26. A building has to be maintained at 21 °C (dry bulb) and 14.5°C(wet bulb). The dew point temperature under these conditions is 10.17°C. The outside temperature is –23°C (dry bulb) and the internal and external surface heat transfer coefficients are 8 W/m2K and 23 W/ m2K respectively. If the building wall has a thermal conductivity of 1.2 W/mK, the minimum thickness (in m) of the wall required to prevent condensation is

1. 1. 0.471

   
   
   

   2. 0.407

   
   
   

   3. 0.321

   
   
   

   4. 0.125

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

27. For a typical sample of ambient air (at 30°C, 75% relative humidity and standard atmospheric pressure), the amount of moisture in kg per kg of dry air will be approximately.

1. 1. 0.002

   
   
   

   2. 0.027

   
   
   

   3. 0.25

   
   
   

   4. 0.75

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given data:
   Dry bulb temperature:
   T_db = 35°C
   Relative humidity
   φ = 75° = 0.75
   At 35°C, P_s = .05628 bar
   

   φ =		Pv	⇒ 0.75 =	Pv
   		Ps		.05628

   
   P_v = 0.75× .05628 =.04221 bar
   Specific humidity,
   

   ω =		.622Pv	=	.622 × .04221
   		P - Pv		1.0135 - .04221

   
   = 0.0270 kg/kg of dry air.

   Correct Option: B

   Given data:
   Dry bulb temperature:
   T_db = 35°C
   Relative humidity
   φ = 75° = 0.75
   At 35°C, P_s = .05628 bar
   

   φ =		Pv	⇒ 0.75 =	Pv
   		Ps		.05628

   
   P_v = 0.75× .05628 =.04221 bar
   Specific humidity,
   

   ω =		.622Pv	=	.622 × .04221
   		P - Pv		1.0135 - .04221

   
   = 0.0270 kg/kg of dry air.

---

---

28. Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is ______ grams.

1. 1. 14.2 gm/kg of dry air

   
   
   

   2. 15.2 gm/kg of dry air

   
   
   

   3. 13.2 gm/kg of dry air

   
   
   

   4. 12.2 gm/kg of dry air

   
   
   

---

1. View Hint View Answer Discuss in Forum

   w = 0.622	Pv
   	P - Pv

   
   

   φ =	Pv
   	Ps

   
   

   ⇒ 0.55 =	Pv
   	Ps

   
   ⇒ P_v = 2235.3
   

   w = 0.622 ×	2235.3	- 2235.3
   	(100 × 103)

   
   w = 0.0142 kg/kg of dry air
   w = 14.2 gm/kg of dry air

   Correct Option: A

   w = 0.622	Pv
   	P - Pv

   
   

   φ =	Pv
   	Ps

   
   

   ⇒ 0.55 =	Pv
   	Ps

   
   ⇒ P_v = 2235.3
   

   w = 0.622 ×	2235.3	- 2235.3
   	(100 × 103)

   
   w = 0.0142 kg/kg of dry air
   w = 14.2 gm/kg of dry air

---

29. For air with a relative humidity of 80%

1. 1. the dry bulb temperature is less than the wet bulb temperature

   
   
   

   2. the dew point temperature is less than wet bulb temperature

   
   
   

   3. the dew point and web bulb temperature are equal

   
   
   

   4. the dry bulb and dew point temperatures are equal

   
   
   

---

1. View Hint View Answer Discuss in Forum

   

   Correct Option: B

   

---

---

30. For air at a given temperature, as the relative humidity is increased isothermally,

1. 1. the wet bulb temperature and specific enthalpy increase

   
   
   

   2. the wet bulb temperature and specific enthalpy decrease

   
   
   

   3. the wet bulb temperature increases and specific enthalpy decreases

   
   
   

   4. the wet bulb temperature decreases and specific enthalpy increases

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   DBT = DBT₂
   DBT →
   φ ↑, Twb ↓
   Relative humidity is increased isothermally.

   Correct Option: A

   
   DBT = DBT₂
   DBT →
   φ ↑, Twb ↓
   Relative humidity is increased isothermally.

---