Definition of efficiency of a thermodynamic cycle

[greypilgrim]

Hi.

Sorting the exchanges of heat and mechanical work in a thermodynamic cycle by the signs and summarizing, I get
##Q_{in}>0##: heat transferred into the system
##Q_{out}<0##: heat transferred to the cold reservoir
##W_{out}<0##: work done by the system
##W_{in}>0##: work done on the system (e.g. by moving back a piston)

The net work is ##W=W_{out}+W_{in}<0## (if it's a heat engine).
Normally, the efficiency is defined by
$$\eta=\frac{|W|}{Q_{in}}=\frac{|W_{out}+W_{in}|}{Q_{in}}\enspace .$$

But wouldn't it be more reasonable to use
$$\eta=\frac{|W_{out}|}{Q_{in}+W_{in}}\enspace ,$$
i.e. only the work actually done by the system divided by all energy put into the system?

 

[mehul mahajan]

usually when we talk about heat engine, it gets all the energy needed in the form of heat, so the formula η= W/Qin is used, but in your case i think that the formula that you gave should be used.(i'm not sure)

 

[DrClaude]

but in your case i think that the formula that you gave should be used.(i'm not sure)

That's not correct. The numerator is always taken as net work.

But wouldn't it be more reasonable to use

η=|Wout|Qin+Win,​

\eta=\frac{|W_{out}|}{Q_{in}+W_{in}}\enspace ,
i.e. only the work actually done by the system divided by all energy put into the system?

While this makes sense in the overall concept of efficiency, it doesn't work very well when applied to heat engines, where one is mostly concerned by their ability to convert heat into work. Using your definition, a transmission belt could be seen as a heat engine with an efficiency of almost unity. Taking net work divided by input heat allows to concentrate on the conversion process.

 

[greypilgrim]

Indeed, most often Win=0W_\mathrm{in} = 0, so both forms would be equivalent.

I'm not sure if we are talking about different things, but I can't think of a single case where this is true. Every compression of a gas (which is present in virtually any thermodynamic cycle) is work done on the gas.

Only because ##W=W_{out}+W_{in}## has a negative part ##W_{out}<0## and a positive part ##W_{in}>0##, its absolute value ##|W|## corresponds to the area enclosed by the cycle in the PV diagram.

 

[DrClaude]

I'm not sure if we are talking about different things, but I can't think of a single case where this is true. Every compression of a gas (which is present in virtually any thermodynamic cycle) is work done on the gas.

I completely misspoke. I have corrected my post.