Handbook of air conditioning and refrigeration / Shan K

9.28 CHAPTER NINE
From Eq. (9.21), the enthalpy of saturated vapor refrigerant at a temperature of 35°F (1.7°C),
point 1, is
h 1 � 104.465 � 0.098445T s � 0.0001226T s 2 � 9.861 � 10 �7 T s 3
� 104.47 � 0.098445(35) � 0.0001226(35) 2 � 9.861 � 10 �7 (35) 3
� 104.47 � 3.445 � 0.150 � 0.042 � 107.72 Btu / lb
Then the refrigeration effect is calculated as
q rf � h 1 � h 4 � 107.72 � 37.71 � 70.01 Btu/lb (162.8 kJ/kg)
2. From Eq. (9.22), the enthalpy differential h 2 � h 1 on the constant-entropy line corresponding to
a saturated temperature differential T s2 � T s1 � 95 �35 � 60°F in the two-phase region is
h 2 � h 1 ��0.18165 � 0.21502(T s2 � T s1) � 0.0012405(T s2 � T s1) 2
That is, the work input W in � h 2 � h 1 � 10.024 Btu/lb (23.73 kJ/kg).
3. According to Eq. (9.16), COP ref of the refrigerating system is calculated as
4. From Eq. (9.17), the mass flow rate of the refrigerant can be calculated as
If the liquid refrigerant is subcooled to a temperature of 90°F, (32.2°C), then
h 3 � h 4 � 10.409 � 0.26851(90) � 0.0001479(90) 2 � 5.3429 � 10 �7 (90) 3
� 10.409 � 24.166 � 1.198 � 0.389 � 36.16 Btu/lb
Refrigeration effect is then increased to
Also COP ref is increased to
�8.1982 � 10 �6 (T s2 � T s1) 3
� � 0.18165 � 0.21502(60) � 0.0012405(60) 2 � 8.1982 � 10 �6 (60) 3
� �0.182 � 12.901 � 4.466 � 1.771 � 10.02 Btu / lb
m˙ r � Q rc
q rf
�
q rf � 107.72 � 36.16 � 71.56 Btu/lb (166 kJ/kg)
Since 1 ton � 200 Btu/min and 1 hp � 42.41 Btu/min, electric power input to the compressor P in
without subcooling is
P in �
COP ref � q rf
W in
500 � 12,000
70.01
COP ref � 71.56
10.024
500 � 200
42.41 � 6.98
� 70.01
10.024
� 85,702 lb / h (38,874 kg / h)
� 7.14
� 6.98
� 337.8 hp (252 kW)