# If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/t1?

1. Sep 24, 2010

### kntsy

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?

2. Sep 24, 2010

### Andrew Mason

Re: If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/

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

3. Sep 24, 2010

### kntsy

Re: If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/

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

4. Sep 25, 2010

### Drakkith

Staff Emeritus
Re: If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/

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.

5. Sep 25, 2010

### kntsy

Re: If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/

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?

6. Sep 25, 2010

### Andrew Mason

Re: If i build a carnot engine in the real world, is its efficiney smaller than 1-t2/

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