CHEMICAL ENGINEERING

MASS TRANSFER 235

The first boundary condition is atyD 0 ,CADCAi, andCAiDA.

Hence: CADCAicoshaxCBsinhax iii

The second boundary condition is that whenyDLand:

NADDdCA/dy

yD 0 D 2 DdCA/dy

yDL

Differentiating equation (iii):

dCA/dyDCAiasinhayCBacoshay

and: dCA/dy
yD 0 DBa

and: dCA/dy
yDLDaB/ 2 DCAiasinhaLCBacoshaL

so that: BD

2 CAisinhaL

1 2coshaL

iv

Substituting equation (iv) into equation (iii):

CADCAicoshayC

2 CAisinhaLsinhay

1 2coshaL

DCAi[coshay 2 coshaycoshaLCsinhaLsinhay

]/ 1 2coshaL



DCAi[coshay2coshayCL

]

The molar flux at the surfaceDNADDdCA/dy

yD 0.

dCA

dy

DCAi[asinhay 2 asinhaaCL

]

dCA/dy

yD 0 D 2 CAia^2 sinhaL

NAD 2 DCAia^2 sinhaL

aD

√

k/D

NAD 2 DCAik/D

sinhL

√

k/D

D 2 CAiksinhL

p

k/D

PROBLEM 10.18

4cm^3 of mixture formed by adding 2 cm^3 of acetone to 2 cm^3 of dibutyl phthalate is
contained in a 6 mm diameter vertical glass tube immersed in a thermostat maintained at
315 K. A stream of air at 315 K and atmospheric pressure is passed over the open top
of the tube to maintain a zero partial pressure of acetone vapour at that point. The liquid
level is initially 11.5 mm below the top of the tube and the acetone vapour is transferred
to the air stream by molecular diffusion alone. The dibutyl phthalate can be regarded
as completely non-volatile and the partial pressure of acetone vapour may be calculated
from Raoult’s law on the assumption that the density of dibutyl phthalate is sufficiently
greater than that of acetone for the liquid to be completely mixed.