Pulse analysis - Principles of Chemical Separations with Environmental Applications - page 293

2 c

D 0
α

∂ V 0

α

∂ c 3(1 − α )

= ∂

c

−

z −

k p ( c − c i R )

∂ z 2 ∂ R α ∂ t

R

D i ∂

2 c

D 0
α

∂ V 0

α

∂ c 3(1 − α ) c i

∂

= ∂

c

−

z −

interstitial fluid equation

∂ z 2 ∂ R α r ∂ t

r ∂

β ∂

c i
∂

D i 1
r

∂
∂

c i
∂

− ρ p ∂

c ads
∂ t =

r r t

2 c i
∂ r − ρ p

D i
β

∂ 2 D i

β

∂ c i

∂

∂ c ads

∂

= ∂

c i
∂

+ intra-particle equation.

r 2 r β t t

Take Laplace transform of each equation and substitute:

R D i 1

R =

D 0
α

d 2 c
d z 2

V 0
α

d c
d z −

3 − α

α

d c i
d r

− s c − 0

s c i + ρ p
β

c ads

d 2 c i
d r 2

D i
β

2 D i
β

d c i
d r −

+ = 0

r

s c i + ρ p
β

d 2 c i
d r 2

D i
β

2 D i
β

d c i
d r −

k a c i

+ = 0

r s + k a / K a

1

r d 2 c i
d r 2

2 d c i

s D i

+ ρ p
β

k a

b 2 r c i = b 2 +

d r −

0; = .

s + k a / K a

Boundary conditions become:

c i ( r = 0 , s ) = finite

r = 0 =

d c i
d r

0 .

To solve the equation for c i , we can use a variable substitution.

u
r

c i =

Let:

u
r −

d c i
d r =

u
r 2

d 2 c i
d r 2

u
r −

u
r 2 −

u
r 2 +

u
r −

2 u
r 2

2 u
r 3

2 u
r 3 .

= = +

Substitute:
u − b 2 u = 0 .

The solution for u is:

k 1 e br k 2 e − br u = +

u
r =

k 1
r

k 2
r

b
r ( k 1 sinh br

e − br k 2 cosh br )

e br

+ = +

(note that we have added a constant ( b ) and changed k to k ).