Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 45.0 MWMW of electricity? This is a typical value for a small hydroelectric dam.

This question involves the concepts of efficiency, potential energy, and electrical energy.

"2.2 x 10⁵ kg/s" of water must pass through the turbines each second, assuming a waterfall height of 26 m.

First, we will use the formula for the efficiency of conversion from potential energy to electrical energy.

assuming h = 26 m:

Learn more about efficiency here:

link

The attached picture shows the pyramid representing energy efficiency.

Explanation:

Let assume that the water dam has a fall of 35 meters. The required instantaneous potential energy is:

The instantaneous potential energy has the following model:

The mass flow rate needed to produced the required energy production:

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow