Connecting two Carnot engions-

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In summary, the conversation is about finding the overall efficiency of a combined engine made up of two Carnot engines in series. The efficiency of the combined engine is expressed in terms of the efficiencies of the individual engines, e1 and e2. The conversation discusses various attempts at manipulating the equations to find the desired expression, with the final result being e = e1 + e2 + (w2/w1). However, this formula is incorrect and the participants continue to try different approaches to find the correct expression for the overall efficiency.
  • #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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  • #2
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
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
Keep trying, it's possible.
 
  • #5
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
Are there any relevant formulas I have not tried in my above attempt?
 
  • #7
Nope, it's just a matter of manipulating the equations you already have.
 
  • #8
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
What's the relationship between the heat input to the first engine and the heat input to the second engine?
 
  • #10
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
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
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
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
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
No, I don't agree;

[tex]\frac{A}{B+C}\neq\frac{A}{B}+\frac{A}{C}[/tex]
 
  • #16
Oh, yeah, my Algebra was wrong afterall. Sorry about this but I do not actually know what else I can try, any further hints?
 

Related to Connecting two Carnot engions-

What is a Carnot engine and how does it work?

A Carnot engine is a theoretical thermodynamic cycle that operates between two heat reservoirs at different temperatures. It consists of four processes - isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. The engine works by converting heat energy into mechanical work, and it operates on the principle of maximum efficiency.

Can two Carnot engines be connected?

Yes, two Carnot engines can be connected in series or in parallel to increase their efficiency. However, the overall efficiency of the connected engines will still be limited by the efficiency of the lower performing engine.

How do you determine the efficiency of connected Carnot engines?

The efficiency of connected Carnot engines can be determined by taking the product of the individual efficiencies of the engines. For example, if two Carnot engines with efficiencies of 50% and 60% are connected in series, the overall efficiency will be 30% (0.5 x 0.6 = 0.3).

What is the maximum possible efficiency of connected Carnot engines?

The maximum possible efficiency of connected Carnot engines is determined by the temperature of the two heat reservoirs. It is given by the Carnot efficiency formula: (Thot - Tcold) / Thot, where Thot is the temperature of the hot reservoir and Tcold is the temperature of the cold reservoir.

Are there any practical applications for connected Carnot engines?

Connected Carnot engines are used in various industrial processes, such as power generation and refrigeration. They are also used in heat pumps for heating and cooling buildings. However, due to their theoretical nature, they are not commonly used in everyday applications.

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