# How to calculate the efficiency from a $p$-$v$ diagram?

Question: Calculate the efficiency of the real and ideal cycles.

The process 1-2 is polytropic, 2-3 is isochoric, 3-4 is polytropic as well, and 4-1 completes the cycle with another isochoric process.

My attempt: I know that to calculate the efficiency of a cycle we must know the work done as output and the heat supplied to the system. As such, the efficiency will be the ratio of these two: $$\eta = \frac{w_{out}}{q_{in}}$$ From this diagram, I am able to calculate $w_{out}$ but I don't see how $q_{in}$ can be determined.

I originally just thought it to be equal to the work done from process 1-2, but I don't see think that is correct at all.

Would the heat in be through the isochoric process 4-1?

I'm not even sure if the heat in can be calculated straight from the p-v diagram.

Any help regarding how to determine $q_{in}$ would be greatly appreciated.

Edit: Can assume a working substance is undergoing this cycle. E.g. air

• This information about the Otto cycle may help you? theory.physics.manchester.ac.uk/~judith/stat_therm/node16.html Commented May 17, 2017 at 7:45
• Do you have any idea what the molar heat capacity of this gas is? Without that, it doesn't seem like you are going to be able to determine the amount of heat in. Commented May 17, 2017 at 11:39
• I was just told that we need to assume a working substance, which helps, but how to calculate the heat in still @ChesterMiller Commented May 17, 2017 at 11:46
• Do you know the relationship for the "equivalent" heat capacity of a polytropic expansion or compression? Commented May 17, 2017 at 12:09

See if you can show that, for a polytropic expansion or compression, $$Q=m\left[C_V-\frac{R}{(n-1)}\right]\Delta T$$where m is the number of moles. So you can get the heat for the polytropic expansion and compression in your process by knowing the temperatures and pressures at the two end points (using the polytropic equation in conjunction with the ideal gas law). So, the first step in your calculation should be to get the temperatures and pressures at the four corners of your cycle.

Of course, for the constant volume changes, $Q=mC_v\Delta T$.

So you can get the heat for all four legs of your cycle.

• Have you been able to derive the equation I presented for the heat Q in the polytropic segments of the process? Even if you haven't, have you been able to use the equation to solve for the efficiency? Commented May 17, 2017 at 23:15
• Assuming my working substance was air, I attempted to calculate the efficiency by finding the work output from the cycle and used the equation you provided to determine whether the polytropic processes were transferring heat in to or out of the system. Using the $[C_v - \frac{R}{n-1}]$ I was able to determine with constants n I have, these would be negative, which would be transferring heat out I believe, so I was then able to determine the heat in, and thus, the efficiency from there. Commented May 18, 2017 at 4:01
• Hmmm. Are you sure that both polytropic heats are negative? Maybe you can provide an edit to show your work? Regarding the equation for the equivalent heat capacity, I think that it's more important to understand how it is derived than to solve this specific problem. Commented May 18, 2017 at 10:48
• I got the following for 1 mole of air: Q12=+2450 J/cycle, Q23= -18154 J/cycle, Q34= -2936 J/cycle, Q41 = +24525 J/cycle. This give a value of W= +5885 J/cycle, a net heat in Qin = + 26975 J/cycle, and an efficiency = 21.8%. Commented May 19, 2017 at 4:04

The $P$-$V$ diagram, in isolation, is only enough to calculate the work done by a cycle. To calculate the efficiency you need to be able to measure the heat flux into and out of the system. Doing that requires something equivalent to the $T$-$S$ (temperature-entropy) diagram. This is not usually stated explicitly because entropy is not easy to measure, so it is normally reconstructed from the $P$-$V$ diagram using an equation of state for the working substance of the engine (e.g. $PV=NkT$). The equation of state gives you $T(V)$ or $T(P)$ (i.e. $T$ as a function of...), when those are combined with some expression for the heat capacity (usually the heat capacity at constant volume, $C_V$) it is possible for the calculation of heat flow in each step of the cycle.

For an ideal monatomic gas the heat capacity is given by: $$C_V = \frac{3}{2} N k.$$ That $3$ in the numerator comes from the number of ways in which the gas atoms can move through space (3 dimensions). It literally comes from the equipartition theorem that any term in the description of the energy of the atoms that make up a substance that is quadratic in position or momentum contributes $k/2$ per atom to $C_V$. So, for a gas of diatomic molecules at low temperatures we get two rotational degrees of freedom, giving $C_V=\frac{5}{2}Nk$. As the temperatures go up the molecules can start vibrating, too, increasing $C_V$ to approximately $\frac{7}{2} Nk$ (1 each for the atoms' vibrational kinetic and potential energies) until the atoms fully disassociate and it drops back down to $\frac{6}{2} Nk = \frac{3}{2} N_{\mathrm{new}}k$.

Bottom line: you need to use the properties of your working substance and the processes you subjected it to (polytropic and isochoric/isometric) to deduce the heat flow during each step.

• So if I'm trying to calculate the heat input for the system, I know there would be a source of heat in from process 4-1, but will there be another heat input from the polytropic process from 1-2? I'm attempting this question with the assumption that the working substance is air. Commented May 17, 2017 at 12:40
• I don't know. Is the polytropic process isentropic/adiabatic? If so, then the heat flow is zero. If not, you'll need to figure out what constraint defines the polytropic process, and go from there. Look at the Wikipedia article on polytropic processes for more info. Commented May 17, 2017 at 12:54
• The process is not adiabatic, I'm trying to determine whether there will be another source of heat for $q_{in}$ from the polytropic process. Commented May 17, 2017 at 13:07