1
$\begingroup$

An ideal Carnot engine is composed of two reservoirs and a working fluid. The hot reservoir and the cold reservoir have temperatures $T_1$ and $T_2$ respectively, with $T_1>T_2$. The working fluid is in a phase transition and has temperature $T_1$ at the start of the Carnot cycle. It undergoes another phase transition at $T_2$ at the end of the cycle to return to its original state.

This is a P-V diagram of the Carnot cycle which proceeds in four steps:

enter image description here

I'm particularly interested in the two stages (from 1 to 2) and (from 3 to 4). They can be described as follows:

1) Stage (from 1 to 2) is a reversible isothermal expansion of the working fluid to transform from the liquid state to the gaseous one. The working fluid is at $T_1$ and it happens to have boiling point at $T_1$. Hence, heat $Q_1$ is supplied to the fluid from the hot reservoir until it transforms to a gas keeping its temperature constant along the whole process. (That the fluid's temperature is constant during the whole process is owing to it being in a phase transition.)

2) Stage (from 3 to 4) is a reversible isothermal compression, and it's similar to what we have just described, with the difference being in this case, heat $Q_2$ is drawn out of the fluid and transfers to the cold reservoir, and the fluid transforms from gas to liquid retaining a constant temperature of $T_2$ throughout the whole process.

I'm puzzled by the mechanism by which the working fluid undergoes phase transition. So, at stage (from 1 to 2), both the fluid and the hot reservoir have the exact same temperature, so that they're in a thermal equilibrium. Hence, there should be no heat or energy exchange between the two bodies. The same can be said of stage (from 3 to 4).

So how is it possible for heat to flow from two bodies having the exact same temperature?

$\endgroup$
  • $\begingroup$ You're also assuming that the liquid phase is compressible (from point 4 to 1)? $\endgroup$ – Soba noodles Sep 28 '15 at 15:25
  • $\begingroup$ @Sobanoodles Yes. $\endgroup$ – Omar Nagib Sep 28 '15 at 17:21
  • $\begingroup$ @OmarNagib There is something wrong with the cycle description. The classic Carnot cycle does involve a phase change. What's more, when a phase change occurs both the fluid temperature and pressure are constant. The pressure of the working fluid is not constant during any stage of the Carnot cycle. The correct answer is given by Hyportnex in the comments below. $\endgroup$ – Bob D Sep 2 at 10:51
1
$\begingroup$

The only reason the reservoir and the fluid in 1-2 are at the same temperature is that the fluid is boiling. Heat is being absorbed by the fluid, and providing the energy required to cause a phase change until all the liquid has been converted to vapor. If the working fluid were not undergoing a phase change, the process would not be isothermal: both the reservoir and the fluid would be increasing in temperature.

Likewise, in 3-4, the reservoir and fluid are losing heat to the environment, and the temperature would be falling if not for the release of energy of fusion.

$\endgroup$
  • $\begingroup$ I know that the fluid is boiling, I already stated that in my question(that the working fluid has a boiling point at $T_1$ and it's in a phase change), my question is: how it is possible for heat to be absorbed by the fluid when it has the same temperature as that of the hot reservoir? $\endgroup$ – Omar Nagib Sep 28 '15 at 17:19
  • 1
    $\begingroup$ for the heat exchange to take place you only need infinitesimally different temperatures between the bodies, the heat exchange will be reversible but very slow. $\endgroup$ – hyportnex Sep 30 '15 at 13:41
  • $\begingroup$ there is always heat exchange..that is the point...even when the temperature is the same. With different temperatures this exchange is more on one side than the other. Isothermal process is one that is very slow, unlike adiabatic, where we assume very quick, so quick so no heat can be transfered.. $\endgroup$ – Žarko Tomičić Jun 24 '18 at 9:13
  • $\begingroup$ In fact, the heat transfer will be so slow as to be totally impractical. Nobody will pay for an engine that produces an infinitesimal amount of power, and takes years or decades to get the job done. $\endgroup$ – David White Sep 2 at 18:03
1
$\begingroup$

The correct answer is given by @hyportnex. The temperatures of the reservoirs and the system temperature are not exactly the same. They are infinitesimally different so that the system and reservoirs are in thermal equilibrium throughout the process. In any case the cycle is not for a Carnot cycle involving two phases (liquid-gas). Phase transition generally occurs at both constant temperature and constant pressure.

The diagram below shows a PV diagram for a 2 phase Carnot Cycle. Note the pressure is constant during the isothermal phase changes.

Hope this helps.

enter image description here

$\endgroup$
1
$\begingroup$

The answer to the question is that there is a very small temperature difference between the substance and the thermal reservoir or environment when heat flows between them. In the limit where the process speed tends to zero, this temperature difference tends to zero.

On a related matter, the question is mixing up two concepts which do not need to be mixed. These concepts are:

  1. Carnot cycle

  2. Phase transition

You can have a Carnot cycle with or without a phase transition. To get an isothermal expansion or compression without a phase transition, just expand or compress the fluid while keeping it in contact with a thermal reservoir at fixed temperature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.