## Pump Power

Calculate the power required for a pump to lift water against a head of 25 meters flowing at 5 cubic meters per second. Neglect all friction losses. Assume a pump efficiency of .77, and the specific weight of water is 9.8 kN/m^3.

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 $$\dot{W}$$ . 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 power:
$$\dot{W}=\frac{(5m^3)(9.8kN)(25m)}{sec(m^3)(.77)}$$$$$\dot{W}=1591\:\frac{kN\cdot m}{sec}=1591\:kW$$$
1591 kW