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Homework Help: Coal fired plant Thermodynamics help

  1. May 16, 2003 #1


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    A coal fired plant generates 600MW of electric power. The plant uses 4.8x10^6kg of coal a day. The heat of combustion is 3.3x10^7 J/kg. The steam that drives the turbines is at temp of 300 degrees C and the exhaust water is at 37 degrees C. Whar is the overall effiecy of the plant for generating electric power?

    I have a formula of and have substituted like so e = W/Q_H = (Q_H - Q_L)/ Q_H = 1- Q_L / Q_H.

    I came up with 37% is that correct?

    Dx :wink:
    Last edited by a moderator: May 18, 2003
  2. jcsd
  3. May 16, 2003 #2
    That's amazing! That plant generates enough power to light 6 bulbs, and uses ONLY 4.8 million kilograms of coal per day.

    Maybe they should consider switching to windmills.

    (I'm guessing it's supposed to be 600 megaWatts. Try again; show YOUR work, not just the formula you copied out of the book...)
    Last edited: May 16, 2003
  4. May 16, 2003 #3

    Tom Mattson

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    I don't have a calculator on me, but your method is right.
  5. May 17, 2003 #4
    It's a good formula, but how are you using it?

    Why not just:
    (600*10^3 kW * 24 h * 3.6 * 10^6 J/kWh)/(4.8*10^6 kg * 3.3*10^7 J/kg) = 32.7%

    (In this problem, aren't the steam and exhaust temperatures red herrings?)
  6. May 18, 2003 #5


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    Good Guess, Gnome!

    Thanks Tom!
    its right.
  7. May 18, 2003 #6
    So please enlighten me. Exactly how did you get 37%?
  8. May 18, 2003 #7


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    I did it just the way you showed me here, gnome.
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