Page 269 - Schaum's Outline of Theory and Problems of Applied Physics
P. 269
254 THERMODYNAMICS [CHAP. 21
Here T 1 =−20 C + 273 = 253 K and T 2 = 40 C + 273 = 313 K. The CP of an ideal refrigerator operating
◦
◦
between these temperatures is
T 1 253 K
CP = = = 4.2
T 2 − T 1 313 K − 253 K
The actual refrigerator is 2.0/4.2 = 0.48 = 48 percent as efficient as the ideal refrigerator.
SOLVED PROBLEM 21.15
At what rate does the refrigerator of Prob. 21.14 remove heat from the freezer compartment?
Since P = W/t and Q 1 = (CP)(W),
Q 1 W
= (CP) = (CP)(P) = (2.0)(1kW) = 2kW
t t
For every joule of energy supplied to the refrigerator, it removes2Jof heat from the freezer compartment.
SOLVED PROBLEM 21.16
A refrigerator which is half as efficient as an ideal refrigerator extracts heat from a storage chamber at
◦
0 F and exhausts it at 100 F. How many foot-pounds of work per Btu extracted does this refrigerator
◦
require?
We begin by finding the coefficient of performance of the ideal refrigerator. Since
T 1 = 0 F = 460 = 460 R T 2 = 100 F + 460 = 560 R
◦
◦
◦
◦
◦
◦
◦
Q 1 T 2 560 R
= = = 5.6
W T 2 − T 1 560 R − 460 R
◦
◦
Here Q 1 = 1 Btu and, since 1 Btu = 778 ft·lb,
778 ft·lb
Q 1
W = = = 139 ft·lb
5.6 5.6
This refrigerator is half as efficient as an ideal refrigerator, so the required work per Btu is twice as great, or 278 ft·lb.
SOLVED PROBLEM 21.17
The British unit of refrigeration capacity is the ton, which is that rate of heat extraction that can freeze 1
◦
ton of water at 32 F to ice at 32 F per day. Since the heat of fusion of water is 144 Btu/lb,
◦
1 refrigeration ton = 12,000 Btu/h
The refrigerator of Prob. 21.16 has a capacity of 2 tons. How much power is required to operate its
compressor?
Since the refrigerator requires 278 ft·lb of work per Btu of heat extracted,
W Btu/h ft·lb 1
P = = (2 tons) 12,000 278 = 1853 ft·lb/s
t ton Btu 3600 s/h
In terms of horsepower,
1853 ft·lb/s
P = = 3.37 hp
550 (ft·lb/s)/hp