Refrigeration and Airconditioning Miscellaneous
 Ambient air is at a pressure of 100 kPa, dry bulb temperature of 30° C and 60% relative humidity, The saturation pressure of water at 30° C is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is (correct to two decimal places).

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P_{atm} = 100 kPa
DBT = t = 30°C → P_{VS} = 4.24 k Pa
φ = 60° = 0.6φ P_{V} ⇒ 0.6 = P_{V} P_{VS} 4.24
P_{V} = 2.544 kPaw = 0.622 × P_{V} P_{atm}  P_{V} w = 0.622 × 2.544 100  2.544
w = 16.24 gms/kg of dry airCorrect Option: B
P_{atm} = 100 kPa
DBT = t = 30°C → P_{VS} = 4.24 k Pa
φ = 60° = 0.6φ P_{V} ⇒ 0.6 = P_{V} P_{VS} 4.24
P_{V} = 2.544 kPaw = 0.622 × P_{V} P_{atm}  P_{V} w = 0.622 × 2.544 100  2.544
w = 16.24 gms/kg of dry air
 In a mixture of dry air and water vapor at a total pressure of 750 mm of Hg, the partial pressure of water vapor is 20 mm of Hg. The humidity ratio of the air in grams of water vapor per kg of dry air (g_{w}/kg_{da}) is ________

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w = 0.622 P_{v} P_{t}  P_{v} = 0.622 × 20 750  20
= 0.017 Kgw/Kgda = 17 gw/KgdaCorrect Option: B
w = 0.622 P_{v} P_{t}  P_{v} = 0.622 × 20 750  20
= 0.017 Kgw/Kgda = 17 gw/Kgda
 The partial pressure of water vapour in a moist air sample of relative humidity 70% is 1.6 kPa, the total pressure being 101.325 kPa. Moist air may be treated as an ideal gas mixture of water vapour and dry air. The relation between saturation temperature (T_{s} in K) and saturation pressure (P_{s} in kPa) for water is given by
in p_{s} = 14.317  5304 p_{0} T_{s}
where p_{0} = 101.325 kPa. The dry bulb temperature of the moist air sample (in °C) is_______.

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φ = 0.7, P_{0} = 101.325
φ = P_{v} ⇒ P_{sat} = P_{v} P_{s}  P_{w} φ P_{sat} = 1.6 = 2.28571 kPa 0.7
Corresponding to saturation pressure whatever is the temperature is the DBT
<In P_{s} = 14.317  5304 P_{0} T_{sat} ln 2.28571 = 14.317  5304  3.7916 = 14.317  5304 101.325 T_{sat} T_{sat} T_{sat} = 5304 = 292.898 K = 19.89°C 18.1086 Correct Option: C
φ = 0.7, P_{0} = 101.325
φ = P_{v} ⇒ P_{sat} = P_{v} P_{s}  P_{w} φ P_{sat} = 1.6 = 2.28571 kPa 0.7
Corresponding to saturation pressure whatever is the temperature is the DBT
<In P_{s} = 14.317  5304 P_{0} T_{sat} ln 2.28571 = 14.317  5304  3.7916 = 14.317  5304 101.325 T_{sat} T_{sat} T_{sat} = 5304 = 292.898 K = 19.89°C 18.1086
 Air in a room is at 35°C and 60% relative humidity (RH). The pressure in the room is 0.1 MPa. The saturation pressure of water at 35°C is 5.63 kPa. The humidity ratio of the air (in gram/kg of dry air) is_________.

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φ = P_{w} = 0.6 = P_{w} P_{s} 5.63
∴ P_{w} = 3.378 kPa
Humidity Ratio,w = 0.622 P_{w} = 0.622 × 3.378 P_{s}  P_{w} 100  3.378
= 0.021745 kg/kg of dry air
or 21.745 g/kg of dry airCorrect Option: A
φ = P_{w} = 0.6 = P_{w} P_{s} 5.63
∴ P_{w} = 3.378 kPa
Humidity Ratio,w = 0.622 P_{w} = 0.622 × 3.378 P_{s}  P_{w} 100  3.378
= 0.021745 kg/kg of dry air
or 21.745 g/kg of dry air
 Moist air at 35°C and 100% relative humidity is entering a psychrometric device and leaving at 25°C and 100% relative humidity. The name of the device is

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Since water content in moist air is reducing. The device is dehumidifer.Correct Option: B
Since water content in moist air is reducing. The device is dehumidifer.