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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.
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
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
Medium SterilizationMedium Sterilization
 Filtration
 Radiation
 Ultrasonic treatment
 Chemical treatment
 Heat
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
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
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
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
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
 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.
E. coli B. stereothermophillus
Time
Ln N
Sensitive organism
Resistant organism
Whole culture
Ln N
Time
Higher no of heat
sensitive organism
Higher no of heat
resistant organism
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]
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
RTE
to etANN −
⋅⋅=ln
to NNln=∇
RTE
etA −
⋅⋅=∇
Combining eqn 2 and 4
[5]
Del factor is the sterilization criterion
( )ARTEt ∇+= lnln
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.
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
ProblemProblem
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
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
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
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
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
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.
TimeTime
TemperatureTemperature
Heating Holdup Cooling
coolingholdupheatingoverall ∇+∇+∇=∇
tooverall NNln=∇
heating∇ cooling∇
coolingheatingoverallholdup ∇−∇−∇=∇
kt holdup∇=
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
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
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
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
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
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
Media Sterilisation
Direct steam spargingDirect steam sparging
T
To
t
h
S
M
Cp
U
A
TH
Q
C’p
W
Tco
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
Electrical HeatingElectrical Heating
T
To
t
h
S
M
Cp
U
A
TH
Q
C’p
W
Tco
Cooling coilsCooling coils
T
To
t
h
S
M
Cp
U
A
TH
Q
C’p
W
Tco
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
Temperature profile during sterlization
0
20
40
60
80
100
120
140
0 50 100 150
Time
Temp
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
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
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
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
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
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 −
=
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
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.
Media Sterilisation
Media Sterilisation
 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.
Media Sterilisation
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
=
Media Sterilisation
Media Sterilisation
Soln to Problem 1
 N1= 1.44 X 1016
 N2/N1=6.9X 10 -17
 u= 254.6m/h
 Re=7.07X103
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
Media Sterilisation
 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
Media Sterilisation
 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.
Media Sterilisation
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.
Media Sterilisation
Media Sterilisation
 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.
Media Sterilisation
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.
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
−±−
=
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
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
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
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
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.
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.
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
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.
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.
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.
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.
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.
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).
Air FilterAir Filter
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
 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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Media Sterilisation

  • 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. 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. 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. Medium SterilizationMedium Sterilization  Filtration  Radiation  Ultrasonic treatment  Chemical treatment  Heat
  • 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. 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. 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. 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. 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.  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. E. coli B. stereothermophillus
  • 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. 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. 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. RTE to etANN − ⋅⋅=ln to NNln=∇ RTE etA − ⋅⋅=∇ Combining eqn 2 and 4 [5] Del factor is the sterilization criterion ( )ARTEt ∇+= lnln
  • 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. 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
  • 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
  • 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. 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. 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. 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.
  • 26. coolingholdupheatingoverall ∇+∇+∇=∇ tooverall NNln=∇ heating∇ cooling∇ coolingheatingoverallholdup ∇−∇−∇=∇ kt holdup∇=
  • 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. 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. 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. 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. 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. 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
  • 34. Direct steam spargingDirect steam sparging T To t h S M Cp U A TH Q C’p W Tco
  • 35. 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
  • 38. 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
  • 39. Temperature profile during sterlization 0 20 40 60 80 100 120 140 0 50 100 150 Time Temp
  • 40. 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
  • 41. 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
  • 42. 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
  • 43. 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
  • 44. 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
  • 45. 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 − =
  • 46. 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
  • 47. 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.
  • 50.  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.
  • 52. 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 =
  • 55. Soln to Problem 1  N1= 1.44 X 1016  N2/N1=6.9X 10 -17  u= 254.6m/h  Re=7.07X103
  • 56. 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
  • 58.  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
  • 60.  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.
  • 62. 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.
  • 65.  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.
  • 67. 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.
  • 68. 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 −±− =
  • 69. 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
  • 70. 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
  • 71. 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
  • 72. 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
  • 73. 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.
  • 74. 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.
  • 75. 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
  • 76. 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.
  • 77. 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.
  • 78. 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.
  • 79. 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.
  • 80. 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.
  • 81. 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).
  • 83. 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
  • 84.  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