CHEMICAL ENGINEERING

280 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

The average mass transfer rate is:

1

te

{

2

√

D



CAt^1 e/^2

}

D 2

√

D

te

CA

In the steady state if ft

is the age distraction function of the surface, then:

surface in age groupttotCdtDft

dt

and: surface in age grouptdttotDftdt
dt

Surface of agetdttotwhich is destroyed is that not entering the next age group

ftdt

dtft

dtDsfftdt

dtgdt

As dt!0, then:f^0 t
dtdtDsft
dtdt

estf^0 t

Cestsft

D 0

Integrating gives: estft
DK

then: ft
DKest i

As the total surfaceD 1 ,K

∫ 1

0

estdtDK

[

est

s

] 1

0

D

K

s

∴KDs

and: ft
Dsest ii

The mass transfer rate into the fraction of surface in the age groupttdtis:

√

D

t

CAsestdt

The mass transfer rate into the surface over the age spantD0totD1is:

D

√

D



CAs

∫ 1

0

t^1 /^2 estdtD

√

D



CA

PuttingDstDˇ^2 , then:t^1 /^2 D

p

s

ˇ

and: sdtD 2 ˇdˇ.

ID

∫ 1

0

p

s

ˇ

eˇ

22 ˇdˇ

s

D

2

p

s

∫ 1

0

eˇ

2

dˇD

2

p

s

Ð

p



2

D

√



s

Thus the mass transfer rate per unit area for the surface as a whole is:

√

D



CAs

√



s

D

p

DsCA

With the new age distribution function:

0 <t<

1

s

, surface renewal rate/areaDs ft

DKest

1

s

<t< 1 , surface renewal rate/areaD 2 s ft

DK^0 e^2 st







from equation (ii)