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Energy of falling water

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    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)?



    3. 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.
     
  2. jcsd
  3. Oct 5, 2009 #2

    mgb_phys

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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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