OPEN CHANNELS

(OPEN CHANNEL FLOW AND HYDRAULIC

MACHINERY)

UNIT – I

AITSCE A.Anil , 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
 2. Uniform and Non-uniform Flow
1.
➢IfSteady and Unsteady
for a given Flow
length of 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

w A
substituteEqn. 2&
V= f sin i →
P 1 3 in Eqn.1,
VA == m
C = Hydraulic
m.sin i Radius → 2
P
forwsmall values of i, sin i = tan i = i
f = C = Chezy's Constant → 3
 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 1 +
Formula + 0.00155  N
23  m
 i
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 =A
1. Cost of construction should be minimum
Q A C m i
2. Discharge should be maximum
1
Q =
Types of channels
K
1. Rectangular
whereK =
based on shape:
A C A i
P
2. Trapezoidal
If P is minimum, Q will be maximum
3. Circular
Rectangular Section
for mosteconomicalsection,
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 mosteconomicalsection,
d  + 2d 
A
dP d  −A
= 0 = 0  P should
2
+ 2 = 0 be
A minimum
= 2d 2  bd = 2d 2
d (d ) d (d ) d
b = 2d or d = b/2 dP
A bd 2d
2
d =0
m = = = =
P b + 2d 2d + 2d
d(d)
2
Trapezoidal Section
for mosteconomicalsection,
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
forshouldbe
for most economicalseciton,P mosteconomicalsection,
minimum
 A P2
d  − nd + 2d n + 1 
should be minimum
dP d  b + 2nd
= 0 = 0  = d n2 + 1
d(d) dP 2
d(d)
=0
m =
d
and θ = 60 0
d(d)
2
Circular Section
d
A
P 
for Max. Velocity, =0
d
 3
d A 
 P 
for Max. D i s c h a r g e , =0
d
 sin 2θ
A = R 2 (θ - )→ 1

P = 2Rθ → 2
2
Circular Section
d 
A R sin 2θ A
m = = (θ - )→ 3
P 2θ 2 P 
dm for Max. Velocity, =0
=0  θ= , d = 0.81 D ,m = 0.3D
'
f o r m a x . v e l o c i t y,
dθ d 
128045
A A3  3 
Q = AC m i = AC i= C i , C a n d i a re cdon s tAa n ts
P P
  P  
  Max. D i s c h a r g e ,
A 3for =0
d  d
 P 
for max.discharge,   = 0
 θ= , d = 0.95D
dθ 0
 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
v 2
To t a l E n e r g y o f f l o w i n g fluid, E = z + h +
2g
w h e r e z = Height o f b o t t o m o f c h a n n e l a b o v e d a t u s ,

I f t h e c h a n n e l b o t t o m is t a k e n a s d a t u m ,

v 2
Es = h + w hi chis called a s S p e c i f i c E n e r g y
2g
 Specific Energy
Q Q
Q =AVV = =
A bh
Q
If dischargeper unit width, q = = constant
b
Q q
V= =
bh h
2 Modified Equation
V2 q
 Es = h + = h + to plot Specific Energy Curve
2g 2g h 2
f orCriticalDepth,
dE
=0 Specific Energy Potential Energy (h)
dh
2
q Es= h + q2/2gh2
where, E = h+
2g h 2

1
q2  3 q
2
3 2
 
3 h .g = q → 1
hc =  h c = c
g  g

Q bh. v
subsitutevalue q = = = h c V c in Eqn. 1
b b

 Vc = g hc
 2
fMinimum
dE
orCriticalDepth,SpecificEnergy
=0 Specific Energy
in termsof Critical Depth; E = h
Potential Energy (h)
+
q
2g h 2
dh
2
when specificenergyis
q minimum,Depthof
E = h + q
flow
2/2gh2 is critical
where, E = h + s
2g h 2 1
q
2 q2  3 q 2
E = h cq 2+ 3
1
su bsti tu 3teh c =2
3 q2
 or h c 3 =
2  g  g
hc =    hcc =  hc .g = q → 1
2g h  
 g  3g
h h c 3h c
E m i n = hc +
c = hc + =
Q bh. v 2 2
subsitute value q = 2g =h c
2 = h c V c in Eqn. 1
b b
2 E min
orVch =c = g hc

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
d1 d1 2q 2
d 2 = − + + → intermsof q
2 4 g d1

2 2
d1 d1 2v 1 d 1
d 2= − + + → i n t e r ms o f V 1
2 4 g1

d1  2 
d2 =  1+ 8 F e −1  → in termso f F e
2  
 Hydraulic Jump

Loss of Energy 2: 2
d1 d1 2q
d 2 = − + + → intermsof q
2 d − d 3
4 g d1
 2 1 
= −
h L E1 E 2 2 = 2
d2=−
d1
+
d1
+
2v
 1 →2i n t e r ms o f V 1
d1 d
14 d
2 4 g1
Length of jump = 5 to 7 timesof (d2 − d1)

d1 
d 2 =  1 + 8 F e 2 −1  → intermsof 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 bi e
−i
= → in terms of Velocity
dx  V
2
1 −  h1

 gh 

h2

dh ib−ie

dx
=

1− ( F e )
2

→in terms of Fe

L =
E 2 - E1

ib - ie
dh
w here representsthe variationof water depthalongthe bottomof the channel
dx
 Gradually Varied Flow (GVF)
IfEquationof
dh/dx = 0,GVF
Free
: Surface of water is Sc or ib Energy Line Slope
parallel to the bed of channel So or ie Bed Slope
dh i b − ie
= → in terms of Velocity
dh/dx > 0, Depth
V 
Ifdx 2 increases in the
1 of
direction  flow (Back Water
− water h1
Curve) gh  h2

dh ib−ie

 
If dh/dx < 0, Depth of water decreases
= 2 → in terms of Fe
indxthe direction of flow (Dropdown
1− ( )
Curve) F e L =
E 2 - E1

ib - ie
dh
where representsthe variationof water depthalongthe 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