Water flows through a semi-circular weir 2 m deep and 4 m wide at the top under a head of 1.50 m. Assuming C = 0.65, determine the discharge through the weir.Please solve it something like the example below. However, the example is parabolic while the problem above is semi circular.Please show step by step solution T.Y. :) %3DQ =through the weir.2 mSolutionQ =1 mBy V2 mActuaActua1.5 m1.5-h dQ = dA 2ghdA = 2x dhExpress x in terms of h by squared property of parabola:1.%D1.5-h2.x = 0.707 /1.5-hdA = 2(0.707 /1.5 -h )dh%DdQ = 2(0.707 /1.5 -h )dh 2gh%3DdQ = 6.263 /1.5-h h dh1.5Q = 6.263V1.5-hh dhBy trigonometric substitution:h= 1.5 sin? 0Vh = 1.2247 sin 0Let%3Ddh = 3 sin 0 cos 0 do%3Dwhen h = 0, 0 = 0°%3Dwhen h 1.5, 0= 90° = T/2%3DtopargeQ = 6.263V1.5 -1.5 sin2 0(1.2247 sin 0)(3 sin @ cos ede)P/2Q= 28.182 sin² 0 cos? Od0By Walli's formula:1(1)Q= 28.1824(2)Q-5.5336 m/s(theoretical discharge)Actual discharge 3.597 m/s

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Answered: %3D through the weir. 2 m Solution 1 m…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Water flows through a semi-circular weir 2 m deep and 4 m wide at the top under a head of 1.50 m. Assuming C = 0.65, determine the discharge through the weir.

Please solve it something like the example below. However, the example is parabolic while the problem above is semi circular.

Please show step by step solution T.Y. :)

%3D
Q =
through the weir.
2 m
Solution
Q =
1 m
By V
2 m
Actua
Actua
1.5 m
1.5-h

dQ = dA 2gh
dA = 2x dh
Express x in terms of h by squared property of parabola:
1.
%D
1.5-h
2.
x = 0.707 /1.5-h
dA = 2(0.707 /1.5 -h )dh
%D
dQ = 2(0.707 /1.5 -h )dh 2gh
%3D
dQ = 6.263 /1.5-h h dh
1.5
Q = 6.263
V
1.5-hh dh
By trigonometric substitution:
h= 1.5 sin? 0
Vh = 1.2247 sin 0
Let
%3D
dh = 3 sin 0 cos 0 do
%3D
when h = 0, 0 = 0°
%3D
when h 1.5, 0= 90° = T/2
%3D
top
arge
Q = 6.263
V1.5 -1.5 sin2 0(1.2247 sin 0)(3 sin @ cos ede)
P/2
Q= 28.182 sin² 0 cos? Od0
By Walli's formula:
1(1)
Q= 28.182
4(2)
Q-5.5336 m/s
(theoretical discharge)
Actual discharge 3.597 m/s

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