In the first law of thermodynamics, We write the change in internal energy in terms of heat exchanged and work done. The answer to the above question doesn't matter if our final aim is to find the change in internal energy. However, When calculating the efficiency of an irreversible cycle, We always write difference of the heat taken in and heat rejected to be the work done by the system. In this, are we also including the work that is being done against friction?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Work done by friction leads to a loss of energy $\endgroup$– QuIcKmAtHsCommented Oct 14, 2018 at 8:06
-
$\begingroup$ It depends on whether the frictional work occurs within the "system" containing the working fluid, or whether it occurs outside of the system, and is part of the work done on the surroundings. $\endgroup$– Chet MillerCommented Oct 14, 2018 at 12:01
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Work done by friction is still work rather than heat. It is worth noting however that it is an irreversible process, not a reversible form of work.
In terms of efficiency, however, we are normally interested in "useful" work done by the system, or work done by the system on some other external system and so would not count work done against friction.
answered Oct 14, 2018 at 11:12
-
$\begingroup$ When we derive clausius inequality we consider the work done in that irreversible cycle as Q(high)-Q(low) for the work done by system, Why don't we also substract the work done by friction? If we only want to take into account the "useful work". $\endgroup$– JayCommented Oct 14, 2018 at 17:46
-
$\begingroup$ I am not totally sure what you are getting at. Could you explain the proof of Clausius' theorem that you are looking at in more detail? $\endgroup$ Commented Oct 14, 2018 at 19:23