1. Sterilization CP504 – Lecture 15 and 16
Learn about thermal sterilization of liquid medium
Learn about air sterilization
Learn to do design calculations
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2. 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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3. 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.
And more….
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4. 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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5. 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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6. 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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7. 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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8. 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; 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
oC; 15 min or longer
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9. Thermal Death Kinetics:
dnt
- kd nt =
(10.1) dt
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)
(0.5 to 5/min)
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10. 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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11. 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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12. ( )   Thermal Death Kinetics (continued): ln nt no = - kd dt (10.2)
Integrating (10.1) using the initial condition n = no at t = 0 gives ln nt no =
- kd dt  t
(10.2) nt no =
- kd dt  t
exp
( )
(10.3)
Survival factor 1 no
Inactivation factor ≡ =
Survival factor nt
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13. 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 ln
- kd t
(10.4) = no nt
exp(- kd t) =
(10.5) no
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14. ( ) Thermal Death Kinetics (non-isothermal operation): - (10.6) kd exp
kd is expressed by the Arrhenius equation given below:
( ) Ed -
(10.6) kd
exp =
kdo RT
where
kdo Arrhenius constant for thermal cell death
Ed is the activation energy for thermal cell death
R is the universal gas constant
T is the absolute temperature
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15. ( )  Thermal Death Kinetics (non-isothermal operation): nt ln - kdo
When kd of (10.6) is substituted in (10.2), we get the following: t
( ) nt  Ed ln
- kdo
exp - dt
(10.7) = no RT
To carry out the above integration, we need to know how the temperature (T) changes with time (t).
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16. ( ) Determining the Arrhenius constants: kd exp - (10.6) = kdo ln(kd)
( ) Ed kd
exp -
(10.6) =
kdo RT
ln(kd)
ln(kdo) = - RT Ed
(10.7)
ln(kdo)
ln(kd) Ed R
1/T
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17. Example 10.1:
A fermentation medium contains an initial spores concentration of 8.5 x The medium is sterilized thermally at 120oC, and the spore density was noted with the progress of time as given below:
a) Find the thermal specific death rate.
b) Calculate the survival factor at 40 min.
Time
(min) 5 10 15 20 30
Spore density (m-3)
8.5 x 1010
4.23 x 109
6.2 x 107
1.8 x 106
4.5 x 104
32.5
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18. Solution to Example 10.1: Data provided: no = 8.5 x 1010
nt versus t data are given
Isothermal operation at 120oC.
Since it is an isothermal operation, thermal specific death rate (kd) is a constant. Therefore, (10.4) can be used as follows: nt
- kd t ln = no
Plotting ln(nt /no) versus t and finding the slope will give the numerical value of kd.
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19. Solution to Example 10.1: kd = -slope = 0.720 per min
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20. Solution to Example 10.1:
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21. Solution to Example 10.1: We know that nt cannot be less than 1.
b) Since kd is known from part (a), the survival factor at 40 min can be calculated using (10.5) as follows: nt
= exp ( per min x t) no
= exp ( per min x 40 min)
= 3.11 x = survival factor
nt = 3.11 x no = 3.11 x 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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22. 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:
a) Calculate the activation energy (Ed) and Arrhenius constant (kdo) of the thermal specific death rate kd.
b) Find kd at 130oC.
Temperature (oC)
115
120
125
kd (min-1)
0.035
0.112
0.347
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23. Solution to Example 10.2:
Data provided: kd versus temperature data are given
Activation energy (Ed) and Arrhenius constant (kdo) of the thermal specific death rate (kd) can be determined starting from (10.7) as follows: Ed
ln(kd)
ln(kdo) = - RT
Plot ln(kd) versus 1/T (taking T in K).
Slope gives (–Ed/R) and intercept gives ln(kdo).
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24. Solution to Example 10.2: Slope = –Ed/R = –35425 K
Intercept = ln(kdo) =
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25. Solution to Example 10.2: Slope = –Ed/R = –35425 K
Ed = (35425 K) (R) = x kJ/kmol = kJ/mol
Intercept = ln(kdo) =
kdo = exp(87.949) = x 1038 per min
= x 1036 per s
Activation energy
Arrhenius constant
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26. ( ) ( ) 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:
( ) Ed - kd =
kdo
exp RT
( )
294.5 x 103 - =
(2.616 x 1036 per s)
exp
8.314 ( ) =
per s =
per min
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27. Solution to Example 10.2: Calculated value at 130oC
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28. ( )    Design Criterion for Sterilization: = kd dt ln no nt =t ln no nt = 
(10.8)
kdo = - RT Ed
exp
( ) dt  t
(10.9)
Del factor (which is a measure of fractional reduction in living organisms count over the initial number present)
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29. Determine the Del factor to reduce the number of cells in a fermenter from 1010 viable organisms to 1: =
kd dt  t ln no nt =
= ln
1010 1 
= 23
Even if one organism is left alive, the whole fermenter may be contaminated.
Therefore, no organism must be left alive. That is, n = 0 =
kd dt 
infinity ln no nt =
= ln
1010 
= infinity
To achieve this del factor, we need infinite time that is not possible.
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30. 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) :
kd dt  t ln no nt =
= ln
1010
0.001 
= 30 =
Using the Arrhenius law, we get
kdo = - RT Ed
exp
( ) dt  t 
= 30
Temperature profile during sterilization must be chosen such that the Del factor can become 30.
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31. Typical temperature profile during sterilization:
heating
holding
cooling
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32. Let us take a look at some sterilization methods and equipment
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33. Batch Sterilization (method of heating):
Direct steam
sparging
Electrical
heating
Steam
heating
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34. Batch Sterilization (method of cooling):
Cold water
cooling
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35. For batch heating by direct steam sparging:
H ms t
T = T0 +
(10.10)
c (M + ms t)
T – final temperature (in kelvin)
T0 – initial temperature (in kelvin)
c – specific heat of medium
M – initial mass of medium
ms – steam mass flow rate
t – time required
H – enthalpy of steam relative to
raw medium temperature
Direct steam
sparging
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36. For batch heating with constant rate of heat flow:q t
T = T0 +
(10.11)
c M
T – final temperature (in kelvin)
T0 – initial temperature (in kelvin)
c – specific heat of medium
M – initial mass of medium
t – time required
q – constant rate of heat transfer
Electrical
heating
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37. ( ) For batch heating with isothermal heat source: U A t
( )
U A t
T = TH + (T0 - TH) exp -
(10.12)
c M
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
Steam
heating
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38. T = TC0 + (T0 - TC0) exp{- [1 – exp( )] } U A c m m t M
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( )] }
U A
c m
m t M
(10.13)
T – final temperature (in kelvin)
T0 – initial temperature of medium (in kelvin)
TC0 – initial temperature of heat sink (in kelvin)
U – overall heat transfer coefficient
A – heat transfer area
c – specific heat of medium
m – coolant mass flow rate
M – initial mass of medium
t – time required
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39. 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 = x 105 kJ / kmol
For the medium: c = kJ/kg.K and ρ = 1000 kg/m3
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40.   kd dt = 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 =
kd dt  t ln n0 nt =
= ln
2x1014
0.001  =
The above integral should give 39.8.
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41.     heat hold cool  = kd dt + = kd dt kd dt + kd dt
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
kd dt  t1
kd dt  t1 t2
kd dt  t2 t3
heating
holding
cooling = + +
heat
hold
cool
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42.   heat hold = + hold + cool = 39.8 = kd dt = kd (t2-t1)
Solution to Example 10.3:
The design problem is therefore, 
heat
+ hold
+ cool = =
(10.14)
Since the holding process takes place at isothermal condition, we get
kd dt  t1 t2
holding
holding
hold =
= kd (t2-t1)
(10.15)
To determine heat and cool, we need to get the temperature profiles in heating and cooling operations, respectively.
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43. 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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44. H = enthalpy of saturated steam at 345 kPa - enthalpy of water at 25oC
Solution to Example 10.3:
Data provided:
T0 = ( ) K = 298 K
c = kJ/kg.K
M = 40 x 1000 kg
ms = 5000 kg/hr
H = enthalpy of saturated steam at 345 kPa
- enthalpy of water at 25oC
= 2,731 – 105 kJ/kg = 2626 kJ/kg
Therefore, we get
78.4 t
T = t
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45. It is the time required to heat the medium from 25oC to 122oC.
Solution to Example 10.3:
For
T = ( ) K = 395 K
78.4 t
395 = t
It is the time required to heat the medium from 25oC to 122oC.
t = 1.46 h
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46. ( ) ( )  heat kd dt = kd kdo = - exp 2.834 x 105 - = 5.7 x 1039 exp
Solution to Example 10.3:
kd dt 
1.46
heating
heat = kd
kdo = - RT Ed
exp
( )
Use
( )
2.834 x 105 - =
5.7 x 1039
exp
8.318 x T
78.4 t
where
T = t
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47. kd dt 
1.46
heating
heat
= 14.8
(10.16) =
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48. T = TC0 + (T0 - TC0) exp{ [1 – exp( )] } U A c m m t M
Solution to Example 10.3:
Cooling is carried out by passing cooling water through vessel jacket.
Temperature profile during cooling using a continuous non-isothermal heat sink is given by (10.13)
T = TC0 + (T0 - TC0) exp{ [1 – exp( )] }
U A
c m
m t M
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49. Solution to Example 10.3: Data provided: T0 = (122 + 273) K = 395 K
TC0 = ( ) K = 293 K
U = 2,500 kJ/hr.m2.K
A = 40 m2
c = kJ/kg.K
m = 100 x 1000 kg/hr
M = 40 x 1000 kg/hr
Therefore, we get
T = exp{ [1 – exp( )] } 1 t
4.187
0.4
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50. It is the time required to cool the medium from 122oC to 30oC.
Solution to Example 10.3:
For
T = ( ) K = 303 K
393 = exp{ [1 – exp( )] } 1 t
4.187
0.4
t = 3.45 h
It is the time required to cool the medium from 122oC to 30oC.
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51. ( ) ( )  cool kd dt = kd kdo = - exp 2.834 x 105 exp - = 5.7 x 1039
Solution to Example 10.3:
kd dt  t2
t2+3.45
cooling
cool = kd
kdo = - RT Ed
exp
( )
( )
2.834 x 105
exp - =
5.7 x 1039
8.318 x T
T = exp{ t}
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52.  cool kd dt = t2 t2+3.45 cooling (10.17) = 13.9
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53. Putting together the results:
kd dt 
1.46
kd dt 
3.45
heating
holding
cooling +
kd Δt + =
heat = 14.8
hold
cool = 13.9
hold = kd Δt = = 11.1
holding
Δt = 11.1 / (kd at 1220C)
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54. ( ) Putting together the results: kd at 1220C 2.834 x 105 exp -
( )
2.834 x 105
5.7 x 1039
exp - =
8.318 x T
T = 395
= per hr
Δt = 11.1 / = hr = 3.37 min
Sterilization is achieved mostly during the heating (14.8 hr) and cooling (13.9 hr)
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55. Putting together the results:
Drawback: Longer heat-up and cool-down time
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