# Civil Engineering - Waste Water Engineering - Discussion

### Discussion :: Waste Water Engineering - Section 4 (Q.No.32)

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.

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.