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Chapter 13
Problem 13.1. Use eq. 9.72 to see if S > 0 at r = 0
3
33
0.438 10 mol s cm
−
= −
( )
( )
22
m
0
e
2
3
06
33
3
r
From 9.72 S S R r ;
6D
0.438 10 0.2
for r 0 S S 6 5 10
S 2 mol cm 0.584 mol cm
S 1.4 mol cm at r 0
−
−
= − −
= = −
= −
= =
the answer is Yes
Problem 13.2. 1st order so rcod = k Ccod We will use eq 9.70; the 1st/order rxn. Rate constant
needs to be based on the whole pellet; Dry wt. is only 5%. The mass of cells/bead is 0.44
mg/bead. Or a cell concentration of 0.44 mg/0.0335 cm3 = 13.1 g/L – thus the rate constant is
k = 3.3 ×10–8 L/g dw s ∙ 13.1 g/L = 4.3 ×10–7 s–1
11
1tanh3 3
1 1 1
0.0306 tan h 0.0918 0.0918
0.9994
= −
= −
=
Clearly diffusional limitations – not important.
Problem 13.3. Use eq 10.1:
( )
2L
Xqo k a C C
=−
( )
22
2
1
2
C 8mg L C 0.10 C 0.8mg L
qo 0.2mmol O g dry wt h
6.4mgO gdry wt h
20h 8 0.8 mg L
X6.4mg O gdrywt. h
X 22.5gdry wt. L
−
= = =
=−
=−
−
=−
=
Problem 13.4. Glucose balance over “dz” yields
So
S
kXS
Qds r dV A d z
kS
− = =
+
Upon integration we obtain
So
K S kXA
ds d z
SQ
+=−
( )
o
SH
So
oo
So
H
S kXA
K ln S S H
SQ
kX
+ − =
=
( )
( )
( )
( )
1H
H
S 30 0.05 1.5g / L
30
0.4 Ln 30 1.5 5d 6 .
1.5
0.99d ld 24h
−
==
+ − =
= = =
( )
( )
( )
3
H
2
02
2
0
V Q 1L h 24h 24L 0.024m
4 0.024
4V
V D H; H 1.17m
4D
0.15
H 1.2m.
= = = =
= = = =
b.
( ) ( )
( )
P S 0
P
P Y S S 0.02 30 1.5 0.57gIA L
Pr QP 1L h 0.57g L 0.57g IA h 13.7g d
= − = − =
= = = =
