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 k 0 = 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 cm 2 /s, from the notes or Eq. 17.1-17 on p. 482, k = 1×10 -2 cm/s = Dκ coth Dκ (k 0 ) 2 Given D = 9×10 -7 cm 2 /s, we have 1x10 !! = 9x10 !! ∗ coth ( 9x10 !! 6x10 !! ! ) 10.54 = coth (2.5 ) By using the solver in MS Excel, κ = 111 s -1 Problem 2 In the kinetic regime: j kin =V cat k 1 c, V cat - volume of catalyst, k 1 - rate constant per unit volume of catalyst Balance equation: c k V dt dc V cat 1 0 − = ; V 0 - volume of pores between catalyst particles. Solution: ) exp( 1 0 0 t k V V c c cat − = . Accounting for 0 V V cat =1, k 1 = ln2/t 1/2 = 7.7 s -1 , In the diffusion-controlled regime, the Thiele modulus
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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
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