Today we will be looking at this biomethanation reaction. Now, what happens in biomethanation
is something like this.
Generally you have insoluble organics say
if you going to a let say a dairy industry
for example, the waste organics lot of solids
are coming. So, insoluble organics they gets
digested by extra cellular enzymes in the
reaction equipment and become soluble organics.
And then volatile acids from in bacteria works
on it; it give you a acetic acid; and then
gasification by methane producing bacteria
it will give you methane and carbon dioxide.
So, is the 3 step reaction that is what you
see in the biomethanation process.
Stage 1 is solubilization, stage 2 is acid
formation, and stage 3 is gasification using
methane bacteria. Now, generally we find that
it is not a bad assumption to assume that
the methane producing step is the slowest
step of the reaction. So, is the mostly design
therefore is done on the basis of this assumption
that methane producing is the slowest step.
Although, there are instances where this solubilization
is quite slow as a result we may have to take
this also into our account. So, as well this
problem is concerned we are looking at production
of methane using acetic acid as a substrate
and that is a chemical reaction that is set
up acetic acid.
So, this is the chemical representation what
is going on acetic acid this is the nitrogen
that is present in the environment to give
you some bacterial growth; which is this is
the chemical representation of bacteria and
methane and carbon dioxide ok. So, you notice
here the roughly equal quantities of methane
and carbon dioxide is produced volume basis;
there is lot of literature in biomethanation
over the last 30, 40 years many many plans
have been set off all over the world. It is
a reasonable well known technology; there
are many problems with it we will talk about
it as we will go along. Now, what is seems
to be an appropriate growth function is a
let me write down the growth function.
The growth function is something that we have
set up it look something like this mu m mu
equal to mu m within brackets 1 by 1 plus
K s by H s and H s by K I; this was a propose
by Graef and Andrews rules bay back in 1974;
there is a lot of material on this what is
H s? H s is the an ionized acid. So, this
is the an ionized part acetic acid which is
the which was is important as far as the a
growth function is concerned ok.
The numbers are given K s is given K i is
given this form of Graef function we have
seen earlier also; this is the form where
substrate inhibits the rate of chemical reaction.
And, therefore we would like to operate the
process at H s value equal to K s times K
i square root. So, the best choice of H s
which is the substrate which is active as
far as the process is concerned is square
root H s times K i we can calculate that is
from here. So, what you want look at is a
very important area dairy waste water just
taken 3000 cubic per day 6000 milligram per
liter 6000 roughly 100 mile mole is not mole
100 mille mole sorry 100 mile mole per liter
is the concentrational acetic acid they will
come out of this 3000 milligram per liter
sorry 3000, 6000 milligrams per liter ok;
it is being treated in a chemostat. Chemostat
is our C S T R equivalent of which we are
talk about all ready; what you want to do
is what just get an idea of the size and what
are the outputs that we can expect?
Now, in go around the world you find that
dairy waste waters what they would do is that
they would first do a biomethenation. And,
as you will see as you go around this problem
you will see the biomethanation is not able
to complete remove or consume this 6000 milligram
per liter; it is only able to do part of it.
And, the other words the effluents come out
of dairy waste water still is vary concentrated
from the point of view of discharge. So, you
will have to do another step of treatment
before it to suitable for discharge. So, we
will have to look at that also find out how
to handle this kinds of problems.
So, if this is the rate function where mu
is mu m times this one; so what is the best
choice of H s that we will take? What is that
value, tell me? What is that please calculate and tell me?
This is the ionization reaction here K a is
given K a is given somewhere I think K a is
4.5 and operating P H is 6.8. So, you can
find out what is the H s value? So, the H
s at which we will operates 0.15; is there
with everybody 0.15 root of k 1 k 2 K s K
i square root is 0.15; is it all right do
we all agree?
This is what I have got H S optimum I got
0.15. So, as per the data given the best choice
mu equal to D correct we must have mu equal
to D chemostat mu equal to D. Therefore, dilution
rate at which we will operate is 0.27 per
day; is it can you please calculate and tell
me whether this mistake in my calculation
0.27 per day is 0.27 is all right? So, what
should be we have?
So, our equipment volume should be 3000 divided
by 0.27 that is about 11000 cubic meter is
what you get no 3000 divided by 0.27 is about
11000; is it all right 11000 is ok. So, what
is the now if it is 11000 what is the value
of S minus I will get is a 1795, S minus comes
from the acetic acid equilibrium here; what
is equilibrium is given all the numbers I
given yes or no?
So, what is the acetic acid that you have
started with stills remains un reacted; that
is acetic acid that is still remains un reacted
1795 is a S minus; how much is H s can you
calculate, what is H s at the end of the process
S minus? What is the H s is this correct;
if there is a mistake tell me I will correct
it 0.15 mile mole per liter 0.15 multiplied
by 60 is 9 is it not, it is it all right?
See our operating the process as optimum value
of H s; H s is optimum is 0.15 mile mole per
liter that is what you told me multiplied
by 60 is a molecular weight of acetic acid.
So, I put it as 9; is it ok?
So, what we saying that we are the process
that we are steadying at 6000 and the biomethanation
reduces it to 2000; is that clear? So, biomethanation
process is able to reduce the organics to
a level of 2000 it is not able to reduces
it to a level that is suitable to discharge.
So, it is 18 and answer is so I have to make
some adjustment in these numbers. So, these
numbers are not exactly correct; please help
me now what is a loss C O D S naught minus of S.
So, this is a s not exactly 6; so you
tell me the actual numbers please so it is
6 minus 1.8 not 2. So, this is not exactly 11000 it be slightly
less slightly more. So, please tell me these
numbers; 12 126.
12600. 12600; is it k is it all right with everybody?
So, what am I saying that so much loss of
C O D as 12000.
Now, if you look back at our Stoichiometry
it says what is it say C O 2 and there roughly
in equal volumes. So, in terms of mass basis
I will calculated here. So, based on the just
look back at this one per every 60 we get
0.92 times methane which is 0.92 point 16
and 0.92 times 44 correct. So, there is Y
axis this is Y Y P S and this Y P carbon dioxide;
so that is what I have done. So, fraction
methane to substrate is 0.24 I get carbon
dioxide substrate 0.67 please verify it is
1 6; how did I get 1795.
So, brilliant question see 29.9 multiplied by…
60. 60 is it ok all right is it with everybody
ok? Now, please tell me if these numbers are
correct 0.92 times 16 is 0.24 and then 0.92
44 is 0.67. so, if I ask you what is the methane
fraction in biogas; what will you tell me
typically weight by weight? Methane fraction
0.24 divided by 0.91 what about that number is.
So, you will find in biogas on weight
basis the methane is not more than about 26,
27 percent; on volume basis it is equal on
weight basis it is only 25, 26 percent is
it with everybody 0.24, 0.67 all right ok.
So, what is the methane production a made
a small mistake here. So, you can tell me
so it is 600; so this number is how much please
tell me? 3025 fine 0.26 is it?
Yes just 2 minutes just 2 minutes there is
some important questions are coming please
I am just what is mistake I made here please
tell me C O D losses 12600 that is this is
the acetic acid loss out of that we said this
is a fraction; that is why you multiplied
by 0.24. What we are saying is for every 60
we are getting p0.92times 16. So, that is
the fraction we are talking about it; is that
is it with everybody what we are saying. So
where are so we have 3000 k g of methane and
so many kilo grams of some slight mistake
here; what is this number please tell me?
12600 8440 fine. Suppose I ask you what is
the thermal value of biogas what will you
tell me; what is the thermal value of biogas?
So, the methane I calculate here just to put
it in the context; methane is about 14000
kilo calories per kilo gram I taken this from
the literature. So, biogas is about close
to 5000 kilo cal 3600 per k g or per cubic
meter you can also calculate that I have done
that. So, 3640 is what we get showing that
biogases is a lean gas is not a very rich
gas it is a lean gas. And, therefore its combustion
in a I c engine will require suitable design
which people done all ready is very popular
around the world; biogas engines are operating
very well. And, what is the problem that we
face biogas in I c engines? The problem with
I c engines is that and this reaction if there
is protein generally you produce hydrogen
sulfide here. And, that creates problem in
the combustion for which in small scale it
is not easy to handle, in large scale you
can remove the hydrogen sulfide. What we find
is by biomethanation gives you biogas; yes
it gives you very good thermal value; but
it produces a waste water which is not ready
for discharge. So, we must look at doing a
system design in which the waste water is
taken care.
And, this waste water we said the fluent that is comes down the biogas is this is 1804 somebody said.
So, we said the yesterday that we our
systems will have dilution rate of this type
correct; where mu m I taken whatever you done
yesterday; so this is a kind of number I get.
So, the second system which will take care
of the waste water is that; it were it has
a dilution rate of 0.5 and the system volume
of 6000 is it please tell me? This is all
right 3000 cubic meter per days coming the
dilution rate is 1 by B times mu m s K s and
all there have taken yesterdays values or
given you yesterday. So, D value is 0.3 is
knows 0.5 and this is what I get please tell
me this is it is I have taken B as 0.3 mu
m as 0.3 per day K s has 10; this is in problem
number one same thing we done this yesterday
same number. Say B is 0.3 everything is what
we have done in yesterday yes or no shall
we go forward all right where have we?
So, we have an equipment size of 6000 cubic
meters; which is what that treatment of that
waste water. And, what is the size of the
equipment the biomethanation? What was biomethanation size we got 11000. So, you can say
11000 plus 6000 17000 cubic meter have we used off. Now, let us look at the real problem?
The real problem is here see what happens
is that is this correct total C O D loss that
in this in this is the biogas; is this correct
what I have written 2000 nah it is 2000. See
we have to remove, see we have to what comes
out off the biogas plant is a 2000 milligrams
per liter 3000 cubic meters. So, the total
amount of C O D that we have to treat in the
fluent treatment plant are the E T P they
call; is this is this correct? Let me repeat
what comes out of the biogas plant is 1804
1.084. So, this is slightly less; so is 540
k g per day ok. So, what comes out of the biogas plant comes
at 1804 I have written 1.80 k g per cubic
meter 3000 cubic meter 5400 k g per day has
be handle in your E T P plant correct; is
this what we have say I am coming to then
I am coming to that ok. Now, how much oxygen
is required to treat to oxidized 5400 k g
of oxygen demand; if it is typically glucose
1 is to 1 is reasonable number. So, to be
able to process 5400 hundred k g; so you will
required 5400 k g of oxygen. Now, you can
ask how this numbers comes from; of course
this number comes from practice if you go
to a any waste treatment plant they will tell
you that every kilogram of oxygen demand you
roughly require 1 kilo water of n h v. That
means you aeration equipment the amount of
energy that you must put in; so that you know
it is sort of bubbles through this is about
1 kilo a tar a every kilogram of oxygen you
have to supply; is this clear?
So, if you have 5400 kilo I mean kilograms
of oxygen to be supplied you will required
5400 kilo tower of electricity. So, how many
kilograms of methane did you produce? 3024.
So, to produce see what we do in this one
that we burned, we burned methane and then
converting into electricity. So, your 3024
k g a methane burning and then generally efficiency
between 0.3 to 0.2 in that range. In the other
words 20 to 30 percent typically 20to 23 percent
of the thermal energies converted to electricity.
So, I will just estimated these numbers are
not correct numbers are not correct; please
tell me now 3024 14000 by 860; what is a number
correct if it is 0.3 what is it? If it is
0.2 what is it?
14768.
14768; I will write 770; if it is 0.2? So,
what we are saying is that if you burned this
you can get so much of electricity; if it
is a 30 percent efficiency so much; if it
is 20 percent efficiency so much. Generally
you would find that I c engines efficiency
is a quite good. So, you have get something
in this range. So, I will get by 860 see kilo
calories a kilo tars conversion see 860 kilocalories
equal to 1 kilo tar this is something that
I have learnt in my school. So, what we are
trying to say here is that see in E T P you
consume so much and you generate so much.
So, there is a slight power surplus, this
is power surplus which is use to various other
purposes; is it clear? So, when you have biogas
plant you have some surplus energy you can
use a various other purposes; part is a reason
why biogas is a very popular in varies places
is that it does give you some power surplus;
very good question that is lets go back to
that question. Where are we; we calculate
its C O D loss for the biogas plant which
is converted to biogas, we calculated C O
D loss in the E T P plant it is converted
into carbon dioxide; in both cases there is
a loss of carbon. In 1 case the carbon is
given comes to you as methane, in another
case the carbon comes to as carbon dioxide;
is it clear? Now, the important point to recognize here
is that in many places you will find when
they do this power production using gas the
wasted energy which is the exhaust from the
engine in many of these dairies they need
lot of hot water. Because dairy requires lot
of washing to be done in varies places. Therefore,
they are able to use that energy to generate
hot water. And, because of that thermal efficiency
of the process because they are able to recover
lot of heat from the waste heat, thermal efficiency
is the very high. And, other words many of
these dairies they are able to integrate biogas,
waste water treatment and therefore that waste
water they are able to converted into hot
water and use it for various kinds of washing.
So, it integrates very nicely in terms of
thermal, in terms of say protecting the environment
and so on. So, it is very popular around the
world; if you can effort the investment on
the biogas plant, investment in the waste
treatment both are very large energy intensive
investment intensive activity.
The second exercise see alcohol fermentation are status in this country.
Of course all the alcohol in this country
is produce from sugarcane molasses. And, in
many parts in the world particularly the world the warmer religious in the world it is from
sugarcane molasses. Now, this data that I
have taken is a rated which the substrate
gets consume is given by this function mu
m S k minus k P and all that; these numbers
by enlarge whole for sugarcane molasses. And,
these numbers 0.17 all these effects are quite.
Now, what happens in a sugar industry is that
the molasses comes to you as 50 percent sugar
44, 45 sometimes 48 percent; 48 percent means
400 to 500 grams per liter as hexose sugar.
Now, industry what they will do is that they
will dilute this to about 10 percent; they
will dilute this from 500 or 450 grams per
liter to 100 grams per liter. The question
is fermentation alcohol industries typically
dilute molasses 200 grams per liter why? Now,
you will you will realize that let me write
this Stoichiometry here C 6 H 1 2 0 6 giving
you twice C 2 H 5 O H plus twice C 0 2; this
is 180, this is 92; this is 88. And, other
words strictly anaerobic conversion hexose
sugar to alcohol is actually roughly 49, 51
percent roughly say 50; is that clear?
Now, if you start with 1 gram of sugar you
will get half a gram of alcohol. Now, if it
is if you start this as 100 grams per liter
then you should get 50 grams per liter of
alcohol correct; is that clear? If you a started
with 500 grams per liter you would have got
250 grams per liter of alcohol. Therefore,
the amount of water you have to remove in
this case it is only 4 times you know 250
to and you have to only remove per every k
g of alcohol you have to remove 250 k m in
750 k g of water; here you will have to remove
950 k g of water. Just see the difference
if you could have used 500 grams per liter
sugar you have removing 750 k g of water per
every k g of alcohol; here you are using per
every 50 you are removing 950. So, 1 to 19
here it is 1 to 3 you can see the advantage
if you can use a higher concentration; is
that clear? Alcohol industries they do not use 500 grams
per liter they use 100 grams per liter; they
are willing to scarifies this advantage why?
Because energy cost of distillation is very
high we all know that but in spite of the
energy cost is so high they diluted to 100
grams per liter and not worth with 500 grams per liter.
So, question is why; is a is a
question clear to all of you why do we dilute
molasses? Say it again; molasses is available
in a sugar industry typically 500 grams per
liter as hexose sugar.
Now, this is diluted to 100 grams per liter
hexose sugar prior to fermentation question
is why? What I am trying to explain here is
that if they could have work with 500 grams
per liter they would have go 250 grams per
liter of alcohol. Therefore, per every 250
grams they have to evaporate 750 grams of
water; 1, 2, 3 per every k g 3 k g of water
has to be evaporated. But if they do this
they have to remove 5 to 95 so 1 to 19 the
amount of water to be removed in this case
is per every k g 19 k g of water has to be
remove; while here it is only 3 k g. So, there
is such great advantage in distillation when
you can use 500 but they only use 100; the
question is why? The answer is there are many
answers many reasons why they do this; first
reason is that molasses come is extremely
viscous; it is so viscous it is not easy to
pump at all. If you want to pump molasses
you will have to dilute; that is the first reason.
Secondly, molasses contains lost of suspended
solids which comes from the process; see what
we do is that we have criticize the sugar.
So, we are just rejecting what we are not
able to recover correct. So, it has lot of
suspended solids and to be able to remove
the suspended solids you have to reduce the
viscosity. So, that is another reason why
they dilute. The third reason they dilutes
this that this is negative effect of alcohol
accumulation in the process which inhibits
the rate of chemical reaction was called product
inhibition. With all these factors combined
make it essential for them to diluted to a
level which they can work with; is that clear
that is why they dilute. Third is this reason
that alcohol accumulation in the process inhibits
the rate of chemical reaction because is this
minus k P can look at this; this reason this
minus k P effect is so that that makes it
worth. So, that is why they would dilute thank
you my friend k let us go forward now.
Now, let us quickly calculate; let us say
we have the typical size of alcohol distilleries
in brazil is very big but ours is only forty
tones per day is typical; and biggest would
be a little larger. So, if you looking at
40 tons per day alcohol factory.
So, just made some calculation a 40 tons per
day alcohol factory; how much should be this
40 tons per day; is this correct? This see
whether what is this calculation is correct
40 thousand k g per day divided by 24 and
then this is the S naught is 100; S is 10.
So, we are taking around the 90 percent conversion;
so I put is 90 ok. And, you can see here Y
P is 0.4 only 40 percent of alcohol is we
get 48 percent is carbon dioxide you see;
is it all right? And, then remaining 40 and
then this is 48; what happens is remaining?
This is 8 it is going l squares on only 8
percent is sell. So, 96 still 4 percent is
unaccounted; there are other products of formed
in fermentation you know let little bit of
glycerol and other product of forms that is
why the data is not there.
So, what is f equal to 46 is this correct
46; 46 cubic meter per hour. So, what is the
concentration alcohol in solution, what is
the concentration of alcohol in solution?
36 k g per cubic meter is 3.6 percent, 3.6
percent is the alcohol in solution. So, in
other words if we start with 1 k g you will
get 36 percent is alcohol the rest is going
to carbon dioxide. See you can see here 8
percent is sell 48 percent is carbon dioxide
and there are some side reactions like glycerol
and all that which takes a way little bit.
So, what the useful alcohol is only 40 percent
not much more in that. And, please recognize
90 percent conversion of hexose sugar are
also not very common you know; the reason
is this inhibition makes it quite difficult
to handle; as a alcohol accumulate this inhibition
becomes worst. So, you will find a process
a very good factory in India would be 80 to
83 percent not much more than that.
So, we are looking at here we getting 36 per
36 k g per cubic meter; but in actuality we
are not looking at much more then about 30,
29, 30 like that; that is what is the common.
So, the amount of energy that is required
to recover the alcohol is very large. So,
for 36 means you have to remove 9, 960 k g
of water has to remove to be able to concentrate
this. So, this is a serious problem with the
problem with alcohol fermentation lies really
in the chemical reaction equipment; we have
been able to figure out what is the way to
take it forward.
Now, an interesting work got done in 1970
is very old research; they spent a lot of
time trying to see whether this problem that
we are facing here very lean concentration of alcohol.
And, very high energy cost of
recovery; can we do something about it 1970’s
1980’s it till goes on. And, then varies
kinds of answers have been given. Let me quickly
run through that because it is context is
the important. But before that let us finish of it.
What is the dilution rate at which we must
operate the plant? What is the dilution rate;
mu m mu m is 0.5 S is 10 is it right S is
10 yes or no? S is 10and K is minus point
0.017 P is 30 things; 0.22 is correct 525
seen all right. So, what is the size of the
equipment? Suppose you have producing 40 tons
per day a alcohol what is the size of the
equipment that you will have?
Size of the equipment is what I have got here
about 200 cubic meter per hour per hour. Now,
this is the term see what is the productivities alcohol D times P is about 8 k g per cubic
meter per hour or 8 gm per liter per hour;
is it clear what we are saying? And, this
is what makes the whole technology little
inviolable because per unit volume of equipment
you are not producing very much in 7.9 k g
per cubic meter per hour. Now, if you look
at the reality of alcohol production in India
what you find is situation is much worse.
Because this is for a continuous process;
why there is no shut down time you know filling
and empty which is very substantial amount
of times required to fill and remove. And,
then in between also you have to clean equipment
and so on; you will find because is the fact
that the see here our residents time is about
5 hours correct. But you will find there the
actual filling time and then cleaning time
etcetera it becomes 18 hours ok.
So, this what we are getting a 7.9 actually
more like a 2 or 3 grams per liter per hour
is what we achieve in a commercial process;
is this point clear to all of you. And, all
the problem arises from the fact that we have
got the chemical reaction technology right;
the real we that engineering of that reaction
is were the problem is; that is why all these
problems are coming.
Therefore, just look at what was done. So,
people have looked at what can we do I mean
from the point of view of reactor design is
there anything we can do; is that clear is
the problem create to all of you? See you
have just done a calculation for continuous
process we said that its productivity is 7.9
correct based on the fact that we have produce
36 k g per cubic meter and our dilution rate
is 0.22. And, therefore our productivity is
7.9; if you have run a batch process then
the actual residence time would have been not 5 hours more like 18 hours, 20 hours sometimes even more.
So, you are getting much much lower productivity
in the batch equipments that you see operating
in this country. Continuous fermentations
are not very common around the world not very
common there only may few factories. If I
ask you why; why is it that continuous fermentations are not very common people prepare a batch?
What would be a our answer, why is it that
people prepare to do alcoholic fermentation
in a batch; why would they prepare batch?
In fact you would see if you go to places
like Madhya Pradesh and Uttar Pradesh and
all that there are much smaller distilleries
5 kilo liters per day, 5, 2 kilo liters per
day small plants are running you see. And,
therefore the productivities are much much
lower.
And, therefore many of these alcohol they
are not going for pharmaceutical not controlled
by government varies issues are there in a
in the commercial sector. But the problem
is that this productivities are very very
low and there are not good answers yet. So,
let us drew you what does being happening
in this fermentation technology alternatives for alcohol.
A very nice book I mean a P H D thesis 1971
Berkley it is in our library; it is only about
120, 130 pages. And, they have just looked
at batch, chemostats, I immobilize cell reactor,
vacuum fermentation and some combinations
also; they have looked at to see whether we
can get around this problem of very poor productivity
of the reaction equipment; what they have
done is that see this batch reactor correct.
Now, if we have to overcome this problem;
what do what can we do? See this inhibition
due to product if you want to knock out this
problem what can we do? Productivity is low
but we want to somehow get rate of this alcohol;
you have to remove the product; how do you
remove a product? So, they said let us do
vacuum. So, what happen if you do a fermentation
and the vacuum? So, you have an equipment
which is operating something like 30 millimeters
30, 35 millimeters of mercury ok; what would
happened to alcohol; it will go into the vapor
phase, alcohol will go into the vapor phase
and water will go also into the vapor phase;
not just the alcohol water would also go into
the vapor phase. So, what will be the net effect of that as
for as product in the in the broth in solution
concentration will come down correct. And,
other words as the fermentation proceeds alcohol
in the broth would go into the vapor phase
we continuously remove it because it is under
vacuum. Therefore, the reaction keep going
forward. So, continuous fermentation essentially
helps you knock out this effect. So, that
reaction is able to proceed in the forward
direction and give you a much higher level
of substrate conversion; this has been proven
in the laboratory for 1980’s lot of research
is there which shows it is a very interesting
technology from the point of view of driving
the sugar tools nearly complete conversion. But a huge problem came many try to commercialize this.
So, what happened was this gas phase which
contains the alcohol and water which at a
pressure of 30 millimeters of mercury; if
it has to be in a taken to the next stage
of the process you have to compress it. So,
you have to compress from 30 mm to atmospheric
pressure. So, suddenly it can be release into
the atmosphere into a tank which contains
alcohol. So, you have to compress from 30
mm to one atmosphere that compression is a
huge cost; you have spent money in a creating vacuum; vacuum also cost money correct there
is a energy costing vacuum. You need power
to compress the product to atmospheric pressure.
So, energy cost of production become very
large. Therefore, while in our technology
of it seem very attractive; it would able
to drive the reaction to completion you see.
And, therefore since here able to drives the
reaction to completion much higher concentration
of s ok. See the your s need not be 100 grams per liter,
it can be 500 grams per liter; it can be anything
there is no great problem. So, you have an
advantage of very highly concentrated alcohol
I mean it is from 3 percent you go to something
like 16, 17 percent; because there is a lot
of alcohol water vapor in the gas phase. So,
you do not get very high concentration 20
percent you will get. So, to that extent you
can say on a energy cost of distillation.
But the net result was that the energy cost
of vacuum operation, energy cost of compression
was so large that the benefits of using the
much higher s was not satisfactory.
So, after may be about 10, 15 years of a lot
of research it had to given up. In fact the
last I mean chapter of this theses I mean
sort of says very very surprisingly it is
says that may be 5000 years ago the understood
this; that is what it says you know that technology
of last so many thousands of years we are
not been able to fundamentally bring about
the improvement in the alcohol fermentation
technology; that is what it says. That is
very interesting to observe they we still
do not seem to have an answer to alcohol fermentation;
not yet. Now, something more interesting also
has happened that means draw attention to that.
See if we see it says you can read here saccharomyces
is the is the organism that is used all over
the world. Saccharomyces is a fungi they work
well under acetic P H. So, the P H of fermentation
is about 4.2, 4.3 and the P H adjustment is
done by alcohol ok.
But in 1970’s the figure out that is a this
fungi works very slowly the we should change
about to bacteria; this particular bacteria
zymomonas it accurse in a naturally in certain
fermentations in a particularly in some of
the farm and juice it occurs. So, they isolated
and all that and then they have try to commercialize
a process using zymomonas as the bacteria
for fermentation.
And, there are several laboratory extremely
in even pilot plants in which this has been
tried out and it is a excellent; performance
is excellent. The reason is zymomonas bacteria
they rated which is able to do the fermentation
is much much superior to fungi bacteria works
faster than fungi. So, the something went
wrong, what went wrong? What went wrong was the bacteria works best at neutral P H. And,
when you want to do anything neutral P H which
means you have sugar solution all nutrients
already and it is neutral P H they invention
by several competing organisms become very high.
Therefore, if to be able to operate
a bacterial fermentation for alcohol requires
that molasses be sterilized; while if you
do with it saccharomyces at P H of 4.2 the
competition from other organisms are very
low.
So, alcoholic fermentation do go through sterilization
if you use fungi. But if you use bacteria
you have to sterilize. So, the energy cost
of sterilization and cooling it to ambient
temperature and so on prohibitive. So, while
the technology worked very well economics
did not favor it is adoption. So, it this
is also fail and there is lot of research
in 1980’s which says that you know this
can be done but it is also not very satisfactory.
I went to a factory I will not name a factory
in 1970’s; they said no no no we have got
a great idea we have go do this; they said
they will do sell recycle. What is the ideal
cell recycle r x is mu times x correct; higher
the cell concentration higher the cell reaction rate.
So, they said I have got a great answer to
this problem. So, they invested on what is
called as a centrifuge in those days and then
did this process and it failed. So, process
did not work; so they were wondering what
they hell went wrong in our most fundamentally wrong mu r x is equal to mu x must be wrong
then correct; they put more cells into the
system. But the process failed; how do you
explain this you understand what I am saying
you have a process in which you have recycle
the cells. So, cell concentration is several
times more than you normally operate but the
process does not quite give you the result
that you expect; you got 36 you got 38 now;
it is not anything very different. So, what
seems to gone wrong is interestingly see when
you run a process particularly and say batch
process or generally about batch process 20,
22 hours 26 hours; as the product accumulate
the bacteria is starts to some sort of denature
the enzyme starts to it starts to die basically.
So, you have lot of cells in the output but
many of them are not viable cells.
So, when you do a centrifuge and recycle you
are recycling cells all right but the viability
of those cells to recycles are very very small.
So, essentially because of viabilities are
very small it did not give you any great result.
And, the other words pure culture fermentation see alcohol is a pure culture fermentation;
where you recycle you have to be sure there
what you recycle is all viable. But in the
in the environment of fermentation of 24,
26 hours many of those cells die; it does
not survive. And, that is part of the reason
why it is the recycle has not been very successful.
But interestingly recycles are very very common
in waste treatment. See in pure culture it
is not very successful but in waste treatment
if you do not recycle your process will fail;
you understand what I am saying how do you
explain that? Why we recycle in a in a waste
treatment plant but we do not have great benefits
in recycling in a pure culture fermentation?
The answer is in pure culture the product
accumulation seems to have very very bad effect
on the organism; while in places like waste
treatment it is poly culture; where there
is predator pray interaction. And., therefore
predator ensures that the pray is always active
because the predator wants to eat correct.
So, this predator pray relationship ensures
that all organisms are active. So, waste treatment
recycles are critical to success of a process
but in pure culture recycle is actually not
very useful. So, recycle see this people have
try recycle in pure culture like alcoholic
fermentation it is not very successful. The
reason is not successful because during the
process of fermentation of 18, 20 hours viability
of many of this organisms starts to go down.
Therefore, when you try to separate this organism
after the fermentation 18, 20 hours and recycle
the viable cells in that recycle stream is
not very large. And, therefore it does not
benefit you.
But in waste treatment what happens it is
a poly cultural there are hundreds of organisms
working; and they all work to feed on the
waste that is existing. So, the predator pray
relationship ensures there is very highly
active environment is established; that is
why recycles are crucial for success of waste
treatment and recycles are not very useful
in pure culture. In waste treatment the predator pray relationship
ensures that all organisms are always active;
in the waste treatment environment will have
spends some time on ecology which you will
do when we meet; we will talk about ecology
shortly in the next class. So, we will explain
that but ecology is what is crucial to success
of what we call as natural processes. All
natural process you need ecology ok this sort
of completes this part. So, we will go to
the most interesting part of which is natural
selection. How recycle affect predator pray
relationship.
Let me draw once again we had a bio reactor
they then we had we put this back correct.
Now, what happens here in the bio reactor
there is lot of growth. And, this growth is
because of what with this lot of nutrients
available, lot of food available. And, therefore
it grows; therefore you have starting with
x naught which is very small. So, there is
a lot of cell growth here x which your concentrative
here; and this cell growth is because of the
poly culture. So, that means this environment
the multiplicity organism present they selected
what is most useful to this environment correct.
And, that you have concentrated here; that
means you have done concentration of the most
useful organism for your process which you
are putting back. But in a alcoholic fermentation you are started with pure culture; and that
pure culture in this environment it was as
it is no ecology. Therefore, it is getting
affected because of the accumulation of the
new coming toxicity or the products. And,
that is why it is start to die when you do
this it starts to die; because so much of
toxicity is there it will die it is clear
to all of us.
Poly culture is what nature is all about and
predator pray is what ensures highly active
population; this is a very beautiful example
of natural selection. See Darwin talked about
natural selection you have read in biology
that natural selection it what happens in
the natural environment.
This is an example where I illustrate you
how this natural selection actually happens;
you can do this numbers to realize how beautifully
it works. The example is here there is a chemostat
in which this organism is growing; it says
it is got specific growth rate mu m is so
much and all that. And, then this chemostat
is working and as it when it is working suddenly
there is a infections that happens; what is
happening that you have this chemostat.
And, then it is coming in and going out S
naught is 100 and it will going out at 80.
So, organism 1; this is organism 1 x 1 is
it correct organism 1 growing, organism 2
is growing, organism 2 is growing. At an instant
when there steady state is establish organism
1 infects the process this what it says is
the problem statement clear to all of you.
Your chemostat is running S naught equal to
100 and 80 is coming out therefore x 2 is
growing. So, you running a process in which
x 2 is growing and when they x 2 growth is
a steady state suddenly a infection occurs
ok.
Infection means what somewhere from where
this x 1 has coming some delta value of x
1 has coming; our question is what will happened
to the process; is this question clear to
all of us? The growth rates of both the organisms
all those data is given here mu m is given
k s everything is given here; is this clear?
So, let us say suppose you make a plot of
mu is a growth rate o verses. And, let us
say this is what this is organism 2 is it
right; organism 2 it is growing. Now, it says
organism 1 is only 0.5 S by 50 plus S this
is mu 1 correct and this 0.3 S divided by
30 plus S this is mu 2 is that clear; 10 is
it this is 10 is it ok.
Now, tell me I mean what is our understanding
of chemostats? Our understanding of chemostat
is a D equal to mu; that means when steady
state is establish when steady state is establish
D was equal to mu or in other words mu 2 was
equal to D when steady state was establish;
do we agree? What is mu 2 do we know mu 2
80 grams per liter is given. So, if you plot
mu 2 verses S let us say this is a curve this
is 80 correct. And, then corresponding to
that what is the we can find out yes or no;
is this clear to all of us?
Now, we say it is there is a infection due
to organism ` we should plot this I plotted
that I will show you in a minute; that curve
looks something like this, this is organism
1 you will plot shortly. So, this is how organism
1 looks like. Now, what will happen you please
tell me we looking at this curve; what will
happen is this question clear to all of us?
We have a steady state of organism 2 what
is mu equal to can you please calculate and tell me; how much is it? 0.26.
0.26. So, it is running at 0.26 per hour is
the dilution rate at which is working. Hence,
that instant organism 1 has come in. Now,
what is this specific growth rate of organism
1corresponding to 80 it is higher than 2 is
it not; it somewhere here yes or no organism
1 it will grow faster than 2. So, it is here
so what will happen in the process substrate
it organism 1 will compete for the same substrate;
both will start eating what is our steady
state d must be equal to mu is our steady
state correct d must be equal to mu is our
steady state. So, what will happened to the
process now? D will become; what is D? D is
F by V does D change our wall setting is not
change correct. So, F by V has not change
when D is not changed; if D is not changed
what will happened to the process? Organism
1 it will grow correct what happens to 2 it
will reduce; then what eventually what will
happened process? Do 2 will get washed out;
you can see here that S will keep coming down
and then this will be the steady state value
of s is it not it. In the steady state value
S organism 2 will be knocked out organism;
well 1 is would have would be growing and
the value of S will reduce from 80 to some
lower value.
See, I plotted at here so you can see here
it comes in a 80 to about 54 some some smaller
value; is that clear is the phenomena clear
what I am trying to bring out see natural
selection happens every minute in the natural
environment. We are talking about bio diversity
and all that today but the importance is write
here ok that simply by adjusting your dilution
rate you have selected a different organism.
For example, it is important to realize and
if you go to any you know nay river which
is starting; as you can see as the velocity
of the river changes the ecology changes;
the whole ecology changes because dilution
rates so different. And, therefore the organism
that can sustain will be also keep on changing.
And, this is a illustration of that; is that
clear to all of you? Why is that organism
1 this is organism this is organism 2 is knocked
out and organism 1 starts to grow you can
see here yeah it is. Finally, at steady state
there is only 1 organism which is organism
2 is knocked out and organism 1starts to growing.
Steady state there is only 1 organism; we
having a chemostat in which our dilution rate
is fixed. Now, organism this infection has
taken place that infection has taken place.
Therefore, eventually what would happened
eventually we said D must be equal to mu;
because that is a representation of steady
state. And, our this graph says there is only
1 point of intersection correct; at that point
of intersection D can only be equal to mu
1correct and mu 2 has to get out organism
2 have no chance of surviving in that environment. What am I saying is that I am running my process
D equal to 0.266; under that environment when
this infection accurse organism 2 will have
to get out; is this clear, is this clear to
all of you? Can now see how in the natural
environment this happens at every minute.
So, that is why you see the organism that
is populating an environment really depends
upon the share rate of the environment. See,
if you look at just yourself diarrhea for
example, what is diarrhea; what is our what
is our system doing? It is increasing the
dilution rate to knock out the organism; diarrhea
is an example where it is it is a phenomena
of increasing the dilution rate. So, there
is knocks out the organisms; it is a natural
response to getting rate of the organism;
there is what happens in the natural environment
is this clear, is that phenomena clear to
all of you? So, you can repeat this there is second part
is same exercise is repeated saying that if
on this side. For example, what I am trying
to put a cross is different organism; for
example in the second in this exercise what
I said is there organism 1 growing at 30 not
80 at 30. And, suddenly this infection accurse
here the selection is for the opposite organism;
you know we can see here here when you are
starting a 30, 30 see here you can see here
that the organism what is what is selected
is organism 2; 1 is knocked out. The important
point here is to recognize that the environment
determines if this substrate concentration
changes at different organism is selected;
see the way nature is organized you see this
it is this organization that we must understand
properly to understand the importance of diversity.
Natural environment is much bigger what I
am saying is much bigger. So, this phenomena
explains how things happen in the natural
world. Now, the question that we want to answer
is how long does it take for the steady state
to get establish? We know that steady state
one organism get knocked out but how long
does it take; can we tell how long does it
take, can we write the differential equations
to tell how long does it take? Let us write
the differential equation you can solve it
at home; let us write the differential equation.
So, I will write for organism 1; first one
is organism 2 is it not it I will write for
2 is it all right yes or no? Why have I knock
this out; there is no the infection is momentary,
it has infection is not coming continuously
it is momentary. So, let me write for the
there are 2 ways infection can occur; one
that you continuously put in the infection
let me write I will come back in a minute.
I am taking Y to be the same for both; is
this ok or material balance; this is our material
balance this is what you have written material
balance is correct; these we have knocked
out this terms. Now, in problem this in 2.1
what are the initial conditions, what is the
initial condition for this problem? It is
at steady state that means steady state value
is what 80 grams per liter is the value of
steady state. That means, at this point when
this happens S naught is 80 what is x 2 and
it going out at 80. I am talking about question
2.1 is coming at hundred going out at 80.
So, what is a cell mass? 100 minus of 80 is
20 you will coefficient this point five 10
grams per liter is cell mass ok ; yes or no
do you all agree is 10 grams per liter 10
mass. Now, at this instant of time this is
0 time an infection occurs what is the value
of x 1? Let us say some delta some some 0.001
grams. So, much has entered so that the density
has become 0.001 grams per liter x 1; so much
of x 1 has entered is this clear? It is a
momentarily now can you solve this problem
now; initial condition is fully specified;
is it clear what we are saying can you solve
this differential equation? because initial
conditions fully specified x 2 is known what
is that not x this is S is 80 S naught is
always 100 is it clear S naught is 100, S
is 80 at the time of infection; x 1 became
0 0 1.
Now, instead of 0 0 1 you can put 0.1 you
can vary that number when I say how long does
it takes it depends upon the magnitude the
inflection; if it is small it will take a
long time if it is large the steady state
the new steady state would be reached quickly;
is this clear? Can you solve this problem
and find out how long it takes for the new
steady state to be achieved yes or no? So,
similarly for q 2.2 also you can do the same
thing what is the initial condition for 2.2.
Our question initial condition for 2.2 S naught
is this is 2.2 S naught is 100, S is 30 correct.
And, what is x 1 is 100 minus of 30 70 is
it is 35 grams per liter. And, x 2 is let
us say point 0 0 1 gram per liter this is
the infection ok. So, this is t equal to 0;
when the infection is just occurred. Similarly,
this is t equal to 0 when the infection is
just occurred 0 plus. So, you can solve forward
march is it can we solve can we finish this
problem now; is it is it yes or no? Now, the
question is I want to ask you is this is the
most important question.
So, how do we use the above phenomena to manage
the infections in a process; after all infections
are something that happens to us it happens
to a process if it is a biological process;
how do we manage these infections? What is
that you and I have learned having known this
phenomena; we have learn this phenomena now
showing that if you change the dilution rate;
you can knock out the organisms. Essentially
what we are saying is that if there is infection
you have to change the dilution rate in an
appropriate direction to knock out the organism
correct. So, we have learned that now that
if you want to knock out an organism you change
your dilution rate. So, that particular organism
just knocked out. In fact industry what they
do they do lot of washing it is essentially
do it this only correct; it is increasing
the dilution rate to knock out the organism.
So, if you are talking about the pure culture
you know fermentations; if you want to knock
out the organism that is the infected the
process you will have to change the dilution
rate. And, that all that substrate will have
to go into your waste treatment plant; you
see this is the problem of biology that you
have to except huge amount of waste that has
to going to waste treatment plant. But that
is something that you have to except there
is an infection or I will stop that.
