What is the Efficiency of a Heat Engine?

In summary: Wait, I just realized that you've been integrating from a point on path 31 back to the same point. So, the integral is not W31, it's -W31. This is because you're doing a clockwise loop: you're moving in the direction of increasing volume, but you're doing negative work: the gas is pushing the piston outwards, but you're defining work as the force exerted by the gas on the piston, and the piston is moving inward. So, on path 31, the gas is doing work on the environment, and W31 is negative. IOW, the gas is expanding, but dV is negative, so W = -p dV is positive. So
  • #36
I am also assuming that the nR can cancel, so:

[tex] P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})} [/tex]

Does this look right?
 
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  • #37
To be honest I have lost track. However, it would be desirable to simplify the term
v1 v1γ-1
The idea is, once you have an expression for the efficiency, for you to manipulate and simplify it algebraically and get

[tex]
e = 1 - \frac{1}{\gamma}\left(\frac{1 - \frac{p_3}{p_1}}{1 - \frac{v_1}{v_3}}\right)
[/tex]​

which, as I had mentioned earlier, has a typo corrected from what was given in Post #1 (v1/v3 instead of v3/v1).
 
  • #38
Oops, my Latex went a bit wrong. it should have been:

[tex] \frac{P_1(v_1\frac{v_1^{\gamma - 1}}{v_3^{\gamma - 1}} - V_3)}{V_3(P_1 - P_3 \frac{V_3^{\gamma - 1}}{v_1^{\gamma - 1}})} [/tex]
 

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