its a topic of fluid mechanics detailing about the most economic section of channels its parameters and so.

1. MOST ECONOMICAL CHANNEL
SECTION:
APPLICABILITY TO
RECTANGULAR AND
TRAPEZOIDAL CHANNELS
MADE BY :
VAIBHAV PATHAK
MECHANICAL B.TECH 3 YEAR

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
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
Area, A
Wetted Perimeter
D
B

4. Empirical Flow Equations for Estimating Normal Flow
Velocities
 a) Chezy Formula (1775):
Can be derived from basic principles. It states that: ;
 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.
SRCV 

5. Definitions
a) Freeboard: Vertical distance between
the highest water 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. Z = e/d = e’/D

6. DESIGN OF CHANNELS FOR
STEADY UNIFORM FLOW
 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.

7. 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
flow depth. For a given discharge, slope and
roughness, the designer aims to minimize the
cross-sectional area A in order to reduce
construction costs

8. 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

9. 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mi
Where m= hydraulic mean depth.
i= bed slope.
m=A/P where P=wetted perimeter.
For m to be maximum ,P minimum

10. A=BD
B=A/D;
P=B+2D ,then
P=A/D+2D
𝑑𝑃
𝑑𝐷
=
−𝐴
𝐷2
+ 2
𝑑𝑃
𝑑𝐷
= 0
−𝐴
𝐵2
+ 2 = 0
𝐷2 =
𝐴
2
=
𝐵𝐷
2
D =
𝐵
2
𝑚 =
𝐵𝐷
𝐵 + 2𝐷
=
2𝐷2
2𝐷 + 2𝐷
=
𝐷
2
𝑚 =
𝐷
2

11. TRAPEZOIDAL 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mi
Where m= hydraulic mean depth.
i= bed slope.
m=A/P where P=wetted perimeter.
For m to be maximum ,P minimum

12. 𝐴 = 𝐵𝐷 + 𝑛𝐷2
𝑃 = 𝐵 + 2𝐷 1 + 𝑛2
𝐵 =
𝐴
𝐷
−
𝑛𝐷2
𝐷
=
𝐴
𝐷
− 𝑛𝐷
𝑃 =
𝐴
𝐷
− 𝑛𝐷 + 2𝐷 1 + 𝑛2
𝑑𝑃
𝑑𝐷
=
−𝐴
𝐷2
− 𝑛 + 2 1 + 𝑛2
𝑑𝑃
𝑑𝐷
= 0, ⇒
−𝐴
𝐷2
− 𝑛 + 2 1 + 𝑛2 = 0
2 1 + 𝑛2 = 𝑛 +
𝐴
𝐷2
2 1 + 𝑛2 = 𝑛 +
𝐵𝐷
𝐷2
+ 𝑛
𝐵
𝐷
= 2( 1 + 𝑛2 − 𝑛
𝐵 + 2𝑛𝐷
2
= 𝐷 1 + 𝑛2
CONCLUSION:
HALF OF THE TOP WIDTH = SIDE WALL LENGTH

13. m = D/2
The best side slope for
Trapezoidal section

𝑃 =
𝐴
𝐷
− 𝑛𝐷 + 2𝐷 1 + 𝑛2
𝑑𝑃
𝑑𝑛
= 0 − 𝐷 +
2𝐷
2 1 + 𝑛2
∗ 2𝑛
𝑑 𝑃 𝑑 𝑛 = 0
𝑛 = 1 √ 3
𝜃 = 600

15. open channel as shown Q=10m3/s, velocity =1.5m/s, for most
economic section. find wetted parameter, and the bed slope n=0.014.
Trapezoidal
Example 4
 
mD
DDDA
DkDBA
m
V
Q
A
BD
DB
D
kDB
kD
78.1
667.6)
2
3
6055.0(
667.6
5.1
10
6055.0
2
2
32
2
31
2
2
1
2
2
2










16. mP
kDDP
kDBP
49.7
2
3
178.12)78.1(6055.0
126055.0
12
2
2
2









To calculate bed Slope
6.1941:1
5.189.0
014.0
1
89.0
49.7
667.6
m49.7
m667.6
1
3
2
2
3
2






S
SV
P
A
R
P
A
SR
n
V
h
h