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Connecting two Carnot engions- please help

  1. May 24, 2009 #1

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    1. The efficiency of a cyclic engine is given by the work done by the engine, divided
    by the heat intake from the surroundings. One Carnot engine drives another in series. Draw a schematic diagram of the combined engine and give an expression for the overall efficiency of this arrangement which contains only the efficiencies of the individual engines.

    3. Hi, I have managed to draw the diagram but am really struggling with the next bit of the question. Let e1, e2 be the efficiencies of engions 1 and 2 respectively and let e be the overall efficiency.

    I know that e[tex]\ 1=w1/Q_H and e2=w2/(Q_h-w1)[/tex] but I am struggling to find e in terms of e1 and e2, please help.
     
    Last edited: May 24, 2009
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  3. May 24, 2009 #2

    Mapes

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    Try writing the efficiency of the two engines in combination, then express the variables in terms of [itex]e_1[/itex] and [itex]e_2[/itex].
     
  4. May 24, 2009 #3

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    What do you mean by writing them "in combination"?

    I get that overall e= (w1+w2)/Qh = e1+w2/Qh but I can't get w2/Qh solely in terms of e2, any ideas?
     
  5. May 24, 2009 #4

    Mapes

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    Keep trying, it's possible.
     
  6. May 24, 2009 #5

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    Ok, here we go!

    w2/Qh= (|Qh|-|w1|-|Qc|)/|Qh|=(-|Qc|-|w1|)/|Qh| +1 =1-e1-|Qc|/|Qh|

    Therefore (w1+w2)/Qh= 1-|Qc|/|Qh| =(|Qh|-|Qc|)/|Qh| = (w1+w2)/Qh and I am back to

    the beggining again. Can you see what I should be doing differently? I can't see any other ways of trying to manipulate the

    expression.
     
    Last edited: May 24, 2009
  7. May 24, 2009 #6

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    Are there any relevant formulas I have not tried in my above attempt?
     
  8. May 24, 2009 #7

    Mapes

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    Nope, it's just a matter of manipulating the equations you already have.
     
  9. May 24, 2009 #8

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    Could you please be a bit more specific? (I Have been stuck on this for a while and so I think I need a push in the right direction). I don't think I will get there otherwise.
     
    Last edited: May 24, 2009
  10. May 24, 2009 #9

    Mapes

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    What's the relationship between the heat input to the first engine and the heat input to the second engine?
     
  11. May 24, 2009 #10

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    let Qh(1), Qh(2) be the heat inputs for engions 1 and 2 respectively.

    Qh(2)=Qh(1)-w1. Is this right?

    I thought I was already using that when I said e2= w2/(Qh-w1).
     
  12. May 24, 2009 #11

    Mapes

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    Right; can't you go straight from

    [tex]e=\frac{w_1+w_2}{Q_{h,1}}[/tex]

    to something containing just [itex]e_1[/itex] and [itex]e_2[/itex]? Where are you getting stuck?
     
  13. May 24, 2009 #12

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    This is what I keep doing and where I am getting stuck.

    e=[tex]\frac{w1+w2}{Qh(1)}[/tex]

    e=e1+[tex]\frac{w2}{Qh(1)}[/tex]

    e=e1+[tex]\frac{w2}{Qh(2)+w1}[/tex]

    e=e1+e2+[tex]\frac{w2}{w1}[/tex]

    This is where I get stuck.

    I can't get rid of the w2/w1 pluss this seems wrong as two engions of efficiency=0.5 would violate the second law if this formula holds.
     
  14. May 24, 2009 #13

    Mapes

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    Then try something else! I tried a few approaches before something worked. I don't mean to be unhelpful, but just giving you the answer would deprive you of the experience of working it out yourself.
     
  15. May 24, 2009 #14

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    Ok, I will try new approaches but just out of interest the comment I made at the end of my last post about the second law do you agree with it? If so what have I done wrong (though it may not show the desired result I still thought my Algebra was correct)?
     
  16. May 24, 2009 #15

    Mapes

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    No, I don't agree;

    [tex]\frac{A}{B+C}\neq\frac{A}{B}+\frac{A}{C}[/tex]
     
  17. May 24, 2009 #16

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    Oh, yeah, my Algebra was wrong afterall. Sorry about this but I do not actually know what else I can try, any further hints?
     
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