Given are four points in a PV-diagram: $(V_1, 3/2P_1)$ is the leftmost point, $(V_2,3/2P_1)$ is the rightmost point, $(\frac{V_1+V_2}{2},P_1)$ is the lowest point and $(\frac{V_1+V_2}{2},2P_1)$ is the highest point. This gives us an ellipse. I have to calculate the efficiency of this process. I started with the definition of the efficieny $e = \frac{W}{Q_h}$. I calculated $W$ by calculating the surface of this ellips what gave me $ W = \frac{\pi}{4}(V_2-V_1)P_1$. Given is that $Q_h$ is the heat included from the expansion from $V_1$ to $V_2$. But I got stuck there, someone who can help me by explaining how to calculate $Q_h$? Thanks!

  • $\begingroup$ There is, of course, an alternative form of the efficiency equation. Using absolute temperatures, you can calculate efficiency with $eff = 1 - T_c/T_h$, where $T_c$ is the temperature of the sink and $T_h$ is the temperature of the source. See en.wikipedia.org/wiki/Thermal_efficiency $\endgroup$ – David White Apr 22 at 20:14
  • $\begingroup$ Definition of the efficiency of a process is $e = 1- \frac{Q_c}{Q_h} $which is always equal or smaller than $1- \frac{T_c}{T_h}$. It is true that is it is equal when it is a Carnot process, but that is not given here. $\endgroup$ – Kabouter9 Apr 23 at 19:09

Along the 'top half' of the ellipse the gas (?) is expanding and heat, $Q_h, $ is flowing in. Now, $$Q_h = \text{work out + rise in internal energy}.$$

You can find the first term on the right by considering the area under the top half of the ellipse down to the $V$ axis (not difficult because you already know the area inside the ellipse, and the ellipse is symmetrical).

You can find the second term on the right in terms of $c_V$ if you assume that the working substance is an ideal gas, as you can calculate the temperature rise from pressure and volume data.

I leave the details to you.

  • $\begingroup$ But how do I calculate the rise in internal energy, nothing is given about the temperature... $\endgroup$ – Kabouter9 Apr 22 at 9:59
  • $\begingroup$ Try using $pV=nRT$ $\endgroup$ – Philip Wood Apr 22 at 10:11
  • $\begingroup$ The number of moles, $n,$ cancels in the calculation of rise in internal energy over the top half of the ellipse. What doesn't cancel (unless I've made some mistake) is the ratio $c_V/R.$ $\endgroup$ – Philip Wood Apr 22 at 10:35
  • $\begingroup$ How did you find the surface under the curve from V1 to V2? :) $\endgroup$ – Kabouter9 Apr 22 at 11:01
  • $\begingroup$ I gave you hints in my answer. Sketch the area involved, and break it down into a rectangle with half an ellipse on top! $\endgroup$ – Philip Wood Apr 22 at 11:18

Applying the first law to this problem (assuming a reversible cycle), we have $$nC_vdT=dq-PdV$$But, from the first law, $nRdT=d(PV)=PdV+VdP$So we have: $$\frac{C_v}{R}(PdV+VdP)=dq-PdV$$or$$dq=\frac{C_p}{R}PdV+\frac{C_v}{R}VdP\tag{1}$$ So, integrating clockwise, in the 2nd quadrant of the ellipse, dP and dV are both positive, so dq is positive, and in the 4th quadrant of the ellipse, dP and dV are both negative, so dq is negative, but, in the 1st and 3rd quadrants, dq changes sign.

From the previous equation dq = 0 when $$\frac{dP}{dV}=-\gamma \frac{P}{V}\tag{2}$$

The equation for the ellipse is:$$\left(\frac{P-\bar{P}}{a}\right)^2+\left(\frac{V-\bar{V}}{b}\right)^2=1\tag{3}$$where $\bar{P}=P_1$, $\bar{V}=\frac{V_1+V_2}{2}$, $a=\frac{P_1}{2}$, and $b=\frac{V_2-V_1}{2}$. From this, it follows that, $$\frac{dP}{dV}=-\left(\frac{a}{b}\right)^2\frac{V-\bar{V}}{P-\bar{P}}\tag{4}$$ The further analysis looks a little complicated, so I'll stop here for now.


If we combine Eqns. 2 and 4, we obtain: $$\left(\frac{a}{b}\right)^2\frac{V-\bar{V}}{P-\bar{P}}=\gamma \frac{P}{V}\tag{5}$$Next, we make the following dimensionless substitutions: $p=P/a$,$\bar{p}=\bar{P}/a$, $v=V/b$, and $\bar{v}=v/b$Eqns. 1, 3, and 5 then become: $$\frac{dq}{ab}=\frac{C_p}{R}pdv+\frac{C_v}{R}vdp\tag{6}$$ $$(p-\bar{p})^2+(v-\bar{v})^2=1\tag{7}$$ and $$\frac{v-\bar{v}}{p-\bar{p}}=\gamma \frac{p}{v}\tag{8}$$ If we next make the substitutions $p-\bar{p}=\sin{\theta}$ and $v-\bar{v}=\cos{\theta}$, Eqn. 7 is satisfied exactly and Eqns. 6 and 8 become: $$\frac{dq}{ab}=\left[-\frac{C_p}{R}(\bar{p}+\sin{\theta})\sin{\theta}+\frac{C_v}{R}(\bar{v}+\cos{\theta})\cos{\theta}\right]d\theta\tag{9}$$ $$(\bar{v}+\cos{\theta})\cos{\theta}=\gamma(\bar{p}+\sin{\theta})\sin{\theta}\tag{10}$$ where $\theta$ is the counterclockwise angle that a line through the center of the ellipse makes with the positive v direction. There are two values of $\theta$ that satisfy Eqn. 10, one value in the 3rd quadrant of the ellipse and the other in the 1st quadrant of the ellipse. Eqn. 9 is integrated between these two values of $\theta$ to get the total heat added to the working fluid during the cycle. Eqn. 9 can be integrated analytically between these two values.

  • $\begingroup$ See the ADDENDUM to my answer for more details. $\endgroup$ – Chet Miller Apr 24 at 17:59

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