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I Definition of efficiency of a thermodynamic cycle

  1. Mar 28, 2017 #1
    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?
     
  2. jcsd
  3. Mar 28, 2017 #2
    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: Mar 28, 2017
  4. Mar 28, 2017 #3

    DrClaude

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

    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: Mar 28, 2017
  5. Mar 28, 2017 #4
    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.
     
  6. Mar 28, 2017 #5

    DrClaude

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    Staff: Mentor

    I completely misspoke. I have corrected my post.
     
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