Proving that a heat engine cannot exceed the carnot efficiency

[handsomecat]

In many textbooks, a proof is provided where the work output of a super-efficient heat engine is provided to a carnot refrigerator, with the net result that a spontaneous heat transfer occurs from the cold reservoir to a hot reservoir.

Let's use some numbers, T_H = 600 K and T_L = 300 K, so that means the carnot efficiency is 50% and the carnot COP is 1.
Between these temperatures, By connecting a heat engine of efficiency 60% to the carnot fridge of COP = 1, then one can show that the impossible occurs.

But, between these temperatures (T_H = 600 K and T_L = 300K) what if we connect a super-efficient heat engine of efficiency 60% to a fridge of COP = 0.5?

Though a super-efficient heat engine can't exist, coupling the two together gives a net heat transfer from the hot reservoir to the cold reservoir, which could occur. What is wrong with this line of argument?

 

[Andrew Mason]

In many textbooks, a proof is provided where the work output of a super-efficient heat engine is provided to a carnot refrigerator, with the net result that a spontaneous heat transfer occurs from the cold reservoir to a hot reservoir.

Let's use some numbers, T_H = 600 K and T_L = 300 K, so that means the carnot efficiency is 50% and the carnot COP is 1.
Between these temperatures, By connecting a heat engine of efficiency 60% to the carnot fridge of COP = 1, then one can show that the impossible occurs.

But, between these temperatures (T_H = 600 K and T_L = 300K) what if we connect a super-efficient heat engine of efficiency 60% to a fridge of COP = 0.5?

Though a super-efficient heat engine can't exist, coupling the two together gives a net heat transfer from the hot reservoir to the cold reservoir, which could occur. What is wrong with this line of argument?

You end up with more mechanical work output from the heat engine than is needed to simply return the original Qh. If you use that excess mechanical work to move more heat from the cold reservoir you end up moving more than Qh heat back to the hot reservoir.

For the heat engine: W/Qh = (Qh-Qc)/Qh = .6; ∴W=.6Qh

For the reverse cycle: Qc/W = Qc/(Qh-Qc) = Tc/(Th-Tc) = 300/(600-300) = 1; ∴W = Qc'; Qh' = W + Qc' = 2W

So let the heat engine produce W = .6Qh of work. This results in .4Qh flowing to the cold reservoir.

Then use that work to move heat back to the hot reservoir: Qh' = 2W = 1.2Qh. This consists of .6Qh coming from the cold reservoir and .6Qh from the work produced from the heat engine. So there is .2Qh more heatflow (.6Qh - .4Qh) out of the cold reservoir than the heat flow into the reservoir.

So there is a net heat flow from cold to hot with no work being added to the system (all work is generated from the heat flow within the system).

AM

 

[russ_watters]

But, between these temperatures (T_H = 600 K and T_L = 300K) what if we connect a super-efficient heat engine of efficiency 60% to a fridge of COP = 0.5?

I think you mean COP of 2.

Plug the numbers into the efficiency equations. You'll see that a 60% efficient heat engine and COP 2.0 refrigerator don't operate at the same temperatures.

 

[Andrew Mason]

But, between these temperatures (T_H = 600 K and T_L = 300K) what if we connect a super-efficient heat engine of efficiency 60% to a fridge of COP = 0.5?

Though a super-efficient heat engine can't exist, coupling the two together gives a net heat transfer from the hot reservoir to the cold reservoir, which could occur. What is wrong with this line of argument?

'
What you are trying to do is see if a heat engine of efficiency greater than (Th-Tc)/Th violates the second law.

You could simply not run the reverse cycle at all and have a net flow of heat to the cold reservoir. But that would prove nothing.

AM

 

[sardar]

I would respond by saying that the proof presented in the textbooks is based on the fundamental laws of thermodynamics, which state that heat will always flow spontaneously from a hot reservoir to a cold reservoir. This is known as the second law of thermodynamics. The Carnot efficiency is derived from these laws and represents the maximum possible efficiency that can be achieved by a heat engine.

In the scenario presented, where a super-efficient heat engine is connected to a Carnot refrigerator, the net result of a spontaneous heat transfer from the cold reservoir to the hot reservoir violates the second law of thermodynamics. This is because a super-efficient heat engine, by definition, would have an efficiency greater than the Carnot efficiency, which implies that it could transfer heat from a cold reservoir to a hot reservoir without any external work being done. This goes against the principle that heat always flows from hot to cold.

Furthermore, the scenario presented of connecting a super-efficient heat engine to a refrigerator with a COP of 0.5 is not physically possible. The COP of a refrigerator is defined as the ratio of heat removed from the cold reservoir to the work input. In this scenario, the COP would be greater than 1, which is not physically achievable.

In conclusion, the proof provided in textbooks is based on the fundamental laws of thermodynamics and cannot be violated. The scenario presented is not physically possible and goes against the principles of thermodynamics. Therefore, it can be concluded that a heat engine cannot exceed the Carnot efficiency.