MASS TRANSFER 243
X eX
2
erfX erfcXX
p
erfcX Right hand side
1 0.368 0.843 0.157 0.278 0.0897
0.9 0.445 0.797 0.203 0.324 0.121
0.97 0.390 0.830 0.170 0.292 0.098
0.96 0.398 0.825 0.175 0.297 0.101
∴ XD 0. 96 Dy^2 / 4 Dt
∴ yD 0. 96 ð 4 ð 10 ^9 ð 60
0.^5 D 4. 8 ð 10 ^4 mor 0.48 mm
PROBLEM 10.23
A chamber, of volume 1 m^3 , contains air at a temperature of 293 K and a pressure of
101 .3kN/m^2 , with a partial pressure of water vapour of 0.8kN/m^2. A bowl of liquid
with a free surface of 0.01 m^2 and maintained at a temperature of 303 K is introduced
into the chamber. How long will it take for the air to become 90% saturated at 293 K
and how much water must be evaporated?
The diffusivity of water vapour in air is 2. 4 ð 10 ^5 m^2 /s and the mass transfer resis-
tance is equivalent to that of a stagnant gas film of thickness 0.25 mm. Neglect the effects
of bulk flow. Saturation vapour pressure of waterD 4 .3kN/m^2 at 303 K and 2.3kN/m^2
at 293 K.
Solution
Moles transferred,
nDDA/L
CAsCA
whereCADconcentration (kmol/m^3 ),CAsis the saturation value ofCAat the surfaceD
is the diffusivity andLis the thickness of the stagnant gas film.
If the saturated vapour pressure at the interface is 4.3kN/m^2 and if at any time the
partial pressure in the air isPAkN/m^2 , then the rate of evaporation is given by:
dn
dt
D
0. 01 ð 2. 4 ð 10 ^5
0. 25 / 1000
[
1
22. 4
ð
273
303
(
4. 3
101. 3
PA
101. 3
)]
D 3. 81 ð 10 ^7 4. 3 PA kmol/s
1m^3 of air at 303 K and 101.3kN/m^2 is equivalent to 1 / 22. 4
273 / 293
D 0 .0416 kmol
Initial moisture contentD 0. 8 / 101. 3
ð 0. 0416 D 3. 29 ð 10 ^4 kmol
Final moisture contentD 0. 9 ð 2. 3 / 101. 3
ð 0. 0416 D 8. 50 ð 10 ^4 kmol
∴ Water evaporatedD 8. 50 3. 29
ð 10 ^4 D 5. 21 ð 10 ^4 kmol
D 5. 21 ð 10 ^4 ð 18
D 9. 38 ð 10 ^3 kg water