Computational Techniques in Civil Engineering Finite Difference Method (FDM) part: Numerical solution - Dr. K.N. Dulal FDM concepts 1. Consider a rectangular channel, 30m wide, with bed slope of 0.015 and Manning’s n = 0.035. The following flow rates are given: =30 m 3 /s, =22 m 3 /s and =20 m 3 /s. Taking ∆ = 1500m and ∆ = 10 min, determine using finite difference scheme for a linear kinematic wave model. Assume lateral flow to be zero. The equation for linear kinematic wave model with no lateral flow is = ∆ ∆ ∆ ∆ . Take wetted perimeter is approximately equal to width of channel. Solution: Width of channel (b) = 30m Bed slope (S) = 0.015 Manning’s n = 0.035 , from Manning’s equation = / / = / / / = / √ / / Comparing to = = / √ . = . / √. . = 1.84 = 0.6 ∆ = 1500m and ∆ = 10 min = 600 s = ∆ ∆ ∆ ∆ = . . . . . . =25.67 m 3 /s
TU Bachelor in Civil Engineering Eight semester Computational Technique Finite differential method
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1. Consider a rectangular channel, 30m wide, with bed slope of 0.015 and Manning’s n = 0.035. Thefollowing flow rates are given: =30 m3/s, =22 m3/s and =20 m3/s. Taking ∆ = 1500m and∆ = 10 min, determine using finite difference scheme for a linear kinematic wave model. Assumelateral flow to be zero. The equation for linear kinematic wave model with no lateral flow is
= ∆∆ ∆∆ . Take wetted perimeter is approximately equal to width of
channel.
Solution:Width of channel (b) = 30mBed slope (S) = 0.015Manning’s n = 0.035, from Manning’s equation= / / = // /= / √ / /Comparing to == /√ . = . /√ . .
= 1.84= 0.6∆ = 1500m and ∆ = 10 min = 600 s
= ∆∆ ∆∆ = . . .. . . =25.67 m3/s
2. Solve exercise 1 by using non-linear kinematic wave model using the equation∆∆ + = ∆∆ + ( )Take value of computed in exercise 1 as initial value and solve by Newton-Raphson iteration.
As the difference in Q in iteration 1 and 2 is small, the iteration is stopped.=25.578 m3/s
3. Consider a rectangular channel, 90m wide and 5km long with bed slope of 0.015 and Manning’s n =0.02. The inflow hydrograph for the channel is given below:Time (min) 0 5 10 15 20Flow (m3/s) 14 19 28 32 40The initial condition is a uniform flow of 14 m3/s and there is no lateral flow.Use the linear kinematic wave model to route the inflow hydrograph through the channel taking ∆ =1000m and ∆ = 5 min . The equation for routing is
= ∆∆ ∆∆ . Take wetted perimeter is approximately equal to width of
channel.
Solution:Width of channel (b) = 90mBed slope (S) = 0.015Manning’s n = 0.02, from Manning’s equation= / / = // /= /√ / /Comparing to == /√ . = . /√ . .
Sample computationFor i = 1 all values are given. For t = 0, all values are given.i = 1, n = 1: = =19 m3/s, = =14 m3/s and = =14 m3/s.
= ∆∆ ∆∆= . .. . =16.17 m3/s
Compute other values in a similar way.
4. The values of flow rate at four points in the space-time gird are given below:=15 m3/s, =15.8 m3/s, =13.9 m3/s and =14.7 m3/. Taking ∆ = 500m, ∆ = 1 hr and =
0.55, calculate the values of and using implicit four-point method.
1. A pipe conveys water from a reservoir as shown in the figure.
Take f = 0.02, c = 1200m/s. The hydraulic grade line (HGL) at the reservoir is given as = 100 +2sin( ). The discharge at the downstream end is zero at all times. By using only one reach, computedischarge from A and elevation of hydraulic grade line at B at 3Sec using discretized equation of theMethod of characteristics in the form of HGL and discharge .
Computation sequence= 100 + 2 sin( )= . where HB = HPB of previous step= + 973.9 − 96.9 | | where HA = HPA of previous step and QA = QPA of previous step