An electric motor (efficiency eta_1 = 0.95) drives a pump…

An electric motor (efficiency $\eta_1 = 0.95$) drives a pump (efficiency $\eta_2 = 0.90$) that pushes water up a pipe whose end is $h = 10\ \mathrm{m}$ above the pump. The flow rate is $Q = 200\ \mathrm{dm^3/s}$. What electrical power does the motor draw?

Explanation of Method

Input power of a motor-driven pump.

Step 1: Useful power

\[ P_{\text{useful}} = \rho Q g h = 1000 \cdot 0.2 \cdot 9.81 \cdot 10 = 19\,620\ \mathrm{W}. \]

Step 2: Input power

\[ P = \frac{\rho Q g h}{\eta_1\eta_2} = \frac{19\,620}{0.855}. \]

\[P \approx 22.9\ \mathrm{kW}\]

$22.95\ \mathrm{kW}$ $\text{electric input}$ $21.8\ \mathrm{kW}$ $\text{motor output}$ $19.62\ \mathrm{kW}$ $\text{water (useful)}$ $\times 0.95$ $\times 0.90$ $P_{\text{useful}} = \rho Q g h = 1000 \cdot 0.2 \cdot 9.81 \cdot 10$

Power along the chain: each machine passes on only its efficiency's share, so the input must be the useful power divided by $0.95 \times 0.90 = 0.855$.

Result

\[\boxed{\,P \approx 22.9\ \mathrm{kW}\,}\]

Remark

Check. The original result is correct, but its formula $P = Qgh/(\eta_1\eta_2)$ is missing the water density $\rho$; with $Q$ in m³/s the density is what turns it into a power in watts.