Civil Engineering - Waste Water Engineering - Discussion

32. 

For a peak discharge of 0.0157 cumec, with a velocity of 0.9 m/sec, the diameter of the sewer main, is

[A]. 10 cm
[B]. 12 cm
[C]. 15 cm
[D]. 18 cm
[E]. 20 cm.

Answer: Option C

Explanation:

No answer description available for this question.

Rahul said: (Jun 3, 2015)  
Q = AV.
0.0157 = Ax0.9.
A = 0.01744.

Pi/4xd2 = 0.01744.
D = 15 cm.

Rajeev Yadav said: (Jul 19, 2015)  
A = Q/V = 0.0157/0.9 = 0.01744 m^2.

So A = (pi/4)*d^2, so d = 0.149 m = 14.9 cm = 15 cm.

Shubham Yadav said: (Jun 16, 2018)  
But what is π?

Raj said: (Aug 19, 2018)  
Thank you @Rahul.

Abir Dey said: (Mar 8, 2019)  
What is the value of π here?

Manoj Suryavanshi said: (Dec 31, 2019)  
Q = AV.
A= Q/V,. 0.0157/0.9 = 0.0174 m^2.

So A = π/4 x d^2,
0.0174m^2 = 3.14/4 x d^2.
0.0174= .785 d^2.

d^2 =0.0174/.785.
d^2 = 0.0221.
d = .148 m.

Leeots said: (Apr 24, 2021)  
Q=0.0157 cum.
A=π*r*r.

V = 0.9m/s
Sol, Q = axv
0.0157=0.9xπ*r*r. where d=2r.
0.0157/0.9= r*r
0.0055527 = r*r
√0.0055527 = r
r = 0.0745
d = 2*r= 0.149 m
d = 15 cm.

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