WEBVTT
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welcome to this second lecture on bioreactors
this is a mock on bioreactors in the last
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lecture previous lecture we looked at module
one which is introduction after mentioning
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the formalities of the course we looked at
some day to day examples such as cancer
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concerns day to day concerns cancer we
looked at what cancer was and how cancer
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is treated allopathically and then we said
that monoclonal antibodies are used to target
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the drugs that kill cancerous cells more directly
to the cancer cells these monoclonal antibodies
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when required in large amounts are produced
through by reactors or bioprocess and that's
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how we started then we looked at other examples
such as alternative liquid fuels bio ethanol
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and bio diesel and saw that they were also
produced through bio processes had given you
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links to some information on ethanol making
ethanol from corn algae to fuels and so on
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i hope your seen them they are interesting
videos then we looked at insulin which is
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a classic drug infact one of the first ones
to be produce large scale one of the first
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new ones to be produced large scale using
bio processes and finally we looked at
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curd making and said curd making is also a
bio process the vessel in which the curd is
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made is nothing but a bioreactor of course
it's a simpler form of a bioreactor a bioreactor
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is nothing but a vessel highly instrumented
and controlled vessel in which bio reactions
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take place and bio products are made normally
with the multiplication of sets then we looked
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at the over view of the bio process we said
that it has two major aspects the bioreactor
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aspect or the upstream aspect and the downstream
processing aspects we said that the raw material
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goes into the bioreactor which contains the
micro organism and nutrients for it to grow
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and the produce the product of interest or
it could be an enzyme which is doing an enzymatic
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conversion and what is harvested from the
bioreactor these liquids stream that is
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harvested from the bioreactor contains the
product of interest along with the micro organism
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and the other material that arose from these
two since we are interested only in the product
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we need to remove that from the other material
and that is what is done through downstream
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processing steps to achieve the final purified
product then we looked at common bioreactor
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types some common bioreactor types we said
that the stirred tank bioreactor is a one
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of the common types it's nothing but a
stirred tank and then an air lift bioreactor
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a packed bed bioreactor fluidized bed bioreactor
which is nothing but a packed bed with
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gigues around there are some advantages there
there was a video which lead you to some
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detail on the fluidized bed bioreactor and
then solid state bioreactors single use bioreactors
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that are helpful in meeting the stringent
demands or stringent regulatory procedures
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for the production of any biological material
then also a photo bioreactor that's typically
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used with micro algae and other photosynthetic
organisms we said there are many other kinds
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of reactors these are the typical ones
there could be other kinds too then
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we said that any bioreactor or many bioreactors
where each bioreactor can be operated in three
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basic modes either a batch operation where
we dump everything in wait for the process
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to complete and dump everything out and proceed
with further processing or a continuous operation
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where there is a continuous input a continuous
output the growth product formation flow of
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nutrients all happens all happen simultaneously
then a fed batch operation which is in between
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batch and a continuous operation where
there could be intermittent input or output
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or both both intermittent output and the intermittent
input at the same time some reactors we said
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nicely fit into all three modes of the
possibilities of operating them in all three
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modes of operation some of them don't this
is think where we finished up the last lecture
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we said that whenever a bio processes involved
we typically need a clean slate we want only
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the organisms of our interest to grow in the
bioreactor and therefore the bioreactor should
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be a clean slate when we begin the ways of
achieving a clean slate is either through
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high temperature or a chemicals vapors or
through radiation such as u v and gamma which
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kills all the organisms that are present in
the bioreactor and then provides us with the
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clean slate and then we introduce the organisms
of our interest which grows a multiplies produces
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the product of interest so let us move further
in this lecture from this point we will look
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at high temperature in some details of achieve
a a clean slate or sterilization as it is
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called using high temperature this is specifically
thermal sterilization the one that uses high
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temperature if we plot the percent viable
cell concentration versus time as is shown
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here this percent viable cell concentration
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we are plotting on a log scale hundred ten
one point one at a temperature of fifty which
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is typically a higher temperature then
many organisms grow at thirty degree c thirty
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seven degree c we sum at twenty twenty seven
oh sorry twenty five twenty seven and so on
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so fifty degree c is some what high for most
organisms there are some thermophiles at grow
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well at high temperatures but let us not consider
them for the time being if it is at fifty
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degree c then the viable cell concentration
decreases linearly with time at fifty degree
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c at sixty degree c again the response is
linear but it is steeper it gets killed faster
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and similarly at seventy degree c the response
is linear but the slope of this curve is larger
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compare to the one side fifty degree c
or sixty degree c this is the typical response
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that is absorbed when viable cells are
exposed to high temperature temperature that
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is higher than their normal operating temperature
this kind of a behavior is applicable for
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a suspension of single cells cells separately
not clump together and so on and so for and
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what this graph tells us is that the logarithm
of the viable cell concentration know this
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is on a log scale so the logarithm of the
viable cell concentration percentage or a
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fraction is you know that fraction times under
is percentage so fraction is directly proportion
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to the time of exposure at the high temperature
so this is the kind of behavior that we see
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when we expose single cell suspensions to
high temperature so how we going to use
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this before we look at how to use this lets
us understand this a little better so that
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it becomes easier to use this at any required
situation in any required situation this
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kind of a relationship that we saw earlier
results when the rate of decrease in viable
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cell concentration is directly proportion
to the viable cell concentration present at
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any time now let me repeat this the rate of
decrease of viable cell concentration is directly
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proportional to the viable cell concentration
present at any time for example this is what
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is what we mentioned here but written
in mathematical terms the rate of decrease
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in concentration of viable cells is proportional
to the viable cell concentration x subscript
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v
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if we call the rate r d the rate of decrease
in concentration as r d since it is proportional
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we can replace the proportionality by equal
to sign and here we need to multiply x v with
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the constant let us call this constant k d
you might be familiar with this kind of a
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first order expression in your earlier courses
so we are using a first order kind of a representation
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here note that this r d is rate of decrease
in concentration which is the concentration
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decrease per unit time mass or number of cells
per unit volume per unit time that is the
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unit here therefore the bases is the concentration
bases we need to remember this now lets us
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light let us write a balance on cells taking
the bioreactor as the system you all gone
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through a course on material balances so we
are going to write a material balance on cells
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and you all know that a balance can be written
over a region of focus and that is called
11:18.860 --> 11:24.920
the system region of focus is called the system
in this case the bioreactor broth the liquid
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part in the bioreactor in which the cells
grow that is what we are going to take as
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the system and we are going to write a balance
on cells on that system you might recall this
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equation here the rate of input of the
cells into the system which is the bioreactor
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broth minus the rate of output of cells from
the bioreactor broth plus the rate of generation
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of cells in the bioreactor broth minus
the rate of consumption of cells in the bioreactor
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broth equals the rate of accumulation of cells
in the bioreactor broth ok this if you
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consider all that can happen to this species
cells this is all that can happen you can
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either comment to the system go out of the
system get generated in the system or get
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consumed in the system and the net result
of all these things must equal the accumulated
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rate or accumulated amount expresses the
rate here because all these are rates this
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also a rate also important to note that this
is on a mass bases this comes from the principle
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of mass conservation and that principle
is what allows us to write this equal to sign
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mass is conserved mass can be either be created
nor destroyed we are of course not considering
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either nuclear reactions or travelling at
the speed of light and in such situations
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the mass of a species is conserved the mass
of the species before a process equals the
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mass of species after the process
so that is the bases for this if it's a
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not clear when you students take a while to
understand this so let us a review the basis
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for this equation you may have done some of
this in your first course and material balances
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so you could take this as a review and try
to understand this to the extent needed to
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be able to manipulate or write balances
according to our need to do that let us consider
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a situation that we are all in chennai
please very familiar with in summers now this
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is the use of water tankers what are unreserved
then water tankers are used to get water
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for daily needs let and the typical size of
a water tanker could be around twelve thousand
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liters know this is my representation of a
water tanker here i would like you to use
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your imagination to see how this is water
tanker this is the back of the lorry on which
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sea water tanker rest and let us say that
this is the filling of the water tanker
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with water at it's source
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so this typically is either eight thousand
liters ten thousand liters twelve thousand
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liters or sixteen thousand liters let us consider
a tank of twelve thousand liter capacity this
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is the case what is the mass of the water
in the tank all of you know that if we know
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the density we can find out the mass of the
water in the tank and the mass happens to
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be twelve thousand kg because the density
of water is one gram per cc or thousand kilogram
14:48.850 --> 14:54.220
per meter cubed one kilogram per liter and
therefore twelve thousand liters comes out
14:54.220 --> 15:03.339
to be twelve thousand kg now the question
is how long would it take to fill a tank to
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fill this tank twelve thousand liter tank
that's the question so people who remember
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their material balance scores or who are
comfortable with this will immediately ask
15:16.360 --> 15:22.279
the question what is the input rate of water
that fills the tank if you know the input
15:22.279 --> 15:30.439
rate of water then you can figure out the
time taken to fill the tank quite easily if
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the input rate happens to be ten kilogram
per second the time would be twelve thousand
15:36.959 --> 15:45.739
sorry thousand two hundred seconds or a twenty
minutes if the input rate is twenty the time
15:45.739 --> 15:51.669
would be six hundred seconds or ten minutes
if the input rate is fifty kilogram per second
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the time would be two forty seconds or four
minutes how did we get at this if we know
15:59.519 --> 16:00.519
the rate of water
16:00.519 --> 16:07.619
input the time by the mass is nothing but
mass divided by the rate and that is what
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this turns out to be for example twelve
thousand kilogram by ten kilogram per second
16:14.179 --> 16:21.230
gives you twelve by thousand two hundred
seconds which is twenty minutes and so on
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a typical time in which the tank gets filled
is about ten minutes and therefore let us
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choose this as the input rate for further
discussion note that we are talking of rates
16:40.809 --> 16:47.709
now let us complicate this process a little
bit suppose there is a hole in the tanker
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which oozes water at the rate of five kilogram
per second how long would it take to fill
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the tank the if you recall your course you
would say if i know the net rate i can very
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easily find the tank so that is the advantage
here if you work in terms of rate all these
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very standard questions become easier to
calculate if you work in terms of volumes
17:17.150 --> 17:22.770
in it becomes rather difficult to see what
is happening and so on so for therefore we
17:22.770 --> 17:28.740
work in terms of rate and that's why rate
is considered some sort of a standard parameter
17:28.740 --> 17:34.950
in a while dealing with the anvex systems
engineering systems it's so on will look that
17:34.950 --> 17:42.030
in a little while in a little while the
net rate in this case there is a ten kilograms
17:42.030 --> 17:48.350
coming in five kilograms going out per second
ten kilograms per second coming in sorry
17:48.350 --> 17:52.360
i think is twenty kilograms per second coming
in five kilograms per second going out so
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the net rate happens to be fifteen kilogram
per second and once you know the rate you
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can find out the time as the volume divided
by the net rate or the in this case the
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mass divided by the net rate this has to
be m twelve thousand kg is divided by fifteen
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kilogram per second or eight hundred seconds
or about thirteen point three minutes is what
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we get
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now let us complicate this in a creative fashion
suppose that in addition to the leak there
18:29.330 --> 18:35.190
is some mechanism inside the tank it'self
that is generating water at one kilogram per
18:35.190 --> 18:41.820
second ok and some other reaction in which
the water is used up inside the tank at point
18:41.820 --> 18:49.730
two five kilogram per second all of which
simultaneously occur ok this is the crucial
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line here they all simultaneously occur if
all these simultaneously occur how long would
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it take to fill the tank here comfortable
with rates by now so you ask me what is the
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net rate when if i know the net rate i can
find it out how do you find the net rate the
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net rate is rate of input minus rate of output
plus rate of generation minus rate of consumption
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rate of input is twenty kilogram per second
rate of output is five kilogram per second
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that we already seen now the rate of generation
here water is being generated at one kilogram
19:24.280 --> 19:30.950
per second and it is being consumed by a reaction
at point two five kilogram per second therefore
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the net rate turns out to be fifteen point
seven five kilogram per second so this is
19:37.660 --> 19:43.620
the rate at which water gets accumulated inside
the system or inside the tank the rate of
19:43.620 --> 19:50.630
change of water mass with time in the tank
so that will turn out to be twelve thousand
19:50.630 --> 19:58.120
kgs divided by the net rate which is fifteen
point seven five kilogram per second which
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turns out to be seven sixty one point nine
seconds or twelve point seven minutes
20:08.030 --> 20:18.410
now i gave you this background two are equation
rate of input minus rate of output plus rate
20:18.410 --> 20:23.120
of generation minus rate of consumption equals
the rate of accumulation and the rate of accumulation
20:23.120 --> 20:31.360
we write as the derivative so that the form
is easy to directly use during heat sterilization
20:31.360 --> 20:37.320
of cells you know you are sterilizing let
say a stirred tank bioreactor containing medium
20:37.320 --> 20:43.240
it which probably has some cells growing in
it already it's not sterilized and we are
20:43.240 --> 20:49.060
going to heat sterilize and kill all the cells
that are present there during that process
20:49.060 --> 20:55.390
the cells are being killed in the bioreactor
broth it is a system nothing as is happening
20:55.390 --> 21:01.380
ok there is no input there is no output and
so on so for therefore the rate of input of
21:01.380 --> 21:11.580
cells mass of cells into the system is zero
there is no output of cells from the system
21:11.580 --> 21:20.270
the rate of output of cells again mass of
cells or number of cells is zero there is
21:20.270 --> 21:26.140
no generation of cells in the system or let
assume that therefore the rate of generation
21:26.140 --> 21:37.110
of cells is zero and therefore the only term
that remains from r i minus r o plus r g minus
21:37.110 --> 21:48.690
r c is this minus r c and minus r c the rate
of consumption of cells r c is equal to d
21:48.690 --> 21:56.820
d t the rate at which mass of cells gets accumulated
in the system this is what we get by considering
21:56.820 --> 22:01.610
it from first principles
22:01.610 --> 22:06.621
the consumption of cells is through the death
of cells as they are getting expose to the
22:06.621 --> 22:15.040
high temperature or this is the heat sterilization
that we are taking about therefore r c which
22:15.040 --> 22:23.220
is the rate of consumption of cells on a mass
basis is r d which was on a volume basis remember
22:23.220 --> 22:28.820
that was a sorry that was on concentration
basis a concentration is nothing but mass
22:28.820 --> 22:34.960
divided by the volume therefore to get mass
from concentration you need to multiply concentration
22:34.960 --> 22:41.740
by volume therefore to get a rate that is
mass based from a rate that is concentration
22:41.740 --> 22:53.120
ra[te]- based you need to multiply the concentration
based rate by volume therefore this r d into
22:53.120 --> 23:01.000
v which is the rate of a consumption of
cells or the rate of death of cells on a mass
23:01.000 --> 23:09.940
basis this minus comes from our balance
equation equals the accumulation rate of the
23:09.940 --> 23:17.910
mass of cells inside the bioreactor when it
is being sterilized when is r d into v equals
23:17.910 --> 23:26.890
d d t of m x now we have everything on a mass
basis which is nice to have now from the definition
23:26.890 --> 23:33.380
of concentration mass is nothing but concentration
times volume or in another words concentration
23:33.380 --> 23:43.640
is mass by volume that's the definition and
therefore m x which we have here is x v the
23:43.640 --> 23:54.430
concentration of viable cells into the volume
if we use this to replace this m x then we
23:54.430 --> 24:04.780
get this minus r d into v equals m x as x
v into v therefore d d t of x v into v and
24:04.780 --> 24:11.330
in this case since a volume is a constant
volume of the broth volume of the system that
24:11.330 --> 24:23.770
is the constant we can take volume out of
the derivative v into d d t of x v thus if
24:23.770 --> 24:29.070
we now we have a volume here we have a
volume here we can cancel them out there is
24:29.070 --> 24:39.490
same volumes therefore minus r d equals minus
k d x v which is what we had seen earlier
24:39.490 --> 24:47.300
we had represented r d as first order
expression k d into x v the rate of death
24:47.300 --> 24:53.720
of cells equals constant times the viable
cell concentration that's the first order
24:53.720 --> 25:02.510
concentration that equals d x v d t since
we have we have cancel the volumes here
25:02.510 --> 25:15.130
ya this is the final expression here d d t
of x v equals minus k d times x v this is
25:15.130 --> 25:22.210
the first order differential equation which
you are all familiar with if we solve this
25:22.210 --> 25:32.650
first order differential equation we get lon
or natural log x v naught by x v equals k
25:32.650 --> 25:41.950
d times t time where x v naught is the concentration
when we began the process of sterilization
25:41.950 --> 25:53.560
heat sterilization and x v is the viable cell
concentration at any time t or the way to
25:53.560 --> 26:02.570
look at this from perspective by design perspective
is the time needed for the viable cell concentration
26:02.570 --> 26:11.090
to go down to x v starting from x v naught
when the process started is nothing but i
26:11.090 --> 26:17.120
am converting the natural log to log base
ten you know that the conversation factor
26:17.120 --> 26:22.070
is two point three naught three therefore
natural log x v naught x v is two point three
26:22.070 --> 26:30.090
naught three log to the base ten x v naught
x v and k d comes to the denominator here
26:30.090 --> 26:36.880
therefore the time required for this amount
of a decrease in viable cell concentration
26:36.880 --> 26:42.400
from x v naught by x v is this expression
two point three naught three k d divided by
26:42.400 --> 26:57.020
k d log to the base ten of x v naught by x
v the time taken for a ten fold reduction
26:57.020 --> 27:03.310
in the viable cell concentration now when
if it was a initially a certain value it is
27:03.310 --> 27:10.700
gone down to one tenth of water towards initially
that is the ten fold reduction in viable cell
27:10.700 --> 27:16.770
concentration at any given temperature this
is an important parameter for the design of
27:16.770 --> 27:21.840
thermal sterilization ok we are figuring out
means by which we collect the information
27:21.840 --> 27:28.990
so that it can be used else where without
much information to design the sterilization
27:28.990 --> 27:37.250
process to design the sterilization equipment
and so on this time taken for ten fold reduction
27:37.250 --> 27:43.450
in viable cell concentration is called the
decimal reduction time typically expressed
27:43.450 --> 27:54.790
as capital d in terms of whatever we have
derived at ten fold reduction in viable cell
27:54.790 --> 28:01.190
concentration essentially means that the initials
viable cell concentration x v naught divided
28:01.190 --> 28:10.110
by the viable cell concentration at that time
t x v must be ten straight forward and if
28:10.110 --> 28:16.380
we substitute this in the expression for time
that we had derived earlier this expression
28:16.380 --> 28:25.920
the time taken for the concentration to go
from x v naught to x v then we would get an
28:25.920 --> 28:33.220
expression for the decimal reduction time
in terms of the variables that we would like
28:33.220 --> 28:39.470
two point three naught three divided by k
d log ten to the base ten you all know log
28:39.470 --> 28:45.680
a to the base a is one and therefore this
reduces to two point three naught three by
28:45.680 --> 28:46.680
k d
28:46.680 --> 29:00.400
recall that the above analysis is applicable
only to a suspension of single cells for which
29:00.400 --> 29:06.050
the rate of kill is directly proportional
to the viable cell concentration at any given
29:06.050 --> 29:15.160
time remember this has been observed and then
this observation we look at representing
29:15.160 --> 29:22.980
mathematically and we said that the first
order decrease would be good way to represent
29:22.980 --> 29:28.790
this that is the rate of decrease of cell
concentration at the high temperature is directly
29:28.790 --> 29:35.270
proportional to the viable cell concentration
at any time and that is what resulted in the
29:35.270 --> 29:40.080
earlier mathematical expression that we had
mathematical expression always generalizes
29:40.080 --> 29:47.440
things it makes it a lot more useful in situations
that has not been seen earlier relevant situation
29:47.440 --> 29:54.960
that have not been seen earlier and thereby
it leads to design of a sterilization processes
29:54.960 --> 30:00.350
in this case
30:00.350 --> 30:07.140
that was for a suspension of single cells
there are other death behaviors that are non
30:07.140 --> 30:15.290
linear let us see whether you are able to
guess this first this is a plot of again percent
30:15.290 --> 30:21.210
viable cell concentration hundred ten one
point one therefore on a log scale versus
30:21.210 --> 30:29.760
time and can you guess what would give a curve
like this earlier it was linear a linear decrease
30:29.760 --> 30:33.940
in this case the decreases something like
this ok it doesnt decrease much for quite
30:33.940 --> 30:42.670
a while and then it starts decreasing can
you guess what kind of content or viable
30:42.670 --> 30:52.990
cells viable cells at what form would give
raise to this the answer is clumped cells
30:52.990 --> 31:00.940
now cells clump with each other then it takes
a while for the killing of the internal
31:00.940 --> 31:06.910
parts of the clump compared to the outer parts
of the clump so that results in this kind
31:06.910 --> 31:14.500
of a behavior for a plot of percent viable
cell concentration versus time let us look
31:14.500 --> 31:22.750
one other non linear behavior this for the
variety this is an another non linear vari[ety]-
31:22.750 --> 31:30.400
behavior as you can see it is represented
as two lines from hundred to lets say point
31:30.400 --> 31:37.000
two or something like that this is one
line and then there is another line here with
31:37.000 --> 31:47.760
a difference slope compare to this when do
you think such a behavior would be seen
31:47.760 --> 31:53.230
in a wave already seen that this happens a
suspension of single cells only thing is that
31:53.230 --> 32:01.309
we have two lines with different slopes ok
so this could arise when you have mixed population
32:01.309 --> 32:09.970
of two different cell types the type differing
the the difference in types arising from their
32:09.970 --> 32:16.320
response to a high temperature this decreases
these type of cells decrease fast these types
32:16.320 --> 32:23.470
of cells decreases slow the viable cell concentration
versus time and this is what this is a
32:23.470 --> 32:28.200
kind of behavior that arises when you have
a mixed population of two different cell types
32:28.200 --> 32:41.310
ok now i am going to present the first practice
problem this is let me call at one point
32:41.310 --> 32:47.200
one because this is module one that is what
the first number indicates and this is problem
32:47.200 --> 32:53.820
one of module one therefore one point one
i am going to assign this problem to you and
32:53.820 --> 33:00.790
you can solve it you need not submit this
problem what we will this will not count
33:00.790 --> 33:07.820
toward evaluation for this mock however i
believe that you need to pick up these skills
33:07.820 --> 33:16.420
of problem solving therefore what i am going
to do is when we begin the next lecture i
33:16.420 --> 33:23.960
am going to solve this first for you and then
go forward ok the practice problem here reads
33:23.960 --> 33:31.809
a bioreactor needs to be sterilized before
use the solution in the bioreactor consist
33:31.809 --> 33:39.850
of single cells with similar thermal response
characteristics ok so it is going to be a
33:39.850 --> 33:48.960
single line that's dropping at seventy degree
c it takes five minutes for the viable cell
33:48.960 --> 33:57.710
concentration to reduce to twenty percent
of it's original value a determine the decimal
33:57.710 --> 34:02.990
reduction time remember we said it was an
important parameter so determine that first
34:02.990 --> 34:11.139
determine the decimal reduction time and be
how long would it take for the viable cell
34:11.139 --> 34:17.139
concentration to reduce to point one percent
of it's original value under the same conditions
34:17.139 --> 34:25.220
ok this is a problem for you you need to
know how to solve problems we will probably
34:25.220 --> 34:30.619
touch up on that when we solve the pro[blem]-
or atleast i will show you some thinking behind
34:30.619 --> 34:41.720
the solution when we begin the next lecture
therefore this is what we did in the introduction
34:41.720 --> 34:48.499
i guess we went through the earlier parts
of the introduction when we began this lecture
34:48.499 --> 34:55.480
essentially some examples of a bioprocesses
and what exactly is a bioprocess and that
34:55.480 --> 34:59.109
the bioreactor is at the heart of the bioprocess
34:59.109 --> 35:05.720
different types of bioreactors are commonly
used some commonly used bioreactors we saw
35:05.720 --> 35:13.180
such as the stirred tank bioreactor the air
lift bioreactor the solid state bioreactor
35:13.180 --> 35:19.489
single used bioreactors photo bioreactors
packed bed bioreactors fluidized spread bioreactors
35:19.489 --> 35:25.099
and so on there are many different types
bioreactor is nothing but an instrumented
35:25.099 --> 35:31.720
controlled vessel in which bioreactions take
place typically mediated by cells or it could
35:31.720 --> 35:37.549
be an enzyme that mediates the reaction that's
exactly what a bioreactor is then we said
35:37.549 --> 35:43.700
that any bioreactor can be operated in different
modes some bioreactors lend them cells to
35:43.700 --> 35:50.019
operation all the three basic modes and some
don't for example a stirred tank bioreactor
35:50.019 --> 35:56.299
can be operated in a batch mode in a continuous
mode as well as in a fed batch mode whereas
35:56.299 --> 36:00.849
may be a packed bed it's a little difficult
to operate only in a batch mode
36:00.849 --> 36:06.740
you can but it takes a lot more effort
rather it it it is normally not operated in
36:06.740 --> 36:14.099
a batch mode it is operated in a continuous
mode then we looked at the need for
36:14.099 --> 36:21.789
a clean slate that is we want only our organisms
to grow and produce the product of interest
36:21.789 --> 36:25.849
and therefore we need to get of all other
organisms with that are normally present we
36:25.849 --> 36:30.809
said that organisms are present in the air
around us all the time and nothing prevents
36:30.809 --> 36:37.510
them from getting into the bioreactor broth
that is open to the atmosphere and therefore
36:37.510 --> 36:44.339
we need to kill all the cells that are initially
present there the main modes of killing
36:44.339 --> 36:52.440
include thermal killing a high temperature
can be used to kill cells or could use a appropriate
36:52.440 --> 36:58.990
vapours or liquids seventy percent ethanol
solution is used to kill cells on your hands
36:58.990 --> 37:08.099
when you when you handle these organisms
cell cultures and so on or a formalin vapour
37:08.099 --> 37:16.369
formaldehyde vapours from formalin solution
is used to kill the organisms in a space such
37:16.369 --> 37:22.380
as a lab and then you could also use gamma
rays and so u v rays gamma rays gamma rays
37:22.380 --> 37:28.980
especially if you are sterilizing lets
say syringes and things like that that cannot
37:28.980 --> 37:33.270
be expose to high temperature there might
of plastics they cannot be expose to high
37:33.270 --> 37:39.069
temperature so those three modes of killing
then we said we will look at the thermal
37:39.069 --> 37:47.450
sterilization and some detail we reaffirmed
the need to be clear about the concept of
37:47.450 --> 37:56.859
rate as a basic parameter in any engineering
system analysis dynamics system analysis the
37:56.859 --> 38:00.270
rate by it'self is
38:00.270 --> 38:05.779
important the way you measure rate could be
very different for example you have speed
38:05.779 --> 38:10.750
which is equivalent of rate and you could
measure an average speed by the distance travelled
38:10.750 --> 38:18.789
by time taken but speed is rather central
to the kind of to answer the kind of questions
38:18.789 --> 38:24.789
that we would be interested in when we deal
with design operation of engineering systems
38:24.789 --> 38:33.269
in this case biological engineering systems
and then using rates we said that the rate
38:33.269 --> 38:39.989
of killing of cells by high temperature
could be directly proportional to the viable
38:39.989 --> 38:46.559
cell concentration at a particular time of
first order kind of expression then we
38:46.559 --> 38:54.130
saw that lead to the definition of a decimal
reduction rate is standard parameter and
38:54.130 --> 39:01.079
then also a more general expression before
that that gives the time of time that it takes
39:01.079 --> 39:06.530
to go from let us say an initial concentration
of x v naught initial viable cell concentrate
39:06.530 --> 39:15.109
of x v naught to certain viable cell concentration
of x v at a particular time t then we looked
39:15.109 --> 39:21.839
at this practice problem here ya ya before
that we did see that this linear killing
39:21.839 --> 39:26.950
is only one kind of a behavior you could have
other kinds of killing such as a curve when
39:26.950 --> 39:34.829
there are clump cells and combination
of two straight lines with different slopes
39:34.829 --> 39:40.950
if you have a mix population of thermally
different organisms then we looked at this
39:40.950 --> 39:46.730
practice problem you might want to spend
some time and solving this it will be helpful
39:46.730 --> 39:52.369
these are the kind of things that engineers
are expected to do and it will be good to
39:52.369 --> 39:59.640
pick up the skills when we meet next we will
solve this problem and start module two
39:59.640 --> 40:01.460
see you in the next lecture