Thermodynamics

506 | ThermodynamicsBoth the Stirling and Ericsson cycles are totally reversible, as is the Carnotcycle, and thus according to the Carnot principle, all three cycles must havethe same thermal efficiency when operating between the same temperaturelimits:h th,Stirling h th,Ericsson h th,Carnot 1 T L(9–14)T HThis is proved for the Carnot cycle in Example 9–1 and can be proved in asimilar manner for both the Stirling and Ericsson cycles.EXAMPLE 9–4Thermal Efficiency of the Ericsson CycleUsing an ideal gas as the working fluid, show that the thermal efficiency ofan Ericsson cycle is identical to the efficiency of a Carnot cycle operatingbetween the same temperature limits.Solution It is to be shown that the thermal efficiencies of Carnot andEricsson cycles are identical.Analysis Heat is transferred to the working fluid isothermally from an externalsource at temperature T H during process 1-2, and it is rejected again isothermallyto an external sink at temperature T L during process 3-4. For areversible isothermal process, heat transfer is related to the entropy change byq T ¢sThe entropy change of an ideal gas during an isothermal process is¢s c p ln T e ¡0 R ln P eR ln P eT i P iThe heat input and heat output can be expressed asP iandq in T H 1s 2 s 1 2 T H aR ln P 2P 1b RT H ln P 1P 2q out T L 1s 4 s 3 2 T L aR ln P 4P 3b RT L ln P 4P 3Then the thermal efficiency of the Ericsson cycle becomesh th,Ericsson 1 q outq in 1 RT L ln 1P 4 >P 3 2RT H ln 1P 1 >P 2 2 1 T LT Hsince P 1 P 4 and P 3 P 2 . Notice that this result is independent ofwhether the cycle is executed in a closed or steady-flow system.Stirling and Ericsson cycles are difficult to achieve in practice becausethey involve heat transfer through a differential temperature difference in allcomponents including the regenerator. This would require providing infinitelylarge surface areas for heat transfer or allowing an infinitely long timefor the process. Neither is practical. In reality, all heat transfer processes takeplace through a finite temperature difference, the regenerator does not havean efficiency of 100 percent, and the pressure losses in the regenerator areconsiderable. Because of these limitations, both Stirling and Ericsson cycles