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Use the Muskingum-Cunge method to route the hydrograph of Example 9.9 for S₁ = 0.0007. Example 9.9 An inflow hydrograph for a river reach has a peak discharge of 4500 cfs (128 m³/s) at a time to peak of 2 hr with a base time of 6 hr. Assume that the inflow hydrograph is triangular in shape with a base flow of 500 cfs (14 m³/s). The river reach has a length of 18,000 ft (5490 m) and a slope of 0.0005 ft/ft. The channel cross section is trape- zoidal with a bottom width of 100 ft (30.5 m) and side slopes of 2:1. The Manning's n for the channel is 0.025. Find the outflow peak discharge and time of occurrence for the river reach using the Muskingum-Cunge method and compare them to the dynamic routing method using the method of characteristics.

Use the Muskingum-Cunge method to route the hydrograph of Example 9.9 for S₁ = 0.0007. Example 9.9 An inflow hydrograph for a river reach has a peak discharge of 4500 cfs (128 m³/s) at a time to peak of 2 hr with a base time of 6 hr. Assume that the inflow hydrograph is triangular in shape with a base flow of 500 cfs (14 m³/s). The river reach has a length of 18,000 ft (5490 m) and a slope of 0.0005 ft/ft. The channel cross section is trape- zoidal with a bottom width of 100 ft (30.5 m) and side slopes of 2:1. The Manning's n for the channel is 0.025. Find the outflow peak discharge and time of occurrence for the river reach using the Muskingum-Cunge method and compare them to the dynamic routing method using the method of characteristics.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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Use the Muskingum-Cunge method to route the hydrograph of Example 9.9 for S = 0.0007. Example 9.9 An inflow hydrograph for a river reach has a peak discharge of 4500 cfs (128 m³/s) at a time to peak of 2 hr with a base time of 6 hr. Assume that the inflow hydrograph is triangular in shape with a base flow of 500 cfs (14 m³/s). The river reach has a length of 18,000 ft (5490 m) and a slope of 0.0005 ft/ft. The channel cross section is trape- zoidal with a bottom width of 100 ft (30.5 m) and side slopes of 2:1. The Manning's n for the channel is 0.025. Find the outflow peak discharge and time of occurrence for the river reach using the Muskingum-Cunge method and compare them to the dynamic routing method using the method of characteristics. Solution. First the kinematic wave speed is calculated based on Manning's equation and a reference discharge of 2500 cfs (70.8 m³/s), which is midway between the base flow and the peak discharge. For the given conditions, the resulting normal depth is 5.71 ft (1.74 m) and the velocity, V, is 3.93 ft/s (1.20 m/s). If we consider the channel to be very wide as a first approximation, then c = (5/3)Vo = 6.55 ft/s (2.00 m/s). The value of At is tentatively chosen to be 0.5 hr based on discretization of the time to peak, which is equal to 2 hr. Then Ax can be estimated from the inequality of (9.73): Ax = 1/ (GAT + B) = 1 (6.55 x=0.5(1 Therefore, two routing reaches, each with a length of 9000 ft (2743 m), can be used. This gives a Courant number of cAt/Ax = 6.55 x 1800/9000 = 1.31, which is slightly greater than unity, and the value of X from (9.67) is BSociAx) = 0,5(1. If the value of either C, or X seems unsatisfactory, further slight adjustments of At and Ax are possible, since the criterion given by (9.73) is conservative and guarantees Co≥ 0.33 rather than Co≥ 0, which is all that really is required to avoid negative outflows. 3.0 3.5 4.0 4.5 5.0 1- 5.5 6.0 6.5 7.0 7.5 6.55 x 0.50 x 3600 + = 9003 ft (2744 m) Time, hr Inflow, cfs Co XI₂ C₁ X 1₁ C₂x0₁ 0.0 0.5 1.0 1.5 2.0 2.5 TABLE 9.7 Muskingum-Cunge routing (C₁ = 0.333; C₁ = 0.540; C₂ = 0.127) of Example 9.6 500 1500 2500 3500 4500 4000 3500 3000 2500 2000 1500 1000 500 500 500 500 500 833 1166 270 810 1350 1499 1890 1332 2430 1166 2160 999 1890 833 1620 666 1350 500 1080 333 810 167 540 167 270 167 270 167 270 2500 122.8 x 0.0005 X 6.55 x 9000/ 64 106 222 348 474 2500 122.8 x 0.0005 × 6.55 538 491 429 366 303 239 176 112 70 64 x = 9000 ft; x = 18000 ft; 0₂, cfs 0₂, cfs 500 833 1748 2738 3736 4236 3864 3380 2882 2382 1882 1382 882 549 506 501 500 611 1110 1997 2976 3806 4058 3727 = 0.155 3258 2763 2264 1764 1264 819 569 512
Transcribed Image Text:Use the Muskingum-Cunge method to route the hydrograph of Example 9.9 for S = 0.0007. Example 9.9 An inflow hydrograph for a river reach has a peak discharge of 4500 cfs (128 m³/s) at a time to peak of 2 hr with a base time of 6 hr. Assume that the inflow hydrograph is triangular in shape with a base flow of 500 cfs (14 m³/s). The river reach has a length of 18,000 ft (5490 m) and a slope of 0.0005 ft/ft. The channel cross section is trape- zoidal with a bottom width of 100 ft (30.5 m) and side slopes of 2:1. The Manning's n for the channel is 0.025. Find the outflow peak discharge and time of occurrence for the river reach using the Muskingum-Cunge method and compare them to the dynamic routing method using the method of characteristics. Solution. First the kinematic wave speed is calculated based on Manning's equation and a reference discharge of 2500 cfs (70.8 m³/s), which is midway between the base flow and the peak discharge. For the given conditions, the resulting normal depth is 5.71 ft (1.74 m) and the velocity, V, is 3.93 ft/s (1.20 m/s). If we consider the channel to be very wide as a first approximation, then c = (5/3)Vo = 6.55 ft/s (2.00 m/s). The value of At is tentatively chosen to be 0.5 hr based on discretization of the time to peak, which is equal to 2 hr. Then Ax can be estimated from the inequality of (9.73): Ax = 1/ (GAT + B) = 1 (6.55 x=0.5(1 Therefore, two routing reaches, each with a length of 9000 ft (2743 m), can be used. This gives a Courant number of cAt/Ax = 6.55 x 1800/9000 = 1.31, which is slightly greater than unity, and the value of X from (9.67) is BSociAx) = 0,5(1. If the value of either C, or X seems unsatisfactory, further slight adjustments of At and Ax are possible, since the criterion given by (9.73) is conservative and guarantees Co≥ 0.33 rather than Co≥ 0, which is all that really is required to avoid negative outflows. 3.0 3.5 4.0 4.5 5.0 1- 5.5 6.0 6.5 7.0 7.5 6.55 x 0.50 x 3600 + = 9003 ft (2744 m) Time, hr Inflow, cfs Co XI₂ C₁ X 1₁ C₂x0₁ 0.0 0.5 1.0 1.5 2.0 2.5 TABLE 9.7 Muskingum-Cunge routing (C₁ = 0.333; C₁ = 0.540; C₂ = 0.127) of Example 9.6 500 1500 2500 3500 4500 4000 3500 3000 2500 2000 1500 1000 500 500 500 500 500 833 1166 270 810 1350 1499 1890 1332 2430 1166 2160 999 1890 833 1620 666 1350 500 1080 333 810 167 540 167 270 167 270 167 270 2500 122.8 x 0.0005 X 6.55 x 9000/ 64 106 222 348 474 2500 122.8 x 0.0005 × 6.55 538 491 429 366 303 239 176 112 70 64 x = 9000 ft; x = 18000 ft; 0₂, cfs 0₂, cfs 500 833 1748 2738 3736 4236 3864 3380 2882 2382 1882 1382 882 549 506 501 500 611 1110 1997 2976 3806 4058 3727 = 0.155 3258 2763 2264 1764 1264 819 569 512
in most cases. The values of the routing coefficients are computed from (9.64) through (9.66) to yield Co= 0.333; C₁ = 0.540; C₂ = 0.127 which sum to unity as required. The routing equation is solved step by step in Table 9.7 for the first subreach of 9000 ft (2743 m). Then the outflow becomes the inflow for the next subreach, for which only the final results are shown in the table without the intermediate computations. The Muskingum-Cunge results are compared with dynamic routing results from the method of characteristics (MOC) in Figure 9.12. It is apparent that the assumption of a constant wave celerity in the Muskingum-Cunge method fails to capture the nonlinear steepening and flattening on the rising and falling sides, respectively, of the outflow hydrograph. However, the peak outflow rates agree very well. The occurrence of the outflow peak is at t = 3.0 hr, but this is limited by the time step of the approximate routing method. The method of charac- teristics gives a peak time of 2.8 hr. Note that the criterion of Equation 9.74 for dif- fusion routing is not met, but reasonable agreement still is obtained between the two methods in this example. 5000 Q, cfs 4000 3000 2000 1000 0 Inflow ● 2 Co= 0.333; C₁ = 0.540; C₂ = 0.127 4 Muskingum-Cunge outflow Time, hr 6 MOC outflow 8 FIGURE 9.12 Inflow and outflow hydrographs for Muskingum-Cunge and MOC routing. 10
Transcribed Image Text:in most cases. The values of the routing coefficients are computed from (9.64) through (9.66) to yield Co= 0.333; C₁ = 0.540; C₂ = 0.127 which sum to unity as required. The routing equation is solved step by step in Table 9.7 for the first subreach of 9000 ft (2743 m). Then the outflow becomes the inflow for the next subreach, for which only the final results are shown in the table without the intermediate computations. The Muskingum-Cunge results are compared with dynamic routing results from the method of characteristics (MOC) in Figure 9.12. It is apparent that the assumption of a constant wave celerity in the Muskingum-Cunge method fails to capture the nonlinear steepening and flattening on the rising and falling sides, respectively, of the outflow hydrograph. However, the peak outflow rates agree very well. The occurrence of the outflow peak is at t = 3.0 hr, but this is limited by the time step of the approximate routing method. The method of charac- teristics gives a peak time of 2.8 hr. Note that the criterion of Equation 9.74 for dif- fusion routing is not met, but reasonable agreement still is obtained between the two methods in this example. 5000 Q, cfs 4000 3000 2000 1000 0 Inflow ● 2 Co= 0.333; C₁ = 0.540; C₂ = 0.127 4 Muskingum-Cunge outflow Time, hr 6 MOC outflow 8 FIGURE 9.12 Inflow and outflow hydrographs for Muskingum-Cunge and MOC routing. 10
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