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April 6, 2008

Gravitational Potential Energy

Filed under: Physics — Administrator @ 7:18 pm

A total of 10^4 kg of water per second flows over a waterfall 25m high!
If half of the power this flow could be converted into electricity, how many 100W light bulbs could be supplied?

This is a classic gravitational potential energy problem where we are asked for electrical power.

The formula for the potential energy of the water is
P=mgh

where
m = 1×10^4 Kg
g=9.8m/s^2
h=25m

the power is simply the amount of energy available in one second or:

P= 2.45 x 10^6 J/s

With 50% efficiency

Power = 1.225 x 10^6 Watts
divide by 100 watts/bulb
12,250 light bulbs

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