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Heat Engine

  1. Sep 14, 2005 #1
    At a steam power plant, steam engines work in pairs; the rejected heat from the first is the heat input to the second. The operating temperatures of the first are 670 C and 440 C. The operating temperatures of the second are 440 C and 290 C. Assume both engines work at 60% of their Carnot efficiency. The heat of combustion of coal is 2.8 x 10^7 J/kg. How many kg of coal must be burned per day if the power output of the plant is to be 1000 MW?

    I know that the QL of the first engine is the QH of the second engine, but how do I get the actual values with only the efficiency and the ratio of the temperatures?
  2. jcsd
  3. Sep 16, 2005 #2
    Anyone have an idea?
  4. Sep 16, 2005 #3


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    You have more than just the ratio of temperatures, you have their actual values. Does this formula look familiar?


    That's the efficiency of a perfect Carnot cycle. Using this and what they gave you in the text, you should be able to figure out the efficiency of the machines themselves.

    Once you have that, it's just a matter of using the definition of efficiency to get your answer. Here's another formula that might come in handy


    Remember, efficiency is just the ratio of the energy you get out to what you put in.
  5. Sep 16, 2005 #4


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    Alright, the both engines operate at 60% of their Carnot efficiency. You know this to be equal to:
    You know the hot and cold sink temperatures, so you can find the actual thermal efficiency of the engines by multiplying the Carnot efficiency by .6. You also know the thermal efficiency to be defined as:
    You know the power output you want to be 1000MW, which is 1000MJ in one second, or 8.64*10^7J per day. This is w. All you need to do is find an expression for the amount of work that comes from a given amount of heat input. To do this, you need to know how much energy is rejected by the first engine and put into the second. This comes from conservation of energy. It is just the difference between the heat supplied and the work performed.
  6. Sep 16, 2005 #5


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  7. Sep 17, 2005 #6


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    Correction on this.

    Wi = QH,i - QL,i is the ideal case with eff = 100%

    if eff. < 100%, the Wi = [itex]\eta_i[/itex] (QH,i - QL,i), where [itex]\eta_i[/itex] is the efficiency.
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