## Pump Work

What is the external work done by a pump that delivers 3,000 cubic meters of water per hour against a total head of 10 m? Ignore all losses and note 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 Convert the volumetric flow rate into $$m^3/s$$ . First, let’s convert the volumetric flow rate into $$m^3/s$$ . $$3,000\frac{m^3}{hr}\cdot \frac{1hr}{60min}\cdot \frac{1min}{60sec}=\frac{3,000m^3}{3600sec}=0.83\:m^3/s$$$
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. Since we are ignoring all losses, we can assume efficiency is 100%. Thus, $$\dot{W}=\frac{(0.83m^3)(9,800N)(10m)}{(sec)(m^3)1}=81,340\frac{N\cdot m}{s}=81.3\:kW$$$
81.3 kW