Pulse analysis - Principles of Chemical Separations with Environmental Applications - page 294

Applying the boundary conditions:

k 2 =

c i ( r = 0) → finite ⇒ 0

k 1
r

c i ( r , s ) = sinh br

k 1 b
r 2

k 1 b 2
r

k 1 b
r 2

d c i
d r =

sinh br + cosh br = ( br cosh br − sinh br ) .

Use L'Hopital's rule to check lim r → 0 :
lim r → 0 d c i b cosh br

k 1 b
2 r

b cosh br

b 2 r sinh br

d r =

+ − = 0

R =

c

sinh br

k 1 b
R

d c i
d r

k p
D i ( c

k p
D i

− − c i R ) = − .

Combine and solve for k 1 :

R 2 k p c

k 1 =

D i b bR cosh bR

k p R
D i −

1 sinh bR .

+

Now we know c i and d c i /

d r .

Go back to:

R =

R D i 1

d c i
d r

d 2 c
d z 2 −

D 0
α

V 0
α

d c
d z −

3 − α

α

s c − 0 .

R

d c i
d r

Substitute for :

d 2 c
d z 2 −

D 0
α

V 0
α

d c
d z +

A c = 0

1 k p

3
R

− α
α

k p R sinh bR
D i bR cosh bR

k p R
D i −

1 sinh bR

A =− s + 1 − .

!

+

h ( s )

Let c ( z , s ) = c 0 ( s )e λ z

V 0
2 D 0

2

D 0

V 0
2 D 0 −

− √ λ = + + .

[ s h ( s )] (note that we only used value)