CHEMICAL ENGINEERING

228 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

and if the resistance to its mass transfer in air is equivalent to that of a 1 mm layer for
conditions of molecular diffusion, what will be the rate of cooling due to evaporation?
The water can be considered as well mixed and the water equivalent of the system is
equal to 10 kg. The diffusivity of water vapour in air may be taken as 0.20 cm^2 /sand
the kilogram molecular volume at NTP as 22.4m^3.

Solution

If subscripts 1 and 2 refer to the water and air side of the stagnant layer and subscriptsA
andBrefer to water vapour and air, then the rate of diffusion through a stagnant layer is:

NADD/RTL

P/PBM

PA 2 PA 1

 (equation 10.34)

where,PA 1 is the vapour pressure of water at 350 KD 41 .8kN/m^2.

PA 2 D0 (since the air currents remove the vapour as it is formed.)

PB 1 D 101. 3  41. 8

 D 59 .5kN/m^2 andPB 2 D 101 .3kN/m^2.

∴ PBMD 101. 3  59. 5
/ln 101. 3 / 59. 5
D 78 .17 kN/m^2.

and:P/PBMD 101. 3 / 78. 17
D 1 .296.

∴ NAD 0. 2 ð 10 ^4 / 8. 314 ð 350 ð 10 ^3

1. 296  0  41. 8

D 3. 72 ð 10 ^4 kmol/m^2 s

D 3. 72 ð 10 ^4 ð 18

 D 6. 70 ð 10 ^3 kg water/m^2 s

Area of bowlD/ 4

0. 32 D 0 .0707 m^2

Therefore the rate of evaporationD 6. 70 ð 10 ^3 ð 0. 0707

 D 4. 74 ð 10 ^4 kg/s

Latent heat of vaporisationD2318 kJ/kg

Specific heat capacity of waterD 4 .187 kJ/kg K

Rate of heat removalD 4. 74 ð 10 ^4 ð 2318

 D 1 .10 kW

If the rate of coolingDd)/dtK/s, then:

water equivalent ðspecific heat capacity ðd)/dt

D 0. 0617

or: 10 ð 4. 187 ðd)/dt
D 1 .10 and d)/dtD 0 .026 deg K/s

PROBLEM 10.12

Show by substitution that when a gas of solubilityCCis absorbed into a stagnant liquid
of infinite depth, the concentration at timetand depthyis:

CCerfc

y

2

p
Dt
Hence, on the basis of the simple penetration theory, show that the rate of absorption
in a packed column will be proportional to the square root of the diffusivity.