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Media Sterilisation

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Media Sterilisation, Batch Sterilisation Design, Thermal Death Kinetics, Filter Sterilisation, Theory of Depth Filters

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Media Sterilisation

  1. 1. If the fermentation is contaminated 1. The medium would have to support the growth of both producer and contaminant organism resulting in loss of productivity. 2. In case of continuous cultivation the contaminant may outgrow the producer organism. 3. In case of biomass as product, the contaminant will result in total loss of product. 4. The contaminant may produce compounds, which make downstream processing difficult. 5. The contaminant may degrade the product; this is common in antibiotics fermentation eg. β Lactamase producing organism degrade β-Lactum antibiotics. 6. Contaminant with phage could result in the lysis of culture.
  2. 2. For avoiding contamination 1. Use pure inoculum to start the fermentation 2. Sterilizing the medium to be employed 3. Sterilizing the fermenter 4. Sterilizing all materials to be added during fermentation 5. Maintaining the aseptic condition during fermentation
  3. 3. Some fermentations are described as protected i.e the medium may not be utilised by the other organisms or the extreme pH are used for cultivation etc., Brewing of Beer falls into this category i.e Hop resins used as the supplement in the medium inhibit growth of most of the organism. In such case of fermentations the medium can be boiled ,the reactors can be cleaned with disinfectants before fermentation not necessarily sterilized. But most of the fermentation are not protected
  4. 4. Medium SterilizationMedium Sterilization  Filtration  Radiation  Ultrasonic treatment  Chemical treatment  Heat
  5. 5. Many cellular materials absorb ultraviolet light, leading to DNA damage and consequently to cell death. Wavelengths around 265 nm have the highest bacterial efficiency. However they have very little ability to penetrate. Use is limited to clean chambers, operation theatre etc., X-rays are lethal and have penetration ability. However due to expensive and safety concerns it is not used RadiationRadiation
  6. 6. Chemical agents are frequently used for disinfection. Major antimicrobial chemical agents are : phenol and compounds (phenol, cresol, orthophenylphenol) alcohol (ethyl,methyl) halogens (iodine, hypochlorites, chloramines) detergents,acids, alkali, gaseous chemosterlizers (ethylene oxide, formaldehyde) Chemical SterilizationChemical Sterilization
  7. 7. Destruction of microorganisms by steam is represented by first order reaction -dN/dt = kN ------------[1] Where N no of viable organisms present t sterlization treatment time k rate constant or Sp. Death rate It is important to note that we are considering the number of organisms and not the concentration – the minimum number of organism required to contaminate the batch is one regardless of the volume
  8. 8. Rearranging the eqn [1] -dN/N = kdt Upon Integration ln (Nt/No) = - kt-------[2] Nt/No = e-kt ------------[3] Nt = No of organisms present after sterlization No = No of organisms present before sterlization
  9. 9. o t N N Time Time o t N N ln Nt/No = e-kt  An infinite time is required to achieve sterile condition i.e Nt=0  After certain period there will be less than one viable cell present i.e Nt<1 ln (Nt/No) = - kt
  10. 10.  Nt<1 is the probability of an organism surviving the treatment  i.e Nt=0.1 is the probability that one organism surviving ten treatments.  In other words one in ten batches may get contaminated due to improper sterlization.  The value of k dependant on temperature, species and its physiological form.  But your sterlization media contains mixture of microorganism.
  11. 11. E. coli B. stereothermophillus
  12. 12. Time Ln N Sensitive organism Resistant organism Whole culture Ln N Time Higher no of heat sensitive organism Higher no of heat resistant organism
  13. 13. As in any first order, The reaction rate increases with increase in temperature i.e k increases with increase in temperature. This relation ship is given by Arrhenius eqn E activation energy R gas constant T absolute temperature A Arrhenius constant RTE Aek /− = [4]
  14. 14. Taking logarithm of eqn 4 [5] Plot of ln k Vs 1/T will give a straight line 1/T Ln k Intercept = ln A Slope = -E/R
  15. 15. RTE to etANN − ⋅⋅=ln to NNln=∇ RTE etA − ⋅⋅=∇ Combining eqn 2 and 4 [5] Del factor is the sterilization criterion ( )ARTEt ∇+= lnln
  16. 16. Thus it is evident from the eqn that the temperature and time are inversely proportional and same degree of sterilization can be achieved by varying time and temperature. Also we have seen that the sterlization criterion depend mainly on heat resistant organism. Bacillus stereothermophilus is the known most heat resistant organism Assume all the contaminants are B.st and calculate the temperature and time. E=67.7 kcal/mole A=1 X 1036.2 1/sec Industrially accepted level of contamination is 0.001. i.e 1 in 1000 batches can get contaminated.
  17. 17. Another common factor used in food industry is decimal reduction time D. It is the time required to reduce the cell population by 10 fold. D=2.303/k Decimal reduction timeDecimal reduction time
  18. 18. ProblemProblem
  19. 19. Step 1 Plot ln No/Nt Vs Time for each temperature and find out the slope. Slope will give k value. Step 2 Plot ln k Vs 1/T from the above values. Slope will give E/R values and Intercept will give you the A value Step 3 From the above graph calculate the k value for 100 o C Step 4 From the k value calculate time required to kill 99% spores ktNN to =ln
  20. 20. 0.00E+00 5.00E-02 1.00E-01 1.50E-01 2.00E-01 2.50E-01 3.00E-01 3.50E-01 0 2 4 6 8 10 Time [min] Y=0.0122 X Y=0.0326 X
  21. 21. 0.00E+00 4.00E+00 8.00E+00 1.20E+01 1.60E+01 2.00E+01 0 2 4 6 8 10 Time[min] Y=1.5943 X Y=9.5859 X
  22. 22. y = -26869x + 70.624 -5 -4 -3 -2 -1 0 1 2 3 0.0025 0.0026 0.0026 0.0027 0.0027 0.0028 0.0028 0.0029 Lnk 1/T
  23. 23. E/R = 26869 o K R = 8.3144 J/K/gmol E = 26869 X 8.3144 = 2.24 X 105 J/gmol=224 KJ/gmol A =4.7 X 1030 min-1 kd= 4.7 X 1030 e-26869/T T=100 +273 =373 o K kd=0.244 min-1 Ln(100/1)=0.244 t t=18.9 min ktNN to =ln
  24. 24. Batch sterilization  Sterilization of the medium in a fermenter can be carried out in batch mode by direct steam sparging, by electrical heaters or by circulating constant pressure condensing steam through heating coil or jacket.  Cooling the medium is carried out by sparging the steam through the cooling coil or jacket.
  25. 25. TimeTime TemperatureTemperature Heating Holdup Cooling
  26. 26. coolingholdupheatingoverall ∇+∇+∇=∇ tooverall NNln=∇ heating∇ cooling∇ coolingheatingoverallholdup ∇−∇−∇=∇ kt holdup∇=
  27. 27. Heating CurveHeating Curve RTE etA − ⋅⋅=∇ 1 11 RTE etA − ⋅⋅=∇ 2 22 RTE etA − ⋅⋅=∇ 3 33 RTE etA − ⋅⋅=∇ TimeTime TemperatureTemperature t1 t2 t3 T3 T2 T1 •••+∇+∇+∇=∇ 321heating
  28. 28. The initial number of contaminants is 1014 and the decimal reduction value at 121o C is 1.5 min. Calculate the time required to bring down the contaminants to industrially acceptable limits. ProblemProblem
  29. 29. D is the time required to reduce the contaminants by 10 fold. Del value is 1017 Time required = 17 X D = 17 X 1.5 Time required = 25.5 min
  30. 30. The initial concentration of contaminants is 106 per ml and the fermenter volume is 10000 litres. The del factor contribution from heating is 9.8 and that of cooling is 10.1. Calculate the holdup time assuming all the contaminants are B.st having kd value at 121o C of 2.54 min-1 ProblemProblem
  31. 31. coolingholdupheatingoverall ∇+∇+∇=∇ 1.10=∇cooling8.9=∇heating 8.36)ln( ==∇ tooverall NN NNoo = 10= 1066 X1000X10000 =10X1000X10000 =101313 NNtt=0.001=0.001 9.161.108.98.36 =−−=∇holdup coolingheatingoverallholdup ∇−∇−∇=∇ kt holdup∇= min65.654.2/9.16 ==t
  32. 32. An autoclave is used for sterilizing 10 litre complex medium containing 105 spores/ml. Due to the problem in autoclave the temperature reached only 110o C. If same holdup time of 20 min is maintained what is the probability of contamination. Compare that with of regular autoclaving at 121o C. ProblemProblem E=2.83 X 105 J/gmol A=1036.2 sec-1 R = 8.3144 J/K/gmol
  33. 33. Direct steam spargingDirect steam sparging T To t h S M Cp U A TH Q C’p W Tco
  34. 34. Steam sparging through the coil or jacketSteam sparging through the coil or jacket T To t h S M Cp U A TH Q C’p W Tco
  35. 35. Electrical HeatingElectrical Heating T To t h S M Cp U A TH Q C’p W Tco
  36. 36. Cooling coilsCooling coils T To t h S M Cp U A TH Q C’p W Tco
  37. 37. Batch sterlization is carriedout in 1 litre fermenter to prevent contamination. The initial no of spores in the medium is 1010 spores/litre. a) Calculate the holding time required to achieve standard contamination levels neglecting heating and cooling periods. b) What is the probability of contamination if a 100 litre fermenter is heated for the same time. c) Redo the calculation for part (a) assuming constant temperature steam heating and constant temperature cooling. Heat exchange coils are used. For steam heating α and β are 0.05 sec-1 and -0.25 and that of cooling 0.065 sec-1 and 0.20. The heating and cooling periods are 3 min and 10 min respectively. d) Redo the calculation for part (a) using the temperature profile
  38. 38. Temperature profile during sterlization 0 20 40 60 80 100 120 140 0 50 100 150 Time Temp
  39. 39. HEATING CURVE Time Temp 0 30 5 35 10 50 15 64 20 86 25 96 30 104 35 111 40 116 45 121 COOLING CURVE Time Temp 0 121 7 110 14 95 21 87 28 75 35 64 42 54 49 45 56 39 64 37
  40. 40. A steam sterilizer is used to sterilize liquid medium for fermentation. The initial concentration of contaminants is 108 per litre. How long 1 m3 medium be treated if the temperature is 80o C, 121o C, 140o C Assume all the contaminants are B.st. E=283 KJ/gmol A=1036.2 sec-1 R = 8.3144 J/K/gmol ProblemProblem
  41. 41. RTE etA − ⋅⋅=∇ E=2.83 X 105 J/gmol A=1036.2 sec-1 R = 8.3144 J/K/gmol No=1011 Nt=0.001 Ln(No/Nt)=34.54 -E/R=-3.4X104 o K T -E/RT exp(-E/RT) A*exp(-E/RT) time (min) 353 -9.64E+01 1.33047E-42 2.10866E-06 272991.8 394 -8.64E+01 3.03152E-38 0.048046344 11.98106 413 -8.24E+01 1.61319E-36 2.556734665 0.225149
  42. 42. It is evident from theIt is evident from the problem that a regime ofproblem that a regime of time and temperature maytime and temperature may now be determined tonow be determined to achieve the desiredachieve the desired sterilization.sterilization. However the fermentationHowever the fermentation media is not an inertmedia is not an inert mixture.mixture. Due to sterilizationDue to sterilization deleterious reactions maydeleterious reactions may occur resulting in loss ofoccur resulting in loss of nutrient qualitynutrient quality Duration of sterilizationProductyield
  43. 43. Loss of nutrient quality is due to two reasons i) Interactions between nutrient components of the medium  Maillard type browning reaction – reactions of the carbonyl group usually from reducing sugar with the amino group of aminoacids, proteins.  Effect of sterilization time on availability of glucose in CSL medium was investigated Time Glucose remaining 60 min 35% 40 min 46% 30 min 64% Problems of this type are normally resolved by sterilizing sugar separately and adding to the medium
  44. 44. II. Degradation of heat labile components  Certain vitamins, amino acids and proteins may be degraded during sterilization regime In certain cases such as animal cell culture media filtration is used In other cases judicious choice of sterilization regime is selected. Nutrient destruction follows first order reaction Activation energy for nutrient destruction – 10 to 30 kcal/mole Activation energy for B.st – 67.7 kcal/mole kt o t e x x − =
  45. 45. Lnk 1/T Spores destruction Nutrient destruction From the graph it isFrom the graph it is evident that sterilizingevident that sterilizing the medium at higherthe medium at higher temperature for lessertemperature for lesser time will result in sporestime will result in spores destruction with lesserdestruction with lesser loss in nutrient quality.loss in nutrient quality. Heating the huge reactorsHeating the huge reactors for higher temperaturefor higher temperature and cooling down in shortand cooling down in short duration is impossibleduration is impossible Alternately continuousAlternately continuous sterilizer can be usedsterilizer can be used
  46. 46. Advantages of continuousAdvantages of continuous sterilizationsterilization Superior maintenance ofSuperior maintenance of medium qualitymedium quality Ease of scale upEase of scale up Easier automatic controlEasier automatic control Reduction in fermenterReduction in fermenter corrosioncorrosion Advantages of BatchAdvantages of Batch sterilizationsterilization Lower capitalLower capital equipment costequipment cost Lower risk ofLower risk of contaminationcontamination Easier to use withEasier to use with media containing solidmedia containing solid suspended matter.suspended matter.
  47. 47.  Performance ofPerformance of continuous sterilizercontinuous sterilizer depends on the nature ofdepends on the nature of fluid flow in the systemfluid flow in the system  Plug flow is an ideal flowPlug flow is an ideal flow where no mixing orwhere no mixing or change in velocity occurchange in velocity occur  Deviation from plug flowDeviation from plug flow is characterized by axialis characterized by axial dispersion the degree todispersion the degree to which mixing occurswhich mixing occurs along the length of thealong the length of the pipe.pipe.  Axial dispersion is theAxial dispersion is the critical factor in designcritical factor in design of continuous sterilizer.of continuous sterilizer.
  48. 48. Axial dispersion and flow through the pipe isAxial dispersion and flow through the pipe is characterized by dimensionless variable PECLETcharacterized by dimensionless variable PECLET NumberNumber u – linear velocityu – linear velocity L- Length of the pipeL- Length of the pipe DDzz- axial dispersion- axial dispersion For perfect plug flow axial dispersion is zero and NFor perfect plug flow axial dispersion is zero and NPePe isis infiniteinfinite NNPePe between 3 – 600 is typical.between 3 – 600 is typical. Once the NOnce the NPePe is known the extent of cell destruction andis known the extent of cell destruction and another dimension less variable Damkohler number cananother dimension less variable Damkohler number can be calculatedbe calculated kkdd- specific death rate- specific death rate zD uL Pe = u Lk Da d =
  49. 49. Soln to Problem 1  N1= 1.44 X 1016  N2/N1=6.9X 10 -17  u= 254.6m/h  Re=7.07X103
  50. 50. 0.650.65 From graph Dz/uD=0.65From graph Dz/uD=0.65 Dz=16.6 sq.m/hDz=16.6 sq.m/h Pe=368Pe=368
  51. 51.  Damkholer number for Pe 368 and N2/N1 6.9 X 10 -17 is 42 (From graph)  k=445.6 1/h  T=-(E/R)/ln(k/A)  T=398.4 K
  52. 52.  For perfect plug flow Dz is zero.  Hence Pe number is infinity.  Dilution rate = Feed rate/reactor volume  For first part calculate Nt/No.  From the graph for this Nt/No value and Pe infinity calculate the Da Number.  Calculate Kd with the given value of Arrhenius constant, activation energy and temperature 130o C  Calculate velocity u from the feed rate and diameter of the steriliser.  Length of the sterliser holding section can be calculated from the above.
  53. 53. Second part:  Calculate NRe.  From the graph calculate Dz/uD  Calculate Dz  Calculate Pe number using the length calculated in the first part.  Now using the Pe number and Nt/No value calculate Da number.  From this calculate length.
  54. 54.  For third part  Using the length of perfect plug flow calculate Da number.  For this Da number and Peclet number calculated in the second part, Calculate Nt/No value from the graph.  From this calculate No and find the contamination rate . i.e 1 organism in how many days.
  55. 55. A continuous steirliser with a steam injector and a flash cooler will beA continuous steirliser with a steam injector and a flash cooler will be employed to sterlize medium continuously with the flow rate of 2 memployed to sterlize medium continuously with the flow rate of 2 m33 /h. The/h. The time for heating and cooling is negligible with this type of steriliser. Thetime for heating and cooling is negligible with this type of steriliser. The typical bacterial count of the medium is about 5 X 10typical bacterial count of the medium is about 5 X 101212 mm-3-3 , which needs to be, which needs to be reduced to such an extent that only one organism can survive during tworeduced to such an extent that only one organism can survive during two months of operation. The heat reesistant bacterial spores in the mediummonths of operation. The heat reesistant bacterial spores in the medium can be characterized by A of 5.7 X 10can be characterized by A of 5.7 X 103939 1/h and E of 2.834X101/h and E of 2.834X1055 KJ/Kmol. TheKJ/Kmol. The steriliser will be constructed with the pipe with an inner diameter of 0.102m.steriliser will be constructed with the pipe with an inner diameter of 0.102m. Steam is available at a temperature of 125Steam is available at a temperature of 125oo C. The physical properties of theC. The physical properties of the medium aremedium are ρρ=1000kg/m=1000kg/m33 andand μμ=4 Kg/m/h R=8.3144 J/K/g mol=4 Kg/m/h R=8.3144 J/K/g mol  What length should the pipe be in the sterliser if you assume ideal plugWhat length should the pipe be in the sterliser if you assume ideal plug flowflow  What length should the pipe be in the sterliser if the effect of axialWhat length should the pipe be in the sterliser if the effect of axial dispersion is considered.dispersion is considered.
  56. 56. The following equations can be used for the design: Dz/uD=2 X 107 (NRe) -1.951 for NRe < 10000 Dz/uD=1.6317(NRe) –0.1505 for NRe > 10000 ln(Nt/No)= -Da + (Da2 /Pe) Solution to ax2 +bx+c=0 is a acbb x 2 42 −±− =
  57. 57. C.s.area 0.008174571 sq.m Vol. flow rate 2 cu.m/h Velocity 244.6611443 m/h Dia 0.102 m density 1000 Kg/cu.m Viscosity 4 Kg/m/h Reynolds no 6238.85918   Dz/UD 0.788460214   Dz 19.67636897 sq.m/h
  58. 58. Nt 1   No 1.44E+16   Nt/No 6.94444E-17   ln(Nt/No) -37.2060046   A 5E+39 1/h E 283400 J/gmol R 8.3144 J/K/gmol T 398 K K 320.0387736 1/h
  59. 59. Peclet infinity   Da=KL/U 37.2   L=DaU/K 28.43841221 m Peclet=UL/Dz 353.6106933   a 1   b -353.61   c 13156.41529   Da 42.25540789   L=DaU/K 32.3031373 m
  60. 60. Filter sterilizationFilter sterilization Suspended solids may be separated from aSuspended solids may be separated from a fluid during sterilization by the followingfluid during sterilization by the following mechanismsmechanisms 1.1. Inertial ImpactionInertial Impaction 2.2. DiffusionDiffusion 3.3. Electrostatic attractionElectrostatic attraction 4.4. InterceptionInterception
  61. 61. Inertial ImpactionInertial Impaction  The fluid will tend to move through the leastThe fluid will tend to move through the least resistance path during filtration.resistance path during filtration.  Suspended particles in the fluid stream haveSuspended particles in the fluid stream have momentum.momentum.  Due to this they tend to move straight andDue to this they tend to move straight and therefore impact on the filtration fibres wheretherefore impact on the filtration fibres where they may be retained.they may be retained.  This mechanism is more prominent in filtration ofThis mechanism is more prominent in filtration of gases than liquids.gases than liquids.
  62. 62. DiffusionDiffusion  Extremely small particles suspended in the fluidExtremely small particles suspended in the fluid are subjected to Brownian movement which isare subjected to Brownian movement which is random movement due to collision of particles.random movement due to collision of particles.  Thus due to this the fluid particles deviate fromThus due to this the fluid particles deviate from the path of movement and impact on the filterthe path of movement and impact on the filter fibres.fibres.  Diffusion is more significant in the filtration ofDiffusion is more significant in the filtration of gases than liquids.gases than liquids.
  63. 63. Electrostatic attractionElectrostatic attraction  Charged particles are attracted byCharged particles are attracted by opposite charges on the surface of theopposite charges on the surface of the filtration mediumfiltration medium
  64. 64. InterceptionInterception  The fibres comprising filter are interwoven to definite openingThe fibres comprising filter are interwoven to definite opening of different sizes eg: 0.22 µm etc.,of different sizes eg: 0.22 µm etc.,  Particles which are larger than these pore sizes are removed byParticles which are larger than these pore sizes are removed by direct interception.direct interception.  Also significant number of smaller size particles are alsoAlso significant number of smaller size particles are also retained due to:retained due to:  More than one particle arriving at a time to the pore.More than one particle arriving at a time to the pore.  An irregularly shaped particle may bridge the pore.An irregularly shaped particle may bridge the pore.  Interception is the equally important mechanism in both liquidsInterception is the equally important mechanism in both liquids and gases.and gases.
  65. 65. Filters are classified into two typesFilters are classified into two types Depth filter Or Non fixed pore filterDepth filter Or Non fixed pore filter In depth filter the pore sizes are larger than theIn depth filter the pore sizes are larger than the particle to be removedparticle to be removed In these filters particles are removed more byIn these filters particles are removed more by impaction, diffusion and electrostatic attractionimpaction, diffusion and electrostatic attraction rather than interceptionrather than interception Absolute filter Or Fixed pore filterAbsolute filter Or Fixed pore filter In this the pore sizes are smaller than the particle toIn this the pore sizes are smaller than the particle to be removed.be removed. Particles are removed by interception.Particles are removed by interception.
  66. 66. In depth filters the removal of microorganism is aIn depth filters the removal of microorganism is a probability and it cannot be absolute.probability and it cannot be absolute. Thus always there is probability that microorganismThus always there is probability that microorganism can pass through the filter immaterial of the depth ofcan pass through the filter immaterial of the depth of the filter.the filter. Fibres are just tightly packed and not fixed in the filter.Fibres are just tightly packed and not fixed in the filter. Hence due to larger pressure movement of materialHence due to larger pressure movement of material and creation of larger channels is possible.and creation of larger channels is possible. Also increased pressure may sometime remove theAlso increased pressure may sometime remove the previously trapped particles.previously trapped particles.
  67. 67. Filter sterilization of mediumFilter sterilization of medium During Animal cell culture media filtration followingDuring Animal cell culture media filtration following precautions have to be taken.precautions have to be taken. Filtered medium should be free of fungal,Filtered medium should be free of fungal, bacterial and mycoplasma contamination.bacterial and mycoplasma contamination. Minimal adsorption of the proteins from theMinimal adsorption of the proteins from the medium during filtrationmedium during filtration Medium should be free of viruses andMedium should be free of viruses and endotoxins.endotoxins.
  68. 68. Absolute filters are used.Absolute filters are used. Pressure drop in these filters are high hence thePressure drop in these filters are high hence the pleated structure membranes are used.pleated structure membranes are used. Pore sizes of these membranes are well controlledPore sizes of these membranes are well controlled during manufacture.during manufacture. To extend the life of the filter prefilters of bigger poreTo extend the life of the filter prefilters of bigger pore sizes are used.sizes are used. Generally polypropylene or nylon membranes are usedGenerally polypropylene or nylon membranes are used Filter should be steam sterilizable in between batches.Filter should be steam sterilizable in between batches.
  69. 69. Filter sterilization of airFilter sterilization of air  For aerobic fermentations air needs to be supplied continuously.For aerobic fermentations air needs to be supplied continuously.  Typical aeration rates are 0.5 to 2 vvm.Typical aeration rates are 0.5 to 2 vvm.  This requires continuous and large amount of air free of microbialThis requires continuous and large amount of air free of microbial contaminants.contaminants.  Sterilization of air by means of heat is economically not viable.Sterilization of air by means of heat is economically not viable.  Most effective technique is filtration by fibrous or membrane filtersMost effective technique is filtration by fibrous or membrane filters  The cotton plug used as closure for flasks during cultivation is good exampleThe cotton plug used as closure for flasks during cultivation is good example of fibrous filter.of fibrous filter.  Simple air filter can be made by packing cotton in the tube.Simple air filter can be made by packing cotton in the tube.  However the pressure drop is huge and wet cotton may become source forHowever the pressure drop is huge and wet cotton may become source for contamination.contamination.  Therefore glass wool is favourable filtration medium.Therefore glass wool is favourable filtration medium.  In the case of absolute filters the filter should be hydrophobic (PTFEIn the case of absolute filters the filter should be hydrophobic (PTFE membranes).membranes).
  70. 70. Air FilterAir Filter
  71. 71. Theory of Depth filters  To reduce the population entering into the filter and exiting out, Expression given as, dN/dx = - KN Where, N is the concentration of particles in the air at a depth x in the filter & K is a constant, On integrating above expression, we get: N/No = e-kt Where, No is the no. of particles entering the filter N is the no. of particles leaving the filter Efficiency of filter is given by E = (No – N) / No But, (No – N) / No = 1 – (N/No) Thus, (No – N) / No = 1 – e-kt
  72. 72.  The log penetration relationship has been used by Humphery and Garden (1955) in the filter design, by using the concept X90 the depth of filter required to remove 90% of the total no. of particles entering the filter,  Thus If, No were 10 and x were X90, then N would be 1, ln (1/10) = - K X90 2.303 log10 (1/10) = - K X90 2.303 (-1) = - K X90 X90 = 2.303 / K

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