Air at a temperature of 25 degree C and pressure 750 mm Hg has a relative humidi

Question

Air at a temperature of 25 degree C and pressure 750 mm Hg has a relative humidity of 80%. Use SI units. Solve the problem using the humidity formulas, and give the graphically solution in the humidity chart. Calculate: Humidity of air. (2) Molal humidity of air. (2) Weight of water condensed from 100 m^3 of original wet air, if the temperature of the air is reduced to 15 degree C and the pressure is increased to 2 bar. At 25 degree C the vapour pressure of the water is 2.5 kPa. (7) Graphical solution. (4) Humid volume = 22.414(H/28.86 + 1/18.02)(T_K/272)(1/P) m^3_wetair/kg_BDA Where, P in bar. H (humidity) in kg H_2O/kg BDA

Explanation / Answer

At 25 C,

Vapor pressure of water, P* = 2.5 kPa

= 2500 / 1 x 105 bar = 0.025 bar

Relative humidity, = 80 /100 = 0.8

= Partial pressure of water / Vapor pressure of water

0.8 = P / P*

P = 0.02 bar

Ptotal = 750 mmHg

= 1 bar

2.

Molal humidity of air

= P / (Ptotal – P) = 0.02 / (1 – 0.02)

= 0.02 / 0.98 = 0.0204

1.

Humidity of air

= Molar humidity * MWwater / MWair

= 0.0204 * 18.02 / 28.86 = 0.013

3.

Water condensed at 15 C

Relative humidity, RH = 100 %

Wet bulb temperature = 15 C

Moisture content from psychrometric chart = 0.0106 kg/kg dry air

Humid volume = 22.414 (0.0106 /28.86 + 1 / 18) (288 / 273) (1/2) = 0.66 m3 / kg dry air

Weight of dry air = 100 m3 / 0.66 m3 / kg dry air = 151.25 kg

Weight of water vapor present = 151.25 * 0.0106 = 1.60 kg

In the original wet air

RH = 80%

Wet bulb temperature = 25 C

Humidity = 0.013 kg/kg dry air

Humid volume = 22.414 (0.013 /28.86 + 1 / 18) (298 / 273) (1/1) = 1.37 m3 / kg dry air

Weight of dry air = 100 m3 / 1.37 m3 / kg dry air = 72.98 kg

Weight of water vapor present = 72.98 * 0.013 = 0.95 kg

Weight of water condensed from 100 m3 of original wet air = 1.60 – 0.95 = 0.65 kg

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