Explain how real machines can never recover the full amount of energy they were supplied with?.

1 Answer
Nov 23, 2016

Sounds like efficiency to me. In general, efficiency is defined as the work output acquired from some work input.

where:

  • #w# is work.
  • #q_H# is the heat flow into the engine from the hot reservoir.
  • #q_C# is the heat flow from the engine into the cold reservoir.
  • #T_H# is the temperature of the hot reservoir.
  • #T_C# is the temperature of the cold reservoir.

The ideal machine can achieve #100%# efficiency, and the work it performs is #100%# reversible. Real machines, however, can never quite achieve #100%# efficiency, and the work they do is nearly #100%# reversible.

As I derived above, a cyclic process has a change in entropy of #DeltaS = 0#, and if heat flow #q# is reversible, then #DeltaS = q_"rev"/T#.

If we separate the cyclic process into two processes, then

#q_H/T_H + q_C/T_C = 0#

Also from the above, I calculated that

#color(blue)(e) = |w|/q_H = (q_H + q_C)/q_H = color(blue)(1 + q_C/q_H)#.

#q_C# flows out of the engine, and thus, #q_C <= 0#. When #q_C = 0#, the efficiency is #100%#.

Or, using the first relation:

#color(blue)(e) = |w|/q_H = (q_H + q_C)/q_H = color(blue)(1 - T_C/T_H)#.

So, we could even say that since all absolute temperatures are positive (i.e. when in #"K"#), as they are in this equation, #0 <= e <= 1#. When #T_C = T_H#, #e = 0#, and when #T_C = "0 K"# (which has yet to be accomplished!), or #T_H# is absurdly large, #e ~~ 1#.

Therefore, the efficiency can never be more than #100%#, and for real machines, the efficiency is effectively never exactly #100%#.