Why can't we plot irreversible processes on a $T$-$S$ diagram?

In the book "Heat and Thermodynamics" by Zemanky, It is stated that

"only reversible processes can be plotted on a $T$-$S$ diagram. Since the definition of entropy involves a reversible change in heat."

But, As $T$ and $S$ are state variables, We should be able to plot the process despite the fact that it is irreversible. If the process is quasi-static we can have the precise set of values of $T$ and $S$ at every instant of the process.

• "If the process is quasi-static we can have the precise set of values of T and S at every instant of the process." But then, by doing that you would have a reversible process again. – myradio Sep 3 '18 at 8:28
• For a process to be reversible, We don't only need it to be quasi-static but also we need the system to be free from dissipative forces. – Jay Sep 3 '18 at 10:34

4 Answers

When we plot a process as a continuous line, curve, etc., of states using thermodynamic variables we are assuming that each point on the chart represents an equilibrium state, i.e., the process taking place is a series of “quasi-static” or “quasi-equilibrium” states. For example, refer to the T-S diagram below for a Rankine refrigeration cycle. The compressor, condenser, and evaporator processes are each reversible. The expansion valve (throttling) process is not. Note how it is represented by a series of hashed lines implying that the states between 3 and 4 are not equilibrium states. The enthalpy at state 3 and state 4 are the same but it is not a constant enthalpy process. Although Zemansky singles out the T-S diagram, the same holds true for other graphs of processes. The following is from an MIT web course on thermodynamics concerning processes.

“…properties define a state only when a system is in equilibrium. If a process involves finite, unbalanced forces, the system can pass through non-equilibrium states, which we cannot treat. An extremely useful idealization, however, is that only infinitesimal unbalanced forces exist, so that the process can be viewed as taking place in a series of quasi-equilibrium states”.

Hope this helps

I think, its because we cannot use a single value of a property for the entire system i.e. if you consider temperature, it will be different at different points in the system.So we cant plot it as a graph(for the entire system).

• In an equilibrium state temperature is well definied and uniform in the whole system, for definition of equilibrium. – Landau Sep 3 '18 at 11:48
• @Landau Ok.Your answer was nice.So in an irreversible process, there will be unaccounted heat transfer due to friction,dissipation or any other irreversible reasons.So thats why,even if we plot the process, the area is meaningless.Is this interpretation right? – user196272 Sep 3 '18 at 12:55
• Yeah, I ve edited my answer, hope it's more clear now. – Landau Sep 3 '18 at 13:03

Here are some characteristics of an irreversible process in a closed system (no mass in or out, but heat exchange and work allowed):

1. The entropy of the system can be changed both by entropy entering or leaving (by heat transfer) through the system boundaries and by entropy generation within the system. (In a reversible process, there is negligible entropy generation within the system, and only entropy exchange with the surroundings is important)

2. The local rate of entropy generation is the sum of that from significant temperature gradients and that from significant velocity gradients within the system. The local rate of entropy generation from temperature gradients is proportional to the square of the local temperature gradient. The local rate of entropy generation from velocity gradients is proportional to the square of the local velocity gradient. These rates are negligible in a reversible/quasi-static process.

3. The total rate of entropy generation within the system at any time is equal to the integral of the local rate of entropy generation, integrated over the mass of the system.

4. If we could solve the dynamic energy balance, momentum balance, and transport equations within the system, we could determine the local temperatures and velocities within the system, and, from that, the local entropy generation rate. This would allow us to determine the overall (cumulative) entropy change at any time. We could then plot this vs the mass average temperature. However, the value of such a plot would be highly questionable, since the entropy change would not just involve exchange of entropy (via heat) with the surroundings.

I think you are right, in an equilibrium state T and S are well definied for a quasi-static process, so you can draw any quasi-static process on a T-S diagram. However in this case the area below the process from the point 1 to the point 2 doesn't have any meaning. While if you consider a reversibile process the area rapresents the heat that the system exchanges to go from 1 to 2. You have: $$ds=\frac{\delta q_{rev}}{T} \rightarrow Q_{rev}=\int_{1}^{2}Tds$$ While $$ds\not=\frac{\delta q_{irrev}}{T} \rightarrow Q_{irrev}\not=\int_{1}^{2}Tds$$

• Yes, The same argument can be made when plotting P-V diagram for an irreversible process, replacing heat with work. – Jay Sep 3 '18 at 18:00
• Of course, I think it's just an inaccuracy done by your book. – Landau Sep 3 '18 at 21:16
• I think my previous comment was wrong. For an Irreversible quasi-static process, the area does in fact indicate the work done by the system. Because Internal pressure can be expressed as the sum of external pressure and any dissipative force (Pressure). – Jay Sep 4 '18 at 7:09
• So, P-V diagram and T-S diagram are not exactly equivalent, When considering the area under the curve. – Jay Sep 4 '18 at 7:11
• Wait... If you have a reversible process you can express the work in the first principle always only as $dW=pdV$. If you don't have a reversible process, then the work is $dW=pdV+dW_{other}$ where $dW_{other}$ can be made by any mechanical device. So in the case of a rev process the area under the P-V diagram is always the work done. While in the case of a not rev process there are two possibilities: 1)It is not quasi-static -> you can't draw it 2)It is quasistatic and in this case you can draw it into the P_V but the area is the work only if there are not any mechanical devices. – Landau Sep 4 '18 at 8:41