Calculating the Efficiency of a Stirling Cycle

Compute the efficiency of a Stirling cycle operating under the same conditions as the Ericsson cycle above.

Question

Calculating the Efficiency of a Stirling Cycle

Compute the efficiency of a Stirling cycle operating under the same conditions as the Ericsson cycle above.

Accepted Answer

The terms for each of the steps in the Stirling cycle are the same as for the Ericsson cycle; however, the properties are slightly different since in this case the heat exchanger (regenerator) operates such that each of its streams is at constant volume, not constant pressure. Therefore, rather than [latex]P_{2}=P_{1}[/latex] as in the Ericsson cycle, here we have (by the ideal gas law) [latex]P_{2}=P_{1} T_{2} / T_{1}[/latex]. Similarly, [latex]P_{4}=P_{3} T_{4} / T_{3}[/latex]. Therefore,

[latex]\begin{aligned}\eta=& \frac{-\dot{W}_{ ext {out }}}{\dot{Q}_{ T }}=\frac{R T_{2} \ln \frac{P_{3}}{P_{2}}+R T_{1} \ln \frac{P_{1}}{P_{4}}}{R T_{2} \ln \frac{P_{3}}{P_{2}}}=\frac{R T_{2} \ln \frac{P_{3}}{P_{2}}+R T_{1} \ln \frac{P_{2} T_{1} / T_{2}}{P_{3} T_{4} / T_{3}}}{R T_{2} \ln \frac{P_{3}}{P_{2}}} \=& \frac{R T_{2} \ln \frac{P_{3}}{P_{2}}+R T_{1} \ln \frac{P_{2}}{P_{3}}}{R T_{2} \ln \frac{P_{3}}{P_{2}}}=\frac{T_{2}-T_{1}}{T_{2}}\end{aligned}[/latex]

Since [latex]T_{2}=T_{3} ext { and } T_{1}=T_{4}[/latex], both the compressor and turbine operate isothermally. Therefore, the efficiency of the Stirling cycle is also equal to that of the Carnot cycle. For the operating conditions here, η = 0.525.

Question 5.2.4: Calculating the Efficiency of a Stirling Cycle Compute the e...

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