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

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 = -1.09.833.0 = -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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