This document discusses the design of open channel sections to convey water flow in the most economical way. It examines rectangular, trapezoidal, triangular, and circular channel cross-sections. For rectangular channels, the most economical section is when the base width is twice the flow depth. For trapezoidal channels, the most economical section is when the side slopes are at an angle of 60 degrees from horizontal and the half top width is equal to the flow depth. Empirical flow equations like Chezy's and Manning's formulas are also presented to estimate normal flow velocities based on hydraulic radius and channel slope.

1. Design of MOST ECONOMICAL
Rectangular, Trapezoidal, Triangular
and Circular
CHANNEL SECTIONS

2. UNIFORM FLOW IN OPEN
CHANNELS
Definitions
a) Open Channel: Duct through which
Liquid Flows with a Free Surface - River,
Canal
b) Steady and Non- Steady Flow: In Steady
Flows, all the characteristics of flow are
constant with time. In unsteady flows, there
are variations with time.

3. Parameters of Open
Channels
a) Wetted Perimeter, P : The Length of contact
between Liquid and sides and base of Channel
P = B + 2 D ; D = normal depth
Area, A
Wetted Perimeter
Hydraulic Mean Depth or Hydraulic Radius (R): If
cross sectional area is A, then R = A/P, e.g. for
rectangular channel, A = B D, P = B+2D
D
B

4. Empirical Flow Equations for Estimating Normal Flow
Velocities
a) Chezy Formula (1775):
Can be derived from basic principles. It states that: ;
V  C R S
Where: V is velocity; R is hydraulic radius and S is
slope of the channel. C is Chezy coefficient and is a
function of hydraulic radius and channel roughness.

5. Empirical Flow Equations for Estimating Normal Flow
Velocities
(b) Manning’s Formula
Q = A
.
𝟏
𝒏
𝟐 𝟏
𝑹𝟑 𝑺
𝟐
Where: V is velocity; R is hydraulic radius
and S is slope of the channel. n is
Manning’s coefficient of roughness depends
up on channel roughness.

6. Definitions
a) Freeboard:
the highest water
Vertical distance between
level anticipated in the
design and the top of the retaining banks. It is
a safety factor to prevent the overtopping of
structures.
b) Side Slope (Z): The ratio of the horizontal
to vertical distance of the sides of the
channel.

7. DESIGN OF CHANNELS
Channels are very important in Engineering projects
especially in Irrigation and, Drainage.
Channels used for irrigation are normally called canals
Channels used for drainage are normally called drains.

8. MOST EFFICIENT SECTION
During the design stages of an open channel, the
channel cross-section, roughness and bottom slope
are given.
The objective is to determine the flow velocity, depth
and flow rate, given any one of them. The design of
channels involves selecting the channel shape and
bed slope to convey a given flow rate with a given
roughness, the designer aims to minimize
flow depth. For a given discharge, slope and
the
cross-sectional area A in order to reduce
construction costs

9. The most ‘efficient’ cross-sectional shape is determined for uniform flow
conditions. Considering a given discharge Q, the velocity V is maximum
for the minimum cross-section A. According to the Manning equation the
hydraulic diameter is then maximum.
It can be shown that:
1.the wetted perimeter is also minimum,
2.the semi-circle section (semi-circle having its centre in the surface)
is the best hydraulic section
Because the hydraulic radius is equal to the water cross section area
divided by the wetted parameter, Channel section with the least wetted
parameter is the best hydraulic section

10. RECTANGULAR SECTION
For a rectangular section
Q=AV, where Q=discharge through the
channel,
A=area of flow. V=velocity
with which water is
flowing in the channel.
For Q to be maximum ,V needs to be
maximum, Since A = constant.
But V=C 𝑅
𝑆
Where R= hydraulic mean radius.
S = bed slope.
R=A/P where P=wetted perimeter.
For R to be maximum ,P minimum

11. As, Area of flow (A) = by
b = 𝐴
𝑦
Wetted Perimeter (P) = b + 2y
P = 𝐴
+ 2y
𝑦
For the section to be most economical /efficient-the wetted perimeter (P) must be
minimum
Therefore, differentiate P w.r.to y
𝑑
𝑃
=0
𝑑
𝑦
𝑑
𝑑
𝑦 𝑦
[𝐴
+ 2y]= 0
- 𝐴
𝑦
2
+ 2 = 0
-
𝐴
𝑦
2 = -2

12. -
𝐴
𝑦
2 = -2
𝐴
𝑦
2 = 2
𝑏
𝑦
=
2
𝑦
2
𝑦
𝑏
= 2 Therefore, b = 2y, y = 𝒃
𝟐
𝑃
R = 𝐴
=
𝑏
𝑦
𝑏
+2
𝑦
=
2
2
𝑦
.
𝑦
= 2
𝑦
2
𝑦
+
2
𝑦 4
𝑦
R = 𝒚
𝟐
The rectangular channel section willbe most economical when:
(i) base width (b) = 2y,and Hydraulic radius (R) = 𝑦
2

13. Problem 1. A rectangular channel is to be dug in the rocky portion of a soil. Find
its most economical cross-section if it is to convey 12 m3/sec of water with an average
velocity of 3 m/sec. Take Chezy’s constant C =50.
Data. Q = 12 m3/sec, V= 3 m/sec,
Chezy’s constant C =50
b = ??,y=d = ??S= ??
Solution. The geometric relations for optimum discharge through a
rectangular channel are: b = 2y, and R = 𝒚
𝟐
As,
Also,
Area (A) = by = 2y.y= 2y2
Q = AxV= 2y2 .V
12 = 2y2 x 3
y= 1.414 m(Ans.)

14. Solution.
Now, b = 2y= 2 x 1.414 = 2.828 m(Ans.)
Now, R = 𝑦
= 1
.
4
1
4
2 2
= 0.707 m (Ans.)
As, V = C R
S
2
S= 𝑉
𝐶2
𝑅
S=
(
3
)
2
(
5
0
)
2x0
.
7
0
7
S =
𝟏
𝟏
𝟗
𝟔
(Ans.)

15. Problem 2. A rectangular channel is to be dug in the rocky portion of a soil. Find
its most economical cross-section if it is to convey 10 m3/sec of water with an average
velocity of 2.5 m/sec. Take Manning’s coefficient of roughness n = 0.223.
Data.
Q = 10 m3/sec,
V= 2.5m/sec,
Manning’s coefficient of roughness n = 0.223
b = ??,
y=d = ??
S= ??

16. Problem 3. Determine the most economical section of a rectangular channel
carrying water at the rate of 0.5 m3/sec, the bed slope of the channel being 1 in 2000. Take
Chezy’s constant C =50.
Data. Q = 0.5 m3/sec,
S= 1 in 2000,
Chezy’s constant C =50
b = ??,
y=d = ??

17. Solution. The rectangular channel section willbe most economical when:
(i) base width (b) = 2y,and
(ii) Hydraulic radius (R) = 𝑦
2
As, Area of flow (A)
And , Q
= by = 2y.y= 2y2
= AxV = 2y2 .V
Q = 2y2 .C R
S
2
Q = 2y2 .C 𝑦
xS
0.5 = 2y2 .5
0
𝑦
x 1
2 2
0
0
0

18. Solution.
0.5 = 2y2 .5
0
𝑦
x 1
2 2
0
0
0
0.5 = 100
1
4
0
0
0
5
𝑦
2
5
0.5 = 1.581.𝑦
2
5
𝑦
2 =
0
.
5
1
.
5
8
1
= 0.316
2
y = 0
.
3
1
65 = 0.63 m (Ans.)
Now, b = 2y= 2 x 0.63 = 1.26 m(Ans.)

19. Problem 4. Determine the most economical section of a rectangular channel
carrying water at the rate of 0.8 m3/sec, the bed slope of the channel being 1 in 2000. Take
Manning’s coefficient of roughness n =0.025.
Data. Q = 0.8 m3/sec,
S= 1 in 2000,
Manning’s coefficient of roughness n =0.025
b = ??,
y=d = ??

20. Most Economical Trapezoidal Channel Section
Consider a trapezoidal channel section as shown in Figure.
Now, area of flow (A) = by + y2z
A= y (b+yz)
y
Or, A
= b+yz
b = A
- yz
y
Wetted Perimeter (P)
P = b+ 2y 1+z
2
y
P = A
- yz + 2y 1+z
2

21. Most Economical Trapezoidal Channel Section
For trapezoidal channel to be most efficient one, P must be minimum, so for P
d
y
minimum, d
P
= 0
− y
z + 2
y 1+z
2 = 0
-Ay−1−1-z +2
d
P d A
=
d
y d
y y
1+z
2= 0
−A
- z +2 1+z
2= 0
y
2
- A
+ z = - 2 1+z
2
y
2
y
2
A
+ z= 2 1+z
2
Put, A= by + y2z = y (b+yz)

22. Most Economical Trapezoidal Channel Section
y
2
y(
b+y
z
)
+ z= 2 1+z
2
y
(
b+y
z
)
+ z= 2 1+z
2
y
b+y
z+y
z
= 2 1+z
2
2
b+2
y
z
= y 1+z
2
Half of top width = length of side slope [1st Condition]

23. Most Economical Trapezoidal Channel Section
For, 2nd Condition
P
As, R = A
=
y(
b
+
y
z
)
b+2
y 1
+
z
2
R =
y(
b
+
y
z
)
b+b
+
2
y
z
R = y(
b
+
y
z
)
2
b
+
2
y
z
R = y(
b
+
y
z
)
= y
2
(
b
+
y
z
) 2
Hydraulic radius = half of flow depth

24. Most Economical Trapezoidal Channel Section
According to given conditions
b
+2
y
z
2
= y 1+z
2 and R = y
2
Consider ∆OFC with θ= OCF, and ∆BCE with θ= CBE
From ∆OFC (right angled triangle)
Sin θ= OF
OC
Therefore, OF = OC Sinθ
But, OC = half of top width = b
+
2
y
z
2
As, half of top width = side slope
2
b+2
y
z
= y 1+z
2

25. Most Economical Trapezoidal Channel Section
∴ OC = y 1+z
2
Now, OF = y 1+z
2 Sin θ (A)
Now, in the triangle ∆BCE
Sin θ= P
= C
E
=
H B
C
y
y 1
+
z
2
1
1
+
z
2
Put this in Eq. (A), we get:
∴Sin θ=
OF = y 1+z
2x
1
1
+
z
2
OF = y (Hence proved that OF = y)
In this way, if circle is drawn with flow depth (y) as its radius, the circle can touch all
the three sides of the trapezoidal channel section. Hence, it can be said that most
economical trapezoidal is half of the hexagonalsection.

26. Most Economical Trapezoidal Channel Section
Best Sloping Angle for the Most Economical trapezoidal Channel Section
From the given Figure, we have:
θ= angle of sloping side with the horizontal,
y = flowdepth
A= by + y2z
A= y (b+yz)
y
Or, A
= b+yz
b = A
- yz
y
Wetted Perimeter (P) = b + 2y 1+z
2
P = A
- yz + 2y 1+z
2
y

27. Most Economical Trapezoidal Channel Section
d
z
Differentiate P w. r.to z, we have: d
P
=𝟎
d
P d A
=
d
z d
z y
− y
z + 2
y 1+z
2 = 0
0 - y + 2y x 1
2
2
−1
(
1+z )2 2 z = 0
- y +
2
y
z
1
(
1
+
z
2
)
2
y =
2
y
z
1
(
1
+
z
2
)
2
2yz = y 1+z
2

28. Most Economical Trapezoidal Channel Section
2 z = 1+z
2 Squaring both sides
4z
2 = 1+z
2
3z
2= 1
z
2= 1
3
1
3
z =
z =
1
3
Now, Tan θ =
𝐵
𝑃
= 1
1
3
Tan θ = 3
θ = t
a
n
−
1 3= 600 (angle with horizontal)
For most economical trapezoidal channel section, z =
1
3
and θ= 600

29. Most Economical Trapezoidal Channel Section
According to given condition
2
b+2
y
z
= y 1+z
2
Put, z =
1
3
b+2
y
1
2
3
= y 1+
1
3
2
b+
2
𝑦
3
= 2
y 1+
1
3
b+
2
𝑦
3
= 2
y
4
3

30. Most Economical Trapezoidal Channel Section
b+
2
𝑦
3
= 4
y
1
3
b= −
4
𝑦 2
𝑦
3 3
b=
4
𝑦− 2
𝑦
b=
3
2
𝑦
3
b = 1.155y

31. Problem. A trapezoidal channel is to carry 5000 m3/minute of water is
designed to have a minimum cross-section. Find the bottom width and flow
depth, if bed slope is 1in 1200, the side slope at 600 and C = 100.
Data.
Q = 5000 m3/minute = 5000/60 = 83.33 m3/sec
b = ??
y = ??
S = 1in1200,
𝜽=600 (side slope with the horizontal = 600) and
C = 100.

32. Solution.
According to Most Economical Conditions,
b
+2
y
z
2
= y 1+z2
,
R = y
2
𝐭
𝐚
𝐧θ= 600 , z =
1
3
b+2
y
1
2
3
= y 1+
1
3
2
b+
2
𝑦
3
= 2
y 1+
1
3
b+
2
𝑦
3
= 2
y
4
3

33. Solution.
b+
2
𝑦
3
= 4
y
1
3
b= −
4
𝑦 2
𝑦
3 3
b=
4
𝑦− 2
𝑦
b=
3
2
𝑦
3

34. Solution.
Area of flow (A) = by + y2z
A= y (b+yz)
3 3
A= y ( 2
𝑦
+ 𝒚
)
3
A= y ( 2𝑦+𝒛
)
3
A= y ( 𝟑𝑦
)
3
A= y ( 3 3𝑦
)
A= 3y2

35. Solution.
As, we know that
Q = AV
Q = AC 𝑅
𝑆
R = 𝒚
𝟐
83.33 = 3y2 x 100
𝒚
𝐱 𝟏
𝟐 𝟏
𝟐
𝟎
𝟎
/
83.33 = 3y𝟓 𝟐
x 100
𝟏
𝟐
𝟒
𝟎
𝟎
83.33 = 𝟑
.𝟓
𝟓y𝟓
/
𝟐
𝟑
.
𝟓
𝟓
y = (𝟖𝟑.𝟑𝟑
)2/5= 3.53m
As, b=
2
𝑦
3
b=
2𝐱𝟑
.𝟓
𝟑
3
b = 4.07m

36. Problem. Apower canal of trapezoidal section has to be excavated through
hard clay at least cost. Determine the dimensions of the canal for the discharge
equal to 14 m3/sec, bed slope is 1 in 2500 and Manning’s coefficient of roughness
is 0.02. Assume side slope 1:1.
Data.
Q = 14m3/sec
b = ??
y = ??
S = 1in2500,
n = 0.02
Z = 1

37. Solution.
According to Most Economical Conditions,
b
+2
y
z
2
= y 1+z2
,
R = y
2
2
b+2
y 𝟏
= y 1+ 𝟏2
b+𝟐
𝐲 = 2
y 𝟐
b+𝟐
𝐲 = 𝟐
.𝟖
𝟑𝐲
b= 𝟐
.𝟖
𝟑𝐲− 𝟐𝐲
b = 0.83y

38. Solution.
Area of flow (A) = by + y2z
A= y (b+yz)
A= y (0.83 y+yx1)
A= 1.83 y2
As, we know that
Q = AV
Q = A.
𝟏
𝒏
𝟐 𝟏
𝑹𝟑 𝑺
𝟐
14= 1.83y2.
𝟎
.
𝟎
𝟐
𝟏 𝒚 𝟐
( )
𝟑(
𝟏
𝟐 𝟐
𝟓
𝟎
𝟎
𝟏
)
𝟐
𝟖
14= 91.5 (
𝒚
)
𝟑
𝟎
.𝟔
𝟑
𝟗𝐱𝟎
.𝟎
𝟐

39. Solution.
𝟖
14= 1.153𝒚𝟑
y = (
𝟏
𝟒
𝟏
.
𝟏
𝟓
𝟑
𝟑
)𝟖
y = 2.55 m
Now, b=0.83 y
b=0.83 x 2.55 = 2.12m
Top width (T) = b+2yz
Top width (T) = 2.12+2 x 2.55 x 1= 7.22 m

40. Most Economical Triangular Channel Section
As, tan𝜽 =
𝑷
= 𝑿
𝑩 𝒀
∴x = ytan𝜽
cos𝜽 =
𝟏
∴ 𝒙 =
𝑩
= 𝒚
𝑯 𝒙
𝟏
𝒚
𝐜
𝐨
𝐬
𝜽
𝒙
𝟏= ysec𝜽

41. Most Economical Triangular Channel Section
As, we know that
Q = AV
Q = AC 𝑅
𝑆 = AC
𝑃
𝐴
𝑆
Now, area of whole section is A
= 2 1
𝑦𝑡
𝑎
𝑛
𝜃𝑥𝑦 = 𝑦
2𝑡
𝑎
𝑛
𝜃
2
𝑦
2=
𝐴
𝑡
𝑎
𝑛
𝜃
y =
𝐴
𝑡
𝑎
𝑛
𝜃

42. Most Economical Triangular Channel Section
As, we know that P = 2x1
P = 2 y𝑠
𝑒
𝑐
𝜃
P = 2
𝐴
𝑡
𝑎
𝑛
𝜃
𝑠
𝑒
𝑐
𝜃
For, maximum Q, P would be minimum, to get P minimum, differentiate P w. r.to 𝜃
.
d
P
d
𝜃
= 0
d
P
=
d
θ d
θ
d 2 𝐴𝑠
𝑒
𝑐
𝜃
𝑡
𝑎
𝑛
𝜃
= 0

43. Most Economical Triangular Channel Section
As,
d 𝑈
d
x 𝑉
Vdu
−𝐔dV
= d
x d
x
d
θ
d
P
=
1
𝑉2
𝑡
𝑎
𝑛
𝜃2 𝐴𝑠
𝑒
𝑐
𝜃𝑡
𝑎
𝑛
𝜃
− 2 𝐴 𝑠
𝑒
𝑐
𝜃 𝑿 2
𝑡
𝑎
𝑛
𝜃
−1
2
2𝑠
𝑒
𝑐 𝜃
= 0
= 2 𝐴 𝑡
𝑎
𝑛
𝜃 𝑡
𝑎
𝑛
𝜃
𝑠
𝑒
𝑐
𝜃
𝑡
𝑎
𝑛
𝜃
−
( 𝑡
𝑎
𝑛
𝜃
)
2
3
𝐴𝑠
𝑒
𝑐 𝜃 𝑡
𝑎
𝑛
𝜃
−1
2
𝑡
𝑎
𝑛
𝜃
= 0
= 2 𝐴 𝑡
𝑎
𝑛
𝜃𝑠
𝑒
𝑐
𝜃 −
𝐴𝑠
𝑒
𝑐
3𝜃
1
𝑡
𝑎
𝑛
𝜃 𝑡
𝑎
𝑛
𝜃 2
= 0
𝐴 𝑠
𝑒
𝑐
𝜃 2 𝑡
𝑎
𝑛
𝜃−
𝑠
𝑒
𝑐
2𝜃
𝑡
𝑎
𝑛
𝜃 𝑡
𝑎
𝑛
𝜃
= 0
d
θ d
θ
d
P d 2 𝐴𝑠
𝑒
𝑐
𝜃
=
𝑡
𝑎
𝑛
𝜃
=0

44. Most Economical Triangular Channel Section
2 𝑡
𝑎
𝑛
𝜃− =
𝑠
𝑒
𝑐
2𝜃 0
2 𝑡
𝑎
𝑛
𝜃−
𝑡
𝑎
𝑛
𝜃 𝑡
𝑎
𝑛
𝜃 𝐴𝑠
𝑒
𝑐
𝜃
𝑠
𝑒
𝑐
2𝜃
𝑡
𝑎
𝑛
𝜃 𝑡
𝑎
𝑛
𝜃
= 0
2 𝑡
𝑎
𝑛
𝜃−
𝑠
𝑒
𝑐
2𝜃
3
(
𝑡
𝑎
𝑛
𝜃
)
2
= 0
Take LCM
2
(
𝑡
𝑎
𝑛
𝜃
)
1
+32
2
− 𝑠
𝑒
𝑐 𝜃
3
(
𝑡
𝑎
𝑛
𝜃
)
2
= 0
2 2
2𝑡
𝑎
𝑛 𝜃
− 𝑠
𝑒
𝑐 𝜃
3
(
𝑡
𝑎
𝑛
𝜃
)
2
= 0

45. Most Economical Triangular Channel Section
3
2𝑡
𝑎
𝑛
2
𝜃− 𝑠
𝑒
𝑐
2𝜃 = 0 x (
𝑡
𝑎
𝑛
𝜃
)
2
2𝑡
𝑎
𝑛
2
𝜃− 𝑠
𝑒
𝑐
2𝜃 = 0
2𝑡
𝑎
𝑛
2
𝜃= 𝑠
𝑒
𝑐
2𝜃
𝑐
𝑜
𝑠
2
𝜃
2 𝑠
𝑖
𝑛
2
𝜃 1
=
𝑐
𝑜
𝑠
2
𝜃
2 𝑠
𝑖
𝑛
2
𝜃= 1
𝑠
𝑖
𝑛
2
𝜃= 1
2
𝑠
𝑖
𝑛
𝜃=
1
2
𝑠
𝑖
𝑛
𝜃=
1
2
θ = s
i
n
−
1 1
2
= 450
2
R = 𝐴
= 𝑦 𝑡
𝑎
𝑛
𝜃
𝑃 2
𝑦 𝑠
𝑒
𝑐𝜃
R =
𝒚𝑡
𝑎
𝑛4
5
0
2 𝑠
𝑒
𝑐4
5
0
R =
𝑦 𝐱1
= 𝑦
2 2 2 2
R = 0.356y
Hence, a triangular channel section will be the most
economical when each of its sloping sides make an angle
of 450 with the vertical and R = 0.356 y.

46. Problem. A triangular channel section, whose sides include anangle of 600,
conveys water at a uniform depth of 0.25 m. If the discharge is 0.04 m3/s ec,
determine the gradient of the trough (bed slope). Use C = 52.
Data.
Angle AOB = 600,
Angle AOC = 300,
y = 0.25 m,
Q = 0.04m3/sec,
S =??

47. Most Economical Triangular Channel Section
Solution.
In ∆ACO, (right angled triangle)
Cos 300 = B
= OC
H AO
Therefore, AO =
OC
C
o
s3
0
0
=
0
.
2
5
0
.
8
5
6
AO = OB = 0.29 m
Tan 300 = P
= AC
B OC
Therefore, AC= 𝑂
𝐶t
a
n3
0
0
AC = 𝑦 t
a
n3
0
0
AC= 𝐵
𝐶=0
.
2
5𝑥0
.
5
7
7=0
.
1
4
4𝑚

48. Most Economical Triangular Channel Section
Now, Area = 2 x 1
𝑥0.25 x 0.144 = 0.036m2
2
P = 2 x 0.29 = 0.58m
Now, Q = AC 𝑅
𝑆
S =
𝑄
𝐴
𝐶 𝑅
S = (
𝑄
𝐴
𝐶 𝑅
)
2
𝑃
R = 𝐴
= 0
.
0
3
6
= 0.062 m
S = (
0
.
5
8
0
.
0
4
0
.
0
3
6𝑥5
2 0
.
0
6
2
)
2
S =
1
1
3
6
.
7
5
= 1in137

49. Problem. Achannel 0.9 m wide has vertical sides and bottom is V-shaped,
the angle of the V being 120o. If the depth of water flowing along the channel
measured from the bottom of the V is 0.6 m. Calculate the flow in liters/sec, when
the bed has a slope of 1in 1200. Find the depth of water in the channel for the same
flow when the V has silted up so that the bottom of the channel is now level and
the slope of the bed remains unchanged. Take C = 55.
Data.
Q = ??
C = 55
Y= 0.6 m
S = 1in1200,
y = ??(when Vis siltedup)

50. Most Economical Circular Channel Section
Consider a circular channel section (shown below) that is not flowing full.
Area of Sector
𝐴
𝑟
𝑒
𝑎𝑜
𝑓𝑆
𝑒
𝑐
𝑡
𝑜
𝑟
= 𝐴
𝑛
𝑔
𝑙
𝑒𝑜
𝑓𝑆
𝑒
𝑐
𝑡
𝑜
𝑟
𝐴
𝑟
𝑒
𝑎𝑜
𝑓𝐶
𝑖
𝑟
𝑐
𝑙
𝑒 𝐴
𝑛
𝑔
𝑙
𝑒𝑜
𝑓𝐶
𝑖
𝑟
𝑐
𝑙
𝑒
=
𝐴
1 𝜃
𝜋 𝑟
2 2𝜋
Now
1
A =
𝜋𝑟
2
𝜃
2𝜋
1
A =
𝑟
2
𝜃
2

51. Most Economical Circular Channel Section
𝑟
2
𝜃
x 2
Area of full section =
Area of full section =
2
𝑟
2
𝜃
2
Area of triangle (∆)= 1
x base x altitude
Sin θ= 𝑃
𝐻
= 𝑃
𝑟
P = rSinθ
Cos θ= 𝐵
= 𝐵
𝐻 𝑟
Base = rCosθ

52. Most Economical Circular Channel Section
2
Area of triangle (∆)= 1
x rx cos θx rx sin θ
Area of triangle (∆) = 1
r2x cos θx sin θ
2
As, sin 2θ = 2 sin θcos θ
2
𝑆
𝑖
𝑛2
θ
= sin θcos θ
2
Area of triangle (∆)= 1
r2 𝑆
𝑖
𝑛2
θ
2 2
Now, area of whole triangle (∆)= r2 𝑆
𝑖
𝑛2
θ
Now,
Area of sector = r2θ- r2𝑆
𝑖
𝑛2
θ
2
Area = r2 𝛉−
𝑺
𝒊
𝒏𝟐
𝛉
𝟐
Wetted Perimeter (P):
𝐴
𝑟
𝑐𝑙
𝑒
𝑛
𝑔
𝑡
ℎ𝑜
𝑓𝑆
𝑒
𝑐
𝑡
𝑜
𝑟
= 𝐴
𝑛
𝑔
𝑙
𝑒𝑜
𝑓𝑆
𝑒
𝑐
𝑡
𝑜
𝑟
𝐴
𝑟
𝑒
𝑎𝑙
𝑒
𝑛
𝑔
𝑡
ℎ𝑜
𝑓𝐶
𝑖
𝑟
𝑐
𝑙
𝑒 𝐴
𝑛
𝑔
𝑙
𝑒𝑜
𝑓𝐶
𝑖
𝑟
𝑐
𝑙
𝑒
=
𝐿 𝜃
2𝜋𝑟 2𝜋
Arc length of required ½ section (L) = r θ
Arc length (Perimeter) of
required full section (L) = 2r θ = d θ

53. Most Economical Circular Channel Section
For Max. Velocity
2
θ = 2
5
7
.
5
0
θ = 1
2
8
.
7
5
0
y = 1.62r
y = 0.81 d
R = 0.305d
For Max. Discharge
2
θ = 3
0
8
0
θ = 1
5
4
0y =
0.95 d R
= 0.29d

54. Problem. A concrete lined circular channel of 3.6 m diameter has a bed
slope of 1in 600.Determine the velocity and flow rate for the conditions of:
Maximum velocity, and Maximum discharge, Take Chezy’s constant, C =50
Data.
Diameter of the circular channel, d = 3.6 m
Bed slope, S = 1/600
Chezy’s constant, C = 50
Let 2θ = Total angle subtended by the water surface at the center of
the channel.
Flow velocity, V=??
Flow rate, Q =??

55. Solution. (a) Maximum velocity condition:
For maximum velocity condition,
2θ = 257.50 = 257.5 x π/180 = 4.49radians
Flow depth = y = 0.81 d = 0.81 x 3.6 = 2.92 m
Area of flow =
𝟐
= (
𝟏
.
𝟖
)
𝟐
𝟒
.𝟒
𝟗− 𝑺
𝒊
𝒏2
5
7
.
5
𝟎 = 8.85m2
Wetted perimeter, P = 2rθ
P = 1.8 x 4.49 = 8.08m
Hydraulic radius, R = A/P = 8.85/8.08 = 1.095 m
Now, Flow velocity, V = C 𝑅
𝑆
V = 50 1
.
0
9
5x
1
6
0
0
= 2.14m/sec
Flow rate, Q = A
V= 8.85 x 2.14 = 18.94 m3/sec (Ans.)
Area = r2 𝛉−
𝑺
𝒊
𝒏𝟐
𝛉
𝟐
Area
=
𝑟
2
2
2
𝛉− 𝑺
𝒊
𝒏𝟐
𝛉
2
𝑟
2
Area = 2
𝛉− 𝑺
𝒊
𝒏𝟐
𝛉

56. (b) Maximum discharge condition:
For maximum discharge condition,
2θ = 3080 = 308 x π /180 = 5.375radians
Flow depth = y = 0.95 d = 0.95 x 3.6 = 3.42 m
Area of flow = (
𝟐
𝟏
.
𝟖
)
𝟐
𝟓
.𝟑
𝟕
𝟓− 𝑺
𝒊
𝒏3
0
8
𝟎 = 9.984 m2
Wetted perimeter, P = 2rθ
P = 1.8 x 5.375 = 9.675m
Hydraulic radius, R = A/P = 9.984/9.675 = 1.032 m
𝑅
𝑆
1
.
0
3
2x
1
6
0
0
= 2.07m/sec
Now, Flow velocity, V= C
V= 50
Flow rate, Q = A
V
Q = 9.984 x 2.07 = 20.66 m3/sec (Ans.)
𝑟
2
Area =
2
2
𝛉− 𝑺
𝒊
𝒏𝟐
𝛉