3
$\begingroup$

With the heat engine

enter image description here

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.

$\endgroup$
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.