ADV. TRANSPORT II 2012, solutions for problem set 8

Corrected solutions are due 5:00 pm, Apr 10 together with your Set 9 solutions.

Problem 1

Absorption on the beads of activated carbon in the absence of reaction is apparently slow, due to slow diffusion with MTC

k0 = 6×10-4 cm/s

Absorption on the beads of resin is much faster, due to reaction-facilitated diffusion. Assuming that the diffusion coefficient of antibiotic within the beads of carbon and resin is the same given as 9*10-7 cm2/s, from the notes or Eq. 17.1-17 on p. 482,

k = 1×10-2 cm/s = Dκ  coth  Dκ(k0)2

Given D = 9×10-7 cm2/s, we have

1x10!! = 9x10!! ∗ 𝜅coth  (9x10!!

6x10!! ! 𝜅)

10.54 = 𝜅coth  (2.5 𝜅)

By using the solver in MS Excel, κ = 111 s-1

Problem 2

In the kinetic regime:

jkin=Vcatk1c, Vcat - volume of catalyst, k1- rate constant per unit volume of catalyst

Balance equation:

ckVdtdcV cat 10 −= ; V0- volume of pores between catalyst particles.

Solution:

)exp( 10

0 tkVV

cc cat−= .  Accounting for  0V

Vcat =1, k1 = ln2/t1/2 = 7.7 s-1,  

In the diffusion-controlled regime, the Thiele modulus

φ = RDk

0

1 = !.!!.!"#$

0.05 = 1.12

Effectiveness factor η=3(φcothφ -1)/φ2=0.93 Balance equation

ckVdtdcV cat 10 η−=

Thus, the apparent rate constant 1kη =7.16 s-1

Problem 3

Here, we are dealing with unsteady diffusion in a slab with the equilibrium absorption of dye into the cells. Equilibrium absorption is characterized by the equilibrium partition coefficient K. Accounting that cell occupy just ε=0.05 of the suspension volume

Assume the equilibrium constant is

K  =  mol  dye/vol  cell*vol  cell/vol  soln

mol  dye/vol  agar  =  c2ε/c1  =  3.5×103ε  

The amount of dye uptake M is

M  =  Aj1t  

where A is the dish area and j1 is the average flux through the solution-suspension interface. Since characteristic diffusion time is much larger than the observation time,

!!

!!"= !.!"!

!∗!.!×!"!!∗!"##= 24.4 ≫ 1

we  assume  that  agar  is  a  semi-­‐infinite  slab,  thus  from  the  notes,  

j1  =    tKD

π)1(4 +c10  

where  c10  =  csol/20  is  equilibrium  concentration  at  the  solution-­‐suspension  interface  

then  𝑀 = 𝜋 ∗ 5! ∗!.!!"""!"

∗ !!∗ 3.2×10−6 ∗ 1+ 3.5�10! ∗ 0.05 ∗ 1800 = 4.46×10−4  mol  

Note:  factor  of  1/2  compensate  for  4  under  the  square  root  

 

Problem  4.  

(1)  Mass  balance  and  boundary  conditions  

Assume  pH  is  high  so  that  no  CO2  initially  presents,  and  catalysis  gives  fast  reaction.  

!!!!"= 𝐷 !!!!

!!!− 𝜅𝑐!    

subject  to  

t  =  0   all  z   c1  =  0  

t  >  0   z  =  0   c1  =  c10  

  z  =  ∞   c1  =  0  

(2)  Two  limits:  

For  fast  reaction,  κ  is  large,  then  

j1|z  =  0  =   Dκ  c1(1  +  0)  =   Dκ  c1  

For  slow  reaction,  κ→0,  then  

j1|z  =  0  =!!"𝑐!"    

which  is  the  same  as  Eq.  2.3-­‐18  with  c1∞  equals  to  zero.