open channel

1. OPEN CHANNELS
(OPEN CHANNEL FLOW AND HYDRAULIC
MACHINERY)
UNIT – I
Rambabu Palaka, Assistant Professor
BVRIT

2. Learning Objectives
1. Types of Channels
2. Types of Flows
3. Velocity Distribution
4. Discharge through Open Channels
5. Most Economical Sections

3. Learning Objectives
6. Specific Energy and Specific Energy Curves
7. Hydraulic Jump (RVF)
8. Gradually Varied Flow (GVF)

4. Types of Channels
 Open channel flow is a flow which has a free
surface and flows due to gravity.
 Pipes not flowing full also fall into the
category of open channel flow
 In open channels, the flow is driven by the
slope of the channel rather than the pressure

5. Types of Channels
 Open channel flow is a flow which has a free
surface and flows due to gravity.
 Pipes not flowing full also fall into the
category of open channel flow
 In open channels, the flow is driven by the
slope of the channel rather than the pressure

6. Types of Flows
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow
4. Sub-critical, Critical and Super-critical Flow

7. 1. Steady and Unsteady Flow
 Steady flow happens if the conditions (flow
rate, velocity, depth etc) do not change with
time.
 The flow is unsteady if the depth is
changes with time

8. 2. Uniform and Non-uniform Flow
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
 If for a given length of channel, the velocity
of flow, depth of flow, slope of the channel
and cross section remain constant, the flow is
said to be Uniform
 The flow is Non-uniform, if velocity, depth,
slope and cross section is not constant

9. 2. Non-uniform Flow
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
Types of Non-uniform Flow
1. Gradually Varied Flow (GVF)
If the depth of the flow in a channel changes gradually
over a length of the channel.
2. Rapidly Varied Flow (RVF)
If the depth of the flow in a channel changes abruptly
over a small length of channel

10. Types of Flows
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow

11. 3. Laminar and Turbulent Flow
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow
Both laminar and turbulent flow can occur in open channels
depending on the Reynolds number (Re)
Re = ρVR/µ
Where,
ρ = density of water = 1000 kg/m3
µ = dynamic viscosity
R = Hydraulic Mean Depth = Area / Wetted Perimeter

12. Types of Flows
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow

13. Types of Flows
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow
4. Sub-critical, Critical and Super-critical Flow
4. Sub-critical, Critical and Super-critical Flow

14. Types of Flows
1. Steady and Unsteady Flow
2. Uniform and Non-uniform Flow
3. Laminar and Turbulent Flow
4. Sub-critical, Critical and Super-critical Flow

15. Velocity Distribution
 Velocity is always vary across channel
because of friction along the boundary
 The maximum velocity usually found just
below the surface

16. Velocity Distribution
 Velocity is always vary across channel
because of friction along the boundary
 The maximum velocity usually found just
below the surface

17. Discharge through Open Channels
1. Chezy’s C
2. Manning’s N
3. Bazin’s Formula
4. Kutter’s Formula

18. Discharge through Open Channels
1. Chezy’s C
2. Manning’s N
3. Bazin’s Formula
4. Kutter’s Formula
Forces acting on the water between sections 1-1 & 2-2
1. Component of weight of Water = W sin i 
2. Friction Resistance = f P L V2 
where
W = density x volume
= w (AL) = wAL
Equate both Forces:
f P L V2 = wAL sin i

19. 3
Constant
s
Chezy'
C
f
w
2
Radius
Hydraulic
m
P
A
1
i
sin
P
A
f
w
V








Chezy’s Formula, mi
C
V 

20. 3
Constant
s
Chezy'
C
f
w
2
Radius
Hydraulic
m
P
A
1
i
sin
P
A
f
w
V








i
m.
C
V
i
i
tan
i
sin
i,
of
values
small
for
i
sin
m.
C
V
1,
Eqn.
in
3
&
2
Eqn.
substitute





Chezy’s Formula, mi
C
V 

21. 1. Manning’s N
Chezy’s formula can also be used with Manning's
Roughness Coefficient
C = (1/n) R1/6
where
R = Hydraulic Radius
n = Manning’s Roughness Coefficient

22. 2. Bazin’s Formula
1. Manning’s N
2. Bazin’s Formula
Chezy’s formula can also be used with Bazins’ Formula
where
k = Bazin’s constant
m = Hydraulic Radius
m
k
1.81
157.6
C



23. Chezy’s Formula,
1. Manning’s N
2. Bazin’s Formula
mi
C
V 

24. 3. Kutter’s Formula
1. Manning’s N
2. Bazin’s Formula
3. Kutter’s Formula
Chezy’s formula can also be used with Kutters’ Formula
where
N = Kutter’s constant
m = Hydraulic Radius, i = Slope of the bed
m
N
i
0.00155
23
1
N
1
0.00155
23
C












25. Chezy’s Formula,
1. Manning’s N
2. Bazin’s Formula
3. Kutter’s Formula
mi
C
V 

26. Problems
1. Find the velocity of flow and rate of flow of water through a
rectangular channel of 6 m wide and 3 m deep, when it is
running full. The channel is having bed slope as 1 in 2000.
Take Chezy’s constant C = 55
2. Find slope of the bed of a rectangular channel of width 5m
when depth of water is 2 m and rate of flow is given as 20
m3/s. Take Chezy’s constant, C = 50

27. Problems
3. Find the discharge through a trapezoidal channel of 8 m
wide and side slopes of 1 horizontal to 3 vertical. The depth
of flow is 2.4 m and Chezy’s constant C = 55. The slope of
bed of the channel is 1 in 4000
4. Find diameter of a circular sewer pipe which is laid at a
slope of 1 in 8000 and carries a discharge of 800 litres/s
when flowing half full. Take Manning’s N = 0.020

28. Problems
5. Find the discharge through a channel show in fig. 16.5.
Take the value of Chezy’s constant C = 55. The slope of
bed of the channel is 1 in 2000

29. Most Economical Sections
1. Cost of construction should be minimum
2. Discharge should be maximum
Types of channels based on shape:
1. Rectangular
2. Trapezoidal
3. Circular

30. Most Economical Sections
1. Cost of construction should be minimum
2. Discharge should be maximum
Types of channels based on shape:
1. Rectangular
2. Trapezoidal
3. Circular maximum
be
will
Q
minimum,
is
P
If
i
A
C
A
K
where
P
1
K
Q
i
m
C
A
V
A
Q





32. Rectangular Section
0
d(d)
dP
minimum
be
should
P
section,
economical
most
for


33. Rectangular Section
0
d(d)
dP
minimum
be
should
P
section,
economical
most
for

2
2
2
2d
2
A
m
b/2
d
or
2d
b
2d
bd
2d
A
0
2
d
A
0
)
(
2
0
)
(
minimum
be
should
P
seciton,
economical
most
for
2
2
2
1
2
2
2
2
d
d
d
d
b
bd
P
d
d
d
d
A
d
d
d
dP
d
d
A
d
b
P
d
A
b
bd
A




































35. Trapezoidal Section
0
d(d)
dP
minimum
be
should
P
section,
economical
most
for


36. Trapezoidal Section
0
d(d)
dP
minimum
be
should
P
section,
economical
most
for

600
θ
and
2
d
m
1
n
d
2
2nd
b
0
d(d)
1
2
n
2d
nd
d
A
d
0
d(d)
dP
minimum
be
should
P
seciton,
economical
most
for
2
1
n
2d
nd
d
A
1
n
2d
b
P
1
nd
d
A
b
nd)d
(b
A
2
2
2

































37. Circular Section
0
d
P
A
3
d
Discharge,
Max.
for
0
d
P
A
d
Velocity,
Max.
for



















38. Circular Section
0
d
P
A
3
d
Discharge,
Max.
for
0
d
P
A
d
Velocity,
Max.
for


















0.95D
d
,
154
θ
0
dθ
P
3
A
d
discharge,
max.
for
constants
are
i
and
C
,
i
P
A
C
i
P
A
AC
i
m
AC
Q
0.3D
m
0.81D,
d
,
45
128
θ
0
dθ
dm
velocity,
max.
for
3
)
2
2θ
sin
-
(θ
2θ
R
P
A
m
2
2Rθ
P
1
)
2
2θ
sin
-
(θ
R
A
0
3
'
0
2




























39. Problems
1. A trapezoidal channel has side slopes of 1 horizontal and 2
vertical and the slope of the bed is 1 in 1500. The area of
cross section is 40m2. Find dimensions of the most
economical section. Determine discharge if C=50
Hint:
 Equate Half of Top Width = Side Slope (condition 1) and find b in terms of d
 Substitute b value in Area and find d
 Find m = d/2 (condition 2)
 Find V and Q

40. Problems
1. A trapezoidal channel has side slopes of 1 horizontal and 2
vertical and the slope of the bed is 1 in 1500. The area of
cross section is 40m2. Find dimensions of the most
economical section. Determine discharge if C=50

41. Problems
1. A trapezoidal channel has side slopes of 1 horizontal and 2
vertical and the slope of the bed is 1 in 1500. The area of
cross section is 40m2. Find dimensions of the most
economical section. Determine discharge if C=50

42. Problems
2. A rectangular channel of width 4 m is having a bed slope of
1 in 1500. Find the maximum discharge through the
channel. Take C=50
3. The rate of flow of water through a circular channel of
diameter 0.6m is 150 litres/s. Find the slope of the bed of
the channel for maximum velocity. Take C=50

43. Non-uniform Flow
In Non-uniform flow, velocity varies at each section of the
channel and the Energy Line is not parallel to the bed of the
channel.
This can be caused by
1. Differences in depth of channel and
2. Differences in width of channel.
3. Differences in the nature of bed
4. Differences in slope of channel and
5. Obstruction in the direction of flow

44. Specific Energy
Energy
Specific
as
called
is
which
2g
v2
h
Es
datum,
as
taken
is
bottom
channel
the
If
datus,
above
channel
of
bottom
of
Height
z
where
2g
v2
h
z
E
fluid,
flowing
of
Energy
Total







45. Specific Energy
h2
2g
q
h
2g
V
h
Es
h
q
bh
Q
V
constant
b
Q
q
,
unit width
per
discharge
If
bh
Q
A
Q
V
V
A
Q
2
2













Modified Equation
to plot Specific Energy Curve

46. Specific Energy
Potential Energy (h)
Es= h + q2/2gh2
hc
g
Vc
1
Eqn.
in
Vc
hc
b
v
bh.
b
Q
q
value
subsitute
q
2
hc
g
q
2
hc
g
q
2 3
1
hc
h
2
2g
q
2
h
E
where,
0
dh
dE
Depth,
Critical
for






















1
.
3
3
g

47. Specific Energy
Potential Energy (h)
Es= h + q2/2gh2
hc
g
Vc
1
Eqn.
in
Vc
hc
b
v
bh.
b
Q
q
value
subsitute
q
2
hc
g
q
2
hc
g
q
2 3
1
hc
h
2
2g
q
2
h
E
where,
0
dh
dE
Depth,
Critical
for






















1
.
3
3
g
3
Emin
2
or
hc
g
q
critical
is
flow
of
Depth
minimum,
is
energy
specific
when
Depth;
Critical
of
terms
in
Energy
Specific
Minimum
hc
2
3hc
2
hc
hc
2
2g
hc
3
hc
Emin
hc
3
or
g
q
2 3
1
hc
substitute
hc
2
2g
q
2
hc
E
h
2
2g
q
2
h
E
2





















48. Specific Energy Curve
Alternate Depths 1 & 2
Hydraulic Jump

49. Problems
1. The specific energy for a 3 m wide channel is to be 3 kg-m/kg. What
would be the max. possible discharge
2. The discharge of water through a rectangular channel of width 6 m, is
18 m3/s when depth of flow of water is 2 m. Calculate: i) Specific Energy
ii) Critical Depth iii) Critical Velocity iv) Minimum Energy
3. The specific energy for a 5 m wide rectangular channel is to be 4 Nm/N.
If the rate of flow of water through the channel us 20 m3/s, determine the
alternate depths of flow.

50. Hydraulic Jump

51. The hydraulic jump is defined as the rise of
water level, which takes place due to
transformation of the unstable shooting flow
(super-critical) to the stable streaming flow
(sub-critical).
When hydraulic jump occurs, a loss of energy
due to eddy formation and turbulence flow
occurs.
Hydraulic Jump

52. Hydraulic Jump
The most typical cases for the location of
hydraulic jump are:
1. Below control structures like weir, sluice are
used in the channel
2. when any obstruction is found in the
channel,
3. when a sharp change in the channel slope
takes place.
4. At the toe of a spillway dam

53. Fe
of
interms
1
Fe
2
8
1
2
d1
d2
V1
of
interms
g1
d1
2v1
2
4
d1
2
2
d1
d2
q
of
interms
d1
g
2q
2
4
d1
2
2
d1
d2




















Hydraulic Jump

54. Fe
of
interms
1
Fe
2
8
1
2
d1
d2
V1
of
interms
g1
d1
2v1
2
4
d1
2
2
d1
d2
q
of
interms
d1
g
2q
2
4
d1
2
2
d1
d2




















 
d1
d2
Jump
Hydrualic
)
d1
(d2
of
times
7
to
5
jump
of
Length
d2
d1
4
d1
d2
3
E2
E1
hL
:
Energy
of
Loss
















Hydraulic Jump

55. Problems
1. The depth of flow of water, at a certain section of a
rectangular channel of 2 m wide is 0.3 m. The discharge
through the channel is 1.5 m3/s. Determine whether a
hydraulic jump will occur, and if so, find its height and loss
of energy per kg of water.
2. A sluice gate discharges water into a horizontal rectangular
channel with a velocity of 10 m/s and depth of flow of 1 m.
Determine the depth of flow after jump and consequent loss
in total head.

56. Gradually Varied Flow (GVF)

57. Gradually Varied Flow (GVF)
In GVF, depth and velocity vary slowly, and the free surface is stable
The GVF is classified based on the channel slope, and the magnitude of
flow depth.
Steep Slope (S): So > Sc or h < hc
Critical Slope (C): So = Sc or h = hc
Mild Slope (M): So < Sc or h > hc
Horizontal Slope (H): So = 0
Adverse Slope(A): So = Negative
where
So : the slope of the channel bed,

58. Gradually Varied Flow (GVF)
In GVF, depth and velocity vary slowly, and the free surface is stable
The GVF is classified based on the channel slope, and the magnitude of
flow depth.
Steep Slope (S): So > Sc or h < hc
Critical Slope (C): So = Sc or h = hc
Mild Slope (M): So < Sc or h > hc
Horizontal Slope (H): So = 0
Adverse Slope(A): So = Negative
where
So : the slope of the channel bed,

59. Flow Profiles
The surface curves of water are called flow profiles (or water surface profiles).
Depending upon the zone and the slope of the bed, the water profiles are classified
into 13 types as follows:
1. Mild slope curves M1, M2, M3
2. Steep slope curves S1, S2, S3
3. Critical slope curves C1, C2, C3
4. Horizontal slope curves H2, H3
5. Averse slope curves A2, A3
In all these curves, the letter indicates the slope type and the subscript indicates the
zone. For example S2 curve occurs in the zone 2 of the steep slope

60. Flow Profiles in Mild slope
Flow Profiles in Steep slope
Critical Depth Line
Normal Depth Line

61. Flow Profiles in Critical slope
Flow Profiles in Horizontal slope

62. Flow Profiles in Adverse slope

63. Gradually Varied Flow (GVF)
 
channel
the
of
bottom
the
along
depth
water
of
variation
the
represents
dx
dh
where
Fe
of
in terms
)
Fe
(
2
1
ie
i b
dx
dh
Velocity
of
in terms
gh
V
2
1
ie
i b
dx
dh
:
GVF
of
Equation
















ie
i -
b
E1
E -
2
L 
Sc or ib Energy Line Slope
So or ie Bed Slope
h2
h1

64. Gradually Varied Flow (GVF)
 
channel
the
of
bottom
the
along
depth
water
of
variation
the
represents
dx
dh
where
Fe
of
in terms
)
Fe
(
2
1
ie
i b
dx
dh
Velocity
of
in terms
gh
V
2
1
ie
i b
dx
dh
:
GVF
of
Equation
















If dh/dx = 0, Free Surface of water is
parallel to the bed of channel
If dh/dx > 0, Depth increases in the
direction of water flow (Back Water
Curve)
If dh/dx < 0, Depth of water decreases
in the direction of flow (Dropdown
Curve)
ie
i -
b
E1
E -
2
L 
Sc or ib Energy Line Slope
So or ie Bed Slope
h2
h1

65. Problems
1. Find the rate of change of depth of water in a rectangular
channel of 10 m wide and 1.5 m deep, when water is
flowing with a velocity of 1 m/s. The flow of water through
the channel of bed slope in 1 in 4000, is regulated in such a
way that energy line is having a slope of 0.00004
2. Find the slope of the free water surface in a rectangular
channel of width 20 m, having depth of flow 5 m. The
discharge through the channel is 50 m3/s. The bed of
channel is having a slope of 1 in 4000. Take C=60

66. Reference
Chapter 16
A Textbook of Fluid Mechanics and
Hydraulic Machines
Dr. R. K. Bansal
Laxmi Publications