A refrigerator using a 0.5 hp motor has a coefficient of performance of 3.05. 2 gallons of tea (model as water) is placed into the refrigerator at 80oF. How long will it take for the refrigerator to cool the tea to 40oF, assuming that this is the only heat load to be considered. Does this seem realistic? If not, what assumptions were made which make the calculation appear unrealistic?

Question

Given: ?= 3.05; W?= -0.5 hp = 0.3534 Btu/s; Water: V= 2 gal = 0.2674 ft3; T1 = 80oF; T2 = 40oF

Assume: The refrigerator is operating in a cyclic fashion. The water behaves as an incompressible substance with constant specific heats: c = 1 Btu/lbm-RSolution:
What will be an ideal response?

Answer 1

The refrigeration cycle heat input rate is
Q?C=?|W?| = 1.078 Btu/s
Take the density of the water to be ? = 62.4 lbm/ft3
Then m = pV = 16.69 lbm
Considering the water to be a closed system, and W = ?KE=?PE = 0 for the cooling water, the First Law for closed systems reduces to
Q = m (u2 – u1) = mc(T2 – T1) = -667.4 Btu
This is heat out of the tea/water and into the cool reservoir, and this heat is removed by the refrigeration cycle. So Qc = 667.4 Btu
The time required to remove this heat is t = QCQ?C = 619 s = 10.3 min.
This appears to be somewhat fast. It is probably unrealistic to neglect the conduction/convection of heat from the center of the containers to the edges where it gets removed to the refrigerator interior. Also, the analysis neglects to consider the movement of the heat from the containers through the air to the walls, where it is removed from the refrigerator. The outside regions of the tea may be cooled to 40oF this quickly, but it is doubtful that the entire mass will be.
However, it should also be noted that the compressor power is rather large for a typical refrigerator, and so while likely still short, the predicted cooling time is probably not too far off.