In 
the last class we started about the biological
process for water and wastewater treatment
and we have seen that the selection of the
treatment processes is based upon the nature
of the pollutant present in the
water either it is biological or physiochemical
processes because when we talk about wastewater
treatment why we are going for biological
processes is the majority of the solids are
organic in nature so it is very
easy to remove the organic matter by the help
of microorganisms. And we were discussing
about what are the different types of reaction
or how can we classify the reactions based
upon the kinetics. We have
seen what is the zero order reaction, first
order reaction and second order reaction.
We also discussed that in biological processes
we assume that most of the reactions are of
first order though many times it may be deviating.
We have also discussed about what are the
different reactor
configurations available. We discussed about
what all are the advantages of plug flow reactors
and CSTR reactor and what CMBR reactor is.
It seems that plug flow reactor needs less
volume compared to a
CSTR reactor if you want to achieve the same
degree of treatment or same degree of removal.
Today we will discuss about the biological
process in detail or we will be starting on
activated sludge process. So,
coming to the secondary treatment we have
already discussed what is the primary treatment.
The primary treatment is based on physical
unit operation. That means the gravity force
is the prominent force there so it will be
removing the solids which will be having high
density or high particle size. We
have also discussed about primary sedimentation
tanks usually employed in wastewater treatment
and it is working on type two settling. We
have already discussed the design procedure
when we were talking
about sedimentation tank.
In wastewater treatment the primary sedimentation
tank removes around 50 to 60% of the suspended
solids. But coming to the soluble BOD the
removal is almost negligible. What we do with
this sludge? This
sludge whatever is collected in the primary
sedimentation tank it is of organic nature
or most of them are organic matter so we cannot
directly dispose them off so usually what
we do this sludge along with the
secondary sedimentation tank sludge and send
it to sludge digesters where the sludge treatment
or sludge stabilization will be taking place
under anaerobic conditions and as a result
we will be getting biogas or
carbon dioxide and methane and stabilized
sludge. This stabilize sludge can be put into
sludge drying beds afterwards can be used
manure or you can use it for land filling
also. So these are the typical values of a
primary sedimentation tank which is commonly
used in wastewater treatment processes.
The detention time varies from 1.5 to 2.5.
As we discussed earlier if the detention time
is higher then whatever is the organic matter
that is settled in the bottom of the tank
starts degrading anaerobically and that
generates gas that gas will be creating turbulence
in the sedimentation tank and it will be affecting
the efficiency of the tank. and our flow rate
varies from 30 to 48 and wier loading is in
the range of 125 to 500 and
these the dimensions of the tank depth of
the tank varies from 3 to 5 meters, length
from 15 to 90 and width 3 to 24 meter and
typical value is 6 to 10 meters and sludge
process speed is around 0.6 to 1.2 meter
per minute and if it is a circular tank the
depth is the same of that of rectangular tank
that is 3 to 5m and diameter 3.6 to 60 meters
and typical value varies from 12 to 45 meters
and bottom slope is 60 to 160 mm
per meter and typical value is 80. So this
is the primary sedimentation tank designed
parameters. It is almost similar to the primary
sedimentation tank of water treatment but
here we are assuming it is type two
settling so we cannot use stock's law for
the design of the sedimentation tank of wastewater
treatment.
Earlier we talked about many secondary treatments
and we have also seen that in certain secondary
treatment units they do not require a primary
treatment at all. for example, oxidation pond
and aerated lagoons
because that system itself will be acting
as a primary unit so everything will be settling
down in the bottom of the tank and we are
giving such a large detention time so the
particles whatever is coming either it is
suspended or it is inert material everything
will be getting settled and that itself it
will be acting as primary and secondary treatment
unit. Now we will see in detail what this
secondary treatment is. Each unit we will
be seeing in detail.
Secondary treatment in domestic wastewater:
mostly biological treatment is being employed.
The reason is the most of the solids are organic
in nature so biological treatment is the most
cost effective one and the
organic matter whatever is present in the
wastewater is highly degradable organic matter
or we take the BOD COD ratio the ratio is
very very high so we can go for biological
treatment. When we talk about the
biological treatment we can classify the biological
treatment into two different categories; one
is suspended growth system and another one
is attached growth system.
In suspended growth system the entire biomass
is in suspension. For example, activated sludge
process is a suspended growth system and another
suspended growth systems are aerated lagoons,
oxidation
ponds and nitrification and denitrification
process. And coming to attached system tricking
filters is an example and another one is rotation
biological contactors.
Anyway we will be discussing each of these
units in detail as what the design parameters
are, how to design, what are the important
factors we have to consider etc. So before
going the biological treatment
systems we should know what are the nutrients
requirements or the microorganisms, what are
the environmental conditions conducive for
the microorganisms and how the metabolism
of microbes are taking
place and what is the relationship between
the microbial growth and substrate utilization
because in any biological treatment process
when we talk the microorganisms are the workers
so we have to create
conducive environments for the workers otherwise
they will not be doing the work properly so
our treatment system will not be performing
better.
First we will discuss about the nutrients.
We have seen the microbial growth curve and
most of the biokinetic parameters in the previous
lectures. So what is the purpose of these
nutrients?
The nutrients are essential for supplying
material for the creation of cell material.
That is very very important. From where they
will be getting the carbon molecule for the
creation of the cell materials, for this
purpose nutrients are required. The microorganisms
require a lot of energy so this energy source
is also from the nutrients and the last one
is how the microorganisms are generating the
energy. They are
generating the energy by a series of biochemical
reactions. Most of them are oxidation reduction
reactions.
If the oxidation reduction reaction has to
take place there should be one electron donor
and electron acceptor. Unless the electron
donor and electron acceptor are there oxidation
reduction reactions are not
possible. Therefore, in an aerobic system
the organic matter will be acting as the electron
donor so the organic matter will be giving
the electrons. So that will be passing through
a series of steps and finally the
electrons should be accepted from the system.
Now, who will be accepting the electrons?
In some cases the nutrients itself will be
accepting the electrons or sometimes it will
be oxygen in an aerobic system.
These are the essential things required for
the microorganisms.
Now, coming to the environmental parameters
the pH temperature etc should be conducive.
We know that microorganisms are not able to
survive either at low pH or at high ph. when
we were discussing about
disinfection we have seen that if the pH is
reduced below 3 or if you increase the pH
above 10 then most of the microorganisms will
die. When you talk about the biological system
we should see that the
microorganisms or the ph conditions in the
system are within this limit and we should
be able to provide the optimum condition.
The optimum condition is most of the time
it will be near the neutral pH.
Coming to the temperature we all know that
the enzymes or the proteins whatever is present
in the cell of microorganisms will be getting
denatured at high temperature in most of the
cases. But there are
microorganisms which can survive very high
temperature and very low temperature. But
most of the microorganisms whatever we are
employing for this wastewater treatment they
come under mesophilic
temperature range or mesophilic microorganisms.
That means their optimum temperature is around
35 to 37 centigrade. So if you reduce the
temperature or increase the temperature the
microbial activity will be
reducing.
It was shown that for every ten degree change
in temperature the activity will be varying
100%. So we can see that how the environmental
condition is going to affect the microorganism.
Now coming back to the
nutrients itself depending upon the nutrients
selectivity we can classify the microorganisms
into various categories. The nutrients will
be acting as a energy source, as an electron
acceptor and as a carbon source.
First we will see the carbon source.
The microorganisms can use either organic
matter as the carbon source or inorganic materials
like carbon dioxide or bicarbonate as the
carbon source. The organisms which are using
inorganic carbon as their
carbon source for cell synthesis are known
as autotrophic microorganisms. Once again
the microorganisms which use inorganic carbon
for their cell synthesis is known as autotrophic
microorganisms. The
microorganisms which use organic matter or
which takes carbon from organic matter are
known as heterotrophic microorganisms.
Now, coming to the energy source they can
derive energy from sunlight, they can derive
energy from chemical reactions. The microorganisms
which derive energy from sunlight is known
as phototrophs and the
microorganisms which derive energy from the
chemical reactions are known as chemotrophs.
In chemotrophs itself the microorganisms which
use organic matter is known as chemoorganotrophs
or
chemoheterotrophs. And the microorganisms
which require chemical reactions for the energy
but they use inorganic components for this
reactions are known as chemoautotrophs.
We have seen different types of microorganisms;
one is autotroph and another one is heterotrophs.
And depending upon the energy requirements
we can see that they are phototrophs, chemoautotrophs
and
chemoheterotrophs. Phototrophs are the ones
which generate energy from photosynthesis.
Just like plants there are many microorganisms
which can get energy from the photosynthesis
reactions. autotrophs the
ones which use inorganic carbon for their
cell synthesis and heterotrophs the ones which
use organic matter for their cell synthesis.
Now, coming to the donor acceptor we have
seen that for any oxidation reduction reaction
there should be one electron donor and one
electron acceptor. So, when we talk about
wastewater treatment in most of
the cases the electron donor is in the wastewater
itself or the organic matter present in the
wastewater but who is the electron acceptor.
The electron acceptor can be either oxygen
or inorganic compounds
containing oxygen like nitrate, sulphate etc
so they can also act as electron acceptors
and get reduced. The third one is the organic
matter itself can act as electron acceptor.
So depending upon the electron
acceptors we can divide the processes into
different categories those are aerobic process
anoxic process and anaerobic process in aerobic
process oxygen molecular oxygen is the final
electron acceptor.
Hence, organic matter releases the electrons
and it will be traveling through a series
of reactions and because of this oxidation
reactions a lot energy will be released that
will be utilized for the metabolisms
purpose of the cell and finally electrons
will be accepted by molecular oxygen and it
will be getting reduced to water. But in certain
cases when oxygen is not available inorganic
compounds like nitrate, sulphate
etc can be accepted by the electrons. For
example, if you talk about nitrification,
denitrification what is happening nitrate
is getting converted to nitrogen this is very
essential when we talk about high ammonia
containing wastewater because there is a limit
for nitrate discharge to the receiving water
so if you have excess nitrate how we can remove
that one.
We can use biological process for the nitrate
removal that process is known as dinitrification.
Denitrification is an example of anoxic process.
Here no molecular oxygen will be available
so nitrate will be acting as
the electron acceptor. Once the compounds
accept the electron it will be getting reduced
so nitrate will be getting reduced to nitrogen
gas. Those processes are coming under anoxic
category. Now we will see
what is antrophic? Anaerobic means either
oxygen or inorganic compounds which is having
oxygen are not present. So organic matter
themselves will be acting as the electron
acceptor.
For example, in anaerobic process of wastewater
the organic matter itself will be acting as
the electron acceptor and gets reduced to
methane gas. That is why we are getting bio
gas or methane gas from
anaerobic processes. These are the various
classifications. These classifications are
very very important because when we select
the microorganisms for a particular treatment
we should know what type of
microorganisms we want and what type of microorganisms
can be used as a particular pollutant and
converted into environmental friendly byproduct.
Hence, the important terms one has to remember
is; what is an autotroph, what is heterotroph,
what are phototrophs, what are chemoautotrophs
and chemoheterotrophs and what is the difference
between
aerobic process and anoxic process and anaerobic
process.
Now we will come back to our activated sludge
process.
This is the most commonly used biological
process for domestic wastewater treatment.
Here what is happening is the wastewater is
coming and we are aerating the wastewater.
The wastewater will be having a lot
of microorganisms so when we supply sufficient
quantity of oxygen the microorganisms will
be utilizing the organic matter present in
the wastewater and they will be converting
it into carbon dioxide, water and
more and more cells. That is what is happening
in activated sludge aeration tank.
In aeration tank the dissolve form of organic
matter is getting converted into colloidal
form of organic matter or new cells so all
these cells are discrete in nature. So if
you let them out like that there will not
be any
COD or BOD removal as such because we are
synthesizing more cells into the system. Thus,
the secondary sedimentation tank of the activated
sludge processes is very very important. And
if you see the
aeration tank of the activated sludge process
what is the nature of the microorganism present
in that?
It is a mixed population. So that mix population
of microorganisms will be utilizing the organic
matter whatever is present in the wastewater
and more and more cells are generated and
it is coming to the
secondary sedimentation tank. But here we
have to separate the cells. Therefore, how
can we separate the discrete cells, it will
be very very difficult. It is seen that when
the microorganism reaches a particular
growth stage
We have discussed about the growth curves.
Are you remembering? We have seen the lag
phase, low growth phase, stationary phase
and endogenous or declining growth phase.
Therefore it was found that if the
microorganisms reach the end of that stationary
phase or if they start coming to the endogenous
phase they start producing some extra cellular
polymers. These exo-cellular polymers are
having some binding
properties. So because of this binding property
all these colloidal cells will aggregate and
they will form flocs. So once they form the
floc the settling will be very very easy.
Hence, in the sedimentation tank the flocs
will be settling and we get clear effluent
that can be removed. We also have to treat
this sludge carefully because this sludge
is containing lot of active biomass so we
cannot just discharge it. Some sludge will
be recycled to the aeration tank and remaining
will be sent to the sludge digesters. What
is sludge and why it is called as activated
sludge process?
This sludge is nothing but microorganisms
having high activity. The reason why it is
known as activated sludge is because the sludge
has an activity. This is the reaction taking
place in the aeration tank; organic
matter plus active sludge or microorganisms
plus dissolved oxygen gives stable compounds
acceptable to biosphere plus increased concentration
of biomass. The stable compounds can be carbon
dioxide,
water, nitrate phosphate depending upon the
compound or depending upon the nature of the
wastewater.
What is happening in the aeration tank is
the soluble organic matter coming to the secondary
treatment unit is converted to settable organic
matter. So if you take the aeration tank alone
we cannot call it as a
treatment system because what is happening
is only the nature of the solid is getting
converted from the soluble form to colloidal
form. So if you want to get removal of the
solids we have to have the
sedimentation tank. The sedimentation tank
or secondary sedimentation tank is an integral
part of activated sludge process.
We have already seen that the settling tank
the secondary sedimentation tank has to be
designed based upon zone settling because
the particles are flocculant in nature and
it is concentrated suspension. So
concentration suspension with flocculant particle
is coming under zone settling and towards
the bottom of the sedimentation tank the concentration
of the particles is so high so they will be
experiencing the
compression settling.
Now we will talk about the ecology of the
activated sludge process. we all know that
it is not a pure culture system, not only
one single type of microorganisms is present
but there are multiple groups of
microorganisms present which are responsible
for the conversion of organic matter into
carbon dioxide and water and more cells. And
if you see the ecology of activated sludge
process we can see that mostly it
is bacteria which are converting the organic
matter into carbon dioxide and water. But
there are protozoa or rotifiers and fungi
present in the activated sludge process.
If you see it is a heterogenous microbial
mixer mainly pseudomonas and zooglea are present.
Both these microorganisms are heterotrophic
in nature. Heterotrophic means they use organic
matter for the cell
synthesis and the energy is derived by oxidation
reduction reaction and here also organic matter
is being used. But in activated sludge process
also there can be autotrophs for example nitrosomomonas
and
nitrobacter. These microorganisms are coming
into the picture whenever we talk about ammonia
oxidation because we know that protein contains
lot of amino acids so naturally the wastewater
will be containing
lot of organic nitrogen.
So, if you aerate the wastewater for long
time, first all the organic matter that will
be getting oxidized and afterwards this organic
nitrogen whatever is present in the wastewater
will be getting converted to
ammonia and that ammonia will be further oxidized
to nitrite and nitrate. So these nitrosomomonas
nitrobacter are responsible for oxidation
of ammonia to nitrite and nitrate. These are
autotrophs; that means they
use only inorganic carbon for their cell synthesis
and their energy. The energy also generated
from chemical reactions but inorganic substances.
Coming to this protozoa, rotifiers and fungi
we have seen that mainly microorganisms or
bacteria are responsible for the removal of
organic matter whatever is present in the
activated sludge present in the
wastewater. But what is the role of this protozoa
rotifiers and fungi?
Protozoas are very very important. They are
known as polishers or scavengers of activated
sludge process. What they do is, this protozoa
are predators of bacteria. In the activated
sludge process we have seen
that the floc formation is there so once the
floc is formed it is very easy to settle the
sludge. But there will be bacteria which are
present in the liquid as discrete particles.
so if the discrete particles are coming to
the sedimentation tank it will not be settling
down so those bacteria will be coming out
along with the effluent so naturally when
there is solid present in the effluent that
too organic solids the BOD and cod of the
effluent will be very high so naturally you
will not be getting the efficiency of whatever
you require to meet the effluent discharge
standard. So what this protozoas does is they
will be eating away the discrete
microorganisms or discrete bacteria whatever
is present in the system so in a way it is
polishing the effluent what is coming out
of the secondary sedimentation tank that is
why those are known as polishers or
scavengers.
Protozoa present here are also an advantageous
thing for activated sludge process to increase
the efficiency.
Now will see the microorganisms are utilizing
the organic matter present in the system.
We are aerating the system and lot of organic
matter either in the colloidal form or dissolve
form is present in the water. We
have also seen that the bacteria are using
the organic matter. So how they are using
the organic matter, what are the steps involved
in this one. It is basically a four step process.
The first one is adsorption of soluble and
colloidal substrate on floc surface. Whatever
is present in the system has to get adsorbed
to the microorganism. Then after the adsorption
the organic material will be
getting transported from outside the cell
to the inside of the cell. It has to pass
through the cell wall and the cell membrane.
If metabolism has to take place it has to
reach the interior of the cell. So once it
reaches
the transport area then metabolism will be
taking place. That is the third process and
the fourth one is solid liquid separation.
We have seen that the wastewater will be containing
dissolved solids as well as colloidal solids.
But if the colloidal solids get attached to
the cell it will be bigger in size so how
they can penetrate into the cell wall,
it will be very very difficult for them to
penetrate into the cell wall. So how the cell
is able to handle these colloidal particles?
It is very interesting. What they do is the
cell will be creating some sort of an enzyme
and that will be hydrolyzing the organic matter
which is in colloidal form. So once it is
hydrolyzed, the hydrolysis is nothing
but adding water molecule to the organic compound
and because of that one it will be splitting
into small parts. Now, once the hydrolysis
is done, the molecules will be smaller in
size so they can easily penetrate
into the cell. Once it penetrates into the
cell metabolism will be taking place depending
upon the condition whether it is aerobic,
anaerobic or anoxic process. Then because
of the metabolism energy will be
generated and more and more cells will be
synthesized.
Therefore, in your system the organic matter
whatever is present in the dissolved form
or colloidal form will be getting converted
to new cells. So the last step is solid liquid
separation because if you allow the
cells to go out of the system the treatment
will not be proper. These are the four steps
that usually takes place;
• Adsorption of soluble and colloidal substrate
on floc surface
• Transport of substrate across the cell
membrane
• Metabolism in the cell and
• Solid liquid separation process
We have already discussed about the agglomeration
of bacteria. Agglomeration of bacteria is
possible because of biopolymers. The biopolymers
secretion depends upon the nature of the system.
If all the
microorganisms are in the low growth phase
they will not be generating any of this biopolymers
so it is very very difficult to separate the
microorganisms from the liquid. But if they
are in stationary phase, the
end of the stationary phase of endogenous
phase the biopolymer secretion will be very
high so it is easy to separate the microorganisms
or flocculent sludge.
This is the typical activated sludge process.
Wastewater is coming here and this is the
aeration tank. Here what is happening is we
are giving lot of oxygen, so under aerobic
condition the organic matter is getting
converted to carbon dioxide water and new
cells and from the aeration tank entire liquid
is going to the secondary sedimentation tank.
So here settling will be taking place because
excess sludge will be generated
and here we will be getting clear effluent
and a portion of the sludge will be wasted
that will be going to anaerobic sludge digester
and a portion will be recycled to the aeration
tank to maintain a constant biomass
concentration in the system. We have to provide
more and more workers then only the process
will be efficient. So the purpose of recycling
activated sludge process is to increase the
biomass concentration in
the aeration tank.
Now we will see how we can design an activated
sludge process. You know what is the amount
of wastewater coming to the system and what
is the amount of ultimate BOD of that wastewater.
Whenever we talk
about biological wastewater treatment processes
we are interested only in the BOD because
BOD gives us the amount of organic matter
that is biodegradable. But if you talk about
COD it will be giving you all
the compounds which is present in the system
which is either biodegradable or non biodegradable.
Therefore if you want to decide whether we
have to go for biological processes or physiochemical
processes what we have to do is first find
out the BOD of the system BOD of the wastewater
and cod of the
wastewater and see what is the BOD by COD
ratio. If BOD by COD ratio is more than 0.5
we can go for biological processes. That means
whatever is the total organic matter present
in that system above 50%
is biodegradable so naturally we can go for
a biological system followed by some other
treatment.
When we talk about the domestic wastewater
this BOD by COD ratio varies from 0.6 to 0.8
depending upon the wastewater. That means
it is highly biodegradable. We will discuss
about the design activated
sludge process based upon kinetic principles.
We have seen what all are the biokinetic parameters
of microorganisms so based upon that one we
can design the activated sludge process. So,
if you want to make a mass balance of liquids,
solids and biomass
we can write it as something like this, this
is Q this is the quantity of wastewater coming
into the system and So is the substrate concentration
or ultimate BOD of the wastewater and we are
assuming that the
microorganism concentration present in the
incoming wastewater is very negligible compared
to the biomass concentration in the system
that is why we are not giving any X value
here and this is our aeration tank
the volume of the aeration tank is Va and
biomass concentration is X and substrate concentration
or BOD concentration is Sc. this aeration
tank we are assuming that it is a CSTR reactor
continuous flow tired
tank reactor.
What is happening here is, as soon as this
So enters in the system instantaneously the
concentration will be changing to Se as what
we have seen yesterday, and the same thing
happens here. And here the flow is
1 plus r into Q and we will be having X and
Se our substrate concentration and biomass
concentration and this is the secondary sedimentation
tank and here we will be having a sludge wastage
of Qw.
So if you want to see here this is Q minus
Qw because Qw will be wasting here, I forgot
to put it here and Qq minus Qw is the flow
coming out and Sc is the substrate concentration
on BOD and Xe is the
biomass concentration and a portion of sludge
will be recycled to the aeration , Rq is the
flow rate, R is the recycling ratio and Se
is the BOD because same Se will be coming
and biomass concentration is Xr
because it is the same sedimentation tank
so the biomass will be getting concentrated
and we are recycling the concentrated sludge.
Always we use this X to the present MLSS.
MLSS is nothing but mix liquid suspended solids.
So if you have the mixed liquid and if you
find out the suspended solid concentration
in that liquid that is known as
MLSS. So we assume that mainly that is biomass.
Or another term is there that is mLVSS mixed
liquid volatile suspended solids. Sometimes
along with the wastewater along with organic
solids the inorganic
solids also will be there so what will happen
is this MLSS will be containing organic and
inorganic solids. But if you want to find
out what is exactly the microbial concentration
then we have to go for mLVSS.
Many studies have shown that the ratio of
MLSS and mLVSS is like this; if you have the
MLSS value, if you multiply by point eight
that is mLVSS or for the total MLSS around
80% is mLVSS that means
volatile suspended solids are fresh cells.
And when we talk about the activated sludge
process another important term is hydraulic
retention time. The amount of time the liquid
is staying in the treatment system is
the hydraulic retention time. That can be
calculated by V by Q.
Therefore, v by Q is the hydraulic retention
time and another important time is Biological
Sludge Retention Time that is BSRT. BSRT is
very very important in any biological treatment
system which is nothing
but biological sludge retention time. The
definition is average time for which a unit
of biomass remains in the system. The biological
retention time is nothing but the total time
the biomass staying in the system
once the biomass enters the system or is generated
in the system.
The BSRT is small in the system. What does
it mean? It means the biomass concentration
is the system is more because more and biomass
is getting generated and it is staying in
the system for a longer time that
is BSRT. It is represented by thetaC and most
of the design purposes this is the most important
criteria thetaC which tells us what is the
biological sludge retention time we are providing
in the system.
This is the mathematical expression of thetaC.
That means what is the time the biomass is
staying in the system. So how can we find
out this one? It is nothing but the total
biomass concentration divided by the
biomass wasted from the system daily. Here
XT is the total active biomass in the system
and dx by dt is the total quantity of active
biomass withdrawn from the system daily so
we will be getting thetaC in terms
of days.
When dx by dt is equal to 0 that means if
you are not withdrawing any sludge from the
system. That means whatever sludge is generated
entire sludge is staying inside the system
so what does it mean? It means
that thetaC or the biological sludge retention
time of that system is infinity because we
know that thetaC is nothing but total X divided
by dx by dt and when the term dx by dt becomes
zero naturally thetaC
become infinity. The process modification
of activated sludge that is extended aeration
system is working on this principle and there,
there will not be any sludge wastage or sludge
recycling theoretically so there
the thetaC will be infinity and all the sludge
whatever is generated in the system will be
undergoing auto oxidation or they will be
in the endogenous respiration phase and will
be auto oxidized.
Therefore, if you want to find out the thetaC
of an activated sludge unit we can find out
like this; X into Va this will give you the
total biomass present in the system that means
in the aeration tank so Va is the
volume of the aeration tank and X is the biomass
concentration per liter or per meter cube.
Now, what is the biomass wasted from the system,
it is nothing but Qw into X where Qw is the
quantity wasted into
what is the concentration of biomass in the
wasted liquid plus Q minus Qw into Xe this
is coming out along with the effluent from
the secondary treatment plant. So it is nothing
but one by mu.
Here we have seen what is mu which is the
specific growth rate. The specific growth
rate is nothing but the growth whatever is
taking place per unit biomass or dx by dt
by X so this is nothing but X by dx by dt
so thetaC is nothing but one by mu. Hence,
if you know the specific growth rate of the
system we can find out what is the thetaC
of the system. We have seen that this is the
most important design parameter of
activated sludge process or any biological
process where we get to know, for how much
time we can retain the sludge in the system.
Hence, if you want to change the thetaC to
get different efficiencies or to modify your
system we have to change the design parameter
thetaC. How can we change the thetaC without
affecting the volume of the
time? Once we design the tank the volume will
be fixed. So if you do not want to change
the volume how can we change the thetaC?
here as per the equation; thetaC is nothing
but x into Va by Qw into X plus Q minus Qw
into Xe so if you change the Qw without changing
either x or Va we can change thetaC and another
one is HRT Hydraulic
Retention Time that v by Q and thetaC is the
design parameter this is very very important.
Now we can see how to find out the thetaC
if you know what is the thetaC we can provide
and what is the influent BOD and effluent
BOD required. We know what is the treatment
required, it should meet the
discharge standard. So in terms of BOD it
is 30 milligrams per liter. So if you want
to find out the volume of the tank how we
can go around?
Therefore we will make the material balance
equation with respect to biomass. So if you
want to take the material balance how can
we do it? The net rate of change in the amount
of biomass within the system is
equal to rate of biomass production in the
system minus rate at which biomass is leaving
the system plus rate of biomass entering with
influent. This is the biomass concentration.
This is the activated sludge
process; we are taking the biomass concentration
in the aeration tank so we have to take the
net rate of change of biomass concentration
here that is equal to the biomass generated
here and the biomass leaving
the system and the biomass entering the system.
If you take all those three into account we
can get the mass balance that is what we have
done here.
So if want to write the mathematical expression
for that one it is like this; Va this is the
volume of the aeration tank into dx by dt
changing microbial concentration net change
that is equal to dx by dt growth into
volume of the tank that means whatever is
generated in the system minus Qw into X plus
Q minus Qw into Xe this is the entire biomass
leaving the system. This is because of the
sludge wastage and this is going
out through the effluent from the secondary
sedimentation tank and this zero is coming
because we are assuming that whatever biomass
is coming to the system is zero. We are assuming
that the biomass
concentration in the wastewater is very very
negligible compared to the biomass concentration
in the aeration tank.
When we discussed about the biokinetic parameters
we have seen that dx by dt by X that is the
specific growth rate so if you multiply that
one with X we will be getting dx by dt growth
equal to YT into dx by dt
growth minus kd into X. that means whatever
is the yield coefficient into dx by dt minus
kd which is the endogenous decay constant
kd into whatever is the biomass present initially.
Therefore, if you substitute
this one for dx by dt in this equation the
final equation will be; Va into dx by dt net
equal to y t into dx by dt g minus kd into
X into volume because volume is coming in
common minus this term which we can
find out from the thetaC equation thetaC is
nothing but x into Va divided by Qw into X
plus Q minus Qw into Xc. So we can write this
term as x into Va by thetaC.
We also know that at steady state the net
biomass generation is equal to zero. Whatever
is generated is leaving from the system or
at steady state Va into dx by dt net is equal
to zero so finally this entire term will
become zero and we will be having zero is
equal to YT into dx by dt g minus k into kd
into X into Va minus x into Va by thetaC or
x is equal to thetaC into YT into Q into So
minus Se by volume of the aeration
tank into 1 plus kt thetaC so we can find
out what is the biomass required for the particular
treatment efficiency.
So minus Se is the substrate removal from
the system that means this is the efficiency
of treatment. For example, if you have a BOD
of 300 initially and se is 20 milligram per
liter So minus Se is the one which is
getting removed from the system and kd we
will be knowing and YT will be knowing because
these are the biokinetic parameters and thetaC
is nothing but 1 by mu. So X we can find out
using this formula and
similarly we can develop an equation for substrate
concentration. We know that mu is equal to
one by thetaC which is equal to YTQ minus
kd or and we know what is the definition for
Q, Q is nothing but
specific substrate utilization rate, what
is the substrate utilized by unit biomass
in the system so Q is equal to ds by dt by
X where dx by dt is the rate of change of
substrate concentration in the system or Q
is
equal to Qmax Se by ks plus Se this is the
maximum substrate utilization rate and this
is the half saturation constant. We have discussed
all these in detail earlier.
Or we can write like this; ds by dt, what
is the rate of change of substrate concentration
in the system that is equal to Qmax into Se
whatever is the substrate concentration present
in the system by ks plus Se into
X. But in activated sludge process what we
assume is the substrate is limiting. That
means as soon as it enters into the system
So will be getting converted to se because
it is the CSTR reactor so the biomass the
substrate or the food available for the biomass
is always under limiting condition. In other
words, compared to this ks value this Se value
will be very very negligible, Se is very very
less compared to ks value. So
how can we write this equation? We can write
it like this.
It is; ds by dt equal to Qmax by ks, we have
neglected Se in the denominator into Se into
X or we can replace this Qmax and ks by a
single constant small k because Qmax is a
constant for the system and ks is
also a constant for a system so two constants
will give another constant so that is why
we are replacing that one with another constant
k. So ds by dt is equal to k into Se into
X where k is Qmax by ks. In other
words we know what Q is in an activated sludge
process. What is ds by dt? It is Q into So
minus Se that is the total substrate concentration
change in the system and the total biomass
concentration in the
system is nothing but Va into X so Q is nothing
but Q into So by Se by Va into X and we know
1 by thetaC is equal to YTq minus kd.
Now, if you substitute this one here we can
get the value for Se, we have to substitute
this one here then we will be getting Se is
equal to 1 plus kd thetaC by k into YT into
thetaC. So what does it mean? The
substrate concentration as well as the biomass
concentration in the system is depending upon
the biokinetic constants of the microbial
population present in the system. So Se is
a function of kd which is the
decay constant and thetaC and thetaC is nothing
but 1 by mu and YT is the yield coefficient
and k is nothing but Qmax by ks
Another important thing in activated sludge
process is the recirculation ratio R. How
this recirculation ratio R is related with
the thetaC the biological sludge retention
time. Again we can take the mass balance dx
by dt is equal to RQ into Xr whatever biomass
that is coming into the aeration tank plus
dx by dt into Va minus 1 plus R into Q into
X this is the biomass mass balance for the
aeration tank alone. So we can
write like this; RQ into Xr this term we are
replacing by YTQ minus kd multiplied by X
into Va minus Q into 1 plus R into X.
At steady state again this term will be becoming
zero so we will be getting another relationship
1 by thetaC equal to Q by Va into 1 plus R
minus R into Xr by X. again we can see that
if the thetaC is fixed, the
volume of the tank is fixed and the flow rate
is fixed and the R is depending upon the concentration
of the recycle sludge and the concentration
of biomass in the aeration tank.
So here Xr is the concentration of sludge
in the re-circulated sludge, Xr by x reflects
the settling property of the sludge that means
how the sludge settling is taking place in
the secondary sedimentation tank that is
very very important for the proper functioning
of an activated sludge process. If the secondary
sedimentation tank of an activated sludge
process is perfectly fine the entire process
will be working perfectly all
right but if it is the other way the process
will not be having the required efficiency.
How can we find out this Xr?
Xr can be obtained from SVI, this term is
also important. This is nothing but sludge
volume index. Sludge volume index is nothing
but the volume occupied by a unit of weight
of biomass and usually SVI is
represented in terms of milliliter per grams.
That means if this much ml of biomass is there
what is the rate of that biomass or what is
the volume occupied by a unit mass of biomass
that is SVI. How can we find
out the SVI?
Before that one we will see how this SVI is
important for activated sludge process.
Small SVI is preferred. If SVI is less than
100 it is a good sludge, if it is hundred
it is desirable, if it is greater than one
fifty it is bulking sludge and if it is greater
than 200 it is not at all desirable. So how
can we
find out the SVI of the system? What we have
to do is take a graduated cylinder and take
one liter of the mixed liquid suspended solids
in that one and allow it to settle for 30
minutes. After 30 minutes we can
find out the volume because it is a graduated
cylinder so we can find out what is the volume
occupied by the sludge and afterwards we find
out the MLSS concentration is for milligrams
or grams per liter and
from that one we can find out the sludge volume
index. And sludge volume index is very important
for the proper functioning of activated sludge
process.
Now we will see what are the things we discussed
today. We have seen that biological processes
are suitable for either water or wastewater
depending upon the pollutant nature. Coming
to the wastewater what is
happening is most of the solids are highly
biodegradable so biological process is the
most desirable treatment process for secondary
treatment of wastewater especially domestic
wastewater.
The biological process is depending upon the
microorganisms for their efficiency. It is
essential to provide sufficient nutrients,
environmental conditions for the microbial
growth and depending upon the nutrient
requirement we can classify the microorganisms
into different categories like autotrophs,
heterotrophs, phototrophs, chemotrophs etc
and depending upon the electron donor we can
classify the process into
aerobic, anoxic or anaerobic.
We were also discussing about activated sludge
process as to what is happening there. We
are providing aeration so the dissolved form
of organic matter is getting converted to
fresher cells in the colloidal form.
And since more and more cells in the colloidal
form is present a solid liquid present process
is essential so secondary sedimentation tank
is an essential part of activated sludge process.
How the sludge settling is taking place in
the secondary sedimentation tank?
It is with the help of some bio polymers which
is excreted by the biomass themselves. Then
we have seen the two parameters; one is hydraulic
retention time and another one is biological
sludge retention time.
For any biological treatment system the design
parameter is thetaC not the HRT. ThetaC is
nothing but it shows how much time the sludge
is remaining in the system and we have seen
the mass balance of
biomass and the substrate of BOD in the entire
system.
We also discussed how to find out the biomass
concentration the volume of the aeration tank
or the efficiency of the treatment unit using
the bio kinetic parameters of the microbial
population present in the
system and another parameter we have discussed
is sludge volume index. This shows the settling
characteristic of the sludge and this is also
very very important.
