R. Shanthini 18 Nov 2011 1 Sterilization CP504 – Lecture 15 and 16 - Learn about thermal...

R. Shanthini 18 Nov 2011

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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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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.

And more….


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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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- Direct flaming

- Incineration

- Hot air oven

-170 °C for 1 hour

-140 °C for 3 hours

.

Thermal sterilization using dry heat


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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

dt = - kd nt

wherent is the number of live organisms presentt is the sterilization timekd 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)

(10.1)


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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.)

Spore germination

Spores

Spore production

Hyphal growth


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A chief difference between spores and seeds as dispersal units is that spores have very little stored food resources compared with seeds.

A spore is a reproductive structure that is adapted for dispersal and surviving for extended periods of time in unfavorable conditions.


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Thermal Death Kinetics (continued):Integrating (10.1) using the initial condition n = no at t = 0 gives

lnnt

no= - kd dt

0

t

nt

no

= - kd dt0

t

exp ( )

(10.2)

(10.3)

Survival factor

Survival factor1

Inactivation factor ≡ =no

nt


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Thermal Death Kinetics (isothermal operation):kd is a function of temperature, and therefore it is a constant for isothermal operations.

(10.2) therefore gives

lnnt

no= - kd t (10.4)

nt

no

= exp(- kd t) (10.5)


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kd kdo= - RT

Edexp( )

wherekdo Arrhenius constant for thermal cell deathEd is the activation energy for thermal cell deathR is the universal gas constantT is the absolute temperature

kd is expressed by the Arrhenius equation given below:

(10.6)

Thermal Death Kinetics (non-isothermal operation):


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Thermal Death Kinetics (non-isothermal operation):

lnnt

no

(10.7)

When kd of (10.6) is substituted in (10.2), we get the following:

- kdo= - RT

Edexp( ) dt

0

To carry out the above integration, we need to know how the temperature (T) changes with time (t).

t


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ln(kd)

1/T

ln(kdo)

EdR

ln(kd) ln(kdo)= - RT

Ed

Determining the Arrhenius constants:

kd kdo= - RT

Edexp( ) (10.6)

(10.7)


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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:

a) Find the thermal specific death rate.b) Calculate the survival factor at 40 min.

Time

(min)

0 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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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

no

= - kd tln

Plotting ln(nt /no) versus t and finding the slope will give the numerical value of kd.


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y = -0.7201x

R2 = 0.9988-25

-20

-15

-10

-5

0

0 10 20 30t (min)

ln(n

t/no)

kd = -slope = 0.720 per min


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0

0.2

0.4

0.6

0.8

1

0 10 20 30t (min)

nt/n

o


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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:

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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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)= - RT

Ed

Plot ln(kd) versus 1/T (taking T in K).

Slope gives (–Ed/R) and intercept gives ln(kdo).


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y = -35425x + 87.949

R2 = 1

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.0025 0.00252 0.00254 0.00256 0.00258 0.0026

1/T (per K)ln

(kd)

Slope = –Ed/R = –35425 K

Intercept = ln(kdo) = 87.949


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Slope = –Ed/R = –35425 K

Ed = (35425 K) (R) = 35425 x 8.314 kJ/kmol = 294.5 kJ/mol

Intercept = ln(kdo) = 87.949

kdo = exp(87.949) = 1.5695 x 1038 per min

= 2.616 x 1036 per s

Activation energy

Arrhenius constant


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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= - RT

Edexp ( )

(2.616 x 1036 per s)= - 8.314 (273+130)

294.5 x 103

exp( )0.0176 per s= 1.0542 per min=


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y = -35484x + 88.101

R2 = 1

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

0.00245 0.0025 0.00255 0.0026

1/T (per K)ln

(kd)

Calculated value at 130oC


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Design Criterion for Sterilization:

= kd dt0

t

lnno

nt=

kdo= - RT

Edexp( ) dt

0

t

Del factor (which is a measure of fractional reduction in living organisms count over the initial number present)

(10.8)

(10.9)


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lnno

nt=

Determine the Del factor to reduce the number of cells in a fermenter from 1010 viable organisms to 1:

= ln1010

1= 23

Even if one organism is left alive, the whole fermenter may be contaminated.

= kd dt0

t

Therefore, no organism must be left alive. That is, n = 0

lnno

nt= = ln

1010

0= infinity

To achieve this del factor, we need infinite time that is not possible.

= kd dt0

infinity


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lnno

nt=

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) :

= 30 = kd dt0

t

= ln1010

0.001

kdo= - RT

Edexp( ) dt

0

t

= 30

Temperature profile during sterilization must be chosen such that the Del factor can become 30.

Using the Arrhenius law, we get


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cooling

holdingheating

Typical temperature profile during sterilization:


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Let us take a look at some sterilization methods and equipment


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Batch Sterilization (method of heating):

Steamheating

Electricalheating

Direct steam sparging


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Batch Sterilization (method of cooling):

Cold water cooling


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Direct steam sparging

T = T0 + H ms t

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

For batch heating by direct steam sparging:

(10.10)


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For batch heating with constant rate of heat flow:

T = T0 + q

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

Electricalheating

t (10.11)


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For batch heating with isothermal heat source:

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

T = TH + (T0 - TH) exp - U A tc M ( )

Steamheating

(10.12)


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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

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):

(10.13)


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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

- For the medium: c = 4.187 kJ/kg.K and ρ = 1000 kg/m3


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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.

lnn0

nt= = 39.8 = kd dt

0

t

= ln2x1014

0.001

n0 = 5 x 1012 per m3 x 40 m3 = 200 x 1012 = 2 x 1014

nt = 1/1000 = 0.001


The above integral should give 39.8.


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kd dt0

t1

kd dtt1

t2

kd dtt2

t3

heating holding cooling

+ +

heat coolhold


Given sterilization process involves heating from 25oC to 122oC, holding it at 122oC and then cooling back to 30oC. Therefore

=

kd dt0

t

=


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= 39.8


Since the holding process takes place at isothermal condition, we get

kd dtt1

t2

holding

= kd (t2-t1)

holding

hold =

heat + cool+ hold

The design problem is therefore,

= (10.14)

(10.15)

To determine heat and cool, we need to get the temperature profiles in heating and cooling operations, respectively.


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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 = T0 + H ms t

c (M + ms t)


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Data provided:

T0 = (25 + 273) K = 298 K c = 4.187 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

T = 298 + 78.4 t

1 + 0.125 t

Therefore, we get


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For T = (122 + 273) K = 395 K

395 = 298 + 78.4 t

1 + 0.125 t

t = 1.46 h It is the time required to heat the medium from 25oC to 122oC.


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heat kd dt0

1.46

heating

=

kd kdo= - RT

Edexp( )

5.7 x 1039= - 8.318 x T

2.834 x 105

exp( )T = 298 +

78.4 t

1 + 0.125 t

Use

where



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heat kd dt0

1.46

heating

= = 14.8

0

50

100

150

200

250

0 0.25 0.5 0.75 1 1.25 1.5

time (in hr)

Temperature (deg C)

kd (per hr)

(10.16)


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T = TC0 + (T0 - TC0) exp{ [1 – exp( )] } U A c m

m t M


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)


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T0 = (122 + 273) K = 395 K TC0 = (20 + 273) K = 293 K U = 2,500 kJ/hr.m2.K A = 40 m2

c = 4.187 kJ/kg.K m = 100 x 1000 kg/hr M = 40 x 1000 kg/hr

Data provided:


Therefore, we get

T = 293 + 102 exp{ [1 – exp( )] } 1 4.187

t 0.4


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T = (30 + 273) K = 303 K

For


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 122oC to 30oC.


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cool kd dtt2

t2+3.45

cooling

=

kd kdo= - RT

Edexp( )

5.7 x 1039= - 8.318 x T

2.834 x 105

exp( )T = 293 + 102 exp{- 0.674 t}



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cool kd dtt2

t2+3.45

cooling

= = 13.9

0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3 3.5

time (in hr)

Temperature (deg C)

kd (per hr)

(10.17)


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Putting together the results:

kd dt0

1.46

kd dt0

3.45

heating holding cooling

+ + = 39.8

heat = 14.8 hold

kd Δt

hold = kd Δt = 39.8 -14.8 -13.9 = 11.1holding

Δt = 11.1 / (kd at 1220C)

cool = 13.9


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Δt = 11.1 / 197.6 = 0.056 hr = 3.37 min

kd at 1220C

5.7 x 1039= - 8.318 x T

2.834 x 105

exp( )= 197.6 per hr

T = 395

Sterilization is achieved mostly during the heating (14.8 hr) and cooling (13.9 hr)


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0

50

100

150

200

250

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

time (in hr)

Temperature (deg C)

kd (per hr)

Drawback: Longer heat-up and cool-down time

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