Answer to Question #193161 in Physics for Davis

Question

if a hydroelectric station is 93% efficient. What mass of water must pass through the station each 1.00s to produce 855kW of electricity from a 34.6 meter drop?

Answer 1

The 855 kW of generated power correspond to

$$P_i=\frac{P_o}{\eta}$$

of input power, or the power of falling water. This power can be found from the work done by water under the influence of gravitation:

$$P_i=\frac{W}{t}=\frac{mgh}{t},\\\space\\ m=\frac{P_it}{gh}=\frac{P_0t}{\eta gh}=2573\text{ kg}.$$