Show that the work performed by an engine during an irreversible cycle operating between two thermal reservoirs at temperatures T 1 and T 2 &lt; T 1 is given by W = W m a x − T 2 Δ S, where Δ S is the increase of entropy of the Universe, and W m a x is the corresponding work performed in a reversible Carnot cycle.

Question

Show that the work performed by an engine during an irreversible cycle operating between two thermal reservoirs at temperatures       T          1       and       T          2        &amp;lt;      T          1       is given by   W  =      W          m      a      x        −      T          2        Δ  S, where   Δ  S is the increase of entropy of the Universe, and       W          m      a      x       is the corresponding work performed in a reversible Carnot cycle.

Evelin Castillo · Accepted Answer

The basis for the model you presented is that (1) the system under consideration is the working fluid and (2) the reservoirs are ideal so that there is no entropy generated within them (and all the entropy due to irreversibility is generated within the working fluid). Since the engine is working in a cycle, the change in internal energy and the change in entropy of the working fluid in each cycle is zero. From this, it follows from the first law of thermodynamics (applied over a cycle) that      Q    1    −      Q    2    =  Wwhere Q1 is the heat received from the hot reservoir and Q2 is the heat rejected to the cold reservoir. From the second law of thermodynamics, it also follows that  0  =            Q      1              T      1        −            Q      2              T      2        +  σwhere   σ is the entropy generated within the working fluid per cycle (and also, by assumption (2), the entropy generated within the universe per cycle). This entropy generation is positive for an irreversible process or zero for a reversible process.If we eliminate Q2 from these equations, and solve for W, we obtain:  W  =      Q    1              (              T        1            −              T        2            )              T      1        −  σ      T    2  Since the entropy generation is positive or zero, the maximum work is if the entropy generation is zero:      W          m      a      x        =      Q    1              (              T        1            −              T        2            )              T      1      Therefore,  W  =      W          m      a      x        −  σ      T    2

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