242 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Solution
As in the previous problem, the basic equation is:
∂CA
∂tDD
∂^2 CA
∂y^2(equation 10.66)which can be solved using the same boundary conditions to give the rate of mass transfer
at depth,y, NA (^) y,tas:
NA (^) y,tDD
dCA
dy
DCAi
√
D
tey(^2) / 4 Dt
At some other value ofyDL, the amount which has been transferred in timetper unit
area is: ∫
t
0
CAi
√
D
tey(^2) / 4 Dt
dt
This integral can be solved by making the substitution:
ˇ^2 Dy^2 / 4 Dt
so that: ˇDy/ 2
p
Dt
and: tDy^2 / 4 Dˇ^2 ,t^1 /^2 Dy/ 2 ˇ
p
D
dtDy^2 / 2 D
ˇ^3 dˇ
The amount transferred at depthLis then:
DCAi
√
D
y
p
D[
2
p
Dt
yey(^2) / 4 Dt
p
erfc
y
2
p
Dt
]
DCAi[
2
√
Dt
ey(^2) / 4 Dt
yerfc
y
2
p
Dt
]
and:
mass transfer atL
mass transfer atyD 0D
[
2
√
Dt
ey(^2) / 4 Dt
yerfc
y
2
p
Dt
]
2
√
Dt
Dey(^2) / 4 Dt
y
2
p
Dt
p
erfc
y
2
p
Dt
DeX
2
X
p
erfcX
where XDy/ 2
p
Dt
Under the conditions in this problem, this ratioD 0 .1.
erfcXD 1 erfX
so that erfcXcan be calculated from Table 13 in the Appendix of Volume 1. Values of
Xwill be assumed and the right hand side evaluated until a value ofXis found such that
the right hand sideD 0 .1.