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68
Chapter 10
Problem 10.1
C* = 7.3 mg/l
Plot DO versus t:
Determination of OUR:
*
LL L 2
dC k a (C C ) qo X
dt = − −
with air off: kLa (C*–CL) = O
L2
dC qo X
dt
= −
Integrating:
O
Ct
L2
CO
O2
dC qo X dt
C C qo X t
=−
− = −
∴ plotting C versus t should yield a straight line with slope = –qo2X = OUR
From least squares regression:
Slope = –1.01 mg/(L min)
Table of values:
t (h) CL (mg/l) dCL/dt (mg/l min)
13 2.9 0.168
14 3.0 0.101
15 3.1 0.042
From least squares regression: slope = − 0.265 min–1
∴ kLa = 0.265 min–1
Note: OUR can be calculated from α, i.e.
2
*
L
1
OUR qo x
k a C
0.265 min (7.3 mg/l) 0.923 mg/l min
OUR 1.01mg/l min (This agrees with the value obtained above).
−
=
= −
= −
=
Problem 10.2
kLa = 30 h–1 at 0.5 v.v.m.
qo2 = 10 mmol/g·h
Ccrit = 0.2 mg/l C* = 7.3 mg/l
mmol 32 mg
10 gh
X 0.67 g/l
=
b) Solubility of O2 in broth varies with the partial pressure of O2 in the gas,
( )
*
*
LL
2
1
1atm
C 7.3 mg/l 34.8 mg/l
0.21atm
ka
x (C C )
qo
30 h mmol
(34.8 0.2 mg/l)
mmol 32 mg
10 gh
x 3.24 g/l
−
= =
=−
=−
=
Problem 10.3
TOTAL GR
lh
But QTOTAL = ṁCP (t2 – t1)
TOTAL
P 2 1
5
43
Q
mC (t t )
9.6 10 Kcal / h
1.0 Kcal/kg C (30 15 C)
m 6.4 10 kg/h 1067 kg/min 1.07 m /min
= −
= −
= = =
broth
12
cooling
water
min
T 35 C T 35 C
t 15 C t 30 C
T 5 C
= ⎯⎯⎯→ =
= ⎯⎯⎯→ =
=
b) dc = 2.5 cm U = 1420 J/S m2°C
TOTAL
Q
Problem 10.4
a. dCL /dt = KL a (C* - CL) – qO2 X
b. KL a = a (P/V)0.4 Vs0.5 N0.5 = a (N3 Di2)0.4 Vs0.5 N0.5
Problem 10.5.
V2/V1 = 100 = 3
Then the scale up factor = 4.64 = D02/D01 =H2/H1 = Di2/Di1
V1 = (3.14/4) D012 H1 = (3.14/4) (2) D013 = 5x 10-3 m3
Then, D01 = 0.147 m , Di1 = 0.049 m , H1 = 0.294 m
a. Scale up by constant KL a
b. Scale up by constant P/V
c. Constant impeller tip speed
d. Mixing time tm is proportional to 4V/ (1.5 N Di3) or 2.67 D02 H/NDi3
Problem 10.6
H (10 l) = 0.43 S–1 K1 = 0.05 S–1
H (104 l) = 0.075 S–1
Probe is in bottom compartment, ∴ C3 = 25 mg/l
HF
( )
1
12
1 1 1
12
1
1
1
22
21 1 1
11
2
KHF
C2
H H K 3K K H
0.05 0.43 4.35 mg/l S
2(0.05 S )
0.43 0.43 0.05 3(0.05S ) 0.43S
C 34.07 mg/l
F 4.35 mg/l S
C3K K H 3(0.05 S ) 0.05S 0.43 S
C 27.92 mg/l
−
−
−
− − −
= + −
++
= + −
+
+
=
==
++
=
Similarly for 104 l tank:
1
2
3
F 7.64 mg/l S
C 86.12 mg/l
C 41.67 mg/l
C 25.0 mg/l
=
=
=
=
Problem 10.7
1 5 1
1
41
1
1
F 5 mg/l S
D 0.2 h 5.556 10 S
H (10 l) 0.43S
H (10 l) 0.075S
K 0.05S
− − −
−
−
−
=
= =
=
=
=
Steady-state mass balances on each compartment:
12 1 1 1 1
21 2 2 3 1 2
32 3 1 3
dC H (C C ) F K C DC 0
dt
dC H (C C ) H (C C ) K C 0
dt
dC H (C C ) K C 0
dt
= − + − − =
= − − − − =
= − − =
Rearranging:
1 1 2
1 1 2 3
2 1 3
(H K D) C HC F
HC (2H K ) C HC 0
HC (H K ) C 0
− + + + = −
− + + =
− + =
Substituting numbers for 10 l tank:
12
1 2 3
23
0.4800556 C 0.43 C 5
0.43 C 0.91C 0.43 C 0
0.43 C 0.48 C 0
− + = −
− + =
−=
Solving:
1
2
3
C 39.15 mg/l
C 32.08 mg/l
C 28.73 mg/l
=
=
=
1
51
F DC
Fraction S consumed F
mg
5 mg/l S 5.56 10 S 39.15 l
5 mg/l S
0.9996 1.000
−−
−
=
−
=
=
∴ Essentially all of the substrate is consumed in the 10 l tank
Problem 10.8.
4
41
0
1
d1
0
V 5 10 l V/F 2 h
N /V 10 l
1min at121 C
k61min at140 C
P (t) 0.99 over 4 weeks
−
−
−
= =
=
=
=
4 1 4
0
1h
N 10 l (5 10 l) (28 days) 24
2h day
−
=
77
1
d
k 61min−
Problem 10.9.
Achieving adequate mixing during batch sterilization is very important. Good
Problem 10.10
N0/V = 105 l–1 Cvit,0 = 30 mg/l
V = 10 l, 104 l 1 – P0 (t) = 0.001
Calculating values of kd at 121°C:
4
36 1 1
d,S
6 10 cal/mol
k 1 10 min exp 0.9027 min
cal
1.987 (394.15 K)
mol K
−−
= − =
4
4 1 1
d,V
1 10 cal/mol
k 1 10 min exp 0.0285 min
cal
1.987 (394.15 K)
mol K
−−
= − =
Calculating sterilization time for 10 l fermenter:
78
N0 = 105 l–1 (10 l) = 106spores 1–P0 (t) = 0.001
From Fig. 10.12: kd,St = 20.78
1
d,S
20.78 20.78
t 23.0 min
k 0.9027 min−
= = =
Assuming first order degradation of vitamin:
1
d,V
kt 0.0285 min (23 min)
vit vit,0
mg
C C e 30 e 15.6 mg/l
l
−
−−
= = =
∴ 52% of the vitamin is active in the 10 l fermenter after sterilization
For 104 l fermenter: N0 = 105 l–1 (104 l) = 109spores
Problem 10.11.
V = 104 l Cvit,0 = 30 mg/l
n0 = 105 l–1
36 1
d,S
65000 cal/mol
1 10 min exp 1.987 cal / mol K(T K)
k−
= −
41
d,V
10000 cal/mol
k 1 10 min exp 1.987 cal / mol K(T K)
−
= −
Polynomial regressions of temperature profile:
a)
d,S
dn kn
dt =−
0
t
0d,S
t
t
n
n=
For the heat-up and cool-down periods:
i) Calculate temperatures for small increments of t using polynomials from
above (used ∆t = 2.5 min for this solution).
ii) Evaluate kd,S for each T.
iii) Use numerical integration (Simpson’s Rule) to evaluate integral.
140
vit,S d,v
60
vit,C
C
ln k dt 0.4716
C==
vit,0
vit,C
C
ln 1.0691
C
=
or
vit,C
vit,0
C
ln 1.0691
C=−
1.0691 1.0691
vit,C vit,0
C C e 30 mg/l e
10.3 mg/l
−−
= =
=
∴ 34.3 % of the vitamin remains undegraded
0.3126 0.3126
−−
21.95 16.51mg/l
Fraction degraded 0.181
30 mg/l
−
==
∴ 18.1 % of the vitamin is degraded during the sterilization period
d) Fraction degraded in heat-up and cool-down periods:
Sh
1
2.5920 l
−
=
Fraction of deactivated spores that is deactivated during heat-up and cool-down:
0 C h S
0C
5 4 1
5
(n n ) (n n )
f(n n )
(10 0.6184) (2.1578 10 2.5920) l
(10 0.6184)
f 0.784
−
− − −
=−
− − −
=−
=
∴ 78.4 % of the deactivated spores are deactivated during heat-up and cool-
down.
Problem 10.12
Note that we are asked for a value at a specific time – i.e. when X = 20 g/L.
Assume over the period of observation dCL/dt ≈ 0 (i.e. CL changes slowly with
time which is a good assumption)
Thus
*2L
L L L
2L
qo max C
dC /dt 0 k a (C C ) x
Ko C
= − − +
or
*2L
LL
2L
qo max C
k a (C C ) x
Ko C
−= +
or
12L
L
L
240 mg O /q-cells-h 20 g/l C
120 h (28 mg/L C ) 0.2 mg / l C
−
−= +
Problem 10.13.
a) With respect to sterilization the heating-up/cooling-down periods make no
measurable contribution. This is because of the high value of Ead for spores. The
41
1 10 min exp[ 10 000 cal/g-mol/1.99 cal/g-mol K 273.15 C T in celsius]
−−
= − − +
So
d
1
1
11
11
t0
– k t
t 15
0.00807 min 15 min
0.029 min 1.25 min
t 40
0.00807 min 10 min
t 50
0.00000018 min 10 min
t 60
C 50 mg/L
C 50 mg/L e @85 C
50 e 44.3 mg/l
@121 C C 44.3 e 21.4 mg/l
@85 C C 21.4 e 19.7 mg/l
@55 C C 19.7 mg/l e 1
−
−
−−
−−
=
=
−
−
=
−
=
−
=
=
=
==
= =
= =
= 9.7 mg/l 1
Final C ≈ 19.7 mg/l or 39 % remains
Problem 10.14.
The solution approach is identical to that in example 10.1 except C* = 7.3 mg/l
is replaced by C* = 35 mg/l. Thus, (at C = 1 mg/l)
L
1
0.25 mg/l-min 0.8 mg/l-min
ka (35 1) mg/l
0.031min−
+
=−
=
C* = 35 mg/L would correspond to a case where the gas used to aerate the
fermenter is highly enriched in O2
Problem 10.15.
2
*
LL L O
dC K a (C C ) Q X
dt = − −
C* = 9 mg/L
11
L
K a 0.388 min 23.3 h
−−
==
Problem 10.16.
43
1
23
2
V 10 L 10 m
V 10 L 0.1m
==
==
3
1
2
V10 100
V 0.1
4.64 Scale down factor
= = =
= − −
a. Geometric Similarity
1
2
1
2
t
11
2 t 2
t
t
1
2
1
122
T1
D
H Di 4.64
H D Di
D2
D 0.43 m.
4.64
Di 0.5
Di 0.108 m
4.64
4V 4 (10)
H 3.185 m
D (2)
= = = =
= = =
= = =
= = =
i1
21
i2
D
N N 100 ( ) 100(4.64) 464 rpm.
D
= = = =
2. Constant impeller Re:
22
i2
2
N 2153 rpm.
=
Problem 10.17.
56
0
N 10 spores/l 10 l 10 spores= =
od
E /KT
d
ke
−
=
1 90 000/1.99.394.16 K
d
@121 C k 1min e
− −
= =
⇒
49 1
3.83 10 min−
=
49 90000/1.99 392.66 K
d
1
@119.5 C k 3.83 10 e
0.65 min
−
−
=
=
If t = 20 min., then
1
d
k t 0.65 min 20 min
13
−
=
=
The sterilization chart is hard to read in this region, but we know
0
6
N
Kdt
13 10
1 P 1 [1 ]
1 [1 e ]
1 P 0.896
e−
−
− = − −
= − −
−=
Probability of success ≈ 0.10
Demonstrates how sensitive system is to small ∆T
Problem 10.18
Note that the number of spores to be killed is:
4 10
10 /L 330,000 / 30 9.9 10 sporesx L d x d x=
With a probability of failure to be 0.001.
b.) If the temperature is 140ºC and
1
32.560min−
Problem 10.19
