Civil Engineering - Hydraulics - Discussion

Power= density*Q*g*h.
= 1000*1*9.81*37.5,
= 367873 watts,
= 367873/746,
= 494 HP.

P = wqh/75.

P = 1000*1*37.5/75.

P = 500.

Why did not we consider the pump efficiency. ! please anyone can explain it. ?

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.

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.

@Harsha.

1 HP =746 Watts.

What is q here?

What is w here?

Consider head loss also, right?

Thank you for explaining the answer.