EFFECT OF SUBCOOLING (OR UNDERCOOLING) OF REFRIGERANT VAPOUR

Consider the T-s and p-h diagrams for the vapour compression cycle shown in Fig. 19.20 in which the refrigerant after condensation process 2′−3′, is cooled below the saturation temperature T₃′ before throttling process to temperature T₃. Such a process is called undercooling or subcooling of the refrigerant. It is generally done along the saturated liquid line. The effect of undercooling is to increase the COP.

Refrigerating effect, R_E = h₁ − h₄ = h₁ − h_f3, where h_f3 = h_f3′−c_pf(T_3′ − T₃)

Work done, w = h₂ − h₁

Figure 19.20 Vapour compression cycle with subcooling of refrigerant: (a) T-s diagram, (b) p-h diagram

Example 19.12

A food storage locker requires a refrigeration system of 42 kW capacity at an evaporator temperature of −5°C and a condenser temperature of 40°C. The refrigerant, R-12, is sub-cooled 5°C before entering the expansion valve, and the vapour is superheated 6°C before leaving the evaporator coil. The compression of the refrigerant is reversible adiabatic. A two-cylinder vertical single acting compressor with stroke equal to 1.5 times the bore is to be used operating at 960 rpm. Determine (a) the refrigerating effect/kg, (b) the mass of refrigerant to be circulated per minute, (c) the theoretical piston displacement per minute, (d) the theoretical power, (e) the co-efficient of performance, (f) the heat removed through condenser/kg and (g) the theoretical bore and stroke of compressor.

Data for R−12 refrigerant is given below:

Solution

Given: Q = 42 kW, T₁ = −5°C or 273−5= 268 K, T_2′ = 40°C or 273 + 40 = 318K, T_3′ −T₃ = 5°C, T₁−T_1′ = 6°C, n = 2, L = 1.5D, N = 960 rpm.

The T-s diagram is shown in Fig. 19.21.

Figure 19.21 T-s diagram

Process l′−1: superheating

h₁ = h_g1′ + c_pgl′(T₁ − T_1′) = 350.44 + 0.6455 × 6

= 354.313 kJ/kg

= 1.5778 kJ/kg.K

Process 1−2: Isentropic compression

T₂ = 326.38 K

h₂ = h_g2′ + c_pg2′(T₂ − T_2′)

= 368.81 + 0.6455 (326.38 − 313)

= 379.00 kJ/kg

Process 3′−3: Subcooling

 

h₂ = h_g3′ + c_pg3′(T₃ − T_3′) = 239.03 − 1.030 × 5 = 233.88kj/kg

Process 3−4: Throttling

h₄ = h₃ =233.88

Specific volume at suction to compressor

1. Refrigeration effect/kg, q₄₋₁ = h₁ − h₄ = 354.313 −233.88 = 120.433kJ/kg
2. Mass of refrigerant required/min, 
3. Theoretical suction volume/min = ṁ × v₁ = 20.924 × 0.06706 = 1.403 m³/min
4. Power, 
5. 
6. Heat removed through the condenser/kg = h₂ − h₃ = 379.00 − 233.88 = 145.12 kJ/kg
7. Theoretical suction volume per cylinder per minute