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Hydroelectric power plant

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data

    upload_2016-10-18_20-24-9.png

    2. Relevant equations


    3. The attempt at a solution

    What I attempted to do is

    .80(1/2)mv2 = Ee

    Where v is the velocity of the falling water, m is the mass of the falling water, and Ee is the electrical power generated by the electric plant.

    So then I solved for m since you are given the Ee and the velocity and I got 3.86 x 10e5 kg/s.
     

    Attached Files:

  2. jcsd
  3. Oct 18, 2016 #2

    kuruman

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    Your number is correct. To be strictly correct you should use dm/dt in your equation instead of m.
     
  4. Oct 18, 2016 #3

    haruspex

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    But then it would be dE/dt on the other side, assuming E stands for energy, not power.
    The OP equation can be justified by prefixing it with "in the time in which a mass m of water passes through".
     
  5. Oct 19, 2016 #4

    kuruman

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    If E stands for energy, .80(1/2)(dm/dt)v2 = dE/dt is correct.
    If Ee stands for power (as indicated by OP in attempt at solution), .80(1/2)(dm/dt)v2 = Ee is correct.
     
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