Answer to Question #113813 in Classical Mechanics for Caylin

Question

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 80% efficiency ( 80% kinetic energy becomes electric energy).

A small hydroelectric plant at the base of the dam generates 50 MW of electric power when the falling water has a speed of 18 m/s.

What is the water flow rate - kilograms of water per second through turbines?

Please assist.

Answer 1

50 MW of electric power corresponds to the power of water of

$$P_W=\frac{P}{\eta}.$$

Power is kinetic energy over time:

$$P_W=\frac{K}{t}.$$

The kinetic energy is

$$K=\frac{1}{2}mv^2.$$

Substitute:

$$\frac{mv^2}{2t}=\frac{P}{\eta},\\ \space\\ \frac{m}{t}=\frac{2P}{v^2\eta}=\frac{2\cdot50\cdot10^6}{18^2\cdot0.8}=385800\text{ kg/s}.$$