Question

Wet leather enters a continuous drier at 45% moisture (wet) and leaves at a rate of 200 lbs/h with a moisture content of 25% moisture (dry). Air at 140°F and 10% RH enters the drier, and leaves the drier at 120°F and 40% RH. Assume atmospheric pressure on both inlet and outlet. Vapor pressures of water at 120°F and 140°C are 87.55 mm Hg and 149.44 mm Hg, respectively.

a.) Calculate the rate (lbs/h) at which water is removed from the leather.

b.) What is the mole fraction of dry air in the air-water vapor mixture entering the dryer?

c.) The volumetric flow rate of inlet air (ft3/h) is?

Answer 1

Given the mass fraction of water in entering wet leather = 0.45

let the mass flow rate of entering leather is L₁.

mass flowrate of leaving leather = L₂ = 200 lbs/h.

mass fraction of water in leaving leather = 0.25

mass flowrate of pure leather = L = 0.75 x 200 = 150 lbs/h

mass flowrate of leather entering = L₁ = 150 / (1-0.45) = 272.73 lbs/h

only water is lost from laether, hence change in leather mass is purely due to water lost into the air.

therefore the rate of water removed from leather = L₁ - L₂ = 272.73 - 200 = 72.73 lbs/h

b)

let the entering gas flowrate = G₁ .

Relative humidity of air entering = 0.1

vapor pressure of water in the air at entering = 149.44 mmHg.

partial pressure of water vapor in the air = 149.44 x 0.1 = 14.944 mmHg.

mol fraction of water vapor in the air entering at 140 F = 14.944 / 760 = 0.01966

mol fraction of dry air in the entering air = 1 - 0.01966 = 0.98034

c)

let the molecular weight of water = 18 lb/lbmol and molecular weight of air = 29 lb/lbmol

mass fraction of water vapor in air entering = 14.944 x 18 / 760 x 29 = 0.012

vapor pressure of water vapor at 120 F = 87.55 mmHg.

relative humidity of air leaving at 120 F = 0.4

partial pressure of water vapor in air leaving = 0.4 x 87.55 = 35.02 mmHg.

mass fraction of water vapor in the air leaving = 35.02 x 18 / 760 x 29 = 0.0286

by mass balance :

air entering + leather entering = air leaving + water leaving

G₁ + L₁ = G₂ + L₂ .

G₁ + 272.73 = G₂ + 200

G₂ - G₁ = 72.73 lbs/h ----------------eq-1

Water vapor balance :

0.012 G₁ + 0.45 x L₁ = 0.0286 x G₂ + 0.25 x L₂

0.0286 x G₂ - 0.012 x G₁ = 0.45 x 272.73 - 0.25 x 200

0.0286 x G₂ - 0.012 x G₁ = 72.73 lbs/h --------------eq-2

solving eq-1 and eq-2

G₂ = 4328.75 lbs/h

G₁ = 4256.02 lbs/h

therefore mass flowrate of entering air = 4256.02 lbs/h

mass flowrate of water vapor alone = 0.012 x 4256.02 = 51.072 lbs/h

mass flowrate of dry air alone = 0.988 x 4256.02 = 4204.95 lbs/h

molar flowrate of entering air = 51.072 / 18 + 4204.95 / 29 = 147.84 lbmol/h = 67.06 kmol/h

volumetric flowrate is obtained by assuming that entering air is ideal

Temperature = 140 F = 333.15 K

pressure P = 760 mmHg = 1.013 x 10⁵ N/m²

Volumetric flowrate V₁ = n RT / P = 67.06 x 8.314 x 333.15 x 1000 / 1.013 x 10⁵ = 1833.6 m³ /h = 64752.9729 ft³/h