Brief discussion of open channel flow.

1.  Chezy’s and Manning’s equations for uniform flow
in open channel,
 Velocity distribution,
 Most efficient channel section,
 Compound channels.
01/07/17 1MODASSAR ANSARI

2.  BY MODASSAR ANSARI
 2nd
Year
 Department of civil Engineering
 SUBJECT- HYDRAULICS & HYDRAULIC
MACHINES
 SUBJECT CODE-NCE 403
01/07/17 2MODASSAR ANSARI

3.  Introduced by the French engineer Antoine Chezy in 1768
while designing a canal for the water-supply system of Paris
01/07/17
h fV C R S=
150<C<60
s
m
s
m
where C = Chezy coefficient
where 60 is for rough and 150 is for smooth also a function of R (like f in
Darcy-Weisbach)
2
f h
g
V S R
λ
=compare
0.0054 > > 0.00087λ
4 hd R=
For a pipe
0.022 > f > 0.0035
3MODASSAR ANSARI

4.  Most popular in U.S. for open channels
01/07/17
(english system)
1/2
o
2/3
h SR
1
n
V =
1/2
o
2/3
h SR
49.1
n
V =
VAQ =
2/13/2
1
oh SAR
n
Q = very sensitive to n
Dimensions ofDimensions of nn??
isis nn only a function of roughness?only a function of roughness?
(MKS units!)
NO!
T /L1/3
Bottom slope
4MODASSAR ANSARI

5.  A section of a channel is said to be most economical when the
cost of construction of the channel is minimum. But the cost
of construction of a channel depends on excavation and the
lining.To keep the cost down or minimum, the wetted
perimeter, for a given discharge, should be minimum. This
condition is utilized for determining the dimensions of
economical sections of different forms of channels. Most
economical section is also called the best section or most
efficient section as the discharge passing through a most
economical section of channel for a given cross-sectional area
A, slope of the bed So and a resistance coefficient, is
maximum. But the discharge
01/07/17 5MODASSAR ANSARI

6. Consider a rectangular section of channel as shown.
Let B = width of channel,
D = depth of flow. B
∴ Area of flow, A = B x D,
Wetted perimeter, P = 2D + B, D
we have
we get P = 2D +
01/07/17
D
A
B =
D
A
6MODASSAR ANSARI

7. we get P =2D+ A/D
For most economical cross section, P should be minimum for a
given area;
dP/ dD = 0
So, dP/dD = 2-A/D2
=0
2=A/D2
=BD/D2
2=B/D
Hence D =B/2
Hydraulic radius Rh=A/P
= BxD/B+2D
 = 2D2
/4D
 = D/2
01/07/17 7MODASSAR ANSARI

8.  Derive P = f(y) and A = f(y) for a trapezoidal channel
 How would you obtain y
01/07/17
z
1
b
y
zyybA 2
+=
2/13/2
1
oh SAR
n
Q =
8MODASSAR ANSARI

9. 01/07/17





 −
=
r
yr
arccosθ
( )θθθ cossin2
−= rA
θsin2rT =
θ
y
T
A
r
θrP 2=
radians
Maximum discharge
when y = ______
0.938d
9MODASSAR ANSARI

10. 01/07/17





 −
=
r
yr
arccosθ
( )θθθ cossin2
−= rA
θsin2rT =
θ
y
T
A
r
θrP 2=
radians
Maximum discharge
when y = ______
0.938d
9MODASSAR ANSARI