Calculation of the efficiency of an engine

  • #1
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Main Question or Discussion Point

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) ?
 

Answers and Replies

  • #2
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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.
 
  • #3
russ_watters
Mentor
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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) ?
It's neither. Work output already does not include the compression energy. You are subtracting it twice.
 
  • #4
CWatters
Science Advisor
Homework Helper
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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.
 
  • #5
Randy Beikmann
Gold Member
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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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