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Efficiency PowerPlant

  1. Oct 30, 2014 #1

    rlc

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    1. The problem statement, all variables and given/known data
    A steam-electric power plant delivers 900 MW of electric power. The surplus heat is exhausted into a river with a flow of 5.51×105 kg/s, causing a change in temperature of 1.35 oC.
    What is the efficiency of the power plant?
    What is the rate of the thermal source?

    2. Relevant equations
    Q=cm(delta T)
    c=4.186 J/g*Celsius

    3. The attempt at a solution
    I know how to answer the first question already:
    Q=(4186 J/kg C)(5.51E5 kg/s)(1.35 C)
    Q=3113756100 J/s (J/s=Watt)
    Convert from Watt to Megawatt --> 3113.76 MW
    900+3113.76=4013.76=Total output
    900/total output=0.2242=22.42 %

    I don't know how to calculate the rate of the thermal source.
    What am I missing?
     
  2. jcsd
  3. Oct 30, 2014 #2

    NTW

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    Input power = 4013,76 MW = 900 MW + 3113,76 MW

    That's the input, i.e., the power delivered by the high-temp source to the machine, and not the output.

    The original, high-temp power (4013,76 MW) is transformed by the machine in 900 MW of electricity and 3113,76 of 'rejected', low-temp power.

    The calculation of the efficiency is right: 900/Total input
     
  4. Oct 30, 2014 #3

    rlc

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    That makes sense, but how do you calculate the rate of the thermal source?
     
  5. Oct 30, 2014 #4

    NTW

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    I don't know what you mean by that. I suspect that it may be the rate of energy delivery of the thermal source. That's the power.
     
  6. Oct 30, 2014 #5

    rlc

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    Yeah, the question was asked in a weird way.
    I tried 4013 MW and it said that that is correct for the rate.

    Thank you for helping me!
     
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