Calculation of the efficiency of an engine

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Discussion Overview

The discussion centers around the calculation of engine efficiency, particularly in the context of the Otto cycle. Participants explore different formulations of efficiency, the definitions of work and heat input, and the implications of these definitions on the understanding of engine performance.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that efficiency can be defined as η = (WO - WC)/WH, while others suggest η = WO/(WH + WC).
  • One participant argues that the only useful definition of efficiency is useful work out divided by work in, emphasizing the subjective nature of what is considered "useful."
  • Another participant challenges the previous claims by stating that work output does not include the compression energy, suggesting that subtracting it leads to double counting.
  • A participant notes that defining input power can be complex, citing examples of gas home heating boilers in the UK that are historically defined as more than 100% efficient.
  • It is mentioned that in the context of the Otto cycle, there is no work input, only heat input (Q_h), and that thermal efficiency can be expressed as (W_e - W_c)/Q_h.

Areas of Agreement / Disagreement

Participants express differing views on the definition and calculation of efficiency, with no consensus reached on a single formulation. Multiple competing perspectives remain regarding the roles of work and heat in determining efficiency.

Contextual Notes

There are limitations in the definitions and assumptions regarding work and heat inputs, as well as the subjective nature of what constitutes useful work. The discussion also highlights the historical context affecting efficiency definitions.

vroy
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We know efficiency (η) = Woutput/Winput

Now in any cycle, for instance, the Otto cycle, let the total work output be WO, total heat added be WH & the total work done on the gas to compress it be WC.

Then, will efficiency be equal to (η) = (WO-WC)/WH ?
Or will it be equal to (η) = WO/(WH+WC) ?
 
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There are lots of internal things that happen in an engine. Compression is one. Friction is another.

The only useful definition of efficiency is useful work out divided by work in. Out and in refer to things external to the engine. Useful is also necessary, because for example heat leaves the engine and warms up the environment. That's energy out, but it is not useful.

In other words, efficiency is a human measure that you define to be useful for your purposes. In that respect, it is less about physics, and more about your opinions.
 
vroy said:
We know efficiency (η) = Woutput/Winput

Now in any cycle, for instance, the Otto cycle, let the total work output be WO, total heat added be WH & the total work done on the gas to compress it be WC.

Then, will efficiency be equal to (η) = (WO-WC)/WH ?
Or will it be equal to (η) = WO/(WH+WC) ?
It's neither. Work output already does not include the compression energy. You are subtracting it twice.
 
Defining the input power in the case of a liquid or gas fuel is not always straight forward. Here in the UK we have gas home heating boilers that are more than 100% efficient because of the way the input power was defined historically.
 
For an Otto cycle, there is no work input. There is heat input (Q_h) from the fuel. Work and heat are thermodynamically different forms of energy.

Taking the work of compression as W_c and the work of expansion as W_e the thermal efficiency is (W_e - W_c)/Q_h.

In words, it is the net output work divided by the heat input.
 

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