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Heat Engine Question

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data


    18. (II) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate ehat input of the second. Th eoperating temeprautres of the first are 680 degrees C and 430 degrees C, and of the second 415 degrees C and 280 degrees C. If the heat of combuston of coal is 2.8 E8 J/kg, at what rate msut coal be burned if the plant is to put out 900 MW of power. Assume the efficiency of the engines is 65 percent of the ideal (Carnot) efficiency.


    2. Relevant equations

    e = |W|/|QH|

    were the H is the subscript

    e ideal = 1-TL/TH

    were L and H are subscripts

    3. The attempt at a solution

    Ok I calculated the efficiency of the first engine to be about 17.05 % efficient and the second one to be about 14.64 % efficient.

    I know that efficiency is

    e = |W|/|QH|

    were the H is the subscript

    I know that 900 MW is a value of power which is not work

    I know that |QH| is 2.8 E8 J/kg but I do not like the units for this, J/kg, as I thought Q was measured in joules only...

    How do I find the work done?
     
  2. jcsd
  3. Dec 16, 2009 #2

    ideasrule

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    Consider one second of time. 900 MJ of energy is produced, so |W|=900 MJ. e is the total efficiency of the entire system, first and second engine combined. Once you get |QH|, remember that this is the energy used per second.
     
  4. Dec 16, 2009 #3
    but it's the power done not the work
    900 MW of power
    hmmmm ok hold up
     
  5. Dec 16, 2009 #4

    ideasrule

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    I know, but you can always consider the energy expended in a single second of time. After all, that's what power is: energy expended in a single second of time!

    If you want, you can rearrange e = |W|/|QH| into |QH| * e = |W|. Take the time derivative of both sides and you get e*d(|QH|)/dt = d|W|/dt: in other words, the rate of heat usage times e is equal to the power generated. If you don't know what a time derivative is or what I'm talking about, don't worry about it; you aren't required to know this.
     
  6. Dec 16, 2009 #5
    ok so I got for |QH| 2816 E 7 J were do I go from here
    hmmm hold up
     
  7. Dec 16, 2009 #6
    Ok so I'm doing this wrong lets see here
     
  8. Dec 16, 2009 #7
    e ideal = 1 - TL/TH = 1 - 703.16 K/953.16 K = .2623 [first engine]
    e = .65 (e ideal) = .65(.2623) = .1705 [first engine]

    e ideal = 1 - TL/TH = 1 - 533.16 K/688.16 K = .2252 [second engine]
    e = .65 (e ideal) = .65 (.2522) = .1639 [second engine]
     
  9. Dec 16, 2009 #8
    e = .1705 + .1639 = .3344 [system]
    P = W/t = 900 MW = W/1 s
    then
    W = 900 MW (1 s)
    seconds cancle out I'm getting
    900 MJ for |W|
     
  10. Dec 16, 2009 #9

    ideasrule

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    e ideal = 1 - TL/TH = 1 - 533.16 K/688.16 K = .2252 [second engine]

    533 K corresponds to 260 C. Is it 260 or 280?
     
  11. Dec 16, 2009 #10
    e = |W|/|QH|
    then
    |QH| = |W|/e = (900 MJ)/.3344 = 2691 J
     
  12. Dec 16, 2009 #11

    ideasrule

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    That's not right. Suppose an amount of energy, Q, goes into the first engine. Q*e_1 gets spat out as work and goes into the second engine. Then, Q*e_1*e_2 gets spat out as work. So the total efficiency is e1*e2, not e1+e2.
     
  13. Dec 16, 2009 #12
    It was 280 in the problem
     
  14. Dec 16, 2009 #13
    ok intersection and union...
    makes sense lets see now then...
     
  15. Dec 16, 2009 #14

    ideasrule

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    OK, so it's 553 K and not 533 K.
     
  16. Dec 16, 2009 #15
    e = e_1 intersection e_2 = (.1705) .1639 = .0279 [system]
    that does not seem correct :O
     
  17. Dec 16, 2009 #16

    ideasrule

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    Well, you have to repeat the calculation for e2 because, as I was saying, you used 533 K for the temperature when it should be 553 K. Other than that, you calculation is correct. You're right that this efficiency is unrealistically low; Google tells me that coal-fired plants typically have an efficiency of 30-50%.
     
  18. Dec 16, 2009 #17
    e = |W|/|Q_H|
    then |Q_H| = |W|/e = 500 MJ/.0279 = 1.792 E 10 J
     
  19. Dec 16, 2009 #18
    I'm lost...

    ok the T_L for the second engine was 280 degrees Celceius

    Kelvin = degrees Celceius + 273.16
    = 280 degrees Celceius + 273.16
    = 533.16 K
    What am I doing wrong here?
     
  20. Dec 16, 2009 #19
    1.792 E 10 J /(2.8 E 7 J/Kg)

    I'm getting 640 Kg of coal every second?
     
  21. Dec 16, 2009 #20
    Chapter 20: 18





    18.First, find the efficiencies of each engine. That’ll make things easier later and help you fill in more information on your diagram. You should find an efficiency of 17.1% for the first engine and an efficiency of 12.8% for the second. Worth noting is the fact that the output of the first engine is the input of the first (or, QL1 = QH2). That means that the work done by the first engine is 17.1% of the energy fed into the first engine, and the work done by the second engine is 12.8% of the wasted heat from the first engine. If your diagram doesn’t make sense at this point, eMail me.

    The next trick is an algebraic one. You already have the work from the first engine in terms of the input heat (it’s the input heat times the efficiency of course). Now you need to express the work done by the second engine in terms of the input heat of the first engine. You can do this by first expressing the heat input of the second engine in terms of the input heat of the first (so relate the exhaust of the first engine to its input; the first exhaust equals the second intake).

    Now you should have two expressions for the output work, both in terms of the input heat of the first engine. That’s good, because you know the output work (per second) is 600MW. From this, you should be able to calculate the input heat. Your answer for this should be in the neighborhood of 3.25 gigajoules per second.

    Knowing the energy density of the fuel, you can easily find out how many kilograms of fuel are needed to supply the appropriate energy per second. It’s a lot, so don’t panic. If you followed all this through, you should have an answer to the tune of 116 kilograms per second.
     
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