6. 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. 1. 1/11

   
   
   

   2. 1/10

   
   
   

   3. 1/9

   
   
   

   4. 1/1.1

   
   
   

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1. View Hint View Answer Discuss in Forum

   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

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7. 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. 1. 8.931 kW
      

   
   
   

   2. 89.31 kW
      

   
   
   

   3. 9.831 kW
      

   
   
   

   4. 9.813 kw

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   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

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8. 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. 1. 5.0
      

   
   
   

   2. 5.5
      

   
   
   

   3. 4.5
      

   
   
   

   4. 1.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   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

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9. The power required for compressor in kW is
   

1. 1. 5.94
      

   
   
   

   2. 1.83
      

   
   
   

   3. 7.9
      

   
   
   

   4. 39.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   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

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10. In a vapour compression refrigeration system, liquids to suction heat exchanger is used to
    

1. 1. Keep the COP constant
      

   
   
   

   2. Prevent the liquid refrigerant from entering the compressor
      

   
   
   

   3. Sub-cool the liquid refrigerant leaving the condenser
      

   
   
   

   4. Sub-cool the vapour refrigerant from the evaporator

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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