Refrigeration and Airconditioning Miscellaneous
Direction: A refrigerator based on ideal vapour compression cycle operates between the temperature limits of –20°C and 40°C. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in the table below.
T  h_{f}  h_{g}  s_{f}  s_{g} 
(°C)  (kJ/kg)  (kJ/kg)  (kJ/kgK)  (kJ/kgK) 
–20  20  180  0.07  0.7366 
40  80  200  0.3  0.67 
 If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to

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h_{1} = h_{4} = 80 kJ/kg , s_{3} = 0.67
h_{3} = 200 kJ/kg , s_{2} = 0.67
h_{1}' = 20 kJ/kg , s_{2}' = 0.7366
h_{2}' = 20 kJ/kg , s_{1}' = 0.07
s_{2} = s_{1}' + x(s_{2}'  s_{1}')
= 0.07 + x + (0.7366 – 0.07)
⇒ 0.67 = 0.07 + 0.6666 x⇒ x = 0.67  0.07 = 0.90 0.6666
h_{2} = h_{1}' + x(h_{2}'  h_{1}')
= 20 + 0.9(180 – 20)
= 20 + 0.9 × 160 = 165 kJ/kg
Refrigerating effect per kg = h_{2} – h_{1} = 164 – 80 = 84 kJ/kg
Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kWCorrect Option: A
h_{1} = h_{4} = 80 kJ/kg , s_{3} = 0.67
h_{3} = 200 kJ/kg , s_{2} = 0.67
h_{1}' = 20 kJ/kg , s_{2}' = 0.7366
h_{2}' = 20 kJ/kg , s_{1}' = 0.07
s_{2} = s_{1}' + x(s_{2}'  s_{1}')
= 0.07 + x + (0.7366 – 0.07)
⇒ 0.67 = 0.07 + 0.6666 x⇒ x = 0.67  0.07 = 0.90 0.6666
h_{2} = h_{1}' + x(h_{2}'  h_{1}')
= 20 + 0.9(180 – 20)
= 20 + 0.9 × 160 = 165 kJ/kg
Refrigerating effect per kg = h_{2} – h_{1} = 164 – 80 = 84 kJ/kg
Hence refrigerating effect for 0.025 kg/sec circulation = 84 × 0.025 = 2.1 kW
 In a vapour compression refrigeration system, liquids to suction heat exchanger is used to

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NA
Correct Option: C
NA
 Round the clock cooling of an apartment having a load of 300 MJ/day requires and air conditioning plant of capacity about

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Capacity= Q = 300 MJ/day
= 300 × 10^{3} 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 × 10^{3} kJ / s 24 × 60 × 60
= 3.47 kW
1 ton = 3.5 kW∴ Q = 3.47 = 0.99 ton = 1 ton 3.5
 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

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Given: p = 1.013 bar, (p_{υ})_{tdp} = 0.02062
Specific humidity = 0.622 P_{υ} p  P_{υ} = 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 = p_{υ} × 0.622 = 0.01292 kg/kg of dry air. p  p_{υ}
Total enthalpy,
h = ^{c}p_{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 P_{υ} p  P_{υ} = 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 = p_{υ} × 0.622 = 0.01292 kg/kg of dry air. p  p_{υ}
Total enthalpy,
h = ^{c}p_{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.
 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 ______.

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ω_{new} = m_{1}ω_{1} + m_{2}ω_{2} m_{1} + m_{2} ⇒ 0.1 × 10.1 × + .1 × 1 = 0.02kg/kg dry air 10.1 + .1
Correct Option: A
ω_{new} = m_{1}ω_{1} + m_{2}ω_{2} m_{1} + m_{2} ⇒ 0.1 × 10.1 × + .1 × 1 = 0.02kg/kg dry air 10.1 + .1