How Much Energy Does Falling Water Generate at Niagara Falls?

AI Thread Summary
The discussion centers on calculating the energy generated by falling water at Niagara Falls, specifically using the formula mgh to determine energy output. It is established that one kilogram of water falling 50 meters produces approximately 490.5 joules of energy. To generate one megawatt of power, around 20,400 kilograms of water must pass through the generators each second. There is confusion regarding the accuracy of this calculation, with the user questioning the correctness of their answer. The conversation highlights the complexities of energy conversion efficiency in hydroelectric power generation.
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Homework Statement



In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 50 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (10e6 watts)?



The Attempt at a Solution



I used the idea that energy of falling water is found with mgh. I found that 1 kg of water falling 50 meters gives off 490.5 joules, so i used (10e6)/(490.5) and got 20387.35984 kg's. The site keeps telling me that answer is wrong and I have no idea what I'm not doing right.
 
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mgh = 1 * 9.8 * 50 = 490J/kg
So 10,000,000 W/ 490J/kg/s = 20 400 kg/s

Seems right, a rather optimistic number of significant figure though,
 
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