Chapter 3:
Steady uniform flow in open
channels
Learning outcomes
• By the end of this lesson, students should be
able to:
– Understand the concepts and equations used in
open channel flow
– Determine the velocity and discharge using
Chezy’s & Manning’s equation
– Able to solve problems related to optimum cross
section in both conduits and open channel
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Introduction
• Comparison between full flow in closed conduit and
flow in open channel
Full flow in closed conduit
Open channel flow
No free surface and pressure in the Existence of free water surface through
pipe is not constant
out the length of flow in the channel.
Pressure at the free surface remains
constant,
with
value
equal
to
atmospheric pressure.
Flow cross sectional area remains Flow cross sectional area may change
constant and it is equal to the cross throughout the length depending on the
sectional area of the conduit (pipe). depth of flow.
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Flow classification based
on fluid particles motion
Turbulent
flow
Laminar
Flow
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Flow classification
• Turbulent flow:
– Characterized by the random and irregular
movement of fluid particles.
– Movement of fluid particles in turbulent flow is
accompanied by small fluctuations in pressure.
– Flows in open channel are mainly turbulent.
– E.g. Hydraulic jump from spillway, Flow in fast
flowing river
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Flow classification
• Laminar flow:
– Flow characterized by orderly movement of fluid
particles in well defined paths.
– Tends to move in layers.
– May be found close to the boundaries of open
channel.
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For flow in pipes
Turbulent
Re > 2000
Laminar
Re < 2000
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Flow in open channel
Turbulent
Re > 500
Laminar
Re < 500
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Flow
classification
wrt space
Uniform flow
Flow parameters (velocity & pressure)
remain constant wrt space at any point
in the flow.
E.g. Flow with a fixed discharge in a
channel of constant geometrical shape
and slope
Non-uniform flow
Flow parameters (velocity &pressure)
change wrt time & space.
E.g. Flow of fixed discharge through
a channel which changes either in
geometrical shape or slope of the
channel
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Flow classification
wrt time
Steady flow
Flow
parameters
remain
constant over a specified time
interval
Unsteady flow
Flow parameters vary over
time
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Steady uniform flow
•Flow parameters do not change wrt space
(position) or time.
•Velocity and cross-sectional area of the stream
of fluid are the same at each cross-section.
•E.g. flow of liquid through a pipe of uniform
bore running completely full at constant
velocity.
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Steady non-uniform flow
• Flow parameters change with respect to space
but remain constant with time.
• Velocity and cross-sectional area of the stream
may vary from cross-section to cross-section,
but, for each cross-section, they will not vary
with time.
• E.g. flow of a liquid at a constant rate through
a tapering pipe running completely full.
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Unsteady uniform flow
• Flow parameters remain constant wrt space but
change with time.
• At a given instant of time the velocity at every
point is the same, but this velocity will change
with time.
• E.g. accelerating flow of a liquid through a
pipe of uniform bore running full, such as
would occur when a pump is started.
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Unsteady non-uniform flow
• Flow parameters change wrt to both time &
space.
• The cross-sectional area and velocity vary
from point to point and also change with time.
• E.g. a wave travelling along a channel.
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Flow classification
• Normal depth – depth of flow under steady
uniform condition.
• Steady uniform condition – long channels with
constant cross-sectional area & constant
channel slope.
– Constant Q
– Constant terminal v
– Therefore depth of flow is constant (yn or Dn)
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Total energy line for flow in open channel
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• From the figure :
– Water depth is constant
• slope of total energy line = slope of channel
• When velocity and depth of flow in an open
channel change, then non-uniform flow will
occur.
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Flow classification
• Non uniform flow in open channels can be divided
into two types:
– Gradually varied flow, GVF
• Where changes in velocity and depth of flow take place
over a long distance of the channel
– Rapidly varied flow, RVF
• Where changes in velocity and depth of flow occur over
short distance in the channel
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Non-uniform flow
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Analysis of flow in
open channel
Continuity
equation
Momentum
equation
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Energy
equation
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Total energy line for flow in open channel
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Continuity equation:
Q1  Q2
A1v1  A2v2
• For rectangular channel:
Q1  Q2
A1v1  A2 v2
B1 D1v1  B2 D2v2
• Express as flow per unit width, q:
q
Q vBD
 vD
B B
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Momentum equation
F  Q(v2  v1 )
• Produced by the difference in hydrostatic forces at
section 1 and 2:
Forcein directionof motionat section1  gA1 x1
Forcein opposingdirectionof motionat section2  gA 2 x2
• Resultant force,
Force Rate of changeof momentum
gA1 x1  A 2 x2   Qv2  v1 
Qv2  v1 
A1 x1  A 2 x2  
g
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Energy equation
p v2
H

 ( z  d  x)
g 2 g
• But hydrostatic pressure at a depth x below free
surface, x  p
g
• Therefore,
v2
H
zd
2g
• Energy equation rewritten as,
2
2
v1
v
 z1  d1  2  z2  d 2  hL
2g
2g
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Energy equation
• For steady uniform
flow, d1  d 2
v1  v2
2
1
2
v1
v
 d1  2  d 2
2g
2g
• Therefore the head loss
is, h  z  z
L
• And energy equation
reduces to,
2
• Known as specific
energy, E, (total energy
per unit weight
measured above bed
level),
v2
E
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2g
D
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Geometrical properties of open channels
• Geometrical properties of open channels:
– Flow cross-sectional area, A
– Wetted perimeter, P
– Hydraulic mean depth, m
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• A – covers the area
where fluid takes place.
D
B
A  BD
• P – total length of sides
of the channel crosssection which is in
contact with the flow.
P  B  2D
A
BD
m 
P B  2D
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Example 3.1
Determine the hydraulic mean depth, m, for the
trapezoidal channel shown below.
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Roughness
coefficient
Chezy, C
Manning, n
v  C mi
1 2 3 12
v m i
n
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Chezy’s coefficient, C
• From chapter 1,
fL v 2
hL 
m 2g
• Rearranging to fit for open channel, the velocity:
v  C mi
• Where Chezy roughness coefficient,
C
2g
f
• For open channel i can be taken as equal to the
gradient of the channel bed slope s. Therefore,
v  C ms
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Manning’s n
• Introduced roughness coefficient n of the
channel boundaries.
1 2 3 12
v m i
n
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Example 3.2
Calculate the flow rate, Q in the channel shown in
Figure 3.5, if the roughness coefficient n = 0.025 and
the slope of the channel is 1:1600.
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Example 3.3
• Determine the flow velocity, v and the flow rate for
the flow in open channel shown in the figure. The
channel has a Manning’s roughness n = 0.013 and a
bed slope of 1:2000.
B
2.75 m
900
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Example 3.4 (Douglas, 2006)
• An open channel has a cross section in the
form of trapezium with the bottom width B of
4 m and side slopes of 1 vertical to 11/2
horizontal. Assuming that the roughness
coefficient n is 0.025, the bed slope is 1/1800
and the depth of the water is 1.2 m, find the
volume rate of flow Q using
a. Chezy formula (C=38.6)
b. Manning formula
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Example 3.5 (Munson, 2010)
• Water flows along the drainage canal having
the properties shown in figure. The bottom
slope so = 0.002. Estimate the flow rate when
the depth is 0.42 m .
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Example 3.6 (Bansal, 2003)
Find the discharge of water through the channel
shown in figure. Take the value of Chezy’s constant =
60 and slope of the bed as 1 in 2000.
A
D
1.2 m
C
2.7 m
B
1.5 m
E
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Example 3.8 (Bansal, 2003)
Find the diameter of a circular sewer pipe which is
laid at a slope of 1 in 8000 and carries a discharge of
800 L/s when flowing half full. Take the value of
Manning’s n = 0.020.
D
d
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Optimum cross-sections for open channels
• Optimum cross section – producing Qmax
for a given area, bed slope and surface
roughness, which would be that with Pmin
and Amin therefore tend to be the
cheapest.
• Qmax : Amin, Pmin
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Example 3.9
Given that the flow in the channel shown in figure is
a maximum, determine the dimensions of the channel.
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Optimum depth for non-full flow in
closed conduits
• Partially full in pipes can be treated same as
flow in an open channel due to presence of a
free water surface.
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Optimum depth for non-full flow in
closed conduits
• Flow cross sectional area,
A  sectorOMNP - triangleOMP

1 2

  r  2   r 2 sin  cos
2

1

2
 r   sin 2 
2



• Wetted perimeter,
P  2 r
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Optimum depth for non-full flow in
closed conduits
• Under optimum condition, vmax,
 P   1  P dA  A dP   0
d A
 2 
d
 P  d
dA
dP
P
A
d
d

d 
• Substituting & simplifying,
2  tan 2
  257.50
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Optimum depth for non-full flow in
closed conduits
• Hence depth, D, at vmax,
D  1.62r
 0.81d
• Using Chezy equation,
1
Q  ACm i
A
 C 
 P
3
2
1
2
1
 2 12
 i

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Optimum depth for non-full flow in
closed conduits
• Qmax occurs when (A3/P) is maximum,
 P   1 3PA dA  A dP   0
3
d A
d
P 
dA
dP
3P
A
d
d
2
2
3
d
d 
• Substituting & simplifying,
2  3080
  1540
• Therefore depth at Qmax,
D  1 .9 r
 0.95d
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Review of past semesters’
questions
OCT 2010
• Analysis of flow in open channels is based on
equations established in the study of fluid
mechanics. State the equations.
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OCT 2010
• Determine the discharge in the channel (n = 0.013) as
shown in Figure Q3(b). The channel has side slopes of 2
: 3 (vertical to horizontal) and a slope of 1 : 1000.
Determine also the discharge if the depth increases by 0.1
m by using Manning's equation.
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OCT 2010
• State the differences between :
i) Steady and unsteady flow
ii) Uniform and non-uniform flow
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OCT 2010
• Figure Q4(b) shows the
channel. Prove that for a
channel, the optimum
cross-section
occurs
when the width is 4
times its depth (B =
4D). (Hint: A = 4/3BD
and P = B + 4D)
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