Question Bank • QB_IDE_EXP_065 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Wells and Pumps

Subtopic: Classification of pumps

Question Statement:

A centrifugal pump discharges $30\text{ L/s}$ ($0.030\text{ m}^3/\text{s}$) of water against a total dynamic head of $25\text{ m}$. If the pump efficiency is $75\%$, calculate the brake power required to drive the pump in $\text{kW}$. (Take $\rho = 1000\text{ kg/m}^3$ and $g = 9.81\text{ m/s}^2$)

Answer Key:

9.81

Step-by-Step Solution & Explanation

Water power (hydraulic power): $P_w = \frac{\rho g Q H}{1000} = \frac{1000 \times 9.81 \times 0.030 \times 25}{1000} = 7.3575\text{ kW}$ Brake power: $P_{\text{brake}} = \frac{P_w}{\eta} = \frac{7.3575}{0.75} = 9.81\text{ kW}$
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