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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; PV Diagram

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!

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

2 Answers 2

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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 $$\alpha=\frac{\left(1-\frac{V_i}{V_f}\right)}{\ln{(V_f/V_i)}}$$

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    $\begingroup$ Hi Chet. I avoided giving a complete answer because the question was a H&E one and I have been slammed in the past for doing so. I find it frustrating because sometimes I get slammed and other times I don't. The judgement aspect is rather vague, in my opinion. Anyway, I agree with you answer and hope you don't get slammed for it. Bob $\endgroup$
    – Bob D
    Apr 13, 2020 at 21:22
  • $\begingroup$ @BobD Well, it was only a small increment on what the OP already had. But, if I get slammed, so be it. $\endgroup$ Apr 13, 2020 at 21:44
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Hint: Heat is added in both the isothermal expansion 2-3 and the isochoric process 1-2. Heat is rejected in the isobaric compression 3-1. The efficiency can also be written as

$$e=\frac{Q_{in}-Q_{out}}{Q_{in}}$$

Hope this helps

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