A pump delivers a discharge of $40\text{ L/s}$ against a total head of $25.0\text{ m}$. If the overall efficiency of the pump set is $70\%$, taking $g = 9.81\text{ m/s}^2$ and density of water $\rho = 1000\text{ kg/m}^3$, the power input required by the electric motor in $\text{kW}$ is ________ (round off to 2 decimal places).
Answer Key:
14.01
Step-by-Step Solution & Explanation
1. Water horsepower / hydraulic power $P_w$:
$P_w = \rho g Q H = 1000 \times 9.81 \times 0.040 \times 25.0 = 9810\text{ W} = 9.81\text{ kW}$
2. Motor input power $P_{in}$:
$P_{in} = \frac{P_w}{\eta} = \frac{9.81\text{ kW}}{0.70} \approx 14.014\text{ kW} \approx 14.01\text{ kW}$