An agricultural centrifugal water pump lifts $30\text{ L/s}$ ($= 0.030\text{ m}^3\text{/s}$) of water against a total head of $25.0\text{ m}$. The density of water is $1000\text{ kg/m}^3$ and $g = 9.81\text{ m/s}^2$. If the overall pump efficiency is $75\%$, the required electric motor brake power in kilowatts (kW) is ________ (round off to 2 decimal places).
Answer Key:
9.81
Step-by-Step Solution & Explanation
1. Water power (hydraulic power output):
$P_w = \rho \cdot g \cdot Q \cdot H$
$P_w = 1000 \times 9.81 \times 0.030 \times 25.0 = 7357.5\text{ W} = 7.3575\text{ kW}$
2. Brake power input to pump:
$P_{brake} = \frac{P_w}{\eta} = \frac{7.3575\text{ kW}}{0.75} = 9.81\text{ kW}$