A river has a flow rate of $250{\text{m}}^3/\text{s}$, a usable head of $30\text{m}$ and the overall efficiency of the hydro‑electric plant is $85\%$. 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 formula for hydraulic power: $P_{\text{hyd}}=\rho gQH$.
2. Substitute the given values: $P_{\text{hyd}}=1000\times 9.81\times 250\times 30$.
3. Calculate the product: $1000\times 9.81\times 7500=73,575,000\text{W}$ (or $73.575\text{MW}$).
4. Include the plant efficiency: $P_{\text{elec}}=\eta \times P_{\text{hyd}}=0.85\times 73.575\text{MW}$.
5. Compute the final value: $0.85\times 73.575=62.53875\text{MW}\approx 62.5\text{MW}$.

Concept tested: Calculation of hydro‑electric power using flow, head and efficiency

Published 28 August 2026 · How this solution was prepared and checked

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