Design of Open Channels
Basic Principles
Channel Flow
Q
Cn
AR 2 / 3 S 01/ 2
n
If depth of flow is unknown the above equation has to be solved by trials for
most of the channel shapes.
Newton’s Method to solve Manning’s equation
Error=
f (y j )  Qj  Q
for Q=constant
dQ j
df

dy j dy j
Assuming n as constant
 df 
d  Cn 1/ 2

  
S 0 A j R 2j / 3 


 dy  j dy j  n
C n 1 / 2  2 AR 1 / 3 dR
dA 


S 0 
 R2/3
n
3
dy
dy  j

Cn
1 dA 
2 / 3 1 / 2  2 dR


A j R j S 0 

n
 3R dy A dy  j
 2 dR 1 dA 

 Q j 

 3R dy A dy  j
1
Now,
0 f (yj )
 df 
  
y j 1  y j
 dy  j
or
y j 1  y j 
f (yj )
 df 
 dy  j
 yj 
1 Q / Q j
 2 dR 1 dA 



 3R dy A dy  j
2
3
Best Hydraulic Section
Cn
AR 2 / 3 S 01/ 2
n
3/ 5
 Qn 
2/5

 A  
P
1/ 2 
C
S
 n 0 
Q
(1)
To minimize the cost of material used in the channel, we need to minimize
the wetted perimeter, for given roughness, discharge and bed slope.
Consider a trapezoidal channel for example.
B=Bw+2zy
Free-board
A
y
1
z
Bw
A  Bw y  zy 2
P  Bw  2 y z 2  1
(2)
Eliminating the base width from above
A  ( P  2 y z 2  1) y  zy 2
(3)
Equating (1) and (3)
( P  2 y z 2  1) y  zy 2  cP 2 / 5
where,
 Qn 

c  
1/ 2 
 Cn S 0 
3/ 5
and Cn=1 for SI units.
4
Holding z constant and differentiating with respect to y and setting
P / y  0 we get,
P  4 y z 2  1)  2 zy
(4)
Similarly holding y constant and differentiating with respect to z and setting
P / z  0 , we get
3
z
 0.577
3
Putting this value in (2), (3) and (4) we get the geometrical properties of the
best trapezoidal section as
P  2 3y
Bw 
2
y
3
P  3Bw
A  3y2
We can observe that
in this condition.
Similarly we can find the geometric characteristics of various shapes as
shown in the following Table.
Shape
Best
A
P
B
Geometry
Trapezoidal
Half hexagon
3.46 y
2.31y
1.73 y 2
Rectangle
Half square
2 y2
4y
2y
Triangle
Half square
y2
2y
Semicircle
-
0.5y 2
2.83 y
y
Parabola
-
1.89 y 2
3.77 y
2.83 y
2y
Minimum Permissible Velocity: 0.6 to 0.9 m/s
Channel Slopes :
Longitudinal Slope should be preferably equal to the ground slope, but if the
permissible velocities are not achievable, the slope should be increased.
Side slopes depend on the type of the material, i.e. angle of repose.
USBR prefers a 1.5:1 (H:V) slope for usual sizes of lined channels.
Recommended side slopes are given in the Table below;
5
Freeboard: Vertical distance between the water surface and top of the
channel.
Freeboard can be estimated as
F  0.55 Cy
Where, the coefficient, C  0.01184Q  1.493
The minimum freeboard is usually 30 cm.
At the channel bends additional freeboard must be provided to account for
the super-elevation of the water surface. Super-elevation of the water surface
can be estimated as
V 2B
hs 
grc
V  average velocity in the channel, B  top width of the channel,
and rc  radius of curvature of the centerline of the channel.
It is also recommended that rc  3B
Where,
Lined Channel Design
According to ASCE all channels carrying supercritical flow should be lined
with concrete reinforced in both longitudinal and lateral directions.
6
Steps in design
1. Estimate the roughness coefficient, n for the lining material and
freeboard coefficient, C for the design discharge.
2. Compute the normal depth of flow by Manning’s equation, with
assumed bed slope and if possible using best hydraulic cross-section.
3. Check the minimum velocity and Froude Number. Repeat step 2 and 3
if necessary to meet the specifications.
4. Calculate the required freeboard and increase the freeboard on the
bends if required using the equation for super-elevation.
Example: Design a lined trapezoidal channel to carry 20 m3/s on a
longitudinal slope of 0.0015. The lining of the channel is to be float-finished
concrete. Consider: (a) the best hydraulic section, and (b) a section with side
slopes of 1.5:1 (H:V).
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Solution
(a) n=0.015,
for best section we compute,
z  0.58,
yn  2.09m
V
 0.66, C  1.7, F  1.04m
gD
(b) yn  1.94m, V  2.53m / s, Fr  0.76, F  1.0m
V  2.6 m / s, Fr 
D
A
B
8