The document provides an overview of uniform flow in open channel flow, including key concepts, basic formulas, and methods for calculating discharge, conveyance, and normal depth. It covers various channel shapes, such as rectangular and trapezoidal, and provides examples and equations to compute flow properties. Additionally, it discusses the implications of different parameters like manning's n and channel slope on flow characteristics.

1. 1
UNIFORM FLOW COMPUTATIONS
Civil Engineering Sem. - 03
OPEN CHANNEL FLOW

2. Contents :
Introduction : uniform flow
Some basic formulas for uniform flow computations
Discharge in uniform flow
Conveyance of the Channel & Section factor
Channels of first & second kind
Computation of Normal Depth
2

3. In the given figure, we have…
L – length between points 1 & 2
V – flow velocity
y0 – Normal depth
𝜃 – Angle between channel bottom of flow & horizontal
g – Acceleration due to gravity
W - weight to fluid in the control volume
Ff - shear force at the boundary
Introduction : Uniform flow
A flow is said to be uniform if its properties (y, A, V ) remain constant with respect to
distance.
Q = AV = constant or
dQ
= 0
dx
Figure 1 :
Uniform flow
(Source : FLOW
IN OPEN
CHANNELS by K
Subramanya,
3rd edition,
Page no - 86)
3

4. Chezy’s Formula :
o
V = C RS
Some basic formulas in uniform flow computations:
Here
V – Flow velocity
C – Chezy coefficient which depends on nature of surface & flow.
-------(2)
k – a coefficient which also depends on nature of surface and flow parameters.
R – hydraulic radius
S0 – Channel bottom slope
A
R =
P
4
------- (1)
------- (3)
𝑪 =
𝜸. 𝟏
𝝆𝒌

5. Manning’s formula
n = Roughness coefficient known as Manning’s n
2 1
3 2
0
1
V = R S
n
Uniform flow computations:
Discharge in uniform flow :
Q = AV
From Manning’s formula (4)
Putting this in Formula of Q
2 1
3 2
0
1
V = R S
n
2 1
3 2
0
1
Q = AV = AR S
n
5
-------(4)
-------(5)

6. Conveyance of the Channel
It expresses Discharge capacity of the channel per unit longitudinal slope.
From equation (5) we have
2 1
3 2
0 0
1
Q = AR S = K S
n
2
3
1
K = AR
n
K is known as conveyance of the channel.
Section factor
The above term is sometimes known as the section factor for uniform flow computations.
2
3
nK = AR
6
-------(6)
-------(7)

7. Variation of φ with y/B in trapezoidal channels
For a given channel, is a function of the depth of flow.
For example, for a trapezoidal section of bottom width = B and side
slope m horizontal : 1 vertical as shown in Figure 2.
Then,
.
2
3
AR
2
P = B + 2y m + 1
A
R =
P
2
(B + my)y
R =
B + 2y m + 1

5 5
2 3 3
3
2
2 3
(B+ my) y
AR =
(B+ 2y m +1 ) 7
From (3) we have
Raising Power both side by 2/3 and then multiplying by A, we get
-------(9)
------(8)
Figure 2 : variation of ∅ with y/b in trapezoidal
channels
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 105)
𝑨 = 𝑩 + 𝒎𝒚 𝒚
f(B,m,y)

8. For a given channel, if B and m are fixed & . The Figure 2 shows relationship of equation (9) in a non-
dimensional manner by plotting vs y/B for different values of m.
8
• For and if n
and S0 are fixed for a
channel ,then there is
unique depth in uniform
flow associated with each
discharge known as normal
depth denoted as y0 .
• This graph is also valid for
any other shape of channel
provided top width is
either constant or increases
with depth.
• The majority of channels
are of this kind having only
one normal depth only,
denoted as channels of 1st
kind.

2 3
8 3
AR
=
B
2
3
0
Qn
AR =
S
Figure 2 : variation of ∅ with y/b in trapezoidal channels
(Source : FLOW IN OPEN CHANNELS by K Subramanya, 3rd
edition, Page no. 105)
𝑨𝑹𝟐 𝟑 = 𝒇 𝒚

9. For the channel of 2nd kind .
• In some channels there may be more than one
value of depth and closing top width such as in
circular, ovoid or triangular channels.
The plot in such type of channels is shown in
figure 3 .
• It can be seen that in some ranges of depths, A𝑹𝟐 𝟑
is not a single valued function of depth. For
Example…
 For circular channel for y/B > 0.82
 For triangular channels it also depends on m
 For m = -0.25, the range y/B > 0.71
 For m = -0.50 , the range y/B > 1.30
For which two depths can be found.
9
Figure 3 : Variation of A𝑹𝟐 𝟑
in channels of second
kind
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 106)

10. Computation of Normal Depth
Rectangular channels
Wide Rectangular channel ( )
For the given rectangular channel
Area A = By0
Wetted perimeter P = B + 2y0
Hydraulic radius R
∴ Using equation (2)
As , the aspect ratio of the channel decreases, R y0 .
Such channels with large bed-widths as compared to their
respective depths are known as wide rectangular channels.
Considering the width as unity.
A = y0 , R = y0 and B = 1
= discharge per unit width =
0 0
0 0
By y
R = =
B + 2y 1+ 2y B
0
y B
Q
= q
B
5 3 1 2
0 0
1
y S
n
 
 
 
 
3 5
0
0
qn
y =
S
0
y
B
< 0.02
10
-------(11)
Figure 4 : Rectangular channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 110)
-------(10)

11. Rectangular channels with
𝒚𝟎
𝑩
≥ 𝟎. 𝟎𝟐
From equation (5)
For this type of rectangular channel we have to calculate from equation (9) .
(Putting m = 0 for rectangle)
A𝑹𝟐/𝟑
=
(𝑩𝒚𝟎)𝟓/𝟑
(𝑩+𝟐𝒚𝟎)𝟐/𝟑 =
(𝒚𝟎 𝑩)
𝟓/𝟑
(𝟏+𝟐𝒚𝟎/𝑩)𝟐/𝟑 𝑩𝟖/𝟑
𝑸𝒏
√𝒔𝟎𝑩𝟖/𝟑 =
𝑨𝑹𝟐/𝟑
𝑩𝟖/𝟑 =
(η𝟎)𝟓/𝟑
𝟏+𝟐η𝟎
𝟐/𝟑 = ∅(η𝟎) ------(12)
Where 𝜼𝟎 = 𝒚𝟎/𝑩
Tables of ∅ 𝜼𝟎 vs 𝜼𝟎 will give a non-dimensional graphical solution aid for general application.
Using equation (12) we can easily find y0/B from the table for any combination of Q, n, S0 and
B.
𝑹 =
𝑨
𝑷
=
𝑩 + 𝒎𝒚𝟎 𝒚𝟎
𝑩 + 𝟐 𝒎𝟐 + 𝟏𝒚𝟎
2 3 1 2
0
1
Q = AR S
n
2 3
0
Qn
= AR
S

11
Figure 4 : Rectangular channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition , Page no. 110)

12. 12
Trapezoidal channel
For the trapezoidal channel of section of side slope m : 1 as shown in
figure 5.
A = (B+m𝒚𝟎)𝒚𝟎
𝑹 =
𝑨
𝑷
=
𝑩+𝒎𝒚𝟎 𝒚𝟎
𝑩+𝟐 𝒎𝟐+𝟏𝒚𝟎
(using equation 5 and 9)
Dividing both side by
𝑸𝒏
𝑺𝟎𝑩𝟖/𝟑
=
𝑨𝑹𝟐/𝟑
𝑩𝟖/𝟑 =
(𝟏+𝒎𝜼𝟎)𝟓/𝟑𝜼𝟎
𝟓/𝟑
(𝟏+𝟐 𝒎𝟐+𝟏𝜼𝟎)𝟐/𝟑
= ∅(𝜼𝟎,m) -------(13)
 
 
5 3 5 3
2 3 0 0
2 3
2
0
0
B+ my y
Qn
= AR =
S B+ 2 m +1y
2
0
P = (B+ 2 m +1y )
8 3
B
Figure 5 : Trapezoidal channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 111)

13. • Where 𝜼𝟎=𝒚𝟎/𝑩
A curve of ∅ vs 𝜼𝟎 with m as the 3rd parameter is used to provide a
general normal depth solution aid.
• When we put m = 0, it will be reduced to a rectangular channel.
Circular channel
For the circular channel let the diameter be D and 2θ be the angle in
radians subtended by the water surface at the centre.
A = area of the flow section = area of sector OMN – area of the
triangle OMN
A =
𝟏
𝟐
𝒓𝟎
𝟐
𝟐𝜽 −
𝟏
𝟐
× 𝟐𝒓𝟎 𝐬𝐢𝐧 𝜽 × 𝒓𝟎 𝐜𝐨𝐬 𝜽
=
𝟏
𝟐
(𝒓𝟎
𝟐
𝟐𝜽- 𝒓𝟎
𝟐
𝐬𝐢𝐧 𝟐𝜽)
=
𝑫𝟐
𝟖
(𝟐𝜽 − 𝒔𝒊𝒏𝟐𝜽)
13
Figure 6 : Circular channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 112)

14. P = wetted perimeter = 2𝒓𝟎𝜽 = 𝑫𝜽
Also 𝐜𝐨𝐬 𝜽 =
𝒓𝟎−𝒚𝟎
𝒓𝟎
= (𝟏 −
𝟐𝒚𝟎
𝑫
)
Hence 𝜽 = 𝒇(𝒚𝟎, 𝑫)
Using Equation (5)
Assuming n = constant for all depths and using equation (3) & (5) and
substituting the above values for A & P we get
𝑸𝒏
𝑺𝟎
=
𝑨𝟓/𝟑
𝑷𝟐/𝟑
=
𝑫𝟏𝟎/𝟑
𝟖𝟓/𝟑
(𝟐𝜽 − 𝒔𝒊𝒏 𝟐𝜽)𝟓/𝟑
(𝑫𝜽)𝟐/𝟑
𝑸𝒏
𝑺𝟎𝑫𝟖/𝟑
=
𝑨𝑹𝟐/𝟑
𝑫𝟖/𝟑 =
𝟏
𝟑𝟐
(𝟐𝜽−𝐬𝐢𝐧 𝟐𝜽)𝟓/𝟑
𝜽𝟐/𝟑 = ∅
𝒚𝟎
𝑫
--------(14)
• The functional relationship of the above equation is evaluated for
different values of and a table is made.
• Using this table, the normal depth for a given D, Q, n and S0 in a circular
channel can be easily derived. Sometimes linear interpolations may be
required.
2 3 1 2
0
1
Q = AR S
n
0
y / D
14
(dividing each side by 𝑫𝟖/𝟑 )
Figure 6 : Circular channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 108)

15. 15
Question on Normal depth
Q. A trunk sewer pipe of 2.0 m diameter is laid on a slope of 0.0004. Find the depth of flow when the discharge is 2
m3/s. (Assume n = 0.014 )
Solution : Given D = 2 m S0 = 0.0004 Q = 2 m3/s n = 0.014
Using equation (14)
From the Table (Figure 7)
at y0/D = 0.62
= 0.22532 at y0/D = 0.63
By interpolation at y0/D = 0.621
Hence the normal depth of flow y0 = 0.621 × 2 = 1.242 m
2 3
8 3 8 3 8 3
0
2 0.014
0.0004 2
0.22049
AR Qn
D S D

 


2 3
8 3
0.22004
AR
D

2 3
8 3
0.22049
AR
D
 Figure 7 : Table for circular channel
(Source : FLOW IN OPEN CHANNELS by K
Subramanya, 3rd edition, Page no. 81)

16. 16
1. A triangular irrigation lined canal carries a discharge of 25 m3/s at bed slope = 1/6000 .If the side slopes of the canal
are 1:1 and Manning’s coefficient is 0.018, the central depth of flow is equal to
a) 2.98 m b) 3.62 m c) 4.91 m d) 5.61 m
Solution – Using equation (5)
and putting the values Q = 25 m3, S0 = 1/6000, n = 0.018
y
y 1
1
m
2. A rectangular channel 6.0 m wide carries a discharge of 16 m3/s under uniform flow condition with normal depth of
1.60 m. Manning’s ‘n’ is 0.015. The longitudinal slope of the channel is
a) 0.000585 b) 0.000485 c) 0.000385 d) 0.000285
Solution – As given Q = 16 m3/s, n = 0.015, B = 6.0 m A = 6.0×1.6 = 9.6m2
2 3 1 2
0
1
Q A R S
n

 
 
2 3 1 2
2
8 3
1 1
25
0.018 6000
2 2
1 1
25
0.018 154.92
4.91
y
y
y
y
   
     
 
 
 
   
 

17. 17
As given Q = 16 m3/s, n = 0.015, B = 6.0 m A = 6.0×1.6 = 9.6m2
Using equation (3)
But Q = AV
Using equation (5)
3. A trapezoidal channel is 10.0 m wide at the base and has a side slope of 4 horizontal to 3 vertical. The bed slope is 0.002.
The channel is lined with smooth concrete (Manning’s n = 0.012). The hydraulic radius(in m) for a depth of flow of 3.0 m is
a)20.0 b)3.5 c)3.0 d)2.1
Solution : m2
3V
m y = 3 m 4H
10 m
 
9.6
1.04
6.0 1.6 2
A
R
P
  
 
2 1
3 2
0
1
Q AV AR S
n
 
 
2 3 1 2
1
16 9.6 1.04
0.015
0.00059
S
S
   
 
2
2
4
( ) 10 3 3 (10 4)3 42
3
4 5
2 1 10 6 1 10 6 20
3 3
42
2.1
20
A B ny y
P B y n
A
R
P
 
        
 
 
 
         
 
 
   

18. 18
4. The normal depth in a wide rectangular channel is increased by 10 %. The % increase in discharge in the channel is
a) 20.1 b) 15.4 c) 10.5 d) 17.2
Solution : Method – 1 Method – 2 (Approximate method)
Using equation (5) for discharge
for
For a wide rectangular channel, B >> y
Let
2 1
3 2
0
1
Q AV AR S
n
 
 
2 3
1 2
0
1
2
By
Q By S
n B y
 
   

 
  2 3 1 2
0
5 3 5 3
1
( )
Q By y S
n
Q y Q ky
 
  
 
   
 
5 3
1
5 3
2
5 3 5 3
2 1
5 3
1
1.1
1.1
100 100
17.21%
Q ky
Q k y
y y
Q Q
Q y




   

 
2 3
1 2
0
5 3
1
2
5 5
10% 17.33%
3 3
By
Q By S
n B y
B y
Q y
dQ dy
Q y
 
  

 


   

19. 19
5. A circular pipe has a diameter of 1 m , bed slope of 1 in 1000, and Manning’s roughness coefficient equal to 0.01. It may
be treated as an open channel flow when it is flowing just full i.e., the water level just touch the crest. The discharge in this
condition is denoted by Qfull . Similarly ,the discharge when the pipe is flowing half-full i.e., with a flow depth of 0.5 m, is
denoted by Qhalf.
The ratio Qfull/Qhalf is
a) 1 b)√2 c) 2 d) 4
Given: D = 1 m Using equation (5)
Hence,
Qfull
Qhalf
=
𝝅𝒅𝟐
𝟒
𝝅𝒅𝟐
𝟖
= 2
2 1
3 2
2 3
0
2 2 3
1
3 2
0
1
1
full full full
half
half
half
AR S
AR
n
Q
Q AR
AR S
n
 
   
   
 
   
 
 
 
2
full half
d
R R
  

20. 20
6. A 4 m wide rectangular channel, having bed slope of 0.001 carries a discharge of 16 m3/s. Considering Manning’s n =
0.012 and g = 10 m/s2, the category of the channel slope is
a)Horizontal b) mild c) critical d) steep
Solution : A = By0
using equation (10)
m
Critical depth,
Flow is subcritical.
Hence channel is mild.
2 1
3 2
0
1
Q AV AR S
n
 
0
0
2
By
R
B y


 
 
 
2 3
1 2
0
0
0
5 3
0
2 3
0
5 3
0
2 3
0
3 5
2 3
0
0
1
16 (0.001)
0.012 2
6.07 4
4 2
1.5178
4 2
1.5178 4 2
3.3
By
By
B y
y
y
y
y
y y
y
 
   
 

 
  

 

 
   
 
 
1 3
2 2
0
4
1.169
10
c
c
q
y
g
y y
   
  
   
   
 

21. References :
1. FLOW IN OPEN CHANNELS by K Subramanya, 3rd edition
2. Open channel flow by VT Chow
3. NPTEL lecture
https://youtu.be/oADqCYG28ps
21