# Efficiency of a reversible thermodinamic cycle

1 mole of ideal monoatomic gas completes a reversible ABC cycle, in which AB is a VOlume constant tranformation, BC an adiabatic expansion(Q=0) and CA an isotherm(Temperature during tranformation is constant). Knowing that $$\Delta S_{AB} = 12\frac{J}{K}$$. $$\Delta S_{AB}$$ is entropy variation.

Calculate:

A) $$\frac{T_B}{T_A}$$

B) the cycle efficiency.

About point A) all is clear. About point B) I don't understand why I can't use thate formula efficiency$$=1-\frac{T_A}{T_B}$$. Infact if I use definition of efficiency: $$\frac{W_{tot}}{Q_{IN}}$$ I get correct number(in relation to the result on the text).

In my case Can I considerate that to make AB transformation my gas is in contact with a source at temperature $$T_A$$. If I can where is the problem?

• I think that if you replace $T_B$ in your equation by the log-mean temperature between A and B, $$T_{lm}=\frac{(T_B-T_A)}{\ln{(T_B/T_A)}}$$you will get the right answer. – Chet Miller Sep 9 at 23:51
• @ChetMiller Good point, but it still won't be the Carnot efficiency, correct? – Bob D Sep 10 at 14:27
• Well, it will be if you regard $T_{lm}$ as the average temperature of heat transfer from the hot reservoir. – Chet Miller Sep 10 at 14:40
• @ChetMiller Hi Chet. If you have a chance, check out the bountied question "What is the meaning of maximum work in this thermodynamics question?" I can't make heads of tails of it. Can you? – Bob D Sep 13 at 22:21
• @BobD Hi. The question makes no sense to me, and the "given solution" makes even less sense. – Chet Miller Sep 14 at 1:40

The general formula for the efficiency of any cycle is

$$η=\frac {W_{net}}{Q_{in}}$$

Where $$W_{net}=Q_{in}-Q_{out}$$

That includes your cycle as well as the Carnot Cycle.

But the Carnot Cycle consists of two reversible isothermal (constant temperature) processes and two reversible adiabatic (constant entropy) processes. For the two isothermal reversible processes

$$Q_{in}=T_{B}\Delta S$$

$$Q_{out}=T_{A}\Delta S$$

Where $$T_A$$ is a low temperature thermal reservoir where heat is rejected, $$T_B$$ is a high temperature thermal reservoir where heat $$Q_{in}$$ is added, and the magnitude of the change in entropy, $$\Delta S$$, is the same for both isothermal processes.

Plugging the last two equations into the general formula for efficiency gives you the Carnot Cycle efficiency of

$$η=1-\frac{T_A}{T_B}$$

Which is the greatest efficiency possible for a heat engine operating between two temperatures.

In my case Can I considerate that to make AB transformation my gas is in contact with a source at temperature 𝑇𝐴. If I can where is the problem? If I'm agree so I can use 1−𝑇𝐴/𝑇𝐵

No you can't. The problem is temperature is not constant during process AB and the Carnot efficiency only applies to an isothermal (constant temperature) processes. For process AB the gas has to be in contact multiple temperature sources each infinitesimally higher than the previous in going from A to B.

For an ideal gas we have

$$\frac{P_{A}V_{A}}{T_{A}}=\frac{P_{B}V_{B}}{T_{B}}$$

Since it is a constant volume process, $$V_{A}=V_{B}$$, therefore

$$\frac{P_A}{T_A}=\frac{P_B}{T_B}$$

Since $$P_{B}>P_{A}$$, we have $$T_{B}>T_{A}$$

The temperature for process AB is only $$T_A$$ at point A. To apply the Carnot cycle efficiency heat has to be added at the same constant temperature throughout the process.

Hope this helps.

• In my case Can I considerate that to make AB transformation my gas is in contact with a source at temperature $T_A$. If I can where is the problem? If I'm agree so I can use $1-T_A/T_B$ – ABC Sep 9 at 21:10
• @ABC That won't work because the temperature is not constant in going from A to B. I have updated my answer to show you why. – Bob D Sep 9 at 22:12
• Thanks for the reply. There is a small thing that does not come back to me. Why does my university textbook write that $1-\frac{T_A}{T_B}$ represents the performance of ALL reversible machines that work with only two sources at temperatures $T_A$ and $T_B$? – ABC Sep 9 at 22:21
• @ABC That is true but only for reversible machines that operate between only TWO temperatures. All heat is added reversibly at $T_B$ and all heat is rejected at $T_A$. For your cycle although all heat is rejected at one temperature $T_A$ in the isothermal process CA, but it is not added at only one temperature in constant volume process AB. That's the difference. – Bob D Sep 9 at 22:29
• Perfect Now I understand!! To get a reversible transformation with constant volume I need about infinite sources. Only with 1 source I get a irreversible transformation! Thanks! – ABC Sep 9 at 22:33