11. Round the clock cooling of an apartment having a load of 300 MJ/day requires and air conditioning plant of capacity about
    

1. 1. 1 ton
      

   
   
   

   2. 5 tons
      

   
   
   

   3. 10 tons
      

   
   
   

   4. 100 tons

   
   
   

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1. View Hint View Answer Discuss in Forum

   Capacity= Q = 300 MJ/day
   

   =	300 × 103	kJ / s
   	24 × 60 × 60

   
   = 3.47 kW
   1 ton = 3.5 kW
   

   ∴ Q =	3.47	= 0.99 ton = 1 ton
   	3.5

   Correct Option: A

   Capacity= Q = 300 MJ/day
   

   =	300 × 103	kJ / s
   	24 × 60 × 60

   
   = 3.47 kW
   1 ton = 3.5 kW
   

   ∴ Q =	3.47	= 0.99 ton = 1 ton
   	3.5

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12. 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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13. If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to
    

1. 1. 2.1 kW
      

   
   
   

   2. 2.5 kW
      

   
   
   

   3. 3.0 kW
      

   
   
   

   4. 4.0 kW

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   h₁ = h₄ = 80 kJ/kg , s₃ = 0.67
   h₃ = 200 kJ/kg , s₂ = 0.67
   h₁' = 20 kJ/kg , s₂' = 0.7366
   h₂' = 20 kJ/kg , s₁' = 0.07
   s₂ = s₁' + x(s₂' - s₁')
   = 0.07 + x + (0.7366 – 0.07)
   ⇒ 0.67 = 0.07 + 0.6666 x
   

   ⇒ x =	0.67 - 0.07	= 0.90
   	0.6666

   
   h₂ = h₁' + x(h₂' - h₁')
   = 20 + 0.9(180 – 20)
   = 20 + 0.9 × 160 = 165 kJ/kg
   Refrigerating effect per kg = h₂ – h₁ = 164 – 80 = 84 kJ/kg
   Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kW

   Correct Option: A

   
   h₁ = h₄ = 80 kJ/kg , s₃ = 0.67
   h₃ = 200 kJ/kg , s₂ = 0.67
   h₁' = 20 kJ/kg , s₂' = 0.7366
   h₂' = 20 kJ/kg , s₁' = 0.07
   s₂ = s₁' + x(s₂' - s₁')
   = 0.07 + x + (0.7366 – 0.07)
   ⇒ 0.67 = 0.07 + 0.6666 x
   

   ⇒ x =	0.67 - 0.07	= 0.90
   	0.6666

   
   h₂ = h₁' + x(h₂' - h₁')
   = 20 + 0.9(180 – 20)
   = 20 + 0.9 × 160 = 165 kJ/kg
   Refrigerating effect per kg = h₂ – h₁ = 164 – 80 = 84 kJ/kg
   Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kW

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14. The COP of the refrigerator is
    

1. 1. 2.0
      

   
   
   

   2. 2.33
      

   
   
   

   3. 5.0
      

   
   
   

   4. 6.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   C.O.P =	h2 – h1
   	h3 – h2

   
   

   C.O.P =	164 – 80	=	84	= 2.33
   	200 – 164		36

   Correct Option: B

   C.O.P =	h2 – h1
   	h3 – h2

   
   

   C.O.P =	164 – 80	=	84	= 2.33
   	200 – 164		36

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15. The vapour compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of the evaporator. The coordinate system used in this figure is
    
    

1. 1. p – h
      

   
   
   

   2. T – s
      

   
   
   

   3. p – s
      

   
   
   

   4. T – h

   
   
   

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1. View Hint View Answer Discuss in Forum

   T-h curve

   Correct Option: D

   T-h curve

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