Derive the Carnot Efficiency without using the Carnot Cycle? [closed]

How can you derive the Carnot efficiency using only properties of reversible cycles?

closed as unclear what you're asking by Whit3rd, Gert, stafusa, Jon Custer, Kyle KanosNov 22 '17 at 11:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What do you mean deriving it without the Carnot cycle? Carnot's formula is the efficiency of the Carnot cycle. – hyportnex Nov 21 '17 at 20:44
• The Carnot Efficiency is the efficiency for ANY reversible heat engine. I want to know if there is a way of deriving it without using the Carnot Cycle, just from the properties of reversible heat engines. – user140323 Nov 21 '17 at 21:15
• maybe this will help: physics.stackexchange.com/questions/300347/… – hyportnex Nov 21 '17 at 21:38
• Yes it is. Carnot's Theorem states that ALL reversible engines have the same efficiency. The Carnot Cycle is just one special, very simple case of a reversible engine. – user140323 Nov 21 '17 at 21:39
• @Gert dude, how right you are! – hyportnex Nov 21 '17 at 21:42

Use the second law of thermodynamics.

Consider as the system a generic reversible engine plus the cold and hot sources. This is a closed and reversible system, therefore its entropy change vanishes. You decompose this change in changes due to the engine and due to the sources. The entropy change of the engine after a cycle vanishes, thus $$\Delta S=\Delta S_{\mathrm{sources}}=0.$$ The hot source loses $|Q_1|$ at constant temperature $T_1$, whereas the cold source gain $|Q_2|$ at temperature $T_2$. Hence, $$\Delta S=-\frac{|Q_1|}{T_1}+\frac{|Q_2|}{T_2}=0,$$ i.e. $$\frac{|Q_2|}{|Q_1|}=\frac{T_2}{T_1}.$$

Plug this into the efficiency $$\eta=\frac{W}{|Q_1|}=1-\frac{|Q_2|}{|Q_1|},$$ and obtain the efficiency of a reversible engine $$\eta=1-\frac{T_2}{T_1}.$$

Notice that we have not made any assumption about the cycle or even the agent responsible by the engine. That is in the Core of Carnot theorem. Any engine, regardless its nature, working between the same sources have the same efficiency, so it is natural that this efficiency can be calculated without any reference to the Carnot cycle which is a specific cycle followed by a specific agent (ideal gas).

• wait a minute- does not the carnot efficiency represent the theoretical maximum possible efficiency of any heat engine operating between two given temperatures, rather than representing the universal efficiency of any and all heat engines operating between those two temperatures? – niels nielsen Nov 21 '17 at 23:04
• It represents the efficiency of any reversible engine operating between two temperatures. That efficiency happens to be the maximum possible efficiency for heat engines. – user140323 Nov 21 '17 at 23:10
• @nielsnielsen It represents the theoretical maximum efficiency of any heat engine between two sources, equivalently, the universal efficiency of all reversible engines operating between two sources. The key point here is the term reversible. – Diracology Nov 21 '17 at 23:15
• Thanks! Is there anyway of deriving it without the use of entropy? Because increasing entropy is derived from the Claudia Inequality, which is in turn derived from Carnot's Theorem. Is there any way? – user140323 Nov 21 '17 at 23:19
• @user140323 The only other way I know is by calculating heats and work explicitly from the Carnot cycle diagram. – Diracology Nov 21 '17 at 23:22