1. 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).

1. 1. 26.24 gms/kg of dry air

   
   
   

   2. 16.24 gms/kg of dry air

   
   
   

   3. 18.24 gms/kg of dry air

   
   
   

   4. 19.24 gms/kg of dry air

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_atm = 100 kPa
   DBT = t = 30°C → P_VS = 4.24 k Pa
   φ = 60° = 0.6
   

   φ	PV	⇒ 0.6 =	PV
   	PVS		4.24

   
   P_V = 2.544 kPa
   

   w = 0.622 ×	PV
   	Patm - PV

   
   

   w = 0.622 ×	2.544
   	100 - 2.544

   
   w = 16.24 gms/kg of dry air

   Correct Option: B

   P_atm = 100 kPa
   DBT = t = 30°C → P_VS = 4.24 k Pa
   φ = 60° = 0.6
   

   φ	PV	⇒ 0.6 =	PV
   	PVS		4.24

   
   P_V = 2.544 kPa
   

   w = 0.622 ×	PV
   	Patm - PV

   
   

   w = 0.622 ×	2.544
   	100 - 2.544

   
   w = 16.24 gms/kg of dry air

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2. 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 ________

1. 1. 10 gw/Kgda

   
   
   

   2. 17 gw/Kgda

   
   
   

   3. 27 gw/Kgda

   
   
   

   4. 25 gw/Kgda

   
   
   

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1. View Hint View Answer Discuss in Forum

   w = 0.622	Pv
   	Pt - Pv

   
   

   = 0.622 ×		20
   		750 - 20

   
   = 0.017 Kgw/Kgda = 17 gw/Kgda

   Correct Option: B

   w = 0.622	Pv
   	Pt - Pv

   
   

   = 0.622 ×		20
   		750 - 20

   
   = 0.017 Kgw/Kgda = 17 gw/Kgda

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3. 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		ps		= 14.317 -	5304
   		p0			Ts

   
   where p₀ = 101.325 kPa. The dry bulb temperature of the moist air sample (in °C) is_______.

1. 1. 15.09°C

   
   
   

   2. 20°C

   
   
   

   3. 19.89°C

   
   
   

   4. 21°C

   
   
   

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1. View Hint View Answer Discuss in Forum

   φ = 0.7, P₀ = 101.325
   

   φ =	Pv	⇒ Psat =	Pv
   	Ps - Pw		φ

   
   

   Psat =	1.6	= 2.28571 kPa
   	0.7

   
   Corresponding to saturation pressure whatever is the temperature is the DBT
   <

   In		Ps		= 14.317 -	5304
   		P0			Tsat

   
   

   ln		2.28571		= 14.317 -	5304	- 3.7916 = 14.317 -	5304
   		101.325			Tsat		Tsat

   
   

   Tsat =	5304	= 292.898 K = 19.89°C
   	18.1086

   Correct Option: C

   φ = 0.7, P₀ = 101.325
   

   φ =	Pv	⇒ Psat =	Pv
   	Ps - Pw		φ

   
   

   Psat =	1.6	= 2.28571 kPa
   	0.7

   
   Corresponding to saturation pressure whatever is the temperature is the DBT
   <

   In		Ps		= 14.317 -	5304
   		P0			Tsat

   
   

   ln		2.28571		= 14.317 -	5304	- 3.7916 = 14.317 -	5304
   		101.325			Tsat		Tsat

   
   

   Tsat =	5304	= 292.898 K = 19.89°C
   	18.1086

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4. 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_________.

1. 1. 21.745 g/kg of dry air

   
   
   

   2. 21g/kg of dry air

   
   
   

   3. 21.745 m/kg of dry air

   
   
   

   4. 21g/kg of dry air

   
   
   

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1. View Hint View Answer Discuss in Forum

   φ =	Pw	= 0.6 =	Pw
   	Ps		5.63

   
   ∴ P_w = 3.378 kPa
   Humidity Ratio,
   

   w = 0.622	Pw	= 0.622 ×	3.378
   	Ps - Pw		100 - 3.378

   
   = 0.021745 kg/kg of dry air
   or 21.745 g/kg of dry air

   Correct Option: A

   φ =	Pw	= 0.6 =	Pw
   	Ps		5.63

   
   ∴ P_w = 3.378 kPa
   Humidity Ratio,
   

   w = 0.622	Pw	= 0.622 ×	3.378
   	Ps - Pw		100 - 3.378

   
   = 0.021745 kg/kg of dry air
   or 21.745 g/kg of dry air

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5. 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

1. 1. Humidifier

   
   
   

   2. Dehumidifier

   
   
   

   3. Sensible heater

   
   
   

   4. Sensible cooler

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   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.

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