Refrigeration and Airconditioning Miscellaneous
 A moist air sample has dry bulb temperature of 30°C and specific humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at 30°C is 4.24 kPa and the total pressure is 90 kPa, then the relative humidity (in%) of air sample is

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Given: DBT = 30°C, w = 11.5 g/kg
∴ Relative humidity= Specific humidity × Total pressure 0.622 saturation vapour pressure φ = w × P_{a} 0.622 P_{vs} φ = 11.5 × 90 = 0.385 0.622 × 1000 × 4.24
φ ≈ 38.5%Correct Option: B
Given: DBT = 30°C, w = 11.5 g/kg
∴ Relative humidity= Specific humidity × Total pressure 0.622 saturation vapour pressure φ = w × P_{a} 0.622 P_{vs} φ = 11.5 × 90 = 0.385 0.622 × 1000 × 4.24
φ ≈ 38.5%
 Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35°C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 35°C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to

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Assuming that compression is isentropic in air compressors, the process can be described on the T  s diagram. The process in the intercooler is constant pressure
p_{2} = p_{3}
But p_{2} = 5_{p1}p_{1} = p_{2} = p_{3} 5 5 p_{1} = 5.628 = 1.13kPa 5
Correct Option: B
Assuming that compression is isentropic in air compressors, the process can be described on the T  s diagram. The process in the intercooler is constant pressure
p_{2} = p_{3}
But p_{2} = 5_{p1}p_{1} = p_{2} = p_{3} 5 5 p_{1} = 5.628 = 1.13kPa 5
 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

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NA
Correct Option: B
NA
 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.

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Given data:
Dry bulb temperature:
T_{db} = 35°C
Relative humidity
φ = 75° = 0.75
At 35°C, P_{s} = .05628 barφ = P_{v} ⇒ 0.75 = P_{v} P_{s} .05628
P_{v} = 0.75× .05628 =.04221 bar
Specific humidity,ω = .622P_{v} = .622 × .04221 P  P_{v} 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 barφ = P_{v} ⇒ 0.75 = P_{v} P_{s} .05628
P_{v} = 0.75× .05628 =.04221 bar
Specific humidity,ω = .622P_{v} = .622 × .04221 P  P_{v} 1.0135  .04221
= 0.0270 kg/kg of dry air.
 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.