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Connecting two Carnot engions- please help

  • Thread starter C.E
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  • #1
C.E
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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.
 
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Answers and Replies

  • #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].
 
  • #3
C.E
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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?
 
  • #4
Mapes
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Keep trying, it's possible.
 
  • #5
C.E
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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.
 
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  • #6
C.E
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Are there any relevant formulas I have not tried in my above attempt?
 
  • #7
Mapes
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Nope, it's just a matter of manipulating the equations you already have.
 
  • #8
C.E
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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.
 
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  • #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?
 
  • #10
C.E
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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).
 
  • #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?
 
  • #12
C.E
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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.
 
  • #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.
 
  • #14
C.E
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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)?
 
  • #15
Mapes
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No, I don't agree;

[tex]\frac{A}{B+C}\neq\frac{A}{B}+\frac{A}{C}[/tex]
 
  • #16
C.E
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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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