A hydroelectric plant has a water flow rate of $300,{\text{m}}^3!!,/\text{s}$, a usable head of $150,\text{m}$ and overall efficiency of $88$. Using $\rho =1000,\text{kg}!!,/{\text{m}}^3$ and $g=9.81,\text{m}!!,/{\text{s}}^2$, calculate the electrical power output in megawatts (MW).

Step-by-step solution

1. Write the power formula: $P=\eta ,\rho ,g,Q,H$.
2. Insert the values: $\eta =0.88$, $\rho =1000$, $g=9.81$, $Q=300$, $H=150$.
3. Compute stepwise: $9.81\times 300=2943$.
4. $2943\times 150=441450$.
5. $441450\times 1000=441,450,000$.
6. $441,450,000\times 0.88=388,716,000$ watts.
7. Convert watts to megawatts: $388,716,000/10^6=388.716$ MW.

Concept tested: Hydroelectric power calculation

Published 4 October 2026 · How this solution was prepared and checked