Question Bank • QB_IDE_ADV_111 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Wells and Pumps

Subtopic: Pump characteristics

Question Statement:

A centrifugal pump lifts an irrigation discharge of $Q = 50.0\text{ L/s}$ ($0.050\text{ m}^3/\text{s}$) against a total dynamic head of $H = 28.0\text{ m}$. The density of water is $\rho = 1000.0\text{ kg/m}^3$ and $g = 9.81\text{ m/s}^2$. If the overall pump efficiency is $\eta = 68.0\%$, the brake shaft power input required to drive the pump in kilowatts ($\text{kW}$) is ________ (round off to one decimal place).

Answer Key:

20.2

Step-by-Step Solution & Explanation

**Method 1: Hydraulic Water Power Formulation** 1. Water power (hydraulic output power, $P_w$): $P_w = \rho g Q H$ Where: - $\rho = 1000.0\text{ kg/m}^3$ - $g = 9.81\text{ m/s}^2$ - $Q = 0.050\text{ m}^3/\text{s}$ - $H = 28.0\text{ m}$ $P_w = 1000.0 \times 9.81 \times 0.050 \times 28.0 = 13,734.0\text{ W} = 13.734\text{ kW}$ 2. Brake power input ($P_{\text{shaft}}$) with $\eta = 0.680$: $P_{\text{shaft}} = \frac{P_w}{\eta} = \frac{13.734\text{ kW}}{0.680} = 20.197\text{ kW} \approx 20.2\text{ kW}$ **Method 2: Direct Single Equation Shortcut** $P_{\text{shaft}} = \frac{9.81 \times 0.050 \times 28.0}{0.680} = \frac{13.734}{0.680} = 20.197\text{ kW} \approx 20.2\text{ kW}$
← Previous (QB_IDE_ADV_103) Next (QB_IDE_005) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test