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Homework Help: Seawater power plant efficiency

  1. Sep 25, 2013 #1
    1. Problem statement
    A power plant generates electricity from the difference in sea temperature. At the surface of the sea the temperature is 27°C. At the bottom of the sea the temperature is 6°C.

    a) What is the highest efficiency that this plant can have?
    b) If the power plant generates 210 W of power, how much heat per time must be released by the hot sea water? How much heat per time is released into the cold sea water?

    2. Equations
    The Carnot efficiency: efficiency = 1-(Tc/Th)
    P[supplied] = P[output]/efficiency
    P[wasted] = P[supplied] – P[output]

    3. Attempt at a solution
    The idea here is that the power plant uses heat from the hot sea water to run a heat engine. The heat engine turns the heat into mechanical work to generate electricity. The heat engine is not efficient enough, so some heat energy is wasted. This wasted heat energy is used to heat the cold sea water.

    In a) I used The Carnot efficiency: efficiency = 1-(Tc/Th)

    Tc = temperature in kelvin of cold reservoir = 273.15+6
    Th = temperature in kelvin of hot reservoir = 273.15+27

    Efficiency = 0.0699 = 7%

    In b) to find heat per time released by the hot sea water I used:

    P[supplied] = P[output]/efficiency

    P[output] = 210 W
    Efficiency = 1-(Tc/Th) = 0.0699

    P[supplied] = 3043.478 W = 3 KW

    To find heat per time released into the the cold sea water I used:

    P[wasted] = P[supplied] – P[output]

    P[wasted] = 3043.478 W – 210 W = 2833.478 W = 2.8 KW
    Last edited: Sep 25, 2013
  2. jcsd
  3. Sep 25, 2013 #2


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    Staff: Mentor

    I think that you typed in a number incorrectly on your calculator, as that number is about 1% off.

    Otherwise, did you have a question?
  4. Sep 25, 2013 #3
    Yes I meant 0.0699. I used 0.069. I'm not sure if my solution is correct, does something seem wrong in my solution? Can I use carnot efficiency equation in this case. I only know the temperatures of the reservoirs, so I'm guessing I have to use the Carnot efficiency formula.
  5. Sep 25, 2013 #4


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    Staff: Mentor

    Everything looks fine. Since you are not given more information, you must assume that the plant works on a Carnot cycle, otherwise you would be inventing numbers! You can also formulate the answer to say that the plant must receive at least 3.0 kW from the hot water and release at least 2.8 kW to the cold water, as these are the minimum values based on an ideal heat engine.
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