Discharge from a centrifugal pump operating at 1000 rpm with a total head of 30 m is 300 L min⁻¹. The pump efficiency is 65%. If speed of the pump is increased to 1200 rpm, the power required to operate the pump in kW is _______________ (rounded off to 2 decimal places). Consider g = 9.81 m s⁻².
Official Answer Key:
3.90 to 4.00
Step-by-Step Solution & Explanation
Pump affinity laws: Q ∝ N, H ∝ N² (efficiency assumed constant).
Q₂ = Q₁×(N₂/N₁) = 300×(1200/1000) = 360 L/min = 0.006 m³/s.
H₂ = H₁×(N₂/N₁)² = 30×(1.2)² = 30×1.44 = 43.2 m.
Power required = ρgQ₂H₂/η = (1000×9.81×0.006×43.2)/0.65 = 2542.2/0.65 = 3912 W = 3.91 kW. Within official range 3.90–4.00.
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