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 (TH) and a constant temperature sink (TL) are the same. They only depend on the high temperature TH=const and the low temperature TL=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 TH and the same temperature sink at constant TL.
A is an irreversible heat engine. It absorbs heat in the amount of QH from the high temperature reservoir (TH) and rejects heat in the amount of QL to the low temperature reservoir (TL). According to the 1st law of thermodynamics, work is done by this heat engine: W=QH-QL.
B is a reversible refrigerator. Heat Q’L is absorbed from the low-temperature reservoir (TL=constant) and heat Q’H is rejected to a high-temperature reservoir (TH=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 1st 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 ηth,heat > ηth,ref, so
which states that QH < Q’H.
We know from the above discussion:
W=QH – QL=Q’H – Q’L
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, ηth,heat > ηth,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 ηth,heat = ηth,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 ηth,heat < ηth,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.