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Homework Help: Waterfall power, work

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    A waterfall is 85ft high, and 3000 cubic ft/sec flow over it driving a water wheel connected to an electric generator. If the overall efficiency is 22%, how many kilowatts does the generator develop?

    2. Relevant equations
    density of water =1000kg/cubic meter
    P.E. = mgh
    weight=density*g*volume

    3. The attempt at a solution
    Firstly convert to SI units.
    85ft = 25.91m
    3000 cubicft/sec = 914.4 cubic meter/sec

    weight of the water being applied = (1000)(9.81)(914.4)
    = 8.97 MN

    P.E.=(8.97 MN)(25.91m)
    =232.42 MJ
    power in = 232.42 MJ /1sec = 232.42 megawatts

    eff =0.22

    0.22 = POUT/232.42 MW

    Are my steps correct?
     
  2. jcsd
  3. Nov 6, 2007 #2

    learningphysics

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    I'm getting 3000 cubicft/sec = 84.95 cubic m/s

    other than this, everything looks right to me. be sure to convert to kW when you finish...
     
  4. Nov 6, 2007 #3
    how do you do a conversion such as 3000 cubic ft to cubic meter?

    Yea, I realized i just did a ft-m conversion instead of cubic ft to cubic meter (a function on my calc).
     
  5. Nov 6, 2007 #4

    learningphysics

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    I actually just used google.com. I typed in 3000 ft^3 and it automatically converted it.

    But we can do it this way:

    we know that 1 ft = 0.3048m

    (1ft)^3 = (0.3048m)^3

    so 1ft^3 = 0.0283168466 m^3

    so 3000 ft^3 = 84.95 m^3
     
  6. Nov 6, 2007 #5
    amazing...thanks a lot learningphysics. You are truly an asset to this board. I have another test this friday. Hope I do well.
     
  7. Nov 6, 2007 #6

    learningphysics

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    thanks so much. I appreciate it! good luck on your test! you'll do great!
     
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