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An ideal diatomic gas undergoes an elliptic cyclic process characterized by the following points in a $PV$ diagram:

$$(3/2P_1, V1)$$ $$(2P_1, (V1+V2)/2)$$ $$(3/2P_1, V2)$$ $$(P_1, (V1+V2)/2)$$

A rough sketch:

enter image description here

This system is used as a heat engine (converting the added heat into mechanical work).

Evaluate the efficiency of this engine

We know that the efficiency is defined as the benefit/cost ratio:

$$e = \frac{W}{Q_h}$$

Let's focus first on the work done by the engine; taking into account the quasistatic approximation, $W=PV$. Then:

$$W = (P_2 - P_1)(V_2 - V_1)$$

Note that from the given points we can guess that $P_2 = 2P_1$. Then:

$$W = P_1(V_2 - V_1)$$

Now let's focus on $Q_h$. I have the following issue here: none of the 4 steps of the cycle has either P or V constants. This means that the strategy of using:

$$Q = nc\Delta T$$

Won't work because you cannot use neither $c_p$ nor $c_p$.

However, when we deal with a rectangular cycle:

enter image description here

It would be really easy to derive expressions for both $Q_a$ and $Q_b$ and then the efficiency for the system would be obtained. That is because in each step $Q_h$ is added, either $P$ or $V$ are constant (and thus $Q = nc\Delta T$ works).

What to do with the elliptical cycle to get $Q_h$?

EDIT

My bad, the work done by the working substance is the area under the PV graph. So as Chet Miller pointed out, the work is:

$$W = \pi (P_2 - P_1)(V_2 - V_1)$$

I have been trying to solve the heat equation so that we get the two angles.

So what I did was deriving $P$ and $V$ wrt the angle:

$$dP = (P_{max} -P_0)\cos \theta d\theta$$

$$dV = (V_{max} -V_0)\sin \theta d\theta$$

And plugging it into 3:

$$0=[-(C_v+R)P(V_{max} -V_0)\sin \theta+C_VV(P_{max} -P_0)\cos \theta]$$

This above equation is satisfied by two $\theta$ angles. But how to solve it?

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    $\begingroup$ Your sign on dV is incorrect. And, please include the d theta's. You also need to substitute for P and V. See my comment after my answer. $\endgroup$ – Chet Miller 2 days ago
  • $\begingroup$ Your equation above for $W$ is still wrong because you are using the full major and minor axes, not the semi-major and semi-minor axes. It's like saying the area of a circle is $A=\pi D^2$ instead of $A=\pi r^2$ $\endgroup$ – user5713492 2 days ago
  • $\begingroup$ @user5713492 I don't see what you mean. The area of an ellipse is $A = \pi a b$ $\endgroup$ – JD_PM yesterday
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    $\begingroup$ Look at your expression for the area of a rectangle. Then look at your expression for the area of an ellipse contained completely within that rectangle. Which is bigger? Which should be bigger? $\endgroup$ – user5713492 yesterday
  • $\begingroup$ @user5713492 I see what you mean now. Btw thank you very much for your answer at MSE; for anyone interested: math.stackexchange.com/questions/3262034/… $\endgroup$ – JD_PM yesterday
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Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle. Then, the work is the area of the ellipse, which is given by: $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-major and semi-minor axes. This differs from the result which you gave.

The shape of the ellipse can be represented parametrically in terms of the angle $\theta$ around the cycle, assuming $\theta$ is measured clockwise from the point $(V_0-(V_0-V_\mathrm{max}),P_0)$: $$P=P_0+(P_\mathrm{max}-P_0)\sin{\theta}\tag{1a}$$ $$V=V_0-(V_\mathrm{max}-V_0)\cos{\theta}\tag{1b}$$ Application of the first law of thermodynamics to the working fluid over a differential portion of the cycle gives us: $$dU=nC_vdT=dQ-PdV\tag{2}$$But from the ideal gas law, $$nRdT=d(PV)$$Substitution of this into Eqn. 2 yields: $$\frac{C_v}{R}d(PV)=dQ-PdV$$ So the differential heat added during an arbitrary portion of the cycle is given by:$$dQ=\frac{1}{R}[(C_v+R)PdV+C_VVdP]\tag{3}$$So the differential heat is zero when dQ = 0, or, equivalently, when $d\ln{P}+ d\ln{V^{\gamma}}=0$, or, equivalently when $PV^{\gamma}=\mathrm{Const}$. This is the equation for an isentropic line tangent to the ellipse.

There are two angles $\theta$ at which adiabats are tangent to the ellipse. These two angles can be obtained by substituting Eqns. 1 into Eqn. 3 with dQ = 0. Once these angles have been determined, the total positive heat added Q during the cycle can then be obtained by integrating Eqn. 3 with respect to $\theta$ between the two angles.

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    $\begingroup$ Nice answer,To my prior comment, I forgot it is not so simple for general p-V diagram. :-) $\endgroup$ – Poutnik Jun 13 at 7:02
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    $\begingroup$ You are aware that the internal energy U of an ideal gas is a function only of temperature, and is equal to nCvdT irrespective of whether the volume is constant, correct? $\endgroup$ – Chet Miller Jun 13 at 15:56
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    $\begingroup$ You also need to substitute for P and V. It then reduces to $$\gamma \frac{P_0}{(P_{max}-P_0)}\sin{\theta}+\frac{V_0}{(V_{max}-V_0)}\cos{\theta}+\gamma \sin^2{\theta}-\cos^2{\theta}=0$$I don't think that this has an analytic solution. $\endgroup$ – Chet Miller 2 days ago
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    $\begingroup$ Definitely has an analytic solution: let $z=\tan\left(\theta/2\right)$ and the above reduces to a quartic equation in $z$. The general quartic equation has a solution, not necessarily a pretty solution, but a solution. $\endgroup$ – user5713492 2 days ago
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    $\begingroup$ Well, @user5713492 indicates a method for solving the equation analytically. I haven't tried this yet, but, if so, you can get the cosines and sines of the two angles, and then integrate to get the heat analytically. Don't bother with the rectangular shape. In my judgment, that will lead to nothing. $\endgroup$ – Chet Miller yesterday

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