Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

The coefficients of performance of actual and reversible refrigerators

operating between the same temperature limits can be compared as follows:

(6–22)

A similar relation can be obtained for heat pumps by replacing all COPR’s

in Eq. 6–22 by COPHP.

The COP of a reversible refrigerator or heat pump is the maximum theo-

retical value for the specified temperature limits. Actual refrigerators or heat

pumps may approach these values as their designs are improved, but they

can never reach them.

As a final note, the COPs of both the refrigerators and the heat pumps

decrease as TLdecreases. That is, it requires more work to absorb heat from

lower-temperature media. As the temperature of the refrigerated space

approaches zero, the amount of work required to produce a finite amount of

refrigeration approaches infinity and COPRapproaches zero.

COPR•

6 COPR,rev¬¬irreversible refrigerator

 COPR,rev¬¬reversible refrigerator

7 COPR,rev¬¬impossible refrigerator

310 | Thermodynamics

EXAMPLE 6–6 A Questionable Claim for a Refrigerator

An inventor claims to have developed a refrigerator that maintains the refrig-

erated space at 35°F while operating in a room where the temperature is

75°F and that has a COP of 13.5. Is this claim reasonable?

Solution An extraordinary claim made for the performance of a refrigerator

is to be evaluated.

Assumptions Steady operating conditions exist.

Analysis The performance of this refrigerator (shown in Fig. 6–52) can be

evaluated by comparing it with a reversible refrigerator operating between

the same temperature limits:

Discussion This is the highest COP a refrigerator can have when absorb-

ing heat from a cool medium at 35°F and rejecting it to a warmer medium at

75°F. Since the COP claimed by the inventor is above this maximum value,

the claim isfalse.



1

175 460 R2> 1 35 460 R 2  1

12.4

COPR,maxCOPR,rev

1

TH>TL 1

EXAMPLE 6–7 Heating a House by a Carnot Heat Pump

A heat pump is to be used to heat a house during the winter, as shown in

Fig. 6–53. The house is to be maintained at 21°C at all times. The house is

estimated to be losing heat at a rate of 135,000 kJ/h when the outside tem-

perature drops to 5°C. Determine the minimum power required to drive

this heat pump.

Cool refrigerated space

at TL = 35°F

Refrigerator

Warm environment

at TH = 75°F

COP = 13.5

FIGURE 6–52

Schematic for Example 6–6.

135,000 kJ/h

Heat loss

House

TH = 21°C

Cold outside air

TL = –5°C

HP

Wnet,in =?

QH

QL

·

·

·

FIGURE 6–53

Schematic for Example 6–7.