96 Computational Modelling in Hydraulic and Coastal Engineering
clc; clear all; close all;
% Input data;
g=9.81;
Dx=1000;
Dt=5;
nx=41;
nt=9000;
dt=2000;
nth=15;
for i=1:nx
h(i)=0;
S(i)=0.001;
b(i)=100;
C(i)=50;
end
Qin=[0,600,400,200,100,50,0,0,0,0,0,0,0,0,0];
k=0;
% Main program;
for k=1:nt
k=k+1;
% Calculation of inflow discharge by interpolation;
jj=k*Dt/dt;
jr=round(jj)+1;
if jr >= nth
2.5
2
1.5
0.5
0 0
1
5
1 0
1 5
2 0
2 5
3 0
3 5
4 0
Longitudinal distance × 1000 (m)
Water depth (m)
t = 1000*Dt
t = 1500*Dt
t = 2000*Dt
t = 3000*Dt
t = 4500*Dt
t = 9000*Dt
Figure 5.7 Water depths over the channel at different times.