# How to get the efficiency of a heat engine which undergoes an elliptical cycle?

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:

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:

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?

• 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. – Chet Miller Jun 14 at 13:56
• 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 $A=\pi r^2$ – user5713492 Jun 14 at 15:23
• @user5713492 I don't see what you mean. The area of an ellipse is $A = \pi a b$ – JD_PM Jun 14 at 19:16
• 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? – user5713492 Jun 14 at 20:37
• @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/… – JD_PM Jun 15 at 10:12

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.
• 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. – Chet Miller Jun 14 at 13:50
• 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. – user5713492 Jun 14 at 15:20