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 3 of 3.
Aslam said:
2 years ago
Consider head loss also, right?
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.
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)
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