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Engine efficiency

  1. Apr 7, 2009 #1
    process A: isothermal compression at low temperature with an input of work of 83 J
    process B: constant volume increase in pressure with an energy input by heating of 200 J
    process C: isothermal expansion at high temperature with work output of 139 J
    process D: constant volume cooling to the original pressure, volume and temperature

    In this cycle, the energy input in process B is the same as the energy rejected in
    process*D.

    Complete the table by applying the first law of thermodynamics to each process and to the whole cycle. Which I have done:

    http://img12.imageshack.us/img12/8733/21144137.jpg [Broken]
    http://g.imageshack.us/img12/21144137.jpg/1/ [Broken]

    The highest and lowest temperatures of the air during the cycle are 500 K and 300 K. Show that the thermal efficiency of the ideal cycle is equal to the maximum possible efficiency for any heat engine working between these temperature limits.


    For this part, I am fine with calculating the maximum possible efficiency using the temperatures given. However, the markscheme for finding the actual thermal efficiency of the ideal cycle says:

    http://img10.imageshack.us/img10/3898/15977913.jpg [Broken]
    http://g.imageshack.us/img10/15977913.jpg/1/ [Broken]


    Can someone please explain the reasons for using 56 and 139 - I would have thought that 139J of work is done, since this much work is done during the isothermal expansion, with the input of 200 + 139 = 339J, since this is the total energy input. Can someone please explain what is wrong with this reasoning.

    Thank you very much.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
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