fluid mechanics

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