CP504 – Lecture 15 and 16

Sterilization

- Learn about thermal sterilization of liquid medium - Learn about air sterilization - Learn to do design calculations R. Shanthini 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.

R. Shanthini 18 Nov 2011 2

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.

R. Shanthini 18 Nov 2011 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) R. Shanthini 18 Nov 2011 4

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 121 o C 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 100 o C)

Destroys pathogens without altering the flavor of the food.

Classic method: 63

o

C; 30 min High Temperature/Short Time (HTST) : 71.7

o

C; 15-20 sec Untra High Temperature (UHT) : 135

o

C; 1 sec

- Boiling (at 100 o C)

killing most vegetative forms microorganisms Requires 10 min or longer time Hepatitis virus can survive for 30 min & endospores for 20 h

- Autoclaving (above 100 o C)

killing both vegetative organisms and endospores 121-132

o

C; 15 min or longer R. Shanthini 18 Nov 2011 8

Thermal Death Kinetics:

dn t dt = - k d n t (10.1) where n t is the number of live organisms present t is the sterilization time k d is the first-order thermal specific death rate k d depends on the type of species, the physiological form of the cells, as well as the temperature.

k d for vegetative cells > k d (10 to 10 10 /min) for spores > k d (0.5 to 5/min) for virus R. Shanthini 18 Nov 2011 9

Hyphal growth Spore germination Spore production 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.) R. Shanthini 18 Nov 2011 10

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 = n o at t = 0 gives ln n t n o n t n o = = 0  t k d dt exp 0 t

( )

(10.2) (10.3)

Survival factor Inactivation factor ≡ 1 Survival factor =

n o n t R. Shanthini 18 Nov 2011 12

Thermal Death Kinetics (isothermal operation) :

k d is a function of temperature, and therefore it is a constant for isothermal operations. (10.2) therefore gives ln n t n o = - k d t (10.4) n t n o = exp(- k d t) R. Shanthini 18 Nov 2011 (10.5) 13

Thermal Death Kinetics (non-isothermal operation) :

k d is expressed by the

Arrhenius equation

given below: E d k d = k do exp (10.6) where k do E d Arrhenius constant for thermal cell death is the activation energy for thermal cell death R is the universal gas constant T is the absolute temperature R. Shanthini 18 Nov 2011 14

Thermal Death Kinetics (non-isothermal operation) :

When k d of (10.6) is substituted in (10.2), we get the following: ln n t n o = - k do 0  t exp E d dt (10.7) To carry out the above integration, we need to know how the temperature (T) changes with time (t). R. Shanthini 18 Nov 2011 15

Determining the Arrhenius constants:

k d = k do exp (10.6) ln(k d ) = ln(k do ) E d RT ln(k d ) ln(k do ) (10.7) R. Shanthini 18 Nov 2011 E d R 1/T 16

Example 10.1:

A fermentation medium contains an initial spores concentration of 8.5 x 10 10 . The medium is sterilized thermally at 120 o C, and the spore density was noted with the progress of time as given below: 15 20 30 Time (min) Spore density (m -3 ) 0 8.5 x 10 10 5 10 4.23 x 10 9 6.2 x 10 7 1.8 x 10 6 4.5 x 10 4 32.5

a) Find the thermal specific death rate.

b) Calculate the survival factor at 40 min.

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Solution to Example 10.1:

Data provided: n o n t = 8.5 x 10 10 versus t data are given Isothermal operation at 120 o C. a) Since it is an isothermal operation, thermal specific death rate (k d ) is a constant. Therefore, (10.4) can be used as follows: ln n t n o = - k d t Plotting ln(n t /n o ) versus t and finding the slope will give the numerical value of k d . R. Shanthini 18 Nov 2011 18

Solution to Example 10.1: t (min) 0 10 0 20 -5 -10 -15 -20 y = -0.7201x

R 2 = 0.9988

-25 30

R. Shanthini 18 Nov 2011 k d = -slope = 0.720 per min 19

Solution to Example 10.1:

1 0.8

0.6

0.4

0.2

0 0 10 t (min) 20

R. Shanthini 18 Nov 2011

30

20

Solution to Example 10.1:

b) Since k d is known from part (a), the survival factor at 40 min can be calculated using (10.5) as follows: n t n o = exp (- 0.720 per min x t) = exp (- 0.720 per min x 40 min) = 3.11 x 10 -13 = survival factor n t = 3.11 x 10 -13 n o = 3.11 x 10 -13 x 8.5 x 10 10 = 0.026

We know that n t 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 ( o C) k d (min -1 ) 115 0.035

120 0.112

125 0.347

a) Calculate the activation energy (E d ) and Arrhenius constant (k do ) of the thermal specific death rate k d .

b) Find k d at 130 o C. R. Shanthini 18 Nov 2011 22

Solution to Example 10.2:

Data provided: k d versus temperature data are given a) Activation energy (E d ) and Arrhenius constant (k do ) of the thermal specific death rate (k d ) can be determined starting from (10.7) as follows: ln(k d ) = ln(k do ) E d RT Plot ln(k d ) versus 1/T (taking T in K). Slope gives ( –E d /R) and intercept gives ln(k do ).

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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 -0.5

-1 -1.5

-2 y = -35425x + 87.949

R 2 = 1 -2.5

-3 -3.5

-4

Slope = –E d /R = –35425 K Intercept = ln(k do ) = 87.949

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Solution to Example 10.2:

Slope = –E d /R = –35425 K E d = (35425 K) (R) = 35425 x 8.314 kJ/kmol = 294.5 kJ/mol Activation energy Intercept = ln(k do ) = 87.949

k do = exp(87.949) = 1.5695 x 10 38 per min = 2.616 x 10 36 per s Arrhenius constant R. Shanthini 18 Nov 2011 25

Solution to Example 10.2:

b) Since activation energy (E d ) and Arrhenius constant (k do ) of the thermal specific death rate (k d ) are known from part (a), k d at 130 o C can be determined using (10.6) as follows: k d = k do exp = (2.616 x 10 36 per s) exp = 0.0176 per s = 1.0542 per min R. Shanthini 18 Nov 2011 26

Solution to Example 10.2: 1/T (per K) 0.0025

0.00255

0.00245

0.5

0 -0.5

-1 -1.5

-2 -2.5

-3 -3.5

-4 0.0026

y = -35484x + 88.101

R 2 = 1

Calculated value at 130 o C R. Shanthini 18 Nov 2011 27

Design Criterion for Sterilization:

 = ln n o n t = 0  t k d dt = k do 0  t exp E d dt (10.8) (10.9)

Del factor

(which is a measure of fractional reduction in living organisms count over the initial number present) R. Shanthini 18 Nov 2011 28

Determine the Del factor to reduce the number of cells in a fermenter from 10 10 viable organisms to 1:  = ln n o n t = ln 10 10 1 = 23 = 0  t k d dt Even if one organism is left alive, the whole fermenter may be contaminated. Therefore, no organism must be left alive. That is, n = 0  = ln n o n t = ln 10 10 0 = infinity = 0  infinity k d dt To achieve this del factor, we need infinite time that is not possible.

R. Shanthini 18 Nov 2011 29

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) :  = ln n o n t = ln 10 10 0.001

= 30 = 0  t k d dt Using the Arrhenius law, we get  = k do 0  t exp E d dt = 30 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

R. Shanthini 18 Nov 2011

Electrical heating Steam heating

33

Batch Sterilization (method of cooling) :

Cold water cooling

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For batch heating by direct steam sparging:

H m s t T = T 0 + c (M + m s t)

(10.10) T T 0 c M m s t H – – – – – – – 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

Direct steam sparging

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For batch heating with constant rate of heat flow:

q T = T 0 + c M t

(10.11) T T 0 c M t q – 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

R. Shanthini 18 Nov 2011 36

For batch heating with isothermal heat source:

T = T H + (T 0 - T H

( )

c M

(10.12)

Steam heating

R. Shanthini 18 Nov 2011 T T H T 0 c M t U A – – final temperature (in kelvin) temperature of heat source – – – – – – (in kelvin) initial temperature (in kelvin) specific heat of medium initial mass of medium time required overall heat transfer coefficient heat transfer area 37

For batch cooling using a continuous non-isothermal heat sink

(eg: passing cooling water through a vessel jacket)

: T = T C0

T T 0 T C0 U A c m M t

+ (T 0 - T C0 ) exp {- [ 1 – exp ( U A c m )] m t M

(10.13)

}

– – – – – 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 R. Shanthini 18 Nov 2011 38

Example 10.3: Estimating the time required for a batch sterilization

- Typical bacterial count in a medium is 5 x 10 12 per m 3 , 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 m 3 in volume and is at 25 o C. 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 122 o C.

- The medium is held for some time at 122 o C. Heat loss during holding time is neglected.

- Medium is cooled by passing 100 m 3 /hr of water at 20 o C through the cooling coil until medium reaches 30 o C. Coil heat transfer area is 40 m 3 , and U = 2500 kJ/hr.m

2 .K.

- For the heat resistant bacterial spores: k do = 5.7 x 10 39 per hr E d = 2.834 x 10 5 kJ / kmol - For the medium: c = 4.187 kJ/kg.K and ρ = 1000 kg/m 3 18 Nov 2011 39

Solution to Example 10.3:

Problem statement:

5 x 10 12 per m 3 Typical bacterial count in a medium is , which is to be reduced by sterilization such that the chance for a contaminant surviving the sterilization is 1 in 1,000.

 n 0 = 5 x 10 12 per m 3 x 40 m 3 = 200 x 10 12 = 2 x 10 14 n t = 1/1000 = 0.001

= ln n 0 n t = ln 2x10 14 = 39.8

0.001

= 0  t k d dt R. Shanthini 18 Nov 2011 The above integral should give 39.8.

40

Solution to Example 10.3:

Given sterilization process involves heating from 25 o C to 122 o C, holding it at 122 o C and then cooling back to 30 o C. Therefore  = 0  t k d dt = 0  t 1 k d dt heating + t 1  t 2 k d dt holding + t 2  t 3 k d dt cooling  heat R. Shanthini 18 Nov 2011  hold  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  hold = t 1  t 2 k d dt holding = k d (t 2 -t 1 ) holding (10.15) To determine  heat and  cool, we need to get the temperature profiles in heating and cooling operations, respectively.

R. Shanthini 18 Nov 2011 42

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): T = T 0 + H m s t c (M + m s t) R. Shanthini 18 Nov 2011 43

Solution to Example 10.3:

Data provided:

T 0 = (25 + 273) K = 298 K c = 4.187 kJ/kg.K M = 40 x 1000 kg m s = 5000 kg/hr H = enthalpy of saturated steam at 345 kPa - enthalpy of water at 25 o C = 2,731 – 105 kJ/kg = 2626 kJ/kg Therefore, we get T = 298 + 78.4 t 1 + 0.125 t R. Shanthini 18 Nov 2011 44

Solution to Example 10.3:

For

T = (122 + 273) K = 395 K 78.4 t 395 = 298 + 1 + 0.125 t t = 1.46 h It is the time required to heat the medium from 25 o C to 122 o C.

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Solution to Example 10.3:

 heat = 0  1.46

k d dt heating Use E d k d = k do exp = 5.7 x 10 39 exp 2.834 x 10 5

( )

8.318 x T where T = 298 + R. Shanthini 18 Nov 2011 78.4 t 1 + 0.125 t 46

 heat = 0  1.46

k d dt heating

= 14.8

250 200 Temperature (deg C) kd (per hr) 150 100 50 0 0 0.25

0.5

R. Shanthini 18 Nov 2011 0.75

time (in hr)

1 (10.16) 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 non isothermal heat sink is given by (10.13) T = T C0 + (T 0 - T C0 ) exp { [ 1 – exp ( U A c m )] m t M } R. Shanthini 18 Nov 2011 48

Solution to Example 10.3:

Data provided:

T 0 = (122 + 273) K = 395 K T C0 = (20 + 273) K = 293 K U = 2,500 kJ/hr.m

2 .K

A = 40 m 2 c = 4.187 kJ/kg.K

m = 100 x 1000 kg/hr M = 40 x 1000 kg/hr Therefore, we get T = 293 + 102 exp { [ 1 – exp ( 1 4.187 )] t 0.4 } R. Shanthini 18 Nov 2011 49

Solution to Example 10.3:

For

T = (30 + 273) K = 303 K 393 = 293 + 102 exp { [ 1 – exp ( 1 4.187 )] t 0.4 } t = 3.45 h It is the time required to cool the medium from 122 o C to 30 o C.

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Solution to Example 10.3:

 cool = t 2  t 2 +3.45

k d dt cooling E d k d = k do exp = 5.7 x 10 39 exp 2.834 x 10 5

( )

8.318 x T T = 293 + 102 exp{- 0.674 t} R. Shanthini 18 Nov 2011 51

 cool

=

t 2  t 2 +3.45

k d dt cooling 250 200 150 100 50 0 0 0.5

1

= 13.9

1.5

2

time (in hr)

2.5

(10.17) Temperature (deg C) kd (per hr) 3 3.5

R. Shanthini 18 Nov 2011 52

Putting together the results:

0



1.46

k d dt

heating

+ k d Δt

holding

+ 0



3.45

k d dt = 39.8

cooling  heat = 14.8

 hold  cool = 13.9

 hold = k d Δt = 39.8 -14.8 -13.9 = 11.1

holding Δt = 11.1 / (k d at 122 0 C) R. Shanthini 18 Nov 2011 53

Putting together the results:

k d at 122 0 C = 5.7 x 10 39 exp 2.834 x 10 5

( )

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)

R. Shanthini 18 Nov 2011 54

Putting together the results:

250 200 150 100 50 0 0 0.5

1 1.5

2 2.5

3

time (in hr)

Temperature (deg C) kd (per hr) 3.5

4 4.5

5 55 18 Nov 2011