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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.