Problem Statement: Consider a substrate-inhibited enzyme that is immobilized via physical entrapment in spherical gel beads. Unless noted otherwise, use the following parameters.
v(s)=vm*s/(Km+s+Ki*s2)
where vm=0.05g/L-sec
Km=3g/L
Ki=8L/g
Diffusion coefficient: De=10-5cm2/sec
Substrate concentration in the bulk solution: sb=2g/L
Radius of the spherical beads: R=0.5cm
Solution:
For substrate diffusing through the polymer gel.
Solution:
differential diffusional mass transfer rate = differential reaction rate
De*d[4*p*r2*(ds/dr)] = 4*p*r2*dr*v(s)
which is equivalent to:
De*[(d2s/dr2)+2/r*(ds/dr)]=v(s)
B.C. s(R)=sb
ds(0)/dr=0
Dimensionless form (normalize s wrt sb, and r wrt R):
(d2s/dr2)+2/r*(ds/dr)=f2*v(s)
B.C. s(1)=1
ds(0)/dr=0
f=R*sqrt(vm/De/Km)
Transform into two coupled 1st-order odes
ds/dr = z
dz/dr = -2/r*z + f2*v(s)
B.C. s(1)=1
ds(0)/dr=0
Solution Approach:
1. Guess s(0)
2. Start from s(0) and z(0)=0, integrate dimensionless odes from r=0 to r=1
3. See if s(1)=1. If yes, stop; otherwise, make another guess of s(0) and
iterate until s(1)=1 is satisfied.
Solution:
In dimensional units, ds/dr=8.08*sb/R=8.08*(2g/L)/(0.5cm)=32.32g/L-cm
Rate of reaction = mass transfer across the surface = area*De*(ds/dr)
= 4*p*(0.5cm)2*(10-5cm2/sec)*(32.32g/L-cm)=1.015*10-6g/sec
Solution:
Rate of reaction /wo mass transfer = volume*v(sb)
= volume*vm*sb/(Km+sb+Ki*sb2)
= 4/3*p*(0.5cm)3*(0.05g/L-sec)*(2g/L)/(3g/L+2g/L+8L/g*(2g/L)2)
= 1.415*10-6g/sec
Solution:
rate /w mass transfer resistance
Effectiveness factor = ---------------------------------
rate /wo mass transfer resistance
= (1.015*10-6g/sec)/(1.415*10-6g/sec)
= 0.718
The magnitude indicates that mass transfer resistance reduces the
reaction rate by about 30%.
Solution:
The effectiveness factor from the preceeding part is for sb=2g/L.
The substrate concentration in the bulk solution for this
part of the problem is se=0.1g/L. Thus, the denominator in the
effectiveness factor, i.e., rate /wo mass transfer resistance, is different.
Furthermore, the solution to the boundary value problem in Part b) also
depends on the value of the substrate concentration in the bulk, which is
half of the boundary conditions. Thus, the numerator in the effectiveness
factor is also different. To calculate the effectiveness factor numerically,
we can numerically solve the boundary value problem, which is beyond what
we can do in this exam.
Since se is small, the reaction rate expression is almost
linear (i.e., first-order) with respect to s. With this linear
approximation, we can approximate the effectiveness factor from the formula
derived for linear rate expression.
v(s)=(vm/Km)*s=(0.05/3sec)*s
f=R*sqrt(vm/De/Km)
=(0.5cm)*sqrt((0.05g/L-sec)/(10-5cm2/sec)/(3g/L))
=15.8
h=3/f*(1/tanh(f)-1/f)=0.178
Solution:
Calculate the volume based on an ideal, fully packed reactor /wo mass transfer resistance.
From the full rate expression in the original problem description, we have:
At se=0.1g/L, v(se)=0.00157g/L-sec
From the simplified, linearized approximation from part f), we have
At se=0.1g/L, v(se)=0.00167g/L-sec
Thus, we confirm that there is very little difference between the
full rate expression and the linear approximation.
Material balance at steady-state around a CSTR.
In-Out=Reaction
F*(sf-se)=V*v(se)
Note that since the reactor is well mixed (irrespective of the shape --
column or otherwise), we use the CSTR equation, not the plug flow equation.
Thus, the ideal volume is:
V = F*(sf-se)/v(se)
= 1(L/hr)(2-0.1g/L)/(0.00167g/L-sec)
= 0.317L
Volume for a column corrected for mass transfer resistance and void fraction
V=0.317L/0.178/(1-0.6)=4.45L
Column length = V/(column cross section area)
= 4.45L/3.14/(5cm)2
= 56.6cm
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