Question Bank • QB_IDE_EXP_046 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Wells and Pumps

Subtopic: Pump characteristics

Question Statement:

For the same pump ($N_1 = 1450\text{ rpm}$, $H_1 = 20.0\text{ m}$), when the speed increases to $N_2 = 1740\text{ rpm}$, the new total head $H_2$ in meters is ________ (round off to 1 decimal place).

Answer Key:

28.8

Step-by-Step Solution & Explanation

From pump affinity laws: $\frac{H_2}{H_1} = \left(\frac{N_2}{N_1}\right)^2$ Given $\frac{N_2}{N_1} = \frac{1740}{1450} = 1.20$: $H_2 = 20.0 \times (1.20)^2 = 20.0 \times 1.44 = 28.80\text{ m}$
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