Modern Engineering Thermodynamics

14.3 Carnot Refrigeration Cycle 537
14.3 CARNOT REFRIGERATION CYCLE
In Chapter 7, we discovered that refrigerators, air conditioners, and heat pumps usually have actual thermal
efficiencies in excess of 100%. This is due simply to the mathematical way in which their thermal efficiencies are
defined (the desired energy output divided by the required energy input) and does not imply the violation of any
physical law. However, claims of thermal efficiency in excess of 100% cause obvious credibility problems in the
public domain, so the term efficiency is not often used with this technology. Instead, we simply rename the
thermal efficiency the coefficient of performance (COP), which is expressed as a pure number, usually between 1
and 10, rather than as a percentage. The COP definitions have been given in Eqs. (7.17) and (7.19) (recall that
work into and heat out of a system are both negative quantities in our sign convention; as in previous chapters, we
use their absolute values in these equations and assign algebraic signs to the symbols to avoid confusion):
and
and it is easily shown that
ðη T Þ heat
hump
ðη T Þ refrigerator or
air conditioner
= COP HP = jQ Hj
jW HP j = j _Q H j
j _W HP j =
= COP R/AC =
Q L
jW R/AC j =
jQ H j
jQ H j − Q L
=
_Q L
j _W R/AC j =
Q L
jQ H j − Q L
=
j _Q H j
j _Q H j − _Q L
(7.17)
_Q L
j _Q H j − _Q L
(7.19)
COP HP = COP R/AC + 1 (14.1)
From Figure 14.1 and Eq. (14.1), it is evident that Eqs. (7.17) and (7.19) can also be written as
and
COP R/AC =
COP HP =
j _Q H j
j _W in j net
_Q L
j _W in j net
= COP HP − 1
(7.17a)
(7.19a)
Note that Eq. (7.17a) is simply the inverse of the general forward-running heat engine thermal efficiency equation
(see Eq. (7.5)); that is,
1
COP HP
ðη T Þ forward-running
(14.2)
heat engine
and Eq. (7.17a) then gives
COP R/AC
1
ðη T Þ forward-running
heat engine
− 1 (14.3)
Therefore, the COP for any of the heat engines discussed in Chapter 13 operating on a reversed thermodynamic cycle
as a heat pump, refrigerator, or air conditioner can be easily obtained through the use of Eqs. (14.2) and (14.3). 2
For example, Eq. (7.16) gives the Carnot thermal efficiency as
ðη T Þ Carnot
= 1 − T L
= T H − T L
T H T L
Then, Eqs. (14.2) and (14.3) can be used directly to give the COP of a Carnot engine running backward as a
heat pump, refrigerator, or air conditioner as
Also, it is easy to show that Eq. (14.1) remains valid for these systems.
COP Carnot HP =
T H
T H − T L
(14.4)
T
COP Carnot R/AC = L
T H − T L
(14.5)
2 However, it is difficult to imagine an internal combustion engine (like the Otto and Diesel cycle) running backward because it would
require a heat-absorbing (endothermic) combustion reaction.