economic channel section
TRANSCRIPT
![Page 1: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/1.jpg)
MOST ECONOMICAL CHANNEL SECTION:APPLICABILITY TO RECTANGULAR AND TRAPEZOIDAL CHANNELSMADE BY :VAIBHAV PATHAK MECHANICAL B.TECH 3 YEAR
![Page 2: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/2.jpg)
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.
![Page 3: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/3.jpg)
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
![Page 4: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/4.jpg)
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
![Page 5: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/5.jpg)
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
![Page 6: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/6.jpg)
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.
![Page 7: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/7.jpg)
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
![Page 8: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/8.jpg)
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
![Page 9: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/9.jpg)
RECTANGULAR SECTIONFor 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=CmiWhere m= hydraulic mean depth.i= bed slope.m=A/P where P=wetted perimeter.For m to be maximum ,P minimum
![Page 10: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/10.jpg)
A=BDB=A/D;P=B+2D ,thenP=A/D+2D
𝑑𝑃𝑑 𝐷=
−𝐴𝐷2 +2
𝑑𝑃𝑑 𝐷=0
−𝐴𝐵2 +2=0
𝐷2= 𝐴2
=𝐵𝐷2
D=𝐵2
𝑚=𝐵𝐷
𝐵+2𝐷=2𝐷2
2𝐷+2𝐷=𝐷2
𝑚=𝐷2
![Page 11: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/11.jpg)
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=CmiWhere m= hydraulic mean depth.i= bed slope.m=A/P where P=wetted perimeter.For m to be maximum ,P minimum
![Page 12: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/12.jpg)
𝐴=𝐵𝐷+𝑛𝐷2
𝑃=𝐵+2𝐷√1+𝑛2
𝐵= 𝐴𝐷 −
𝑛𝐷2
𝐷 = 𝐴𝐷−𝑛𝐷
𝑃= 𝐴𝐷 −𝑛𝐷+2𝐷√1+𝑛2
𝑑𝑃𝑑𝐷=
− 𝐴𝐷2 −𝑛+2√1+𝑛2
𝑑𝑃𝑑𝐷=0 ,⇒ −𝐴
𝐷2 −𝑛+2√1+𝑛2=0
2√1+𝑛2=𝑛+𝐴𝐷2
2√1+𝑛2=𝑛+ 𝐵𝐷𝐷2 +𝑛
¿
CONCLUSION:HALF OF THE TOP WIDTH = SIDE WALL LENGTH
![Page 13: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/13.jpg)
m = D/2The best side slope for Trapezoidal section
(𝑃= 𝐴
𝐷 −𝑛𝐷+2𝐷√1+𝑛2
𝑑𝑃𝑑𝑛=0−𝐷+ 2𝐷
2√1+𝑛2∗2𝑛
𝑑𝑃 /𝑑𝑛=0𝑛=1/√3𝜃=600
)
![Page 14: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/14.jpg)
![Page 15: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/15.jpg)
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
mVQA
BD
DBD
kDBkD
78.1
667.6)236055.0(
667.65.1
106055.0
22
322
31
221
2
2
2
![Page 16: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/16.jpg)
mP
kDDP
kDBP
49.723178.12)78.1(6055.0
126055.0
12
2
2
2
To calculate bed Slope
6.1941:1
5.189.0014.01
89.049.7667.6
m 49.7m 667.6
1
32
2
32
S
SV
PAR
PA
SRn
V
h
h
![Page 17: economic channel section](https://reader036.vdocuments.us/reader036/viewer/2022062523/58f9b6fc1a28abf5148b45b7/html5/thumbnails/17.jpg)
THANK YOU