Page 324 - Numerical Methods for Chemical Engineering

Problems 313

You use an immobilized enzyme system in which the enzyme is encapsulated in beads of
radius R of a polymer hydrogel at a mass loading density ρ E = 10 −2 mg /ml of gel. The
E
2
substrate diffusivity in the hydrogel is 10 −6 cm /s. The bulk substrate concentration is 1
M. Neglect external mass transfer resistance. Define an internal effectiveness factor, and
compute its value as a function of R in the range 10 −4 –10 −2 m. Since the enzyme is
expensive, what radius would you use, and why?
6.B.3. Consider a system in which a charged surface, maintained at a constant electric
potential 0 , is in contact with an aqueous solution of NaCl, with a bulk salt concentration
−
c NaCl .If 0 > 0, the surface selectively attracts the Cl anions, else if 0 < 0 it attracts the
Na cations. In either case, near the surface there is a region of charge imbalance in the salt
+
solution that counteracts the surface potential. As the distance z from the surface increases,
the electric potential decays to zero, as the surface electric charge is screened by the salt
solution.
We present here a model for the electric potential near a charged planar surface known as
Gouy–Chapman theory. Let c + (z) and c − (z) be the ion molar concentrations as functions
of the distance z from the surface. The charge density, also a function of z, is then

ρ(z) = q e N av [c + (z) − c − (z)] (6.233)
23
q e = 1.602×10 −19 C is the charge of an electron and N av = 6.022×10 . The electric
potential is governed by the Poisson equation
2
d ρ(z)
− = (6.234)
dz 2 ε 0 ε r
2
ε 0 = 8.854×10 −12 C /(J m) is the permittivity of free space and ε r = 78 is the dielectric
constant of water.
We obtain a closed-form equation for (z) at equilibrium using statistical mechanics. The
potential energy of a cation is E + (z) = q e (z), and that of an anion is E − (z) =−q e (z).
Then, with the far-field condition (z) → 0as z →∞, the Boltzmann distribution predicts
the ion concentration fields
q e (z) q e (z)

c + (z) = c NaCl exp − c − (z) = c NaCl exp + (6.235)
k b T k b T
The charge density is then
1
q e (z) q e (z)
ρ(z) = q e N av c NaCl exp − − exp + (6.236)
k b T k b T
The electric potential field at equilibrium is obtained from the following BVP, involving
the nonlinear Poisson–Boltzmann equation,
2
d ρ(z) q e N av c NaCl q e (z) q e (z) 1
− 2 = = exp − − exp +
dz ε 0 ε r ε 0 ε r k b T k b T
(6.237)
BC 1 (0) = 0
BC 2 (∞) = 0
The quantity q e N av c NaCl /ε 0 ε r on the right-hand side must have the same units as the left-
hand side (electric potential divided by length squared). Likewise, from the argument of