- #1
davidbenari
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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 the system because of 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}##.
In case it's not clear yet, I'm saying ##Q_c+Q_h## is the same as ##-(W{out}+W{in})## where ##W_{out}<0## and ##W_{in}>0##.
Consider the 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.
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 the system because of 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}##.
In case it's not clear yet, I'm saying ##Q_c+Q_h## is the same as ##-(W{out}+W{in})## where ##W_{out}<0## and ##W_{in}>0##.
Consider the 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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