Refrigeration and Airconditioning Miscellaneous
 A R12 refrigerant reciprocating compressor operates between the condensing temperature of 30º and evaporator temperature of –20ºC. The clearance volume ratio of the compressor is 0.03. Specific heat ratio of the vapour is 1.15 and the specific volume at the suction is 0.1089 m^{3}/kg. Other properties at various states are given in the figure. To realize 2 tons of refrigeration, the actual volume displacement rate considering the effect of clearance is

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Given: Clearance volume ratio, C = 0.03
Specific volume at suction, υ_{1} = 0.1089 m^{3} /kg
Net refrigerating effect = 2 ton = 2 × 3.516 kJ = 7.032 kJ/s
Specific heat ratio, c = 1.15
∴ Net refrigerating effect = ṁ(h_{1} – h_{4})
or 7.032 = ṁ(176 – 65)
or ṁ = 0.063 kg/s
Volume = 0.063 × 0.1089 = 6.89 × 10^{3} m^{–3} /sVolumetric efficiency = 1 + C  C p_{2} ^{1 / n} p_{1} Volumetric efficiency = 1 + 0.03  0.03 7.45 ^{1 / 1.15} 1.50
= 0.909
∴ Volume displacement rate considering clearance effect = 6.89 × 10^{–3} × 0.909 = 6.26 × 10^{–3} m^{3} /sCorrect Option: A
Given: Clearance volume ratio, C = 0.03
Specific volume at suction, υ_{1} = 0.1089 m^{3} /kg
Net refrigerating effect = 2 ton = 2 × 3.516 kJ = 7.032 kJ/s
Specific heat ratio, c = 1.15
∴ Net refrigerating effect = ṁ(h_{1} – h_{4})
or 7.032 = ṁ(176 – 65)
or ṁ = 0.063 kg/s
Volume = 0.063 × 0.1089 = 6.89 × 10^{3} m^{–3} /sVolumetric efficiency = 1 + C  C p_{2} ^{1 / n} p_{1} Volumetric efficiency = 1 + 0.03  0.03 7.45 ^{1 / 1.15} 1.50
= 0.909
∴ Volume displacement rate considering clearance effect = 6.89 × 10^{–3} × 0.909 = 6.26 × 10^{–3} m^{3} /s
 In the window air conditioner, the expansion device used is

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The capillary tube is used as expansion device
Correct Option: A
The capillary tube is used as expansion device
Direction: A refrigerator based on ideal vapour compression cycle operates between the temperature limits of –20°C and 40°C. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in the table below.
T  h_{f}  h_{g}  s_{f}  s_{g} 
(°C)  (kJ/kg)  (kJ/kg)  (kJ/kgK)  (kJ/kgK) 
–20  20  180  0.07  0.7366 
40  80  200  0.3  0.67 
 The COP of the refrigerator is

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C.O.P = h_{2} – h_{1} h_{3} – h_{2} C.O.P = 164 – 80 = 84 = 2.33 200 – 164 36 Correct Option: B
C.O.P = h_{2} – h_{1} h_{3} – h_{2} C.O.P = 164 – 80 = 84 = 2.33 200 – 164 36
 If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to

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h_{1} = h_{4} = 80 kJ/kg , s_{3} = 0.67
h_{3} = 200 kJ/kg , s_{2} = 0.67
h_{1}' = 20 kJ/kg , s_{2}' = 0.7366
h_{2}' = 20 kJ/kg , s_{1}' = 0.07
s_{2} = s_{1}' + x(s_{2}'  s_{1}')
= 0.07 + x + (0.7366 – 0.07)
⇒ 0.67 = 0.07 + 0.6666 x⇒ x = 0.67  0.07 = 0.90 0.6666
h_{2} = h_{1}' + x(h_{2}'  h_{1}')
= 20 + 0.9(180 – 20)
= 20 + 0.9 × 160 = 165 kJ/kg
Refrigerating effect per kg = h_{2} – h_{1} = 164 – 80 = 84 kJ/kg
Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kWCorrect Option: A
h_{1} = h_{4} = 80 kJ/kg , s_{3} = 0.67
h_{3} = 200 kJ/kg , s_{2} = 0.67
h_{1}' = 20 kJ/kg , s_{2}' = 0.7366
h_{2}' = 20 kJ/kg , s_{1}' = 0.07
s_{2} = s_{1}' + x(s_{2}'  s_{1}')
= 0.07 + x + (0.7366 – 0.07)
⇒ 0.67 = 0.07 + 0.6666 x⇒ x = 0.67  0.07 = 0.90 0.6666
h_{2} = h_{1}' + x(h_{2}'  h_{1}')
= 20 + 0.9(180 – 20)
= 20 + 0.9 × 160 = 165 kJ/kg
Refrigerating effect per kg = h_{2} – h_{1} = 164 – 80 = 84 kJ/kg
Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kW
 In a vapour compression refrigeration system, liquids to suction heat exchanger is used to

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NA
Correct Option: C
NA