Question

A room contains 50 kg of dry air and 06 kg of water vapor at 25°C and 95 kPa total pressure. The saturation p ressure of water at 25 "C is 3.1698 kPa. The molar masses for air and water are 29 kg/kmol and 18 kg/kmol, respectively Determine (a) The specific humidity of air in the room; (b) The relative humidity of air in the room.

Answer 1

a) w = 0.12 kg/ kg of dry air

b) d = 484.69 %

Step-by-step explanation:

Given:

Mass of dry air, m = 50 kg

mass of water vapor = 6 kg

saturation pressure of water = 3.1698 kPa

Molar mass of the air = 29 kg/mol

Molar mass of the water = 18 kg/mol

a) The specific humidity (w) is calculated as:

w = (mass of water vapor) / (mass of air)

on substituting the values, we get

w = 6 kg / 50 kg

or

w = 0.12 kg/ kg of dry air

b) Now,

w = (0.622Pv) / (Pt - Pv)

where,

Pv is the actual vapor pressure

Pt is the vapor pressure at room temperature

on substituting the values, we have

0.12 = (0.622 × Pv) / (95 - Pv)

or

11.4 - 0.12Pv = 0.622Pv

or

Pv = 15.36 kPa

also,

Relative humidity is given as:

d = (actual vapor pressure) / (saturated vapor pressure)

saturated vapor pressure for water = 3.1698 kPa

on substituting the values, we have

d = (15.36 / 3.1698 ) × 100

or

d = 4.846 × 100

or

d = 484.69 %