The efficiency of the cycle shown in the figu | vedclass

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(B) Let the cycle be $A \rightarrow B \rightarrow C \rightarrow A$. $A \rightarrow B$ is isobaric,$B \rightarrow C$ is adiabatic,and $C \rightarrow A$

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$x^2 = e^{x-1}$

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(B) Let the cycle be $A \rightarrow B \rightarrow C \rightarrow A$. $A \rightarrow B$ is isobaric,$B \rightarrow C$ is adiabatic,and $C \rightarrow A$ is isothermal. Let $T_A = T_{min}$ and $T_B = T_{max}$. The ratio is $x = T_B / T_A$. For isobaric process $A \rightarrow B$: $Q_{AB} = n C_p (T_B - T_A) = n \left( \frac{\gamma R}{\gamma - 1} \right) (T_B - T_A)$. For isothermal process $C \rightarrow A$: $Q_{CA} = n R T_A \ln(V_A / V_C)$. Since $B \rightarrow C$ is adiabatic,$T_B V_B^{\gamma-1} = T_C V_C^{\gamma-1}$. Since $T_C = T_A$,$T_B V_B^{\gamma-1} = T_A V_C^{\gamma-1}$. Thus,$(V_C / V_B)^{\gamma-1} = T_B / T_A = x$,so $V_C / V_B = x^{1/(\gamma-1)}$. Also,for isobaric process $A \rightarrow B$,$V_B / V_A = T_B / T_A = x$. So,$V_C / V_A = (V_C / V_B) \cdot (V_B / V_A) = x^{1/(\gamma-1)} \cdot x = x^{\gamma/(\gamma-1)}$. Efficiency $\eta = 1 - |Q_{CA}| / Q_{AB} = 1 - [n R T_A \ln(V_C / V_A)] / [n R \frac{\gamma}{\gamma-1} (T_B - T_A)]$. Substituting values: $\eta = 1 - [T_A \cdot \frac{\gamma}{\gamma-1} \ln x] / [\frac{\gamma}{\gamma-1} T_A (x - 1)] = 1 - \frac{\ln x}{x - 1}$. Given $\eta = 0.5$,so $0.5 = 1 - \frac{\ln x}{x - 1} \Rightarrow \frac{\ln x}{x - 1} = 0.5 \Rightarrow \ln x = 0.5(x - 1) \Rightarrow \ln x = \ln(e^{0.5(x-1)})$. Thus $x = e^{0.5(x-1)} \Rightarrow x^2 = e^{x-1}$.

The efficiency of the cycle shown in the figure (consisting of one isobar,one adiabat,and one isotherm) is $50 \%$. The ratio $x$ between the highest and lowest temperatures attained in this cycle obeys (the working substance is an ideal gas):

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