Can a Real-World Carnot Engine Achieve Carnot Efficiency?

AI Thread Summary
The discussion centers on the Carnot cycle's efficiency, emphasizing that it is derived from a reversible process, not friction. It is noted that the Carnot cycle represents the theoretical maximum efficiency, which cannot be achieved in practice. The conversation also addresses the Stirling cycle, clarifying that it is not reversible and cannot reach Carnot efficiency, despite claims in some literature. The key distinction is that the Carnot cycle maintains infinitesimally small temperature differences, allowing for zero net entropy change, while the Stirling engine does not. Ultimately, the Carnot cycle remains an ideal benchmark for thermodynamic efficiency that real-world engines cannot attain.
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I think it is due to friction and the carnot efficiency is derived in a reversible cycle.





Also,can i build just a little bit irreversible but still carnot? But if irreversible how can i draw the graph? Still smooth line?
 
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kntsy said:
I think it is due to friction and the carnot efficiency is derived in a reversible cycle.Also,can i build just a little bit irreversible but still carnot? But if irreversible how can i draw the graph? Still smooth line?
The thermodynamic inefficiency of the Carnot cycle is NOT due to friction. It has nothing to do with friction and everything to do with the second law of thermodynamics.

The Carnot cycle is an ideal. It represents the theoretical limit of thermodynamic efficency. It can be approached but never reached in the real world.

AM
 


Andrew Mason said:
The thermodynamic inefficiency of the Carnot cycle is NOT due to friction. It has nothing to do with friction and everything to do with the second law of thermodynamics.

The Carnot cycle is an ideal. It represents the theoretical limit of thermodynamic efficency. It can be approached but never reached in the real world.

AM

If i use a stirling cycle in a reversible process, can the efficiency reach (1-T2/T1)? It is reversible so can?
 


kntsy said:
If i use a stirling cycle in a reversible process, can the efficiency reach (1-T2/T1)? It is reversible so can?

I'm not familiar with whatever equation you are trying to use so i can't comment on that.

However, the stirling engine is very far from 100% efficient. At best i believe it is around 40%. Since you cannot reverse the stirling engine without using work, it isn't reversible in a thermodynamic standpoint.
 


So a reversible stirling cycle is always less efficient to carnot efficiency? BOOK says ANY reversible cycle can reach carnot effieciency! Is the BOOK wrong?
 


kntsy said:
So a reversible stirling cycle is always less efficient to carnot efficiency? BOOK says ANY reversible cycle can reach carnot effieciency! Is the BOOK wrong?
A Stirling cycle is not reversible. You cannot simply make any cycle reversible. The Carnot cycle is reversible because heat flows into and out of the engine with the engine and reservoirs at the virtually the same temperature (in infinitessimally small difference in temperature). Since the temperature of the reservoir and system gas are the same, the change in entropy of the reservoir is equal and opposite to that of the engine (sum of entropy changes = 0). The other parts of the cycle are quasistatic adiabatic processes which could be reversed with an infinitessimal change in pressures (since they are adiabatic, dQ = 0 -> dS = dQ/T = 0). So there is no change in entropy of the system and surroundings during the Carnot cycle.

In the Stirling engine, heat does not flow into or out of the engine due to infinitessimally small differences in temperature. So there is a net increase in entropy during flows into or out of the engine and the process is not reversible. The Stirling engine cannot be made to equal the efficiency of the Carnot cycle no matter how well it is designed.

AM
 
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