Refrigeration and Airconditioning Miscellaneous
 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.

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w = 0.622 P_{v} P  P_{v} φ = P_{v} P_{s} ⇒ 0.55 = P_{v} P_{s}
⇒ P_{v} = 2235.3w = 0.622 × 2235.3  2235.3 (100 × 10^{3})
w = 0.0142 kg/kg of dry air
w = 14.2 gm/kg of dry airCorrect Option: A
w = 0.622 P_{v} P  P_{v} φ = P_{v} P_{s} ⇒ 0.55 = P_{v} P_{s}
⇒ P_{v} = 2235.3w = 0.622 × 2235.3  2235.3 (100 × 10^{3})
w = 0.0142 kg/kg of dry air
w = 14.2 gm/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
 The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30°C and 70%, respectively. If the saturated steam pressure at 30°C is 4.25 kPa, the specific humidity of the room air in kg of water vapour/kg dry air is

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RH = 0.7 = P_{v} and w = 0.622p_{v} 4.25 100  p_{v}
Correct Option: C
RH = 0.7 = P_{v} and w = 0.622p_{v} 4.25 100  p_{v}
 A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room is

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m_{a} = 35 kg, m_{v} = 0.5 kg
w = m_{v} = 0.5 = 0.01428 m_{v} 35 also w = 0.622 P_{v} P_{a}
P_{t} = 100 kP_{a}∴ w = 0.622 × P_{v} P_{t}  P_{v} Now, 0.01428 = 0.622 × P_{v} P_{t}  P_{v}
P_{v} = 2.238 kPa
Relative humidityφ = P_{v} = 2.238 × 100 P_{s} 3.17
φ = 70.6% = 71%Correct Option: D
m_{a} = 35 kg, m_{v} = 0.5 kg
w = m_{v} = 0.5 = 0.01428 m_{v} 35 also w = 0.622 P_{v} P_{a}
P_{t} = 100 kP_{a}∴ w = 0.622 × P_{v} P_{t}  P_{v} Now, 0.01428 = 0.622 × P_{v} P_{t}  P_{v}
P_{v} = 2.238 kPa
Relative humidityφ = P_{v} = 2.238 × 100 P_{s} 3.17
φ = 70.6% = 71%
 If a mass of moist air in an airtight vessel is heated to a higher temperature, then

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R.H. decreases
Correct Option: D
R.H. decreases