Refrigeration and Airconditioning Miscellaneous
 Refrigerant vapor enters into the compressor of a standard vapor compression cycle at –10°C (h = 402 J/kg) and leaves the compression at 50°C (h = 432 kJ/kg). It leaves the condenser at 30°C (h = 237 kJ/kg). The COP of the cycle is ________.

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Work done = 432 – 402 = 30 kJ/kg
Refrigerating effect = 402 – 237 = 165 kJ/kg.COP = 165 = 5.5 30 Correct Option: B
Work done = 432 – 402 = 30 kJ/kg
Refrigerating effect = 402 – 237 = 165 kJ/kg.COP = 165 = 5.5 30
 Clearance volume of a reciprocating compressor is 100 ml, and the volume of the cylinder at bottom dead centre is 1.0 litre. The clearance ratio of the compressor is

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Clearance ratio = Clearance volume Swept volume = 100 = 1 1000  100 9 Correct Option: C
Clearance ratio = Clearance volume Swept volume = 100 = 1 1000  100 9
 A refrigerator uses R134a as its refrigerant and operates on a ideal vapourcompression refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, the rate of heat rejection to the environment is _______ kW.
Given data:
At P = 0.14 MPa, h = 236.04 kJ/kg , s = 0.9322 kJ/kgK
At P = 0.8 MPa, h = 272.05 kJ/kg (superheated vapour)
At P = 0.8 MPa, h = 93.42 kJ/kg (saturated liquid)

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h_{1} = 236.04 kJ/kg
h_{2} = 272.05 kJ/kg
h_{3} = 93.42 kJ/kg
s_{1} = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h_{2} – h_{3}) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kWCorrect Option: A
h_{1} = 236.04 kJ/kg
h_{2} = 272.05 kJ/kg
h_{3} = 93.42 kJ/kg
s_{1} = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h_{2} – h_{3}) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kW
 ListI
A. Liquid to suction heat exchange
B. Constant volume heat addition
C. Normal shock
D. Ammoniawater
ListII
1. Vapour absorption refrigeration
2. Vapour compression refrigeration
3. Diesel cycle
4. Otto cycle
5. Converging nozzle
6. Converging diverging nozzle

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A – 2, B – 4, C – 6, D – 1
Correct Option: A
A – 2, B – 4, C – 6, D – 1
 The power required for compressor in kW is

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Power required by compressor p = ṁ(h_{2} – h_{1})
= 0.2 (276.45 – 237) = 7.9 kWCorrect Option: C
Power required by compressor p = ṁ(h_{2} – h_{1})
= 0.2 (276.45 – 237) = 7.9 kW