## A Pump’s Efficiency

If a pump requires 2,000 kW of power to lift against a total water head of 30 m at 5 cubic meters per second, what is the pump’s efficiency? Assume no friction losses and the specific weight of water is 9.8 kN/m^3.

Expand Hint
The pump power equation:
$$\dot{W}=\frac{Q\gamma h}{\eta_t}$$$where $$Q$$ is the volumetric flow, $$h$$ is the fluid head needed to be lifted, $$\eta_t$$ is the total efficiency ( $$\eta_{pump} \times \eta_{motor}$$ ), $$\dot W$$ is the power, and $$\gamma$$ is the specific weight of the fluid. Hint 2 Solve for $$\eta_t$$ . The pump power equation: $$\dot{W}=\frac{Q\gamma h}{\eta_t}$$$
where $$Q$$ is the volumetric flow, $$h$$ is the fluid head needed to be lifted, $$\eta_t$$ is the total efficiency ( $$\eta_{pump} \times \eta_{motor}$$ ), $$\dot W$$ is the power, and $$\gamma$$ is the specific weight of the fluid.

Solving for efficiency:
$$2,000kW=\frac{(5m^3)(9.8kN)(30m)}{sec(m^3)\eta_t}$$$$$\eta_t=\frac{(5)(9.8kN)(30m)}{sec(2,000\frac{kN\cdot m}{sec})}=\frac{1,470}{2,000}=0.74$$$
0.74
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