Bioreactor Design and Operation - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 940

We obtain L ¼ 49.1 m. Therefore, a 50-m long section of sterilization can meet the needs.

Example 18-3. Continuous sterilizer.
A fermentor of 10 m 3 is to be used to carry out the bioprocess. Medium at a flow rate of
1.2 m 3 /h is to be sterilized by heat exchange with steam in a continuous sterilizer. The liquid
contains bacterial spores at a concentration of 2.5 10 11 m 3 . The sterilization temperature is

to be conducted at 125 C, at which the death rate of the bacterial spores is k d ¼

500/h.
The sterilizer tube has an inner diameter of 1.2 m; the density of the medium is 1100 kg/m 3
and the viscosity is 0.00125 Pa $ s. How long of the tube is needed to sterilize the medium
for a risk of one viable cell contamination in 6 months? What is the required dimensionless
sterilization time?
Solution: Before we answer the question, we need to decide what the question asks us to
do. A risk of one viable cell contamination in 6 months seems to imply that

1
C X 0 Qt ¼

1

1:2630:524 ¼ 7:5910 16
which is a relative condition for sterilization. In the problem statement, we were told the
volume of the fermentor V

pð s Þ¼

2:510 11

10 m 3 but not whether the fermentation is continuous or batch.
One can also use a more stringent (absolute) condition, that is, the probability of unsuccessful
fermentation

¼

102:510 11
2:510 11

P 1 ð s Þ¼ VC X 0

C X 0 Qt ¼ 1:2630:524 ¼ 0:001897

for each reactor full of medium. We shall compare both cases.
The sterilization process is continuous in a tubular reactor. The flow conditions are

4 1100 3 : 333 10 4
p0:00125 1:2 ¼ 311:24 <
2000 , i.e. laminar flow. Therefore, Eqns (18.47) through (18.49) apply.

4rQ
pm v D R ¼

1.2 m 3 /h 10 4 m 3 /s; Q ¼ ¼ 3.333 Re ¼

exp

4ðk d s þ2Þ
k d s ðk d s þ8Þ

k d s
2

pð s Þ z (18.47)

exp

4 ðk d s þ2Þ
k d s ðk d s þ8Þ

1
2 k d s

C X 0 V

P 1 ð s Þ z

exp

(18.48)

2 ðC X 0 V 1 Þðk d s þ 2 Þ
k d s ðk d s þ8Þ

1
2 k d s

1þ

and

1
2 k d s ln

4ðk d s þ2Þ
k d s ðk d s þ8Þ lnðC X0 VÞ

t S ¼ (18.49)

The kinetic condition to use with these equations is given by
k d s ¼ k d 4Q D

500
3600

p
43:33310 4 1:2 2

R L ¼

L ¼ 471:24 L

with the tube length L being in meters.