Thermodynamics

310 | ThermodynamicsWarm environmentat T H = 75°FRefrigeratorCOP = 13.5The coefficients of performance of actual and reversible refrigeratorsoperating between the same temperature limits can be compared as follows:COP R •6 COP R,rev irreversible refrigerator COP R,rev reversible refrigerator7 COP R,rev impossible refrigerator(6–22)A similar relation can be obtained for heat pumps by replacing all COP R ’sin Eq. 6–22 by COP HP .The COP of a reversible refrigerator or heat pump is the maximum theoreticalvalue for the specified temperature limits. Actual refrigerators or heatpumps may approach these values as their designs are improved, but theycan never reach them.As a final note, the COPs of both the refrigerators and the heat pumpsdecrease as T L decreases. That is, it requires more work to absorb heat fromlower-temperature media. As the temperature of the refrigerated spaceapproaches zero, the amount of work required to produce a finite amount ofrefrigeration approaches infinity and COP R approaches zero.EXAMPLE 6–6A Questionable Claim for a RefrigeratorCool refrigerated spaceat T L = 35°FFIGURE 6–52Schematic for Example 6–6.HouseT H = 21°CHPQ HQ·L·Cold outside airT L = –5°C135,000 kJ/hHeat loss·W net,in = ?FIGURE 6–53Schematic for Example 6–7.An inventor claims to have developed a refrigerator that maintains the refrigeratedspace at 35°F while operating in a room where the temperature is75°F and that has a COP of 13.5. Is this claim reasonable?Solution An extraordinary claim made for the performance of a refrigeratoris to be evaluated.Assumptions Steady operating conditions exist.Analysis The performance of this refrigerator (shown in Fig. 6–52) can beevaluated by comparing it with a reversible refrigerator operating betweenthe same temperature limits:COP R,max COP R,rev 1T H >T L 11175 460 R2>135 460 R2 1 12.4Discussion This is the highest COP a refrigerator can have when absorbingheat from a cool medium at 35°F and rejecting it to a warmer medium at75°F. Since the COP claimed by the inventor is above this maximum value,the claim is false.EXAMPLE 6–7Heating a House by a Carnot Heat PumpA heat pump is to be used to heat a house during the winter, as shown inFig. 6–53. The house is to be maintained at 21°C at all times. The house isestimated to be losing heat at a rate of 135,000 kJ/h when the outside temperaturedrops to 5°C. Determine the minimum power required to drivethis heat pump.