In a hydroelectric dam, water falls 33.0m and then spins a turbine to generate electricity.

A. What is delta U of 1.0 kg of water?

B. 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 50.0 MW of electricity? This is a typical value for a small hydroelectric dam.

### 1 Answer

you have to watch out if you dont define the variables I almost thought of U as being initial velocity..

U = potential energy

since the m = 1.0kg

h = 33.0 m

g = 9.8 m/s^2

delta U = -mgh = -1.0

*9.8*33.0 = -323.4 Jsince the surface is assumed to be 0 the potential lowers as it gets close to the surface.

B) 50 MW = 50

*10^6 W = 50*10^6 J/ssince the change in potential is the energy we use to rotate the turbines

each kg will produce 323.4*0.8 = 258.72 J of electricity

to find the amount of water needed we divide this into the total Joules needed per second.

mass of water = 50*10^6/258.72 = 193259 kg of water per second..