OPEN CHANNELS
(OPEN CHANNEL FLOW AND HYDRAULIC
MACHINERY)
UNIT – I
BVRIT Rambabu Palaka, Assistant Professor
Learning Objectives
1. Types of Channels
2. Types of Flows
3. Velocity Distribution
4. Discharge through Open Channels
5. Most Economical Sections
Learning Objectives
6. Specific Energy and Specific Energy Curves
7. Hydraulic Jump (RVF)
8. Gradually Varied Flow (GVF)
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
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
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
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
2. Uniform and Non-uniform Flow
1.
Steady and Unsteady
If for a given length of Flow
channel, the velocity
of Uniform
2. flow, depth
andofNon-uniform
flow, slope of the channel
Flow
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
2. Non-uniform Flow
Types
1. Steadyofand
Non-uniform
Unsteady FlowFlow
1. Uniform
2. Gradually Varied
and Flow (GVF)
Non-uniform Flow
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
Types of Flows
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
1. Steady and Unsteady Flow
depending on the Reynolds number (Re)
2. Uniform and Non-uniform Flow
Re = ρVR/µ
3. Laminar and Turbulent Flow
Where,
ρ = density of water = 1000 kg/m3
µ = dynamic viscosity
R = Hydraulic Mean Depth = Area / Wetted Perimeter
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
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
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
Velocity Distribution
Velocity is always vary across channel
because of friction along the boundary
The maximum velocity usually found just
below the surface
Velocity Distribution
Velocity is always vary across channel
because of friction along the boundary
The maximum velocity usually found just
below the surface
Discharge through Open Channels
1. Chezy’s C
2. Manning’s N
3. Bazin’s Formula
4. Kutter’s Formula
Discharge through
Forces acting Open
on the water between Channels
sections 1-1 & 2-2
1. Component of weight of Water = W sin i
2. Friction Resistance = f P L V2
1. Chezy’s C
where
2. Manning’s N W = density x volume
= w (AL) = wAL
3. Bazin’s Formula Equate both Forces:
f P L V2 = wAL sin i
4. Kutter’s Formula
Chezy’s Formula, V C mi
V w A sin i 1
f P
A m Hydraulic Radius 2
P
w C Chezy's Constant 3
f
Chezy’s Formula, V C mi
substitute
V w A Eqn. 2&
sin i 1 3 in Eqn.1,
f P
VA m
C Hydraulic
m. sin i Radius 2
P
forwsmall values of i, sin i tan i i
C Chezy's Constant 3
f
V C m. i
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
2. Bazin’s Formula
Chezy’s formula can also be used with Bazins’ Formula
1. Manning’s N
2. Bazin’s Formula
C 157.6
1.81 k
where m
k = Bazin’s constant
m = Hydraulic Radius
Chezy’s Formula, V C mi
1. Manning’s N
2. Bazin’s Formula
3. Kutter’s Formula
Chezy’s formula can also be used with Kutters’ Formula
1. Manning’s N
1
2. Bazin’s Formula23 0.00155
C N
3. Kutter’s Formula 0.00155 N
1 23
i m
where
N = Kutter’s constant
m = Hydraulic Radius, i = Slope of the bed
Chezy’s Formula, V C mi
1. Manning’s N
2. Bazin’s Formula
3. Kutter’s Formula
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
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
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
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
Most Economical Sections
V
1. Cost of construction should be minimum
Q A A C m i
2. Discharge should be maximum
1
Q
Types of channels
K
1. Rectangular
where K
based on shape:
A C A i
P
2. Trapezoidal
If P is minimum, Q will be maximum
3. Circular
Rectangular Section
for most economicalsection,
P should be minimum
dP
0
d(d)
A
A bd b
Rectangular Section
d
1
A
P b 2d 2d 2
d
for most economicalseciton, P should be minimum
for most economicalsection,
A
d 2d
dP
0
d 0 Pshould
A
2 be
0 minimum
A 2d 2
bd 2d 2
d (d ) d (d ) d2
b 2d or d b/2 dP
A bd 2d
2
d 0
m d(d)
P b 2d 2d 2d 2
Trapezoidal Section
for most economicalsection,
P should be minimum
dP
0
d(d)
A
Trapezoidal Section
A (b nd)d b nd 1
d
A
P b 2d n 1 nd 2d n 2 1 2
2
d
for most economical
for most economicalseciton,P should be minimum
section,
A P2 be minimum
should
d nd 2d n 1
dP d b 2nd
0 dP 0 d n 2
1
d(d) d(d)
0 2
d d(d)
m and θ 600
2
Circular Section
d
A
P
for Max. Velocity, 0
d
3
A
d
P
for Max. Discharge, 0
d
sin 2θ
A R (θ -
2
) 1
P 2Rθ 2
2
Circular Section
m
A
R
(θ -
sin 2θ
)3 d
A
P 2θ 2 P
dm for Max. Velocity, 0
for max. velocity, 0 θ 128 45 , d 0.81D,
0 '
d m 0.3D
dθ
A A3 3
Q AC m i AC i C d
A
i , C and i are constants
P P
P
Max. Discharge,
3for 0
A
d d
P
for max. discharge, 0 θ 1540 , d 0.95D
dθ
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
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
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
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
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
Specific Energy
v2
Total Energy of flow ing fluid, E z h
2g
w here z Height of bottom of channel above datus,
If the channel bottom is taken as datum,
v2
Es h w hich is called as Specific Energy
2g
Specific Energy
Q Q
QAVV
A bh
Q
If dischargeper unit width, q constant
b
Q q
V
bh h
2 2 Modified Equation
V q
Es h h to plot Specific Energy Curve
2g 2g h 2
f or CriticalDepth,
dE
dh
0 Specific Energy Potential Energy (h)
2
q Es= h + q2/2gh2
where, E h
2g h 2
1
q2 3 q
2
hc
3 3 2
hc hc .g q 1
g g
Q bh. v
subsitute value q h c V c in Eqn. 1
b b
Vc g hc
2
dE
or CriticalDepth,Specific
fMinimum
dh
0 Specific Energy
Energy in termsPotential Energy (h)
of Critical Depth; E h
q
2g h 2
2
when specific q energyis minimum,Depth
E = hof
+ q2/2gh2 is critical
flow
where, E h s
2g h 2 1
1 q
2 q2 3 3 q
2
E hc substitute h or h
q 32g 23 q
2 2 c
2 g
c
g
hc hc
h c 3
hc .g q 1
g 3g
hc h 3hc
Emin h c hc c
Q bh.2v 2 2
subsitute value q 2ghc
hc Vc in Eqn. 1
b b
2 Emin
orVch c g h c
3
Alternate Depths 1 & 2
Hydraulic Jump
Specific Energy Curve
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.
Hydraulic Jump
Hydraulic Jump
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
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
Hydraulic Jump
2 2
d1 d1 2q
d2 interms of q
2 4 g d1
2 2
d1 d1 2v1 d1
d 2 interms of V1
2 4 g1
d1 2
d2 1 8 F e 1 interms of F e
2
Hydraulic Jump
Loss of Energy 2: 2
d1 d1 2q
d2 interms of q
2 d d 3
4 g d1
h L E1 E 22 2
2 1
d1 d1
2v14 dd11 d2interms
of V
d 2
2
4 g
1
1
Length of jump 5 to 7 times of (d2 d1)
d1 2
d2 1 8 F e 1 interms of F e
2 Jump d 2 d1
Hydrualic
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.
Gradually Varied Flow (GVF)
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,
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,
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
Normal Depth Line
Flow Profiles in Mild slope
Critical Depth Line
Flow Profiles in Steep slope
Flow Profiles in Critical slope
Flow Profiles in Horizontal slope
Flow Profiles in Adverse slope
Gradually Varied Flow (GVF)
Equationof GVF : Sc or ib Energy Line Slope
So or ie Bed Slope
dh ib ie
in terms of Velocity
dx V
2
1 h1
gh
h2
dh i b ie
dx 1 ( F )
e
2
in terms of Fe
L
E 2 - E1
i b - ie
dh
where representsthe variationof water depth along the bottomof the channel
dx
Gradually Varied Flow (GVF)
IfEquation
dh/dx =of0,GVF
Free
: Surface of water is Sc or ib Energy Line Slope
parallel to the bed of channel So or ie Bed Slope
dh ib ie
in terms of Velocity
dh/dx > 0, Depth
Ifdx 2 increases in the
V
1 of
direction flow (Back Water
water h1
Curve) gh
h2
dh/dxi b<0,
Ifdh i e Depth of water decreases
indxthe direction
1 ( )
Curve) F e
in terms of Fe
2 of flow (Dropdown
L
E 2 - E1
i b - ie
dh
where representsthe variationof water depth along the bottomof the channel
dx
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
Reference
Chapter 16
A Textbook of Fluid Mechanics and
Hydraulic Machines
Dr. R. K. Bansal
Laxmi Publications
