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The thermodynamic efficiency $\eta$ is calculated by $\eta= \frac{W_{out}}{Q_{in}}$

Using the first law of thermodynamics we usually say that $W_{out}$ is $Q_c+Q_h$, where $Q_c$ is the heat dissipated into a cold reservoir, and $Q_h$ is the heat absorbed by a hot reservoir. Both are measured within the system, such that $Q_c<0$ and $Q_h>0$

However I object to that. $W_{out}$ is not $Q_c+Q_h$. That calculation is simply the magnitude of net energy in the process due to work. Namely, its considering the $W_{input}$, done by the surroundings on the gas, to calculate the $W_{output}$.

Consider the a Carnot engine. I would say that the work output is the area underneath the expansion isotherm, and the expansion adiabat. All other works are done on the system and are not "outputs".

You might say that the other works are "negative outputs". But although this makes sense in a mathematical sense, it doesn't make practical sense that this should be a $W_{output}$.

I'm obviously wrong about this, so I'd like someone to clear things up for me. Thanks.

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  • $\begingroup$ For starters, you can tell us what $Q_c$ and $Q_h$ are. $\endgroup$ – garyp Feb 11 '15 at 18:33
  • $\begingroup$ Ok, I've done it on my 2nd paragraph. thx $\endgroup$ – DLV Feb 11 '15 at 18:40
  • $\begingroup$ Hmm. Not sure I follow ... Is the problem the terminology, and what exactly is meant by "work output"? Does it help to consider $W_{out}$ as being the net work, which for an engine is "output"? I think that's the intended meaning. $\endgroup$ – garyp Feb 11 '15 at 23:04
  • $\begingroup$ So I'm saying Qc+Qh is not the work coming out of the system. This calculation is considering work done on the system. The work coming out of the system or "output", shouldn't have anything to do with the work coming into it if it really is an "output". $\endgroup$ – DLV Feb 11 '15 at 23:36
  • $\begingroup$ Well, it is the net output, and I think that's the sense of the word that is intended. $\endgroup$ – garyp Feb 12 '15 at 2:45
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The Carnot cycle is ... well ... a cycle. In each turn of the cycle, the system returns to exactly the same state over and over again. Pick a point, call it the beginning of the cycle. After one turn of the cycle, the system is where it started, with exactly the same energy it had when it started. During the cycle, it has done some work. Where has that energy come from? It can't have come from the system, the system has exactly the same energy it had when it started. The energy can only have come from the heat reservoirs. $Q_h$ is removed from the hot side, $Q_c$ is deposited the cold side. The net energy, $Q_h + Q_c$ has been converted to work. The Carnot engine provides the mechanism for the transfer of the energy from heat to work.

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  • $\begingroup$ Hey, I don't think this is my question. My question regards the definition of $\eta$ and why $Qh+Qc$ is not a valid calculation of $W_{out}$ Thanks. $\endgroup$ – DLV Feb 11 '15 at 21:28

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