In a hydroelectric dam, water falls 33.0m and then spins a turbine to generate electricity.
In a hydroelectric dam, water falls 33.0 meters and spins a turbine to generate electricity.
A. What is the change in potential energy (ΔU) of 1.0 kg of water during its fall?
B. If the dam operates with an efficiency of 80% in converting the water's potential energy into electrical energy, how many kilograms of water must flow through the turbines each second to produce 50.0 MW of electrical power? This value is typical for a small hydroelectric dam.
1 Answers
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 = -1.09.833.0 = -323.4 J
since the surface is assumed to be 0 the potential lowers as it gets close to the surface.
B) 50 MW = 5010^6 W = 5010^6 J/s
since 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..
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