Posted by **Zach** on Monday, November 30, 2009 at 2:07pm.

In a hydroelectric dam, water falls 26.0 m and then spins a turbine to generate electricity. What is ∆U of 1.0 kg of water? 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 54.0 MW of electricity? This is a typical value for a small hydroelectric dam.

- Physics -
**bobpursley**, Monday, November 30, 2009 at 2:15pm
potentialenergy= mgh=1kg*9.8N/kg*26m

massflow*g*h/time * .80=54*10^6

Flowrate=54^10^6/gh * 1/.80

- Physics -
**drwls**, Monday, November 30, 2009 at 2:19pm
If what you are calling ∆U includes the gravitational potential energy*, then it is -(density*g*(delta H)

= -(10^3 kg/m^3)*26 m*9.81 m/s^2

The product will be in Joules.

For P = 54*10^6 Watts, you need a mass flow rate such that

P = (mass flow rate)*(∆U)

*Usually, in thermodynamics, U represents internal energy only, which has nothing to do with elevation.

- Physics -
**drwls**, Monday, November 30, 2009 at 2:22pm
I forgot to include the 80% efficiency factor, but BobPursley did it right

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