CP504 – Lecture 15 and 16
Sterilization
- Learn about thermal sterilization of liquid medium
- Learn about air sterilization
- Learn to do design calculations
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18 Nov 2011
1
Sterilization is a process to kill or inactivate all viable
organisms in a culture medium or in a gas or in the
fermentation equipment.
This is however not possible in practice to kill or
inactivate all viable organisms.
Commercial sterilization is therefore aims at reduce risk
of contamination to an acceptable level.
Factors determining the degree of sterilization include
safety, cost and effect on product.
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Reasons for Sterilization:
• Many fermentations must be absolutely devoid of foreign
organisms.
• Otherwise production organism must compete with the
foreign organisms (contaminants) for nutrients.
• Foreign organisms can produce harmful (or unwanted)
products which may inhibit the growth of the production
organisms.
• Economic penalty is high for loss of sterility.
• Vaccines must have only killed viruses.
• Recombinant DNA fermentations - exit streams must be
sterilized.
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And more….
3
Sterilization Methods:
• Thermal:
preferred for economical large-scale sterilizations
of liquids and equipment
• Chemical:
preferred for heat-sensitive equipment
→ ethylene oxide (gas) for equipment
→ 70% ethanol-water (pH=2) for equipment/surfaces
→ 3% sodium hypochlorite for equipment
• Irradiation:
→ ultraviolet for surfaces
→ X-rays for liquids (costly/safety)
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Sterilization Methods continues:
• Sonication (sonic / ultrasonic vibrations)
• High-speed centrifugation
• Filtration:
preferred for heat-sensitive material and filtered air
Read pages 213 to 214 of J.M. Lee for more
on sterilization methods
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Thermal Sterilization:
• Dry air or steam can be used as the heat agent.
• Moist (wet) steam can also be used as the heat agent
(eg: done at 121oC at 2 bar).
• Death rate of moist cells are higher than that of the dry
cells since moisture conducts heat better than a dry
system.
• Therefore moist steam is more effective than dry
air/steam.
• Thermal sterilization does not contaminate the medium
of equipment that was sterilized (as in the case of use
of chemical agent for sterilization).
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Thermal sterilization using dry heat
- Direct flaming
- Incineration
- Hot air oven
-170 °C for 1 hour
-140 °C for 3 hours
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Thermal sterilization using moist heat
- Pasteurization (below 100oC)
Destroys pathogens without altering the flavor of the food.
Classic method: 63oC; 30 min
High Temperature/Short Time (HTST) : 71.7oC; 15-20 sec
Untra High Temperature (UHT) : 135oC; 1 sec
- Boiling (at 100oC)
killing most vegetative forms microorganisms
Requires 10 min or longer time
Hepatitis virus can survive for 30 min & endospores for 20 h
- Autoclaving (above 100oC)
killing both vegetative organisms and endospores
121-132oC; 15 min or longer
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Thermal Death Kinetics:
dnt
= - kd nt
dt
(10.1)
where
nt is the number of live organisms present
t is the sterilization time
kd is the first-order thermal specific death rate
kd depends on the type of species, the physiological form of
the cells, as well as the temperature.
kd for vegetative cells > kd for spores > kd for virus
(10 to 1010/min)
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18 Nov 2011
(0.5 to 5/min)
9
Hyphal growth
Spore
production
Spore
germination
Spores
Spores form part of the life cycles of many bacteria, plants,
algae, fungi and some protozoa. (There are viable bacterial
spores that have been found that are 40 million years old on
Earth and they're very hardened to radiation.)
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A spore is a reproductive structure that is adapted for
dispersal and surviving for extended periods of time in
unfavorable conditions.
A chief difference between spores and seeds as dispersal
units is that spores have very little stored food resources
compared with seeds.
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Thermal Death Kinetics (continued):
Integrating (10.1) using the initial condition n = no at t = 0 gives
t
nt
ln
no
= -

(10.2)
kd dt
0
t
( )
nt
= exp no
kd dt
(10.3)
0
Survival factor
Inactivation factor ≡
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18 Nov 2011
1
Survival factor =
no
nt
12
Thermal Death Kinetics (isothermal operation):
kd is a function of temperature, and therefore it is a constant
for isothermal operations.
(10.2) therefore gives
nt
- kd t
ln
=
no
nt
= exp(- kd t)
no
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18 Nov 2011
(10.4)
(10.5)
13
Thermal Death Kinetics
(non-isothermal operation):
kd is expressed by the Arrhenius equation given below:
( )
kd = kdo exp -
Ed
RT
(10.6)
where
kdo
Ed
R
T
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18 Nov 2011
Arrhenius constant for thermal cell death
is the activation energy for thermal cell death
is the universal gas constant
is the absolute temperature
14
Thermal Death Kinetics
(non-isothermal operation):
When kd of (10.6) is substituted in (10.2), we get the following:
nt
ln
-k
no = do

0
t
( )
exp -
Ed
RT
dt
(10.7)
To carry out the above integration, we need to know how the
temperature (T) changes with time (t).
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Determining the Arrhenius constants:
( )
Ed
kd = kdo exp -
RT
Ed
ln(kd) = ln(kdo) ln(kd)
RT
(10.7)
ln(kdo)
Ed
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18 Nov 2011
(10.6)
R
1/T
16
Example 10.1:
A fermentation medium contains an initial spores concentration
of 8.5 x 1010. The medium is sterilized thermally at 120oC, and
the spore density was noted with the progress of time as given
below:
Time
(min)
Spore density
(m-3)
0
5
10
15
20
30
8.5 x
1010
4.23 x
109
6.2 x
107
1.8 x
106
4.5 x
104
32.5
a) Find the thermal specific death rate.
b) Calculate the survival factor at 40 min.
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18 Nov 2011
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Solution to Example 10.1:
Data provided:
no = 8.5 x 1010
nt versus t data are given
Isothermal operation at 120oC.
a) Since it is an isothermal operation, thermal specific death rate
(kd) is a constant. Therefore, (10.4) can be used as follows:
nt
ln
= - kd t
no
Plotting ln(nt /no) versus t and finding the slope will give
the numerical value of kd.
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18 Nov 2011
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Solution to Example 10.1:
0
10
t (min)
20
30
0
ln(nt/no)
-5
-10
-15
-20
y = -0.7201x
R2 = 0.9988
-25
kd = -slope = 0.720 per min
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Solution to Example 10.1:
1
nt/no
0.8
0.6
0.4
0.2
0
0
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18 Nov 2011
10 t (min) 20
30
20
Solution to Example 10.1:
b) Since kd is known from part (a), the survival factor at 40 min
can be calculated using (10.5) as follows:
nt
no
= exp (- 0.720 per min x t)
= exp (- 0.720 per min x 40 min)
= 3.11 x 10-13 = survival factor
nt
= 3.11 x 10-13 no = 3.11 x 10-13 x 8.5 x 1010 = 0.026
We know that nt cannot be less than 1.
The above number is interpreted as the chances of
survival for the living organism is 26 in 1000.
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Example 10.2:
The thermal death kinetic data of Bacillus stearothermophilus
(which is one of the most heat-resistant microbial type) are as
follows at three different temperatures:
Temperature (oC)
kd (min-1)
115
120
125
0.035
0.112
0.347
a) Calculate the activation energy (Ed) and Arrhenius constant
(kdo) of the thermal specific death rate kd.
b) Find kd at 130oC.
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Solution to Example 10.2:
Data provided: kd versus temperature data are given
a) Activation energy (Ed) and Arrhenius constant (kdo) of the
thermal specific death rate (kd) can be determined starting
from (10.7) as follows:
ln(kd)
= ln(kdo) -
Ed
RT
Plot ln(kd) versus 1/T (taking T in K).
Slope gives (–Ed/R) and intercept gives ln(kdo).
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Solution to Example 10.2:
1/T (per K)
0.0025 0.00252 0.00254 0.00256 0.00258 0.0026
0
ln(kd)
-0.5
-1
y = -35425x + 87.949
R2 = 1
-1.5
-2
-2.5
-3
-3.5
-4
Slope = –Ed/R = –35425 K
Intercept = ln(kdo) = 87.949
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Solution to Example 10.2:
Slope = –Ed/R = –35425 K
Ed = (35425 K) (R) = 35425 x 8.314 kJ/kmol = 294.5 kJ/mol
Activation energy
Intercept = ln(kdo) = 87.949
kdo = exp(87.949) = 1.5695 x 1038 per min
= 2.616 x 1036 per s
Arrhenius constant
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18 Nov 2011
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Solution to Example 10.2:
b) Since activation energy (Ed) and Arrhenius constant (kdo) of
the thermal specific death rate (kd) are known from part (a), kd
at 130oC can be determined using (10.6) as follows:
kd = kdo exp
( )
-
Ed
RT
= (2.616 x 1036 per s) exp
= 0.0176 per s
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18 Nov 2011
(
)
294.5 x 103
8.314 (273+130)
= 1.0542 per min
26
Solution to Example 10.2:
ln(kd)
1/T (per K)
0.00245
0.5
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
0.0025
0.00255
0.0026
y = -35484x + 88.101
R2 = 1
Calculated value at 130oC
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Design Criterion for Sterilization:
t

=
no
ln
nt
t
= kdo

0
=

(10.8)
kd dt
0
( )
exp -
Ed
RT
dt
(10.9)
Del factor (which is a measure of
fractional reduction in living organisms
count over the initial number present)
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Determine the Del factor to reduce the number of cells in a
fermenter from 1010 viable organisms to 1:
t

=
no
ln
nt
=
ln
1010
1
= 23 =

kd dt
0
Even if one organism is left alive, the whole fermenter may be
contaminated.
Therefore, no organism must be left alive. That is, n = 0

=
no
ln
nt
=
ln
1010
0
= infinity =
infinity

kd dt
0
To achieve this del factor, we need infinite time that is not
possible.
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18 Nov 2011
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Therefore n should not be 1, and it cannot be 0.
Let is choose n = 0.001
(It means the chances of 1 in 1000 to survive) :
t

=
no
ln
nt
=
ln
1010
= 30
0.001
=

kd dt
0
Using the Arrhenius law, we get
t
=
kdo
 ( )
exp -
Ed
RT
dt
= 30
0
Temperature profile during sterilization must be chosen
such that the Del factor can become 30.
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Typical temperature profile during sterilization:
heating
holding
cooling
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Let us take a look at some sterilization
methods and equipment
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Batch Sterilization (method of heating):
Direct steam
sparging
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Electrical
heating
Steam
heating
33
Batch Sterilization (method of cooling):
Cold water
cooling
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For batch heating by direct steam sparging:
T = T0 +
T
T0
c
M
ms
t
H
Direct steam
sparging
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18 Nov 2011
–
–
–
–
–
–
–
H ms t
(10.10)
c (M + ms t)
final temperature (in kelvin)
initial temperature (in kelvin)
specific heat of medium
initial mass of medium
steam mass flow rate
time required
enthalpy of steam relative to
raw medium temperature
35
For batch heating with constant rate of heat flow:
q
t
T = T0 +
cM
T
T0
c
M
t
q
–
–
–
–
–
–
(10.11)
final temperature (in kelvin)
initial temperature (in kelvin)
specific heat of medium
initial mass of medium
time required
constant rate of heat transfer
Electrical
heating
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For batch heating with isothermal heat source:
(
T = TH + (T0 - TH) exp - U A t
cM
Steam
heating
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)
(10.12)
T – final temperature (in kelvin)
TH – temperature of heat source
(in kelvin)
T0 – initial temperature (in kelvin)
c – specific heat of medium
M – initial mass of medium
t – time required
U – overall heat transfer coefficient
A – heat transfer area
37
For batch cooling using a continuous non-isothermal
heat sink (eg: passing cooling water through a vessel jacket):
T = TC0 + (T0 - TC0) exp{- [1 – exp(
UA
cm
)]
mt
M
}
(10.13)
T
T0
TC0
U
A
c
m
M
t
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18 Nov 2011
–
–
–
–
–
–
–
–
–
final temperature (in kelvin)
initial temperature of medium (in kelvin)
initial temperature of heat sink (in kelvin)
overall heat transfer coefficient
heat transfer area
specific heat of medium
coolant mass flow rate
initial mass of medium
time required
38
Example 10.3: Estimating the time required for a batch sterilization
- Typical bacterial count in a medium is 5 x 1012 per m3, which is to
be reduced by sterilization such that the chance for a contaminant
surviving the sterilization is 1 in 1,000.
- The medium is 40 m3 in volume and is at 25oC. It is to be sterilized
by direct injection of saturated steam in a fermenter. Steam available
at 345 kPa (abs pressure) is injected at a rate of 5,000 kg/hr, and will
be stopped once the medium reaches 122oC.
- The medium is held for some time at 122oC. Heat loss during
holding time is neglected.
- Medium is cooled by passing 100 m3/hr of water at 20oC through
the cooling coil until medium reaches 30oC. Coil heat transfer area is
40 m3, and U = 2500 kJ/hr.m2.K.
- For the heat resistant bacterial spores:
kdo = 5.7 x 1039 per hr
Ed = 2.834 x 105 kJ / kmol
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- For
the medium: c = 4.187 kJ/kg.K and ρ = 1000 kg/m3
18 Nov 2011
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Solution to Example 10.3:
Problem statement: Typical bacterial count in a medium is
5 x 1012 per m3, which is to be reduced by sterilization such
that the chance for a contaminant surviving the sterilization
is 1 in 1,000.
n0 = 5 x 1012 per m3 x 40 m3 = 200 x 1012 = 2 x 1014
nt = 1/1000 = 0.001

=
n0
ln
nt
=
t
ln
2x1014
0.001
= 39.8
=

kd dt
0
The above integral should give 39.8.
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Solution to Example 10.3:
Given sterilization process involves heating from 25oC to 122oC,
holding it at 122oC and then cooling back to 30oC. Therefore
t


=
kd dt
0
t2
t1
=

0
heat
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18 Nov 2011
+
kd dt
heating

t1
t3
+
kd dt
holding
hold

t2
kd dt
cooling
cool
41
Solution to Example 10.3:
The design problem is therefore,
 = heat + hold + cool
= 39.8
(10.14)
Since the holding process takes place at isothermal condition,
we get
t2
hold
=

kd dt
= kd (t2-t1)
t1
holding


(10.15)
holding
To determine heat and cool, we need to get the
temperature profiles in heating and cooling operations,
respectively.
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Solution to Example 10.3:
Heating is carried out by direct injection of saturated steam in a
fermenter.
Temperature profile during heating by steam sparging is given
by (10.10):
H ms t
T = T0 +
c (M + ms t)
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Solution to Example 10.3:
Data provided:
T0
c
M
ms
H
= (25 + 273) K = 298 K
= 4.187 kJ/kg.K
= 40 x 1000 kg
= 5000 kg/hr
= enthalpy of saturated steam at 345 kPa
- enthalpy of water at 25oC
= 2,731 – 105 kJ/kg = 2626 kJ/kg
Therefore, we get
T = 298 +
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18 Nov 2011
78.4 t
1 + 0.125 t
44
Solution to Example 10.3:
For
T
= (122 + 273) K = 395 K
395 = 298 +
78.4 t
1 + 0.125 t
t = 1.46 h
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18 Nov 2011
It is the time required to heat the
medium from 25oC to 122oC.
45
Solution to Example 10.3:
1.46
heat
=

kd dt
0
Use
heating
( )
(
kd = kdo exp -
Ed
RT
= 5.7 x 1039 exp -
where
)
2.834 x 105
8.318 x T
78.4 t
T = 298 +
1 + 0.125 t
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1.46

heat =

(10.16)
= 14.8
kd dt
heating
0
250
Temperature (deg C)
200
kd (per hr)
150
100
50
0
0
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18 Nov 2011
0.25
0.5
0.75
time (in hr)
1
1.25
1.5
47
Solution to Example 10.3:
Cooling is carried out by passing cooling water through vessel
jacket.
Temperature profile during cooling using a continuous nonisothermal heat sink is given by (10.13)
T = TC0 + (T0 - TC0) exp{
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18 Nov 2011
[1 – exp(
UA
cm
)]
mt
}
M
48
Solution to Example 10.3:
Data provided:
T0
TC0
U
A
c
m
M
= (122 + 273) K = 395 K
= (20 + 273) K = 293 K
= 2,500 kJ/hr.m2.K
= 40 m2
= 4.187 kJ/kg.K
= 100 x 1000 kg/hr
= 40 x 1000 kg/hr
Therefore, we get
T = 293 + 102 exp{
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18 Nov 2011
[1 – exp(
1
4.187
)]
t
0.4
}
49
Solution to Example 10.3:
For
T
= (30 + 273) K = 303 K
393 = 293 + 102 exp{
t = 3.45 h
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18 Nov 2011
[1 – exp(
1
4.187
)]
t
0.4
}
It is the time required to cool the
medium from 122oC to 30oC.
50
Solution to Example 10.3:
t2+3.45
cool
=

kd dt
t2
cooling
( )
(
kd = kdo exp -
Ed
RT
= 5.7 x 1039 exp -
)
2.834 x 105
8.318 x T
T = 293 + 102 exp{- 0.674 t}
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51
t2+3.45
cool
=

kd dt
t2
(10.17)
= 13.9
cooling
250
Temperature (deg C)
200
kd (per hr)
150
100
50
0
0
0.5
1
1.5
2
2.5
3
3.5
time (in hr)
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Putting together the results:
1.46

3.45
+ kd Δt
kd dt
0
heating
heat = 14.8
hold = kd Δt
+
holding

0
hold
kd dt
= 39.8
cooling
cool = 13.9
= 39.8 -14.8 -13.9 = 11.1
holding
Δt = 11.1 / (kd at 1220C)
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Putting together the results:
kd at 1220C
(
= 5.7 x 1039 exp -
)
2.834 x 105
8.318 x T
T = 395
= 197.6 per hr
Δt = 11.1 / 197.6 = 0.056 hr = 3.37 min
Sterilization is achieved mostly during
the heating (14.8 hr) and cooling (13.9 hr)
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Putting together the results:
Temperature (deg C)
250
kd (per hr)
200
150
100
50
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time (in hr)
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Drawback:
Longer heat-up and cool-down time
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