This document discusses uniform open channel flows and how to calculate water depth and discharge in open channels. It provides the momentum balance equations for open channel flows and defines parameters like hydraulic diameter, roughness, and slope. It also presents MATLAB code to solve these equations numerically for different channel geometries like rectangular and trapezoidal channels using variables like discharge, roughness, and slope. The code can calculate the normal water depth given other channel parameters.

1. Uniform open channel flows
Prof. Dr. Andreas Malcherek

2. The fundamental problem of open channel flows
Euroluftbild.de, Grahn
For a given discharge Q the water level h needs to be determined!

3. Hydraulic diameter of open channels
Example: Rectangular cross section

4. Rivers are wide open channels
Wide rectangular cross section

5. The momentum balance for a pipe

6. Momentum balance for an uniform open channel
Same water depth everywhere
Stationary conditions

7. MATLAB Program
• function riverflow
• rho=1000;
• h=1.5; %Water depth within the river
• B=50; % Width of the river
• A=B*h;
• U= B+2*h;
• dhyd=4*A/U;
• z1=20; % Upper height of the river
• z2=0; % Loweder height of the river
• L=10000; % Length of the river
• J=(z1-z2)/L; % Slope
• Ks=0.03; %Roughness of the river bed
•
• [T v]=ode45(@RHS,[0 1000],0);
• plot(T,v*A,'Color',[0 0 0],'LineWidth',2)
• xlabel('Time[s]','FontWeight','bold')
• ylabel('Discharge[m^3/s]','FontWeight','bold')
• grid on
•
• function dvdt=RHS(t,v)
• Re=v*dhyd/nu;
• lambda=colebrook_white(Re,Ks/dhyd);
• dvdt=(p1-p2)/(rho*L)+g*(z1-z2)/L-lambda/8*U/A*v*abs(v);
• end
• end
7

8. Open channel and river slopes
Example: Weser river in
Germany: At the source
z1=123m, at the mouth z0=0
m, lenght 300km

9. Balance between driving force and friction
Discharge

10. The friction coefficient for open channel flows

11. Roughness of open channel beds
d
l
d
D

12. Grain Roughness

13. Ripple and dune roughness







 D
D

r
r
r
r
r
s
k
l

20  
d
d
e
k d
d
d
s
l
 /
25
1
77
.
0 D


D

d
l
d
D

14. Rectangular channel
Bh
A 
Given: Q, ks, J B, l
B
2h
4Bh
dhyd



15. Rectangular channel
4gJB3h3 - ʎQ2h-
ʎ
2
Q2B = 0

16. Solving roots of polynomials in MATLAB
g=9.81;
B=10;
Q=8;
J=0.001;
lambda=0.03;
poly=[4*g*B^3 0 -lambda –Q^2];
h=max(real(roots(poly)));
4gJB3h3 - ʎQ2h-
ʎ
2
Q2B = 0

17. Polynomials in MATLAB

18. Discharge characteristics of the river

19. Manning formula

20. Mannings coefficient

21. Calculation with Manning formula

22. Trapezoida channels
2
mh
Bh
A 

 
B
m
1
2h
mh
Bh
4
d
2
2
hyd




Rectangular
profile for m=0

23. Trapezkanäle
2g
Q
I
1
λ
)
)A(h
(h
d
2
So
2
N
N
hyd 
    2g
Q
I
1
λ
mh
Bh
B
m
1
2h
mh
Bh
4 2
So
2
2
N
N
2
N
2
N
N





   
B
m
1
2h
2g
Q
I
1
λ
mh
Bh
4 2
N
2
So
3
2
N
N 




24. Normal water depth in trapezoidal channel
 
B
2h
2g
Q
I
1
λ
h
4B N
2
So
3
N
3


B=100m
Iso=1:10000
Ks=6mm