OPEN CHANNEL FLOW Numerical Methods and Computer Applications
OPEN CHANNEL FLOW Numerical Methods and Computer Applications Roland Jeppson
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Contents List of Figures....................................................................................................................................xi List of Tables..................................................................................................................................... xv Preface............................................................................................................................................xvii Computer Programs with Listings of Code.................................................................................. xxiii Chapter 1 Dimensions, Terminology, and Review of Basic Fluid Mechanics...............................1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11
Introduction........................................................................................................1 One, Two, and ThreeDimensional Flows.......................................................1 Steady versus Unsteady Flow.............................................................................2 Uniform versus Nonuniform Flow.....................................................................3 Prismatic versus Nonprismatic Channels...........................................................4 Subcritical, Critical, and Supercritical Flows.....................................................4 Turbulent versus Laminar Flow.........................................................................5 Review of Basic Fluid Mechanics Principles.....................................................5 Physical Properties of Fluids and Their Effects on OpenChannel Flows.........6 Conservation of Mass, or Continuity Equations.............................................. 19 Energy Principle............................................................................................... 22 1.11.1 Kinetic Energy Correction Coefficient, α........................................... 27 1.12 Momentum Principle in Fluid Flow................................................................. 30 1.12.1 Momentum Flux Correction Coefficient, β......................................... 34 Problems...................................................................................................................... 38 Chapter 2 Energy and Its Dissipation in Open Channels............................................................ 43 2.1 2.2
Introduction...................................................................................................... 43 Approaches to Frictional Resistance................................................................ 43 2.2.1 Friction Factors in Open Channels......................................................44 2.3 Combining the Chezy and the Chezy C Equations.......................................... 52 2.4 Empirical Formula: Use of Manning’s Equation............................................. 56 2.5 Channels with Varying Wall Roughness, but Q = Constant............................ 74 2.6 Specific Energy, Subcritical and Supercritical Flows...................................... 75 2.7 Flumes............................................................................................................ 100 2.8 Delivery Diagrams......................................................................................... 103 2.9 Graphical Aids to Solving Critical Flow Problems........................................ 131 2.10 Upstream Depth When Critical Conditions Occur at Reduced Downstream Section....................................................................................... 145 2.11 Dimensionless Treatment of Upstream Trapezoidal Channel to Downstream Rectangular Channel............................................................ 146 2.11.1 Upstream Channel Also Rectangular................................................ 150 2.12 Hydraulically Most Efficient Section............................................................. 153 2.12.1 Nondimensional Variables for This Section...................................... 155 Problems.................................................................................................................... 157 References................................................................................................................. 182
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Chapter 3 The Momentum Principle Applied to Open Channel Flows..................................... 183 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
The Momentum Function............................................................................... 183 Characteristics of the Momentum Function................................................... 185 Rectangular Channels and Momentum Function per Unit Width.................. 188 Polynomial Form for Momentum Function................................................... 194 Dimensionless Momentum Functions............................................................ 196 Celerity of Small Amplitude Gravity Waves..................................................208 Constant Height Waves................................................................................... 212 Open Channel to Pipe Flow...........................................................................224 Multiple Roughness Coefficient for Channel Section— Compound Sections.................................................................... 233 Problems.................................................................................................................... 236 Problems to Solve Using Program CHANNEL........................................................ 253
Chapter 4 Nonuniform Flows.................................................................................................... 255 4.1 4.2
4.3
4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19
Types of Nonuniform Flows........................................................................... 255 Ordinary Differential Equation for Gradually Varied Flow.......................... 255 4.2.1 Bulk Lateral Outflow......................................................................... 256 4.2.2 Lateral Inflow.................................................................................... 257 4.2.3 Generalization of Gradually Varied Flow Equations........................ 258 Gradually Varied Flow in Prismatic Channels without Lateral Inflow or Outflow........................................................................................... 259 4.3.1 Classification of Gradually Varied Profiles....................................... 259 4.3.2 Sketching GVF Profiles in Prismatic Channels................................264 4.3.3 Alternative Forms of the ODE That Describe GVF Profiles............ 268 Numerical Methods for Solving ODEs..........................................................280 Canal Systems................................................................................................ 294 Simultaneous Solution of Algebraic and Ordinary Differential Equations.............................................................................301 Flow into a Mild Channel with a Downstream Control.................................304 Different Modes of Gate Operation................................................................ 315 Hydraulic Jump Downstream from a Gate in a Finite Length Channel........ 316 Nonprismatic Channels.................................................................................. 341 Culverts..........................................................................................................346 4.11.1 Solutions When Upstream Control Exists......................................... 347 4.11.2 Solutions When Downstream Control Exists.................................... 359 GVF Profiles in Nonprismatic Channels........................................................ 368 GVF Profiles in Branched Channel Systems.................................................. 374 GVF Profiles in Parallel Channels................................................................. 390 Solutions to Spatially Varied Flows............................................................... 399 4.15.1 Outflow from Side Weirs...................................................................400 4.15.2 ClosedForm Solution to Side Weir Outflow....................................404 Spatially Varied Inflows.................................................................................408 4.16.1 Algebraic Solution............................................................................. 413 Spatially Varied Flow in Nonprismatic Channels.......................................... 414 Tile Drainage.................................................................................................. 429 Downstream Controls in Nonprismatic Channels.......................................... 433
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4.20 Gutter Flow and Outflow through Grates.......................................................449 4.20.1 Gutter Flow.......................................................................................449 4.20.2 Flow into Grates at Bottom of Channel............................................ 457 4.20.3 Combined Problem: Gutter Inflow and Grate Outflow..................... 474 4.20.4 Lateral Inflow over Grate Length......................................................484 4.20.5 No Negative Flow Rates....................................................................484 4.20.6 Last Grate at End of Gutter............................................................... 492 4.20.7 Subcritical Flow through n Gutter–Grates........................................ 498 4.21 Multiple Branched Channel Systems............................................................. 510 4.22 Other Dependent Variables in GVF Computations........................................540 4.23 Varied Flow Function.....................................................................................544 4.24 Moving Waves................................................................................................ 549 4.24.1 QuasiUnsteady Analysis.................................................................. 549 4.24.2 Downstream Controlled Waves......................................................... 549 4.24.2.1 Explicit Method................................................................. 550 4.24.2.2 Implicit Method................................................................. 559 4.24.3 Upstream Controlled Waves.............................................................. 568 4.25 Moving Hydraulic Jump................................................................................. 578 Problems.................................................................................................................... 593 References................................................................................................................. 656 Chapter 5 Common Techniques Used in Practice and Controls................................................ 657 5.1 5.2 5.3
5.4
5.5
5.6
Introduction.................................................................................................... 657 Resistance to Flow in Natural Streams and Rivers........................................ 657 Techniques Used for Solving Steady Flows in Irregular Channels................660 5.3.1 Defining Irregular Channel Properties............................................. 661 5.3.2 Use of Cubic Splines to Define Cross Section.................................. 667 5.3.3 Solving Manning’s and Energy Equations in Natural Channels....... 671 5.3.4 Hand Solution to GVFFlow by the Standard Step Method............. 681 5.3.5 Using an ODE Solver to Compute GVFProfile in Irregular Channels........................................................................ 685 5.3.6 Multiple Parallel Channels and Flood Plain Storage........................ 696 5.3.7 Implementation of Solution for Compound Channel........................ 698 5.3.8 System of Natural Streams and Rivers............................................. 709 5.3.9 HEC2 Water Surface Profiles.......................................................... 709 Water Measurement in Channels.................................................................... 710 5.4.1 Current Meters.................................................................................. 710 5.4.2 Weirs................................................................................................. 711 5.4.3 Parshall Flumes and Cutthroat Flumes............................................. 713 5.4.4 Other Critical Flow Flumes............................................................... 718 Design of Transitions...................................................................................... 720 5.5.1 Subcritical Transitions....................................................................... 721 5.5.1.1 Design of Transitions by Hand Computations................... 721 5.5.1.2 Transition Design by Solving an ODE Problem................ 724 5.5.2 Supercritical Transitions.................................................................... 727 5.5.2.1 Channel Contractions........................................................ 728 5.5.3 Design of Supercritical Transitions................................................... 729 5.5.4 Channel Enlargements...................................................................... 735 Gates............................................................................................................... 741
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5.7
Submerged Flow Downstream from Vertical Gates...................................... 744 5.7.1 Dimensionless Forms of Equations................................................... 748 5.8 Series of Submerged Gates............................................................................. 763 5.9 Design of Side Weirs...................................................................................... 772 5.10 Optimal Design of Trapezoidal Channels Considering Total Costs.............. 779 5.10.1 Additional Optional Analyses........................................................... 788 Problems.................................................................................................................... 793 References................................................................................................................. 822
Chapter 6 Unsteady Flows......................................................................................................... 823 6.1 6.2
When Should Flow Be Handled as Unsteady?............................................... 823 Basic OneDimensional Equations for Unsteady Channel Flows (The St. Venant Equations).............................................................................824 6.2.1 QY Set of St. Venant Equations.......................................................824 6.2.2 VY Set of St. Venant Equations....................................................... 826 6.2.3 Derivation Based Directly on Newton’s Second Law of Motion...... 826 6.2.4 No Lateral Inflow or Outflow and Prismatic Channels..................... 827 6.2.5 Changes in Dependent Variables...................................................... 828 6.3 Determination of Mathematical Type of St. Venant Equations..................... 828 6.3.1 SecondOrder PDEs.......................................................................... 829 6.4 Taking Advantage of the Equation Characteristics........................................ 831 6.5 Solution to Unsteady Flows That Deviate Only Slightly from Uniform Conditions............................................................................... 833 6.6 Boundary Conditions...................................................................................... 836 6.6.1 Depth at Origin Specified, Y(0, t)—Known..................................... 836 6.6.2 Velocity at Origin Specified, V(0, t)—Known.................................. 837 6.6.3 Flow Rate q at the Origin Specified, q(0, t)—Known...................... 837 6.6.4 Upstream Reservoir Water Surface Specified, H(0, t)—Known....... 837 6.6.5 Understanding Characteristics Better............................................... 851 6.7 Maximum Possible Flow Rates...................................................................... 863 6.7.1 Maximum Point Outflow, Δq............................................................866 6.8 Extending the Methods to Nonrectangular Channels.................................... 867 6.8.1 Trapezoidal Channels........................................................................ 867 6.8.2 Dimensionless Variables................................................................... 868 6.8.3 Circular Channels.............................................................................. 883 6.9 Maximum Flow Rates in Nonrectangular Channels...................................... 890 6.9.1 Maximum Outflow ΔQ That Can Be Taken at an Intermediate Position................................................................ 897 6.10 Positive Waves................................................................................................ 898 6.11 Control Structures.......................................................................................... 901 6.12 Partial Instant Opening of Gates in Rectangular Channels........................... 910 6.13 Partial Instant Closing of Gates in Trapezoidal Channels............................. 915 6.14 Partial Instant Closure Followed by Slow Movement Thereafter.................. 919 6.14.1 Rectangular Channels....................................................................... 919 6.14.2 Nonrectangular Channels..................................................................926 6.15 Dam Break Problem....................................................................................... 937 Problems.................................................................................................................... 938
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Chapter 7 Numerical Solution of the St. Venant Equations....................................................... 963 7.1 7.2 7.3
Background..................................................................................................... 963 Method of Characteristics.............................................................................. 965 Boundary Conditions...................................................................................... 971 7.3.1 Downstream End............................................................................... 971 7.3.2 Upstream End.................................................................................... 972 7.3.3 Solution to Boundary Unknowns...................................................... 973 7.4 Using Characteristics with Specified Time Increments................................. 987 7.4.1 Based on SecondOrder Approximations......................................... 987 7.4.2 Upstream Boundary Condition Equations........................................ 989 7.4.3 Downstream Boundary Condition Equations...................................990 7.5 Iterative Solution Technique......................................................................... 1001 7.6 Explicit Evaluation of Variables at Points L and R...................................... 1003 7.7 Accuracy of Numerical Solutions................................................................. 1008 7.8 Implicit Methods.......................................................................................... 1013 7.8.1 Direct Implicit Finite Differencing..................................................1014 7.8.1.1 Boundary Conditions........................................................1017 7.8.1.2 Solving the Difference Equations.................................... 1019 7.8.1.3 Computer Code to Implement Direct Implicit Method..........................................................1020 7.8.2 Alternative Approach to Boundary Condition Equations............... 1028 7.9 Gauss–Seidel or SuccessiveOverRelaxation (SOR) Iterative Solution Techniques...................................................................................... 1030 7.10 Crank–Nicolson Newton Iterative Implicit Method..................................... 1043 7.11 Weighting Current and Advanced Time Steps Differently.......................... 1059 7.12 The Preissmann Implicit Method................................................................. 1064 7.12.1 Double Sweep Method of Solution.................................................. 1067 7.12.1.1 Boundary Conditions....................................................... 1068 7.12.1.2 Downstream Boundary Conditions................................. 1068 7.12.1.3 Upstream Boundary Conditions...................................... 1069 7.13 Solving Preissmann Difference Equations Using the Newton Method....... 1075 7.14 TwoDimensional Free Surface Flows......................................................... 1084 7.14.1 TwoDimensional St. Venant Equations......................................... 1084 Problems.................................................................................................................. 1087 References............................................................................................................... 1100
Appendix A...................................................................................................................................1101 A.1 Open Channel Geometry and Properties..................................................... 1105 A.1.1 Classification of Channels............................................................... 1105 A.1.2 Geometric Properties of Common Prismatic Channels.................. 1105 A.1.2.1 Rectangle......................................................................... 1105 A.1.2.2 Trapezoid......................................................................... 1106 A.1.2.3 Circle................................................................................ 1107 A.1.2.4 Obtaining Geometric Properties of Irregular Cross Sections...................................................................1114 Problems (For You to Work)....................................................................................1121
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Appendix B Numerical Methods............................................................................................... 1123 B.1 Newton Method............................................................................................ 1123 B.1.1 How the Newton Method Works..................................................... 1123 B.1.2 Solving Systems of Equations......................................................... 1125 B.1.3 Implementing the Newton Method in Solving Manning’s Equation........................................................................ 1129 B.1.3.1 Implementation of the Newton Method in Solving Manning’s Equation in a Circular Channel..................... 1130 B.2 Lagrange’s Interpolation Formula................................................................ 1132 B.2.1 Implementation of Lagrange’s Formula in a Computer Program..............................................................1134 B.3 Numerical Integration.................................................................................. 1136 B.3.1 Trapezoidal Rule............................................................................. 1136 B.3.1.1 Simpson’s Rule................................................................. 1139 B.4 Spline Functions........................................................................................... 1146 B.4.1 Background..................................................................................... 1146 B.4.2 Quadratic Splines.............................................................................1147 B.4.3 Cubic Splines................................................................................... 1148 B.5 Linear Algebra.............................................................................................. 1159 B.5.1 Use of Subroutine LAGU That Implements Laguerre’s Method....... 1162 Appendix C ODESOL: Subroutine to Solve ODEs................................................................... 1165 C.1 Background to Algorithm............................................................................. 1165 C.2 Using FORTRAN......................................................................................... 1165 C.3 How Do You Use ODESOL......................................................................... 1166 C.4 Subroutine That You Must Supply................................................................1167 C.5 odesolcCFunction to Solve ODE’s..........................................................1174 C.6 How Do You Use odesolc..........................................................................1175 C.7 Procedure That You Must Supply.................................................................1176 C.8 DVERK: ODE Solver from the International Statistical Mathematical Libraries, ISML............................................................................................ 1180 C.9 Runge–Kutta Method....................................................................................1183 C.9.1 Background......................................................................................1183 C.9.2 Description of Method.....................................................................1183 C.9.2.1 Illustrative Use of Routines............................................. 1188 C.10 Runge–Kutta Using C....................................................................................1191 C.11 Runge–Kutta–Fehlberg Method................................................................... 1193
List of Figures FIGURE 2.1 Diagram for Chezy’s C for use in determining the flow rate, velocity and slope of the energy line, or head loss in open channels................................... 47 FIGURE 2.2 Relationship between n parameter and relative roughness to establish relation of n to e/R h................................................................................................ 61 FIGURE 2.3 Plot of dimensionless Manning’s equation in trapezoidal channels (including rectangular when m = 0)....................................................................... 73 FIGURE 2.4 Plot of dimensionless Manning’s equation in circular channels............................ 74 FIGURE 2.5 Sketches of specific energy diagrams in (a) a general channel and (b) in a rectangular channel in which the bottom width changes, but Q = constant.......... 76 FIGURE 2.6 Relationship of dimensionless depth to dimensionless flow rate parameters under “critical flow” conditions............................................................................ 132 FIGURE 2.7 Dimensionless specific energy diagram trapezoidal channels. (Individual curves apply for Q′ = m3Q2/(gb5).)........................................................................ 133 FIGURE 2.8 Dimensionless specific energy diagrams for circular channels. (Individual curves apply for Q′ = Q2/(gD5).)........................................................................... 133 FIGURE 2.9 Critical condition in a trapezoidal channel........................................................... 136 FIGURE 2.10 Critical condition in a circular channel................................................................ 136 imensionless critical depth at the beginning of a trapezoidal channel feed FIGURE 2.11 D by a reservoir........................................................................................................ 139 imensionless critical depth at the beginning of a circular channel feed FIGURE 2.12 D by a reservoir........................................................................................................ 139 ight graphs for different upstream side slopes that provide the solution FIGURE 2.13 E of Equation 2.40 of dimensionless depth Y1′ as a function of b′ and b1′ for upstream trapezoidal channels to downstream rectangular channels with critical flow........................................................................................................... 147 olution of Equation 2.43 of dimensionless depth Y′ as a function FIGURE 2.14 S of b′ and b1′ for rectangular channels both upstream and downstream and critical flow in the downstream channel..................................................... 151 Figure 3.1 Dimensionless momentum function diagrams for trapezoidal sections. (Individual curves apply for Q′ = m3Q2/(gb5).)..................................................... 197 Figure 3.2 Dimensionless momentum function diagrams for circular sections. (Individual curves apply for Q′ = Q2/(gD5).)........................................................ 198 Figure 3.3 Dimensionless momentum diagram for rectangular sections..............................205
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Figure 3.4 R elationship between ratio of depths r = Y2/Y1 with the upstream Froude number Fr1 = q /(gY12 )1/ 2 = V1 /(gY1 )1/ 2 for rectangular channels and with the upstream Froude number Fr1 = {Q 2 T1 /(gA13 )}1/ 2 and the dimensionless downstream depth Y1′ = mY1 / b for trapezoidal channels.......................................208 elationship between ratio of depths r = Y1/Y2 with the downstream Froude Figure 3.5 R number Fr2 = q /(gY23 )1/ 2 = V2 /(gY2 )1/ 2 for rectangular channels and with the downstream Froude number Fr2 = {Q 2 T2 /(gA 32 )}1/ 2 and the dimensionless downstream depth Y2′ = mY2 / b for trapezoidal channels...................................... 210 radually varied profiles caused by a single control. (a) The weir causes flow Figure 4.1 G above the normal depth in a mild channel and this is an M1 GVF. (b) The small dam causes the same M1 GVF upstream as the weir. (c) The flow upstream from a gate is always subcritical and if flow emerges as “free flow” from the gate it is supercritical downstream. Thus upstream an M1 GVF occurs, and an M3 GVF since the channel is mild. (d) If the contraction is severe enough, it will cause critical flow at this reduced section with an M1 GVF upstream. If the contraction is small and the channel enlarges again downstream, or is steeper, then the upstream flow may remain at normal depth. (e) In a steep channel, a dam will cause a jump to take the flow from supercritical to subcritical conditions, and an S1 GVF will exist downstream there from. Upstream of the jump the depth will be normal. Thus the S1 GVF will start at the depth conjugate to Y0 and end at a depth with a specific energy that is equal to the critical specific energy at the crest of the dam plus the height of the dam, or E = Ec + Δz. (f) The depth in the mild downstream channel will be normal so an H2 GVF will occur downstream from the hydraulic jump and an H3 GVF upstream there from. If the horizontal channel is short, and the water has a very large velocity as it passes under the gate the hydraulic jump may be pushed into the mild channel in which case the H3 GVF will exist over the entire length of the horizontal channel and continue as an M3 GVF in the downstream channel until the jump occurs. (g) A gate in a steep channel will force the flow to be subcritical upstream from it so an S1 GVF will occur that has its beginning at the depth conjugate to the normal depth in the steep channel. Downstream as S3 GVF will occur. (h) In a critical channel, a gate will cause an C1 GVF upstream that starts at critical depth so no jump occurs, and an C3 GVF will occur downstream. (i) No jump occurs downstream from a gate in a steep channel unless something downstream there from forces the flow to be subcritical. (j) A steep channel abruptly changes to a mild channel will either cause an S1 GVF upstream from the break in grade as shown if the momentum function associated with the downstream normal depth is greater than that for the upstream normal flow, Mo1 < Mo2. If Mo1 > Mo2 then the jump will occur downstream with an M3 GVF starting at the break in grade and ending with the depth conjugate to Yo2. (k) A reduction in the bottom slope in a mild channel will cause an M1 GVF upstream from the break in grade. (l) Adverse or horizontal channels have profiles much like those in mild channels. The difference is they are designated with an A or H letter.......................................... 265 Figure 4.2 Examples of GVF profiles...................................................................................... 267 Figure 5.1 Typical irregular channel cross section.................................................................. 661 Figure 5.2 Plan view of a small river between position x = 5,000 ft and x = 11,000 ft............ 682
List of Figures
xiii
Figure 5.3 Transition shapes for supercritical expansion....................................................... 737 Figure 5.4 Solution of dimensionless depth Y immediately downstream from gate based on dimensionless depth y3 and y2. (Assumption is that jet flow occurs beneath gate so that special use of energy and momentum equations are valid.)................ 749 Figure 5.5 Special dimensionlessspecific energy and momentum functions....................... 750 Figure 5.6 Solution of dimensionlessspecific energy across a gate...................................... 752 Figure 5.7 Limiting values for submergence to first occur.................................................... 753 Figure 5.8 Solution of dimensionless parameter equations that separate submerged and free flow past a gate. (Submerged flow will occur for Y3 ≥ Y1y3.)................. 755 Figure 5.9 Dimensionless solution to loss of energy past submerged gate, Δe = e1 − e3, and submergence depth, y = Y/Y1........................................................................ 759 ariations of upstream Froude number squared with dimensionless depth Figure 5.10 V y2 and y3................................................................................................................ 760 imensionless variables displayed as functions of the upstream Froude Figure 5.11 D number squared..................................................................................................... 761 plot that shows how the critical side weir height varies with the flow rate Figure 5.12 A in the upstream channel and the bottom width of a rectangular channel. For this plot Manning’s n = 0.016 and the channel has a bottom slope So = 0.0005, and the discharge coefficient for the side weir is Cd = 0.45............. 774 plot that shows how the critical side weir height varies with the flow rate Figure 5.13 A in the upstream channel and the bottom width of a rectangular channel. For this plot Manning’s n = 0.016 and the channel has a bottom slope So = 0.001, and the discharge coefficient for the side weir is Cd = 0.45.................................. 774 engths of side weir needed to discharge varying ratios of downstream to Figure 5.14 L upstream channel flow, under varying upstream flow rates. The following were specified in obtaining this graph: the channel is rectangular with a constant width b = 2 m, the side weir height is 10% above the critical height, the bottom slops is constant, So = 0.0005, with Manning’s n = 0.016, and the discharge coefficient Cd = 0.45............................................................................. 779 Figure 5.15 Variables that define trapezoidal channel’s crosssections................................... 780 ive graphs that show how optimal b, Yo, m, V, and C vary with lining Figure 5.16 F and excavation costs. The following parameters are specified for the series of solutions that gives these graphs: Q = 20 m3/s, n = 0.016, eb = er = 0.15 m, R = 0.5 m, Ce = $15/m and So = 0.0002................................................................. 786 ive graphs that show how optimal b, Yo, m, V, and C vary with lining Figure 5.17 F and excavation costs. The following parameters are specified for the series of solutions that gives these graphs: Q = 20 m3/s, n = 0.016, eb = er = 0.15 m, R = 0.5 m, Ce = $15/m and So = 0.002.................................................................. 788 Figure 6.1 Function of celerity and flow rate per unit width, q, at origin for a rectangular channel...................................................................................... 865 Figure 6.2 Graphs that give the dimensionless function f(c′,q′) for a rectangular channel. (Roots are where the curves cross the zero horizontal axis. Only the smaller root has physical significance.) ..............................................................................866
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List of Figures
Figure 6.3 (a) Dimensionless critical depth and stage variable related to initial w′o − Vo′ for a trapezoidal channel. (b) Dimensionless critical depth and stage variable related to initial w′o − Vo′ for a circular channel ..................................................... 892 Figure 6.4 (a) Dimensionless critical values (Yc′, Vc′, and Q′c ) related to initial dimensionless depth, Yo′ and velocity, Vo′ for a trapezoidal channel. (b) Dimensionless critical values (Yc′, Vc′, and Q′c ) related to initial dimensionless depth, Yo′ and velocity, Vo′ for a circular channel ........................... 893 FIGURE 7.1 Intersection of C+ with C− characteristic lines to define grid points at which unknown will be solved.........................................................................................966 FIGURE 7.2 Characteristics uses to solve values at new grid m from values at known grids L and R......................................................................................................... 967 FIGURE 7.3 Sketch of C+ and C− characteristics in the xt plane with the numbering starting at downstream end.................................................................................... 977 FIGURE 7.4 Finite difference grid in the xt plane at which values for the dependent variables Y and Q will be solved..........................................................................1014 FIGURE 7.5 Linear system of finite difference equations represented in matrix notation for constant upstream reservoir supplying the channel and the flow rate given as a function of time at the downstream end of the channel............................... 1020 Figure A.1 Geometric properties of a circular section ...........................................................1111 Figure A.2 G eometric properties of a circle, including the dimensionless conveyance, K′ = A′(A′/P ′)2/3 .....................................................................................................1111
List of Tables Table 1.1 Properties of Water Related to Temperature.................................................................7 TABLE 2.1 Summary of Equations That Define Chezy’s Coefficient, C...................................... 47 TABLE 2.2 Value of Wall Roughness, e, for Different Channel Materials................................... 49 TABLE 2.3 Typical Values for Manning’s n.................................................................................. 57 ariations of Manning’s n with a Fixed Value of e = 0.004 ft TABLE 2.4 V (v = 1.217 × 10−5 ft2/s) and Other Variables Changed in a Trapezoidal Channel........59 ariations of Manning’s n with a Fixed Value of e = 0.001219 m TABLE 2.5 V (v = 1.31 × 10 −6 m2/s) and Other Variables Changed in a Circular Channel..............60 atio of Downstream to Upstream Depth, Y2/Y1, across a Hydraulic Jump TABLE 3.1 R in Rectangular and Trapezoidal Channels................................................................209 Table 3.2 R atio of Upstream to Downstream Depth, Y1/Y2, across a Hydraulic Jump in Rectangular and Trapezoidal Channels................................................................ 211 Table 4.1 Gradually Varied Profiles in Prismatic Channels.....................................................260 Table 5.1 CrossSection Data for the Six Stations along the River Shown in Figure 5.2......... 682 ables Giving Areas, A and Perimeter, P as a Function of the Depth Y Table 5.2 T for the Six Sections of River..................................................................................... 683 preadsheet Solution to Previous GVFProfile in the River of Figure 5.1 Table 5.3 S (EXCEL RIVER1.XLS)............................................................................................684 imensionless Variables Associated with Stage Variable, w, Table 6.1 D for Trapezoidal Channels.......................................................................................... 870 TABLE 6.2 D imensionless Variables Associated with Stage Variable, w, for Circular Sections.............................................................................................. 884 Table A.1 Geometric Properties of Cross Sections Often Used for Channels.........................1110 Table A.2 Geometric Properties for Irregular Channel Given Above.....................................1118
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Preface This book was developed over several years while teaching courses in open channel flow to graduate students. Initially, while on the quarter system, two open channel graduate‑level courses were taught: the first dealing with steadystate flow was a fourcredit course, including one credit for a laboratory; and the second dealing with unsteady flow, that is, numerical solutions of the St. Venant equations, was a threecredit course. When the university switched to the semester system, these two courses were both expanded into threecredit graduate‑level semester courses, and the amount of material covered was essentially that in the current book. Before undertaking the second course, most of the students had also taken a course dealing with numerical methods in engineering. While the material was developed and intended to complement lectures in this subject area, it should also be useful to the practicing engineer. There are numerous example problems throughout the book that elucidate principles, formulate and set up problems, and/or apply techniques of problem solutions. Thus, the book is also intended for selfstudy for those who have taken courses in fluid mechanics and hydraulics. The basic principles of conservation of mass, energy, and momentum are emphasized in the hope that this will help students master this important subject rather than just learn routine techniques in solving the large host of open channel applications. In so doing, students will enhance and enlarge their understanding of the fundamental principles of fluid mechanics and apply them in solving complex real problems. This emphasis is accomplished by devoting an entire chapter (Chapter 2) to the energy principle as it applies to open channel flow. Chapter 3 is devoted to the momentum principle, but since energy and conservation of mass have been covered previously, all three principles are used in setting up and solving problems. (Since the principle of conservation of mass is relatively easily implemented in solving open channel problems, a separate chapter is not devoted to it.) Many of these equations are nonlinear, and therefore numerical means for solving them are covered in addition to the open channel hydraulics. Real channels generally do not consist of a single channel of constant size, but rather a series of channels with different sizes and control structures, and/or parallel systems. Therefore, in dealing with these principles, they are applied repeatedly to link the equations together, which must be solved simultaneously to obtain depths, velocities, and flow rates throughout channel systems. Again, numerical means for accomplishing the solution for a system of nonlinear equations are covered, and the techniques for accomplishing such solutions are documented through computer codes and program listings. The progression from a single channel to a multichannel system is a distinguishing feature that sets this book apart from other books on this subject. Seldom is the flow in real channels uniform, that is, the depth varies with position along the channel. Except near control structures, these variations in depth can be handled as gradually varied flow, that is, the flow is assumed to be onedimensional, or the dependent variables are only a function of the position along the channel. Such gradually varied flows are described by an ordinary differential equation (ODE), for which closedform solutions are only possible using very restrictive assumptions and, therefore, seldom apply in practice. The longest chapter in the book, Chapter 4, deals with gradually varied flows. It begins by deriving the general gradually varied flow equation, and documents numerical methods for solving this firstorder ODE. Computer codes based on mathematical numerical methods rather than the traditional standard step method for solving a single ODE are provided, and these are then applied to solve a variety of problems associated with upstream and downstream controls, side weirs, etc. Again, as in previous chapters, after the student is thoroughly familiar with how gradually varied flow in single channels can be solved, he or she is shown how to set up and solve gradually varied flows in a system of channels.
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This instruction involves numerical methods for solving systems of ordinary differential and nonlinear (with some linear) algebraic equations simultaneously. To assist in the instruction, computer codes and/or computer programs are provided. Following the linebyline instructions for the computer to follow is in fact a most effective means of learning how such complex problems can be solved. The setting up of simultaneous algebraic equations in Chapters 2 and 3 provides the basis for setting up the equations that govern gradually varied flows in channel systems. The extension is that now, not only are there algebraic equations involved, but also ODEs. Commonly available software packages are not capable of solving combined systems of ODEs and algebraic equations without “addin.” Thus, much of the material in Chapter 4 is not available in other textbooks on this subject. The variety of these complex problems is almost unlimited, but in an attempt to provide the student with the tools needed to set up and solve a particular problem that he or she may encounter, a large number of example problems are provided as an integral part of the text. The solutions to these example problems generally contain numerous computations and therefore require the use of a computer. Thus, many of the example problems contain computer programs. There are also a large number of homework problems at the end of the chapters. These problems provide the student not only with experience to solve problems somewhat similar to the example problems, but require him or her to also apply the principles to solve problems that expand upon the text material. The material covered in Chapters 1 through 4 assumes that the channel’s geometry is rectangular, trapezoidal, circular, or can be defined by simple parameters such as bottom width, side slope, diameter, etc. Chapter 5, entitled “Common techniques used in practice and controls,” describes how the geometry of natural channels can be defined using a table of xy values for its bottom shape, and how quantities such as areas, perimeters, and top widths can be obtained from this data rather than just by solving an equation. It then covers water measurements in open channels, gates, and transitions, and concludes with a section dealing with total least cost design of channels. The last two chapters, Chapters 6 and 7, deal with unsteady flow. Chapter 6 derives the various forms of unsteady flow equations for onedimensional flow, that is, the St. Venant equations, and describes their characteristics. By assuming that the difference between the slopes of the energy line and channel bottom is the same, these unsteady flow equations can be solved along characteristic lines; this concept can then be used to obtain solutions to a variety of problems with upstream and downstream controls. Such simplified solutions provide the student with a good understanding of unsteady channel flow, and become almost indispensable in setting up the complete unsteady flow equations for a variety of problems, as described in Chapter 7. Initially, the material also included a followon chapter dealing with solutions of the two and threedimensional unsteady flow equations, but that has now been deleted. These two and threedimensional equations are now only derived. I would like to express my heartfelt gratitude to Dr. Oulhaj Ahmed at the Institut Agronomique et Véterinaire Hassan II, Rabat, Morocco, who has translated this book into French and used it as course material for years now. He had translated material years ago, and recently updated that translation to include this book. The French version of this book is available at the Institut Agronomique et Véterinaire Hassan II, Rabat, Morocco. An electronic “User’s Manual” is also available on the CRC Press Web site (www.crcpress.com) that contains the solutions to the homework problems that are located at the end of each chapter. To obtain these solutions, please contact the publisher (Taylor & Francis Group). I wish to thank the many students who have participated in this course and in other courses. The satisfaction and joy associated with teaching are truly enormous, and the enthusiasm of students adds much thereto. It is difficult to think of any profession that is as rewarding as teaching at a university. I sincerely hope that this book will contribute to the important subject of open channel flow and the use of numerical methods in engineering practice.
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For MATLAB® and Simulink® product information, please contact The MathWorks, Inc. 3 Apple Hill Drive Natick, MA, 017602098 USA Tel: 5086477000 Fax: 5086477001 Email:
[email protected] Web: www.mathworks.com
Guidelines for Studying This Book The details of learning and internalizing a new subject until it becomes an integral part of your working knowledge cannot be described here. Rather, as a university student, or a professional, by now you should know how you learn best. Unless you are genius, it is unlikely that you will learn the subject of open channel flow by just reading, as you would read a newspaper, a magazine such as National Geographic, or a novel. Rather you will have to acquire the ability to set up and solve problems. Therefore, in these guidelines, I will suggest objectives and goals that you should set, and establish a means for measuring whether these goals are being systematically met. In other words, you will not really understand the subject of open channel flow unless you are able to set up mathematical equations that properly describe an open channel flow system and then solve those equations. To help in this process, this book contains numerous example problems, and an even larger number of homework problems. Some of these problems are what one might call routine, for example, problems that simply deal with solving a specified equation for a specified unknown variable. Other problems require that you identify the basic principles that apply, and how the proper application of these principles produces the system of equations that needs to be solved simultaneously to get the numerical values that describe how the channel system will perform under given conditions. It is hoped that this course will help you appreciate how the mathematics and mathematical methods that were developed mainly for purely theoretical reasons in your previous courses on linear algebra, calculus, and differential equations, suddenly assume great importance when dealing with engineering problems. In your previous math courses, practically all of the learning was associated with linear equations and differential equations for which “closedform” solutions are possible. Real engineering problems are, most frequently, governed by nonlinear and differential equations that must be solved using numerical techniques. This is certainly true for open channel flow. In brief, engineering mathematics comes down, ultimately, to numerical results, for example, numbers that define a problem’s properties, such as depth, velocity, flow rate, force, etc., and the variations of these quantities in space and time. Consequently, much of what you will learn during this course deals with numerical methods. You will find that the material in the appendixes, especially Appendixes B and C, will need to be studied in detail, and fully mastered. Since numerical methods require a large amount of number crunching, beyond what can be practically accomplished by hand, it will be vital that you use a computer to solve many problems. Thus, plan on making your computer a heavily relied upon workhorse in studying this subject. It is for good reason that the title of this book contains not only open channel flow, but also numerical methods and computer applications. Throughout the text there are listings of computer programs in Fortran, C (or C++), and math applications software such as Mathcad and TKSolver that are used to solve example problems and to illustrate concepts. A folder on the CDROM on the back cover of this book contains MATLAB® programs that accomplish the same tasks of many of the Fortran programs. You might not be thoroughly familiar with these languages and/or applications for obtaining the numerical solutions, but it is relatively easy to gain sufficient understanding of Fortran (or C++) to follow the logic and computations needed to implement solutions. You will discover that it is often much easier to fully
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understand how principles and equations are used to solve problems by studying these programs than by just reading the text. The listings of these programs will be a very important tool for you to accomplish the goals (as listed below) that you should establish for yourself during the courses you take using this book. Most of our bodies of knowledge in engineering, and its practice, are based on, relatively, a few fundamental principles and/or laws. In open channel flow, the three underlying principles are (1) conservation of mass, (2) conservation of energy, and (3) conservation of momentum. A thorough and comprehensive understanding of these principles is vital to properly understanding open channel flows and to solving complex channel systems. The proper application of these relatively simple concepts often entails considerable insight; therefore, devote time and effort in understanding these principles completely, and how they are used in setting up the equations that provide solutions to a wide variety of open channel situations. The problems solved in Chapters 1 and 2 use only the first two of these principles and gets you acquainted with equations such as Manning’s equation and Chezy’s equation, which describe how energy is dissipated due to fluid friction. Chapter 3 adds the momentum principle to your working tools. Chapter 4 and Appendix C devote much space to solving firstorder ODEs, because the vast majority of steadystate open channel flows are mathematically described by such equations. The ability you acquire in numerically solving ODEs related to open channel flows will enhance your ability to cope with many other engineering problems, since ODEs are a most important body of knowledge in engineering. The first five chapters assume that flow conditions do not change with time, that is, they deal with the subject of steadystate open channel flow. The last two chapters are devoted to unsteady flows in open channels. Initial conditions, or what the flow consists of at time zero, require that steadystate solutions be obtained. Thus, you will need to have a good understanding of Chapters 1 through 5 before beginning Chapter 6. Since there is a semester’s amount of study (or more) in Chapters 1 through 5, to study the material in this book will take at least two semesters of graduatelevel course work. In fact, the last time I taught the last two chapters to PhD students in Fluid Mechanics Hydraulics Program at Utah State University, Logan, Utah, it took me two semesters, the second at the request of the students, so they could solve more general unsteady open channel problems. Now, let us define goals you should set for yourselves. Below, only the goals for the first portion of the book will be outlined, that is, for steadystate flows. After studying this portion of the book, you should be able to set your goals associated with solving unsteady open channel flow problems. (You should repeatedly read these goals and assess your progress in completing them.) Goal 1: Develop computer software that will solve for any of the variables associated with Manning’s equation for (a) rectangular channels, (b) trapezoidal channels, (c) circular channels, and (d) natural, or irregularly shaped, channels. A vital part of this goal is to understand why the Newton method works and how it is implemented to solve for variables that cannot be placed on the left side of the equal sign by manipulating an equation, for example, solving implicit equations. Goal 2: Develop a similar software that solves the combined Chezy and ChezyC equations. The simultaneous solution of these two equations can be considered a more fundamentally sound approach to open channel flow than using the empirical Manning’s equation. A similar comparison in pipe flow is use of the Darcy–Weisbach equation (with the friction factor therein being a function of the relative roughness of the pipe wall, and the Reynolds number associated with the flow) versus use of the empirical Hazen–Williams equation. Goal 3: Develop computer software that will solve the energy equation (and energies equate at two different positions) for any of the variables associated therewith for (a) rectangular channels, (b) trapezoidal channels, (c) circular channels, and (d) natural, or irregularly shaped, channels.
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Goal 4: Develop computer software that will solve the critical flow equation for any of the variables associated therewith for (a) rectangular channels, (b) trapezoidal channels, (c) circular channels, and (d) natural, or irregularly shaped, channels. Goal 5: Develop computer software that will solve the momentum equation (and momentum functions equated at two different positions) for any of the variables associated therewith for (a) rectangular channels, (b) trapezoidal channels, (c) circular channels, and (d) natural, or irregularly shaped, channels. Goal 6: Combine the solution capabilities of Goals 1 through 5 into a single software package that is easy to use by allowing you to select the type of problem you are solving, and what variable(s) is (are) to be solved. Upon completing these first six goals, you will have developed a program similar to program CHANNEL that is available on the CDROM on the back cover of this book. You might wish to use your software to solve the problems given at the end of Chapter 3 “Problems to solve using program CHANNEL.” In addition to, or in conjunction with, the achievement of the above goals you should establish and complete the following goals: Goal 7: Using the energy equations and/or critical flow equations associated with channel systems that consist of branched and parallel channels, write out the system of equations that describe the flow rates, velocities, and depths throughout the system. Goal 8: Use linear algebra in combination with the expanded Newton method to solve the system of equations from Goal 7. Goal 9: Use the momentum principle, in addition to the energy and critical flow equations, to define and solve problems involving branched and parallel channel systems that have controls such as gates that cause hydraulic jumps to occur. The above goals deal with flow situations in which the depth does not vary with position along the channel, except in the immediate position of control structures, such as gates, intakes, etc., or, in other words, uniform flows occur. Similar goals need to be set to handle gradually varied flows (GVFs), or situations in which depths, velocities, and possibly flow rates vary with position along the channel. These latter types of flows are governed by ODEs, if steady state, or, if unsteady, by partial differential equations. Goal 10: Become thoroughly familiar with the general ODE that defines GVFs, which allows for lateral inflow/outflow and changing channel size and shape, and how this equation simplifies depending upon the conditions. Goal 11: When possible solve the GVF equation by numerical integration, otherwise learn and obtain numerical solutions of this equation using techniques designed to solve ODEs, and apply these numerical methods to solving GVF problems in single channels. Part of this goal should be to develop computer software that implements the solutions. Goal 12: Be able to write out the system of equations that define channel systems that involve both algebraic equations and ODEs. Goal 13: Solve these combined systems of algebraic equations and ODEs using the Newton method in combination with numerical solutions of ODEs. Again, part of this objective is to develop computer software to obtain the solutions. This later software (or computer program), because of the variety of equations involved, will not be a general program, but will need to be modified to handle different given situations. Chapter 4 is devoted to helping you achieve Goals 10 through 13. Because the subject of GVF is more complex, and the type of problems more varied, Chapter 4 is longer than Chapters 1 through 3 combined.
Computer Programs with Listings of Code Name of Program Chapter 1 EXPRB1_5.FOR EXPRB1_5.C EPRB1_9.FOR EPR1_15.FOR EXPR1_15.C Chapter 2 CHEZYC.FOR CHEZYC.C CH2PR2.FOR CHEZYCTC.FOR MANNING.PAS MANNING.FOR MANNING.C MANNTC.FOR MANNTC.C ALTDEP.FOR ALTDEP.CPP ROOTSE.FOR ROOTSE.C E_UN.FOR E_UN.C E_UN1.FOR E_UN1.C UENCHEZ.FOR UENCHEZ.C UENCHEZ1.FOR
What Program Does? Numerical integration using Simpson’s rule and cubic splines for natural channel Numerical integration using Simpson’s rule and cubic splines for natural channel (solves Example Problem 1.5) Numerical integration using Simpson’s rule and cubic splines (Example Problem 1.9) Solves Example Problem 1.15—Cubic splines and Simpson’s rule Solves Example Problem 1.15—Cubic splines and Simpson’s rule Solves Chezy’s C from transitional equation Solves Chezy’s C from transitional equation Solves Chezy’s equation and Chezy’s C in the transitional zone, simultaneously. (Example Problem 2.2) Solves Chezy’s equation for any of the variables as unknown in both trapezoidal and circular channels Solves Manning’s equation for any of the variables in both trapezoidal and circular channels Solves Manning’s equation for any of the variables in both trapezoidal and circular channels Solves Manning’s equation for any of the variables in both trapezoidal and circular channels Completely solves Manning’s equation with the Newton method Completely solves Manning’s equation with the Newton method Solves energy equation for alternative depth Solves energy equation for alternative depth Uses cubic equation to solve alternative depth Uses cubic equation to solve alternative depth Simultaneous solution of energy and Manning’s equations (Example Problem 2.12) Simultaneous solution of energy and Manning’s equations (Example Problem 2.12) Solves energy and Manning’s equations by substitution Solves energy and Manning’s equations by substitution Uniform flow Simultaneous solution of energy and Chezy’s equations Simultaneous solution of energy and Chezy’s equations
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UENCHEZ1.C E_CR1.FOR E_CR1.C E_CR.FOR E_CR.C E_UNTC.FOR E_UNTC.C THREEC.PAS THREECH.FOR THREECH.C BRANCHCHL.FOR BRANCHCH.C
Chapter 3 ROOTSM.FOR LAGU.FOR LAGU.CPP LAGU5.FOR GATE_WAVE.PAS GWAVE.FOR WAVE.FOR WAVE.C WAVETR.FOR WAVETR.C PRB3_12.FOR PRB3_12.C Chapter 4 GVFXY1.FOR GVFXY1.PAS GVFXY1.C EPRB4_2.FOR EPRB4_2.C EPRB4_2.FOR EPRB4D.FOR EPRB4R.FOR GVFDYXK.C RUKUY4.FOR
Computer Programs with Listings of Code
Simultaneous solution of energy and Chezy’s equations Simultaneous solution of energy and critical flow equations (Example Problem 2.13) Simultaneous solution of energy and critical flow equations (Example Problem 2.13) Simultaneous solution of energy and critical flow equations (Example Problem 2.13) Simultaneous solution of energy and critical flow equations (Example Problem 2.13) Entrance trapezoidal channel to circular channel—Energy and Manning’s equations Entrance trapezoidal channel to circular channel—Energy and Manning’s equations Solves three channel branch Solves three channel branch Solves three channel branch General branched channels—Trapezoidal or circular (Example Problem 2.18) General branched channels—Trapezoidal or circular (Example Problem 2.18) (also see: BRANCHCR.FOR, BRANCHC1.FOR, and BRANCHCH.FOR)
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Solves conjugate depth as cubic equation Uses Laguerre’s method to find roots of momentum equation Uses Laguerre’s method to find roots of momentum equation Uses Laguerre’s method to find roots of momentum equation Solves moving wave problems Solves moving wave problems Solves moving waves (Example Problem 3.9) Solves moving waves (Example Problem 3.9) Moving wave (Example Problem 3.10) Moving wave (Example Problem 3.10) Jump in steep to mild channel (Example Problem 3.12) Jump in steep to mild channel (Example Problem 3.12)
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Solves dx/dY—Numerical integration (Example Problem 4.2) Solves dx/dY—Numerical integration (Example Problem 4.2) Solves dx/dY—Numerical integration (Example Problem 4.2) Solves dx/dY by calling on SIMPR (Simpson’s rule) (Example Problem 4.2) Solves dx/dY by calling on SIMPR (Simpson’s rule) (Example Problem 4.2) Solves dY/dx = (So − Sf )/(1 − Fr2 ) by calling on ODESOL Solves dY/dx = (So − Sf )/(1 − Fr2 ) by calling on DVERK Solves dY/dx = (So − Sf )/(1 − Fr2 ) by calling on RUKUST Solves dY/dx = (So − Sf )/(1 − Fr2 ) by calling on RUKUST To solve ODE dY/dx (Example Problem 4.5)
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96 97 97 99 107 108 112 113 115 121
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Computer Programs with Listings of Code
RUKUY4.C RUKUX4.PAS EPRB4_6.FOR EPRB4_6A.FOR RUKUSTS.C EPRB4_6B.FOR EPRB4_8.FOR EPRB4_8.C SOLGVF.FOR SOLGVF.C SOLGATE.FOR GVFJMP.FOR GVFJMP.C GVFJMP2.FOR EPRB4_14.FOR EPRB4_14.C SOLGVFDF.FOR
SOLGVFKF.C
SOLINF.FOR SOLINF.C SOLINF2.FOR EPRB4_18.FOR CULVERTU.FOR CULVERTD.FOR CULVERD1.FOR EPR4_29.FOR EPR4_29A.FOR
SOLBRA.FOR SOLJMP2.FOR SOLPARG.FOR EPRB4_35.FOR
To solve ODE dY/dx (Example Problem 4.5) To solve ODE dY/dx (Example Problem 4.5) Solves dY/dx Manning’s equation—Calls on RUKUY4 (Example Problem 4.6) Solves dY/dx Manning’s equation—Calls on RUKUST Solves dY/dx Manning’s equation—Calls on RUKUST Solves dY/dx—Chezys equation Solves gate at end of channel (Example Problem 4.9) Solves gate at end of channel (Example Problem 4.9) Solve GVF in mild channel, three simultaneous equations Solve GVF in mild channel, three simultaneous equations GVF—Gate downstream (Example Problem 4.10) Hydraulic jump downstream from gate Hydraulic jump downstream from gate Hydraulic jump downstream from gate (Example Problem 4.12) Solves three equations—Gaussian elimination (Example Problem 4.14) Solves three equations—Gaussian elimination (Example Problem 4.14) Gradually varied flow from reservoir to steep rectangular channel at end. Solves three equations simultaneously (Example Problem 4.15) Gradually varied flow from reservoir to steep rectangular channel at end. Solves three equations simultaneously (Example Problem 4.15) Lateral inflow—Solves four equations (solves Example Problem 4.16) Lateral inflow—Solves four equations (solves Example Problem 4.16) Lateran inflow—Solves five equations (solves Example Problem 4.16) Solves GVF (Example Problem 4.18) Solves culvert flow with upstream control Solves culvert flow with downstream control Solves culvert flow with downstream control in which the flow rate Q is specified. GVF in nonprismatic channel (Example Problem 4.29) calls on ODESOL (EPR4_29K.FOR calls on RUKUSTF) GVF in nonprismatic channel (Example Problem 4.29) calls on ODESOL (EPR4_29K.FOR calls on RUKUSTF) calls on DVERK Solves upstream channel branching into N downstream channels (Example Problem 4.30) Solves branched channel with hydraulic jump (Example Problem 4.33) Solves N parallel channels with an upstream and downstream channels (Example Problem 4.35) Spatially varied outflow from side weir (Example Problem 4.37)
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EPRB4_44.FOR EQUNOM.FOR EPRB4_46.FOR TDRAIN.FOR SOLSID.FOR SOLWEIM.FOR SOLWEIS.FOR GUTTER.FOR GUTTER.C GUTTERHZ.FOR GRATE.FOR GRATE.C GRATMILD.FOR GRATMILD.C GRATE2E.FOR GUTTER4T.FOR GUTTER3I.FOR GUTGRAT1.FOR GUTGRTN.FOR GUTGRTBA.C SOLGBR.FOR SOLGBRO.FOR EPRB4_49.FOR EPRB4_66.FOR VARIFUN.FOR WAVEMOV.FOR WAVEMOVR.C WAVEMOVI.FOR WAVEMVRI.C WAVEUCE.FOR WAVEUCE.C WAVEUCI.FOR WAVEMJP7.FOR WAVEMJP7.C Chapter 5 CHTABL.FOR CHTABL.C SPLINENA.FOR NATURAL.FOR NATREG.C GVFNAT.FOR GVFNATN.C GVFNATM.FOR TRANSIT.FOR
Computer Programs with Listings of Code
Solves spatially varied flow in nonprismatic channel (Example Problem 4.44) Newton method to solve Nequations (Example Problem 4.45) Modified spatially varied flow (Example Problem 4.46) Solves tile drain problems Lateral outflow with gate downstream and reservoir upstream Outflow from side weir (Example Problem 4.50) Jump in side weir length Lateral inflow to gutter with inlet drains (Example Problem 4.52) Lateral inflow to gutter with inlet drains (Example Problem 4.52) Gutter flow (Example Problem 4.53) Lateral outflow through bottom grates (Example Problem 4.54) Lateral outflow through bottom grates (Example Problem 4.54) Grates in mild triangular channel—Supercritical flow Grates in mild triangular channel—Supercritical flow Modification of previous program Combined gutter grates Gutter–grate system with inflow over gutter lengths Gutter–grate—No flow out end Supercritical gutter–grate flow Supercritical gutter–grate flow Multiple branched channel Multiple branched channel with outflow/inflow Solves dz/dx for change in bottom elevation through transition Design channel for outflow Uses varied flow function Quasiunsteady moving waves Quasiunsteady moving waves Design channel for outflow—Implicit method Design channel for outflow—Implicit method Design channel for outflow—Upstream gate instantly raised Design channel for outflow—Upstream gate instantly raised As in previous program with uniform flow downstream Moving hydraulic jump Moving hydraulic jump Makes table of geometry of natural channel Makes table of geometry of natural channel Obtains A, P, V, α, and β in a natural channel using cubic splines (Example Problem 5.1) Completely solves Manning’s equation for natural channels Completely solves Manning’s equation for natural channels Solves gradually varied flows in natural channels Solves gradually varied flows in natural channels Handles compound section in natural channels Design of subcritical transition (Example Problem 5.2)
415 419 423 431 434 440 444 451 454 456 458 459 462 465 470 476 486 493 500 505 513 526 541 543 547 552 556 560 564 570 573 574 581 587 663 665 669 673 678 686 691 699 723
Computer Programs with Listings of Code
EXPR43.FOR EXPR5_5.FOR SUBMESER.FOR SUBMESER.C LENSIDEW.FOR OPTIMAES.FOR Chapter 6 EPRB6_2.FOR EPRB6_2A.FOR EPRB6_2B.FOR UNTYPE4.FOR EPRB6_5.FOR EPRB6_5A.FOR HVARYX.FOR YTIMEX.FOR YTIMEX.PAS YTIMEX.C MAXFLOW.FOR MAXFLOW.C UNSTGT.FOR UNSTGT.PAS SURGMO6.FOR GATETR.FOR GATETR.C GATEDWU.FOR GATETRDU.FOR Chapter 7 REGUL.FOR BOUNDY.FOR UNSCHG.FOR HARTREE.FOR SOLVE.FOR IMPLICIT.FOR IMPLICBC.FOR IMPLICAL.FOR
Transition from trapezoidal to rectangular channel (Example Problem 5.3) Obtains series of solution of transition problem (Example Problem 5.5) Solves flow through series of submerged gates Solves flow through series of submerged gates Find the length of side weir (ups and downs conditions given) Obtains least cost design for trapezoidal channel (total cost) Unsteady flow—Solves Example Problem 6.1 Obtains roots Obtains roots Determines unsteady depths and velocity—method of characteristics Obtains time‑dependent solution (Example Problem 6.5) Obtains time‑dependent solution (Example Problem 6.5) Solves dispersive waves in which H decreases (Example Problem 6.6) Time‑dependent solution at several positions Time‑dependent solution at several positions Time‑dependent solution at several positions Solves for the maximum flow rate in trapezoidal and circular channels Solves for the maximum flow rate in trapezoidal and circular channels Solves surge movement downstream from gate Solves surge movement downstream from gate Solves six equations that describe wave movement from closing gate Solves partial instant closing of gates in trapezoidal channels Solves partial instant closing of gates in trapezoidal channels Solves instant drop of gate followed by slow movement (Example Problem 6.24) Solves instant drop – dQ/dt thereafter, solves 11 variables Subroutine for regular grid points Subroutine for upstream and downstream boundary grid points Solves the St. Venant equations using the method of characteristics Solves the St. Venant equations Using the Hartree method, i.e., characteristics with fixed time steps Subroutine for solving equations Solves the St. Venant equations using the direct implicit method Solves the St. Venant equations using the Gauss–Seidel successiveoverrelaxation iterative method Solves the St. Venant equations using the Crank–Nicolsen Newton implicit method (Example Problem 7.6)
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725 739 765 769 776 782 838 841 841 844 845 847 849 877 878 879 894 896 903 904 908 916 918 921 928 969 973 979 993 1006 1021 1037 1049
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PREISDBS.FOR PREISM6.FOR Appendix A GEONAT.PAS Appendix B NEW3EQ.FOR MANNINGC.PAS LAGR.PAS TRAPR.FOR TRAPR.C SIMPR.FOR SIMPR.C TRAP1.FOR TRAP1.C SIMP1.FOR SIMP2.FOR PRESFOR.FOR APPB4.FOR SPLINE3.FOR SPLINE3.C SPLINE3.PAS SPLINESU.FOR CIRCSPLI.FOR MAINSOL.FOR MAINSOL1.FOR LAGUSUB.FOR Appendix C APPEC1.FOR APPEC2.FOR APPC3.FOR CHANRE2.C CHANRE2O.C APPDVERK.FOR APPERUK.FOR APPCRKT.FOR RUKUST.FOR APPCR4A.FOR APPCRKT.FOR RUKUST.C RUKUSTS.C
Computer Programs with Listings of Code
Solves the St. Venant equations using the Preissmann double sweep method Solves the St. Venant equations using the Newton method to solve the Preissmann equations
1070 1078
Generates geometric properties of an irregular channel
1117
Solves three simultaneous equations Solves Manning’s equation Implements the Lagrange’s formula Implements the trapezoidal rule with subroutine Implements the trapezoidal rule with subroutine Implements Simpson’s rule with subroutine Implements Simpson’s rule with subroutine Uses subroutine TRAPR.FOR to solve problem Uses subroutine TRAPR.FOR to solve problem Integrated pressure above gate (Example Problem B.2) Integrated pressure above gate (Example Problem B.2) Solves ODE to integrate pressure above gate (Example Problem B.3) Combined solution of dY/dx and Simpson’s rule (Example Problem B.4) Implements interpolation with cubic splines Implements interpolation with cubic splines Implements interpolation with cubic splines Subroutine that uses cubic splines Examines how cubic spline duplicates circle (Example Problem B.6) Program that demonstrates use of subroutine SOLVEQ Program that demonstrates use of subroutine SOLVEQL Subroutine that implements Laguerre’s method
1127 1130 1134 1138 1138 1139 1140 1140 1141 1142 1142
Uses subroutine ODESOL (Example Problem C.1) Uses subroutine ODESOL (Example Problem C.2) Uses subroutine ODESOL (Example Problem C.3) Solves problem of flow from reservoir into trapezoidal channel Solves problem using procedure solodesc Uses DVERK to solve problem of flow from reservoir to breakingrade Subroutine that uses Runge–Kutta method Subroutine that uses Runge–Kutta method Subroutine that implements the Runge–Kutta method Uses subroutine RUKU4SA to solve a problem Uses subroutine RUKUST to solve a problem Procedure to use Runge–Kutta method Uses RUKUST.C to solve a problem
1168 1170 1172 1178 1179
1143 1145 1151 1152 1153 1155 1156 1160 1162 1163
1182 1184 1185 1186 1188 1190 1191 1192
Computer Programs with Listings of Code
RKFEHL.FOR RKFEHLF.C RKFEHLF1.FOR RKFEHL1.C MRKFEHL.FOR RKFEHLSG.FOR RKFEHLSC.C PJECTILE.FOR PJECTILF.FOR PJECTILE.C PJECTILF.C
Subroutine that implements the Runge–Kutta–Fehlberg method Subroutine that implements the Runge–Kutta–Fehlberg method Uses subroutine RKFEHL.FOR to solve problem Uses procedure RKFEHL.C to solve problem Uses procedure RKFEHL.C to solve problem Subroutine with alternative arguments Subroutine with alternative arguments Solves projectile problem using RUKUST Solves projectile problem using FKFEHL Solves projectile problem using rukust Solves projectile problem using RKFEHLSC
xxix
1194 1194 1195 1196 1198 1199 1200 1202 1202 1203 1204
Terminology, 1 Dimensions, and Review of Basic Fluid Mechanics 1.1 Introduction Openchannel flow is distinguished from closedconduit flow by the presence of a free surface, or interface, between two different fluids of different densities. The two most common fluids involved are water and air. The presence of a free surface makes the subject of openchannel flow more complex, and more difficult to compute commonly needed information about the flow, than closedconduit flow, or pipe flow. In pipeflow problems, the crosssectional area of the flow is known to equal the area of the pipe. In openchannel flow, the area depends upon the depth of flow, which is generally unknown, and must be determined as part of the solution processes. Coupling this added complexity with the fact that there are more openchannel flows around us than there are pipe flows, emphasizes the need for engineers, who plan work in waterrelated fields, to acquire proficiency in openchannel hydraulics. The wide use of computers in engineering practice reduces the need for graphical, table lookup, and other techniques learned by engineers who received their training a decade ago. Technical fields apply very specific meanings to words. With a knowledge of these meanings it is possible for individuals educated in that field to communicate much more effectively and, with fewer words, convey clear and concise information. Often these words have a more general, less concise, meaning in their general use and, therefore, have a less concise meaning for the public at large. Other words are coined especially for a technical discipline. This chapter introduces the terminology used in openchannel flow. It is important that some terminology be fully mastered to effectively read and understand the remainder of this book, and to converse verbally, or in writing with openchannel hydraulic engineers.
1.2 One, Two, and ThreeDimensional Flows The dimensionality of a flow is defined as the number of independent space variables that are needed to describe the flow mathematically. If the variables of a flow change only in the direction of one space variable, e.g., in the direction along the channel x, then the flow is described as one dimensional. For such flows, variables such as depth Y and velocity V are only functions of x, i.e., Y(x) and V(x). If the variables of the flow change in two directions, such as the position along the channel x, and the position from the bottom of the channel y, or the position across the channel z, then the flow is described as two dimensional. For twodimensional flows the mathematical notation of variables contains two arguments such as V(x,y) and Y(x,y), or V(x,z) and Y(x,z). If the variables of the flow change in three directions, such as the position along the flow, with the vertical position within the flow, and the horizontal position across the flow, then the flow is three dimensional. If a Cartesian coordinate system with axes x, y, and z (note here that lower case y is not the depth of flow, Y) is used, then threedimensional flows are described mathematically by noting that the variables of the flow are a function of all three of these independent variables, or the velocity, for example, is denoted as V(x,y,z), to indicate that its magnitude varies 1
2
Open Channel Flow: Numerical Methods and Computer Applications
with respect to x, y, and z in space, and since velocity is a vector, its direction also depends on x, y, and z. If the velocity also changes with time, this additional dependency will be denoted by V(x,y,z,t). A twodimensional flow that changes in time would have its velocity described as V(x,y,t) or V(x,z,t), and thus depends upon three independent variables and, from a mathematical point of view, is three dimensional. However, in fluid mechanics such flows are called two dimensional, unsteady. A flow with V(x,z) is called two dimensional, steady. The notation for the variables of a onedimensional, unsteady flow consists of Y(x,t), V(x,t), etc. Since the computations needed to solve a onedimensional problem are much simpler than a twodimensional steady problem, we wish to define the flow as one dimensional, if assuming this does not deviate too much from reality. The equations describing one dimensional, nontimedependent or steadystate flows, are either just algebraic equations, or ordinary differential equations, whereas equations needed to describe twodimensional flows are almost without exception partial differential equations. A onedimensional flow that does change with respect to time must be described mathematically by partial differential equations also. Furthermore, because of the complex nature of the two and threedimensional flow equations that describe real flows, a very small number of closed form solutions to the mathematical boundary value problems governed by these partial differential equations are available for theoretically simple flows. Therefore, it is necessary to resort to approximate numerical methods to solve general two and threedimensional flows. The term hydraulics of open channel flow is often used for onedimensional free surface flow. The assumption made, that allows the flow in an open channel to be defined as one dimensional, is that the average velocity at a cross section can be used, and it is not necessary to be concerned with variations of the velocity with depth, or position across the flow. Based on this assumption the velocity, V(x) at any position along the channel equals the flow rate Q at this section divided by the crosssectional area. Since the velocity varies from the bottom of a channel to the top at any position, and this velocity distribution may vary across the channel, correction factors are sometimes utilized to provide more accurate values of the kinematic energy per unit weight, and the momentum flux when the average velocity is used in the appropriate formula. However, the assumption is retained that the flow is one dimensional. These correction factors are discussed later in this chapter.
1.3 Steady versus Unsteady Flow A fluid flow is steady if none of the variables that can be used to describe the flow change with respect to time. Mathematically, steady flow is described by having partial derivatives of such variables as the depth of flow, the velocity, the crosssectional area, etc., with respective to time all equal to zero, e.g., ∂Y/∂t = 0, ∂V/∂t = 0, ∂A/∂t = 0. A fluid flow is unsteady if any of the variables that describe the flow changes with respect to time. Thus a flow is unsteady if the depth at a given position in an openchannel flow changes with respect to time. Mathematically, a flow would be determined to be unsteady if any of the partial derivatives of any variable that describe the flow such as the depth, the velocity, the crosssectional area, etc., with respect to time is different from zero. Generally, if one variable changes with respect to time all variables do. Since for steady flows the variables that describe the flow are not a function of time, the independent variable t is not included in describing the flow, mathematically. If the flow is one dimensional in space, then the variables of the flow are only a function of that space variable. Thus the depth varies only as a function of the position variable x, or Y(x). If the flow is unsteady then the dependent variable Y varies with x and t, and this is denoted as Y(x,t). For a threedimensional timedependent flow, the depth is a function of the Cartesian coordinate system x,y, and z as well as time, and therefore the depth is denoted mathematically as Y(x,y,z,t).
Dimensions, Terminology, and Review of Basic Fluid Mechanics
3
1.4 Uniform versus Nonuniform Flow A flow is uniform if none of the variables that describe the flow vary with respect to position, x along the channel. Therefore, the flow is uniform if the depth, velocity, and crosssectional area are all constant. When dealing with onedimensional openchannel hydraulics, uniform flows are not dependent on x. In fact, there is not such a thing as one, two, or threedimensional uniform flow since the variables of the flow do not change with respect to x, y, or z. In theory, there could be a timedependent uniform openchannel flow, but in practice such flows never actually exist. To have a uniform unsteady flow, the depth would have to increase (or decrease) at the same rate throughout the entire length of channel so that at all times neither the depth nor the velocity changes with respect to position, but is constantly changing with time. Mathematically, a flow is uniform if ∂Y/∂x = 0, ∂V/∂x = 0, ∂A/∂x = 0, etc., throughout the flow. Setting partial derivatives with respect to the position x equal to zero, and having this zero occur throughout the flow is synonymous with stating that there is no dependency upon x. A flow is nonuniform if any of the variables of the flow vary from position to position. Thus, in a onedimensional channel flow, if the depth either increases or decreases from one position to another in a channel, the flow is nonuniform. A nonuniform flow may be one, two, or three dimensional, and may be either steady or unsteady. The flow over the crest of a dam’s spillway is nonuniform but steady if the flow rate does not change. If, however, the flow rate is either increasing or decreasing with time then the flow is nonuniform and unsteady. Nonuniform flows will be further subdivided into gradually varied, rapidly varied, and spatially varied. When the radius of curvature of the streamlines is large, e.g., the streamlines are nearly straight, such that the normal component of acceleration can be ignored, a nonuniform flow will be referred to as gradually varied. In a gradually varied openchannel flow the pressure will increase in the vertical direction just as it does in the same fluid for a uniform flow. This variation of pressure with depth is hydrostatic. A rapidly varied flow occurs when the change in depth is too rapid to ignore the normal acceleration component of the flow, and the pressure distribution is not hydrostatic. Another way of looking at the difference between gradually varied and rapidly varied flow is that it is possible to use onedimensional hydraulic equations for gradually varied flows, but rapidly varied flows are two dimensional, or three dimensional. Because of the complexities involved in solving two and threedimensional openchannel flows, often rapidly varied flows are handled by utilizing onedimensional openchannel equations that are modified by experimental coefficient that account for the deficiencies in the onedimensional assumption. This utilization of experimental coefficients distinguished onedimensional hydraulics from pure fluid mechanics. There is no parameter of measurement of the radius of curvature, or other characteristic of the flow with a threshold value that separates a gradually varied from a rapidly varied flow. Rather the distinction is subjective. Almost everyone would agree that the flow over a dam’s spillway crest is rapidly varied, whereas the flow in the channel upstream from the dam is gradually varied. Likewise the flow immediately downstream from a sluice gate, where the flow is contracting rapidly from the gate height, is rapidly varied whereas the flow both upstream and further downstream from the gate is gradually varied until the position downstream from the gate where a hydraulic jump occurs, should this be the case. The flow through the hydraulic jump is rapidly varied, again. Whether a flow through a transition between two channels of different sizes is rapidly or gradually varied may be debatable. The classification will depend upon how rapidly the transition changes the channel’s cross section, and how accurate it is necessary that the computed results correspond to the actual flow characteristics. An abrupt enlargement or an abrupt contraction will cause a small section of rapidly varied flow to occur. Whereas the solution of onedimensional hydraulic equations for gradually varied flow may represent an accurate method for solving the variation of depth across a 50 ft long smoothly formed transition. If so, the flow is gradually varied. Spatially varied flows are those portions of the main channel flow over which either lateral inflow or lateral outflow occurs. Therefore, in a spatially varied flow, the flow rate changes with
4
Open Channel Flow: Numerical Methods and Computer Applications
position along the channel. Strictly speaking, the joining of two channels creates a spatially varied flow for a short distance. Generally, however, spatially varied flows occur where the lateral inflow or outflow is over some length of channel. A side weir that runs parallel to the direction of the channel over which discharge occurs creates a section of spatially varied flow. In this case spatially varied flow has distributed outflow from the channel, and the flow rate decreases in the direction of the main channel flow. Water accumulating from rainfall over a roadway surface and flowing into the gutters along the sides of the roadway causes a spatially varied openchannel flow in the gutters. In this case the channel flow in the gutter has a lateral inflow, and the flow rate increases in the direction of flow. As this gutter flow crosses the grates of a storm drain a spatially varied outflow occurs in the gutter, but in the storm drain that receives the flow from the grates, a spatially varied inflow occurs. If all the gutter flow can enter the storm drain then the channel flow in the gutter terminates at the end of the spatially varied flow.
1.5 Prismatic versus NonPrismatic Channels Definitions that are closely associated with uniform and nonuniform flows, but apply to the channel rather than the flow in the channel, are prismatic and nonprismatic channels. A prismatic channel has the same geometry throughout its length. This may consist of a trapezoidal section, a rectangular section, a circular section, or any other fixed section. If the shape and/or size of the section changes with position along the channel, the channel is referred to as a nonprismatic channel. In theory, it is possible for a natural channel created by nature to be prismatic. However, in practice natural channels are nonprismatic.
1.6 Subcritical, Critical, and Supercritical Flows An openchannel flow is classified according to how the average velocity, V, of the flow compares with the speed, c, of a small amplitude gravity wave in that channel. If V is less in magnitude than c, then the flow is subcritical. If V is greater in magnitude than c, then the flow is supercritical, and if V = c, then the flow is critical. The speed of a small amplitude gravity wave is given by
c=
gA = gYd T
(1.1)
where g is the acceleration of gravity A is the cross section of the flow T is the top width of the flow Yd is the hydraulic depth, A/T Subcritical flows behave differently from supercritical flow because in a subcritical flow the effect of downstream changes are noted by the fluid and it adjusts in anticipation of that downstream occurrence. Thus, if an obstruction exists in a subcritical flow the depth will gradually increase to the depth needed to pass by the obstruction. This signal that something exists downstream is propagated continuously to the upstream flowing fluid by gravity waves. These gravity waves can travel upstream because the velocity of flow is smaller than their speeds. In supercritical flows the effect of changes cannot travel upstream because the velocity in the channel exceeds the propagation speed of gravity waves. Therefore, flow does not adjust itself for downstream conditions. For example, if a channel containing a supercritical flow ends abruptly, the depth at the end of the channel at the free overfall will be the same as if the channel had continued. If the flow were subcritical, the depth would decrease toward critical depth at the end of a free overfall, however.
Dimensions, Terminology, and Review of Basic Fluid Mechanics
5
As a consequence of whether gravity waves can move upstream or not, subcritical flows have their control “downstream,” whereas supercritical flows are “upstream controlled.” An example of both downstream and upstream control exists at a gate in a channel. Upstream from the gate the flow must be subcritical because the gate, which is downstream, controls the depth, velocity, area, etc., of the flow. Downstream from the gate the flow will be supercritical, and the gate determines the magnitude of the variables of the flow. If the gate is lowered, for example, it will decrease the downstream depth while increasing the downstream velocity, and has the opposite effect on the upstream flow. The Froude number, Fr, is the ratio of the velocity in a channel divided by the speed of propagation, or celerity of a small amplitude gravity wave c, or
Fr =
V = c
V gA/T
=
Q2T gA3
(1.2)
Therefore, the determination of whether a flow is subcritical, critical, or supercritical is commonly accomplished by computing the Froude number of the flow. If this value is less than unity then the flow is subcritical. If the Froude number is exactly equal to one, then the flow is critical, and if the Froude number is larger than unity, then the flow is supercritical. It turns out that the Froude number is also the ratio of inertia to gravity forces. A more indepth treatment of subcritical and supercritical flows and their associated Froude numbers is given in subsequent chapters.
1.7 Turbulent versus Laminar Flow The Reynolds number, or the ratio of inertia to viscous forces, is used to distinguish whether a flow is laminar or turbulent. For openchannel flows the Reynolds number is defined by using the hydraulic radius, R h, or the crosssectional area A divided by the wetted perimeter P as the length variable. Thus the Reynolds number is defined by
Re =
VR h Q ρVR h = = ν νP µ
(1.3)
where V is the average velocity of the flow Q is the volumetric flow rate μ is the absolute viscosity ν is the kinematic viscosity of the fluid If the Reynolds number is less than 500, the flow is laminar. Otherwise, the flow is turbulent. Often 4Rh is used as the length parameter in Reynolds number for channel flows because this is equivalent to the diameter of a pipe, e.g., the hydraulic radius of a pipe is Rh = (πD2/4)/(πD) = D/4. When using this latter definition, the numerators on the left side of Equation 1.3 should be multiplied by 4. Laminar flows are rare in open channels if the fluid is water. Examples of laminar flow might be the sheet flow over the surface of a watershed produced by precipitation, or the lateral flow over the crest of a roadway as it moves toward the side gutter. To have laminar water flow in an open channel, the depth generally has to be very small, in conjunction with a not too large velocity, since the kinematic viscosity of water is about 1.2 × 10 −5 ft2/s.
1.8 Review of Basic Fluid Mechanics Principles The rest of this chapter provides basic theory upon which the book is based. The presentation assumes that you are familiar with fluid mechanics, and therefore the remainder of this chapter should be considered a review; but this review is slanted toward openchannel hydraulics. Books dealing with engineering fluid mechanics will contain a more thorough treatment of this subject material.
6
Open Channel Flow: Numerical Methods and Computer Applications
The application of fluid mechanics in solving engineering problems involves a thorough understanding of the following four related items: (1) Physical properties of fluids, e.g., density, specific weight, viscosity, surface tension, and how these cause pressures to change, resistance to motion, etc. (2) The conservation of mass, or the continuity principle. (3) The conservation of energy and its dissipation into nonrecoverable forms. (4) Utilization of momentum fluxes as vector quantities to deal with external forces on fluid in motion. A section for each of these four important subjects follows as the rest of this chapter. The specific application of these subjects to the flow of water in open channels constitutes the remaining chapters of this book. The review in this chapter will introduce the symbols that will be used through the rest of the book, and subsequent chapters are written assuming that you are acquainted with these symbols and their meanings. The problems at the end of this chapter have been listed under four similar headings. Problems given under subsequent headings generally also require an understanding of the principles involved in the previous headings.
1.9 Physical Properties of Fluids and Their Effects on OpenChannel Flows In solid mechanics, since the object being dealt with generally stays together its mass or weight is used when dealing with the effects it has on its environment. With fluids, however, total mass or total weight, generally, have no significance since these totals are directly related to the length of time the flow has been occurring. Rather, mass per unit volume or weight per unit volume are used. These quantities are the density ρ and the specific weight γ, or the fluid respectively. In the SI (International System of Units) the density ρ is given in kilograms per cubic meter, e.g., ρ (kg/m3). In ES units (English System of Units) the density is in slugs per cubic foot, ρ (Slug/ft3). Specific weight γ is related to density through Newton’s second law of motion,
Force = Mass × Acceleration.
Weight is a force due to resisting gravity, and therefore γ = ρg in which g is the acceleration of gravity and equals 32.2 fps2 in ES units or equals 9.81 m/s2 when using SI units. (With more digits of precision g = 32.174049 fps2 in ES units, and g = 9.80685 m/s2 in SI units.) In SI units the specific weight is generally given as kilonewtons per cubic meter, e.g., γ (kN/m3) and in ES units γ is given in pounds per cubic foot, e.g., γ (lb/ft3). The density and specific weight for water vary moderately with its temperature as shown in Table 1.1. The weight of a fluid causes pressure to increase with depth. If z is taken as the vertical coordinate, positive upward against gravity, from a selected datum and there is no motion in the fluid, then the pressure varies according to the hydrostatic law,
dp = − γ = −ρg dz
(1.4)
and if the fluid is incompressible (which is another way of indicating that γ and ρ are constant v alues), then this equation integrates to
p2 − p1 = γ (z 2 − z1 ) = ρg(z 2 − z1 )
(1.4a)
When the fluid is a liquid with a free surface it is convenient to use a coordinate h (e.g., independent variable) that has its origin on this surface and is positive downward. h is related to z by dh = −dz, and, therefore, Equations 1.4 and 1.4a become as following with p considered a function of h:
32.0 41.0 50.0 59.0 68.0 77.0 86.0 95.0 104.0 113.0 122.0 131.0 140.0 149.0 158.0 167.0 176.0 185.0 194.0 203.0 212.0
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
999.8 1000.0 999.7 999.1 998.2 997.0 995.7 994.1 992.2 990.2 988.0 995.7 983.2 980.5 977.8 974.9 971.8 968.6 965.3 962.2 958.4
3
1.940 1.940 1.940 1.939 1.937 1.934 1.932 1.929 1.925 1.921 1.917 1.932 1.908 1.902 1.897 1.892 1.886 1.879 1.873 1.867 1.860
Slug/ft
kg/m
°F
°C
3
Density, ρ (Mass/Vol.)
Temperature
9.805 9.807 9.804 9.798 9.789 9.777 9.764 9.749 9.730 9.711 9.689 9.764 9.642 9.615 9.589 9.561 9.530 9.499 9.466 9.436 9.399
kN/m 3
62.42 62.43 62.41 62.37 62.32 62.24 62.16 62.06 61.94 61.82 61.68 62.16 61.38 61.21 61.04 60.86 60.67 60.47 60.26 60.07 59.83
lb/ft 3
Specific Weight (Weight/Vol.)
Table 1.1 Properties of Water Related to Temperature
1.785 1.518 1.307 1.139 1.002 0.890 0.798 0.720 0.653 0.596 0.547 0.504 0.466 0.433 0.404 0.378 0.354 0.333 0.315 0.294 0.282
2
(N · s/m )
μ × 10 3
3.728 3.170 2.730 2.379 2.093 1.859 1.667 1.504 1.364 1.245 1.142 1.053 0.973 0.904 0.844 0.789 0.739 0.695 0.658 0.614 0.589
(lb · s/ft ) 2
5
μ × 10
Abs. Viscosity, μ
1.785 1.518 1.307 1.140 1.004 0.893 0.801 0.724 0.658 0.602 0.554 0.506 0.474 0.442 0.413 0.388 0.364 0.344 0.326 0.306 0.294
(m /s) 2
v × 106
1.922 1.634 1.407 1.227 1.080 0.961 0.863 0.780 0.708 0.648 0.596 0.545 0.510 0.475 0.445 0.417 0.392 0.370 0.351 0.329 0.317
(ft /s) 2
v × 105
Kinematic Viscosity, v
0.61 0.87 1.23 1.70 2.34 3.17 4.24 5.62 7.38 9.58 12.33 15.73 19.92 25.00 31.16 38.53 47.34 57.58 70.10 79.36 101.33
kN/m2 (abs)
0.09 0.13 0.18 0.25 0.34 0.46 0.61 0.82 1.07 1.39 1.79 2.28 2.89 3.63 4.52 5.59 6.87 8.35 10.17 11.51 14.70
psia
Vapor Pressure, pv
Dimensions, Terminology, and Review of Basic Fluid Mechanics 7
8
Open Channel Flow: Numerical Methods and Computer Applications
dp = γ = ρg dh
(1.4b)
p = γ h = ρgh
(1.4c)
and
In the last equation, the pressure p assumes that atmospheric pressure is the reference base pressure, and therefore p is gage pressure. Atmospheric pressure must be added to gage pressure to get absolute pressure. A few example problems follow that illustrate the use of Equations 1.4; however, it is worthwhile to examine how the pressure varies with depth in a channel with a steep bottom slope that contains a flow, before leaving the subject of pressure within a fluid caused by gravity. If the angle that the channel bottom makes with the horizontal is denoted by θ (note that the tan θ = So the slope of the channel bottom), then the normal depth Yn is related to the vertical depth Yv by the cosine of this angle, or Yn = Yvcosθ, as shown in the sketch below. Since gravity acts on fluid elements in the vertical direction, the height water would rise in a piezometer with its opening pointing downward, and located at any position will equal γhncosθ = gρhncosθ = γhvcos2θ, where hn is the normal distance from the water surface down to the point being considered, and hv is the vertical distance from the water surface to the point as shown on the sketch. Thus, the pressure at the bottom of the channel equals γYncosθ = γYvcos2θ.
Yv
θ
Yn
hn
hv
p = γY c n osθ tan(θ) =
So
gn
g
1
An alternative, to the above method for determining the pressure distribution, is to note that the acceleration of gravity g acts in the vertical direction on any fluid element. The component of the acceleration in the normal direction is gn = gcosθ, and therefore p = ρgnhn = ρghncosθ = γhvcos2θ. Should the bottom of the channel be curved instead of having a constant bottom slope, then the normal acceleration in the fluid due to the curvature of the channel bottom will affect the pressure distribution as illustrated in the sketches below. The fluid at the very bottom of the channel will have a normal acceleration equal to v2/r and will add to the gravitational acceleration when the curvature is concave upward and subtract from g when the curvature is concave downward. Note that since the fluid is not free falling under gravity’s acceleration, upward accelerations add to g and downward accelerations subtract from g. The component of acceleration on a fluid particle in the direction of the radius of curvature equals g cosθ + v2/r, where θ now is the angle between the vertical and the radius of curvature, and therefore at any point in the fluid dp/dr = ρ(g cosθ + v2/r). Since the radii of curvature of streamlines other than the bottom streamline will not, in general, equal the radius of curvature of the channel, it is not possible to determine the pressure distribution without making assumptions related to the radius of curvature between the channel bottom and the free surface.
Dimensions, Terminology, and Review of Basic Fluid Mechanics
9
Concave downward Concave upward
The bulk modulus Ev– is a similar quantity in dealing with fluids, as is the modulus of elasticity when dealing with solids. It has the same units as pressure and is the reciprocal of the compressibility. The bulk modulus equals the change in pressure needed to cause a decrease in volume divided by that decreased volume divided by the volume of fluid involved, or Ev = −
∆p dp dp =− = ∆V / V dv/v dρ/ρ
(1.5)
The second and third parts of Equation 1.5 are obtained by taking one unit mass of a fluid so V becomes the specific volume (volume/unit mass) v which is the reciprocal of the density ρ, and therefore dv/v = −dρ/ρ. The bulk modulus of pure water equals 320,000 psi. A small amount of free air entrained in the water can significantly reduce the bulk modulus of the mixture. If x is taken as the fraction of free air (not dissolved) mixed in water, then the density of the air–water mixture is given by ρm = xρa + (1 − x)ρw in which ρa and ρw are the densities of air and water, respectively. To obtain the bulk modulus of this mixture note from Equation 1.5 that ΔV = −V Δp/Ev–. The change in volume of the mixture ΔVm will be the sum of the changes in volume of the air and the water or, ΔVm = ΔVa + ΔVw. By taking an original volume of a unit amount (V = 1), this last expression for the change in volume becomes the following upon substituting for the ΔV’s: −
(1 − x)∆p ∆p x∆p =− =− Em Ea Ew
or E m =
Ea E w (1 − x)E a + xE w
where Ea is the bulk modulus of air Ew denotes the bulk modulus of water Since the air mass in a water–air mixture generally is an extremely small fraction of the water mass, the air’s temperature will remain constant when it is compressed. The compression of a gas at constant temperature is referred to as an isothermal process. The bulk modulus of a gas undergoing an isothermal process equals its absolute pressure, or Ea = pabs. Therefore, the bulk modulus for a small fraction x of air entrained in water becomes
Em =
pabs E v (1 − x)Pabs + xE v
(1.6)
where the symbol Ev has been used for the bulk modulus of water again instead of Ew. The table below indicates how the bulk modulus is affected by a small fraction of air. Note that a small amount of air entrainment reduces the density of the mixture very modestly, but reduces the bulk modulus by orders of magnitude. These values assume air at atmospheric pressure of 14.7 psia, and at a temperature of 60°F, so its density equals 0.00237 slugs/ft3.
10
Open Channel Flow: Numerical Methods and Computer Applications Fraction of air x Density of mixture ρm Bulk modulus of mixture Em × 10−2 (psi)
0.0000 1.940 3,200
0.0001 1.940 1,007
0.0005 1.939 26.927
0.001 1.938 7.115
0.005 1.930 2.913
0.01 1.921 1.463
0.10 1.746 0.147
Example Problem 1.1 A sensitive pressure transducer is used to record the pressure at the bottom of a river that carries a sediment load that causes the density to increase linearly from 1.945 slug/ft3 on the surface at a rate of 0.012 slug/ft3/ft of depth. Develop the equation that gives the depth of flow from this pressure reading. What is the error if the sediment load is ignored, and the density taken equal to 1.94 and a pressure of 3.5 psi is recorded? Solution The problem is solved by defining ρ = 1.945 + 0.012h, substituting this into Equation 1.4b, separating variables, and integrating. The result is p = g(1.945h + 0.006h 2 ),
where p is in pound per square foot h is in feet
Applying the quadratic formula gives the depth h as a function of the pressure reading as p h = −162.083 + 83.333 3.783 + 0.024 g
1/ 2
.
Substituting p = 3.5 × 144 into the above equation gives h = 7.856 ft. If ρ = 1.94 (constant), then p = 62.4h, or h = 8.077 ft, or an error of +0.221 ft. Example Problem 1.2 Water is being drawn from a well whose water level is 1500 ft below the ground surface. Determine the significance of the compressibility of water in determining the head the pumps must supply if the friction loss in the well pipe is 25 ft, and the pump must supply 80 psi of pressure at the ground surface. Assume the bulk modulus for water remains constant and equal to 320,000 psi. Solution First, it is necessary to determine the relationship between density and pressure since the effects of increasing density of the fluid are to be taken into account. The definition of a fluid’s bulk modulus Ev– provides this relationship since Ev– = −Δp/(ΔV – /V – ) = ρdp/dρ, in which V – is fluid volume and ΔV is the change in this volume due to the pressure increase Δp. Separating variables – in this equation and integrating the density from ρo (the density at atmospheric pressure, which will be taken as 1.94 slugs/ft3) to ρ and integrating the pressure from 0 (atmospheric) to p gives
p ρ = Εxp ρo Εv
(An alternative to integrating between the two limits is to just integrate and add a constant to the resulting equation. This constant can then be determined from a known condition. In this case the known condition is that the density equals ρo = 1.94 slugs/ft3 when the pressure, p = 0.) Substituting this expression into the hydrostatic Equation 1.4b gives
p dp = gρ = gρo Εxp dh Εv
Dimensions, Terminology, and Review of Basic Fluid Mechanics Again separating variables and integrating gives
gρ h p = − E v Ln 1 − o Εv
The value of h to use in this equation is the sum 1500 ft, the pressure head needed at the surface, 80 × 144/(32.2 × 1.94) = 184.41 ft, and the frictional loss of 25 ft, or h = 1709.41 ft. Substituting this value for h in the above equation gives a pressure of 742.41 psi at the pump in the well. If the compressibility of the water is ignored, the pressure is obtained from p = gρh/144 = 741.55 psi. This small difference of about 1 psi points out that the compressibility of water is not very significant for most engineering applications. An exception is whenever the speed of pressure waves are concerned, because for such applications a small amount of free air in water can dramatically change the speed of this wave as noted above. Example Problem 1.3 Water is flowing at a rate of 800 cfs in a 6 ft wide rectangular channel at a constant depth of 4 ft. The bottom of the channel changes slope by means of a circular arc of radius 50 ft. The depth of flow through this arc remains constant at 4 ft. Assuming that the streamlines of this flow are concentric circles (have the same center of curvature), determine the pressure along a radial line at the beginning of the arc where it connects to the straight upstream channel that has a bottom slope So = 0.15, and also determine the pressure distribution along a vertical radial line. Solution Using the upstream bottom slope θ = arctan(0.15) = 8.531°. The velocity in the channel equals 33.333 fps and therefore,
dp (33.333)2 = ρ g cos(8.531°) + dr r
or after integrating p from 0 to p as r is integrated between 46 and r, this results in
r p = ρ 31.844r + 1111.11 Ln − 1464.81 46
to give the pressure distribution as a function of r. On the bottom r = 50, and the pressure here is p = 426.75 psf = 2.964 psi, at the channel bottom, where r = 50 ft. Ignoring the added pressure due to the radius of curvature, the pressure at the bottom of the channel is p = 4γ/144 = 1.73 psi. This latter amount is 1.234 psi too small.
r
Flipbucket spillways cause highvelocity water to have a considerable normal component of acceleration, as illustrated in this example. To find the added forces on such structures caused by changing the direction of the fluid can be handled easier through the use of the momentum principle that will be discussed in detail in Chapter 3. Example Problem 1.4 A circular tank with a 3 m diameter (or radius ro = 1.5 m) initially containing water to a depth of ho = 4 m is rotated at an angular velocity ω = 5 rad/sec for a long time until the water is brought into solid body rotation. Determine the shape of the water surface in the tank, and its depth at the center and outside walls of the tank.
11
12
z
Zmax
Water surface
Zmin
4m
Static equilibrium
Open Channel Flow: Numerical Methods and Computer Applications
r ω = 5 rad/s D=3 m
Solution In this problem the fluid (water) is being accelerated. The equation for fluid statics ∂p/∂z = − ρg (with ∂p/∂x = 0 and ∂p/∂y = 0) which accounts only for gravitational acceleration in the zdirection (vertical) can be generalized to the following: ∂p/∂z = −ρ(g + az), ∂p/∂x = −ρax, ∂p/∂y = −ρay. When applied to the rotating tank, with z as the vertical coordinate, and r the radial coordinate, these equations become ∂p/∂z = −ρg and ∂p/∂r = −ρa r = ρrω2. From the definition of a differential dp = (∂p/∂r)dr + (∂p/∂z)dz. Along any constant pressure surface, such as the free surface, dp = 0, and therefore the slope of the water surface at any radial position r is dz/dr = − ar/(g + az) = rω2/g. Separating variables and integrating gives the following parabolic relationship between z and r for the water surface: z=
r 2ω 2 +C 2g
The constant C can be evaluated by noting that the same amount of water exists in the tank after rotation as before, or ro
∫ 0
ro
ω 2 ro4 ro2 ω2r2 + 2πrz dr = 2π r + C dr = 2π 2 8g 2g
∫ 0
C = πh o ro2
or solving for C
(1.5 × 5) = 4 − 1.433 = 2.567m. ro 2 ω 2 =4− 4g 4(9.81) 2
C = ho −
At the center axis of the tank the water depth will be zin = C = 2.567 m, and at the outside zmax = C + (ro × ω)2/(2g) = 2.567 + 2.867 = 5.433 m. As a check on the computations, etc., we might note that if the volume of the cylinder formed by zmax as its height has the volume of the paraboloid subtracted from it, then the original volume of the cylinder with a height ho should occur. The volume of a paraboloid equals 1/2 the area of the base time the height. Therefore, Aho = A{zmax − 0.5(zmax − C)} = A(zmax + C)/2, or ho = (zmax + C)/2 = (5.433 + 2.567) = 4 m. Before leaving this problem, it is worth noting that the Bernoulli equation cannot be applied across the streamlines in this rotation tank, e.g., from the inside radius to the outside. If this were done then the total head on the free surface would be the sum of the elevation and velocity heads (since p = 0) or
z+
V2 (rω)2 =z+ =H 2g 2g
13
Dimensions, Terminology, and Review of Basic Fluid Mechanics or z=H−
r 2ω 2 , 2g
but this equation has a constant minus the velocity head whereas the above equation indicated that z was equal to a constant plus the velocity head. The reason is that the Bernoulli equation is based on irrotational flow, whereas this is rotational flow. A free vortex represents irrotational flow, but its equation indicates that the transverse component of velocity v t = Constant/r. The forced rotation in the tank gives vt = rω = r × Constant. Example Problem 1.5 Water is flowing in a natural channel, and the flow rate Q is to be determined by measuring the velocity on the surface and the depth of flow at various positions x across the channel. These measurements have produced the values in the table below. Position, x (m) Depth, Y (m) Velocity, Vs (m/s)
0.0 0.0 0.0
4 4.7 0.35
8 8.3 0.45
12 11.2 0.47
16 12.7 0.49
20 13.3 0.50
24 13.4 0.49
28 12.8 0.50
32 12.0 0.48
36 11.8 0.48
40 11.5 0.48
44 11.7 0.48
48 12.3 0.48
52 13.3 0.49
56 14.5 0.50
60 15.0 0.50
64 14.8 0.45
68 12.7 0.35
72 8.0 0.20
75 0.0 0.0
It has also been determined that the velocity varies according to the same dimensionless profile from the bottom of the channel at any position x according to the data given in the table below. (In this table y′ = y/Y is the dimensionless depth, with y beginning at the bottom at this position x, and likewise v′ = v/Vs is the dimensionless velocity, in which Vs is the velocity on the surface at this position x.) Dimensionless depth y′ Dimensionless velocity v′
0.0 0.04 0.08
0.12
0.20
0.28
0.40
0.52
0.64
0.72
0.80 0.90
1.0
0.0 0.22 0.375 0.500 0.675 0.815 0.955 1.040 1.070 1.068 1.06 1.03
1.0
Solution There are a number of methods that could be used to solve this problem, including plotting the data from the above tables and determining area under the curves. However, since in this book we wish to emphasize the use of the computer to do numerical computations, the problem will be solved by writing a computer program. First, however, let us note that a dimensionless flow rate per unit width q′ can be obtained by integrating the dimensionless velocity profile given in the second table, or q′ = ∫v′dy′, with limit from 0 to 1. The flow rate per unit width q equals the dimensionless q′ multiplied by the depth Y and the surface velocity at any position, or q = ∫v dy = VsY∫v′dy′ = VsYq′. Thus, the flow rate Q, which is the integral of the unit flow rate q times dx, or Q = ∫qdx, can be determined from Q = q′∫VxY dx, with the limits of this integration from 0 to the total width of the section, or 75 m. These integrations will be accomplished by using the subroutine SIMPR, which is described in Appendix B, and implements Simpson’s rule to numerically evaluate integrals. To provide the integrand as a continuous function of the variable being integrated, a cubic spline function will be used, i.e., subroutine SPLINESU, also described in Appendix B is utilized. The program EXPRB1_5 (both in FORTRAN and C) is given below to provide the solution. The following are the key components of this program: (1) The first two READS store the data from the above two tables in arrays, with arrays YP and VP storing the dimensionless velocity profile data in the second table, and arrays X, Y, and VS storing the data
14
Open Channel Flow: Numerical Methods and Computer Applications in the first table. (2) The cubic spline function SPLINESU is called three times to provide (a) the second derivatives d2v′/dy′ 2, (b) the second derivatives d2Y/dx2, and (c) the second derivatives d2Vs/dx2 corresponding to the points in the tables. (3) The subroutine SIMPR is called next to integrate the dimensionless velocity profile. It calls on function subprogram VPROF to provide v′ corresponding to any dimensionless depth y′, and accomplishes this by using the d2v′/dy′ 2 supplied by the first call to SPLINESU. (4) The subroutine SIMPR is called again to provide the integral ∫(VsY)dx, and subprogram Dq provides the arguments for this integration. (5) Finally by multiplying q′ (the result from the first integration) by the result from the second integration the flow rate Q is printed out. (Note that the C program contains the function that supplies what is integrated as argument 1, as mentioned in Appendix B if the name is not equat used; thus allowing the two different arguments vprof (for v′) and qp (for Y*Vs).) Listing of program EXPRB1_5.FOR EXTERNAL VPROF,Dq REAL DUM(30) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX I1=1 I2=2 READ(2,*) NP,(YP(I),VP(I),I=1,NP) READ(2,*) NX,(X(I),Y(I),VS(I),I=1,NX) CALL SPLINESU(NP,YP,VP,D2VP,DUM,0) CALL SPLINESU(NX,X,Y,D2Y,DUM,0) CALL SPLINESU(NX,X,VS,D2VS,DUM,0) CALL SIMPR(VPROF,0.,1.,qPRIM,1.E6,20) WRITE(*,*)' Integral of dimensionless'/ &' velocity profile=',qPRIM I1=1 I2=2 CALL SIMPR(Dq,0.,X(NX),Q,1.E4,20) WRITE(*,*)' Flowrate, Q =',Q*qPRIM END FUNCTION Dq(XX) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX 1 IF(XX.LT.X(I2) .OR. I2.EQ.NX) GO TO 2 I1=I2 I2=I2+1 GO TO 1 2 IF(XX.GE.X(I1) .OR. I1.EQ.1) GO TO 3 I2=I1 I1=I11 GO TO 2 END FUNCTION VPROF(YY) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX 1 IF(YY.LT.YP(I2) .OR. I2.EQ.NP) GO TO 2 I1=I2 I2=I2+1 GO TO 1 2 IF(YY.GE.YP(I1) .OR. I1.EQ.1) GO TO 3 I2=I1 I1=I11 GO TO 2
Dimensions, Terminology, and Review of Basic Fluid Mechanics 3
3
DYP=YP(I2)YP(I1) A=(YP(I2)YY)/DYP B=1.A VPROF=A*VP(I1)+B*VP(I2)+((A*A1.)*A*D2VP(I1)+ &(B*B1.)*B*D2VP(I2))*DYP**2/6. RETURN DX=X(I2)X(I1) A=(X(I2)XX)/DX B=1.A AA=A*(A*A1.)*DX*DX/6. BB=B*(B*B1.)*DX*DX/6. DEPTH=A*Y(I1)+B*Y(I2)+AA*D2Y(I1)+BB*D2Y(I2) VSURF=A*VS(I1)+B*VS(I2)+AA*D2VS(I1)+BB* &D2VS(I2) Dq=DEPTH*VSURF RETURN END
Listing of program EXPRB1_5.C #include <stdio.h> #include <stdlib.h> #include <math.h> float yp[20],vp[20],x[30],y[30],vs[30],d2vp[20],d2y[30],d2vs[30]; int i1,i2,np,nx; extern float simpr(function equat,float xb,float xe,\ float err,int max); extern void splinesu(int n,float *x,float *y, float *d2y,\ float *d,int ity); float vprof(float yy){float a,b,dyp; while((yy>=yp[i2])&&(i2
1)){i2=i1;i1−−;} dyp=yp[i2]−yp[i1]; a=(yp[i2]−yy);b=1.−a; return a*vp[i1]+b*vp[i2]+((a*a−1.)*a*d2vp[i1]+(b*b−1.)*b*d2vp[i2])\ *dyp*dyp/6.; } // End of vprof float qp(float xx){float a,b,dx,aa,bb,depth; while((xx>=x[i2])&&(i11)){i2=i1;i1−−;} dx=x[i2]−x[i1];a=(x[i2]−xx)/dx;b=1.−a; aa=a*(a*a−1.)*dx*dx/6.;bb=b*(b*b−1.)*dx*dx/6.; depth=a*y[i1]+b*y[i2]+aa*d2y[i1]+bb*d2y[i2]; return depth*(a*vs[i1]+b*vs[i2]+aa*d2vs[i1]+bb*d2vs[i2]); } // End of qp void main(void){FILE *fili; char filnam[20];int i; float dum[30];qprim,q; i1=1;i2=2; printf("Give input file name\n"); scanf("%s",filnam); if((fili=fopen(filnam,"r"))==NULL){ printf("Cannot open file\n");exit(0);} fscanf(fili,"%d",&np); for(i=0;i
15
16
Open Channel Flow: Numerical Methods and Computer Applications printf("Integral of dimensionless velocity profile =%f\n",qprim); printf("Flowrate, Q =%f\n",qprim*simpr(qp,0.,x[nx−1],1.e−4,20)); } The following is the input file needed to solve the problem (EXPRB1_5.DAT): 13 0. 0. .04 .22 .08 .375 .12 .5 .2 .675 .28 .815 .4 .955 .52 1.04 .64 1.07 .72 1.068 .8 1.06 .9 1.03 1 1 20 0 0 0 4 4.7 .25 8 8.3 .45 12 11.2 .47 16 12.7 .49 20 13.3 .5 24 13.4 .49 28 12.8 .48 32 12. .48 36 11.8 .48 40 11.5 .48 44 11.7 .48 48 12.3 .48 52 13.3 .49 56 14.5 .5 60 15. .5 64 14.8 .45 68 12.7 .35 72 8. .2 75 0. 0. The solution produced is Integral of dimensionless velocity profile = 8.726063e − 01 Flowrate, Q = 342.505700
Viscosity is the fluid property that defines its resistance to motion. By definition, the absolute viscosity μ of a fluid is defined as the coefficient that relates the internal shearing stress τ within the fluid to the velocity gradient at this point in the flowing fluid, or ∂v τ = −µ ∂n
(1.7)
where n is a direction normal to velocity v. Fluids, such as water, whose absolute viscosities are not dependent upon the rate of shear, ∂v/∂n, are referred to as Newtonian fluids. The viscosity of water does depend upon its temperature, as shown in Table 1.1. If the viscosity of a fluid does change with the rate of shear (μ is a function of ∂v/∂s or τ, i.e., μ(∂v/∂s) or μ(τ)) then the fluid is referred to as a nonNewtonian fluid. Silly putty is a nonNewtonian fluid that behaves similar to an elastic solid if sheared rapidly (i.e., bounces as a rubber ball), but flows as a very viscous fluid if given sufficient time to do this. Mud and debris flows, which are openchannel flows, of a fluid that is a mixture of water and soil (and possibly debris from the channel) that have mobilized are other examples involving nonNewtonian fluids. The kinematic viscosity v equals the absolute viscosity divided by the fluid density, or v=
µ ρ
(1.8)
Effects that viscosity has on fluid motion can be examined by considering a small section of uniform, steady flow in a channel in which an element of fluid with a differential length dx, and a height dy is taken as shown in the sketch below.
(τ + dτ)dx Y
dY τ dx
γdx dy So
τo
Dimensions, Terminology, and Review of Basic Fluid Mechanics
17
The channel has a bottom slope equal to So and is wide enough that only the shearing stress on the bottom is important (the side shears can be neglected). Since the flow is steady and uniform, the summation of forces on the differential fluid element in the xdirection must equal zero. This summation produces the following:
− τ dx + ( τ + d τ)dx + So γ dx dy = 0
In this equation the slope of the channel bottom is small so that the sin θ = tan θ = So. Simplifying gives dτ = −Soγ dy, which can be integrated between the bottom of the channel to any depth y, or τ
∫
Y
∫
dτ = − γ So dy
τo
0
which gives
τ = τ o − So γ y
An examination of the section of this flow from the bottom to the top with a length dx shows that τo = γYSo. For any arbitrary section in which the shear stress on the side is considered, it can be shown that
τo =
γASo = γ R hS o P
(1.9)
where A is the crosssectional area P is wetted perimeter R h is the hydraulic radius, A/P which reduces to the above for the wide section being considered here. Substituting for τo gives the following equation that defines the shearing stress as a function of the position from the bottom y:
τ = γ So (Y − y)
(1.10)
which shows the shearing stress varies linearly from γSoY at the bottom to zero at the top of the channel flow. Substituting Equation 1.10 into Equation 1.7 allows for the velocity distribution from the bottom to the top of the channel to be determined. Making this substitution gives
γS dv = o (Y − y)dy µ
which integrates to give the parabolic relationship,
v=
γ So µ(Yy − y 2 /2)
(1.11)
It turns out that Equation 1.11 is correct only for laminar flows in which the viscous forces dominate over the inertia forces. For a turbulent flow the fluid near the walls has its velocity increased, whereas the velocity near the top of the flow away from the solid boundaries has its velocity decreased from the velocity given by Equation 1.11 by mixing of the flows. In other words, the
18
Open Channel Flow: Numerical Methods and Computer Applications
large velocity portion of the flow is continually being mixed with the small velocity portion of the flow tending to make the velocity distribution more nearly the same throughout the flow than that described by Equation 1.11. This mixing is referred to as momentum transfer within the flow, and results in the need to modify Equation 1.7 for turbulent flows. This momentum transfer causes the inertial forces to play a more important role in determining the distribution of the velocity with depth. For turbulent flows it is necessary to multiply the velocity gradient in Equation 1.7 by an additional eddy viscosity term ε to get the shearing stress, or μ is replaced by (μ + ε). This eddy viscosity varies with y in a manner that cannot be defined easily for different flow situations, and therefore has limited practical value. Even though Equations 1.7 and 1.11 hold only for laminar flows, Equation 1.10 is valid for turbulent flows as well. Example Problem 1.6 A trapezoidal channel with a bottom width b = 3 m, and a side slope m = 2 has a bottom slope So = 0.0008 and is 5000 m long. This channel contains a uniform flow of Q = 20 m3/s at a depth Y = 1.6 m. Determine the total shear force on the sides of this channel caused by the flowing water. Solution The shear stresses on the boundaries of this channel are given by Equation 1.9. Computing the area and wetted perimeter gives A = (b + mY), Y = 9.920 m2, P = 10.155 m. Substituting into Equation 1.9 gives
9.920 τ o = 9800 (0.0008) = 7.658 N/m 2 . 10.115
The area over which this shear stress acts equals the perimeter times the length of channel or 10.155 × 5000 = 50,775 m2. Therefore, the total shear force equals 50,775 × 7.658 = 388,848 N. Alternatively, this shear force equals the component of the weight in the channel in the direction of the flow, or weight XSo. This computation gives 9800(9.92 × 5000)(0.0008) = 388,848 N. It should be noted in this problem that the shear force does not involve using the flow rate. In actuality, it is the flow rate and the channel roughness that determines the depth of flow. Example Problem 1.7 Water at a rate of Q = 25 m3/s flows around a circular bend in a rectangular channel with a width of 4 m. The inside radius of this bend is 4 m. If the depth upstream from the bend is Yo = 3 m, determine the velocity distribution through a radial line through this bend under the following assumptions: (1) an average velocity through the depth of flow can be used so that the velocity can be considered only a function of r (distance from the center of the bend); (2) the shear stress is proportional to the velocity gradient dv/dr; and (3) the shear stress is zero along the circle midway between the two channel walls and varies linearly to τo at the two walls. Solution The solution starts by noting from the problem’s description that τ = a(dv(r)/dr) = τo(r − 6)/2, in which a is a constant. Integrating this equation and evaluating the constant of integration gives the following equation for the velocity distribution:
6r − r 2 v = C − 16 (Note that v = 0 when r = 4 and r = 8) 2
To evaluate the constant C, the integral of vYdr is equated to the flow rate Q, or
∫
Q = vY dr = 25
19
Dimensions, Terminology, and Review of Basic Fluid Mechanics If the depth is assumed constant at Y = 3 m across the circular portion of the channel then 6r 2 r 3 r2 − − 16 r 25 = 3C 6r − − 16 dr = 3C 2 2 6
∫
8
= 3C[5.33], or C = 1.563 4
The assumption of constant depth is not good since the centrifugal forces of the flow around the bend will cause the depth at its outside to be larger than at the inside. The slope of the water surface will be normal to the total acceleration vector, or dY/dr = v2/(gr). Substituting for v and integrating gives C2 r 4 Y = − 2r 3 + 26r 2 − 192r + 256 Ln(r) + C1 g 16
Since the depth is assumed to equal 3 m when r = 6 m (the midradius of the bend), C1 can be evaluated as C2 C1 = 3 + (108.31). g
The above expressions for Y and v are again substituted in 25 = ∫ vY dr and this integrated from 4 to 8 with the result, 25 = 16C − 0.015982C3
The three roots of this equation are 1.5663, −32.395, and 30.828. The first root is the desired one, i.e., C = 1.5663. The depths and velocities through the bend are given in the table below. r
4
4.5
5
5.5
6
6.5
7
7.5
8
Y V
2.800 0.000
2.808 1.371
2.847 2.350
2.917 2.937
3.000 3.133
3.077 2.937
3.131 2.350
5.157 1.371
3.162 0.000
1.10 Conservation of Mass, or Continuity Equations A basic principle of the mechanics is that mass is conserved. When this principle is applied under steadystate flow conditions to onedimensional fluid flows it is generally expressed that the mass flow rate or flux past a second section equals the mass flow rate past the first section plus the mass flow rate entering between the two sections, or
∫
∫
(ρVA)2 = (ρVA)1 + (ρVA)1− 2 = (ρVA)1 + ρvY dx = ρq dx
(1.12)
where the Vs represent the average velocities at the designated positions the As represent the corresponding crosssectional areas normal to these velocities When dealing with an incompressible fluid the density ρ is constant and can be divided out of Equation 1.12 to give a volumetric flow equation. The volumetric flow rate with dimensions of L3/t will be given the symbol Q, so Equation 1.12 can be written in either of the following two forms:
∫
Q 2 = Q1 + q(x)dx
20
Open Channel Flow: Numerical Methods and Computer Applications
or
∫
(VA)2 = (VA)1 + q(x)dx
(1.13)
where q(x) is the lateral inflow (negative for outflow) in units of flow rate per unit length (L2/t) being equal to viY, and the integrations are over the length between the two sections 1 to 2. If the lateral inflow is constant, then the last term in Equation 1.13 can be simplified to qL1–2 where L1–2 is the distance between section 1 and 2. The average velocity V in Equations 1.12 and 1.13 represents the point velocity integrated over the crosssectional area of the flow divided by this area, or
V=
∫ v dA
(1.14)
A
When dealing with two or threedimensional flows, or an unsteady onedimensional flow, the continuity equation becomes a partial differential equation rather than one of the above algebraic equations. The general equation that applies for threedimensional unsteady flow of a compressible fluid is ∂(ρu) ∂(ρv) ∂(ρw) ∂(ρ) + + =− ∂x ∂y ∂z ∂t
(1.15)
where u, v, and w are the velocity components in the x, y, and z coordinate directions, respectively. For either steady flow, or the flow of an incompressible fluid the righthand side of this equation becomes zero. For an incompressible fluid the continuity equation reduces to ∂u ∂v ∂w + + =0 ∂x ∂y ∂z
(1.16)
The continuity equation for an unsteady onedimensional flow in an open channel needs to apply for the entire cross section of the channel rather than at a point as is the case with Equations 1.15 and 1.16. To develop this equation consider a differential length of a channel flow as shown in the sketch below whose depth, etc. changes with time, as well as the position x along the channel. The principle of conservation of mass requires that the following, in relationship to this control volume, CV, be true:
(Vol. in CV in time ∆t) − (Vol. leaving CV in time ∆t) = Vol. change in ∆t, q(x,t) T
(∂Y/∂t)dt dx
Q
Y
Q+ dx
∂Q dx ∂x
x
Y
21
Dimensions, Terminology, and Review of Basic Fluid Mechanics
or substituting the appropriate quantities into this equation gives
∂Q ∂Y Q − Q + ∂x dx + q(x,t)dx dt = T ∂t dt dx
Simplifying and dividing by dxdt gives ∂Q ∂Y − q(x,t ) + T =0 ∂x ∂t
(1.17)
This equation, coupled with a second partial differential equation of motion, will be the basis for solving unsteady channel flows in Chapter 6. It is worthwhile noting that if ∂Q/∂x is of the same positive magnitude as the magnitude of lateral inflow q(x,t) at all points x and at all times, then ∂Y/∂t = 0, i.e., the flow is steady state, which means at any position x the depth and velocity are constant. However, it is important to note that q and ∂Q/∂x represent two different quantities. q(x,t) is the lateral inflow that can in general vary along the channel as well as vary with time. It is generally known and not part of what is solved for. Should lateral outflow occur, then q(x,t) is a negative quantity. The term ∂Q/∂x represents the change in flow rate in the channel direction x for a given instant in time, and in general varies from instant to instant. Sometimes an additional subscript is applied to partial derivatives to denote what is being held constant. In Equation 1.17 the two independent variables are x and t. Therefore, it is understood that t is constant when dealing with partial derivatives with respect to x and x is held constant when the partial derivative is with respect to t. Q is unknown and therefore its variation with x and t is a part of the solution. Since the volumetric flow rate Q equals the average velocity V times the crosssectional area A, the first term in Equation 1.17 can be expanded to give the following alternative continuity equation for onedimensional unsteady flow in a channel:
A
∂V ∂A ∂Y +V − q(x,t ) + T =0 ∂x ∂x ∂t
(1.17a)
Equation 1.17 generally cannot be solved by itself because it involves two unknowns, or dependent variables, Q and Y. The principle that as many independent equations must exist as there are unknowns, dictates that another equation must be obtained. This second equation comes from Newton’s second law of motion, and will be given later in Chapter 6. Equation 1.17 and this second equation are often referred to as the StVernant equations that govern onedimensional unsteady openchannel hydraulic problems. Example Problem 1.8 A 10 m wide rectangular channel has a 10 m long section of side weir that discharges a volumetric flow rate per unit length as given by the equation q = 0.5 / 2g H 3/2 in which H is the depth of water in the channel above the weir crest. The weir crest is 2 m above the channel bottom, and the depth in the channel at the beginning of the side weir is 2.5 m, and the flow rate here equals 50 m3/s. If the depth of flow through the side weir is such that V2/2g + Y = constant, determine the flow rate in the channel over the length of the side weir.
2.5 m 50 m3/s
2m
22
Open Channel Flow: Numerical Methods and Computer Applications Solution Letting Qo and Yo be the flow rate and depth, respectively, at the beginning of the side weir the constant referred to can be determined from constant = Q o 2 /(2gA o2 ) + Yo = (50/25)2/19.62 + 2.5 = 2.704. Thus, the following implicit equation provides the depth Y: Y = 2.704 −
Q2 19.62(10Y)2
This equation will converge with Q given if the Y computed on the left of the equal sign is iteratively used for the Y on the right of the equal sign. After determining Y the lateral outflow can be determined by q = 2.215(Y − 2)3/2
and the flow rate determined from Equation 1.13, or
∫
Q = 50 − q dx
Since this integration cannot be completed in closed form, it must be carried out using numerical techniques. The small FORTRAN program listed below does this under the assumption that if small enough steps are used for dx then the Q at the current position x can be used to compute the depth, etc. Techniques designed to solve ordinary differential equations (ODEs) provide a better alternative for solving this problem than the crude numerical integration used in the above program. The ODE for this problem is dQ/dx = 0.5/ 2g(Y − 2)3 / 2. Numerical methods for solving ODEs will be described in Chapter 4 and are extensively utilized in solving gradually varied flow problems. Listing of FORTRAN program that numerically integrates the above equation in the s olution for Example Problem 1.8 Q=50. Y=2.5 WRITE(6,100) Q,Y 100 FORMAT(' x',7X, 'Q',7X, 'Y',/,19('_'), &/,' 0',2F8.3) DO 30 I=1,1000 Y1=2.704−(Q/Y)**2/1962. 10 IF(ABS(Y1−Y).LT. .000001) GO TO 20 Y=Y1 GO TO 10 20 Q=Q−.02215*(Y1−2.)**1.5 IF(MOD(I,100).EQ.0) WRITE(6,110) I/100,Q,Y1 110 FORMAT(I3,2F8.3) 30 CONTINUE END
Solution from Progaram X
Q (cfs)
Y (ft)
0 1 2 3 4 5 6 7 8 9 10
50.000 49.208 48.397 47.569 46.723 45.859 44.977 44.078 43.161 42.227 41.275
2.500 2.500 2.515 2.523 2.530 2.537 2.545 2.552 2.559 2.556 2.573
1.11 Energy Principle The equations of motion are of extreme importance to all mechanics, whether dealing with fluids or solids. The equations of motion are based on Newton’s second law, that states that force = mass × acceleration. In this section the scalar application of this fundamental law of mechanics is discussed, as it applies to fluids in motion. The section is entitled “Energy Principle” because this scalar application gives equations similar to the energy equation in Thermodynamics. In the next section, the vector application of Newton’s second law to fluid motion is discussed under the heading “Momentum Principle.” The application of Newton’s second law to a general threedimensional unsteady fluid
23
Dimensions, Terminology, and Review of Basic Fluid Mechanics
flow results in the Navier Stokes equations, whose development can be found in books dealing with fluid mechanics. Applying the same law to an ideal fluid that is inviscid (viscous shear stresses are ignored) produces the Euler equations of motion. Neither of these equations is presented herein. Rather simplified versions of these equations are given for onedimensional and special flows, and these are most useful in handling problems of onedimensional openchannel hydraulics. It will be necessary to define acceleration in fluid motion, first. Acceleration is a vector that represents the time rate of change of the velocity vector, or a = dV/dt. Since velocity in a fluid is a function of the position as well as time in general, the full derivative dV/dt is obtained by the chain rule in calculus, or using Cartesian coordinates, dV DV ∂V ∂x ∂V = = + dt dt ∂x ∂t ∂y
The quantity dV/dt is called the substantial derivative and is often denoted by DV/Dt to make it clear that the partial derivative with respect to time is not intended. Since dx/dt, dy/dt, and dz/dt equal the velocity components u, v, and w in the x, y, and zdirections, respectively, the above equation becomes
dV DV ∂V ∂V ∂V ∂V = =u +v +w + dt dt ∂x ∂y ∂z ∂t
The terms u∂V/∂x, v∂V/∂y, and w∂V/∂z are called the convective accelerations and can be visualized as the acceleration that occurs as streamlines within a flow converge together. The term ∂V/∂t is the local acceleration, and represents the variation in velocity that occurs at a point in the fluid with time. Also note that the above equations are vector equations since every term contains the velocity vector V. If the flow is one dimensional, then the magnitude of the velocity is only a function of the coordinate in that direction and time, if the flow is unsteady, or V(s,t). s will be used for the space coordinate rather than x, because the onedimensional flow will be allowed in the curved direction along a streamline, and not restricted to just the direction of the channel. For such onedimensional flows the magnitude of the acceleration is given by a=
dV ∂V ∂V =V + dt ∂s ∂t
Consider an element of fluid of length ds and crosssectional area dA along a streamline in the Fs = ma s to this direction s, as shown in the sketch below. Applying Newton’s second law element results in (note that only magnitudes are used since only one direction s is involved)
∑
ds
p+
p dA
θ
τ
∂p ∂s
ds dA
Stream line
s
gρdA ds
∂p ∂V ∂V p dA − p + ds dA − τdP ds − ρg dA ds × sin(θ) = ρ dA ds V + ∂s ∂t ∂s
24
Open Channel Flow: Numerical Methods and Computer Applications
where P is the perimeter of the element. Noting that sinθ = ∂z/∂s, dividing by (−ρ dA ds) and simplifying results in
1 ∂p τ dP ∂z ∂V ∂V + +g +V + =0 ρ ∂s ρ dA ∂s ∂s ∂t
It is helpful in understanding equations to follow their dimensions. In the equation immediately above, the dimensions are force divided by mass, since every term started out as a force which was divided by the ρdAds, or the mass of the differential element. If flow is steady then ∂V/∂t = 0, and the variables are only a function of s and, therefore, the above equation can be integrated along a streamline to give
p τ dp v2 + ds + gz + = constant ρ ρ dA 2
∫
Now the dimensions of the equation are force times length divided by mass. The product of force and length is work or energy. Thus, upon integrating along the streamline each term in the equation represents a form of fluid energy per unit mass. Dividing by g results in the Bernoulli Equation, that has units of energy per unit weight of fluid, or
p v2 +z+ + γ 2g
τ dp
∫ γ dA ds = H(constant)
(1.18)
The integral in Equation 1.18 cannot be evaluated in general since τ is not a known function of s, and therefore its value over the distance between a couple of points in the flow is designated as the headloss hL1–2 and Equation 1.18 is commonly applied between the two points in the following form:
p1 V2 p V2 + z1 + 1 = 2 + z 2 + 2 + h L1−2 γ 2g γ 2g
(1.18a)
Each term in Equation 1.18 is considered a head because they all have the dimensions of length. In ES units this length is given in feet, and in SI units this length is in meters. It is important to note that terms are actually energy per unit weight of fluid. That is, when using ES units each term represents some form of energy in ftlbs that each pound of fluid contains, and when using SI units each term is this energy in Nm that each Newton weight of fluid contains. p/γ is referred to as the pressure head, and represents the potential energy due to pressure per unit weight. z is the elevation head and is the potential energy due to the vertical position of the fluid per unit weight, and V2/2g is the kinetic energy per unit weight. The sum of p/γ and z is called the piezometric head. To visualize these energies first consider the elevation head z. If a unit weight of fluid were a distance z above a datum then it would be capable of doing z times its weight of work in dropping under gravity to the datum. Since energy is the ability to do work, the potential energy that this unit weight of fluid has equals z. To visualize the potential energy due to the pressure of a fluid consider an infinitely large reservoir of liquid. At the bottom of this reservoir there is a cylinder that can extract work from the fluid due to its constant pressure here. This pressure acts over the area of the piston within the cylinder, and if the piston is permitted to move through a distance L, the amount of work done on it by the fluid equals pAL. The weight of fluid involved in doing this work equals γAL. Therefore, the energy per unit weight of this fluid is pAL/(γAL) = p/γ. Note from the hydrostatic equations that p/γ = h the height of the liquid above the cylinder. Thus, at the top of the reservoir the fluid contains elevation head, z = h, and at the bottom the liquid contains pressure head and at neither position does the liquid possess any velocity head since it is
25
Dimensions, Terminology, and Review of Basic Fluid Mechanics
not moving, but these sum to a constant value. To understand that the velocity head V2/2g is the kinetic energy per unit weight, note that kinetic energy equals 1/2 the mass times the velocity squared, or ρ(Vol)V2/2. Dividing this by the weight of this mass of fluid gives K.E./(unit weight) = {ρ(Vol)V2/2}/{gρ(Vol)} = V2/2g. P+
∂P ∂n
dn dAn
V
ds
dn s
V + dV
ds
+
dV n
z θ
g
p dAn
r
ρg dAndn
Some applications in openchannel flow call for looking at what happens across streamlines. Flow over the crest of a dam is one such application. Therefore, consider applying Newton’s second law normal to the streamlines, as shown in the sketch. If forces on this element due to viscous shear are ignored then the summation of forces in the direction of an outward normal equated to the mass times outward normal acceleration results in
V2 ∂Vn ∂p pdAn − p + dn dA − ρg dAn cos(θ) = ρdAn + ∂n ∂t r
The cosθ = ∂z/∂n and the last quantity within parenthesis is the acceleration in the normal direction. To see the latter, note that Vn = f(s,t), and therefore by the chain rule,
an =
∂Vn ∂V V+ n ∂s ∂t
where V =
∂s ∂t
From the small vector diagram in the sketch above it can be seen from similar triangles that
dVn ds = V r
and therefore
∂Vn V = r ∂s
dividing the above force equation by ρdAndn, and simplifying gives
1 ∂p ∂z V2 ∂Vn +g + + =0 ρ ∂n ∂n r ∂t
Special cases are worth examining. If the flow is steady and streamlines are straight then the last two terms in the last equation are zero and upon integrating in the ndirection, the following is obtained:
p + z = constant γ
26
Open Channel Flow: Numerical Methods and Computer Applications
which indicates that the pressure distribution in the normal direction is hydrostatic in a moving flow provided the streamlines are straight. Since this is the assumption made to classify flows as one dimensional, pressure distributions are generally considered hydrostatic unless the channel is steep as discussed previously. Before the above equation can be integrated with curved streamlines, it is necessary to have a relationship between V and r. Two special cases will be considered; irrotational flow, and solid body, or forced vortex, rotational flow. Real flows are between these two cases. Fluid viscosity must be absent for a flow to be irrotational, whereas the rotation of a forced vortex follows the solid body rotation law that the velocity increases with the radius of rotation, V = rω. In general, a flow is irrotational if the vorticity or crossproduct of the differential del operator and the velocity vector equal zero, or ξ = ∉XV = 0 where (using the Cartesian coordinate system)
v=
∂ ∂ ∂ i+ j+ k ∂x ∂y ∂z
and i, j, and k are unit vectors in the x, y, and zdirections respectively.
∇=
∂ ∂ ∂ i+ j+ k ∂x ∂y ∂z
and i, j, and k are unit vectors in the x, y, and z. The component of vorticity for the plane of the curved flow above in the natural coordinate system is ∂Vn/∂s − ∂V/∂n and if this flow is irrotational then this quantity equals zero. Since ∂Vn/∂s = V/r, the term V2/r in the above equation can be written as V∂V/∂n, permitting the equation to be integrated in the normal direction, giving
p v2 +z+ = H (constant) 2g γ
(1.18b)
or identically the same as Equation 1.18, without the viscous loss term; but since viscous forces were not included one would expect this. Therefore, if the flow is irrotational the Bernoulli equation applies across, as well as, along streamlines. The usual concept is that the Bernoulli equation is applied for the entire section of a onedimensional flow. If the radius of curvature of the streamline is constant, it can be shown that for an irrotational flow the velocity times the radius is constant, or Vr = constant; thus, as r becomes zero V must become infinite. For a forced vortex, the vorticity equals twice the actual rotation about an axis, or
∂v n ∂v 2v − = 2ω = r ∂s ∂n
Therefore, ∂V/∂n = −V/r and integration of the above equation in the normal direction gives
p v2 +z− = H (constant) 2g γ
(1.18bc)
(Note the minus sign in front of the velocity head.) A centrifugal pump causes the flow within the housing of the pump to nearly follow this forced vortex law. The exception is that there must be a radial component of velocity for the fluid to pass onto and leave the pump impeller. This radial component will be governed by the usual Bernoulli equation and, therefore, the head produced by a pump will be approximated by (with zi = zo)
27
Dimensions, Terminology, and Review of Basic Fluid Mechanics
hp =
p2 − p1 Vi2 − Vo2 Vo2 − Vi2 + + γ 2g r 2g t
where subscripts i and o denote inlet and outlet, respectively, and subscripts r and t denote radial and tangential directions, respectively.
1.11.1 Kinetic Energy Correction Coefficient, α The Bernoulli equation developed above is generally applied for an entire section of an open channel flow, and the average velocity for the cross section is used to compute the velocity head, or kinetic energy per unit weight. Since the velocity is not constant throughout the entire section, the actual or exact kinetic energy per unit weight, K.E./Wt, can be defined by multiplying the velocity head by a correction coefficient α, or V2 K.E. = α Wt 2g
The evaluation of α involves determining the exact kinetic energy and dividing this by the velocity head. The exact, or correct, kinetic energy K.E., passing a cross section is the integral over the area of the kinetic energy through a differential area dA, or
∫ ρv dQ = ∫ ρv dA K.E. = 2
3
2
2
Dividing K.E. by gρQ = gρVA gives K.E./Wt and, therefore,
α=
∫
∫
v2dQ v2dA K.E./Wt = = v 2 / 2g v2Q v2 A
(1.19)
For a natural channel, or when it is not possible to define v and A as functions of the depth, the integral sign is replaced by a summation, and Equation 1.19 becomes
K.E./Wt α= 2 = v /(2g)
∑ v ∆Q = ∑ v ∆A 2
V2Q
2
V2 A
(1.19a)
The value of α is only slightly larger than unity for typical turbulent openchannel flows. Furthermore, when the flow is subcritical, as it always is in natural rivers and streams, the velocity head will be smaller than the other terms in the Bernoulli equation, and therefore it is common practice to take α as 1. Example Problem 1.9 Water flows over the crest of a small dam that is 4 m high at a rate of 48.2 m3/s. The depth of water over the dam’s crest is 6 m, and the channel is rectangular and 10 m wide. Determine the depth and velocity in the channel downstream from the dam. Ignore any frictional losses. Solve the problem twice; first, under the assumption that at both sections the kinetic energy correction coefficient is unity, and second, the αs are not unity, and that the following dimensionless velocity profiles are known.
28
Open Channel Flow: Numerical Methods and Computer Applications Over the Dam’s Crest y/Y v/V
0.00 0.00
0.07143 0.3529
0.2143 0.8235
0.3571 1.0353
0.5357 1.1764
0.7143 1.2823
0.8571 1.2235
y/Y v/V
0.00
0.0357
0.0714
0.1429
0.3571
0.7143
1.0000
0.00
0.5743
0.8614
0.9763
1.0452
1.0567
1.0452
1.0000 1.0823
In the Channel at the Dam’s Toe
Solution For the first part with the α’s = 1, the upstream velocity is determined from V1 = Q/A1 = 48.2/60 = 0.8033 m/s. Apply the Bernoulli equation, Y2 +
Q2 V2 = Y1 + 1 = 6.033 m 2 2g 2g(10Y2 )
This is a cubic equation that must be solved by trial, or an iterative technique such as the Newton Method described in Appendix B. The solution is Y2 = 0.461 m. Another solution to this cubic equation is 6 m, but this is obviously not the wanted root since it is the upstream depth. The corresponding velocity is V2 = 48.2/4.61 = 10.46 m/s. For the second part of the problem q = 4.82 = ∫v dy, or 4.82V = ∫(v/V)dy, in which the limits of integration are 0–6. If dy is replaced by the dimensionless depth y/Y, then the upper limit changes to 1. There are many methods that can be used to evaluate this integral including plotting the profile and determining the area by counting squares. We will use a cubic spline as described in Appendix B to interpolate between the given values and Simpson’s rule (described in Appendix C) to obtain a numerical integral. A FORTRAN program EPRB1_9.FOR listing to accomplish this task is given below. 1 2 3
EXTERNAL EQUAT COMMON /TRA/Y(8),V(8),DQV(8),I1,IP,N I1=1 IP=2 WRITE(*,*)' Give no. and then this may pairs' READ(*,*) N,(Y(I),V(I),I=1,N) CALL SPLINESU(N,Y,V,DQV,D,0) CALL SIMPR(EQUAT,0.,Y(N),FLOW,1.E−5,30) WRITE(*,*) FLOW END FUNCTION EQUAT(YY) COMMON /TRA/Y(8),V(8),DQV(8),I1,IP,N IF(YY.LT.Y(IP).OR.IP.EQ.N) GO TO 2 I1=IP IP=IP+1 GO TO 1 IF(YY.GT.Y(I1).OR.I1.EQ.1) GO TO 3 IP=I1 I1=I1−1 GO TO 2 DY=Y(IP)−Y(I1) AY=(Y(IP)−YY)/DY BY=1.−AY EQUAT=AY*V(I1)+BY*V(IP)+((AY*AY−1.)*AY*DQV(I1)+(BY*BY−1.)* &BY*DQV(IP))*DY*DY/6. RETURN END
Dimensions, Terminology, and Review of Basic Fluid Mechanics In the program, the subroutine SPLINESU is called to provide the values of the second derivatives at the points. The common block passes the arrays of dimensionless depths y/Y and corresponding dimensionless velocities, v/V, i.e., the values in the above table over the crest of the weir. The result of this numerical integration is 1.000 and (meaning that the above data truly give a proper dimensionless velocity profile), therefore, the average velocity V1 is the same as above or V1 = 0.8033 m/s. Next, the last two lines of the function subprogram EQUAT can be changed to the following: EQUAT=(AY*V(I1)+BY*V(IP)+((AY*AY−1.)*AY*DQV(I1)+(BY*BY−1.)* &BY*DQV(IP))*DY*DY/6.)**3 Giving both the above tables of dimensionless values provides the solution for the kinetic energy correction coefficients: α1 = 1.2557, and α2 = 1.0555. Now the implicit equation that must be solved is 0.80332 4.82 2 Y2 + 1.0555 2 = 6.0413 = 6 + 1.2557 19.62 Y2 19.62
The solution is Y2 = 0.472 m. The average velocity V2 = 4.82/0.472 = 10.21 m/s. Example Problem 1.10 A pump increases the pressure from its inlet to its outlet pipe by 52 psi. The inlet and outlet pipes are 10 and 12 in. in diameter, respectively, and the flow rate is Q = 6 cfs. Determine the amount of energy per unit weight that the pump supplies to the water. If the pump has an efficiency of 85%, determine the horsepower required to drive the pump. Solution In applying the energy equation to this problem the head of the pump needs to be added to the left side of the Bernoulli equation, or V2 p V2 p + = h + p 2g + γ 2g γ 1 2
Upon substituting the known values into this equation gives hp = 120.97 ft. To get energy from the head given by Bernoulli’s equation multiply by the weight flow rate γQ, and therefore the horsepower is
HP =
γ Qh p = 96.88 hp (550 is the conversion for hp from energy in ftlb/s.) 550e
Example Problem 1.11 Assume that the water at the surface of a river flowing around a bend follows the free vortex law. The radius of the inside of the bend is 30 m, and at the outside of the bend the radius is 45 m. If the velocity at the inside of the bend is measured as 0.8 m/s, what is the velocity on the surface at the outside of the bend? Solution The free vortex law indicates that Vr = constant. This constant equals 0.8(30) and, therefore, the velocity at the outside of the bend is Vo = (0.8)(30)/45 = 0.533 m/s. Since at the bottom of the river the velocity will be smaller than at the surface, there is a net flow outward on the surface and an inflow at the bottom. This movement creates a secondary motion in a river that moves sediment toward the downstream inside of the bend. A pumping station here would bring in river sediments; therefore, such a pumping station is best located on the outside of the bend.
29
30
Open Channel Flow: Numerical Methods and Computer Applications Example Problem 1.12 The velocity distribution in a wide open channel can be defined by the following logarithmic function between the channel bottom and a position y = 0.1 ft above the bottom: v = C/Ln(y), and a seconddegree polynomial for the rest of the depth. When the depth of flow is 2.6 ft, velocities are measured at three positions with the following results: Depth, ft Velocity, fps
0.1 2.3
2.1 2.45
2.6 2.35
Determine the kinetic energy correction coefficient, α. Solution To obtain the equation for the velocity distribution over the bottom 0.1 ft substitute 2.3 in for v, and 0.1 for y in the given equation and solve for C. This gives C = 2.3 Ln(0.1) = −5.296. The equation for the rest of the profile is obtained by evaluating a,b, and c in v = a + by + cy2 by Lagrange’s interpolation equation described in Appendix B, or solving a, b, and c from the three equations obtained by substituting the y and v from above into this polynomial equation to give v = 2.2694 + 0.317y − 0.110y 2
The flow rate q per unit width of channel is found first from Q = ∫v dA with the integration carried out in two parts, or −0.1
q = −5.296
∫ 0
dy + Ln(y)
2.4
∫ (2.2694 + 0.317y − 0.11y )dy = 0.1715 + 6.0989 = 6.2705 cfs/ft. 2
0.1
The first integral was evaluated numerically because of the complexity of a closedform integration. Next, evaluate ∫v3dA as shown below.
∫
0.1
v3dA = −5.296
∫ 0
dy + Ln(y)3
2.6
∫ (2.2694 + 0.317y − 0.11y ) dy = 0.581 + 36.356 = 36.937 2 3
0.1
To evaluate α substitute into α = ∫v3dA/(qV2) = 36.937/[6.2705(2.4117)2] = 1.013.
1.12 Momentum Principle in Fluid Flow An equally important skill for solving a fluidflow problem to that associated with the use of the energy principle is the application of the momentum principle. The momentum principle must generally be utilized to solve problems dealing with external forces acting on the fluid. Its use occurs when dealing with the overall flow picture rather than a detailed examination of individual fluid particles, and their associated internal processes. The momentum principle produces vectors, and therefore is the natural choice when external forces are involved. It can be applied to one, two or threedimensional flows, but when dealing with openchannel hydraulics the common use of the momentum principle will be for one and twodimensional flows. The momentum principle in fluid mechanics is based on Newton’s second law of motion, F = ma, where the force F and the acceleration a are vector quantities. Remember, to define a vector it is necessary that its direction as well as its magnitude are given, and that a change in direction (with the magnitude constant) is equally important to a magnitude change (with the direction remaining constant). Since acceleration is the time rate of change of velocity, Newton’s second law can be written as
F=m
dV m = dV dt dt
31
Dimensions, Terminology, and Review of Basic Fluid Mechanics
In writing the term after the last equal sign in the above equation the concept is that rather than dealing with a derivative we are dealing with a differential change in the velocity vector dV within the differential time dt. In solid mechanics dt is moved to the other side of the equal sign, associated with vector F and the product FΔt is called the linear impulse, and this is equated to the change in linear momentum that is defined as Δ(mV). In fluid mechanics, however, we deal with flow rates, and dm/dt will be denoted as m, the mass flow rate, which equals ρQ. The interpretation of m/dt is the change in mass per time or a mass flow rate ρQ. Therefore, for fluid applications the above equation can be written as r = ρQ∆V
where R is the resultant force on the CV. When applied to a control volume, CV, of fluid between the two sections 1 and 2, as shown in the sketch below, it is written as r = ρQ(V2 − V1 )
(1.20)
The forces on the CV include the body forces, i.e., the weight of the CV, as well as external forces that may be applied by adjacent fluid and structures, etc., that act against the CV of fluid. F P2 FP1 V1
V2
ρQ
ρQ
2 F
1
Weight
If the flow divides so that it leaves through sections 2 and 3 as shown below in the sketch, then the momentum equation would need to be written as r = (ρQV)2 + (ρQV)3 − (ρQV)1
F p2 )2
(ρQ
2
V2
Fp1 Fx
(ρQ)1
V1
Fy
)3
Weight
(ρQ
1
F p3 V3
3
The quantity (ρQV) is called the momentum flux. A more generalized expression of the momentum principle states that the resultant external force acting on a CV of fluid equals the difference between the momentum flux vectors leaving the CV minus the momentum flux vectors entering the CV. Expressed as an equation the momentum principle becomes
r=
∑ (ρQV)
leaving
−
∑ (ρQV)
entering
(1.21)
32
Open Channel Flow: Numerical Methods and Computer Applications
Since Equation 1.21 is a vector equation, for two and threedimensional problems it can be written as two or three scalar equations, respectively, in coordinate directions of that space. For a onedimensional problem the coordinate direction will coincide with the directions of the vectors on both sides of the equal sign in Equation 1.21 and the vectors become effectively scalar quantities. For a twodimensional problem in Cartesian coordinates Equation 1.21 can be written as the two scalar equations,
∑ F = (ρQV
− ρQVx1 )
(1.21a)
∑ F = (ρQV
− ρQVy1 )
(1.21b)
x
x2
and
y
y2
where the subscripts x and y represent the x and y coordinate directions, respectively the 1 and 2 subscripts represent entering and leaving the control volume, respectively The summation of forces in the x and ydirections include all external and body forces acting on this CV. The application of the momentum principle to most problems includes the following five steps:
1. Construct a control volume of the fluid. The upstream section(s) of this CV is (are) upstream from the occurrence of interest to a section of the flow where the velocity and other variables of the problem can be defined. Likewise, the downstream section(s) is (are) taken where fluid variables can be defined. Adjacent fluid and structures are removed from this CV. 2. Replace all removed fluid and structural components by the equivalent forces that they apply to the fluid in the CV. 3. Apply Equations 1.21a and 1.21b. In applying these equations it is vital to watch signs. Should the component of force or velocity be in the direction opposite to the positive direction taken for either x or y, then it is negative in these equations. Likewise, any vector not in either the x or ydirection must be multiplied by the cos or sin of the appropriate angle to get the component in the coordinate direction. 4. Solve for the unknowns. Since Equations 1.21a and 1.21b are two equations, two unknowns can be obtained. For a onedimensional problem only one unknown may exist, and for a threedimensional problem three unknowns may exist. These unknowns may be the components of one unknown vector, such as the force. 5. If the unknown(s) solved for in step 4 consists of a vector, note that its sense will be opposite on the structure than on the fluid. The value obtained by the above procedure is the force on the CV of fluid.
In implementing step 2 above, the fluid adjacent to the CV of fluid is replaced by an equivalent force. In applying this to openchannel flows, this adjacent fluid will often have a hydrostatic pressure distribution, and the force will be a hydrostatic force. A hydrostatic, Fs, force equals the crosssectional area times the pressure at the centroid of this area, or
Fs = pc A = γ h c A = γ Ah c
(1.22)
33
Dimensions, Terminology, and Review of Basic Fluid Mechanics
where hc is the distance from the free surface to the centroid of the area. The last form given in Equation 1.22 contains the term Ahc which is the first moment of area about the free surface. For rectangular and trapezoidal cross sections, this first moment of area is easily obtained. For a rectangle it equals the area times the distance from the surface to the center or bY2/2. For a trapezoid it can be obtained by the method of composite areas, e.g., the rectangle and the two triangles and is Ahc = bY2/2 + mY3/3. For other cross sections such as an arc of a circle, this first momentum of area must be obtained by integration. See Appendix A and Table A.1 for this first moment of area for a circular cross section. Example Problem 1.13 A trapezoidal channel with a bottom width b1 = 14 ft and a side slope m1 = 1.5 divides into a rectangular channel with b2 = 8 ft whose direction is 30° from the direction of the upstream channel, and a pipe with a 4 ft diameter whose direction is 45° from the direction of the upstream channel. The top of the pipe is below the water surface so it flows full. The depth in the downstream channel is 5.2 ft. The pipe drops in elevation by 110 ft over its length of 1500 ft, and delivers water at a pressure of 40 psi. The wall roughness for the pipe is e = 0.02 ft. Determine the resultant force on this structure for the design flow rate of Q1 = 450 cfs. Solution First applying the Bernoulli equation between sections 1 and 2 and then sections 1 and 3 (or 2 and 3) along with the continuity equation and equations for pipe flow give the following five simultaneous equations to solve for the five unknowns Q2, Q3, f, Y1, and pt (pressure on top of pipe): H=
Q12 Q 22 Q 22 + Y1 = + 5.2 2 2 2gA1 2gA 2 64.4(38.4)2
(1)
Q 32 pt + + 4.0 2gA 32 γ
(2)
H=
Q 2 + Q 3 = 450
hf =
(3)
f ( L/D ) Q 32 40 (144 ) = 110 − + H (the Darcy − Weisbach equation from pipe flow) (4) γ 2gA 32
and
e 1 9.35 = 1.14 − 2 log + (the ColeBrook − White equation from pipe flow) D Re f f
(5)
The solution gives Q2 = 322.40 cfs, Q3 = 127.60 cfs, f = 0.03036, Y1 = 5.965 ft and pt = 33.17 psf. (To solve these five equations you need to use the techniques described in Appendix B as the Newton method, or Mathcad or MATLAB on your PC.) From these the forces on the CV become 14Y12 1.5Y13 = 22,162 lbs, F2 = γA 2 h c2 = γ 41.6(22.6) = 6,749.2 lbs, F1 = γA1h c1 = γ + 3 2
p F3 = γ A 3 t +2 = 1,985.1 lbs. γ
34
Open Channel Flow: Numerical Methods and Computer Applications Applying the momentum equation in the x and ydirections, respectively, gives the following two equations: xdirection 22,162 − 6,749.18 cos 30 − 1,985.1 cos 45 − R x = 1.94[(322.4)7.75 cos 30 + (127.6)10.15 cos 45 − 450(3.288)] = 3,105.1
and solving R x = 11,808 lbs
ydirection −6,749.2 sin 30 + 1,985.1 sin 45 + R y = 1.94[(322.40)7.75 sin 30
− (127.6)10.154 sin 45] = 646.3 and solving R y = 2,617 lbs
which results in R = 12,095 lbs at an angle of 12.5° from the horizontal with its direction downward to the right on the structure.
1.12.1 Momentum Flux Correction Coefficient, β When the momentum principle is applied as if a threedimensional problem is a twodimensional problem, as in Example Problem 1.13, or as if a twodimensional problem is a onedimensional problem, i.e., when an average velocity is used for a section of flow, then ρQV does not represent the exact momentum flux, EMF, passing the section. The EMF can be obtained by multiplying by a correction coefficient, β in a manner similar to the use of α as a correction for the velocity head. By definition this coefficient, β, is
∫
∫
v2dA EMF ρ v dQ β= = = V2 A ρQV ρQV
(1.23)
For a natural channel the integral is replaced by a summation and Equation 1.23 becomes
∑
v∆Q EMF ρ β= = = ρQV ρQV
∑ v ∆A 2
V2 A
(1.23a)
and Equation 1.20 becomes
r = (ρβQV)2 − (ρβQV)1 Example Problem 1.14 The velocity distribution with depth in a wide rectangular channel is given by v = Ln(1 + 10y) where y is the distance from the channel bottom to the position where the velocity v is given. If the depth of flow Y = 5 ft, determine the momentum correction coefficient β. Solution Per unit width β = ∫v2dy/(qV) in which q is the volumetric flow rate per unit width and equal the depth time the average velocity V. To solve for β it is necessary to first solve q = ∫vdy = ∫Ln(1 + 10y)dy with limits of 0 to 5. This can be integrated by letting x = 1 + 10y, and dy = dx/10; thus q = ∫Ln(x)dx/10 with limits from 1 to 51, or q = [x Ln(x) − x]/10 evaluated between 51 and 1; giving q = 15.0523107. You will find it instructive to have your HP48 calculator obtain this answer by numerically integrating the above equation, and also have it carry out the symbolic integration. Since you will be using computer programs throughout this course, how about
(1.20)
Dimensions, Terminology, and Review of Basic Fluid Mechanics gaining some experience with this problem, but utilizing the Simpson’s numerical integration described in Appendix B? The main program that calls SIMPR is listed below; the first column obtains q and also prints out the average velocity, and the second column integrates ∫v2/dy between 0 and 5, and also prints out β. EXTERNAL EQUAT CALL SIMPR(EQUAT,0.,5.,VALUE) WRITE(*,*) VALUE,VALUE/5. END FUNCTION EQUAT(Y) EQUAT=ALOG(1.+10.*Y) RETURN END
EXTERNAL EQUAT CALL SIMPR(EQUAT,0.,5.,VALUE) WRITE(*,*) VALUE,VALUE/45.312708 END FUNCTION EQUAT(Y) EQUAT=ALOG(1.+10.*Y)**2 RETURN END
The results are q = 15.05205 cfs/ft, V = 3.0104 fps from the first integration, and ∫v2dy = 48.73822 and β = 1.0756. Example Problem 1.15 Determine α and β for the natural channel in Example Problem 1.5. Solution The numerator in Equation 1.19 can be viewed as a double integral since the differential area dA = dy dx, or this numerator is
∫∫
Xf
v3dy dx =
∫ 0
1 Xf 1 Vs3Y V′ 3dy ′ dx = v′ 3dy ′ Vs3Y dx 0 0 0
∫
∫
∫( )
Note that the quantity within { } after the last equal sign is constant, since the dimensional velocity profile does not change from position to position and, therefore, it needs to be evaluated only once. Thus, the argument of numerical integration with respect to x (across the channel) involves the surface velocity Vs cubed, multiplied by the depth Y at this section, i.e., Vs3Y, and both Vs and Y can be evaluated using the cubic spline interpolation of the data in the first table in Example Problem 1.5. The same applies for the numerator in Equation 1.23 to evaluate β; with the surface velocity squared rather than cubed. The program EXPR1_15 (in both FORTRAN and C) is given below to provide the numerator for both Equations 1.19 and 1.23. It also provides the area of the cross section, which can be done by evaluating the last integral in the above equa0 tion with the exponent of the velocity equal to zero, since Vs = 1. Note that this program contains the additional variable NE that is used as the exponent of the surface velocity. When NE = 0, the area of the cross section is evaluated; when NE = 2, the numerator for β is evaluated, and when NE = 3, the numerator for α is evaluated. Listing for Program EPR1_15.FOR EXTERNAL VPROF,Dq REAL DUM(30) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX,NE READ(2,*) NP,(YP(I),VP(I),I=1,NP) READ(2,*) NX,(X(I),Y(I),VS(I),I=1,NX) CALL SPLINESU(NP,YP,VP,D2VP,DUM,0) CALL SPLINESU(NX,X,Y,D2Y,DUM,0) CALL SPLINESU(NX,X,VS,D2VS,DUM,0) NE=0 CALL SIMPR(Dq,0.,X(NX),AREA,1.E6,20) WRITE(*,*)' Area=',AREA NE=2 5 I1=1
35
36
Open Channel Flow: Numerical Methods and Computer Applications 1 2 3 1 2 3
I2=2 CALL SIMPR(VPROF,0.,1.,qPRIM,1.E6,20) WRITE(*,*)' Integral of dimensionless', &velocity**',NE,' profile=',qPRIM I1=1 I2=2 CALL SIMPR(Dq,0.,X(NX),Q,1.E4,20) WRITE(*,*)' Integral with respect to x=' &,Q*qPRIM NE=NE+1 IF(NE.LT.4) GO TO 5 END FUNCTION VPROF(YY) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX,NE IF(YY.LT.YP(I2) .OR. I2.EQ.NP) GO TO 2 I1=I2 I2=I2+1 GO TO 1 IF(YY.GE.YP(I1) .OR. I1.EQ.1) GO TO 3 I2=I1 I1=I1–1 GO TO 2 DYP=YP(I2)YP(I1) A=(YP(I2)YY)/DYP B=1.A VPROF=(A*VP(I1)+B*VP(I2)+((A*A1.)*A* &D2VP(I1)+(B*B1.)*B*D2VP(I2))*DYP**2/6.)**NE RETURN END FUNCTION Dq(XX) COMMON YP(20),VP(20),X(30),Y(30),VS(30), &D2VP(20),D2VS(30),D2Y(30),I1,I2,NP,NX,NE IF(XX.LT.X(I2) .OR. I2.EQ.NX) GO TO 2 I1=I2 I2=I2+1 GO TO 1 IF(XX.GE.X(I1) .OR. I1.EQ.1) GO TO 3 I2=I1 I1=I1−1 GO TO 2 DX=X(I2)X(I1) A=(X(I2)XX)/DX B=1.A AA=A*(A*A1.)*DX**2/6. BB=B*(B*B1.)*DX**2/6. DEPTH=A*Y(I1)+B*Y(I2)+AA*D2Y(I1)+BB*D2Y(I2) VSURF=A*VS(I1)+B*VS(I2)+AA*D2VS(I1)+ &BB*D2VS(I2) Dq=DEPTH*VSURF**NE RETURN END
Listing for Program EXPR1_15.C #include <stdio.h> #include <stdlib.h> #include <math.h>
Dimensions, Terminology, and Review of Basic Fluid Mechanics float yp[20],vp[20],x[30],y[30],vs[30],d2vp[20],d2y[30],d2vs[30]; int i1,i2,np,nx,ne; extern float simpr(float (*equat)(float xx),float xb,float xe,\ float err,int max); extern void splinesu(int n,float *x,float *y, float *d2y,float *d,\ int ity); float vprof(float yy){float a,b,dyp; while((yy>=yp[i2])&&(i20)){i2=i1;i1–;} dyp=yp[i2]yp[i1]; a=(yp[i2]yy)/dyp;b=1.a; return pow(a*vp[i1]+b*vp[i2]+((a*a1.)*a*d2vp[i1]+(b*b1.)\ *b*d2vp[i2])*dyp*dyp/6.,ne); } // End of vprof float qp(float xx){float a,b,dx,aa,bb,depth; while((xx>=x[i2])&&(i10)){i2=i1;i1––;} dx=x[i2]x[i1];a=(x[i2]xx)/dx;b=1.a; aa=a*(a*a1.)*dx*dx/6.;bb=b*(b*b1.)*dx*dx/6.; depth=a*y[i1]+b*y[i2]+aa*d2y[i1]+bb*d2y[i2]; return depth*pow(a*vs[i1]+b*vs[i2]+aa*d2vs[i1]+bb*d2vs[i2],ne); } // End of qp void main(void){FILE *fili; char filnam[20];int i; float dum[30],qprim,q; printf("Give input file name\n"); scanf("%s",filnam); if((fili=fopen(filnam, "r"))==NULL){printf("Cannot open file\n"); exit(0);} fscanf(fili, "%d",&np); for(i=0;i
37
38
Open Channel Flow: Numerical Methods and Computer Applications
Problems Terminology 1.1 Define the flows in the following situations according to (1) steady or unsteady; (2) uniform or nonuniform and if nonuniform, whether rapidly or gradually varied; (3) subcritical or supercritical; and (4) turbulent or laminar. If insufficient information is provided for a classification, indicate what other information is needed. (a) A flow in a river that produces a storm hydrograph at the site of observation. (b) The flow downstream from a gate supplying water to a prismatic channel from a constant water surface elevation reservoir. (c) The same for as in (b) except the gate is slowly being closed. (d) A constant flow rate entering a trapezoidal canal of constant cross section that supplies water to three turnouts (consider: (1) the flow upstream from the first turnout, (2) the flow between the turnouts, and (3) the section downstream from the last turnout). A depth of water is required for the canal to supply the last turnout. (e) Water flow into a canal from a large lake with the control gate fully open. The canal has a constant cross section and a small bottom slope. 1.2 Water at a temperature of 55°F is flowing at a rate Q = 550 cfs in a trapezoidal channel with a bottom width of b = 12 ft, and a side slope m = 1.5. If the depth of flow is 4.5 ft, compute (a) the Froude number, (b) the Reynolds number, and (c) the speed of a small amplitude gravity wave in this flow. Classify the flow in this channel. 1.3 A cross section of a natural channel has the following transect data: x (ft) y (ft)
0 0
2 0.6
4 1.5
5 2.5
6 4.5
8 5.0
10 4.3
12 2.8
14 1.2
15 0.3
17 0
For a flow rate of Q = 280 cfs at a depth of 4.6 ft compute (a) the crosssectional area, (b) the Froude number, and (c) the Reynolds number if the water temperature is T = 50°F. 1.4 The sketch below shows a long canal system whose bottom slope changes at several points, and is controlled by gates. Indicate the reaches where the flow is subcritical (downstream controlled) and where it is supercritical (under upstream control). Under uniform flow a steep channel will sustain a supercritical flow whereas a mild channel will sustain a subcritical flow.
1.5 Starting with the perfect gas law ρ = p/(RT) and the definition of bulk modulus, prove that a gas undergoing an isothermal compression or expansion has a bulk modulus equal to its absolute pressure. 1.6 The speed of sound propagates through a fluid with a velocity given by c = E v /ρ. Compare the speed of sound in pure water to that of water that contains 1% free air by volume. 1.7 Compute the pressure variation in the ocean to a depth of 4000 m if the compressibility is not ignored, and then compute it on the basis of an incompressible fluid. In this computation assume both g and Ev are constant. Take the density of ocean water as ρ = 1020 kg/m3. 1.8 Water is flowing down a steep spillway with slope So = 0.30 at a depth of 0.8 m. Determine the pressure at the bottom of this channel. 1.9 A pressure transducer records 4.00 psi of pressure on the bottom of a steep channel with a bottom slope of two to one (horizontaltovertical distances). What are the normal and vertical depths of this flow? The bottom slope is constant. 1.10 At the base of a dam spillway the bottom of the channel suddenly changes from having a constant slope of So = 0.40 to a circular arc with a radius of 30 m. The normal depth of flow at
39
Dimensions, Terminology, and Review of Basic Fluid Mechanics
this point of change is 0.75 m and the velocity V = 10 m/s. Assume the radius of curvature of the streamlines changes linearly from 30 to 20 m through this depth, and compute the pressure distribution through the depth of flow where the radius is 20 m, and also the pressure distribution along the bottom of the channel through the length where the radius of curvature varies. What is the force against a section of fluid per unit width of flow at the section where the circular arc first begins? 1.11 Water is being rotated in a cylindrical drum of 2 m radius and height 5 m at a rotational speed of (a) ω = 1200 rpm and (b) ω = 40 rpm. The axis of rotation is vertical. Compute the pressure distribution in the vertical direction at the following three radii: r1 = 0, r2 = 1 m, and r3 = 2 m. When rotated, water is to the top of the tank. 1.12 Determine the average velocity, V, for a laminar flow that has a velocity profile given by Equation 1.11. How is the average velocity related to the velocity on the surface? How does this result relate to the area under a parabola? 1.13 Solve illustrative Example Problem 1.7 if the flow rate is Q = 20 m3/s and Y still equals 3 m in the upstream channel.
Continuity 1.14 Prove whether the following twodimensional flow fields are (1) continuous (satisfy the continuity equation), and (2) are rotational or irrotational. (a) u = U (constant) and v = 0 (b) u = 10x/(x2 + y2) and v = 10y/(x2 + y2). For this flow determine the tangential and radial components of velocity. (c) u = sin(x)cosh(y) and v = cos(2x)sinh(y) (d) u = 3x2 − 2y2 and v = 6xy (e) u = sinh(y)cos(x) and v = cosh(y)sin(x) (f) u = 5(x2 − y2) and v = 10xy 1.15 The velocity and depth of water flowing in a 2 m diameter pipe are 3 m/s and 1.4 m at one section. At another section the depth equals 0.6 m. What is the velocity at this second section? 1.16 The flow from a trapezoidal channel with b = 10 m and m = 2 goes through a transition section and enters a circular culvert with an 8 m diameter. If the flow rate is 350 m3/s and the depths in these two channels equal 4 m and 6.4 m respectively, what are the average velocities at the two sections. 1.17 A river whose cross section is defined by the following x y data from the left bank (y is positive downward from the top of the left bank) has been gaged by a current meter giving the velocities shown at the 0.2 and 0.6 depth at five position across the river, when the river stage was 2 ft. Under the assumption that the average of these two velocities gives the average velocity for this incremental position across the river, determine the flow rate in the river. Crosssection data for the river x (ft) y (ft)
0 0
2.0 0.8
4.0 1.1
6.0 2.3
8.0 2.1
10.0 1.0
12.0 0.1
Current meter measurements Position
1
2
3
4
5
x (ft) v0.2 (fps) v0.8 (fps)
1.2 1.1 1.3
3.6 1.3 1.5
6.0 1.6 1.9
8.4 1.4 1.5
10.8 1.2 1.3
40
Open Channel Flow: Numerical Methods and Computer Applications
1.18 Water in a rectangular channel flows around a 90° bend. The channel is 10 ft wide, the inside radius of the bend is r i = 30 ft and the radius of the outside of the bend is ro = 40 ft. A flow rate of Q = 400 cfs is in the channel when the depth of water at the inside of the bend is measured at Yi = 5 ft. Make the following assumptions: (a) there is no secondary flow; (b) the angular momentum is constant, i.e., the flow around the bend follows the free vortex law; (c) the velocity does not change with position between the bottom of the channel to the water surface at any give radial distance. Determine the following: (1) The depth at the outside of the bend, (2) The velocity at the inside of the bend, Vi, and (3) the velocity at the outside of the bend. (To accomplish these tasks you should first develop the equation that gives the depth through the bend as a function of the radius r, and the free vortex constant, and then evaluate the free vortex constant by making sure that 400 cfs passes through the bend.)
=4 r0
30
ft
t
b = 10 ft
=
0f
r1
1.19 Water comes off a spillway with a velocity of V = 100 fps at its toe. The spillway flip bucket at its end consists of a circular arc with a radius R = 20 ft. The depth of flow through the flip bucket is 2 ft. Assume that the velocity distribution through the depth of flow is constant. Derive the equation that gives the pressure distribution in the water on the spillway bucket through a vertical section passing through the center of the circle, i.e., the section vertically above point A on the sketch. What is the pressure at the bottom of the flow at this position? 10
0f
ps
2f
t
V=
R=
A
20
ft
p=?
Energy 1.20 Prove that for an irrotational flow that if the radius of curvature for a particle of fluid remains constant that its velocity is given by v = C/r (C = constant). Therefore, as rϖ0, then vϖ∞. 1.21 Obtain the equation y = f(x) that describes the top surface of water flowing over a sharp crested weir in a wide channel under the following assumptions: (a) the top surface remains horizontal for all x < 0 (where x has its origin at the weir crest), (b) there is no friction between this top streamline and the fluid below it, and (c) there is no pressure gradient in the normal direction at the top streamline. The average velocity and depth upstream from the weir are Vo and Yo, respectively. What will cause the real flow to have a top surface that deviates from this mathematical description.
41
Dimensions, Terminology, and Review of Basic Fluid Mechanics
8
7.5
1.22 A weir at the end of a wide rectangular channel has its crest 1.2 m above the channel bottom, and the depth of water above the weir crest is 0.8 m. Under the assumption that the flow is inviscid, compute the thickness of the falling water at a point 2.5 m below the weir crest. 1.23 An irrotational flow occurs around a 90° circular bend in a rectangular channel. The channel is 10 ft wide, and the inside radius of the bend is 60 ft. If velocity at the inside of the bend is 8 fps, and the velocity does not vary in the vertical direction, determine the flow rate in this channel if the depth at the inside of the bend is 4.5 ft. What is the difference between the water surface elevation at the inside and outside of the bend under this ideal situation? How would the real flow around this bend deviate from this behavior? 1.24 Assume the velocity distribution is parabolic in a shallow depth of flow in a wide channel. Determine the value of the kinetic energy correction coefficient, α. 1.25 Lines of constant velocity are shown in cross sections of a flow in three channels. Determine the kinetic energy correction coefficient for each.
5
6
7.5
7
7
2
6
4
1.26 Lateral outflow at a rate of 2 cfs/ft is taking place over a 20 ft long side weir. There is zero flow downstream from this weir, and the width of the channel is 5 ft. Develop the equation that describes the depth of flow as a function of x across the length of the side weir. How would this solution be complicated if the discharge from the side weir depended upon the depth of water above its crest to the 3/2 power? 1.27 The velocity distribution in a wide rectangular channel is as given in illustrative Problem 1.13, v = Ln(1 + 10y). Determine the kinetic energy correction coefficient, α. 1.28 Assume the velocity distribution with depth in a trapezoidal channel with a bottom width b = 3 m and a side slope m = 1.2 is given by the equation v = 0.5 Ln(1 + 3y) (m/s). The depth of flow Y = 2 m. What is the volumetric flow rate? What is the kinetic energy correction coefficient, α? What is the average energy; per unit weight in this channel flow?
Momentum 1.29 Solve Example Problem 1.13 if the pressure delivered at the end of the pipe is 45 psi instead of 40 psi. 1.30 Compute the resultant force on a spillway bucket that turns a flow rate of q = 300 cfs/ft from its direction down the spillway at a slope of So = 0.4 to the horizontal direction. The depth of flow at the beginning of this bucket is 5 ft and at the end of the bucket is 4.6 ft. 1.31 If the spillway bucket of the previous problem turned through a total angle of 33° what would the force on the bucket be? How high would the jet of water rise above the bottom of the bucket and what would its velocity be at this highest point? 1.32 A structure takes water in the vertical direction from a position 10 ft below the bottom of a channel under atmospheric pressure. The area of this outlet is 50 ft2. The upsteam channel is of trapezoidal shape with b = 10 ft and m = 1.5, and the depth in this channel just upstream from this structure is 4 feet. Compute the flow rate and the resultant force on this structure. Ignore frictional losses.
42
Open Channel Flow: Numerical Methods and Computer Applications
4 ft b = 10 ft m = 1. 5
10 ft
50 ft2
2f
t
v=
Cl
n(1
35°
35° rb=12 ft
rs
=1
0f
t
1.33 Determine the momentum flux correction coefficient, β for the velocity v = 0.5 Ln(1 + 3y) given for the flow in the trapezoidal channel of Problem 1.28 (b = 3 m, m = 1.2 and Y = 2 m). 1.34 A flip bucket at the toe of a spillway has a central portion consisting of a circle with radius of rb = 12 ft. After a short smooth transition from the constant slope spillway face, the circular portion of the bucket begins at an angle of 35° to the left of the vertical and it ends at an angle of 35° to the right of the vertical as shown in the sketch. After the end of the circular portion a constant section 2 ft in length ends the bucket. A flow rate of q = 160 cfs/ft is coming down the spillway, and the depth normal to the bottom at the beginning of the circular arc is 2 ft. Assume that the velocity distribution can be defined by the following logarithmic function of the radius r. v = C ln(13 − r). Do the following: (1) Determine the constant C (in integral tables you will find Iln(x)dx = x ln(x) − x). (2) Write the expression that provides the pressure gradient dp/dr through the circular portion of the spillway bucket. How would you obtain the pressure distribution at the end of the circular portion of the bucket at the plus 35° angle?
q=
1
c 60
fs/
ft
3r
)
1.35 Repeat Example Problems 1.5 and 1.15, except use the following dimensionless velocity profile, rather than the one given in Example Problem 1.5. Dimensionless depth, y′ Dimensionless velocity, V′
0.00
0.03
0.05
0.10
0.15
0.20
0.60
0.80
0.90
1.00
0.00
0.30
0.55
0.75
0.95
0.98
0.985
1.00
1.01
1.00
1.36 The program EXPRB1_5 that was written to solve Example Problem 1.5 integrated the dimensionless velocity profile to obtain a dimensionless unit flow rate q′. Modify this program so the actual velocity profile at each position x is integrated to get the actual unit flow rate q. (In doing this you can evaluate q for each unit of width of channel, and sum these q’s to get the total flow rate Q.) What flow rate, channel area, and average velocity do you get with this process? 1.37 Using the approach used in the previous problem, determine the values of the energy correction coefficient α, and the momentum correction coefficient β for the natural channel in Example Problem 1.5.
and Its Dissipation 2 Energy in Open Channels 2.1 Introduction In applying the energy principle to the flow of liquids in open channels, it is necessary at the outset to be able to describe the resistance of fluid to motion. This resistance occurs because all real fluids have the property called viscosity, which causes internal shearing stresses to exist within the fluid as a consequence of a velocity gradient as it passes over a solid boundary. Viscosity is defined in Chapter 1. At the solid boundary, the velocity in the fluid must be zero, or agree with the velocity of the boundary, and increase therefrom for movement to take place. Thus it takes work, or energy, to cause motion of fluids relative to the boundaries that contain the fluids. This resistance and its effects are given several names such as “frictional resistance,” “viscous shear,” “friction factors,” “friction losses,” “friction energy dissipation,” “frictional head loss,” etc. Often, friction is omitted. In the case of liquids, the internal processes associated with fluid friction involve the conversion of useful fluid energy, which is originally in the form of a potential or kinetic energy, into a non recoverable energy, i.e., increase in temperature of the liquid. It is appropriate, therefore, to consider it a head loss, or dissipation of energy per unit weight of fluid. The increase in temperature is very small and of minor, if any, significance. The first part of this chapter deals with how this head loss can be determined practically in computing depths of flow, and velocities that will occur if a given volumetric flow rate is to occur in a channel of a given size. These losses will be restricted to uniform flows. Nonuniform flows will be dealt with in Chapter 4. After being able to determine this loss, the second part of this chapter deals with the application of the energy principle to open channel flows. In this part, concepts associated with specific energy in open channel flow will be covered, as well as the differences between subcritical and supercritical flows.
2.2 Approaches to Frictional Resistance The problem of frictional resistance in open channel flows is complex and depends on sound engineering judgment in selecting appropriate coefficients. The subject is made more complex by the fact that many channels are unlined, and therefore may have a moveable bed under some, or all flow conditions, that exist in that channel. The effects of bed movement are not considered in this chapter and are dealt with in articles and volumes dealing with sediment transport, its scour, and deposition and flow in alluvial channels. This chapter is restricted to flow resistance in fixed bed channels. The different means for handling the resistance to motion in channels can be roughly classified as (1) the more fundamentally sound friction factor approach similar to that used for pipe flows when utilizing the Darcy–Weisbach equation and (2) based on tested and widely used empirical equations. In engineering practice, the use of empirical equations dominates in computing frictional losses in open channels. The reason for this is associated with the complexities involved with the friction factor approach. The same occurs in pipe flow computations in which the Hazen–Williams equation is used more extensively than is the Darcy–Weisbach equation. The use of empirical equations should be limited to ranges of situations for which they give good answers. A more fundamentally 43
44
Open Channel Flow: Numerical Methods and Computer Applications
sound approach generally does not have these restrictions. As computers take over the task of doing the arithmetic, there will likely be a trend in engineering practice to switch to the friction factor approach. The friction factor approach will be described first and it will be followed by a section dealing with the use of Manning’s equation. Before beginning this discussion of fluid frictional losses in open channel flows, it should be pointed out that other complex phenomena may operate in dissipating fluid energy that are not covered by this theory, and using the results from this theory for these channel flows may produce erroneous answers. Such a flow, for example, exists in a steep mountain stream with very large bed elements that extend up to, or above, the water surface in many locations. Data taken from such a river/stream that flows through the Rocky Mountain Hydraulics Laboratory (Jarrett, 1991) indicate that energy dissipated is proportional to the velocity raised to the 8.3 power, whereas frictional theory for turbulent flows has the energy dissipation proportional to the velocity squared. Empirical equations, such as Manning’s equation, are not appropriate for such flows. In fact, neither is Manning’s equation appropriate to determine flow depths, etc. on steep spillways with slopes 0.2 or greater even though they are made of relatively smooth concrete, because it has not been developed to describe such large velocity flows that have largescale turbulence associated with them. Neither is Manning’s equation suited for extremely lowvelocity flows through tall grass, for example.
2.2.1 Friction Factors in Open Channels In 1768, a French engineer, Antoine Chezy, reasoned that the resistance to flow in an open channel would vary with the wetted perimeter and with the square of the velocity and that the force to balance this resistance would vary with the area of the cross section and with the slope of the channel. Therefore, he proposed that
V2 P V2 = = Constant (AS) (R hS)
and would be the same for any similar channel. He used this in designing a canal for the Paris water supply. Years later, in 1897 his manuscript was published, and as his method became known and adopted by other engineers, the square root of the constant became know as the Chezy coefficient, and the equation
Q = CA(R hS)1/2
(2.1a)
V = C(R hS)1/2
(2.1b)
or
became known as Chezy’s equation. While Chezy took C as a constant for a channel with a fixed wall roughness, we now know that C is a function of the Reynolds number (Re = 4VR h/ν = 4Q/(νP); also note as above that the hydraulic radius R h is the area divided by the wetted perimeter, or R h = A/P) of the flow as well as the relative roughness (e/R h) of the channel wall, or C = f(Re,e/R h). In other words, not only does C depend on the type of channel, but its value also depends on the flow conditions in that channel. The processes associated with fluid resistance in open channel flows are similar to those that cause head losses in pipeline flows. Therefore, it is possible to get considerable insight into channel resistance by utilizing what is known from experimentation and theory, and has been adopted into practice in pipe fluid friction. The Darcy–Weisbach equation, hf = f(L/D)(V2/2 g) defines the
45
Energy and Its Dissipation in Open Channels
frictional head loss in a pipe flow, in which f is a friction factor whose magnitude depends upon the relative roughness of the pipe wall, e/D, and the Reynolds number of the flow in the pipe. The Darcy–Weisbach equation is accepted as the fundamentally sound and best method for computing head losses or pressure drops in pipelines due to known flow rates. Therefore, this equation will be compared with Chezy’s equation. The Darcy–Weisbach equation indicates that the slope of energy line, or the head loss, hf, divided by the length, L, of pipe over which this loss occurs equals a friction factor, f, divided by the pipe diameter (which is a representative parameter for the pipe size with dimensions of length), multiplied by the velocity head, V2/2 g. If the hydraulic radius R h multiplied by 4 is used in place of the pipe diameter, then the Darcy–Weisbach equation can be written as S=
hf fV2 fQ 2 = = L 4R h (2g) 8R h (gA2)
(2.2)
in which the friction factor, f, is a function of Reynolds number and the relative roughness, e/D, of the pipe wall (e is the roughness of the pipe wall and D is the diameter of the pipe). The substitution of 4R h in place of D can be justified by noting that for a pipe, the hydraulic radius that equals the area divided by the perimeter is Rh = (πD2/4)/(πD) = D/4. It is worth noting that if f is dimensionless, then the terms separated by equal signs in Equation 2.2 are all dimensionless. Thus, Equation 2.2 can be obtained from dimensional analysis, and this analysis leads to the conclusion that f does depend on the two dimensionless parameters, Re (Reynolds number) and e/D. The friction factor f will have the same value regardless of whether ES or SI units are used. A comparison of Equation 2.1a and b with Equation 2.2 gives the following relationship between the Darcy–Weisbach friction factor, f, and Chezy’s coefficient, C: C=
8g f
(2.3)
For pipe flow, it is known that the friction factor f can be defined well by a Moody diagram that is a plot of f as the ordinate against Reynolds number as the abscissa with different curves for different values of relative roughness e/D. This diagram is defined by the following equations for the following four different types of flow that occur:
1. Laminar flow f=
64 Re
2. Hydraulically smooth flow (the wall roughness are well embedded within the laminar sublayer, less than 1/5 its size, and consequently have no influence on the magnitude of f) 1 = 2Log10 (R e f ) − 0.8 f 3. Transitional zone (in which both Re and e/D determine f’s magnitude; this is the Colebrook– White equation) e e 1 2.52 9.35 = −2Log10 + = 1.14 − 2Log10 + 3.7D R e f D R e f f
46
Open Channel Flow: Numerical Methods and Computer Applications
4. Wholly rough zone (in which Re no longer effects f’s magnitude)
1 e 3.7D = 1.14 − 2Log10 = 2Log10 D e f
Equations that define Chezy’s C might be determined by substituting Equation 2.3 into the above four equations. When doing this for laminar flow, the following equation results:
C=
gR e 8
(2.4)
Equation 2.4 applies for flows in which the Reynolds number is less than about 2100. Experimentation has shown that the value of 8 in Equation 2.4 does not hold constant for all channel, but no fully accepted range of values has been established. Since laminar flows are rare when dealing with water as the fluid, the value of 8 can probably be used, at least for a reasonable first approximation. Substituting Equation 2.3 into the hydraulically smooth equation gives
gR e C = 32gLog10 0.887C
(2.5)
There is some question about whether the value 0.887 should be modified slightly to fit available experimental data. However, since very limited quality experimental data are available for hydraulically smooth flow, the derived value will be used in this book. For flows that occur in the transitional zone in which both the relative roughness of the channel wall and the Reynolds number have an influence on the magnitude of the frictional resistance, the constants that are obtained from substituting Equation 2.3 into the above transitional equation might be modified slightly to give a better fit of some experimental data. (See ASCE Task Force, 1963.) With these modified coefficients the equation is
e 0.884C C = − 32g Log10 + 12R h R e g
(2.6)
If the original coefficients that come from the Colebrook–White or other experimentally based equation are preferred, then Equation 2.6 can be written in the more general form
e bC C = −cLog10 + aR h R e g
(2.6a)
in which the a, b, and c can be given slightly different values. For example, when using the Colebrook–White equation directly a = 12, b = 0.887, and c = (32 g)1/2. For the wholly rough zone substitution of Equation 2.3 into the wholly rough equation that defines the Moody diagram gives
e 12R h C = − (32g) Log10 = (32g) Log10 e 12R h
(2.7)
These equations are summarized in Table 2.1, and Equations 2.5 through 2.7 are plotted on Figure 2.1 to give “Chezy C” diagram for turbulent flows. If Equation 2.4 were plotted on a left addition to Figure 2.1, it would result in a straight line with a slope of 1/2, since power equations such as Equation 2.4 plot as straight lines on log–log graph paper with the exponent in the equation giving the slope on this plot.
47
Energy and Its Dissipation in Open Channels
TABLE 2.1 Summary of Equations That Define Chezy’s Coefficient, C Type of Flow
Equation Giving C
Equation No.
Range of Application
Laminar
C = (gRe/8)
2.4
Re < 2100
Hydraulically smooth
Re g C = 32g log10 0.887C
2.5
Re < 2100
Transition
e 0.884C + C = − 32g log10 12 R Re g h
2.6
2100 < Re < Ve/(Cν) = 100
Wholly rough
e C = − 32glog10 12R h
2.7
Re > Ve/(Cν) = 100
12R h or C = 32g log10 e
100
180
Chezy’s C = V/√RhS (SI units)
80
Material PVC Very smooth concrete Concrete (smooth froms) Ordinary concrete Untreated gunite Rough concrete Clean dirt ditch
70
e (mm)
e (ft)
0.0015 0.30 0.49 1.2 2.0 4.3 10.0
0.000005 0.0010 0.0016 0.0038 0.0067 0.014 0.03
mo
lly s
ca auli
r Hyd
160
e/Rh = 0.0004 e/Rh = 0.0006 e/Rh = 0.0008
140
oth
e/Rh = 0.002
e/Rh = 0.001
120
Relative roughness, e/Rh = 0.004
60
e/Rh = 0.006
e/Rh = 0.008
100
e/Rh = 0.01
e/Rh = 0.015
50
e/Rh = 0.02
80
e/Rh = 0.05
40
e/Rh = 0.1 e/Rh = 0.15
60
e/Rh = 0.2
30
Chezy ’s C = V/√RhS (ES units)
90
0005 = 0.0 e/R h 001 0.0 = e/R h 2 e/R h = 0.000
2
4
6
8 104
2
4
6 8 105 2 4 Reynold’s number, Re = 4V Rh/ν
6
8 106
2
4
6
8 107
50
FIGURE 2.1 Diagram for Chezy’s C for use in determining the flow rate, velocity and slope of the energy line, or head loss in open channels.
The friction factor f in the Darcy–Weisbach equation has a value that is independent of the units used. However, since C is not dimensionless, but has the dimensions of the square root of gravity, e.g., L1/2/T, its value will be different when using SI units than when using ES units. In Figure 2.1, the values for C when using SI units are the leftside ordinate, whereas the right ordinate applies when using ES units. An alternative to having different values of C for different units would be to modify Chezy’s equation to include g, or define the Chezy equation as
V = C1 gR hS
(2.1c)
However, historical developments have not done this, and therefore you must be sure that the appropriate value of C is used to the system of units that you are using. Since many open channel flows do fall within the transitional type of flow, some discussion of the characteristics of Equation 2.6 are in order. It should be noted that C occurs on both sides
48
Open Channel Flow: Numerical Methods and Computer Applications
of Equation 2.6 (the same is true of Equation 2.5). It is not possible to rearrange this equation so that C is on one side of the equation all by itself. Equations of this type are referred to as implicit equations, since an explicit solution of them is not possible. Use of general iterative techniques, such as the Newton method described in Appendix B can be used. However, because of the nature of Equations 2.5 and 2.6 they can be solved by a simple feedback iteration, called a Gauss–Seidel iteration, in which the C solved for on the left side of the equal sign is used for C on the right side of the equal sign for the next iteration. The value produced by the explicit Equation 2.7 can be used as the initial starting value for this iterative solution or easier, just start with a reasonable guess. The following few lines of FORTRAN and C code illustrate implementation of this iterative solution for a trapezoidal channel. For other types of cross sections, the two function statements at the top of this listing that define the area, A(B,FM,Y) and the wetted perimeter, P(B,FM,Y) need to be modified. Program CHEZYC.FOR A(B,FM,Y)=(B+FM*Y)*Y P(B,FM,Y)=B+2*Y*SQRT(FM*FM+1.) 5 WRITE(6,*)' Give:B,FM,Y,G,VISC,E,Q' READ(5,*,ERR=30) B,FM,Y,G,VISC,E,Q AR=A(B,FM,Y) V=Q/AR G8=.884/SQRT(G) SQG=SQRT(32.*G) C1=SQG*ALOG10(12.*AR/P(B,FM,Y)/E) 10 RH=AR/P(B,FM,Y) RE=4.*V*RH/VISC C=SQG*ALOG10(E/(12.*RH)+G8*C1/RE) IF(ABS(CC1).LT. 1.E8) GO TO 20 C1=C GO TO 10 20 WRITE(6,*)' CHEZYS COEF=',C GO TO 5 30 STOP END Program CHEZYC.C #include <stdlib.h> #include <stdio.h> #include <math.h> float a(float b,float m,float y){return (b+m*y)*y;} float p(float b,float m,float y){return b+2.*y*sqrt(m*m+1.);} void main(void){float b,m,y,g,visc,e,q,ar,v,g8,rh,re,c,c1,sqg; printf("Give: b,m,Y,g,Visc,e,Q\n"); scanf("%f %f %f %f %f %f %f",&b,&m,&y,&g,&visc,&e,&q); ar=a(b,m,y); v=q/ar; g8=.884/sqrt(g);sqg=sqrt(32.*g); c=sqg*log10(12.*ar/p(b,m,y)/e); do {c1=c; rh=ar/p(b,m,y); re=4.*v*rh/visc; c=sqg*log10(e/(12.*rh)+g8*c1/re);}while(fabs(cc1)<1.e8); printf("CHEZYS COEF =%f\n",c);} A typical problem involves considerably more than solving one of the equations in Table 2.1 for Chezy’s coefficient. The types of problems associated with steady flow can be categorized as follows:
49
Energy and Its Dissipation in Open Channels
1. The flow rate, or velocity is unknown, and all other variables are known. 2. The depth is unknown and all other variables are known. This type of problem typically asks a question like: With this given channel what will the depth of flow be if the flow rate equals Q. 3. One of the variables associated with the channel size is unknown, but the flow rate is known. This type of problem can be consider a design problem in which the size of channel needed to convey a specified flow rate is wanted. For a trapezoidal channel the unknown may be the bottom width, or the side slope, and for a circular section the diameter is the unknown. 4. All variables are known except the wall roughness. 5. The slope of the channel bottom is unknown, and all other variables are known. Of all problems this is the easiest, since its solution can be obtained most directly, by (a) solving the appropriate equation for Chezy’s coefficient, and (b) solving Chezy’s equation for S. (When the slope, S, refers to the channel bottom subscript o will be used and when the slope refers to the energy line subscript f will be used e.g., in most subsequent equations So or Sf will appear in Chezy’s as well as Mannings equation.)
Problems in any of these categories might be view, as a mathematical problem of solving two nonlinear simultaneous equations; Chezy’s equation and the appropriate equation from Table 2.1 that defines Chezy’s coefficient. Appendix B describes the Newton method for solving systems of nonlinear equations. Alternately software packages, such a Mathcad, TKSolver, or Matlab can be used, or math packs for pocket calculators might be used. In addition to the procedure described above for case (5) problems under category (1) can be solved by cycling through the following steps: (a) solving for C (Equation 2.6), (b) using this C compute Q, or V from Equation 2.1a and b, and (c) based on this Q, or V update Reynolds number and repeat steps (a) through (c) until a small enough change occurs between consecutive values of Q or V that you are willing to accept the results. Convergence of this procedure will be rapid because changes in flow rate have a relatively small effect on the value of C. Should the flow be in the wholly rough zone then only one cycle of steps (a) and (b) above completes the solution since Chezy’s coefficient is independent of the flow rate. Typical values for wall roughnesses that are appropriate are given in Table 2.2. Example Problem 2.1 What is the flow rate in a trapezoidal channel with b = 10 ft, m = 1.5, and a bottom slope of So = 0.0005 if the depth of flow is measured equal to 5 ft. Solution This problem falls in Category (1) above. The solution might begin by solving for C from Equation 2.7, which gives C = 127.49. Next the implicit Equation 2.6 is solved by the Gauss–Seidel iteration based on an assumed Reynolds number, i.e., 1.0E6 (or if one wishes the above C could be used in
TABLE 2.2 Value of Wall Roughness, e, for Different Channel Materials Material PVC Very smooth concrete Concrete (smooth forms) Ordinary concrete Untreated gunite Rough concrete Clean dirt ditch
e (m)
e (ft)
0.0000015 0.00030 0.00049 0.0012 0.0020 0.0043 0.0100
0.000005 0.0010 0.0016 0.0038 0.0067 0.014 0.03
50
Open Channel Flow: Numerical Methods and Computer Applications Chezy’s equation to compute the Reynolds number). The result is C = 125.15. Now using this C the velocity is V = 4.944 fps from Chezy’s equation, and the associated Reynolds number equals 5,019,874 (assuming v = 1.23E−5). To get the solution the steps of (a) solving C from Equation 2.6 based on the most recent Reynolds number, (b) solve V from the Chezy equation (Equation 2.1a and b) and updating Re = 4R h/v. The results are iteration # 2, C = 126.98, V = 5.017 fps, Re = 5,093,390; iteration # 3, C = 126.99, which is close enough giving a flow rate Q = 439.0 cfs. Example Problem 2.2 A trapezoidal channel with b = 8 ft, m = 1.2, and a bottom slope of 0.0006 is to convey a flow rate of Q = 300 cfs. The wall roughness of the channel is e = 0.004 ft. Determine the depth of flow in this channel. Solution An effective way to solve this problem is to use the Newton method (Appendix B) to solve Chezy equation and Equation 2.6 simultaneously for C and Y. For the Newton method these equation can be written as
F1 = C + (32g )
1/ 2
e 0.884C =0 Log10 + 12R h R e g
(
)
F2 = Q − CA(R hS)1/2 = 0
(1)
(2)
Using the Newton method the unknown vector, consisting of the two values C and Y, is updated by the following iterative equation:
C Y
( m +1)
C = Y
(m )
z1 − z 2
in which the vector z is the solution to the linear system of equations:
∂F1 ∂C ∂F2 ∂C
∂F1 ∂Y z1 F1 = ∂F2 z 2 F2 ∂Y
Instead of actually taking the partial derivatives shown above a numerical approximation can be used. This numerical approximation evaluates the function (i.e., equation) twice with the second evaluation based on decreasing (or increasing) the variable that the derivative is taken with respect to by a small increment and then dividing the two values of the function by this increment. The following FORTRAN program can be used to obtain this solution. It is designed to solve for C as well as any of the following variables: e, b, m, Y, Q, or S in a trapezoidal channel. Input to this program as well as the values to solve this problem consist of: 1 the number of the unknown variable which is 4 for Y for this problem, the list of known, including a guess for Y, in the following order: e = 0.004, b = 8, m = 1.2, Y = 4(guess), Q = 300, S = 0.0006, g = 32.2, and kinematic viscosity = 1.23E–5. The solution produced by the program is Y = 4.46 ft. The problem can be solved by using Figure 2.1 in connection with Equation 2.1a and b. You should at least verify Chezy’s C from Figure 2.1. The value of C from the above procedure is C = 125. Listing of FORTRAN program to solve problem using the Chezy equation (CH2PR2): REAL X(6),D(2,2),F(2) EQUIVALENCE (E,X(1)),(B,X(2)),(FM,X(3)),(Y,X(4)),(Q,X(5)), &(S,X(6)) A(B,FM,Y)=(B+FM*Y)*Y RH(AR,B,FM,Y)=AR/(B+2.*Y*SQRT(FM*FM+1.)) F1(C,SQG,RH1,RE,G8,E)=C+SQG*ALOG10(E/(12.*RH1)+G8*C/RE) F2(Q,C,AR,RH1,S)=QC*AR*SQRT(RH1*S)
Energy and Its Dissipation in Open Channels 1
20
100
WRITE(6,*)' Give: No. of UNK,1e,2b,3m,4Y,5Q,6S,g & Vis' READ(5,*) IUN,X,G,VISC IF(IUN.LT.1 .OR. IUN.GT.6) GO TO 1 G8=.884/SQRT(G) SQG=SQRT(32.*G) C=SQG*ALOG10(12.*RH(A(B,FM,Y),B,FM,Y)/E) AR=A(B,FM,Y) RH1=RH(AR,B,FM,Y) RE=4.*Q*RH1/(VISC*AR) F(1)=F1(C,SQG,RH1,RE,G8,E) F(2)=F2(Q,C,AR,RH1,S) C=.98*C D(1,1)=(F(1)F1(C,SQG,RH1,RE,G8,E))/(.02040816*C) D(2,1)=(F(2)F2(Q,C,AR,RH1,S))/(.02040816*C) C=C/.98 X(IUN)=.95*X(IUN) AR=A(B,FM,Y) RH1=RH(AR,B,FM,Y) RE=4.*Q*RH1/(VISC*AR) D(1,2)=(F(1)F1(C,SQG,RH1,RE,G8,E))/(.05263158*X(IUN)) D(2,2)=(F(2)F2(Q,C,AR,RH1,S))/(.05263158*X(IUN)) X(IUN)=X(IUN)/.95 FAC=D(2,1)/D(1,1) D(2,2)=D(2,2)FAC*D(1,2) F(2)=F(2)FAC*F(1) DIF=F(2)/D(2,2) X(IUN)=X(IUN)DIF DIF1=(F(1)DIF*D(1,2))/D(1,1) C=CDIF1 IF(ABS(DIF)+ABS(DIF1).GT. .001) GO TO 20 WRITE(6,100) C,X(IUN),X FORMAT(' C=',F8.2,' Unknown=',F10.3,/,' e=',F10.5,/, &' b=',F8.2,/,' m=',F8.2,/,' Y=',F8.3,/,' Q=',F10.2,/, &' S=',F8.5) WRITE(6,*)' Give 1 to solve another problem; otherwise 0' READ(5,*) IUN IF(IUN.EQ.1) GO TO 1 STOP END
Example Problem 2.3 Determine the depth of flow in a circular channel with D = 10 ft, if its bottom slope equals 0.0005 and its wall roughness equals 0.004 ft and it is to convey a flow rate of Q = 150 cfs. Solution The above program can be modified by changing the function statement to obtain the area by a function subprogram, and function statement for the hydraulic radius to apply for a circular channel. These statements could consist of and
FUNCTION A(D,BETA,Y) BETA=ACOS(1.2.*Y/D) A=D*D/4.*(BETACOS(BETA)*SIN(BETA)) RETURN END RH(AR,D,BETA,Y)=AR/(D*BETA)
51
52
Open Channel Flow: Numerical Methods and Computer Applications The above listing might be altered with BETA replacing FM, and or the present FORTRAN names used with different meaning. The solution gives Y = 4.613 ft, and Chezy’s C = 123.1. Example Problem 2.4 You are to size a trapezoidal channel that is to carry a flow rate of Q = 50 m3/s. The slope of the channel is 0.0012 and it is to be made of formed concrete. For stability of the channel sides their slopes are to be 1.5, and the depth is not to exceed 2 m. What should the bottom width be? Solution The computer program of Example Problem 2.2 will solve this problem. If this program is used the input consists of: 2,.00049,3,1.5,2,50,.0012,9.81,1.14E−6. The solution is b = 4.94 m. If Mathcad is available to you it would be a excellent experience to use it, or some other software package to get the same solution.
2.3 Combining the Chezy and the Chezy C Equations An alternative to solving Chezy’s equation and the Chezy C equation for the transitional zone simultaneously for any of the variables is to eliminate C first by solving for it from Chezy’s equation, and then substituting this in the equation that applies within the transitional zone where it occurs on both sides of the equal sign. By eliminating C between these two equations there is one equation and one of the variables can be solved as the unknown. The resulting equation is implicit for all variables except e, and so in general must be solved by an iterative technique. The HP48 calculator’s SOLVE capability can be used; however, because A and P are functions of the depth and size variables the resulting equation becomes long and complex. However, this approach is readily implemented in a computer program, in TKSolver, or Mathcad models where separate statements can be used to define A and P. Solving Chezy’s equation for C gives C=
Q P/(AS0 ) A
or C = V
P AS0
The transitional zone equation that gives C can be written as
eP 0.221νCP F(ξ) = C + 32g Log10 + =0 Q g 12A
If the first of the two equations above, that contains Q, is used that gives C using Q then the following equation results: F (ξ ) =
eP 0.221νP 23 Q P/(AS0 ) + 32g Log10 + =0 12A A 23 gS0 A
and if the second of the above two equations that gives C using the velocity V then the following equation results: F (ξ ) = V
eP 0.221νP 23 P + 32g Log10 + =0 12 A A 23 gS0 (AS0 )
Energy and Its Dissipation in Open Channels
53
After solving either of these two equations depending on whether the flow rate Q is given (or the unknown) or whether the velocity is given (or the unknown), then Chezy’s equation is solve for the coefficient C if this value is desired. The program CHEZYCTC.FOR, which is given below, implements such a solution using the Newton method to solve for the selected unknown from the variables: m, b, S, Y, e, Q, or V if the channel is trapezoidal, and D, S, Y, e, Q, or V if the channel is circular. The technique used to accommodate both trapezoidal and circular channels is to not use X(1) (for m) and change V(2) to D (the Character string) if the channel is circular. When IC = 1, for a circular channel, then the equation that gives A and P for a circular channel are used; otherwise those that give these quantities for a trapezoidal channel are used. (See Statements starting with label 55.) The approach is very similar to that used in solving the DW and CW equations. This program does not contain the logic, however to generate a guess for the unknown, that is needed in the Newton method. Rather the user must supply this guess as well as the known values. The variable and rule sheets from TKSolver are listed below the program listing that handle first the equation that assumes that the flow rate Q is the flow variable, which is in the combined equation with the other channel property variables, and the second is for the equation that involves the V as the flow variable. Listing of program CHEZYCTC.FOR for solving the Chezy equation for any of the variables for both a trapezoidal and a circular channel
1
100
10 12
REAL X(7) CHARACTER*19 FMT/'(1X,A2,3H = ,F10.4)'/ CHARACTER*1 V(7)/'m','b','S','Y','e','Q','V'/ WRITE(*,*)' Give 1=ES or 2=SI(or 0/=STOP),', &'0=trap or 1=cir. & Visc' READ(*,*) II,IC,VISC IF(II.LT.1) STOP G=32.2 IF(II.GT.1) G=9.81 VISC2=.221*VISC/SQRT(G) G32=SQRT(32.*G) IF(IC.GT.1) THEN I2=2 V(2)='D' ELSE I2=1 V(2)='b' ENDIF WRITE(*,100)(I,V(I),I=I2,7) FORMAT(' Give No. of Unknown',/(I2,'  ',A2)) READ(*,*) IU IF(IU.GT.5) GO TO 10 WRITE(*,*)' Give 1 if Q will be given or 2',= if V is known' READ(*,*) IV GO TO 12 IV=IU5 WRITE(*,*)' Give values to knowns &=,= GUESS for unknown' I3=7 IF(IV.EQ.1) I3=6 DO 20 I=I2,I3
54
20 50 52 55
60
70
Open Channel Flow: Numerical Methods and Computer Applications
IF(IV.EQ.2 .AND. I.EQ.6) GO TO 20 WRITE(*,"(A2,' = ',\)") V(I) READ(*,*) X(I) CONTINUE M=0 XX=X(IU) X(IU)=1.005*X(IU) DX=X(IU)XX IF(IC.EQ.1) THEN COSB=1.2.*X(4)/X(2) BETA=ACOS(COSB) A=.25*X(2)**2*(BETACOSB*SIN(BETA)) P=X(2)*BETA ELSE A=(X(2)+X(1)*X(4))*X(4) P=X(2)+2.*X(4)*SQRT(X(1)**2+1.) ENDIF ADL=G32*ALOG10(X(5)*P/(12.*A)+VISC2*(P/A)**1.5/SQRT(X(3))) IF(IV.EQ.1) THEN F=X(6)*SQRT(P/(A*X(3)))/A+ADL ELSE F=X(7)*SQRT(P/(A*X(3)))+ADL ENDIF M=M+1 IF(MOD(M,2).EQ.0) GO TO 60 X(IU)=XX F1=F GO TO 55 DIF=DX*F/(F1F) X(IU)=XXDIF IF(ABS(DIF).GT. .00001 .AND. M.LT.30) GO TO 52 IF(IV.EQ.1) THEN X(7)=X(6)/A ELSE X(6)=A*X(7) ENDIF DO 70 I=1,7 FMT(18:18)='3' IF(I.EQ.3 .OR. I.EQ.5) FMT(18:18)='6' WRITE(*,FMT) V(I),X(I) WRITE(*,FMT) 'C',X(7)*SQRT(P/(A*X(3))) GO TO 1 END
═══════════ VARIABLE SHEET ════════════ St Input──── Name─── Output─── Unit──── A 75 10 b 1 m 5 Y
Energy and Its Dissipation in Open Channels
55
P 24.142136 Q 478.55564 .001 S 32.099844 g32 .01 e .0000141 v 32.2 g C 114.47967 ═══════════════ RULE SHEET ═════════════════ S Rule────────────────────────────────────── A=(b+m*Y)*Y P=b+2*Y*sqrt(m^2+1) Q*sqrt(P/(A*S))/A+g32*Log(e*P/(12*A)+.221*v/ sqrt(g*S)*(P/A)^1.5)=0 C=Q*sqrt(P/(A*S))/A ═══════════ VARIABLE SHEET ════════════ St Input──── Name─── Output─── Unit───── A 75 10 b 1 m 5 Y P 24.142136 V 6.3807419 .001 S 32.099844 g32 .01 e .0000141 v 32.2 g C 114.47967 ══════════════ RULE SHEET ════════════════ S Rule──────────────────────────────────── A=(b+m*Y)*Y P=b+2*Y*sqrt(m^2+1) V*sqrt(P/(A*S))+g32*Log(e*P/(12*A)+.221*v/ sqrt(g*S)*(P/A)^1.5)=0 C=V*sqrt(P/(A*S)) For laminar flow or water to occur in open channels either the depth or the velocity must be very small. For example, assume water at 15.6°C (60°F) so its kinematic viscosity is v = 1.123 × 10 −6 m2/s (1.217 × 10 −5 ft2/s), and that the largest value of Reynolds number, VR h/v allowed for laminar flow is 500, and that the channel is very wide so that R h = Y, then the product of the velocity time the depth VY ≤ 500(1.123 × 10 −6) = 0.0005615 for SI units or VY ≤ 0.0061 for ES unit. The tables below show these limiting values. The Froude number Fr in the third column of these tables is defined by Fr = V gY . Notice that as the depth becomes very small, in the order of 0.01 ft, or 0.003 m, and the velocity consequently larger that the flow may becomes supercritical (Fr greater than 1). To have conditions right to allow a flow to be simultaneously laminar and supercritical are not common. Furthermore, when the depth becomes small enough for Froude numbers to be larger than unity then the surface tension of the water becomes a significant factor. When water flows in very thin sheets on steep surfaces, it tends to form thread like streamlets. Also most channel surfaces, such as gutter, or road way beds, where very small depths of open channel water flows may be found in
56
Open Channel Flow: Numerical Methods and Computer Applications
practice, are not smooth enough and the water will actually be seen to flow in the lower indentations. The sheet flow over watershed surfaces that occurs from rainfall, which does not infiltrate into the soil, tends to erode the smaller particles and by so doing forms a system of mini channel, which will grow in size in time especially if the rainfall is intense. The last columns in these tables give the bottom slope of the channel that would be required for the flow to take place as computed by the Chezy equation. Note that for super critical flows the bottom slopes must also be very large. Limiting Depths, Velocities and Froude Numbers for Laminar Flow of Water in Wide Open Channels. T = 15.6°C (60°F) v = 1.123 × 10 m2/s (1.217 × 10 ft2/s) V (ft/s)
Y (ft)
Fr
So
V (m/s)
Y (m)
Fr
So
0.020 0.040 0.0625 0.125 0.250 0.500 0.010 0.025 0.050 0.100 0.200 0.300
0.304 0.152 0.097 0.049 0.024 0.012 0.609 0.243 0.122 0.061 0.030 0.020
0.006 0.018 0.035 0.100 0.282 0.799 1.072 0.271 0.096 0.034 0.012 0.007
0.000015 0.000117 0.000447 0.003577 0.028619 0.228955 0.412688 0.026412 0.003302 0.000413 0.000052 0.000015
0.0060 0.0120 0.0200 0.0500 0.1000 0.1500 0.0030 0.0050 0.0100 0.0200 0.0500 0.1000
0.0936 0.0468 0.0281 0.0112 0.0056 0.0037 0.1872 0.1123 0.0561 0.0281 0.0112 0.0056
0.0063 0.0177 0.0381 0.1506 0.4261 0.7828 1.0910 0.5071 0.1793 0.0634 0.0160 0.0057
0.000008 0.000062 0.000288 0.004495 0.35962 0.121372 0.235792 0.050931 0.006366 0.000796 0.000051 0.000006
2.4 Empirical Formula: Use of Manning’s Equation Before the turn of the twentieth century, systematic research was underway to define better fluid resistance in open channel flow. It was recognized then that C in the Chezy equation was not constant under all flow conditions in a given channel. Bazin proposed the following formula that found use in the past to better define C: C=
157.6 1 + m Rh
(for ES units)
in which m takes on a different value depending on the roughness of the channel wall. In 1868, Gauckler proposed that for flat slopes in open channel flow, C varies as the sixth root of the hydraulic radius. Others came to the same conclusion and the result has now been widely accepted throughout the world and is known in the United States as Manning’s formula even though the name Manning is a misnomer and it has been proposed that the formula be called Gauckler–Manning’s equation in part to undo the incorrect naming. In Europe the same formula is called the Strickler formula. Because of its wide use, this will be the formula used in this text book as an alternative to Chezy’s formula. Manning’s equation can be written as V=
Cu 2 / 3 R h So n
or
Q=
C A2 /3 C A5/3 Cu AR 2h / 3 So = u A 2 / 3 So = u 2 / 3 So n n P n P
(2.8)
57
Energy and Its Dissipation in Open Channels
in which Cu equals 1 when using SI units, and Cu = 1.486 (the cube root of the number of feet per meter) when using ES units, and n is the roughness coefficient of the channel wall. In this equation the following symbols apply: A = the cross section of the flow in ft2 when using ES units and in m2 when using SI units; P is the wetted perimeter and is the length of contact between the water and the channel when viewed in a direction normal to the flow direction, and has units of ft in the ES system, and m in the SI system; R h is the hydraulic radius, which is defined as the area A divided by the wetted perimeter P, with units of ft in ES units and m in SI units; and So is the slope of the channel bottom, which for uniform flow equal the slope of both the water surface in the channel as well as the energy line of the flow. The slope of the energy line equal the head loss divided by the length over which this loss occurs. Therefore So might be thought of as hL/L for uniform flow. Since three digits of precision cannot be maintained in the selection of n, it is common to use Cu = 1.49 in ES units. Typical values for use in Manning’s equation for different material that channel are constructed from are given in Table 2.3. If n value larger than 0.05, the largest value in Table 2.3, is needed to describe a given channels flow, other mechanisms than just fluid friction are likely involved and Manning’s equation is probably inappropriate, e.g., a constant value of n will not describe depths, etc. over much of a range of flow rates. This condition exists for very small velocity flows that may be approaching flow through a porous media, or very large velocity flows. It is worth noting that Manning’s formula indicates that the head loss is proportional to the square of the velocity, or flow rate. On the Chezy C diagram this corresponds to the wholly rough zone. When the Reynolds number becomes small with a magnitude of 100,000 or less then according to the Chezy equation the head loss is proportional to the velocity to a power less than 2, but greater than 1. For laminar flow, Equation 2.4 indicates that the head loss is proportional to the velocity to the first power. Thus the use of Manning’s equation will essentially duplicate the results obtained from Chezy’s formula for very large Reynolds numbers when the flow lies in the wholly rough zone, provided corresponding roughness values are selected. On the other hand, the use of Manning’s equation is questionable for lowvelocity flows in small smooth channels. A natural question is: What is the relationship between the values of Manning’s n and the equivalent sand roughness e used in connection with Chezy’s formula? The answer is that there is no direct
TABLE 2.3 Typical Values for Manning’s n Channel Material Lined channels Smooth brass, glass, lucite, PVC Cement plaster Planed lumber, unpainted steel, trowelled concrete Unplaned lumber, smooth asphalt, vitrified clay, brick in cement mortar, cast iron Asphalt (rough), untreated gunite Rough concrete Corrugated metal Natural channels Clean excavated earth Earth (good condition), rock excavation, gravel (straight chan.) Earth (straight with some grass) Earth (winding, no grass), clean natural beds Gravel beds (plus large boulders) Earth (winding), weedy streams
n 0.010 0.011 0.012 0.013 0.016 0.020 0.023 0.023 0.025 0.026 0.030 0.040 0.050
58
Open Channel Flow: Numerical Methods and Computer Applications
relationship, or simple equation that can give n from e, or e from n. For any given e, the value of n is different depending on the channel type and size, and the flow rate, i.e., Reynolds number of this flow. To get an n that corresponds to an e for any given situation, one must solve both equations. For example, if one wanted to determine what n corresponds to e = 0.004 ft, it is first necessary to decide what channel and flow rate this correspondence is to apply for. To illustrate, assume a flow rate of 400 cfs occurs in a rectangular channel with a bottom width of 10 ft and bottom slope of 0.0005 has a known e = 0.004 ft. First the depth is solved from Chezy’s equation. The depth is Y = 8.04 ft. Next Manning’s equation is solved for n, giving n = 0.0141. Thus for this channel containing this flow rate the corresponding n = 0.0141 for an e = 0.004 ft. For this e, the value of n will be different for each different flow rate and each channel. Table 2.4 illustrates this variation of n with several different variables that can be changed in a trapezoidal channel. Since a rectangular channel is a special trapezoidal channel with side slope m = 0, this table includes a couple of rectangular channels. All of the values for n given in this table correspond to a equivalent sand roughness of e = 0.004 ft in the Chezy equation. The first three columns in Table 2.4 are for the rectangular channel used in the above illustration with a bottom width of 10 ft and a bottom slope of So = 0.0005. The flow rate was varied from 8 to 400 cfs in this channel. The second columns gives the depths that are obtained by solving Chezy’s formula, and the third column indicates the value of n that is solved by Manning’s equation for this column 2. This third column of n values shows that n increases with Q and Y from 0.0131 to 0.0141, or a 7.6% change over this range of conditions. The next three columns, i.e., columns 4, 5, and 6 in Table 2.4, were obtained similarly except for a trapezoidal channel with a side slope of 1.5 (with b = 10 ft and So = 0.0005 as for the rectangular channel). With the flow rate changing from 12 to 600 cfs, the variation in Manning’s n is again 7.6%. The next four groups of three columns each were obtained by holding the flow rate constant, letting the slope of the channel bottom vary, solving Chezy’s equation for the depth, and based on this computed depth and other variables solving Manning’s equation for n. The third column of each group of 3 shows the variation of n. These variations in n are from 3.9% to 5.1%. The last three columns in Table 2.4 were obtained for a rectangular channel with a bottom slope of So = 0.0005 and the depth held constant at 2 ft. The first column of this group that gives the bottom width b was varied from 0.5 to 25.0 ft, and the flow rate in column 2 was obtained by solving Chezy’s formula. The last column represents the solution of Manning’s equation for this flow rate and channel. Note again a variation in the n value of 5.4%. Table 2.5 represents the results from similar calculations to those used to get the values in Table 2.4 with the exception that they apply to circular cross sections, and the units are SI instead of ES. Approximately the same variations in n occur. Both of these tables represent common channel sizes, and the value of e = 0.004 ft is a typical value for manmade lined channel. The variation of 3%–8% shown in these tables is within the accuracy of selecting n for a given application. Therefore, we might conclude that for most practical problems it is satisfactory to use Manning’s equation. Should the Reynolds number of the flow be less than 100,000, then it is probably best to use Chezy’s formula even if it does entail a little more arithmetic. The accuracy of predicting flow rates and/or other variables in open channels cannot be expected to be better than the percentages shown in Tables 2.4 and 2.5. If one assumes Manning’s equation applies only in the wholly rough zone, then equating Chezy’s and Manning’s equations, V = C(R hSo)1/2 = (Cu/n)R h2/3So1/2, with C defined by Equation 2.7 gives
Cu 1 6 12R h R h = 32g Log10 n e
If we take n as dimensionless, then this equation indicates Cu has dimensions of L1/3/t. Making the assumption that n is dimensionless allows the same values of n to be used for both ES and SI units. This is what has occurred in practice, i.e., Cu = 1 for SI units, and for ES units Cu is the cubic root
7.6%
Y
0.64 2.46 3.62 4.50 5.88
n
7.6%
0.0132 0.0137 0.0140 0.0141 0.0142
Note: Q is in cfs and Y is in ft, b is in ft.
Variation
12 120 240 360 600
Q
8 80 160 240 400
n
Y
0.52 2.39 3.95 5.37 8.04
Q
0.0131 0.0137 0.0139 0.0140 0.0141
b = 10 ft, m = 1.5 So = 0.0005
b = 10 ft, m = 0 So = 0.0005
Variables Held Constant
0.0001 0.0005 0.0010 0.0025 0.0050
So 12.39 6.39 4.88 3.46 2.69
Y
n
4.4%
0.0143 0.0141 0.0140 0.0138 0.0137
b = 10 ft, m = 0 Q = 300 cfs 0.0001 0.0005 0.0010 0.0025 0.0050
So 8.05 5.35 4.46 3.48 2.87
Y
5.1%
0.0145 0.0142 0.0141 0.0139 0.0138
n
b = 10 ft, m = 1.5 Q = 500 cfs So 0.0001 0.0005 0.0010 0.0025 0.0050
2.77 1.42 1.08 0.765 0.594
Y
n
3.9%
0.0134 0.0132 0.0131 0.0130 0.0129
b = 2 ft, m = 0 Q = 5 cfs 0.0001 0.0005 0.0010 0.0025 0.0050
So
2.58 1.76 1.49 1.18 0.988
Y
4.6%
0.0137 0.0134 0.0133 0.0132 0.0131
n
b = 2 ft, m = 1.5 Q = 20 cfs
TABLE 2.4 Variations of Manning’s n with a Fixed Value of e = 0.004 ft (v = 1.217 × 10−5 ft2/s) and Other Variables Changed in a Trapezoidal Channel
b 0.5 1.0 5.0 10.0 25.0
0.937 2.758 26.462 61.932 174.04
Q
n
5.4%
0.0130 0.0131 0.0135 0.0136 0.0137
m = 0, Y = 2 ft So = 0.0005
Energy and Its Dissipation in Open Channels 59
Y
0.58 1.29 2.87 3.77 5.38
n
8.8%
0.0135 0.0140 0.0145 0.0146 0.0147
Q
0.073 0.438 1.02 1.75 3.43
0.204 0.492 0.770 1.05 1.83
Y
n
5.3%
0.0131 0.0134 0.0136 0.0137 0.0138
D = 2 m So = 0.0005
Note: Q is in m3/s and Y is in m, and D is in m.
Variation
1.20 6.00 26.4 40.0 60.0
Q
D = 6 m So = 0.0005
Variables Held Constant
Q 0.0007 0.0054 0.0122 0.0204 0.0306
0.0359 0.0987 0.1526 0.2087 0.3086
Y
n
2.3%
0.0134 0.0131 0.0131 0.0131 0.0131
D = 0.33333 m So = 0.0005 So
5.8%
0.0001 0.0003 0.0010 0.0020 0.0050
0.722 0.498 0.354 0.294 0.238
n
4.6%
0.0136 0.0134 0.0132 0.0131 0.0130
Y
n 0.0147 0.0145 0.0144 0.0141 0.0139
Y 3.70 2.97 2.28 1.57 1.24
D = 1 m Q = 0.20 m3/s
D = 6 m Q = 25 m3/s 0.0002 0.0004 0.0010 0.0020 0.0050
So
0.0001 0.0005 0.0010 0.0020 0.0050
So
TABLE 2.5 Variations of Manning’s n with a Fixed Value of e = 0.001219 m (v = 1.31 × 10 −6 m2/s) and Other Variables Changed in a Circular Channel
3.23 7.26 10.3 14.6 21.8
Q
D = 3 m Y = 2 m n
1.4%
0.0142 0.0141 0.0141 0.0141 0.0140
So 0.0001 0.0005 0.0010 0.0020 0.0050
0.025 0.056 0.080 0.113 0.179
Q
n
3.1%
0.0134 0.0131 0.0131 0.0130 0.0130
D = 0.5 m Y = 0.3 m
60 Open Channel Flow: Numerical Methods and Computer Applications
61
Energy and Its Dissipation in Open Channels
of the number of feet per meter, or Cu = (1/.3048)1/3 = 1.486. However, it is generally accepted that n is not dimensionless. Often when working in one system of units (either ES or SI), the quantity on the right of the above equation, which is Chezy’s C, as well as Cu are taken as dimensionless, and therefore the literature will often suggest that n has dimensions of L1/6. A log–log plot of the above equation is given below. If n is assumed to have dimensions of L1/6 then one would expect n to vary as the onesixth root of the wall roughness size, e, or n = Ke1/6 (where K is a constant). Such a relationship is shown on the graph as a dashed line, but any other line parallel to this line could be used depending on what n one selects to correspond to e for a given hydraulic radius. For the given dashed line R h is taken equal to 3 ft and Cu R1h/6 /{n(32g)1/2} = 2.079 corresponding to e/R h = 0.1 (Figure 2.2). Thus for e = 0.3 ft, n = 0.0267, or n = 0.03268e1/6. This gives n = 0.0182 corresponding to e = 0.03 ft, and n = 0.0124 corresponding to e = 0.003 ft. Note from this graph that the dashed line would fit the curve closer if its slope were flatter; suggesting that the dimensions of n might be closer to L1/7. Let us now focus attention on solving Manning’s equation, which can be solved directly if the flow rate Q, the roughness coefficient n, or the slope of the channel bottom So is the unknown. To solve for n, its place is interchanged in Equation 2.8 with Q. In solving for So Manning’s equation becomes 2
2 2 nQ P 2 / 3 nQ P 2/3 nV 2/3 = = So = R 5 /3 Cu Cu A Cu A A
(2.8a)
If one of the variables that goes into defining the crosssectional area A, the wetted perimeter P, i.e., the hydraulic radius R h is unknown, then Manning’s equation becomes implicit in that variable, and must be solved by an iterative method such as the Newton’s method, or by trial and error. Consider a trapezoidal section for example. For a trapezoid, the area and wetted perimeters are 5
4
n(32g)1/2
CuRh1/6
3
2
1 0.0001
n= 1/6
e
CuRh
Rh
n(32g)1/2
0.00010 0.00015 0.00020 0.00025 0.00030 0.00040 0.00050 0.00060 0.00070 0.00080 0.00090 0.00100 0.00150 0.00200 0.00250 0.00300 0.00400
5.079181 4.903090 4.778151 4.681241 4.602060 4.477121 4.380211 4.301030 4.234083 4.176091 4.124939 4.079181 3.903090 3.778151 3.681241 3.602060 3.477121
2
3
0.00500 0.00600 0.00700 0.00800 0.00900 0.01000 0.01500 0.02000 0.02500 0.03000 0.04000 0.05000 0.06000 0.07000 0.08000 0.09000 0.10000 5
3.380211 3.301030 3.234083 3.176091 3.124939 3.079181 2.903090 2.778151 2.681241 2.602060 2.477121 2.380211 2.301030 2.234083 2.176091 2.124939 2.079181
0.001
0 .0
326
8e 1/6
1
Relative roughness, e/Rh
6
0.01
FIGURE 2.2 Relationship between n parameter and relative roughness to establish relation of n to e/R h.
0.1
62
Open Channel Flow: Numerical Methods and Computer Applications
defined respectively by A = (b + mY)Y, and P = b + 2Y m 2 + 1, and Manning’s equation written as a function of this unknown equal to zero for use in the Newton iterative Equation B.2 becomes
F(ξ) = nQ b + 2Y m 2 + 1
2 /3
− [(b + mY)Y]5 / 3 Cu So = 0
(2.8b)
in which variable ξ represent b, Y, or m depending respectively whether the bottom width, the depth or the side slope is the unknown. The Newton method for solving implicit equations such as Equation 2.8b is discussed in Appendix B, and is illustrated by example problems below. If the channel is circular, then it is best to introduce the additional angle β = cos−1(1 – 2Y/D) as defined in Appendix A. Thus the depth Y is related to this angle (in radians) by Y = D (1 − cos β)/2, and Manning’s equation can be written as
F(ξ) = nQ(βD)
2 /3
D2 (β − cos β sin β) − 4
5/ 3
Cu So = 0
(2.8c)
in which variable ξ now represents either β or D depending respectively whether the depth, or the diameter is unknown. Should the depth be unknown then β is first obtained by solving for it from Equation 2.8c, and thereafter the depth Y is computed by the equation above Equation 2.8c. Example Problem 2.5 A flow rate of Q = 450 cfs is taking place in a trapezoidal channel with the following properties: b = 10 ft, m = 1, and So = 0.0006. Determine the uniform depth of flow in this channel if the appropriate value for Manning’s n is, n = 0.013. Solution This problem can be solved using Equation 2.8b and the Newton method. The Newton method can easily be solved using a programmable pocket calculator using the following steps:
1. Assign the variables, Y, b, m, Q, So, and n to storage registers, and place the values for these variable in the assigned registers (this includes a guess for the unknown Y in this case). 2. Put the calculator in program mode and program Equation 2.8b into it with the value of the equation F(ξ) = F(Y) being displayed upon completion of the equation. 3. Press the operate button on the calculator, and when it is complete store the value displayed in an unused register. 4. Retrieve the value from the register that hold Y, and increase it by a small amount such as 0.001. 5. Store this increased value in the same register for Y again. 6. Press the operate button again. 7. Recall the last value of the equation, subtract it from the current value and divide this difference by 0.001. 8. Recall Y, subtract the result from step 7, and also subtract 0.001 from it and store Y back in its register. 9. Repeat steps 4 through 8 until convergence has occurred. If you want you can also program these steps into your pocket calculator. The implementation of these steps for this problem results in the following: Step 1 Register Value
1(Y)
2(b)
3(m)
4(Q)
5(So)
6(n)
5
10
1
450
0.0005
0.013
63
Energy and Its Dissipation in Open Channels from step 3 F(Y) = 2.902; from step 6 F(Y) = 2.738, after step 8 register # 1 contains 5.018 for Y. After iteration # 2 Y = 5.018, which represents no change to three digits beyond the decimal point, and the solution is terminated. If you have a calculator with the capability to solve implicit equations with a SOLVE key such as the HP 48xs, then all that is needed is to define the equation and give values to the variables. Example Problem 2.6 A flow rate of Q = 30 cfs is to be carried by a pipe with a bottom slope of So = 0.00028, and a Manning’s roughness coefficient of 0.013. If the depth is not to exceed 3/4 of the diameter, what size pipe should be used? Solution Since the depth is not to exceed 3/4 of the diameter, then cos β = 1 – 3/2 = − 1/2, and the equation that must be solved is F(D) = nQ (βD )
2/3
D2 − (β − sin β cos β ) 4
5/3
C u So
for the pipe diameter D. Using the Newton method, the solution to this equation is D = 4.49 ft. Taking the next standard pipe size would call for using a pipe with a 54 in. diameter. Example Problem 2.7 A pipe of diameter 2 m and Manning’s n = 0.013 has a bottom slope of So = 0.00112. Generate a table that gives the depths of flow that would be expected in this pipe under uniform flow conditions for flow rates of Q = 0.5 m3/s to Q = 4.5 m3/s in increments of 0.5 m3/s. Solution The solution for each entry in this table requires that Equation 2.8c be solved by an iterative method such as the Newton method. The following BASIC program implements such a solution for the nine different flow rates giving the depth shown to the right of the BASIC listing. BASIC program listing to solve Example Problem 2.7 10 INPUT "Give:n,So,D,Q1,DQ,est. for Y & No ",N,S,D,Q,DQ,Y,NO% 20 FOR I%=1 TO NO% 30 NCT%=0 40 ARG=12*Y/D 50 TANA=SQR(1ARG*ARG)/ABS(ARG) F ARG>0 THEN BETA=ATN(TANA) ELSE BETA=3.14159265#ATN 60 I (TANA) 70 A=.25*D*D*(BETAARG*SIN(BETA)) Q(m3/s) Y(m) _______ _____ 80 P=BETA*D 90 F=QA/N*(A/P)^.6666667*SQR(S) 0.5 0.42 0.60 100 IF NCT%>0 THEN GOTO 150 1.0 0.74 110 F1=F 1.5 0.87 120 Y=Y.001 2.0 0.99 130 NCT%=1 2.5 140 GOTO 40 3.0 1.10 1.22 150 DY=.001*F1/(F1F) 3.5 1.33 160 Y=YDY+.001 4.0 1.46 170 IF ABS(DY) > .00001 THEN GOTO 30 4.5 180 PRINT Q,Y 190 Q=Q+DQ 200 NEXT I 210 END
64
Open Channel Flow: Numerical Methods and Computer Applications Example Problem 2.8 A natural canal that has a bottom slope of 0.002, and a Manning’s n = 0.018 has the cross section defined by the following transect data for a long distance. Determine the depth of flow in this canal if the flow rate is Q = 90 cfs. x (ft) y (ft)
0 0
2 0.8
4 1.5
7 3.0
9 3.3
11 2.5
14 1.2
18 0.0
Solution From this data it is possible to interpolate along both sides of the canal to generate the following data giving the top width, the area, and perimeter for equal increments of the depth y. Depth, Y 0.17 0.33 0.50 0.66 0.83 0.99 1.16 1.32 1.49 1.65 1.82 1.98 2.15 2.31 2.48 2.64 2.81 2.97 3.14 3.30
Top Width
Area
Perimeter
1.58 3.01 4.30 5.44 6.44 7.29 7.99 8.55 8.96 9.22 9.41 10.21 11.07 12.03 13.05 14.03 15.01 15.99 16.99 18.00
0.13 0.51 1.11 1.92 2.90 4.03 5.29 6.65 8.10 9.62 11.18 12.80 14.56 16.46 18.53 20.77 23.16 25.72 28.44 31.33
1.62 3.10 4.44 5.63 6.68 7.59 8.37 9.02 9.61 10.22 10.88 11.75 12.66 13.68 14.75 15.78 16.82 17.86 18.91 19.98
The solution might proceed by selecting a depth, and then by interpolation in the above table determine the corresponding area and perimeter, substitute these into Manning’s equation, and compute the flow rate. Based on the difference between the computed flow rate and the wanted value of 90 cfs, adjust the depth and repeat the process. A more formal procedure would be to implement the Newton method in the above table interpolation to obtain a better estimate for the next depth to use. The answer is Y = 2.60 ft, with A = 20.21 ft2 and P = 15.56 ft. See Appendix B.5 for details related to how areas and wetted perimeters can be defined from x y crosssectional coordinates, etc.
Solving for the depth, or diameter, in circular sections by the Newton method requires that a good guess be supplied to start the iterative solution process, or else the method will fail. For circular sections, this guess must be much better than for trapezoidal channels. In the small BASIC program given under Problem 2.7 above it is left to the user to provide a satisfactory starting value. However, requiring the user to supply an initial guess is not always desirable. In most cases, an adequate initial guess can be easily generated within a computer program for solving for either the depth or diameter in a circular section, the two variables in Manning’s equation for which an explicit solution is not possible. If the area and wetted perimeter for a circular section are replaced in Manning’s equation by functions of the auxiliary angle β that define these quantities, and terms rearranged, then Manning’s equation can be written as
65
Energy and Its Dissipation in Open Channels
(β − cos β sin β) nQ Q′ = = 8/3 10.079368β2/3 Cu D So
5/ 3
= F(β)
(2.9)
It should be noted that Q′ is dimensionless if Cu has units of the cube root of length per time (with n taken as dimensionless). A close approximation of the above dependency of Q′ on β is given by the following power equation:
Q′ = 0.0313β 4.0984
(2.10a)
β = 2.3286(Q′)0.244
(2.10b)
or the inverse of this equation is,
Thus if the depth of flow is the unknown it is possible to compute Q′ = nQ / (Cu D8 / 3 So ), and then from Equation 2.10b obtain a starting value for β. The reasonableness of this approximation is shown by the values in the small table below that gives the approximate β, the actual β and the difference corresponding to several values of Q′. Note that as the depth approaches the diameter, i.e., β approaches π the difference get larger. Q′ 1.00E−06 1.00E−05 1.00E−04 1.00E−03 1.00E−02 0.100 0.200 0.300 0.330 0.335
βact
βappr
Difference
0.0822 0.1400 0.2388 0.4098 0.7160 1.3478 1.7364 2.1842 2.4564 2.5955
0.0800 0.1403 0.2461 0.4316 0.7570 1.3277 1.5723 1.7359 1.7767 1.7832
0.0022 −0.0003 −0.0072 −0.0218 −0.0410 0.0201 0.1641 0.4484 0.6797 0.8123
Manning’s equation will need to be solved frequently throughout the chapters of this book. It will be useful for you to develop a computer program, or program your pocket calculator so that it will be possible to readily obtain a solution to any variable that may be unknown in Manning’s equation for either a trapezoidal or a circular section. Below is a listing of such a program in TURBO PASCAL, which utilizes the clear screen CLRSCR and the GOTOXY capabilities to give the program a little “user friendliness.” You should study over the program carefully. The array element X[IU] contains the unknown. For a trapezoidal channel the correspondence between the X’s and the variables are: X[1] = Q, X[2] = n, X[3] = S, X[4] = Y, X[5] = b, and X[6] = m. For a circular section the variables are the same through X[4], but then X[5] = D (the pipe diameter). A good way for you to understand how this program solves Manning’s equation completely in either a circular or a trapezoidal section, and allows either ES or SI units to be used is for you to translate the program into FORTRAN, C or BASIC depending upon what you are most familiar with. Listing of PASCAL program MANNING.PAS that completely solves Manning’s equation in both trapezoidal and circular channels Program Manning; Const NX:array[1..7] of char=('Q','n','S','Y','b','m','D'); VNAM:array[1..4] of string[12]=('flow rate','coefficient', 'bottom slope','depth');
66
Open Channel Flow: Numerical Methods and Computer Applications
JW:array[1..4] of integer=(2,4,6,3); Var X:array[1..7] of real; F1,P,Beta,DF,DX,C,AA:real; ITY,I,IU,m:integer; Function Expn(a,b:real):real; Begin if a<0 then Writeln('error in power',a,b) else Expn:=Exp(b*Ln(a)) End; {raises a to the power b} Function A:real; Begin {computes area A & AA and perimeter P} If ITY=1 then begin P:=X[5]+2*X[4]*sqrt(sqr(X[6])+1); AA:=(X[5]+X[6]*X[4])*X[4]; end else Begin P:=12*X[4]/X[5]; if P=0 then Beta:=Pi/2 else begin Beta:=sqrt(1sqr(P))/abs(P); if P>0 then Beta:=arcTan(Beta) else Beta:=PiarcTan(Beta) end;AA:=sqr(X[5])/4*(BetaP*sin(Beta)); P:=Beta*X[5] End; A:=AA End; Function F:real; Begin {Defines Manning's Equation for Newton Method} F:=X[2]*X[1]C*A*Expn(AA/P,0.666667)*SQRT(X[3]) End; Var Ch:Char; Label L1; BEGIN {Start of program} L1:ClrScr; GoToXY(1,10);Writeln('Do you want to use:'); Writeln('1  ES units, or'); Writeln('2  SI units?'); repeat Readln(Ch); until Ch in ['1','2','E','e','S','s']; If Ch in ['1','E','e'] then C:=1.486 else C:=1; ClrScr;GoToXY(1,10); Writeln('Is section:'); Writeln('1  Trapezoidal, or');Writeln('2  Circular?'); repeat Readln(Ch); until Ch in ['1','2','T','t','C','c']; if Ch in ['1','t','T'] then ITY:=1 else ITY:=2; ClrScr; Writeln('Give no. of unknown'); For I:=1 to 4 do Writeln(I:2,'  ',NX[I]:1,' (',VNAM[I]); if ITY=2 then Writeln(' 5  D (diameter)') else begin Writeln(' 5  b (bottom width)'); Writeln(' 6  m (side slope)') end; repeat Readln(IU); until IU in [1..7]; ClrScr; Writeln('Give values for knowns');For I:=1 to 4 do if I<>IU then begin GoToXY(1,I+1);Write(NX[I]:1,' = '); Readln(X[I]) end; If (ITY=2) and (IU<>5) then begin GoToXY(1,6);Write('D = '); Readln(X[5]) end; If ITY=1 then Begin I:=5; if IU<>5 then begin I:=I+1; GoToXY(1,I); Write('b = ');Readln(X[5]) end; if IU<>6 then begin I:=I+1; GoToXY(1,I); Write('m = ');Readln(X[6]) end; End; Case IU of {1 thru 3 solve explicit eqs, 4,5 & 6 use Newton Method} 1:X[1]:=C/X[2]*A*Expn(AA/P,0.6666667)*sqrt(X[3]); 2:X[2]:=C/X[1]*A*Expn(AA/P,0.6666667)*sqrt(X[3]); 3:X[3]:=sqr(X[1]*X[2]/(C*A*Expn(AA/P,0.6666667))); 4,5,6:Begin If ITY=2 Then Begin If IU=4 then begin Beta:=2.3286*Expn(X[1]*X[2]/(C*Expn(X[5], 2.6666667)*sqrt(X[3])),0.244); X[4]:=X[5]/2*(1cos(Beta)) end else X[5]:=2*X[4] End Else
Energy and Its Dissipation in Open Channels
67
case IU of 4:X[4]:=X[5]/2; 5:X[5]:=2*X[4]; 6:X[6]:=1; end; m:=0; repeat F1:=F; DX:=X[IU]/100; X[IU]:=X[IU]DX; DF:=DX*F1/(F1F); X[IU]:=X[IU]+DXDF; m:=m+1; until (m>20) or (abs(DF)<0.0001); End; End; ClrScr; Writeln('Solution to unknown'); If (ITY=2) and (IU=5) then Write('D = ') else Write(NX[IU]:1,' = '); Writeln(X[IU]:11:JW[IU]+1); Writeln; Writeln('Variables of Problem:'); For I:=1 to 4 do Writeln(NX[I]:1,' = ',X[I]:10:JW[I]); If ITY=2 then Writeln('D = ',X[5]:10:3) else begin Writeln('b = ',X[5]:10:3); Writeln('m = ',X[6]:10:4) end; GoToXY(1,24); Write('Do You want to solve another problem?(Y or N) '); Readln(Ch); If (Ch='Y') or (Ch='y') then GoTo L1; END. Listing of FORTRAN program MANNING.FOR designed to solve Manning’s equation in either trapzoidal or circular channels LOGICAL REPT REAL X(6),n,m CHARACTER*1 V(6) COMMON X,BETA EQUIVALENCE (Q,X(1)),(n,X(2)),(S,X(3)),(Y,X(4)), &(b,X(5)),(m,X(6)) DATA V/'Q','n','S','Y','b','m'/ WRITE(6,*)' Give 1.49 for ES units or 1. for SI' &,= units.' READ(5,*) C WRITE(6,*)' Give 1 for trap. sec., or 2 for cir.' &,= sec.' READ(5,*) ITYP IF(ITYP.EQ.1) THEN No=6 ELSE No=5 V(5)='D' ENDIF 1 WRITE(6,100) (I,V(I),I=1,No) 100 FORMAT(' Give No. of unknown:'/6(I2,'',A1)) READ(5,*) IUNK DO 10 I=1,No IF(I.LT.4.AND.I.EQ.IUNK) GO TO 10 WRITE(6,101) V(I) 101 FORMAT(3X,A1,' = ',$) READ(5,*) X(I) 10 CONTINUE NCT=0 20 AA=A(ITYP)
68
21 22 23 24 25 40 105
Open Channel Flow: Numerical Methods and Computer Applications
GO TO(21,22,23,24,24,24), IUNK Q=C/n*AA*(AA/P(ITYP))**.6666667*SQRT(S) GO TO 40 N=C/Q*AA*(AA/P(ITYP))**.6666667*SQRT(S) GO TO 40 S=(n*Q/C*(P(ITYP)/AA)**.6666667/AA)**2 GO TO 40 REPT=.TRUE. F=n*QC*AA*(AA/P(ITYP))**.6666667*SQRT(S) IF(REPT) THEN REPT=.FALSE. F1=F X(IUNK)=X(IUNK).001 AA=A(ITYP) GO TO 25 ENDIF DIF=.001*F1/(F1F) X(IUNK)=X(IUNK)+.001DIF NCT=NCT+1 IF(NCT.LT.20 .AND. ABS(DIF).GT..00001) * GO TO 20 IF(NCT.EQ.20)WRITE(6,*)' FAILED TO CONVERGE' *,DIF WRITE(6,105)(V(I),X(I),I=1,No) FORMAT(' SOLUTION:',6(A2,' =',F12.6)) WRITE(6,*)' Give 1 for another prob.; else 0' READ(5,*) NCT IF (NCT.EQ.1) GO TO 1 STOP END FUNCTION A(ITYP) COMMON X(6),BETA IF(ITYP.EQ.1) THEN A=(X(5)+X(6)*X(4))*X(4) ELSE BETA=ACOS(1.2.*X(4)/X(5)) A=.25*X(5)*X(5)*(BETA.5*SIN(2.*BETA)) ENDIF RETURN END FUNCTION P(ITYP) COMMON X(6),BETA IF(ITYP.EQ.1) THEN P=X(5)+2.*X(4)*SQRT(X(6)**2+1.) ELSE P=X(5)*BETA ENDIF RETURN END
Energy and Its Dissipation in Open Channels
69
Listing of C program designed to solve Manning’s equation in trapezoidal or circular channels /* Solves Manning's Equation in Trapezoidal & Circular Channels */ #include <stdio.h> #include <math.h> int type; float beta; float A(float b,float m,float Y) { if (type==1) return((b+m*Y)*Y); else {beta=acos(1.2.*Y/b); return(0.25*b*b*(beta0.5*sin(2.*beta)));}} float P(float b,float m,float Y) { if(type==1) return(b+2.*Y*sqrt(m*m+1.)); else return(b*sin(beta));} main () { float x[6],cc,AA,F,F1,DIF,dumm; int IUNK,nct,No,II,i; char u,V[7]; dumm=sqrt(.04); /* drag floating pt lib. into the linker */ strcpy(V,"QnSYbm\n"); puts("Give 1.49 for ES units or 1. for SI units."); scanf("%f",&cc); puts("Give 1 for trapezoidal section, or 2 for circular section."); scanf("%d",&type); if (type==1) No=6; else {V[4]='D'; No=5;} a2:puts("Give No. of unknown:"); for (i=0;printf(" %d  %c,",i,V[i]),i2)) { printf(" %c = ",V[i]); scanf("%f",&x[i]);} switch (IUNK) { case 0: x[0]=cc/x[1]*pow(A(x[4],x[5],x[3]),1.6666667)/pow(P(x[4],\ x[5],x[3]),0.66667)*sqrt(x[2]); break; case 1: x[1]=cc/x[0]*pow(A(x[4],x[5],x[3]),1.6666667)/pow(P(x[4],\ x[5],x[3]),0.66667)*sqrt(x[2]); break; case 2: AA=A(x[4],x[5],x[3]); DIF=x[1]*x[0]/cc*pow(P(x[4],\ x[5],x[3])/AA,0.666667)/AA; x[2]=DIF*DIF; break; default: nct=0; do {II=0; a1:AA=A(x[4],x[5],x[3]); F=x[0]*x[1]cc*AA*pow(AA/P(x[4],x[5],x[3]),0.66667)*sqrt(x[2]); if (II==0) {II=1; F1=F; x[IUNK]=x[IUNK]0.001; goto a1;} DIF=0.001*F1/(F1F); x[IUNK]=x[IUNK]+0.001DIF; } while (abs(DIF)>0.00001 && ++nct<15);} for (i=0;i
70
Open Channel Flow: Numerical Methods and Computer Applications
Listing of FORTRAN program that uses the Newton method to solve all variables (MANNTC. FOR) PARAMETER (EX=.6666667,EX1=1.6666667) CHARACTER*1 VAR(6)/'Q','n','S','y','b','m'/,UNK CHARACTER*33 FMT/"(' Solution for ',A1,' = ',F10.3)"/ REAL X(6),ns,neq,C/1.486/ C X(1)=Q, X(2)=n, X(3)=S, X(4)=y, X(5)=b, X(6)=m LOGICAL R /.FALSE./ WRITE(*,*)'Give 0 if ES; 1 if SI units' READ(*,*) IUNIT IF(IUNIT.EQ.1) C=1. WRITE(*,*)'Is the section: 1 trapzoidal, or 2 circular?' READ(*,*) ISECT IF(ISECT.EQ.2) THEN NVAR=5 VAR(5)='D' ELSE NVAR=6 ENDIF WRITE(*,*)'Give value to variables' DO 10 I=1,NVAR WRITE(*,"(2X,A1,' = '\)") VAR(I) IF(I.EQ.2 .AND. ISECT.EQ.1) THEN WRITE(*,"(' (1st for sides=) ')") READ(*,*) ns WRITE(*,"(' (now for bottom=) ')") ENDIF 10 READ(*,*) X(I) 15 WRITE(*,*)'Give symbol for unknown' READ(*,'(A1)') UNK IUNK=1 DO 20 WHILE (VAR(IUNK).NE.UNK.AND.IUNK.LE.NVAR) 20 IUNK=IUNK+1 IF(IUNK.GT.NVAR) GO TO 15 DO 30 I=1,20 25 IF(ISECT.EQ.2) THEN COSB=1. 2.*X(4)/X(5) B=ACOS(COSB) A=.25*X(5)*X(5)*(B COSB*SIN(B)) P=B*X(5) F=X(2)*X(1) C*SQRT(X(3))*A*(A/P)**EX ELSE P=X(5)+2.*X(4)*SQRT(X(6)**2+1.) neq=X(5)/P*X(2)+2.*X(4)*SQRT(X(6)**2+1.)/P*ns F=neq*X(1)*P**EX C*SQRT(X(3))*((X(5)+X(6)*X(4))*X(4))**EX1 ENDIF IF(R) GO TO 28 XX=X(IUNK) X(IUNK)=1.005*X(IUNK) F1=F
Energy and Its Dissipation in Open Channels
28 30 40
R=.TRUE. GO TO 25 DIF=F1*(X(IUNK) XX)/(F F1) X(IUNK)=XX DIF IF(ABS(DIF).LT. .000001) GO TO 40 CONTINUE IF(IUNK.EQ.2)FMT(32:32)='4' IF(IUNK.EQ.3)FMT(32:32)='7' WRITE(*,FMT) VAR(IUNK),X(IUNK) END
Listing of C program that uses the Newton method to solve all variables (MANNTC.C) #include <stdio.h> #include <math.h> #include <stdlib.h> main() {int units,sect,i,r,iunk,nvar=5,nct=0; float c,x[6],f,f1,xx,dif,cosb,a,b,p,ns,neq,ex=.6666667,\ ex1=1.6666667; char var[7]="QnSybm\0",unk; clrscr(); printf("Give 0 if ES; 1 if SI units ");scanf("%d",&units); if (units) c=1; else c=1.486; printf("Is the section: 1  trapezoidal, or 2  circular? "); scanf("%d",§);if(sect==2) {nvar=4;var[4]='D';} printf("Give value to variables\n");for(i=0;i<=nvar;i++){ printf(" %c = ",var[i]); if((i==1)&&(sect==1)){ printf(" (1st for sides =) ");scanf("%f",&ns); printf(" (now for bottom=) ");} scanf("%f",&x[i]);} do {printf("Give symbol for unknown ");scanf("%s",&unk);iunk=0; while (var[iunk]!=unk && iunk<=nvar) iunk++;} while(iunk>unk); do {r=1; L1:if(nvar==4){cosb=12*x[3]/x[4];b=acos(cosb); a=.25*x[4]*x[4]*(bcosb*sin(b)); p=b*x[4]; f=x[1]*x[0]c*sqrt(x[2])*a*pow(a/p,ex);} else { p=x[4]+2.*x[3]*sqrt(x[5]*x[5]+1.); neq=x[4]/p*x[1]+2.*x[3]*sqrt(x[5]*x[5]+1.)/p*ns; f=neq*x[0]*pow(p,ex)c*sqrt(x[2])*pow((x[4]+x[5]*x[3])*x[3],\ ex1);} if(r) {xx=x[iunk];x[iunk]=1.005*x[iunk];f1=f;r=0;goto L1;} dif=f1*(x[iunk]xx)/(ff1); x[iunk]=xxdif; nct++; } while (nct<20 && fabs(dif)>.000001); printf("\nSolution for %c = %f",var[iunk],x[iunk]);} Example Problem 2.9 Determine the maximum flow rate that can be accommodated in the channel shown below without causing the depth in the upstream channel to rise above its normal depth. The upstream channel has a bottom width b1 = 10 ft, a side slope m1 = 2, a Manning’s roughness coefficient, n1 = 0.014, and a bottom slope, So1 = 0.0002. The downstream channel is steep, i.e., under uniform flow conditions the depth will be less than critical depth and is rectangular in shape with a bottom width b2 = 8 ft. Investigate the relationship between the slope of bottom of the upstream channel on the flow rates and the depths that are possible in this channel under uniform flow conditions.
71
72
Open Channel Flow: Numerical Methods and Computer Applications
ion
Q
b2 =
Tr an sit
b1 = 10 ft, m1 = 2, n1 = 0 .014, So1 = 0.0002
8 ft S teep c hanne l
Solution Since the downstream channel is steep, critical flow will occur at the end of the transition to 8 ft wide rectangular channel, the specific energy here is given by q2 E c = 1.5Yc = 1.5 2 g
1/ 3
(Q / b 2 ) = 1.5 g
1/ 3
= 0.11787Q 2 / 3
The specific energy in the larger upstream channel must equal Ec or Y1 +
(Q /A)12 = E c = 0.11787Q 2 / 83 (2 g )
If the depth upstream is to be uniform, then 2/3
Q=
1.486 A A1 S1o/12 P1 1 n
These three equations allow for the variables Ec, Y1, and Q to be solved, or if one wishes to eliminate the first equation then the latter two equations will solve for Y1 and Q. Their solution is Y = 4.146 ft, Q = 218.42 cfs.
In general, a problem of this nature that is to determine what the maximum flow rate is that can occur in a channel with a transition from an upstream mild channel to a downstream steep channel requires the simultaneous solution of (1) the critical flow equation in the downstream channel; (2) the energy equation, which equates the specific energy in the upstream channel to the critical specific energy in the downstream channel; and (3) the uniform flow equation. To investigate the relationship between the bottom slope of the upstream channel and the maximum uniform flow possible requires that the latter two above equations be solved for different values of So1. The table below shows the results from several such solutions: Bottom slope So1 Flow rate Q (cfs) Depth Y1 (ft)
0.002 0.457 0.0616
0.001 6.703 0.376
0.0009 9.466 0.476
0.0008 13.59 0.610
0.0007 19.88 0.792
0.0006 29.66 1.094
0.0005 45.37 1.406
0.0003 119.4 2.734
0.0002 218.4 4.146
0.00018 251.4 4.566
0.00014 344.1 5.662
0.00012 412.1 6.404
0.0001 504.7 7.352
0.00008 638.6 8.627
0.0004 71.72 1.922
It is interesting to note that the flow rate decreases to extremely small values as the bottom slope of the upstream channel becomes larger. Thus it is very easy to “choke” the upstream flow, i.e., cause it to be above normal depth by reducing the size of a downstream channel that will have critical flow in it.
If you don’t have a computer available, nor a programmable calculator, then it is possible to use graphical means for solving Manning’s equation. To develop such a graphical solution let us define
73
Energy and Its Dissipation in Open Channels
a dimensionless depth Y′ = Y/b for trapezoidal channels. (When defining a dimensionless depth for the specific energy later we will let Y′ = mY/b because this eliminates m from the resulting dimensionless equation.) Then after some algebraic manipulation the following equation can be obtained: nQ 8
Cu b 3 S0
5
=
(
(Y′ + mY′ 2 ) 3
1 + 2Y′ 1 + m 2
)
2 3
= Q′
The parameter on the left of the equal sign might be taken as a dimensionless flow rate, denote as Q′. (Note that Q′ is dimensionless if n is taken as dimensionless and Cu is assumed to have dimensions of L1/3/t.) This equation shows that the flow rate parameter Q′ is a function of the dimensionless depth Y′ = Y/b, and the side slope m of the trapezoidal channel, and when m = 0 the channel is rectangular. This relationship is given in Figure 2.3 using several different curves for different m values. The main graph is a linear plot, and the insert gives the same graph except using log–log paper. This graph can be used to solve Manning’s equation for several different unknowns (Figure 2.3). For example, to find the normal depth Yo, one would first compute the flow rate parameter Q′ from the known values; then enter this value on the ordinate of the graph, and read the corresponding dimensionless depth Y′ on the abscissa; and finally compute Yo as the product of Y′ and the bottom width b. Manning’s equation can also be solved graphically for a circular section. To develop such a graphical solution Equation 2.9 can be used to define the relationship between the angle β and 10
10
Q΄ =
9
=2
=
1.
5
m
nQ
CuSo1/2b8/3
0.1
5
m
nQ
CuSo1/2b8/3
(1 + 2 √ m2 + 1 Y΄)2/3
1
7
m
4 0.01
3
m=
0.1
2
0.2 0.3 0.4 0.6 0.8 1 Dimensionless depth, Y΄ = Y/b
2
=1
5
0.7
m=
0.5
Rectangle
1 0
=
Cu√Sob8/3
(Y΄ + mY΄2)5/3
8
6
nQ
0
0.2
0.4
0.6
0.8 1 1.2 Dimensionless depth, Y΄ = Y/b
1.4
1.6
1.8
2
FIGURE 2.3 Plot of dimensionless Manning’s equation in trapezoidal channels (including rectangular when m = 0).
74
Open Channel Flow: Numerical Methods and Computer Applications 0.35 Q΄ =
0.3
nQ CuD8/3√So
nQ
(β – cos β sin β)5/3 10.079368 β2/3
β = cos–1(1 – 2Y΄)
0.25 Cu√SoD8/3
=
0.2 0.15 0.1 0.05 0
0.1
0.2
0.3
0.4 0.5 0.6 0.7 Dimensionless depth Y΄= Y /D
0.8
0.9
1
FIGURE 2.4 Plot of dimensionless Manning’s equation in circular channels.
the flow rate parameter Q′ = (β − cos β sin β)5/3/(10.0793684β2/3), and the angle β in turn can be obtained from the dimensionless depth Y′ = Y/D from β = cos−1(1 − 2Y′). Such a graphical solution of Manning’s equations for circular channels is given in Figure 2.4. Table D.2 provides greater precision than can be obtained from the graph, and the example problems at the end of Table D.2 illustrate how the table can be used.
2.5 Channels with Varying Wall Roughness, but Q = Constant Uniform flow cannot exist in a channel in which the wall roughness varies in the direction of the channel unless the bottom slope varies in precisely the correct manner so the velocity and depth remain constant with x. This combination of n and So would create uniform flow only for one flow rate. However, uniform flow can occur in long channels with constant bottom slopes when the wall roughness varies along the position of the channel cross section. It is not uncommon to have a channel’s sides with a different roughness than its bottom. An example is a laboratory flume whose sides are Plexiglas and its bottom is filled with gravel. A modification to Manning’s equation for channels with varying roughness coefficients in different portions of its cross section might be to compute an equivalent Manning’s n that weights the individual n values according to the portion of the perimeter to which they apply. For a trapezoidal channel with a different Manning’s n along the bottom, nb, than that for the sides, ns, the equivalent roughness coefficient would then be
n eq =
b 2Y 1 + m 2 nb + ns P P
The last two program listings are designed to handle a trapezoidal channel that has a different roughness coefficient along the bottom than the sides of the channel using an equivalent roughness coefficient computed by this equation. The validity of using an equivalent roughness coefficient would need to be verified by field or laboratory measurements for a given channel. The need for verification is that Manning’s equation is empirical and therefore it is not possible to use theory alone to derive an “equivalent” Manning’s equation for channels with varying roughnesses along
75
Energy and Its Dissipation in Open Channels
the cross section. The above formula will produce an neq that equals nb and ns when these are the same, whereas other methods will not. For example, one might be inclined to associate n with P2/3 n i Pi2/3 . Using this approach for a for which then n applies. Then nP2/3 would be replaced by trapezoidal channel with a different bottom roughness than side roughness would use one of the following Manning’s equation:
∑
Q=
Cu A 5/3 So 2 /3 2 /3 2 n b b + 2Y m + 1 n s
or
{
}
2/3 2 /3 2 (Y/n b ) + mY 2n g m +1 Q = Cu A So {n b + n s}1/ 3
or
{Y/n + mY 2n Q=C A S b
u
o
s
m 2 + 1
{n b + ns }1/ 3
}
2 /3
The problem with any of these latter formulas is that they will not produce the same results as the original Manning’s equation does when n = nb = ns.
2.6 Specific Energy, Subcritical and Supercritical Flows When dealing with open channel flow it is convenient to reference the energy per unit weight from the channel bottom. Thus instead of having a horizontal datum from which the energy is referenced, a sloping data is used. The result is call the “specific energy,” and it consists of the depth of flow in the channel plus the velocity head, or
E = Y+α
V2 Q2 = Y+α 2g 2gA 2
(2.11)
in which α is the kinetic energy correction coefficient defined in Chapter 1. In Equation 2.11 Y is the depth of flow, and if the pressure distribution is hydrostatic, then this depth Y will equal the pressure head p/γ on the bottom of the channel. Therefore, regardless of the position within the channel flow the sum of the two terms in Equation 2.11 represent the distance between the channel bottom and the energy line. For a uniform flow the specific energy will be constant. If the channel is on a steep slope then it is necessary to adjust the depth as described in Chapter 1, that is, if Y represents the vertical distance through the fluid, then it needs to be multiplied by the cosine squared of the angle of the bottom slope. It is common in practice to ignore the kinetic energy correction coefficient α (i.e., assume α = 1) and then the specific energy becomes
E = Y+
V2 Q2 = Y+ 2g 2gA 2
(2.11a)
76
Open Channel Flow: Numerical Methods and Computer Applications r ,o Q 1 nnel 1 a Ch Q2 > Q1, or Channel2
Y
°
45
Y
(a)
V2/2g
E=Y+
Q2 2gA2
Subcritical flow
Rectangular channel, q = Q/b
Y
q2
Yc
Supercri ti
V2c /2g
cal flow
Specific energy
(b)
°
45
Subcritical flow q1
V2/2g
Depth
Depth
l flow
a Critic
Y
q3
q1
E=Y+
Supercritical flow q2 Specific energy 2gy2
FIGURE 2.5 Sketches of specific energy diagrams in (a) a general channel and (b) in a rectangular channel in which the bottom width changes, but Q = constant.
For a rectangular channel, it is convenient to deal with the flow rate per unit width of channel, q = Q/b. For a rectangular channel the specific energy can be written as follows if α is assumed equal to 1, and the bottom slope of the channel is small enough so that the cos θ = 1:
E = Y+
V2 q2 = Y+ 2g 2gY 2
(2.11b)
A plot of the depth Y as the ordinate, and the specific energy E as the abscissa is referred to as a specific energy diagram. In Figure 2.5 two sketches of specific energy diagrams are given. The first applies for any channel, and the second is specific for a rectangular channel. The following should be observed:
1. If the specific energy is held constant (i.e., a vertical lines is drawn on the specific energy diagram) then there are two depths. These depths are called alternative depths. The flow associated with the larger of these two depths produces subcritical flow, and the smaller depth is associated with supercritical flow. 2. As these two depths merge into a single depth, a minimum value for the specific energy occurs that can exist for any given flow rate. This depth is called critical depth, and the flow associated with it is called critical flow and will be denoted by Yc. The specific energy associated with critical depth will be denoted by Ec. 3. As the flow rate in a given channel increases the specific energy curve is shifted to the right and upward, i.e., the critical depth is increased, and both the subcritical and supercritical depths are closer to the critical depth. In dealing with a rectangular channel the flow rate per unit width q increases (with Q constant) when the bottom width of the channel becomes less, i.e., a channel transition reduces the width of the channel.
That there are generally two depths associated with any given value of the specific energy can be understood best by examining Equation 2.11b. By multiplying this equation by Y2 one notes that Equation 2.11b is a cubic equation. A cubic equation has three roots, generally, and if there are imaginary roots they occur in pairs. The third root of Equation 2.11b gives a negative value for Y, but since a negative depth is physically not possible, this root is ignored. From a mathematical view point, if the specific energy is reduced to a value less than the critical value, Ec, for a given flow rate in a given channel, then the alternative depths become imaginary, or complex roots of Equation
77
Energy and Its Dissipation in Open Channels
2.11b. For a general channel it is not obvious what type of equation (Equation 2.11b) represents, but two real positive roots of Y exist for values of E > Ec and these are called alternative depths. A simple explicit equation exists for computing the alternate depths in a rectangular channel. This equation can be used to obtain the depth upstream from a gate if the downstream depth is known or the downstream depth if the upstream depth is known, for example. To obtain this equation, equate E1 to E2 or Y1 +
q2 q2 = + Y 2 2gY12 2gY22
(2.11c)
Let the specific energy computed from the known depth be denoted by Ek and the other depth be given without a subscript. The Equation 2.11c becomes the following cubic equation: Y3 − E k Y 2 +
q2 =0 2g
(2.11d)
Since one depth Yk is known this equation can be reduced to a quadratic equation by synthetic division or −E k Yk
1
Yk − E k Yk (Yk − E k )
1
0 Yk (Yk − E k )
q 2 / 2g Yk 0
giving Y 2 + (Yk − E k )Y + (Yk − E k )Yk = 0
Solving for Y gives
Y=
{
}
1 E k − Yk + (E k − Yk )2 + 4Yk (E k − Yk ) 2
(2.11e)
Since Ek − Yk is the velocity head Vk2 /(2g) = Vh associated with the known depth, Yk, Equation 2.11e can be written as
Y=
(
)
1 Vh + Vh2 + 4Yk Vh 2
(2.11f)
Note that Equations 2.11e and f are valid only for situations in which E1 = E2 and the flow rate per unit width in the rectangular channel is the same at the two positions 1 and 2. To find the minimum value of the specific energy, i.e., Ec and the corresponding critical depth that is associated with critical flow, the well known principle of calculus can be employed of setting the first derivative of E with respect to Y equal to zero. This principle will first be applied to Equation 2.11b, and later to 2.11a. Differentiation of Equation 2.11b with respect to Y and equating dE/dY to zero gives (with q held constant for a given specific energy curve)
dE q2 = 1− = 0 or q 2 = gY 2 dY gY 2
or q = gY
78
Open Channel Flow: Numerical Methods and Computer Applications
which can be rewritten in several different ways as given by the following equations that define critical flow (the subscript c has been added to emphasize that these equations define critical flow conditions):
q c = gYc3
q2 3 Yc = c g
(2.12)
1
Vc2 Yc = 2g 2
(2.12a)
(velocity head equal 1/2 the critical depth)
(2.12b)
2 Yc = E c 3
(2.12c)
E c = 1.5Yc
(2.12d)
It needs to be noted that Equations 2.12 apply only for rectangular channels. Since the energy equation for a rectangular channel is a cubic equation (Equation 2.11d) in terms of depth Y, then if the specific energy E, and the flow rate per unit width q are known then the general solution of a cubic equation can be used (see CRC standard Math. Tables for methods to solve cubic equations), to solve it. This solution procedure consists of first substituting x + E/3 for Y in Equation 2.11d to eliminate the squared term. This substitution produces 1 q2 2 3 − x3 − E2x + E =0 3 2g 27
Let a = −E2/3 and b = q2/(2g) – 2E3/27. Then the three roots for x are solved for next by the following three equations:
x1 = A + B, x 2 = 0.5(i 3 )(A − B) − 0.5x1, and x 3 = 0.5(i 3 )(B − A) − 0.5x1
in which 1/ 3
b2 a 3 1 / 2 A = + − 0.5b 4 27
1/ 3
b2 a 3 1 / 2 and B = − + − 0.5b 4 27
If b2/4 + a3/27 > 0. there will be one real root and two conjugate imaginary roots, i.e., the specific energy E < Ec and physically only a meaningless negative solution for Y exists. If b2/4 + a3/27 = 0, there will be three real roots of which at least two are equal, i.e., critical flow occurs in which the two real roots gives the critical depth, and the other real root is a meaningless negative solution for Y. If b2/4 + a3/27 > 0 there will be three real and unequal roots, i.e., the two positive roots are the alternative depths and the third root is a meaningless negative depth Y.
Energy and Its Dissipation in Open Channels
79
After the solution for x values have been obtained the final step is to obtain the depths Y from
1 1 1 Y1 = x1 + E; Y2 = x 2 + E; Y3 = x 3 + E 3 3 3
The program ALTDEP, listed below, implements this method for solving for the alternative depths. Generally to use the program one would first solve the specific energy equation to get E for the known depth, and then select the other depth from the output from the program as the alternative depth. However, the program lets us know what portion of the specific energy diagram the problem as been specified in. If two of the root are imaginary and the program gives the message “E < Ec so negative real root and 2 imaginary roots” we know that physically the energy must be increased for the specified flow rate to be possible. An alternative to using the complex arithmetic involved in the above procedure for solving a general cubic equation is to use an implicit solution method, such as Newton’s method, to solve for the depth desired by providing an appropriate guess, or software such as TKSolver or Mathcad (or an HP calculator or a spreadsheet) that has the capability to solve implicit equations. Listing of program ALTDEP.FOR COMPLEX AA,BB,C3,ARG,X1,X2,X3 C3=.5*CSQRT(CMPLX(3.,0.)) 1 WRITE(*,*)' Give q, E & g or 0/=STOP' READ(*,*) Q,E,G IF(Q.LT.1.E8) STOP A=E*E/3. B=Q*Q/(2.*G)+2.*A*E/9. BH=.5*B BS4=B*B/4. YC=(Q**2/G)**.33333333 ARG=CSQRT(CMPLX(BS4,0.)+CMPLX(A,0.)**3/27.) WRITE(*,*)' Yc =',YC,' Ec =',1.5*YC AARG=BS4+A**3/27. IF(AARG.GT.0.) THEN WRITE(*,*)' E<Ecsonegative real root and' &,'2',' imaginary &roots' ELSEIF(AARG.GT.1.E5 .AND. ARG.LT.1.E5) THEN WRITE(*,*)' Critical condition' ELSE WRITE(*,*)'Alternative Depths&negative Y' ENDIF AA=(ARGCMPLX(BH,0.))**.33333333 BB=(ARGCMPLX(BH,0.))**.33333333 X1=AA+BB X2=(AABB)*C3.5*X1 X3=(BBAA)*C3.5*X1 WRITE(*,100) X1+E/3.,X2+E/3.,X3+E/3. 100 FORMAT(3(2F9.3,3X)) GO TO 1 END
80
Open Channel Flow: Numerical Methods and Computer Applications
Listing of Program ALTDEP.CPP #include #include <stdlib.h> #include #include void main(void){float a,b,q,e,g,bh,bs4,yc,aarg; complex aa,bb,c3,arg,x1,x2,x3; c3=.5*sqrt(3.)*complex(0.,1.); L1:cout <<"Give q, E & g or 0 0 0 =STOP"<< endl; cin >>q>>e>>g; if(q<1.e5) exit(0); a=e*e/3.; b=q*q/(2.*g)+2.*a*e/9.; bh=.5*b;bs4=b*b/4.; yc=pow(q*q/g,.3333333); arg=sqrt(complex(bs4,0.)+pow(complex(a,0.),3.)/27.); cout <<"Yc ="<0.) cout<<"E<Ec so negative real root and 2 imaginary roots"<<endl; else if((aarg>1.e5)&&(aarg<1.e5))\ cout <<"Critical condition"<< endl; else cout <<"Alternative Depths & negative Y"<< endl; aa=pow(argcomplex(bh,0.),.3333333); bb=pow(argcomplex(bh,0.),.3333333); x1=aa+bb;x2=(aabb)*c3.5*x1;x3=(bbaa)*c3.5*x1; cout.width(9); cout<<setprecision(3)<<x1+e/3.<<" "<<x2+e/3.<<" "<<x3+e/3.<< endl; goto L1;} Example use of program Input: 5 2 32.2 Output: Yc = 9.190971E01 Ec = 1.378646 Alternative depths and negative Y 1.891 .000 .511 .000 .402 Input: 5 1.378646 32.2 Output: Yc = 9.190971E01 Ec = 1.378646 Critical condition .920 .000 .919 .000 .460 Input: 5 1 32.2 Output: Yc = 9.190971E01 Ec = 1.378646 E<Ec so negative real root and 2 imaginary roots .754 .444 .754 .444 .507 Input: 0/
.000
.000
.000
The above procedure can be simplified by using the arc cosine (and subsequently the cosine). To use this alternate method, first compute the angle θ from
(
)
6.75q 2 /g − E 3 / 27 θ = cos −1 3 E / 3 ( )
Energy and Its Dissipation in Open Channels
81
The three roots are then obtained from
E θ Y1 = 1 − 2 cos (negative depth) 3 3
θ + 2π E Y2 = 1 − 2 cos 3 3
θ + 4π E Y3 = 1 − 2 cos 3 3
Of course, if E = Ec the two latter depths become equal, or become the critical depth Yc associated with the given q. If the given E is less than the critical depth, then Y2 and Y3 are imaginary and the above procedure will not work because the argument for the arc cosine is not within the allowable range of −1 to +1. Thus if this alternative method is implemented in a computer program, as in ROOTSE listed below, the critical specific energy Ec, as described below, should be computed and if E < Ec the computation of the roots should not be attempted. Program ROOTSE.FOR PARAMETER (PI=3.14159265) REAL X(3) PI2=2.*PI PI4=4.*PI 1 WRITE(*,*)' Give: q,E,g' READ(*,*) q,E,g IF(q.LT.1.E5) STOP Yc=(q*q/g)**.33333333 Ec=1.5*Yc IF(E.LT.Ec) THEN WRITE(6,110) q,E,g,Yc,Ec 110 FORMAT(' q=',F8.3,' E=',F8.3,' g=',F8.2,/' Only 1 real &root(neg.)',' Yc=',F8.3,' Ec=',F8.3) GO TO 1 ENDIF E3=E/3. THETA=ACOS(((6.75*q*q/gE**3)/27.)/E3**3) X(1)=E3*(1.2.*COS(THETA/3.)) X(2)=E3*(1.2.*COS((THETA+PI2)/3.)) X(3)=E3*(1.2.*COS((THETA+PI4)/3.)) WRITE(6,100) q,E,g,Yc,Ec,X 100 FORMAT(' q=',F8.3,' E=',F8.3,' g=',F8.2,/,' Yc=',F8.3,' &Ec=',F8.3,' Roots are:',/,3F9.3) GO TO 1 END
82
Open Channel Flow: Numerical Methods and Computer Applications
Program ROOTSE.C #include <stdlib.h> #include <stdio.h> #include <math.h> void main(void){float pi=3.14159265,q,g,e,yc,ec,theta,e3,y1,y2,y3; L1:printf("Give: q,E,g\n"); scanf("%f %f %f",&q,&e,&g); if(q<1.e5) exit(0); yc=pow(q*q/g,.33333333); ec=1.5*yc; if(e<ec) {printf("q=%8.3f E=%8.3f g=%8.2f \nOnly 1 real \ root(beg.)\n Yc=%8.3f Ec=%8.3f\n",q,e,g,yc,ec); goto L1;} e3=e/3.; theta=acos(((6.75*q*q/gpow(e,3.))/27.)/pow(e3,3.)); y1=e3*(1.2.*cos(theta/3.)); y2=e3*(1.2.*cos((theta+2.*pi)/3.)); y3=e3*(1.2.*cos((theta+4.*pi)/3.)); printf("q=%8.3f E=%8.3f g=%8.3f\nYc=%8.3f \ Ec=%8.3f\nRoots are:\n%9.3f %8.3f \ %8.3f\n",q,e,g,yc,ec,y1,y2,y3); goto L1;} For any channel Equation 2.11a is differentiated, and dE/dY is set to zero to find the minimum value of E. This procedure leads to
Q 2 dA =1 gA 3 dY T dA
dY
Y
From the accompanying sketch it is clear that dA/dY equals the top width T for all channels and therefore the critical flow equation for all channels becomes
Q 2c Tc V2 2 F = = =1 r gA 3c c2
(2.13)
in which Fr is the Froude number which is defined as the ratio of inertia to gravity forces. The speed at which a small amplitude gravity wave travel in an open channel is given by
c=
gA 3 T
(2.14)
83
Energy and Its Dissipation in Open Channels
Proof of this equation is given in Chapter 3. From Equations 2.13 and 2.14, we see that the Froude number is also the ratio of the average velocity in the channel to the speed of a small amplitude gravity wave. From Equation 2.13, it can be concluded that critical flow occurs when the average velocity V of a channel flow exactly equals the speed c of a small amplitude gravity wave. For subcritical flows the Froude number Fr will be less than unity, since the denominator of Equation 2.13 increases with increasing depth, and for supercritical flows the Froude number will be larger than unity. In a subcritical flow, the fact that the velocity in the channel is less than the speed of a small amplitude gravity wave allows these waves to move upstream against the flowing fluid. Thus in a sense the fluid can receive a signal that things are to change downstream, and it adjust gradually. This adjusting to downstream conditions is referred to as downstream control. Subcritical flows are always controlled by downstream conditions, i.e., an obstruction to the flow such as a partly closed gate will cause the depth upstream from the gate to increase, and the velocity to decrease. On the other hand if the flow is supercritical, the velocity in the channel exceeds the speed of a small amplitude gravity wave, and the flow does not receive a signal about downstream conditions. Therefore, if the flow is supercritical it will not change its depth or velocity to meet downstream conditions. For example, should the channel abruptly end, a supercritical flow would continue to the end of the channel at the same depth and velocity as it has upstream there from. Supercritical flows are, therefore, said to be upstream controlled. An example of a control that affects both the upstream flow as well as the downstream flow conditions is a sluice gate in a channel flow. Since subcritical flows are controlled downstream and supercritical flows are controlled by an upstream device we can conclude that the flow upstream from the gate must be subcritical, and the flow downstream from the gate must be supercritical.
6
5
3
Sub
Ec
Y Ec
1
0
2
V /(2g) Yc 1
cri
tic
al f
3.5
low
w l flo ica crit Q3 = 300 per cf
Ec
2
E = 4 ft Subcritical flow
4
1 1.25
Y 10
4
Yc
u Yc S Q 2 = 200 cfs Q1 = 100 cfs
2 3 4 5 Specific energy, E = Y + V2/(2g)
Depth, Y (ft)
Depth, Y (ft)
5
0
E = 5 ft
4.5
3
Qmax
2.5
s
Yc
1.5 Yc
0.5 7
0
Yc
Qmax
2 1
6
Qmax
E = 3 ft
0
50
100
150
Supercritical flow
200 250 300 Flow rate, Q (cfs)
10 350
400
Y
1 1.25 450
500
Above a specific energy diagram and a depthdischarge diagram for a trapezoidal channel are given. The separate curves on the depthdischarge diagram show the relationship of the flow rate to changes in depth for a constant specific energy. The maximum point on these curves corresponds to critical conditions, i.e., the depth is critical and the Qmax is the maximum flow rate that could be obtained into this trapezoidal if it were supplied by a reservoir with a water surface elevation equal to the E for that curve. This assumes there is no entrance loss. You might note that the minimum point on the specific energy diagram corresponds to the maximum point on the depthdischarge diagram. For example, the curve for E = 3 ft on the depth–discharge diagram gives a maximum flow rate Qmax = 200 cfs, and the minimum specific energy on the specific energy diagram of the curve on the for Q = 200 cfs is Ec = 3 ft.
Open Channel Flow: Numerical Methods and Computer Applications
Subcritical flow
(Downstream controlled)
Control
84
Supercritical flow
(Upstream controlled)
To further illustrate the meanings of these diagrams assume a reservoir with a head H = 5 ft supplies a steep trapezoidal channel with b = 10 ft, and m = 1.25, the flow rate that will enter the channel is Q = 494 cfs (the maximum flow rate possible on the depthdischarge diagram of the curve for E = 5 ft), and the depth at the entrance will be Yc = 3.62 ft, the ordinate of the maximum point of this curve. These values are at the extreme, Qmax position of the curve for E = 5 ft, but these values could also be obtained by solving the critical flow equation Q2T/(gA3) = 1 and the specific energy equation 5 = Y + (Q/A)2/(2 g) simultaneously. Likewise if the reservoir level were 4 ft above the channel bottom Qc = Qmax = 331.5 cfs and Yc = 2.87 ft, or if H = 3 ft, Qc = Qmax = 201.2 cfs and Yc = 2.12 ft. It is not possible for the flow from a reservoir into a steep channel to be in the lower (the supercritical) portion of the depthdischarge diagram, because the flow rates would be less than the amount Qmax that can be supplied. However, if the channel is not steep, but mild, then the flow into the channel will be reduced by downstream conditions, i.e., the fluid frictional resistance. For example, for the trapezoidal channel for which the depthdischarge diagram is made if the reservoir supplied a specific energy of 5 ft, and the flow rate were 400 cfs, the depth would be 4.5 ft in the channel. If the channel has an n = 0.014, then Manning’s equation could be solved to show that it would have to have a bottom slope of So = 0.000701 for this uniform flow to occur. The need to determine critical depth, or critical flow conditions, occurs frequently in open channels. The illustrative problems, as well as the problems at the end of this chapter, point out a few situations where computation of critical depth is needed. Therefore, it is worth discussing how the above critical flow Equations 2.12 and 2.13 can be effectively solved. Should the channel be rectangular the solution of critical flow conditions is very easy since Equations 2.12 are all explicit. However, for a nonrectangular channel Equation 2.13 must be solved for the critical depth Yc, and since this equation is implicit, the solution requires an iterative technique such as the Newton method. Alternatively graphical solution can be utilized as discussed below to get answers with adequate precision for most applications. To solve Equation 2.13 by the Newton method, it can be rewritten as
F = Q 2c Tc − gA 2c = 0
(2.15)
in which Tc and Ac are functions of the critical depth Yc and the parameters that define the given channel. Example Problem 2.10 Water enters a steep rectangular channel that has a bottom width of 8 ft from a reservoir whose water surface is 5 ft above the channel bottom. Determine the flow rate into the channel, and the depth of flow at the channel entrance. Neglect the entrance loss coefficient. A long distance downstream a gate exists that produces a depth Y2 = 1.5 ft downstream from it. What is the depth immediately upstream from the gate?
85
Energy and Its Dissipation in Open Channels Solution Equations 2.12 apply to this problem because the channel is rectangular. Since the channel is steep, and a steep channel will contain a supercritical flow under uniform flow conditions, and the water in the reservoir (where V = 0) is subcritical, the flow must pass through critical depth at the entrance of the channel. Therefore, Yc = (2/3) E = (2/3)5 = 3.333 ft, and the flow rate per unit width q = (gYc3 )1/ 2 = (32.3(3.333)3 )1/ 2 = 34.53 cfs /ft, or Q = bq = 8(34.53) = 276.3 cfs. The second part of the problem requires that the specific energy downstream from the gate be solved, or E 2 = Y2 + q 2 /(2gY22 ) = 1.5 + (34.53 /1.5)2/ 64.4 = 9.729 ft . Now the alternative depth to 1.5 is sought with E = 9.729 ft. One way is to extract the root 1.5 from the cubic equation, another is to solve the cubic equation with the Newton method starting with a “subcritical” guess for Y1, or use the program ALTDEP. The solution for the depth upstream from the gate is Y1 = 9.525 ft. A hydraulic jump (discussed in Chapter 3) will occur somewhere upstream from the gate changing the flow from super to subcritical. Example Problem 2.11 Instead of the rectangular channel a steep pipe of 8 ft diameter is used. What is the flow rate and depth of flow at the entrance? A gate downstream causes a depth downstream from it of Y2 = 2.8 ft. What is the depth immediately upstream from the gate? Solution Since this is a circular channel, the implicit Equation 2.13 applies. Since there are two unknowns, Y and Q, a second equation must be obtained. This second equation is the specific energy Equation 2.11a. If angle β is used initially as a substitute for Y, the two equations that must be solved simultaneously are D2 F1 = Q 2 (D sin β) − g (β − cos β sin β) 4
3
and F2 =
{1 − K L } Q2 D2 (β − cos β sin β) E−D − 2(1 − cos β) 2g 4
2
The solution to these equations by the Newton method uses the following iterative equation
Q β
( m +1)
Q = β
(m )
z1 − z 2
in which the Z values are obtained by solving the following linear system of two equations:
∂F1 ∂Q ∂F2 ∂Q
∂F1 ∂β z1 F1 = ∂F12 z 2 F2 ∂β
in which F1 and F2 and its derivatives are evaluated using the current values, i.e., those with an m superscript. The solution gives Q = 207.2 cfs, and β = 1.474 rad, and from this angle Yc = D(1 − cos β)/2 = 3.615 ft. This solution assumed K L = 0, e.g., ignores entrance losses. If the entrance loss coefficient is taken as 0.1, then the solution is Q = 197.2 cfs, and Y = 3.523 ft. To solve part two, E2 = Y2 + (Q/A2)2/(2 g) = 5.512 ft. Now the energy equation must be solve using an implicit method because extracting a root or using the general solution to a cubic equation is not available. In fact since the variable β relates Y to A, it also needs to be solved. The solution is β = 1.787 rad, and Y1 = 4.859 ft.
86
Open Channel Flow: Numerical Methods and Computer Applications Example Problem 2.12 Instead of the rectangular channel (with b = 8 ft) of Example Problem 2.10 having a steep slope this problem deals with this channel on a “mild slope” with So = 0.00075. The channel has a Manning’s n = 0.014, and is long. What are the depth and the flow rate? The entrance loss coefficient is Ke = 0.1. As a second part solve for the flow rate and the uniform depth if the channel is trapezoidal with b = 8 ft, and m = 1.2. (So is still 0.00075, n = 0.014 and H = 5 ft). Also solve these problems using Chezy’s, rather than, Manning’s equation, if e = 0.01 ft. As a fourth part of this problem assume the bottom width b of the trapezoidal channel is sought that will supply a flow rate Q = 400 cfs. Solution For these mild channels the two equations that govern are: (1) a uniform flow equation which is Manning’s equation for the first part of the problem or
Q=
Cu AR 2h / 3S1o/ 2 n
or F1 = nQP 2 / 3 − C u A 5 / 3S1o/ 2 = 0
and (2) the energy equation H=Y+
1 − K L Q2 2g λ 2
or F2 = H − Y −
1 − K L Q2 =0 2g λ 2
Using the Newton method, the solution to these two equations for the rectangular channel is: Q = 177.72 cfs, and Y = 4.602 ft, and for the trapezoidal channel is: Q = 336.03 cfs, and Y = 4.451 ft. The program E_UN listed below is designed to solve Manning’s and energy equations simultaneously using the Newton method. You should study either the FORTRAN or C versions of these programs to understand how this solution is accomplished. The subroutine FUN (void function in the C program) supplies the values to the two above equations whenever it is called. The main program numerically evaluates the four derivatives in the Jacobian matrix using ∂Fi/∂x = {Fi(x + Δx) − Fi(x)}/Δx, in which Δx is obtained by multiplying the current value of the unknown x by 1.005 and then subtracting its current value or Δx = 1.005x − x = 0.005x. Carefully study the listing starting with the statement 30 SUM=0. to the statement IF(NCT.LT.30 .AND. SUM. GT.1.E5) GO TO 30 (the do {} while; in the Cprogram) to see how the Newton method is implement in solving a system of simultaneous equation because this approach will be used repeatedly. The subroutine SOLVEQ is called on to solve the linear system of equation even though a 2 × 2 matrix such as occurs in this program could be solved with a few lines of code. Notice that the program is designed to solve for any two of the first 7 variables 1 = b, 2 = m, 3 = So, 4 = n, 5 = Q, 6 = H, 7 = Y, and 8 = K in the array X() and that the array ID() is given a value of 1 to identify these two unknown variables and a value of 0 if the variable is known. Also notice that the program accommodates either a trapezoidal (which includes a rectangular channel with m = 0) and a circular channel by giving ITYPE a 0 or a 1 respectively. Program E_UN.FOR C Solves Manning's (uniform flow) and Energy simultaneously C for any 2 unknowns C See E_UN1 to solve Q & Y and method that can be used C with calculator CHARACTER*17 FMT/'(1X,A1,'' ='',F9.3)'/ CHARACTER*1 CX(8)/'b','m','S','n','Q','H','Y','K'/ CHARACTER*5 CH(0:1)/'value','guess'/ READ(*,*) G,FKE,ITYPE 1 DO 10 I=1,7 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 10 WRITE(*,'(I2,2X,A1)') I,CX(I) 10 ID(I)=0
Energy and Its Dissipation in Open Channels 2
100 20 30
35 40
50
WRITE(*,*)' Give two numbers for 2 unknown variables' READ(*,*) I1,I2 IF(I1.LT.1.OR.I1.GT.7.OR.I2.LT.1.OR.I2.GT.7) GO TO 2 ID(I1)=1 ID(I2)=1 DO 20 I=1,7 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 20 WRITE(*,100) CH(ID(I)),CX(I) FORMAT(' Give ',A5,' for ',A1,' = ',\) READ(*,*) X(I) CONTINUE NCT=0 SUM=0. CALL FUN(F) I1=0 DO 40 I=1,7 IF(ID(I).EQ.0) GO TO 40 XX=X(I) I1=I1+1 X(I)=1.005*X(I) CALL FUN(F1) DO 35 J=1,2 IF(ITYPE.EQ.1) THEN CX(1)='D' CX(2)=' ' ENDIF IF(G.GT.15.) THEN Cu=1.486 ELSE Cu=1. ENDIF X(8)=FKE FKE=(1.+FKE)/(2.*G) REAL F(2),F1(2),D(2,2) INTEGER*2 ID(8),INDX(2) COMMON X(8),G,FKE,Cu,ITYPE WRITE(*,*)' Give:g,entr. loss C. & 0=TRAP or 1=CIRLCE' D(J,I1)=(F1(J)F(J))/(X(I)XX) X(I)=XX CONTINUE CALL SOLVEQ(2,1,2,D,F,1,DD,INDX) I1=0 DO 50 I=1,7 IF(ID(I).EQ.0) GO TO 50 I1=I1+1 SUM=SUM+ABS(F(I1)) X(I)=X(I)F(I1) CONTINUE NCT=NCT+1 WRITE(*,*)' NCT=',NCT,' SUM=',SUM IF(NCT.LT.30 .AND. SUM.GT.1.E5) GO TO 30 WRITE(*,*)' Solution:' DO 60 I=1,8 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 60 IF(I.EQ.3) THEN FMT(16:16)='6'
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Open Channel Flow: Numerical Methods and Computer Applications
60
ELSEIF(I.EQ.4) THEN FMT(16:16)='3' ELSE FMT(16:16)='3' ENDIF WRITE(*,FMT) CX(I),X(I) CONTINUE WRITE(*,*)' Give 1 to solve another prob. (0=STOP)' READ(*,*) I2 IF(I2.EQ.1) GO TO 1 END SUBROUTINE FUN(F) REAL F(2) COMMON X(8),G,FKE,Cu,ITYPE IF(ITYPE.EQ.1) THEN BETA=ACOS(1.2.*X(7)/X(1)) A=.25*X(1)**2*(BETACOS(BETA)*SIN(BETA)) P=X(1)*BETA ELSE A=(X(1)+X(2)*X(7))*X(7) P=X(1)+2.*X(7)*SQRT(X(2)**2+1.) ENDIF F(1)=X(4)*X(5)*P**.6666667Cu*A**1.6666667*SQRT(X(3)) F(2)=X(6)X(7)FKE*(X(5)/A)**2 RETURN END
Program E_UN.C // Solves Manning's (uniform flow) and Energy simultaneously for // any 2 unknowns // See E_UN1 to solve Q & Y and method that can be used with // calculator #include <stdio.h> #include <stdlib.h> #include <math.h> float x[8],g,fke,cu; int itype; extern void solveq(int n,float **d,float *f,int itype,float *dd,\ int *indx); void fun(float *f){float beta,a,p; if(itype){beta=acos(1.2.*x[6]/x[0]); a=.25*x[0]*x[0]*(betacos(beta)*sin(beta));p=x[0]*beta;} else {a=(x[0]+x[1]*x[6])*x[6]; p=x[0]+2.*x[6]*sqrt(x[1]*x[1]+1.);} f[0]=x[3]*x[4]*pow(p,.6666667)cu*pow(a,1.6666667)*sqrt(x[2]); f[1]=x[5]x[6]fke*pow(x[4]/a,2.);return;} //End of fun void main(void){char *fmt=" %c =%9.3f\n",*ch[]={"value","guess"},\ *cx="bmSnQHYK"; float f[2],f1[2],xx,sum,**d,*dd; int id[8],indx[2],i,j,i1,i2,nct; d=(float**)malloc(2*sizeof(float*)); for(i=0;i<2;i++)d[i]=(float*)malloc(2*sizeof(float)); printf("Give: g,entrance loss C. & 0=TRAP or 1=CIRLCE\n"); scanf("%f %f %d",&g,&fke,&itype); if(itype){stpcpy(cx[0],"D");stpcpy(cx[1]," ");} if(g>15.) cu=1.486; else cu=1.; x[7]=fke;fke=(1.+fke)/(2.*g); L1: for(i=0;i<7;i++){if((!itype)  ((itype)&&(i != 1)))\ printf("%2d %c\n",i+1,cx[i]);id[i]=0;}
Energy and Its Dissipation in Open Channels do{printf(" Give two numbers for 2 unknown variables\n"); scanf("%d %d",&i1,&i2);}while((i1<1)(i1>7)(i2<1)(i2>7)); id[i11]=1;id[i21]=1; for(i=0;i<7;i++){if((!itype)  ((itype)&&(i != 1))) {printf("Give %s for %c =",ch[id[i]],cx[i]); scanf("%f",&x[i]);}} nct=0; do{sum=0.; fun(f); i1=1; for(i=0;i<7;i++){if(id[i]){xx=x[i];i1++;x[i]*=1.005;fun(f1); for(j=0;j<2;j++) d[j][i1]=(f1[j]f[j])/(x[i]xx); x[i]=xx;}} solveq(2,d,f,1,dd,indx); i1=1; for(i=0;i<7;i++){if(id[i]){sum+=fabs(f[++i1]);x[i]=f[i1];}} printf("nct= %d SUM=%f\n",nct,sum);}while((++nct<30)&&(sum>1.e5)); printf("Solution:\n");for(i=0;i<8;i++){ if((!itype)  ((itype)&&(i != 1))){ if(i==2)stpcpy(fmt[8],"6"); else stpcpy(fmt[8],"3"); printf(fmt,cx[i],x[i]);}} printf("Give 1 to solve another prob. (0=STOP)\n"); scanf("%d",&i2); if(i2) goto L1;} The above listing of the program will solve the fourth part of the problem in which a flow rate of Q = 400 cfs is given and b is wanted along with Y. The difference is for parts 1 and 2 variables 5 and 7 (Q and Y) are given as the unknowns and for part 3 variables 1 and 7 (b and Y) are identified as the unknowns. The solution for part 3 is b = 10.194 ft and Y = 4.416 ft. If the flow rate Q and depth Y are always the two unknowns, then it is not necessary to solve the two equations simultaneously. Rather Manning’s equation is substituted into the energy to eliminate the flow rate Q, or F = H − Y − (1 + Ke)(Cu/n)2(A/P)4/3So/(2g) = 0 and solve this equation for the depth Y. Thereafter solve for Q from either of the original equations. This is the approach you would use to solve parts 1 and 2 with a pocket calculator such as an HP with a SOLVE function. The program E_UN1, listed below use this latter approach to solve for Q and Y. Program E_UN1.FOR C Solves Manning's and Specific Energy for Y & Q by substituting Manning's equation C into Energy and first solving for Y and thereafter Q. (This is the procedure C to use with a pocket calculator) See program E_UN to solve any 2 variables. COMMON B,FM,D,FN,SO,H,C,ITYPE WRITE(*,*)' Give: g,(0=trap or 1=cir)' READ(*,*) G,ITYPE IF(ITYPE.EQ.0) THEN WRITE(*,*)' Give: b,m,n,So,H,Ke' READ(*,*) B,FM,FN,SO,H,FKE ELSE WRITE(*,*)' Give: D,n,So,H,Ke' READ(*,*) D,FN,SO,H,FKE ENDIF Y=.8*H Cu=1. IF(G.GT.15.) Cu=1.486 C=(Cu/FN)**2*SO*(1.+FKE)/(2.*G) NCT=0 10 F=FUN(Y) DIF=.05*F/(FUN(Y+.05)F) Y=YDIF NCT=NCT+1
89
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Open Channel Flow: Numerical Methods and Computer Applications 100
WRITE(*,*) NCT,DIF IF(NCT.LT.30 .AND. ABS(DIF).GT. 1.E5) GO TO 10 CALL AREAP(Y,A,P) Q=CU/FN*A*(A/P)**.6666667*SQRT(SO) WRITE(*,100) Y,Q FORMAT(' Solution: Y =',F8.3,' Q =',F10.2) END SUBROUTINE AREAP(Y,A,P) COMMON B,FM,D,FN,SO,H,C,ITYPE IF(ITYPE.EQ.1) THEN COSB=1.2.*Y/D BETA=ACOS(COSB) P=D*BETA A=.25*D*D*(BETACOSB*SIN(BETA)) ELSE P=B+2.*Y*SQRT(FM**2+1.) A=(B+FM*Y)*Y ENDIF RETURN END FUNCTION FUN(Y) COMMON B,FM,D,FN,SO,H,C,ITYPE CALL AREAP(Y,A,P) FUN=HYC*(A/P)**1.3333333 RETURN END
Program E_UN1.C /* Solves Manning's and Specific Energy for Y & Q by substituting Manning's Eq.into Energy and first solving for Y and thereafter Q. (This is the procedure to use with a pocket calculator) See program E_UN to solve any 2 variables. */ #include <stdio.h> #include <stdlib.h> #include <math.h> float b,m,d,n,so,h,c; int itype; void areap(float y,float *a,float *p){float cosb,beta; if(itype){cosb=1.2.*y/d;beta=acos(cosb); *p=d*beta;*a=.25*d*d*(betacosb*sin(beta));} else {*p=b+2.*y*sqrt(m*m+1.); *a=(b+m*y)*y;return;}} float fun(float y){float *a,*p; areap(y,a,p); return(hyc*pow((*a)/(*p),1.333333));} void main(void){float g,dif,ke,y,f,q,cu=1.,*a,*p; int nct=0; printf("Give: g(0=trap or 1=cir)\n"); scanf("%f %d",&g,&itype); if(itype){printf("Give: D,n,So,H,Ke\n"); scanf("%f %f %f %f %f",&d,&n,&so,&h,&ke);} else {printf("Give: b,m,n,So,H,Ke\n"); scanf("%f %f %f %f %f %f",&b,&m,&n,&so,&h,&ke);} y=.8*h; if(g>15.) cu=1.486; c=pow(cu/n,2.)*so*(1.+ke)/(2.*g); do{f=fun(y); dif=.05*f/(fun(y+.05)f); y=dif; printf("%d %f %f\n",++nct,dif,y); }while((nct<30) && (fabs(dif)>1.e5)); areap(y,a,p); q=cu/n*(*a)*pow((*a)/(*p),.6666667)*sqrt(so); printf("Solution: Y = %8.3f Q = %10.2f\n",y,q); }
91
Energy and Its Dissipation in Open Channels To solve the third part of the problem that requests that Chezy’s equation be used instead of Manning equation, requires that 3 (rather than 2) equations be solved simultaneously for Q, Y, and C. These equations are as follows:
F1 = Q − CA(R hSo )1/2 = 0 (Chezy’s equation)
e 0.884C F2 = C + (32g)1/ 2 Log10 + = 0 (Chezy ’s C equation) R e g1/ 3 (12R h )
(1 + K e ) Q 2 F3 = H − Y − = 0 (Energy equation) ( 2g ) A
(
)
Program UENCHEZ is designed to solve these three equations simultaneously. Notice that now the subroutine FUN provides values to 3 F values, and the part of the program that implements the Newton method has the DO loops changed from 2 to 3. The program UENCHEZ1 reduces the Newton solution to the depth by substituting Chezy’s equation into the energy equation, as has been done previous when using Manning’s equation as the uniform flow equation. Now however it is necessary to use a Gauss–Seidel type iteration to solve C associated with the current value of Q. This technique can be used with a calculator with a SOLVE function. Program UENCHEZ.FOR CHARACTER*17 FMT/'(1X,A1,'' ='',F9.3)'/ CHARACTER*1 CX(9)/'b','m','S','e','Q','H','Y','C','K'/ CHARACTER*5 CH(0:1)/'value','guess'/ REAL F(3),F1(3),D(3,3) INTEGER*2 ID(9),INDX(3) COMMON X(9),G,FKE,G32,CG,ITYPE WRITE(*,*)' Give:g,k.vis.,entrance loss C.', &' & 0=trap,1=cir' READ(*,*) G,VIS,FKE,ITYPE IF(ITYPE.EQ.1) THEN CX(1)='D' CX(2)=' ' ENDIF IF(G.GT.15.) THEN X(8)=100. ELSE X(8)=60. ENDIF X(9)=FKE FKE=(1.+FKE)/(2.*G) G32=SQRT(32.*G) CG=.884*VIS/(4.*SQRT(G)) 1 DO 10 I=1,7 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 10 WRITE(*,'(I2,2X,A1)') I,CX(I) 10 ID(I)=0 2 WRITE(*,*)' Give 2 num. for 2 unknown var.', ' in addition to C' READ(*,*) I1,I2 IF(I1.LT.1.OR.I1.GT.8.OR.I2.LT.1.OR.I2.GT.8) GO TO 2 ID(I1)=1 ID(I2)=1
92
Open Channel Flow: Numerical Methods and Computer Applications
100 20 30
35 40
50
60
ID(8)=1 DO 20 I=1,7 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 20 WRITE(*,100) CH(ID(I)),CX(I) FORMAT(' Give ',A5,' for ',A1,' = ',\) READ(*,*) X(I) CONTINUE NCT=0 SUM=0. CALL FUN(F) I1=0 DO 40 I=1,8 IF(ID(I).EQ.0) GO TO 40 XX=X(I) I1=I1+1 X(I)=1.005*X(I) CALL FUN(F1) DO 35 J=1,3 D(J,I1)=(F1(J)F(J))/(X(I)XX) X(I)=XX CONTINUE CALL SOLVEQ(3,1,3,D,F,1,DD,INDX) I1=0 DO 50 I=1,8 IF(ID(I).EQ.0) GO TO 50 I1=I1+1 SUM=SUM+ABS(F(I1)) X(I)=X(I)F(I1) CONTINUE NCT=NCT+1 WRITE(*,*)' NCT=',NCT,' SUM=',SUM IF(NCT.LT.30 .AND. SUM.GT.1.E5) GO TO 30 WRITE(*,*)' Solution:' DO 60 I=1,9 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 60 IF(I.EQ.3 .OR. I.EQ.4) THEN FMT(16:16)='6' ELSE FMT(16:16)='3' ENDIF WRITE(*,FMT) CX(I),X(I) CONTINUE WRITE(*,*)' Give 1 for another pb. (0=STOP)' READ(*,*) I2 IF(I2.EQ.1) GO TO 1 END SUBROUTINE FUN(F) REAL F(3) COMMON X(9),G,FKE,G32,CG,ITYPE IF(ITYPE.EQ.1) THEN BETA=ACOS(1.2.*X(7)/X(1)) A=.25*X(1)**2*(BETACOS(BETA)*SIN(BETA)) P=X(1)*BETA ELSE A=(X(1)+X(2)*X(7))*X(7) P=X(1)+2.*X(7)*SQRT(X(2)**2+1.)
Energy and Its Dissipation in Open Channels ENDIF Rh=A/P F(1)=X(5)X(8)*A*SQRT(X(3)*RH) F(2)=X(8)+G32*ALOG10(X(4)/(12.*Rh)+CG*X(8)*P/X(5)) F(3)=X(6)X(7)FKE*(X(5)/A)**2 RETURN END Program UENCHEZ.C #include <stdio.h> #include <stdlib.h> #include <math.h> float x[9],g,fke,g32,cg; int itype; extern void solveq(int n,float **d,float *f,int itype,\ float *dd,int *indx); void fun(float *f){float beta,a,p,rh; if(itype){beta=acos(1.2.*x[6]/x[0]); a=.25*x[0]*x[0]*(betacos(beta)*sin(beta));p=x[0]*beta;} else {a=(x[0]+x[1]*x[6])*x[6];p=x[0]+2.*x[6]*sqrt(x[1]*x[1]+1.);} rh=a/p; f[0]=x[4]x[7]*a*sqrt(x[2]*rh); f[1]=x[7]+g32*log10(x[3]/(12.*rh)+cg*x[7]*p/x[4]); f[2]=x[5]x[6]fke*pow(x[4]/a,2.);return;} //End of fun void main(void){char *fmt=" %c =%9.3f\n",*ch[]={"value","guess"},\ *cx="bmSeQHYCK"; float f[3],f1[3],xx,sum,**d,*dd,vis; int id[9],indx[3],i,j,i1,i2,nct; d=(float**)malloc(3*sizeof(float*)); for(i=0;i<3;i++)d[i]=(float*)malloc(3*sizeof(float)); printf("Give: g,k.viscosity,entrance loss C. & 0=trap or 1=cir\n"); scanf("%f %f %f %d",&g,&vis,&fke,&itype); if(itype){stpcpy(cx[0],"D");stpcpy(cx[1]," ");} if(g<15.) x[7]=100.; else x[7]=60.; x[8]=fke; fke=(1.+fke)/(2.*g);g32=sqrt(32.*g);cg=.884*vis/(4.*sqrt(g)); L1: for(i=0;i<7;i++){if((!itype)  ((itype)&&(i != 1))) printf("%2d %c\n",i+1,cx[i]);id[i]=0;} do{printf(" Give two numbers for 2 unknown variables\n"); scanf("%d %d",&i1,&i2); }while((i1<1)(i1>7)(i2<1)(i2>7)); id[i11]=1;id[i21]=1; for(i=0;i<7;i++){if((!itype)  ((itype)&&(i != 1))) {printf("Give %s for %c =",ch[id[i]],cx[i]); scanf("%f",&x[i]);}} nct=0; do{sum=0.; fun(f); i1=1;for(i=0;i<8;i++){if(id[i]){xx=x[i];i1++; x[i]*=1.005;fun(f1); for(j=0;j<3;j++) d[j][i1]=(f1[j]f[j])/(x[i]xx); x[i]=xx;}} solveq(3,d,f,1,dd,indx); i1=1; for(i=0;i<8;i++){if(id[i]){sum+=fabs(f[++i1]);x[i]=f[i1];}} printf("nct= %d SUM=%f\n",nct,sum); }while((++nct<30)&&(sum>1.e5)); printf("Solution:\n"); for(i=0;i<9;i++){if((!itype)  ((itype)&&(i != 1))){ if((i==2)(i==3))stpcpy(fmt[8],"6"); else stpcpy(fmt[8],"3"); printf(fmt,cx[i],x[i]);}} printf("Give 1 to solve another prob. (0=STOP)\n"); scanf("%d",&i2); if(i2) goto L1;}
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Open Channel Flow: Numerical Methods and Computer Applications Using this program to solve for the flow rate Q and the depth Y gives Q = 164.18 cfs, Y = 4.67 ft, and C = 109.32 for the rectangular channel, and Q = 368.70 cfs, Y = 4.52 ft, and C = 113.45 for the trapezoidal channel. When part 4 is solved for the bottom width b = 11.14 ft, Y = 4.51 ft, and C = 113.77. Program UENCHEZ1.FOR C Solves Chezys and Specific Energy for Y & Q by substituting Chezys Eq. C into Energy and first solving for Y and thereafter Q. The needed Chezy's C C is solved using Gauss_Seidel iteration (This is the procedure C to use with a pocket calculator) See program E_UN to solve any 2 variables. COMMON B,FM,D,H,CC,FMS,C,ITYPE WRITE(*,*)' Give: g,viscosity,(0=trap or 1=cir)' READ(*,*) G,VIS,ITYPE IF(ITYPE.EQ.0) THEN WRITE(*,*)' Give: b,m,e,So,H,Ke' READ(*,*) B,FM,e,SO,H,FKE FMS=2.*SQRT(FM*FM+1.) ELSE WRITE(*,*)' Give: D,e,So,H,Ke' READ(*,*) D,e,SO,H,FKE ENDIF G32=SQRT(32.*G) CG=.884*VIS/(4.*SQRT(G)) CC=SO*(1.+FKE)/(2.*G) IF(G.GT.15.) THEN C=100. ELSE C=60. ENDIF Y=.8*H NCT=0 10 F=FUN(Y) DIF=.05*F/(FUN(Y+.05)F) Y=YDIF CALL AREAP(Y,A,P) Rh=A/P Q=C*A*SQRT(Rh*SO) MCT=0 12 C1=C C=G32*ALOG10(e/(12.*Rh)+CG*C1*P/Q) MCT=MCT+1 IF(ABS(CC1).GT. 5.E6 .AND. MCT.LT.30) GO TO 12 IF(MCT.EQ.30) WRITE(*,*)' Did not converge',C,C1 NCT=NCT+1 WRITE(*,*) NCT,DIF,Y,Q IF(NCT.LT.30 .AND. ABS(DIF).GT. 1.E5) GO TO 10 WRITE(*,100) Y,Q,C 100 FORMAT(' Solution: Y =',F8.3,' Q =',F10.2,' C=',F9.2) END SUBROUTINE AREAP(Y,A,P) COMMON B,FM,D,H,CC,FMS,C,ITYPE IF(ITYPE.EQ.1) THEN
Energy and Its Dissipation in Open Channels COSB=1.2.*Y/D BETA=ACOS(COSB) P=D*BETA A=.25*D*D*(BETACOSB*SIN(BETA)) ELSE P=B+FMS*Y A=(B+FM*Y)*Y ENDIF RETURN END FUNCTION FUN(Y) COMMON B,FM,D,H,CC,FMS,C,ITYPE CALL AREAP(Y,A,P) FUN=HYCC*(A/P)*C*C RETURN END Program UENCHEZ1.C /* Solves Chezys and Specific Energy for Y & Q by substituting Chezys Eq.into Energy and first solving for Y and thereafter Q. The needed Chezy's C is solved using Gauss_Seidel iteration (This is the procedure to use with a pocket calculator) See program E_UN to solve any 2 variables. */ #include <stdio.h> #include <stdlib.h> #include <math.h> float b,m,d,h,cc,fms,c; int itype; void areap(float y,float *a,float *p){float cosb,beta; if(itype){cosb=1.2.*y/d;beta=acos(cosb);*p=d*beta; *a=.25*d*d*(betacosb*sin(beta));} else {*p=b+fms*y; *a=(b+m*y)*y;return;}} float fun(float y){float *a,*p; areap(y,a,p); return(hycc*((*a)/(*p))*c*c);} void main(void){float g,dif,ke,y,f,q,e,so,vis,g32,cg,rh,c1,*a,*p; int mct,nct=0; printf("Give: g,viscosity,(0=trap or 1=cir)\n"); scanf("%f %f %d",&g,&vis,&itype); if(itype){printf("Give: D,e,So,H,Ke\n"); scanf("%f %f %f %f %f",&d,&e,&so,&h,&ke);} else {printf("Give: b,m,e,So,H,Ke\n"); scanf("%f %f %f %f %f %f",&b,&m,&e,&so,&h,&ke); fms=2.*sqrt(m*m+1.);} if(g>15.) c=100.; else c=70.; g32=sqrt(32.*g); cg=.884*vis/(4.*sqrt(g)); cc=so*(1.+ke)/(2.*g); y=.8*h; do{f=fun(y); dif=.05*f/(fun(y+.05)f); y=dif; printf("%d %f %f\n",++nct,dif,y);areap(y,a,p); rh=(*a)/(*p); q=c*(*a)*sqrt(rh*so); mct=0; do{c1=c;c=g32*log10(e/(12.*rh)+cg*c1*(*p)/q); }while((fabs(cc1)>5.e6) && (++mct<30)); }while((nct<30) && (fabs(dif)>1.e5)); printf("Solution: Y = %8.3f Q = %10.2f C= %9.2f\n",y,q,c); } Example Problem 2.13 Rather than having a trapezoidal channel as in the previous example problem the channel is circular with a diameter D = 10 ft. Its Manning’s n = 0.014 and its bottom slope is So = 0.00075 as in the previous problem. (Ke = 0.1) Find the flow rate and depth. As a second part assume this
95
96
Open Channel Flow: Numerical Methods and Computer Applications circular channel has a steep slope. Now solve for Q and Y. As a third part the channel is a steep trapezoidal channel with b = 10 ft, and m = 1.2. Now solve for Q and Y. As a fourth part find the size of steep circular channel requires to convey a flow rate Q = 300 cfs. For all part the reservoir head is H = 5 ft and the entrance loss coefficient is Ke = 0.1. Solution For the first part either the program E_UN or E_UN1, given in the previous problem, can be used to solve the problem identifying the type of channel as circular. The solve gives: Q = 178.58 cfs and Y = 4.549 ft. You should verify these results with you calculator, using TKSolver and/or Mathcad. For the second part in which the channel is steep the critical flow equation F1 = Q2T − gA3 = 0 replaces the uniform flow equation of the previous example problem. However, when the unknowns are Q and Y, then Q can be solve from this equation and substituted into the energy equation to give the single equation F = H − Y − (1 + Ke)A/(2T) = 0 to solve for Y and thereafter solve the critical flow equation for Q. The program E_CR1 uses this later approach to solve for Y and then Q, and the program E_CR solves the critical and energy equations simultaneously. The answers to second part of the problem are: Y = 3.56 ft and Q = 230.20 cfs. For the third part of the problem in which the steep channel has a trapezoidal cross section the answers are: Y = 3.51 ft and Q = 474.37 cfs. For the fourth part of the problem in which the diameter is want that will convey Q = 300 cfs, the two equations must be solved simultaneously for Y and D (or program UCR must be used.) The answers are Y = 3.60 ft and D = 15.413. Program E_CR1.FOR C Solves Critical and Specific Energy for Y & Q by substituting C Critical Eq. into Energy and first solving for Y and thereafter Q. C This is the procedure to use with a pocket calculator) See program E_CR to solve any 2 variables. COMMON B,FM,D,H,C,ITYPE WRITE(*,*)' Give: g,(0=trap or 1=cir)' READ(*,*) G,ITYPE IF(ITYPE.EQ.0) THEN WRITE(*,*)' Give: b,m,H,Ke' READ(*,*) B,FM,H,FKE ELSE WRITE(*,*)' Give: D,H,Ke' READ(*,*) D,H,FKE ENDIF C=.5*(1.+FKE) Y=.7*H NCT=0 10 F=FUN(Y) DIF=.05*F/(FUN(Y+.05)F) Y=YDIF NCT=NCT+1 WRITE(*,*) NCT,DIF,Y IF(NCT.LT.30 .AND. ABS(DIF).GT. 1.E5) GO TO 10 CALL AREAP(Y,A,T) Q=SQRT(G*A**3/T) WRITE(*,100) Y,Q 100 FORMAT(' Solution: Y =',F8.3,' Q =',F10.2) END SUBROUTINE AREAP(Y,A,T) COMMON B,FM,D,H,C,ITYPE IF(ITYPE.EQ.1) THEN COSB=1.2.*Y/D
Energy and Its Dissipation in Open Channels BETA=ACOS(COSB) T=D*SIN(BETA) A=.25*D*D*(BETACOSB*SIN(BETA)) ELSE T=B+2.*Y*FM A=(B+FM*Y)*Y ENDIF RETURN END FUNCTION FUN(Y) COMMON B,FM,D,H,C,ITYPE CALL AREAP(Y,A,T) FUN=HYA*C/T RETURN END Program E_CR1.C /* Solves Critical and Specific Energy for Y & Q by substituting Critical Eq.into Energy and first solving for Y and thereafter Q. (This is the procedure to use with a pocket calculator) See program E_CR to solve any 2 variables. */ #include <stdio.h> #include <stdlib.h> #include <math.h> float b,m,d,h,c; int itype; void areap(float y,float *a,float *t){float cosb,beta; if(itype){cosb=1.2.*y/d;beta=acos(cosb);*t=d*sin(beta); *a=.25*d*d*(betacosb*sin(beta));} else {*t=b+2.*y*m; *a=(b+m*y)*y;return;}} float fun(float y){float *a,*t; areap(y,a,t); return(hyc*(*a)/(*t));} void main(void){float g,dif,ke,y,f,q,*a,*t; int nct=0; printf("Give: g(0=trap or 1=cir)\n"); scanf("%f %d",&g,&itype); if(itype){printf("Give: D,H,Ke\n"); scanf("%f %f %f",&d,&h,&ke);} else {printf("Give: b,m,H,Ke\n"); scanf("%f %f %f %f",&b,&m,&h,&ke);} y=.7*h; c=.5*(1.+ke); do{f=fun(y); dif=.05*f/(fun(y+.05)f); y=dif;printf("%d %f %f\n",++nct,dif,y); }while((nct<30) && (fabs(dif)>1.e5)); areap(y,a,t); q=sqrt(g*pow((*a),3.)/(*t)); printf("Solution: Y = %8.3f Q = %10.2f\n",y,q); } Program E_CR.FOR C Solves critical flow and Energy simultaneously C for any 2 unknowns CHARACTER*17 FMT/'(1X,A1,'' ='',F9.3)'/ CHARACTER*1 CX(6)/'b','m','Q','H','Y','K'/ CHARACTER*5 CH(0:1)/'value','guess'/ REAL F(2),F1(2),D(2,2) INTEGER*2 ID(6),INDX(2) COMMON X(6),G,FKE,ITYPE WRITE(*,*)' Give: g,entrance loss C. & 0=TRAP',' or 1=CIRLCE' READ(*,*) G,FKE,ITYPE IF(ITYPE.EQ.1) THEN
97
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Open Channel Flow: Numerical Methods and Computer Applications
1 10 2
100 20 30
35 40
50
60
CX(1)='D' CX(2)=' ' ENDIF X(6)=FKE FKE=(1.+FKE)/(2.*G) DO 10 I=1,5 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 10 WRITE(*,'(I2,2X,A1)') I,CX(I) ID(I)=0 WRITE(*,*)' Give two numbers for 2 unknown variables' READ(*,*) I1,I2 IF(I1.LT.1.OR.I1.GT.7.OR.I2.LT.1.OR.I2.GT.7) GO TO 2 ID(I1)=1 ID(I2)=1 DO 20 I=1,5 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 20 WRITE(*,100) CH(ID(I)),CX(I) FORMAT(' Give ',A5,' for ",A1,' = ',\) READ(*,*) X(I) CONTINUE NCT=0 SUM=0. CALL FUN(F) I1=0 DO 40 I=1,5 IF(ID(I).EQ.0) GO TO 40 I1=I1+1 X(I)=1.005*X(I) XX=X(I) CALL FUN(F1) DO 35 J=1,2 D(J,I1)=(F1(J)F(J))/(X(I)XX) X(I)=XX CONTINUE CALL SOLVEQ(2,1,2,D,F,1,DD,INDX) I1=0 DO 50 I=1,5 IF(ID(I).EQ.0) GO TO 50 I1=I1+1 SUM=SUM+ABS(F(I1)) X(I)=X(I)F(I1) CONTINUE NCT=NCT+1 WRITE(*,*)' NCT=',NCT,' SUM=',SUM IF(NCT.LT.30 .AND. SUM.GT.1.E5) GO TO 30 WRITE(*,*)' Solution:' DO 60 I=1,6 IF(ITYPE.EQ.1 .AND. I.EQ.2) GO TO 60 WRITE(*,FMT) CX(I),X(I) CONTINUE WRITE(*,*)' Give 1 to solve another prob. (0=STOP)' READ(*,*) I2 IF(I2.EQ.1) GO TO 1 END SUBROUTINE FUN(F) REAL F(2)
Energy and Its Dissipation in Open Channels COMMON X(6),G,FKE,ITYPE IF(ITYPE.EQ.1) THEN BETA=ACOS(1.2.*X(5)/X(1)) A=.25*X(1)**2*(BETACOS(BETA)*SIN(BETA)) T=X(1)*SIN(BETA) ELSE A=(X(1)+X(2)*X(5))*X(5) T=X(1)+2.*X(2)*X(5) ENDIF F(1)=X(3)**2*TG*A**3 F(2)=X(4)X(5)FKE*(X(3)/A)**2 RETURN END Program E_CR.C // Solves critical flow and Energy simultaneously for any 2 // unknowns #include <stdio.h> #include <stdlib.h> #include <math.h> float x[6],g,fke; int itype; extern void solveq(int n,float **d,float *f,int itype,\ float *dd,int *indx); void fun(float *f){float beta,a,p,t; if(itype){beta=acos(1.2.*x[4]/x[0]); a=.25*x[0]*x[0]*(betacos(beta)*sin(beta)); t=x[0]*sin(beta);} else {a=(x[0]+x[1]*x[4])*x[4];t=x[0]+2.*x[1]*x[4];} f[0]=x[2]*x[2]*tg*pow(a,3.); f[1]=x[3]x[4]fke*pow(x[2]/a,2.);return;} //End of fun void main(void){ char *fmt=" %c =%9.3f\n",*ch[]={"value","guess"},*cx="bmQHYK"; float f[2],f1[2],xx,sum,**d,*dd; int id[6],indx[2],i,j,i1,i2,nct; d=(float**)malloc(2*sizeof(float*)); for(i=0;i<2;i++)d[i]=(float*)malloc(2*sizeof(float)); printf("Give: g,entrance loss C. & 0=TRAP or 1=CIRLCE\n"); scanf("%f %f %d",&g,&fke,&itype); if(itype){stpcpy(cx[0],"D");stpcpy(cx[1]," ");} x[5]=fke;fke=(1.+fke)/(2.*g); L1: for(i=0;i<5;i++){if((!itype)  ((itype)&&(i != 1)))\ printf("%2d %c\n",i+1,cx[i]);id[i]=0;} do{printf(" Give two numbers for 2 unknown variables\n"); scanf("%d %d",&i1,&i2); }while((i1<1)(i1>7)(i2<1)(i2>7)); id[i11]=1;id[i21]=1; for(i=0;i<5;i++){if((!itype)  ((itype)&&(i != 1)))\ {printf("Give %s for %c =",ch[id[i]],cx[i]); scanf("%f",&x[i]);}} nct=0; do{sum=0.; fun(f); i1=1; for(i=0;i<5;i++){if(id[i]){xx=x[i];i1++;x[i]*=1.005;fun(f1); for(j=0;j<2;j++) d[j][i1]=(f1[j]f[j])/(x[i]xx); x[i]=xx;}} solveq(2,d,f,1,dd,indx); i1=1; for(i=0;i<5;i++){if(id[i]){sum+=fabs(f[++i1]);x[i]=f[i1];}} printf("nct= %d SUM=%f\n",nct,sum); }while((++nct<30)&&(sum>1.e5)); printf("Solution:\n");
99
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Open Channel Flow: Numerical Methods and Computer Applications for(i=0;i<6;i++){if((!itype)  ((itype)&&(i != 1))){ if(i==2)stpcpy(fmt[8],"6"); else stpcpy(fmt[8],"3"); printf(fmt,cx[i],x[i]);}} printf("Give 1 to solve another prob. (0=STOP)\n"); scanf("%d",&i2); if(i2) goto L1;}
2.7 Flumes Flow rate measurements in open channels can be made by devices called flumes, which reduce the size of a throat section by reducing its width and/or raising its bottom. If the channel downstream from where the device is installed does not back the water up, so “free flow conditions” exist, then critical flow occurs within the throat length. To help assure that critical flow occurs the section diverges downstream from the throat and the bottom may have a steep slope through the throat or possibly downstream therefrom. Such devices are called “critical flow” meters because their means of measuring the flow is based on causing the depth to be critical in the throat section, or the raised section. If only a hump in the bottom occurs the device is referred to as a broadcrested weir. Chapter 5 gives more information about open channel flow measurement devices and how they are constructed. Using the theory associated with critical flow to determine how the flow rate is determined by a measured depth in such devices is covered here. First let us deal with the simple case of a flat bottom device with contracting sides in a rectangular channel upstream from the device and diverging walls downstream therefrom to a larger, or steeper sloping downstream channel that can readily convey the water away from the device, so critical flow occurs within the throat as shown in the sketch below.
b2
b1
Venturi flume
Y1
Y2 = Yc
Flat bottom
With the depth critical within the throat the flow rate Q equals unit flow rate q2 times the throat width b2. And from Equation 2.12 q c = q 2 = (gYc3 )1 / 2, so
Q = b2
2 gY = 3 3 c
1.5
gb2 E1.5 c
(2.12e)
(The number 12 is used as a reminder that the equation is restricted to rectangular channels.) The latter of which comes from Equation 2.12c Yc = 2Ec/3. Because the depth at section 2 is less stable with the larger velocities here, the upstream depth Y1 is generally measured rather than Y2 = Yc. So the typical flow equation equates E1 = Y1 + Q2/(2gb12Y12) to Ec = E2, and the flow rate equation becomes the following implicit equation that gives Q as a function of the upstream depth Y1:
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Energy and Its Dissipation in Open Channels
2 Q= 3
1.5
1.5
QZ gb2 Y1 + 2gb1Z Y1Z
(2.12f)
Rather than solve this implicit equation practice replaces Ec by Y1 and introduces a coefficient, Cv, called the velocity coefficient, since it accounts for the upstream velocity head and approaches unity as the upstream velocity head becomes negligible, or 2 Q = Cv 3
1.5
gb2 Y11.5
(2.12g)
To obtain an equation that gives Cv let’s relate Y1 to the critical depth at section 2, or Y1 +
Q2 = 1.5Yc (2gb12 Y12 )
(2.12h)
By dividing by Yc and defining the dimensionless upstream depth Y1′ = Y1/Yc we can, with a little algebraic manipulation, obtain the following cubic equation relating Y1′ to the ratio of the throat to the upstream width b2/b1. 2
b 2Y1′3 − 3Y1′2 + 2 = 0 b1
(2.12i)
The table below gives the three roots to this equation for several width ratios. Table Giving Roots of Equation 2.12i for Several Width Ratios b2/b1
0.95
0.90
0.85
0.80
0.75
0.70
0.65
0.60
root 1 root 2 root 3
1.1708 0.8069 −.4777
1.2341 0.7211 −.4551
1.2793 0.6531 −.4324
1.3149 0.5945 −.4094
1.3444 0.5418 −.3862
1.3693 0.4933 −.3627
1.3908 0.4481 −.3389
1.4094 0.4055 −.3149
0.55
0.50
1.4256 0.3651 −.2906
1.4397 0.3264 −.2660
0.45 1.4520 0.2892 −.2411
0.40 1.4626 0.2533 −.2159
0.35 1.4717 0.2186 −.1904
0.30 1.4794 0.1850 −.1644
0.25 1.4910 0.1523 −.1381
Of the three roots of this equation only the larger of the two positive roots occurs in practice because the negative root is physically impossible, and the smaller of the positive roots is associated with supercritical flow upstream. The largest root, on the other hand, allows us to conclude that for any flow rate in a rectangular flume with a flat bottom there is a fixed upstream depth Y1 associated with any Q. To find this Y1 first solve Yc = {q2/g}1/3 = {(Q2/b2)/g}1/3 (Equation 2.12). Next solve Equation 2.12i for Y1′, and finally compute Y1 = Y1′Yc. In order for the flume to make its measurement properly (i.e., have critical flow at its throat) the depth of flow (which is often normal depth) at the position where the flume is to be installed must be equal to or less than this value of Y1. If it is larger than Y1 then the flow through the flume’s throat will be subcritical. For example, assume a venturi flume with a width ratio of 0.8 is to be installed in a 5 m wide rectangular channel with
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Open Channel Flow: Numerical Methods and Computer Applications
n = 0.013 to measure a maximum flow rate Q = 40 m3/s. What is the maximum bottom slope this channel can have for the flume to work properly? From the above table Y1′ = 1.3149 corresponding to b2/b1 =0.8, and b2 = 0.8(5) = 4 m, so q2 = 40/4 = 10 m2/s and Yc = 2.168 m. Thus the maximum normal depth is Y1 = Y1′Yc = 2.851 m. Solving Manning’s equation we get a bottom slope of So = 0.000908. If the bottom slope were 0.0008 for example then its normal depth would be Yo1 = 2.992 with Eo1 = 3.357, and setting E2 = 3.357 and solving for Y2 = 2.607. The associated Froude number is 0.758, i.e., the flow is not critical at section 2. Let us assume the flow rate is Q = 20 m3/s. Then the normal depth, with So = 0.000908 is Yo1 = 1.709 m and Eo1 = 1.988 m. The critical depth is 1.366 m and Ec = 2.049 m. Thus the flume will cause the depth upstream to be larger than Yo1 so E1 = Ec = 2.049 m. Also Equation 2.12i can be used to get an equation for the velocity coefficient Cv. By dividing Equation 2.12g by Equation 2.12e we get 1 = Cv(Y1/E1)1.5, or Y1 / E1 = Cv−2 / 3, but Y1′ = Y1 / Yc = 1.5Y1/ E1, so Y1′ = 1.5Cv−2 / 3, which when substituted into Equation 2.12i gives 2
4 b2 2 2 /3 27 b Cv − Cv + 1 = 0 1
(2.12j)
Alas! Another implicit equation, but you can readily let your pocket calculator solve it for you. The table below provides solutions to Cv for several width ratios b2/b1. Table That Provides the Solution of the Velocity Coefficient Cv from Equation 2.12j for Several Width Ratios b2/b1 Cv
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.002233 1.009064 1.020918 1.038597 1.063487 1.097979 1.146490 1.218388 1.340083 1.835955
Thus you can choose whether you solve Equation 2.12f for Q given an upstream depth Y1, or solve Equation 2.12j and substitute the Cv obtained therefrom into Equation 2.12 g to solve Q. A final option is to solve the energy equation E1 = E2c + Δz and the critical flow equation Q 2 T 5 /(2gA 32 ) = 1 simultaneously for Yc and Q. If the channel at either position 1 or 2 is not rectangular, the bottom is not flat, and Y1 is known then only this later option exists. A broad crested weir measures the flow rate by having a hump, with a height Δz, in the bottom of a rectangular channel, which is high enough to cause critical flow over it as shown in the sketch below.
Y1
Yc ΔZ
Flat bottom, So = 0
Equation 2.12e applies equally to this device as it does to the venturi flume. Now, however, Ec = E1 – Δz, and therefore the implicit equation that gives Q is as follows, rather than Equation 2.12f:
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Energy and Its Dissipation in Open Channels
2 Q= 3
1.5
1.5
Q2 gb2 Y1 − ∆ Z + 2gA12
(2.12k)
Following the same procedure used to obtain the dimensionless cubic Equation 2.12i, that gives the dimensionless upstream depth Y1′ = Y1/ Yc as a function of the width ratio yields 2
b 2Y1′3 − (3 + 2∆ Z )Y1′2 + 2 = 0 b1
(2.12l)
Equation 2.12l also assumes that the channel is rectangular upstream from the broad crested weir, but Equation 2.12k applies for a non rectangular upstream channel whose crosssectional area is A1. (Likewise Equation 2.12f could be used for situations in which the channel is not rectangular if A1 replaces b1Y1 in it.) It is not practical, however, to introduce a velocity coefficient into Equation 2.12k and there after develop an equation for Cv. The difference between Equations 2.12l and 2.12i is that now Y1′ depends upon two parameters, the width ratio b2/b1 and (3 + 2Δz/Yc). Thus the upstream dimensionless depth Y1′ is fixed by b2/b1 but this value will be different for each different height of weir divided by the critical depth over the weir Δz/Yc, and since Yc varies with Q therefore Y1′ varies with b2/b1, Δz, and Q.
2.8 Delivery Diagrams A delivery diagram is a graph that gives the flow rate Q that will occur in a given channel as a function of the head H of the reservoir that supplies it. Often only one channel is involved and this channel is long so that until critical flow occurs at its entrance the flow will be uniform throughout its length. Under these conditions the data for Q versus H to plot the delivery diagram can be obtained by solving the energy and uniform flow equations simultaneously, as has been done in the previous example problems, for a series of H values until the Froude number becomes unity, and thereafter solve the energy and critical flow equations. For a given channel (its size variables, roughness, and bottom slope do not change with depth and flow rate), the Froude number associated with the flow will increase as the head H of the supply reservoir increases. Thus in obtaining the series of solutions for the delivery diagram it is best to start with the smallest reservoir head H that is anticipated and increase with a specified increment toward the largest H anticipated. To illustrate how delivery diagrams are created let’s use a trapezoidal channel with b = 10 ft, m = 1.2, n = 0.014. Delivery diagrams will be created for this channel for bottom slopes varying from So = 0.0005 to So = 0.0028 to demonstrate that as the bottom slope of the channel increases and H increases that critical flow at the entrance can limit the flow rate that the channel will carry. The table below provides the solution for the depth Y and the flow rate Q from the simultaneous solution of the Energy and Manning’s equations for the reservoir head H varying from 1 to 8 ft, for five different bottom slopes. The Froude number squared has been added as a third column to each of these solutions. The last two columns in this table contain the critical depth Yc and the critical flow rate Qc, i.e., the simultaneous solution of the energy and critical flow equations associated with the same H values. Notice that for bottom slopes of 0.0005, 0.001, and 0.002 that all the Froude numbers are less than one. Thus all these Q versus H data do represent the delivery diagram. However with bottom slopes of 0.0025 and 0.0028 the Froude
Y
.929 1.112 1.295 1.479 1.662 1.845 2.028 2.212 2.395 2.578 2.762 2.945 3.129 3.312 3.496 3.679
H
1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0
21.0 28.5 36.9 46.3 56.5 67.6 79.7 92.6 106.4 121.2 136.8 153.4 170.9 189.3 208.7 229.0
Q
So = 0.0005
Fr
0.15 0.16 0.17 0.17 0.18 0.18 0.18 0.19 0.19 0.19 0.19 0.20 0.20 0.20 0.20 0.21
2
Y
0.868 1.037 1.207 1.376 1.545 1.714 1.883 2.052 2.221 2.391 2.560 2.729 2.899 3.068 3.238 3.407
26.5 35.9 46.3 57.9 70.5 84.3 99.1 115.0 132.0 150.1 169.2 189.5 210.9 233.4 257.0 281.8
Q
So = 0.001 Fr 0.30 0.31 0.33 0.34 0.34 0.35 0.36 0.37 0.37 0.38 0.38 0.39 0.39 0.40 0.40 0.41
2
Y 0.771 0.918 1.065 1.212 1.358 1.505 1.651 1.798 1.944 2.091 2.237 2.384 2.531 2.677 2.824 2.971
30.8 41.3 53.0 65.9 80.1 95.3 111.8 129.3 148.1 167.9 189.0 211.2 234.5 259.1 284.8 311.7
Q
So = 0.002 Fr 0.58 0.61 0.63 0.65 0.67 0.68 0.70 0.71 0.72 0.74 0.75 0.76 0.76 0.77 0.78 0.79
2
Y 0.731 0.869 1.007 1.145 1.282 1.420 1.557 1.694 1.831 1.969 2.106 2.244 2.381 2.519 2.656 2.794
31.5 42.1 54.0 67.0 81.2 96.5 113.0 130.6 149.3 169.2 190.2 212.3 235.6 260.0 285.6 312.4
Q
So = 0.0025 Fr 0.72 0.75 0.78 0.80 0.83 0.84 0.86 0.88 0.89 0.91 0.92 0.93 0.94 0.95 0.97 0.97
2
Y 0.709 0.843 0.976 1.109 1.241 1.374 1.506 1.638 1.770 1.903 2.035 2.168 2.300 2.433 2.565 2.698
31.6 42.3 54.1 67.1 81.3 96.6 112.9 130.5 149.1 168.8 189.6 211.6 234.7 258.9 284.2 310.7
Q
So = 0.0028 Fr
0.80 0.83 0.87 0.89 0.92 0.94 0.96 0.98 1.00 1.01 1.02 1.04 1.05 1.06 1.07 1.08
2
0.661 0.797 0.934 1.071 1.210 1.349 1.489 1.630 1.771 1.913 2.056 2.199 2.343 2.488 2.632 2.778
Yc
Qc 31.7 42.3 54.1 67.1 81.2 96.4 112.9 130.4 149.1 168.9 189.9 212.1 235.4 259.8 285.4 312.2
Critical
Table that Solves the Energy and Mannings Equations Simultaneously for Varying Reservoir Heads H and Bottom Slopes, and also the Energy and Critical Flow Equations for Yc and Qc for these Same H’s for a Trapezoidal Channel with b = 10 ft, m = 1.2, n = .014 and an Entrance Loss Coefficient Ke = 0.1
104 Open Channel Flow: Numerical Methods and Computer Applications
4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0 6.2 6.4 6.6 6.8 7.0 7.2 7.4 7.6 7.8 8.0
3.863 4.046 4.230 4.413 4.597 4.781 4.964 5.148 5.331 5.515 5.698 5.882 6.065 6.249 6.432 6.615 6.799 6.982 7.166 7.349
250.3 272.5 295.8 320.0 345.2 371.4 398.6 426.9 456.2 486.6 518.0 550.5 584.1 618.8 654.6 691.5 729.5 768.7 809.0 850.5
0.21 0.21 0.21 0.21 0.21 0.22 0.22 0.22 0.22 0.22 0.22 0.22 0.23 0.23 0.23 0.23 0.23 0.23 0.23 0.23
3.577 3.746 3.916 4.086 4.255 4.425 4.594 4.764 4.934 5.103 5.273 5.442 5.612 5.781 5.951 6.120 6.290 6.459 6.629 6.798
307.7 334.7 362.9 392.3 422.9 454.7 487.6 521.9 557.3 594.0 631.9 671.1 711.5 753.3 796.3 840.7 886.4 933.4 981.7 1031.4
0.41 0.41 0.42 0.42 0.42 0.43 0.43 0.43 0.43 0.44 0.44 0.44 0.45 0.45 0.45 0.45 0.46 0.46 0.46 0.46
3.118 3.265 3.412 3.560 3.707 3.854 4.001 4.149 4.296 4.443 4.590 4.738 4.885 5.032 5.179 5.326 5.474 5.621 5.768 5.915
339.8 369.1 399.6 431.3 464.3 498.5 534.0 570.7 608.7 647.9 688.5 730.3 773.5 818.0 863.8 910.9 959.4 1009.3 1060.5 1113.1
0.80 0.80 0.81 0.82 0.82 0.83 0.84 0.84 0.85 0.85 0.86 0.86 0.87 0.87 0.88 0.88 0.89 0.89 0.90 0.90
2.932 3.070 3.208 3.346 3.484 3.622 3.760 3.898 4.036 4.174 4.312 4.450 4.589 4.727 4.865 5.003 5.141 5.279 5.417 5.555
340.3 369.4 399.6 431.1 463.8 497.6 532.7 569.0 606.6 645.3 685.4 726.7 769.2 813.1 858.2 904.6 952.3 1001.4 1051.7 1103.4
0.98 0.99 1.00 1.01 1.02 1.02 1.03 1.04 1.05 1.05 1.06 1.07 1.07 1.08 1.08 1.09 1.10 1.10 1.11 1.11
2.831 2.964 3.097 3.230 3.363 3.496 3.629 3.762 3.895 4.028 4.161 4.295 4.428 4.561 4.694 4.827 4.961 5.094 5.227 5.360
338.3 367.1 397.0 428.1 460.4 493.9 528.5 564.3 601.4 639.7 679.1 719.8 761.8 805.0 849.4 895.2 942.2 990.4 1040.0 1090.8
1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.16 1.17 1.18 1.19 1.19 1.20 1.21 1.21 1.22 1.22 1.23 1.24
2.924 3.070 3.216 3.363 3.511 3.658 3.806 3.955 4.103 4.252 4.401 4.550 4.700 4.850 5.000 5.150 5.301 5.451 5.602 5.753
340.2 369.4 399.8 431.5 464.3 498.4 533.7 570.3 608.2 647.3 687.8 729.5 772.5 816.9 862.6 909.6 958.0 1007.7 1058.8 1111.3
Energy and Its Dissipation in Open Channels 105
106
Open Channel Flow: Numerical Methods and Computer Applications
numbers becomes larger than unity when H becomes larger than 4.6 and 2.6 ft, respectively. In other words, for these latter two bottom slopes critical conditions at the channel entrance limit the flow rates as the head of the reservoir rises above these values. What occurs is critical depth at the entrance of the channel, and then the depth gradually decreases downstream therefrom toward the normal depth associated with this critical flow rate. Such gradually varied flow profiles will be dealt with in Chapter 4. To plot the delivery diagram the critical depth Yc, and the critical flow rate Qc from the last column are substituted in place of the computed Y and Q whenever the Froude number is larger than unity. The delivery diagrams for this channel for the five different bottom slopes are given in the graph below, with the depths plotted using the right ordinate. 6
1000
5
28
800 600 De
400
at pth
t en
S for ce ran
28
.00
=0 o
2,
=
So
0 0.0
= ,So
25
= So
0 0.0
1
.00
=0
4
So
3
.00
=0
So
05
0 0.0
2 1
200 0
Depth at entrance, Y (ft)
Flowrate into channel, Q (cfs)
1200
1
2
3
5 6 4 Reservoir head, H (ft)
7
8
0
In the design of entrances from reservoirs to channels the size, or shape of the entrance may be different than the channel. If this is the case then solving the flow into the channel will involve the solution of three simultaneous equation for the three unknowns, Q, Y1, and Y2, in which Y1 is the depth in the entrance geometry and Y2 is the normal depth in the channel downstream therefrom. In other words in place of a single normal depth two depths are added to the list of unknowns. Therefore an additional equation is needed, and this equation comes from writing the energy equation across the transition from the entrance geometry to the channel geometry downstream therefrom. These three equations are
F1 = nQP22 / 3 − Cu A 52 / 3S1o/ 2 = 0 (Manning ’s, or a Uniform F, Equation) F2
2 1 + K e )( Q /A1 ) ( = H−Y − 1
F3 = Y +
H
(2 g )
= 0 (Energy at enterance)
(Q /A1 )2 (1 + K L )(Q /A 2 )2 − Y2 − − ∆z = 0 (Energy across transition) (2 g ) (2 g )
Y1 Δz Y1 b
Y2 D, n, S o
m
Energy and Its Dissipation in Open Channels
107
in which Δz is the rise in the bottom from the entrance to the beginning of the channel, as shown above in the sketch that shows a trapezoidal section at the entrance from the reservoir and a transition to a circular channel downstream therefrom. Program E_UNTC is designed to solve for any 3 of the 12 variables involved if the entrance from a reservoir to a circular channel is trapezoidal. Generally this program would be used to solve for variables 6 = Q, 8 = Y1, and 9 = Y2. For example, if a reservoir with H = 5 ft supplies a circular channel with a diameter D = 12 ft, n = 0.014, and So = 0.001 and the entrance consists of a trapezoidal section with b = 10 ft, m = 1.2, and the entrance loss coefficient is Ke = 0.1, and the loss coefficient K L = 0.1, and there is no change in bottom elevation across the transition from the trapezoidal to the circular sections then this program produces the following: b = 10 ft, m = 1.2, D = 12 ft, So = 0.001, n = 0.014, Q = 223.7 cfs (solution), H = 5.0, Y1 = 4.86 ft (solution), Y2 = 4.37 ft (solution), Ke = 0.1, K L = 0.1 and Dz = 0. E_UNTC.FOR olves flow in circular channel that has a trapezoidal entrance C S C at the reservoir. C Equations are: Manning's, Energy at entrance, & Energy between C 2 shapes. CHARACTER*17 FMT/'(1X,A2,'' ='',F9.3)'/ CHARACTER*2 CX(12)/'b ','m ','D ','So','n ','Q ','H ', &'Y1','Y2','Ke','KL','Dz'/ CHARACTER C1(12)/'2','2','2','5','3','1','2','2','2','3', &'3','2'/ CHARACTER*5 CH(0:1)/'value','guess'/ REAL F(3),F1(3),D(3,3) INTEGER*2 ID(12),INDX(3) COMMON X(12),G,G2,Cu WRITE(*,*)' Give: g' READ(*,*) G G2=2.*G Cu=1. IF(G.GT.15.) Cu=1.486 1 DO 10 I=1,12 WRITE(*,'(I3,2X,A2)') I,CX(I) 10 ID(I)=0 2 WRITE(*,*)' Give three numbers for 3 unknown','variables' READ(*,*) I1,I2,I3 IF(I1.LT.1.OR.I1.GT.12.OR.I2.LT.1.OR.I2.GT.12 &.OR.I3.LT.1.OR.I3.GT.12) GO TO 2 ID(I1)=1 ID(I2)=1 ID(I3)=1 DO 20 I=1,12 WRITE(*,100) CH(ID(I)),CX(I) 100 FORMAT(' Give ',A5,' for ',A2,' = ',\) READ(*,*) X(I) 20 CONTINUE NCT=0 30 SUM=0. CALL FUN(F) I1=0
108
35 40
50
60
Open Channel Flow: Numerical Methods and Computer Applications
DO 40 I=1,12 IF(ID(I).EQ.0) GO TO 40 XX=X(I) I1=I1+1 X(I)=1.005*X(I) CALL FUN(F1) DO 35 J=1,3 D(J,I1)=(F1(J)F(J))/(X(I)XX) X(I)=XX CONTINUE CALL SOLVEQ(3,1,3,D,F,1,DD,INDX) I1=0 DO 50 I=1,12 IF(ID(I).EQ.0) GO TO 50 I1=I1+1 SUM=SUM+ABS(F(I1)) X(I)=X(I)F(I1) CONTINUE NCT=NCT+1 WRITE(*,*)' NCT=',NCT,' SUM=',SUM IF(NCT.LT.30 .AND. SUM.GT.3.E5) GO TO 30 WRITE(*,*)' Solution:' DO 60 I=1,12 FMT(16:16)=C1(I) WRITE(*,FMT) CX(I),X(I) CONTINUE WRITE(*,*)' Give 1 to solve another prob.','(0=STOP)' READ(*,*) I2 IF(I2.EQ.1) GO TO 1 END SUBROUTINE FUN(F) REAL F(3) COMMON X(12),G,G2,Cu BETA=ACOS(1.2.*X(9)/X(3)) A2=.25*X(3)**2*(BETACOS(BETA)*SIN(BETA)) P2=X(3)*BETA A1=(X(1)+X(2)*X(8))*X(8) FKe=(1.+X(10))/G2 FKL=(1.+X(11))/G2 F(1)=X(5)*X(6)*P2**.6666667Cu*A2**1.6666667*SQRT(X(4)) F(2)=X(7)X(8)FKE*(X(6)/A1)**2 F(3)=X(8)+(X(6)/A1)**2/G2X(9)FKL*(X(6)/A2)**2X(12) RETURN END
E_UNTC.C // Solves Manning's (uniform flow) and Energy simultaneously for // any 2 unknowns // See E_UN1 to solve Q & Y and method that can be used with // calculator #include <stdio.h>
Energy and Its Dissipation in Open Channels
109
#include <stdlib.h> #include <math.h> float x[12],g,g2,cu; extern void solveq(int n,float **d,float *f,int itype,\ float *dd,int *indx); void fun(float *f){float beta,a1,a2,p2,fke,fkl; beta=acos(1.2.*x[8]/x[2]); a2=.25*x[2]*x[2]*(betacos(beta)*sin(beta));p2=beta*x[2]; a1=(x[0]+x[1]*x[7])*x[7];fke=(1.+x[9])/g2;fkl=(1.+x[10])/g2; f[0]=x[4]*x[5]*pow(p2,.6666667)cu*pow(a2,1.6666667)*sqrt(x[3]); f[1]=x[6]x[7]fke*pow(x[5]/a1,2.); f[2]=x[7]+pow(x[5]/a1,2.)/g2x[8]fkl*pow(x[5]/a2,2.)x[11]; return;} //End of fun void main(void){char *fmt=" %s =%9.3f\n",*ch[]={"value","guess"},\ *c1="222531222332"; char *cx[]={"b ","m ","D ","So","n ","Q ","H ","Y1","Y2","Ke","KL","Dz"}; float f[3],f1[3],xx,sum,**d,*dd; int id[12],indx[3],i,j,i1,i2,i3,nct; d=(float**)malloc(3*sizeof(float*)); for(i=0;i<3;i++)d[i]=(float*)malloc(3*sizeof(float)); printf("Give: g\n");scanf("%f",&g); g2=2.*g; if(g>15.) cu=1.486; else cu=1.; L1: for(i=0;i<12;i++){printf("%2d %s\n",i+1,cx[i]);id[i]=0;} do{printf(" Give three numbers for 3 unknown variables\n"); scanf("%d %d %d",&i1,&i2,&i3); }while((i1<1)(i1>12)(i2<1)(i2>12)(i3<1)(i3>12)); id[i11]=1;id[i21]=1;id[i31]=1; for(i=0;i<12;i++){printf("Give %s for %s =",ch[id[i]],cx[i]); scanf("%f",&x[i]);} nct=0; do{sum=0.; fun(f); i1=1;for(i=0;i<12;i++){if(id[i]){xx=x[i]; i1++;x[i]*=1.005;fun(f1); for(j=0;j<3;j++) d[j][i1]=(f1[j]f[j])/(x[i]xx); x[i]=xx;}} solveq(3,d,f,1,dd,indx); i1=1; for(i=0;i<12;i++){if(id[i]){sum+=fabs(f[++i1]);x[i]=f[i1];}} printf("nct= %d SUM=%f\n",nct,sum); }while((++nct<30)&&(sum>3.e5)); printf("Solution:\n");for(i=0;i<12;i++){fmt[8]=c1[i]; printf(fmt,cx[i],x[i]);} printf("Give 1 to solve another prob. (0=STOP)\n"); scanf("%d",&i2); if(i2) goto L1;} In developing a delivery diagram of flow into a channel with a different geometry at entrance than the channel, critical flow in the channel or its entrance geometry may control. To check which controls the Froude numbers associated with both Y1 and Y2 can be computed, and should either of these become larger than unity, then the critical flow equation in this section will replace the uniform flow equation above. Problems associated with water entering canal systems from reservoirs can become more complex than those illustrated above in which a single channel is taking water from the reservoir. Often several channels may take water from the same inlet works, with the different channels conveying the water away in different directions, at different slopes, and each may have its own controls. Problems may be further complicated by not having uniform flow in some of these channels because
110
Open Channel Flow: Numerical Methods and Computer Applications
controls in them may exist downstream quite some distance from the inlet works creating gradually varied flows in these channels. Problems in which gradually varied flows exist will be discussed in a later Chapter 4. However, if the flow is uniform in the channels, or critical at the entrances in some of the channels, or if the controls in the separate channels are short distances downstream from where the water is taken from the reservoir, then the tools we have covered in this chapter suffice in solving the problems. The explanation and illustrative examples below will help in understanding how to identify what variables need to be solved, and what the governing equations are for a given multiply branched channel system. At this juncture in the course the governing equations come from the following four principles: (1) Conservation of mass, e.g., that the flow out from a junction equals that into the junction. In brief the junction continuity equation applies at each junction. (2) The energy is the same (with local losses accounted for if desired) in upstream and downstream channels branching therefore. Also the head in a reservoir equals the specific energy in the channel it supplies (minus the entrance loss), and the energy line is the same upstream and downstream of gates. (3) For channels that are long after the junction, and do not contain gates, the uniform flow equation (Manning’s or Chezy’s equation is available). (4) If a situation results in critical flow, the critical flow equation is available. Later, after the tools associated with the momentum principle (Chapter 3) and means for solving gradually varied flows (Chapter 4) are covered, we will allow different depths to exist at the beginning and end of each of the channels in the system, with hydraulic jumps occurring if the conditions allow this. Based on the current tools there will generally be two unknowns in each channel, the flow rate Qi and the depth Yi. Thus if five channels are in the branched system, there will be 10 unknowns and 10 equations need to be written to solve for these 10 variables. The junction continuity equations take on the form Fj = Q u − ΣQ di = 0
in which subscript u is the upstream channel number (generally one, but may be more), and subscript d stands for the downstream channels with the extra i indicating there will generally be several of these. The energy equation at the reservoir that supplies the system is of the form Fj = H − Y1 +
(1 + K e )Q12 2gA12
The energy equation at the junctions is of the form
Fj = Yu +
Q 2u Q2 − Ydi − (1 + K e ) di2 − ∆z ui = 0 2 2gA u 2gA di
The number of these equations is one less than the number of channels at this junction. If desired energy between two of the downstream channels can replace an equation requiring the energy at the upstream channel be the same as in the downstream channel. The energy equations across gates are of the form
Fj = Yu +
Q 2u Q2 − Ydi − (1 + K e ) di2 − ∆z G = 0 2 2gA u 2gA di
in which subscripts u and d denote upstream and downstream of the gate, respectively in channel i. For any channel that contains uniform flow an equation of the form
Fj = n iQ i Pi2 / 3 − Cu A 5i / 3 So = 0
111
Energy and Its Dissipation in Open Channels
is available (or Chezy’s equation might be used, alternatively, in which case an additional Chezy’s C is added to the list of unknown variables for each channel, and a additional Chezy’s C equation is added to the list of equations for each channel.) One must have insight in deciding if critical flow occurs. If so then the critical flow equation in conjunction with the energy equation can be used to solve for a limiting flow rate. For example, if the downstream channels have greater capacity than the upstream channel can get from the reservoir head H, then the simultaneous solution of the energy equation and the critical flow equation for Yc and Qc dictate what the flow rate is. Under these circumstances the specific energy in the upstream channel at the junction, Hj, is unknown and is less than the reservoir head H. Example Problem 2.14 Water is taken from a reservoir by means of a rectangular inlet channel that is 10 m wide. A short distance downstream therefrom the channel divides into a trapezoidal section with b2 = 4 m, m2 = 1.5, n2 = 0.015, and So2 = 0.0008, and a pipe with a diameter D3 = 3 m, n3 = 0.013, and So3 = 0.0014. The bottom of the pipe is 1.8 m above the bottom of the rectangular channel, and the trapezoidal and rectangular channels have the same bottom elevation. When the water surface elevation in the reservoir is 4.5 m above the bottom of the channel determine the depths and flow rates in all three channel (six unknowns). Assume the entrance loss coefficient equals 0.12. Solution To solve for the six unknowns in this problem it is necessary to write six equations, and solve them simultaneously. The equations are
F1 = Q1 − Q 2 − Q 3 = 0 (continuity)
F2 = H – Y1 –
F3 = Y1 – Y2 +
F4 = Y1 – Y3 – ∆z3 +
C F5 = Q 2 – u A 2 R 2h2/ 3 So2 = 0 (uniform flow) n2
(5)
C F6 = Q 3 – u A 3R 2h3/ 3 So3 = 0 (uniform flow) n3
(6)
(1 + K L )(Q1/A1 )2 2g
{(Q /A ) 1
1
2
2g 1
1
2
(energy )
}=0
4m
008 0.0 5 = .01 S o2 m, n 2= 0 4 , b 2= = 1 .5 m2 Q2
b1 = 10 m Δz Q1 3 =1 .8
D=
}=0
– (Q 3 /A 3 )2 2g
Δz2 = 0
H = 4.5 m KL = 0.12
(energy entrance)
– (Q 2 /A 2 )2
{(Q /A )
(1)
3 3m Q3 ,S n3 = o3 = 0.0 0 0.01 14 3
(energy )
(2)
(3)
(4)
112
Open Channel Flow: Numerical Methods and Computer Applications in which H = 4.5 m, and Cu is 1 for SI units. A systematic method such as the Newton method described in Appendix B is needed to solve these equations for the six unknowns: Q1, Q2, Q3, Y1, Y2, Y3. It is relatively easy to provide reasonable guesses to these unknowns since the velocity heads will be small for this problem in comparison to the depths. Therefore one would expect Y1, Y2, and Y3 to be only slightly less than the water surface elevation in the reservoir. The solution provides the following values: Q1 = 136.433 m3/s, Q2 = 121.127 m3/s, Q3 = 15.306 m3/s, Y1 = 3.741 m, Y2 = 3.920 m, Y3 = 2.249 m. The following TURBO PASCAL computer program implements a solution to this problem with the following input data: 4.5 .12 10 0 0 1.8 4 1.5 .015 .0008 3 .013 .0014
140 3.6 125 3.8 15 1.25 {est. for unknown}
The FORTRAN and C programs are written so the input consist of g, H, K L and then for each channel thereafter: a 0 for trapezoidal and 1 for circular channel, b, m, n, So and Δz for each channel including #1. When giving the data for channel 1 its n, So, and Δz can be given as any values since they are not used. The input for these programs is 9.81 4 .12 0 10 0 .013 .001 0 0 4 1.5 .015 .0008 0 1 4 0 .013 .0014 1.8 140 3.6 125 3.8 15 1.25 A TKSolver model and a Mathcad model are also given to solve three channel problems. The given variables in these models solve this problem. PASCAL program for solving the above problem (THREEC.PAS) Program ThreeCh; Const NC=3;NU=6;g=9.81;g2=19.62;C=1.0; Var H,KL,P,AA: real; b,m,n,So,Y,Q,Z:array[1..NC] of real; EQ:array[1..NU] of real; D:array [1..NU] of array [1..NU] of real; Cir:array [1..NC] of boolean; Function EXPN(a,b:real):real;Begin EXPN:=Exp(b*Ln(a)) End; Function A(K:integer):real; Begin If Cir[K] then begin P:=12*Y[K]/b[K]; if abs(P)<0.00001 then m[K]:=Pi/2 else begin AA:=sqrt(1sqr(P))/abs(P); if P<0 then m[K]:=PiArcTan(AA) else m[K]:=ArcTan(AA); AA:=sqr(b[K])/4*(m[K]P*sin(m[K]));P:=m[K]*b[K]; A:=AA end; end else begin P:=b[K]+2*Y[K]*sqrt(sqr(m[K])+1); AA:=(b[K]+m[K]*Y[K])*Y[K];A:=AA end; End; Function F(K:integer):real; Begin Case K of 1: F:=Q[1]Q[2]Q[3]; 2: F:=HY[1]KL*sqr(Q[1]/A(1))/g2; 3,4:F:=Y[1]Y[K1]+(sqr(Q[1]/A(1))sqr(Q[K1]/A(K1)))/g2Z[K1]; 5,6:F:=Q[K3]C/n[K3]*A(K3)*EXPN(AA/P,0.6666667) *sqrt(So[K3]) end; End; Var I,J,KI:integer; ADIF,FAC,SUM:real; Dif:array [1..NU] of real; BEGIN Cir[1]:=false;Cir[2]:=false;Cir[3]:=true; Writeln('Give:H,KL,b1,m1,Z2,Z3,b2,m2,n2,So2,D3,n3,So3'); Readln(H,KL,b[1],m[1],Z[2],Z[3],b[2],m[2],n[2],So[2],b[3], n[3],So[3]); Writeln('Give est. for:Q1,Y1,Q2,Y1,Q3,Y3');
Energy and Its Dissipation in Open Channels Readln(Q[1],Y[1],Q[2],Y[2],Q[3],Y[3]); KL:=1+KL; { Defines equations and Jacobian matrix for Newton Solution} repeat For I:=1 to NU do Begin EQ[I]:=F(I); For J:=1 to NU do begin If J>NC then Y[JNC]:=Y[JNC]0.001 else Q[J]:=Q[J]0.001; D[I,J]:=(EQ[I]F(I))/0.001; If J>NC then Y[JNC]:=Y[JNC]+0.001 else Q[J]:=Q[J]+0.001 end; End; { Solves linear equations} For KI:=1 to NU1 do Begin For I:=KI+1 to NU do if D[I,KI]<>0 then begin FAC:=D[I,KI]/D[KI,KI]; For J:=KI+1 to NU do D[I,J]:=D[I,J]FAC*D[KI,J]; EQ[I]:=EQ[I]FAC*EQ[KI] end; End; Dif[NU]:=EQ[NU]/D[NU,NU]; Y[NC]:=Y[NC]Dif[NU]; ADIF:=abs(Dif[NU]); For I:=NU1 downto 1 do Begin SUM:=0; For J:=I+1 to NU do SUM:=SUM+Dif[J]*D[I,J]; Dif[I]:=(EQ[I]SUM)/D[I,I]; If I>NC then Y[INC]:=Y[INC]Dif[I] else Q[I]:=Q[I]Dif[I]; ADIF:=ADIF+abs(Dif[I]) End; until (ADIF < 0.0001); Write('Unknown flow rates: '); For I:=1 to NC1 do Write('Q(',I:1,')=',Q[I]:10:3,','); Writeln('Q(',NC:1,')=',Q[NC]:10:3); Write('Unknown Depths: '); For I:=1 to NC1 do Write('Y(',I:1,')=',Y[I]:10:3,','); Writeln('Y(',NC:1,')=',Y[NC]:10:3); END FORTRAN program to solve problem (THREECH.FOR) PARAMETER (NC=3,NU=6) INTEGER*2 Cir(NC) CHARACTER*49 FMT/'(3(3H Q(,I1,2H)=,F10.3),/,3(3H Y(,I1,2H)=, &F10.4))'/ COMMON g,g2,C,H,FKL,P,b(NC),FM(NC),FN(NC),So(NC),Y(NC),Q(NC), &Z(NC),EQ(NU),D(NU,NU),Dif(NU),Cir FMT(2:2)=CHAR(48+NC) FMT(27:27)=CHAR(48+NC) WRITE(*,90) NC 90 FORMAT(' Give g,H,KL & for',I2,' Channels:0=Trap or &1=Cir,b,m,n,So & z(or Cir D & 0 for b,m)') READ(*,*) g,H,FKL,(Cir(I),b(I),FM(I),FN(I),So(I),z(I),I=1,NC) WRITE(*,*)' Give Est. of Q and Y for',NC,' Channels' READ(*,*)(Q(I),Y(I),I=1,NC) g2=2.*g C=1.486 IF(G.LT.20.) C=1. FKL=FKL+1 10 DO 20 I=1,NU EQ(I)=F(I) DO 20 J=1,NU IF(J.GT.NC) THEN Y(JNC)=Y(JNC).001 ELSE
113
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Open Channel Flow: Numerical Methods and Computer Applications
20
30 40
50
60
Q(J)=Q(J).001 ENDIF D(I,J)=(EQ(I)F(I))/.001 IF(J.GT.NC) THEN Y(JNC)=Y(JNC)+.001 ELSE Q(J)=Q(J)+.001 ENDIF CONTINUE DO 40 KI=1,NU1 DO 40 I=KI+1,NU IF(ABS(D(I,KI)).LT.1.E7) GO TO 40 FAC=D(I,KI)/D(KI,KI) DO 30 J=KI+1,NU D(I,J)=D(I,J)FAC*D(KI,J) EQ(I)=EQ(I)FAC*EQ(KI) CONTINUE Dif(NU)=EQ(NU)/D(NU,NU) Y(NC)=Y(NC)Dif(NU) ADIF=ABS(Dif(NU)) DO 60 I=NU1,1,1 SUM=0. DO 50 J=I+1,NU SUM=SUM+Dif(J)*D(I,J) Dif(I)=(EQ(I)SUM)/D(I,I) IF(I.GT.NC) THEN Y(INC)=Y(INC)Dif(I) ELSE Q(I)=Q(I)Dif(I) ENDIF ADIF=ADIF+ABS(Dif(I)) IF(ADIF.GT. .0001) GO TO 10 WRITE(6,FMT) (I,Q(I),I=1,NC),(I,Y(I),I=1,NC) END FUNCTION A(K) PARAMETER (NC=3,NU=6) INTEGER*2 Cir(NC) COMMON g,g2,C,H,FKL,P,b(NC),FM(NC),FN(NC),So(NC),Y(NC),Q(NC), &Z(NC),EQ(NU),D(NU,NU),Dif(NU),Cir IF(Cir(K).GT.0) THEN COSB=1.2.*Y(K)/b(K) FM(K)=ACOS(COSB) P=FM(K)*b(K) A=.25*b(K)**2*(FM(K)COSB*SIN(FM(K))) ELSE P=b(K)+2.*Y(K)*SQRT(FM(K)**2+1.) A=(b(K)+FM(K)*Y(K))*Y(K) ENDIF RETURN END FUNCTION F(K) PARAMETER (NC=3,NU=6) INTEGER*2 Cir(NC) COMMON g,g2,C,H,FKL,P,b(NC),FM(NC),FN(NC),So(NC),Y(NC),Q(NC), &Z(NC),EQ(NU),D(NU,NU),Dif(NU),Cir IF(K.LT.2) THEN
Energy and Its Dissipation in Open Channels
2
QQ=Q(2) DO 2 J=3,NC QQ=QQ+Q(J) F=Q(1)QQ ELSE IF(K.EQ.2) THEN F=HY(1)FKL*(Q(1)/A(1))**2/g2 ELSE IF(K.LT.NC+2) THEN F=Y(1)Y(K1)+((Q(1)/A(1))**2(Q(K1)/A(K1))**2)/ &g2Z(K1) ELSE AAA=A(KNC) F=Q(KNC)C*AAA*(AAA/P)**.6666667*SQRT(So(KNC))/FN(KNC) ENDIF RETURN END
Program THREECH.C #include <stdlib.h> #include <stdio.h> #include <math.h> #define NC 3 #define NU 6 float g,g2,c,h,fkl,p,b[NC],fm[NC],fn[NC],so[NC],y[NC],q[NC],\ z[NC],eq[NU];int cir[NC]; float a(int k){float cosb; if(cir[k]){cosb=1.2.*y[k]/b[k]; fm[k]=acos(cosb);p=fm[k]*b[k]; return .25*b[k]*b[k]*(fm[k]cosb*sin(fm[k]));} else {p=b[k]+2.*y[k]*sqrt(fm[k]*fm[k]+1.); return (b[k]+fm[k]*y[k])*y[k];}} // End of a float f(int k){float qq,aaa; int j; if(k<1){qq=q[1];for(j=2;jNC) y[jNC]=.001; else q[j]=.001; d[i][j]=(eq[i]f(i))/.001; if((j+1)>NC) y[jNC]+=.001; else q[j]+=.001;}} for(ki=0;ki1.e7){fac=d[i][ki]/d[ki][ki]; for(j=ki+1;j
115
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Open Channel Flow: Numerical Methods and Computer Applications dif[NU1]=eq[NU1]/d[NU1][NU1]; y[NC1]=dif[NU1]; adif=fabs(dif[NU1]); for(i=NU2;i>1;i){sum=0.; for(j=i+1;jNC) y[iNC]=dif[i]; else q[i]=dif[i]; adif+=fabs(dif[i]);}}while (adif>.0001); for(i=0;i
Q12 2·g·(b1·Y1)2
Y1+
Q12 Q22 = Y2+ 2 2·g·(b1·Y1) 2·g·((b2+ m2·Y2)·Y2)2
117
Energy and Its Dissipation in Open Channels
cos(Beta)=12·
Q2=
Y3 D
Y1+
Q32 Q12 = dz+ Y3+ 2 2 2·g·(b1·Y1) 2·g· D·D /4·(Betacos(Beta)·sin(Beta))
((b2+ m2·Y2)·Y2)1.666667
(b2+2·Y2
m2·m2+1
)
.6666667
·
So2 n2
D·D /4·(Betacos(Beta)·sin(Beta)) Q3 = (D·Beta).6666667
1.6666667
·
So3 n3
136.433 121.127 15.307 Find(Q1,Q2,Q3,Y1,Y2,Y3,Beta)= 3.741 3.92 2.249 2.094
The PASCAL, FORTRAN, and C programs are written in a general way so that with minor modifications they would solve problems in which more than two channels branch from a single channel. The constants NC (number of channels) and NU (number of unknowns) would need to be increased to 4 and 8, respectively if four channels were involved. The “Case of” statement to 3, 4 and 5, 6 would need to read 3, 4, 5 and 5, 6, 8 respectively, and the values subtracted from K would also need to be changed. Also the input and output statements would need modification to handle a larger problem, but the portion of the program that generates the Jacobian for the Newton method, and the solution by this method would need no change. D is a twodimensional array for the Jacobian matrix used in the Newton method, and its derivative elements are numerically evaluated by subtracting 0.001 from the value of the variable that the derivative is being taken with respect to, evaluating the equation again, and then dividing the difference between the two values of the equation by 0.001. The program is also written general enough that any of the three channels may be of circular or trapezoidal cross sections, by setting the boolean Cir equal to true for any circular section. When a section is circular b[I] holds the diameter of the section instead of the bottom width, and upon obtaining the solution m[I] will contain the angle β. It should be noted that if the entrance loss coefficient were zero, with the short length for the upstream channel so that there is no variation of depth from its beginning to end, then it would be possible to solve for the flow rate and depth in each of the downstream branched channels separately because the head at the beginning of each of these would be H then. However, with a nonzero entrance loss coefficient Ke the specific energy becomes dependent upon the flow rate in the upstream main channel, which in turn depends upon the flow rate in the two downstream channels; thus the system of six equations must be solved simultaneously. A good estimate of the solution could be obtained, however by solving each downstream channel separately. Example Problem 2.15 Modify the above program to handle similar problem but in ES units. Solution The only modification needed is to change the constants g, g2 and C to 32.2, 64.4, and 1.486, respectively. Example Problem 2.16 Consider the following variation of the previous problem. A channel takes the water from a reservoir and then divides into two channels as in the previous problem, but in this problem gates in the two downstream channels are used to control the flows in each. In channel 2 the gate is set so
118
Open Channel Flow: Numerical Methods and Computer Applications its bottom is 1.0 m above the channel bottom, and in channel 3 the gate is 1.5 m above the channel bottom. All channels have the same bottom elevation at the Y section. If the water surface elevation in the reservoir is 3.5 m above the channel bottom, and the minor loss coefficient is 0.09 determine the flow rates in the three channels. Solution In this problem it is a reasonable assumption that the depths upstream from the gates in all three channels are the same, i.e., Y1, since the gates will restrict the flow the velocity heads in all channels upstream from the gates will be relatively small. Under this assumption the unknowns are: Q1, Q2, Q3, and Y1. If one is not willing to live with this assumption then two additional unknowns, Y2 and Y3 must be added. Based on the above assumption the four equations whose simultaneous solution provides the solution to the problem are
F1 = Q1 − Q 2 − Q 3 = 0
F2 = H – (1 + K L )(Q12 / (2gA12 )) = 0 F3 = Y1 +
(Q 2 /A 2u )2 (Q /A )2 − Y2d − 2 2d = 0 2g 2g
F4 = Y1 +
(Q 3 /A 3u )2 (Q /A )2 − Y3d − 3 3d = 0 2g 2g
e Gat
H = 3.5 m
Q1
Q2
m =6 , b2 m . t c Re = 1.0 yG
E1, Y1
b1 = 10 m, m1 = 1.5
Gat e
KL = 0.09 Q3 Rec t., yG = b3 = 4 m 1.5 m
in which H = 3.5 m, K L = 0.09, and the added second subscripts u and d stand for upstream and downstream from the gates. The depth downstream from the gate will equal the gate position times its contraction coefficient, Cc = 0.6. The solutions to these four equations are Y1 = 3.316 m, Q1 = 46.348 m3/s, Q2 = 13.855 m3/s, Q3 = 32.493 m3/s. Example Problem 2.17 A flow rate of 15 cfs/ft passes under a sluice gate whose tip is placed at the distance of 1 ft above the channel bottom. The contraction coefficient for the gate is Cc = 0.6, and the channel is rectangular. Determine the depth of flow upstream from this gate. Solution This problem is relative easy to solve because it deals with a rectangular channel. The depth a very short distance downstream from the gate will equal the height of the gate times the contraction coefficient, or Y2 = 0.6(1) = 0.6 ft. The problem now consists of finding the alternate depth to this downstream supercritical depth. Substituting this depth into the specific energy Equation 2.11c gives E2 = 10.305 ft. There is very small energy loss past a gate and therefore E1 = E2, or writing this out gives the following equation: Y+
q2 = 10.305 2gY 2
119
Energy and Its Dissipation in Open Channels Rearranging results in the cubic equation,
q2 Y 3 – 10.305Y 2 + 3.494 = 0 or in general Y 3 – EY 2 + = 0 2g
Since the downstream depth Y2 = 0.6 ft is a known root of this cubic equation it can be reduced to a quadratic equation by extracting this root. The use of synthetic division as follows is a convenient means of reducing the equation to a quadratic: 1 − 10.305 1
0
3.494  .6
− 5.823 .6 − 9.705 − 5.823
− 3.49 0

and therefore the quadratic equation is Y2 – 9.705Y – 5.823 = 0, which can be solved by the quadratic formula giving Y = 10.272 ft. (Note the negative root obtained from the quadratic equation can be discarded as physically impossible.) This technique works only if E1 = E2. Example Problem 2.18 Three channels branch from a main channel a short distance downstream from where it receives water from a reservoir as shown in the sketch below. The properties of the channels are as follows: b1 = 15 ft, m1 = 2, b2 = 8 ft, m2 = 1, n2 = 0.013, So2 = 0.0008, b3 = 5 ft, m3 = 1, n3 = 0.013, So3 = 0.0008, b4 = 5 ft, m4 = 1, n4 = 0.013, So4 = 0.0008, H = 5 ft. All minor loss coefficients, K’s = 0.2, which include the entrance loss coefficient from the reservoir into main channel, and from this main channel into each of the three branch channels. All bottom elevations at the junction are the same. Channel # 3 contains a gate. Determine the flow rates and depth in all of the channels as the gate is closed from a wide open position to a completely closed position in channel # 3. The gates contraction coefficient is Cc = 0.6.
Q4
K14
Y4
Δz 14n H
Ke
Y1 Q1
Gate
K 13
Y3u
Δz13
Q3
Y3d
b3, m3, So3
Δz
12
b1, m1
, S o4 m4 b 4,
K12
Y2
b2 , m
2 , So 2
Q2
Solution In this problem there are eight unknown, namely the flow rates and the depths in all four channels. These unknowns are: Q1, Y1, Q2, Y2, Q3, Y3, Q4, Y4. The eight equations needed to solve these eight unknowns are
F1 = Q1 − Q 2 − Q 3 − Q 4 = 0 (1 + K e ) {Q1/A1 } =0 ( 2g ) 2
F2 = H – Y1 –
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Open Channel Flow: Numerical Methods and Computer Applications
F3 = Y1 +
{Q1/A1}2 − Y
F4 = Y1 +
{Q1/A1}2 − Y
F5 = Y1 +
{Q1/A1}2 − Y
2g
2g
2g
2
(1 + K13 ) {Q 3 /A 3u } − ∆z13 = 0 2g 2
3u
4
(1 + K12 ) {Q 2 /A 2 } − ∆z12 = 0 2g 2
−
−
(1 + K14 ) {Q 4 /A 4 } − ∆z14 = 0 2g 2
−
F6 = n 2 Q 2 P2 2 / 3 − CA 2 5 / 3So21/ 2 = 0
F7 = Y3u +
{Q3 /A3u }2 − Y
3d
2g
(1 + K g ) {Q 3 /A 3d } =0 2g 2
−
F8 = n 4 Q 4 P42 / 3 − CA 54 / 3S104/ 2 = 0
The listing below is a FORTRAN program that is designed to solve for 2 × Nc unknowns in which Nc is the number of channels at the branch. The unknown might be the width of the channel, if the flow rate is specified, for example. This program has been modified slightly in PRB18.FOR (and PRB18.C) to solve the above problem repeatedly for different depths Yd3, in channel # 3 downstream from the gate to produce the table given below. These depth and flow rates are plotted in the figures below. Notice from these figures (and the table) that as the gate in channel # 3 is closed only a very modest increase in depths and flow rates in channel # 2 and # 3 occur. This is the case since the velocity heads are relatively small in comparison to the depths. For steeper channels the effects would be larger. Table Giving Depth and Flowrate in All Four Channels in Problem 2.18 as They Change with the Depth Downstream from the Gate in Channel # 3. The Other Channels Do Not Contain Gates
Height Gate (ft) 0.01 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80
Channel # 1
Channel # 2
Channel # 3
Channel # 4
Q (cfs)
Y (ft)
Q (cfs)
Y (ft)
Q (cfs)
Y (ft)
Q (cfs)
Y (ft)
545.1 549.2 553.7 558.3 563.0 567.6 572.3 577.0 581.8 586.5 591.3 596.0 600.8 605.6 610.3 615.0 619.8 624.5 629.1
4.535 4.526 4.514 4.503 4.491 4.478 4.466 4.452 4.438 4.424 4.409 4.394 4.378 4.362 4.345 4.327 4.309 4.290 4.270
315.5 315.3 315.1 314.8 314.6 314.3 314.1 313.8 313.6 313.3 313.0 312.7 312.4 312.1 311.7 311.4 311.0 310.6 310.3
4.229 4.228 4.226 4.225 4.223 4.221 4.219 4.217 4.216 4.213 4.211 4.209 4.207 4.205 4.202 4.200 4.197 4.194 0.191
0.5 4.9 9.9 14.9 19.9 25.0 30.2 35.3 40.5 45.8 51.1 56.3 61.7 67.0 72.3 77.7 83.0 88.4 93.8
4.923 4.921 4.918 4.916 4.913 4.909 4.905 4.900 4.896 4.890 4.884 4.878 4.871 4.863 4.855 4.847 4.838 4.828 0.817
229.1 229.0 228.8 228.7 228.5 228.3 228.1 227.9 227.7 227.5 227.2 227.0 226.8 226.5 226.3 226.0 225.7 225.4 25.1
4.318 4.317 4.315 4.314 4.312 4.310 4.308 4.306 4.304 4.302 4.300 4.298 4.295 4.293 4.290 4.288 4.285 4.282 4.80
121
Energy and Its Dissipation in Open Channels 700
5
Flow rate in channels (cfs)
1
4.8
500
4.7
400
4.6
Channel # 2
300
4.5
Channel # 4
200
Chann el # 1
4.4
Channel # 4
4.3
100 0
Channel # 3
Depths in channels (ft)
Channel #
600
4.9
4.2
Channel # 3
0
0.2
0.4 0.6 0.8 1 1.2 1.4 Depth downstream, gate 3 (ft)
Channel # 2
1.6
1.8
4.1
0
0.2
0.4 0.6 0.8 1 1.2 1.4 Depth downstream, gate 3 (ft)
1.6
Program BRANCHCHL.FOR solves general branched channels REAL X[ALLOCATABLE](:),F[ALLOCATABLE](:), &F1[ALLOCATABLE](:),D[ALLOCATABLE](:,:) &,DZ[ALLOCATABLE](:),KLOS[ALLOCATABLE](:) INTEGER*2 IT[ALLOCATABLE](:),INDX[ALLOCATABLE](:), &IV[ALLOCATABLE](:) LOGICAL*1 STEEP[ALLOCATABLE](:),KEYB CHARACTER*2 CU[ALLOCATABLE](:) CHARACTER*6 S(2)/'SQYbmn','SQYDn '/ CHARACTER*1 CH CHARACTER*38 FMT/"(1X,A1,' =',F8.6,/,(1X,A1,' =',F8.3))"/ COMMON CMAN,G,G2,YGA(5),EYG(5),IYG(5),IGATE(6),NGATE WRITE(*,*)' Give: (1) 1=ES, 2=SI;', (2) No. channels;', &' (3) No. gates (4) INunit;',= (5) OUTunit' KEYB=.FALSE. ! Trapezoidal Circular READ(*,*)II,NC,NGATE,IN,IOUT! IGATE(NGATE+1)=NC+1 ! H = x(1) H = x(1) IF(IN.EQ.0.OR.IN.EQ.5)KEYB=.TRUE.!Q1 = x(2) Q1 = x(2) IF(II.EQ.2) THEN ! Y1 = x(3) Y1 = x(3) G=9.81 ! b1 = x(4) D1 = x(4) CMAN=1. ! m1 = x(5) n1 = x(5) ELSE ! n1 = x(6) G=32.2 ! So2= x(7) So2= x(6) CMAN=1.486 ! Q2 = x(8) Q2= x(7) ENDIF ! Y2 = x(9) Y2 = x(8) G2=2.*G ! b2 = x(10) D2 = x(9) N=2*NC ! m2 = x(11) n2 = x(10) ALLOCATE(IT(NC),KLOS(NC),DZ(NC),F(N),F1(N),D(N,N), &CU(N),STEEP(NC)) IF(KEYB)WRITE(*,*)' For each channel give 1 = trap.', &', or 2 = cir.' NV=0 DO 10 I=1,NC IF(KEYB) WRITE(*,"(' Channel #',I2,' = ',\)") I READ(IN,*) IT(I) 10 NV=NVIT(I)+7 ALLOCATE(X(NV),IV(N),INDX(N)) II=0 DO 30 I=1,NC STEEP(I)=.FALSE.
1.8
122
Open Channel Flow: Numerical Methods and Computer Applications
20
30
120
35 40 42
45 50
IF(KEYB) THEN WRITE(*,"('Give variables for Channel',I2)") I DO 20 J=1,7IT(I) CH=S(IT(I))(J:J) IF(I.EQ.1 .AND. J.EQ.1) CH='H' WRITE(*,"(5X,A1,' = ',\)") CH READ(*,*) X(J+II) WRITE(*,"(' Loss Coef =',\)") READ(*,*) KLOS(I) IF(I.GT.1) THEN WRITE(*,"(' Change in bottom position =',\)") READ(*,*) DZ(I) ENDIF ELSE IF(I.EQ.1) THEN READ(IN,*)(X(J),J=1,7IT(I)),KLOS(I) ELSE READ(IN,*)(X(J+II),J=1,7IT(I)),KLOS(I),DZ(I) IF(X(II+1).GT. .008) STEEP(I)=.TRUE. ENDIF ENDIF II=II+7IT(I) IF(NGATE.GT.0)READ(IN,*) (IGATE(I),YGA(I),I=1,NGATE) IF(KEYB) WRITE(*,*)' Give symbols for',N, &' Unknowns, i.e. Q1 Y1 b2 etc.' READ(IN,120) (CU(I),I=1,N) FORMAT(26(A2,1X)) IU=0 IPOS=0 II1=1 DO 40 I=1,N II=ICHAR(CU(I)(2:2))48 DO 35 J=1,7IT(II) IF(S(IT(II))(J:J).NE.CU(I)(1:1)) GO TO 35 IF(II.NE.II1) THEN IPOS=IPOS+7IT(II1) II1=II ENDIF IU=IU+1 IV(IU)=IPOS+J GO TO 40 CONTINUE CONTINUE NCT=0 CALL FUNCT(NC,N,NV,IT,STEEP,X,F,DZ,KLOS) DO 50 J=1,N XX=X(IV(J)) X(IV(J))=1.005*X(IV(J)) CALL FUNCT(NC,N,NV,IT,STEEP,X,F1,DZ,KLOS) DO 45 I=1,N D(I,J)=(F1(I)F(I))/(X(IV(J))XX) X(IV(J))=XX CALL SOLVEQ(N,1,N,D,F,1,DD,INDX) NCT=NCT+1 SUM=0. DO 60 I=1,N X(IV(I))=X(IV(I))F(I)
Energy and Its Dissipation in Open Channels 60
63
64
65
70
SUM=SUM+ABS(F(I)) WRITE(*,*)' Iteration=',NCT,' SUM=',SUM IF(NCT.LT.20 .AND. SUM.GT. .0005) GO TO 42 DO 65 I=1,NGATE NCT=2 II=IYG(I) FF=EYG(I)X(II)(X(II1)/((X(II+1)+X(II+2)* &X(II))*X(II)))**2/G2 IF(MOD(NCT,2) .NE. 0) GO TO 64 NCT=NCT+1 F11=FF XX=X(II) X(II)=1.005*X(II) GO TO 63 NCT=NCT+1 DIF=(X(II)XX)*F11/(FFF11) X(II)=XXDIF WRITE(*,*)' NCT=',NCT,' SUM=',SUM IF(NCT.LT. 40 .AND. ABS(DIF).GT. .0001) GO TO 63 IF(NCT.GE.40) WRITE(*,*)' Failed to converge', &' with gates',DIF CONTINUE II=0 S(IT(1))(1:1)='H' FMT(16:16)='3' DO 70 I=1,NC WRITE(IOUT,"(' For Channel',I2)") I WRITE(IOUT,FMT)(S(IT(I))(J:J),X(J+II),J=1,7IT(I)) IF(I.EQ.1) S(IT(1))(1:1)='S' FMT(16:16)='6' II=IIIT(I)+7 END SUBROUTINE FUNCT(NC,N,NV,IT,STEEP,X,F,DZ,KLOS) REAL X(NV),F(N),DZ(N),KLOS(N) INTEGER IT(NC) LOGICAL*1 STEEP(NC) COMMON CMAN,G,G2,YGA(5),EYG(5),IYG(5),IGATE(6),NGATE II=0 IG=1 DO 20 I=1,NC IF(IT(I).EQ.1) THEN A=(X(II+4)+X(II+5)*X(II+3))*X(II+3) IF(STEEP(I)) THEN T=X(II+4)+2.*X(II+5)*X(II+3) ELSE P=X(II+4)+2.*SQRT(X(II+5)**2+1.)*X(II+3) ENDIF ELSE ARG=1.2.*X(II+3)/X(II+4) BETA=ACOS(ARG) A=.25*X(II+4)**2*(BETASIN(BETA)*ARG) IF(STEEP(I)) THEN T=X(II+4)*SIN(BETA) ELSE P=BETA*X(II+4) ENDIF ENDIF
123
124
Open Channel Flow: Numerical Methods and Computer Applications
10
20
IF(I.EQ.1) THEN E1=X(3)+(X(2)/A)**2/G2 F(1)=X(1)E1KLOS(1)*(X(2)/A)**2/G2 F(2)=X(2) JJ=7IT(1) DO 10 J=2,NC F(2)=F(2)X(JJ+2) JJ=JJ+7IT(J) ELSE IF(I.EQ.IGATE(IG)) THEN F(2*I1)=E1YGA(IG)(1.+KLOS(I))*(X(II+2)/((X(II+4) &+X(II+5)*YGA(IG))*YGA(IG)))**2/G2DZ(I) EYG(IG)=E1 IYG(IG)=II+3 IG=IG+1 ELSE F(2*I1)=E1X(II+3)(1.+KLOS(I))*(X(II+2)/A)**2/G2DZ(I) ENDIF IF(STEEP(I)) THEN F(2*I)=T*X(II+2)**2G*A**3 ELSE F(2*I)=X(II+7IT(I))*X(II+2)CMAN*A*(A/P)** &.66666667*SQRT(X(II+1)) ENDIF ENDIF II=II+7IT(I) RETURN END
Input data file to solve Example Problem 2.18 1 1 1 1 5. 531 3.961 15 2 .013 .2 .0008 206.5 4.589 8 1 .013 .2 0. .0008 161.8 4.572 5 1 .013 .2 0. .0008 161.8 4.752 5 1 .013 .2 0. 3 1.701 Q1 Y1 Q2 Y2 Q3 Y3 Q4 Y4 Input from keyboard 1 4 1 2 3 Program BRANCHCH.C #include <stdio.h> #include <stdlib.h> #include <math.h> float cman,g,g2,yga[5],eyg[5]; int ngate,iyg[5],igate[6]; extern void solveq(int n,float **a,float *b,int itype,\ float *dd,int *indx); void funct(int nc,int *it,int *steep,float *x,float *f,\ float *dz,float *klos){ int ii,ig,j,jj,i; float a,t,p,arg,beta,e1; ii=0; ig=0;
Energy and Its Dissipation in Open Channels for(i=0;i
125
126
Open Channel Flow: Numerical Methods and Computer Applications x=(float *)calloc(nv,sizeof(float)); iv=(int *)calloc(n,sizeof(int)); indx=(int *)calloc(n,sizeof(int)); for(i=0,ii=0;i.008)steep[i]=1;}} ii+=6it[i];} // end for i if(ngate>0){ if(keyb) printf("Give pairs of values: channel no\ and depth downs. from gate(s)\n"); for(i=0;i.0005)); for(i=0,nct=1;i
127
Energy and Its Dissipation in Open Channels printf("NCT= %d DIF=%f\n",nct,dif); if((nct<40) && (fabs(dif)>.0001)) goto L63; if(nct==40) printf("Not converge with gates\n");} s[it[0]][0]='H'; for(i=0,ii=0;i
In solving the system of equations above that describe the flow rates and depths in several channels that branch off from a short main channel, it was assumed that critical depth at the entrance of the main channel does not occur and limit the flow rate into it. If the composite carrying capacities of the channels that branch off from this main channel exceed the capability of the main channel, then critical flow will occur at the entrance. When this occurs a valid solution to the system of equations does not exist. It is generally not easy to determine that a valid solution to the system of equations does not exist, however. An easier means for determining whether critical flow limits the flow rate is to complete the following steps:
1. Solve for the critical depth and flow rate from the reservoir into the main channel by using the energy and critical flow equations. By substituting the critical flow equation into the energy equation gives
F = Y+
(1 + K e )A −H = 0 2T
(2.16)
from which Y = Yc can be solved. With Yc solved, the critical flow rate can be solved from
Qc =
gA 3 T
(2.17)
2. Solve for the critical specific energy Ec from E c = Yc +
Q 2c 2gA 2c
If the entrance loss coefficient is zero (Ke = 0), then Ec = H and this equation does not need to be solved. 3. Using the specific energy solve for the normal depths and flow rates in all of the branched channels from the two equations,
F1 = E c − Yi −
(1 + K L1− i )Q 2i − ∆z1i = 0 2gA 2i
F2 = n iQ i Pi2 / 3 − Cu A 5i / 3 So = 0
(2.18) (2.19)
4. Sum the flow rates that the downstream branched channel can carry by Qbranches = ΣQi, with i = 2 to n. If Qbranches is larger than Qc then no valid solution to the equations exist, and Qc limits the flow rate into the channels. With the flow rate limited into the downstream
128
Open Channel Flow: Numerical Methods and Computer Applications
branched channels they will not flow at normal depths even if no downstream control influences their upstream flow conditions based on the critical specific energy. Rather the limiting flow rate will force specific energy to be less. The solution to branched channels whose flow is limited by critical depth in the main upstream channel at the reservoir can be based on the assumption that energy is lost in the main channel such that the specific energy at the junction with the branched channels is that needed to simultaneously satisfy all the energy equations at this junction and the uniform flow equations, in addition to the continuity equation. What happens in the real branched channel system is that just downstream from the entrance the flow will become supercritical with a hydraulic jump occurring in the main channel before the junction of the branched channels, or oblique jumps will occur at the junctions where the direction of the flow must change to enter the individual branch channels. In writing these equations the main channel will have a subscript 1, the first branched channel a subscript 2, etc. to the last with a subscript n. There will be 2(n − 1) + 1 equations needed to solve for n − 1 flow rates, n − 1 depths in the downstream channel, and the specific energy or head Hj at the junction. The head Hj will be less than the reservoir head H reduced by the local entrance loss K eQ12 /(2gA12 ). These equation are Q 2 + Q 3 + + Q n − Q c = 0 (Continuity)
(1 + K L1− 2 )Q 22 − ∆z1− 2 = 0 (Energy) 2gA 22
H j − Y2 −
A n 2Q 2 − A2 2 P2 Cu
H j − Y3 −
A n 3Q 3 − A3 3 P3 Cu
H j − Yn −
A n n 3Q n − An n Pn Cu
(2.21a)
2 /3
So 2 = 0 ( Uniform flow)
(2.22a)
(1 + K L1− 3 )Q 32 − ∆z1− 3 = 0 (Energy) 2gA 22
(2.21b)
2 /3
So3 = 0 ( Uniform flow)
(2.22b)
(1 + K L1− n )Q 2n − ∆z1− n = 0 (Energy) 2gA 2n
(2.21c)
2 /3
Son = 0 ( Uniform flow)
Q4
All minor losses KL = 0.2 H = 5 ft
(2.20)
Q1 Y1
b1 = 10 ft, m1 = 2
Y4
3
.01
=0 4 1, n
= 4 t, m 5 f .0008 = b4 =0 S o4 Y3 Q3
b3 = 5 ft, m3 = 1, n3 = 0.013 So3 = 0.0008 Y2 b2 = 10 f t, m So Q2 2 =0 .000 2 = 2, n 8 2 =0 .013
(2.22c)
129
Energy and Its Dissipation in Open Channels Example Problem 2.19 A main channel with b1 = 10 ft, and m1 = 2 takes water from a reservoir with a head H = 5 ft above the channel bottom. A short distance downstream from its entrance this channel branches into three channels with the following properties: b2 = 8 ft, m2 = 1, n2 = 0.013, So2 = 0.0008, b3 = 5 ft, m3 = 1, n3 = 0.013, So3 = 0.0008, b4 = 5 ft, m4 = 1, n4 = 0.013, So4 = 0.0008. The bottoms of all channel at the junction are at the same elevation, i.e., all Δz values are zero, and the entrance loss coefficient Ke and loss coefficients between the main and the branched channel K L1−i are all 0.2. Determine the flow rates and depths in the three branched channels. Solution Solving the energy and critical flow equations simultaneously at the entrance to the main channel from the reservoir gives: Qc = 531.4 cfs, Yc = 3.509 ft, and because of the loss coefficient Ke = 0.2, the specific energy associated with this critical flow is Ec = 4.74 ft. Next using a head of H = Ec = 4.74 the energy equation and Manning’s equation are solve simultaneously for the three branched channels with the results: Yo2 = 4.071 ft, Q2 = 292.5 cfs, Yo3 = 4.157 ft, Q3 = 211.4 cfs, Yo4 = 4.157 ft, Q4 = 211.4 cfs. These flow rates sum to 715.4 cfs, which exceeds the critical flow into the main channel. Therefore the flow will be limited to Qc = 531.4 cfs. To determine the flow rates and depths in the branched channels the seven equations below are solved for the following seven unknowns: Hj, Y2, Q2, Y3, Q3, Y4, Q4 (Hj is the head at the junction).
F1 = Q 2 + Q 3 + Q 4 − 531.4 = 0 F2 = H j – Y2 –
F3 =
F7 =
(So2 )1/ 2 = 0
(1 + K L1−3 )(Q 3 /A 3 )2 – ∆z1−3 = 0 2g
A n 3Q 3 – A3 3 P3 Cu
F6 = H j – Y4 –
2/3
A n 2Q 2 – A2 2 P2 Cu
F4 = H j – Y3 –
F5 =
(1 + K L1−2 )(Q 2 /A 2 )2 – ∆z1−2 = 0 2g
2/3
(So3 )1/ 2 = 0
(1 + K L1− 4 )(Q 4 /A 4 )2 – ∆z1− 4 = 0 2g
A n 4Q 4 – A4 4 P4 Cu
2/3
(So4 )1/ 2 = 0
The solution to these seven equations gives: Hj = 4.04 ft, Q2 = 219.6 cfs, Y2 = 3.467 ft, Q3 = 155.9 cfs, Y3 = 3.540 ft, Q4 = 155.9 cfs, Y4 = 3.540 ft. Thus the energy loss in the main channel from its entrance to the branched channel must equal ΔH = 4.74 − 4.04 ft = 0.70 ftlb/lb. This solution was obtained using TKSolver. It can also be obtained with the program above by specifying the flow rate in the upstream channel as the unknown, and indicating that S1 is unknown for the upstream main channel. The reason for specifying S1 as unknown, rather than H1, is that the program is written so that for all channels, except the first, variable So, the bottom slope, is in the first position of the array x for the variables for that channel. For the first, or main, channel the head H is in this first position, but the logic to note this does not exist in the program. Thus S1 actually
130
Open Channel Flow: Numerical Methods and Computer Applications is interpreted as H when given for channel # 1. The input to the above program BRANCHCH and its solution are given below. The real occurrence for a situation like this for which critical conditions limit the flow rate is that this limiting flow reduces the energy available at the junction below that of the reservoir. Input to BRANCHCH for critical flow at entrance 1 1 1 1 4.6 531.41 3.882 10 2 .013 .2 0.0008 217.3 4. 8 1 .013 .2 0. 0.0008 157.05 4. 5 1 .013 .2 0. 0.0008 157.05 4. 5 1 .013 .2 0. S1 Y1 Q2 Y2 Q3 Y3 Q4 Y4 Solution for critical flow at entrance For Channel 1 H = 4.038 Q = 531.410 Y = 3.875 b = 10.000 m = 2.000 For Channel 2 S = .000800 Q = 219.610 Y = 3.467 b = 8.000 m = 1.000 n = .013 For Channel 3 S = .000800 Q = 155.900 Y = 3.540 b = 5.000 m = 1.000 n = .013 For Channel 4 S = .000800 Q = 155.900 Y = 3.540 b = 5.000 m = 1.000 n = .013 In obtaining this solution with program BRANCHCH you will notice that difficulties exist in achieving convergence. As soon as the residual becomes small, it get large the next iteration. This numerical action results because BRANCHCH also attempts to solve Y1 = Yc associated with the critical flow rate Q1, which is specified. The infinite derivative of the depth in the upstream channel causes the Newton method to move off the desired solution as Y1 approaches Yc1. To solve this problem adequately BRANCHCH should be modified so that Y1 is not included as an unknown, i.e., solve 7 rather than 8 equations simultaneously.
131
Energy and Its Dissipation in Open Channels
2.9 Graphical Aids to Solving Critical Flow Problems Since the critical flow equation (with the exception for rectangular channels), and the specific energy equation are implicit when solving them for the depth, or channel size graphical methods for solving these problems are widely used in practice. The need for such graphical solutions is rapidly diminishing with the wide spread use of programmable pocket calculators, and PC computers. However, even with these modern tools it is often advantageous to obtain a “quick” graphical solution, in examining proposed alternatives. To examine how graphs for solving critical flow equations might be developed the critical flow equation will be written specifically for a trapezoidal channel as
Q 2 (b + 2mYc ) Q 2 b(1 + 2mYc /b) = g(bYc + mYc 2 )3 gb6 Yc / b + m(Yc / b)2
{
{Q /(gb )} (1 + 2mY /b) = 1 } {Y / b + m(Y /b) } 5
2
3
c
=
c
c
2
3
(2.23)
In the last form of this equation it should be noted that Q2/(gb5) is a dimensionless parameter, which can be denoted as Q′. Likewise the ratio Yc/b is dimensionless and can be denoted by Y′. Thus the last part of the above equation can be written as
Q′ =
{Y′ + mY′ 2}3 1 + 2mY′
(2.24)
or the dimensionless flow rate Q′ is a function of the dimensionless depth Y′, and the side slope of the trapezoidal channels. This equation is in a form in which it is relative easy to generate the values needed to plot Q′ versus Y′ for selected values of m. Such a plot is given in Figure 2.6. This figure also includes a similar curve for circular channels. To use Figure 2.6 to determine the critical depth the value of Q′ = Q2/(gb5) is computed from the flow rate, and bottom width of the channel. This value is entered on the abscissa of the graph, and projected vertically upward to the curve for the known side slope of the channel. The dimensionless critical depth Y′ = Yc/b is next read from the graph, and from this value the critical depth determined. It was convenient to leave the side slope m out of the definition of Q′ in Equation 2.24, so different curves could be plotted on Figure 2.6. Note, however, if the definition for dimensionless depth Y′ = mY/b and Q′ = m3Q2/(gb5) for a trapezoidal channel then Equation 2.23 can be written as
Q′c =
mYc m 3Q c2 with and Y = Q = ′ ′ c c b gb5
Yc′ + Yc′2 1 + 2Yc′
(2.24a)
(These latter definitions for dimensionless variables will be used below.) To obtain the curve for circular channels on Figure 2.6 write the critical flow equation for a circle using the auxiliary angle β = cos−1(1 − 2Yc/D) as the variable to give
Q 2 D sin β = 1 gD (β − sin β cos β)3 /64 6
Letting Q′ = Q2/(gD5) be a dimensionless flow rate, this equation becomes
Q′ =
(β − sin β cos β)3 64 sin β
(2.25)
132
Open Channel Flow: Numerical Methods and Computer Applications
ββ
m
=
5
1.
.0 m =
.0 4 =3 = m m
5.0
1.5
m
=0
Yc
0.7
m = m 0.5 =1 .0
1.0
.0
=2
2.0
m
120 140 180
D D
100
1.2
3.0
1.5
ap Tr
1.0
C
al
id
o ez
Yc
0.2
Angle, β (rad)
ul ng
0.3
ta
r
la
u irc
60
D
ar
b or 0.4 Yc
Angle, β(°)
80
0.5
Re c
Yc
m
40
b
0.1 10–3
10–2
1 10–1 Q2/gD5 or Q2/gb5
10
102
FIGURE 2.6 Relationship of dimensionless depth to dimensionless flow rate parameters under “critical flow” conditions.
This equation is in the form in which it is relatively easy to generate a table of values for Q′ corresponding to the dimensionless β and since Y′ = Yc/D = (1 − cos β)/2 this table can be expanded to give values of Q′ versus Y′. Thus a single curve defines critical flow for all circular sections. You should note that the first portion of this curve on the log–log plot of Figure 2.6 is nearly a straight line. Thus Y′ or β can be approximated quite closely by a power function of Q′ in the form a Q′b. This power function was discussed earlier for providing starting values to solve for critical depth in circular sections by the Newton method. It is also possible to develop dimensionless specific energy diagrams that provide solutions for critical flow conditions as well as other problems associated with the use of the specific energy principle in open channel flow. Figures 2.7 and 2.8 provide such dimensionless plots for trapezoidal and circular channels respectively. If one wants to determine the critical depth by using these figures, one needs only compute the value of Q′ (which you should note is defined differently, as is Y′, than in Figure 2.6 for a trapezoidal channel to include m) and read the ordinate Y′ corresponding to the minimum value for the curve for that Q′. The flow rate can be determined by reading the value from the abscissa corresponding to the minimum value of this curve. In addition, it is relatively simple to solve for an alternate depth. For example, if the supercritical depth is known and its alternate subcritical depth is desired, one need only compute Q′ and Y′ for the supercritical depth, enter this Y′ on the ordinate, and project horizontally to the curve corresponding to Q′. Next project vertically upward to this same curve, and then horizontally over the ordinate where the subcritical value for Y′ is read.
133
Energy and Its Dissipation in Open Channels
1.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
0.9
0.9 y
0.8
m
I
0.8
Q΄
0.5 0.4 0.3 0.2 0.1
Q΄ =
0.0
0.1
Q΄
=0
0.00
.00
25
1
0.2
Q΄ = 0.0 15 Q΄ = 0.01 Q΄ = 0. 005
0.3
0.4
Q ΄= Q΄ Q΄ = 0.18 = 0 0.1 Q΄ .14 6 = Q΄ Q΄ 0 = . = 0 .12 Q΄ .1 09 =0 .07 Q΄ = Q΄ 0.08 =0 .05 Q΄ = 0.0 Q΄ = 4 0.03 Q΄ = 0.02
Q΄ Q΄
=0
8 0.
Q΄ Q΄
0.9
΄=
Q
0.6
0.7
Q
Dimensionless flowrate Q΄ = m3Q2/(gb3)
΄=
Dimensionless depth, Y΄ = mY/b
b
0.7
0.0
1.0
0.6
=0
.6
=0
.5
0.5
.4
=0
.3
0.4
=0
.20
0.3 0.2 0.1
0.5 0.6 0.7 0.8 0.9 1.0 Dimensionless specific energy, E΄= mE/b
1.1
1.2
1.3
1.4
0.0
FIGURE 2.7 Dimensionless specific energy diagram trapezoidal channels. (Individual curves apply for Q′ = m3Q2/(gb5).)
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.4
1.5
1.6
1.7
1.8
ca
l
0.9
1.3
Q
Su b
cr
iti
΄=
Dimensionless depth, Y΄ = Y/D
Q΄
0.6
Q΄
0.4 0.3
Q΄
=0
.01
.18
0.5
=0
Dimensionless flow rate Q΄ = Q2/(gD5)
Q΄
Q΄ =0 =0
.25
Q΄ =0
Q΄
.30
=0
Q΄
.4
Q΄
=0
=0
=0
0.9 0.8
8
y
D
9
0.
Q΄
β
0.7
0.
1.0
΄=
Q
0.8
Q΄= 1 .0
1.0
.7
.6
0.7
.5
0.6
.20
=0 Q΄ Q΄ = 0 .1 .16 =0 4 .12 Q Q΄ ΄= 0 .1 0 Q΄ Q΄= = 0 .0 9 =0 0 .07 .08 Q Q΄ ΄= 0 .0 = 6 0 .0 Q΄ 5 =0 Q΄ .04 =0 .03 Q΄ =0 .02
Sup e
rcrit
0.5
ical
0.4 0.3
0.2
0.2
0.1
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Dimensionless specific energy, E΄= E/D
1.3
1.4
1.5
1.6
1.7
1.8
0.0
FIGURE 2.8 Dimensionless specific energy diagrams for circular channels. (Individual curves apply for Q′ = Q2/(gD5).)
134
Open Channel Flow: Numerical Methods and Computer Applications
To determine how Figure 2.7 is developed write the specific energy equation specifically for a trapezoidal channel as shown below. E = Y+
Q2 Q2 = Y+ 2 2 4 2g(bY + mY ) 2gb [(y/b) + m(Y/b)2 ]2
(2.26)
If this equation is multiplied by m and divided by b, and the dimensionless specific energy mE/b defined as E′, and mY/b denoted as a dimensionless depth Y′ then the following results: E′ = Y′ +
m 2Q 2 Q′ = Y′ + 2 2 2gb (mY/b + (mY/b) ) 2(Y′ + Y′ 2 )2
(2.27)
5
in which Q′ = m3Q2/(gb5). Thus it is clear that dimensionless specific energy E′ can be defined as a function of dimensionless depth Y′ with dimensionless flow rate Q′ held constant, i.e., for any Q′ the dimensionless specific energy curve can be defined, as has been done in Figure 2.7. Figure 2.8 for a circular section can be obtained in a similar manner. The difference is that the specific energy equation is written specifically for a circular channel using angle β, as follows: E=
D(1 − cos β) Q2 + 4 2 gD (β − sin β cos β)2/8
(2.28)
by dividing this equation by D, and defining Q′ = Q2/(gD5), the above equation becomes, E′ =
1 − cos β 8Q′ 8Q′ = Y′ + + 2 2 (β − sin β cos β) (β − sin β cos β)2
(2.28a)
and since cos β = 1 − 2Y/D = 1 − 2Y′ it is possible to develop a dimensionless specific energy curve for any dimensionless value of Q′. Figure 2.8 represents such a graph. Example Problem 2.20 A transition occurs from a 2 m diameter pipe to one of 1.5 m diameter. The transition keeps the centerlines of the two pipes lined up. For a flow rate of Q = 2 m3/s, and a bottom slope of So = 0.00095, and n = 0.013 for the downstream 1.5 m diameter pipe, determine the depth of flow just upstream from the transition. Solve this problem graphically, and then verify the solution. Solution The solution must begin by solving a uniform flow equation for the downstream 1.5 m diameter channel. Using Manning’s equation the uniform, or normal, depth in the downstream channel is Yo2 = 1.131 m. Dividing this by the downstream diameter of 1.5 m gives Y′ = 0.754. Next compute Q ′2 = 2 2 /(9.81(1.5)5 ) = 0.054. Entering the ordinate of Figure 2.8 with 0.754 and finding the intersection with the curve Q′ = 0.054 gives E′ = 0.82 on the abscissa. Therefore E2 = 1.5(0.82) = 1.23 m. The specific energy in the larger pipe just upstream from the transition must be E1 = E2 + Δz = 1.23 + 0.25 = 1.48 m, and therefore E1′ = 0.74. Also Q1′ = 2 2 /(9.81(2)5 ) = 0.013. Entering Figure 2.8 with these latter two values produces , Y′ = 0.725 or Y1 = 1.45 m. A numerical solution to the problem consists of solving the following implicit equation for Y1: E1 = E 2 , or Y1 +
Q2 Q2 = Y2 + + ∆z 2 2gA1 2gA 22
in which A2 can be evaluated from the known depth of 1.131 m, and the area A1 must be defined in terms of Y1 and/or the auxiliary angle β. Solving this implicit equation by the Newton method gives: Y1 = 1.446 m.
Energy and Its Dissipation in Open Channels
135
Example Problem 2.21 What is the smallest pipe that could be used downstream in Example Problem 2.20 if the slope were sufficient to carry the flow that could pass through the transition and upstream conditions are not to be changed? Solution Critical flow will occur at the end of the transition in the entrance of the smaller downstream pipe. If the depth upstream is to remain the same as in Example Problem 2.20, then the critical specific energy for the downstream channel will be Ec = E1 − (D1 − D2)/2 = 1.48 − (2 − D2)/2 = 0.48 + D2/2. This equation involves two unknowns; D2 and Y2 since the critical specific energy Ec is the sum of the depth and the velocity head. The second needed equation to solve the problem is the critical flow equation applied to the downstream channel, or Q2T/(gA3) = 1. Solving these two implicit equations simultaneously gives D2 = 1.237 m and Y2 = 0.770 m.
The above dimensionless equations that were used to plot Figures 2.7 and 2.8 for trapezoidal and circular channels, respectively, can be used to develop dimensionless equations that relate the critical dimensionless flow rate, Q′c to the critical dimensionless depth Yc′ and the critical dimensionless energy Ec′. These equations can be used to solve more readily for critical flow conditions than the critical flow equation. Furthermore, explicit equations can be developed that closely approximate these equations, and these approximate equations can be used for rough answers, or can be used to provide starting guesses for the Newton method to solve the critical flow equation. First consider the dimensionless specific energy in a trapezoidal channel, i.e., E′ = Y′ + Q″/(Y′ + Y′2)2 (Note that Q″ is 1/2 the Q′ used in Figures 2.7 and 2.8, or Q″ = Q′/2). Taking the derivative of E′ with respect to Y′, and setting this derivative to zero produces the following dimensionless equation for Q″ after a little mathematical rearrangement: Q′′c = 0.5
(Yc′ + Yc′ 2 )3 0.5Yc′ 3 (1 + Yc′)3 = 1 + 2Yc′ 1 + 2Yc′
(2.29)
the subscript c has been added to Y′ and Q″ because setting the derivative to zero produces critical flow conditions. Figure 2.9 is a log–log plot of this equation, and also the above dimensionless critical specific energy equation. Note that the curves on this graph can be approximated by straight lines. The equations for these approximate straight line fits are:
Yc′ = 0.925(Q′′c )0.284
(2.30)
E′c = 1.238(Q′′c ) 0.272
(2.31)
and
The same can be accomplished for a circular open channel. Taking the derivative of the equation E′ = Y′+ 8Q′/(β − cos β sin β)2 with respect to Y′ and setting this derivative to zero gives the following equation after some algebraic manipulations:
Q′c =
(β − cos β sin β)3 (64 sin β)
(Also given below Equation 2.25)
(2.32)
in which β is the angle associated with the dimensionless critical depth Yc′ = Yc / D, or β = cos −1 (1 − 2Yc′ ). This equation as well as E′c are plotted on Figure 2.10. Again note that except as Qc and Yc approach unity these relationships are approximated by a straight lines on this log–log graph. The equations for such straight lines fits are
136
Open Channel Flow: Numerical Methods and Computer Applications 10
E΄c = m E c/b
Dimensionless critical depth and specific energy
Approximate equations Y΄c = 0.925(Q˝c )0.284 = 0.76(Q΄c)0.284 E΄c = 1.238(Q˝c )0.272 = 1.025(Q΄c )0.272 Q΄c = 2.628(Y΄c )3.521
2
Q΄c = 0.913(E΄c )3.676
1 0.8 0.6 0.5 0.4
Y΄c = mYc/b
E΄c=
Y΄c=
0.3
/b mE c
/b mY c
0.2 Q˝ = Q΄/2 0.1 0.001
2
3
5
0.01 2 3 5 0.10 2 3 5 1.00 Dimensionless critical flow rate, Q˝c = m3Qc2/(2gb5)
2
3
5
10.00
FIGURE 2.9 Critical condition in a trapezoidal channel. 3.142 = π 2.498 2.214 1.982 1.772 1.571 1.361
Approximate equations Y΄c = (Q ˝c )0.254
Dimensionless critical depth and specific energy Y΄c = Y/D E΄c = E/D
E΄c = 1.4(Q ˝c )0.256 Q ΄c =(Y΄c)3.3937
1.159
Q ΄c = 0.2687(E΄c)3.906
E
0.927
/D
E ΄c=
1
0.1
/D
=Y Y΄c
Dimensionless critical depth and specific energy
0.95
= E΄c
0.9 0.85
D
E/
Approx. equations Ec΄ = 1.70 + 0.9Log10Q΄c
/D
=Y Y΄c
0.7
0.348
0.65
Approx. equations
0.6
1E – 06
0.284
Y΄c = 0.97 + 0.395Log10Q΄c Q΄c = 0.0035e 5.829Yc΄
0.55
0.01 1E – 07
0.574 0.536 0.495 0.451 0.403
Q΄c = .0129e 2.588Ec΄
0.8 0.75
0.5
0.644
0.1
0.2 0.3 0.4 0.5 0.6 0.7 0.8 Dimensionless critical flowrate, Q΄c = Qc z/(gD5)
1E – 05 0.0001 0.001 0.01 Dimensionless critical flowrate, Q΄c = Qcz/(gD5)
FIGURE 2.10 Critical condition in a circular channel.
0.1
1
1
0.200
Angle, β
1
Energy and Its Dissipation in Open Channels
137
Yc′ = (Q′c )0.254
(2.33)
E′c = 1.4(Q′c )0.256
(2.34)
and
The insert in the lower right hand corner of Figure 2.10 shows Yc′ and E′c plotted against Q′c on semilogarithmic paper that starts with a value of Q′c = 0.1. On this insert a straight line plot for these curves are given by the following equations:
Yc′ = 0.97 + 0.395 log10 (Q′c ) or Q′c = 0.0035e 5.829Yc ′
(2.35)
E′c = 1.70 + 0.9 log10 (Q′c ) or Q′c = 0.0129e 2.558Ec ′
(2.36)
and
If the channel is rectangular then another definition of the dimensionless depth needs to be used since multiplication of Y by m will always produce a zero dimensionless depth. A useful dimensionless depth is defined by dividing the depth by the critical depth Yc or defining Y′ = Y/Yc. If the dimensionless specific energy E′ is also defined by dividing E by Yc, then the dimensionless specific energy equation for a rectangular channel becomes,
E′ = Y′ + 0.5/Y′2 or the cubic equation Y′3 − E′ Y′2 + 0.5 = 0
The methods described earlier can be used to solve this resulting cubic equation for the three real roots that exist when E′ is greater, or equal, to 1.5. Using the alternate method that uses the arc cosine (and cos) results in defining the angle θ as
(6.75 − E′ 3 )/ 27 θ = cos −1 3 3 ( E / ) ′
with the three roots given by
E′ θ Y1′ = 1 – 2cos 3 3
E′ [θ + 2π] Y2′ = 1 – 2cos 3 3
E′ [θ + 4 π ] Y3′ = 1 – 2cos 3 3
The program ROOTSED, given below obtains these three roots for any given value for the dimensionless depth Y′. By modifying this program to include a DO loop one could easily generate data to provide a graph of the alternate dimensionless depths as a function of the dimensionless specific energy.
138
Open Channel Flow: Numerical Methods and Computer Applications
Program ROOTSED.FOR PARAMETER (PI=3.14159265) 1 WRITE(*,*)' Give: E''=E/Yc must be 1.5 or greater' READ(*,*) Ep IF(Ep.LT. 1.E5) STOP E3=Ep/3. THETA=ACOS(((6.75Ep**3)/27.)/E3**3) Y1=E3*(1.2.*COS(THETA/3.)) Y2=E3*(1.2.*COS((THETA+2.*PI)/3.)) Y3=E3*(1.2.*COS((THETA+4.*PI)/3.)) WRITE(*,100) Y1,Y2,Y3 100 FORMAT(' Y1=',F8.4,' Y2=',F8.4,' Y3=',F8.4) GO TO 1 END For rectangular channels, it was useful to know that under critical flow the depth equals twothirds of the critical specific energy, or Yc = 2Ec/3. What this equation represents is the flow rate eliminated between the critical flow equation and the specific energy equation. While not as simple, the same can be done for nonrectangular channels by solving for Q from the critical flow equation and substituting for it into the energy equation, giving Equation 2.16, or H c = Yc +
(1 + K e )A c (2Tc )
(2.37)
For a trapezoidal channel, let the dimensionless critical head be defined as H′c = mH c /b (the same definition as for the dimensionless specific energy) and Yc′ = mYc /b , then the dimensionless form of the critical flow equation that relates head to dimensionless critical depth is H′c = Yc + (1 + K e )
Y′ + Y′2 2(1 + 2Y′)
(2.38)
This equation could also be obtained by eliminating Q′ between the dimensionless critical flow and energy equations for a trapezoidal channel. Figure 2.11 shows this relationship with different curves on it for different values for the entrance loss coefficient, Ke. If Ke = 0 the distance between the E′c and Yc′ curves could be added to Yc′. Or stated differently the results on Figure 2.11 along the Ke = 0 curve can be obtained from Figure 2.9, by entering the ordinate for Yc′ and then move horizontally over to this curve; then move vertically up to the E′c curve, and finally read the ordinate for E′c. For circular channels define the critical dimensionless head as H′c = H c / D (and as before Yc′ = Yc / D = (1 − cosβ)/ 2 , and then the dimensionless version of Equation 2.37 is,
H′c =
1 − cos β (1 + K e )(β − cos β sin β (1 + K e )(β − cos β sin β + = Yc′ + 2 8 sin β 8 sin β
(2.39)
Figure 2.12 shows this dimensionless relationship. Generally problems involving the use of Equations 2.38 or 2.39 have the total H head given. So the first step is to determine the dimensionless H′c, and from this value find the corresponding value for the critical depth Yc′ . Whether the channel is trapezoidal or circular the implicit Equations 2.38 or 2.39 can be solved for Yc′ (with H′c known), rather than using the graphs. Once Yc′ is determined then the dimensionless flow rate can be computed from Equation 2.29 (or Equation 2.24) for trapezoidal channels, or Equation 2.32 for circular channels.
139
Energy and Its Dissipation in Open Channels 2 Dimensionless critical head, H΄c = mHc/b
1.8 1.6
.5 = 1 .0 K e e= 1 K
1.4 1.2 1 0.8 0.6
Lo
0.4
l
ca
s los
ff
e co
t, K en ici
.0
=2 e
.2
=0 0 Ke K e=
.4
0 K e=
K e=
.8
=0 e K 6 . 0
0.2 0
0
0.1
0.2 0.3 0.4 0.5 0.6 0.7 0.8 Dimensionless critical depth, Y΄c = mYc/b
0.9
1
FIGURE 2.11 Dimensionless critical depth at the beginning of a trapezoidal channel feed by a reservoir.
Dimensionless critical head, H΄c = Hc/D
2.5
2
K e= Ke = 1.0
1.5
0.8
Ke = 1.5
.0
1
c Lo
0.5
0
0
0.1
al
0.2
los
oe sc
ffi
cie
=2 e t, K
n
K e=
0
K e= Ke = 0.4 Ke = 0.2
0.6
0.3 0.4 0.5 0.6 0.7 0.8 Dimensionless critical depth, Y΄c = Yc/D
0.9
1
FIGURE 2.12 Dimensionless critical depth at the beginning of a circular channel feed by a reservoir.
Often problems whose solutions are based on the energy principle are design problems that call for sizing the channel that will carry a specified flow rate. The bottom slope is known, or dictated by the general slope of the land over which the channel is to carry the water, and the roughness coefficient is determined by the type of material the channel will be built from. Thus if the channel is trapezoidal in shape the bottom width and/or the side slope are/is the unknown variable(s), or if the channel is circular then it is the diameter that is sought after. Basically such problems are not different than those we have dealt with previously, e.g., they involve the simultaneous solution to a couple of implicit equations. The difference is that for this type of problem Q is known, and the size variable, and the depth of flow are generally the unknown variables. The two equations that need to be solved are either (1) a uniform flow equation, and the energy equation if the flow is subcritical, or (2) the critical flow equation and the energy equation if the situation will create a critical flow section.
140
Open Channel Flow: Numerical Methods and Computer Applications Example Problem 2.22 It is desired to find the pipe diameter that would be carry a flow rate of 30 m3/s from a reservoir whose water surface elevation will be 4.5 m above the bottom of the channel at its entrance. The slope of the channel is So = 0.001, and it Manning’s roughness coefficient is n = 0.013. The entrance loss coefficient is estimated to equal 0.12. Solution Since this is a mild slope the governing equations are the energy equation and Manning’s (or Chezy’s) equation, or F1 = H – Y –
(1 + K L )Q 2 =0 2gA 2
and F2 = nQP 2 / 3 − A 5 / 3 S = 0
in which the area A and the perimeter P involve the second unknown D. The solution gives Y = 3.047 m and D = 4.152 m. You should verify these answers, by at least substituting into the above two equations, but better still solve these two equations using the Newton method, or by utilizing an available software package such as TK solver, MATLAB, or Mathcad. Example Problem 2.23 A pipe has its bottom filled at section 2 to a depth of Δz, as shown in the sketch. Using dimensionless variables generate a graph that gives the maximum dimensionless height of hump for given upstream flow conditions. The diameter of the pipe remains constant between section 1 and 2, only the rise in the bottom reduces the size of the pipe at section 2.
D Y1
Yr 2
1
Δz
Solution The specific energy and continuity equations are written in dimensionless form below by dividing length variables by the pipe diameter D and area by the diameter squared D2.
F 2 (A ′r − A ′b ) Y′ Y Y1′ + Fr12 d1 = Yr′ + Fr2 2 d2 = Yr′ + r2 2 2 (2Tr′)
in which subscript 1 denotes section 1, subscript r denotes section 2 as if the step did not exist, and subscript b denotes the bottom step. A is area and Yd is the hydraulic depth, or area divided by the top width (A′ = A/D2 and Yd′ = Yd / D). The continuity equation becomes: A1′Fr1Yd1 ′ 0.5 =
Fr2 (A ′r − A ′b )1.5 Tr′ 0.5
Define the angle β = cos−1(1 − 2Y′), then A′ = (β − cos β sin β)/4, T′ = sin β. (These apply to define the dimensionless area at section 1, for the bottom step, or the regular area, sub r at section 2 ignoring the step.) The TKSolver Variable and Rule Sheets are given below.
════════════════════ RULE SHEET ═════════════════════════ S Rule─────────────────────────────────────────────────── * A1*sqrt(Yd1*Fr2)=(ArAb)*sqrt((ArAb)/sin(Br)) * Y1+Fr2*Yd1/2=Yr+.5*(ArAb)/sin(Br) * Br=acos(12*Yr) * Ar=.25*(Brcos(Br)*sin(Br)) * Bb=acos(12*Yb) * Ab=.25*(Bbcos(Bb)*sin(Bb)) * r6=Yb/Y1 * Fr=sqrt(Fr2)
════════════════════ VARIABLE SHEET ══════════ (Model PAPER6.TK or PAPER*.TK for other Y1') St Input──── Name─── Output─── Unit───── Comment───────── .6 Y1 Upstream dimensionless depth 1.7721542 B1 Upstream angle β .49202836 A1 Upstream Area .50217434 Yd1 Upstream Hydraulic Depth L .01 Fr2 Upstream Froude No.squared (Fr(2) = 1) LG .54897512 Yr Depth from bot. circle to water at sect. 2 LG .44221115 Yb Depth from bot. circle to rise LG 1.6689039 Br LG .44159578 Ar LG 1.4549598 Bb LG .33503915 Ab L r6 L Fr
Energy and Its Dissipation in Open Channels 141
Table Sheet ════════════════════ TABLE: Title: Element Yr──── Yb──── 1 .54898 .44221 2 .53184 .38071 3 .5229 .34336 . . . 24 .5198 .1015 25 .52166 .09599 . . . 42 .56063 .02814 43 .56321 .02536 . . . 59 .59067 .0036 60 .59144 .00319 61 .59221 .00281 Br──── 1.6689 1.6345 1.6166 . 1.6104 1.6141 . 1.6923 1.6975 . 1.7531 1.7547 1.7563
Ar──── .4416 .42452 .41559 . .4125 .41436 . .45318 .45574 . .48287 .48363 .48439
Bb──── 1.455 1.3299 1.2522 . .6485 .63 . .3371 .31983 . .12 .1131 .106
Ab──── .33504 .27455 .23867 . .04178 .03849 . .00624 .00534 . .00029 .00024 .0002
════════════════════════════ Fr──── .1 .16673 .21354 . .64761 .66121 . .86012 .8704 . .96954 .97211 .97468
r6──── .73702 .63452 .57227 . .16917 .15998 . .0469 .04226 . .00599 .00532 .00468
r7──── .91496 .88641 .87149 . .86634 .86944 . .93438 .93868 . .98444 .98573 .98702
142 Open Channel Flow: Numerical Methods and Computer Applications
143
Energy and Its Dissipation in Open Channels A similar Mathcad model EXPRB2∼1.MCD is listed below (with only a top portion of the range variables given by changing Fr2:=0.01,0.03…0.12) Variables
Y1:= 6
Yb:=.4
B1:= 1.7721542
Br:= 1.7
Ar:=.4
A1:=.49202836
Bb:= 1.5
Yd1:=.5
Yr :=.5
Ab:=.3
Given Ar  Ab sin(Br) Yd1 Ar  Ab Y1+Fr2 · = Yr +.5· 2 sin(Br) cos(Br)= 12·Yr Ar =.25·(Br cos(Br)·sin(Br)) cos(Bb)= 12·Yb Ab =.25·(Bb cos(Bb)·sin( (Bb)) F(Fr2):= Find(Yr,Yb,Br,Ar,Bb,Ab) Fr2:= 0.01,0.03..0.12 A1 (Yd1·Fr2)=(Ar  Ab)·
(Fr2)0 F(Fr2)1 F(Fr2)2 F(Fr2)3 F(Fr2)4 F(Fr2)5 F( 0.549 0.531 0.521 0.516 0.512 0.509
1.669 1.632 1.613 1.602 1.594 1.589
0.442 0.376 0.336 0.307 0.283 63 0.26
0.442 0.423 0.414 0.408 0.404 0.402
1.455 1.319 1.237 1.174 1.122 1.076
F(Fr2)1 Y1
Fr2
0.1 0.173 0.224 0.265 0.3 0.332
0.335 0.27 0.232 0.204 83 0.18 0.165
0.737 0.626 0.56 0.511 0.472 0.438
The variables listed are used to obtain the curve for an upstream dimensionless depth of 0.6. To get the separate curves a series of solutions were obtained for different dimensionless upstream depths, 1/2 Y1′. The dimensionless flow rate can be defined as Q′ = {Q2/(gD5)}1/2. Then Q ′ = Fr1/ A1′3/ T1′ . The graphs below were obtained by plotting the solutions for Y b/Y1 and Yr/Y1, respectively as the ordinate, and the Froude number Fr1 as the abscissa for Y1′ = 0.1, 0.4, 0.6, 0.7, 0.8, and 0.9. Ratio of depth Yr to upstream depth
Ratio of step height to upstream depth
{
0.8 0.7 0.6
Dimensional upstream depth Y1/D = 0.1 0.4
0.5 0.4
0.6 0.7 0.8 0.9
0.3 0.2
1 0.98 0.96 Dimensional upstream depth Y1΄ = Y1/D = 0.1
0.94 0.92 0.9
0.4
0.88 0.86
0.6
0.7
0.8
0.84
0.1 0
}
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Upstream Froude number, Fr1
0.9
1
0.82 0.8
0.9 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Upstream Froude number, Fr1
Example Problem 2.24 Develop a special algorithm, and implement it in a subroutine, based on the Newton method to solve for the bottom width b and the critical depth Yc of a trapezoidal channel that will result in critical flow for a given flow rate Q and specific energy E (or head H).
1
144
Open Channel Flow: Numerical Methods and Computer Applications Solution The two equations needed to solve for these two unknowns Yc (used as Y in equations below) and b are the critical flow and energy equations, or, F1 = Q 2 T − gA 3 = 0
F2 = E – Y –
Q2 =0 2gA 2
Notice that the derivative ∂F2/∂Y in the Jacobian for the Newton method consists of ∂F2 Q2T = −1 + ∂Y gA 3
and since we want the flow to be critical, we will set ∂F2/∂Y = 0, even though when it is evaluated for guesses for b and Y it may not be zero. Thus the Jacobian becomes
∂F1 ∂Y D = ∂F2 ∂Y
∂F1 2mQ 2 − 3gA 2 T ∂b = ∂F2 0 ∂b
Q 2 − 3gA 2 Y D11 Q2Y = 0 (gA 3 )
D12 D22
Notice that the Jacobian is already in upper triangular form, so the solution vector {z} in [D]{z} = {F} can be obtained from back substitution, or z2 = F2/D22, and z1 = (F1 – D12z2)/D11. The subroutine SOLYB given below implements this special Newton iteration. SOLYB.FOR SUBROUTINE SOLYB(G,FM,E,Q,Y,B) FM2=2.*FM QS=Q*Q FMQ=FM2*QS QSG=QS/G G3=3.*G M=0 1 A=(B+FM*Y)*Y T=B+FM2*Y AS=A*A F1=QS*TG*A*AS F2=EYQSG/(2.*AS) D11=FMQG3*AS*T D12=QSG3*AS*Y Z2=F2/(QS*Y/(G*A*AS)) Z1=(F1D12*Z2)/D11 B=BZ2 Y=YZ1 M=M+1 IF(ABS(Z1)+ABS(Z2).GT.1.E5 .AND. M.LT.20) GO TO 1 IF(M.EQ.20) WRITE(*,*) ' Didnot converge' RETURN END
Energy and Its Dissipation in Open Channels
145
A main program that calls on this subroutine can consist of the following. If the input: 32.2 1.5 5 500 3.5 10 is given then the solution is: Y = 3.666, b = 9.216. 1
100 99
WRITE(*,*)' Give:g,m,E,Q, and guesses for Y & b' READ(*,*,ERR=99) G,FM,E,Q,Y,B CALL SOLYB(G,FM,E,Q,Y,B) WRITE(*,100) Y,B FORMAT(' Solution Y =',F8.3,' b =',F8.3) GO TO 1 STOP END
2.10 Upstream Depth When Critical Conditions Occur at Reduced Downstream Section This section will expand upon the problems of determining upstream conditions from a reduced section that causes critical flow to exist. The principles have already been covered: First the critical flow equation Q2T/(gA3) = 1 is solved at the downstream section and using the solved critical depth Yc, the specific energy is computed E2 = Ec = Yc + (Q/Ac)2/(2g), and finally the energy upstream is equated to the downstream critical specific energy accounting for any change in the channel bottom and/or local loss, or E1 = Y1 + (Q/A1)2/(2g) = Ec + Δz + K(Q/Ac)2/(2g). In order for a solution to exist for the upstream depth Y1 the specific energy E1 must be larger than the upstream critical specific energy Ec1 (E1 > Ec1) associated with this flow rate. This generally means that the upstream channel must have a larger crosssectional area for this depth Y1, than the crosssectional in the downstream channel associated with the critical depth here, but for special geometries this may not be true. Furthermore, of the two real solutions available for Y1 if E1 > Ec1 the one that produces a Froude number less than one, i.e., the subcritical depth, is selected. Otherwise a hydraulic jump would take place, as discussed in Chapter 3. A common design problem that involves this procedure is to size the upstream channel so that for the design flow rate the upstream channel will flow under uniform flow conditions. The terrain over which the downstream channel occurs may slope significantly so that critical flow will occur at its beginning and thereafter the depth will decrease toward the normal supercritical depth supported by its steep slope. It is since the bottom slope increases that the size of the channel is reduce. The design of transitions from the upstream to downstream channel is covered in Chapter 5, in which the transition’s design may result in normal depths in both the upstream and downstream channels for the selected flow rate. In this section we will assume that there is no change in bottom elevation across the transition, and that the local loss caused by the transition can be ignored, or perhaps a more realistic statement is that the energy loss through the transition equals the drop in the channel bottom so E1 = Ec2. The following are some of the possible combinations of upstream and downstream channels: (1) upstream trapezoidal channel to smaller downstream trapezoidal channel, (2) upstream trapezoidal channel to downstream rectangular channel, (3) upstream rectangular channel to smaller downstream rectangular channel, (4) upstream trapezoidal (or rectangular) channel to downstream circular channel, (5) upstream circular channel to downstream trapezoidal (or rectangular) channel, and (6) upstream circular channel to smaller downstream circular channel. The approach to the solution to any of these six combinations follows the procedure illustrated in Example Problem 2.25. Example Problem 2.25 A channel system is to be designed for a flow rate Q = 400 cfs. The slope of the land changes so that the upstream channel will have So1 = 0.00035, and the downstream channel So2 = 0.015, the channels are of concrete with n = 0.013. The downstream channel will consist of a circular section with a diameter D2 = 8 ft. Determine the bottom width b1 of an upstream trapezoidal channel with a side slope m1 = 0.8 that will flow under uniform conditions. (The bottom remains horizontal through the transition from the trapezoidal to the circular channels.)
146
Open Channel Flow: Numerical Methods and Computer Applications Solution Step 1: Since the downstream channel is steep, critical flow will occur at its entrance so solve Q2T/(gA3) = 1 in the 8 ft diameter pipe, for Yc = 5.084 ft. Step 2: Compute the specific energy associated with this Yc, or Ec = E2 = Yc + (Q/Ac)2/(2g) = 7.272 ft. Step 3: Equate the upstream energy to that downstream, or Y1 + (Q/A1)2/(2g) = Ec = 7.272 ft and solve this equation simultaneously with the uniform flow equation (Manning’s equation) Q = (C u / n)A15 / 3 (So1 )1/ 2/ P12 / 3 for Y1 and b1 to give: Yo1 = Y1 = 6.906 ft and b = 6.407 ft. The Froude number associated with this solution is Fr1 = 0.39. Using b1 = 6.407 ft, another solution for Y1 is 2.722 ft, but the Froude Number associated therewith is 2.047, obviously not the sought for solution. (The normal depth in the downstream circular channel is Yo2 = 3.308 ft (Eo2 = 9.757 ft), so the depth will be critical at its beginning and gradually thereafter decrease to this latter uniform depth.)
2.11 Dimensionless Treatment of Upstream Trapezoidal Channel to Downstream Rectangular Channel Generalization of the solution to the problem of finding conditions in an upstream trapezoidal channel when critical flow exists in a downstream rectangular channel can be handled by introducing dimensionless variables. (This is combination (2) mentioned above.) The following three dimensionless variables will be introduced: Y1′ = m1Y1/ b1 (the dimensionless depth in the upstream channel) b1′ = b1/b2 (the ratio of the upstream bottom width to the downstream bottom width), and b′ = b1/ Yc (the ratio of the upstream bottom width to the critical depth in the downstream rectangular channel). Since the downstream channel is rectangular the critical depth here can be computed from Yc = {q 22 / g}1/ 3 = {(Q / b2 )2 / g}1/ 3. We should also note, since the downstream Froude number is unity, that Q 2 /(gYc3b22 ) = 1 and Ec = E2 = 1.5Yc. Multiplying the energy equation E1 = Ec by m1/Yc gives,
m1Y1 m1Q 2 + = 1.5m1 Yc 2gYc (b1Y1 + m1Y12 )2
or written in terms of the dimensionless variables this energy equation becomes, Y1′b1 +
0.5m13 /(b1′) 2
(
b′ 2 Y2′ + Y2′ 2
)
2
= 1.5m1
(2.40)
Equation 2.40 can be written as the following fifth degree polynomial.
1.5m1 4 3m1 3 1.5m1 0.5m13 = 0 Y1′ 5 + 2 – Y1′ + 1 – Y1′ – + b′ b′ b′Y1′2 b1′2 b′ 3
(
)
(2.41)
By introducing these dimensionless variables in the upstream Froude number squared, Fr12 = Q 2 (b1 + 2m1Y1 )/{g(b1Y1 + m1Y12 )3} after some manipulation, it can be expressed by the following equation as a function the upstream side slope m1 and these dimensionless variables:
Fr21 =
(m1 b′ )3 (1 + 2Y1′ ) m13 (1 + 2Y1′ ) = b1′ 2 b′ 3 (Y1′ + Y1′ 2 )3 b1′ 2 (Y1′ + Y1′ 2 )3
(2.42)
Since a fifth degree polynomial has five roots, it is clear that some precautions be used in solving either Equations 2.40 or 2.41 to insure that the sought after subcritical root for Y1′ is obtain.
147
Energy and Its Dissipation in Open Channels
If m1 and b′1 are large enough then two positive real positive roots, one the subcritical and the other the supercritical depth, one negative root, and two complex conjugate roots are available. The values of these roots will depend on the ratio b′ = b1/Yc. It will be a worthwhile exercise for you to write a computer program to produce several tables for different m1 values of solutions Y1′ as a function of b′1 and b′. (See the homework problems at the end of this chapter.) The solutions for dimensionless depth Y1′ from Equations 2.40 or 2.41 for eight different upstream side slopes m1 = 0.25, 0.5, 0.75, 1.0, 1.25, 1.5, 1.75 and 2.0 are given in the eight graphs in Figure 2.13. On these 1.6
Dimensionless depth, Y1΄ = m1Y1/b1
Dimensionless depth, Y1΄ = m1Y1/b1
0.8
1.4
0.7
1.2
b΄=
0.2
1
b/ 1
0.1 0.5
1
0.2
b= 2 0 .8
1.5
2
2.5
3
3.5
4
Dimensionless width, b΄= b1/Yc
4.5
0
5
Dimensionless depth, Y1΄ = m1Y1/b1
1.5
2
2.5
3
3.5
4
4.5
5
4
4.5
5
4
4.5
5
Dimensionless width, b΄= b1/Yc
.0
b΄ 1 = 2
b΄ 1 = 1
0.5
m1 = 0.75
1
1.5
2
2.5
1
0.5
b1΄ = b /b 1 2 = 0.8
0.5
.0 b΄ 1 = 1.2 b΄ 5 1 = 1.5 b΄ 1 = 1. 75 b΄ 1 = 2.0
1.5
b΄ 1 = 1
1
2
b΄ 1 = 1
.0 b΄ 1 = 1
.25 b΄ 1 = 1.5 .75
3
3.5
4
Dimensionless width, b΄= b1/Yc
4.5
0
5
4
m1 = 1.0
b1΄ = b1/b2 = 0.8 0.5
1
1.5
2
2.5
3
3.5
Dimensionless width, b΄= b1/Yc
Dimensionless depth, Y1΄ = m1Y1/b1
4.5
3.5
4
3.5
3 .0 b΄ 1 = 1.2 5 .5
b΄ 1 = 1
b΄ 1 = 1.7 5 b΄ 1 = 2.0
b΄ 1 = 1
2
1.5 1
0.5 1
1.5
2
2.5
3
2
1.5
m1 = 1.25
3.5
4
Dimensionless width, b΄= b1/Yc
4.5
5
m1 = 1.5
1
0.5
b1΄ = b1/b2 = 0.8 0.5
3
2.5
b΄ 1 = 1.2 b΄ 1 = 1.5 5 b΄ 1 = 1.7 5 b΄ 1 = 2.0
2.5
.0
Dimensionless depth, Y1΄ = m1Y1/b1
1
2.5
1.5
Dimensionless depth, Y1΄ = m1Y1/b1
0.5
3
2
0
m1 = 0.5
b1΄ = b1/b2 = 0.8
0.2
2.5
0
.0
0.6
b΄ 1 = 1
0
0.8
m1 = 0.25
.25 b΄ 1 = 1.5 b΄ 1 = 1.7 5 b΄ 1 = 2.0
0.3
b΄ 1 = 1
0.4
1
b΄ 1 = 1
b΄ 1 = 1.5 b΄ 1 = 1.7 b΄= 5 1 2 .0
b΄ 1 = 1
b΄ 1 = 1
0.5
.25
.0
0.6
0
b1΄ = b1/b2 = 0.8 0.5
1
1.5
2
2.5
3
3.5
Dimensionless width, b΄= b1/Yc
FIGURE 2.13 Eight graphs for different upstream side slopes that provide the solution of Equation 2.40 of dimensionless depth Y1′ as a function of b′ and b1′ for upstream trapezoidal channels to downstream rectangular channels with critical flow. (continued)
148
Open Channel Flow: Numerical Methods and Computer Applications
Dimensionless depth, Y1΄ = m1Y1/b1
6 5
5
2
.0 b΄ 1 = 1
3 2
m1 = 1.75
m1 = 2.0
1
1 0
b΄ 1 = 1
3
b΄ .2 1 = 1.5 5 b΄ 1 = 1.7 5 b΄ 1 = 2.0
4
4 b΄ 1 = 1. 0 b΄ 1 = 1. 2 b΄ 5 1 = 1.5 b΄ 1 = 1.7 5 b΄ 1 = 2.0
Dimensionless depth, Y1΄ = m1Y1/b1
6
b1΄ = b1/b2 = 0.8 0.5
1
1.5
2
2.5
3
3.5
Dimensionless width, b΄= b1/Yc
4
4.5
5
0
b1΄ = b1/b2 = 0.8 0.5
1
1.5
2
2.5
3
3.5
4
Dimensionless width, b΄= b1/Yc
4.5
5
FIGURE 2.13 (continued)
graphs the ratio of upstream bottom width b1 to the downstream critical depth b′ = b1/Yc are the abscissas, and Y1′ are the ordinates, with different lines for the ratio b1′ = b1/b2 . The bottom curve on each graph is for b′1 = 0.8, the next curve up for b′1 = 1.0, etc. with the top curve for b′1 = 2.0 . The lower curves end when a real root to Equation 2.41 is not available, i.e., the downstream critical specific energy is not larger than the upstream critical specific energy because b1 is too small. (Conditions of no solution will occur only for b′1 < 1.) In other words the upstream flow has also become critical, or Fr1 = 1 as well as Fr2 = 1. The dimensionless depth Y1′ and ratio b′, associated with any b′1 can be solve from Equations 2.40 (or 2.41) and the equation obtained by setting the Froude number, Equation 2.42, to unity. A TKSolver model providing such solutions for several different values of b′1 less than 1 and for six side slopes m1 is given below. (The “Solve List” capability of TKSolver was used separately for each of the m1 as given and these separate solutions were combined into the single table given below.) The dimensionless upstream depth Y1′ does not change with m1 as can be observed from the separate graphs on Figure 2.13, or by examining the nature of the equations, but of course the actual depth Y1 will change. Notice how b′ becomes larger with both m1 and b′1. TKSolver model to solve for b′ = b1/Yc and Y1′ = m1Y1/ b1 for different values of b1′ = b1/b2. (Solves dimensionless Fr12 = 1 and E1 = Ec2 simultaneously.) ══════════════ VARIABLE SHEET ══════════ St Input──── Name─── Output─── Unit───── LG .25707709 Y1p .25 m LG 1.0308088 bp L .5 b1p ══════════════ RULE SHEET ════════════ S Rule──────────────────────────────── (m/bp)^3/b1p^2*(1+2*Y1p)/(Y1p+Y1p^2)^3=1 Y1p*bp+.5*m^3/(b1p*bp*(Y1p+Y1p^2))^2=1.5*m b′ = b1/Yc b1′ 0.500 0.550 0.600 0.650
Y1′
m1 = 0.2
m1 = 0.25
m1 = 0.30
m1 = 0.40
m1 = 0.50
m1 = 0.75
1.05510900 0.86045400 0.69857550 0.56197330
0.213722 0.259824 0.317100 0.390313
0.267152 0.324780 0.396375 0.487891
0.320582 0.389736 0.475650 0.585470
0.427443 0.519648 0.634200 0.780626
0.534304 0.649560 0.792750 0.975783
0.801456 0.974341 1.189124 1.463674
149
Energy and Its Dissipation in Open Channels (continued) b′ = b1/Yc b1′ 0.700 0.750 0.800 0.850 0.900 0.920 0.930 0.940 0.950 0.960 0.970 0.971 0.972 0.973 0.974 0.975
Y1′
m1 = 0.2
m1 = 0.25
m1 = 0.30
m1 = 0.40
m1 = 0.50
m1 = 0.75
0.44529950 0.34464220 0.25707710 0.18037750 0.11282030 0.08804428 0.07609940 0.06443842 0.05305317 0.04193582 0.03107887 0.03000722 0.02893809 0.02787148 0.02680738 0.02574579
0.487402 0.622656 0.824647 1.160043 1.828846 2.329720 2.687304 3.163939 3.831054 4.831504 6.498624 6.728554 6.974905 7.239501 7.524447 7.832185
0.609253 0.778320 1.030809 1.450054 2.286057 2.912150 3.359131 3.954925 4.788817 6.039380 8.123280 8.410693 8.718632 9.049376 9.405559 9.790231
0.731104 0.933984 1.236971 1.740065 2.743268 3.494580 4.030957 4.745909 5.746581 7.247256 9.747935 10.092830 10.462360 10.859250 11.286670 11.748280
0.974805 1.245312 1.649294 2.320086 3.657691 4.659440 5.374609 6.327879 7.662107 9.663008 12.997250 13.457110 13.949810 14.479000 15.048890 15.664370
1.218506 1.556640 2.061618 2.900108 4.572114 5.824300 6.718261 7.909849 9.577634 12.078760 16.246560 16.821390 17.437260 18.098750 18.811120 19.580460
1.827759 2.334960 3.092426 4.350162 6.858171 8.736450 10.077390 11.864770 14.366450 18.118140 24.369840 25.232080 26.155900 27.148130 28.216670 29.370690
Example Problem 2.26 An upstream trapezoidal channel is to have a bottom width b1 = 9 ft and a side slope m1 = 0.5, and transitions to a rectangular channel with a bottom width b2 = 10 ft at the beginning of a steep slope. What flow rate will result in critical depth in both the trapezoidal and rectangular channels? What are the depths, Y1 and Y2? Solution From the above table on the line corresponding to b1′ = 9/10 = 0.9 and the column for m1 = 0.50 we read b′ = 4.572114 and therefore Yc2 = b1/b′ = 9/4.572114 = 1.968 ft (Ec2 = 1.5Yc2 = 2.952 ft). The flow rate per unit width in the rectangular channel is q2 = (gYc23)1/2 = (32.2 × 1.9683)0.5 = 15.672 cfs/ft, so Q = 10(15.672) = 156.72 cfs. The depth in the upstream channel will be Y1 = b1Y1′ / m1 = 9(0.11282030)/ 0.5 = 2.031 ft (E1 = E1c = 2.953 ft ), i.e., just 0.067 ft greater than the downstream depth, but the specific energies are equal. If Manning’s n = 0.013, this critical flow will occur in the upstream channel if So1 = 0.002641. Example Problem 2.27 An upstream trapezoidal channel with b1 = 15 ft and m1 = 1, changes to a rectangular channel with b2 = 10 ft at the head of a steep slope. For a flow rate of Q = 400 cfs what are the depths in the trapezoidal and rectangular channels and the Froude number of the upstream flow? Solution b1′ = 15/10 = 1.5, and from the critical flow equation in the rectangular channel Yc2 = (402/32.2)1/3 = 3.676 ft so b′ = b1/Yc2 = 15/3.676 = 4.080. Next solve Equation 2.40 (or 2.41), or read the graph in Figure 2.13, or use the table you generated for the home problem for Y1′ = 0.35333 (the graph will produce at most 2 digits of accuracy), so Y1 = 15(0.35333)/1 = 5.300 ft, and from Equation 2.41 Fr1 = 0.316. Example Problem 2.28 An upstream trapezoidal channel is to have a side slope m1 = 1.5, and a bottom width of 1.5 times that of a downstream rectangular channel at the head of a steep slope. If the upstream channel has n = 0.013, and a bottom slope So1 = 0.0003, what should the bottom widths of the two channels be for a design flow rate of Q = 15 m3/s if uniform flow is to take place in the upstream
150
Open Channel Flow: Numerical Methods and Computer Applications channel. Solve the problem using dimensionless variables, and verify this solution by solving the problem using dimensioned variables. What occurs when the bottom slope of the upstream channel increases, i.e., say to So1 = 0.0005? Solution A good approach involves solving Manning’s equation with Y1 replaced by b1Y1′/ m1 and Equation 1/ 3 2 2.40 with b′ = b1/ Yc = b1 ( b1′Q / b1 ) / g = b11/ 3/ 3.723 simultaneously for Y1′ and b1. The solution is Y1′ = 0.0525 and b1 = 20.092 m. The upstream depth Y1 = b1Y1′ / m1 = 0.704 m . A general TKSolver model (TRARECS1.TK) designed to solve problems of this type is given below. ═══════════ VARIABLE SHEET ═══════════ St Input──── Name─── Output─── Unit─── Y1p .05254506 b1 20.092804 Y1 .70385174 bp 39.887384 1.5 m 1.5 b1p 15 Q 1 Cu .0003 So1 .013 n 9.81 g ═════════════ RULE SHEET ═══════════ S Rule────────────────────────────── Y1p=m*Y1/b1 Q=Cu*So1^.5*((b1+m*Y1)*Y1)^1.666667/(n*(b1+2.*(m*m+1)^. 5*Y1)^.6666667) bp=b1/(((b1p*Q/b1)^2/g)^.3333333) Y1p*bp+.5*m^3/(b1p*bp*(Y1p+Y1p^2))^2=1.5*m To solve the problem using dimensioned variables requires that Y1, Ec, and b1 be solved using: (1) Manning’s equation in the upstream channel, (2) The critical specific energy equation
{
}
E c = 1.5 ( b1′Q / b1 ) /g 2
1/ 3
, and (3) the energy equation Y1 + (Q/A1)2/(2g) = Ec. The solution is the
same as above. As the upstream bottom slope increases the width of channel increases rapidly to be unrealistically wide, i.e., the above solution with b1 = 20.1 m is a very wide channel already. A better solution would be to use a smaller ratio b1′, i.e., let b1′ = 1.2 or smaller. If b1′ = 1.2 and So1 = 0.0005, then the above model gives: 0.979Y1′ = 0.0979, Y1 = 0.805 m, b1 = 12.333 m, and b′ = 20.52.
2.11.1 Upstream Channel Also Rectangular The above dimensionless variables are not valid for a rectangular channel since the dimensionless depth is obtained by multiplying the actual depth by the side slope m, and this is zero. For a rectangular channel let
Y1′ = Y1/b1 with b1′ = b1/b2 and b′ = b1/Yc2 as before
With this redefined Y1′ the dimensionless specific energy between the upstream rectangular channel, and the downstream rectangular channel where critical flow occurs is,
b′Y1′ +
1 = 1.5 2b′ b1′ 2 Y1′2 2
(2.43)
151
Energy and Its Dissipation in Open Channels
or written as a third degree polynomial
0.5 1.5 Y1′3 – Y1′2 + 3 2 = 0 b′ (b′ b1′ )
(2.44)
1 3 = −2 (b1′2 b′ 3Y1′3) (b′Y1′)
(2.45)
and the Froude number squared,
Fr12 =
When both upstream and downstream channels are rectangular, and the depth is to be critical in the downstream channel it is clear that the width ratio b′1 must be larger than unity for a real solution to be available from Equation 2.44 (or Equation 2.43). When this condition is met then there will be at least one positive real root. Figure 2.14 provides graphical solutions (one using linear paper and the other using log–log paper) of this larger positive dimensionless depth Y1′ as a function of the ratio b′ = b1/Yc for several width ratios b′1 starting with 1.05 and ending with 2.5. The lowest curve applies for b′1 = 1.05, and the upper curve for b′1 = 2.5 , but note again as was the case when the upstream channel is trapezoidal, that the spread between the curves is small and gets smaller with increasing values of b′1 and b′. You will find it useful to generate a table of values that would allow Figure 2.14 to be plotted to be used in solving problems. (See one of the homework problems at the end of this chapter.) Because whenever Y1′ or b = occur in Equation 2.43 they form a product b′Y1′ = c, which must be a constant for any given ratio b′1 = b1 / b2 . This fact is apparent also from the log–log plot on Figure 2.14 with straight lines with a slope of −1. Straight lines on a log–log plot conform to the equation y = cxb in which b is the slope and c is the intercept. (b′Y1′ is not only constant for the subcritical dimensional depth Y1′ , but also is constant for the supercritical dimensionless depth Y1′ and the nonphysical negative root Y1′ .) Therefore a much simpler means for evaluating Y1′ for any b′1 is to find c = b′Y1′ (the constant, which is given in the table below on the row of Y1′ ) and then solve for Y1′ = c / b′.
2
1.3 1.2
1.6 1.5 1.4
5 1.0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 Dimensionless width b΄= b1/Yc
0.5 0.3 0.2
0.5
2
= 2.5 8
1.
7 1. .6 1
1.5
1.4
1.0 5
0.5
2
1
b 1/b 2 5 2.2
1.2
1
.25
b΄1=
1.1
1.5
b΄1
1. 1. 8 7
2
.5
=2 b2
/ =b1
2
1.3
2.5
Dimensionless depth Y΄1 = Y1/b1
3
1.1
Dimensionless depth Y΄1 = Y1/b1
3
1 2 5 3 Dimensionless width b΄= b1/Yc
10
FIGURE 2.14 Solution of Equation 2.43 of dimensionless depth Y′ as a function of b′ and b1′ for rectangular channels both upstream and downstream and critical flow in the downstream channel.
152
Open Channel Flow: Numerical Methods and Computer Applications
The second means for evaluating the Froude number given by Equation 2.45 shows that Fr1 (or Fr12), does not change with the ratio b′1 = b1/b2. Thus the upstream Froude number for any upstream rectangular channel with a width b1 that reduces to a rectangular channel with width b2 in which the flow is critical can be computed by finding this constant c = b′Y1′ and then using Fr1 = {3/c – 2}1/2. The table below gives the subcritical solution Y1′ corresponding to b′ = 1 for different ratios b1′ , so this Y1′ equals this constant c, and the corresponding upstream Froude numbers. Table of Subcritical Roots Y1′ of the Third Degree Polynomial for a RectangulartoRectangular Channel Reduction that Produces Yc at b2, Corresponding to b′ = 1 b1′
1.05
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
2.00
2.25
2.50
Y1′
1.1670
1.2243
1.2920
1.3337
1.3626
1.3840
1.4004
1.4134
1.4239
1.4397
1.4532
1.4626
Fr1
0.755
0.671
0.567
0.494
0.449
0.409
0.377
0.350
0.327
0.289
0.254
0.226
Example Problem 2.29 A flow rate Q = 400 cfs occurs in a upstream rectangular channel with a width of b1 = 15 ft, which reduces to a rectangular channel with a width b2 = 10 ft at the head of a steep slope. Determine the upstream depth and Froude number. If n = 0.013 what bottom slope of this upstream channel will result in uniform flow? Solve for the same unknowns when Q = 200 cfs. Solution The critical depth in the smaller channel is Yc = {(Q/b2)2/g}1/3 = {402/32.2}1/3 = 3.676 ft, so b′ = b1/Yc = 15/3.676 = 4.08. With b1′ = 1.5 the larger positive solution to Equation 2.43 (or 2.44), which may be read to two digits from Figure 2.14 but can be solved easiest by dividing c by b′, is Y1′ = 1.385/4.08 = 0.3392, so Y1 = 5.088 ft. From Equation 2.45 Fr1 = 0.4095. Solving Manning’s equation gives So1 = 0.000479. For Q = 200 cfs, Yc = 2.314 ft, b′ = 6.477, Y1′= 1.384 / 6.477 = 0.2137, so Y1 = 3.205 ft, and Fr1 = 0.4095 (the same since b1′ is the same), and So1 = 0.000449. Example Problem 2.30 What width b2 of steep downstream rectangular channel will result in uniform flow in a b1 = 4 m wide upstream rectangular channel with n = 0.013 and a bottom slope So1 = 0.0005 if the flow rate is Q = 10 m3/s. Solution Depth ratio b′ can be given as the following function of b1′: b′ = 4/ Yc = 4 /{[(10 / 4)b1′ ]2 /g}1/ 3 = 4.6486086 / b1′2 / 3. With this substituted into Equation 2.43, it can be solved simultaneously with Manning’s equation for b1′ and Y1′. The solution is b1′ = 1.52135 and Y1′ = 0.3949 (b′ = 3.514), so b2 = 4/1.52135 = 2.629 m, and Y1 = 4(.3949) = 1.580 m. The Froude number 0.5 0.5 Fr1 = 3 / ( b′Y1′) − 2 = 3 / (3.514 × 0.3949) − 2 = 0.402. Critical depth in the downstream channel can now be computed from Yc = {(10/2.629)2/9.81}1/3 = 1.138 m, or Yc = b1/b′ = 4/3.514 = 1.138 m. The Froude number can also be computed from Fr1 = {(10/4)2/(9.81 × 1.583}0.5 = 0.402. A TKSolver model for the above solution is,
{
}
{
}
══════════════ VARIABLE SHEET ═══════════ St Input──── Name─── Output─── Unit───── bp 3.5142794 b1p 1.5213531 Y1p .39491513 Y1 1.5796605 10 Q .0005 So .013 n
153
Energy and Its Dissipation in Open Channels ══════════════ RULE SHEET ═══════════ S Rule──────────────────────────────── bp=4.6486086/b1p^.6666667 bp*Y1p+.5/(bp*b1p*Y1p)^2=1.5 Y1=4*Y1p Q=(4*Y1)^1.666667*So^.5/(n*(4+2*Y1)^.666667)
2.12 Hydraulically Most Efficient Section A major consideration in the design of a channel should be to keep total costs to a minimum. The costs associated with a channel lining are directly proportional to its wetted perimeter. The fluid frictional losses are directly related to the perimeter also. Thus a key factor in designing a channel is to maximize its conveyance capabilities while minimizing its perimeter. The hydraulically most efficient section is defined as a channel with the least wetted perimeter for a given conveyance capability. Therefore, the costs of a channel will generally be near minimum if it is designed as the “hydraulically most efficient section. See the Section 5.9 for designing least cost trapezoidal channels accounting for lining and excavation costs. If Manning’s equation is assumed to properly account for the frictional losses then to determine the relationship between the geometric variables that will result in the least wetted perimeter the flow rate, Q, the roughness coefficient, n and the bottom slope, So are held constant. Combining these variables in a single constant c results in
nQ = A5/3 (CuS1/2 )P 2 / 3
or cP 2 / 3 = A 5 / 3
which can be written with c redefined as
A = cP 2/5 (in which c is redefined as c = c3/5 )
(2.46)
First, consider a circular section because it is generally known that a circle is the shape that has the largest possible area for a given perimeter. For a circle the area is given by A = D2/4(β − sin β cos β) and Equation 2.46 can be written as
D2 (β − sin βcos β) = cP 2 / 5 4
Differentiating this equation with respect to β and setting ∂P/∂β = 0 gives,
1 + sin 2 β − cos2 β = 0
replacing cos2 β with 1 − sin2 β gives sin β = 0, or β = 0. Clearly this is a maximum. Replacing sin2 β with 1 − cos2 β gives,
1 + 1 – cos2 β − cos2 β = 2 – 2cos2 β = 0 or cos β = 1 or β =
Thus a circular section is hydraulically most efficient when flowing half full.
π 2
154
Open Channel Flow: Numerical Methods and Computer Applications
Don’t confuse this section with the depth in a circular section that will maximize the flow rate it can carry. For a given diameter, wall roughness, and bottom slope the latter depth equals Y = 0.938D, or β = 2.6391 rad (151.21°), as shown in Figure A.2. To get this depth, the conveyance K = A5/3/P2/3 is differentiated with respect to Y and the set to zero. After some algebra 5 sin β = 1 − cos β sin β/β results. Its solution is β = 2.6390536 radians, and Y = D(1 − cos β)/2 = 0.93818122D. Next, consider a trapezoidal channel. For a trapezoidal channel its crosssectional area is given by A = (b + mY)Y, and its perimeter is given by P = b + 2Y(m2 + 1)1/2. Solving b from the last equation and substituting this for b in the equation for the area and finally using this area in Equation 2.46 gives the following equation that we wish to find the extreme value of: (P − 2Y(m 2 + 1)1/ 2 )Y + mY 2 = cP 2 / 5
(2.47)
Taking the partial derivative with respect to m, with Y held constant, results in,
2Y 2 m ∂P 2 ∂P Y – + Y 2 = cP −3 / 5 2 1/2 ∂m (m + 1) 5 ∂m
setting ∂P/∂m = 0 results in
2m 1 3 = 1 or 4m 2 = m 2 + 1 or m = = (m 2 + 1)1/2 3 3
(2.48)
A side slope of m = √ 3/3 result in an angle of 30° from the vertical or an angle of 120° from the horizontal. Next, differentiate Equation 2.47 partially with respect to Y, with m held constant, gives,
∂P 2 ∂P Y + P – 4Y(m 2 + 1)1/2 + 2mY = cP −3 / 5 ∂Y 5 ∂Y
Setting ∂P/∂Y = 0 results in, P − 4(2 m)Y + 2 mY = 0 since (m2 + 1)1/2 = 2 m from above, or
3 P = 6mY = 6Y =2 3
( 3 ) Y = 3
2 Y = 3b 3
(2.49)
since the hypotenuse of a right triangle with sides 1 and m = √ 3/3 is 2/√ 3. Thus with P = 3b and Y = (√ 3/ 2) b, it is apparent that the hydraulically most efficient section is 1/2 of a hexagon. This result is not surprising since a hexagon is inscribed within a circle. Substitution of Y = (√ 3/2)b into the equation for the area of a trapezoidal channel gives A = (bY + mY 2 ) = (√ 3/2)b2 + (1 /√ 3)(3 / 4)b2 = √ 3(1 / 2 + 1/ 4)b2 = 1.299038b2 .
155
Energy and Its Dissipation in Open Channels
30° 60° 1 √3
2
The area for this section for a trapezoidal channel is 2 2 3 2 1 2 A= = Y + Y = 3Y and the wetted perimeter by, 2 3Y 3 3 2 P = 3 Y = 2 3Y and the hydraulic radius by, 3
( 3Y ) Y (2 3Y) 2 2
Rh =
(2.50)
The last equation when applied to a rectangular section indicates that the channel is operating hydraulically most efficient when the depth of flow is 1/2 the width of the channel.
2.12.1 Nondimensional Variables for This Section It is interesting to note the values of dimensionless variables coming out of the hydraulically most efficient cross section, or onehalf a hexagon, described in the previous section. The dimensionless depth Y′ = mY/ b = (1/ 3 )[( 3 / 2)b]/b = 1/ 2 or a constant for all sizes. Defining the dimensionless area A′ = mA/b2, it also is a constant or,
A′ =
3 mA = Y′ + Y′ 2 = 0.5 + 0.52 = 0.75 = , or substituting A = 3Y 2 2 4 b
and A′ =
3 3 3 = 3 4 4
As is noted above P = 3b, thus if the dimensionless wetted perimeter is define by P′ = P/b = 3. The hydraulic radius R h = A/P = [(b2/m)A′]/(bP′) = (b/m)(A′/P′), or letting
1 R A′ 3 1 R′h = m h = = = 0.25 = 4 b P′ 4 3
The same result can be obtained from Equation 2.50, namely R h = Y/2 = (bY′/m)/2, or R ′h = mR h / b = Y′/2 = 1/ 4.
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Open Channel Flow: Numerical Methods and Computer Applications
Since Manning’s equation is not dimensionally homogeneous (unless dimensions are assigned to Cu and/or n), it makes little sense to introduce a dimensionless flow rate Q into it. However, by substituting the above into Manning’s equation the following relationships between Q and Cu, n, and So, and b, or Y can be obtained for the hydraulically most efficient trapezoidal channel: C C Q = 0.7435137 u So b8 / 3 = 1.0911236 u So Y8 / 3 n n
The graph below provides this relationship of nQ /(Cu So ) and the bottom width b and the depth Y.
Bott
o
th b
th Y
D ep
9 8 7
7
6
6
5
tto m
3
Bo
4
4
wi
ep
th
5
dt h
Y
b
8
D
Bottom width, b
9
id mw
2
3 2
Depth, Y
10
0
50
100
150 200 nQ/(Cu√So)
250
300
1 350
The relationship of the optimal bottom width, and the optimal depth with Manning parameter nQ /(Cu So ) trapezoidal channels Example Problem 2.31 A flow rate of Q = 500 cfs is to be conveyed in a channel of maximum hydraulic efficiency with n = 0.014 and So =0.0008. What width of trapezoidal channel b, and what diameter of circular channel D should be used? If the cost of channel per foot of length is a linear function of its perimeter, i.e., $ = 300P, compare the costs of these two channels. Also determine the perimeter, and cost of a circular section in which the conveyance is a maximum. For a circular section, make a graph similar to the one above that shows how the optimal diameter D (and/or the optimal depth Y) varies with the parameter nQ /(C u So ). Solution When P = 3b and A = 1.299038b2 are substituted into Manning’s equation and b solved for
{
}
3/8
the following equation is found: b = Q / (0.7435137C u So / n) = {500/(0.7435137(1.486) (0.0008)1/2/0.014)}3/8 = 7.609 ft for the trapezoidal channel. For the hydraulically most efficient circular section, i.e., onehalf a circle, β = π/2, and from Manning’s equation 500 = 1.486(0.0008)1/2/0.014[D2/4(π/2)]5/3/(Dπ/2)2/3. Solving for D = 13.672 ft. For the circular channel with the maximum conveyance β = 2.6391 rad 500 = 1.486(0.0008)1/2/0.014[D2/4(β − cos β sin β)]5/3/[Dβ]2/3 = 1.0065746D8/3, or D = 10.2576 ft. For the trapezoidal section P = 3b = 3(7.609) = 22.828 ft, and for the circular sections P = (π/2)13.672 = 21.475 ft, and P = βD = 2.6391(10.2576) = 27.071 ft when K is a maximum, or the costs are: trapezoidal section $/ft = $6848.40, and circular section $/ft = $6442.57/ft and $/ft = $8121.30 when K is a maximum, or $1678.83 more than for the onehalf circle.
157
Energy and Its Dissipation in Open Channels For the last part of this problem that require a graph to be produced that shows how the optimal diameter varies with nQ /(C u √ So ) , we start with nQ /(C u √ So ) = A5/3/P2/3, and note that for the optimal β = 2.6390536 rad, the area which is A = (D2/4)(β − sin β cos β) that the optimal area is A = 0.7652888D2, and the optimal wetted perimeter is P = βD = 2.6390536D or the optimal nQ /(C u √ So ) = 0.335282D8 / 3 or D = 1.506507[ nQ / (C u √ So )]3 / 8 . Sine the optimal depth Y = 0.93818122D, we get Y = 1.41338 [ nQ / (C u √ So )]3 / 8 . The graphs below provide plots of these latter two equations; the first graph uses linear axes, whereas the second graph uses logarithmic axes. Because of the nature of the equations, they plot as straight lines on the log–log graph. 14 Optimal diameter, D
10
10
8
8 6
6
4
4
2
2
0
100
50
0
200 150 Parameter, nQ/(Cu√So)
250
Optimal depth, Y
12
12
0 350
300
14
Optimal diameter, D Optimal depth, Y
10
9 8 7 6 5
ter me Dia
4
De
3
pth
2
1 1
2
3
4
5 6 7 89 20 30 40 10 Parameter, nQ/(Cu√So)
60
80
100
110
120
Relationship the optimal diameter D, and optimal depth Y to the parameter nQ /(Cu √ So ) for a circular channel. The first graph uses linear axes, and the second graph uses logarithmic axes.
Problems 2.1 Determine what the uniform flow rate will be in a circular channel with a diameter of 2 m, if the bottom slope is 0.0004, and its wall roughness is 0.0003 m, if the depth of flow is 1.3 m. The kinematic viscosity of the water flowing in this pipe is v = 1.519 × 10 −6 m2/s. 2.2 What depth of flow would be expected in a 5 ft diameter channel that has a bottom slope of 0.004, if the wall roughness is 0.0005 ft and 200 cfs is flowing in this channel under uniform conditions. (v = 1.41 × 10 −5 ft2/s)
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Open Channel Flow: Numerical Methods and Computer Applications
2.3 What is the depth of flow in Problem 2.2 if the flow rate is 100 cfs? 2.4 Determine the depth of flow that would occur in a trapezoidal channel with b = 10 ft, m = 1.5, So = 0.0008 if it is carrying 400 cfs. The wall roughness for this channel is e = 0.001 ft. What is Chezy’s C? What is the corresponding value of Manning’s n? (v = 1.41 × 10 −5 ft2/s) 2.5 What size pipe should be used to carry 6 m3/s down a mild slope of 0.0015 if the pipe is not to flow more than 3/4 full (e = 0.0015 m and temperature = 20°C). 2.6 A 6 ft circular conduit has a bottom slope So = 0.001, and a wall roughness of e = 0.004 ft. What will the depth be for a flow rate of 80 cfs (temperature = 50°F). Using this depth, compute the corresponding value for Manning’s coefficient. 2.7 Use Chezy’s formula to fill in the blank in the table below that applies for uniform flow in channels of trapezoidal shapes (v = 1.41 × 10 −5 ft2/s). Flow Rate Q (cfs) — 324.5 550.0 400.0 450.0 600.0
Rough. e (ft)
Bot. Slope So
Bot. Width b (ft)
Side Slope m
Depth Y (ft)
0.005 — 0.003 0.004 0.004 0.005
0.0006 0.0008 — 0.0015 0.0009 0.00075
6.0 8.0 10.0 — 7.0 10.0
1.6 0.0 1.3 1.5 — 1.2
4.2 5.0 4.8 4.0 4.1 —
2.8 The table below applies for uniform flow in circular channels. Fill in the mission blanks (v = 1.317 × 10 −6 m2/s). Flow Rate Q (m3/s)
Rough. e (m)
Bot. Slope So
Diameter D (m)
Depth Y (m)
— 20.0 25.0 30.0 60.0
0.005 — 0.003 0.0025 0.0045
0.00080 0.00068 — 0.00040 0.00045
3.0 5.0 6.0 — 8.0
2.3 2.52 2.9 3.0 —
2.9 Assume the ratio of Y/D (depth to diameter) is the same as in Problem 2.6 but the conduit is of 1 ft diameter. What flow rate is occurring? What is Manning’s n for this flow? Why is n not the same as in Problem 2.6? 2.10 A flow rate of 40 m3/s is flowing in a trapezoidal channel under uniform conditions at a depth of Y = 2 m. The channel has a bottom width of b = 3 m, and a side slope of m = 1.5. Its wall roughness e = 0.002 m. What is the bottom slope of this channel? (v = 1.519 × 10 −6 m2/s). 2.11 Water is to enter a circular channel from a reservoir so that the depth in the channel will be 3.5 ft. The channel will have a bottom slope of So = 0.0009, and a wall roughness of e = 0.002 ft. What size pipe must be used to carry a flow rate of 120 cfs? 2.12 Use the following example of an open channel to show that the selection of an inappropriate n causes a much larger variation in the computed flow rate, than does a corresponding percent error in the selection of e. D = 3 m, Y = 2 m, So = 0.0008. Assume the correct value of n = 0.013, but you use n = 0.015 (a 15% error), and that the correct e = 0.00057 m, but you use e = 0.00066 m (also a 15% error) (v = 1.519 × 10 −6 m2/s). 2.13 Determine the appropriate bottom width that should be used in the design of an open channel with a trapezoidal cross section and a side slope of m = 1.6, if the depth of flow is to be 5.8 m when the flow rate is 550 m3/s. The slope of the channel bottom is So = 0.00113, and its wall roughness is e = 0.0015 m.
159
Energy and Its Dissipation in Open Channels
2.14 Water enters a trapezoidal canal from a reservoir at the uniform depth of 5.2 ft. The canal is to convey 450 cfs over a long distance at a slope of 0.0005, and stability considerations dictate that the side slope should be 2 to 1. If the wall roughness e = 0.012 ft, determine what the bottom width of the canal should be. 2.15 The following data defines the cross section of an earthen canal, in which x represents the distance in feet from the left bank when looking downstream, and y represents the corresponding vertical distance from this bank elevation to the elevation of the canal bottom. x (ft)
0
3
5
7
10
12
14
18
21
25
27
29
y (ft)
0
1.4
2.3
2.8
3.4
3.5
3.3
2.8
1.5
0.9
0.2
0.0
Estimate the flow rate in the canal if its bottom slope is 0.001, and the depth of flow is 3.2 ft. 2.16 Use Manning’s formula to fill in the blank in the table below that applies for uniform flow in channels of trapezoidal shapes. Flow Rate Q (cfs) — 324.5 550.0 400.0 450.0 600.0
Coeff. n
Bot. Slope So
Bot. Width b (ft)
Side Slope m
Depth Y (ft)
0.013 — 0.014 0.013 0.013 0.013
0.0006 0.0008 — 0.0015 0.0009 0.00075
6.0 8.0 10.0 — 7.0 10.0
1.6 0.0 1.3 1.5 — 1.2
4.2 5.0 4.8 4.0 4.1 —
2.17 Fill in the missing blank in the table below using Manning’s formula. These are trapezoidal channel, and the flow in them is uniform. Flow Rate Q (m3/s) — 50.0 60.0 100.0 80.0 20.0
Coeff. n
Bot. Slope So
Bot. Width b (m)
Side Slope m
Depth Y (m)
.014 — .013 .015 .014 .0135
0.0015 0.0012 — 0.0008 0.00075 0.00068
3.0 3.0 2.5 — 6.0 2.0
1.0 1.5 2.0 2.0 — 1.5
2.3 2.2 2.5 3.0 2.9 —
2.18 For each entry in the table for Problem 2.17 compute the value of Chezy’s C coefficient, and from this C determine what the wall roughness e is. 2.19 Write a computer program that is designed to solve Manning’s formula completely for a trapezoidal section. It should display a menu of variables, allow the user to select the unknown, then request values for the knowns, and finally provide the solution to the unknown. 2.20 The table below applies for uniform flow in circular channels. Fill in the mission blanks. Flow Rate Q (cfs) — 300.0 250.0 200.0 100.0
Coeff. n
Bot. Slope So
Diameter D (ft)
Depth Y (ft)
.0142 — .013 .013 .013
0.00055 0.00065 — 0.03000 0.00100
10.0 10.0 5.0 — 8.0
6.8 7.2 3.2 2.5 —
160
Open Channel Flow: Numerical Methods and Computer Applications
2.21 The table below applies for uniform flow in circular channels. Fill in the mission blanks. Flow Rate Q(m3/s) — 20.0 25.0 30.0 60.0
Coeff. n
Bot. Slope So
Diameter D (m)
Depth Y (m)
.015 — .013 .0125 .0145
0.00080 0.00068 — 0.00040 0.00045
3.0 5.0 6.0 — 8.0
2.3 2.52 2.9 3.0 —
2.22 Compute the normal depth in a trapezoidal channel with b = 10 ft, m = 1.5, So = 0.0008 if Q = 400 cfs is flowing in the channel and it has a bottom roughness coefficient for Manning’s equation of nb = 0.05, and a side roughness coefficient ns = 0.013. 2.23 A rectangular laboratory flume has Plexiglas sides (ns = 0.010) and gravel on its bottom (nb = 0.035). The flume is 3 ft wide has a bottom slope of So = 0.001. If the depth of flow is 1.5 ft estimate the flow rate in the flume. 2.24 A 8 ft wide rectangular channel made of concrete (n = 0.013) is filled to a depth of 1 ft with gravel (n = 0.04). The slope of the channel is So = 0.0015. For a flow rate of Q = 90 cfs predict the height of water in the channel with and without the gravel. 2.25 Determine the alternate depth in a rectangular channel to a depth Y2 = 0.2 m if the flow rate is q = 3 m2/s. 2.26 Determine the alternate depth to 1.5 ft in a trapezoidal channel with b = 5 ft, and m = 1.2 if the flow rate is Q = 150 cfs. 2.27 Determine the depth that will occur upstream of gate in a trapezoidal channel with b = 6 ft, and m = 1.2 if the flow rate is Q = 200 cfs, and the channel bottom rises in elevation by 1.5 ft across the gate, and the depth of flow downstream of the gate is Y2 = 2.0 ft. 2.28 Determine the alternate depth to 4 m in a trapezoidal channel with b = 3 m, and m = 1.5 if the flow rate is Q = 30 m3/s. 2.29 The depth of flow downstream from a sluice gate in a rectangular channel is 1.2 ft. If the flow rate per unit width under the gate is q = 20 cfs/ft determine the depth upstream of the gate. 2.30 The depths upstream and downstream from a sluice gate in a rectangular channel are measured to be 5.2 and 1.1 ft, respectively. If the channel is 15 ft wide, determine the flow rate passing under the gate. 2.31 A specific energy diagram has the specific energy E along the abscissa and the depth Y along the ordinate with different curves for constant values of flow rate Q. Generate tables of values, and then plot these to make a depth–discharge diagram for constant specific energies, i.e., graph Q on the abscissa, Y on the ordinate and different curves apply for constant values of the specific. Make this graph specific for a trapezoidal channel with b = 10 ft, m = 1.25, and use three curves for E = 3 ft, E = 4 ft, and E = 5 ft. Prove that the maximum flow rate for any constant E corresponds to the critical depth Yc. 2.32 The depth upstream from a sluice gate in a rectangular channel 4.1 m. The gate is set at a distance 0.5 m above the bottom of the channel, and the contraction coefficient for the gate is 0.58. Determine the flow rate passing the gate per unit width. 2.33 A contraction reduces a rectangular channel from b = 4 m to b = 3 m. The downstream channel has a bottom slope So2 = 0.0015, and n2 = 0.013. For a flow rate Q = 10 m3/s determine the upstream depth and the change in water surface elevation. The bottom elevation of the channel does not change through the contraction. 2.34 In Problem 2.33, the bottom rises 0.2 m. Now determine the upstream depth and change in water surface elevation. 2.35 In Problem 2.33 the bottom fall 0.2 m. Now determine the upstream depth and the change in water surface elevation.
161
Energy and Its Dissipation in Open Channels
2.36 How much must the bottom rise or fall in Problem 2.33 so that critical flow occurs at the smallest section of the contraction, but the water level upstream is the same as in Problem 2.33? 2.37 A transition from a trapezoidal channel with b1 = 10 ft, m1 = 1.5, and So1 = 0.0001 to a circular channel with D2 = 8 ft, n2 = 0.012, and So2 = 0.058. What are the depths at the upstream and downstream ends of this transition if the flow rate Q = 450 cfs? 2.38 Water from a reservoir with a water surface elevation 5 m above the bottom of a 10 m wide and steep rectangular channel enters it unrestricted. Determine the flow rate. 2.39 A gate exists in Problem 2.38 with a contraction coefficient of 0.8 to control the flow rate. If the gate is 0.3 m above the channel bottom what is the flow rate? 2.40 A transition occurs between an upstream trapezoidal channel with a bottom width b1 = 3 m, and a side slope m1 = 1.5 to a rectangular section with a bottom width b2 = 2.5 m. Simultaneously with the reduction of width the bottom drops by Δz = 0.2 m from the trapezoidal to the rectangular sections. If the bottom slope of the trapezoidal channel is So1 = 0.0006 and has a Manning’s roughness coefficient n = 0.013, determine the maximum flow rate that can pass through the transition without increasing the depth upstream therefrom above the uniform depth. What is the maximum flow rate if the bottom remains horizontal across the transition? 2.41 A steep and a mild rectangular channel with a bottom width of b = 10 ft take water from a reservoir whose water surface is 5 ft above the channel bottom. The mild channel has a bottom slope So = 0.00075, and a Manning’s roughness coefficient n = 0.013. Determine the flow rate in both of these channels. Assume the entrance loss coefficient is 0.
H = 5 ft H = 5 ft
Stee p
b = 10
ft
b = 10 ft, n = 0.013, So = 0.00075
2.42 A transition from a trapezoidal channel with b1 = 4 m, m1 = 1 to a rectangular channel with b2 = 3 m occurs. The bottom of the channel remains at the same elevation, and both channels have a Manning’s roughness coefficient, n1 = n2 = 0.014. The upstream channel has a bottom slope So1 = 0.0005, and the downstream channel has a bottom slope So2 = 0.009. If a flow rate of Q = 24 m3/s is occurring what are the depths at the beginning of the transition and at its end? What change in elevation of the bottom of the channel would be necessary so the upstream depth remains at uniform depth? What should the width of the downstream rectangular channel be if uniform flow to exist throughout the upstream channel if the bottom elevation does not change? 2.43 Generate the data needed to plot a dimensionless specific energy graph that has the subcritical dimensionless depth Y1′ = mY1/b as the abscissa and the supercritical depth Y2′ = mY2 / b as the ordinate that applies for trapezoidal channels, and create this graph. The graph should have curves for Q′ = m3Q2/(gb5) the same as Figure 2.4. 2.44 Repeat the graph of the previous problem except make it apply for circular channels. 2.45 Develop a relationship for the diameter D in a circular channel with the bottom width b and the side slope m in a trapezoidal channel so the dimensionless flow rates Q′ are the same in these two sections. Assume that a circular section exists downstream from a gate with a diameter that satisfied the above relationship, and that a trapezoidal section with b and m exists upstream from the gate. Develop the data and plot three graphs that apply for m = 0.5, m = 1.0, and m = 1.5 that provide the relationship between the downstream supercritical depth in the circular section to
162
2.46
2.47 2.48 2.49 2.50
2.51 2.52 2.53 2.54
Open Channel Flow: Numerical Methods and Computer Applications
the upstream subcritical depth in the trapezoidal section, assuming that the above relationship of D to b and m applies so the dimensionless depths are the same in the two sections. Develop a graphical solution that provides the depth downstream from a gate if the depth upstream is known, or provides the depth upstream from the gate if the depth downstream therefrom is known. Have this graphical solution apply for flow rates per unit width of rectangular channel from q = 1 to 10 m2/s. Note that if a log–log plot is used that the curves for the different unit flow rates are not nearly as curved as when the plot uses a linear horizontal and vertical axis. The graphical solution can be developed rather easily using a spreadsheet since the equations that provide the alternative depths are explicit for a rectangular channel. Also write a computer program that provides a table of these solutions for different flow rates, q. (You should also generate the above tables using CHANNEL.) Repeat the previous problem for an upstream trapezoidal channel with a bottom width b1 = 3 m and a side slope m = 1.5 and a rectangular channel at the position of the gate with a bottom width b2 = 2.5 m. Use flow rates of Q = 3 to 25 m3/s. A pipe with a diameter D = 6 m and on a steep slope takes water from a reservoir with a water surface elevation 5 m above the bottom of the pipe. What flow rate might be expected to enter the pipe? A trapezoidal channel with b = 7 ft, and m = 1.5 and a bottom slope of So = 0.08 gets water directly from a reservoir with its water surface 4.5 ft above the channel bottom. What flow rate would you predict will enter this channel? A transition takes a trapezoidal channel with b = 10 ft, and m = 1.5 to a rectangular section with b = 8 ft. Through the transition the bottom of the channel rises 0.2 ft. The slope of the very long upstream trapezoidal channel is So = 0.00015, and its n = 0.014. For a flow rate of Q = 330 cfs determine the following: (a) the depth upstream from the transition, (b) the depth immediately downstream from the transition, (c) the change in water surface through the transition, and (d) the Froude numbers associated with both the upstream flow and that immediate downstream from the transition. If the depth in the rectangular channel is not to change downstream from the transition what should its slope be? Its n = 0.014 also. If its slope is less than this amount what will happen? If the water surface is not to change through a transition from a trapezoidal to a rectangular channel with the dimensions in the previous problem how much must the bottom rise or fall as a tabular function of the upstream depth? Plot this function. What is the maximum rise that can take place in the bottom elevation through the transition in Problem 2.50 for the upstream flow to be possible? Everything is the same as in Problem 2.50 except that the bottom rises by 0.8 ft through the transition. What is the depth upstream? A broad crested weir consists of a rounded rectangular hump placed in the bottom of a channel that is sufficiently high to cause critical flow over its top, and is a common flow measurement device. If the depth Y1 upstream from a broad crested weir, that is 0.6 m high, is 2 m, and the rectangular channel containing the weir is b = 4 m wide, what is the flow rate Q? What is the depth over the crest of the weir?
Yc
Y1 = 2 m
Δz = 0.6 m b=4m
Q=?
Energy and Its Dissipation in Open Channels
163
2.55 For the flow over the broad crested weir of the previous problem plot the function F(q) on the ordinate, and q on the abscissa of a graph and show that two solutions exist, and demonstrate that one of these is associated with an upstream supercritical flow, and the other subcritical. However, since physically the supercritical condition cannot exist that the only physically viable solution is the larger of the two roots. (Note that if the upstream flow were supercritical that a hydraulic jump would form upstream so that the flow is subcritical in front of the weir so that the flow depth decreases as the velocity head increases to pass the reduced flow section.) 2.56 Water enters a trapezoidal channel with b = 8 ft, m = 2.0, and a bottom slope of So = 0.000115 from a reservoir whose water surface is 8 ft above the channel bottom. If the entrance loss coefficient is K L = 0.04 determine the flow rate and depth of flow in this channel. The channel is very long, and n = 0.013. 2.57 Gates control flow in a trapezoidal channel with b = 6 m, m = 1.5, n = 0.014, and So = 0.00025. There are three rectangular gates each 1.5 m wide with contraction coefficients of Cc = 0.8. Two gates are set 0.5 m above the channel bottom and the third gate is set 0.2 m above the channel bottom. The depth upstream from the gates is 4.2 m. Determine the flow rate in the channel. How much is the upstream depth above or below the uniform (normal) depth in this channel? 2.58 The land slopes at a rate of 0.0016 ft per foot of length and you are to design a trapezoidal channel that will carry 400 cfs of water when the depth of water in the reservoir is 5 ft above its bottom. The channel is to have a side slope of m = 1.4, and a Manning’s n = 0.013. What should its bottom width be? What is the Froude number associated with this flow? 2.59 Develop the stage discharge curve for the channel you designed in the previous problem with the reservoir water surface elevation varying from 0 to 5.5 ft above the channel bottom. 2.60 Determine the size of pipe that should be used to convey a flow rate of Q =16 m3/s from a reservoir whose water surface elevation is 3 m above the pipe’s bottom. The pipe is to have a bottom slope of So = 0.00035, and a Manning’s n = 0.014. The entrance loss coefficient K L = 0.1. What will the depth of flow be in this channel? What is the Froude number associated with this flow. 2.61 A 3 m diameter pipe with a steep bottom slope is taking water from a reservoir with a depth of 2.8 m above the pipes bottom. What flow rate would you estimate is entering the pipe, and at what depth? 2.62 You are to size a pipe that will take 30 m3/s from a reservoir whose water surface elevation is 3 m above the pipe’s bottom, and the depth at its entrance in the pipe is not to exceed 2 m. Determine the size of pipe to use. After a very short distance a transition occurs to a trapezoidal section with b = 6 m, and m = 1.2. What is the depth in this trapezoidal section? What must the bottom slope of this trapezoidal channel be if this depth is not to change, and its Manning’s roughness coefficient n = 0.013? 2.63 Water is to be taken from a reservoir with H = 6 ft using a trapezoidal channel with b = 8 m and m = 1.5. After a very short distance the section changes to that of a pipe with a diameter D = 6 m. The entrance loss coefficient including the transition is K L = 0.15. What will the flow rate be if the roughness of the pipe wall is n = 0.013 and its slope is So = 0.0008? What will the depth be at the entrance of the pipe? What will the depth be at the entrance of the trapezoidal channel? 2.64 Repeat the previous problem except the pipe is made of corrugated steel with a roughness of n = 0.035. 2.65 Water enters a trapezoidal channel from a reservoirs whose bottom has a Manning’s roughness coefficient of nb = 0.035, and whose sides have a Manning’s roughness coefficient ns = 0.012. The channels bottom width is b = 5 m, and its side slope m = 1.5 and its bottom slope is So = 0.00085. If the depth of water in the reservoir is 3.2 m above the channel bottom and the entrance loss coefficient is Ke = 0.08 determine the discharge into the channel.
164
2.66 2.67 2.68
2.69
2.70
2.71
2.72 2.73
2.74 2.75
Open Channel Flow: Numerical Methods and Computer Applications
(Solve this problem by using an equivalent neq based on weighting according to the fraction of the perimeter each roughness applies for.) Repeat the previous problem except assume nb = 0.012 and ns = 0.035. Also solve Problem 2.65 assuming that nP2/3 in Manning’s equation should be replace by (nP2/3)b plus (nP2/3)s. Why is the depth obtained by an equivalent Manning’s n different? Develop the delivery diagrams for a trapezoidal channel with b = 3 m, m = 1.4, n = 0.013 for bottom slopes varying from So = 0.001 to So = 0.003. The entrance loss coefficient is Ke = 0.12 and the reservoir head varies from H = 0.2 m to H = 2.2 m. A 10 ft diameter pipe that has a Manning’s n = 0.012 takes water from a reservoir whose head varies between H = 1 ft and H = 10 ft. The entrance loss coefficient is Ke = 0.12. Obtain delivery diagrams for this long channel for bottom slopes So = 0.001, 0.002, 0.0025, 0.0028, and 0.003. For the 12 ft diameter pipe with n = 0.012, that has a trapezoidal entrance section through which water from the reservoir passes before entering the pipe, and that was solved in the text with Program E_UNTC, develop the delivery diagrams for H varying from 1 to 8 ft (b = 10′, m = 1.2, Ke = 0.1, K L = 0.1). Program E_UNTC assumes uniform flow occurs in the downstream circular channel. Modify this program so critical depth may occur in either the upstream entrance (trapezoidal section), or at the beginning of the circular channel. Use this program to solve the problem solved in the text except that the channel is steep. In the text the program E_UNTC is designed to solve flow rate into a circular channel that has a trapezoidal entrance where it receives water from a reservoir. Write a computer program, or develop a computer model, to solve problems in which a trapezoidal channel has a circular section at its entrance to the supply reservoir. Use this program (or model) to solve for Q, Y1 and Y2 from H = 3 m, D = 5 m, Ke = 0.1, b = 3 m, m = 1.4, n = 0.013, So = 0.001, K L = 0.1 and Δz = 0.8 m. For the channel with the transitional entrance of the previous problem develop the delivery diagrams with H varying from 0.25 to 5 m for bottom slopes varying from So = 0.0002 to So = 0.0015. In developing the delivery diagrams in the previous problem you should have computed supercritical depths at the beginning of the trapezoidal section when critical flow occurs in the circular entrance. To have smooth flow from this position on, when this condition of critical flow occurs, the channel must have a bottom slope sufficient to maintain uniform flow at this depth, or a steep slope. Compute these slopes and compare them with Sc needed to separate a mild from s steep channel if the trapezoidal channel receives water directly from the reservoir. For the trapezoidal channel with a circular entrance of Problem 2.71 compute the change in bottom elevation Δz needed so that critical flow occurs in both the circular section and at the beginning of the trapezoidal channel when the reservoir head is H = 2.5 m. Develop an iterative solution for either the subcritical, or supercritical depth corresponding to a given value of the specific energy E in a trapezoidal channel based on being able to solve the cubic equation for the three real roots if the channel is rectangular. The development of this iterative solution can be based on the follow two observations. (1) When the channel is rectangular, the root Y2 (or the root obtained by adding 2π to θ to make the argument for the cosine) is the subcritical depth, and root Y3 (obtained by adding 4π to θ to make the argument of the cosine) is the supercritical depth. (2) A mean flow rate per unit width q in a trapezoidal channel can be define by q = 2Q/(T + b) = Q/(mY + b) = Q/(A/Y). Use this iterative approach to solve both the sub and supercritical depth in a 10 ft wide trapezoidal channel with m = 1 if E = 5 ft, and the flow rate is Q = 400 cfs.
165
Energy and Its Dissipation in Open Channels
2.76 Modify the iterative solution method developed to solve for the depths associated with a given specific energy E in a trapezoidal channel of the previous problem to find the subcritical depth, or the supercritical depth in a circular channel with a known diameter, and a specified value of the specific energy E. 2.77 Water is taken from a reservoir by means of a rectangular inlet channel that is 10 ft wide. A short distance downstream therefrom the channel divides in a trapezoidal section with b2 = 4 ft, m2 = 1.5, n2 = 0.015, and So2 = 0.0008, and a pipe with a diameter D3 = 3 ft, n3 = 0.013, and So3 = 0.0014. The bottom of the pipe is 1.8 ft above the bottom of the rectangular channel, and the trapezoidal and rectangular channel have the same bottom elevation. When the water surface elevation in the reservoir is 4.5 ft above the bottom of the channel determine the depths and flow rates in all three channel (six unknowns). Assume the entrance loss coefficient equals 0.12.
Q2
Δz2 = 0
b1 = 10 ft Q1 H = 4.5 ft KL = 0.12
Δz
4 ft
008 0.0 5 = 1 2 0.0 t, S o 4 f , n 2= = .5 b2 =1 m2
3 = 1.
8 D=
3 Q3 3 ft ,S o3 = 0.00 n3 = 14 0.01 3
2.78 Modify the program THREECH to accommodate any number of channels branching from the upstream main channel, i.e., allow the number of channels to be 4, 5, etc. Also make the following changes: (1) In place of the FUNCTION F make this a subroutine (a void function in C) that supplies all of the equations each time it is called, and (2) Rather than have a built in linear algebra solver when implementing the Newton method, call on a linear algebra subroutine such as SOLVEQ. Use this modified program to solve the following problem. A reservoir with a head H = 5 ft, and an entrance loss coefficient, Ke = 0.1, supplies a main trapezoidal channel with b1 = 10 ft and m = 1.5. A short distance downstream this channel branches into three identical long trapezoidal channels with b = 4, m = 1, n = 0.013 and So = 0.0005. Solve for the flow rates and depths in the four channels. 2.79 For the four channel system of the previous problem develop the delivery diagram for reservoir heads varying from H = 1 ft to H = 8 ft in increments of ΔH = 0.25 ft. 2.80 A trapezoidal channel with b1 = 10 ft, and m1 = 1.5 takes water from a reservoir with a water surface elevation 5.5 ft above its bottom. As short distance downstream therefrom it divides into two rectangular channels with the following properties: b2 = 6 ft, and n2 = 0.014, and b3 = 4 ft, and n3 = 0.013. A short distance downstream in the two rectangular channels there are gates to control the flow rate. The gates are set at yG2 = 1.5 ft and yG3 = 0.8 ft above the bottom of the channel, respectively. The bottom of all three channels is at the same elevation, and the contraction coefficients for the gates are both 0.6. If the entrance loss coefficient equals K L = 0.09. Determine the depths in all three channels upstream from the gates, and the flow rates in each of the three channels.
Open Channel Flow: Numerical Methods and Computer Applications
te Ga
166
H = 5.5 ft
Q1
Q2
ft =6 t., b 2 5 ft c e R 1. y G=
E1, Y1
b1 = 10 ft, m1 = 1.5
Rec
Gat e
KL = 0.09
t., b yG = 3 = 4 ft 0.8 ft
Q3
2.81 Write a computer program, or use an available software program capable of solving system of nonlinear simultaneous equations to develop the depths and discharges that will occur in all three channels of the previous problem with the water surface elevation in the reservoir at 5.5 ft, but with different gate setting in channel two varying from yG2 = 0 ft to wide open. This second rectangular channel has a bottom slope of So2 = 0.0005, and extends downstream for a very long distance. 2.82 Assume the gate in channel 2 of Problem 2.80 is wide open and that this channel has a steep bottom slope. Also its bottom is 1.5 ft above the bottom of the other two channels. What will the depths and flow rates be in all three channels now. 2.83 Two long channels are joined by a smooth transition. The upstream channel has the following properties: Its bottom width is b1 = 10 ft; its side slope is m1 = 2; its Manning’s roughness coefficient is n1 = 0.014, and its bottom slope is, So1 = 0.0002. The downstream channel has the following properties: b2 = 8 ft, m2 = 1.0, n2 = 0.014, and So2 = 0.001. The bottom rises by Δz = 0.5 ft through the transition. The design flow rate is Q = 450 cfs. Determine the depths both immediately upstream and downstream from the transition.
sit
Tr an
b1 = 10 ft, m1 = 2, n1 = 0.014, So1 = 0.0002
ion
Q = 450 cfs ΔZ
b2 = 8 ft, m2 = 1, n2 = 0.014, So2 = 0.001 =0 Very long .5 f t
2.84 Write a computer program capable of solving any of the variables that may be unknown through a transition between two trapezoidal channels. The variables of the problem are: Y1, b1, m1, Δz, Y2, b2, m2, and Q. 2.85 Write a computer program capable of solving any of the variables that may be unknown through a transition between two circular channels. The variables of the problem are: Y1, D1, Δz, Y2, D2, and Q. 2.86 Write a computer program to solve the problem of water entering a trapezoidal channel from a reservoir at uniform flow. This program should be able to solve any of the variables in the following list in addition to determine the flow rate Q: H, b, m, K L, So, or n. (H is the head of water in reservoir above the channel bottom.) 2.87 Write a computer program to solve the problem of water entering a circular channel from a reservoir at uniform flow. This program should be able to solve any of the variables in the following list in addition to determine the flow rate Q: H, D, K L, So, or n. 2.88 A trapezoidal channel is to be designed to carry a flow rate Q = 400 cfs. The single channel divides into two trapezoidal channels with the first having a bottom width b2 = 6 ft, a side slope m2 = 1.5, a Manning’s n2 = 0.013, and a bottom slope So2 = 0.0005, and the second having a bottom width b3 = 8 ft, a side slope m3 = 1.2, a Manning’s n3 = 0.013 and a bottom
167
Energy and Its Dissipation in Open Channels
slope So3 = 0.0006. The bottom of the channel with a bottom width of 8 ft is 0.5 ft above the other divided channel. The upstream single channel is to have a bottom slope of So1 = 0.0008, m1 = 2.0, and a Manning’s n1 = 0.014. Determine the depths and flow rates in the downstream divided channels, and the bottom width and depth in the upstream channel so that uniform flow will occur in it when the above design flow rate occurs.
Solution: Q2 = 202.61 cfs, Q3 = 197.39 cfs, Y2 = 3.87 ft, Y3 = 3.28 ft, b1 = 12.13 ft, and Y1 = 3.69 ft.
ft, m 2 b 2= 6
Q = 400 cfs b 3= m1 = 2.0, n1 = 0.014, So1 = 0.0008 8 Δz = 0.5΄ ft, m
3
=1
.2,
n
3
=0
.01
3,
S
o3
= 1.5,
=0
n2
3, = 0.01
S o2 =
5
0.000
.00
06
2.89 Four channels branch from the upstream main channel as shown in the sketch below. The sizes, etc. of the channel are as shown on the sketch, including the proposed width of 16 ft for the upstream main channel. A gate exists a short distance downstream in channel 4, with a setting that produces a depth downstream from it of 0.6 ft. Also notice that the bottom of channel 2 is 0.3 ft above the bottoms of the other channels. Is it possible to have this upstream channel 16 ft and not restrict the flow that the other channels can carry under the conditions given? Solve for the flow rates and depths in all of the channels if the upstream trapezoidal channel has a bottom width b1 = 20 ft, and a side slope m1 = 2. 1
.00
Δz
=
.01
0.3
= , m2
5 b 2=
Gate
H = 5 ft
b1 = 16, m1 = 2, n = 0.013 1
=0 2 3, S o
b3 =
=0 2 1, n
b3 = 6, m3 = 1.5, n3 = 0.013, So3 = Yd4 = 0.6
ft
b4 = 8, m 6, m
5 = 1,
4 = 0,
n5 =
0.01
3, S
n4 = 0.0
o5 = 0
13
.000
5
0.0007
168
Open Channel Flow: Numerical Methods and Computer Applications
2.90 Determine the minimum width b1 that the upstream channel of the previous channel system must have so that critical flow will not occur at its entrance, thus limiting the flow rate that the downstream channels can carry. You might use one of the models you used to solve the previous problem in which you attempt to successively reduct the width from 20 ft until it fails to produce a solution to get an idea how the upstream depth decrease with decreasing b1 until critical depth occurs. Why are the flow rates and depths of the channels downstream from the junction not affected by changing widths of the upstream channel until critical conditions occur? 2.91 Fix the upstream width of channel #1 in the system of the previous two problems to b1 = 19 ft, and obtain a series of solutions in which the reservoir water surface H increases from 5 ft until a solution no longer exists. What caused this situation with rising reservoir heads? 2.92 Enlarge the downstream steep rectangular channel in illustrative Example Problem 2.9 from b2 = 8 ft to b2 = 10 ft, and repeat this problem including solving values of Q and Y for a number of different bottom slopes for the upstream channel. 2.93 A trapezoidal channel with b1 = 8 ft, and m1 = 1.5 is laid on a bottom slope of So1 = 0.0005, and has a Manning’s roughness coefficient, n1 = 0.013. This channel conveys the water into a smooth transition to a circular channel with a diameter D2 = 18.2 ft. The circular channel has a steep slope. Determine the maximum flow rate that can exist, and not increase the depth in the upstream channel above its normal depth, through the transition if the bottom elevation does not change between the trapezoidal and circular channels. Determine what the relationship is between the maximum flow rate possible and the slope of the upstream channel by solving the problem for several different values of So1. 2.94 Determine the maximum flow rate that can be accommodated in the channel shown below without causing the depth in the upstream channel to rise above its normal depth. The upstream channel has a bottom width b1 = 8 ft, a side slope m1 = 2, a Manning’s roughness coefficient, n1 = 0.014, and a bottom slope, So1 = 0.00025. The downstream channel is steep, i.e., under uniform flow conditions the depth will be less than critical depth, has a bottom width b2 = 7 ft, and a side slope m2 = 1. Solve the problem with the upstream slope changing and plot the maximum flow rate, and the depths upstream and at the head of the steep channel versus So1. Can you explain these trends? (Increase So to about 0.001.) K Q
L =0
Y1
.08
Y2
b1 = 8 ft, m1 = 2, n1 = 0.14, So1 = 0.0004 (varying) b 2 = 7 ft,
m2 = 1 (s
leep)
2.95 A transition between an upstream mild channel and a downstream steep channel is smooth with no change in the bottom position. The upstream channel has b1 = 12 ft, m1 = 2, n1 = 0.014, and a bottom slope So1 = 0.0002. The downstream steep channel has a bottom width b2 = 8 ft and a side slope m2 = 1. Determine the maximum flow rate possible in this channel if the flow in the upstream channel is to be uniform. What is the depth of this upstream uniform flow, and what is the depth at the beginning of the steep channel? 2.96 In the previous problem obtain the solutions for the same unknowns for the downstream channel having a side slope of m2 = 0.5 and m2 = 0, respectively. The upstream mild channel has the same size, slope, etc. as in the previous problem and the bottom width of the steep downstream channel is b2 = 8 ft. 2.97 Solve Problem 2.95 for several width of downstream rectangular channel with bottom widths varying from b2 = 6 ft to b2 = 12 ft (m2 = 0). Note from these solution how significant the choking effect is as the size of the downstream steep channel is reduced in size. 2.98 Water is taken from a reservoir with a water surface elevation 5 ft above a 12 ft wide, trapezoidal channel with a side slope m1 = 1.5. A short distance downstream from the channel
169
Energy and Its Dissipation in Open Channels
entrance it divides into two trapezoidal channels with b2 = 8 ft, m2 = 1.0, and b3 = 4 ft, and m3 = 2. The second channel has a gate in it a short distance downstream from the branch that causes the depth of flow downstream from it to be at Y22 = 0.8 ft. The third channel is long, has a Manning’s n3 = 0.015, and a bottom slope So3 = 0.0005. The bottom rises by Z13 = 0.3 ft between the upstream channel and the third channel. Just upstream from the junction a 12 in. diameter pipe takes water from near the bottom of the channel. This pipe is 2000 ft long, has an equivalent sand roughness e = 0.012 in., and delivers water at it end with a head He = 0 ft. Determine the flow rates in all channels and the pipe, as well as the depths in the channels. Solution: Q1 = 367.6 cfs, Q2 = 114.9 cfs, Q3 = 250.6 cfs, Q4 = 2.11 cfs, Y1 = 4.683, Y2 = 4.885, Y3 = 4.639.
b1 = 12 ft, m1 = 1.5
= 3
m ft ,
.8 2 =0
Q1
Gat e
H= 5 ft
b
3
=
z1
t
4
.3 f
=0 3
2. 0,
S
o3
=
0. 00 05
Y2
D= 12 ft
b2 = 8 ft ,m He = 0 ft 2 =1 .0 L = 2000 ft e = 0.012 ft
2.99 Example Problem 2.18 obtains a series of solution for a four channel branched system in which the height of the gate in channel 3 varies from 1.7 ft to being closed. Solve this same channel system except the upstream channel # 1 has a width of 12 ft rather than 15 ft. The gate’s contraction coefficient is Cc = 0.6. Obtain this series of solutions with the gate’s position starting at 0.1 ft. 2.100 A reservoir with a water surface elevation 4.8 ft above the bottom of a trapezoidal channel with b1 = 20 ft, m1 = 1, n1 = 0.014, supplies water to a branched channel system as shown, in which this upstream channel divides into three channels; channels 2 and 3 consisting of pipes with D2 = 5 ft, n2 = 0.013, So2 = 0.0006, and D3 = 6 ft, n2 = 0.013, So3 = 0.0005. Channel 4 is trapezoidal with b4 = 8 ft, m4 = 1, n4 = 0.015. This channel contains a gate that produces a depth of 1.7 ft downstream from it. The entrance loss coefficient is Ke = 0.15, and the loss coefficients to the 3 branched channels are all 0.1. The bottoms of the pipes for channels 2 and 3 are Δz12 = Δz13 = 0.2 ft above the bottom of channel 1. Solve for the flow rates and depths in these four channels.
(2)
Δz12 = 0.2 ft
D 2=
(1)
Δz 13 = 0.2 KL = 0.1
Ke = 0.15 b1 = 20 ft, m1 = 1, n1 = 0.014
06
0.00
D3 = 6 ft, n2 = 0.013 , So4 = 0.0006
(4) b4 = 8
n 2=
S o4=
(3)
ft
ft , m
4 = 1,
Gate
H = 4.8 ft
5 ft ,
3, 0.01
n4 = 0
Yd4 =
.015,
1.7 ft
So4 =
0.001
2
170
Open Channel Flow: Numerical Methods and Computer Applications
2.101 Example Problem 2.19 gave the answers that were obtained by solving the problem using, but did not actually provide the sheets of this solution. Using TKSolver, Mathcad, or other software capable of solving a system of equations verify the given answers. Obtain these answers: (1) by using the seven equations given in this example problem (you may add areas and wetted perimeters to the list of unknown variables if you want) but specify the value of Q1 = 531.4, and (2) Also include Q1 as an unknown and include the additional equations needed. 2.102 Water enters a rectangular channel with a 20 ft bottom width from a reservoir whose water surface is 9 ft above the channel bottom. The channel has a Manning’s roughness coefficient n = 0.014. (a) If the slope of the channel bottom is So = 0.055 what is the flow rate into the channel? (b) If the slope of the channel bottom is So = 0.0005 what is the flow rate into the channel? (c) A vertical gate is placed a short distance downstream from the channel entrance with its tip 2 ft above the channel bottom. The gates contraction coefficient is Cc = 0.6. Now what is the flow rate? 2.103 A trapezoidal channel with b1 = 5 m, m1 = 1.2, and So1 = 0.0008 changes to a rectangular channel with a bottom width of b2 = 4.5 m and a bottom slope So2 = 0.015. Both channels have n = 0.014 and the bottom elevation remains constant through the transition. For a flow rate Q = 50 m3/s what will the depths be immediately upstream and at the end of the transition? What is the change in water surface through the transition? As a (b) part to this problem, what will the depths be if the bottom drops by 1 m through the transition, i.e., Δz = −1. Q = 50
b1 = 5 m , m
1 = 1.2, So1 = 0.00 08
b2 = 4.5
m 3/s
m, S
o2 = 0.01
5
2.104 Give the equations that need to be solved to determine the flow rate in each of the four channels shown in the sketch below. The upstream main channel is short and is supplied by a reservoir with a water surface elevation of H = 5 ft. The bottom elevation of channel 2 is 0.3 ft above the bottom of the main channel and the other channel’s bottoms are at the same elevation. Channel 4 contains a sluice gate with a contraction coefficient, Cc = 0.6, and its gate’s position above the channel bottom is YG = 1.0 ft. Ignore all minor losses YG =
1 ft
Q4
(C
c =0
.6) Y d4
Y u4
1 b 4=
Q1
H= 5 ft (1)
Y1
b1 = 15 ft, m1 = 2
Δz
b2 =
0.3
6 ft
ft
,m
=0
4
Q3
Y3
(3)
12 =
,m 0 ft
b3 = 8 ft, m3 = 1.5, n3 = 0. 013, So3 = 0.0007
(2)
2 =1
Y
2
,n
2 =0
.01
3, S
o2 =
0.0
01
Q2
171
Energy and Its Dissipation in Open Channels
Q1 = 10 m3/s
Y1= ? YG
Pier
Gate #3
Pier Gate #2
Gate #1
2.105 Write a computer program, or use a software package such as Mathcad, or TKSOLVER to obtain the solution to Example Problem 2.19. 2.106 Solve the same problem as given as Example Problem 2.19 with the exception that the bottom slopes of the three channels downstream from the branch are: So2 = So3 = So4 = 0.0005. 2.107 Solve the same problem as given as Example Problem 2.19 with the exception that the bottom slopes of the 3 channels downstream from the branch are: So2 = So3 = So4 = 0.0003. What are the flow rates and depths in the four channels when So2 = So3 = So4 = 0.0002. 2.108 The program BRANCHCH.FOR will solve problems in which the critical depth at the entrance of the channel from the reservoir restricts the flow rate. However, to obtain such a solution it is necessary that you previously determine what this restrictive flow rate is. Modify the program so that if it fail to converge, or if you specify that critical flow governs that it will solve the problem without the necessity of giving what this restricting flow rate is. 2.109 At the location of sluice gates in a trapezoidal channel with b = 5 m and m = 1.5, n = 0.013 and So = 0.0005, the bottom of the channel rises by 0.3 m and the crosssection changes to a rectangle that contains 3 gates each 1 m wide as shown in the sketch below. The two outside gates are set 0.5 m above the channel bottom and the center gate is set 0.7 m above the channel bottom. All three gates have contraction coefficients equal to Cc = 0.6, and a pier 0.5 m wide exists between the gates to support them. If the flow rate in the upstream channel is Q1 = 10 m3/s, write the system of equations, and then solve them, that give the upstream depth, Y1, and the flow rates past each of the gates.
.6
=0 Cc
0.3 m
YG2 = 0.7 m YG1 = 0.5 m Y 3 = 0.5 m G
2.110 Write a computer program that is capable of completely solving the critical flow equation for a trapezoidal channel and that uses the dimensionless form of this equation in the computations. By completely solving the equation is meant that any of the variables Yc, Ec, Qc, b, or m may be unknown with the other variables known. Also accomplish this with a software package such as TKSolver or Mathcad. 2.111 Write a computer program that is capable of completely solving the critical flow equation for a circular channel and that uses the dimensionless form of this equation in the computations. By completely solving the equation is meant that nay of the variables Yc, Ec, Qc, or D may be unknown with the other variables known. Also accomplish this with a software package such as TKSolver or Mathcad. 2.112 Write a computer program that provides a table of values relating the depth upstream from a sluice gate to the depth downstream therefrom. Write this program so it only needs to evaluate explicit equations. Execute the program for a flow rate Q = 40 m3/s in a 4 m wide channel with the downstream depth varying from Y2 = 0.2 m to Y2 = 1.8 m in increments of 0.05 m and plot the results. Note if this were done for several flow rates, q, the graph could be used to solve the specific energy across the gate. Verify the results from your program with CHANNEL. Also use a spreadsheet, TKSolver, Mathcad or similar software to solve the problem, and plot the results.
172
Open Channel Flow: Numerical Methods and Computer Applications
2.113 An upstream channel receives water from a reservoir with a head of H = 5 ft, and shortly downstream therefrom divides into three branch channel, with the following sizes: Channel (2) is rectangular with a bottom width of 8 ft, and it contains a gate a short distance downstream from its beginning that cause the depth of flow downstream therefrom to be Yd2 = 1.0 ft; Channel (3) is trapezoidal with b3 = 8 ft, a side slope m3 = 1, and a bottom slope of So3 = 0.001; Channel (4) is circular with a diameter D4 = 8 ft, and has a bottom slope So4 = 0.0008. The bottom drops by Δz12 = 0.2 ft between channels (1) and (2) and also by Δz14 = −0.2 ft between channels (1) and (4). All Manning’s roughness coefficients are n = 0.013, and the entrance loss to the reservoir is Ke = 0.2. If the upstream main channel has a side slope m1 = 1.5, and a bottom width b1 = 16 ft, what are the flow rates in the four channels and the depths? If the width of the upstream channel is reduced to b1 = 14 ft what are the flow rates and the depths? What is the critical flow rate, i.e., the maximum flow rate, for channel (1)? What will occur if the width of channel (1) is reduced to 14 ft?
g
Lon
Δz 14 =
H = 5 ft
(4) =8 D4
b1 = 16 ft, m1 = 1.5
ft ,
Δz1 = – 2 0.2 ft
all n = 0.013
(2)
08
.00
=0
S o4
Long
(3)
(1)
b3 = 8 ft, m3 = 1, So3 = 0.001 Gat e
Ke = 0.2
–0.2 ft
Yd
2 =1
b2 =
.0 f
8 ft
t
rec t
.
2.114 The plan view of the inlet portion of a channel system is shown in the sketch below. It consist of two parallel channels (1) and (2) that receive water from a reservoir with a water surface elevation H = 6 ft above the bottom of these channels. Channels (1) has a bottom width b1 = 8 ft, and a side slope m1 = 1.5, and channel (2) has b2 = 10 ft, and m2 = 1.2. A short distance downstream these two channels join to form channel (3), but shortly thereafter branch into three separate channels denoted by (4), (5), and (6) on the sketch. A gate controls the flow in channel (4) and is set so it produces a depth Yd4 = 1.2 ft immediately downstream from the gate. Channel (4) is rectangular with b4 = 8 ft. Channel (5) is a pipe with a diameter D5 = 8 ft, and its bottom is 1 ft below the other channels. Channel (5) has a bottom slope So5 = 0.00095, and a Manning’s n5 = 0.012. Channel (6) has b6 = 10 ft, m6 = 1.5, So6 = 0.0006, and n6 = 0.013. The minor loss coefficients are Ke = 0.05 (entrance from reservoirs to both channels (1) and (2)), K13 = K23 =0.08, K34 = K36 = 0.1, and K35 = 0.15, in which the double subscripts denote from which channel to what channel the flow occurs. Do the following: (a) Determine the variables that are unknown and list them, (b) Write out the system of equations that will provide the solution to these unknowns, (c) Solve this system of equations. If instead of dropping 1 ft to the bottom of channel (5) its bottom were 1 ft above the other channels what would the flow rates and depths in the six channels be?
173
36
0.08
(6)
K
K 13=
b1 = 8 ft, m1 = 1.5
=0
.1
Energy and Its Dissipation in Open Channels
(1) H = 6 ft
(3)
b6
ft
.05
K
K2
=
b3
25
ft ,
m
34 =
.5
=1
3
0.1
,
.08
=0 3
095
D5 = 8 ft, n5 = 0.012, So5 = 0.00 Kg ate = 0 (4) b4 =
K
b2 = 10 ft, m2 = 1.2
0.0 n 6= 1.5, = 6 ft, m K = 0.15 35 = 10
6
000
= 0.
(5)
35 Δz
(2)
e =0
1 =–
S o6 13,
8 ft
,m
4 =0
Yd = 4 1. 2 ft
2.115 The sketch below shows two channels that take water from separate reservoirs, and combine this flow into a single channel a short distance downstream. Using the sizes of trapezoidal channels shown on the sketch below solve for the flow rates and depth in the three channels. (The loss coefficient at both entrances, as well as between the channels is K = 0.05.) Q1
H1 = 2 m
Q3
Y1
b1 = 3 m, m1 = 1.5
Y3
Q2
H
=2 2
m
Y2
.5 =3
,m
m
.0 =1
b3 = 6 m, m3 = 1.5, n3 = 0 .014, So3 = 0.0006
1
b2
Q1
H1 = 2 m
Y1
Y3
b1 = 3 m, m1 = 1.5
Q2
m =2
H2
.5 =3
Y2 ,m
m
.0
=1 2
Gate
2.116 The same channel system as in the previous problem, except that the depth of the water surface in reservoir 2 is H2 = 2.04 m, or 0.04 m above that in reservoir 1. How much has this effected the flow. Why do you think the flow from the reservoir into channel 2 approaches critical conditions so rapidly as H2 increases? What is the limiting value of H2 for critical conditions to occur. What would happen if this water surface rose to 2.1 m? 2.117 The same channel system as in Problem 2.115, except that a gate is placed in channel 3 that causes a depth of Y3d downstream from it equal to 0.7 m, solve for the depths and flow rates in all channels. Q3 Y3d = 0.7 m
b3 = 6 m, m3 = 1.5
b2
2.118 A system of six channels are involved in branches as shown in the sketch below. Assume all minor loss coefficients between an upstream channel and its branch downstream, as well as
174
Open Channel Flow: Numerical Methods and Computer Applications
the entrance loss coefficient equal 0.05, i.e., Ke = Kij = 0.05. Channel (3), which contains a gate that produces a depth of Yd3 = 0.7 m downstream therefrom, is rectangular with a bottom width of b3 = 3 m. The sizes and properties of the other channels, which are trapezoidal, are: b1 = 5 m, m1 = 1.5, b2 = 2 m, m2 = 1, n2 = 0.013, So2 = 0.0008, b4 = 2 m, m4 = 1, b5 = 1.5 m, m5 = 1, n5 = 0.013, So5 = 0.0005, and b6 = 1 m, m6 = 1, n6 = 0.013, So6 = 0.001. The bottoms of all channels are at the same elevation at the junctions. Write out the system of equations whose solution will provide the depths and flow rates in all six channels, and obtain a solution to these equations. g
Lon
H = 2 .5 m
All minor losses between channels Kij = 0.05
Gate
(1)
Ke = 0.05 b1 = 5 m, m1 = 1.5
,
013
= 0. 1, n 2 8 = , m 2 000 2 m 2= 0. So b 2= Yd3 = 0.7 m (3)
(2)
b3 = 3 m
(4) 4= 2m ,m
b
4
=1 =1 b6
,m
m
b5
.5
m,
m
5
Lo = 1 ng ,n 5= 0
.01
3,
So
5
=0
.00
05
1
.00
=0 ng , S o6 3 .01
Lo
=0 ,n6
=1
6
(6)
(5 ) =1
2.119 Solve for the flow rates and depths in the various channels shown in the sketch below if the head of water from both of the supply reservoirs equals 5 ft, i.e., H1 = 5 ft and H2 = 5 ft. The flow in channel 6 is controlled by a vertical gate, so that the depth downstream from it is Yd6 = 0.6 ft, and Channels 3 and 4 are very long both with bottom slopes of So3 = So4 = 0.0008, and Manning’s roughness coefficients n3 = n4 = 0.014 so that uniform flow exist in these channels. The entrance loss coefficients are Ke = 0.05, and the loss coefficients from each upstream to downstream channel is Kj = 0.04. There is no loss across the gate.
b 2=
H1 = 5΄
b1 = 10 ft, m1 = 1.5 (1)
Ke = 0.05
.5
Q4
.04 K =0 j
(3)
Q2
Y3
b3 = 8 ft, m
3=1
Q3 S = 0. o3 0008, n
3 = 0.014
= H2
5΄
=0 4 t, m f Y4 =4 b4 (4)
t
Y2 (2)
Q1
Y1
10 f
So5 = 0.0008, n5 = 0.014 Q5 4 ft 5) b 5= Y5 ( Gate Y6 Q6 (6) Kj = 0.04 b6 = 5 ft
Ke = 0.05
2.120 In the previous problem solve for the flow rates and depths if the height of the water surface of the reservoir that supplies channel 4 increases a small amount to H2 = 5.05 ft. Solve for the flow rates and depths if this reservoir water surface fall a small amount to 4.98 ft. How
175
Energy and Its Dissipation in Open Channels
do you explain why the depth in channel 4 decreases as the reservoir water surface elevation rises, and vice versus? What is the minimum H2 possible to have the flow in channel 4 not reverse itself and flow into reservoir 2? What is the maximum head H2 of reservoir 2 to have the condition in the other channels as specified in the previous problem? What is the minimum H2 possible to have the other conditions as specified in the previous problem with the flow reversing itself and flowing into reservoir 2 from channel 4? 2.121 The same channel system as in Problem 2.119 except that a gate is placed in channel 3 that causes a depth of Y3d downstream from it equal to 0.7 m. Solve for the depths and flow rates in all channels.
Y5
0 ft
b 2= 1
(1)
=5 H2
ft
=4 b4
(3)
.5
=0 m4 ft ,
Y4 (4)
Kj = 0.04
Q6
Kj = 0.04
Y3
4
Ke = 0.05
Q1
Y1
4 ft b 5= Gate
(5) Y6
Q2
Y2 (2)
Q
H1 = 5 ft
b1 = 10 ft, m1 = 1.5
So5 = 0.0008, n5 = 0.014
Q5
(6)
b6 = 5 ft
Q3
b3 = 8 ft, m
3=1
Ke = 0.05
2.122 Water is taken from a reservoir whose water surface is 6 ft above a large channel bottom that branches into several channels as shown in the plan view of the system sketched below. The properties of the channels are given in the table below. The distances between branches are small enough that losses can be ignored, except that the entrance loss from the reservoir which has a loss coefficient Ke = 0.2. A gate exists downstream in channel 5 that produces a depth of Yd = 0.8 ft downstream from it. An 8 in. diameter pipe also takes water from channel 1 near where it branches and delivers the water at 40 psi at its end where the elevation is 300 ft. The pipe is 2000 ft long and has an equivalent sand roughness for use in the Darcy– Weisbach, Colebrook–White equations of e = 0.002 in. The elevation of the bottoms of all the channels where they branch is 400 ft. Lo
H = 6 ft
All n = 0.014 Elev = 400 ft
(2)
(1)
(3)
γ = 1.41 × 10–5 ft2/s
ow ipe fl Full p 02 f t e = 0.0 000΄
8΄–2
Ke = 0.2
(5)
(6)
Elev = 300 ft
p = 40 psi
ng
Lo
ng
Long
(4) Y
d5 =
0.8
ft
176
Open Channel Flow: Numerical Methods and Computer Applications Channel 1 2 3 4 5 6
b (ft)
m
So
20 6 8 3 3 5
1.8 1.0 0.0 0.0 0.0 1.0
0.001 0.001 0.0005 0.0005 0.0007 0.0007
Qout = 50 c fs
2.123 In the previous six channel pipe system, it is desired that channel 4 delivers 100 cfs. What size should it be? In addition the pipe size is to be increased to 2 ft diameter (same e, etc.) and the gate in channel 5 is raised so the depth downstream from it is 1.1 ft. Now what are the flow rates and depths in the channels (and the width need for channel 4)? 2.124 A trapezoidal channel with b1 = 10 ft and m1 = 1 receives water from a reservoirs whose water surface is H = 5 ft above the channel bottom. A short distance downstream from the entrance the channel branches into channels 2, 3, and 4. Channels 2 and 4 have gates a short distance downstream that cause downstream depths of Yd2 = 1.5 ft and Yd4 = 1.2 ft, respectively. These channels are rectangular with b2 = 4 ft, and b4 = 3 ft, respectively. Channel 3 consists of a pipe with a diameter D3 = 8 ft, a Manning’s n3 = 0.014, and a bottom slope So3 = 0.0008. There is a diversion at the junction of the channels of Qout = 50 cfs. At the branch all channel have the same elevation. Write out the equations whose solution will provide the flow rates and depths in all channels.
H = 5 ft
(2)
.5 f
=1 Y d2
t
D3 = 8 ft, n3 = 0.014, So3 = 0.0008
(1)
b1 = 10 ft, m1 = 1
4 ft
te Ga
b 2=
(3)
(4) Gate
b4 = 3 ft Yd = 4 1.2 ft
2.125 For the branched channel system of the previous problem make up a table that gives the discharges and depths in the four channels as a function of the gate setting YG2 in channel 2. The contraction coefficient for this gate is Cc = 0.6, and the channel downstream from the gate is rectangular with b2 = 4 ft, n2 = 0.014, and So2 = 0.0009. 2.126 For the branched channel system of Problem 2.124 make up a table that gives the discharges and depths in the four channels as a function of the gate setting YG4 in channel 4. The contraction coefficient for this gate is Cc = 0.6, and the channel downstream from the gate is rectangular with b4 = 3 ft, n4 = 0.014, and So4 = 0.0012. 2.127 Repeat Problem 2.124 with all four channel with the same sizes and gate settings, etc. but the bottom of channel 3 (the pipe) at its beginning is 1 ft below the bottom of the other channels at the junction position, i.e., Δz13 = −1.0 ft. 2.128 In the branched channel system of the previous problems, what is the maximum flow rate that can be obtained through channel 3 if its diameter were to be increased? What are the flow rates in the other channels when this condition occurs and the gates are both completely
177
Energy and Its Dissipation in Open Channels
2.129 2.130
2.131 2.132
open? What is the minimum diameter that the pipe can have for this maximum flow rate to occur? Solve the flow rates and depths for both Δz13 = 0 and Δz13 = −1.0 ft. Modify the model used to obtain the dimensionless graph of Example Problem 2.23 so that it can solve the specific energy between section 1 and 2 (with the filled bottom) of a pipe for other variables without the flow being critical as section 2. Develop a model that is capable of solving problems involving the specific energy between section 1 and 2 in a pipe such as in Example Problem 2.23 in which the bottom of the pipe is filled to a depth Δz, but do this so that the actual, rather than the dimensionless, variables are solved. Make a graph similar to that given in Example Problem 2.23 except relate the dimensionless flow rate to the Froude number. A complex branching channel system takes water from a reservoir with a water surface elevation of H = 2 m above the channel bottom as shown. The sizes of the channels are shown on the sketch below. All channels are rectangular except number 3, which is trapezoidal. Channel 2 contains two gates each 0.6 m wide that both have contraction coefficients Cc = 0.6. The first gate is set so its tip is 0.2 m above the channel bottom and the other gate is set 0.15 m above the channel bottom. Channels 3 and 4 are long with Manning’s roughness coefficients, and bottom slope as shown on the sketch. Identify what unknowns you would solve for and then give the equations that need to be solved. Then solve these equations.
=
Y 2a
0.2
Q 2a
m
.6 =0
b 2a
H=2 m
Q2
.6
=0
b 2b
Q 2b
m
= b2
m Y2b = 0.2 m
2m
Q1 Q3
Long
b3 = 1.5 m, m3 = 1, So3 = 0.0005, n3 = 0.014 Ke = 0.05
b1 = 4 m
b4 =
1.5 m
,m
4 = 0,
So = 4 0. 000
4, n
4 = 0.
014
Q4
Long
2.133 The same channel exists as in the previous problem except now channel 4 contains a pier over a length of it that divides the flow into two parts Q5 and Q6. Set up and solve the equations that now govern the problem. (Note: Under the assumptions that we do not need to solve the gradually varied flow in the channel upstream and near the junctions, and that the pier’s length is short so that the energy at it upstream and downstream ends are the same we end up with one less equation than unknowns unless some condition is established across the pier. Since under our assumptions one would expect the flow in the two channels through the pier length to be close to the ratio of the channel widths, this might be used as the condition.)
178
Open Channel Flow: Numerical Methods and Computer Applications
= Y 2a
0.2
Q 2a
m
.6 =0
b 2a
H=2m
Q2
.6
=0
b 2b
Q 2b
m
= b2
m
2m
Y2b = 0.2 m
Q1 Q3
Long
b3 = 1.5 m, m3 = 1, So3 = 0.0005, n3 = 0.014 Ke = 0.05
b1 = 4 m
b4 =
1.5 m
,m
4 = 0,
So = 4 0. 0004 ,n
4 = 0.
014
Q4
Long
2.134 Parshall and Cutthroat flumes that are widely used to measure flow rates in open channels and ditch make use of the energy principle, and if free flow conditions exist critical depth occurs close to the throat. A Cutthroat flume always has its throat width, W, 2 ft less than its upstream and downstream width, and its bottom is flat. The upstream stilling well is 2.11 ft upstream from the throat whereas the upstream convergence sides are 3.16 ft long. (The length of the flume downstream from the throat is 3 ft, and flume beginning is 2 ft upstream from the throat, so the diverging portion is 1 on 6 and the upstream converging portion of the flume converges 1 on 3.) The laboratory calibration equation for Cutthroat flumes is Q (cfs) = Cha1.56 (ha in ft) in which C = 3.50 W1.025 (with the width W in ft). Use this free flow equation for a Cutthroat flume to locate the position where critical flow will occur in a 6 ft wide Cutthroat flume for a range of flow rates. Use one dimensional hydraulics in this determination, and assume there is no energy loss between the upstream stilling well position where depth ha is measured and the critical flow section. 2.135 Repeat the previous problem except for a 2 ft wide Cutthroat flume. 2.136 A 6 ft wide Cutthroat flume is installed in a trapezoidal channel with a bottom width of 12 ft, and a side slope of m = 2. The slope of the downstream channel bottom is So = 0.00287. When the flume was first installed the downstream channel consisted of a large sand grain materials that had a Manning’s roughness coefficient n = 0.018. However because this material was transported downstream the downstream channel was lined to a depth of 6 in. around it entire cross section with riprap rock with a Manning’s n = 0.035. Assuming the downstream channel is very long so it will flow at uniform depth, and that the local loss from the downstream stilling well into the wider trapezoidal channel equals the difference in velocity heads between these two section, or that the normal depth in the downstream channel equals the depth in stilling well hb determine the effect of the riprap in changing the flow from free flow to submerged flow through the Cutthroat flume. The transition submergence ratio, St = hb/ha equals 0.88 for a 6 ft wide Cutthroat flume.
179
Energy and Its Dissipation in Open Channels Stilling well for hb
6.08 ft
W + 2 ft
5 ft
2.11
T
Q
1
T
ft
W
3 3.16
6 ft
W + 2 ft
Stilling well for ha
1
Plan
(c) (b)
Q ha
hb
(a)
2.137 Example Problem 2.23 examined how the ratio of step height in the bottom of a circular channel divided by the upstream depth that produced critical flow varied as a function of the Froude number. Derive the dimensionless equations, whose solution provided this relationship. 2.138 A circular channel with a diameter D = 3 m, with a Manning’s n = 0.014 is laid on a slope of So = 0.0008. If the flow rate being carried by this channel is Q = 10 m3/s, what is the maximum height of hump that can fill the bottom of the channel (but the diameter at this section is still the same, i.e., D2 = 3 m) if the upstream depth is not to be effected. How does this height compare with a reduction in radius that produces critical flow? 2.139 A flow rate Q = 400 cfs is taking place in a circular channel with a diameter D = 15 ft. The bottom slope of the channel is So = 0.0005, and its Manning’s roughness is n = 0.013. Concrete has been placed in the bottom of this channel to a depth of 1.5 ft at a section, so that its bottom is flat here. The filled in portion of the channel is 1200 ft long and then it discharges into a reservoir whose water surface is 7.0 ft above the channel bottom. Determine the depths immediately upstream and downstream from where the filled in bottom occurs, and the GVF profiles upstream from this position and to the downstream reservoir. 2.140 Write a program, or a computer model, that will generate tables of values to plot the graphs in Figure 2.13, i.e., make tables for different side slopes m1 that solve Equation 2.41 (or Equation 2.40). It will be instructive for you to use the Laguerre method described in Chapter 3 and extract all roots from this fifth degree polynomial, at least for smaller values of m1. 2.141 An upstream trapezoidal channel has a bottom width b1 = 3 m, and a side slope m1 = 0.75, and transitions to a rectangular channel with b2 = 3.5 m. What flow rate will result in critical depths in both the trapezoidal and rectangular channels? What are these depths and their corresponding specific energies? (Solve the problem using dimensionless variables.) 2.142 An upstream trapezoidal channel with m1 = 1.5 and b1 = 5 m smoothly changes to a rectangular channel with b2 = 5 m wide, and critical flow occurs here. If the flow rate is Q = 12 m3/s, what are the depths upstream and downstream from the transition? What is the upstream
180
2.143
2.144
2.145
2.146 2.147
2.148
2.149 2.150
2.151 2.152
2.153
Open Channel Flow: Numerical Methods and Computer Applications
Froude number? If n = 0.014, what bottom slope of the upstream channel will result in uniform flow? (Solve this problem using both dimensionless and dimensioned variables.) An upstream trapezoidal channel has m1 = 1.5, n = 0.013, and So1 = 0.00025, and a bottom width ratio to a downstream rectangular channel of b1′ = b1 / b2 = 1.5. For a flow rate Q = 500 cfs, solve for b1 and b2 so critical depth occurs in the rectangular channel, and uniform flow occurs upstream. Write a program, or a computer model, that will generate a table of values to plot Figure 2.14, i.e., make a table that solves Equation 2.44 (or Equation 2.43). It will be instructive for you to use the Laguerre method described in Chapter 3 and extract the three roots from this third degree polynomial. A flow rate Q = 11.5 m3/s occurs in a rectangular channel that reduces its width by 2/3 through a transition. The upstream channel width is b1 = 5 m. Determine the upstream depth and Froude number. If n = 0.014 what bottom slope of upstream channel will result in uniform flow? What width b2 of a steep downstream rectangular channel will result in uniform flow upstream if the upstream rectangular channel’s width is b1 = 12 ft, n = 0.015 and So1 = 0.0005 for a flow rate Q = 350 cfs. Determine the value of c in the equation Y1′ = c / b′ that defines the upstream dimensionless depth Y1′ = Y1 / b1 as a function of the width ratio b′ = b1/Yc2 for a transition from an upstream rectangular channel with b1 = 6 ft and a downstream rectangular channel with b2 = 4.8 ft. What is the upstream Froude number? What ratio of upstream to downstream widths b1/b2 of rectangular channels should be used if the upstream width b1 should equal twice the critical depth in the downstream channel, and the depth Y1 is to be 10% larger than the critical depth. For a flow rate Q = 20 m3/s and the upstream width b1 = 5 m, what are these depths? What is the upstream Froude number? Write a computer program, or develop a computer model, that solves Equation 2.43 for any of the variables: b1′, b′, or Y1′ given the other two and use this model to solve Problems 2.145 and 2.148. The eight graphs in Figure 2.13 use linear graph paper to display the relationship of the dimensionless variables, Y1′, b1′, and b′ for different values of m1. Plot these relationships on eight loglog graphs and obtain an approximate equation of the form Y1′ = a(b′ )b for each b1′ and m1. From these equations can you suggest one equation that may be used to provide a “guess” to start the Newton solution to Equation 2.40. Write a computer program, or develop a computer model, that solves Equation 2.40 (or Equation 2.41) for any of the variables: b1′ , b′, or Y1′, and use this program (model) to solve Problem 2.142. An upstream trapezoidal channel with a bottom slope of So1 = 0.0002, a side slope m1 = 1, and n = 0.013 has a smooth transition to a steep downstream circular channel with D = 10 ft, (a) If the upstream channel has a width b1 = 10 ft, what flow rate will result in uniform flow upstream and what will this uniform depth be, and what will the critical depth be at the beginning of the circular channel? Repeat this solution for larger downstream pipe diameters up to 12 ft, and examine how the flow rates and depths change. (b) Determine how the upstream bottom width must vary for uniform flow to occur in the upstream channel for specified flow rates varying from 400 to 800 cfs in increments of 50 cfs. (Also determine the upstream uniform depth Y1, and the critical depth Yc2 associated with each of these flow rates.) (c) Determine how the upstream bottom slope So1 necessary for uniform flow varies as the flow rates vary from 400 to 800 cfs. It has been determined that the side slope of a trapezoidal channel must be m = 1.5. What should the bottom width, b, be to have the hydraulically most efficient section, if the channel is to be designed for a flow rate Q = 500 cfs, its Manning’s roughness coefficient is n = 0.015, and it has a bottom slope So = 001?
181
Energy and Its Dissipation in Open Channels
2.154 The most efficient trapezoidal sections is to be used to convey 500 cfs of water from a reservoir whose head H = 5 ft above the channel bottom. The channel will have a bottom slope So = 0.0008 and its Manning’s n = 0.013. Find the bottom width of channel to use. The entrance loss coefficient is Ke = 0.15. 2.155 Determine the hydraulically most efficient trapezoidal channel # 2 in the three channel system shown below. This # 2 channel has a Manning’s n = 0.013 and a bottom slope of So2 = 0.00085. Channel # 1 is trapezoidal also with a bottom width b1 = 5 m and a side slope m1 = 1.3, and receives its water from a reservoir with a head H = 3 m. The entrance loss coefficient is Ke = 0.15. Channel # 3 is a pipe with a diameter D = 1.5 m and is steep. Its bottom is 1.2 m above the level of the other two channels.
Q2
H=3m
b=?
yd. st h o M
b1 = 5 m, m1 = 1.3
Steep D
085
.00
=0 2
3, S o
.01
0 n 2=
Q1
Ke = 0.15
eff.
Q3
= 1.5 m
2.156 If the elevation of the bottom of a channel changes simultaneously with a contraction of its size, critical flow may not occur at the channel’s throat, or at its end, if it ends in a free overfall. The position where this control (or critical depth) occurs will be where the sum of the critical specific energy E c, and the elevation of the bottom z is a maximum, or where Hc = E c + z is maximum. A rectangular channel that is b = 10 ft wide and is carrying a flow rate Q = 250 cfs end in a free overfall. However, before the termination of the channel the width of the channel reduces to 5 ft at its end over a 50 ft length, while simultaneously the bottom of the channel rises and then fall again according to the second degree polynomial z = 3(2 − x/25)(x/25) as shown in the sketch below. Find the position xc where critical depth will occur and this critical depth. To help understand why critical depth occurs where H is maximum, make up a table that gives z, E, Y, and Fr2 as a function of the position x. Rectangular b = 10 ft
Plan view
Convergent section
L = 50 ft
z = 3 (2 – x/25)(x/25) Profile view
b = 5 ft
End free overfall
182
Open Channel Flow: Numerical Methods and Computer Applications
2.157 For the channels of the previous problem investigate how the position, and magnitude of the critical depth varies with the flow rate, using values of Q from 50 to 400 cfs. 2.158 Rather than have the bottom of the channels in the previous two problems rise according to a second degree polynomial the bottom rises 3 ft linearly over the first 10 ft of the transition, and then falls 3 ft over the last 40 ft of the transition. Now where does critical depth occur? 2.159 A trapezoidal channel, which is b = 5 m wide with a side slope m = 1, except near its end carries a flow rate Q = 25 m3/s. Before ending in a free overfall at its end the channel has a 25 m long transition in which the bottom width reduces to 2.5 m, and the side slope m to 0, both linearly. Simultaneously the bottom of the channel rises according to the second degree polynomial, z = (x/12.5)(2 − x/12.5). Find the position where critical depth occurs, and then make up a table as in Problem 2.156 that gives the Froude number as a function of x. 2.160 Figure A.2 provides a plot of several dimensionless variables as functions of the dimensionless depth Y′ = Y/D for circular channels. Prove that the maximum flow rate Q will occur in a given circular channel (D, So and n fixed) according to Manning’s equation when β = 2.6391 rad (a corresponding dimensionless depth Y′ = Y/D = 0.93818. Also prove that the dimensionless hydraulic radius R ′h = R h / D2 occurs when β = 2.2468 rad, and the maximum dimensionless conveyance K′ = A′ = (A′/P′) = 0.5015, i.e., verify some of the values given on Figure A.2. What value of A′ is associated with the maximum Q? Give a physical explanation why the maximum Q is associated with a larger β than the maximum hydraulic radius. 2.161 As in the previous problem determine the angle β associated with the maximum flow rate Q that will occur in a circular channel based on Chezy’s Equation, rather than Manning’s equation, assuming that Chezy’s C is constant. Compute A′, Ad, R′h, and K′ associated with this condition. Give a physical explanation why this β is larger than the β of the previous problem. 2.162 In the previous problem you were to determine angle β associated with the maximum Q in a circular channel based on Chezy’s C being constant. Determine this angle for several values of the relative roughness e/R h if the flow is wholly rough.
References ASCE Task Force. 1963. Friction factors in open channels. Journal of the Hydraulic Division, ASCE, 89(2), 97–143. Jarrett, R. D. 1991. Hydrologic and hydraulic research in mountain rivers, Water Resources Bulletin, American Water Resources Association, 26(3), 419–429.
Momentum 3 The Principle Applied to Open Channel Flows 3.1 The Momentum Function Use of the momentum principle is needed when forces control the direction or conditions associated with fluid motions, or when it is not possible to define what is happening to the fluid on a small element basis but a large picture of a mass of fluid within a control volume is possible. If a vector quantity, such as force or velocity with both magnitude and direction, is the unknown, or is one of the important known variables of the problem, then it will most likely be necessary to use the momentum principle in solving the problem. In order to introduce and develop the momentum function for use in open channel hydraulics, an interesting phenomena, the hydraulic jump will be analyzed. A hydraulic jump abruptly takes the flow in an open channel from supercritical flow to subcritical flow, so that through a hydraulic jump the depth of flow rather abruptly increases. Downstream from a hydraulic jump there will be a control, which may be a flatter channel, a gate, or dam, etc. that requires the flow to be at a subcritical depth. Upstream from the hydraulic jump something will cause the flow to be supercritical, such as a gate, or steep channel. The supercritical flow rushes down the channel with a velocity in excess of the speed of small amplitude gravity waves, and consequently receives no signal from the downstream flow. Since it must change to subcritical flow at some position because a downstream control dictates this, the change occurs in the form of a hydraulic jump. The sketch below illustrates these conditions resulting in a hydraulic jump. A hydraulic jump actually takes place over a finite length of several feet, but since sketches herein have an enlarged vertical to the horizontal scale the hydraulic jump is shown as a near vertical line.
Supercritical flow Subcritical flow
To apply the momentum principle to the hydraulic jump, the steps outlined in Chapter 1 will be followed: Step # 1: A control volume of the hydraulic jump will be created by moving a short distance upstream and downstream from it and removing the fluids. Hydraulic jump hc1
Y1
Fp1
γhc1A1
Y2
Y1
Fp2 γhc2A2
hc2 Y2
dA
h Y
Control volume
183
184
Open Channel Flow: Numerical Methods and Computer Applications
Step # 2: These columns of removed fluid will be replaced by the hydrostatic forces that they apply to the control volume fluid. A hydrostatic pressure force equals the pressure at the centroid of the area times the area, and the pressure at the centroid equals the specific weight γ times the distance from the water surface to the centroid, hc. Therefore, on the upstream side this force is, Fp1 = γhc1A1 and on the downstream side this force is, Fp2 = γhc2A2, in which the subscripts denote upstream and downstream respectively. The quantity hcA is also the first moment of area about the fluid surface. Methods for its evaluation for common cross sections are given in Table A.1, but in general it is given by hcA =
∫ hdA
(3.1)
in which h is the distance from the fluid surface down to the differential element dA of the crosssectional area. Only momentum fluxes in the direction of flow will be considered. Neither the normal nor shear force from the bottom of the channel will be included. Nor will the weight of the fluid in the control volume. In steep channels, the component of the fluid weight in the direction of flow will be a significant force and possibly also the shear force. However, since the weight will depend upon the size of the control volume, its inclusion into the analysis must be based on experimental data. Our intent here is to develop the momentum principle for use in open channel flow analysis, and inclusion of the component of weight, and shear forces in the flow direction complicate the problem so that basic principles cannot be developed. Should a hydraulic jump occur in a steep channel, then the results that follow will need modification. Step # 3: The application of the momentum equation in the direction of flow results in γ h c1A1 − γ h c2 A 2 = ρQ(V2 − V1 )
Note that every term in the above equation has dimensions of force, since γ is dimensionally force per length cubed, or (F/L3), and this is multiplied by (L3) in hcA, and the momentum flux terms have dimensions of (Ft2/L4)(L3/t)(L/t) = (F). Since ρ = γ/g, the above equation can be divided by the specific weight, which will result in each term having the dimensions of L3. Upon rearranging terms with the same subscript on the same side of the equal sign the following equation results: h c1A1 +
Q2 Q2 = h c2 A 2 + (gA1 ) (gA 2 )
(3.2)
The sum of the two term Ahc + Q2/(gA) is called the momentum function, and will be given the symbol M. Thus an alternative way of expressing Equation 3.2 is that the momentum function on the two ends of the hydraulic jump are equal, or
M1 = M2
(3.3)
Q2 V2 A = Ah c + (gA) g
(3.4)
in which the M’s are defined by M = Ah c +
Table A.1 gives the result from integrating ∫hdA for trapezoidal and circular shapes. For a trapezoidal channel, Ah c =
bY 2 mY 3 + 2 3
(for a trapezoid )
185
The Momentum Principle Applied to Open Channel Flows
and for a circular channel one form in which the results can be given is Ah c =
D3 D D2 2 (3 sin β − 3β cos β − sin 3β) (for a circle) sin β − A cos β = 2 6 24
3.2 Characteristics of the Momentum Function A plot of M = Ahc + Q2/(gA) as the abscissa, and the depth Y as the ordinate is called a momentum diagram (or some books may refer to it as a thrust diagram) just as E versus Y is referred to as a specific energy diagram. A sketch of a momentum diagram is shown below. From an examination of the two terms that sum to give M, it is clear that if a fixed channel is to convey a specified flow rate Q, then as the velocity in that channel approaches zero the term Ahc becomes very large, and the term Q2/(gA) becomes very small since A becomes large; thus M becomes large. On the other hand, as the depth approaches zero (the velocity becomes large), the area must become very small and the term Q2/(gA) becomes very large, even if Ahc is small. Therefore, a momentum function diagram will have a minimum value for M at some intermediate depth, much as is the case with a specific energy diagram. 6 Y
5
1 1.25
10 ft
Depth, Y (ft)
4 3
Q
3 =3
2
Q2 = 1 0
al
itic
cr Sub
Q1 = 100
0
200
00
cfs
ical
it Supercr
cf s
cfs
50
100
150
Momentum function M = hc
200
A + Q2/(gA)
(ft3)
250
300
To find this minimum value the principle of calculus of setting the first derivative to zero can be used, that is dM/dY will be equated to zero. In order to differentiate Ahc it is necessary to utilize Leibniz’s rule since Ahc is defined by the above integral. Leibniz’s rule is d dα
φ2 (α )
∫
φ1 ( α )
φ2 (α )
F(x, α )dx =
∫
φ1 ( α )
dφ dφ ∂F1 dx + F(φ1, α ) 1 − F(φ2 , α ) 2 ∂α dα dα
Applying this rule to the derivative d(Ahc)/dY gives d(Ah c ) = dα
0
∂h
∫ ∂Y dA + (0)1 − Y(0) = A Y
186
Open Channel Flow: Numerical Methods and Computer Applications
and therefore dM Q 2 ∂A Q2T = A− = A − =0 dY gA 2 ∂Y gA 2 or Q2T = 1 or Fr2 = 1 gA 3
(3.5)
This result is also the “critical flow” equation that indicates that the average velocity of the flow equals the speed of propagation of a small amplitude gravity wave. Therefore the minimum value for the momentum function M exactly coincides with the minimum value for the specific energy for this same flow rate in the same channel. Thus a hydraulic jump will always take the flow from a supercritical condition to a subcritical condition. Furthermore, because of the shape of the momentum function the height of the hydraulic jump will be greater as the depth upstream is further below critical depth. Should the depth upstream from a jump be only slightly below Yc then the downstream depth will only be slightly above Yc. This is referred to as a mild hydraulic jump, and in observing such occurrences in a channel one would note only a waviness in the water surface. A strong hydraulic jump on the other hand will have a highly supercritical flow upstream (e.g., Fr1 is much larger than unity), a slow moving flow downstream, with a large change in the water surface elevation. Such a strong hydraulic jump will occur, for example, on the apron of a dam spillway where the water after flowing down the steep spill way of the dam has a very high velocity. Strong hydraulic jumps dissipate a large fraction of the total energy per unit weight that the upstream fluid possesses. The two depths across a hydraulic jump are referred to as conjugate depths. Thus the momentum function M is constant for conjugate depths, one of which is associated with supercritical flow and the other is associated with subcritical flow. Conjugate depths are not equal to alternate depth for which the specific energy is constant for a given flow rate. Example Problem 3.1 Find the first moment of area about the water surface of a flow in a 6 ft diameter pipe that is 2 ft deep. Determine this value for Ahc by the following four procedures: (1) use your calculator to numerically integrate the appropriate function, (2) use the Simpson’s rule SIMPR described in Appendix B, (3) and (4) use the formula given in Appendix A. Solution First solve for β = cos−1(1 − Y/R) = cos−1(1 – 2/3) = 1.23096 rad, and the area A = R2(β − cos β sin β) = 8.25021 ft2. The first moment of area is shown in the sketch below to be
D
h
dA
θ
y
Y
β
The Momentum Principle Applied to Open Channel Flows Y
∫
Ah c = hdA with cos θ = 1 − 0
y R
h = Y − R(1 − cos θ), and dA = 2R sin θ dy so dy = R sin θ dθ β
∫
Ah c = [ Y − R(1 − cos θ)][2R sin θ](R sin θ dθ) 0
1.23096
∫
Ah c = 2R 3
[ Y /R − 1 + cos θ]sin 2 dθ
0
1.23096
Ah c = 54
∫ 0
1 2 cos θ − 3 sin θ dθ
1. Integrating with the HP48G calculator produces Ahc = 6.8347 ft3 2. The computer program that calls on SIMPR can consist of the following: PARAMETER (NMAX=21,A=0,B=1.23096,ERR=1.E5) EXTERNAL EQUAT CALL SIMPR(EQUAT,A,B,VALUE,ERR,NMAX) WRITE(*,*) 54.*VALUE END FUNCTION EQUAT(T) EQUAT=(COS(T).33333333)*SIN(T)**2 RETURN END
As a solution it prints out 6.83474 for the solution. 3. Using the equation Ahc = D3/24{3sin β − 3β cos β − sin3 β} = 6.8347 ft3 4. Using the equation Ahc = 0.5D{D2/6sin3 β − A cos β} = 6.8347 ft3 Example Problem 3.2 A hydraulic jump occurs in a circular channel with a diameter D = 4 m. The flow rate Q = 22 m3/s and the depth upstream from the jump is 0.8 m. What is the depth downstream from the hydraulic jump, and how much energy per unit weight of fluid is dissipated by the jump? Solution From the known conditions upstream from the jump the momentum function can be calculated here, or M = (D/2){(D2/6)sin3 β − A cos β} + Q2/(gA) in which A = (D2/4)(β − cos β sin β) and β = cos−1(1 – 2Y/D). Substituting the known values into this equation produces, M1 = 28.16 m3. Across the hydraulic jump M2 = M1, and therefore the above implicit equation (Equation 3.2) needs to be solved in which M = 28.16 and D = 4. Using the Newton method, the solution is 2.871 rad, from which Y2 = 3.927 m. The Froude number associated with the upstream flow is Fr1 = 5.25, and that associated with the downstream flow conditions is Fr2 = 0.16. The specific energy upstream is E1 = 8.51 m and that associated with the downstream conditions is E2 = 4.08 m. The difference represents the head loss through the hydraulic jump, or the energy per unit weight of fluid dissipated by the jump. The head loss equals 4.43 m, or more energy is dissipated than remains in the downstream flow. The most difficult part of this solution is solving the implicit momentum equation for the depth of flow.
187
188
Open Channel Flow: Numerical Methods and Computer Applications
3.3 Rectangular Channels and Momentum Function per Unit Width Simplification occurs if the channel is rectangular in shape. For a rectangular channel it is useful to deal with the momentum function per unit width of channel. This unit momentum function will be denoted by lower case m, or m = M/b much the same as q = Q/b. For a rectangle Ahc = bY2/2, and dividing Equation 3.4 by b gives m=
M Y2 q2 Y 2 V2 Y = + = + b 2 (gY) 2 g
(3.6)
Therefore a hydraulic jump in a rectangular channel is defined by
Y12 q2 Y2 + = 2 + q 2 (gY2 ) 2 2 (gY1 )
(3.7)
It can be noted that when Equation 3.7 is multiplied by either Y1 or by Y2 that a cubic equation results just as a cubic equation defined the alternative depths when dealing with specific energy in a rectangular channel. To investigate further the characteristic of Equation 3.7 collect the terms with Y squared on one side of the equal sign, and the terms contain q on the other side so that Equation 3.7 becomes
Y12 Y22 (Y1 + Y2 )(Y1 − Y2 ) q 2 1 1 q 2 (Y1 − Y2 ) − = = − = g Y2 Y1 gY1Y2 2 2 2
(3.8)
Upon dividing this equation by (Y1 – Y2) and then multiplying it by Y1 the following quadratic equation results:
Y12 + Y2 Y1 −
2q 2 =0 (gY2 )
It should also be noted that the above cubic equation has been reduced to a quadratic equation by the division of (Y1 – Y2). Using the quadratic formula to solve this equation and then dividing the result by Y2 gives the following useful equation to computer the depth upstream of a hydraulic jump if the downstream depth is known:
Y1 −1 + 1 + 8q 2/(gY23 ) −1 + 1 + 8Fr22 = = Y2 2 2
(3.9)
The possible minus in front of the square root is ignored since it produces a physically impossible negative depth. If upon dividing Equation 3.8 by (Y1 – Y2) the result had been multiplied by Y2 instead of Y1, then a quadratic equation in Y2 will result. After using the quadratic formula the following equation results which is identical to Equation 3.9 with the subscripts reversed:
Y2 −1 + 1 + 8q 2 /(gY13 ) −1 + 1 + 8Fr21 = = Y1 2 2
(3.10)
189
The Momentum Principle Applied to Open Channel Flows
Thus regardless of whether the depth is known upstream, or downstream the other conjugate depth in a rectangular channel can be obtained by use of the explicit Equations 3.9 or 3.10. Another useful form for the hydraulic jump equation, (or the equation resulting from equating m1 to m2, i.e., m1 = m2 can be obtained from Equation 3.8 by solving it for the Froude number squared, or Fr12 =
1 Y2 Y2 q2 = + 1 = 0.5Y′(Y′ + 1) gY13 2 Y1 Y1
(3.11)
or Fr22 =
1 Y1 Y1 Y′ + 1 q2 = +1 = 3 gY2 2 Y2 Y2 2Y′ 2
(3.12)
in which the latter parts of Equations 3.11 and 3.12 contain the dimensionless depth Y′ = Y2/Y1. The momentum function per unit width for critical flow mc can be evaluated either from the critical depth, Yc or the flow per unit width qc. For critical flow q 2c /g = Yc3 or q c2 /(gYc ) = Yc2. Substituting this in Equation 3.6 gives
q2 Y2 m c = c + Yc2 = 1.5Yc2 = 1.5 c 2 g
2/3
It is interesting that the constant is also 1.5 = 3/2 in the relationship between critical specific energy and critical depth for a rectangular channel (Ec = 1.5Yc); the difference is now Yc is squared, as needed, so the dimensions are the same on both sides of the equation since the dimensions of m are L2. In the case of Ec twothirds of its value is the depth (potential energy/unit weight) and onethird the velocity head Vc2 /(2g), whereas twothirds of mc comes from the momentum flux divided by the specific weight, or ρqV / γ = q 2c /(gYc ) = Yc2 and onethird from the hydrostatic force divided by γ. Writing Equation 3.6 as a cubic equation gives F(Y) = Y 3 − 2mY +
2q 2 =0 g
Any general cubic equation x3 + px2 + qx + r = 0 can be put in the form y3 + ay + b = 0 by substituting y = p/3 for x (see CRC Standard Math. Tables). Our momentum function per unit width cubic equation is already in this form. The three roots are y1 = A + B, y 2 = 0.5(i 3 )(A − B) − 0.5y1, and y 3 = 0.5(i 3 )(B − A) − 0.5y1 in which 1/ 3
b2 a 3 1 / 2 A = + − 0.5b 4 27
1/ 3
b2 a 3 1 / 2 and B = − + − 0.5b 4 27
190
Open Channel Flow: Numerical Methods and Computer Applications
If b2/4 + a3/27 > 0 there will be one real root and two conjugate imaginary roots. If b2/4 + a3/27 = 0 there will be three real roots of which at least two are equal. If b2/4 + a3/27 < 0 there will be three real and unequal roots. For the cubic momentum equation a = –2m and b = 2q2/g. The first condition occurs for values of m < mc and the one real root gives a negative depth (which of course has no physical meaning); the second condition represents critical flow conditions; and the third condition where three real unequal roots occur is when m > mc and there are two positive real roots that are the two conjugate depths, and the third root gives a negative depth. The graph below is a plot obtained by solving the cubic momentum equation for m ≥ mc for q = 1, 2.5, 5.0, 7.5 and 10.0 cfs/ft. One can perform the above complex arithmetic to find the conjugate depths, but it is much easier to use Equations 3.9 or 3.10.
q= q 1 cf q = = 2.5 s / f t c 5 cf fs/f s/ ft t
8 6
q=
Depth, Y (ft)
4 2
q=
0
1
cfs
7.5
/f t
/ft
fs 0c
q=
2
/ft
fs 0c
q
0 =3
/ft
cfs
q = 1 cfs/ft q = 2.5 cfs/ft q = 5 cfs/ft
–2
–8
0
/f t cfs
q = 20
10
–6
q=
q = 7.5 cfs/
ft
–4 cfs/ f t
q = 30 cfs/ft
5 10 15 Momentum function, m (ft2)
20
If one only knows the momentum function per unit width for a rectangular channel, and the unit flow rate q, and not one of the conjugate depths so Equations 3.9 or 3.10 cannot be used, it is slightly easier to use the solution for the above cubic F(Y) that involves the arc cosine (and cosine), as was used in solving the alternate depths for a given specific energy in the previous chapter. To use this alternative method for solving the cubic equation for its three roots first compute the angle θ from the first root Y1 is the negative depth, the second root Y2 is θ = cos[(q2/g)/(2m/3)1.5] and thereafter solve for the three roots from the subcritical depth and the third root Y3 is the supercritical depth. 1/ 2
2m Y1 = −2 3
1/ 2
θ 2m cos , Y2 = −2 3 3
1/ 2 ( θ + 4π ) ( θ + 2π ) 2m cos cos , Y3 = −2 3 3 3
The first root Y1 is the negative depth, the second root Y2 is the subcritical depth and the third root Y3 is the supercritical depth. The program ROOTSM, listed below is designed to provide these three depths given the unit flow rate q, the unit momentum function m, and the acceleration of gravity.
191
The Momentum Principle Applied to Open Channel Flows
Note that it computes critical depth and the associated unit momentum function, and tells you if you give an m less than mc. Program ROOTSM.FOR PARAMETER (PI=3.14159265) 1 WRITE(*,*)' Give: q,m (momentum/width),g' READ(*,*) q,F,g IF(q.LT.1.E4) STOP qg=q*q/g Yc=qg**.3333333 Fc=.5*Yc**2+qg/Yc IF(F.LT.Fc) THEN WRITE(*,100) Fc,F,Yc 100 FORMAT(' No real roots exist since m<mc m(crit)=',F8.3, &' m=',F8.3,/,' Yc=',F8.3) GO TO 1 ENDIF FF=2.*sqrt(.66666667*F) THETA=ACOS(qg/(.6666667*F)**1.5) Y1=−FF*COS(THETA/3.) Y2=−FF*COS((THETA+2.*PI)/3.) Y3=−FF*COS((THETA+4.*PI)/3.) WRITE(*,*) Y1,Y2,Y3 GO TO 1 END An explicit equation giving the head loss across a hydraulic jump in a rectangular channel can be obtained as a function of the upstream and downstream depths, Y1 and Y2. The head loss caused by a hydraulic jump is given by h L = E1 − E 2 = Y1 − Y2 +
q2 1 1 − 2g Y12 Y22
(3.13)
in which the expression after the second equal sign applies only for rectangular channels. From Equation 3.12
q2 1 = (Y1Y2 )(Y1 + Y2 ) (2g) 4
which when substituted into Equation 3.13 gives
h L = Y1 − Y2 +
(Y1 + Y2 )(Y22 − Y12 ) 4Y1Y2
Upon multiplying (Y1 – Y2) by 4Y1Y2 to give it the same denominator, the numerator becomes the cube of (Y1 – Y2) and the head loss is given by
hL =
(Y2 − Y1 )3 4Y1Y2
(3.14)
192
Open Channel Flow: Numerical Methods and Computer Applications
By combining Equation 10, which gives the ratio of the conjugate depths Y2/Y1 as a function of the upstream Froude number, the head loss across a hydraulic jump (Equation 3.14) can be expressed by the following equation as a function of the upstream Froude number Fr1 and the upstream depth Y1:
hL
{ 1 + 8F − 3} = 16 { 1 + 8F − 1} 3
2 r1
Y1
(3.14a)
2 r1
(
)
(
)
and as a fraction of the upstream specific energy E1 = Y1 + q 2/ 2gY12 = Y1 1 + Fr12 / 2 by the following equation that shows that hL/E1 is only a function of the upstream Froude number, Fr1.
hL E1
{(1 + 8F ) − 3} = 8 ( 2 + F ) {(1 + 8F ) 2 r1
3
1/ 2
2 r1
2 r1
1/ 2
}
−1
(3.14b)
Note from the plot of hL/E1 versus Fr1 that the fraction of total upstream energy lost through a hydraulic jump in a rectangular channel is relatively small for upstream Froude numbers only modestly larger than unity, and increases rapidly for large upstream Froude numbers. 0.9
0.7 0.6 0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0
Fraction of upstream energy lost
Fraction of upstream energy lost
0.8
0.5 0.4 0.3 0.2 0.1 0
1
2
3
4
5
6
1
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 Upstream Froude number
7 8 9 10 Upstream Froude number
11
12
13
14
2 15
If Equation 3.14 is written using the dimensionless depth Y′ = Y2/Y1 then the following dimensionless equation results:
h′L =
h L (Y′ − 1)3 = 4Y′ Y1
(3.14c)
193
The Momentum Principle Applied to Open Channel Flows
A plot of h′L versus Y′ below shows how the dimensionless headloss in a rectangular channel rapidly increases from 0, when Y′ = 1 to 18.225, when Y′ = 10. When Y′ = 100, h′L = 2425.75. Another dimensionless relationship giving the headloss in a rectangular channel can be obtained by defining the following dimensionless alternative depths by dividing by the critical depth Yc : Y1c′ = Y1 / Yc and Y2c′ = Y2 / Yc . Then Equation 3.14 becomes h′Lc =
h L (Y2′c − Y1′c )3 = Yc 4Y1′c Y2′c
(3.14d)
If Yc = (q2/g)1/3 is substituted in the middle part of this equation and both numerator and denominator of the result divided by Y1, then h L / Yc = (h L /Y1 )/ Fr12/ 3 = h′L / Fr12 / 3 and Equation 3.14d becomes h′L =
(Y2′c − Y1′c )3 2 / 3 Fr1 4Y1′c Y2′c
(3.14e)
Since Equation 3.10 gives the dimensionless depth Y′ = 0.5{(1 + 8Fr12 )1/ 2 − 1}, and Y1c′ = Y1 / Yc = (Y1 / Y2 ) / q 2 /(gY23 )
1/ 3
= 1/(Y′Fr22 / 3 ) and Y2c′ = Y2 / Yc = (Y2 / Y1 ) / q 2 /(gY13 )
Fr22 =
1/ 3
= Y′ / Fr12 / 3 and
q2 q 2 /(gY13 ) Fr21 = = 3 gY23 Y′ 3 Y′
it is possible to express all of these dimensionless variables as a function of the upstream Froude Number, Fr1. The table below provides these dimensionless variables, and the accompanying figure plots them. Fr1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
y′
Fr2
y′1c
y′2c
h′L
1.000 1.6794 2.3723 3.0707 3.7720 4.4749 5.1789 5.8836 6.5887 7.2942 8.0000 8.7060 9.4121 10.1184 10.8248 11.5312 12.2377 12.9443 13.6510
1.000 0.6892 0.5474 0.4646 0.4095 0.3697 0.3394 0.3153 0.2956 0.2792 0.2652 0.2530 0.2424 0.2330 0.2246 0.2171 0.2102 0.2040 0.1983
1.000 0.7631 0.6300 0.5429 0.4807 0.4338 0.3969 0.3669 0.3420 0.3209 0.3029 0.2871 0.2733 0.2610 0.2500 0.2401 0.2311 0.2229 0.2154
1.0000 1.2817 1.4944 1.6670 1.8134 1.9412 2.0553 2.1586 2.2533 2.3410 2.4228 2.4996 2.5721 2.6408 2.7062 2.7686 2.8284 2.8858 2.9410
0.0000 0.0467 0.2723 0.7229 1.4117 2.3442 3.5228 4.9489 6.6233 8.5465 10.7188 13.1403 15.8113 18.7319 21.9022 25.3221 28.9918 32.9114 37.0807
194
Open Channel Flow: Numerical Methods and Computer Applications 20 Dimensionless headloss, hL΄
18 16 14 12 10 8 6 4 2 0
2
1
3
4
5 6 7 Dimensionless depth, Y΄
8
9
10
Graph showing how the dimensionless head loss h′L = h L / Y1 varies with the dimensionless depth Y′ = Y2/Y1. 40
io
ss h ΄
Dim
lo
20 15 Dim
10
wn stre am
5 0
ensio
Do
nless
e sh es l n Y1 = Y 2/ sio th Y΄ p en e d m nless Di ensio Dim
Depth Y1c ΄ = Y1/Yc Froude number, Fr2 = q2/(gY 3) 2
1
2
2
L
ens
25
th
dep
L /Y 1
ss nle
2.5
Yc Y 2/
=h
30
= Y΄2c
ad
Dimensionless depth, Y΄ = Y2/Y1 and hL΄
35
3
4 5 6 7 8 9 10 Upstream Froude number, Fr1 Figure relationship of dimensionless variables to upstream Froude number
1.5
1
0.5
Dimensionless depths, Y΄1c Y΄2c and Fr2
3
0
3.4 Polynomial Form for Momentum Function A natural question to raise is whether it is possible to simplify the hydraulic jump equation to an explicit form for other cross sections besides a rectangle. The answer is no, but it is useful to go through similar attempts for a trapezoidal section because doing so provides insights into the characteristics of the hydraulic jump equation. To simplify the notation let r = Y2/Y1 and t = (area of the rectangular part)/(area of the triangular parts) = b/(mY1) (which you might note is the reciprocal of the dimensionless depth Y′ used in defining Figure 2.4, the dimensionless specific energy diagram). After some algebraic manipulation of the momentum equation M1 = M2 for a trapezoidal section the following fifth degree polynomial can be produced.
The Momentum Principle Applied to Open Channel Flows
195
r 5 + 2.5tr 4 + 1.5t 2 r 3 − (1.5t + 3F 2 (t + 1) + 1)r 2 − (1.5t 2 + t + 3F 2 t(t + 1))r + 3F 2 (t + 1)2 = 0 (3.15)
in which F 2 = V12 /(gY1 ). Note F is not the Froude number for a trapezoidal channel. This fifth degree polynomial can be reduced to the following fourth degree polynomial by extracting the root r = 1:
r 4 + (2.5t + 1)r 3 + (1.5t + 1)(t + 1)r 2 + (0.5t 2 + (t − 3F 2 )(t + 1))r − 3F 2 (t +11)2 = 0
(3.16)
These equations are also reversible, i.e., r could be defined as Y1/Y2 and the subscript for t and F could be 2, and identically the same equations would result. For a triangular section, (i.e., b = 0, t = 0) the above fourth degree polynomial reduces to: r 4 + r 3 + r 2 − 3F 2 (r + 1) = 0
(3.17)
In general a fourth degree polynomial will have four roots, or solutions. Should there be imaginary roots these must occur as complex conjugate pairs. Since two real roots exist that have the same value for the momentum function in a trapezoidal channel, e.g., the conjugate depths, we can conclude that two complex, or imaginary roots exist, or two additional real but negative roots exist. The root for the known real depth was eliminated by reducing the fifth degree polynomial to a fourth degree polynomial. In solving problems involving finding a conjugate depth, the only advantage in solving Equation 3.16 over solving Equation 3.3 is that much is known about finding roots of polynomials. However, Equation 3.16 is restricted to trapezoidal channels and Equation 3.3 is general for all cross sections. Later in this chapter general methods for finding roots of a polynomial will be used to solve a dimensionless form of the momentum function equation. Example Problem 3.3 The depth upstream from a hydraulic jump in a rectangular channel is 0.6 m. If the channel has a bottom width b = 5 m and contains a flow rate of Q = 20 m3/s, what depth of flow would be expected downstream from the hydraulic jump? How much head loss occurs through the jump, and how much power in horsepower is dissipated? Solution The solution for the conjugate depth after the hydraulic jump can be computed directly from Equation 3.10 or Y2 = 0.6 −1 + 1 + 8q 2 /(gY3 ) 2 = 2.05 m. The head loss hL = E1 − E2 = 2.87 − 2.24 = 0.63 m. The power dissipated = γQhL = 9.8(20)(0.63) = 123.48 kW or hp = 123.48/0.746 = 165.5 hp.
Should an external force exist on a control volume of fluid, then Equation 3.3 can be modified to allow for this by noting that in the development of the momentum function M, it is equal to force divided by the specific weight γ of the fluid or M(L3) = F/γ(F/(F/L3)) and therefore Equation 3.3 might be generalized to M1 = M2 +
F γ
and likewise if a rectangular channel is involved the momentum equation per unit width becomes, m1 = m 2 + in which f = F/b.
f γ
196
Open Channel Flow: Numerical Methods and Computer Applications Example Problem 3.4 A sluice gate is positioned 0.6 ft above the channel bottom. Its contraction coefficient is Cc = 0.6. If a flow rate of q = 10 cfs/ft is passing the gate determine the force on the gate. How does this actual force compare with the force computed assuming that the pressure on the gate were hydrostatic? Solution The solution of the energy equation E1 = E2 (with Y2 = (0.6)(0.6) = 0.36 ft) gives Y1 = 12.33 ft. Therefore the force per unit width is obtained from f = γ (m1 − m 2 ) = γ {Y12 / 2 − Y22 / 2 + q 2 / g(1/ Y1 − 1/ Y2 )} = 62.4(75.95 − 8.38) = 4216.4 lb / ft . If the force is computed from a hydrostatic pressure distribution its magnitude equals γ(E − yG)2/2 = 62.4(12.34 − 0.6)2/2 = 4300.4 lb/ft, or 2% larger. The reason for the larger hydrostatic force is that the actual pressure distribution goes to zero at the tip of the gate. Example Problem 3.5 Baffles exist at the toe of a dam to stabilize the hydraulic jump on the apron of the dam. The total cross section of these baffles normal to the direction of flow is 20 ft2, and they exist in a trapezoidal channel with b = 10 ft, and m = 1.2. This channel extends downstream from the dam for a long distance at a slope of So = 0.0012, and its roughness coefficient is n = 0.014. If the drag coefficient for the baffles has been determined equal to 0.75 based on the downstream dynamic pressure, determine what depth of the upstream flow should be if the flow rate is Q = 500 cfs. Solution Uniform flow exists downstream. Therefore solving Manning’s formula gives the depth downstream from the hydraulic jump as Y2 = 4.425 ft and M2 = 247.17 ft3. The velocity downstream is V2 = 7.38 fps, and therefore the drag force against the baffles can be obtained from Drag = CDAb(ρV2/2) = 803.05 lb. The upstream depth can now be solved from the momentum equation M1 = M2 − Drag/γ = 260.04 ft3. Solution of this equation by the Newton method gives Y1 = 1.924 ft. Without the baffles the upstream depth must be larger for the downstream depth to be possible, i.e., the conjugate depth to 4.425 ft, which is 2.973 ft.
3.5 Dimensionless Momentum Functions Dimensionless momentum functions have utility similar to dimensionlessspecific energy curves. Dimensionless momentum functions will be developed for both trapezoidal and circular sections in the follow few paragraphs. For a trapezoidal section the momentum function is M=
1 1 Q2 bY 2 + mY 3 + 2 3 g(bY + mY 2 )
This equation can be rearranged as M=
b3 m 2 Y 2 m 3Y 3 Q 2 /g + + 2 2 2 3 m 2b 3b b / m(mY/b + m 2 Y 2 / b2 )
If the following dimensionless parameters are defined: M′ = m2M/b3, Y′ = mY/b and Q′ = m3Q2/ (gb5), then the above equation can be written as the following dimensionless momentum function:
M′ =
1 2 1 3 Q′ Y′ + Y′ + 2 3 Y′ + Y′ 2
(3.18)
Dimensionless momentum function curves for a number of values of Q′ are plotted on Figure 3.1. The two dimensionless depths associated with a constant value of M′ on this graphs for a given curve Q′ are the dimensionless conjugate depths. Dimensionless critical depth Yc′ is also shown
197
The Momentum Principle Applied to Open Channel Flows 0.0 1.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
0.9
0.9 y
l 0.8
Q΄ = m3 Q2/(g b5)
0.7 0.6
Q΄
0.5
Q΄ Q΄
0.4
Q΄
Q Q ΄ = Q΄ ΄ = 0 ΄ .2 = 0 Q = 0. .16 0.1 8 ΄= 14 Q 0.1 ΄= 2 1
Q
=. Q΄
0.3 .001 d=0
0.1
Q΄
05
Q .0 ΄ 3 Q΄ =.02 Q΄ =.01 =.0 5 1 Q΄ Q΄ =.0 =0 05 .00 25
=0
Q΄ =0
Q΄ =0
.6
=0
0.7 0.6
=0
.9
.8
0.5
.5
.4
0.4
=0
.3
0.3
09
=.
΄=
07
Q΄
΄=
0.0
=.
Q
Q
0.1 0.0
m
b
Q΄
Dimensionless depth, y΄ = my/b
0.8
0.2
1.0
0.
=.
08
0.2
4
0.2
0.1
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2 2
1.3 3
Dimensionless momentum function, M΄ = m M/b
1.4
1.5
1.6
1.7
1.8
0.0
Figure 3.1 Dimensionless momentum function diagrams for trapezoidal sections. (Individual curves apply for Q′ = m3Q2/(gb5).)
in the graph. The graph can be used to solve problems associated with the momentum function. In addition to having the dimensionless critical depth read as the minimum point on the curves on Figure 3.1, the critical flow equation Q2T/(gA3) = 1 can be written using the same dimensionless variables. For a trapezoidal section the critical flow equation becomes Q2(b + 2mY) = Q2b(1 + 2Y′) = (b2/m)3g(mbY/b2 + m2Y2/b2)3, or letting Q′ = m3Q2/(gb5), the dimensionless critical flow equation for a trapezoidal section becomes Q′ =
(Y′ + Y′ 2 ) (1 + 2Y′ )
(3.18a)
and this equation gives Y′ corresponding to the minimum M′ for any Q′ on Figure 3.1, or the position on the Y′ ordinate where critical flow occurs for any Q′ curve. A similar graph for a circular channel is given in Figure 3.2. Dimensionless momentum functions for a circular channel can be obtained by defining the dimensionless depth in a circle as Y′ = Y/D. Then with auxiliary angle defined from β = cos−1(1 − 2Y′) the momentum function is M=
Q 2/g 11 2 3 1 D sin β − D2 (β cos β − sin β cos2 β) + 1 2 26 4 D (β − cos βsin β) 4
If M′ is defined as M′ = M/D3 and Q′ = Q2/(gD5), then the following dimensionless momentum function occurs:
M′ =
11 3 1 4Q′ sin β + cos β(sin β cos β − β) + β − sin β cos β 2 6 4
(3.19)
198
Open Channel Flow: Numerical Methods and Computer Applications 0.1
0.0 1.0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
0.9
0.9 0.8
Q
Dimensionless depth, y΄ = y/D
΄=
0.7
Q΄ Q΄
0.6 Q΄
΄=
0.3 Q
΄=
0.2
Q΄
.0
15
Q΄
0.1 Q΄= .001
0.0
0.1
Q΄ =.
Q΄ =.0 2
Q΄
=.
03
=.
04
.0
Q
7
΄=
=.
Q΄ 3
Q
΄=
.8
0.8 .9
0.7 =.
6
=.
0.6
5
=.
4
0.5
8 .1
.08
΄=
΄=
4
Q
=0
Q
Q
.1
Q΄
.1
=0
0.4
= Q΄
Q΄
= Q΄
0.5
0.0
1.0
.1
6
Q΄
=.
20
0.4
.1
2
.0
9
0.3
05
0.2
=. 01 Q΄= .005 Q΄= .002 5
0.2
0.1
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
Dimensionless momentum function,
1.2
1.3 M΄ = M/D3
1.4
1.5
1.6
1.7
1.8
0.0
Figure 3.2 Dimensionless momentum function diagrams for circular sections. (Individual curves apply for Q′ = Q2/(gD5).)
You should note that Y′ and M′ are defined differently for a trapezoidal section than for a circular section. The dimensionless critical flow equation for a circular section is
Q′ =
(β − cos β sin β)3 64 sin β
(3.19a)
and this equation gives the ordinate Y′ of the minimum position on any Q′ curve on Figure 3.2. There are often advantages in solving a dimensionless equation since much of the variation due to the size of the variables no longer exists. If a hydraulic jump occurs in a trapezoidal channel then the dimensionless momentum function M′1 as defined by Equation 3.18 equals this value evaluated downstream from the hydraulic jump, or 1 2 1 3 Q′ 1 1 Q′ Y1′ + Y1′ + = Y2′ 2 + Y2′ 3 + 2 3 3 Y1′ + Y1′ 2 2 Y2′ + Y2′ 2 If this equation is manipulated into the form of a polynomial in Y1′ , it is possible to eventually produce the following equation that assumes that Y2′ is known: 3Q′ 3Q′ =0 Y1′ 4 + (2.5 + Y2′ )Y1′ 3 + (Y2′ 2 + 2.5Y2′ + 1.5)Y1′ 2 + Y2′ 2 + 1.5Y2′ − Y1′ − 2 (Y Y2′ + Y2′ ) Y2′ This equation is a fourth degree polynomial, and therefore would be expected to possibly have four real roots. In the process of obtaining Equation 3.20 the value (Y1′ − Y2′ ) was divided out, and
The Momentum Principle Applied to Open Channel Flows
199
therefore the known root downstream from the hydraulic jump is no longer a solution to Equation 3.20. It turns out that two of the possible roots are generally complex conjugate roots, but may become real negative roots, and one of the remaining roots always gives a negative value for Y1′, and since negative roots are generally complex conjugate roots, but may become negative roots, and one of the remaining roots always gives a negative value for Y1′ , and since negative roots are physically not possible, it is the remaining real root with a positive value for Y1′ that is sought. The trick is how do you find this root best? A host of iterative methods might be used, including the Newton method, which has been used extensively to this point. It is informative to utilize a method capable of finding all roots including the complex roots. Laguerre’s method is well suited for this purpose. However, since Laguerre’s method is quite involved, it is not described herein. For a description of this method for finding roots of a polynomial consult a book that deals with numerical methods. The following FORTRAN program is designed to extract all four roots from Equation 3.20, print them out, and then compute the depth which is conjugate to the given depth. Appendix B describes the arguments of Subroutine LAGU. The results from the program are applicable only for trapezoidal channels. Listing of FORTRAN program LAGU.FOR that uses Laguerre’s method of finding all roots of dimensionless momentum function M1′ = M2′ . PARAMETER (ND=4,EPS=1.E−6) REAL m COMPLEX C(ND+1),ROOTS(ND),AD(ND+1),Z1,Z2,Z3 EPS1=2.*EPS*EPS 10 WRITE(6,*)' Give:b,m,Q,g & Y or one of conjugate depths' READ(5,*) b,m,Q,G,Y QP=(Q/b)**2*(m/b)**3/G YP=m*Y/b C(5)=CMPLX(1.,0.) C(4)=CMPLX(2.5+YP,0.) YP1=YP*(YP+1.) C(3)=CMPLX(YP1+1.5*(YP+1.),0.) C(2)=CMPLX(YP1+.5*YP−3.*QP/YP1,0.) C(1)=CMPLX(−3.*QP/YP,0.) DO 20 J=1,ND+1 20 AD(J)=C(J) DO 30 J=ND,1,−1 Z1=CMPLX(0.,0.) CALL LAGU(AD,J,Z1,EPS) IF(ABS(AIMAG(Z1)).LE.EPS1*ABS(REAL(Z1))) & Z1=CMPLX(REAL(Z1),0.) ROOTS(J)=Z1 Z2=AD(J+1) DO 30 JJ=J,1,−1 Z3=AD(JJ) AD(JJ)=Z2 30 Z2=Z1*Z2+Z3 DO 50 J=2,ND Z1=ROOTS(J) DO 40 I=J−1,1,−1 IF(REAL(ROOTS(I)).LE.REAL(Z1)) GO TO 50 40 ROOTS(I+1)=ROOTS(I) I=0
200
50 60 70 80 10
Open Channel Flow: Numerical Methods and Computer Applications
ROOTS(I+1)=Z1 DO 60 I=1,ND IF(AIMAG(ROOTS(I)).NE. 0.) GO TO 60 IF(REAL(ROOTS(I)).LT. 0.) GO TO 60 YP=REAL(ROOTS(I)) WRITE(6,70) ROOTS(I) FORMAT(2F12.4) WRITE(6,80) Y,YP*b/m FORMAT(' Conjugate depth to',F10.3,' equals ',F10.3,//) WRITE(6,*)' Give 1 if to solve another problem,', &' otherwise 0' READ(5,*) I IF(I.EQ.1) GO TO 10 STOP END SUBROUTINE LAGU(C,ND,Z1,EPS) COMPLEX C(ND+1),Z1,DX,ZO,Z2,Z3,Z4,Z5,DZ,SS,Z6,Z7,Z8, &ZERO,XX,FF ZERO=CMPLX(0.,0.) DO 20 ITER=1,50 Z2=C(ND+1) Z3=ZERO Z4=ZERO DO 10 J=ND,1,−1 Z4=Z1*Z4+Z3 Z3=Z1*Z3+Z2 Z2=Z1*Z2+C(J) IF(CABS(Z2).LE.1.E8) THEN DX=ZERO ELSE IF(CABS(Z3).LE.1.E8.AND.CABS(Z4).LE.1.E8) THEN &DX=CMPLX(CABS(Z2/C(ND+1))**(1./FLOAT(ND)),0.) ELSE Z5=Z3/Z2 Z8=Z5*Z5 DZ=Z8−2.*Z4/Z2 XX=(ND−1)*(ND*DZ−Z8) YY=ABS(REAL(XX)) ZZ=ABS(AIMAG(XX)) IF(YY.LT.1.E12.AND. ZZ.LT.1.E12) THEN SS=ZERO ELSE IF (YY.GE.ZZ) THEN FF=(1./YY)*XX SS=SQRT(YY)*CSQRT(FF) ELSE FF=(1./ZZ)*XX SS=SQRT(ZZ)*CSQRT(FF) ENDIF Z6=Z5+SS Z7=Z5−SS IF(CABS(Z6).LT.CABS(Z7)) Z6=Z7 DX=FLOAT(ND)/Z6
The Momentum Principle Applied to Open Channel Flows
20
201
ENDIF ZO=Z1−DX IF(Z1.EQ.ZO) RETURN Z1=ZO IF(CABS(DX).LE.EPS*CABS(Z1)) RETURN CONTINUE WRITE(6,*)' FAILED TO CONVERGE' RETURN END
Listing of program LAGU.CPP #include #include <math.h> #include <stdio.h> #include <stdlib.h> void lagu(complex *c,int nd,complex *z1,float eps){ complex dx,zo,z2,z3,z4,z5,dz,ss,z6,z7,z8,zero=complex(0.,0.),xx,ff; int iter,j;float yy,zz; for(iter=1;iter<=50;iter++){z2=c[nd]; z3=zero;z4=zero; for(j=nd;j>0;j−−){z4=(*z1)*z4+z3;z3=(*z1)*z3+z2; z2=(*z1)*z2+c[j−1];} if(abs(z2)<=1.e−20) dz=zero; else if((abs(z3)<=1.e−8)&&(abs(z4)<=1.e−8))\ dx=complex(pow(abs(z2/c[nd]),1./(float) nd),0.); else {z5=z3/z2;z8=z5*z5;dz=z8−2.*z4/z2; xx=(float)(nd−1)*((float)nd*dz−z8); yy=fabs(real(xx)); zz=fabs(imag(xx)); if((yy<1.e−12)&&(zz<1.e−12)) ss=zero; else if(yy>=zz){ ff=(1./yy)*xx;ss=sqrt(yy)*sqrt(ff);} else {ff=(1./zz)*xx;ss=sqrt(zz)*sqrt(ff);} z6=z5+ss;z7=z5−ss;if(abs(z6)0;j−−){z1=complex(0.,0.);*zz1=z1; lagu(ad,j,zz1,eps); z1=(*zz1); if(fabs(imag(z1))<=eps1*fabs(real(z1))) z1=complex(real(z1),0.); roots[j−1]=z1; z2=ad[j];for(jj=j;jj>0;jj−−){z3=ad[jj−1]; ad[jj−1]=z2;z2=z1*z2+z3;}}
202
Open Channel Flow: Numerical Methods and Computer Applications
for(j=2;j<=nd;j++){z1=roots[j−1]; for(i=j−1;i>0;i−−){ii=i−1; if(real(roots[ii])<=real(z1)) goto L50; roots[i]=roots[ii];}ii=−1; L50:roots[ii+1]=z1;} for(i=1;i<=nd;i++){ii=i−1;if(imag(roots[ii])!=0.) goto L60; if(real(roots[ii])<0.) goto L60;yp=real(roots[ii]); L60:printf("%12.4f %11.4f\n",roots[ii]);} printf("Conjugate depth to %10.3f equals %10.3f \n\n",y,yp*b/m); printf("Give 1 if to solve another problem, otherwise 0\n"); scanf("%d",&i); if(i>0) goto L10;} Equation 3.20 is written so that Y1′ is unknown, and Y2′ is known. By developing the equation slightly differently, it is possible to obtain an equation identical to Equation 3.20 except that the 1 and 2 subscripts are interchanged. In other words Equation 3.20 is valid if we consider Y2′ the depth upstream from the hydraulic jump and Y1′ the depth downstream from the hydraulic jump. Thus while one of the purposes associated with examining dimensionless momentum functions was to obtain graphical solutions for problems controlled by the momentum principle, we see that their usefulness extends also into solving these problems numerically. The Newton method might be used in solving Equation 3.20 in place of Laguerre’s method. It is generally easier to supply a good enough starting value to a dimensionless equation, than to the original equation. An alternative form of the dimensionless momentum equation to Equation 3.20 that contains both real positive conjugate roots plus the roots that are physically not possible is to use Equation 3.18 directly so that M′ = m2M/b3 is evaluated using the known depth. Upon multiplying Equation 3.18 through by the denominator under Q′ the following fifth degree polynomial equation results:
Y′5 5 4 Y′3 Y′ + − M′Y′2 − M′Y′ + Q′ = 0 + 2 3 6
(3.18a)
The subscripts are left off from the dimensionless variables, but Q′ and M′ will be evaluated using the known dimensionless depth, and the roots of the equation for Y′ will include this value as one of the possibilities. To extract the roots from this equation using the above program that implements the Laguerre’s method the degree of polynomial needs to be increased by 1 by changing ND to 5 and defining the coefficient with the following statements rather than those in the listed program. Fortran QP=(Q/b)**2*(m/b)**3/G YP=m*Y/b MP=(.5+YP/3.)*YP**2+QP/(YP*(1.+YP)) C(6)=CMPLX(.3333333,0.) C(5)=CMPLX(.8333333,0.) C(4)=CMPLX(.5,0.) C(3)=CMPLX(−MP,0.) C(2)=CMPLX(−MP,0.) C(1)=CMPLX(QP,0.) C qp=pow(q/b,2.)*pow(m/b,3.)/g; yp=m*y/b; mp=(.5+yp/3.)*yp*yp+qp/(yp*(1.+yp)); c[5]=complex(.3333333,0.);
203
The Momentum Principle Applied to Open Channel Flows
c[4]=complex(.8333333,0.); c[3]=complex(.5,0.); c[2]=complex(−mp,0.);c[1]=complex(−mp,0.); c[0]=complex(qp,0.); The other change desirable is to bring the WRITE statement within the DO 60 loop so that both real positive roots are written out. The same procedure and program can be used to obtain all roots from the dimensionless specific energy Equation 2.27. Rewriting this equation as a fifth degree polynomial results in Y′ 5 + (2 − E′ )Y′ 4 + (1 − 2E′)Y′ 3 − E′Y′ 2 +
Q′ =0 2
and the statements that define the polynomial coefficients in the program to find the roots of this equation can be modified to the following (LAGUE.FOR):
QP=(Q/b)**2*(m/b)**3/g YP=m*Y/b EP=.5*QP/(YP+YP**2)**2+YP C(6)=CMPLX(1.,0.) C(5)=CMPLX(2.−EP,0.) C(4)=CMPLX(1.−2.*EP,0.) C(3)=CMPLX(−EP,0.) C(2)=CMPLX(0.,0.) C(1)=CMPLX(.5*QP,0.)
To handle any of these problems, the main program can be modified to read in the degree of the polynomial and its coefficients as given below. Listing of program LAGU5.FOR designed to extract roots of any polynomial PARAMETER (EPS=1.E−6) REAL CO(10) COMPLEX C(11),ROOTS(10),AD(11),Z1,Z2,Z3 EPS1=2.*EPS*EPS 10 WRITE(6,*)' Give: ND, C0,C1,C2,..CND' READ(5,*) ND,(CO(I),I=1,ND+1) DO 15 I=1,ND+1 15 C(I)=CMPLX(CO(ND+2−I),0.) DO 20 J=1,ND+1 20 AD(J)=C(J) DO 30 J=ND,1,−1 Z1=CMPLX(0.,0.) CALL LAGU(AD,J,Z1,EPS) IF(ABS(AIMAG(Z1)).LE.EPS1*ABS(REAL(Z1)))Z1=CMPLX(REAL(Z1),0.) ROOTS(J)=Z1 Z2=AD(J+1) DO 30 JJ=J,1,−1 Z3=AD(JJ) AD(JJ)=Z2 30 Z2=Z1*Z2+Z3 DO 50 J=2,ND
204
40 50 65 60 70
Open Channel Flow: Numerical Methods and Computer Applications
Z1=ROOTS(J) DO 40 I=J−1,1,−1 IF(REAL(ROOTS(I)).LE.REAL(Z1)) GOTO 50 ROOTS(I+1)=ROOTS(I) I=0 ROOTS(I+1)=Z1 DO 60 I=1,ND IF(AIMAG(ROOTS(I)).NE. 0.) GO TO 60 IF(REAL(ROOTS(I)).LT. 0.) GO TO 60 YP1=REAL(ROOTS(I)) WRITE(6,65) YP1 FORMAT(' A positive real root is',F10.3) WRITE(6,70) ROOTS(I) FORMAT(2F14.8) END Example Problem 3.6 Using the dimensionless momentum Equation 3.20 find the depth conjugate to 6 ft in a trapezoidal channel with a bottom width b = 10 ft, and a side slope m = 1.5 for a flow rate Q = 450 cfs. Solution With Y2 = 6 ft, Y2′ = 0.9 , and Equation 3.20 becomes, Y1′ 4 + 3.4Y1′ 3 + 4.00Y1′ 2 + 1.788Y1′ − 0.707 = 0
The roots for this equation, as determined by the above program are: root # 1 (−1.261,0.) real but negative, root # 2 (−1.186,−1.004) complex, root # 3 (−1.186,1.004) complex, and root # 4 (0.2324,0.) real and positive or the root that is sought. It is informative to solve this problem for different flow rates and depths downstream from the hydraulic jump. The small table below shows a few such solutions. Note that only for the problem in which the flow rate was reduced to 50 cfs, and the downstream depth Y2 = 3 ft did the two complex conjugate roots become two additional real negative roots. Table giving four solution to M1′ = M 2′ in trapezoidal with b = 10 ft, and m = 1.5 Flow Rate Q (cfs) Y2 (ft) 600 450 100 50
6 6 3 3
Root # 1
Root # 2
Root # 3
Root # 4
Real
Imaginary
Real
Imaginary
Real
Imaginary
Real
Imaginary
Conjugate Depth Y1 (ft)
−1.356 −1.261 −1.284 −1.258
0.000 0.000 0.000 0.000
−1.198 −1.185 −0.867 −0.873
−1.094 −1.004 −0.232 0.000
−1.198 −1.185 −0.867 −0.838
1.094 1.004 0.232 0.000
0.3523 0.2324 0.0676 0.0190
0.000 0.000 0.000 0.000
2.349 1.550 0.450 0.127
It will be instructive for you to obtain these same answers by using the modification of the program that solve the fifth degree polynomial.
For a rectangular section the above definition of Y′, when dealing with dimensionless momentum functions, cannot be used because m = 0. However by redefining the dimensionless depth as the depth of flow divided by the critical depth for rectangular channels some interesting characteristics of this special dimensionless momentum function will be seen. If Equation 3.6 is divided by the critical depth squared and this quantity denoted as a dimensionless momentum function per unit width, then the following results:
205
The Momentum Principle Applied to Open Channel Flows
m′ =
1 m Y2 q2 Ym′ 2 = + = + 2 2 3 2Yc (Y / Yc )gYc 2 Yc Ym′
(3.21)
in which the definition for Yc comes from setting the Froude number squared for a rectangular channel to unity, or Fr2 = q 2 /(gYc3 ) = 1 and the dimensionless depth is defined as Ym′ = Y / Yc. The subscript m used to define this dimensionless depth denotes that it applies for the momentum function per unit width. Below YE′ will be used for a similar dimensionless depth but applicable to specific energy per unit width. Since m′ is a function of the single variable Ym′ Equation 3.21 applies for all rectangular channels and any flow rate in any of these channels. In other words momentum functions for rectangular channels can be reduced to a single dimensionless function per unit width. Figure 3.3 gives this dimensionless momentum function, and it can be used to graphically solve any problem involving momentum in a rectangular channel. Since the Froude number squared Fr2 = q 2 /(gY 3 ) = (Yc / Y)3 for a rectangular channel, it can be noted that either the upstream supercritical, or downstream subcritical Froude number squared can be obtained by taking the reciprocal of the appropriate dimensionless depth cubed, or the Froude number, Fr = 1/(Ym′ )1.5.
5.0
0
1
2
3
4
5
6
7
8
9
10
4.5
4.5
er, F
r1
4.0
Froude numb
3.5 3.0
D
m
n ow
ea str
de
, pth
2.5 2.0 1.5 1.0 0.5
Downstre
e number , Fr2
0 0
1
Upstrea m
am Froud
2
3
4.0
Y΄
Upstre am
Dimensionless depth, Y΄m = Y/Yc
5.0
depth, Y ΄
Subcritical
Supercritical
m΄
Y΄
Fr
Y΄
Fr
1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 3.10 3.20 3.30 3.40 3.50 3.60 3.70 3.80 3.90 4.00 4.10 4.20 4.30 4.40 4.50 4.60 4.70 4.80 4.90 5.00
1.0000 1.2793 1.4064 1.5078 1.5959 1.6751 1.7481 1.8163 1.8806 1.9416 2.0000 2.0560 2.1100 2.1622 2.2127 2.2618 2.3096 2.3561 2.4015 2.4459 2.4893 2.5318 2.5735 2.6144 2.6545 2.6940 2.7328 2.7710 2.8085 2.8455 2.8820 2.9180 2.9534 2.9884 3.0230 3.0571
1.000 0.6911 0.5996 0.5401 0.4960 0.4612 0.4327 0.4085 0.3878 0.3696 0.3536 0.3392 0.3263 0.3145 0.3038 0.2940 0.2849 0.2765 0.2687 0.2614 0.2546 0.2482 0.2422 0.2366 0.2312 0.2262 0.2214 0.2168 0.2125 0.2083 0.2044 0.2006 0.1970 0.1936 0.1903 0.1871
1.000 0.7648 0.6812 0.6226 0.5768 0.5392 0.5073 0.4796 0.4553 0.4337 0.4142 0.3966 0.3806 0.3659 0.3524 0.3399 0.3283 0.3175 0.3074 0.2980 0.2892 0.2809 0.2730 0.2656 0.2586 0.2520 0.2457 0.2397 0.2340 0.2286 0.2235 0.2185 0.2138 0.2093 0.2050 0.2008
1.0000 1.4952 1.7786 2.0356 2.2826 2.5257 2.7678 3.0106 3.2550 3.5017 3.7511 4.0036 4.2593 4.5183 4.7808 5.0468 5.3164 5.5896 5.8664 6.1469 6.4309 6.7186 7.0098 7.3047 7.6030 7.9050 8.2104 8.5193 8.8316 9.1474 9.4666 9.7892 10.1152 10.4444 10.7770 11.1128
3.5 3.0 2.5 2.0 1.5 1.0 0.5 0
4
5
6
7
8
Dimensionless momentum function, m΄ = Y΄m2/2 + 1/Y΄m
Figure 3.3 Dimensionless momentum diagram for rectangular sections.
9
10
206
Open Channel Flow: Numerical Methods and Computer Applications
Cubic Equation 3.21 can readily be solved for its three roots. First solve θ from 1.5 1.5 θ = cos−1 m′ Then the three roots are given by θ ′ = −1.632993m′1 / 2cos negative root Ym1 3 (θ + 2π) Ym2 ′ = −1.632993m ′1/ 2cos subcritical root 3 (θ + 4π) Ym1 ′ = −1.632993m ′1/ 2cos supercritical root 3 The minimum value for m′ is 1.5 corresponding to critical conditions when m′ = m′c . Since a single dimensionless momentum function diagram can be developed for rectangular channels, let us attempt to accomplish the same for the specific energy by dividing E, applicable per unit width, by Yc. The result of this division is E Y q2 Y Yc3 = + = + Yc Yc 2gY 2 Yc Yc 2Y 2 Yc by defining E′ = E/Yc, and YE′ = Y / Yc , this equation becomes the following dimensionless specific energy for a rectangular channel.
E′ = YE′ +
1 (2YE′ 2 )
(3.22)
It needs to be noted that if YE′ = 1/ Ym′ is substituted into Equation 3.22, then Equation 3.21 results. Thus Figure 3.3 can be used to solve specific energy problems in rectangular channels as well as momentum problems. To use Figure 3.3 for specific energy problem the ordinate becomes 1/ YE′ and the abscissa is interpreted as E′. Example Problem 3.7 A hydraulic jump occurs immediately downstream from a sluice gate in a rectangular channel. If the depth upstream in this channel is 4.5 ft, and the flow rate per unit width is q = 5 cfs/ft, then what is the depth downstream from the hydraulic jump, and how much energy is dissipated? Solution This problem can be solved using the appropriate equations, or by utilizing Figure 3.3. If equations are used, then the energy equation E1 = E2 is applied across the gate first to determine the depth downstream from the gate. Solution gives Y2 = 0.30 ft. Next the momentum equation m1 = m2 is applied across the hydraulic jump to find the depth downstream. Therefore, this depth Y3 = 2.13 ft from this last solution. The energy loss due to the hydraulic jump is h L = E1 − E3 = 4.52 − 2.22 = 2.30 ftlb/lb. Using Figure 3.3 requires that the critical depth be computed first
207
The Momentum Principle Applied to Open Channel Flows from Yc = (q2/g)1/3 = 0.919 ft. Therefore Y1′ = 4.5 / 0.919 = 4.896 , or the YE′ for use with the figure equals 1/4.896 = 0.204. Entering with this value gives E′ = 5.0 and therefore 1/ YE2 ′ = 3.06 , which is the value for Ym2 ′ on the ordinate of Figure 3.3. Thus YE2 ′ = 1/ 3.06 = 0.327 , and Y2 = 0.327(0.919) = 0.300 ft. Also Ym2 ′ = 0.327(or m ′m = (0.327)2 / 2 + 1/ 0.327 = 3.112). Then from Figure 3.3 Ym3 ′ = 2.32 and therefore Y3 = 2.13 ft.
An alternative to the dimensionless relationships for a hydraulic jump (i.e., M′1 = M′2 ) is to express the ratio of depths r = Y2′ / Y1′ as a function of the upstream Froude number, Fr1 (or r = Y1′ / Y2′ as a function of Fr2). For a trapezoidal channel the area is A = bY + mY2 = (b2/m)/(Y′ + Y′2) and the Froude number squared is Fr2 = Q 2 T /(gA 3 ) = m 3Q 2 (1 + 2Y′ ) / gb5 (Y′ + Y′ 2 )3 = Q′(1 + 2Y′ )/(Y′ + Y′ 2 )3 in which, as before, Q′ = m3Q2/(gb5). Thus Q′ is related to the Froude number by Q′ = {(Y′ + Y′ 2 )3 /(1 + 2Y′ )}Fr2 . The dimensionless momentum function M′ as given by Equation 3.18 can be written across a hydraulic jump in a trapezoidal channel, (i.e., M1′ = M′2 ) as
{
}
Y1′2 Y1′3 (Y1′ + Y1′2 )2 2 Y2′2 Y2′3 Q′ + + + + Fr1 = (Y 2 3 (1 + 2Y1′) 2 3 Y2′ + Y2′2 ) But since Q′ on the right side of the equation can also be expressed in terms of the upstream dimensionless depth and the upstream Froude number, the last term in this equation can be written as (Y1′ + Y1′2 )3 / {(Y2′ + Y2′ )(1 + 2Y2′ )} Fr1 . Upon dividing the above equation by Y12 and defining the ratio r = Y2′ / Y1′ , and rearranging terms the following cubic equation results: 3 2 Y1′(r 3 − 1) (r 2 − 1) (1 + Y1′ ) /(r(1 + rY1′ )) − (1 + Y1′) 2 + + Fr1 = 0 3 2 (1 + 2Y1′)
One root of this equation is r = 1 and generally one of the two remaining roots will be positive (the one being sought), and the other is negative (a physical impossible root). Note from this equation that the ratio r of downstream to upstream depth in a trapezoidal channel is a function of the upstream Froude number Fr1 and the dimensionless upstream depth Y1′ = mY1 / b . This relationship is shown in Figure 3.4, and also given in Table 3.1. If the ratio r is defined as upstream depth divided by the downstream depth, r = Y1/Y2, and the same procedure followed except that the downstream Froude number squared, Fr22 , is retained, then an identical equation with the subscripts reversed will be obtained. This equation is 3 2 Y2′ (r 3 − 1) (r 2 − 1) (1 + Y2′ ) /(r(1 + rY2′ )) − (1 + Y2′ ) 2 + + Fr2 = 0 3 2 (1 + 2Y2′ )
For a rectangular channel the arithmetic is simpler. Equating the dimensionless momentum function per unit width m1′ = m′2 , dividing by Y1′2 , and rearranging terms gives r 3 − (2Fr12 + 1)r + 2Fr12 = 0 and upon extracting the known root r = 1 reduces this equation to the quadratic equation, r 2 + r − 2Fr12 = 0
208
Open Channel Flow: Numerical Methods and Computer Applications
Downstream depth/upstream depth, Y2/Y1
30
25
= Y΄1
pez Tra
15
10
Re
c
g tan
c
5
0.0
0.1
0.2
1.5
5
0
r ula
l ne
n ha
1
2
3
4
5
6
7
ls
nne
ha al c
oid
20
1
0.0
0.3 0.4
0.5
0.8 0.6 0.7 0.9 1 1.25
8 9 10 11 12 13 14 15 16 17 18 19 20 Upstream Froude number, Fr1
Figure 3.4 Relationship between ratio of depths r = Y2/Y1 with the upstream Froude number Fr1 = q /(gY12 )1/ 2 = V1 /(gY1 )1/ 2 for rectangular channels and with the upstream Froude number Fr1 = {Q 2 T1 /(gA13 )}1/ 2 and the dimensionless downstream depth Y1′ = mY1 / b for trapezoidal channels.
Applying the quadratic formula to this equation produces Equation 3.10, as would be expected, since r = Y2′ / Y1′ = Y2 / Y1 . Figure 3.4 and Table 3.1 contains the result for a rectangular channel, as well as for trapezoidal channels with different values for the dimensionless upstream depth Y1′. Figure 3.5 and Table 3.2 are similar to Figure 3.4 and Table 3.1, except the ratio is now for the upstream depth divided by the downstream depth, or r = Y1 / Y2 = Y1′/ Y2′ , and the downstream Froude number Fr2 is given in the first column.
3.6 Celerity of Small Amplitude Gravity Waves An application of the momentum principle is the determination of the speed of movement, or celerity, c, of a small amplitude gravity wave on the surface of a body of water. Should one throw a stone in a still pool of water, the entry of the stone through the water surface causes the depth to increase by a very small amount, and an increase in depth propogates radially outward with a celerity c. Should a stick be held in a stream of flowing water its existence in the flow causes the depth to rise, and this effect is propogated outward also. Should the velocity of flow be larger than c, e.g., the flow is supercritical flow, then the effect would not proprogate upstream but be seen as a wedgeshaped wave downstream from the stick. For larger velocities the angle of the wedge becomes smaller. The Froude number is the ratio of the average velocity of the flow in a channel divided by the celerity of a small amplitude gravity wave. If the amplitude, or height of the wave, is more than a very small fraction of the depth, then the speed of this wave is not defined by the variable c. Consider a small amplitude gravity wave as shown in the sketch below in which the increase in depth at the wave is dY, and the depth in this channel, which we will assume is not flowing, to
Rect.
1.00 2.37 3.77 5.18 6.59
8.00 9.41 10.82 12.24 13.65
15.06 16.48 17.89 19.31 20.72
22.13 23.55 24.96 26.37 27.79
Fr
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
20.80 22.04 23.28 24.51 25.74
14.43 15.72 17.00 18.28 19.54
7.82 9.16 10.50 11.82 13.13
1.00 2.36 3.74 5.11 6.47
0.01
17.70 18.64 19.57 20.49 21.39
12.74 13.77 14.78 15.77 16.74
7.26 8.41 9.53 10.63 11.70
1.00 2.32 3.61 4.87 6.08
0.05
0.1
15.68 16.47 17.24 18.00 18.75
11.51 12.38 13.23 14.07 14.88
6.78 7.78 8.76 9.70 10.62
1.00 2.27 3.49 4.63 5.73
0.2
13.55 14.19 14.82 15.44 16.05
10.11 10.84 11.54 12.22 12.89
6.15 7.00 7.82 8.61 9.37
1.00 2.20 3.29 4.30 5.25
12.36 12.93 13.49 14.04 14.58
9.30 9.94 10.57 11.18 11.77
5.75 6.51 7.25 7.96 8.64
1.00 2.14 3.15 4.08 4.94
0.3
11.57 12.10 12.61 13.12 13.62
8.75 9.34 9.92 10.48 11.03
5.46 6.18 6.86 7.51 8.14
1.00 2.09 3.05 3.91 4.71
0.4
11.00 11.50 11.98 12.46 12.93
8.35 8.91 9.45 9.98 10.50
5.25 5.93 6.57 7.19 7.78
1.00 2.05 2.96 3.78 4.54
0.5
10.58 11.05 11.51 11.96 12.41
8.05 8.58 9.10 9.60 10.09
5.09 5.73 6.35 6.93 7.50
1.00 2.02 2.90 3.69 4.41
0.6
Dimensionless Upstream Depth Y1′ = mY1/b
10.24 10.69 11.13 11.57 12.00
7.81 8.32 8.82 9.30 9.77
4.96 5.58 6.17 6.73 7.28
1.00 1.99 2.84 3.60 4.30
0.7
9.96 10.40 10.83 11.25 11.67
7.61 8.11 8.59 9.06 9.52
4.85 5.45 6.02 6.57 7.10
1.00 1.97 2.80 3.54 4.22
0.8
9.74 10.16 10.58 10.99 11.40
7.45 7.93 8.40 8.85 9.30
4.76 5.35 5.90 6.44 6.95
1.00 1.95 2.76 3.48 4.14
0.9
9.54 9.96 10.37 10.77 11.17
7.31 7.78 8.24 8.68 9.12
4.68 5.26 5.80 6.32 6.82
1.00 1.94 2.73 3.43 4.08
1.0
TABLE 3.1 Ratio of Downstream to Upstream Depth, Y2/Y1, across a Hydraulic Jump in Rectangular and Trapezoidal Channels
9.18 9.57 9.96 10.35 10.72
7.04 7.49 7.93 8.35 8.77
4.53 5.08 5.60 6.10 6.58
1.00 1.90 2.66 3.34 3.96
1.25
8.91 9.29 9.67 10.04 10.40
6.85 7.28 7.70 8.11 8.52
4.42 4.95 5.45 5.94 6.40
1.00 1.88 2.61 3.27 3.87
1.5
The Momentum Principle Applied to Open Channel Flows 209
210
Open Channel Flow: Numerical Methods and Computer Applications 1
Upstream depth/downstream depth, Y1/Y2
0.9 0.8 0.7
Trapezoidal channels, Y΄2
0.6 0.5 0.4
Y΄2 = 1.5
1.25
1
0.3 Re
0.2
ct
0.8 0.9
an
l gu
0.6 0.7
ar
0.5
el nn
0.4
0.01 0.05 0.2 0.3
a ch
0.1 0
0
0.1
0.6 0.7 0.3 0.4 0.5 Downstream Froude number, Fr2
0.2
0.8
0.9
1
Figure 3.5 Relationship between ratio of depths r = Y1/Y2 with the downstream Froude number Fr2 = q /(gY23 )1/ 2 = V2 /(gY2 )1/ 2 for rectangular channels and with the downstream Froude number Fr2 = {Q 2 T2 /(gA 32 )}1/ 2 and the dimensionless downstream depth Y2′ = mY2 / b for trapezoidal channels.
simplify the analysis but will not affect the magnitude determine for c, is Y. The channel is prismatic and has a crosssectional area, A. To analyze this wave consider an observer moving with the celerity of the wave so that from his point of view a steady state flow exists. A stationary observer would see unsteady flow in the channel as the wave passed. c Y
dY
v
Y + dY
γA(hc + dY)
A + dA
dY
c
A γ Ahc
To the moving observer the continuity equation is, v(A + dA) = cA
in which v is the average velocity this moving observer sees at the section with the incrementally larger depth Y + dY and dA is the incremental increase in this area due to the rise, dY, of the liquid surface. Rewriting this equation so that v appears by itself on one side of the equal sign gives
v=c
A A + dA
(continuity equation as seen by a moving observer )
Rect.
.0050 .0196 .0431 .0745 .1124 .1557 .2036 .2550 .3093 .3660 .4247 .4849 .5464 .6091 .6726 .7369 .8019 .8675 .9335 1.00
Fr2
.05 .10 .15 .20 .25 .30 .35 .40 .45 .50 .55 .60 .65 .70 .75 .80 .85 .90 .95 1.00
0.01
.0050 .0197 .0433 .0747 .1127 .1562 .2041 .2556 .3100 .3668 .4255 .4857 .5473 .6099 .6733 .7376 .8025 .8679 .9337 1.00
0.05
.0051 .0200 .0439 .0757 .1142 .1581 .2065 .2584 .3131 .3701 .4290 .4893 .5508 .6132 .6764 .7403 .8047 .8695 .9346 1.00
0.1
.0052 .0204 .0448 .0771 .1162 .1608 .2097 .2621 .3173 .3746 .4336 .4940 .5553 .6176 .6804 .7437 .8075 .8715 .9357 1.00
0.2 .0054 .0213 .0467 .0804 .1208 .1667 .2168 .2702 .3261 .3840 .4432 .5036 .5647 .6263 .6883 .7506 .8129 .8753 .9377 1.00
0.3 .0057 .0223 .0489 .0839 .1257 .1729 .2241 .2785 .3351 .3933 .4527 .5129 .5736 .6346 .6958 .7569 .8180 .8789 .9396 1.00
.0060 .0235 .0512 .0875 .1307 .1792 .2315 .2867 .3438 .4023 .4617 .5217 .5819 .6423 .7025 .7626 .8225 .8820 .9412 1.00
0.4 .0063 .0246 .0536 .0913 .1358 .1854 .2387 .2945 .3521 .4108 .4701 .5298 .5895 .6492 .7086 .7677 .8264 .8848 .9426 1.00
0.5 .0066 .0258 .0559 .0950 .1408 .1915 .2456 .3020 .3599 .4186 .4778 .5372 .5964 .6554 .7140 .7722 .8300 .8872 .9439 1.00
0.6 .0069 .0270 .0583 .0987 .1456 .1973 .2522 .3090 .3671 .4259 .4849 .5439 .6026 .6609 .7188 .7762 .8330 .8893 .9449 1.00
0.7
Dimensionless Downstream Depth Y2′ = mY2/b .0072 .0282 .0607 .1023 .1503 .2029 .2584 .3156 .3739 .4326 .4914 .5499 .6082 .6659 .7232 .7798 .8358 .8912 .9459 1.00
0.8 .007 .029 .0631 .1058 .1549 .2082 .2642 .3218 .3801 .4387 .4973 .5555 .6132 .6704 .7270 .7830 .8382 .8928 .9467 1.00
0.9
1.0 .0079 .0306 .0654 .1092 .1593 .2133 .2697 .3275 .3859 .4444 .5026 .5605 .6178 .6745 .7305 .7858 .8404 .8943 .9475 1.00
Table 3.2 Ratio of Upstream to Downstream Depth, Y1/Y2, across a Hydraulic Jump in Rectangular and Trapezoidal Channels 1.25 .0087 .0336 .0711 .1174 .1694 .2249 .2821 .3402 .3985 .4566 .5142 .5713 .6275 .6831 .7378 .7918 .8450 .8974 .9490 1.00
.0096 .0365 .0764 .1249 .1785 .2349 .2927 .3509 .4090 .4667 .5237 .5799 .6353 .6899 .7436 .7965 .8485 .8998 .9503 1.00
1.5
The Momentum Principle Applied to Open Channel Flows 211
212
Open Channel Flow: Numerical Methods and Computer Applications
Since this observer sees a steadystate flow, the momentum at both sections must be equal or M1 = M 2, or,
A(h c + dY) +
v2 (A + dA) c2A = Ah c + g g
(momentum equation as seen by a moving observer)
In the above equation, it should be noted that since the water surface has risen an amount dY at the wave that a firstorder approximation (i.e., ignoring products of two differential amounts) is obtained by adding dY to hc. If v, as defined by the continuity equation, is substituted into this momentum equation, the result rearranged, the second order term dY/dA deleted, and dA/dY replaced by T (the top width) the following equation is obtained: c=
gA = gYh T
(3.23)
in which Yh = A/T is called the hydraulic depth. For a rectangular channel the hydraulic depth is the same as the depth Y, but for other channels this is not the case. For a rectangular channel the celerity of a small amplitude gravity wave is given by c = gY
(3.24)
3.7 Constant Height Waves General unsteady flow problems in channels is the subject of Chapter 6 (and the numerical solution to such problems in Chapter 7), but certain special situations occur that allow an unsteady problem to be solved by having the observer move such that the flow is steady to him. For these types of problems, the momentum principle often provides one of the equations needed for a solution. A class of these problems involves the rapid operation of gates in a channel that contains uniform, or near uniform flow, prior to the change in the gate opening. For example, assume that a gate is suddenly closed down further. Since this incremental closure will reduce the flow rate past the gate, a surge will form upstream from the gate across which the prior depth changes to the new depth established by energy for the new gate setting. If the upstream flow is uniform, then the speed, v, at which this wave moves will be constant, and to an observer moving with this wave a steadystate flow will exit. In fact, he will see a hydraulic jump in front of him, with an incoming velocity equal to the sum of the previous velocity and his velocity in the opposite direction. This combined velocity produces a supercritical flow in his eyes. Downstream he sees a subcritical flow. In other words from his view point the momentum equation M1 = M2 is valid; however, in place of the upstream flow rate Q1 the equation would contain (V1 + v)A1 and the downstream flow rate Q2 = (V2 + v)A2, in which V1 is the original velocity upstream from the surge, and V2 is the new velocity in the channel downstream from the surge. Making these substitutions the momentum equation across this moving surge becomes
(Ah c )1 +
(v + V1 )2 A1 (v + V2 )2 A 2 = (Ah c )2 + g g
(3.25)
Likewise the continuity equation to this moving observer is
(V1 + v)A1 = (V2 + v)A 2
(3.26)
213
The Momentum Principle Applied to Open Channel Flows
v Y1
V1
V2
Y2
Y3 = CcyG yG
The unknown variables of a problem of this type are v, Y2, and V2. The third equation that is required comes from energy at the gate from the viewpoint of a stationary observer. As the gate closes rapidly, it will cause a reduction in depth downstream from the gate. If an observer moves with this resulting surge again a steadystate flow will be seen if the original downstream depth is constant. The above Equations 3.25 and 3.26 are valid from his view point if the + signs within the ( ) are changed to − signs. The reason for the sign change is that the observer is now moving in the same direction as the flow.
Y1
b1 = 10 ft m1 = 1 n1 = 0 .014, So1 = 0.0005
.82
=0 c
C
V2 Y2
G=
V1
Y
v
0.5
ft
Example Problem 3.8 A gate in a channel is suddenly closed from an entirely open position to a distance of 0.5 ft from the channel bottom. Its contraction coefficient is Cc = 0.82. Upstream from the gate the channel is trapezoidal with b1 = 10 ft, m1 = 1, n1 = 0.014, and So1 = 0.0005. Downstream from the gate the width of the channel is 8 ft, and it is rectangular here. Prior to the gate closure uniform flow existed throughout the trapezoidal channel. Determine the flow rate past the gate after it is partly closed, the depth on the upstream side of the gate and the speed at which the surge moves upstream if prior to closure the flow rate was Q = 250 cfs.
Y3 V3
Rectangular b3 = 8 ft
Solution The assumption made in solving this problem is that to an observer moving with the speed of the surge a steady state flow occurs, and he sees a hydraulic jump in front of him. First, uniform flow is solved in the trapezoidal channel with the results: V1 = 4.504 fps, and Y1 = 3.973 ft. Now the unknowns to this problem are: v, Y2, and V2. The three available equations are: From moving observer
F1 = (V1 + v)A1 − (V2 + v)A 2 = 0 (continuity) Q2 1 1 − F2 = h c1A1 − h c2 A 2 + = 0 (momentum) g A1 A 2 Stationary observer at gate
F3 = Y2 − Y3 +
V22 ( V A )2 − 2 22 = 0 (energy ) (2g) (2gA 3 )
The solution to these three equation produces the following results: Y2 = 5.202′, V2 = 0.729 fps, v = 8.101 fps.
214
Open Channel Flow: Numerical Methods and Computer Applications The listing of the following PASCAL and FORTRAN programs illustrate how problems of this type can be solved utilizing the Newton method. PASCAL computer program to solve wave problems Program Gate_Wave; Var Y3,vw,b,m,A3S,A1:Real; Y,V: array [1..2] of real; EQ:array [1..3] of real; D: array[1..3] of array[1..3] of real; Const g=32.2;g2=64.4; Function A(K:integer):real; Begin A:=(b+m*Y[K])*Y[K] End; Function MOM(K:integer):real; Begin MOM:=(b/2+m*Y[K]/3)*sqr(Y[K])+sqr((V[K]+vw))*A(K)/g; End; Function F(K:integer): real; Begin Case K of 1:F:=(V[1]+vw)*A1−(V[2]+vw)*A(2); 2:F:=MOM(1)−MOM(2); 3:F:=Y[2]−Y3+(sqr(V[2])−sqr(V[2]*A(2))/A3S)/g2 end; End; Var I,J,N:integer; FAC,Cc,yG,b3,m3,D1,D2,D3:real; BEGIN Writeln('Give: b,m,V1,Y1,yG,Cc,b3,m3 & est. for vw,Y2,V2'); Readln(b,m,V[1],Y[1],yG,Cc,b3,m3,vw,Y[2],V[2]); Y3:=Cc*yG; A3S:=sqr((b3+m3*Y3)*Y3); A1:=A(1); repeat For I:=1 to 3 do Begin EQ[I]:=F(I); For J:=1 to 3 do begin Case J of 1:vw:=vw−0.001; 2:Y[2]:=Y[2]−0.001; 3:V[2]:=V[2]−0.001;end; D[I,J]:=(EQ[I]−F(I))/0.001; Case J of 1:vw:=vw+0.001; 2:Y[2]:=Y[2]+0.001; 3:V[2]:=V[2]+0.001;end; end; End; {Solves equations} For N:=1 to 2 do Begin For I:=N+1 to 3 do begin FAC:=D[I,N]/D[N,N]; For J:=N+1 to 3 do D[I,J]:=D[I,J]−FAC*D[N,J]; EQ[I]:=EQ[I]−FAC*EQ[N];end; End; D1:=EQ[3]/D[3,3]; V[2]:=V[2]−D1; D2:=(EQ[2]−D1*D[2,3])/D[2,2]; Y[2]:=Y[2]−D2; D3:=(EQ[1]−D2*D[1,2]−D1*D[1,3])/D[1,1]; vw:=vw−D3; until (abs(D1)+abs(D2)+abs(D3))<0.0001; Writeln('Wave speed=',vw:10:3,' Depth=',Y[2]:10:3, ' Velocity=',V[2]:10:3); END. FORTRAN computer program for wave prob., GWAVE.FOR LOGICAL LDV COMMON b,Fm,Y(2),V(2),EQ(3),D(3,3),vw,A1,Y3,A3S,g2,g,FMOM1 g=32.2 g2=64.4 WRITE(6,*)' Give: b,m,V1,Y1,yG,Cc,b3,m3, & est. vw,Y2,V2' READ(5,*) b,Fm,V(1),Y(1),yG,Cc,b3,Fm3,vw,Y(2),V(2) Y3=Cc*yG A3S=((b3+Fm3*Y3)*Y3)**2 A1=A(1)
The Momentum Principle Applied to Open Channel Flows FMOM1=FOM(1) 10 DO 30 I=1,3 EQ(I)=F(I) DO 30 J=1,3 DV=−.001 LDV=.FALSE. 21 IF(J−2) 22,23,24 22 vw=vw+DV GO TO 25 23 Y(2)=Y(2)+DV GO TO 25 24 V(2)=V(2)+DV 25 IF(LDV) GO TO 30 D(I,J)=(EQ(I)−F(I))/.001 DV=.001 LDV=.TRUE. GO TO 21 30 CONTINUE DO 50 N=1,2 DO 50 I=N+1,3 FAC=D(I,N)/D(N,N) DO 40 J=N+1,3 40 D(I,J)=D(I,J)−FAC*D(N,J) 50 EQ(I)=EQ(I)−FAC*EQ(N) D1=EQ(3)/D(3,3) V(2)=V(2)−D1 D2=(EQ(2)−D1*D(2,3))/D(2,2) Y(2)=Y(2)−D2 D3=(EQ(1)−D2*D(1,2)−D1*D(1,3))/D(1,1) vw=vw−D3 IF(ABS(D1)+ABS(D2)+ABS(D3).GT..0001) GO TO 10 WRITE(6,100) vw,Y(2),V(2) 100 FORMAT(' Wave speed=',F10.3,' Depth=',F10.3, &' Velocity=',F10.3) STOP END FUNCTION F(K) COMMON b,Fm,Y(2),V(2),EQ(3),D(3,3),vw,A1,Y3,A3S,g2,g,FMOM1 GO TO (1,2,3),K 1 F=(V(1)+vw)*A1−(V(2)+vw)*A(2) RETURN 2 F=FMOM1−FOM(2) RETURN F=Y(2)−Y3+(V(2)**2−(V(2)*A(2))**2/A3S)/g2 3 RETURN END FUNCTION A(K) COMMON b,Fm,Y(2),V(2),EQ(3),D(3,3),vw,A1,Y3,A3S,g2,g,FMOM1 A=(b+Fm*Y(K))*Y(K) RETURN END FUNCTION FOM(K) COMMON b,Fm,Y(2),V(2),EQ(3),D(3,3),vw,A1,Y3,A3S,g2,g,FMOM1 FOM=(.5*b+Fm*Y(K)/3.)*Y(K)**2+(V(K)+vw)**2*A(K)/g RETURN END
215
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Open Channel Flow: Numerical Methods and Computer Applications
The previous application creates a constant depth upstream moving wave by instantly closing a gate some incremental amount. This will be referred to as a downstream controlled wave (DCW), because it is created by a downstream control. By instantly increasing the flow rate, or depth, at the upstream end of a channel a constant height wave is created that moves downstream. This will be referred to as an upstreamcontrolled wave (UCW), because it is created by an upstream control. In both DCW’s and UCW’s the depths, velocities and flow rates, both upstream and downstream from the waves are constant so that an observer moving with the velocity v of the wave sees a steadystate hydraulic jump. We will now examine an UCW, and thereafter manipulate the resulting equations into various other forms. The equivalent DCW equations will also be given. You should verify these equations. Assume uniform flow exists in a channel with depth Y1, and velocity V1 (or flow rate Q1) constant, when suddenly at its upstream end the velocity (or flow rate) is suddenly increased to V2 (or Q2). Then a constant height wave will result that will move down the channel with a velocity v, and the depth upstream from this wave will be constant at Y2 as shown on the sketch below. v Y2
V2(Q2) V1(Q1)
Y1
From the viewpoint of an observer traveling downstream with the wave velocity v, the continuity equation is (v − V1 )A1 = (v − V2 )A 2 (Cont. Equation)
(3.27)
and equating the momentum function upstream to that downstream (M1 = M2) gives
A1h c1 +
(v − V1 )2 A1 (v − V2 )2 A 2 = A2hc2 + g g
(Mom. Equation)
(3.28)
Notice in both Equations 3.27 and 3.28 that the velocity seen by the moving observer is v–V, rather than v + V as in the DCW. Equations 3.27 and 3.28 allow for two variables to be solved. Typically Y1 and V1(Q1) are known, so if a new velocity V2(Q2) is given then the depth Y2 and v are solved. However, another variable might be specified, such as v, and then V2 might be solved. Depending upon the type of upstream control and what is known, a third equation might be added to Equations 3.27 and 3.28, as was done with a gate having its opening changed in the DCW. For an UCW the gate’s opening would be increased, however, rather than decreased. By examining Equations 3.27 and 3.28 further we can learn more about constant height waves. First let us eliminate V2 by substituting Equation 3.27 into Equation 3.28 to give
A A g(A 2 h c 2 − A1h c1 ) = (v − V1 )2 A1 1 − 1 = (v − V1 )2 1 (A 2 − A1 ) A2 A2
or UCW v = V1 +
gA 2 (A 2 h c 2 − A1h c1 ) g(h c 2 A 2 / A1 − h c1 ) = V1 + A1 ( A2 − A1 ) 1 − A1 / A 2
(3.29)
217
The Momentum Principle Applied to Open Channel Flows
DCW v=
gA 2 (A 2 h c 2 − A1h c1 ) − V1 = A1 (A 2 − A1 )
g(h c 2 A 2 / A1 − h c1 ) − V1 1 − A1 / A 2
(3.29a)
Of course if one also knows what V2 is by having solved it then the wave velocity can be solved from only the continuity equation, or UCW v=
V2 A 2 − V1A1 Q 2 − Q1 = A 2 − A1 A 2 − A1
(3.30)
v=
V1A1 − V2 A 2 Q1 − Q 2 = A 2 − A1 A 2 − A1
(3.30a)
DCW
For a rectangular channel Equations 3.27 and 3.28 simplify to UCW
(v − V1 )Y1 = (v − V2 )Y2 (Rect. Cont. Equation)
(3.31)
(v + V1 )Y1 = (v + V2 )Y2 (Rect. Cont. Equation)
(3.31a)
DCW
and equating the momentum per unit width upstream to that downstream (m1 = m2 by a moving observer) gives UCW
Y12 (v − V1 )2 Y1 Y22 (v − V2 )2 Y2 + = + g g 2 2
(Rect. Mom. Equation)
Y12 (v + V1 )2 Y1 Y22 (v + V2 )2 Y2 + = + g g 2 2
(Rect. Mom. Equation)
(3.32)
DCW
(3.32a)
Equation 3.29 simplifies to the following equations for a rectangular channel: UCW v = V1 +
1 2
gY1
Y22 Y12 − 1 Y22 Y12 − 1 = V1 + c1 1 − Y1 Y2 2 − 2Y1 Y2
(3.33)
or manipulated to another form
2 Y 12Y v = V1 + ( gY1 2 ) 22 + 2 Y 1 Y1
12
= V1 +
c1 2
Y22 Y2 Y 2 + Y 1 1
12
gY = V1 + 2 ( Y2 + Y1 ) 2Y1
12
(3.33a)
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Open Channel Flow: Numerical Methods and Computer Applications
DCW v=
1 2
gY1
Y22 Y12 − 1 Y22 Y12 − 1 − V1 = c1 − V1 1 − Y1 Y2 2 − 2Y1 Y2
(3.33b)
or manipulated to another form
gY v = 1 2
12
Y22 Y2 Y 2 + Y 1 1
12
− V1 =
c1 2
Y22 Y2 Y 2 + Y 1 1
12
gY − V1 = 2 ( Y2 + Y1 ) 2Y1
12
− V1
(3.33c)
or just using the Continuity Equation, UCW v=
V2 Y2 − V1Y1 q 2 − q1 = Y2 − Y1 Y2 − Y1
(3.34)
v=
V1Y1 − V2 Y2 q1 − q 2 = Y2 − Y1 Y2 − Y1
(3.34a)
DCW
As the height of the wave becomes very small the constant height wave becomes a small amplitude wave, and the equations should reflect this. For a small amplitude wave, Y2 approaches the depth Y1, or Y2/Y1 → 1. Let the velocity V1 = 0 and note that if Y2/Y1 = 1 in Equations 3.33a and 3.33c that v = (gY1)1/2 = (gY)1/2 = c, e.g., indicates that the wave velocity equals the celerity of a small amplitude gravity wave as given by Equation 3.24 for a rectangular channel. Notice also that if the wave is moving upstream due to downstream control that the wave velocity is v = c − V1, or the speed with which a small change in depth moves upstream equals the celerity of a small amplitude gravity wave minus the channel velocity that carries it downstream. Likewise a small disturbance in depth will move downstream with a velocity v = c + V1 according to Equation 3.33a as Y2/Y1 = 1. The same results can be obtained from Equations 3.33 and 3.33b, but since they produce the indeterminate form 0/0 L’Hospital’s rule must be applied. We might define the celerity of a standing wave as the speed it moves in a channel minus the velocity in the channel, e.g., the speed it moves as seen by an observer moving with the velocity in the channel. Thus for a UCW this celerity is c = v + V1 and for a DCW this celerity is c = v + V1. From Equations 29 see this celerity is given by v=
gA 2 (A 2 h c 2 − A1h c1 ) = A1 (A 2 − A1 )
g(h c 2 A 2 / A1 − h c1 ) 1 − A1 / A 2
(3.35)
and for a rectangular channel from Equations 3.33,
gY v = 1 2
12
Y22 Y2 Y 2 + Y 1 1
12
=
c1 2
Y22 Y2 Y 2 + Y 1 1
12
gY = 2 ( Y2 + Y1 ) 2Y1
12
(3.35a)
219
The Momentum Principle Applied to Open Channel Flows
For problems in which a third equation is not available so that only the continuity and momentum equations are used to solve for v and Y2, it is convenient to eliminate v from these two equations and solve Y2 from one implicit equation. Thereafter v can be obtained from Equation 3.30 (or 3.30a), or if the channel is rectangular, from Equation 3.34 (or 3.34a). Substituting v from the continuity equation into the momentum equation gives UCW 2
Q − Q1 g(h c 2 A 2 / A1 − h c1 ) V2 A 2 − V1A1 = − V1 = 2 − V1 A 2 − A1 A 2 − A1 1 − A1 / A 2
2
(3.36)
DCW 2
Q − Q1 g(h c 2 A 2 / A1 − h c1 ) V2 A 2 − V1A1 = + V1 = 2 + V1 A 2 − A1 A 2 − A1 1 − A1 / A 2
2
(3.36a)
For a rectangular channel Equations 3.36 simplify to UCW 2
gY2 ( Y2 + Y1 ) = V2YY2 −− YV1Y1 − V1 = Yq 2 −− qY1 − V1 2Y1 2 1 2 1
2
(3.37)
An alternative to this equation is
g(Y2 − Y1 ) 2 (Y2 − Y12 ) = (V1 − V2 )2 2Y1Y2
(3.37a)
DCW 2
V Y − V1Y1 q −q gY2 (Y2 + Y1 ) = 2 2 + V1 = 2 1 + V1 Y2 − Y1 Y2 − Y1 2Y1
2
(3.37b)
With the same alternative Equation 3.37a. Example Problem 3.9 Obtain a series of solutions that provide the depth and velocity and wave speed that will occur in a rectangular channel if the flow rate is instantly increased at its beginning from uniform conditions of q = 2 m2/s and Y1 = Yo = 2 m. Solution V1 = q/Yo = 2/2 = 1 m/s. For each solution in this series Equation 3.37 is solved first for Y2. With Y2 known, Equation 3.34 is used to solve the wave speed v. V2 = q/Y2 in which q is the new specified flow rate per unit width. A TKSolver model is given below to solve this problem. In this model Equation 3.37 (using Y22 for Y2 and V22 for V2) and Equation 3.37a are solved to verify that identical results are produced by these two equations. A Fortran and C program that solves this problem follows the TKSolver model. This program uses the Newton method with a numerical evaluation of the derivative of Equation 3.37a.
220
Open Channel Flow: Numerical Methods and Computer Applications TKSolver Model ================================================VARIABLE SHEET=================================================== St LG
LG LG L LG LG
Input
Name
1 1.3793999 9.81 2 2.1748587 2.1748587 3 1.3793999 6
V1 V2 g Y1 Y2 Y22 q2 V22 v
Output
Unit
================================================RULE SHEET================================================ S Rule (V1−V2)∧2=g*(Y1−Y2)*(Y1∧2−Y2∧2)/(2.*Y1*Y2) ((V22*Y22−V1*Y1)/(Y22−Y1)−V1)∧2=.5*g*Y22/Y1*(Y22+Y1) V2=q2/Y2 V22=q2/Y22 v=(V2*Y2−V1*Y1)/(Y2−Y1) ===================================================================TABLE: Title: Element q2 — V2—————1 2.01 1.00407586 2 2.2 1.0803217 3 2.4 1.15825367 · · · 38 9.4 3.09385488 39 9.6 3.13633146 40 9.8 3.17837366 41 10 3.21999374
wave=========================================================== V22————10.6695871 1.0803217 1.27939865 · 3.09385488 3.13633146 3.17837366 3.21999374
Y2——————2.00184077 2.03643044 2.07208496 · 3.03828084 3.06090097 3.08333791 3.1055961
Y22————.188385922 2.03643044 1.87588129 · 3.03828084 3.06090097 3.08333791 3.1055961
Fortran Program WAVE.FOR WRITE(*,*) 'Give: Y1,V1,g,q2,N' READ(*,*) Y1,V1,G,q2,N q1=V1*Y1 dq=(q2−q1)/FLOAT(N−1) Y2=Y1+.05 v=SQRT(G*Y2)+V1 DO 10 I=1,N q=q1+dq*float(I−1) IF(I.EQ.1) q=q+.01 M=0 1 F=(V1−q/Y2)**2−g*(Y1−Y2)*(Y1**2−Y2**2)/(2.*Y1*Y2) M=M+1 IF(MOD(M,2).EQ.0) GO TO 2 F1=F Y22=Y2 Y2=1.005*Y2 GO TO 1 2 DIF=(Y2−Y22)*F1/(F−F1) Y2=Y22−DIF
v———————5.4325044 5.48991413 5.54900801 · 7.12716607 7.16372235 7.19996959 7.23591551
The Momentum Principle Applied to Open Channel Flows 10 100
IF(ABS(DIF).GT.1.E−5.AND.M.LT.30)GO TO 1 IF(M.EQ.30)WRITE(*,*)'Failed to conv.',I,DIF WRITE(3,100) q,q/Y2,Y2,(q−q1)/(Y2−Y1) Y2=1.05*Y2 FORMAT(F5.2,3F8.3) END
CProgram WAVE.C #include <math.h> #include <stdlib.h> #include <stdio.h> void main(void){int n,i,m; FILE *filo; char fnam[20]; float y1,v1,v,g,q2,q1,dq,y2,q,f,f1,dif,hc1,hc2a,y22,a; printf("Give: Y1,V1,g,Q2,N\n"); scanf("%f %f %f %f %d",&y1,&v1,&g,&q2,&n); printf("Give output filename\n");scanf("%s",fnam); if((filo=fopen(fnam,"w"))==NULL) {printf("Cannot open output file\n");exit(0);} q1=v1*y1;dq=(q2−q1)/(float)(n−1); y2=y1+.05; for (i=1;i1.e5)&&(m<20)) goto L1; if(m==30) printf("Failed to converge %d %f/n",i+1,dif); fprintf(filo,"%5.2f %7.3f %7.3f %7.3f\n",q,q/y2,y2,\ (qq1)/(y2y1)); y2*=1.05;}} Input Data: 2 1 9.81 10 41 q
V2
Y2
v
2.01 2.20 2.40 2.60 · 9.40 9.60 9.80 10.00
1.004 1.080 1.158 1.234 · 3.094 3.136 3.178 3.220
2.002 2.036 2.072 2.107 · 3.038 3.061 3.083 3.106
5.306 5.490 5.549 5.607 · 7.127 7.164 7.200 7.236
Example Problem 3.10 A uniform flow of Q = 300 cfs exists in a trapezoidal channel with b = 10 ft, m = 1.5, n = 0.014, and So = 0.0004. Compare the velocity of the constant height waves that will occur from increasing the flow rate in increments of 50 to 1000 cfs with the celerities of small amplitude gravity waves in the channel upstream from the wave.
221
222
Open Channel Flow: Numerical Methods and Computer Applications
16 14 Velocity (fps)
8
y, v velocit Wave l
ma
12
r No
10
7.5 7
, Yo
th
p de
6.5
Celerity, c2
6
,Y Depth 2 ity, V 2 Veloc
8
5.5 5
6 4 300
Depth (ft)
18
4.5 400
500
600 700 800 Flow rate, Q (cfs)
900
4 1000
Solution First the initial condition is obtained by solving Manning’s equation giving Y1 = 4.327 ft and V1 = 4.2043662 fps (the accuracy seven digits is used so Q =300 cfs). For each new increment of Q Equation 3.36 needs to be solved for the new depth Y2, i.e., Equation 3.36 replaces Equation 3.33c used in the previous problem. Thereafter the wave velocity v is solved using Equation 3.30, and the celerity is computed from c2 = (gA2/T2)1/2. The Fortran program WAVETR.FOR given below solves the problem for 15 increments, and the solution table is given thereafter. Because of the nature of Equation 3.36 for depth Y2 close to Y1 (or Q2 close to 300) guesses very close to the sought after solution must be provided, or else a solution for Y2 less than Y1, “the supercritical root,” or another root will be obtained. For flow rates close to 300 cfs the following table provides the desired roots for Y2. Q(cfs) Y2 (ft)
301 302 303 4.3301 4.3331 4.3362
304 4.3392
305 310 315 320 330 340 350 4.3422 4.3574 4.3725 4.3875 4.4172 4.4464 4.4755
Notice from the solution table (and the graph) that even with the increasing area of this trapezoidal channel with m = 1.5 with depth Y, that the normal depth Yo given in the last column of the solution table is greater than Y2. Therefore, in practice a constant height wave would not actually occur because the depth upstream of the wave would be gradually varied. Program WAVETR.FOR WRITE(*,*) 'Give: Y1,V1,g,Q2,N,b,m,n,So' READ(*,*) Y1,V1,G,Q2,N,B,FM,FN,So CU=1.486 IF(G.LT.20) CU=1. So=SQRT(So) FMS=2.*SQRT(FM*FM+1.) A1=(B+FM*Y1)*Y1 Q1=V1*A1 dQ=(Q2Q1)/FLOAT(N1) Y2=4.48 Yo=Y2 FM2=2.*FM FM3=FM/3. B2=B/2. T=B+FM2*Y1 C1=SQRT(G*A1/T) v=C1V1 WRITE(3,110) C1,v,V1,Q1
The Momentum Principle Applied to Open Channel Flows 110 FORMAT(' Celerity c1=',F8.3,' Initial Wave Vel=',F8.3,' & V1=',F8.3,' Q1=',F8.3) HC1=(B2+FM3*Y1)*Y1**2/A1 WRITE(3,100) Q1,V1,Y1,C1+V1,C1,Y1 DO 10 I=2,N Q=Q1+dQ*float(I1) M=0 1 A=(B+FM*Y2)*Y2 HC2A=(B2+FM3*Y2)*Y2**2 F=g*(HC2A/A1HC1)/(1.A1/A)((QQ1)/(AA1)V1)**2 M=M+1 IF(MOD(M,2).EQ.0) GO TO 2 F1=F Y22=Y2 Y2=1.005*Y2 GO TO 1 2 DIF=(Y2Y22)*F1/(FF1) Y2=Y22DIF IF(ABS(DIF).GT. 1.E5 .AND. M.LT.30) GO TO 1 IF(M.EQ.30) WRITE(*,*)' Failed to converge',I,DIF A=(B+FM*Y2)*Y2 C2=SQRT(g*A/(B+FM2*Y2)) M=0 3 Ao=(B+FM*Yo)*Yo F=FN*QCU*Ao*(Ao/(B+FMS*Yo))**.66666667*So M=M+1 IF(MOD(M,2).EQ.0) GO TO 4 F1=F Yoo=Yo Yo=1.005*Yo GO TO 3 4 DIF=(YoYoo)*F1/(FF1) Yo=YooDIF IF(ABS(DIF).GT. 1.E5 .AND. M.LT.30) GO TO 1 IF(M.EQ.30) WRITE(*,*) ' Manning failed',I,DIF WRITE(3,100) Q,Q/A,Y2,(QQ1)/(AA1),C2,Yo 10 Y2=1.05*Y2 100 FORMAT(F8.2,5F8.3) END Program WAVETR.C #include <math.h> #include <stdlib.h> #include <stdio.h> void main(void){int n,i,m; FILE *filo; char fnam[20]; float y1,v1,g,q2,b,fm,fn,so,a1,q1,dq,y2,yo,yoo,fms,fm2,fm3,b2,t,\ c1,q,f,f1,dif,hc1,hc2a,y22,a,c2,ao,cu=1.486; printf("Give: Y1,V1,g,Q2,N,b,m,n,So\n"); scanf("%f %f %f %f %d %f %f %f %f",&y1,&v1,&g,&q2,&n,&b,&fm,&fn,\ &so); printf("Give output filename\n");scanf("%s",fnam); if((filo=fopen(fnam,"w"))==NULL){ fprint("Cannot open output file\n");exit(0);} if(g<20.) cu=1; so=sqrt(so); fms=2.*sqrt(fm*fm+1.); a1=(b+fm*y1)*y1; q1=v1*a1;dq=(q2q1)/(float)(n1);
223
224
Open Channel Flow: Numerical Methods and Computer Applications y2=4.48;yo=y2;fm2=2.*fm;fm2=fm/3.;t=b+fm2*y1;c1=sqrt(g*a1/t); fprintf(filo,"Celerity c1= %8.3f Initial Wave Vel= %8.3f\ V1= %8.3f Q1 = %8.3f\n",q1,v1,y1,c1+v1,c1,y1); hc1=(b2+fm3*y1)*y1*y1/a1;fprintf(filo,"%8.2f %7.3f %7.3f\ %7.3f %7.3f %7.3f/n",q1,v1,y1,c1+v1,c1,y1); for (i=1;i1.e5)&&(m<20)) goto L1; if(m==30) printf("Failed to converge %d %f/n",i+1,dif); a=(b+fm*y2)*y2; c2=sqrt(g*a/(b+fm2*y2));m=0; L3: ao=(b+fm*yo)*yo; f=fn*qcu*ao*pow(ao/(b+fms*yo),.66666667)*so; if((++m)%2){f1=f;yoo=yo;yo*=1.005;goto L3;} dif=(yoyoo)*f1/(ff1); yo=yoodif; if((fabs(dif)>1.e5)&&(m<30)) goto L3; if(m==30)printf("Manning failed %d %f\n",i,dif); fprintf(filo,"%8.2f %7.3f %7.3f %7.3f %7.3f %7.3f\n",q,q/a,y2,\ (qq1)/(aa1),c2,yo); y2*=1.05;}} Input data: 4.327 4.2043662 32.2 1000 15 10 1.5 .014 .0004 Output table: Celerity c1= 9.999 Initial Wave Vel= 5.795 V1= 4.204 Q1= 300.000 Q 300.00 350.00 400.00 450.00 500.00 550.00 600.00 650.00 700.00 750.00 800.00 850.00 900.00 950.00 1000.00
V2
Y2
v
c2
Yo
4.204 4.679 5.121 5.535 5.926 6.296 6.647 6.983 7.304 7.612 7.909 8.195 8.472 8.739 8.998
4.327 4.475 4.615 4.748 4.874 4.994 5.110 5.220 5.327 5.430 5.529 5.625 5.719 5.809 5.897
14.203 14.514 14.810 15.091 15.360 15.618 15.867 16.106 16.338 16.563 16.782 16.994 17.200 17.401 17.598
9.999 10.140 10.270 10.391 10.504 10.611 10.712 10.808 10.898 10.985 11.068 11.147 11.224 11.297 11.368
4.327 4.688 5.021 5.332 5.624 5.900 6.162 6.412 6.651 6.880 7.101 7.314 7.520 7.720 7.913
3.8 Open Channel to Pipe Flow If the flow is under supercritical conditions at a depth such that its conjugate depth is above the top of the conduit, then a modified hydraulic jump will change the flow from open channel to closed conduit flow. The momentum principle allows this situation to be analyzed. Following the usual steps needed in utilizing the momentum principle, the first step produces a control volume such as that above. The force upstream on this control volume is the same as that upstream from a hydraulic jump, namely the hydrostatic force from the removed upstream fluid, e.g., γAhc1. Downstream in the closed conduit section the force from the removed fluid can best be represented by two forces;
225
The Momentum Principle Applied to Open Channel Flows
that due to the head of water above the top of the conduit, and that due to a hydrostatic pressure distribution of liquid depth at the top of the conduit. These two downstream forces are: Fp21 = γAtd and Fp22 = γAt(D/2), respectively. At = πD2/4 for a pipe conduit, or the entire crosssectional area of a noncircular conduit, and (D/2) is the distance from the top of the conduit to its centroid, and d is the depth water would stand in a standpipe attached to the top of the conduit just downstream from where the conduit flows full. The pressure on the top of the conduit is p = γd. HGL
d Fp2 = γAtd Fp1 = γ(Ahc)1
Fp3 = γAt(D/2)
Y1 Closed conduit
Applying the equation that the summation of forces on the control volume of fluid must equal the momentum flux leaving the control volume minus the momentum flux entering the control volume in the x direction gives γA1h c1 − γA t d − γ
D γ Q2 1 1 At = − g A t A1 2
or solving for the head d on the top of the conduit at the downstream section
Q2 1 1 1 − A1h c1 − 2 DA t + g A1 A t d= At
(3.38)
An alternate form of Equation 3.38 involves writing it in terms of the momentum functions, or d=
(M1 − M t ) 4(M1 − M t ) = At (πD2 )
(3.28a)
in which Mt is the momentum function based on the pipe flowing full and is given by Mt = At(D/2) + Q2/(gAt). Example Problem 3.11 A 12in. diameter PVC sewer line suddenly changes grade from So1 = 0.0496 to So2 = 0.0005. The flow rate is Q = 1.5 cfs. Downstream from the change in grade at a distance of 5060 ft the pipe discharges into a pond with a water surface elevation at the top of the pipe. Determine where the flow changes from open channel to pipe flow. Solution The normal depth for Q = 1.5 cfs equals Yo1 = 0.230 ft from Manning’s equation. A computation of the slope Sf of the HGL for the downstream pipe flowing full gives, Sf = [nQP2/3/(1.49A5/3)]2 = 0.0006677 (using n = 0.008 for PVC pipe). Since this value is greater than the slope of the downstream pipe, it will flow full from its end to the position where the modified hydraulic jump takes place. Substituting into Equation 3.38 gives,
226
Open Channel Flow: Numerical Methods and Computer Applications 1 π π 1 1.52 d = 0.013 − + − 4 2 32.2 0.136 4
π = 0.070 4
From the slope of the downstream HGL it is 0.849 ft above the top of the pipe at the position of the break in grade. Therefore the position where the modified jump will occur upstream from the break in grade is
L=
d2
0.849 = 15.9 ft (0.0174 − 0.0006677)
0.849 ft
HGL, S = 0.00 06677 f
So1 =
D = 1.0 ft So2 = 0.0005, L = 5060 ft
Q = 1.5 cfs
Example Problem 3.12 Water is flowing down a steep pipe with a diameter, D1 = 3.5 ft, a Manning’s roughness coefficient, n1 = 0.012, and a bottom slope of So1 = 0.15. At the end of the pipe the water is directed into a trapezoidal channel with b2 = 2 ft, m2 = 0.8, n2 = 0.013, and So2 = 0.000006. It is observed that 3 ft from the end of the pipe the flow changes from open channel to pipe flow. For the pipe flow the equivalent sand roughness is e = 0.002 in. Determine what the flow rate Q is. Solution There are actually six unknowns in this problem. Q, Sf (the slope of the energy line in the pipe flow), d, Y1, Y2, and f. The equations needed to solve this problem are
Q= Q=
1.486 A1 n1A1 P1
1.486 A 2 n 2 A 2 P2
S1o1/ 2
(Manning’s equation: upstream channel)
(1)
2/3
S1o2/ 2
(Manning’s equation: downstream channel)
(Momentum)
(3)
(So1 − Sf ) x = Y2 − D − d (From geometry of slopes) fQ 2 (2g(π / 4)2 D5 )
(2)
M1 − A t (D / 2) − Q 2 /(gA t ) d= At
Sf =
2/3
(4)
(Darcy −Weisbach for pipe flow)
e 7.343473vD 1 = 1.14 − 2Log + f Q/ f D
(5)
(Colebrook −White)
(6)
These equations are used in the TKSolver variable and rule sheets as shown below to obtain the solution. Thereafter a similar Mathcad model is given. Notice in these models that separate equations have been used to define β, the areas, the wetted perimeters and the upstream momentum
227
The Momentum Principle Applied to Open Channel Flows function so that 6 (and if one wants to include At in this list 7) additional variables, β1, A1, P1, A2, P2, M1, and At, have been added to the list of unknowns, 12 (or 13) equations are being solved simultaneously. The flow rate is Q = 141.2 cfs. Notice in this model, since separate equations have been used for areas, wetted perimeters, the upstream momentum function, and the total area, that 7 additional variables have been added in the list of unknown variables, so 13 equations are being solved for 13 unknowns. ============================================================================VARIABLE SHEET =========================================================================== St
Input
Name
Output
Unit
Comment
Q 141.28483 1.486 C .012 n1 A1 3.5758985 P1 4.7823839 .15 So1 .013 n2 A2 177.62387 P2 37.096815 .000006 So2 beta 1.3663954 Y1 1.394784 3.5 D 2 b2 .8 m2 Y2 13.703004 M1 175.44462 d 9.788304 At 9.6211275 32.2 g Sf .01176663 3 x f .01229892 .00033333 e .00001217 v =====================================================================================RULE SHEET==================================================================================== S Rule— Q=C/n1*A1*(A1/P1)∧.66666667*sqrt(So1) Q=C/n2*A2* (A2/P2)∧.66666667*sqrt(So2) cos (beta)=1.−2*Y1/D A1=D*D/4* (beta−cos (beta)*sin(beta)) P1=beta*D A2=(b2+m2*Y2)*Y2 P2=b2+2*Y2*sqrt (m2*m2+1) M1=.5*D*(D*D/6.*sin(beta)∧3−A1*cos (beta))+Q*Q/g/A1 d=(M1−.5*D*At−Q*Q/(g*At))/At At=.25*pi()*D*D (So1−Sf)*x=Y2−D−d Sf=f*Q*Q/(1.23370055*g*D∧5) 1/sqrt(f)=1.14−2*log(e/D+7.3434728*v*D/abs(Q)/sqrt(f))
Mathcad model to solve Problem 3.12. Variables Cu: = 1.486 nl: = .012 n2: = .013 Sol: = .15 So2: = .000006 D: = 3.5 b2: = 2 m2: = .8 g: = 32.2 x: = 3 e: = .0003333333 v: = .00001217 Q: = 130 A1: = 3.5 P1: = 4.5 A2: = 180 P2: = 35 β: = 1.5 Y1: = 1.4 Y2: = 13.7 Ml: = 170 d: = 9.8 At: = 9.6 Sf: = .0117 f: = .013
228
Open Channel Flow: Numerical Methods and Computer Applications Given
Q=
Cu A1 ⋅ A1 ⋅ P1 nl
.6666667
⋅ So1 Q =
Cu A2 ⋅ A2 ⋅ P 2 n2
.6666667
⋅ So2 cos(β) = 1. − 2 ⋅
Y1 D
Al = 25 ⋅ D 2 ⋅ (β − cos(β) ⋅ sin(β)) P1 = β ⋅ D A2 = (b2 + m2 ⋅ Y2) ⋅ Y2 P 2 = b2 + 2 ⋅ Y2 ⋅ m2 ⋅ m2 + 1 D2 Q2 M1 = 5 ⋅ D ⋅ ⋅ sin(β)3 − A1 ⋅ cos(β) + 6 g ⋅ A1 M1 − D ⋅ At = .25 ⋅ π ⋅ D2 d =
At Q2 − f ⋅ Q2 2 g ⋅ At (Sol − Sf ) ⋅ x = Y2 − D − d Sf = At 1.2337005 ⋅ g ⋅ D5
e 1 D = 1.14 − 2 ⋅ log + 7.3434728 ⋅ v D f Q ⋅ f
0 1 2 3 4 5 Find (Q, Sf, d, Y1, Y2, f, β, A1, P1, A2, P2,M1, At) = 6 7 8 9 10 11 12
0 141.2848 0.0118 9.7883 1.3948 13.703 0.0123 1..3664 3.5759 4.7824 177.6239 37.0968 175.4446 9.6211
The logic needed to program a solution of simultaneous nonlinear equations has been discussed earlier. A FORTRAN and a C program to solve this problem are given below. In these programs the Colebrook–White equation is solved by a Gausel iteration based on the current values of Q and thus only five simultaneous equations are solved for the variables Q, Sf, d, Y1, and Y2, with the array X containing these unknowns in this order. The Jacobian matrix for the Newton method is evaluated numerically by calling on the subroutine (function) FUN with each variable increment as well as without this increment, as described previously. It does not take big changes in the variable that describe this problem in order for a solution not to exist, i.e., the jump is washed out from the pipe. Program PRB3_12.FOR REAL F(5),F1(5),DJ(5,5) INTEGER*2 INDX(5) COMMON G,CU,SO1,SO11,SO2,D,B2,FM2,ED,VIST,RF,DS4,FMS,AT,DG5, &XX,X(5)
The Momentum Principle Applied to Open Channel Flows WRITE(*,*)' Give: g,n1,n2,So1,So2,D,b2,m2,e,X' READ(*,*) G,FN1,FN2,SO11,SO2,D,B2,FM2,E,XX WRITE(*,*)' Provide guesses for: Q,Sf,d,Y1,Y2' READ(*,*) X RF=8. CU=1.486 VIST=1.217E5 IF(G.GT.20.) GO TO 1 CU=1. VIST=1.31E6 1 ED=E/D VIST=7.3434728*VIST*D DS4=.25*D*D FMS=2.*SQRT(FM2*FM2+1.) SO1=CU*SQRT(SO11)/FN1 SO2=CU*SQRT(SO2)/FN2 AT=.78537816*D**2 DG5=1.23370055*G*D**5 M=0 10 SUM=0. CALL FUN(F) DO 20 J=1,5 XT=X(J) X(J)=1.01*X(J) CALL FUN(F1) DO 18 I=1,5 18 DJ(I,J)=(F1(I)F(I))/(X(J)XT) 20 X(J)=XT CALL SOLVEQ(5,1,5,DJ,F,1,DD,INDX) DO 30 I=1,5 X(I)=X(I)F(I) 30 SUM=SUM+ABS(F(I)) M=M+1 IF(SUM.GT. 1.E4 .AND. M.LT.20) GO TO 10 WRITE(*,100) X 100 FORMAT(' Solution:',/' Q =',F8.1,/' Sf =',F8.6,/ &' d =',F8.3,/' Y1=',F8.3,/' Y2 =',F8.3) END SUBROUTINE FUN(F) REAL F(5) COMMON G,CU,SO1,SO11,SO2,D,B2,FM2,ED,VIST,RF,DS4,FMS,AT, &DG5,XX,X(5) COSB=1.−2.*X(4)/D BETA=ACOS(COSB) A1=DS4*(BETACOSB*SIN(BETA)) F(1)=X(1)SO1*A1*(A1/(D*BETA))**.66666667 A2=(B2+FM2*X(5))*X(5) F(2)=X(1)SO2*A2*(A2/(B2+FMS*X(5)) )**.66666667 F(3)=X(3)(.5*D*(D*D/6.*SIN(BETA)**3A1*COSBAT)+ &X(1)**2/G*(1./A1–1./AT))/AT 1 RF1=RF RF=1.14–2.*ALOG10(ED+VIST*RF1/X(1)) IF(ABS(RFRF1).GT. 1.E6) GO TO 1 SF=(X(1)/RF)**2/DG5
229
230
Open Channel Flow: Numerical Methods and Computer Applications F(4)=(SO11SF)*XXX(5)+D+X(3) F(5)=X(2)SF RETURN END Program PRB3_12.C #include <stdio.h> #include <stdlib.h> #include <math.h> float g,cu,so1,so11,so2,d,b2,m2,ed,vist,rf,ds4,fms,at,dg5,xx,x[5]; void fun(float *f){float cosb,sinb,beta,a1,a2,rf1,sf; cosb=1.2.*x[3]/d; sinb=sqrt(1.cosb*cosb); beta=atan(sinb/cosb); if(beta<0.) beta+=3.14159265; a1=ds4*(betacosb*sinb); f[0]=x[0]so1*a1*pow(a1/(d*beta),.6666667); a2=(b2+m2*x[4])*x[4]; f[1]=x[0]=so2*a2*pow(a2/(b2+fms*x[4]),.6666667); f[2]=x[2](.5*d*(d*d/6.*pow(sinb,3.)a1*cosbat)+x[0]\ *x[0]/g*(1./a1–1./at))/at; do{rf1=rf; rf=1.14−2.*log10(ed+vist*rf1/x[0]) } while(fabs(rfrf1)<1.e6); sf=pow(x[0]/rf,2.)/dg5; f[3]=(so11sf)*xxx[4]+d+x[2]; f[4]=x[1]sf;} // end fun void solveq(int n,float **a,float *b,int itype,float *dd,\ int *indx); void main(void){float e,n1,n2,xt,sum,f[5],f1[5],*dd,**dj; int i,j,m,indx[5]; dj=(float**)malloc(5*sizeof(float*)); for(i=0;i<5;i++)dj[i]=(float*)malloc(5*sizeof(float)); printf(" Give: g,n1,n2,So1,So2,D,b2,m2,e,x\n"); scanf("%f %f %f %f %f %f %f %f %f %f",&g,&n1,&n2,&so11,&so2,&d,\ &b2,&m2,&e,&xx); printf(" Provide guesses for: Q,Sf,d,Y1,Y2\n"); scanf("%f %f %f %f %f",&x[0],&x[1],&x[2],&x[3],&x[4]); rf=8.,cu=1.486;vist=1.217e5; if(g<20.) {cu=1.;vist=1.31e6;} ed=e/d; vist*=7.3434728*d; ds4=.25*d*d;fms=2.*sqrt(m2*m2+1.); so1=cu*sqrt(so11)/n1; so2=cu*sqrt(so2)/n2; at=.78537816*d*d; dg5=1.23370055*g*pow(d,5.); m=0; do{sum=0.; fun(f); for(j=0;j<5;j++){xt=x[j]; fun(f1); for(i=0;i<5;i++) dj[i][j]=(f1[i]f[i])/(x[j]xt); x[j]=xt;} solveq(5,dj,f,1,dd,indx); for(i=0;i<5;i++){x[i]=f[i]; sum+=fabs(f[i]);} } while ((sum<1.e4) && (++m<20)); printf("Solution:\n Q =%8.1f\n Sf =%8.6f\n d =%8.3f\n\ Y1 =%8.3f\n Y2 =%8.3f\n", x[0],x[1],x[2],x[3],x[4]);}
Example Problem 3.13 A steep open channel pipe with D = 5 ft, So1 = 0.0482, and n = 0.012 has a change in its bottom slope to So2 = 0.01 at a point where a 1.5 ft diameter pipe takes water out from the bottom of the larger pipe as shown in the sketch. At a distance 500 ft downstream from the breakingrade the pipe discharges into a stilling basin that discharges the water over a sharp crested weir that is 50.6 ft long. The crest of the weir is Zw = 4 ft above the bottom of the pipe, and its discharge coefficient is Cd = 0.6. The 1.5 ft diameter pipe is 1200 ft long and discharges at an elevation 103 ft
231
The Momentum Principle Applied to Open Channel Flows below the bottom of the larger pipe. It is to supply water with a pressure head p/γ = 90 ft. For use in the Darcy–Weisbach equation use an equivalent sand roughness e = 0.0045 in. for both pipes. If the flow rate coming down the 5 ft diameter pipe is Q1 = 600 cfs, determine the discharge in the smaller pipe Qo, the flow rate entering the spillway Q2, and the position x where a modified hydraulic jump will occur.
Sf1
HGL
1
d
z
Sf 2
1 H
.012
,S
o1 = 0
D2=
2
Q3
L = 1.5 ft, 2
Q
1
1200
ft
e = 0.0045 in. D = 5 ft
Zw
Q2
So2 = 0.01, L2 = 500 ft
Stilling basin
Weir
Δp
/γ΄
=9
0 ft
Δz = 103 ft
.048
ft .6 50 0.6 L = c= C
n=0
Q2
Solution First from Q1 = 600 cfs and the slope and n of the upstream pipe the normal depth can be solved from Manning’s equation as Yo1 = 3.961 ft. Next from the Darcy–Weisbach and Colebrook–White equations, the slope of the EL (or HGL) after the modified jump to the break in grade can be solved as Sf1 = 0.0336853. There are seven unknowns: Q2, Qp, x, d, Sf2, H, and z. The seven available equations are:
1/ 2 2 Q 2 = Cd ( 2g ) L w H1.5 3
(1)
z + D + So2 L 2 = Z w + H + Sf2 L 2
(2)
d + So1x = z + Sf1x
(3)
d=
Q 2 = Q1 − Q p
z + D + ∆elev −
(M1 − M t ) At
(4)
L (Q p / A p ) 2 p = fp p γ D p (2g)
f (Q 2 / A t ) 2 Sf2 = D (2g)
Plus 2 Colebrook–White Equations for f and fp.
(5)
(6)
(7)
232
Open Channel Flow: Numerical Methods and Computer Applications A TKSolver model to solve this problem is given below. ================================== VARIABLE SHEET =============================== St InputNameOutputQ2 577.62052 Qp 22.379478 x 532.04304 H 2.3298401 z 11.87257 Sf2 .03108546 d 4.1501248 f .01156612 fp .01499386 5 D .01 So2 .6 Cd 32.2 g 50.6 Lw 500 L2 4 Zw .0482 So1 .0336853 Sf1 M1 699.97395 beta 2.1954298 A1 16.68629 3.962 Y1 At 19.634954 Mt 618.48644 600 Q1 103 Delev 90 phead 1200 Lp 1.5 Dp .000375 e .0000141 vis ========================================= RULE SHEET ====================================== S Rule——* * * * * * * * * * * * * *
Q2=Cd*(2/3)*sqrt(2.*g)*Lw*H∧1.5 z+D+So2*L2=Zw+ H+Sf2*L2 d+D+So1*x=D+z+Sf1*x cos(beta)=1.−2*Y1/D A1=D∧2/4* (beta−cos(beta)*sin(beta)) M1=.5*D* (D∧2/6*sin (beta)∧3−A1*cos (beta))+Q1*Q1/(g*A1) At=pi()/4*D∧2 Mt=.5*D*At+Q1∧2/(g*At) d=(M1−Mt)/At Q2=Q1−Qp z+Delev+D−phead=fp*Lp/Dp*(Qp/(pi()/4*Dp∧2))∧2/(2.*g) Sf2=f/D*(Q2/At)∧2/(2.*g) 1/sqrt(f)=1.14−2*log(e/D+7.34347283*vis*D/(Q2*sqrt(f))) 1/sqrt(fp)=1.14−2*log(e/Dp+7.34347283*vis*Dp/(Qp*sqrt(fp)))
233
The Momentum Principle Applied to Open Channel Flows
3.9 Multiple Roughness Coefficient for Channel Section—Compound Sections Under flood flow conditions, the depth of water may exceed the height of the main river bank and then the flood plain areas on one or both sides of the main river channel become part of the section that conveys water. The side flood plains generally cause significantly larger resistance to flow than the main river channel. Simulation of such flows in natural streams and rivers will be dealt with in Chapter 6. A channel with several values for the roughness coefficients for different portions of the channel is commonly referred to as a compound channel. The flow in such channels can be visualized as different channel flows that have become combined because a wall between them is missing. Such compound channel flows are different than those discussed in Chapter 2, in which an equivalent Manning’s roughness coefficient was used that was obtained by weighting the separate n′ values according to the fraction of the wetted perimeter to which each applied. The approach in Chapter 2 is applicable for a concrete channel with gravel deposits in its bottom, for example. The approach described in this section for compound channels is applicable for situations in which a main channel is overtopped and the areas adjacent to its sides convey the spilled water in the direction of the main channel flow. We will now examine several interesting hydraulic properties of compound sections. Let us consider an example in which the main channel has a bottom width of b = 9 m, and a side slope of m = 0.5, and when the depth reaches 5 m then the water flows out of the main channel onto both sides with expanded widths of br = 15 m and bl = 15 m, as shown in the sketch. The main channel has a Manning’s roughness of nm = 0.018 and both overflow sides have roughness values of nr = 0.055 and nl = 0.055. The bottom slope of all portions of the channel are the same, with So = 0.0008. For this channel, its geometric properties need to be defined by equations that apply when the depth of flow is less and greater than the capacity of the main channel as follows: For Y ≤ 5 m
1
y
n1 = 0.045 m1 = 0.5
Y
b1 = 50 ft
Y1 = 10 ft nm = 0.013
nr = 0.045 1 mm = 1
bm = 20 ft
br = 40 ft
1 mr = 0.5
So = 0.0008
Sketch of channel section with several roughness coefficients A = (b + mY)Y, P = b + 2Y(m 2 + 1)1/2 ,
T = b + 2mY
(the usual equations for a trapezoidal channel)
For Y > 5 m
A = 57.5 + (b r + b l + 14.0)(Y − 5),
P = 20.18 + b r + b l + 2(Y − 5),
T = 14.0 + b r + b l
in which 57.5, 20.18, and 14.0 are the area, perimeter, and top width for the main channel when full, respectively. These equations will need to be modified according to the geometry of the main channel and overflow channels for other compound sections. If uniform flow were to occur at the top of the main channel then the flow rate would be Qo = 181.6 m3/s, and the Froude number would be
234
Open Channel Flow: Numerical Methods and Computer Applications
Fr = {Q2T/(gA3)}1/2 = 0.50. The specific energy diagram for this normal flow rate is shown as one of the curves on the specific energy diagram below. Note this diagram looks similar to such diagrams in regular sections with the exception that the curve breaks toward the 45° line, especially for larger flow rates in which velocities are larger, as the flow leaves the main channel into the side channels because the rapid expansion in width at this depth reduces the average velocity. In reality, the flow in the main channel will not be reduced, but rather the smaller velocities in the side channels reduce the average velocity. A graph below shows how Manning’s equation defines the relationship between the flow rate and the depth. (The channel bottom slope would need to change if these were to occur, of course.) For this compound channel, when the depth Y = 7.8 m, the main channel contains 431.9 m3/s whereas the side channel contributes 76.6 m3/s or 15% of the total flow even though the area of the side channels is 44.3% of the total. These computations are based on having the main channel continue upward with a side slope of 0.5, but not adding any to the wetted perimeter above the 5 m depth since the flow in the side channels will exist here rather than a channel wall. This graphs also shows the flow rate–depth relationship if the main channel actually did have side walls that extended upward to the 7.8 ft depth. Such an upward extending real walled main channel would carry only 385.4 m3/s instead of 431.9 m3/s because of the added wall resistance.
600 500
Flow rate, Q (m**3/s)
Depth of flow, Y (m)
8 7 6
400
5 4
Q=
3 2 Q
1 0
0
1
2
Q= =
100
Q= 250 Q = 181. 6 (norm Q= al) 150
200 100
50
3
4
5
6
7
8
9
Specific energy, E = Y + Q*Q/(2gA*A), (m) Specific energy diagram
10
0
bi
m
Co
300
300
d ne
ow
fl
ain
*
,Q
te
ra
l, Q 431.9 ne
an
ch
5.4 38 ain m f d s o ende .5 t ide If s nel ex m = 0 h n cha rd wit a w p u 76.6
M
* nel, Q chan Side
1
2
3
4
5
6
7
8
Depth of flow in channel, Y (m) Flow rates computed based on Manning’s equation
In a compound channel, it is possible that critical flow may occur for three different depths. These three critical flow conditions will occur for a given flow rate if critical depth occurs for that flow rate at a depth modestly less than the top of the main channel. When this occurs the Froude number becomes less than 1 as the depth rises to the top of the main channel. As the water rises above the main channel into the side channels, the top width of the compound channel immediately increases thus rapidly increasing the Froude number so it becomes unity again. Further increases in depth reduce the Froude number until it equals unity for a third depth. The graph below shows that for a flow rate of Q = 300 m3/s critical flows occur at Y = 4.436 m, 5.0 + m, and 5.37 m. Again these results are based on computations that assume onedimensional hydraulics. In actuality, especially for compound channels of the size used in this example, the main channel portion will be supercritical flow at certain depths while the flow in the side channels will still be subcritical. In the table below, the assumption is made that the velocities are different in the main channel than the side channels and the total flow rate is Q = 300 m3/s. To determine what portion of this total flow rate occurs in the main and side portions of the channel, the assumption is used that the slopes of the energy lines are the same for both. Letting K be the conveyance, K = A(A/P)2/3/n, gives that Qm /K m = Qs/Ks, or Qm = 300K m /(K m + Ks). In determining the areas it is assumed that the main portion of the channel extends upward with a side slope of 0.5.
235
The Momentum Principle Applied to Open Channel Flows 5
Momenmtum functions, M (m**3)
600
4.5
3
΄/s
m 00 =1
600
Q
2
΄/s
1
Q
0.5 0
Q = 500
=2
Q= 1
2
4
3
00
100
100
5
6
7
Depth of flow in compound channel, Y (m) Variation of Froude number, Fr with depth for several flow rate
Mcompound
200
Q = 400 Q = 300
3
1.5
300
Q=
Q
2.5
400
M
3
* =M ** + M *
Q=
3.5
m 00 =2
Froude number, Fr
500
3 /s m 600 3 /s Q= m 3 s / 500 Q = 400 m 3 s / Q= m 300
4
8
0
nel han in c sion a in m xten M * d its e an
in M*
1
2
3
4
5
e sid
6
ls nne
cha
7
8
Depth of flow, Y (m) Variations of individual and compound momentum functions with depth for a flow rate Q = 300 m3/s
Computations based on assuming the flow in the main portion of the channel is separated from that in the side portions of the channel. Values are based on a total flow rate Q = 300 m3/s. Main Channel
Depth Y (m)
Qm (m /s)
Vm (m/s)
Am (m2)
Pm (m)
Tm (m)
Frm
4.436 4.60 4.80 5.00 5.10 5.20 5.30 5.40 5.50 5.60
300.000 300.000 300.000 300.000 299.471 298.383 296.942 295.251 293.377 291.371
6.029 5.771 5.482 5.217 5.084 4.947 4.809 4.673 4.540 4.409
49.763 51.980 54.720 57.500 58.905 60.320 61.745 63.180 64.625 66.080
18.919 19.286 19.733 20.180 20.404 20.628 20.851 21.075 21.298 21.522
13.436 13.600 13.800 14.000 14.100 14.200 14.300 14.400 14.500 14.600
1.000 .943 .879 .822 .794 .766 .739 .712 .687 .662
Depth Y (m)
Qs (m /s)
Ts (m)
Fr
4.436 4.60 4.80 5.00 5.10 5.20 5.30 5.40 5.50 5.60
3
Side Channels 3
.529 1.617 3.058 4.749 6.623 8.629
Vs (m/s)
.177 .270 .341 .398 .445 .484
As (m2)
2.995 5.980 8.955 11.920 14.875 17.820
Ps (m)
29.976 29.953 29.929 29.906 29.882 29.858
29.900 29.800 29.700 29.600 29.500 29.400
.178 .193 .199 .200 .200 .199
Compound Froude No. 1.000 .943 .879 .822 1.305 1.177 1.069 .976 .896 .827
The first depth in this table is for critical flow. Note under the assumption that the flow is actually two different flows rather than flow in a single channel that the Froude number in the main channel continues to decrease, and of course the Froude number for the side channel flows is less than one. The last column in this table gives the Froude number computed as if this were a single compound channel. Note that Fr for the compound channel is larger than one whereas if one divides the flow into separate channels then both Fr′ values are less then unity. Therefore, we might conclude that
236
Open Channel Flow: Numerical Methods and Computer Applications
only for smaller sized cross sections are the computations of single compound channel cross section applicable for real occurrences. Values of momentum function in the main channel (extended upward with m = 0.5), the side channel and the compound channel are plotted in the above graph. The values of the momentum functions in the side channels are small compared to that in the main channel. There are two reasons for this: first, the depth of flow is small in comparison to that in the main channel, and therefore the term Ahc is small, and second, the flow rate in the side channels is small for a given area and therefore the term Q2/(gA3) is also small. The computations of the momentum function for the compound channel becomes smaller for depths just larger than 5 m as shown. All these facts indicate that caution is called for when using onedimensional hydraulic equations for compound channels.
Problems
500 cfs
Gate
3.1 A hydraulic jump occurs in a rectangular channel with a bottom width of b = 8 m. For a flow rate Q = 80 m3/s the depth upstream from the jump is 0.3 m. What is the depth downstream from the jump? 3.2 A flow rate of Q = 70 m3/s exists in a trapezoidal channel with b = 3 m, and a side slope m = 1.5. If the depth downstream from the jump is measured as Y2 = 6 m, what is the depth upstream from the jump? 3.3 A gate in a rectangular channel with b = 10 ft causes the depth of water downstream from it to be 2 ft. At the gate section the bottom of the channel rises abruptly by 2.0 ft. For a flow rate of Q = 450 cfs determine what the force is on the gate. The minor loss coefficient K L = 0.2. 3.4 A hydraulic jump occurs at the end of a transition between a trapezoidal channel with b1 = 8 ft, and m1 = 1.5 and a rectangular channel with b2 = 7 ft. For a flow rate of Q = 400 cfs, the depth in the downstream channel is Y2 = 8 ft. What is the force against the transition? (Assume head loss across the hydraulic jump is given by the formula for a rectangular channel.) 3.5 A trapezoidal channel with a bottom width of 10 ft, and a side slope of 1.0 has a diversion from its side in a 6 ft wide rectangular channel as shown in the sketch. Upstream the flow rate is 500 cfs, and 200 cfs leave from the diversion. At a gate, the main channel changes to a rectangular channel with a bottom width of 8 ft. The gate is set 3 ft above the channel bottom and has a contraction coefficient of Cc = 0.85.
b = 10 ft
8 ft
m=1 6 ft
Cc = 0.85
200 cfs
Determine the depth just upstream from the gate, the depth upstream in the main channel where the flow rate is 500 cfs, and the force on the gate and diversion structure, and the combined force on both. 3.6 A flow rate of Q = 2220 cfs comes down the spillway of a dam, and at the toe of the dam on its apron the depth is 2 ft. The channel here has a bottom width b = 20 ft, and side slope m = 2. Baffles are to be placed in the apron to keep the hydraulic jump on the apron. If the downstream channel has a bottom slope So = 0.00015, and a roughness coefficient n = 0.013 determine what force should exist on these baffles to keep the jump immediately downstream from the dam. How much crosssectional area should these baffles have if their drag coefficient based on the velocity upstream from the jump is CD = 0.73?
237
The Momentum Principle Applied to Open Channel Flows
2 ft Q = 2200 cfs b = 20 ft
So = 0.00015
m=2
3.7 A trapezoidal channel with a bottom width of b = 20 ft, and a side slope m = 1.5 contains stop logs in a rectangular contraction of this channel to a width of 20 ft, and downstream the flow depth is considerably less than the upstream depth. The stop logs extend upward from the bottom of the channel a distance of 4 ft. If the flow rate in the channel is Q = 1300 cfs, determine the force against the stop logs and its contracting structure. Now assume that the downstream depth is greater, creating a depth above the stop logs of 5.5 ft and determine this force. Also what is the upstream depth Y1 for this latter situation? m = 1.5 Q = 1300 cfs
b = 20
20
4 ft
3.8 A gate valve is used to control the discharge from a 3 ft diameter pipe. Open channel flow exists downstream from the valve. If the valve causes a depth of Y2 = 0.8 ft downstream of the valve, and the flow rate is Q = 30 cfs, determine the force against the gate valve. Ignore minor losses.
Q = 30
8 ft
3 ft
3.9 Stop logs in the bottom of a trapezoidal channel with b1 = 10 ft, and m = 1.5 are used to divert water into a side channel with a bottom width of b2 = 4 ft, and a side slope of m2 = 1. The bottom slope of this channel is So2 = 0.001, and its Manning’s n2 = 0.015. This channel is very long, and its bottom is 2 ft above the bottom of the main channel. The section of the main channel where the stop logs exist contracts to a rectangular section with a bottom width b3 = 10 ft. The stop logs are 3 ft above the channel bottom and cause the flow to be critical over their top. If the flow rate coming into the main channel is Q1 = 600 cfs, determine (a) the flow rates going by the stop logs and into the side channel, and (b) the force against the stop logs, and their hold structure. The side channel runs at 90° from the main channel. m = 1.5 Q1 = 600 cfs b1 = 10
b3 = 10 Δz = 2
So2 = 0.001
4
n2 = 0.015 n2 = 1
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Open Channel Flow: Numerical Methods and Computer Applications
3.10 Derive Equation 3.15. 3.11 Extract the root r = 1 from Equation 3.15 and prove that Equation 3.16 results. Explain what the difference is between Equations 3.15 and 3.16. 3.12 Starting from the dimensionless depth Y′ for a trapezoidal channel, equate M1′ to M′2 and derive Equation 20. 3.13 Solve for the upstream depth in Example Problem 3.5 if the baffles did not exist. 3.14 If the upstream depth is Y1 = 1.924 ft as determined in Example Problem 3.5 with the baffles in place, what would the downstream depth be without the baffles in place? 3.15 Type up the FORTRAN program, or utilize a similar program that will extract roots from a polynomial, including the complex roots, and investigate what range of flow rates, and/or depths upstream from a hydraulic jump will result in three real but negative depths, in addition to the real positive root being solved for instead of two complex roots. The channel you should do this investigation for has a bottom width of b = 4 m and a side slope m = 2. 3.16 Using the dimensionless momentum function find all roots of Equation 3.20 that are associated with a depth of 2.2 ft, and a trapezoidal channel with a bottom width of b = 20 ft, and a side slope m = 2 for a flow rate Q = 1500 cfs. Select the correct root, and determine the conjugate depth that would exit downstream of a hydraulic jump if the 2.2 ft were upstream from the hydraulic jump. Verify the result using Figure 3.1. 3.17 Determine the roots of the dimensionless momentum function associated with a flow rate of (a) 2000 cfs, and (b) 2500 cfs, if the channel is the same size as in the previous problem and Y1 = 2.2 ft as in the previous problem. 3.18 Modify the FORTRAN program used to extract the roots from the dimensionless momentum function from Equation 3.20 so that it will extract the roots from Equations 3.15 and 3.16. Then to verify that your programs work for these two equations solve the problem to get the conjugate depth to Y1 = 2 ft, and a trapezoidal channel with b = 10 ft and m = 1, with a flow rate Q = 400 cfs. 3.19 A gate discharges flow into a trapezoidal channel with a bottom width of b = 5 m, and a side slope m = 2, under submerged conditions. The channel has a bottom slope So = 0.0013, and a Manning’s n = 0.015. The gate is set 1.5 m above the channel bottom. If the depth downstream from the gate is measured as Y2 = 2.5 m, determine the following: (a) the flow rate, Q, (b) the depth upstream assuming that the velocity head under the gate is dissipated, and (c) the force against the gate. How does this force compare with the force computed if the pressure distribution were hydrostatic over the entire gate area wetted by the water?
Y2 = 2.5 m
1.5 m b=5 m
So = 0.0013
m=2
3.20 Using dimensionless values obtained by dividing by the critical depth (or critical depth squared for m′) generate a table that contains the following columns for across a hydraulic jump: (1) the dimensional depth upstream, Yu′ . (2) The Froude number corresponding to
The Momentum Principle Applied to Open Channel Flows
239
(1), Fru. (3) The value of the dimensionless momentum function m′. (4) The corresponding value of the dimensionless specific energy E′u. (5) The dimensionless depth downstream, Yd′. (6) The Froude number corresponding to (5), Frd. (7) The dimensionless specific energy downstream, E′d . (8) The loss of dimensionless specific energy across the hydraulic jump. (9) Using E′d from (7) use the solution of a cubic equation that produces the conjugate dimensionless depths of Figure 3.3 to obtained the downstream depth Yd, i.e., duplicate the values in column 5, except use E′d in place of m′ for the abscissa, and the ordinate is 1 / Ye′ rather than Ym′ . Start this table with Yu′ equal to 1 in column 1 and decrement Yu′ by 0.05 and end with Yu′ = 0.1 in the final row of column 1 of the table. 3.21 As in the previous problem use dimensionless values divided by the critical depth to generate a table that contains the following columns across a vertical gate, rather than a hydraulic jump. (1) the dimensionless depth, Yu′ . (2) The Froude number corresponding to (1), Fru. (3) The value of the dimensionless specific energy E′u . (4) The corresponding value of the dimensionless momentum function m′u . (5) The dimensionless depth downstream, Yd′ . (6) The Froude number corresponding to (5), Frd. (7) The dimensionless downstream specific energy computed from (5), just to verify that E′d = E′u . (8) The dimensionless momentum function downstream, m′d . (9) The dimensionless force per unit width on the gate, i.e., the difference in dimensionless momentum across the gate. (10) Using m′d from (8) use the solution of a cubic equation that produces the alternate dimensionless depths of Figure 3.3 to obtained the downstream depth Yd, i.e., duplicate the values in column 5, using m′d for the abscissa. (11) Provide the alternate depth (subcritical depth) to that given in column 10. Start this table with Yu′ equal to 1 in column 1 and increment Yu′ by 0.05 and end with Yu′ = 2.5 in the final row of column 1 of the table. 3.22 The dimensionless cubic equations for energy and momentum for a rectangular channel have the reciprocal properties that when 1/ Ye′ is substituted into the energy equation the momentum equation is produced, and if 1/ Ym′ is substituted into the momentum equation the energy equation is produced. In this process m′ is treated as E′ and vice versa. The dimensionless values of depth and energy are obtained by dividing by the critical depth, and m′ by dividing by the critical depth squared. Verify these reciprocal properties by generating two tables of values. In the first column of Table 3.1 let m′ vary from 1.5 to 2.5 in increments of 0.05, as the next three columns provide the three roots obtained from solving the cubic momentum equation, and as the final three columns provide the three dimensionless depths associated with the first column of m′ values but obtained from solving the three roots of the energy equation. Note in this table that columns 5 through 7 contain the same values for the dimensionless momentum depths as columns 2 through 4. In the first column of Table 3.2 let E′ vary from 1.5 to 2.5 in increments of 0.05, as the next three columns provide the three dimensionless depths associated with the first column of E′ obtained from solving the three roots of the cubic energy equation, and as the final three columns provide the three dimensionless values associated with the first column of E′ values but obtained from solving the cubic momentum equation. Note in this second table that columns 5 through 7 contain the same values for the dimensionless energy depths as columns 2 through 4. 3.23 An iterative method for solving either the subcritical or supercritical depth associated with a given value for the momentum function in a trapezoidal channel is to use the cubic equation that applies for a rectangular channel, i.e., F(Y) = Y3 − 2cmoY + 2cq2/g = 0, in which mo is the average momentum function per unit width, or M/bav, and q is the average flow rate per unit width or q = Q/bav, in which bav = A/Y = b + mY. Do the following: (1) Write the momentum function equation for a trapezoidal channel in the form of the above cubic equation, and in this process define how c is computed in this equation (note that as the side slope of the trapezoidal channel goes to 0 that c becomes 1). (2) Write a program that will solve for either the subcritical or supercritical depth associated with a specified value for the momentum function M. (3) Verify that your program works to solve the subcritical and supercritical depths if Q = 400 cfs, M = 300 ft3, b = 10 ft, and m = 1.
240
Open Channel Flow: Numerical Methods and Computer Applications
3.24 As requested in the previous problem develop an iterative solution for the subcritical or supercritical depth associated with a given value for the momentum function M, except have it apply for a circular section, rather than a trapezoidal channel. In other words obtain iterative solutions to the cubic equation for a rectangular channel, so that each subsequent iteration causes this solution to give the appropriate depth for a circular channel. Test you solution method by solving for both the sub and supercritical depth associated with a flow rate of Q = 200 cfs in a D = 10 ft diameter pipe if the momentum function M = 100 ft3. 3.25 The dimensionless momentum equation (Equation 3.21) for a rectangular channel is obtained by dividing the depth by the critical depth Yc rather than by the bottom width b and multiplying by m as was done to obtain the dimensionless equation (Equations 3.15 or 3.16) for a trapezoidal channel. Obtain a dimensionless equation for a trapezoidal channel by dividing Y by Yc and define the dimensionless momentum function by dividing by the critical depth cubed, or M′ = M / Yc3 . Note this dimensionless equation contains the side slope m and the bottom with b, as well as Yc and therefore fails to satisfy one of the purposes for nondimensionalizing an equation, namely to remove channel sizes from the equation. Therefore, this equation lacks the practical applications of Equations 3.15 or 3.16. Prove that with the side slope m = 0 in this equation that it reduces to the momentum equation for a rectangular channel. Also develop a solution to this fifth degree polynomial equation that extracts all five of its roots. What are the roots of this equation if Q = 400 cfs, b = 10 ft, m = 1 and the momentum function M = 300 ft3? 3.26 Modify Equation 3.19, which gives the dimensionless momentum function M′ for a circular channel so that rather than involving the dimensionless flow rate Q′ = Q2/(gD5) it contains the angle associated with critical depth, i.e., βc = cos −1 (1 − 2Yc′) . Use this equation to generate a series of tables for different values of βc (which of course has a critical depth associated with it since Yc′ = 0.5(1 − cos βc )) that provide the dimensionless depths Y′ = Y/D (and associated values of angle β) and the associated values of the dimensionless momentum functions M′ = M/D3. If the data from these separate tables were plotted, a graph similar to Figure 3.2 would be produced, the difference being that rather than the separate curves being associated with a Q′, they would be associated with a given critical value of angle βc or a dimensionless critical depth Yc′ . 3.27 A wavemaking device in a rectangular channel of 2 ft width consists of a vertical plate that can be moved against the still water in the channel by a mechanical drive mechanism. If the plate moves at a speed of 2 fps and the depth of water in the channel is 6 ft, determine the height of the wave and the speed of its movement. What force is required to drive the plate forward? 3.28 At the end of a circular channel with a diameter D = 8 m, that carries a discharge of Q = 18 m3/s, there are three identical rectangular side channels that take an equal amount of water from the circular channel at right angles from the direction of its flow. The gate in the first channel is wide open. The side channels are 2 m wide, and have a bottom slope of 0.001, and a Manning’s n = 0.013. The bottoms of the side channels are 1 m above the bottom of the pipe. What are the depths in the main channel between each of the side channels, and the depth upstream from the first side channel and downstream from the last side channel. What force is needed to cause the diversion into each side channel? Explain how this force is developed. (Ignore minor losses.)
2m
D=4 m 18 m3/s
Gate
Gate
241
The Momentum Principle Applied to Open Channel Flows
3.29 If the depth of the water in the Problem 3.27 is 4 ft, determine the speed at which the plate should move to create a wave with a speed of 6 fps. 3.30 A smooth hump of 0.5 ft exists in the bottom of a 10 ft wide rectangular channel. If the upstream depth is 5 ft and the flow rate is Q = 100 cfs, determine the force on the hump. Develop an equation that gives this force in general. 3.31 Flow upstream from a gate in a rectangular channel with b = 15 ft is at a depth of 4 ft and a velocity of 3 fps. Suddenly the gate is completely closed. What is the depth at the gate and the speed of the surge? 3.32 Flow in a very wide channel with So = 0.0013, and n = 0.012 is suddenly decreased from q = 12 cfs/ft to q = 3 cfs/ft. What is the new depth of flow and the velocity of the wave? 3.33 A gate is used to control the flow rate in a channel. Upstream from the gate the channel has a bottom width b = 10 ft, and a side slope m = 1.5. At the gate a smooth transition changes the section to rectangular with b = 10 ft. The gate has been set 4 ft above the channel bottom for a long time, and the flow rate under this setting has been measured to be Q = 400 cfs. Suddenly the gate is lowered to a height 3 ft above the channel bottom. Under the assumption that the depth upstream of the gate is constant compute (a) the new depth upstream from the gate, (b) the speed at which the surge will move upstream, (c) the flow rate past the gate, (d) the force on the gate prior to being closed to the new position, and (e) the force on the gate after being closed to the new position. The contraction coefficient for the gate is Cc = 0.6. Ignore minor losses. 3.34 Initial conditions are as in the previous problem, but the gate is suddenly closed to a position 1 foot above the channel bottom. Compute the same quantities asked for in the previous problem. 3.35 Initial conditions are as in Problem 3.33, but the gate is suddenly completely closed. Compute the same quantities asked for in Problem 3.33. 3.36 A pipe with a diameter D = 3 m is laid on a slope of So = 0.0006, and has a Manning’s n = 0.013. If the flow rate in the pipe is Q = 5 m3/s, find the wave speed v and the depth Y2 and velocity V2 in the pipe upstream from a valve at its end if (a) the valve is instantly closed completely, (b) the valve is instantly closed to a position so Y3 = 0.5 m above the bottom of the pipe, and (c) just before the valve a smooth transition changes the pipe to a square section with a width equal to D = 3 m, and the valve is closed to a position so Y3 = 0.2 m above the bottom of the pipe.
v Q = 5 m3/s
Y1 = Yo
V1 = Vo
D = 3 m, n = 0.013, So = 0.0006
Y2
V2
Y3
Part (a) Valve completely closed Part (b) Valve closed so Y3 = 0.5 m Part (c) Square section at valve with b = 3 m and Y3 = 0.2 m
3.37 A channel with a bottom width of b = 4 ft, and a side slope m = 1.2 discharges into a pipe with a diameter D = 6 ft. The pipe has a bottom slope So = 0.02, and an equivalent sand roughness for use in the Darcy–Weisbach equation of e = 0.0006 ft. At a distance of 1500 ft a valve controls the flow and creates a pressure head of 29 ft on the top of the pipe. The loss coefficient is K L = 0.5 between the channel and the pipe. If the flow rate is Q = 200 cfs determine what the depth is in the channel upstream of the pipe. Under the assumption that the depth you determine is constant throughout the channel determine: (a) the flow rate Q n, (b) the new depth Y1, and (c) the velocity v of a surge if the valve is adjusted so as to create a pressure head of ph = 30 ft. Repeat the solution for these variables if ph = 31 and 32 ft.
242
Open Channel Flow: Numerical Methods and Computer Applications Sf
v Y1 Q = 200 cfs
L = 1500 ft
V1
Vo
ph = 30 ft ph = 31 ft ph = 32 ft p = 29 ft h
D = 6 ft
b = 4 ft, m = 1.2
So = 0.0
2, e = 0
.0006 ft
3.38 Two gates exist at the end of a trapezoidal channel with a bottom width b1 = 3 m, and a side slope m1 = 1.0. Each gate is 1.5 m wide, has a contraction coefficient Cc = 0.6, and initially both gates are open with their tips YG1 = YG2 = 0.9 m above the bottom of the channel. Suddenly one of the gates is completely closed. Determine how much the depth will increase upstream from the gates, and the speed of the surge that will result. The flow coming into the channel is Q = 10 m3/s. v Vo
V1
Yo
Y1
Cc YG = 0.9
YG
b1 = 3 m, m1 = 1.0
1.5 m
YG 1.5 m
3.39 A gate is releasing a flow rate of Q = 30 m3/s into a trapezoidal channel with b = 2.5 m and m = 1.0, and So = 0.0008 from a reservoir and the channel downstream causes the gate to be submerged so the depth downstream of the gate equals the normal depth in the downstream channel. Suddenly the gate is raised so that the flow rate is doubled, e.g., is now Q = 60 m3/s. If the gate remains submerged what is the new depth downstream of the gate, and what is the speed of the surge that will travel downstream in the channel.
v Y1
Qn = 60 m3/s
Yo
Vo
Q = 30 m3/s
V1
b = 2.5 m, m = 1.0, So = 0.0008, n = 0.014
3.40 Derive Equation 3.33a from Equation 3.33, and thus show that the wave velocity v equals the sum of the upstream velocity V1, plus the celerity c2 times the square root of the average depth (Y1 + Y2)/2, i.e., v increases more rapidly than c2 does with Y2. 3.41 A flow rate per unit width of q = 20 cfs/ft occurs in a 10 ft wide rectangular channel with n = 0.013 and a bottom slope So = 0.0005 under uniform conditions. Suddenly a gate is closed completely at the downstream end of the channel. Determine the depth upstream from the gate and the velocity of the wave. Solve this problem by (a) simultaneously solving the continuity and momentum equations from the viewpoint of a moving observer, and (b) first solving only one implicit equation, Equation 3.37 (or 3.37a).
243
The Momentum Principle Applied to Open Channel Flows
3.42 Resolve the previous problem with the gate shut only part way so that flow past the gate is onehalf the original flow rate, or q2 = 10 cfs/ft. 3.43 Solve Example Problem 3.9 using two simultaneous equations rather than one implicit equation followed by the explicit continuity equation to solve v. 3.44 Solve Example Problem 3.10 using two simultaneous equations rather than one implicit equation followed by the explicit continuity equation to solve v. 3.45 A gate at the upstream end of a trapezoidal channel with b = 3 m, m = 1, n = 0.014, and So = 0.0003, is initially set with its tip 0.5 m above the bottom of the channel. The flow behind the gate is submerged, and the gate causes a headloss equal to 1.1 times the velocity head in the jet coming from under the gate, where this velocity is computed by dividing the flow rate by the area corresponding to the gate’s height times its contraction coefficient, which is Cc = 0.60. The water that flows into the channel comes from a constant head reservoir whose water surface elevation is 5 m above the bottom of the channel. Obtain a series of solutions that provides the submerged depth Y2 downstream from the gate, the constant height wave speed v, and the velocity V2 that will occur if the gate’s height is suddenly increased from the initial setting of 0.5 to 0.75, 1.00, 1.25, 1.50, etc. m. 3.46 Flow in a steep pipe of 3 ft diameter is controlled downstream by a valve, so that at the valve the pressure at the top of the pipe is 40 psi. The slope of the pipe is So = 0.0855, its Manning’s n = 0.013, and it contains a flow rate of 32.0 cfs. Determine the position upstream from the valve where the flow changes from open channel to closed conduit flow. 3.47 A square box conduit with a width and height of 2 ft contains a flow rate of 60 cfs. The roughness coefficient for this conduit is n = 0.014. The bottom slope changes from So1 = 0.102 to So2 = 0.00085. Determine where the flow becomes closed conduit flow if the channel ends with water at its top 1000 ft downstream from the change in grade. 3.48 A smooth transition takes place from a 4 m diameter pipe to a 0.5 m diameter pipe. The larger pipe has a bottom slope of So1 = 0.015, and the smaller pipe has a slope So2 = 0.250, and a length of 1000 m. At its end the water flows into a tank with a water surface elevation 10 m above the ground surface. (a) Determine where the flow changes from open channel to pipe flow if the flow rate Q = 0.3 m 3/s. (n = 0.013 for both pipes and for use in the Darcy– Weisbach equation for pipe flow use an equivalence roughness e = 0.001m.) (b) What flow rate will just cause the smaller diameter downstream pipe to flow full to its beginning? (c) If the flow rate is Q = 0.325 m 3/s where will the flow change from open channel to pipe flow? 3.49 Two vertical gates, each 3 ft wide are in a rectangular channel that is 9.5 ft wide. The pier between the gates is 1.5 ft wide. When a flow rate of Q = 150 cfs is occurring, the depth immediately downstream from gate #1 is 1.5 ft, and from gate #2 is 2.0 ft. Determine the following: (a) The amount of flow passing each gate. (b) The force against gate #1. (c) The force against gate #2. (d) The force against the middle pier. (State clearly any assumptions used to obtain this force.)
1 ft
Q = 150 cfs 9.5 ft
3 ft
Yg2 = 2.0 ft
3 ft
Yg1 = 1.5 ft
1.5 ft
1 ft
244
Open Channel Flow: Numerical Methods and Computer Applications
3.50 Write a program, model or spreadsheet to generate data so that a graph can be constructed that provides the supercritical dimensionless depth Y1′ = mY1 / b on the ordinate as a function of the subcritical depth Y2′ = mY2 / b on the abscissa for a trapezoidal channel, and then construct this graph. Have a curve on this graph for the following values of the dimensionless flow rate Q′ = m3Q2/(gb5): 0.01,0.02,0.03,0.04,0.05,0.075,0.10,0.15,0.2,0.3,0.4,0.5,0.6,0.8 and 1. 3.51 Repeat the previous problem except make the graph for a circular channel with Y1′ = Y1 / D, Y2′ = Y2 / D and Q′ = Q2/(gD5). 3.52 A gate in the main channel is used to control the amount of flow that leaves the main channel, and is conveyed away by a side channel that runs at an angle 90° from the main channel. The main channel is trapezoidal in shape with b = 5 m, and m = 1.8. The side channel is circular with D = 3 m, and has a bottom slope So2 = 0.00125, and a Manning’s n = 0.014 and is very long. Its bottom is 2 m above the bottom of the main channel, If the main channel is carrying a flow rate of Q = 100 m3/s, and it is desirable to divert a flow rate 10 m3/s in the side channel determine how far above the bottom of the channel the gate should be set. Its contraction coefficient is Cc = 0.56. What is the force against the gate? What is the force against the side channel wall that is required to divert the water? 3.53 A sewer pipe of 12 in. diameter and a roughness coefficient e = 0.005 in. for use in the Darcy– Weisbach equation discharges into a sewerage treatment pond whose water surface elevation is 2 ft above the top of the pipe. The pipe has a constant slope of So = 0.025 for a long distance. For use in Manning’s formula assume n = 0.013. Determine the position where the flow changes from open channel to pipe flow for flow rates of: (a) Q = 3 cfs, (b) Q = 4 cfs, and (c) Q = 5 cfs. What is the major factor that causes the flow to move upstream with increasing flow rates? 3.54 For Example Problem 3.12, determine how the flow rate Q, the upstream depth Y1, and the downstream depth Y2, etc. vary as the position x changes where the flow changes from open channel to closed pipe flow. Change the position x from 0 ft to just beyond 20 ft where a solution of the governing system of equations is no longer possible. (Note: The solution fails at about 21.875 ft.) 3.55 Compare the headlosses that occur in a rectangular and circular channel across a hydraulic jump (or if the jump causes pipe flow in the circular channel), if the flow rate is Q = 400 cfs, and the diameter of the pipe is 8 ft. For this comparison let the rectangular channel have a width so that it has the same area as the circle when the depth is 8 ft in the rectangle. Have the upstream depth vary from 5 to 2 ft. Also note when the jump hits the top of the circular channel. 3.56 Repeat the previous problem but increase the flow rate from Q = 400 cfs to Q = 800 cfs, and have the upstream depths vary from 7.8 to 2.4 ft. 3.57 At a certain position in a trapezoidal channel it has two side channels that take water in directions 90° from the direction of the main channel. The first such side channel is rectangular with a bottom width of b4 = 6 ft, n4 = 0.014, and So4 = 0.0008. This channel is very long. The second side channel is also rectangular, is controlled by a gate that produces a depth of Y52 = 1.5 ft downstream from the gate. Upstream the main channel has a bottom width, b1 = 10 ft, a side slope m1 = 1.5, and after the first branch channel the width reduces to 8 ft so that here b2 = 8 ft, m 2 = 1.5. Downstream from the second side channel the main channel is controlled by a gate that produces a depth downstream from the gate of Y32 = 1.5 ft. The second side channel has its bottom raised by Δz25 = 0.5 ft. The flow rate in the upstream main channel is Q1 = 500 cfs. Ignoring all minor loss coefficient determine the following four forces: The force on the channel structure between the upstream main channel and the first side channel, Fs4, the force on the channel structure upstream and downstream from the second side channel, Fs5, the force on the gate in channel 3, FG3 and the force on the gate in channel 5, FG5. Before obtaining these forces you will find it necessary to compute the flow rates Q2, Q3, Q4, Q5, and the depths Y1, Y2, Y31, Y4, and Y51.
245
The Momentum Principle Applied to Open Channel Flows
Q2
Y1
F
n4 = 0.014
b4 = 6 ft
3.59 3.60
3.61 3.62
(4)
Y4 Q4
(2)
(3)
F
s5
b2 = 8 ft, m2 = 1.5
(5)
Δz
25
Y51 b5 = 6 ft
s4
b1=10 ft, m1 = 1.5
3.58
Y2
F0.5
(1)
So4 = 0.0008
Q1
=
Y31
FG3
Q3 Y32 = 1.5 ft
b3 = 8 ft, m3 = 1.5
0.
5
ft Y52 = 1.5 ft
Q3
Solution: Fs4 = 6324.2 lb to the right and upward at 63.13° from the horizontal, Fs5 = 4848.1 lb to the right and upward at 78.19° from the horizontal, FG3 = 4071 lb, and FG5 = 722.0 lb. If the gate is raised in channel # 3 of the previous problem so that it produces a depth of 1.7 ft downstream from it, what will the forces, and flow rates become in this problem if the upstream flow rate Q1 is maintained at 500 cfs. What occurs if the gate in channel 3 is raised further so that it attempts to produce a depth of 2.0 ft downstream? If the gate is raised in channel # 5 of Problem 3.57 so that it produces a depth of 2.0 ft downstream from it, what will the forces, and flow rates become in this problem if the upstream flow rate Q1 is maintained at 500 cfs. For the same channel configuration as given in Problem 3.57 the depth of flow immediately upstream from the gate in channel # 3 is measured to equal 5.40 ft. Determine the same forces asked for in Problem 3.57 as well as all the flow rates and depth, including the flow rate Q1 in the main upstream channel. Obtain a series of solutions to Example Problem 3.12 in which the bottom slope So1 of the upstream channel varies. A D1 = 12 ft diameter pipe which is laid on a slope of So1 = 0.20 (with n1 = 0.013 and e1 = 0.0005 ft) connects into a pipe with a diameter D2 = 8 ft, which is laid on as lope of So2 = 0.04 (with e2 = 0.0005 ft). This second pipe is L2 = 500 ft long and discharges into a reservoir with a water surface elevation H2 = 13 ft above the bottom of the pipe. The local loss coefficient KL = 0.5 for the transition between the two pipes. If the flow rate is Q = 1800 cfs, determine the position x upstream from the junction of the two pipes where a modified hydraulic jump will occur.
v Q = 5 m3/s
Y1 = Yo
V1 = Vo
D = 3 m, n = 0.013, So = 0.0006
Y2
V2
Y3
Part (a) Valve completely closed Part (b) Valve closed so Y3 = 0.5 m Part (c) Square section at valve with b = 3 m and Y3 = 0.2 m
3.63 Solve for the flow rate Q in the previous problem if the modified hydraulic jump occurs at x = 50 ft upstream from the position where the pipe diameters change. 3.64 Solve the previous problem, except that the upstream pipe has a diameter D1 = 10 ft. Can you explain why reducing the pipe diameter allows a larger flow rate? 3.65 A flow rate per unit width of q = 10 m2/s is coming down a steep rectangular spillway that is 15 m wide, and has a bottom slope of So1 = 0.25, and n1 = 0.013. A box culvert occurs at the end of the spillway with a width of 15 m, and a height of 5 m, as shown in the sketch, with two vertical partitions spaced at a distance of 5 m. This box culvert is 500 m long, when it changes into a 6 m diameter pipe.
246
Open Channel Flow: Numerical Methods and Computer Applications
So1 =
m 2/s
0.25,
x d
Yo
n1 = 0
1
.013,
b1 = 1
5m
L2 = 500 m
h2 = 5 m
D3 = 6
So2 = 0.0015, n2 = 0.013, b2 = 15 m e2 = 0.0006 m
So3 = 0.0
03
m
e3 = 0.000
3m
L3 = 4000
ph
Valv e
q =10
m
5m 5
5 5 Cross section box culvert
The box culvert has a bottom slope So2 = 0.0015, (with n2 = 0.013 and e2 = 0.0006 m), and the pipe has a bottom slope So3 = 0.003 (with e3 = 0.0004 m). The pipe is 4000 m long where its flow is controlled by a valve. If the pressure head created by the valve on the top of the pipe is ph = 2 m, where will a modified hydraulic jump occur within the box culvert? (Ignore that a gradually varied flow will actually increase the depth in the first portion of the box culvert where the flow is supercritical upstream from the jump, e.g., assume up to the jump in the box culvert the depth remains at Yo1.) The box culvert has a bottom slope So2 = 0.0015, (with n2 = 0.013 and e2 = 0.0006 m), and the pipe has a bottom slope So3 = 0.003 (with e3 = 0.0004 m). The pipe is 4000 m long where its flow is controlled by a valve. If the pressure head created by the valve on the top of the pipe is ph = 2 m, where will a modified hydraulic jump occur within the box culvert? (Ignore that a gradually varied flow will actually increase the depth in the first portion of the box culvert where the flow is supercritical upstream from the jump, e.g., assume up to the jump in the box culvert the depth remains at Yo1.) 3.66 Solve Example Problem 3.13 except that the smaller diameter pipe has a 2.0 ft diameter (rather than 1.5 ft), and is 1000 ft long (rather than 1200 ft). It delivers the same pressure head of 90 ft at its end. 3.67 Solve Example Problem 3.13 except in place of the stilling basin and overflow weir there is a long trapezoidal channel with b = 8 ft, m = 1, n = 0.014, and So = 0.0007 into which the 5 ft diameter pipe discharges. 3.68 For a compound channel verify that the flow rate in the main channel is given by, Qm =
QtK m (K m + K s )
in which K are conveyances defined from Manning’s formula as K = A(A/P)2/3/n and subscript m and s apply for the main channel and side channels, respectively. Qt is the total flow rate. The assumption in the verification of the equation is that the slope of the energy lines Sf of the main and side channels are the same. 3.69 If a flow rate of Q = 300 m3/s occurred in the compound channel used as an example in the text (bm = 9 m, br = 15 m, bl = 15 m, mm = 0.5, nm = 0.018, and nr = nl = 0.055) what will the flow rate be in the main channel and in the side channels under the assumption that the side of the main channel portion of the flow extends upward with the side slope of 0.5 for a depth of Y = 5.4 m. (In other words verify the results provided in the table given in the text.) 3.70 Repeat the previous problem but assume that the main channel portion of the flow extends vertically upward from where its banks end. 3.71 For the compound channel in the text (bm = 9 m, br = 15 m, bl = 15 m, mm = 0.5, nm = 0.018, and nr = nl = 0.055) for a flow rate of Q = 300 m3/s and a downstream depth of 5.6 m determine the
The Momentum Principle Applied to Open Channel Flows
247
depth upstream and the upstream Froude number that would result in a hydraulic jump. Carry out these computations: (a) assuming onedimensional hydraulic equations are valid, and the channel downstream from the hydraulic jump can be handled as a compound channel, and (b) if the hydraulic jump occurred only in the main channel portion. 3.72 Determine the normal and critical depths in a compound channel with the following measurements if the flow rate in this channel is Q = 400 cfs, bm = 4 ft, mm = 1.2, br = 2 ft, mr = 0.5, bl = 3 ft, ml = 0.5, nm = 0.013, nr = nl = 0.022, So = 0.00085 and the height of the main channel is 4 ft. 3.73 Determine the normal and critical depths in a compound channel with the following measurements if the flow rate in this channel is Q = 3400 cfs, bm = 15 ft, mm = 1.0, nm = 0.020, br = 35 ft, mr = 0.5, bl = 40 ft, ml = 0.5, nm = 0.015, nr = nl = 0.042, So = 0.00085, and the height of the main channel banks before the water flow into the side channels is 10 ft. Based on both the assumptions that the sides of the main channel are vertically above the top bank, and extend upward with a side slope of 1, determine what portion of the total flow rate will be within the main channel extended to the water surface and what portion will be in the side channels. Compute the kinetic energy correction coefficient under the assumption that the velocities in the three separate portions of the channel are constant. Compute the momentum functions under normal depth for: (a) the compound section, (b) the main channel flow, and (c) the side channel flow. 3.74 Write a computer program that will solve for any desired four variables associated with the flow in a compound channel. Assume the compound channel consists of a main channel and right and left side channels, all of which are trapezoidal in shape. The four equations available to solve the four unknowns are: (1) the continuity equation that requires that the total flow Q equal the sum of the flows in the main channel; (2) the right side channel and (3) the left side channel, or F1 = Q − Qm − Qr − Ql = 0; and (4) Manning’s equations for the three component channels. Solution: The listing for a FORTRAN program that does this is given below: This program is designed to first prompt for the input of all variables. For those variables that are later selected as the unknowns these given values will be used as the initial values to start the Newton iterative solution. The names used to prompt for the variables are as follows: Q = total flow rate; Qm = flow rate in main channel; Qr = flow rate in right side channel; Ql = flow rate in left side channel; y = depth above top of main channel; bm = bottom width; m m = side slope of main channel; nm = Manning’s n for main channel; br = bottom width of right side channel; m r = side slope of right side channel; nr = Manning’s n for right side channel; bl = bottom width of left side channel; ml = side slope of right side channel; nl = Manning’s n for right side channel; So = bottom slope of channels; Y1 = depth to top of main channel, i.e., the depth is the sum of y and Y1; g = acceleration of gravity to determine with SI or ES units are to be used; and Iv = parameter to determine whether the main channel is to extend upward to the surface vertically from its top width, or whether the main channel is to extend upward with the same side slope to the water surface. If vertically upward Iv = 1, otherwise IV = 0. Listing of program, MANNCO1.FOR INTEGER*2 INDX(4),IUNK(4) REAL D(4,4),X(18),F(4),F1(4),mm,mr,ml,nm,nr,nl CHARACTER*2 CH(18)/'Q ','Qm','Qr','Ql','y','bm','mm','nm', &'br','mr','nr','bl','ml', 'nl','So','Y1','g ','Iv'/ COMMON X,CC,A1,P1,Ivert EQUIVALENCE (Q,X(1)),(Qm,X(2)),(Qr,X(3)),(Ql,X(4)), (Y,X(5)), &(bm,X(6)),(mm,X(7)),(nm,X(8)),(br,X(9)), (mr,X(10)), &(nr,X(11)),(bl,X(12)),(ml,X(13)), (nl,X(14)),(So,X(15)), &(Y1,X(16)) WRITE(*,*)' Give value to each variable' DO 5 I=1,18 WRITE(*,110) I,CH(I) 110 FORMAT(I3,2X,A2,' = ',\)
248
Open Channel Flow: Numerical Methods and Computer Applications 5 6 120 10 15 20 30 100
IF(I.EQ.18) WRITE(*,"(' 1 for vertically up ',\)") READ(*,*) X(I) Ivert=X(18)+.1 WRITE(*,*)' Give number of 4 unknowns' DO 6 I=1,16 WRITE(*,120) I,CH(I),X(I) FORMAT(I3,2X,A2,F10.3) READ(*,*) IUNK IF(X(17).LT.30.) THEN CC=1. ELSE CC=1.486 ENDIF Qm=.75*Q Qr=.125*Q Ql=Qr A1=(bm+mm*Y1)*Y1 P1=bm+2.*Y1*SQRT(1.+mm*mm) NCT=0 CALL FUNCT(F) DO 20 I=1,4 XX=X(IUNK(I)) X(IUNK(I))=1.005*X(IUNK(I)) CALL FUNCT(F1) DO 15 J=1,4 D(J,I)=(F1(J)−F(J))/(X(IUNK(I))−XX) X(IUNK(I))=XX CALL SOLVEQ(4,1,4,D,F,1,DD,INDX) DIF=0. DO 30 I=1,4 X(IUNK(I))=X(IUNK(I))−F(I) DIF=DIF+ABS(F(I)) NCT=NCT+1 WRITE(*,*)' NCT=',NCT,' DIF=',DIF IF(NCT.LT.20 .AND. DIF.GT. .0004) GO TO 10 WRITE(*,100) (IUNK(I),I=1,4),(I,CH(I),X(I),I=1,16),Y1+X(5) FORMAT(' Solution to:'4I3,/,1X,24('−'),/, 16(2X,I2,1X,A2,'=’, &F12.5,/),' 17 Depth =',F10.2) END SUBROUTINE FUNCT(F) REAL F(4),X(18),mm,mr,ml,nm,nr,nl COMMON X,CC,A1,P1,Ivert EQUIVALENCE (Q,X(1)),(Qm,X(2)),(Qr,X(3)),(Ql,X(4)),(Y,X(5)), &(bm,X(6)),(mm,X(7)),(nm,X(8)),(br,X(9)), (mr,X(10)), &(nr,X(11)),(bl,X(12)),(ml,X(13)), (nl,X(14)),(So,X(15)), &(Y1,X(16)) IF(Ivert.EQ.1) THEN Am=A1+(bm+2.*mm*Y1)*Y Ar=.5*(2.*br+mr*Y)*Y Al=.5*(2.*bl+ml*Y)*Y ELSE Am=A1+(bm+2.*mm*Y1+mm*Y)*Y Ar=.5*(2.*br+(mr−mm)*Y)*Y Al=.5*(2.*bl+(ml−mm)*Y)*Y ENDIF Pr=br+Y*SQRT(1.+mr*mr)
249
The Momentum Principle Applied to Open Channel Flows
Pl=bl+Y*SQRT(1.+ml*ml) F(1)=Q−Qm−Qr−Ql F(2)=nm*Qm−Am*(Am/P1)**.666666667*CC*SQRT(So) F(3)=nr*Qr−Ar*(Ar/Pr)**.666666667*CC*SQRT(So) F(4)=nl*Ql−Al*(Al/Pl)**.666666667*CC*SQRT(So) RETURN END
3.75 For the compound channel shown in the sketch below with a main channel bottom width bm = 20 ft, and side slope m m = 1, and a Manning’s nm = 0.013, with a right side channel with br = 40 ft, m r = 0.5, and nr = 0.045, a left side channel with bl = 50 ft, ml = 0.5, and nl = 0.045, and a bottom slope So = 0.0008, determine the flow rates in all component channels for depths varying from 0.5 ft above the top of the main channel to a depth of 10 ft above the main channel, in increments of 0.5 ft. The height of the main channel is Y1 = 10 ft. Solve for these flow rates: (a) assuming that the main channel extends vertically upward from its outer banks to the water surface, and (b) that it continues with the side slope of m m = 1 to the water surface. Plot the 4 flow rates against the depth of flow in the compound channel. 1
y
n1 = 0.045 Y
b1 = 50 ft m1 = 0.5
Y1 = 10 ft nm = 0.013
bm = 20 ft
nr = 0.045 1 mm = 1
br = 40 ft
1 mr = 0.5
So = 0.0008
Solution: Solving Manning’s equation for the main channel with a depth of 10 ft gives Q = 3278.1 cfs. The solution obtained by using a program such as that in the previous problem are given in the tables below. Main Channel Consist of Area Vertically Above y (ft)
Y (ft)
Q (cfs)
Qm (cfs)
Qr (cfs)
Ql (cfs)
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0 16.5 17.0 17.5 18.0 18.5 19.0 19.5 20.0
3676.72 4121.92 4605.58 5124.16 5675.32 6257.34 6868.87 7508.76 8176.07 8869.95 9589.68 10334.59 11104.09 11897.64 12714.76 13554.98 14417.89 15303.10 16210.26 17139.01
3650.34 4038.45 4442.09 4860.97 5294.81 5743.35 6206.37 6683.64 7174.95 7680.10 8198.90 8731.18 9276.77 9835.50 10407.24 10991.82 11589.11 12198.97 12821.28 13455.92
11.72 37.06 72.57 116.78 168.78 227.90 293.65 365.62 443.48 526.94 615.76 709.72 808.64 912.36 1020.71 1133.57 1250.82 1372.33 1498.02 1627.79
14.66 46.40 90.92 146.41 211.74 286.09 368.84 459.50 557.64 662.92 775.02 893.69 1018.68 1149.78 1286.81 1429.59 1577.97 1731.80 1890.95 2055.30
250
Open Channel Flow: Numerical Methods and Computer Applications Main Channel Continues with mm = 1 y (ft)
Y (ft)
Q (cfs)
Qm (cfs)
Qr (cfs)
Ql (cfs)
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0 16.5 17.0 17.5 18.0 18.5 19.0 19.5 20.0
3681.23 4140.20 4647.47 5200.17 5796.71 6436.16 7117.96 7841.82 8607.62 9415.41 10265.33 11157.62 12092.59 13070.63 14092.17 15157.70 16267.74 17422.86 18623.66 19870.79
3655.09 4058.27 4488.46 4946.55 5433.41 5949.93 6497.02 7075.57 7686.50 8330.71 9009.14 9722.71 10472.36 11259.03 12083.67 12947.23 13850.67 14794.96 15781.06 16809.96
11.60 36.30 70.34 112.01 160.21 214.09 273.00 336.38 403.78 474.80 549.07 626.29 706.18 788.48 872.96 959.41 1047.65 1137.48 1228.75 1321.30
14.54 45.63 88.68 141.61 203.10 272.14 347.94 429.86 517.34 609.90 707.12 808.62 914.05 1023.12 1135.54 1251.05 1369.42 1490.42 1613.86 1739.53
Flow rates in component channels, cfs (thousands)
20 18
Main flow continues upward with m = 1 side slope
16 14 12
Q
)
tal
(to
Q
Qm
)
tal
(to
Qm
10 8 6 4
Main flow extends vertically upward
2 0
Main flow extends vertically upward Main flow continues upward with m = 1 side slope Ql Ql Qr
10
12
16 14 Depth in compound channel (ft)
18
20
3.76 Repeat the previous problem except the side channels have ½ their widths, i.e., br = 20 ft, and bl = 25 ft. 3.77 Same as Problem 3.75 except solve for the flow rates in the main channel, and the two side channels, and the total depth of flow if the total flow rate is Q = 6000 cfs. The properties of the channels are as given in Problem 3.75. 3.78 Repeat the previous problem except for side channels having ½ the widths, i.e., br = 20 ft and bl = 25 ft. 3.79 A gate controls the flow of water into a long 5 m wide rectangular channel that has a Manning’s n = 0.012, and a bottom slope So = 0.0013, as shown. The gate is a short distance downstream
251
The Momentum Principle Applied to Open Channel Flows
C
c
Y
Short distance
G= = 0 2.0 .6 m
from the reservoir that supplies the flow with a head H = 3 m. If the gate is 2.0 m above the bottom of the channel, and its contraction coefficient is Cc = 0.6, and the entrance loss coefficient is Ke = 0.06 determine: (1) the flow rate Q, (2) the depths upstream and downstream from the hydraulic jump, (3) the force on the gate, (4) the force on the gate if the pressure distribution on it were hydrostatic, and (5) a channel roughness that would cause the gate to be submerged.
3m
Long
K
e
=
0.
06
b = 5 m, n = 0.012, So = 0.0013
3.80 A steep 10 ft wide rectangular channel receives water from a reservoir with a water surface elevation that is 5 ft above the channel’s bottom. This channel is steep, and it is observed that at the end of this channel where it branches into three channels that the depth upstream from the hydraulic jump that occurs here is Yu = 1 ft. The downstream channels consist of: Channel # 2 is rectangular with a bottom width b2 = 4 ft, a roughness coefficient n2 = 0.014, and a bottom slope So2 = 0.0008; Channel # 3 is rectangular with a bottom width b3 = 5 ft and it contains a gate a short distance downstream from the junction; Channel # 4 is circular with a diameter D = 5.0 ft, has a roughness coefficient n4 = 0.012, a bottom slope So4 = 0.0015, and its beginning is Δz14 = 4 ft above the bottom of the other channels at the junction. Channels # 2 and # 4 each branch off at an angle of 30° from Channel # 1, and Channel # 3 is in the same direction as Channel # 1. Determine the flow rates in all of the channels, and the depth downstream from the gate in Channel # 3. Also determine the force the fluid applies against the junction structure (including the gate). (Ignore all local losses.)
Hyd. jump
b1 = 10 ft
ft =4
b2
,
4, S o
.01
0 n 2=
H = 5 ft
Gate
30°
b3 = 5 ft (3)
Mu = Md
,n
4 =0
=
4
ft
Yu = 1 ft
D = 4 5 .0 f t
4
Steep
1.
Ke
g
Lon
(2)
(1)
Δz
=0
08
.00
=0 2
30°
(4)
.01
2, S
o4 =
0.0
015
252
Open Channel Flow: Numerical Methods and Computer Applications
3.81 A trapezoidal channel with b = 3 m, m = 1.4, n = 0.013, and So = 0.0008 has a circular section with a diameter D = 5 m at its entrance where it receives water from a reservoir with a head H = 4 m. The entrance loss coefficient is Ke = 0.1. Determine the flow rate Q into the channel, and if critical flow controls at its circular entrance section, determine how much headloss a hydraulic jump dissipates in the transition between the circular and trapezoidal sections, if this occurs. As a second part to this problem locate the position of the hydraulic jump based on the following assumptions: (1) The specific energy in the transition upstream from the jump is constant and equals Ec at the entrance to the reservoir, (2) The specific energy in the transition downstream from the jump is constant and equals Eo, the specific energy associated with the normal depth Yo in the trapezoidal channel, (3) The shape of the transition is such that the area and the first moment of area Ahc vary linearly from the circular to the trapezoidal shapes, i.e., A = (1 − x)Acir + xAtrap in which x is the fraction of the distance across the transition, and (4) To account for the force on the enlarging control volume for the jump add 4.2 times the depth immediately downstream of the jump, Yd, to the downstream side of the momentum equation, or (Ahc)u + Q2/(gAu) = (Ahc)d + Q2/(gAd) + 4.2Yd. Where will the hydraulic jump occur if the area and first moment of area vary as the square root of the fraction of the distance across the transition, i.e., A = (1 − √x)Acir + √xAtrap. 3.82 In Problem 2.72 you were to develop the delivery diagrams for several bottom slopes for a trapezoidal channel with b = 3 m, m = 1.4, and n = 0.013 with a circular section with D = 5 m at its entrance to a reservoir, with Ke = 0.1 (the channel of the previous problem), for several different bottom slopes So. Take the delivery diagram for So = 0.0008 (with H varying from 0.25 to 5 m), and for each of these H versus Q entries compute: (1) whether a hydraulic jump will occur in the transition from the circular to the trapezoidal sections and if a jump occurs what headloss it must create for uniform flow to exit in the downstream trapezoidal channel, and (2) the slope the downstream trapezoidal channel must have so the flow will remain supercritical through the transition and into the trapezoidal channel. 3.83 The reservoir head H that supplies the 4 channel system shown below varies from H = 0.5 ft to H = 10 ft. Do the following for a series of H′ values: (1) Solve for the flow rates and depths in all four channels (i.e., develop the delivery diagrams). If critical flow occurs at the entrance, i.e., in Channel 1, assume that the specific energy of the three downstream channels are the same. (2) For those H′ values for which critical flow occurs in Channel 1 determine the energy loss between the upstream Channel 1 and the downstream channels. (3) For the H′ values involved in (2) assume that the supercritical flow exists into the beginning of each of the three downstream channels, and compute this depth under the assumption that the specific energy within the supercritical region is constant. (4) Compute the force that the branching structure must apply to the fluid within the region of the hydraulic jump for each of the downstream branched channels.
KL = 0.1
2 .4, S o
8
000
= 0.
=1 , m2 4 ft = All n = 0.013, all bottom elevation are the same b2
K
e
=0
.1
b3 = 4.5 ft, m2 = 1.5, So3 = 0.001 b1 = 10 ft, m1 = 1.7
b4 = 4 ft,
All channels are long except 1 which is very short
m4 = 1.4
, So4 = 0.0
01
The Momentum Principle Applied to Open Channel Flows
253
Problems to Solve Using Program CHANNEL Now that you developed your skills in solving the implicit equations that are associated with Open Channel flow, you will be provided computer program, CHANNEL that will solve these problems for you. In using CHANNEL it is the basic principle from open channel flow, such as UNIFORM Flow, ENERGY, MOMENTUM, CRITICAL, that identifies the type of problem you want solved. The first time you use CHANNEL ask for Help, study the document produced, and then solve the following 10 problems using channel. You should verify enough answers to be confident that the answers are correct. For the following 10 problems write down the answer(s) or have the answer screen from CHANNEL printed by using the Print Screen key on your PC. (The program CHANNEL will either be provided to you by your instructor, or it will be on the diskette that was distributed with this book. Ask your instructor about the details in how to execute program CHANNEL if he provides it for your use.)
1. Find the uniform depth of flow that will exist in a pipe with an 8 ft diameter if its bottom slope is 0.001, its Manning’s roughness coefficient is 0.012, and it contains a flow rate of 100 cfs. 2. A transition from a pipe with a diameter of 2 m to one with a diameter of 2.5 m occurs. If the depth upstream is 1.0 m, and the flow rate is 1.5 m3/s, what is the depth downstream and what is the change in the water surface elevation? The transition loss coefficient equals 0.1. 3. A transition takes a trapezoidal channel with b1 = 10 ft, and a side slope m1 = 1.2 to a rectangular section with b2 = 8 ft, and the bottom rises 0.3 ft. What is the depth downstream and the change in the water surface elevation if the flow rate is Q = 250 cfs, and the upstream depth is Y1 = 5 m. The transition loss coefficient equal 0.1. 4. Water enters a steep trapezoidal channel with b = 10 ft and m = 1.5 from a reservoir with a water surface elevation of 4 ft above the channel bottom. What is the flow rate, and the depth of flow at the entrance (Ke = 0.12). 5. Water enters a mild channel with the size of Problem 3.4. The slope of the channel bottom is So = 0.0005, and Manning’s n = 0.013. (Ke = 0.12). Find the flow rate and the depth of flow. 6. A gate valve exists in an 8 ft diameter pipe that is flowing as an open channel with Q = 90 cfs. Downstream from the valve the depth of flow is Y2 = 2 ft. Determine the force of the water against the valve. 7. Water is entering a long pipe with a bottom slope of So = 0.0006, n = 0.012, a diameter D = 6 ft from a reservoir whose water surface is H = 5 ft above the bottom of the pipe. The entrance loss coefficient Ke = 0.2. Determine the flow rate and the depth in this pipe. Solve the same problem except the pipe is laid on a steep slope. 8. Determine the bottom width that a trapezoidal channel should have if it is to carry a flow rate of Q = 300 cfs, is to have a bottom slope So = 0.0009, a roughness coefficient n = 0.013, and a side slope m = 1.5, and if the water surface of the reservoir that supplies the channel is H = 5 ft above its bottom (Ke = 0.02). Repeat this problem except that the channel has a bottom slope that is steep. 9. Generate the stage discharge relationship that gives the depth and flow rate that a reservoir will supply a trapezoidal channel if b = 10 ft, m = 1.5, So = 0.001, and n = 0.015 for depths of water in the reservoir from H = 2 ft to H = 8 ft in increments of 0.5 ft (Ke = 0.2). 10. Generate a table of values that gives the bottom width of a trapezoidal channel that will carry flow rates ranging from 200 to 1000 cfs in increments of 100 cfs if the following hydraulic properties of the channel are to be maintained: the side slope m = 1.3, the roughness coefficient n = 0.013, and the slope of the channel bottom So = 0.00085. For all of these cases the water is supplied by a reservoir whose water surface is H = 5 ft above the channel bottom, and the entrance loss coefficient is Ke = 0.09.
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Open Channel Flow: Numerical Methods and Computer Applications
Whenever you need to solve a problem that CHANNEL is capable of solving during the rest of this course you should feel free to use CHANNEL, but remember you are still responsible for knowing whether the results are correct. 11. Use Chezy’s equation to determine the flow rate that a trapezoidal channel with a bottom width b = 10 ft, a side slope m = 1.5, a bottom slope So = 0.001, and a equivalent sand roughness of e = 0.012 ft, if it is at a depth of Y = 5 ft under uniform flow conditions. (Also what is Chezy’s C and the Reynolds number of the flow?) 12. Determine the depth of flow needed to convey a flow rate Q = 400 cfs in a trapezoidal channel with b = 8 ft, m = 1.3, So = 0.0005 if its equivalent wall roughness is e = 0.01 ft. 13. Determine the depth Y in a circular channel with a diameter D = 12 ft that will occur when 300 cfs is flowing. The channel has a bottom slope So = 0.0006, and an equivalent wall roughness e = 0.014 ft. (Also what is Chezy’s C and the Reynolds number of the flow?) 14. What size circular channel with e = 0.009 ft is needed to carry a flow rate of Q = 20 m3/s if the bottom slope So = 0.00085, and the equivalent wall roughness is e = 0.011 ft. 15. The cross section of a natural channel is defined by the transect data in the table below. The wall roughness is defined by Manning’s n = 0.015, and the bottom slope is So = 0.00085. If the depth of flow is Y = 5 ft, what is the flow rate under uniform conditions? Pt
x (ft)
y (ft)
1 2 3 4 5 6 7 8 9
0 5 8 10 13 17 22 25 30
8.0 6.5 5.0 3.2 2.8 3.1 5.3 7.0 8.0
16. For the natural channel defined in the previous problem determine the depth of water under uniform flow conditions if the flow rate is Q = 350 cfs. Determine this depth using both linear and quadratic interpolation of the crosssectional data. 17. Using the crosssectional data of the previous problem determine the depth if rather than using Manning’s equation, it is modified so that the exponent e1 of the hydraulic radius is 0.715, and the exponent e2 of the slope of the channel bottom is 0.35. For this problem use a bottom slope So = 0.0018 and a Manning’s n = 0.045. (Use Cu = 1.486.) 18. Use e1, e2, So and n as in the previous problem, the difference is that the channel has a trapezoidal cross section with b = 10 ft, and m = 1. Determine the flow rate Q if the depth of flow is Y = 5 ft. 19. Using the geometry solving capability of channel to solve the variables left blank in the table below if the other variables have the values given for a trapezoidal channel. Depth (ft) 5 4
Bottom Width (ft)
Side Slope
10 8
1.2 1.5 1.0
Perimeter (ft)
Top Width (ft)
Area (ft2) 70 50
1st Mom. of A (ft3)
4 Nonuniform Flows 4.1 Types of Nonuniform Flows In Chapter 1, it was noted that nonuniform flows in open channels are subdivided into gradually varied and rapidly varied on the basis of whether normal accelerations of fluid moving along the curved streamlines can be ignored or not. Situations, such as the flow directly underneath a gate, and over the crest of a weir or a dam are examples of rapidly varied flows. For such situations, the stream lines curve rapidly and normal accelerations have a significant influence on the flow pattern, i.e., the flow needs to be handled as a two or threedimensional problem. When the water depth and the velocity change over long distances as water impounds behind a gate, or the depth gradually decreases in a mild channel upstream from a break in grade to a steeper channel, then normal accelerations are insignificant, and the flow can be considered onedimensional. In this chapter, only gradually varied flows will be dealt with. Furthermore, only steadystate flows are considered in this chapter. If the flow rate Q increases or decreases in the direction of the channel, then this special type of gradually varied flow is referred to as a spatially varied flow. The adjective “spatially” denotes that the flow rate changes in space, e.g., in the direction of the x coordinate. Alternative terminology calls for situations in which the flow rate changes along the channel, i.e., lateral inflow or lateral outflow. An example of lateral inflow is the gutter flow along a roadway during a rain storm. The runoff from the road crest feeds the gutter flow so that it increases in the flow direction until the gutter passes over a storm drain grate at which time a lateral outflow occurs from the gutter. A side weir, or overflow spillway along a canal side is a lateral outflow. In the previous three chapters, the equations used to describe the problems were algebraic. When dealing with gradually varied flows, the governing equation is an ordinary differential equation (an ODE). Since this ODE can be solved in the closed form for only a limited number of simplified cases, much of the material in this chapter deals with numerical methods for obtaining approximate solutions to this governing equation. Rather than dealing with the numerical methods per se, the emphasis will be on the use of standard algorithms implemented into computer programs. The widespread use of computers in recent years has made some of the hand techniques that engineers have used in the past to solve gradually varied flow problems obsolete. The general gradually varied flow equations will be developed first. The general case will include spatially varied flows in nonprismatic channels. After the general equations have been developed, solutions will be obtained for the simpler cases for gradually varied flows with no lateral inflow or outflow in a prismatic channel. For these simpler cases, gradually varied flows are classified, and this classification system will be described. After dealing with the simpler cases, more complex situations will be handled.
4.2 Ordinary Differential Equation for Gradually Varied Flow In developing differential equations for a gradually varied flow, the case of lateral outflow and lateral inflow will be different. Lateral outflow will be dealt with first.
255
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Open Channel Flow: Numerical Methods and Computer Applications
4.2.1 Bulk Lateral Outflow In dealing with a gradually varied flow in which there may be a lateral outflow from the channel, it is assumed that all the liquid at a given position x in the channel has the same energy per unit weight. Therefore, each unit weight of liquid that leaves the main channel flow carries with it the same energy as each unit weight of liquid that remains in the channel. The energy principle is the appropriate principle to utilize in describing this situation. From any selected horizontal data, the total head H will be constant at any position along the channel, or H =z+Y+
Q2 2gA 2
(4.1)
For a nonuniform flow, this total head will vary from position to position, and this variation is defined as the derivative of H with respect to x, or
dH dz dY Q dQ Q 2 dA = + + − dx dx dx gA 2 dx gA 3 dx
If the channel is nonprismatic, then the cross section A is both a function of the position x and the depth Y at this position, i.e., A = f(x, Y), and therefore in general, by the chain rule of calculus, dA ∂A ∂A dY = + dx ∂x Y ∂Y Y dx
in which the subscripts emphasize which variables are being held constant when the derivative is taken. The partial derivative ∂A/∂Yx equals the top width T. Should the channel be prismatic, e.g., the crosssectional area is only a function of the depth Y, then ∂A/∂x = 0. The term dQ/dx is the lateral outflow per unit length along the channel, and will have a negative magnitude. It will be defined as q*o = −dQ/dx. The derivative dH/dx is the negative slope of the energy line, and this slope will be denoted by Sf, or Sf = −dH/dx. Also, the derivative dz/dx is the negative slope of the channel bottom, or So = −dz/dx. Substitution of these results into the above equation produces the following:
−Sf = −So +
dY Q 2 T Q 2 ∂A Q * 1− − − qo 3 3 dx gA gA ∂x Y gA 2
(4.2)
The quantity of interest is the change in depth with respect to position, so Equation 4.2 will be solved for dY/dx, and Fr2 (Froude number squared) will be substituted in place of Q2T/(gA3) giving
dY = dx
So − Sf +
Q 2 ∂A Qq*o + 3 gA ∂x Y gA 2 1 − Fr2
(4.3)
Equation 4.3 is the general equation that describes a gradually varied flow in a channel that may have bulk lateral outflow q*o . If the channel is prismatic, and no lateral outflow occurs, then this equation reduces to
dY So − Sf = dx 1 − Fr2
(4.4)
257
Nonuniform Flows
Before examining the lateral inflow case, it is well to examine the term ∂A/∂xY for a nonprismatic channel. An example of a nonprismatic channel is a trapezoidal channel going through a transition from b1 and m1 at section 1 to b2 and m2 at section 2. For this transitional part of the trapezoidal channel, the area is defined by A = b(x)Y + m(x)Y 2 = (b(x) + m(x)Y)Y
taking the partial derivative with respect to x gives the following: ∂A db dm =Y + Y2 ∂x dx dx
The full derivatives of b and m are used above because these variables depend only on x, and therefore full and partial derivatives are identical. In other words, for a trapezoidal channel, the change in area with respect to x equals the depth times the change in the bottom width with respect to x plus the depth squared times the change in the side slope with respect to x. For a circular section, the diameter may change through a transition in which case, the area A = D2(β − cos β sin β)/4 is differentiated with respect to x to give
∂A 1 ∂D 1 ∂β = D (β − sin β cos β) + D2 (1 − cos2 β + sin 2 β) ∂x 2 ∂x ∂x 4
in which the partial derivative of β with respect to x is obtained by differentiation of cos β = 1 − 2Y/D to give ∂β 2Y dD =− 2 ∂x D sin β dx
or the substitution of this result into the above equation gives the following for ∂A/∂x:
∂A 1 dD = D(β − cos β sin β) − Y sin β ∂x 2 dx
4.2.2 Lateral Inflow In the case of a lateral inflow, it is necessary to use the momentum principle rather than the energy principle because the amount of energy dissipated as the incoming flow impacts with the main channel flow is unknown. The sketch below shows a control volume of a small length of channel Δx, which can be reduced to a differential length dx, with the forces shown on it. q* Uq
dx γAhcl
γhcdA
γASodx τo = γRHSf
γAhc2
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Open Channel Flow: Numerical Methods and Computer Applications
Since these forces were not included in the development of the momentum function M, and the momentum function was obtained by dividing the summation of force equation by the specific weight γ of the fluid, a summation of forces in the x direction produces the following:
γ
dM dx = γASodx − γ R hSf Pdx + ρq* U q dx − γ h cdA dx
in which Uq is the component of velocity of the inflowing liquid in the direction of the main channel flow V is the average velocity of the main channel flow Differentiation of the momentum function M = Ahc + Q2/(gA) gives dM d 2Q dQ Q 2 dA = (Ah c ) + − dx dx gA dx gA 2 dx
in which dA/dx = T(dY/dx) + ∂A/∂xY and by Leibniz’s rule d(Ahc)/dx = A(dY/dx). Upon substitution of these and q* = dQ/dx into the above equation, and solving dY/dx gives the following spatially varied flow equation for a lateral inflow:
dY = dx
So − Sf +
Q 2 ∂A 2Qq* q* U q h c ∂A − + − gA 3 ∂x gA 2 gA A ∂x Y 2 1 − Fr
(4.5)
4.2.3 Generalization of Gradually Varied Flow Equations There is considerable similarity between Equations 4.3 and 4.5. The sign in front of the term containing Qq* is different and a 2 now multiplies this term, but q* = −qo*. Equation 4.5 contains a couple of additional terms in the numerator; one that accounts for the momentum flux for the incoming liquid q*Uq/(gA), and the other, (hc/A)(∂A/∂x) that accounts for the hydrostatic force of the fluid against the expanding or the contracting cross section for a nonprismatic channel. In addition, there is one additional case that has not been considered, that does occur in channel flow, that of seepage outflow or inflow. If some of the fluid from a channel is lost from seepage into (or out from) the soil that forms the canal, then this fluid will leave from the boundary layer where it has a zero velocity. Therefore, fluid lost by seepage will have an energy per unit weight less by the velocity head than the fluid that remains in the channel. The following equation will accommodate all these cases:
dY = dx
So − Sf +
Q 2 ∂A Qq* − − Fq gA 3 ∂x gA 2 1 − Fr2
in which Fq = 0 for bulk lateral outflow Vq* Qq* Fq = = for seepage flow 2gA 2gA 2 (V − U q )q* h c ∂A for bulk lateral inflow Fq = + gA A ∂x Y
(4.6)
259
Nonuniform Flows
in which Uq is the velocity component of the inflow in the direction of the main channel flow, and q* represents the lateral inflow or outflow per unit length of main channel with q* negative for lateral outflow, and positive for lateral inflow. For a prismatic channel, the terms containing ∂A/∂xY are zero, and Equation 4.6 reduces to
dY = dx
Qq* − Fq gA 2 1 − Fr2
So − Sf −
(4.7)
If there is no lateral inflow or outflow, then note that Equation 4.7 reduces to Equation 4.4. It is Equation 4.4 that we will begin with, and after gaining experience in solving it, more general problems will be discussed in which the channel may be nonprismatic, and lateral inflow or outflow may also occur.
4.3 Gradually Varied Flow in Prismatic Channels without Lateral Inflow or Outflow 4.3.1 Classification of Gradually Varied Profiles For the simplest case of gradually varied flow in a prismatic channel in which the flow rate does not change with the position along the channel, a classification of the flow profiles has been adopted and is understood by hydraulic engineers. This classification consists of an upper case letter that denotes whether the slope of the channel bottom, under uniform flow, will produce subcritical, critical, or supercritical flow, and a subscript to this letter that denotes the relationship of the actual depth to the two reference depths, the normal depth Yo, and the critical depth Yc. The letter designation is as follows: M is used if under uniform flow, the slope of the channel is such that the flow will be subcritical. The M stands for mild channel. S is used if under uniform flow, the slope of the channel is such that the flow will be supercritical. The S stands for steep channel. C is used if under uniform flow, the slope of the channel will produce a critical flow. The C stands for critical. H is used if the slope of the channel bottom is zero, i.e., the channel is horizontal. A is used if the slope of the channel bottom is negative, i.e., the bottom of the channel increases in elevation in the direction of the flow. The A stands for adverse. The subscript to this letter will be
1. If the actual depth is above both the reference depths 2. If the actual depth is between the two reference depths 3. If the actual depth is below both the reference depths
Horizontal and adverse channels cannot have normal depths in them, i.e., Yo approaches infinity. There are only two gradually varied profiles possible in these channel, H2 and H3, and A2 and A3, respectively. Since, in a critical channel, Yo and Yc coincide, the only two possible profiles in this very special channel are C1 and C3. Table 4.1 shows the possible gradually varied profiles, which will be referred to as the GVF profiles,hereinafter. For example, if a channel is mild, i.e., for the given flow rate this channel would contain a subcritical flow if under uniform conditions, and a dam backs the water up so the actual depth is above the normal (or uniform) depth, then this GVF profile is referred to as an M1 profile,
H3
dY − = =+ dx −
H3
M3
H2
S1
Yc
Yc
Yo
Yc
Yo
dX
dY
=0
dX
dY
= Sc
dX
dY
=0
= Sc
dY = Sc dX
dX
dY
dX
dY
=∞
H2
S2
S1
H3
S3
M3
M2
M1
= So
= So
–∞
dY = ∞ dX
dY dX =
dY =0 dX
dX
dY
dY = ∞ dX
dY dX = –∞
dX
dY
Magnitude of dY/dx at Ends of Profile
∞
dY − = =− dx +
S3
S3
S2
M2
M1
Sketch of Profiles and Reference Lines
=– X d
H2
dY − = =+ dx −
S2
dY + = =+ dx +
dY − = =+ dx −
dY − = =− dx +
dY + = =+ dx +
Sign Associated with Equation 4.4
dY + = =− dx −
S1
M3
M2
M1
Profile Designation
dY
Horizontal
Steep
Mild
Slope of Channel B
Table 4.1 Gradually Varied Profiles in Prismatic Channels
260 Open Channel Flow: Numerical Methods and Computer Applications
Adverse (or negative slope)
A3
A2
C3
C1
dY − = =+ dx −
dY − = =− dx +
dY − = =+ dx −
dY + = =+ dx + C3
A3
C1
A2
Yc
Yo = Y c
dX
dY
dX
= Sc
= So
= Sc
dY
dX
dY
dY = Sc dX C3
A3
A2
C1
dY
–∞
= Sc
dY dX =
dX
dX
dY = So
dY dX = ∞
Critical
Nonuniform Flows 261
262
Open Channel Flow: Numerical Methods and Computer Applications
or back water curve. If on the other hand, a gate in this mild channel causes a supercritical flow downstream from it, then the GVF profile is denoted by M3. It will be advantageous to examine the shape of the GVF profiles by examining the signs associated with both the numerator and the denominator of Equation 4.4, which is repeated below:
dY So − Sf = dx 1 − Fr2
(4.4)
The numerator consists of the difference between the slope of the channel bottom and the slope of the energy line, So − Sf, and will be positive if the actual depth Y, is above the normal depth Yo. The reason is that for greater than normal depth, the velocity will be less than the uniform velocity, and consequently less frictional head loss will occur, and therefore Sf is smaller then So. On the other hand, the numerator will be negative whenever the actual depth Y is less than Yo, based on this same reasoning. The denominator will be positive if the actual depth Y is greater than the critical depth Yc, and will be negative whenever the actual depth is less than the critical depth. Reasons for this is that, since the denominator consists of 1 − Fr2, it will have a negative sign if the Froude number is greater than 1, and positive if the Froude number is less than 1, and the Froude number is greater than unity for supercritical flows, and less than unity for subcritical flows. If the numerator and the denominator have the same sign, dY/dx is positive. A positive value for dY/dx means that the depth will increase in the downstream direction. On the other hand, if the numerator and the denominator of the last part of Equation 4.4 are of opposite signs, then dY/dx is negative. A negative value for dY/dx indicates that the depth decreases in the downstream direction. Table 4.1 indicates the sign of dY/dx in its third column for the various GVF profiles. As the following GVF profiles approach the critical depth, the slopes of the water surfaces approach infinity: M2, M3, H2, H3, S1, S2, A2, and A3. For these profiles, Equation 4.4 indicates that dY/dx becomes extremely large as the critical depth is approached because the denominator of Equation 4.4, 1 − Fr2 approaches zero, and the numerator remains finite. In an actual channel flow, the change in depth becomes more rapid as the critical depth is approached, but dY/dx does not approach infinity. The results from Equation 4.4 violate the assumption that was used in its development, namely that the flow be gradually varied and normal acceleration components can be ignored. Therefore, the ODEs for gradually varied flows, Equations 4.2 through 4.7 do not apply near the critical depth. In solving the GVF profile, it is therefore necessary to start with a depth slightly below the critical depth, or slightly above the critical depth. Another problem exists for the ends of the GVF profiles as they approach the normal depth Yo, for here the numerator So − Sf approaches zero, and therefore dY/dx approaches zero, indicating that no change occurs in the depth with a given change in x. No such change is equivalent to saying that the GVF profile has an infinite length in approaching the uniform depth. From a practical viewpoint, however, gradually the varied flow becomes a uniform flow when the depth is within about 1% of the normal depth. To assist in avoiding numerical difficulties in solving Equation 4.4, it is well to examine, and thereafter be aware of, the extreme or the limiting values that dY/dx can have at the ends of the GVF profiles. These limiting values are shown in the last column of Table 4.1. In other words, the range of changes in depth is with respect to the position dY/dx, for each profile is restricted to values between its extreme ends. One might question the need to be concerned with the magnitude of a derivative like dY/dx, when it is the depth that is ultimately sought. The answer is that if we are not aware of when dY/dx becomes zero, or infinite, we may expect the impossible from the numerical solution, and by being aware, we can avoid problems. Whenever the GVF profile approaches the critical depth, the denominator of Equation 4.4 becomes zero, since Fr2 approaches 1 and depending upon whether Sf > So or Sf < So, the magnitude of dY/dx will approach negative, or positive infinity, respectively. GVF profiles whose ending derivative values are negative infinity are M2, H2, and A2. Those with dY/dx equal to positive infinity at their end are M3, H3, and A3. The S1 profile has a positive infinite dY/dx value at its upstream end,
263
Nonuniform Flows
and the S2 a negative infinite value at its upstream end. For all these profiles, where dY/dx becomes infinite, the curvature of the streamlines becomes too large for the onedimensional hydraulics to be valid as critical depth is approached, and therefore Equation 4.4 does not describe the actual GVF profile. When solving GVF profiles, judgement must be applied in starting or ending the numerical computations about 5% above or below the critical depth. For GVF profiles that approach the uniform depth, the numerator of Equation 4.4 approaches zero, since Sf approaches So. Such profiles approach normal depth asymptotically, and according to the mathematics, they are infinitely long. In practice, when the depth is within 1% or 2% of the normal depth, the GVF profile ends and the flow is uniform thereafter. The upstream end of the M1 and the M2 profiles and the downstream ends of the S2 and the S3 profiles are in this category where dY/dx approaches 0. The downstream end of the M1 and the S1 GVF profiles have dY/dx approach the slope of the channel bottom, because as the depth becomes large Sf approaches 0 and Fr2 approaches 0, and thus dY/dx approaches So/1. The same idea applies to the upstream end of the H2 and the A2 GVF profiles. In the case of an H2 GVF profile, So approaches 0, so its upstream dY/dx becomes 0, and for the adverse slope, So is negative, so dY/dx = −So. To determine the values that dY/dx approach at the upstream limit of the M3, the S3, the H3, the A3, and the C3 GVF profiles, let us introduce the critical slope. By definition, the critical slope will produce a uniform flow at the critical depth. Therefore, the critical slope is the bottom slope computed from the uniform flow equation (either Manning’s or Chezy’s equation) when the critical flow equation is satisfied, or 2
nQP 2 / 3 n 2Q 2 P 4 / 3 Sc = = 5/3 Cu A10 / 3 Cu A
Q2T gA 3 = 1, or Q 2 = 3 gA T
and
Cu is used for 1.486 or 1 here to prevent confusion with Chezy’s C. Substituting Q from the critical flow equation into Manning’s equation gives
Sc =
n 2gP 4 / 3 S gP S = f2 or from Chezy ’s equation Sc = = f2 1/ 3 2 u C TA Fr TC Fr
The result of Sf/Fr2 is obtained by regrouping the terms as shown below when using the Manning equation, and thereafter by multiplying the numerator and the denominator by Q2/A10/3.
(n P = (TA 2
Sc
43 13
)( g )( Q
) = n Q P (C A ) = S F ) (Q T gA )
C2u Q 2 A10 3 2
A10 3
2
2
43 2
2 u
10 3
3
f 2 r
or using Chezy’s equation,
Sc =
(P /C2 )(Q 2/A3 ) Q 2 P / (C2 A3 ) S = 2 = f2 2 3 3 (T /g)(Q /A ) Q T / (gA ) Fr
Thus, we see that the critical slope is also obtained by the equation that gives the slope of the energy line divided by the Froude number and by setting the Froude number at unity. Consider what happens as the depth becomes very small. The slope of the channel bottom becomes very small in comparison to Sf, and therefore the numerator of Equation 4.4 approaches −Sf. Likewise, 1 is small in comparison to Fr2, and so the denominator of Equation 4.4 approaches −Fr2 and dY/dx approaches Sf/Fr2 = Sc. In other words, profiles whose upstream depths can approach zero (M3, S3, H3, A3, and C3), have these end values of dY/dx approach the critical slope Sc. The remaining question is what values for dY/dx exist at the ends of the C1 and the C3 profiles as these approach the coincident
264
Open Channel Flow: Numerical Methods and Computer Applications
normal and the critical depths. Here, dY/dx takes on the indeterminate form 0/0, which requires that L’Hôpital’s rule of taking the derivative of the numerator and the denominator be followed. An alternative approach that eliminates the mathematics associated with taking these derivatives, etc., is to substitute Sf = ScFr2 in Equation 4.4 and to note So = Sc for a critical channel, as shown below:
dY So − Sc Fr2 Sc (1 − Fr2 ) = = = Sc dx 1 − Fr2 1 − Fr2
by dividing out (1 − Fr2 )
Thus, a C3 GVF profile has the same value of dY/dx at its two extreme ends. To avoid numerical problems associated with getting too close to the critical depth and the normal depth, we will define the GVF profiles as the water surface to within 1% of the normal depth, and within 5%–10% of the critical depth.
4.3.2 Sketching GVF Profiles in Prismatic Channels The first step in solving gradually varied flow problems in open channels is to sketch in and identify the type of profile that will occur. It is not possible to define an exact stepbystep procedure by which this can be done, but the following are usually important considerations or properties of the flow to be identified: (As you read this it will be helpful to look at the sketches in Figure 4.1.)
1. Identification of the “control points” of the flow. A gate in the channel controls both the flow upstream from it, as well as downstream from it. A gate is therefore, always a good point from which to sketch in GVF profiles, both upstream and downstream from it. The upstream flow will be subcritical, and its depth will be determined by the gate setting. If the channel has a bottom slope that will support a uniform supercritical depth for this flow rate, then the upstream GVF profile will be an S1. If the uniform depth is subcritical, this upstream GVF profile will be an M1. In a mild channel, the GVF profile downstream from the gates will be of the M3 type, and if the channel is steep, then the downstream GVF profile will be of the S3 type. An exception occurs if the flow downstream from the gate is submerged. A break in grade in an upstream mild channel to a steep slope is also a control point. At the break in grade, the depth will be critical with the flow downstream therefrom supercritical unless the water is “backedup” by another control further downstream. A reservoir at the head of a steep channel is a control point, and the depth here will be critical. For a mild channel, this is not the case. 2. A hydraulic jump will terminate an M3 GVF profile in a mild channel, and may be one end of an M1 or an M2 GVF profile on its subcritical side, or if the channel is long, the subcritical side of a hydraulic jump may be at normal depth. If the hydraulic jump occurs in a steep channel, then there will either be no GVF profile upstream from it, or the GVF profile will be an S2 (or S3) profile on its way to approaching the normal depth in this steep channel. Hydraulic jumps are therefore, other key points used in deciding how to sketch in the GVF profile. Often, however, the position of a hydraulic jump must be determined as part of the solution to the GVF problem. If it is not known whether the hydraulic jump exists upstream or downstream from a break in grade, then both possibilities should be shown on the sketch, so that later computations can determine which of these is correct. 3. Downstream reservoirs will create an M2 or an M1 profile depending upon whether the elevation of the water surface in the reservoir is below or above the normal depth (for this flow rate) in the channel, respectively. When a steep channel terminates in a downstream reservoir, there are two possible situations that can occur: (a) If the reservoir elevation level is above the critical depth, then an S1 GVF profile will exist in the upstream channel, and a hydraulic jump will exist upstream from it. This hydraulic jump will be caused by the high water level in the reservoir. (b) If the reservoir water surface is below the critical depth in
265
Nonuniform Flows M1 Yo Yc
(a)
H2 Jump
Mild slope Y > Yo > Yc
(b)
Horizontal
(f )
Yo
Mild slope Y > Yo > Yc
M3
(c)
Yo
d
Steep slope Y>Y c>Y o
(k)
Yo >Y
c
Mild slope Y = Yo > Yc
M1
Yo Yc
Steep slope Y>
Yo Yc
Jump
(j)
Jump
(g)
Yc
Mil
S1
Yc Yo
S1
Yc Yo
M1 Jump
Yo
H3
M1 Yc
Yc
Yo
Mild sl o Y > Y pe o>Y c
Milder slope Y > Yo > Yc
Mild slope
Yo
Yo
Yc
(d)
Yo = Y
Yc
(e)
Critic
(h)
Channel wall contraction on a mild slope
Yc Yo
A2
C1
No jump
M1
A3
c
al slop
S1
e Y>Y
o>Y c
(l)
Yc
ope
sl Adverse
S1
Jump
Steep slope Y>
Yo > Y
Yo
S3
c
(i)
Yc
Steep slope
Yo
Figure 4.1 Gradually varied profiles caused by a single control. (a) The weir causes flow above the normal depth in a mild channel and this is an M1 GVF. (b) The small dam causes the same M1 GVF upstream as the weir. (c) The flow upstream from a gate is always subcritical and if flow emerges as “free flow” from the gate it is supercritical downstream. Thus upstream an M1 GVF occurs, and an M3 GVF since the channel is mild. (d) If the contraction is severe enough, it will cause critical flow at this reduced section with an M1 GVF upstream. If the contraction is small and the channel enlarges again downstream, or is steeper, then the upstream flow may remain at normal depth. (e) In a steep channel, a dam will cause a jump to take the flow from supercritical to subcritical conditions, and an S1 GVF will exist downstream there from. Upstream of the jump the depth will be normal. Thus the S1 GVF will start at the depth conjugate to Y0 and end at a depth with a specific energy that is equal to the critical specific energy at the crest of the dam plus the height of the dam, or E = Ec + Δz. (f) The depth in the mild downstream channel will be normal so an H2 GVF will occur downstream from the hydraulic jump and an H3 GVF upstream there from. If the horizontal channel is short, and the water has a very large velocity as it passes under the gate the hydraulic jump may be pushed into the mild channel in which case the H3 GVF will exist over the entire length of the horizontal channel and continue as an M3 GVF in the downstream channel until the jump occurs. (g) A gate in a steep channel will force the flow to be subcritical upstream from it so an S1 GVF will occur that has its beginning at the depth conjugate to the normal depth in the steep channel. Downstream an S3 GVF will occur. (h) In a critical channel, a gate will cause an C1 GVF upstream that starts at critical depth so no jump occurs, and an C3 GVF will occur downstream. (i) No jump occurs downstream from a gate in a steep channel unless something downstream there from forces the flow to be subcritical. (j) A steep channel abruptly changes to a mild channel will either cause an S1 GVF upstream from the break in grade as shown if the momentum function associated with the downstream normal depth is greater than that for the upstream normal flow, Mo1 < Mo2. If Mo1 > Mo2 then the jump will occur downstream with an M3 GVF starting at the break in grade and ending with the depth conjugate to Yo2. (k) A reduction in the bottom slope in a mild channel will cause an M1 GVF upstream from the break in grade. (l) Adverse or horizontal channels have profiles much like those in mild channels. The difference is they are designated with an A or H letter.
266
Open Channel Flow: Numerical Methods and Computer Applications
this steep channel, then the water depth in the channel will be unaffected by the reservoir, i.e., no GVF profile will exist unless it is in the lower portion of the varied flow from an upstream cause. If the reservoir water surface level is above normal but below critical, then a surge just off the end of the channel will form in the reservoir, but this surge (or standing wave) will be unable to move into the channel because the channel velocity of the channel flow is larger than the speed of its movement. 4. A break in grade, e.g., a change in the bottom slope, will always cause a length of gradually varied flow to occur and often GVF profiles will exist upstream and downstream therefrom. If the change in grade is from a mild to a steep channel, then an M2 profile occurs upstream, and an S2 profile occurs downstream with a critical depth at the break in grade. If the break in grade is from a mild to a milder slope, then an M1 profile will occur upstream, and if the downstream channel is long enough, then the depth in the downstream milder channel will be uniform. Likewise, a change from a milder to a steeper, but still mild slope will cause a portion of an M2 profile in the upstream channel. If both the channel upstream and downstream from a break in grade are steep, then the two cases are: (a) to a steeper slope an S2 GVF profile will exist in the downstream channel, and (b) to a less steep slope, an S3 will occur in the downstream channel. However, no GVF profiles exist upstream from the grade change.
The various possibilities described above are illustrated in the sketches shown in Figure 4.1, which depict situations in which a GVF profile(s) occurs as a result of a single control. It is assumed in these situations that the channel is very long both upstream and downstream from the control. Figure 4.2 depicts situations where this is not the case and the effects of several controls can cause a different type profile to begin where another type ends. Example Problem 4.1 Sketch in the GVF profiles for the four situations shown. You should note that two reference lines are given, one for the normal depth and the other for the critical depth. In a mild channel, the normal depth line is above the critical depth line and in a steep channel the reverse is true, the normal depth is below the critical depth line. Therefore, if the type of channel is not given and reference lines are, you can determine whether the channel is mild or steep, etc. (1) A break in grade occurs from a mild to a steep channel, but the channel’s size does not change. Solution to (1): Since the flow will be supercritical downstream from the break in grade and the subcritical upstream, the flow must pass through the critical depth at the break in grade. Thus upstream, the depth will be between the normal and critical depths, or Yo > Y > Yc and dYdx will be negative, because of a negative numerator and the GVF will be an M2, that passes through the critical depth at the break in grade and asymptotically approaches the normal depth at its other end. Downstream of the break in grade, an S2 GVF will occur. The channel is steep and the depth is still between the two reference depths. dY/dx is also negative here since the denominator is negative. The beginning of the S2 is at the critical depth and at its other end, it asymptotically approaches the normal depth in the downstream steep channel. + –
M
2
Yo1
+
Yc Mild
Yo > Yc
S2
Steep
Yo2
Yo < Yc
267
Nonuniform Flows Ra p var idly ied Yc flo w
S1
H yd ju rau m lic p
S2
Yo
Steep
Rapidly varied flow
M2
Mild
Mild
M1
Dam M1
M3
Hy d ju raul m ic p
M2
Yc
S1
Hy d ju raul m ic p
Mild
S3
S1 Yo
Steep
Yc
Yo
M1
M3
Rapidly varied flow
S1
Hydraulic jump
Mild
Yo
S2
S1
Steep
Yo
Ra va pidly Yc r flo ied w
lic au dr p y H jum
S2
M1 Yo
Yo
Rapi dl varie y d flow S2
M2
Mild
A2
S1
M2
Steep
M2 S2
se Adver ulic dra Hy mp ju
Mild
Yc
Mild S1 S1
H2 H2
Steep Horizontal
Figure 4.2 Examples of GVF profiles.
(2) A mild channel has its bottom slope suddenly reduced, but the channel size remains the same. Solution to (2): Assuming that the downstream channel is very long, fluid friction will control and a uniform flow will exist in it. Therefore, at the break in grade, the depth will equal the normal depth for the downstream channel, or Y = Yo2. Since this depth is above the normal depth in the upstream channel, and this is also a mild channel, an M1 GVF will occur as shown below.
268
Open Channel Flow: Numerical Methods and Computer Applications M1
+ –
Yo2
+
Yo1
Yc
Yc
Mild Yo > Yc
Milder
Yo > Yc
(3) A steep channel suddenly changes to a mild channel but the channel size remains constant. Both the upstream and the downstream channels are very long. Solution to (3): Upstream from the break in grade, the flow will be supercritical, and therefore cannot be controlled by downstream conditions. However, the flow will be controlled by downstream conditions, and in this case, by fluid friction and will be at normal depth. Thus, the flow must change from the supercritical to the subcritical, and this can occur only by means of an abrupt hydraulic jump. The jump may either occur upstream or downstream from the break in grade depending respectively upon whether the momentum function associated with the downstream normal depth is greater or less than that for the upstream normal depth. A jump will occur upstream if Mo1 < Mo2 and downstream if Mo1 > Mo2. If the jump occurs upstream, then an S1 GVF will start at a depth that is conjugate to Yo1 and end at the break in grade with Yo2. If the jump occurs downstream, then an M3 GVF will occur that starts at Yo1 and ends at a depth conjugate to Yo2.
Jump
Cas Yo1 Steep
S1
e2
Case 1
Y2
Yo1 <
Jump
Yo2
M3
Yc
Mild
Yo2
Yo > Yc
(4) A steep channel suddenly has its bottom slope increased but its size remains constant. Solution to (4): Being supercritical, the flow in the upstream channel cannot gradually react to the steeper downstream slope, so it will remain at normal depth to the break in grade. At this position, the normal depth will be below the actual depth, and an S2 GVF will exist that gradually becomes less approaching the downstream normal depth asymptotically.
Yc
Yo1
Steep
Yo2 < Yc
S2
Yo
2
Steep er
Yc Yo1 <
Yc
4.3.3 Alternative Forms of the ODE That Describe GVF Profiles In Equation 4.4, the depth Y is considered the dependent variable (or unknown) and x is considered the independent variable. Alternative forms of ODEs for describing GVF profiles can be developed. The specific energy E may be introduced as the dependent variable in place of the depth Y. To accomplish this change of dependent variables take the differential of the specific energy equation, E = Y + Q2/(2gA2), to give
269
Nonuniform Flows
dE = dY +
Q2T dY = dY + Fr2dY gA 3
Substituting this result into Equation 4.4 gives the following: dE = So − Sf dx
(4.8)
In Equation 4.8, the specific energy E is considered the dependent variable and x the independent variable. Since both E and Sf depend on the depth, Y must be used as an auxiliary variable. While it may not be apparent, Equation 4.8 allows for a prismatic as well as a nonprismatic channel. That is, Equation 4.8 can be used to solve the same problems as Equation 4.6 can (with the term ∂A/∂xY included) provided q* is zero. This can be shown by noting that dA = ∂A/∂xY + T dY for a nonprismatic channel. Another alternative form of Equation 4.4, that still considers Y the dependent variable, can be obtained by solving Sf from Manning’s equation, (or from Chezy’s equation, as another possibility) giving Sf = n2Q2P4/3/(Cu2A10/3). Multiplying and dividing this by gT allows the Froude number squared to be isolated as follows:
Sf =
gn 2 P 4 / 3 Q 2 T gn 2 P 4 / 3 = Fr2 3 10 / 3 gA Cu TA Cu TA10 / 3
A critical slope Sc is defined by letting the Froude number equal unity or
Sc =
gn 2 P 4 / 3 C2u TA10 / 3
so Sf = ScFr2 and the gradually varied ODE becomes
dY So − Sc Fr2 = dx 1 − Fr2
(4.9)
In solving Equation 4.9 versus solving Equation 4.4, the need to repeatedly compute Sf may appear to have been replaced by solving Sc once. However, this is not true because Sc is not constant, but varies with the depth as can be seen in the equation above Equation 4.9. Thus, the computation of Sf is replaced with the computation for Sc. A fundamental theorem of calculus is that when dealing with full derivatives of continuous variables, dx/dy = 1/(dy/dx). Therefore, to interchange the role of the variables, all that is needed is to take the reciprocal of both sides of the expressions that define an ODE. Thus, if the desire is to have an ODE that considers x the dependent variable (i.e. the unknown), and the specific energy the independent variable, then the reciprocal of both sides of Equation 4.8 can be taken to give
dx 1 = dE So − Sf
(4.10a)
If an ODE is desired that considers Y the independent variable, and x the dependent variable, then the reciprocal of both sides of Equation 4.4 can be taken to give
dx 1 − Fr2 = dY So − Sf
(4.10b)
270
Open Channel Flow: Numerical Methods and Computer Applications
It is well to examine the characteristics of these alternative forms of ODEs that describe GVF profiles in a prismatic channel with no lateral inflow or lateral outflow. First note that both the Froude number Fr, and the slope of the energy line Sf, are functions of the depth Y and not directly functions of the position along the channel x. Only indirectly are these variables related to x because for GVF profiles, the depth Y changes with x. The Froude number is defined by its definition or Fr2 = Q2T/(gA3), and since both T and A are functions of Y once the shape of the channel is defined, it is a function of Y. The slope of the energy line Sf is defined by a uniform flow equation, either Chezy’s equation or Manning’s equation, because these equations define the frictional loss due to fluid motion. The difference between using them to define Sf and to solve the slope of the channel bottom (with all other variables known) in a uniform flow is that Sf will have a different value depending upon the depth of flow at a point, and therefore will indirectly vary with x. If Manning’s equation is used, Sf is given by 2
nQP 2 / 3 Sf = 5/3 C A
(4.11)
u
in which Cu = 1.486 when using ES units, and Cu = 1.00 when using SI units. If Chezy’s equation is used, then Sf is given by Sf =
Q2 A2 Rh C2 (e/R h , R e )
(4.12)
in which C is Chezy’s coefficient that is a function of the relative roughness of the channel wall e/R h and the Reynolds number Re as described in Chapter 2. Since the quantities on the right side of the equal sign for both Equation 4.10a and b depend only upon the independent variables Y, they can be solved by multiplying the equation by the differential of this dependent variable and by performing a numerical integration. In other words, Equation 4.10a can be written as
∫
L = x 2 − x1 = dx =
E2
1
∫ S − S dE
E1
o
(4.10c)
f
and Equation 4.10b can be written as
∫
L = x 2 − x1 = dx =
Y2
∫
Y1
1 − Fr2 dY So − Sf
(4.10d)
Any valid numerical integration method may be used for carrying out the integration of Equation 4.10c and/or d including the trapezoidal rule, or Simpson’s rule. In order to solve Equation 4.10b and c, it is necessary that the depth Y be known at both ends of the GVF profile. In other words, in order to solve the GVF equation with Y the independent variable, its beginning and ending value must be known, and a starting value of the dependent variable must be given. This requirement can be stated as a general rule. “To solve an ODE, it is necessary that the two end values of the independent variable be given, and that one value, the starting value, for the dependent variable be given.” Often, this requirement dictates which form of the GVF equation must be solved; one that considers x independent, or one that considers Y independent. Reasons why what is known dictate which form of the equation must be solved is illustrated in some of the example problems that follow.
271
Nonuniform Flows
If you have a calculator that does numerical integrations such as an HP48X, or HP48G, then it can readily solve problems in which Y is considered the independent variable and x the dependent variable, i.e., solve Equation 4.10d. The ultimate way that you would want to write Equation 4.10d is to use the variables, i.e., evaluate the Froude number Fr2 = Q2T/(gA3), and the slope of the energy line Sf = [(nQ/C)/A(P/A)2/3]2 in terms of the variable of the channel and the depth. For a trapezoidal channel, the variables are b, m, n, So, and Y, and for a circular channel they are D, n, So, and Y. To solve any problem, the variables are given the appropriate values for that problem. Before writing these general equations and storing them in your calculator, you may wish to have the calculator solve a simple problem by carrying out the numerical integration for that problem. For this purpose, consider a flow rate Q = 200 cfs in a rectangular channel with b = 10 ft, n = 0.015, and So = 0.0005. A gate creates a depth of Y = 6 ft. The solution of Manning’s equation indicates that the normal depth is 4.93 ft. Therefore, the gate causes an M1GVF backwater curve with a depth change from 5 to 6 ft. After substituting the known values into Equation 4.10d, the length of this profile is given by 6
L=
∫ 5
1 − 12.42236 /Y 3 0.040757 1 + 0.2Y 0.0005 − Y 2 Y
4 /3
dY
Your calculator should give a solution L = 10,379 ft. Now, develop two numerical integration equations that use the variable names, one for a trapezoidal channel, and one for a circular channel, and use the first one to verify the above result from the numerical integration for this specific problem. Doing such numerical integrations with the calculator are a good means for solving problems in which you know the depths at both ends of the GVF profile, i.e., problems in which Y is the independent variable, but not for solving problems in which Y is the dependent variable. Even if you have a calculator that integrates Equation 4.4, you should consider several other alternatives that are presented below. For useinhand computations of the length of GVF profiles, Equation 4.10a is often written as
∆x =
∆E So − Sf
(4.10e)
and the solution is obtained by carrying out a trapezoidal ruletype numerical integration using a table (see the following problem) in which the depth is incremented in the first column, and the subsequent columns contain the numerator and the denominator of Equation 4.10e. The last column performs the division that computes Δx. The summation of this column of Δx gives the length of the GVF profile corresponding to the first and the last depth in the table. Example Problem 4.2 A small dam in a trapezoidal channel with b = 10 ft and m = 1.4 cause the depth to increase to 6 ft. If the bottom slope of this channel is So = 0.0008, and it has a Manning’s n = 0.015 determine how far upstream from the dam the water depth is increased if the flow rate is Q = 450 cfs. Solution Solving Manning’s equation for the depth indicates that the normal depth in this channel is Yo = 4.69 ft. The GVF in this channel is of the M1 type, and takes the depth from 1% above normal depth or 1.01(4.69) = 4.74 (4.75 ft will be used) to 6 ft just upstream of the dam. Since the depths are known at both ends of this GVF profile, one of the equation that considers Y the independent variable, can be solved. The table below gives a solution to Equation 4.10e based
272
Open Channel Flow: Numerical Methods and Computer Applications on using a form of the trapezoidal rule to carry out the numerical integration as done in the FORTRAN, PASCAL, or C programs listed below. There is also a listing below of a program that calls on the more precise Simpson’s rule SIMPR algorithm described in Appendix B. You should also solve this problem using your HP48 calculator, or another calculator that has the capability of doing numerical integrations. Engineers have justified using crude trapezoidal ruletype integrations in the past, but with current tools available, these methods should only be used when solution results do not need to be very accurate. Y
E
6.000
delE
Sf
6.258 −0.216
5.750
6.042 5.832 5.629 5.435 5.253
Yo = 4.69 ft b = 10 ft
0.000403
−521.6
0.000325
−623.2
0.000229
−847.5
0.000107
−1713.1
−464.0 −986.0 −1609.0 −2457.0
0.000762
−4170.0
Q = 450 cfs m = 1.4
x 0.0
0.000625 −0.182
4.750
−464.3
0.000518 −0.194
5.000
0.000465
0.000432 −0.203
5.250
del x
0.000363 −0.210
5.500
So – Sf
0.000307
6 ft n = 0.015
This solution indicates that this M1 GVF profile has a length of 4170 ft upstream from the dam, based on five steps and the linear approximation used in the numerical integration. A more precise numerical integration gives this length as 4417 ft or a difference of 6%. The computer program listings below will carry out the solution as given above. FORTRAN Listing of GVFXY1.FOR to Solve This Problem WRITE(6,*)' Give:Q,b,m,n,So,Y1,Y2,N,g' READ(5,*)Q,b,Fm,Fn,So,Y1,Y2,N,g C=1. If(g.GT.30.) C=1.486 Q2G=Q*Q/(2.*g) QnC=Fn*Q/C A=(b+Fm*Y1)*Y1 E1=Y1+Q2G/(A*A) X=0. SF1=(QnC*((b+2.*Y1*SQRT(Fm*Fm+1.))/A)**.6666667/A)**2 WRITE(3,100)Y1,E1,SF1,X 100 FORMAT(5X,'Y',7X,'E',' delE',6X,'Sf',' SoSf', &1'del x',7X,'x',/,1X,56(''),/2F8.3,8X,F8.6,16X,F9.0) DY=(Y2Y1)/N DO 10 I=1,N Y=Y1+DY*FLOAT(I) A=(b+Fm*Y)*Y E=Y+Q2G/(A*A) SF=(QnC*((b+2.*Y*SQRT(Fm*Fm+1.))/A)**.6666667/A)**2 SFOAV=So.5*(SF+SF1)
Nonuniform Flows
10
DELX=(EE1)/SFOAV X=X+DELX WRITE(3,'(16X,F8.3,8X,F8.6,F8.1)') EE1,SFOAV,DELX WRITE(3,'(2F8.3,8X,F8.6,16X,F9.0)') Y,E,SF,X E1=E SF1=SF STOP END
PASCAL Listing of Program GVFXY1.PAS to Solve This Problem Program GVFXY1; Function Expn(a,b:real):real;Begin if a<0 then Writeln('error in power',a,b) else Expn:=Exp(b*Ln(a)) End; Var Q,b,m,n,So,Y1,Y2,g,Q2G,C,QnC,E,E1,SF,SF1,A,DY,Y,SFOAV,X, DELX:real; I,No:integer; BEGIN Writeln('Give:Q,b,m,n,So,Y1,Y2,N,g'); Readln(Q,b,m,n,So,Y1,Y2,No,g); C:=1; If g>30 then C:=1.486; Q2G:=Q*Q/(2*g); QnC:=n*Q/C; A:=(b+m*Y2)*Y2; E1:=Y2+Q2G/sqr(A); X:=0; SF1:=sqr(QnC*Expn((b+2*Y2*sqrt(sqr(m)+1))/A,0.666667)/A); Writeln(' Y E delE Sf SoSf del x x'); Write(''); Writeln(''); Writeln(Y2:8:3,E1:8:3,' ',SF1:8:6,' ',X:9:0); DY:=(Y1Y2)/No; For I:= 1 to No do Begin Y:=Y2+DY*I; A:=(b+m*Y)*Y; E:=Y+Q2G/sqr(A); SF:=sqr(QnC*Expn((b+2*Y*sqrt(sqr(m)+1))/A,0.666667)/A); SFOAV:=So0.5*(SF+SF1); DELX:=(EE1)/SFOAV; X:=X+DELX; Writeln (' ',E1E:8:3,' ',SFOAV:8:6,DELX:8:1); Writeln (Y:8:3,E1:8:3,' ',SF:8:6,' ',X:9:0); E1:=E; SF1:=SF End; END. C Listing of Program GVFXY1.C to Solve This Problem #include <stdio.h> #include <stdlib.h> #include <math.h> #define sqr(x) x*x void main(void){ int i,no; float q,b,m,n,so,y1,y2,g,c=1,q2g,qnc,a,e1,sf1,sf,dy,y,e,\ sfoav, delx,x=0; printf("Give: Q,b,m,n,So,Y1,Y2,N,g\n"); scanf("%f %f %f %f %f %f %f %d %f",\ &q,&b,&m,&n,&so,&y1,&y2,&no,&g); if(g>30.) c=1.486; q2g=q*q/(2.*g); qnc=n*q/c; a=(b+m*y1)*y1; e1=y1+q2g/(a*a);
273
274
Open Channel Flow: Numerical Methods and Computer Applications sf1=sqr(qnc*pow((b+2.*y1*sqrt(m*m+1.))/a, .6666667)/a); printf(" Y E delE Sf SoSf del \ x x\n"); for(i=1;i<57;i++) printf("");printf("\n"); printf("%8.3f %7.3f %8.6f %9.0f\n",\ y1,e1,sf1,x); dy=(y2y1)/(float)no; for(i=1;i<=no;i++){ y=y1+dy*(float)i; a=(b+m*y)*y; e=y+q2g/(a*a); sf=sqr(qnc*pow((b+2.*y*sqrt(m*m+1.))/a,.6666667)/a); sfoav=so.5*(sf+sf1); delx=(ee1)/sfoav; x+=delx; printf(" %8.3f %8.6f %7.1f\n",\ ee1,sfoav,delx); printf("%8.3f %7.3f %8.6f %9.0f\n" \, y,e,sf,x); e1=e; sf1=sf; }} Listing of EPRB4_2.FOR That Calls on SIMPR for a Numerical Solution EXTERNAL EQUAT CALL SIMPR(EQUAT,6.,4.75,X,1.E6,21) WRITE(*,*) X END FUNCTION EQUAT(Y) A=(10.+1.4*Y)*Y P=10.+3.440935*Y EQUAT=(1.6288.82*(10.+2.8*Y)/A**3)/(.000820.633359 &*((P/A)**.6666667/A)**2) RETURN END Returns as the solution −4417.068 Listing of EPRB4_2.C #include <stdio.h> #include <stdlib.h> #include <math.h> extern float simpr(float (*equat)(float xx),float xb,\ float xe, float err,int max); float equat(float y){float a,p; a=(10.+1.4*y)*y; p=10.+3.440935*y; return (1.6288.82*(10.+2.8*y)/pow(a,3.))/\ (.000820.633359*pow(pow(p/a,.66666667)/a,2.));} void main(void){ printf("Length of GVF profile =%f\n", simpr(equat,6.,4.75,\1.e6,21));} Example Problem 4.3 A gate exists in a channel with b = 10 ft and m = 1.5 that causes a depth of Y1 = 1.5 ft immediately downstream from it. This channel has a bottom slope of So = 0.0005, and its n = 0.013 and it is carrying a flow rate Q = 500 cfs. Determine the location of the hydraulic jump. Solution First, Manning’s equation is used to solve the normal depth in the channel downstream from the hydraulic jump. This normal depth is Yo2 = 5.118. The depth just upstream of the hydraulic jump will be the conjugate depth to this. Therefore, next the momentum equation M1 = M2 must be solved to find the depth upstream from the jump. For this solution, the momentum function downstream of the jump M2 is based on the normal depth of 5.118 ft. This solution gives the depth upstream of the jump as 2.297 ft. An M3 GVF profile will take the depth from 1.5 ft immediately
275
Nonuniform Flows behind the gate to 2.297. The table below gives this solution. Note that since the depths were known at both ends of this GVF profile, it was most appropriate to consider x the dependent variable in this problem also. The solution indicates that the jump will occur 302 ft downstream from the gate. A more precise numerical integration gives this distance as 305 ft. If you are familiar with a spreadsheet (Lotus 123, Quattro, or Excel) you should solve this problem and Problem # 1 using it. Y
E
1.500
delE
12.997
Sf
10.698 9.062 7.878
−0.019406
61.0
−0.014433
60.2
−0.010930
58.7
0.016975
7.009
183.0
0.012891
6.367
244.0
0.009970
1.5 ft
M3 G
So = 0.0005
61.3 122.0
−0.642 2.297
−0.026671 0.022836
−0.869 2.138
61.1
x
61.0
−1.185 1.978
−0.037658 0.031506
−1.635 1.819
del x
0.0
−2.300 1.659
So – Sf
0.044810
302.0
Q = 500 cfs
VF
Yo2 = 5.118
b = 10 ft
m = 1.5 n = 0 .013
Example Problem 4.4 A sluice gate causes a depth of 2 ft in an 8 ft wide rectangular channel immediately downstream from it. The channel has a bottom slope of So = 0.0011, a Manning’s n = 0.013, and 800 ft downstream from the gate the channel terminates in a free overfall. If the flow rate in this channel per unit width is q = 60 cfs/ft, determine the location of the hydraulic jump. 800 ft 2 ft
M2 G
M3 GVF profile
q = 60 cfs/ft
VF pr ofile
4.9 ft 8 ft wide
So = 0.0011
n = 0.013
Solution Since the distance between the sluice gate and the end of the channel is known, and a hydraulic jump, if it occurs, must be somewhere within this distance, it is more convenient to consider x the independent variable, and Y the dependent variable. Thus, Equation 4.4 will be selected. Because this channel is relative short, it is unlikely that a normal depth will exist downstream from the hydraulic jump. If a hydraulic jump does occur, then the depth at the end (free overfall) of the channel will be critical, and an M2 GVF profile will occur between the jump to this critical depth (Yc = (q2/g)1/3 = 4.82 ft), and an M3 GVF profile will occur upstream from the hydraulic
276
Open Channel Flow: Numerical Methods and Computer Applications jump. The hydraulic jump will, therefore, likely occur between these two GVF profiles, at a point where they match the conjugate depths. The three unknowns are as follows: (1) the distance from the gate to the hydraulic jump, (2) the depth upstream from the hydraulic jump, and (3) the depth downstream from the hydraulic jump. The three equations needed to solve these three unknowns are (a) the momentum equation across the jump M1 = M2, (b) the GVF equation for the M3 profile, and (c) the GVF equation for the M2 profile. Methods for solving systems of algebraic and ordinary differential equations will be discussed later. However, for this problem, an easy approach will be to solve both the M2 and the M3 GVF profiles past where it is anticipated that the hydraulic jump will occur, and then find the x position where depths from these two solutions are the two conjugate depths across the hydraulic jump. The tables below shows these two GVF solutions and also the values of the momentum functions M, associated with these depths. Note, the solution for the M2 GVF profile starts at the end of the channel with a depth slightly above Yc, since Y = Yc would result in a division by zero in Equation 4.4, e.g., an infinite value for dY/dx, and proceeds upstream. The position where the depths satisfy the conjugate depth equation is where the values of the momentum functions Ms (the last column in the solution tables) and x’s are both equal in the two solution tables. This position is x = 354 ft, where an upstream depth of Y1 = 3.459 ft exists and a downstream depth of Y2 = 6.494 ft exists. Depending upon the size of the Δx interval used in the solution, it may be necessary to interpolate the distance x between a couple of table entries. The above solutions to the GVF profiles were obtained using the differential equation solvers, ODESOL RUKUST and/or DVERK described in Appendix C. More will be said about using these solvers later. Below are two FORTRAN listings that call these solvers. Note, these programs consist of a main program that defines the problem and appropriately call on the solver, and a subroutine that defines the derivative dY/dx for the solver, so it knows what the differential equation to be solved is. You should read Appendix C now as you go through these program listings and understand how they solve the ODE for a gradually varied flow. The next section discusses what these solver do. M3 GVF–Profile
M2 GVF–Profile
x (ft)
Y (ft)
M (ft )
x (ft)
Y (ft)
M (ft3)
0.000 2.000 4.000 · 352.000 354.000 356.000 358.000 · 400.000
2.000 2.007 2.014 · 3.448 3.459 3.470 3.480 · 3.720
463.2 461.7 460.3 · 306.9 306.4 305.9 305.4 · 295.8
800.000 798.000 · 358.000 356.000 354.000
4.900 4.982 · 6.488 6.491 6.494
278.6 278.8 · 306.2 306.3 306.4
352.000 350.000
6.497 6.500
306.5 306.6
3
FORTRAN Listing of Program EPRB4_2.FOR That Call ODESOL REAL Y(1),DY(1),XP(1),YP(1,1),WK1(1,13) EXTERNAL DYX COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRANS/B,FM,FN,SO,Q2,FNQ,FMS WRITE(6,*) 'GIVE IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,g' READ(5,*) IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,G H1=.01 HMIN=1.E5 Y(1)=YB FMS=2.*SQRT(FM*FM+1.) C=1. IF(G.GT.30.) C=1.486 FNQ=FN*Q/C
Nonuniform Flows
2
100
8 1
20
40
Q2=Q*Q/G X=XBEG A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON XZ=X+DELX CALL ODESOL(Y,DY,1,X,XZ,TOL,H1,HMIN,1,XP,YP,WK1,DYX) X=XZ A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON FORMAT(6X,2F10.3,F12.2) IF(DELX .LT. 0.) GO TO 8 IF(X .LT. XEND) GO TO 2 GO TO 1 IF(X .GT. XEND) GO TO 2 IF(IOUT.NE.6) CLOSE(UNIT=IOUT) STOP END SUBROUTINE DYX(X,Y,DY) REAL Y(1),DY(1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRANS/B,FM,FN,SO,Q2,FNQ,FMS A=(B+FM*Y(1))*Y(1) T=B+2.*FM*Y(1) P=B+FMS*Y(1) SF=(FNQ*(P/A)**.66666667/A)**2 FR2=Q2*T/A**3 DY(1)=(SOSF)/(1.FR2) RETURN END
Input data for above M3 GVF profile 3 .0001 2 2 480 .013 .0011 8 0 0 400 32.2 FORTRAN Listing of Program EPRB4D.FOR That Call DVERK REAL Y(1),CC(24),WK(2,9) EXTERNAL DYX COMMON /TRANS/B,FM,FN,SO,Q2,FNQ,FMS WRITE(6,* ) 'GIVE IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,g' READ(5,*) IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,G IER=0 IND=1 Y(1)=YB FMS=2.*SQRT(FM*FM+1.) C=1. IF(G.GT.30.) C=1.486 FNQ=FN*Q/C Q2=Q*Q/G X=XBEG A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON 2 XZ=X+DELX CALL DVERK(1,DYX,X,Y,XZ,TOL,IND,CC,2,WK) IF(IER.NE.0) THEN
277
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Open Channel Flow: Numerical Methods and Computer Applications
100
8 1
20
40
WRITE(6,*) IER,' ERROR TERMINATION' STOP ENDIF A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON FORMAT(6X,2F10.3,F12.2) IF(DELX .LT. 0.) GO TO 8 IF(X .LT. XEND) GO TO 2 GO TO 1 IF(X .GT. XEND) GO TO 2 IF(IOUT.NE.6) CLOSE(UNIT=IOUT) STOP END SUBROUTINE DYX(N,X,Y,DY) REAL Y(N),DY(N) COMMON /TRANS/B,FM,FN,SO,Q2,FNQ,FMS A=(B+FM*Y(1))*Y(1) T=B+2.*FM*Y(1) P=B+FMS*Y(1) SF=(FNQ*(P/A)**.66666667/A)**2 FR2=Q2*T/A**3 DY(1)=(SOSF)/(1.FR2) RETURN END
Input data for above M2 GVF profile 3 .0001 −2 4.9 480 .013 .0011 8 0 800 350 32.2 FORTRAN Listing of Program That Solves Equation 4.9 rather than Equation 4.4 REAL Y(1),DY(1),XP(1),YP(1,1),WK1(1,13) EXTERNAL DYX COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRANS/B,FM,FN,SO,Q2,FMS,SCN WRITE(6,*) 'GIVE IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,g' READ(5,*) IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND,G H1=.01 HMIN=1.E5 Y(1)=YB FMS=2.*SQRT(FM*FM+1.) C=1. IF(G.GT.30.) C=1.486 Q2=Q*Q/G SCN=G*(FN/C)**2 X=XBEG A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON 2 XZ=X+DELX CALL ODESOL(Y,DY,1,X,XZ,TOL,H1,HMIN,1,XP,YP,WK1,DYX) X=XZ A=(B+FM*Y(1))*Y(1) FMON=B*Y(1)**2/2.+FM*Y(1)**3/3.+Q2/A WRITE(IOUT,100) X,Y,FMON 100 FORMAT(6X,2F10.3,F12.2) IF(DELX .LT. 0.) GO TO 8
Nonuniform Flows
8 10
IF(X .LT. XEND) GO TO 2 GO TO 10 IF(X .GT. XEND) GO TO 2 IF(IOUT.NE.6) CLOSE(UNIT=IOUT) STOP END SUBROUTINE DYX(X,Y,DY) REAL Y(1),DY(1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRANS/B,FM,FN,SO,Q2,FMS,SCN A=(B+FM*Y(1))*Y(1) T=B+2.*FM*Y(1) FR2=Q2*T/A**3 SC=SCN*(B+2.*Y(1)*SQRT(FM*FM+1.))**1.3333333/T/A**.3333333 DY(1)=(SOSC*FR2)/(1.FR2) RETURN END
Listing of Program That Call on the Runge–Kutta Method Described in Appendix C REAL YY(1),YTT(1) COMMON B,FM,FM2,TWOM,FNQ,QG2,SO EQUIVALENCE (Y,YY(1)) WRITE(6,*)'Give:IOUT,Q,b,m,So,n,Xbeg',Xend,DX,Ybeg,g,xstart' READ(5,*) IOUT,Q,B,FM,SO,FN,XBEG,XEND,DX,YBEG,G,DXS C=1. IF(G.GT.30.) C=1.486 N=ABS(XENDXBEG)/ABS(DX)+.5 QG2=Q*Q/G TWOM=2.*FM FNQ=FN*Q/C FM2=2.*SQRT(FM*FM+1.) Y=YBEG WRITE(IOUT,90) XBEG,Y,(.5*B+FM/3.*Y)*Y*Y+QG2/((B+FM*Y)*Y) 90 FORMAT(F10.1,F10.3,F10.1) X1=XBEG DO 10 I=1,N X=X1+DX CALL RUKUST(1,DXS,X1,X,1.E5,YY,YTT) WRITE(IOUT,100)X,Y,(.5*B+FM/3.*Y)*Y*Y+QG2/((B+FM*Y)*Y) 10 X1=X END SUBROUTINE SLOPE(X,Y,DYX) REAL Y(1),DYX(1) COMMON B,FM,FM2,TWOM,FNQ,QG2,SO A=(B+FM*Y(1))*Y(1) SF=(FNQ*((B+FM2*Y(1))/A)**.66666667/A)**2 DYX(1)=(SOSF)/(1.QG2*(B+TWOM*Y(1))/A**3) RETURN END Listing of Program GVFDYXK.C (a C Program) That Calls on the Runge–Kutta Method #include <stdio.h> #include <stdlib.h> #include <math.h> float b,m,m2,twom,fnq,qg2,so; void rukust(int n,float *dxs,float xbeg,float xend,\ float err,float *y,float *ytt);
279
280
Open Channel Flow: Numerical Methods and Computer Applications void slope (float x,float *y, float *dy){ float a,sf; a=(b+m*y[0])*y[0]; sf=pow(fnq*pow((b+m2*y[0])/a,.66666667)/a,2.); dy[0]=(sosf)/(1.qg2*(b+twom*y[0])/pow(a,3.)); return;} //end slope void main(void){ int i,no; float q,n,xbeg,xend,dx,ybeg,g,xstart,*dxs,x,x1,c=1.,y[1],ytt[1]; FILE *fil; char filnam[20],fmt[]="%10.1f %9.3f %9.1f\n"; printf("Give name of output file\n");scanf("%s",filnam); if((fil=fopen(filnam,"w"))==NULL){ printf("File cannot be opened\n"); exit(0);} printf("Give:Q,b,m,So,n,xbeg,xend,DX,Ybeg,g,xstart\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f", &q,&b,&m,&so,&n,&xbeg,&xend,&dx,&ybeg,&g,&xstart); if(g>30.) c=1.486; no=fabs(xendxbeg)/fabs(dx)+.5; qg2=q*q/g; twom=2.*m;fnq=n*q/c; *dxs=xstart; m2=2.*sqrt(m*m+1.); y[0]=ybeg; x1=xbeg; fprintf(fil,fmt,x1,y[0],(.5*b+m/3.*y[0])*y[0]*y[0]+qg2/ ((b+m*y[0])*y[0])); for(i=0;i<no;i++){x=x1+dx; rukust(1,dxs,x1,x,1.e5,y,ytt); fprintf(fil,fmt,x,y[0],(.5*b+m/3.*y[0])*y[0]*y[0]+qg2/\ ((b+m*y[0])*y[0])); x1=x;} fclose(fil);}
4.4 Numerical Methods for Solving ODEs In obtaining the solution to Example Problem 4.4, it was assumed that a solution to Equation 4.4 was readily available in the form of a tabular list of depths Y as a function of the positions x. This tabular solution was obtained by utilizing an available numerical differential equation solver, such as described in Appendix C. Problems involving ODEs are common in all scientific and engineering fields, and therefore much is known about obtaining closed forms and numerical solutions to them. Solving an ODE involves more than just a numerical integration because the derivative(s) of the dependent variable(s) are functions of both the independent and the dependent variables, whereas if the equation can be separated so that each side of the equation involves only one variable, then both sides of the equation can be integrated, if not in the closed form then numerically. It is because the right side of Equation 4.4 is only a function of Y was it possible to reverse the role of the variables, and solve Example Problems 4.2 and 4.3 by a numerical integration. In the more general ODE for gradually varied flow, Equation 4.6, the derivative dY/dx, is a function of both x and Y. The varied flow equation is special in that it is a first order ODE, i.e., only involves a first derivative. The emphasis in this book is on the utilization of algorithms that have been developed for numerically solving problems governed by ODEs. However, it is desirable to give a brief overview of the general techniques employed for their solution. Readers interested in more detail should consult the appropriate portions of one of the many available books dealing with the subject of numerical analysis. The starting of the numerical solutions to ODEs generally utilizes the Runge–Kutta or the Euler method. After the solution has been obtained over a few intervals of Δx, it can be continued using predictor–corrector methods. Runge–Kutta type methods can start and continue the solution over an interval of as many Δx’s as desired. The fourthorder Runge–Kutta method is commonly used by engineers and scientists and is described in Appendix C. You should read this appendix in conjunction with the following pages. The Euler method is illustrated in the sketch below. The dependent variable Yo, is known at the starting point xo (referred to as an initial value problem), and therefore the derivative dY/dx can be evaluated at xo. By multiplying dY/dx by the interval Δx, an estimate of Y1 at xo + Δx = x1 can be obtained by
281
Nonuniform Flows
dY Y1 = Yo + ∆x dx 0
Predictor
Corrector
Y dY/dx
dx
dx
x
Now that an estimate of Y1 is available, a better estimate can be obtained for the derivative at x1 by using the average of the derivatives at xo and x1 or
Y1(1) = Yo +
dY dY 1 ∆x + 2 dx 0 dx 1
since Y1(1) is an improvement over Y1, the above equation can be repeatedly applied giving Y1(2), Y1(3). . Y1(i) always using the most recent Y1 to evaluate (dY/dx)1. This second equation is called Euler’s corrector. In practice, Euler’s corrector is generally only applied once since it is more numerically efficient to use a smaller Δx than to apply the corrector more than once. This process could be applied repeatedly for the entire range of x for which the dependent variable Y is sought. However, once Y’s are known corresponding to several x’s at a spacing of Δx, then higher order polynomials can be used to both predict ahead and correct this prediction. These are referred to as predictor–corrector methods. Milne’s method is one such means of continuing the solution. The predictor equation of Milne’s method is
Yi +1 = Yi − 3 +
dY 4 dY dY ∆x 2 − + 2 dx i 3 dx i − 2 dx i −1
in which the i subscripts refer to the ends of the different intervals for which Y has been obtained, with subscript i at the end of the last such interval, and this equation extrapolates to the end of the next interval, i.e., to i + 1. The corrector for Milne’s method is Yi +1 = Yi −1 +
dY 1 dY dY ∆x + 4 + dx i dx i +1 3 dx i −1
The Hamming method is quite widely used and provides more mathematical stability in the numerical solution than the Milne method, as it contains a predictors, a modifier, and a corrector, and a final value equation as given below:
282
Open Channel Flow: Numerical Methods and Computer Applications
Predictor YI(+01) = Yi − 3 +
dY dY 4 dY ∆x 2 − + 2 dx i − 2 3 dx i dx i −1
Modifier Yi(+11) = Yi(+01) −
(
112 ( 0 ) Yi − Yi( j) 121
)
Corrector Yi(+j1) =
dY 1 dY dY + 2 − 9Yi − Yi − 2 + 3∆x dx i dx i −1 8 dx i +1
Final value Yi +1 = Yi +
(
9 Yi(+01) − Yin+1 121
)
Since such predictor–corrector methods rely upon starting the solution with a method, such as Euler’s method, and the overall accuracy of the solution cannot be greater than that at its start, it can be argued that continuing a solution with sophisticated methods is not justified. Runge–Kutta methods are such a compromise. Consider using a predictor like the above Euler equation to predict Y at the midpoint of the interval, or at Δx/2. Then use the value of x and Y at this midpoint to compute the derivative over the interval Δx. Written in equation form, this becomes dY Yi +1 = Yi + ∆x dx i + 1
2
in which the evaluation of the derivative is accomplished as shown below:
1 1 dY dY dx = f x i + 2 ∆x, Yi + 2 ∆Y ; ∆Y = ∆x dx 1 i+ i 2
in which f( ) represents the function on the right side of the equal sign, e.g., it is a simpler way of writing dY/dx especially when the arguments for its evaluation are given. Because of symmetry, the firstorder term is canceled, and the method becomes a secondorder method versus the Euler predictor being only a firstorder method. By a judicious evaluation of dY/dx, i.e., f( ), it is possible to increase the order of the approximation even further. Such a widely used evaluation gives the following fourthorder Runge–Kutta method (for more details see Appendix C): 1 1 ∆Y1 = ∆xf (x i , Yi ); ∆Y2 = ∆xf x i + ∆x, Yi + ∆Y1 2 2
1 1 ∆Y3 = ∆xf x i + ∆x, Yi + ∆Y2 ; ∆Y4 = ∆xf (x i + ∆x, Yi + ∆Y3 ) 2 2
283
Nonuniform Flows
and evaluating Y at i + 1 with
1 1 Yi +1 = Yi + (∆Y1 + ∆Y4 ) + (∆Y2 + ∆Y3 ) + O(∆x 5 ) 6 3
in which the quantity O(Δx5) indicates that fifthorder terms are being ignored. The following programs are designed to implement this fourthorder Runge–Kutta method. They use a constant interval Δx. For an accuracy of the solution, the interval should be responsive to the properties of the function being solved, etc., the computer code should implement means for sizing this interval for the accuracy requested. Such a code is given in Appendix C in the subroutine ODESOL, DVERK, and RUKUST, which are described in Appendix C. FORTRAN Listing of Program RUKUY4.FOR to Solve ODE Equation (Equation 4.4) COMMON B,FM,FN,SO,Q,Q2G,FM2,FNQ,FMS WRITE(6,*)'Give:IOUT,Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g' READ(5,*) IOUT,Q,B,FM,SO,FN,XBEG,XEND,DX,YBEG,G Q2G=Q*Q/G FNQ=FN*Q IF(G.GT.30.) FNQ=FNQ/1.486 FMS=2.*SQRT(FM*FM+1.) FM2=2.*FM N=ABS(XENDXBEG)/ABS(DX) Y=YBEG WRITE(IOUT,100)XBEG,Y 100 FORMAT(2F10.3) DO 10 I=1,N CALL RUKU4(X,DX,Y) X=XBEG+DX*FLOAT(I) 10 WRITE(IOUT,100) X,Y IF(ABS(XXEND).LT. .001) STOP CALL RUKU4(X,XENDX,Y) WRITE(IOUT,100) XEND,Y STOP END SUBROUTINE RUKU4(X,DX,Y) XH=X+.5*DX DY1=DX*SLOPE(X,Y) DY2=DX*SLOPE(XH,Y+DY1/2.) DY3=DX*SLOPE(XH,Y+DY2/2.) DY4=DX*SLOPE(XH,Y+DY3) Y=Y+(DY1+DY4)/6.+(DY2+DY3)/3. RETURN END FUNCTION SLOPE(X,Y) COMMON B,FM,FN,SO,Q,Q2G,FM2,FNQ,FMS A=(B+FM*Y)*Y A2=A*A FR2=Q2G*(B+FM2*Y)/(A*A2) SF=(FNQ*((B+Y*FMS)/A)**.66666667/A)**2 SLOPE=(SOSF)/(1.FR2) RETURN END
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Open Channel Flow: Numerical Methods and Computer Applications
Listing of Program RUKUY4.C #include #include <stdio.h> #include <math.h> #include <stdlib.h> #define sqr(x) x*x float q,b,m,so,n,xbeg,xend,dx,ybeg,g,q2g,fm2,fnq,fms,x,y; float slope(float xx,float yy){ float a,a2,fr2,sf; a=(b+m*yy)*yy; a2=a*a; fr2=q2g*(b+fm2*yy)/(a*a2); sf=sqr(fnq*pow((b+yy*fms)/a,0.66666667)/a); return (sosf)/(1.fr2);} void ruku4(void){ float xh,dy1,dy2,dy3,dy4; xh=x+0.5*dx; dy1=dx*slope(xh,y); dy2=dx*slope(xh,y+dy1/2.); dy3=dx*slope(xh,y+dy2/2.); dy4=dx*slope(x+dx,y+dy3); y=y+(dy1+dy4)/6.+(dy2+dy3)/3.;} void main(){ int i,no; cprintf("Give:Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g\r\n"); scanf("%f %f %f %f %f %f %f %f %f %f", &q,&b,&m,&so,&n,&xbeg,&xend,&dx,&ybeg,&g); q2g=q*q/g; fnq=n*q; if (g > 30) fnq=fnq/1.486; fms=2*sqrt(m*m+1); fm2=2.*m; x=xbeg; no=fabs(xendxbeg)/fabs(dx); y=ybeg; cprintf("rn%10.3f %10.3f\r\n",xbeg,y); x=xbeg; for(i=1;i<=no;i++){ ruku4(); x+=dx; cprintf("%10.3f %10.3f\r\n",x,y);}} The above program contain a function subprogram SLOPE that can be modified to solve other ODEs. When Equation 4.4 is being solved, it would not be necessary to pass X as an argument of SLOPE, since the dY/dx does not depend upon x. Example Problem 4.5 Modify one of the above programs to solve problems, such as Example Problems 4.2 and 4.3 in which Y was considered the independent variable and x the dependent variable. Solution If one is willing to interpret the input prompt so that Xbeg and Xend are the beginning and the ending values for the depth Y, and Ybeg is the beginning value of x, then the only changes needed is to rewrite SLOPE so its arguments are reversed, i.e., SLOPE(Y,X) and the statement that computes SLOPE be changed to SLOPE=(1.FR2)/(SOSF). The PASCAL program has been modified below (including changes in the prompt). Thus, it is possible to use a differential equation solver to perform a numerical integration with a modest amount of additional computer effort.
Nonuniform Flows Program RuKuX4; Var I,No,IOUT:integer; Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g,Q2G,FM2,FNQ,FMS,X,Y:real; Function Expn(a,b:real):real;Begin if a<0 then Writeln ('error in power',a,b) else Expn:=Exp(b*Ln(a)) End; Function SLOPE(yy,xx:real):real; Var A,A2,FR2,SF:real; Begin A:=(b+m*yy)*yy; A2:=A*A; FR2:=Q2G*(B+FM2*yy)/(A*A2); SF:=sqr(FNQ*Expn((b+yy*FMS)/A,0.66666667)/A); SLOPE:=(1.FR2)/(SoSF); End; Procedure RUKU4; Var XH,DY1,DY2,DY3,DY4:real; Begin XH:=X+0.5*DX; DY1:=DX*SLOPE(XH,X); DY2:=DX*SLOPE(XH,X+DY1/2); DY3:=DX*SLOPE(XH,X+DY2/2); DY4:=DX*SLOPE(XH,X+DY3/2); Y:=Y+(DY1+DY4)/6+(DY2+DY3)/3 End; BEGIN Writeln('Give:Q,b,m,So,n,Ybeg,Yend,DY,Xbeg,g'); Readln(Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g); Q2G:=Q*Q/g; FNQ:=n*Q; If g > 30 then FNQ:=FNQ/1.486; FMS:=2*sqrt(m*m+1); FM2:=2*m; X:=Xbeg; No:=Trunc(abs(XendXbeg)/abs(DX)); Y:=Ybeg; Writeln(Xbeg:10:3,Y:10:3); X:=Xbeg; For I:=1 to No do Begin RUKU4; X:=X+DX; Writeln(X:10:3,Y:10:3) End; If abs(XXend) > 0.001 Then begin DX:=XendX; RUKU4; Writeln(Xend:10:3,Y:10:3) end; END. Example Problem 4.6 Water flows at a rate Q = 20 m 3/s in a 2.5 m wide, with m = 1 trapezoidal channel. The upstream channel is mild, and changes to a steep channel with a bottom slope of So = 0.012 and n = 0.014. At a distance L = 100 m from the break in grade, the channel discharges into a reservoir with a water surface elevation H 2 = 3.0 m above the channel bottom. Locate the position of the hydraulic jump. Also, solve the problem using Chezy’s equation (e = 0.0012 m and ν = 1.003 × 10 −6 m 2/s). Solution First computing critical conditions gives Yc = 1.516 m, with Ec = 2.066 m and Mc = 10.73 m3. Therefore, the problem consists of solving: (1) the S2GVF profile downstream from the break in grade, (2) the S1GVF profile upstream from the reservoir, and (3) determining the position where the momentum functions for these two GVF profiles are equal. Portions of these two solutions are shown below. The jump occurs at a position x = 22 m downstream from the break in grade.
285
286
Open Channel Flow: Numerical Methods and Computer Applications S2 GVF with Q = 20.00, So = .012000, n = .0140, B = 2.5, m = 1.00. Y
E
M
0.000 2.000 4.000 · 18.00 20.00 22.00
1.500 1.399 1.357
2.066 2.085 2.101
10.7 10.8 10.9
1.225 1.213 1.203
2.205 2.218 2.230
24.00 26.00
1.194 1.185
2.243 2.255
x
S1 GVF with Q = 20.00, So = .012000, n = .0140, B = 2.5, m = 1.00. x
Y
E
M
3.000 2.974
3.075 3.051
22.7 22.3
11.4 11.5 11.5
100.0 98.00 · 28.00 26.00 24.00 22.00
1.980 1.944 1.906 1.866
2.239 2.217 2.195 2.173
12.1 11.9 11.7 11.5
11.6 11.6
20.00 18.00
1.823 1.777
2.151 2.130
11.3 11.2
The above solutions were obtained using the programs that called on ODESOL as were used to solve Example Problem 4.3. Essential identical solutions will be obtained using the following program that calls on the Runge–Kutta fourthorder method described in Appendix C, if the following input is used: 3 20 2.5 1 .012 .014 0 100 2 1.5 9.81 5 & 3 20 2.5 1 .012 .014 100 12 2 3. 9.81 5 Program EPRB4_6.FOR COMMON B,FM,FN,SO,Q,Q2G,FNQ,FMS,G,C WRITE(6,*)'Give:IOUT,Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g,Nsetp' READ(5,*) IOUT,Q,B,FM,SO,FN,XBEG,XEND,DX,YB,G,NSTEP C=1. IF(G.GT.30.) C=1.486 FNQ=FN*Q/C Q2G=Q*Q/G FMS=2.*SQRT(FM*FM+1.) N=ABS(XENDXBEG)/ABS(DX)*FLOAT(NSTEP) DX=DX/FLOAT(NSTEP) Y=YB FMOM=(.5*B+FM/3.*Y)*Y*Y+Q2G/((B+FM*Y)*Y) WRITE(IOUT,100)XBEG,Y,FMOM 100 FORMAT(F10.1,F10.3,F10.2) DO 10 I=1,N CALL RUKU4(X,DX,Y) X=XBEG+DX*FLOAT(I) FMOM=(.5*B+FM/3.*Y)*Y*Y+Q2G/((B+FM*Y)*Y) 10 IF(MOD(I,NSTEP).EQ.0) WRITE(IOUT,100)X,Y,FMOM END SUBROUTINE RUKU4(X,DX,Y) XH=X+.5*DX DY1=DX*SLOPE(X,Y) DY2=DX*SLOPE(XH,Y+DY1/2.) DY3=DX*SLOPE(XH,Y+DY2/2.) DY4=DX*SLOPE(X+DX,Y+DY3) Y=Y+(DY1+DY4)/6.+(DY2+DY3)/3. RETURN END FUNCTION SLOPE(X,Y)
287
Nonuniform Flows COMMON B,FM,FN,SO,Q,Q2G,FNQ,FMS,G,C A=(B+FM*Y)*Y SF=(FNQ*((B+FMS*Y)/A)**.66666667/A)**2 SLOPE=(SOSF)/(1.Q2G*(B+2.*FM*Y)/A**3) RETURN END
In this program, that uses the fixedstep Runge–Kutta method, the number of intermediate computation steps consists of the value given to NSETP, which duplicates the above solutions when equal to 5. If NSETP = 1, then the solution of the S2 GVF profile is as follows: x 0.0 2.0 4.0
Y
M
1.500 1.301 1.281
10.73 11.10 11.17
The difference of Y = 1.301 m to Y = 1.399 m is significant over the first 2 m increments used in the computations. The reason is that dY/dx is infinite at the critical depth and starting at a depth of 1.5 m is too close to the critical depth for a fixedstep size algorithm to provide accurate answers unless the step is a fraction of 2 m. A better approach is to use what is called an “Adaptive Step Size” ODESolver. ODESOL, DVERK, and RUKUST fall in this category, i.e., they find a step size that is consistent with the error condition specified. For example, if RUKUST is used, then there is no need for the NSETP (number of intermediate steps) to be greater than 1 to achieve accuracy. Program EPRB4_6 is modified below to call on RUKUST. Program EPRB4_6A.FOR COMMON B,FM,FN,SO,Q,Q2G,FNQ,FMS REAL Y(1),YTT(1) WRITE(*,*)' Give:IOUT,Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,g' READ(*,*) IOUT,Q,B,FM,SO,FN,XBEG,XEND,DX,YB,G C=1 IF(G.GT.30.) C=1.486 FNQ=FN*Q/C Q2G=Q*Q/G FMS=2.*SQRT(FM*FM+1.) N=ABS(XENDXBEG)/ABS(DX) Y(1)=YB FMOM=(.5*B+FM/3.*YB)*YB*YB+Q2G/((B+FM*YB)*YB) WRITE(IOUT,100) XBEG,Y,FMOM 100 FORMAT(F10.1,F10.3,F10.2) DXS=.1 X1=XBEG DO 10 I=1,N X2=X1+DX CALL RUKUST(1,DXS,X1,X2,1.E5,Y,YTT) FMOM=(.5*B+FM/3.*Y(1))*Y(1)**2+Q2G/((B+FM*Y(1))*Y(1)) WRITE(IOUT,100) X2,Y,FMOM
288
10
Open Channel Flow: Numerical Methods and Computer Applications
X1=X2 END SUBROUTINE SLOPE(X,Y,DYX) REAL Y(1),DYX(1) COMMON B,FM,FN,SO,Q,Q2G,FNQ,FMS A=(B+FM*Y(1))*Y(1) SF=(FNQ*((B+FMS*Y(1))/A)**.66666667/A)**2 DYX(1)=(SOSF)/(1.Q2G*(B+2.*FM*Y(1))/A**3) RETURN END
The input needed to solve Example Problem 4.2 is 3 450 10 1.4 .0008 .015 6 4.75 −.25 0 32.2 with Solution = −4508.8 ft Listing of Program RUKUSTS.C #include #include <math.h> #include <stdlib.h> #include <stdio.h> extern void rukust(int neq,float *dxs,float xbeg,float xend,\ float error,float *y, float *ytt); float b,m,n,so,q,c=1,q2g,fnq,fms; void slope(float x,float *y,float *dy){ float a,fr2,sf; a=(b+m*y[0])*y[0]; fr2=q2g*(b+2.*m*y[0])/(a*a*a); sf=pow(fnq*pow((b+fms*y[0])/a,.66666667)/a,2.); dy[0]=(sosf)/(1.fr2); /* End of function slope */ void main(void){float g,xbeg,xend,dx,ybeg,x,*y,*ytt,*dxs; int i,nm; char fnam[20]; FILE *filo; printf("Give output file\n"); scanf("%s",fnam); if((filo=fopen(fnam,"w"))==NULL){ printf("Can not open output file %s",fnam);exit(0);} cprintf("Give: Q,b,m,So,n,Xbeg,Xend,DX,Ybeg,gr\n"); scanf("%f %f %f %f %f %f %f %f %f %f", &q,&b,&m,&so,&n,&xbeg,&xend,\&dx,&ybeg,&g); y=(float *)calloc(1,sizeof(float)); ytt=(float *)calloc(1,sizeof(float)); dxs=(float *)calloc(1,sizeof(float)); if(g>30.) c=1.486; q2g=q*q/g; fnq=n*q/c;fms=2.*sqrt(m*m+1.); nm=fabs(xendxbeg)/fabs(dx); y[0]=ybeg;x=xbeg; fprintf(filo,"%10.1f %10.3f\n",xbeg,ybeg); cprintf("%10.1f %10.3f\r\n",xbeg,ybeg);*dxs=.1*dx; for(i=1;i<=nm;i++){rukust(1,dxs,x,x+dx,1.e4,y,ytt); x+=dx; fprintf(filo,"%10.1f %10.3f\n",x,*y); cprintf("%10.1f %10.3f\r\n",x,*y);}}
Nonuniform Flows
289
The Program EPRB4_6B.FOR is a modification of the above Program EPRB4_6A.FOR that is designed to solve the problem using Chezy’s equation, rather than Manning’s equation. Notice in the input that n is replaced by e and ν. Listing of Program EPRB4_6B.FOR COMMON B,FM,e,SO,Q,Q2G,FMS,SQG,QV4,SG,C REAL Y(1),YTT(1) WRITE(*,*)' Give:IOUT,Q,b,m,So,Chezy e,VISC,Xbeg, &Xend,DX,Ybeg,g' READ(*,*) IOUT,Q,B,FM,SO,e,VISC,XBEG,XEND,DX,YB,G e=e/12. FMS=2.*SQRT(FM*FM+1.) SG=SQRT(G) SQG=SQRT(32.*G) QV4=4.*Q/VISC C=SQG*ALOG10(e/(((B+FM*YB)*YB)/(B+FMS*YB))) Q2G=Q*Q/G N=ABS(XENDXBEG)/ABS(DX) Y(1)=YB FMOM=(.5*B+FM/3.*YB)*YB*YB+Q2G/((B+FM*YB)*YB) WRITE(IOUT,100) XBEG,Y,FMOM 100 FORMAT(F10.1,F10.3,F10.2) DXS=.1 X1=XBEG DO 10 I=1,N X2=X1+DX CALL RUKUST(1,DXS,X1,X2,1.E5,Y,YTT) FMOM=(.5*B+FM/3.*Y(1))*Y(1)**2+Q2G/((B+FM*Y(1))*Y(1)) WRITE(IOUT,100) X2,Y,FMOM 10 X1=X2 END SUBROUTINE SLOPE(X,Y,DYX) REAL Y(1),DYX(1) COMMON B,FM,e,SO,Q,Q2G,FMS,SQG,QV4,SG,C A=(B+FM*Y(1))*Y(1) P=B+FMS*Y(1) Rh=A/P 1 C1=C C=SQG*ALOG10(E/Rh+.884*C/(SG*QV4/P)) IF(ABS(CC1).GT. 1.E5) GO TO 1 SF=(Q/(C*A))**2/Rh DYX(1)=(SOSF)/(1.Q2G*(B+2.*FM*Y(1))/A**3) RETURN END Inputs to solve upstream and downstream GVFs: 3 20 2.5 1 .012 .0012 1.003e−6 0 100 1.5 9.81 and 3 20 2.5 1 .012 .0012 1.003e−6 100 12 3 9.81
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Open Channel Flow: Numerical Methods and Computer Applications
Solution outputs: x (m)
Y (m)
M (m3)
.0 2.0 · 20.0 22.0
1.500 1.399 · 1.213 1.202
10.73 10.83 · 11.49 11.55
24.0 26.0 · 98.0 100.0 100.0 98.0 · 26.0 24.0 22.0
1.193 1.184 · 1.040 1.039 3.000 2.974 · 1.946 1.908 1.869
11.60 11.65 · 12.80 12.82 22.72 22.33 · 11.90 11.72 11.54
20.0 · 12.0
1.827 · 1.571
11.36 · 10.7
Mathcad: This problem is solved below using Mathcad’s fourthorder fixedstep size Runge–Kutta numerical differential equation solver. Note, it also does not provide a close solution near the critical depth. In other words, in order to obtain a better solution using a fixedstep size algorithm smaller increments of x are required to be used for the computations of the S2 GVF profile. You should modify the Mathcad solution to accomplish this. EPRB4_6.MCD FQN := n ⋅
m:= 1 So:=.012 Q:= 20 g:= 9.81 n:=.014 Cu:= 1. Y0 := 1.5 P(Y):= b + 2⋅ Y⋅ m 2 + 1
b: = 2.5 A(Y): = (b + m ⋅ Y)⋅ Y D(x, Y): =
.666666 67 P(Yo) So − FQN ⋅ A(Yo) A(Yo) 1− Q ⋅ 2
Q Cu
b + 2 ⋅ m ⋅ Yo g ⋅ A Yo
3
z: = rkfixed(Y, 0, 100, 50, D) 2 m Q2 m1i: = ⋅5 ⋅ b + ⋅(z< 1 >)i ⋅ (z< 1 >)1 + 3 g ⋅ A (z< 1 >)i
2
i: = 0..50
291
Nonuniform Flows
Z =
Y0: = 3
0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28
1 1.5 1.301 1.281 1.264 1.249 1.235 1.223 1.212 1.202 1.192 1.184 1.176 1.168 1.161 1.154
m1 =
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 10.733 11.097 11.171 11.241 11.308 11.372 11.433 11.492 11.549 11.603 11.655 11.706 11.755 11.802 11.847
z1:= kfixed(Y,100,0,50,D) j: = 0..50
2 m Q2 m2j: = .5⋅b + ⋅(z1< 1 >)j ⋅ (z1< 1 >)j + 3 g⋅ A (z1< 1 >)j
z1 =
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72
1 3 2.974 2.949 2.923 2.897 2.871 2.845 m2 = 2.819 2.793 2.766 2.74 2.714 2.687 2.66 2.633
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 22.721 22.333 21.951 21.574 21.202 20.835 20.474 z1 = 20.118 19.767 19.422 19.082 18.747 18.417 18.092 17.773
36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
0 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0
1 1.98 1.944 1.906 1.866 1.823 1.777 1.725 1.662 1.556 1.536 1.39 1.464 1.361 1.405 1.566
m2 =
36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
0 12.085 11.891 11.704 11.523 11.349 11.183 11.025 10.877 10.743 10.734 10.85 10.751 10.916 10.824 10.749
TKSolver: This problem has also been solved below, using the TKSolver. Its variable, rule, function, and output table sheets are shown. Solution for the S2 GVF profile ════════════════════ VARIABLE SHEET ═════════════════════ St Input──── Name─── Output─── Unit───── Comment───────── 1.5 y0 L 'yyy Y .014 n 2.5 b 1 m 1 C 20 Q .012 So 9.81 g
292
Open Channel Flow: Numerical Methods and Computer Applications
════════════════════ RULE SHEET ════════════════════════ S Rule────────────────────────────────────────────────── place(Y,1) = y0 call RK4_se('DYDx,Y,'x) The call RK4_se is a library in the TKSolver package (that uses a fixedstep size) and it requires that the user supply another function that defines the ODE that is being solved, which is given as FUNCTION DYDx below. ════════════════ PROCEDURE FUNCTION: RK4_se (DIFFEQ\RungeKutta) lassical FourthOrder RungeKutta method, Comment: C single eqn Parameter Variables: Input Variables: EQ,y,x Output Variables: S Statement─────────────────────────────────────────────────────── ; Notation: EQ name of a function with the 1storder equation y'=f(x,y) ; x independent variable (list) ; y dependent variable, y=F(x), (list) ; K RungeKutta coefficients (list) ; Description: This procedure represents an implementation of a ; classical 4thorder RungeKutta procedure for numerical ; integration of a single ordinary differential equation ; y'=f(x,y). Given a function name passed as a symbolic ; value of the Input Variable EQ, list of values of the ; independent variable x, and an initial condition as the value of ; the 1st element of the list y, the procedure generates the ; solution in the rest of the list y. xi:= x[1] yi:= y[1] for i=2 to length(x) ye:= yi h:= (x[i]xi)/2 for j=1 to 3 'K[j]:= apply(EQ,xi,ye) if mod(j,2) then xi:= xi + h if j=3 then h:= 2*h ye:= yi + h*'K[j] next j 'K[4]:= apply(EQ,xi,ye) yi:= yi + dot('K,1,2,2,1)*h/6 y[i]:= yi next i call delete('K) ════════════════════ PROCEDURE FUNCTION: DYDx ═══════════ Comment: Given differential equation Parameter Variables: n,b,m,C,Q,So,g Input Variables: x,y Output Variables: y'
Nonuniform Flows
293
S Statement────────────────────────────────────────────── A:=(b+m*y)*y NUM:= So(n*(b+2*sqrt(m*m+1)*y)^.66666667*Q/(C*A^1.666667))^2 DEN:=1Q*Q*(b+2*m*y)/(g*A^3) y':=NUM/DEN To solve the problem, the following list sheet was created and the x sheet filled in with values starting with 0 and running through 100 in increments of 2 m. ════════════════════ LIST SHEET ════════════════════════ Name───── Elements── Unit───── Comment────────────────── x 51 independent variable yyy 51 dependent variable ames of dependent variables, Y 1 n set of solutions In order to obtain the values of the momentum function, the rules that solve the ODE were taken out, and the additional rule that computes the momentum function added, as shown below, ════════════════════ RULE SHEET ════════════════════════ S Rule────────────────────────────────────────────────── C place(Y,1) = y0 C call RK4_se('DYDx,Y,'x) M=(.5*b+m/3*YY)*YY^2+Q^2/(g*(b+m*YY)*YY) and the solution for the depth copied into the list variable YY before pressing L10 for a list solve. ════════════════════ VARIABLE SHEET ════════════════════ St Input──── Name─── Output─── Unit───── Comment──────── 1.5 y0 L 'yyy Y .014 n 2.5 b 1 m 1 C 20 Q .012 So 9.81 g L 3 YY LG 10 M ════════════════════ TABLE: Solution ═══════════════════ Title: y = f(x) Element x───────── y───────── M───────── ─────── ─────── 1 0 1.5 10.7332866 2 2 1.30055004 11.0968582 3 4 1.28091077 11.1707535 . . . . 10 18 1.19248127 11.6029712 11 20 1.18370612 11.6553916 12 22 1.17552185 11.7059391
294
13 14 .
Open Channel Flow: Numerical Methods and Computer Applications
24 26 .
1.16786187 11.7547211 1.16067065 11.8018339 . .
Jump
Following the same procedure, except using a value of y0 = 3, and filling in the list table for x starting with 100 and moving back toward 0 results in the following solution for the S1 gradually varied profile beyond the hydraulic jump. The momentum functions are equal at a position x between 22 and 24 m according to these solutions, which agree closely with the above solution at 22 m, despite the fact that the fixed step algorithm does not provide a good solution close to critical, but since the depth varies rapidly over a small distance here, the computed position of the jump, from a practical viewpoint, is not so very different. ════════════════════ TABLE: Solution ═══════════════════ Title: y = f(x) Element x───────── y───────── M───────── 1 100 3 22.7211951 2 98 2.9743262 22.3333936 3 96 2.94859016 21.9509297 . . . . 37 28 1.98014744 12.0854955 38 26 1.94377088 11.8913888 39 24 1.905841 11.7037361 Jump 40 22 1.86593079 11.5228252 41 20 1.82338687 11.3490387 In summary, one should note that an ODE solver can be used to solve dx/dY (or dx/dE) in which x is the dependent variable, as well as dY/dx, in which Y is the dependent variable. However, a numerical integrator can only be used to solve the first form of the ODE in which x is the dependent variable. A numerical integrator requires that the derivatives be only a function of the independent variable. An ODE solver allows the derivative to be a function of both the dependent and independent variables, or just one of these.
4.5 Canal Systems Most of the previous considerations have dealt with an individual channel. As the slope of the land changes, channels are built with changing bottom slopes. To keep the costs near a minimum, the size of the channel will be reduced where the bottom slope increases, and vice versa. With a subcritical flow in such systems, it is easy for a transition from a larger to a smaller channel to cause an M1 GVF profile even if the normal depth in the smaller channel is less than the normal depth in the larger channel, unless the position of the bottom is changed. This choking effect is very likely to occur especially if the smaller channel is steep, and the critical flow occurs at its beginning. Gates are used to control the flow rates and the water depths in canal systems. The effects of these gates must extend to the supply reservoirs of the system if they are to control the flow rates, as well as the depths. To determine if a gate does control the flow into the channel, and how much it will reduce the flow rate depending upon its setting requires that the GVF profile be solved from the gate up to the supply reservoir through whatever the canal system between the two consists of. The actual solution to such problems involving downstream gate control are governed by the ODE for the GVF, plus the algebraic equations that define energy, critical flow, and possibly momentum, depending upon what is involved in the given situation. The only means available for solving problems defined by simultaneous ODEs and algebraic equations is by trialanderror or iterative methods such as the Newton method. Thus, canal systems with control gates need to be solved by (1) guessing the flow
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rate into the system, (2) solving the algebraic equations involved in the problem, (3) solving the GVF profiles and (4) upon completing these solutions, verifying whether the guessed flow rate satisfies all conditions. If not, adjust the assumed flow rate and repeat the above steps until all conditions are satisfied. In the next section, use of the Newton method will be discussed to automate the solution. Example Problem 4.7 Determine the depth, etc., through a channel system consisting of an upstream steep channel followed by a mild channel, and this channel followed by another mild channel but with a larger slope than the middle channel. Channel # 1 and Channel # 3 are very long and the middle channel has a length of L2 = 1000 m. The flow rate is Q = 50 m3/s, and all channels have a Manning’s n = 0.014. Consider (a) the case where all three channels are trapezoidal and have the same size, i.e., b = 5 m, and a side slope m = 1.2, and (b) the case where the channels have different sizes, i.e., b1 = 5 m (rectangular), b2 = 5 m, m2 = 1.5, and b3 = 5 m, m3 = 1.2. The middle and the downstream channel have bottom slopes of So2 = 0.0001 and So3 = 0.001. Consider two different slopes for the upstream channel, So1 = 0.09 and So1 = 0.01. Solution The solutions are shown on the sketches below. First, the normal depths are computed for all channels for all cases. Since the downstream long channel will cause the depth at its beginning to always be at its normal depth, an M2 GVF will exist in the downstream portion of the middle channel. Therefore, the next step is to solve this M2 GVF profile. Should the momentum function M2 associated with the depth Y2 thus computed at the beginning of the middle channel be less than Mo1, then the hydraulic jump will occur in the middle channel, and its location is determined where the momentum functions from the M3 and M2 GVFs are equal. Should M2 > Mo1, then an S1 GVF will exist in Channel # 1. When So1 = 0.09, the jump occurs downstream for both the constant size channel and the one with transitions. When So1 = 0.01, the jump occurs upstream in the steep channel since M2 = 40.2 > Mo1 = 37.6 m3. However, when there are transitions in the channel, the specific energy E2 = 2.935 m associated with Y2 = 2.774 m is less than the critical specific energy in the upstream rectangular channel. Thus, even though M2 = 42.81 > Mo1 = 40.03 m3, the jump cannot occur upstream from the transition, and therefore must be within the transition.
Long
M2 GVF
Yo1
Yo1 = 0.628 m Eo1 = 10.389 m Mo1 = 71.62 m3 Fro1 = 5.929
Yo2 Yd = 2.8 5m M3 GVF
So1 = 0.0
9
Yu = 1.10 m
b = 5 m, mSo2 = 0.0001, L2 = 1000 m = 1,2, n=0.01 4 Yo2 = 4.085 m Eo2 = 4.162 m Mo2 = 75.27 m3 Fro2 = 0.239
Yo3 Long
Q = 50 m3/s So3 = 0.001
Yo3 = 2.252 m Eo3 = 2.676 m Mo3 = 31.94 m3 Fro3 = 0.713
Solve ODE for M2 – > Y2 = 2.919 m, M2 = 41.51 m3 Jump occurs in middle channel since Mo1 > M2 Maching Mu = Md from solutions of ODE for M3 and M2 GVF gives x = 180 m, with Yu = 1.10 m and Yd = 2.85 m (M = 40.2 m3) M3—GVF solution x (m) Y (m) E (m)
M (m3)
M2—GVF solution Y (m) E (m) x (m)
M (m3)
.628 .680 . .991 1.044 1.097 1.152 1.209
71.615 65.683 . 44.397 42.228 40.299 38.578 37.039
1000.0 980.0 . 220.0 200.0 180.0 160.0 140.0
31.939 32.245 . 39.927 40.078 40.227 40.376 40.522
.0 20.0 . 140.0 160.0 180.0 200.0 220.0
10.388 8.817 . 4.379 4.036 3.749 3.507 3.304
Jump
2.252 2.287 . 2.836 2.844 2.852 2.860 2.868
2.675 2.693 . 3.060 3.067 3.073 3.079 3.085
Jump
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Open Channel Flow: Numerical Methods and Computer Applications
S1 GVF
Yo1
Yd =
So1 =
Yo1 = 1.189 m Eo1 = 3.371 m Mo1 = 37.56 m3 Fro1 = 2.118
M2 GVF
Y2 = 2.919 m
2.70
Yo2
0m
0.01
Q = 50 m3/s
So2 = 0.0001, L2 = 1000 m b = 5 m, m = 1.2, n = 0 .014 Y = 4.085 m o2
Eo2 = 4.162 m Mo2 = 75.27 m3 Fro2 = 0.239
Yo3 Long
So3 = 0.001 Y = 2.252 m o3 Eo3 = 2.676 m Mo3 = 31.94 m3 Fro3 = 0.713
Now equate Md = Mo1 = 37.56 and solve Yd = 2.700 m Solve dx/dY = (1–F2r )/(So – Sf) between Yd and Y2 = 2.919 m to get x = –17.6 m
Long
M2 GVF
Yo1
Yo1 = 0.628 m Eo1 = 10.389 m Mo1 = 71.62 m3 Fro1 = 5.929
Yo2 Yd = 2.8 5m M3 GVF
So1 = 0.0
9 b = 5 m, m
Yu = 1.10 m
So2 = 0.0001, L2 = 1000 m = 1.2, n = 0. 014 Yo2 = 4.085 m Eo2 = 4.162 m Mo2 = 75.27 m3 Fro2 = 0.239
Yo3 Long
Q = 50 m3/s So3 =0.001
Yo3 = 2.252 m Eo3 = 2.676 m Mo3 = 31.94 m3 Fro3 = 0.713
Solve ODE for M2 –> Y2 = 2.919 m, M2 = 41.51 m3 Jump occurs in middle channel since Mo1 > M2 Matching Mu = Md from solutions of ODE for M3 and M2 GVF gives x = 180 m, with Yu = 1.10 m and Yd = 2.85 m (M = 40.2 m3) M3—GVF solution x (m) Y (m) E (m)
M (m3)
M2—GVF solution Y (m) E (m) x (m)
M (m3)
.628 .680 . .991 1.044 1.097 1.152 1.209
71.615 65.683 . 44.397 42.228 40.299 38.578 37.039
1000.0 980.0 . 220.0 200.0 180.0 160.0 140.0
31.939 32.245 . 39.927 40.076 40.227 40.376 40.522
.0 20.0 . 140.0 160.0 180.0 200.0 220.0
10.388 8.817 . 4.379 4.036 3.749 3.507 3.304
Jump
S1 GVF
Yo1
Long
Yd =
So1 =
Yo1 = 1.189 m Eo1 = 3.371 m Mo1 = 37.56 m3 Fro1 = 2.118
0.01
2.675 2.693 . 3.060 3.067 3.073 3.079 3.085
Jump
M2 GVF
Y2 = 2.919 m
2.70
2.252 2.287 . 2.836 2.844 2.852 2.860 2.868
Yo2
0m
Q = 50 m3/s
So2 = 0.0001, L2 = 1000 m b = 5 m, m = 1.2, n = 0.014 Y = 4.085 m o2
Eo2 = 4.162 m Mo2 = 75.27 m3 Fro2 = 0.239
Now equate Md = Mo1 = 37.56 and solve Yd = 2.700 m Solve dx/dy = (1 – Fr2)/(So – Sf) between Yd and Y2 = 2.919 m to get x = –17.6 m
Yo3 Long
So3 = 0.001 Y = 2.252 m o3 Eo3 = 2.676 m Mo3 = 31.94 m3 Fro3 = 0.713
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Nonuniform Flows Example Problem 4.8 A gate exists 500 m downstream from a supply reservoir. The channel between the reservoir and the gate has a bottom slope So = 0.0009, a bottom width b = 3 m, a Manning’s n = 0.013, and a side slope m = 2. A short transition with a minor loss coefficient of K L = 0.09 changes the channel to a rectangular shape at the gate with a bottom width of b = 2.5 m. The water elevation in the upstream reservoir is 2 m above the bottom of the channel, and the entrance loss coefficient is Ke = 0.12. The contraction coefficient for the gate is constant with its setting above the channel bottom and equals Cc = 0.75. Determine the discharge through the canal system as a function of the gate setting under the assumption that free flow occurs downstream of the gate, i.e., obtain the discharge for several gate settings allowing this information to be plotted to give a discharge curve of Q versus YG. 500 m
H=2 m
.12 b = 3 m, m = 2, S = 0.0009, n = 0.013 1 1 o
=0
Ke
Tran siti KL = on 0.09
Yg= 2
m, C c
= 0.75
b2 = 2.5 m
Solution The solution might begin by solving Manning’s equation simultaneously with the energy equation at the upstream reservoir to determine what the uniform depth and discharge would be. Using this flow rate, the critical depth in the reduced channel at the gate can be solved to provide some guidance about what gate setting to start the requested series of solutions with. These solutions provide Yo = 1.688 m, Qo = 25.06 m3/s, and Yc2 = 2.168 m, with the corresponding specific energy at the end of the trapezoidal channel E2 = 3.25 m. Note that this specific energy is well above the specific energy supplied by the reservoir; therefore an M1 GVF profile will exist between the reservoir and the gate. The solution will be obtained by writing a FORTRAN program that calls on the ODE solver ODESOL described in Appendix C. This program, whose listing is shown below, is designed to determine the depth of flow downstream from the gate by multiplying the gate setting by the contraction coefficient, and then computing the specific energy upstream from the gate. It reads in the necessary parameters to define the problem, as well as the gate setting for which solutions are to be obtained. The specific energy establishes the depths immediately upstream from the gate, as well as at the end of the trapezoidal channel before the transition reduces its size by using the losses coefficient of 0.09. The program then obtains the solution of the GVF profile to the reservoir and displays a prompt indicating what the depth and the specific energy are at the entrance of the channel with a request that a new guessed flow rate be supplied. If the previous given flow rate is correct, the user supplies a negative flow rate value, the absolute value of which will be used for the next gate setting. This program solves the ODE dE/dx = So − Sf, and as such gives an example of an alternative solution methodology. The equation is simpler than the one giving dY/dx, but involves more arithmetic on the part of the computer because Sf is a function of the depth Y, requiring that the specific energy equation E = Y + Q2/(2gA2) be solved for Y every time the ODE solver requests that dE/dx be evaluated. For natural channels, there is some advantage in using the ODE in the form of dE/dx, since it automatically takes care of changes in the crosssectional area without the need for defining a term that involves ∂A/∂x. More information related to natural channels is included in Chapter 5. (You should solve this problem with one of the previously given programs that solves dY/dx = (So − Sf )/(1 − Fr2).) The gate setting flow rate relationship obtained from this solution process is given in the table below. It should be noted from this solution, that with the gate set at 2 m above the channel bottom, which is only modestly below the critical depth in the rectangular channel here, the gate
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Open Channel Flow: Numerical Methods and Computer Applications has reduced the flow rate about 10 m3/s over what would occur in the trapezoidal channel under uniform flow conditions. You should duplicate this solution for the experience gained in guessing appropriate flow rates. Solution to Illustrative Problem 45 Gate (m) 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2
Flow Rate (m3/s) 14.93 14.57 13.90 12.95 11.73 10.28 8.61 6.73 4.66 2.41
Depth at Beg. (m) 1.93 1.93 1.94 1.95 1.96 1.97 1.98 1.99 1.99 1.99
FORTRAN Listing of Program EPRB4_8.FOR to Assist in Solving Illustrate Example Problem 4.8 CHARACTER*2 UNIT LOGICAL SWITCH REAL E(1),EPRIME(1),XP(1),YP(1,1),WK1(1,13),YG(20),KE1,KE2 EXTERNAL DEX COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRANS/QN,B1,FM1,B,FM,FN,Y,Q2G,SO,X1,DB,DFM,XBEG,SWITCH DATA SWITCH/.FALSE./ WRITE(6,*)=GIVE: UNIT,IOUT,TOL,DELX,YB,Q,FN,SO,B1,FM1,''B2, &FM2,XBEG,XEND,H,CV,KE1,KE2' READ(5,*)UNIT,IOUT,TOL,DELX,YB,Q,FN,SO,B1,FM1,B2,FM2,XBEG, &XEND,H,CV,KE1,KE2 WRITE(6,*)' Give N & gate positions' READ(5,*) NSET,(YG(I),I=1,NSET) H1=.1 IF(DELX.GT.0.) H1=ABS(H1) CC=1. G2=19.62 IF(UNIT .EQ. 'SI') GO TO 10 CC=1.486 G2=64.4 10 DO 70 I=1,NSET YY=CV*YG(I) Y=YB 12 QN=(FN*Q/CC)**2 Q2G=Q*Q/G2 E(1)=YY+(1.+KE2)*Q2G/(((B2+FM2*YY)*YY)**2) B=B1 FM=FM1 X=XBEG WRITE(IOUT,15) YG(I),Q,SO,FN,B1,FM1,B2,FM2 15 FORMAT(/,' SOLUTION TO GRADUALLY VARIED=, = FLOW WITH'' &Ygate='F8.2,/,' Q=',F10.2,' SO=',F10.6,n=',F7.4,/,'B1=', &F8.1,' m1=',F8.2,/, B2=',F8.1,' m2=',F8.2,/1X,30(''),/, &3X, DIST. DEPTH E',/1X,30('')) CALL YSOL(E,Y)
Nonuniform Flows
20
40
80
70 100
10 20 30
40
50
WRITE(IOUT,40) X,Y,E XZ=X+DELX CALL ODESOL(E,EPRIME,1,X,XZ,TOL,H1,HMIN,1,XP,YP,WK1,DEX) X=XZ WRITE(IOUT,40) X,Y,E FORMAT(1X,3F10.3) IF(X.GT.XEND) GO TO 20 EE=Y+(1.+KE1)*Q2G/(((B+FM*Y)*Y)**2) WRITE(*,80) Y,H,EE,Q FORMAT(' Y,H,E,Q',4F12.3,/,' Give new Q (neg O.K.)') READ(5,*) QQ IF(QQ.GT.0.) THEN Q=QQ GO TO 12 ENDIF WRITE(4,100) YG(I),Q,Y Q=ABS(QQ) FORMAT(3F12.4) STOP END SUBROUTINE DEX(X,E,EPRIME) LOGICAL SWITCH REAL E(1),EPRIME(1) COMMON /TRANS/QN,B1,FM1,B,FM,FN,Y,Q2G,SO,X1,DB,DFM,XBEG,SWITCH CALL YSOL(E,X) P=B+2.*SQRT(FM*FM+1.)*Y A=(B+FM*Y)*Y SF=QN*((P/A)**.6666667/A)**2 EPRIME(1)=SOSF RETURN END SUBROUTINE YSOL(E,X) REAL E(1) LOGICAL SWITCH COMMON /TRANS/QN,B1,FM1,B,FM,FN,Y,Q2G,SO,X1,DB,DFM, &XBEG,SWITCH IF(Y.LT.E(1).AND.Y.GT..6*E(1)) GO TO 10 Y=.9*E(1) NCT=0 NT=0 IF(SWITCH) GO TO 40 XX=1.(XX1)/(XBEGX1) B=B1+DB*XX FM=FM1+DFM*XX A=(B+FM*Y)*Y F=E(1)YQ2G/A**2 IF(NT.GT.0) GO TO 50 F1=F Y=Y.001 NT=1 GO TO 30 Y=Y+.001 DIF=.001*F1/(F1F) NCT=NCT+1 Y=YDIF IF(NCT.LT.15.AND.ABS(DIF).GT. .00001) GO TO 20
299
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Open Channel Flow: Numerical Methods and Computer Applications IF(NCT.EQ.15) WRITE(6,*)' DID NOT CONVERGE' RETURN END
Input to start solution: 'SI' 3 .000001 −50 2.1 16 .013 .0009 3 2 2.5 0 0 2 .75 .12 .009 10 2 1.8 1.6 1.4 1.2 1. .8 .6 .4 .2
500
and two file names, one for the data giving the individual profiles, and one for the final solution results given in the previous table. The input not shown above consists of a new guess for the correct flow rate Q, so that the two values of specific energy displayed by the program on the monitor (the second and third values) are equal (or nearly so.) When they are equal, the next flow rate given is preceded by a minus sign and is the first guess for the flow rate that will be used in solving the problem with the next gate position read into the array YG(I). The C program listed below requires the same input, but calls for both file names before the input for the gate settings, whereas the FORTRAN program utilizes the MSFORTRAN extension of standard FORTRAN 77 to promote a file name when a write first occurs using a logical unit that has not been attached to a file that has been open with the OPEN statement. Listing of Program EPRB4_8.C Designed to Solve Example Problem 4.8: #include <stdio.h> #include <stdlib.h> #include <math.h> int swith=0; float qn,b1,fm1,b,fm,fn,y,q2g,so,x1,db,dfm,xbeg; extern void rukust(int neq,float *dxs,float xbeg,float xend,\ float error,float *y,float *ytt); void ysol(float *e, float x){int nct,nt; float xx,a,f,f1,dif; if((y>e[0])  (y<.6*e[0])) y=.9*e[0]; nct=0; L20: nt=0; L30: if(!swith){xx=1.(xx1)/(xbegx1);b=b1+db*xx;fm=fm1+dfm*xx;} a=(b+fm*y)*y;f=e[0]yq2g/(a*a); if(nt==0){f1=f;y=.001,nt=1; goto L30;} y+=.001; dif=.001*f1/(f1f);nct++; y=dif; if((fabs(dif)>.00001)&&(nct<15)) goto L20; if(nct==15) printf("DID NOT CONVERGE\n");} //End of ysol void slope(float x,float *e,float *eprime){float p,a,sf; ysol(e,x); p=b+2.*sqrt(fm*fm+1.)*y; a=(b+fm*y)*y; sf=qn*pow(pow(p/a,.6666667)/a,2.); eprime[0]=sosf;} // End of slope void main(void){char unit[2],fname[20]; FILE *out,*ou1; int i,j,iout,nset; float tol,delx,yb,q,b2,fm2,xend,h,cv,ke1,ke2,yg[20],x,xz,ee,\ qq,*h1,cc,g2,yy,e[1],ew[1]; printf("GIVE:UNIT,IOUT,TOL,DELX,YB,Q,FN,SO,B1,FM1,B2,FM2,\ XBEG, XEND,H,CV,KE1,KE2\n"); scanf("%s %d %f %f %f %f %f %f %f %f %f %f %f %f %f %f %f %f",\ &unit,&iout,&tol,&delx,&yb,&q,&fn,&so,&b1,&fm1,&b2,&fm2,\ &xbeg,&xend,&h,&cv,&ke1,&ke2); if(iout!=6){printf("Give output file name\n"); scanf("%s",fname);out=fopen(fname,"wt");} printf("Give file for summary sol.\n");
Nonuniform Flows
301
scanf("%s",fname);ou1=fopen(fname,"wt"); printf("Give: N & gate positions\n"); scanf("%d",&nset); for(i=0;i<nset;i++)scanf("%f",&yg[i]); *h1=.1; if(delx>0.) *h1=fabs(*h1); cc=1.; g2=19.62; if((unit=="es")(unit=="ES")){cc=1.486;g2=64.4;} for(i=0;i<nset;i++){yy=cv*yg[i]; y=yb; L12:qn=pow(fn*q/cc,2.); q2g=q*q/g2; e[0]=yy+(1.+ke2)*q2g/pow((b2+fm2*yy)*yy,2.); b=b1; fm=fm1;x=xbeg; if(iout==6){ printf("nSOLUTION TO GRADUALLY VARIED FLOW WITH Yg=\ %8.2f\nQ=%10.2f, So=%10.3f, n=%7.4f\n B1=%8.1f,m1=%8.2f\ \n B2=%8.1f, m2=%8.2f\n",yg[i],q,so,fn,b1,fm1,b2,fm2); for(j=0;j<31;j++)printf("");printf("\n"); printf(" DIST. DEPTH E\n"); for(j=0;j<31;j++)printf(""); printf("\n");} else {fprintf(out,"nSOLUTION TO GRADUALLY VARIED FLOW WITH \ Yg=%8.2f\nQ=%10.2f, So=%10.3f, n=%7.4f\n B1=%8.1f,\ m1=%8.2f\n\B2=%8.1f, m2=%8.2f\n",yg[i],q,so,fn,b1,fm1,b2,fm2); for(j=0;j<31;j++)fprintf(out,"");fprintf(out,"\n"); fprintf(out," DIST. DEPTH E\n"); for(j=0;j<31;j++)fprintf(out,"");fprintf(out,"\n");} ysol(e,x); if(iout==6)printf(" %10.3f %9.3f %9.3f\n",x,y,e[0]); else fprintf(out," %10.3f %9.3f %9.3f\n",x,y,e[0]); do{ xz=x+delx; rukust(1,h1,x,xz,tol,e,ew); x=xz; if(iout==6) printf(" %10.3f %9.3f %9.3f\n",x,y,e[0]); else fprintf(out," %10.3f %9.3f %9.3f\n",x,y,e[0]); } while(x>xend); ee=y+(1.+ke1)*q2g/pow((b+fm*y)*y,2.); printf("Y,H,E,Q\n%12.3f %11.3f %11.3f %11.3f\n",y,h,ee,q); printf("Give new Q (neg. O.K.)\n"); scanf("%f",&qq); if(qq>0.){q=qq;goto L12;} fprintf(ou1,"%12.4f %11.3f %11.3f\n",yg[i],q,y); q=fabs(qq);} // end of for(i fclose(ou1);if(iout!=6)fclose(out);} Note that the C program calls on the ODE solver RUKUST, whereas the FORTRAN program calls on the ODESOL. It will be a worthwhile exercise for you to modify the program in the language you are most familiar with to call on the other solver.
4.6 Simultaneous Solution of Algebraic and Ordinary Differential Equations We will now consider methods that will allow algebraic and ordinary differential equations to be solved simultaneously. Implementation of these methods in computer code will remove the burden of repeatedly trying values of flow rate, as in Example Problem 4.8, until all equations are satisfied. Instead, the solution can be turned over entirely to the computer. It will be necessary to use
302
Open Channel Flow: Numerical Methods and Computer Applications
some type of iterative process. The Newton method is a good candidate for obtaining this iterative solution since it converges rapidly. As a first step in solving equations by the Newton method, all equations are written as functions of the unknowns equal to zero. Each ODE might be considered a function of unknowns. When the correct combination of these unknowns are used, then each ODE will produce, as its solution, the correct value of its dependent variable. This variable is one of the variables in the algebraic equations. For example, if the variable that the ODE is solving is the depth Y1 at the beginning of the channel, then let Y1 be the variable used in the algebraic equations, and let Yode1 be the solution of the upstream depth produced by the solution of the ODE. This solution will be based on a downstream starting depth Y2 and the other variables that define the GVF problem. Then the equation associated with the ODE for use in the Newton method can be defined as Fi = Y1 − Yode1 = 0. This equation will not be satisfied, e.g., equal zero until the solution of the ODE equals the depth Y1, the value of which is used in the other equations. In other words, as is the case with an algebraic equation, Fi = Y1 − Yode1 will not be zero, until the correct solution vector of unknowns is used in evaluating it. If you have difficulties in comprehending how Fi = Y1 − Yode1 = 0, that comes from an ODE can be included with algebraic equations to form a system of simultaneous equations, consider a simple ODE (with an appropriate boundary, or initial condition) for which a closedform solution can be obtained. This closedform solution is an algebraic equation that can be written as a function of the unknown variables equal to zero, i.e., F(x1, x2, …. xn) = 0, in which the x’s are the unknown variables. Since in general, only numerical solutions and not closedform solutions to ODEs are available,, and since our ODE has a beginning depth that is known at one end of the channel, and the solution produces the depth at the other end of the channel, a convenient way of expressing a function that must equal zero is to subtract the numerical solution of the ODE from the depth that is being sought, as one of the unknown variables. The implementation of the Newton method in solving combined algebraic equations and ODEs can be illustrated by taking a general example, such as Example Problem 4.8, in which the water is supplied from a reservoir with a known head H at the upstream end of the channel, and the flow is controlled by a gate at the downstream end. The equations that define this problem are
F1 = H − Y1 − (1 + K e )
Q2 = 0 (Energy applied between reservoir annd channel) 2gA12
F2 = Y3 +
Q2 Q2 − Yd + = 0 (Energy across downstream gate) 2 2gA 3 2gA 2d
F3 = Y2 +
Q2 Q2 − Y3 − (1 + K L ) =0 2 2gA 2 2gA 32
(Energy across transition upstream from gate)
F4 = Y1 − Yode1 = 0 (From solving an ODE problem from the beginninng of the transition to the beginning of a channel)
In these equations, the subscripts of the areas, Ai, correspond to the subscripts of the depths. Subscript d denotes immediately downstream from the gate; subscript 3 denotes immediately upstream from the gate and at the end of the transition; subscript 2 denotes the beginning of the transition; and subscript 1 denotes the beginning of the channel. Yd is known to be equal to the height of the gate above the channel bottom times the contraction coefficient, or Yd = CcYG, and therefore the unknowns are: the flow rate Q, the depth at the beginning of the channel Y1, the depth at the beginning of the transition Y2, and the depth at the end of the transition Y3. The solution of the above four simultaneous equations provides the solution to these four unknowns.
303
Nonuniform Flows
The implementation of the Newton method in solving these equations involves the iterative equation:
{Y}( m + 1) = {Y}( m ) − {z}
in which {z} is the unknown vector in the linear system of equations, [D]{z} = {F} in which [D] is the Jacobian matrix and {F} is the equation vector represented by the above four equations, e.g., its elements are obtained by evaluating these equations with the current values of the unknown vector {Y}(m). ({Y}T = [Q, Y1, Y2, Y3]). The Jacobian D consists of the following derivative matrix:
∂F1 ∂Q ∂F2 ∂Q D= ∂F3 ∂Q ∂F4 ∂Q
∂F1 ∂Y1 ∂F2 ∂Y1 ∂F3 ∂Y1 ∂F4 ∂Y1
∂F1 ∂Y2 ∂F2 ∂Y2 ∂F3 ∂Y1 ∂F4 ∂Y1
∂F1 ∂Y3 ∂F2 ∂Y3 ∂F3 ∂Y3 ∂F4 ∂Y3
It is not possible to algebraically determine the derivatives of the fourth equation F4, but these can be evaluated numerically. For example,
∂F4 1 ≈ {F4 (Q + ∆Q, Y1, Y2 , Y3 ) − F4 (Q, Y1, Y2 , Y3 )} ∂Q ∆Q
Likewise, derivatives with respect to Y1, Y2, and Y3 are determined by evaluating the equation twice, once with their values incremented and once without their values incremented, taking the difference between these two values of the equation and then dividing this difference by the increment. If xi denotes the unknowns, then the general equation to evaluate any element (with subscript j, i.e., in column j and row i) in the Jacobian is
∂Fi Fi (x1, x 2 ,…, x j + ∆x j , x j+1,…, x n ) − Fi (x1, x 2 ,…, x j , x j+1,…, x n ) = ∂x j ∆x j
The evaluation of each element of the last row of the above Jacobian D entails solving the ODE that describes the GVF profile twice, once with the flow rate, or the depths incremented, and once without this increment. Consider as another example, the problem involving an M1GVF profile upstream from a control due to a critical depth that exists at a distance x = L downstream from the channel’s entrance. The depth at this control is the beginning value of the dependent variable Y2, and the solution YGVF at the entrance where x = 0 is the solution produced upon solving the M1. Then the function Fi might be defined by
Fi = Y1 − YGVF = 0
in which Y1 is the current value of the depth at the entrance of the channel. To carry this example further, assume that the control consists of a short smooth transition to a steep rectangular channel. Then at the beginning of the rectangular channel, the flow will be critical and the critical depth will be defined by Yc =3 /(Q / b)2 / g , and if the minor loss due to the transition is defined by a loss
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Open Channel Flow: Numerical Methods and Computer Applications
coefficient K L that multiples the downstream velocity head, then the specific energy at the end of the upstream channel is given by E2 = (1.5 + K L/2)Yc. At the channel entrance, the energy equation applies. Thus, the following three equations are available: F1 = Y2 +
Q2 1 − 1.5 + K L 3 Q /b)2 / g = 0 2 2gA 2 2
F2 = H − Y1 − (1 + K e )
Q2 =0 2gA12
F3 = Y1 − YGVF = 0
The unknowns in these three equations are: Q, Y1 (depth at channel entrance), and Y2 (depth at the end of the upstream channel) Programming a computer to solve the combination of algebraic and ODE equations that govern Example Problem 4.8 is not difficult. While only derivatives of the equations involving the ODEs are evaluated numerically, the other equations might also be evaluated numerically. Means for doing this are implemented in the computer programs that solve the following problems. With the FUNCTION FUN(II) designed to provide the evaluation of equation II (see following listing of FUN) this code consists of 1
10
DO 10 I=1,N F(I)=FUN(I) DO 10 J=1,N DX=.005*X(J) X(J)=X(J)+DX D(I,J)=(FUN(I)F(I))/DX X(J)=X(J)DX
in which D hold the elements of the N by N Jacobian matrix and F is the equation vector.
4.7 Flow into a Mild Channel with a Downstream Control As a relatively simple example of solving combined algebraic and ODEs simultaneously, consider the flow into a reservoir feed channel that has a GVF in it due to a downstream effect. Such a situation results if a break in grade to a steep channel occurs a relatively short distance L downstream from the reservoir, as shown in the sketch below. Let us consider the case where the channel from the reservoir to the break in grade is trapezoidal, and the downstream channel is rectangular with a different width b2 than the upstream channel.
H = 7 ft
Y1
Q
05 L = 1000 ft, b1 = 12 ft, m1 = 1, n = 0 .02, So1 = 0.0009
0. K e=
Y2
YC 03
0. K L=
b2 Stee p
There are three variables that are to be solved: (1) the flow rate Q, (2) the depth at the upstream end of the channel Y1, and (3) the depth at the downstream end of the upstream channel Y2. One
305
Nonuniform Flows
might also consider the critical depth Yc at the beginning of the downstream rectangular channel an unknown, but since this channel is rectangular, the critical depth will be eliminated by using the explicit equations for the critical flow in rectangular channels. Thus, three equations are needed. Two of these are algebraic, the energy equation at the reservoir and the energy equation across the two channels, and the third is the ODE that describes the GVF in the channel. These equations are
F1 = H − Y1 − (1 + K e )
Q2 =0 2gA12 1
Q2 K ( Q /b ) 2 3 F2 = Y2 + − 1.5 + L =0 2 2gA 2 2 g
F3 = Y1 − Y1,ode (Y2 )
The notation used in the third equation has the following meanings: Y1,ode(Y2) is the depth at the beginning of the channel obtained by solving the ODE for a GVF starting with the depth Y2 at its downstream end. The depth thus obtained should match the depth Y1 that occurs in the other equations. The second equation equates the specific energy at the end of the trapezoidal channel to the critical specific energy in the rectangular channel, plus the local loss here. The local loss equals h L = K LVc2/(2g). Noting that Vc2/(2g) = Yc/2 and Yc = {q2/g}1/3 = {(Q/b}2/g}1/3 and that Ec = 1.5Yc, provides the last term in F2. The Newton method, as described above, can solve these three equations simultaneously. Program SOLGVF is designed to accomplish such solutions. Program SOLGVF.FOR C Solves problem of flow into a mild channel from a reservoir that C has a steep rectangular channel at a position L downstream C from the reservoir. REAL F(3),D(3,3),X(3),KL2,KE1,KL,KE COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN,Q2G,X EQUIVALENCE (Q,X(1)),(Y1,X(2)),(Y2,X(3)) WRITE(*,*)' GIVE:IOUT,TOL,ERR,FN,SO,B1,FM1,B2,H,L,g,KL,KE' READ(*,*) IOUT,TOL,ERR,FN,SO,B1,FM1,B2,H,FL,G,KL,KE IF(G.GT.30.) THEN CC=1.486 ELSE CC=1. ENDIF G2=2.*G KL2=.5*KL+1.5 KE1=1.+KE WRITE(*,*)' GIVE guess for: Q,Y1,Y2' READ(*,*) X NCT=0 1 DO 10 I=1,3 F(I)=FUN(I) DO 10 J=1,3 DX=.005*X(J) X(J)=X(J)+DX
306
10
100
1
2 3
Open Channel Flow: Numerical Methods and Computer Applications
D(I,J)=(FUN(I)F(I))/DX X(J)=X(J)DX FAC=D(3,1)/D(1,1) D(3,2)=D(3,2)FAC*D(1,2) D(3,3)=D(3,3)FAC*D(1,3) F(3)=F(3)FAC*F(1) FAC=D(2,1)/D(1,1) D(2,2)=D(2,2)FAC*D(1,2) D(2,3)=D(2,3)FAC*D(1,3) F(2)=F(2)FAC*F(1) FAC=D(3,2)/D(2,2) D(3,3)=D(3,3)FAC*D(2,3) F(3)=F(3)FAC*F(2) DIF1=F(3)/D(3,3) Y2=Y2DIF1 DIF=(F(2)DIF1*D(2,3))/D(2,2) Y1=Y1DIF SUM=ABS(DIF1)+ABS(DIF) DIF=(F(1)D(1,2)*DIFD(1,3)*DIF1)/D(1,1) SUM=SUM+ABS(DIF) Q=QDIF NCT=NCT+1 IF(NCT.LT.30 .AND. SUM.GT. ERR) GO TO 1 WRITE(IOUT,100) X FORMAT(' Q =',F10.2,' Y1 =',F10.2,' Y2 =',F10.2) END FUNCTION FUN(II) EXTERNAL DYX REAL X(3),W(1,13),KL2,KE1,Y(1),DY(1),XP(1),YP(1,1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN, &Q2G,X GO TO (1,2,3),II Q2G=X(1)*X(1)/G2 FUN=X(3)+Q2G/((B1+FM1*X(3))*X(3))**2KL2*((X(1)/B2)**2/G) &**.333333333 RETURN FUN=HX(2)KE1*Q2G/((B1+FM1*X(2))*X(2))**2 RETURN Y(1)=X(3) H1=.05 HMIN=.001 XX=FL XZ=0. QN=(FN*X(1)/CC)**2 Q2G=X(1)*X(1)/G CALL ODESOL(Y,DY,1,XX,XZ,TOL,H1,HMIN,1,XP,YP,W,DYX) FUN=X(2)Y(1) RETURN END
Nonuniform Flows
307
SUBROUTINE DYX(XX,Y,DY) REAL Y(1),DY(1),KL2,KE1,X(3) COMMON/TRAS/B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN, &Q2G,X EQUIVALENCE (Q,X(1)),(Y1,X(2)),(Y2,X(3)) P=B1+2.*SQRT(FM1*FM1+1.)*Y(1) A=(B1+FM1*Y(1))*Y(1) SF=QN*((P/A)**.666666667/A)**2 T=B1+2.*FM1*Y(1) DY(1)=(SOSF)/(1.Q2G*T/A**3) RETURN END SOLGVF.C #include <stdlib.h> #include <stdio.h> #include <math.h> float b1,fm1,fm12,fm1s,b2,h,g,g2,kl2,ke1,fl,tol,hmin=.001,fn,so, cc,\ fnq,q2g,x[3]; #include "odesol.h" void slope(float x,float *y,float *dyx){float a,sf; a=(b1+fm1*y[0])*y[0];sf=fnq*pow(pow((b1+fm1s*y[0])/a,.6666667)/a,2.); dyx[0]=(sosf)/(1.q2g*(b1+fm12*y[0])/(a*a*a));} // end slope float fun(int ii){float h1,y[1]; h1=.05; q2g=x[0]*x[0]/g2; if(ii==1) return (x[2]+q2g/pow((b1+fm1*x[2])*x[2],2.)\ kl2*pow(pow(x[0]/b2,2.)/g,.3333333)); else if(ii==2) return(hx[1]ke1*q2g/pow((b1+fm1*x[1])*x[1],2.)); else { y[0]=1.005*x[2]; fnq=pow(fn*x[0]/cc,2.); q2g=x[0]*x[0]/g; odesolc(y,fl,0.,tol,h1,hmin); return (x[1]y[0]);} } // End fun void main(void){int nct,i,j; float f[3],dx,fac,dif1,dif,err,sum,**d; d=(float**)malloc(3*sizeof(float*)); for(i=0;i<3;i++)d[i]=(float*)malloc(3*sizeof(float)); printf("Give: TOL,ERR,n,So,b1,m1,b2,H,L,g,KL,Ke\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f %f",\ &tol,&err,&fn,&so,&b1,&fm1,&b2,&h,&fl,&g,&kl2,&ke1); if(g>20.) cc=1.486; else cc=1.; g2=2.*g;kl2=.5*kl2+1.5;ke1+=1.; fm12=2.*fm1;fm1s=sqrt(fm1*fm1+1.); printf("Give guess for: Q,Y1,Y2\n"); scanf("%f %f %f",&x[0],&x[1],&x[2]);nct=0; do{ for(i=1;i<4;i++){f[i1]=fun(i); for(j=0;j<3;j++){ dx=.005*x[j]; x[j]+=dx;d[i1][j]=(fun(i)f[i1])/dx; x[j]=dx;}} fac=d[2][0]/d[0][0];d[2][1]=fac*d[0][1];d[2][2]=fac*d[0][2]; f[2]=fac*f[0]; fac=d[1][0]/d[0][0];d[1][1]=fac*d[0][1]; d[1][2]=fac*d[0][2]; f[1]=fac*f[0]; fac=d[2][1]/d[1][1];d[2][2]=fac*d[1][2];f[2]=fac*f[1]; dif1=f[2]/d[2][2]; x[2]=dif1;dif=(f[1]dif1*d[1][2])/d[1][1];
308
Open Channel Flow: Numerical Methods and Computer Applications
x[1]=dif;sum=fabs(dif1)+fabs(dif); dif=(f[0]d[0][1]*difd[0][2]*dif1)/d[0][0];sum+=fabs(dif); x[0]=dif; printf(" NCT=%d SUM=%f ",++nct,sum); for(i=0;i<3;i++)printf(" %f",x[i]);printf("\n"); }while((nct<30) && (sum>err)); printf("Q = %10.2f Y1 = %10.2f Y2 = %10.2f\n",x[0],x[1],x[2]);} This program contains the following: (1) a main program that defines the problem and implements the Newton solution, (2) a function subprogram, FUN, that defines any of the three equations (which you should write out) when called upon to do this, and when called on to provide the value of the equation that involves the GVF profile, it calls on the ODE solver appropriately, and (3) a subroutine that defines the derivative dY/dx. To obtain values for the equations that are stored in the FORTRAN array F(3), the main program calls the function subprogram FUN with an argument I = 1, 2, and 3 for the different equation numbers. When I = 1 or I = 2, then FUN evaluates the equation with a single FORTRAN line. However, when I = 3, the ODE solver is called upon to supply the solution to YGVF at the entrance of the channel. The difference between the current value Y1, which is being adjusted by the Newton method, and this value defines the third equation and FUN=X1(2)Y(1). Note that I becomes II, the first argument in FUNCTION FUN and the different equations are selected with the computed GO TO(1,2,3),II statement. (The FORTRAN variables are: X1(1)=Q, X1(2) = Y1, and X1(3)=Y2, as can be seen by the equivalence statement.) After the main program fills the elements for the equation vector, F(I), and the Jacobian matrix D(I,J), it solves the resulting linear system of equations, adjusts the unknown vector, X(I) which consists of: Q, Y1, and Y2, and repeats another Newton iteration if the convergence error ERR is not satisfied. The error parameter TOL applies for the ODE solver. Since part of this solution involves the numerical solution of the GVF profile, the error for the Newton solution ERR should not be too small. ERR = 0.1 is probably a good value to use. The subroutine DYX supplies dY/dx to ODESOL or DVERK whenever requested. A more refined solution might be obtained by eliminating the loss coefficient, and in its place solving the GVF profile through the transition using the term that involves ∂A/∂x for the nonprismatic channel effect. To implement this approach, a small amount of additional logic is needed in the subroutine DYX that properly evaluates the change in area with respect to x while in the transition and adds this term in defining dY/dx, and sets ∂A/∂x = 0 when upstream from the transition. The first equation, which now consists of the energy equation between the beginning of the transition and its end, is replaced by just the critical flow equation applied at the downstream end of the transition. The main program reads in the data for the problem that is to be solved, and implements the Newton method in obtaining the solution. To obtain the equation vector and the Jacobian matrix, it calls on the function subroutine FUN to evaluate the equation number corresponding to its argument. When the third equation is to be evaluated, then the function subprogram FUN calls on the ODE solver ODESOL, which in turn call on the subroutine DYX to evaluate the derivative dY/dx = (So − Sf)/ (1 − Fr2). After obtaining the equation vector and the Jacobian matrix, this main program uses the Gaussian elimination method to solve this linear system of equations. However, these statements could be replaced by a call to a subroutine such as SOLVEQ that solves linear systems of equations. You should study this listing over carefully to fully understand how the solution is accomplished. Example Problem 4.9 Using program SOLGVF obtain a series of solutions for the flow rate into an upstream trapezoidal channel with b1 = 5 m, m1 = 1.2, n = 0.015, and So1 = 0.00075. This channel is supplied by a reservoir whose water surface is 3.5 m above the channel bottom. The entrance loss coefficient is Ke = 0.05. At a distance 300 m downstream from the channel’s beginning, there is a transition to a steep rectangular channel with a bottom width b2. The loss coefficient for the transition is
309
Nonuniform Flows K L = 0.025. Solve the problem with the bottom width of this downstream channel varying from b2 = 3 m to b2 = 8 m. Solution If the flow were uniform, then the solution of the energy equation at the entrance and Manning’s equation give: Qo = 73.743 m3/s, Yo = 3.10 m, Eo = 3.48 m, and Mo = 56.41 m3. Solving the critical flow equation simultaneously with the energy equation at the entrance gives: Q = 86.26 m3/s, Yc = 2.54 m, Ec = 3.45 m, and Mc = 59.79 m3. This latter flow rate represents the maximum that can be obtained for the given reservoir head, and would occur only if the length of the upstream channel approached zero, and the downstream rectangular channel were wide enough. The solutions to the other cases for varying b2 are given below. The program failed to produce a solution when the bottom width of the downstream channel was given as b2 = 8.0 m, since the downstream Froude number approaches unity. The following should be observed: (1) With the smaller bottom widths, the flow is considerably smaller than if the flow in the trapezoidal channel were uniform. It is not until the bottom width of the downstream channel gets wider than 6.5 m, that the flow rate exceeds this amount. (2) The GVF profile in the upstream trapezoidal channel is an M1 for the bottom width of the downstream channel less than 6.5 m. This can be observed by the fact Input to SOLGVF No. IOUT 1 2 3 4 5 6 7 8 9 10 11
6 6 6 6 6 6 6 6 6 6 6
Solution
TOL
ERR
FN
SO
B1
FM1
B2
H
L
g
KL
KE
Q
Y1
Y2
1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6 1.e − 6
0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005
0.015 0.015 0.015 0.015 0.015 0.015 0.015 0.015 0.015 0.015 0.015
0.00075 0.00075 0.00075 0.00075 0.00075 0.00075 0.00075 0.00075 0.00075 0.00075 0.00075
5 5 5 5 5 5 5 5 5 5 5
1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2
3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 7.8
3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5
300 300 300 300 300 300 300 300 300 300 300
9.81 9.81 9.81 9.81 9.81 9.81 9.81 9.81 9.81 9.81 9.81
0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025
0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05
35.81 41.54 47.15 52.61 57.88 62.91 67.62 71.92 75.68 78.66 79.93
3.43 3.40 3.37 3.34 3.29 3.25 3.19 3.13 3.07 3.00 2.97
3.63 3.60 3.56 3.50 3.44 3.37 3.28 3.17 3.03 2.82 2.61
that Y2 is larger than Y1. Thus, for a width less than the 7.0 entry in the solution table, the change to the steep downstream rectangular channel reduces the flow from the reservoir. (3) As the widths of the downstream channel increase, the depths throughout the channel decrease. Example Problem 4.10 Resolve Example Problem 4.8 by writing a program that solves the ODE and the algebraic equations simultaneously. Solution For this problem, one could write three algebraic equations plus one ODE that describes the GVF from the gate up to the reservoir. The three algebraic equation could consist of (1) energy from the reservoir to the beginning of the channel, (2) energy across the gate, and (3) energy across the transition upstream from the gate. However, (2) and (3) can be combined into one equation that equates the specific energy immediately upstream from the transition to the specific energy downstream from the gate. Thereafter, the energy across the transition (or across the gate) can be solved to get the depth immediately upstream from the gate. The three simultaneous equations needed to solve the three unknowns, Q, Y1, and Y2 are
F1 = H − Y1 − (1 + K e )Q 2 / (2gA12 ) = 0
F2 = Y2 + Q 2 / (2gA 22 ) − Yd − (1 + K L )Q 2 / (2gA d2 ) = 0
310
Open Channel Flow: Numerical Methods and Computer Applications F3 = Y1 − Y1,ode (Y2 ) = 0 ODE is dY/dx = (So − Sf ) / (1 − Fr2 )
in which Y1 and Y2 are the depths at the beginning and the end of the trapezoidal channel, respectively, and Yd is the depth downstream of the gate and is given by Yd = CcYG. An alternative is to replace F2 with
F2 = Y2 + Q 2 / (2gA 22 ) − Yu − (1 + K L1 )Q 2 / (2gA 2u ) = 0 and
F3 = Yu + Q 2 / (2gA 2u ) − Yd − (1 + K L2 )Q 2 / (2gA d2 ) = 0 and the previous F3 becomes F4, in which now K L1 is the loss coefficient for the transition, and K L2 is the loss coefficient for the gate, whereas earlier F2 K L was the combined loss coefficient for the transition and the gate. A third alternative is to use the above F3 that equates the specific energy immediately upstream of the gate Eu to that downstream of the gate Ed, and to solve the ODE starting at the end of the transition (immediately upstream of the gate) and to include the nonprismatic term in the ODE while solving the GVF through the transition. The program SOLGATE solves this problem using the first of these approaches. In this program, the array X, with dimensions of 3, contains the three unknowns with X(1) = Q, X(2) = Y1, and X(3) = Y2. In the program, the subroutine FUN defines the three equations, i.e., supplies values to F(1) = F1, F(2) = F2, and F(3) = F3, and the Main program implements the Newton method in solving the equations by numerically evaluating the nine elements of the Jacobian matrix. It calls on the ODE solver RUKUST, whose use is described in Appendix C, to solve the GVF in the trapezoidal channel. In order to find the depth Yu in the rectangular channel upstream of the gate, but at the end of the transition, the function subprogram YGATE uses the Newton method to solve the specific energy equation across the gate after Q is solved. Likewise, to provide the normal depth Yo corresponding to this flow rate, the subprogram YNORM uses the Newton method to solve Manning’s equation. The subroutine OPEN (qopen in the Cprogram) solves the energy equation at the entrance and Manning’s equation simultaneously to provide Yo and Qo, as if there were no gate in the channel and it were long so that a uniform flow would occur in the upstream trapezoidal channel. The input to solve the problem with the 10 given gate settings used in Example Problem 4.8 is 3 2 .013 .0009 2 .12 .09 9.81 500 2.5 .0000001 .0005 .75 2 −.2 10 15 1.9 2.2 The solution provided is H = 2.0, Ke = .120, b1 = 3.0, m1 = 2.00, n = .0130, So = .000900, b2 = 2.5, KL = .090, L = 500.0 Nogate Uniform Flow: Yo = 1.688, Qo = 25.17 YG 2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20
Yd
Q
Y1
Y2
Yu
Yo
Yc
1.50 1.35 1.20 1.05 0.90 0.75 0.60 0.45 0.30 0.15
14.76 14.43 13.79 12.87 11.68 10.25 8.59 6.72 4.66 2.41
1.929 1.933 1.939 1.948 1.957 1.968 1.978 1.987 1.994 1.998
2.325 2.332 2.344 2.359 2.377 2.395 2.412 2.427 2.439 2.447
1.830 1.890 1.977 2.069 2.158 2.239 2.310 2.368 2.412 2.440
1.299 1.284 1.255 1.212 1.154 1.080 0.986 0.868 0.714 0.498
1.526 1.503 1.458 1.392 1.306 1.197 1.064 0.903 0.707 0.456
311
Nonuniform Flows 500 m
YG= 2
.12 b1 = 3 m, m1 = 2, So = 0.0009, n = 0.013
=0 Ke
Tran sit KL = ion 0.09
H=2 m
m, C c
= 0.75
b2 = 2.5 m
Listing of program SOLGATE.FOR PARAMETER (N=3) INTEGER*2 ITYP(N) REAL F(N),F1(N),D(N,N),KL1,KE1 COMMON /TRAS/B1,FM1,B2,H,G2,KL1,KE1,FL,TOL,FN,SO,CU,QN,Q2G, &Yd,Ad,FMS,FM2,X(3),DXS WRITE(*,*)' Give:b1,m1,n,So,H,Ke,KL,g,L,b2,TOL,ERR,Cc,Yg1, &dYg,N' READ(*,*) B1,FM1,FN,SO,H,KE1,KL1,G,FL,B2,TOL,ERR,CC, &YG1,DYG,NGATE CU=1.486 DXS=1. FM2=2.*FM1 FMS=2.*SQRT(FM1*FM1+1.) IF(G.LT. 30.) CU=1. G2=2.*G KL1=1.+KL1 KE1=1.+KE1 WRITE(*,*)' Give guess for: Q,Y1 & Y2' READ(*,*) X Yo=.98*X(2) CALL OPEN(Yo,Qo) WRITE(3,110) H,KE11.,B1,FM1,FN,SO,B2,KL11.,FL,Yo,Qo 110 FORMAT(' H =',F7.1,', Ke =',F6.3,', b1 =',F8.1,', m1 =', &F7.2,', n =',F8.4,',',/,' So =',F9.6,', b2 =',F8.1,', &KL =',F6.3,', L =',F8.1,/,' No gateUniform Flow: Yo =', &F8.3,', Qo =',F8.2,/1X,61(''),/,4X,'YG',4X,'Yd',7X,'Q', &6X,'Y1',6X,'Y2',6X,'Yu',6X,'Yo',6X,'Yc',/,1X,61('')) DO 50 KK=1,NGATE YG=YG1+DYG*FLOAT(KK1) Yd=CC*YG Ad=(B2*Yd)**2 NCT=0 1 SUM=0. CALL FUN(F1) DO 10 J=1,N XX=X(J) X(J)=1.005*X(J) CALL FUN(F) DO 5 I=1,N 5 D(I,J)=(F(I)F1(I))/(X(J)XX) 10 X(J)=XX CALL SOLVEQ(N,1,N,D,F1,1,DET,ITYP) DO 15 I=1,N X(I)=X(I)F1(I)
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Open Channel Flow: Numerical Methods and Computer Applications 15
50 100
1
1
SUM=SUM+ABS(F1(I)) NCT=NCT+1 IF(SUM.GT.ERR .AND. NCT.LT.20) GO TO 1 IF(NCT.EQ.20) WRITE(*,*)' Did not Converge',SUM WRITE(3,100) YG,CC*YG,X,YGATE(X(1)),YNORM(X(1)), &((X(1)/B2)**2/G)**.33333333 FORMAT(2F7.2,F8.2,5F8.3) END SUBROUTINE FUN(F) REAL F(3),KL1,KE1,Y(1),YW(1) COMMON /TRAS/B1,FM1,B2,H,G2,KL1,KE1,FL,TOL,FN,SO,CU, &QN,Q2G,Yd,Ad,FMS,FM2,X(3),DXS QN=(FN*X(1)/CU)**2 Q2G=X(1)**2/G2 F(1)=HX(2)KE1*Q2G/((B1+FM1*X(2))*X(2))**2 F(2)=X(3)+Q2G/((B1+FM1*X(3))*X(3))**2YdKL1*Q2G/Ad Y(1)=X(3) CALL RUKUST(1,DXS,FL,0.,TOL,Y,YW) F(3)=X(2)Y(1) RETURN END SUBROUTINE SLOPE(XX,Y,DY) REAL Y(1),DY(1),KL1,KE1 COMMON /TRAS/B1,FM1,B2,H,G2,KL1,KE1,FL,TOL,FN,SO,CU, &QN,Q2G,Yd,Ad,FMS,FM2,X(3),DXS AREA=(B1+FM1*Y(1))*Y(1) SF=QN*(B1+FMS*Y(1))**1.33333333/AREA**3.3333333 DY(1)=(SOSF)/(1..5*Q2G*(B1+FM2*Y(1))/AREA**3) RETURN END FUNCTION YNORM(Q) COMMON /TRAS/B1,FM1,B2,H,G2,FKL1,FKE1,FL,TOL,FN,SO,CU, &QN,Q2G,Yd,Ad,FMS,FM2,X(3),DXS QQ=CU*SQRT(SO)/FN Y=X(2) M=0 F=QQQ*((B1+FM1*Y)*Y)**1.6666667/(B1+FMS*Y)**.666667 Y1=1.005*Y DIF=(Y1Y)*F/(QQQ*((B1+FM1*Y1)*Y1)**1.6666667/ &(B1+FMS*Y1)**.666667F) Y=YDIF M=M+1 IF(ABS(DIF).GT. .000001 .AND. M.LT.20) GO TO 1 IF(M.EQ.20) WRITE(*,*) ' YNORM failed to converge' YNORM=Y RETURN END FUNCTION YGATE(Q) COMMON /TRAS/B1,FM1,B2,H,G2,FKL1,FKE1,FL,TOL,FN,SO,CU, &QN,Q2G,Yd,Ad,FMS,FM2,X(3),DXS M=0 Y=.9*X(3) QQ=Q**2/G2 ED=FKL1*QQ/Ad+Yd F=Y+QQ/(B2*Y)**2ED Y1=1.005*Y DIF=(Y1Y)*F/(Y1+QQ/(B2*Y1)**2EDF)
Nonuniform Flows
1
313
Y=YDIF M=M+1 IF(ABS(DIF).GT. .000001 .AND. M.LT.20) GO TO 1 IF(M.EQ.20) WRITE(*,*) ' YGATE failed to converge' YGATE=Y RETURN END SUBROUTINE OPEN(Yo,Qo) COMMON /TRAS/B1,FM1,B2,H,G2,FKL1,FKE1,FL,TOL,FN,SO,CU, &QN,Q2G,Yd,Ad,FMS,FM2,X(3),DXS M=0 QQ=Cu*SQRT(SO)/FN AREA=(B1+FM1*Yo)*Yo Qo=QQ*AREA**1.66666667/(B1+FMS*Yo)**.66666667 F=HYoFKE1*(Qo/AREA)**2/G2 Y1=1.005*Yo A=(B1+FM1*Y1)*Y1 DIF=(Y1Yo)*F/(HY1FKE1*(QQ*A**1.6666667/(B1+FMS*Y1) &**.6666667/A)**2/G2F) Yo=YoDIF M=M+1 IF(ABS(DIF).GT. .000001 .AND. M.LT.20) GO TO 1 IF(M.EQ.20) WRITE(*,*) ' OPEN failed to converge' Qo=QQ*AREA**1.66666667/(B1+FMS*Yo)**.66666667 RETURN END
#include <stdio.h> #include <stdlib.h> #include <math.h> float b1,m1,b2,h,g2,kl1,ke1,l,tol,n,so,cu,qn,q2g,yd,ad,fms,\ fm2,dxs[1],x[3]; extern void solveq(int n,float **a,float *b,int itype,float *dd,\ int *indx); extern void rukust(int neq,float *dxs,float xbeg,float xend,\ float error,float *y,float *ytt); void fun(float *f){ float y[1],yw[1]; qn=pow(n*x[0]/cu,2.); q2g=x[0]*x[0]/g2; f[0]=hx[1]ke1*q2g/pow((b1+m1*x[1])*x[1],2.); f[1]=x[2]+q2g/pow((b1+m1*x[2])*x[2],2.)ydkl1*q2g/ad; y[0]=x[2];rukust(1,dxs,l,0.,tol,y,yw); f[2]=x[1]y[0]; } //end fun void slope(float xx,float *y, float *dy){ float area,sf; area=(b1+m1*y[0])*y[0]; sf=qn*pow(b1+fms*y[0],1.33333333)/pow(area,3.3333333); dy[0]=(sosf)/(1..5*q2g*(b1+fm2*y[0])/pow(area,3.)); }// end slope float ynorm(float q) {float qq,y,f,y1,dif; int m; qq=cu*sqrt(so)/n; y=x[1]; m=0; do {f=qqq*pow((b1+m1*y)*y,1.66666667)/pow(b1+fms*y,.66666667); y1=1.005*y; dif=(y1y)*f/(qqq*pow((b1+m1*y1)*y1,1.6666667)/\ pow(b1+fms*y1,.66666667)f); y=dif; }while ((fabs(dif)>.000001) && (++m<20)); if(m==20) printf("YNORM failed to converge\n");
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return y;} // end ynorm float ygate(float q){float y,qq,ed,f,y1,dif; int m; m=0; y=.9*x[2]; qq=q*q/g2; ed=kl1*qq/ad+yd; do {f=y+qq/pow(b2*y,2.)ed; y1=1.005*y; dif=(y1y)*f/(y1+qq/pow(b2*y1,2.)edf); y=dif; } while ((fabs(dif)>.000001) && (++m<20)); if(m==20) printf("YGATE failed to converge\n"); return y;} // end ygate void qopen(float *yo, float *qo){ float qq,area,f,y1,a,dif,yyo,qqo; int m; m=0; yyo=*yo; qq=cu*sqrt(so)/n; do {area=(b1+m1*yyo)*yyo; qqo=qq*pow(area,1.6666667)/pow(b1+fms*yyo,.6666667); f=hyyoke1*pow(qqo/area,2.)/g2; y1=1.005*yyo;a=(b1+m1*y1)*y1; dif=(y1yyo)*f/(hy1ke1*pow(qq*pow(a,1.6666667)/\ pow(b1+fms*y1,.6666667)/a,2.)/g2f); yyo=dif; }while ((fabs(dif)>.000001) && (++m<20)); if(m==20) printf("OPEN failed to converge\n"); qqo=qq*pow(area,1.6666667)/pow(b1+fms*yyo,.66666667); *yo=yyo;*qo=qqo; } // end qopen void main(void){ float yo[1],qo[1],*det,sum,xx,err,cc,yg1,dyg,g,yg, f1[3],f[3],**d; int nn,kk,nct,j,i,indx[3];FILE *fil; d=(float**)malloc(3*sizeof(float*)); for(i=0;i<3;i++)d[i]=(float*)malloc(3*sizeof(float)); dxs[0]=1.; cu=1.486; printf("Give: b1,m1,n,So,H,Ke,KL,g,L,b2,TOL,ERR,Cc,Yg1,dYg,nn\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f %f %f %f %f %d",\ &b1,&m1,&n,&so,&h,&ke1,&kl1,&g,&l,&b2,&tol,&err,&cc,&yg1,&dyg,&nn); fm2=2.*m1; fms=2.*sqrt(m1*m1+1.); if(g<30.) cu=1; g2=2.*g; kl1+=1.;ke1+=1.; printf("Give guess for Q,Y1 & Y2\n"); scanf("%f %f %f",&x[0],&x[1],&x[2]); yo[0]=.98*x[1]; qopen(yo,qo); if((fil=fopen("SOLGATE.OUT","w"))==NULL){ printf("Can not open SOLGATE.OUT\n");exit(0);} fprintf(fil, H=%7.1f, Ke=%6.3f, b1=%8.1f, m1=%7.2f, n=%8.4f, n\ So=%9.6f, b2=%8.1f, KL=%6.3f, L=%8.0f\n",\ h,ke11.,b1,m1,n,so,b2,kl11.,l); fprintf(fil," No gateUniform Flow:Yo=%8.3f, Qo=%8.2f\n",yo[0],qo[0]); for(i=0;i<61;i++)fprintf(fil,""); fprintf(fil,"\n"); fprintf(fil," YG Yd Q Y1 Y2 Yu\ Yo Yc\n"); for(i=0;i<61;i++)fprintf(fil,""); fprintf(fil,"\n"); for(kk=0;kk
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Nonuniform Flows
solveq(3,d,f1,1,det,indx); for(i=0;i<3;i++){x[i]=f1[i];sum+=fabs(f1[i]);} } while((sum>err) && (++nct<20)); if(nct==20)printf("Did not converge %f\n",sum); fprintf(fil,"%7.2f %6.2f %7.2f %7.3f %7.3f %7.3f %7.3f %7.3f\n",\ yg,yd,x[0],x[1],x[2],ygate(x[0]),ynorm(x[0]), pow(pow(x[0]/b2,2.)/g,.333333333));} fclose(fil);}
4.8 Different Modes of Gate Operation The operation of gates can be classified into underflow or overflow, free flowing downstream or submerged downstream. The types of gates we have used to illustrate the principles of open channel flow are, underflow gates. The flow under a gate is also often referred to as orifice flow, since the equation describing the discharge past the gate is of the form of an orifice equation that gives the velocity or discharge, proportional to the square root of the head. Overflow gates control the depth and the flow rate by having a plate, etc., rise from the channel bottom. For such overflow gates, the flow rate (discharge) is proportional to the head raised to the 1.5 power. The flow past a gate is classified as either free flowing or submerged depending upon whether the downstream effects the flow past the gate, or not. For example, if an underflow gate is submerged, then the depth immediately downstream from the gate is above the tip of the gate, whereas if a free flow occurs, the depth immediately downstream from the gate is below the tip of the gate. Submerged gate flows will be dealt with in Chapter 5. Often, the structure that holds the gate in place results in a constriction in the channel size, and therefore even though the tip of the gate is above the water depth, the gate site will affect the flow. Such a flow can be classified as a nonorifice flow. The effect is similar to that of a flume placed in the channel. Flumes and weirs are also dealt with in Chapter 5. Because the depth “contracts” below the tip of the gate if free flow occurs to Yd = CcYG, it is possible to have gate settings above the normal depth associated with the flow rate that is occurring in a channel. As the settings of gates are changed, an unsteady flow is initiated in which the volume of water stored in the channel upstream of the gate is either increased or decreased, until another steadystate condition is eventually established. Changes in gate setting may result in the flow changing from free to submerged, or vice versus, or from having the gate’s tip above to below the water surface, e.g., from nonorifice to orifice flow, or vice versa, etc. If a vertical gate is lowered into a channel flow, the results can be quite different, than if it is gradually raised above the water surface. The various possibilities can only be adequately dealt with by solving the unsteady flow equations, which are the subject of Chapters 6 and 7. Example Problem 4.11 Obtain a series of steadystate solutions to the flow into a channel from a reservoir past a gate in which the gate is raised until it rises above the water surface. For these solutions, hold the upstream reservoir head constant with H = 5 ft (and an entrance loss coefficient Ke = 0.2). The gate is 1000 ft downstream from the reservoir. Upstream of the gate, the channel is trapezoidal with b1 = 10 ft, m1 = 1, n = 0.013, and So = 0.0008. The gate is vertical and is b2 = 10 ft wide, and has an assumed constant contraction coefficient Cc = 0.6. L = 1000 ft
H
Y1
Q .2
K
GVF
=0 e
b1 = 10 ft, m1 = 1, n = 0.013, So = 0.0008
Y2
Yu
Cc Yd = CcYG
YG
on b = 10 ft 2 KL = 0
siti
n Tra
Free flow
316
Open Channel Flow: Numerical Methods and Computer Applications Solution The program developed in the previous problem can be used to solve this problem. The last two columns in the solution table below, that provides the volume and the change in volume of water in the channel, between the reservoir and the gate, have been added to provide an idea about what volumes are involved during the unsteady change from one steady state to the new steadystate condition. Also, the Froude numbers immediately upstream and downstream of the gate have been added. The input is: 10 1 .013 .0008 5 .2 0 32.2 1000 10 .0000001 .0005 .6 3 .2 16, and 300 4.7 5.3 as a second line. The table provides the solution, and the graph below the table plots the flow rate on the right ordinate, and several of the depths in the table on the left ordinate are against the gate height as the abscissa. Notice that when the gate’s tip is between 5.6 and 5.8 ft above the channel bottom, the solution of the equations shows Yu = Yd. Both are critical depths. When this occurs, the flow rate is between 356.41 and 356.31 cfs, which is slightly less than what would occur under uniform conditions if the trapezoidal channel were very long, i.e., Qo = 384.29 (with Yo = 4.250 ft). In other words, the tip of the gate is actually 5.80 − 3.485 = 1.315 ft above the water surface immediately upstream of the gate. Of course, this is nonsense; but illustrates that judgement must be used in solving equations. Actually, the gate will rise above the water surface between gate settings 4.2 and 4.4 ft provided the gate is raised extremely slowly so transient effects are insignificant. However, if the gate is raised rapidly, it could be much higher before clearing the water surface. Notice the last solutions in the table with YG = 5.8 and 6.0 ft produce identical depths of Yu = Yd = 3.480 and 3.600 ft, e.g., no longer are alternate depths being solved across the gate. H = 5.0, Ke = 0.200, b1 = 10.0, m1 = 1.00, n = 0.0130, So = 0.000800, b2 = 10.0, KL = 0.000, L = 1000. No gate–Uniform Flow: Yo = 4.250, Qo = 384.29. YG 3.00 3.20 3.40 3.60 3.80 4.00 4.20 4.40 4.60 4.80 5.00 5.20 5.40 5.60 5.80 6.00
Yd
Frd
Q
Y1
Y2
Yu
Fru
Yo
Yc
Volume
Del Vol.
1.80 1.92 2.04 2.16 2.28 2.40 2.52 2.64 2.76 2.88 3.00 3.12 3.24 3.36 3.48 3.60
2.02 1.91 1.81 1.72 1.63 1.55 1.47 1.40 1.32 1.26 1.19 1.13 1.07 1.02 .97 .92
276.58 288.57 299.58 309.59 318.60 326.59 333.58 339.58 344.61 348.71 351.91 354.23 355.73 356.41 356.31 355.44
4.702 4.669 4.637 4.605 4.574 4.545 4.518 4.493 4.472 4.453 4.438 4.427 4.419 4.416 4.416 4.421
5.284 5.223 5.161 5.099 5.037 4.977 4.920 4.867 4.819 4.778 4.743 4.717 4.700 4.692 4.693 4.704
4.989 4.886 4.778 4.667 4.551 4.432 4.311 4.188 4.065 3.940 3.816 3.693 3.570 3.449 3.480 3.600
0.44 0.47 0.51 0.54 0.58 0.62 0.66 0.70 0.74 0.79 0.83 0.88 0.93 0.98 0.97 0.92
3.534 3.620 3.697 3.766 3.827 3.881 3.927 3.966 3.999 4.026 4.046 4.061 4.071 4.075 4.074 4.069
2.875 2.957 3.032 3.099 3.159 3.212 3.257 3.296 3.329 3.355 3.375 3.390 3.400 3.404 3.404 3.398
67755 67297 66842 66395 65964 65555 65174 64828 64521 64259 64045 63882 63771 63713 63722 63793
0 −457 −454 −445 −430 −408 −380 −346 −306 −261 −213 −162 −110 −57 10 70
4.9 Hydraulic Jump Downstream from a Gate in a Finite Length Channel Now, let us consider the problem of finding the position of a hydraulic jump downstream from a gate where the channel terminates at some relatively short distance downstream therefrom either as a free overfall or by discharging into a reservoir with a known depth. With the length of the channel not too long, there will be a GVF downstream from the hydraulic jump, as well as the M3 GVF upstream therefrom. Thus, the downstream depth Y2 is unknown. If a free overfall occurs at the end of the channel, this will be an M2 GVF. When the water level in the reservoir rises above the normal depth, then the downstream GVF is an M1. In this problem, the flow rate and the depth immediately downstream from the gate are known, and the unknowns are as follows: (1) the distance x where the
317
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jump occurs downstream from the gate, (2) the depth Y1 immediately upstream from the jump, and (3) the depth Y2 immediately downstream from the jump. The depth at the end of the channel will be considered to be known either from the water surface in the reservoir if this is given larger than the critical depth, or equal to the critical depth otherwise. Since the flow rate is known, this critical depth can be calculated when necessary. M2 or M
1
Yb
M3 x
WS
Q
Y2
Yc
Y1 b, m, n, So, L
For this problem, there are three equations: the momentum equation across the hydraulic jump, and two ODE equations for the two GVF profiles upstream and downstream from the hydraulic jump. If the channel is trapezoidal, then these three equations are
Q2 b m 2 Q2 b m − + Y2 Y2 + =0 F1 = + Yl Y12 + gA1 2 3 gA 2 2 3
F2 = Y1 − Y1.ode (Y2 ) = 0 Solving the M3 GVF
F3 = Y2 − Y1.ode (Y3 ) = 0 Solving the GVF from the end of the chann nel up to the downstream side of the jump
There are two approaches to solving this problem. The first is a computer implementation of the methodology used in Example Problem 4.4 by hand, i.e., solving the two gradually varied flow profiles, but rather than printing out the solution of depths Y and M’s corresponding to the various positions x, these are stored in memory, and the position where the two momentum fuctions are equal is determined. In other words, the position of the hydraulic jump is where the x’s and the M’s are simultaneously equal from the two GVF solutions. The second approach is to use the Newton method to solve the above three equations simultaneously. First, consider a computer program GVFJMP that uses the first approach. The program will need to solve two GVF profiles; the M3 upstream of the hydraulic jump, and the one from the end of the channel, at least up to the hydraulic jump. Both these solutions should extend beyond where the jump actually occurs so that the solutions can be examined to determine the actual position. In addition to solving Y, at each increment values of the momentum functions need to be computed and stored in arrays. After these solutions are obtained, an interpolation algorithm needs to determine where the two values of x and the two values of the momentum function are simultaneously equal. Program GVFJMP.FOR is designed for this purpose. After reading in the problem specification, this program solves the critical depth. Should the downstream reservoir water surface level WS be less than this Yc, then it sets WS = 1.03Yc. Thus, the user does not need to computer Yc, but rather just give a small value for WS to solve the case of a free overfall. If WS is given larger than Yc, then the case of the channel discharging into a reservoir is accommodated. The ODE solver RUKUST, described in the appendix is used to solve the two GVF profiles. The program variables associated with the upstream GVF contain a U, i.e., XU, YU, and MU that are the upstream position, the corresponding depths and the momentum functions, and those associated with the downstream GVF contain D, i.e., XD, YD, MD, etc. Starting with the statement labeled 14, the interpolation of the two tables (i.e., arrays of values) begins to find the location where the two values of the two momentum functions are equal. The approach taken in this
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algorithm is to start at the upstream end of the channel, and find the index ID of the downstream position XD corresponding with XU(IU). Thereafter, the values of the two momentum functions are compared. If MU(IU) is larger than MD(ID), then the next solution entry is checked in the same manner, etc. After two entries in both the upstream solution table and the downstream solution table have been made, Md = ad + bdx, in which the b’s are evaluated from (M1 − M2)/(x1 − x2) and the a’s from M1 − bx1. Then setting the two M’s equal provides the following interpolation equation for x: x=
(a u − a d ) (bd − b u )
The second approach uses the Newton method to solve the one algebraic and two ODEs, simultaneously. Most differential equation solvers, such as ODESOL, DVERK, and RUKUST are designed to solve systems of simultaneous equations but only over the same interval of the independent variable. Since the upstream and downstream GVF profiles need to be solved in different directions to solve the problem of locating the hydraulic jump between a gate and the downstream end of the channel, it will be necessary to call on the solver two different times. The upstream solution will be from 0 to x (the position of the jump), and the downstream GVF solution will be from L (the length of the channel) up to x. If the solver itself does not find an appropriately small interval to get the solution started near the critical depth, it may also be necessary to call on this solver within a DO loop to solve the downstream GVF. Fixed step Runge–Kutta methods fall in to this category. More sophisticated solvers, such as ODESOL and DVERK, however, need to be called on only once for the entire interval. Program GVFJMP2.FOR implements this second approach to solve the position of the hydraulic jump. Program GVFJMP.FOR LOGICAL DOWNS REAL YY(1),YTT(1) REAL XU[ALLOCATABLE](:),YU[ALLOCATABLE](:),MU[ALLOCATABLE](:), &XD[ALLOCATABLE](:),YD[ALLOCATABLE](:),MD[ALLOCATABLE](:) COMMON B,FM,FM2,TWOM,FNQ,QG2,SO EQUIVALENCE (Y,YY(1)) WRITE(6,*)'Give:IOU,Q,b,m,So,n,L,WS,Yb,dX1,dX2,g' READ(5,*) IOUT,Q,B,FM,SO,FN,FL,WS,YB,DX1,DX2,G IF(FM.LT.1.E5) THEN YC=((Q/B)**2/G)**.333333 GO TO 2 ENDIF QP=FM**3*Q**2/(G*B**5) YC=.925*(.5*qp)**.284 NCT=0 1 F=(YC+YC**2)**3/(1.+2.*YC)QP YC1=1.01*YC X1=(YC1YC)*F/((YC1+YC1**2)**3/(1.+2.*YC1)QPF) YC=YCX1 NCT=NCT+1 IF(ABS(X1).GT.1.E5 .AND. NCT.LT.20) GO TO 1 YC=B*YC/FM 2 WRITE(IOUT,*)' Critical Depth =',YC IF(WS.LT.YC) THEN WRITE(IOUT,*)' Depth at end of channel is critical' WS=1.03*YC
Nonuniform Flows
100 5
10
ENDIF YC1=.95*YC DXS=.05 C=1. IF(G.GT.30.) C=1.486 NUE=FL/ABS(DX1)+.5 NDE=FL/ABS(DX2)+.5 ALLOCATE(XU(NUE),YU(NUE),MU(NUE),XD(NDE),YD(NUE),MD(NDE)) QG2=Q*Q/G TWOM=2.*FM FNQ=FN*Q/C FM2=2.*SQRT(FM*FM+1.) Y=WS A=(B+FM*Y)*Y X1=FL DX=ABS(DX2) N=NDE DOWNS=.TRUE. XD(1)=FL YD(1)=Y MD(1)=(.5*B+FM/3.*Y)*Y*Y+QG2/A WRITE(IOUT,100) X1,Y,MD(1) FORMAT(F10.2,F10.3,F10.1) DO 10 I=2,N X=X1+DX CALL RUKUST(1,DXS,X1,X,1.E5,YY,YTT) A=(B+FM*Y)*Y IF(DOWNS) THEN XD(I)=X YD(I)=Y MD(I)=(.5*B+FM/3.*Y)*Y*Y+QG2/A WRITE(IOUT,100) X,Y,MD(I) ELSE XU(I)=X YU(I)=Y MU(I)=(.5*B+FM/3.*Y)*Y*Y+QG2/A WRITE(IOUT,100) X,Y,MU(I) IF(Y.GT.YC1) THEN NUE=I GO TO 14 ENDIF ENDIF X1=X IF(DOWNS) THEN N=NUE Y=YB DX=ABS(DX1) DOWNS=.FALSE. XD(1)=0. YD(1)=Y A=(B+FM*Y)*Y
319
320
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50
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Open Channel Flow: Numerical Methods and Computer Applications
MD(1)=(.5*B+FM/3.*Y)*Y*Y+QG2/A X1=0. WRITE(IOUT,100) X1,Y,MD(1) GO TO 5 ENDIF ID=NDE4 IDM=NDE2 IU=2 IU1=IU1 DO 20 WHILE (XD(ID).GT.XU(IU).AND.ID.LT.IDM) ID=ID+1 ID1=ID+1 DO 30 WHILE (XD(ID1).LT.XU(IU1).AND.ID1.GT.1) ID1=ID11 ID=ID11 IF(MU(IU).LT.MD(ID1) .OR. IU.EQ.NUE) GO TO 50 IU=IU+1 GO TO 15 BU=(MU(IU)MU(IU1))/(XU(IU)XU(IU1)) WRITE(*,*) IU,IU1,ID,ID1 AU=MU(IU)BU*XU(IU) BD=(MD(ID)MD(ID1))/(XD(ID)XD(ID1)) AD=MD(ID)BD*XD(ID) X=(AUAD)/(BDBU) F=YU(IU1)+(XXU(IU1))/(XU(IU)XU(IU1))* &*(YD(IU)YD(IU1)) QP=YD(ID)+(XXD(ID))/(XU(ID1)XU(ID))*(YD(ID1)YD(ID)) WRITE(*,110) X,F,QP WRITE(IOUT,110) X,F,QP FORMAT(' Position x=',F10.2,' Upst Depth Y1 =',F10.3,' &Downstr. Depth Y2 =',F10.3) DEALLOCATE(XU,YU,MU,XD,YD,MD) END SUBROUTINE SLOPE(X,Y,DYX) REAL Y(1),DYX(1) COMMON B,FM,FM2,TWOM,FNQ,QG2,SO A=(B+FM*Y(1))*Y(1) SF=(FNQ*((B+FM2*Y(1))/A)**.66666667/A)**2 DYX(1)=(SOSF)/(1.QG2*(B+TWOM*Y(1))/A**3) RETURN END
Program GVFJMP.C #include <stdlib.h> #include <stdio.h> #include <math.h> #define sqr(x) x*x float b,fm,fm2,twom,fnq,qg2,so; extern void rukust(int neq,float *dxs,float xbeg,float xend,\ float error,float *y,float *ytt);
Nonuniform Flows
321
void slope(float x,float *y,float *dyx){float a,sf; a=(b+fm*y[0])*y[0];sf=pow(fnq*pow((b+fm2*y[0])/a,.6666667)/a,2.); dyx[0]=(sosf)/(1.qg2*(b+twom*y[0])/(a*a*a));} // end slope void main(void){int downs,nct,i,id,id1,idm,iu,iu1,nde,nue,n; char fmt[]="%10.2f %9.3f %9.1f\n"; float q,fn,fl,ws,yb,dx1,c,*dxs,dx2,g,yc,qp,f,yc1,a,x,x1,dx,bu,au,bd,ad; float y[1],ytt[1],*xu,*yu,*mu,*xd,*yd,*md; FILE *fil;char fnam[20]; printf("Give:Q,b,m,So,n,L,WS,Yb,dX1,dX2,g\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f",\ &q,&b,&fm,&so,&fn,&fl,&ws,&yb,&dx1,&dx2,&g); if(fm<1.e5) yc=pow(sqr(q/b)/g,.333333); else {qp=pow(fm,3.)*sqr(q)/(g*pow(b,5.)); yc=.925*pow(.5*qp,.284);nct=0; do{ f=pow(yc+sqr(yc),3.)/(1.+2.*yc)qp; yc1=1.01*yc; x1=(yc1yc)*f/(pow(yc1+sqr(yc1),3.)/(1.+2.*yc1)qpf);yc=x1; }while((++nct<20) &&(fabs(x1)>1.e5)); yc=b*yc/fm;} printf("Give name for output file\n");scanf("%s",fnam); if((fil=fopen(fnam,"w"))==NULL){ printf("Failed to open file\n"); exit(0);} fprintf(fil,"Critical Depth =%f\n",yc); if(ws20.) c=1.486; nue=fl/fabs(dx1)+.5; nde=fl/fabs(dx2)+.5; xu=(float *)calloc(nue,sizeof(float)); yu=(float *)calloc(nue,sizeof(float)); mu=(float *)calloc(nue,sizeof(float)); xd=(float *)calloc(nde,sizeof(float)); yd=(float *)calloc(nue,sizeof(float)); md=(float *)calloc(nde,sizeof(float)); qg2=q*q/g;twom=2.*fm;fnq=fn*q/c;fm2=2.*sqrt(fm*fm+1.); y[0]=ws; a=(b+fm*y[0])*y[0]; x1=fl; dx=fabs(dx2); n=nde;downs=1;xd[0]=fl; yd[0]=y[0];md[0]=(.5*b+fm/3.*y[0])*y[0]*y[0]+qg2/a; fprintf(fil,fmt,x1,y[0],md[0]); L5: for(i=1;iyc1){nue=i; goto L14;}} x1=x;} // end for if(downs){n=nue;y[0]=yb;dx=fabs(dx1);downs=0;xd[0]=0.;yd[0]=y[0]; a=(b+fm*y[0])*y[0]; md[0]=(.5*b+fm/3.*y[0])*y[0]*y[0]+qg2/a; x1=0.; fprintf(fil,fmt,x1,y[0],md[0]); goto L5;} L14:id=nde5;idm=nde3;iu=1; L15:iu1=iu1; while((xd[id]>xu[iu])&&(id0))id1–;id=id11;
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Open Channel Flow: Numerical Methods and Computer Applications
if((mu[iu]<md[id1])  (iu==(nue1))) goto L50; iu++;goto L15; L50: bu=(mu[iu]mu[iu1])/(xu[iu]xu[iu1]); printf("%d %d %d %d\n",iu,iu1,id,id1); au=mu[iu]bu*xu[iu];bd=(md[id]md[id1])/(xd[id]xd[id1]); ad=md[id]bd*xd[id];x=(auad)/(bdbu); f=yu[iu1]+(xxu[iu1])/(xu[iu]xu[iu1])*(yu[iu]yu[iu1]); qp=yd[id]+(xxd[id])/(xd[id1]xd[id])*(yd[id1]yd[id]); printf("Position x=%10.2f Upst Depth Y1=%10.3f Downstr Depth\ Y2=%10.3f\n",x,f,qp); fprintf(fil,"Position x=%10.2f Upst Depth Y1=%10.3f Downstr Depth\ Y2=%10.3f\n",x,f,qp); free(xu);free(yu);free(mu);free(xd);free(yd);free(md);}
Example Problem 4.12 Solve Example Problem 4.4 using program GVFJMP. Solution The input needed is, 3 480 8 0 .0011 .013 800 4.9 2 20 −20 32.2 and the solution is, x = 353.7, Y1 = 3.459, and Y2 = 6.497. Program GVFJMP2.FOR INTEGER*2 INDX(3) REAL F(3),D(3,3) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON/TRAS/ B,FM,FM2,TWOM,FNQ,QG2,SO,C,G,FL,YB,WS,X(3) WRITE(6,*)'Give:IOUT,Q,b,m,So,n,L,WS,Yb,g,',,'Est x,Y1 & Y2' READ(5,*) IOUT,Q,B,FM,SO,FN,FL,WS,YB,G,X IF(FM.LT.1.E5) THEN YC=((Q/B)**2/G)**.3333333 GO TO 2 ENDIF QP=FM**3*Q**2/(G*B**5) YC=.925*(.5*qp)**.284 YC1=1.01*YC NCT=0 1 FF=(YC+YC**2)**3/(1.+2.*YC)QP X1=(YC1YC)*FF/((YC1+YC1**2)**3/(1.+2.*YC1)QPFF) YC=YCX1 NCT=NCT+1 IF(ABS(X1).GT.1.E5 .AND. NCT.LT.20) GO TO 1 YC=B*YC/FM 2 WRITE(IOUT,*)' Critical Depth =',YC IF(WS.LT.YC) THEN WRITE(IOUT,*)' Depth at end of channel',' is critical' WS=1.03*YC ENDIF C=1. IF(G.GT.30.) C=1.486 QG2=Q*Q/G TWOM=2.*FM
Nonuniform Flows
3
10
12
100
1 2
3
FNQ=FN*Q/C FM2=2.*SQRT(FM*FM+1.) NCT=0 DO 10 I=1,3 F(I)=FUN(I) DO 10 J=1,3 DX=.005*X(J) X(J)=X(J)+DX D(I,J)=(FUN(I)F(I))/DX X(J)=X(J)DX CALL SOLVEQ(3,1,3,D,F,1,DD,INDX) SUM=0. DO 12 I=1,3 SUM=SUM+ABS(F(I)) X(I)=X(I)F(I) NCT=NCT+1 WRITE(*,*) NCT,SUM,X IF(NCT.LT.30 .AND. SUM.GT. 2.E3) GO TO 3 WRITE(IOUT,100) X FORMAT(' x =',F10.2,' Y1 =',F10.2,' Y2 =',F10.2) END FUNCTION FUN(II) EXTERNAL DYX REAL Y(1),W(1,13),DY(1),XP(1),YP(1,1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON/TRAS/ B,FM,FM2,TWOM,FNQ,QG2,SO,C,G,FL,YB,WS,X(3) HMIN=.0001 H1=1. GO TO (1,2,3),II FUN=(.5*B+FM/3.*X(2))*X(2)**2+QG2/((B+FM*X(2))*X(2))&(.5*B+FM/3.*X(3))*X(3)**2QG2/((B+FM*X(3))*X(3)) RETURN Y(1)=YB CALL ODESOL(Y,DY,1,0.,X(1),1.E6,H1,HMIN,1,XP,YP,W,DYX) FUN=X(2)Y(1) RETURN Y(1)=WS H1=1. CALL ODESOL(Y,DY,1,FL,X(1),1.E6,H1,HMIN,1,XP,YP,W,DYX) FUN=X(3)Y(1) RETURN END SUBROUTINE DYX(XX,Y,DY) REAL Y(1),DY(1) COMMON/TRAS/ B,FM,FM2,TWOM,FNQ,QG2,SO,C,G,FL,YB,WS,X(3) YY=ABS(Y(1)) A=(B+FM*YY)*YY SF=(FNQ*((B+FM2*YY)/A)**.66666667/A)**2 DY(1)=(SOSF)/(1.QG2*(B+TWOM*YY)/A**3) RETURN END
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Open Channel Flow: Numerical Methods and Computer Applications Example Problem 4.13 A flow rate Q = 400 cfs passes under a gate with a depth Y b = 1.5 ft downstream from the gate. The channel downstream from the gate is trapezoidal with b = 10 ft, m = 1, n = 0.014, and a bottom slope So = 0.001. At a length 1000 ft downstream from the gate, the channel ends in a free overfall. Locate the position of the hydraulic jump using program GVFJMP2, i.e., solving the momentum equation and the two GVF equations simultaneously. Solution The input to program GVFJMP2 consists of 6 400 10 1 .001 .014 1000 0 1.5 32.2 300 2.5 4.1 The solution is: x = 319.0 ft, Y1 = 2.493 ft, and Y2 = 4.106 ft. Example Problem 4.14 Solve Example Problem 4.8 writing a computer program that implements the Newton method in simultaneously solving the three algebraic and one ODE equations that govern the problem. Solution The listing of a FORTRAN program that solves Example Problem 4.8 iteratively by means of the Newton method follows. The basic difference between it and SOLGVF is that it solves four simultaneous equations rather than three. The equations are (1) energy at the entrance, (2) energy across the gate, (3) energy across the transition, and (4) the GVF from the downstream gate to the reservoir. Program EPRB4_14.FOR PARAMETER (N=4) REAL F(N),D(N,N),X(N),KL1,KE1,KL,KE COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B1,FM1,B2,H,G,G2,KL1,KE1,FL,TOL,FN,SO,CC,QN, &Q2G,X,Y4,A4,QSG EQUIVALENCE (Q,X(1)),(Y1,X(2)),(Y2,X(3)),(Y3,X(4)) WRITE(*,*)' GIVE:IOUT,TOL,ERR,n,So,b1,m1,b2,H,L,YG,g,KL, &KE,Cc,Ns,YG2' READ(*,*) IOUT,TOL,ERR,FN,SO,B1,FM1,B2,H,FL,YG,G,KL,KE, &CCO,NS,YG2 WRITE(IOUT,200) 200 FORMAT(' Gate Flow rate Depths',' (meters)',/,' (m) &(m**3/s) at Beg.C.',' at End C. at End T.') DYG=(YGYG2)/FLOAT(NS1) Y4=CCO*YG A4=(Y4*B2)**2 IF(G.GT.30.) THEN CC=1.486 ELSE CC=1. ENDIF G2=2.*G KL1=1.+KL KE1=1.+KE WRITE(*,*)' GIVE guess for: Q,Y1,Y2,Y3' READ(*,*) X 50 NCT=0 1 DO 10 I=1,N F(I)=FUN(I) DO 10 J=1,N DX=.005*X(J) X(J)=X(J)+DX D(I,J)=(FUN(I)F(I))/DX
Nonuniform Flows 10 X(J)=X(J)DX C Solves system of equations using Gaussian C Elimination DO 12 J=1,N1 DO 12 I=J+1,N FAC=D(I,J)/D(J,J) F(I)=F(I)FAC*F(J) DO 12 K=J+1,N 12 D(I,K)=D(I,K)FAC*D(J,K) F(N)=F(N)/D(N,N) X(N)=X(N)F(N) SUM=ABS(F(N)) DO 16 I=N1,1,1 FAC=0. DO 14 J=I+1,N 14 FAC=FAC+D(I,J)*F(J) F(I)=(F(I)FAC)/D(I,I) X(I)=X(I)F(I) 16 SUM=SUM+ABS(F(I)) NCT=NCT+1 WRITE(*,110) NCT,SUM,X 110 FORMAT(' NCT=',I2,F12.2,4F10.4) IF(NCT.LT.30 .AND. SUM.GT. ERR) GO TO 1 WRITE(IOUT,100) YG,X 100 FORMAT(F6.2,4F10.3) YG=YGDYG IF(YG.LT.YG2.01) STOP Y4=CCO*YG A4=(Y4*B2)**2 GO TO 50 END FUNCTION FUN(II) EXTERNAL DYX REAL X(4),W(1,13),KL1,KE1,Y(1),DY(1),XP(1),YP(1,1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B1,FM1,B2,H,G,G2,KL1,KE1,FL,TOL,FN,SO,CC,QN, &Q2G,X,Y4,A4,QSG H1=.05 HMIN=.001 Q2G=X(1)*X(1)/G2 GO TO (1,2,3,4),II 1 FUN=HX(2)KE1*Q2G/((B1+FM1*X(2))*X(2))**2 RETURN 2 FUN=X(3)+Q2G/((B1+FM1*X(3))*X(3))**2X(4)KL1*Q2G/ &(B2*X(4))**2 RETURN 3 FUN=X(4)+Q2G/(B2*X(4))**2Y4Q2G/A4 RETURN 4 Y(1)=X(3) XX=FL XZ=0. QN=(FN*X(1)/CC)**2 QSG=X(1)*X(1)/G CALL ODESOL(Y,DY,1,XX,XZ,TOL,H1,HMIN,1,XP,YP,W,DYX) FUN=X(2)Y(1) RETURN
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Open Channel Flow: Numerical Methods and Computer Applications END SUBROUTINE DYX(XX,Y,DY) REAL Y(1),DY(1),KL1,KE1,X(4) COMMON/TRAS/B1,FM1,B2,H,G,G2,KL1,KE1,FL,TOL,FN,SO,CC,QN, &Q2G,X,Y4,A4,QSG EQUIVALENCE (Q,X(1)),(Y1,X(2)),(Y2,X(3)),(Y3,X(4)) P=B1+2.*SQRT(FM1*FM1+1.)*Y(1) A=(B1+FM1*Y(1))*Y(1) SF=QN*((P/A)**.666666667/A)**2 T=B1+2.*FM1*Y(1) DY(1)=(SOSF)/(1.QSG*T/A**3) RETURN END Input data to solve the above problem consists of 3 .00001 .001 .013 .0009 3 2 2.5 2 500 2 9.81 .09 .12 .75 10 .2 with guess for the unknown as: 15 2 2.35 1.6 Solution to problem:
Gate (m) 2.00 1.80 1.60 1.40 1.20 1.00 0.80 0.60 0.40 0.20
Depths (m)
Flow Rate (m**3/s)
at Beg. C.
at End C.
at End T.
14.993 14.795 14.228 13.332 12.137 10.671 8.955 7.012 4.861 2.518
1.926 1.929 1.935 1.943 1.954 1.965 1.976 1.985 1.993 1.998
2.345 2.348 2.357 2.370 2.385 2.401 2.416 2.430 2.440 2.447
1.586 1.740 1.879 2.004 2.114 2.211 2.292 2.358 2.408 2.43
The following observations will help you understand this program: there are two subroutines; one, DYX, that defines the derivative dY/dx for the solver ODESOL as usual, and the other a function subprogram, FUN, that evaluates any of the four equations when asked to do so. When asked to evaluate F4, this function calls on the solver ODESOL to provide the solution over the entire length of the GVF profile. The entire length can be used in a single call to ODESOL since we are not interested in having the depths corresponding to the x’s over the length. If desired, a table of x and Y values could be obtained after the flow rate is determined. The main program defines the problem by reading in appropriate variables, defines the Jacobian matrix D (which is stored in the array D(N,N)), and the equation vector (which is stored in the array F(N)). Thereafter, it solves this linear system of equations and implements the Newton method. Finally, it decreases the gate height and solves the new problem. The program listed below uses the same logic as the above program, but is programmed using Borland’s CLanguage. CLanguage program EPRB4_14.C designed to solve Example Problem 4.14 #include <stdio.h> #include <math.h> #include <stdlib.h> #include "odesol.h" #define sqr(a) (a*a)
Nonuniform Flows
327
#define N 4 float f[N],d[N][N],x[N],kl1,ke1,kl,ke,b1,fm1,b2,h,g=9.81,g2=19.62,fn,\ so,cc=1.,qn,q2g,y4,a4,qsg,tol,err,fl; float fun(int ii,float dx,int j){ float h1=.05,hmin=.001,x1[N]; int i;float *y; y=(float *)calloc(nv,sizeof(float)); for(i=0;i<4;i++) x1[i]=x[i]; if(j > 1) x1[j]+=dx;q2g=sqr(x1[0])/g2; switch (ii){ case 0:return(hx1[1]kl1*q2g/sqr((b1+fm1*x1[1])*x1[1])); case 1:return(x1[2]+q2g/sqr((b1+fm1*x1[2])*x1[2])x1[3]kl1*q2g/\ sqr(b2*x1[3])); case 2:return(x1[3]+q2g/sqr(b2*x1[3])y4q2g/a4); case 3: y[0]=x1[2]; qn=sqr(fn*x1[0]/cc); qsg=sqr(x1[0])/g; odesolc(y,fl,0.,tol,h1,hmin,nstor); return(x1[1]y[0]);}} /* end of fun*/ void slope(float x,float *y,float *dydx) { float p,a,sf,t; p=b1+2.*sqrt(fm1*fm1+1.)*y[0]; a=(b1+fm1*y[0])*y[0]; sf=sqr(qn*pow(p/a,0.666666667)/a); t=b1+2.*fm1*y[0]; dydx[0]=(sosf)/(1.qsg*t/(a*sqr(a))); return;} /* end of slope */ void main(void) { int iout,ns,step,nct,k,i,j; float yg,yg2,dyg,cco,dx=0.,fac,sum; char filena[12]; FILE *fil; cprintf("GIVE:IOUT,TOL,ERR,n,So,b1,m1,b2,H,L,YG,g,KL,KE,Cc,Ns,YG2r\n"); scanf("%d %f %f %f %f %f %f %f %f %f %f %f %f %f %f %d %f",&iout,\ &tol,&err,&fn,&so,&b1,&fm1,&b2,&h,&fl,&yg,&g,&kl,&ke,&cco,&ns,&yg2); if(iout !=6){ cprintf("Give filename for solution data\r\n");scanf("%s",filena); fil=fopen(filena,"w"); fprintf(fil," Gate Flow rate Depths (meters)\n"); fprintf(fil," (m) (m**3/s) at Beg.C. at End C. at End T.\n");} dyg=(ygyg2)/(ns1); if(g>30.) cc=1.486; g2=2.*g; kl1=1.+kl; ke1=1.+ke; cprintf("Give guess for:Q,Y1,Y2,Y3r\n"); scanf("%f %f %f %f",&x[0],&x[1],&x[2],&x[3]); for(step=1;step<=ns;step++) { y4=cco*yg;a4=sqr(y4*b2); nct=0; do { for(i=0;i=0;i–){fac=0.;for(j=i+1;jerr); if(iout!=6) fprintf(fil,"%6.2f %9.3f %9.3f %9.3f \ %9.3f\n",yg,x[0],x[1],x[2],x[3]); else cprintf("%6.2f %9.3f %9.3f %9.3f \
328
Open Channel Flow: Numerical Methods and Computer Applications
%9.3f\r\n",yg,x[0],x[1],x[2],x[3]); yg=dyg; } /* end of step */ if(iout!=6) fclose(fil);} Example Problem 4.15 A trapezoidal channel that is 1200 ft long receives its water from a reservoir with a water surface elevation H = 4.5 ft above its bottom. The bottom slope of the channel is So = 0.0014, n = 0.014 b = 10 ft, and m = 1.5. The entrance loss coefficient is Ke = 0.2. At the downstream end of this channel there is a rectangular channel with a bottom width of b2 = 6 ft, and this downstream rectangular channel is steep. Assume that the loss coefficient for this downstream transition is K L = 0.12.
H = 4.5 ft b = 10 ft, m = 1.5, n = 0.014, So = 0.0014
Rec
tang b2 = ular 6 ft
ion
L = 1200 ft
sit
=0
an Tr
.2
Ke
Solution This problem can be solved using program SOLGVF. It is solved using the program SOLGVFDF, whose listing is provided below. This program illustrates the following alternative techniques to those used in SOLGVF: (1) it calls on DVERK (the FORTRAN program, and the C program calls on rukust), rather than ODESOL as the ODEsolver; (2) it calls on the linear algebra equation solver SOLVEQ, rather than having the Gaussian elimination method built into the code: and (3) it calls on the subroutine FUN(F), in which the argument F is an array containing the three equation values upon return, rather than calling on FUN(II) once to evaluate each equation, and thereafter again with each unknown incremented to evaluate the elements of the Jacobian as is done in SOLGVF. Program SOLGVFDF calls subroutine FUN(F) four times; the first time to obtain the values of the equation vector F, and then three times thereafter with each of the unknowns incremented by multiplying their current value by 1.005. Note this alternative means that the Jacobian matrix evaluation requires two arrays (F and FF) to hold the values of the equations and evaluates the rows within the inner loop with the columns as the outer loop, and results in N + 1 calls to the subroutine FUN(F) that evaluates the equations, rather than N + N*N calls to the function FUN(II), in which N is the number of equations. Study the nested DO 12 loop to understand how the elements of the Jacobian are evaluated. The input data to solve this problem consists of (The same input works for SOLGVF.) 6 1.e−5 .0001 .014 .0014 10 1.5 6 4.5 1200 32.2 .12 .2 250 4 6 and the solution is Q = 250.2 cfs, Y1 = 4.26 ft, Y2 = 5.83 ft. Listing of Program SOLGVFDF.FOR enerates Jacobian by columns by calling subroutine FUN that C G supplies all eqs. C Also calls on linear algebraic equation solver SOLVEQ INTEGER*2 INDX(3) REAL F(3),FF(3),D(3,3),X(3),KL2,KE1,KL,KE COMMON B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN,Q2G,X WRITE(*,*)' GIVE:IOUT,TOL,ERR,FN,SO,B1,FM1,''B2,H,L,g,KL,KE' READ(*,*) IOUT,TOL,ERR,FN,SO,B1,FM1,B2,H,FL,G,KL,KE IF(G.GT.30.) THEN
Nonuniform Flows
1
10 12
14
100
CC=1.486 ELSE CC=1. ENDIF G2=2.*G KL2=.5*KL+1.5 KE1=1.+KE WRITE(*,*)' GIVE guess for: Q,Y1,Y2' READ(*,*) X NCT=0 CALL FUN(F) DO 12 J=1,3 XX=X(J) X(J)=1.005*X(J) CALL FUN(FF) DO 10 I=1,3 D(I,J)=(FF(I)F(I))/(X(J)XX) X(J)=XX CALL SOLVEQ(3,1,3,D,F,1,DET,INDX) SUM=0. DO 14 I=1,3 X(I)=X(I)F(I) SUM=SUM+ABS(F(I)) NCT=NCT+1 IF(NCT.LT.30 .AND. SUM.GT. ERR) GO TO 1 WRITE(IOUT,100) X FORMAT(' Q =',F10.2,' Y1 =',F10.2,' Y2 =',F10.2) END SUBROUTINE FUN(F) EXTERNAL DYX REAL X(3),C(24),W(1,9),KL2,KE1,Y(1),F(3) COMMON B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN,Q2G,X Q2G=X(1)*X(1)/G2 F(1)=X(3)+Q2G/((B1+FM1*X(3))*X(3))**2KL2*((X(1)/ &B2)**2/G)**.333333333 F(2)=HX(2)KE1*Q2G/((B1+FM1*X(2))*X(2))**2 Y(1)=X(3) XX=FL XZ=0. IND=1 QN=(FN*X(1)/CC)**2 Q2G=X(1)*X(1)/G CALL DVERK(1,DYX,XX,Y,XZ,TOL,IND,C,1,W) F(3)=X(2)Y(1) RETURN END SUBROUTINE DYX(N,XX,Y,YPRIME) REAL Y(N),YPRIME(N),KL2,KE1,X(3) COMMON B1,FM1,B2,H,G,G2,KL2,KE1,FL,TOL,FN,SO,CC,QN,Q2G,X P=B1+2.*SQRT(FM1*FM1+1.)*Y(1) A=(B1+FM1*Y(1))*Y(1) SF=QN*((P/A)**.666666667/A)**2 T=B1+2.*FM1*Y(1) YPRIME(1)=(SOSF)/(1.Q2G*T/A**3) RETURN END
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Open Channel Flow: Numerical Methods and Computer Applications Listing of Program SOLGVFKF.C #include <stdlib.h> #include <stdio.h> #include <math.h> float b1,fm1,fm12,fm1s,b2,h,g,g2,kl2,ke1,fl,tol,fn,so,cc,\ fnq,q2g,x[3]; extern void rukust(int neq,float *dxs,float xbeg,float xend,\ float error,float *y,float *ytt); extern void solveq(int n,float **a,float *b,int itype,\ float *dd,int *indx); void slope(float x,float *y,float *dyx){float a,sf; a=(b1+fm1*y[0])*y[0]; sf=pow(fnq*pow((b1+fm1s*y[0])/a,.6666667)/a,2.); dyx[0]=(sosf)/(1.q2g*(b1+fm12*y[0])/(a*a*a));} // end slope void fun(float *f){float xx,xz,y[1],ytt[1],*dxs; q2g=x[0]*x[0]/g2; f[0]=x[2]+q2g/pow((b1+fm1*x[2])*x[2],2.)\ kl2*pow(pow(x[0]/b2,2.)/g,.3333333); f[1]=hx[1]ke1*q2g/pow((b1+fm1*x[1])*x[1],2.); y[0]=x[2]; fnq=pow(fn*x[0]/cc,2.);q2g=2.*q2g;*dxs=.02; rukust\(1,dxs,fl,0.,tol,y,ytt);f[2]=x[1]y[0]; } // End fun void main(void){int nct,i,j,indx[3]; float xx,err,sum,f[3],ff[3],*dxs,*det,**d; d=(float**)malloc(3*sizeof(float*)); for(i=0;i<3;i++)d[i]=(float*)malloc(3*sizeof(float)); printf("Give: TOL,ERR,n,So,b1,m1,b2,H,L,g,KL,Ke\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f %f",\ &tol,&err,&fn,&so,\&b1,&fm1,&b2,&h,&fl,&g,&kl2,&ke1); if(g>20.) cc=1.486; else cc=1.; g2=2.*g;kl2=.5*kl2+1.5;ke1+=1.; fm12=2.*fm1;fm1s=sqrt(fm1*fm1+1.); printf("Give guess for: Q,Y1,Y2\n"); scanf("%f %f %f",&x[0],&x[1],&x[2]); nct=0; do{ fun(f); for(j=0;j<3;j++){xx=x[j]; x[j]+=1.005; fun(ff); for(i=0;i<3;i++) d[i][j]=(ff[i]f[i])/(x[j]xx); x[j]=xx;} solveq(3,d,f,1,det,indx); sum=0.; for(i=0;i<3;i++){x[i]=f[i]; sum+=fabs(f[i]);} } while((++nct<30) && (sum>err)); printf("Q = %10.2f, Y1 =%10.2f, Y2 =%10.2f\n",x[0],x[1],x[2]);} A simultaneous solution of the energy equation and Manning’ equation at the entrance of the channel produces the following normal flow rate and depth: Qo = 385.8 cfs and Yo = 3.56 ft. The input data to these programs consists of 6 .000001 .001 .014 .0014 10 1.5 6 4.5 1200 32.2 .12 .2 250 4 6 and the solution produced consists of
Q = 250.16 cfs, Y1 = 4.26 ft, and Y2 = 5.83 ft. Software packages such as Mathcad, TKSolver, etc., can also be utilized to handle problems involving the simultaneous solution of algebraic and ODEs. For example, the above problem of flow from a reservoir into a channel that changes to a steep channel after a relative short distance
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Nonuniform Flows is solved by TKSolver with the following “Rule” and “Variable” sheets. The library function RK4_se that comes with this package is utilized to solve the GVF equation. This library requires that a function be added that defines the ODE that is to be solved. This function is DYDx and consists of a statement that evaluates dY/dx in Equation 4.4 (see listing below). The list of x’s defines where the solution for Y is to be obtained at. The “rules” basically include four equations: (1) the critical flow at the beginning of the steep channel, (2) the energy across the transition, (3) the GVF ODE equation starting with the depth Y2 at the end of the upstream mild channel to the entrance where it produces Y1, and (4) the specific energy at the entrance of the channel. This TKSolver model can be operated in two ways. The first way is to give a guess for the flow rate and to give the variable Y2 the status of G (for guess), and then press the F9 key to obtain the solution that includes H. The H from this solution is then compared with the water surface elevation and the flow rate is adjusted until the two compare. The second, and most advantageous method, is to let the TKSolver’s iterative solver do the entire job, by giving H a value, and indicating that the variables, Y1, Y2, and Q are guesses as shown below. TKSolver model to solve from into channel from short upstream reservoir to break to a steep grade. Before solution VARIABLE SHEET St Input 'yyy L
Name Y
n b m C So g Ke b2 H Q Y2 Y1 Yc Ec
.014 10 1.5 1.486 .0014 32.2 .2 6 4.5 G 200 G 6 G 4.3
After solution VARIABLE SHEET St Input Name
Output
L 'yyy .014 10 1.5 1.486 .0014 32.2 .2 6 4.5
265.73 5.81 4.22 3.93 5.90
Y n b m C So g Ke b2 H Q Y2 Y1 Yc Ec
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Open Channel Flow: Numerical Methods and Computer Applications ════ RULE SHEET════════ S Rule───────────────── Yc=((Q/b2)^2/g)^.3333333 Ec=1.5*Yc Y2+(Q/((b+m*Y2)*Y2))^2/(2*g)=Ec place(Y,1)=Y2 call RK4_sen('DYDx,Y,'x) Y1=ELT(Y,21) H=Y1+(1+Ke)*(Q/((b+m*Y1)*Y1))^2/(2*g) ════ FUNCTION SHEET═════ Name────────── Type───── Arguments── Comment── DYDx Procedure 2;1 Given differential equation RK4_sen Procedure 3;0 Classical 4thOrder RungeKutta ════════════ PROCEDURE FUNCTION: DYDx ═════════ Comment: Given differential equation Parameter Variables: n,b,m,C,Q,So,g Input Variables: x,y Output Variables: y' S Statement────────────────────────────────────────────────────── y':=(So(n*(b+2*sqrt(m*m+1)*y)^.66666667*Q/(C*((b+m*y)*y) ^1.666667))^2)/(1Q*Q/g/((b+m*y)*y)^3) ════════════ PROCEDURE FUNCTION: RK4_sen ════════ Comment: Classical FourthOrder RungeKutta method, single eqn Parameter Variables: b,m,g,Q Input Variables: EQ,y,x Output Variables: S Statement───────────────────────────────────────────────────── name of a function with the 1storder equation ; Notation: EQ y'=f(x,y) ; x independent variable (list) ; y dependent variable, y=F(x), (list) ; K RungeKutta coefficients (list) ; Description: This procedure represents an implementation of a ; classical 4thorder RungeKutta procedure for numerical ; integration of a single ordinary differential equation ; y'=f(x,y). Given a function name passed as a symbolic value ; of the Input Variable EQ, list of values of the independent ; variable x, and an initial condition as the value of the 1st ; element of the list y, the procedure generates the solution ; in the rest of the list y. xi:= x[1] yi:= y[1] for i=2 to length(x) ye:= yi h:= (x[i]xi)/2 for j=1 to 3 'K[j]:= apply(EQ,xi,ye) if mod(j,2) then xi:= xi + h if j=3 then h:= 2*h ye:= yi + h*'K[j] next j
Nonuniform Flows 'K[4]:= apply(EQ,xi,ye) yi:= yi + dot('K,1,2,2,1)*h/6 y[i]:= yi next i call delete('K) ════ LIST: x ════════ independent variable Element── Value────── 1 1200 2 1140 3 1080 4 1020 5 960 6 900 7 840 8 780 9 720 10 660 11 600 12 540 13 480 14 420 15 360 16 300 17 240 18 180 19 120 20 60 21 0 ═════ TABLE: Solution ═════ Title: y = f(x) Element x y 1 1200 5.809 2 1140 5.727 3 1080 5.646 4 1020 5.564 5 960 5.483 6 900 5.402 7 840 5.321 8 780 5.240 9 720 5.160 10 660 5.079 11 600 5.000 12 540 4.920 13 480 4.841 14 420 4.762 15 360 4.683 16 300 4.605 17 240 4.528 18 180 4.451 19 120 4.374 20 60 4.299 21 0 4.224
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Open Channel Flow: Numerical Methods and Computer Applications Example Problem 4.16 A channel receives its water over a 30 ft length. The flow is free to move in both directions from this inflow section. The inflow is constant at a rate q* = 25 cfs/ft, has an incoming velocity of U = 8 fps, and is directed at an angle of 45° downward and to the right as shown. From this inflow section, the channel slopes downward in both directions with So = 0.0005. The channel has a bottom width of b = 10 ft, a side slope of m = 1.5, and a Manning’s n = 0.018 in both directions. Solve the flow rates and depths in both directions. q* = 25 cfs/ft 30 ft Q1
Y1 = Yo1
So = –0.0005
Y2 = Yo2
b=10 ft, m = 1.5, S
o = 0.0005,
Q2
n = 0 .018
Solution In this problem, there are four unknowns: Q1 (flow rate to the left), Q2 (flow rate to the right), and the depths Y1 and Y2 in the channels flowing to the left and right respectively. The four equations needed to be solved for these four unknowns consist of
F1 = Q1 + Q 2 − Lq* = 0
F2 = Q1 − C u /n A1R 2h1/ 3 /So = 0
F3 = Q 2 − C u /n A 2 R 2h2/ 3 /So = 0
F4 = Yo2 − Ygvf = 0 (in which the GVF solution will begin at the beginning of the inflow section with Y = Yo1.) The solution to the problem gives Y1 = 4.60 ft, Y2 = 5.76 ft, Q1 = 293.5 cfs, and Q2 = 456.5 cfs. The program that obtains this solution is listed below. It calls on a linear algebra solver SOLVEQ that returns the solution in the vector F. Notice that the subroutine DYX that supplies dY/dx to the ODESolver shows that the flow rate (variable QQ) is negative upstream from where the flow divides, and if (Sf )1/2 is solved from Manning’s equation with Q negative, a negative value is produced, but when a negative is squared, a positive value results. Therefore, Sf is computed from Sf = (Sf )1/2 S1f / 2 , so it is negative in the numerator of the ODE. FORTRAN listing SOLINF.FOR to solve above four equations. REAL F(4),D(4,4),X(4) INTEGER*2 INDX(4) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,TOL,FN,SO,CC,X WRITE(*,*)' GIVE:IOUT,TOL,ERR,FN,SO,B,FM,L,g,qs,Uq,Angle' READ(*,*) IOUT,TOL,ERR,FN,SO,B,FM,FL,G,QS,Uq,Angle Uqx=Uq*COS(.017455329*Angle) FM2=2.*FM FMS=2.*SQRT(FM*FM+1.) SS=SQRT(SO) IF(G.GT.30.) THEN CC=1.486 ELSE CC=1. ENDIF WRITE(*,*)' GIVE guess for: Q1,Q2,Y1,Y2'
Nonuniform Flows
1
10
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1 2 3 4
READ(*,*) X NCT=0 DO 10 I=1,4 F(I)=FUN(I) DO 10 J=1,4 DX=.005*X(J) X(J)=X(J)+DX D(I,J)=(FUN(I)F(I))/DX X(J)=X(J)DX CALL SOLVEQ(4,1,4,D,F,1,DD,INDX) WRITE(6,222) F NCT=NCT+1 DIF=0. DO 20 I=1,4 X(I)=X(I)F(I) DIF=DIF+ABS(F(I)) IF(NCT.LT.30 .AND. DIF.GT. ERR) GO TO 1 WRITE(IOUT,100) X FORMAT(' Q1 =',F10.2,' Q2 =',F10.2,' Y1 =',F10.2,' Y2 =', &F10.2) END FUNCTION FUN(II) EXTERNAL DYX REAL X(4),W(2,13),Y(1),DY(1),XP(2),YP(2,2) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,TOL,FN,SO,CC,X GO TO (1,2,3,4),II FUN=X(1)+X(2)FL*QS RETURN FUN=FN*X(1)*(B+FMS*X(3))**.666667CC*((B+FM*X(3))* &X(3))**1.666667*SS RETURN FUN=FN*X(2)*(B+FMS*X(4))**.666667CC*((B+FM*X(4))* &X(4))**1.666667*SS RETURN Y(1)=X(3) H1=.05 HMIN=.001 XX=0. XZ=FL CALL ODESOL(Y,DY,1,XX,XZ,TOL,H1,HMIN,1,XP,YP,W,DYX) FUN=X(4)Y(1) RETURN END SUBROUTINE DYX(XX,Y,DY) REAL Y(1),DY(1),X(4) COMMON /TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,TOL,FN,SO,CC,X YY=ABS(Y(1)) P=B+FMS*YY A=(B+FM*YY)*YY A2=A*A*G QQ=QS*XXX(1) VD=QQ/AUqx Q2=QQ*QQ SF=FN*ABS(QQ)/CC*(P/A)**.666666667/A SF=SF*ABS(SF)
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Open Channel Flow: Numerical Methods and Computer Applications DY(1)=(SOSFQQ*QS/A2VD*QS/(G*A))/(1.Q2*(B+FM2*YY)/(A*A2)) RETURN END C listing SOLINF.C to solve the above four equations #include <stdlib.h> #include <stdio.h> #include <math.h> float b,fm,fms,fm2,ss,qs,g,uqx,fl,tol,fn,so,cc,x[4]; extern void solveq(int n,float **d,float *f,int itype,\ float *dd, int *indx); extern rukust(int n,float *dxs,float xb,float xe,float err,\ float*y,float *ytt); void slope(float xx,float *y,float *dy){ float yy,p,a,a2,qq,vd,q2,sf; yy=fabs(*y);p=b+fms*yy;a=(b+fm*yy)*yy;a2=a*a*g; qq=qs*xxx[0]; vd=qq/auqx;q2=qq*qq;sf=fn*fabs(qq)/cc*pow(p/a,.6666667)/a; sf*=fabs(sf); *dy=(sosfqq*qs/a2vd*qs/(g*a))/(1.q2*(b+fm2*yy)/(a*a2));} // End of slope float fun(int ii){float dxs[1],y[1],ytt[1],xx,xz,arg; if(ii==0) arg=x[0]+x[1]fl*qs; else if(ii==1) arg=fn*x[0]*pow(b+fms*x[2],.666667)\ cc*ss*pow((b+fm*x[2])*x[2],1.666667); else if(ii==2) arg=fn*x[1]*pow(b+fms*x[3],.666667)\ cc*ss*pow((b+\fm*x[3])*x[3],1.666667); else{y[0]=x[2];dxs[0]=.05;xx=0.,xz=fl; rukust(1,dxs,xx,xz,tol,y,ytt);arg=x[3]y[0];} return arg;} // End fun void main(void){int i,j,nct,indx[4]; float dx,dif,uq,angle,err,f[4],dd[1],**d;char fname[20]; FILE *filo; d=(float**)malloc(4*sizeof(float*)); for(i=0;i<4;i++)d[i]=(float*)malloc(4*sizeof(float)); printf("Give name of output file\n");scanf("%s",fname); if((filo=fopen(fname,"w"))==NULL){ printf("Cannot open file"); exit(0);} printf("GIVE:TOL,ERR,FN,SO,B,FM,L,g,qs,Uq,Angle\n"); scanf("%f %f %f %f %f %f %f %f %f %f %f",\ &tol,&err,&fn,&so,&b,\&fm,&fl,&g,&qs,&uq,&angle); uqx=uq*cos(.017455329*angle); fm2=2.*fm; fms=2.*sqrt(fm*fm+1.); ss=sqrt(so); if(g>30.) cc=1.486; else cc=1.; printf("Give guess for: Q1,Q2,Y1,Y2\n"); for(i=0;i<4;i++)scanf("%f",&x[i]);nct=0; do{for(i=0;i<4;i++){ f[i]=fun(i); for(j=0;j<4;j++){ dx=.005*x[j];x[j]+=dx; d[i][j]=(fun(i)f[i])/dx; x[j]=dx;}} solveq(4,d,f,1,dd,indx); dif=0.; for(i=0;i<4;i++){ x[i]=f[i];dif+=fabs(f[i]);} printf("nct=%d, dif=%f\n",++nct,dif); }while((nct<30) && (dif>err)); fprintf(filo," Q1 =%10.2f, Q2 =%10.2f, Y1 =%10.3f, Y2=%10.3f\n",\ x[0],x[1],x[2],x[3]); fclose(filo);}
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Nonuniform Flows
In solving the above problem, it was assumed that the slope of the channel upstream from the beginning of the lateral inflow was equal, but in the opposite direction. If the channel has a constant slope in the same direction, then a uniform flow cannot exit upstream, and eventually the depth will become zero for a wedge of water upstream from the inflow that is stationary under a steadystate flow. For this situation, all the lateral inflow will contribute to the downstream flow rate Q2 = q*L. To solve this case, since Q2 is known, first solve the downstream uniform depth Yo2, and as a second step solve the GVF (with the lateral inflow term included) starting at the downstream end with Yo2 for Y1. For this example problem, Q2 = 750 cfs, Yo2 = 7.366 ft, and Y1 = 6.855 ft. Another case is that the slope So continues upstream for an additional distance L1 as shown in the sketch below. At the upstream end, there will be a boundary condition that establishes the depth Yu. The three possible conditions are (1) the channel ends in a free overfall, (2) the channel discharges into a reservoir with a known water surface elevation WS, or (3) there is a channel with a bottom slope in the opposite direction and it contains a uniform flow. Notice that we have now added one more unknown Yu so that Q1, Q2, Y1, Y2 = Yo2, and Yu are five unknowns. The five equations needed for case (1) are L1
Yu
Q1
L
q*
θ Y1
Y
Y2 = Yo2
Q2
b, m, n, So
F1 = Q1 + Q 2 − Lq* = 0
F2 = Q12 Tu − gA 3u = 0 (the critical flow equation)
F3 = Q 2 − Cu /n A 2 R 2h2/ 3 /So = 0 (Manning’s equation downstream)
F4 = Y1 − Y1ode (Yu ) = 0 (ODE solution over L1 starting with Yu ) F5 = Yo2 − Y2ode (Y1 ) = 0 (ODE sol. over L starting with Y1 and including lateral inflow terms)
If the second boundary condition with a known reservoir water surface WS applies, then Yu = WS, and again only four variables are unknown. An alternative for case (2) is to replace the second equation with F2 = Yu − WS = 0 and keep Yu as unknown. For the third case, the second equation would be replaced by Manning’s equation writing the function of Yu equal to zero. For example, F2 = nQ Pu2 / 3 − Cu A 5u / 3S1o/22 = 0 . The program SOLINF2, whose listing is given below, is designed to solve problems with an upstream channel with a length L1 upstream from the lateral inflow, and any of the three above mentioned upstream boundary conditions. Listing of Program SOLINF2.FOR EXTERNAL DYX REAL F(5),D(5,5),X(5) INTEGER*2 INDX(5) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON/TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,FL1,TOL,FN,SO, &CC,X,So2,WS,IBC,INFLOW
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WRITE(*,*)' GIVE:IOUT,TOL,ERR,FN,SO,B,FM,L,L1,g,qs,Uq,Angle, &BC(1=Yc,2=WS or 3=Channel)' READ(*,*) IOUT,TOL,ERR,FN,SO,B,FM,FL,FL1,G,QS,Uq,Angle,IBC IF(IBC.EQ.2) THEN WRITE(*,*)' Give upstream wselevation' READ(*,*) WS ENDIF IF(IBC.EQ.3) THEN WRITE(*,*)' Give slope of upstream channel' READ(*,*) So2 So2=SQRT(So2) ENDIF Uqx=Uq*COS(.017455329*Angle) FM2=2.*FM FMS=2.*SQRT(FM*FM+1.) SS=SQRT(SO) IF(G.GT.30.) THEN CC=1.486 ELSE CC=1. ENDIF WRITE(*,*)' GIVE guess for: Q1,Q2,Y1,Y2,Yu' READ(*,*) X NCT=0 DO 10 I=1,5 F(I)=FUN(I,DX,0) DO 10 J=1,5 DX=.005*X(J) D(I,J)=(FUN(I,DX,J)F(I))/DX CALL SOLVEQ(5,1,5,D,F,1,DD,INDX) NCT=NCT+1 DIF=0. DO 20 I=1,5 X(I)=X(I)F(I) DIF=DIF+ABS(F(I)) WRITE(*,200) NCT,DIF,X FORMAT(' NCT=',I2,' DIF=',E12.5,/,2F10.2,3F10.3) IF(NCT.LT.30 .AND. DIF.GT. ERR) GO TO 1 WRITE(IOUT,100) X FORMAT(' Q1 =',F10.2,' Q2 =',F10.2,' Y1 =',F10.2,' Y2 =', &F10.2,' Yu =',F10.2) END FUNCTION FUN(II,DX,J) EXTERNAL DYX REAL X(5),W(2,13),Y(1),DY(1),XP(2),YP(2,2) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON/TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,FL1,TOL,FN,SO,CC,X, &So2,WS,IBC,INFLOW IF(J.GT.0) X(J)=X(J)+DX GO TO (1,2,3,4,5),II
Nonuniform Flows
1 2
3 4
5
10
FUN=X(1)+X(2)FL*QS GO TO 10 IF(IBC.EQ.1) THEN FUN=X(1)**2*(B+FM2*X(5))G*((B+FM*X(5))*X(5))**3 ELSE IF(IBC.EQ.2) THEN FUN=WSX(5) ELSE FUN=FN*X(1)*(B+FMS*X(5))**.666667CC*((B+FM*X(5))*X(5)) &**1.666667*So2 ENDIF GO TO 10 FUN=FN*X(2)*(B+FMS*X(4))**.666667CC*((B+FM*X(4))*X(4)) &**1.666667*SS GO TO 10 Y(1)=X(5) IF(IBC.EQ.1) Y(1)=1.1*Y(1) INFLOW=0 H1=.01 HMIN=.00001 XX=0. XZ=FL1 CALL ODESOL(Y,DY,1,XX,XZ,TOL,H1,HMIN,1,XP,YP,W,DYX) FUN=X(3)Y(1) GO TO 10 Y(1)=X(3) INFLOW=1 H1=.01 XZ=FL CALL ODESOL(Y,DY,1,XX,XZ,TOL,H1,HMIN,1,XP,YP,W,DYX) FUN=X(4)Y(1) IF(J.GT.0) X(J)=X(J)DX RETURN END SUBROUTINE DYX(XX,Y,DY) REAL Y(1),DY(1),X(5) COMMON/TRAS/B,FM,FMS,FM2,SS,QS,G,Uqx,FL,FL1,TOL,FN,SO, &CC,X,So2,WS,IBC,INFLOW YY=ABS(Y(1)) P=B+FMS*YY A=(B+FM*YY)*YY A2=A*A*G IF(INFLOW.EQ.1) THEN QSS=QS QQ=QS*XXX(1) VD=QQ/AUqx ELSE QQ=X(1) VD=0. QSS=0. ENDIF
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Q2=QQ*QQ SF=FN*QQ/CC*(P/A)**.666666667/A SF=SF*ABS(SF) DY(1)=(SOSFQSS*(QQ/A2+VD/(G*A)))/(1.Q2*(B+FM2*YY)/(A*A2)) RETURN END Let us solve the previous problem with an upstream channel with a length L1 = 500 ft using the three boundary conditions. The input to solve case (1) for a free overfall is 6 1.e−5 .001 .018 .0005 10 1.5 30 500 32.2 25 8 45 1 300 450 4.6 5.7 2.7 with the output for the solution as Q1 = 308.91 Q2 = 441.09 Y1 = 4.33 Y2 = 5.66 Yu = 2.69
For case (2), let us specify a reservoir water surface elevation 4 ft above the channel bottom. The input now becomes 6 1.e−5 .001 .018 .0005 10 1.5 30 500 32.2 25 8 45 2 4 300 450 4.6 5.7 4. with the output:
Q1 = 289.44 Q2 = 460.56 Y1 = 4.67 Y2 = 5.79 Yu = 4.00
For case (3), let us give the channel upstream a bottom slope So2 = 0.001; then the input to INFSOL2 is 6 1.e−5 .001 .018 .0005 10 1.5 30 500 32.2 25 8 45 3 .001 300 450 4.6 5.7 4.4 with the output:
Q1 = 294.51 Q2 = 455.49 Y1 = 4.59 Y2 = 5.76 Yu = 3.84
A couple of comments are in order. First, it was assumed that the long channel upstream from Yu had the same size as the channel downstream, therefore only its bottom slope was in the opposite direction. Should this channel be of a different size, or there be a difference in the bottom elevation, then Yu would need to be replaced by two depths Yu1 and Yu2 in these two different channels. Thus, there would be six unknown variables. The additional sixth equation would be the energy equation across the transition between the two channels, or F6 = Yu2 + Q12/(gAu22) − Yu1 − (1 + K L) Q12/(gAu12) − Δz = 0. The second comment is that if the GVF is solved from the beginning of the channel where the depth is Yu to the end of the lateral inflow where the depth is Y2 = Yo2, then Y1 could be eliminated as an unknown. The solution to the ODE would need to turn to the lateral inflow terms when the solution is with the length L, but have these terms zero when the solution is within L1. If this alternative is used, then equations F4 and F5 would be replaced with the single equation F4 = Yo2 − Y2ode(Yu) = 0.
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4.10 Nonprismatic Channels Transitions between channels of different sizes can alter the type of GVF profile that exists in a channel. The GVF sketches that were previously shown in this chapter assumed that the channel is of constant size and shape throughout its entire length. Such channels are called prismatic channels. If the slope of the land changes, it would not be economical to use the same large channel for steep slopes, that are needed for very mild slopes. The size of the channel will in practice, therefore, be changed to reflect the changing land slopes. When a channel changes from one size to another, or from one shape to another, it is referred to as a nonprismatic channel. The following examples illustrate how effects due to nonprismatic channels can alter the type of GVF that may exist in different portions of the channel system. Example Problem 4.17 Sketch in the GVF profile in the channel shown below assuming it is prismatic throughout its entire length.
H Mild Stee p
Horizontal
Ste ep
Solution Since this channel is prismatic, i.e., neither its shape nor size changes throughout its entire length, the GVF profiles are as given below.
M2
S1
H Yc Mild
S1
S2 Steep
H2
S2
Yc Horizontal
S2 Stee p
Example Problem 4.18 The channel in the previous problem consists of (1) a 5000 ft long trapezoidal channel, followed by; (2) a 6000 ft long rectangular channel that contains a sluice gate, which is followed by; (3) a horizontal trapezoidal channel with a 3000 ft length; and (4) a 5000 ft long steep circular channel as shown in the sketch below. Determine the flow rate entering this channel if the upstream reservoir level is 10 ft above the channel bottom. Also, locate any hydraulic jumps and determine the water surface profiles throughout the channel. Smooth transitions occur between the different shapes of the channel, and the bottom elevation does not change through the transitions. The gate with a contraction coefficient of Cc = 0.6 has its tip at YG = 5 ft above the channel bottom so the depth downstream from the gate is 3 ft.
342
Open Channel Flow: Numerical Methods and Computer Applications S1
10 ft 9.86 ft Q = 1012 cfs
.2 =0
Ke
Trapezoidal mild b = 15 ft m=2 So= 0.0001 L =5000 ft n = 0.012
S
2
14.50 ft 10.11 ft 6.83 ft L=
289
0 ft
20.28 ft
Recta ngula r stee b= p So = 010 ft .0 n=0 5 L1 =30.012 L = 60 00 ft 00 ft
Jump occurs in transition
M1
H2 10.39 ft Horizontal trapezoidal b = 15 ft m=2 n = 0.012 L = 3000 ft
S 10.35 ft 2 7.29 ft
3.79
Circ u D = lar stee So = 012 ft p .0 n= 0 4 L = 50.015 00 ft
Solution The transitions can alter the types of GVF profiles considerably from the previous problem. The question, “What is the flow rate?” raises several other questions, such as (a) Does the critical flow section between the upstream mild trapezoidal channel and the steep rectangular channel “choke” the flow causing an M1 GVF profile to exist in the upstream channel rather than an M2 GVF profile? (b) Does the sluice gate increase the depth at the head of the rectangular channel above the critical depth? (c) If the S1 GVF profile upstream from the gate does not extend up to the break in grade, where will the hydraulic jump occur? (d) Where will the hydraulic jump downstream from the gate occur? (e) What will the depth be at the end of the horizontal channel before the transition changes the cross section to that of a circle? Unfortunately, the answers to all of these questions depend upon knowing the flow rate, and the flow rate will depend upon whether the M2 or M1 GVF profile, which ever it is, extends to the reservoir or not. To arrive at the solution, begin by assuming that a normal depth does exist in the upper portion of the first trapezoidal channel. If so, a simultaneous solution of Manning’s equation and the energy equation will give the flow rate and the normal depth. These equations are
F1 =
nQP 2 /3 − 1.486A 5/3 =0 So
(1)
and
F2 = H − Yo − (1 + K L)Q2(2gA2) = 0
(2)
The solution gives Q = 1320.3 cfs and Yo = 9.707 ft. With this flow rate, the critical depth at the head of the next rectangular channel can be calculated from Yc = 3 q 2 /g = 8.150 ft (q = 132.03 cfs/ft), and Ec = 1.5 Yc = 12.225 ft. If the losses through this transition are ignored, then the depth in the trapezoidal channel at the beginning of the transition can be obtained from
Y2 +
Q2
(2gA ) 2 2
= 12.225
giving Y2 = 12.105 ft. Since this depth is above the normal depth Yo = 9.697 ft, the transition has “choked” the flow causing an M1 GVF profile to exist rather than an M2 as sketched in for the previous problem. Starting from this downstream depth in the trapezoidal channel, the M1 GVF profile will be solved. Since the depth may not be normal at the entrance of the channel, it will be more convenient to use an ODE for this computation that assumes Y is the dependent variable and x is the independent variable. The FORTRAN listing below utilizes ODESOL described in Appendix C to obtain this solution. The input data to this program consists of 3 .001 −500 12.105 1320.3 .0001 .012 15 2 5000 0
ft
Nonuniform Flows FORTRAN listing that utilizes ODESOL (EPRB4_18.FOR) REAL Y(1),DY(1),XP(1),YP(1,1),WK1(1,13) EXTERNAL DYX COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE,B,FM,FN,SO,Q2,FNQ WRITE(6,* )'GIVE IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND' 1 READ(5,*) IOUT,TOL,DELX,YB,Q,FN,SO,B,FM,XBEG,XEND H1=.01 Y(1)=YB FNQ=FN*Q/1.49 Q2=Q*Q/32.2 X=XBEG WRITE(IOUT,100) X,Y 2 XZ=X+DELX CALL ODESOL(Y,DY,1,X,XZ,TOL,H1,HMIN,1,XP,YP,WK1,DYX) X=XZ WRITE(IOUT,100) X,Y 100 FORMAT(6X,2F10.3) IF(DELX .LT. 0.) GO TO 8 IF(X .LT. XEND) GO TO 2 GO TO 1 8 IF(X .GT. XEND) GO TO 2 STOP END SUBROUTINE DYX(X,Y,DY) REAL Y(1),DY(1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE,B,FM,FN,SO,Q2,FNQ 20 A=(B+FM*Y(1))*Y(1) T=B+2.*FM*Y(1) P=B+2.*SQRT(FM*FM+1.)*Y(1) SF=(FNQ*(P/A)**.66666667/A)**2 A3=A**3 FR2=Q2*T/A3 40 DY(1)=(SOSF)/(1.FR2) RETURN END Solution M1 GVF profile x(ft) Y(ft) ————––– ——–––– 5000 12.105 4500 12.073 4000 12.042 3500 12.011 3000 11.980 2500 11.949 2000 11.918 1500 11.888 1000 11.858 500 11.828 0 11.798 The depth of 11.798 at the beginning of the trapezoidal channel is well above the depth computed based on uniform flow conditions, and is even above the water surface elevation in the reservoir. Therefore, the flow rate must be reduced and the entire process, given above, repeated. This flow rate must be selected by trial. After several trials, the flow rate Q = 1012 cfs is used. Based on this flow rate, the critical depth at the end of the first transition is Yc = 6.826 ft, Ec = 10.239 ft, and Y2 = 10.114 ft. The solution to the M1 GVF profile gives
343
344
Open Channel Flow: Numerical Methods and Computer Applications x (ft)
Y (ft)
5000.000 4500.000 4000.000 3500.000 3000.000 2500.000 2000.000 1500.000 1000.000 500.000 .000
10.114 10.088 10.061 10.035 10.010 9.985 9.959 9.935 9.910 9.886 9.862
If the upstream depth of Y1 = 9.862 ft is substituted into the energy equation H = Y1 + Q 2/(2gA 2 ) = 10.02 ft
which is close enough to the 10 ft of head on the reservoir. The flow rate is now known and the rest of the problem can be solved. This solution is given below, but you should verify the details, because getting them right will ensure that you understand what needs to be done. (You should also solve this problem using the Newton method as has been done in the last Example Problems 4.6 and 4.7.)
Having a critical section at the end of a transition, as in Example Problem 4.18, that reduces the flow rate in the channel from 1320.3 to 1012 cfs, does not constitute a very effective design. Elevation changes through the transitions are needed to prevent such occurrences. This will often require some earth work to change the slope of the channel instead of having the bottom of the channel on the same grade as the ground. However, since ground elevations seldom change abruptly, this will be needed anyway. The subject of the design of transitions is covered in Chapter 5. Example Problem 4.19 Redesign the channel in the previous problem by changing the bottom elevation through the transition, so that a uniform depth occurs throughout the entire length of the upstream trapezoidal channel. Also change the bottom elevation through the transition just upstream from the circular section. Solution The sketch below shows this redesigned channel. Q = 1320.3 cfs 10 ft
.0 =3
9.70 ft Trapezoidal b = 15 ft m=2 So = 0.0001 L = 5000 ft n = 0.012
32.82 ft
20.13 ft 3.17 ft
Rectangular b = 10 ft So = 0.0466 L = 6000 ft n = 0.012
ft)
ft n = 5 , Y dow .6 0 = (C c 33.07 ft YG
33.0 ft
3.17 ft Trapezoidal b = 15 ft m=2 So = 0.000 L = 3000 ft n = 0.012
9.75 ft
4.43 ft Circular D = 12 ft So = 0.038 L = 5000 ft n = 0.015
345
Nonuniform Flows Example Problem 4.20 In the sketch below, a channel system is shown that receives water from a reservoir with a water surface elevation 2.5 m above the channel bottom. At its beginning, the channel is rectangular with b1 = 4 m, So1 = 0.001, and a length L1 = 20 m, then there is a short transition to a trapezoidal channel with b2 = 5 m, m2 = 1.5, So2 = 0.0008, and L2 = 2000 m, then again a short transition to a circular section with D3 = 3 m, So3 = 0.05, and L3 = 4000 m, and finally a third transition to a trapezoidal channel with b4 = 4 m, m4 = 1.5, So4 = 0.0012, and L 4 = 8000 m. Manning’s n = 0.013 for all channels. The last trapezoidal channel contains two gates, one at its end, and the other 4000 m upstream therefrom. Immediately in front of each gate, a diversion of 10 m3/s takes place and each gate has a contraction coefficient of Cd = 0.58. The gates are set at YG1 = 0.86 m and YG2 = 0.30 m, respectively, above the channel bottoms. Determine the flow rate into this system and compute the water surface profiles throughout the system’s length. (The entrance loss coefficient is Ke = 0.15.) M
2
.86
b4 = 4m m4 = 1.5 S04 = 0.0012 L4 = 8000 m
m
M3
2 = 0.3
=0
YG
D3 = 3 So3 = 0 m L3 = 4 .05 m 000 m
m3 /s
G1
10
Y
0m
4000 m M1
Q=
b1 = 4 m So1 = 0.001 L1 = 20 m
b2 = 5 m, m2 = 1.5 So2 = 0.008 L2 = 2000 m
M1
0 m3 /s
K
e
=0
.15
S3
Q=1
2.5 m
Solution Note that channel 3, which is the pipe, is steep since So3 = 0.05, and therefore a critical depth is expected at its beginning. The solution will be accomplished without the aid of a computer program that solves a system of algebraic and ordinary differential equations, simultaneously. Thus, the solution must begin by estimating what the flow rate will be. Limiting values for Q are likely between (a) that would occur if the upstream rectangular channel were very long, and (b) if the channel began with the second trapezoidal channel, and it were very long. The simultaneous solution of the energy equation and Manning’s equation for these two cases gives 21.13 and 44.69 m3/s, respectively, with corresponding normal depths of Yo1 = 2.138 and Yo2 = 2.085 m, respectively. The flow rate might be outside of these bounds if the transition to the circular section caused an M1 GVF profile that extended to the reservoir, or if an M2 GVF profile from this position extended to the reservoir. The latter seems unlikely. Start by assuming Q = 25 m3/s, then the critical depth and the specific energy at the beginning of the circular section are Yc3 = 2.192 and Ec3 = 3.23 m, respectively. Equating E2 at the downstream end of the 5 m wide trapezoidal channel to 3.23 m and solving the depth gives Y2end = 3.198 m (the normal depth in this trapezoidal channel for a flow rate of 25 m3/s is Yo2 = 1.521 m). Next, solving the GVF profile for the beginning of this trapezoidal channel gives Y2beg = 1.801 with a corresponding specific energy E2beg = 1.967. However, if we check the specific energy in the upstream rectangular channel associated with a critical flow of Q = 25 m3/s, we find that Ec1 = 2.378 m (Yc1 = 1.585 m), (Ec1 + KeVc12/(2g) = 2.507 m), which is larger than E1beg, and therefore the backwater curve in the trapezoidal channel will have no effect on the flow rate since Ec1 = 2.378 m is greater than E2beg = 1.967 m. The upper limiting flow rate rather than being 44.69 m3/s, as determined above, will be that obtained by the simultaneous solution of the energy, and the critical flow equation for channel 1, or 24.82 m3/s (with Yc1 = 1.587 m). The actual flow rate will be slightly less than this because the frictional losses in the upstream 20 m long channel will reduce this. After a few trial flow rates, shown below, followed by solving the M2 GVF profile to the reservoir, it is found that Q = 24.8 m3/s (e.g., the upstream 20 m long channel frictional loss has had an insignificant effect in reducing the flow rate).
346
Open Channel Flow: Numerical Methods and Computer Applications Trial 1 2 3 4 5
Q (m3/s)
Yc (m)
Y1beg (m)
E (m)
23 24 24.3 24.6 24.8
1.499 1.542 1.555 1.568 1.577
1.684 1.731 1.746 1.762 1.771
2.361 2.435 2.456 2.478 2.594
The normal depths associated with this flow rate of Q = 24.8 m3/s in each of the channels are Yo2 = 1.515 m, Yo3 = 1.016 m, Yo4 = 1.496 m, respectively, and the normal depths associated with the reduced flow rates of 14.8 m3/s downstream from the first gate is 1.132 m. The difference between the critical specific energy in channel 1 (Ec1 = 2.37 m) and that associated with the normal depth in channel 2 (Eo2 = 1.77 m) will be dissipated in the expansion. The solution to the two M1 GVF profiles in front of the two gates are shown below, and because of the small differences in the normal depths between channels 3 and 4, the hydraulic jump that takes the supercritical flow from channel 3 to the subcritical flow in channel 4 will take place in the enlarging transition. M1 GVF profile upstream from gate # 2 x (m)
Y (m)
E (m)
M (m3)
8000.000 7600.000 7200.000 6800.00 6400.00 6000.00 5600.00
2.271 1.809 1.391 1.159 1.134 1.133 1.132
2.310 1.885 1.547 1.411 1.401 1.401 1.400
17.50 11.34 7.85 6.82 6.76 6.75 6.75
M1 GVF profile upstream from gate # 1 x (m)
Y (m)
E (m)
M (m3)
4000.000 3600.000 3200.000 2800.000 2400.000 2000.000
2.394 1.957 1.615 1.505 1.497 1.496
2.489 2.127 1.906 1.858 1.856 1.855
21.77 16.02 13.37 12.89 12.87 12.86
4.11 Culverts An application that involves applying the principles covered thus far, including solutions to GVFs, is the prediction of flow rates and depths throughout culverts. The term culvert is used for circular, oval, rectangular, square or other shaped conduits that pass beneath highways, railroads, or other embankments. The flow through a culvert can be controlled by upstream or downstream conditions. If the culvert has a steep bottom slope, its flow rate (and entrance depth) may be controlled upstream by having a critical flow at its entrance. But if the GVF caused by the downstream depth extends to its entrance, it will not have a critical flow at it entrance even if the bottom slope is steep. It is easier for a downstream control to exist in a short steep culvert than in longer culverts. If the bottom slope is mild, then the flow through the culvert will always be downstream controlled. Generally, the velocities upstream and downstream from culverts are small. Therefore, we will assume reservoirs exist upstream and downstream with water surface elevations H1 and H2, respectively, above the culverts invert (bottom). Since the velocity downstream of the culvert is small, the velocity head in the culvert will be assumed to be dissipated as it enters the downstream reservoir.
347
Nonuniform Flows
4.11.1 Solutions When Upstream Control Exists
ra tio n
The sketch below depicts the flow patterns that may occur when an upstream control exists. For Case I, GVFs exist throughout the culvert’s length. If H2 is less than the
H1
H1
Yc
D
S2
y .h od p M jum
Yd
Yc Yc
Yu
Case IV Top of Culvert S1
Yu
Case II
S2 Yu
H2
Start of Case IV Case III Hyd. jump
H1
p/γ
d.
Hyd. jump
H1
Se
pa
Case V Transfer to downstream control
S1
Case II
Yd
Yu
Steep bottom slope, So > Sc
H2
Case I
H2
Start of Case II
S2 S2
Y2
H2 H2 = conj. to
–H2
downstream depth Y2 at the end of the S2 GVF profile, then the flow ignores it because it is supercritical. As the downstream depth rises above Y2, a roller will appear at the end of the culvert, but in the reservoir. This roller will not penetrate the culvert until the depth H2 becomes larger than the conjugate depth to depth Y2. With downstream depths above the conjugate depth, a hydraulic jump will move into the culvert, and this represents the start of Case II shown in the above sketch. The higher H2 the further this jump will move up the culvert. When a hydraulic jump occurs within the culvert but does not close the top of the culvert, the profiles are denoted as Case II on the above sketch. Since the slope of the channel bottom So is larger than Sf, and the flow is subcritical, the depth will increase beyond the jump in the downstream direction along an S1 GVF profile. Therefore, the hydraulic jump will not be able to close the top of the culvert unless the depth H2 is larger than the height of the culvert. Cases III and IV on the sketch show these possibilities, in which the downstream portion of the flow in the culvert is a closed conduit flow (also commonly called pipe flow even if the culvert is not circular), and the upstream portion contains a supercritical open channel flow. Case III is distinguished from IV in that the conjugate depth to the depth Yu at the end of the S2 GVF is less than D, rather than equal to or above D. Thus, for Case III there is an S2 GVF starting at the entrance, a hydraulic jump, an S1 GVF that ends at the top of the culvert, and then a pipe flow to the culvert’s end. In Case IV a modified jump occurs in which the top of the culvert downstream therefrom has a pressure head on it equal to p/γ = d = (M1 − Mt)/At, the equation derived in Chapter 3 in the section “Open Channel to Pipe Flow.” Finally, as the downstream depth H2 rises, it will eventually push the modified hydraulic jump up to the entrance. When this occurs, control is transferred downstream. The sketch does not show all the possibilities, and the cases shown are not all possible in a given culvert. For example, if the bottom slope of the culvert is large so that the Froude number associated with the flow at its end is large, then the subcritical conjugate depth to this depth may be larger than the diameter of the culvert; in which case there can be no Case II because the roller will be kept in the downstream reservoir for H2 above the top of the culvert. In order for Case IV to occur, the momentum function associated with the supercritical flow must be larger than the momentum function for a full conduit flow with no pressure head on the top, or Mt = At(D/2) + Q2/(gAt). The position where the momentum function of the supercritical flow equals Mt represents the “Start of Case IV.” For a given size culvert and a specified flow rate, or a specified upstream reservoir head H1, Case IV will generally be downstream of what is shown as “Start of Case IV” on the sketch as the bottom slope So is increased more than needed for this limiting case. In other words, the cases shown on the sketch do not necessarily occur in the sequence shown as H2 is increased, but rather depend upon all the variables involved in describing the flow through a culvert with a critical flow at its entrance. In other words, you need to visualize the bottom slope, Manning’s n, etc., being varied, as well as H1 and H2 to understand the different cases shown on the sketch.
348
Open Channel Flow: Numerical Methods and Computer Applications
For Cases I, II, and III, the solution consists of first simultaneously solving the energy and the critical flow equations. For Case I, the gradually varied S2 GVF profile is solved to the end of the channel. For Case II, the governing equations are
F1 = H1 − Yc −
(1 + K g )Q 2 = 0 Variables 2gA 2
F2 = Q 2 T − gA 3 = 0 Q = Flow rate, Yc = Critical depth at entrance F3 = Yu − Yuode = 0 F4 = (Ah c )u +
dY So − Sf = dx 1 − Fr2
for S2 GVF Yu = Depth upstr. H. J., Yd = Depth D. H. J.
Q2 Q2 − (Ah c )d − = 0 x = Position of H. J. gA u gA d
F5 = Yd − Ydode = 0
dY So − Sf = dx 1 − Fr2
for S1 GVF
The unknowns are Q, Yc, Yu, Yd, and x, but since a critical flow at the entrance governs, equations F1 and F2 are first solved simultaneously (or solved using Equation 2.16) for Q and Yc, as for Case I, and thereafter equations F3, F4, and F5 are solved simultaneously for the depths upstream and downstream from the jump and the position of the jump, or Yu, Yd, and x. When the jump hits the top of the culvert as shown in Case IV (and beyond), then Yd is no longer an unknown, but the pressure head d = p/γ on the top of the culvert becomes an unknown. Also, in place of solving the downstream GVF, the slope Sf of the energy line (and slope of the HGL) must be solved. On might use the Darcy–Weisbach equation (and the Colebrook–White equation) for this purpose. However, since the Reynolds number is generally very large for most culvert flows, and f from the Darcy–Weisbach equation is in the wholly rough zone on a Moody diagram, Manning’s equation can be used to solve the slope of the energy line for these pipe flows. The Darcy–Weisbach equation gives the slope of the energy line by Sf = hf/L = fQ22/(2gAt2D), or if Manning’s equation is used, then Sf = [nQ/CuAt(D/ 4)2/3]2, in which At is the full area of the culvert. This slope is utilized to find where the HGL intersects with the pressure head d = p/γ on the top of the culvert. Thus, the last two equations above are replaced by F4 = (Ah c )1 +
Q2 1 Q2 − D A t − gA1 2 gA t
F5 = D + d + (S0 − Sf )(L − x) − H 2 = 0
and these with F3 solve Yu, p/γ and x. The assumption in the last equation F5 is that the velocity head in the culvert is lost as the flow enters a much slower moving flow outside the exit of the culvert. The determination of the case that may exist under given conditions is often more complex than might appear from examining the above sketch, especially for culverts that gradually close in on their tops, such as circular culverts. This difficulty is compounded by having the Froude number approach zero as the depth in the culvert approaches its diameter. Thus, for example, solutions exist for Yc and Q based on solving the upstream energy and the critical flow equations for upstream depths H1 far larger than the diameter D of circular culverts. However, in practice, when H1 is greater than about 1.2D, the flow into the culvert becomes governed by an orificetype equation, rather than the critical flow equation. Another complication is that as the upstream depth rises and causes larger flow rates into the culvert, the value of the critical bottom slope Sc increases. How the flow rate and
349
Nonuniform Flows
Dia = 5 ft
Flow rate, Q (cfs)
180
itic
160
= t en K e ici eff = 0.25 o s c Ke los .1 0 ce an K e = r t n
140 120 100
E
80
=0
.0
Ke
60 40 20
2
oe sc los
ffic
Ke
al
3.5 3 w
t
ien
e nc tra K = 0.1 En K e = 0.0 e
2.5
0.5
Cr
.25
4
h Ye
t dep
0.5 Flo Ke=
, te ra
Q
2.5 2
=0
3 3.5 4 4.5 5 Upstream head, H1 (ft)
1.5 5.5
6
1
0.04 Dia = 5 ft Ke = 0.0
0.035 Critical slope, Sc
200
Critical depth an entrance, Yc (ft)
the critical depths, and also how the critical slope of the culvert vary with H1 for a 5 ft diameter culvert are shown in the figures below. Since Sc increases with H1, a culvert that has a steep slope for smaller flow rates may become a mild culvert for larger flow rates. When this occurs, the control is also passed from upstream to the downstream water level as the upstream head H1 increases.
0.03
0.025
ghn rou g’s 35 n i nn 0.0 Ma n =
ess
.03
n=0
0.02
0.015 2
0.01
n = 0.0
n = 0.15 n = 0.009
0.005 0
2
2.5
3 3.5 4 4.5 5 Upstream head, H1 (ft)
5.5
6
For example, take a 5 ft diameter culvert with a bottom slope of So = 0.02, n = 0.03, and Ke = 0. If the upstream head H1 = 2 ft, then the flow is upstream controlled (unless H2 is above D) since the critical bottom slope is Sc = 0.0175, and therefore at the beginning of the culvert, the depth will be Yc = 1.47 ft and Q = 28.1 cfs. However, as H1 rises to the top of the culvert, or H1 = 5 ft, then the critical bottom slope is Sc = 0.0244, and the downstream depth controls because the culvert has a mild slope. (You should solve the appropriate equations to verify these values.) In solving conditions in culverts under upstream control, either the upstream head H1 or the flow rate Q will be specified along with the downstream head H2. Whether H1 or Q is specified, the upstream energy equation F1 and the critical flow equation F2 are to be solved. If Q is specified, first solve the critical depth from the critical flow equation F2, and thereafter the energy equation F1 can be solved explicitly for H1. If H1 is given, it is best to eliminate Q from F1 and F2 by substituting Q from F2 into F1 and solving the resulting equation. Using the dimensionless depth Y′ = Yc /D and H′ = H1/D, this equation is F = 0.5(1 – cos β) + (1 + Ke){β – cos β sin β}/(8 sin β) − H′ = 0, in which β = cos−1(1 − 2Y′). After solving β, Q is obtained from Q = [gD5(β − cos β sin β)3/(64 sin β)]1/2. For Cases II, III, or IV, if the momentum function associated with the S1 is larger than that associated with the S2 GVF for all positions up to the entrance, then no longer will a critical depth occur at the entrance, and the flow is no longer controlled by upstream conditions. Rather, the effects of the depth H2 downstream for the culvert are felt to its entrance and the downstream control occurs. These conditions are not shown on the above sketch, but are handled in the next sections. The computer program CULVERTU.FOR listed below is designed to solve culvert flow problems when upstream control exists. The approach taken in this program is to first solve critical flow conditions at the entrance. It allows for the flow rate Q, or the upstream reservoir water surface elevation H1 to be given. In the input, this is done by giving a 0 to the one that is unknown. Then starting just below the critical depth, the S2GVF profile is solved throughout the length of the culvert. If the given depth H2 at the downstream end of the culvert is less than this, then Case I occurs, i.e., the flow is supercritical throughout the entire culvert. The conjugate depth to this downstream depth is next computed. Should H2 be less, then this conjugate depth Case I still occurs with a roller occurring just outside the end of the culvert. The program contains logic to distinguish between Cases II, III, and IV and also determines whether a hydraulic jump (or a modified jump for Case IV) can occur, and if not, prints out a message that downstream control exists. It divides the culvert into 30 stations at which the depths, the areas and the momentum functions are computed. In computing the S1 GVF profile at these stations, the program terminates when a computed depth becomes less
350
Open Channel Flow: Numerical Methods and Computer Applications
than (1.5 + So)Yc, but if the depth before the next station reaches critical depth, then ODESOL may terminate because it cannot achieve the specified tolerance, and stops with the message that the minimum step size has been reached. To make CULVERTU handle all possible situations without failure, considerably more logic needs to be added to the program. Listing of program CULVERTU.FOR designed to solve problem in which critical flow occurs at the culvert’s entrance C Program to solve Flow in Culvert with upstream control. PARAMETER (N=3,NP=31) EXTERNAL DYX CHARACTER*3 CASE REAL FMU(NP),FMD(NP),YU(NP),YD(NP) REAL W(1,13),Y(1),DY(1),XP(1),YP(1,1) COMMON NGOOD,NBAD,KMAX,KOUNT,DXSAVE COMMON /TRAS/D,DH,DS6,D25,So,G,QN,Q2G,H1,H2,A,YC,IOUT WRITE(*,*)' GIVE:IOUT,TOL,D,Q,H1,H2,Ke,L,n,So,g' READ(*,*) IOUT,TOL,D,Q,H1,H2,FKE,FL,FN,So,G IF(Q.EQ.0. .OR. H1.EQ.0.) GO TO 1 WRITE(*,*)' Both Q and H1 cannot be given. Which is unknown?' WRITE(*,*)' Give 1 = Q, or 2 = H1' READ(*,*) M IF(M.EQ.1) THEN Q=0. ELSE H1=0. ENDIF 1 IF(G.GT.30.) THEN CC=1.486 ELSE CC=1. ENDIF DH=.5*D DS6=D*D/6. D25=.25*D*D AT=.78539816*D*D FKE1=(FKE+1.)/(2.*G) FKE=(FKE+1.)/8. M=0 IF(Q.EQ. 0.) THEN IQ=0 H1D=H1/D Q=.2687*H1D**3.906 YC=Q**.254 COSB=1.2.*YC IF(COSB.LT. 1.) COSB=.98 BETA=ACOS(COSB) 2 F=.5*(1.COS(BETA))+FKE*(BETACOS(BETA)*SIN(BETA))/ &SIN(BETA)H1D M=M+1 IF(MOD(M,2).EQ.0) GO TO 3 F1=F
Nonuniform Flows
351
BET=BETA BETA=1.01*BETA GO TO 2 3 DIF=(BETABET)*F1/(FF1) BETA=BETDIF IF(ABS(DIF).GT. 1.E6 .AND. M.LT.30) GO TO 2 IF(M.GE.30) WRITE(*,*)' Did not comverge. DIF=',DIF,BETA Q=(BETACOS(BETA)*SIN(BETA))**3/(64.*SIN(BETA)) Q=SQRT(G*Q*D**5) YC=DH*(1.COS(BETA)) Q2G=Q*Q/G ELSE IQ=1 Q2G=Q*Q/G BETA=1.5 Q2GD=D*Q2G 4 F=Q2GD*SIN(BETA)(D25*(BETACOS(BETA)*SIN(BETA)))**3 M=M+1 IF(MOD(M,2).EQ.0) GO TO 5 F1=F BET=BETA BETA=1.01*BETA GO TO 4 5 DIF=(BETABET)*F1/(FF1) BETA=BETDIF IF(ABS(DIF).GT.1.E6 .AND. M.LT.30) GO TO 4 YC=DH*(1.COS(BETA)) H1=YC+4.*FKE*Q2G/(D25*(BETACOS(BETA)*SIN(BETA)))**2 ENDIF WRITE(IOUT,100) Q,YC,H1,H2,D,FL,FN,So 100 FORMAT(' Q=',F8.2,', Yc=',F8.3,', H1=',F8.3,', H2=',F8.3,/ &' D=',F7.1,', L=',F8.0, ', n=',F8.4,', So=',F8.6) FMT=DH*AT+Q2G/AT YCP=(1.15+SO)*YC HMIN=.000001 DELX=FL/FLOAT(NP1) C Solves S2GVF from entrance to end of culvert XX=0. XZ=DELX H11=.05 Y(1)=.95*YC YU(1)=Y(1) FMU(1)=FMOM(Y(1)) WRITE(IOUT,115) 'S2',0.,YU(1),A,FMU(1) 115 FORMAT(/,A3,'  GVF Profile',/,4F10.3) 118 FORMAT(4F10.3) DO 10 I=2,NP CALL ODESOL(Y,DY,1,XX,XZ,TOL,H11,HMIN,1,XP,YP,W,DYX) YU(I)=Y(1) FMU(I)=FMOM(Y(1)) WRITE(IOUT,118) XZ,YU(I),A,FMU(I)
352
Open Channel Flow: Numerical Methods and Computer Applications
XX=XZ XZ=XZ+DELX Y2=2.*YCYU(NP) IF(Y2.GT.D) Y2=.9*D FMOM1=FMOM(YU(NP)) C Solves Conjugate Depth at end of culvert M=0 12 F=FMOM1FMOM(Y2) M=M+1 IF(MOD(M,2).EQ.0) GO TO 13 F1=F Y22=Y2 Y2=1.01*Y2 GO TO 12 13 IF(ABS(FF1).LT.1.E20) THEN DIF=0. ELSE DIF=F1*(Y2Y22)/(FF1) ENDIF Y2=Y22DIF IF(ABS(DIF).GT.1.E6 .AND. M.LT.30) GO TO 12 IF(M.GE.30) WRITE(*,*)' Did not converge for conj. Y',DIF,Y2 IF(H2.GT.Y2) THEN IF(Y2.GT.D) WRITE(IOUT,106) Y2 106 FORMAT(' Conjugate depth at end is above top',' of culvert', &F8.2) GO TO 15 ENDIF WRITE(IOUT,110) H1,H2,Q,YC,YU(NP),Y2 110 FORMAT(' Case I occurs with S2GVF thru entire length' &' of culvert',/,' H1 =',F8.2,' H2 =',F8.2,' Q =',F8.2, &'Yc =',F8.2,' Depth at end =',F8.2,/,' Conjugate &depth =',F8.2) IF(H2.LE.Y2) WRITE(IOUT,120) Y2YU(NP) 120 FORMAT(' Roller exists at end of culvert with',' height =', &F8.2) STOP 15 IF(H2.GT.D) GO TO 40 C Case II  Hyd. Jump in culvert XX=FL H11=1. XZ=XXDELX IB=1 Y(1)=H2 YD(NP)=H2 FMD(NP)=FMOM(Y(1)) WRITE(IOUT,115)'S1',FL,YD(NP),A,FMD(NP) IE=NP CASE='II ' C Solves S1GVF profile 19 DO 20 I=IE1,1,1 10
Nonuniform Flows
20 22 160
25 26
130
30 150 32
40
155
353
CALL ODESOL(Y,DY,1,XX,XZ,TOL,H11,HMIN,1,XP,YP,W,DYX) YD(I)=Y(1) FMD(I)=FMOM(Y(1)) WRITE(IOUT,118) XZ,YD(I),A,FMD(I) IF(Y(1).LT.YCP) THEN IB=I GO TO 22 ENDIF XX=XZ XZ=XZDELX IF(IB.EQ.1 .AND. FMD(1).GT.FMU(1)) THEN WRITE(IOUT,160) H1,H2,Q FORMAT(' Effects extend to entrance so downstream', control &occurs.',/,' Change specifications.',/,' H1=',F7.2,' H2=', &F7.2,' Q=',F8.2) STOP ENDIF DO 25 I=IB,NP IF(FMU(I).LT.FMD(I)) GO TO 26 CONTINUE IP=I1 FAC=(FMU(IP)FMD(IP))/(FMD(IP+1)FMD(IP)FMU(IP+1)+FMU(IP)) XX=DELX*(FLOAT(IP1)+FAC) WRITE(IOUT,130) CASE,Q,YC,YU(IP)+FAC*(YU(IP+1)YU(IP)), &YD(IP)+FAC*(YD(IP+1)YD(IP)),XX FORMAT(' Case ',A3,' Jump occurs within culvert:', &' Q =',F8.2,' Yc =',F7.2,/,' Depth upst. jump=',F6.2, &' Depth downst. jump =',F6.2,' Jump position x=',F7.1) STOP XX=(H2D)/(SOSF) WRITE(IOUT,150) XX FORMAT(' Case III  Culvert full at downstr. end', &' for a distance of',F8.2) IE=NPXX/DELX1.9 Y(1)=.98*D YD(IE)=D FMD(IE)=FMOM(Y(1)) XX=DELX*FLOAT(IE1) XZ=XXDELX H11=1. IB=1 WRITE(IOUT,115)'S1',XX,Y(1),A,FMD(IE) CASE='III' GO TO 19 SF=(FN*Q/(CC*AT*(.25*D)**.6666667))**2 XX=(H2D)/(SOSF) IF(XX.GE.FL) THEN WRITE(IOUT,155) (XXFL)*(SOSF)+FKE1*(Q/AT)**2 FORMAT(' Case V  Downstream H2 causes depth',' at &entrance of',F9.2,/,' Flow is downstream controlled.') STOP
354
41 42
180
C
140
50 170
Open Channel Flow: Numerical Methods and Computer Applications
ENDIF DO 41 I=NP,1,1 IF(FMU(I).LT.FMT) GO TO 42 CONTINUE WRITE(IOUT,160) H1,H2,Q STOP IE=I IF(IE.EQ.NP) GO TO 30 X24L=DELX*(FLOAT(IE1)+(FMTFMU(IE))/(FMU(IE+1)FMU(IE))) H24L=D+(SOSF)*(FLX24L) WRITE(IOUT,180) X24L,FLX24L,H24L FORMAT(' Start of Case IV at x =',F8.2,' (from end=',F8.2,') &Downstr. H2 required=',F8.2) DO 50 I=IE1,NP PHEAD=H2D(SOSF)*(FLDELX*FLOAT(I1)) WRITE(IOUT,*) I,PHEAD,PHEAD*AT+FMT,FMU(I) IF(PHEAD.LT.0.) GO TO 50 IF(PHEAD*AT+FMT