Question Bank • QB_IDE_011 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Wells and Pumps

Subtopic: Pump characteristics

Question Statement:

An agricultural centrifugal water pump operating at $1450\text{ rpm}$ discharges $0.04\text{ m}^3\text{/s}$ against a total operating head of $16.0\text{ m}$. The specific speed ($N_s = \frac{N \sqrt{Q}}{H^{3/4}}$) of the pump in SI metric units is ________ (round off to 1 decimal place).

Answer Key:

36.3

Step-by-Step Solution & Explanation

Specific speed formula: $N_s = \frac{N \sqrt{Q}}{H^{3/4}}$ Given: • $N = 1450\text{ rpm}$ • $Q = 0.04\text{ m}^3\text{/s} \implies \sqrt{Q} = \sqrt{0.04} = 0.20$ • $H = 16.0\text{ m} \implies H^{3/4} = (16)^{3/4} = (2^4)^{3/4} = 2^3 = 8.0$ $N_s = \frac{1450 \times 0.20}{8.0} = \frac{290}{8.0} = 36.25 \approx 36.3$
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