# Computing the efficiency of a heat engine

With the heat engine

we have an ideal gas working the substance. I want to show that the efficiency of this heat engine is

$$\eta = 1 - \frac{1}{\gamma}\frac{\bigg(\frac{V_3}{V_1}\bigg)^\gamma - \bigg(\frac{V_2}{V_1}\bigg)^\gamma}{\bigg(\frac{V_3}{V_1} - \frac{V_2}{V_1}\bigg)}$$

where $\gamma = \dfrac{C_p}{C_v}$ is the adiabetic index.

So we have that $Q_1$ enters during the expansion $B \to C$ and $Q_2$ leaves during the compression $D \to A$. Therefore,

$$\eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$

I'm kind of stuck at this point. Any further help would be appreciated.

Now you just need to compute $Q_1$ and $Q_2$ in terms of state variables.

Let's consider $Q_1=Q_{CB}$. The First Law of Thermodynamics tells us that for the process $B\to C$ we have \begin{align} E_{CB} = Q_{CB} - W_{CB} \end{align} Since the pressure is a constant, say $P_2$, the work done is \begin{align} W_{CB} = \int_{V_2}^{V_3}P\, dV = P_2(V_3-V_2). \end{align} On the other hand, the change in internal energy is \begin{align} E_{CB} = C_V(T_3-T_2) \end{align} so from the First Law, we get \begin{align} Q_1 = C_V(T_3-T_2) + P_2(V_3-V_2) \end{align} Now, you do an analogous thing to determine $Q_2$ in terms of state variables, and do some simplifications using, among other things, the ideal gas law.

• I'm having a bit of trouble getting it into the required format. For instance how does $T_4$ just go away? – Jon Sep 24 '13 at 23:58
• @Jon It can be eliminated in favor of $T_3, V_3$, and $V_1$ since $TV^{\gamma-1}$ is constant for an adiabatic process. – joshphysics Sep 25 '13 at 0:02
• Ok thanks. Also for $Q_2$, $E_{AD}$ is computed the same way, but how do I compute $W_{AD}$? – Jon Sep 25 '13 at 0:35
• @Jon Since $dV=0$ along that path, the work done is zero. – joshphysics Sep 25 '13 at 1:45
• That's what I suspected. It just didn't seem intuitive (physically) – Jon Sep 25 '13 at 2:01