Question

In a hydroelectric dam, water falls 31.0 m and then spins a turbine to generate electricity. Part A What is ΔU of 1.0 kg of water?

Part 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 47.0 MW of electricity? This is a typical value for a small hydroelectric dam.

Answer 1

Given
m = 1.0kg
h = 31.0 m
g = 9.8 m/s^2
Where U = potential energy
delta U = -mgh = -1.0*9.8*31.0 = -303.8J

since the surface is assumed to be 0 the potential lowers as it gets close to the surface.

B) 47 MW = 47*10^6 W = 47*10^6 J/s

since the change in potential is the energy we use to rotate the turbines

each kg will produce 303.8*0.8 = 243.04J of electricity

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

mass of water = 50*10^6/243.04 = 205727 kg of water per second..