Question 9.4: Thermal Efficiency of the Ericsson Cycle Using an ideal gas ...

It is to be shown that the thermal efficiencies of Carnot and Ericsson cycles are identical.
Analysis Heat is transferred to the working fluid isothermally from an external source at temperature T_{H} during process 1-2, and it is rejected again isothermally to an external sink at temperature T_{L} during process 3-4. For a reversible isothermal process, heat transfer is related to the entropy change by

q=T \Delta s

The entropy change of an ideal gas during an isothermal process is

\Delta s=c_{p} \ln \frac{{T_{e}}^{\nearrow ^0} }{T_{i}}-R \ln \frac{P_{e}}{P_{i}}=-R \ln \frac{P_{e}}{P_{i}}

The heat input and heat output can be expressed as

q_{ in }=T_{H}\left(s_{2}-s_{1}\right)=T_{H}\left(-R \ln \frac{P_{2}}{P_{1}}\right)=R T_{H} \ln \frac{P_{1}}{P_{2}}

and

q_{\text {out }}=T_{L}\left(s_{4}-s_{3}\right)=-T_{L}\left(-R \ln \frac{P_{4}}{P_{3}}\right)=R T_{L} \ln \frac{P_{4}}{P_{3}}

Then the thermal efficiency of the Ericsson cycle becomes

\eta_{\text {th, Ericsson }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{R T_{L} \ln \left(P_{4} / P_{3}\right)}{R T_{H} \ln \left(P_{1} / P_{2}\right)}=1-\frac{T_{L}}{T_{H}}

since P_{1}=P_{4} \text { and } P_{3}=P_{2} . Notice that this result is independent of whether the cycle is executed in a closed or steady-flow system.