Sterilization CP504 – Lecture 15 and 16 Learn about thermal sterilization of liquid medium Learn about air sterilization Learn to do design calculations R. Shanthini 18 Nov 2011

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

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…. R. Shanthini 18 Nov 2011

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

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 R. Shanthini 18 Nov 2011

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). R. Shanthini 18 Nov 2011

Thermal sterilization using dry heat Direct flaming Incineration - Hot air oven -170 °C for 1 hour -140 °C for 3 hours . R. Shanthini 18 Nov 2011 7

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 R. Shanthini 18 Nov 2011 8

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) R. Shanthini 18 Nov 2011

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.) R. Shanthini 18 Nov 2011

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.     R. Shanthini 18 Nov 2011

( )   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 R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

( ) 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 R. Shanthini 18 Nov 2011

( )  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). R. Shanthini 18 Nov 2011

( ) 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 R. Shanthini 18 Nov 2011

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) 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 R. Shanthini 18 Nov 2011

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. R. Shanthini 18 Nov 2011

Solution to Example 10.1: kd = -slope = 0.720 per min R. Shanthini 18 Nov 2011

Solution to Example 10.1: R. Shanthini 18 Nov 2011

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 (- 0.720 per min x t) no = 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. R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

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). R. Shanthini 18 Nov 2011

Solution to Example 10.2: Slope = –Ed/R = –35425 K Intercept = ln(kdo) = 87.949 R. Shanthini 18 Nov 2011

Solution to Example 10.2: 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 R. Shanthini 18 Nov 2011

( ) ( ) 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 (273+130) = 0.0176 per s = 1.0542 per min R. Shanthini 18 Nov 2011

Solution to Example 10.2: Calculated value at 130oC R. Shanthini 18 Nov 2011

( )    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) R. Shanthini 18 Nov 2011

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. R. Shanthini 18 Nov 2011

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. R. Shanthini 18 Nov 2011

Typical temperature profile during sterilization: heating holding cooling R. Shanthini 18 Nov 2011

Let us take a look at some sterilization methods and equipment R. Shanthini 18 Nov 2011

Batch Sterilization (method of heating): Direct steam sparging Electrical heating Steam heating R. Shanthini 18 Nov 2011

Batch Sterilization (method of cooling): Cold water cooling R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

( ) 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 R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

  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  = 39.8 The above integral should give 39.8. R. Shanthini 18 Nov 2011

    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 R. Shanthini 18 Nov 2011

  heat hold = + hold + cool = 39.8 = kd dt = kd (t2-t1) 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 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. R. Shanthini 18 Nov 2011

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) R. Shanthini 18 Nov 2011

H = enthalpy of saturated steam at 345 kPa - enthalpy of water at 25oC Solution to Example 10.3: 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 Therefore, we get 78.4 t T = 298 + 1 + 0.125 t R. Shanthini 18 Nov 2011

It is the time required to heat the medium from 25oC to 122oC. Solution to Example 10.3: For T = (122 + 273) K = 395 K 78.4 t 395 = 298 + 1 + 0.125 t It is the time required to heat the medium from 25oC to 122oC. t = 1.46 h R. Shanthini 18 Nov 2011

( ) ( )  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 = 298 + 1 + 0.125 t R. Shanthini 18 Nov 2011

kd dt  1.46 heating heat = 14.8 (10.16) = R. Shanthini 18 Nov 2011

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 R. Shanthini 18 Nov 2011

Solution to Example 10.3: Data provided: 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 Therefore, we get T = 293 + 102 exp{ [1 – exp( )] } 1 t 4.187 0.4 R. Shanthini 18 Nov 2011

It is the time required to cool the medium from 122oC to 30oC. Solution to Example 10.3: For T = (30 + 273) K = 303 K 393 = 293 + 102 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. R. Shanthini 18 Nov 2011

( ) ( )  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 = 293 + 102 exp{- 0.674 t} R. Shanthini 18 Nov 2011

 cool kd dt = t2 t2+3.45 cooling (10.17) = 13.9 R. Shanthini 18 Nov 2011

Putting together the results: kd dt  1.46 kd dt  3.45 heating holding cooling + kd Δt + = 39.8 heat = 14.8 hold cool = 13.9 hold = kd Δt = 39.8 -14.8 -13.9 = 11.1 holding Δt = 11.1 / (kd at 1220C) R. Shanthini 18 Nov 2011

( ) 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 = 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

Putting together the results: Drawback: Longer heat-up and cool-down time R. Shanthini 18 Nov 2011