Sterilization

1. CP504 – Lecture 15 and 16 Sterilization • Learn about thermal sterilization of liquid medium • Learn about air sterilization • Learn to do design calculations

2. Sterilization isa 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.

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….

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)

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

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).

7. Thermal sterilization using dry heat • Direct flaming • Incineration - Hot air oven • -170 °C for 1 hour • -140 °C for 3 hours • .

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; 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

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 (0.5 to 5/min) (10 to 1010/min)

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.)

11. A sporeis 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.

12. t  nt - kd dt ln = no 0 t ( )  nt exp - kd dt = no 0 • Thermal Death Kinetics (continued): Integrating (10.1) using the initial condition n = no at t = 0 gives (10.2) (10.3) Survival factor 1 no Inactivation factor ≡ = nt Survival factor

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 (10.4) - kd t ln = no nt exp(- kd t) (10.5) = no

14. Thermal Death Kinetics • (non-isothermal operation): kd is expressed by the Arrhenius equation given below: ( ) Ed (10.6) - exp kd 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 Tis the absolute temperature

15. Thermal Death Kinetics • (non-isothermal operation): When kd of (10.6) is substituted in (10.2), we get the following: t ( )  Ed nt (10.7) - dt exp ln - kdo = no RT 0 To carry out the above integration, we need to know how the temperature (T) changes with time (t).

16. Ed - ln(kd) ln(kdo) = RT • Determining the Arrhenius constants: ( ) Ed - (10.6) exp kd kdo = RT (10.7) ln(kdo) ln(kd) Ed R 1/T

17. 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.

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.

19. Solution to Example 10.1: kd = -slope = 0.720 per min

20. Solution to Example 10.1:

21. 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 = 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.

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.

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).

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

25. 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

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 - exp kd kdo = RT ( ) 294.5 x 103 - exp (2.616 x 1036 per s) = 8.314 (273+130) 1.0542 per min 0.0176 per s = =

27. Solution to Example 10.2: Calculated value at 130oC

28. t  no kd dt ln = = nt 0 ( ) t  Ed - dt exp kdo = RT 0 Design Criterion for Sterilization: (10.8)  (10.9) Del factor(which is a measure of fractional reduction in living organisms count over the initial number present)

29. t infinity   no no ln ln = = kd dt kd dt nt nt = = 0 0 1010 1010 = ln = ln 0 1 Determine the Del factor to reduce the number of cells in a fermenter from 1010 viable organisms to 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  = infinity To achieve this del factor, we need infinite time that is not possible.

30. no ln = nt t  ( ) t  Ed kd dt - dt exp kdo = RT 0 0 1010 = ln 0.001 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  = Using the Arrhenius law, we get = 30  Temperature profile during sterilization must be chosen such that the Del factor can become 30.

31. Typical temperature profile during sterilization: heating holding cooling

32. Let us take a look at some sterilization methods and equipment

33. Batch Sterilization (method of heating): Electrical heating Steam heating Direct steam sparging

34. Batch Sterilization (method of cooling): Cold water cooling

35. For batch heating by direct steam sparging: H ms t (10.10) T = T0 + 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

36. For batch heating with constant rate of heat flow: q t (10.11) T = T0 + 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

37. For batch heating with isothermal heat source: ( ) 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

38. m t U A T = TC0 + (T0 - TC0) exp{- [1 – exp()]} M c m For batch cooling using a continuous non-isothermal heat sink (eg: passing cooling water through a vessel jacket): (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

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 = 2.834 x 105 kJ / kmol • For the medium: c = 4.187 kJ/kg.K and ρ= 1000 kg/m3

40. t  n0 kd dt ln = = nt 0 2x1014 = ln 0.001 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  = 39.8 The above integral should give 39.8.

41. t1 t2 t3    kd dt kd dt kd dt 0 t1 t2 cooling heating holding 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 + + = heat hold cool

42. t2  kd dt t1 holding holding 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 = = kd (t2-t1) (10.15) To determine heat and cool, we need to get the temperature profiles in heating and cooling operations, respectively.

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)

44. 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

45. 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

46. ( ) Ed - exp kd kdo = RT 1.46  kd dt 0 heating Solution to Example 10.3: heat = Use ( ) 2.834 x 105 - exp 5.7 x 1039 = 8.318 x T 78.4 t where T = 298+ 1 + 0.125 t

47. 1.46  kd dt heating 0 heat (10.16) = 14.8 =

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

49. 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 1 T = 293 + 102 exp{ [1 – exp()]} t 4.187 0.4

50. Solution to Example 10.3: For T = (30 + 273) K = 303 K 1 393 = 293 + 102 exp{ [1 – exp()]} t 4.187 0.4 t = 3.45 h It is the time required to cool the medium from 122oC to 30oC.