Calculation of the efficiency of an engine

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SUMMARY

The efficiency (η) of an engine, specifically in the context of the Otto cycle, is defined as the net output work divided by the heat input. The correct formula for calculating thermal efficiency is (W_e - W_c)/Q_h, where W_e represents the work of expansion, W_c is the work of compression, and Q_h is the heat added from the fuel. It is crucial to understand that work output does not include compression energy, and defining input power can vary, as seen in UK gas home heating boilers that can exceed 100% efficiency due to historical definitions. Efficiency is fundamentally a measure of useful work output versus work input, emphasizing its subjective nature.

PREREQUISITES
  • Understanding of thermodynamics principles, particularly the Otto cycle
  • Familiarity with the concepts of work output (W_o), work of compression (W_c), and heat input (Q_h)
  • Knowledge of energy forms, specifically the distinction between work and heat
  • Awareness of efficiency definitions and their implications in engineering contexts
NEXT STEPS
  • Research the principles of thermodynamic cycles, focusing on the Otto cycle and its efficiency calculations
  • Explore the concept of energy forms and their implications in mechanical systems
  • Investigate the historical definitions of efficiency in various applications, including heating systems
  • Learn about advanced efficiency metrics and their relevance in modern engine design
USEFUL FOR

Engineers, mechanical designers, and students studying thermodynamics who seek to understand engine efficiency calculations and the implications of energy definitions in practical applications.

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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