# COP of Reversible Refrigeration Cycles

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1. Aug 19, 2015

### Soumalya

Considering two thermal energy reservoirs, one at a higher temperature of TH compared to the other at TL, if we operate a reversible refrigeration cycle between the reservoirs then its coefficient of performance is given by,

COPR,rev=TL/TH-TL

This is the basically the maximum COP a refrigeration cycle can achieve operating between two thermal energy reservoirs at TH and TL.Since a Carnot refrigerator is a reversible refrigeration cycle the above expression for COPR,rev is valid for the reversed Carnot cycle.

If we consider any other reversible refrigeration cycle operating between the same two temperature limits it should have a similar COP as given by the above expression.Now considering a reversed Brayton cycle which is a reversible cycle (all processes are internally reversible) operating between two TERs at TH and TL,it should be given similarly as,

COPreversed brayton cycle=TL/TH-TL=COPreversed carnot cycle

But my textbook outlines that the COP of a reversed Brayton cycle is less than that of a reversed Carnot cycle operating under similar temperature limits.(Please refer to the attachment)

This seems contradictory as all reversible refrigerators should have the same COP operating between the same temperature limits.

Any thoughts?

2. Aug 19, 2015

### Andrew Mason

The explanation given seems to be a bit careless with the use of the term "reversible". When it speaks of the Brayton power cycle consisting of processes each of which are "internally reversible" it appears to be using that term in the sense that each of the processes can be run in the opposite direction. That is a confusing use of the term "reversible". It does not mean that each of the processes are "reversible" in the thermodynamic sense (i.e. that the system and surroundings are always infinitesimally close to thermodynamic equilibrium so that the direction can be reversed by an infinitesimal change of conditions). The Brayton power cycle is not a reversible cycle. Like any thermodynamic cycle, it can be run in the opposite or reverse direction to make heat flow to occur from the cold reservoir to the hot reservoir by doing mechanical work on the system.

AM

3. Aug 21, 2015

### Soumalya

Well I guessed it correctly a cycle that is truly "reversible" in thermodynamic sense should be both internally and externally reversible.Probably Carnot cycle is the best known thermodynamic cycle that is totally reversible in nature.Reversed Brayton cycle involves heat transfer through a finite temperature difference and thus is externally irreversible.

Indeed books should be more careful and specific in descriptions.