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
400 H.P.
450 H.P.
500 H.P.
550 H.P.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 1 of 3.

Haidar Ullah said:   8 months ago
To calculate the power generated, we can use the formula:

Power (P) = ρ * g * Q * H/1000.

where:

ρ = density of water (approximately 1000 kg/m³),
g = acceleration due to gravity (approximately 9.81 m/s²),
Q = discharge (1 cumec or 1 m³/s),
H = total head of the nozzle (37.5 m).

First, let's calculate the power in Watts:

P = 1000 * 9.81 * 1 * 37.5 ≈ 367,437 W.

To convert Watts to Horsepower (HP), we divide by 746 (since 1 HP ≈ 746 W):
P ≈ 367,437/746 ≈ 493 HP.

Rounding to the nearest answer, we get:

The correct answer is 500 H.P.
(1)

Aslam said:   2 years ago
Consider head loss also, right?

Chaman said:   3 years ago
Thank you for explaining the answer.

Ghamand said:   5 years ago
@Rishi.

Very nice explanation, thanks.

Ganesh said:   5 years ago
Thanks for explaining @Varsha.

Dheeraj kataria said:   6 years ago
Why did not we consider the pump efficiency. ! please anyone can explain it. ?
(2)

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.
(1)

Manjunath S Munnolli said:   6 years ago
@Harsha.

1 HP =746 Watts.
(1)

Divya teja said:   7 years ago
What is q here?
(1)

Paul said:   7 years ago
Here, w is the specific weight.


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