## Refrigeration and Air-conditioning Miscellaneous

#### Refrigeration and Air-conditioning

1. 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 ________.

1. 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

1. 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

1.  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

1. A refrigerator uses R134a as its refrigerant and operates on a ideal vapour-compression 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)

1. h1 = 236.04 kJ/kg
h2 = 272.05 kJ/kg
h3 = 93.42 kJ/kg
s1 = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h2 – h3) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kW

##### Correct Option: A

h1 = 236.04 kJ/kg
h2 = 272.05 kJ/kg
h3 = 93.42 kJ/kg
s1 = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h2 – h3) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kW

1. List-I
A. Liquid to suction heat exchange
C. Normal shock
D. Ammonia-water
List-II
1. Vapour absorption refrigeration
2. Vapour compression refrigeration
3. Diesel cycle
4. Otto cycle
5. Converging nozzle
6. Converging diverging nozzle

1. A – 2, B – 4, C – 6, D – 1

##### Correct Option: A

A – 2, B – 4, C – 6, D – 1

1. The power required for compressor in kW is

1. Power required by compressor p = ṁ(h2 – h1)
= 0.2 (276.45 – 237) = 7.9 kW

##### Correct Option: C

Power required by compressor p = ṁ(h2 – h1)
= 0.2 (276.45 – 237) = 7.9 kW