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## Chapter 3: Steady uniform flow in open channels

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### 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. – PowerPoint PPT presentation

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Title: Chapter 3: Steady uniform flow in open channels

1
Chapter 3 Steady uniform flow in open channels
2
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
Chezys Mannings equation
• Able to solve problems related to optimum cross
section in both conduits and open channel

3
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 pipe is not constant Existence of free water surface through 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 constant and it is equal to the cross sectional area of the conduit (pipe). Flow cross sectional area may change throughout the length depending on the depth of 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

6
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.

7
For flow in pipes
8
Flow in open channel
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• 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.

12
• 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.

13
• 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.

14
• 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.

15
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)

16
• Total energy line for flow in open channel

17
• 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.

18
Flow classification
• Non uniform flow in open channels can be divided
into two types
• 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

19
Non-uniform flow
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23
• Total energy line for flow in open channel

24
Continuity equation
• For rectangular channel
• Express as flow per unit width, q

25
Momentum equation
• Produced by the difference in hydrostatic forces
at section 1 and 2
• Resultant force,

26
Energy equation
• But hydrostatic pressure at a depth x below free
surface,
• Therefore,
• Energy equation rewritten as,

27
Energy equation
• Therefore the head loss is,
• And energy equation reduces to,
• Known as specific energy, E, (total energy per
unit weight measured above bed level),

28
Geometrical properties of open channels
• Geometrical properties of open channels
• Flow cross-sectional area, A
• Wetted perimeter, P
• Hydraulic mean depth, m

29
• A covers the area where fluid takes place.
• P total length of sides of the channel
cross-section which is in contact with the flow.

30
Example 3.1
• Determine the hydraulic mean depth, m, for the
trapezoidal channel shown below.

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Chezys coefficient, C
• From chapter 1,
• Rearranging to fit for open channel, the
velocity
• Where Chezy roughness coefficient,
• For open channel i can be taken as equal to the
gradient of the channel bed slope s. Therefore,

33
Mannings n
• Introduced roughness coefficient n of the channel
boundaries.

34
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 11600.

35
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 Mannings roughness n 0.013
and a bed slope of 12000.

36
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 (C38.6)
• b. Manning formula

37
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 .

38
Example 3.6 (Bansal, 2003)
• Find the discharge of water through the channel
shown in figure. Take the value of Chezys
constant 60 and slope of the bed as 1 in 2000.

C
39
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 Mannings n 0.020.

40
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

41
Example 3.9
• Given that the flow in the channel shown in
figure is a maximum, determine the dimensions of
the channel.

42
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.

43
Optimum depth for non-full flow in closed conduits
• Flow cross sectional area,
• Wetted perimeter,

44
Optimum depth for non-full flow in closed conduits
• Under optimum condition, vmax,
• Substituting simplifying,

45
Optimum depth for non-full flow in closed conduits
• Hence depth, D, at vmax,
• Using Chezy equation,

46
Optimum depth for non-full flow in closed conduits
• Qmax occurs when (A3/P) is maximum,
• Substituting simplifying,
• Therefore depth at Qmax,

47
Review of past semesters questions
48
OCT 2010
• Analysis of flow in open channels is based on
equations established in the study of fluid
mechanics. State the equations.

49
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.

50
OCT 2010
• State the differences between