Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : 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
Discussion:
7 comments Page 1 of 1.
Leeots said:
4 years ago
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.
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.
(2)
Manoj Suryavanshi said:
6 years ago
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.
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.
(3)
Abir dey said:
6 years ago
What is the value of π here?
Raj said:
7 years ago
Thank you @Rahul.
Shubham Yadav said:
7 years ago
But what is π?
Rajeev yadav said:
1 decade ago
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.
So A = (pi/4)*d^2, so d = 0.149 m = 14.9 cm = 15 cm.
(2)
Rahul said:
1 decade ago
Q = AV.
0.0157 = Ax0.9.
A = 0.01744.
Pi/4xd2 = 0.01744.
D = 15 cm.
0.0157 = Ax0.9.
A = 0.01744.
Pi/4xd2 = 0.01744.
D = 15 cm.
(2)
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