All Problems

Second law of thermodynamics, entropy

Problem 2.128

Making use of the Clausius inequality, demonstrate that all cycles having the same maximum temperature Tmax⁡T_{\max } and the same minimum temperature Tmin⁡T_{\min } are less efficient compared to the Carnot cycle with the same Tmax⁡T_{\max } and Tmin⁡T_{\min }

Reveal Answer
2.128. The inequality ∫δQ1T1−∫δQ2′T2⩽0\int \frac{\delta Q_{1}}{T_{1}}-\int \frac{\delta Q_{2}^{\prime}}{T_{2}} \leqslant 0 becomes even stronger when T1T_{1} is replaced by Tmax⁡T_{\max } and T2T_{2} by Tmin⁡T_{\min }. Then Q1/Tmax⁡−Q_{1} / T_{\max }- −Qa′/Tmin⁡<0.-Q_{\mathrm{a}}^{\prime} / T_{\min }<0 . Hence \frac{Q_{1}-Q_{2}^{\prime}}{Q_{1}}<\frac{T_{\max }-T_{m \ln }}{T_{\max }}, \text { or } \eta<\eta_{\text {Carnot }}
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