Calculating Efficiency from a $pV$ Diagram

I Am trying the calculate the efficiency of this engine however, I'm not sure whether my result is making intuitive sense. The $$pV$$ diagram of the engine is as follows;

Here we note that the process $$2\to3$$ is an isothermal expansion of the engine. So, the efficiency of the engine is defined as; $$\epsilon=\frac{W}{Q_H}$$ Where $$W$$ is the net work. So, to then determine the efficiency of the engine, we must determine the net work of the system and the expression for $$Q_H$$. Namely, we note that the net work of the system is the area enclosed by the cycle. This can in turn be given by; $$W_{net}=NkT_h\int_{V_i}^{V_f}\frac{1}{V}dV-P_i\int_{V_i}^{V_f} dV=NkT_h\ln{(\frac{V_f}{V_i})}-P_i\Delta V=NkT_h\ln{(\frac{V_f}{V_i})}-Nk\Delta T$$ And since $$Q_H$$ is the heat added during the isochoric process, $$Q_H=C_V\Delta T$$ No, then we can substitute these expressions into our from for $$\epsilon$$; $$\epsilon=\frac{NkT_h\ln{(\frac{V_f}{V_i})}-Nk\Delta T}{C_V\Delta T}$$ And, given that $$\frac{V_f}{V_i}=\frac{T_h}{T_c}$$ for this process, where $$T_c$$ is the temperature of the engine at $$(1)$$, we can rewrite the efficiency; $$\epsilon=\frac{NkT_h\ln{(\frac{T_h}{T_c})}-Nk\Delta T}{C_V\Delta T}$$ Also, if we assume the gas to me monatomic, $$C_V=\frac{3}{2}Nk$$ which again simplifies the expression to; $$\epsilon=\frac{2}{3}\left(\frac{T_h\ln{(\frac{T_h}{T_c})}-\Delta T}{\Delta T}\right)=\frac{2}{3}\left(\frac{T_h\ln{(\frac{T_h}{T_c})}-(T_h-T_c)}{(T_h-T_c)}\right)$$ Is this process correct? When I consider the limiting case where $$\frac{T_h}{T_c}\to \infty$$ these seems to be no maximum efficiency as I would expect. Any help would be greatly appreciated!

• @user8736288 Reversing the process will just make $Q_H$ negative I believe. So, for the reverse process, $$Q_H=-C_V\Delta T$$
– JayP
Commented Apr 13, 2020 at 9:53
• Is it an ideal gas? Commented Apr 13, 2020 at 10:26
• @BobD Yes it is
– JayP
Commented Apr 13, 2020 at 10:45
• Heat also comes in during the isothermal expansion, not just the isochoric heating. So $$Q_H=NC_v\Delta T+NKT\ln{(V_2/V_1)}$$ Commented Apr 13, 2020 at 12:05

Your equation for the heat added is incorrect. There is also heat added during isothermal expansion. So heat added is $$Q=C_v(T_h-T_l)+nkT_h\ln{(V_f/V_i)}=C_vT_h\left(1-\frac{V_i}{V_f}\right)+nkT_h\ln{(V_f/V_i)}$$and the work done is $$W=nkT_h\ln{(V_f/V_i)}-nkT_h\left(1-\frac{V_i}{V_f}\right)$$So the efficiency is: $$\epsilon=\frac{\ln{(V_f/V_i)}-\left(1-\frac{V_i}{V_f}\right)}{\ln{(V_f/V_i)}+\frac{C_v}{nK}\left(1-\frac{V_i}{V_f}\right)}=\frac{1-\alpha}{1+\frac{C_v}{nK}\alpha}$$with $$\alpha=\frac{\left(1-\frac{V_i}{V_f}\right)}{\ln{(V_f/V_i)}}$$
$$e=\frac{Q_{in}-Q_{out}}{Q_{in}}$$