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

Question

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?

Answer 1

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.