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. 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.

Solution:

RHS=c = ∆∆ + ( ) = 30 + 1.84 20 . = 23.1

Residual error is = ∆∆ + − = + 1.84 . − 23.1=0.4 + 1.84 . − 23.1′ = 0.4 + 1.1 .

Newton-Raphson formula

= −= 25.67 m3/s (computed in 1)

Iteration 1=0.4 25.67 + 1.84(25.67) . − 23.1 = 0.0645′ = 0.4 + 1.1(25.67) . =0.7= 25.67 − . . = 25.577 m3/s

Iteration 2=0.4 25.577 + 1.84(25.577) . − 23.1 = -0.0007′ = 0.4 + 1.1(25.577) . =0.7= 25.577 − . . = 29.578 m3/s

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 == /√ . = . /√ . .

= 2= 0.6∆ = 1000m and ∆ = 5 min = 300 s

Routing

Distance (m)

RemarksTime(min)

Time index(n) 0 1000 2000 3000 4000 5000

i = 1 2 3 4 5 60 1 14 14 14 14 14 14 Given5 2 19 16.17 14.92 14.39 14.16 14.0710 3 28 21.65 17.91 15.91 14.91 14.4215 4 32 26.64 21.96 18.62 16.52 15.3220 5 40 33.37 27.50 22.77 19.34 17.09

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.

Solution:

n+1dt

n

= ∆ + (1 − ) ∆= 0.55 ( . . )+ (1 − 0.55) ( . ) = -0.0022

= ∆ + ∆= . . + . = 0.00022

Method of characteristics (MOC)

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 .

C+C-

BA

L= 600m, D = 400mm100m

θ

0.50.5

1-θi i+1dx

Solution:f = 0.02, c = 1200m/s, L = 500m, D = 0.4m∆ =L=600m∆ = L/c = 600/1200=0.5SA = 0.4 =0.1256m2= = . . = 973.9= ∆ = .. . . = 96.9

QB = 0= 100 + 2sin( )for C+= − ( − ) − | | = 100 − 973.9(0 − ) − 96.9 | |= + 973.9 − 96.9 | | (I)for C-= + ( − ) + | |= + 973.9( − 0) + 0 0= . (II)

At t = 0, QA = 0, QB = 0, HA = HB = 100m

Computation table

t (sec) HPA QPA HPB

0 100 0.000 100.00.5 102 0.002 100.01 100 0.000 104.01.5 98 -0.006 100.02 100 0.000 92.02.5 102 0.010 100.03 100 0.000 112.0

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