Definition of efficiency of a thermodynamic cycle

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Discussion Overview

The discussion revolves around the definition of efficiency in thermodynamic cycles, particularly in the context of heat engines. Participants explore different formulations of efficiency based on the work done by the system and the energy inputs, debating the implications of these definitions.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant proposes a definition of efficiency as $$\eta=\frac{|W_{out}|}{Q_{in}+W_{in}}$$, suggesting it accounts for all energy input into the system.
  • Another participant agrees that the proposed formula might be applicable but expresses uncertainty about its correctness in general contexts.
  • A different participant asserts that the numerator in efficiency calculations is typically taken as net work, emphasizing the focus on heat conversion in heat engines.
  • It is noted that in many cases, $$W_{in}$$ is zero, which could make both proposed efficiency forms equivalent.
  • Concerns are raised about the applicability of the new definition, particularly regarding its implications for understanding heat engines and their efficiency.
  • One participant reflects on their earlier statement, admitting a mistake and acknowledging the complexity of the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate definition of efficiency, with no consensus reached. Some agree on the traditional definition while others advocate for the alternative proposed formula, highlighting the ongoing debate.

Contextual Notes

Participants acknowledge that the definitions of work and energy inputs can vary based on the specific thermodynamic cycle being considered, and that assumptions about work done on or by the system may influence the discussion.

greypilgrim
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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?
 
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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)
 
Last edited by a moderator:
mehul mahajan said:
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.

greypilgrim said:
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.
 
Last edited:
DrClaude said:
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.
 
greypilgrim said:
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.
 

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