16. 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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17. 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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18. 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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19. Dew point temperature of air at one atmospheric pressure (1.013 bar) is 18°C. The air dry bulb temperature is 30°C. The saturation pressure of water at 18°C and 30°C are 0.02062 bar and 0.04241 bar respectively. The specific heat of air and water vapour respeetively are 1.005 and 1.88 kJ/kgK and the latent heat of vaporization water of 0°C is 2500 kJ/kg. The specific humidity (kg/kg of dry air) and enthalpy (kJ/kg of dry air) of this moist air respectively, are

1. 1. 0.01051, 52.64

   
   
   

   2. 0.01291, 63.15

   
   
   

   3. 0.01481, 78.60

   
   
   

   4. 0.01532, 81.40

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: p = 1.013 bar, (p_υ)_tdp = 0.02062
   

   Specific humidity = 0.622	Pυ
   	p - Pυ

   
   

   =	0.622 × 0.02062	= 0.01291 kg/kg of dry air
   	1.013 - 0.02062

   
   h = 1.022 t_d + w(h_fgdp + 2.3t_dp)
   = 63.15 kJ/kg of dry air.
   Alternative method
   p_ε = Saturation pressure = 0.02062 bar
   p_a = Partial pressure of air = 0.04241 bar
   t_d = Dry bulb temperature = 30º
   t_w = Wet bulb temperature = 18º
   Specific humidity,
   

   W =	pυ	× 0.622 = 0.01292 kg/kg of dry air.
   	p - pυ

   
   Total enthalpy,
   h = ^cp_a t_d + wh_s + w_c p_s (t_d – t_ω)
   = 1.005 × 30 + 0.0129 × 2500 + 0.01292 × 1.88(30 – 18)
   = 63.15 kJ/kg of dry air.

   Correct Option: B

   Given: p = 1.013 bar, (p_υ)_tdp = 0.02062
   

   Specific humidity = 0.622	Pυ
   	p - Pυ

   
   

   =	0.622 × 0.02062	= 0.01291 kg/kg of dry air
   	1.013 - 0.02062

   
   h = 1.022 t_d + w(h_fgdp + 2.3t_dp)
   = 63.15 kJ/kg of dry air.
   Alternative method
   p_ε = Saturation pressure = 0.02062 bar
   p_a = Partial pressure of air = 0.04241 bar
   t_d = Dry bulb temperature = 30º
   t_w = Wet bulb temperature = 18º
   Specific humidity,
   

   W =	pυ	× 0.622 = 0.01292 kg/kg of dry air.
   	p - pυ

   
   Total enthalpy,
   h = ^cp_a t_d + wh_s + w_c p_s (t_d – t_ω)
   = 1.005 × 30 + 0.0129 × 2500 + 0.01292 × 1.88(30 – 18)
   = 63.15 kJ/kg of dry air.

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20. A stream of moist air (mass flow rate = 10.1 kg/s) with humidity ratio of 0.01 kg/kg dry air mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper and uniform mixing with no condensation, the humidity ratio of the final stream in (kg/kg dry air) is ______.

1. 1. 0.02 kg/kg dry air

   
   
   

   2. 0.01 kg/kg dry air

   
   
   

   3. 0.03 kg/kg dry air

   
   
   

   4. 0.5 kg/kg dry air

   
   
   

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1. View Hint View Answer Discuss in Forum

   ωnew =	m1ω1 + m2ω2
   	m1 + m2

   
   

   ⇒	0.1 × 10.1 × + .1 × 1	= 0.02kg/kg dry air
   	10.1 + .1

   
   

   Correct Option: A

   ωnew =	m1ω1 + m2ω2
   	m1 + m2

   
   

   ⇒	0.1 × 10.1 × + .1 × 1	= 0.02kg/kg dry air
   	10.1 + .1

   
   

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