Thermal Efficiency of the Ericsson Cycle

Using an ideal gas as the working fluid, show that the thermal efficiency of an Ericsson cycle is identical to the efficiency of a Carnot cycle operating between the same temperature limits.

Question

Thermal Efficiency of the Ericsson Cycle

Using an ideal gas as the working fluid, show that the thermal efficiency of an Ericsson cycle is identical to the efficiency of a Carnot cycle operating between the same temperature limits.

Accepted Answer

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 [latex] T_{H} [/latex] during process 1-2, and it is rejected again isothermally to an external sink at temperature [latex] T_{L} [/latex] during process 3-4. For a reversible isothermal process, heat transfer is related to the entropy change by

[latex] q=T \Delta s [/latex]

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

[latex] \Delta s=c_{p} \ln \frac{{T_{e}}^{ earrow ^0} }{T_{i}}-R \ln \frac{P_{e}}{P_{i}}=-R \ln \frac{P_{e}}{P_{i}} [/latex]

The heat input and heat output can be expressed as

[latex] q_{ in }=T_{H}\left(s_{2}-s_{1} ight)=T_{H}\left(-R \ln \frac{P_{2}}{P_{1}} ight)=R T_{H} \ln \frac{P_{1}}{P_{2}} [/latex]

and

[latex] q_{ ext {out }}=T_{L}\left(s_{4}-s_{3} ight)=-T_{L}\left(-R \ln \frac{P_{4}}{P_{3}} ight)=R T_{L} \ln \frac{P_{4}}{P_{3}} [/latex]

Then the thermal efficiency of the Ericsson cycle becomes

[latex] \eta_{ ext {th, Ericsson }}=1-\frac{q_{ ext {out }}}{q_{ ext {in }}}=1-\frac{R T_{L} \ln \left(P_{4} / P_{3} ight)}{R T_{H} \ln \left(P_{1} / P_{2} ight)}=1-\frac{T_{L}}{T_{H}} [/latex]

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

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

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