Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 38)
38.
If the total head of the nozzle of a pipe is 37.5 m and discharge is 1 cumec, the power generated is
Discussion:
22 comments Page 2 of 3.
Keshav said:
7 years ago
Power= density*Q*g*h.
= 1000*1*9.81*37.5,
= 367873 watts,
= 367873/746,
= 494 HP.
= 1000*1*9.81*37.5,
= 367873 watts,
= 367873/746,
= 494 HP.
(8)
Megha said:
7 years ago
What is w here?
Paul said:
7 years ago
Here, w is the specific weight.
Divya teja said:
7 years ago
What is q here?
(1)
Manjunath S Munnolli said:
6 years ago
@Harsha.
1 HP =746 Watts.
1 HP =746 Watts.
(1)
Jinish Patel said:
6 years ago
W is a specific weight taken from g = 9.810 gravity weight.
Q is discharged through the pipe.
H is a head loss in the pipe.
And power= w*Q*h.
w is in horsepower.
1 horsepower = 736 watts in metric.
Q is discharged through the pipe.
H is a head loss in the pipe.
And power= w*Q*h.
w is in horsepower.
1 horsepower = 736 watts in metric.
(1)
Dheeraj kataria said:
6 years ago
Why did not we consider the pump efficiency. ! please anyone can explain it. ?
(2)
Ganesh said:
5 years ago
Thanks for explaining @Varsha.
Ghamand said:
5 years ago
@Rishi.
Very nice explanation, thanks.
Very nice explanation, thanks.
Chaman said:
3 years ago
Thank you for explaining the answer.
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