Irrigation and Drainage Engineering · Wells and Pumps · 2 marks · Numerical answer
A centrifugal pump at 1000 rpm delivers300 L/min at 30 m head and 65% efficiency. Raise speed to 1200 rpm. Find input power in kW, to two decimals;use g=9.81 m/s².
Initial discharge is .005 m³/s, so input power is $P_1=1000(9.81)(.005)(30)/.65=2263.846$ W. For dynamically similar operation with approximately unchanged efficiency, pump affinity gives $P\propto N^3$. Thus $P_2=2.263846(1.2)^3=\mathbf{3.91}$ kW.
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