download the script: Carnot Principles

Carnot principles can be expressed as the following two statements:

1. The efficiencies of all reversible heat engines operating between a constant temperature source (T_H) and a constant temperature sink (T_L) are the same. They only depend on the high temperature T_H=const and the low temperature T_L=const.

2. Between the same two reservoirs, the efficiency of a reversible heat engine is always greater than the efficiency of an irreversible one.

We can use Reductio ad absurdum to prove the statements of Carnot principles.

Consider that we have two engines operating between the same temperature source at constant T_H and the same temperature sink at constant T_L.

A is an irreversible heat engine. It absorbs heat in the amount of Q_H from the high temperature reservoir (T_H) and rejects heat in the amount of Q_L to the low temperature reservoir (T_L). According to the 1^st law of thermodynamics, work is done by this heat engine: W=Q_H-Q_L.

B is a reversible refrigerator. Heat Q’_L is absorbed from the low-temperature reservoir (T_L=constant) and heat Q’_H is rejected to a high-temperature reservoir (T_H=constant). In this case work input is required to achieve this process. Assume that the work input here equals to W from the irreversible heat engine A. That means, the work output from irreversible heat engine A is directly applied to the reversible refrigerator. And we know from the 1^st law of thermodynamics, the required work can be determined in W=Q’_H-Q’_L.

We can also calculate the thermal efficiencies of both engines:

For heat engine: and for refrigerator:

Pay attention here, we did not calculate the COP of refrigerator, because COP and thermal efficiency are two different concepts, so that we cannot compare with each other. So we assume that refrigerator here operates like a ‘heat engine’ and, hence, has a thermal efficiency.

If the efficiency of an irreversible engine is greater than the efficiency of a reversible one, that means η_th,heat > η_th,ref, so

which states that Q_H < Q’_H.

We know from the above discussion:

W=Q_H – Q_L=Q’_H – Q’_L

Then ΔQ=Q’_H-Q_H=Q’_L-Q_L>0

So now imagine that a so-called heat-engine-refrigerator combination which consists of these two engines (shown as a box in the figure below). The work transferred between heat engine and refrigerator is internal force which will not be taken into consideration in this new combination.

Since no work is done on or by this engine-combination, heat in the amount of ΔQ is transferred from the temperature sink to the temperature source. The heat-engine-refrigerator-combination plays no role. And according to the second law of thermodynamics, heat cannot flow spontaneously in the direction of rising temperature. Therefore, η_th,heat > η_th,ref is impossible. So now we proved that the efficiency of an irreversible engine cannot be greater than the efficiency of a reversible one

Now let’s consider η_th,heat = η_th,ref . In this case, work from the irreversible heat engine is also applied to the reversible refrigerator, ΔQ=0 and no change of state occurs during this process, because the work fluid returns to its initial state. This is clearly a violation of irreversibility.

So the only possibility is that η_th,heat < η_th,ref. We can use the similar way to prove that in this case heat (ΔQ>0) flows spontaneously in the direction of sinking temperature. It meets the second law of thermodynamics perfectly.