Question

A water pump driven by petrol rises water at a rate of 0.5 m³ min^–1 from a depth of 30 m. If the pump is 70% efficient, what power is developed by the engine?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{h}=30\mathrm{m} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=500\mathrm{kg}/\min \\ \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{500}{60}\mathrm{kg}/\mathrm{s}=\frac{25}{3}\mathrm{kg}/\mathrm{s} \\ \frac{70}{100}\times \mathrm{P}=\frac{\mathrm{mgh}}{\mathrm{t}} \\ \mathrm{P}=\frac{\mathrm{mgh}\times 100}{70\times \mathrm{t}} \\ \mathrm{P}=100\times \frac{25}{3}\times 9.8\times 30\times \frac{1}{70} \\ \mathrm{P}=3500\mathrm{W} \end{gathered}$$