Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 6 (Q.No. 39)
39.
A circular sewer 2 m diameter has to carry a discharge of 2 m3/s when flowing nearly full. What is the minmum required slope to initiate the flow ? (Assume Manning's n = 0.015)
Discussion:
10 comments Page 1 of 1.
Lee said:
8 years ago
Yes, option A is the right answer.
Agree @Amit.
Agree @Amit.
(1)
Phani said:
7 years ago
C is correct.
V = 1/n x r^2/3 x s^1/2.
Q = AxV.
2 = pai/4x4 x 1/0.015 x 1^2/3 x s^1/2,
S = 9*10^-4 = 0.00009.
V = 1/n x r^2/3 x s^1/2.
Q = AxV.
2 = pai/4x4 x 1/0.015 x 1^2/3 x s^1/2,
S = 9*10^-4 = 0.00009.
(1)
Amit padiyar said:
9 years ago
Q = (1/n) * R^(2/3) * s^.5 * d2/4.
S = .00023 ans.
S = .00023 ans.
Kunti said:
7 years ago
You are right @Phani.
Shivaprasad sajjan said:
7 years ago
.00009 is correct.
You are right @Phani.
You are right @Phani.
Vivek mishra said:
6 years ago
@Phani.
Hydraulic mean depth for circular section is 28D for max discharge and 3D for maximum velocity then how could you take m= 1?
Hydraulic mean depth for circular section is 28D for max discharge and 3D for maximum velocity then how could you take m= 1?
Prem said:
6 years ago
I think A is correct one.
Mahesh sinh said:
5 years ago
According to me, A is the correct answer.
Ali said:
5 years ago
r=1/2 not 1,
Then the answer is 0.00023.
Then the answer is 0.00023.
Meraj said:
5 years ago
Q=A*V, here Q is given, from here we can find.
V=2/3, and we know V=1/n*R^2/3*S^1/2.
we can find R=1/2, by putting all the value we get,
S = 0.00023 so A is the correct answer.
V=2/3, and we know V=1/n*R^2/3*S^1/2.
we can find R=1/2, by putting all the value we get,
S = 0.00023 so A is the correct answer.
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