A hydroelectric plant has a dam of height h = 70 m . At the…

A hydroelectric plant has a dam of height $h = 70\ \mathrm{m}$. At the bottom, the water leaves the turbine at $v = 22\ \mathrm{m/s}$. What is the efficiency of the plant (neglecting other losses)?

Explanation of Method

Efficiency of a hydroelectric plant.

Step 1: Energy per kilogram

\[ \text{available: } gh = 686.7\ \mathrm{J/kg}, \]

\[ \text{carried away: } \tfrac12 v^2 = 242\ \mathrm{J/kg}. \]

Step 2: Efficiency

\[ \eta = 1 - \frac{v^2}{2gh} = 1 - \frac{484}{1373.4}. \]

\[\eta \approx 0.65 = 65\%\]

$h$ $v = 22\ \mathrm{m/s}$ $\text{turbine}$ $65\%$ $35\%$ $gh = 686.7\ \mathrm{J/kg}$ $\text{used}$ $\tfrac{v^2}{2} = 242$

Each kilogram of water brings $gh$ of energy. What it still carries away as kinetic energy when it leaves the turbine, $v^2/2$, is not used.

Result

\[\boxed{\,\eta \approx 0.65 = 65\%\,}\]

Remark

Check. The original result is correct.