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Heat: heat input

  1. Jun 28, 2003 #1

    Dx

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    A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

    is 1.8Kw right?

    e=W/q_h
    .30 * 600

    TY
    dx:wink:
     
  2. jcsd
  3. Jun 28, 2003 #2

    Tom Mattson

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    No, that would correspond to an efficiency of 0.33...
     
  4. Jun 28, 2003 #3

    Dx

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    Is my equation wrong, what am i doing incorrectly tom?
    Dx
     
  5. Jun 28, 2003 #4

    Tom Mattson

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    I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input.
     
  6. Jun 28, 2003 #5

    Dx

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    yourt the man Tom! Thats what my symbols mean. e= efficiency, W=work ouot/Q_h=work in. so why did i get this wrong, tom.
    Im lost in the math and dont know where? plz help?
    Dx
     
  7. Jun 29, 2003 #6

    HallsofIvy

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    You might be better off to take Tom's answers with a grain of salt and check them carefully. He seems to be doing the same thing you are: taking the numbers given and putting them together in the simplest way without regard for the correct formula.


    In this case, the formula you give is: e=W/q_h and the values are efficiency= e= 0.30, Work out= W= 600.

    In other words, 0.30= 600/q_h. Solving that, 0.30*q_h= 600 so
    q_h= 600/0.30= 2000 Joules.

    It should occur to you that, since efficiency is always less than 1, there must be more energy put in that you get out.

    .30 * 600
     
  8. Jun 30, 2003 #7

    Tom Mattson

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    No, Dx was quoting himself (not me) in his post above. The only "answer" I gave was...

    "I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input."

    ...which is no different from what you posted.
     
  9. Jun 30, 2003 #8
    Speaking of grains of salt,
    600 watts/30% = 2,000 watts, not joules. :smile:
     
  10. Jun 30, 2010 #9
    can anybody please tell me the formula for heat rate for fuels?
     
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