Computational Techniques in Civil Engineering Water Resources part: Tutorial solution

Civil Engineering BCE | Computational Technique
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FDM
1. Values of discharge (Q) in m3/s at three points in the space-time grid are given as , and . Taking and = 5 minute, compute using following schemes: (a) backward difference (b) forward difference, (c) central difference. Also compute .
Solution:
(a) backward difference
(b) forward difference
(c) central difference
( ) [( ) ( ) ] [ ]
( ) ( )
2. Values of discharge (Q) in m3/s at three points in the space-time grid are given as , and . Taking =1 km and = 30 minute, compute using following schemes: (a) backward difference (b) forward difference, (c) central difference
t n+1 n i-1 i i+1 X
Solution:
(a) backward difference
(b) forward difference
(c) central difference
3. 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.
Solution:
( ) ( )( )
( ) ( )( ) = -0.0022
( )
( ) = 0.00022
0.5 0.5 θ t n+1 t n n-1 i-1 i i+1 X n+1 n i i+1 X 1-θ
4. 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 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. 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
= 1500m and = 10 min = 600 s
( ) ( ( ) ) ( ) ( ( ) ) =25.67 m3/s
5. Solve exercise 4 by using non-linear kinematic wave model using the equation
( ) ( )
Take value of computed in exercise 4 as initial value and solve by Newton-Raphson iteration.
Solution:
RHS=c = ( ) = = 23.1
Residual error is ( ) ( ) = ( )
( )= ( )
( ) ( )
Newton-Raphson formula
( ) ( ) ( ) ( )
( ) = 25.67 m3/s (computed in 1)
Iteration 1
( )= ( ) = 0.0645
( ) ( ) =0.7
( ) = 25.577 m3/s
Iteration 2
( )= ( ) = -0.0007
( ) ( ) =0.7
( ) = 29.578 m3/s
As the difference in Q in iteration 1 and 2 is small, the iteration is stopped.
=25.578 m3/s
6. 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 20
Flow (m3/s) 14 19 28 32 40
The 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) = 90m
Bed slope (S) = 0.015
Manning’s n = 0.02
from Manning’s equation
( √ )
Comparing to
( √ ) ( √ ) = 2
= 1000m and = 5 min = 300 s
Routing
Distance (m)
Remarks
Time (min)
Time index (n)
0
1000
2000
3000
4000
5000
i = 1
2
3
4
5
6
0
1
14
14
14
14
14
14
Given
5
2
19
16.17
14.92
14.39
14.16
14.07
10
3
28
21.65
17.91
15.91
14.91
14.42
15
4
32
26.64
21.96
18.62
16.52
15.32
20
5
40
33.37
27.50
22.77
19.34
17.09
Sample computation
For i = 1 all values are given for all t. For n = 1, all values are given for all i.
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.

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