Civil Engineering May 2021
C by Bazin:
OPEN CHANNELS
-
V by Chezy formula:
v  C RS
S = slope of the egl; hydraulic slope;
energy gradient (as in pipes)
HL hf
S

L
L
R = hydraulic radius (m)
A
P
(as in pipes)
P = wetted perimeter (m)
 m1/2 
C = Chezy coefficient 
 s 


Theoretically,
C
Hydraulics 12
Most Efficient Sections (MES):
87
C
m
1
R
conduits where liquid flows with free or
atmospheric surface.
Q  Av (as in pipes)
R
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C by Kutter:
1
0.00155
 23 
n
S
C

n
0.00155 
1
 23 

S
R

In channels of different shapes/ proportions but
of the same A, S and n, the one that gives the
biggest Q or the smallest P is the “mes” (also
called optimum channel section, most
economical section, most hydraulically efficient
section, most advantageous section, most
favorable section, best hydraulics section and
the like.)
m,n = roughness coefficient
If C is not specified, use the Manning’s C in the
Chezy’s V. The result is the Manning or the
Chezy-Manning formula for V:
1
v  R 2/3S1/2
n
A. UNIFORM FLOW
- the same depth, cross sectional area and
velocity for a length of flow
- the channel bed is parallel to the egl.
Generally,
8g
f
C by manning:
C
1 1/6
R
n
width at the top = sum of the sides
d
R
2
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Civil Engineering May 2021
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c. Calculate the boundary shearing
stress.
If the cross section of the most efficient
trapezoidal section is not specified, assume
1. A rectangular channel 1.8 m wide conveys
water at a depth of 1.2 m.
a. What is the velocity of flow if it is laid
on a slope of 0.002 and n= 0.012?
b. What is the discharge if the channel
slopes 0.1° and n = 0.016? Use
Kutter’s formula for C.
c. What is the velocity of flow if S =
0.000263 and f = 0.02?
2. The figure shows a natural channel flow
consisting of the main channel and two
equal flood plains on the sides. Assume the
slope of the channel is 0.0002, n=0.02 for
the main and n=0.04 for the flood plains.
a. What is the estimated discharge in
the main channel?
b. What is the total discharge?
c. If the total discharge is to be
contained in an asphalt-lined
semicircular channel with n=0.016,
what diameter is required?
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Hydraulics 12
3. An open channel is to be designed to carry
2.6 m3/s of water at a slope of 0.0009. The
channel material has n = 0.014.
a. Compute the depth for a semicircular section.
b. Compute the depth for a rectangular
section.
c. Compute the depth for a trapezoidal
section.
4. For a discharge of 30 m3/s and a velocity of 3
m/s, calculate the depth and bottom width
of the trapezoidal channel shown in order
to have minimum seepage. n = 0.017.
5. A trapezoidal canal has bottom width of 5 m
and side slopes of 1H to 2V. The flow is 30
m3/s when the depth of flow is 1.20 m. Use
n = 0.013.
a. Calculate the specific energy.
b. What is the slope of the channel
bed?
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6. Water flows in a brickwork rectangular
channel (n = 0.015) 2 m wide, on a slope of
5 m/km.
a. Find the flow rate when the normal
depth is 50 cm.
b. If the normal depth remains 50 cm,
find the channel width which will
triple the flow rate.
7. The trapezoidal channel is made of
brickwork (n = 0.015) and slopes at 1:500.
Determine the flow rate if the normal depth
is 80 cm.
6) 2.266 m3/s
5.153 m
7) 5.23 m3/s
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