Good day everyone, and welcome to this lecture
on Water Quality Monitoring. We will be continuing with the previous lectures,
where we studied different parameters for water quality monitoring. And in the previous lecture like we studied
the various chemical parameters which are used for estimating various properties of
the wastewater or water. Now, continuing in the same time, we will
be now going to the third parameters, third set of parameters which are called as like
biological or biochemical parameters. And they are essential parameters, because
they estimate virtually what is the oxygen demand that will be there with respect to
organics which are present in the water and when their natural degradation occurs in the
natural requirement. So, all those aspects will be discussing in
this lecture. One of the foremost important parameters with
respect to these biochemical parameters, it is called as biochemical oxygen demand. As such, whenever we are discharging any water
into another aqueous stream, and that water contains some amount of organics, so, that
organic will always naturally decay, and that decay happens because of the presence of microorganisms,
which use that organic material as a nutrient and convert it into basic forms like carbon
will be converted into CO2, etc. And during that degradation process, they
require a lot of oxygen. This type of decay we observe commonly on
any soil surface also. So, any organic material if it is thrown on
soil surface, so, its natural decay always happens. We do not foresee that, but always some amount
of oxygen is required during the decaying process, which happens because of the presence
of biological organisms in that system. However, if that organic material is present
in the water and that whole water is being discharged into another Aqueous stream. So, that means, the water which is there,
it will be having a maximum oxygen availability depending upon the solubility of oxygen at
that temperature. And if the oxygen demand is more than the
what is present in that aqueous stream, then there will be anaerobic conditions will prevail. And that decay process will be hampered, also
oxygen level will also go down and it will cause effect on the aquatic life of both animals
as well as plants. So, oxygen availability in the water is very,
very important for the sustaining the life. And for doing this we must understand that
any material which we throw into any water body, if it has some oxygen demand, that means
the oxygen present in the water will be diminished and that will cause problem with respect to
that Aqueous stream. So, we always want the oxygen demand to be
minimal, whenever we are discharging any water into another stream. So, that oxygen demand we need to determine
beforehand. So, biochemical oxygen demand is one of the
methods which actually helps in determining how much oxygen will be consumed during the
degradation of all the organic material present in the water in the natural environment, when
different types of biological organisms are present in that aquatic system. So, this BOD estimation helps in determining
the oxygen demand in general. Now, it is any water stream it is having very
high BOD value. That means, if it is mixed with another water
stream or if that water will be discharged into another stream, so, the demand of oxygen
will be more so, that means that the present water stream which is having high BOD, its
water quality is bad. That means it contains a lot of organic material
and which, during degradation will require a lot of oxygen. So, we do not want BOD values to be higher,
we always want BOD values to be minimum possible. So, if BOD value is very less, that means
that the quality of water is good. There are no organics present in the water
and there will be no requirement or degradation of those organics present in that water. So, this BOD is generally expressed in milligrams
of oxygen consumed per liter of the whatever water sample. And there are two methods, earlier method
it was in India, we used to do determine the BOD values by keeping the water sample for
5 days at 20 degrees centigrade. But now, since last few years, the new method
has come under that we incubate the water sample for 3 days at 27 degrees centigrade
for determining the BOD value. Now, during the BOD test, there are or during
any natural degradation of organic matter, there are two stages of decomposition. In the one stage, in the first stage, carbonaceous
stage it is called. During that stage whatever oxygen demand is
there, that is due to conversion of organic carbon into carbon dioxide. And in the second stage, which is also called
as nitrogenous stage, a combined demand is there which is of carbonaceous plus nitrogenous
organic matter and because of their conversion into various other like carbon dioxide, ammonia,
nitrite etc. So, during this nitrogenous demand, there
will be always be organic nitrogen will be converted into ammonia nitrite further into
nitrate. So, nitrogenous demand usually occurs much
later than the carbonaceous demand and it may start occurring early as possible around
6 days. But, in the BOD test, we are usually inclined
towards finding out the carbonaceous demand only we are not finding the nitrogenous demand
in the water. Under some conditions, there is a possibility
that ammonia nitrite and nitrifying bacteria if they are present, this nitrification of
nitrogenous organic matter may start occurring less than 5 days also. So, we do not want because in the BOD, we
want to estimate only the carbonaceous oxygen demand only. So, under that condition, we need to add some
inhibitors during the BOD estimation, so that the only we can determine the carbonaceous
demand, there is no nitrogenous demand, which comes into the BOD estimation. If we do not add that inhibitor, actually,
our BOD values may be 10 to 40 percent higher than the carbonaceous oxygen demand, and that
will be wrong. So, the results of BOD tests are reported
as carbonaceous BOD only and it may be BOD for 3 days, it may be 5 days when a inhibitor
is added. Now, how do we determine the BOD? So, that we will, we are going to learn now. So, BOD is the amount of oxygen which is like
dissolved oxygen required for the biological decomposition of organic matter, so this is
very simple definition this is very simple definition. Now, the oxygen consumed is related to the
amount of biodegradable organic, so, if organic matter is less. So, amount of oxygen demand will be less and
so, it depends upon the how much amount of organic substance is present in that water. And that organic carbon or organic substances
which are present they may be because of the presence of protein, carbohydrates, fats or
in the industrial wastewater, many other organic materials which are actually being used in
that industry as a raw material or they are being produced. So, are they are intermediate during the whole
process. So, they may also come into the water and
they may incur some oxygen demand. So, we need to find out the oxygen demand
in the water. So, this is the BOD dilution method. Now, what we do is that measurement of BOD
is done like finding out the initial oxygen and then finding out the final oxygen either
after 5 days at 20 degrees centigrade or 3 days at 27 degrees centigrade. In India now, we prefer 3 day at 27 degrees
centigrade, which is the standard. What we do is that we keep a number of standard
300 ml bottles are there, which are used during the BOD testing and they like 300 ml BOD bottles
are filled completely with wastewater. Now, if the overall demand is very less or
the amount of organic matter present in the water is less than that what will happen? The oxygen content of one of the bottle is
determined in immediately and that the DO value can be determined using a DO meter. There are many, many simple instruments which
can determine the oxygen or we can determine the DO value using titration as well. Now, the other bottle is incubated at 20 degrees
centigrade for 5 days or at 27 degrees centigrade for 3 days in total darkness. So, it is kept in total darkness, so that
there is no algal growth. And naturally it we are assuming here that
some amount of microorganism is present already present in that wastewater. So, that microorganism will degrade the organic
matter. And for that it will consume the oxygen which
is already present in that water. So, after 5 days or 3 days, we determine the
oxygen content again. And the difference between the 2 DO values
is the amount of oxygen consumed by microorganisms during the 5 or 3 days, and it is reported
as BOD5 or BOD3. So, the simplest method, case first is when
the amount of organic matter present in the water sample is very less we can determine
by just finding out the initial DO and final DO after incubating the wastewater sample
in a 300 ml bottles for 3 days or 5 days. Now, these are the simple DO bottles and we
can use this instrument for finding out the DO very easily. Now, there is a certain limit, we can easily
calculate this and in fact, this we will be doing that what is the saturated value of
dissolved oxygen for water at any temperature. So, this we can determine very easily for
the sake here it is reported that the saturated state value of DO for water at 20 degrees
centigrade is approximately 9.1 milligram per liter. So, that means, any water at 20 degrees centigrade
will generally not contain more than this amount of oxygen. So, any extra oxygen will go out of the surface
into the atmosphere. And if the water contains less DO some amount
of oxygen will go into the water and so that the solubility limit is met again. Now, the oxygen demand for wastewater it is
possible that it is more than this requirement it may be several 100 milligram per liter. So, under that condition, this method which
was given in the previous slide just finding the initial DO and final DO may not work because
the oxygen demand is more, so the oxygen may become 0. So, we do not know within 3 days whether whatever
the actual demand. So, under that condition, what is done is
that that we dilute the water with oxygen with the organic matter free water. So, that the overall demand overall amount
of organic matter in the water sample becomes less and so that we can find out. So, always the dilution is done of the water
sample itself. And that we may know whether we require dilution
or not. There are 2 methods for doing this. Any BOD test is considered to be okay. For that there are 2 conditions. One is that the final DO in the BOD test should
always be less than or greater than 2 milligrams per liter, what does it mean? It means that after 3 days if the DO value
in that water sample bottle is less than 2 milligram per liter we assume that we are
not sure and it may be possible that oxygen demand was more, but it has not been properly
given because availability was not there. So, it is possible that it may become 0 milligram
per liter also. So, if this happens that BOD at the DO value
after 3 days or 5 days is less than 2 milligram per liter, we assume that the BOD test is
not okay, we need to dilute the water further. So, that the final BOD or final DO value should
always be greater than 2 milligram per liter. Also, it is possible that we might dilute
the water so much that actually that we are not sure whether the change in DO which has
happened is correct or not. So, if the change in DO from initial and final
values is less than 2 milligram per liter, the under that condition also, we assume that
our BOD test has failed. So, we have to redo the BOD test. So, under this condition, actually, the dilution
has been more. So, that is, why the demand of oxygen was
very less, and there is absolutely virtually no change. So less, if at least 2 milligram per liter
change is not there, we assumed the BOD test to be failed. And similarly final DO should always be greater
than 2 milligram per liter, then only the BOD test is assumed to be correct. So, these are the methods. Now, this dilution if when we are doing the
dilution, we can find out the BOD using this method it may be BOD5 or BOD3 also depending
upon whether it is 3 day test or 5 day test, so, it is found out by using this formula
BOD is equal to DO initial minus DO final divided by P where P is the dilution factor. And which is defined as the ratio of volume
of sample of wastewater to the total volume of wastewater that means wastewater plus dilution
water. So, P is like P is equal to volume of sample,
volume of sample and it is volume of sample plus dilution water. So, and generally this whole thing is at the
bottom it is 300 ml. So, and this is the volume of the sample,
so, this is how P is defined. So, because we always do the test in 300 ml
BOD bottles, which are also called as Meyer flask, so, we use them. So, for this it is there. Now, in this formula there is presumption. Presumption is that, that dilution wastewater
which has been used, it has no oxygen demand of itself. But it may be possible that dilution water
that we are using it has some oxygen demand. And also there is a second case that sometimes
it is possible that water samples that we are using it has no microorganism. So, sometimes what we do is that we use the
dilution water which is seeded water, dilution water which is seeded water what does it mean? That dilution water that we are using is already
sseeded with some microorganisms. So, if they are seeded with some microorganisms,
they will also incur some demand. And this is done, so that we always have some
micro exam present during the BOD test. So, this is done. And during that condition, it is most likely
that, that dilution water also has some BOD demand. So, under that condition, we use this formula
which is given here for finding out the BOD. Again, it may be BOD5, or BOD3, depending
upon the number of days for which the BOD bottles are kept. Most of the times micro-organisms are added
in the dilution water and it is referred to as seeded water. So, as to have enough micro oxygen for carrying
out biodegradation of organic waste. So, in this case, the oxygen demand of seeded
water has to be subtracted from the demand of mixed the samples of waste plus dilution
water. So, and Bi and Bf are the initial and final
DO concentration. So, in this case, we keep at least 3 bottles,
one is initially will find out that DO of the sample. Second case, will find that DO of the sample. So, one bottle, bottle one will contain wastewater
plus seeded water. So, both will be there and then there will
be at least one another bottle, which will be containing only seeded water. And it will also be kept for that many days,
3 days or 5 days inside the incubator at that particular temperature. So, if it is kept it for 3 days it will be
holding incubator will be kept at 3 days and at 27 degrees centigrade and everything will
be in the dark. So, this is the method and we can find out
the BOD via this method. Now, these there are some other important
things, DO can be modelled as a first order reaction. And this is very important, because many times
we need to determine the carbonaceous oxygen demand for maybe 8 days, 10 days or at any
other temperature also. So, we have performed the test suppose that
27 degrees centigrade, but actual degradation is going to occur at 15 degrees centigrade. So, under that condition, the values will
change, the demand will change. So, for doing all these studies, it is very
important to model the BOD. Now, as usual, since most of the biological
reactions can be modelled as first order reaction here also BOD is also modelled as first order
reaction assuming that the rate of decomposition of organic waste it is proportional to the
whatever is the waste left. So, it is the traditional simple concepts,
so we are putting dL by dt is directly proportional to Lt and where Lt is the amount of oxygen
demand left. So, after time t, and k is the BOD rate constant
and its unit is time inverse. So, if it is kept in days it is will be day
inverse. Now, if you solve it assuming that at time
t is equal to 0, the demand left is 0. So, under that condition we can easily solve
and this equation we will get. And here, there are few things more that will
understand may be in the. Now at any time, whatever is the total oxygen
demand? So, that demand is equal to the BOD already
used up and whatever is left. So, that means L0 is equal to BODt plus Lt.
So, always L0 at time t, at initial BOD and it is also sometimes that what will be the
ultimate oxygen demand this is equal to L0, they both are same and they are equal to BODt
plus Lt. Now, if this is there, then we can find out
the BODt very easily because BODt will be at any time t will be equal to L0 minus Lt
and Lt is already known, because it is Lt is like L0 e raise to minus kt. So, we have BODt is equal to L0 1 minus e
raise to minus kt. So, this is there. Now, here the these parameters the k value
is highly dependent upon the temperature and as the temperature increases, we know very
well that metabolism increases, so, utilization of DO also increases. So, that is why k is a function of temperature
in degree centigrade and its relationship with k20. So, whatever is the value of constant at 20
degrees centigrade, it is related to at any other temperature and theta is one of the
constants which is used and remember the t which is given here it has to be in degrees
centigrade, so t minus 20. And this theta value have been reported differently
and they may vary with respect to water also, type of water whether it is municipal wastewater,
industrial wastewater, etc, but generic values are given here. So, this is theta. So, theta is an empirical constant here and
theta value may be 1.135 if temperature is between 4 to 20 degrees centigrade. It may be 1.056 if the temperature is between
20 to 30 degrees centigrade. The, this is very, very important and this
is used many times for various uses and it has a lot of value. Now, we are taking few problems for better
understanding, so how to solve these problem. First problem is like we perform a test and
we use a 6 ml of wastewater and it is diluted to 300 ml of distilled water in standard BOD
bottle. So, what we have taken? We have taken the volume of sample which has
been taken here is 6 ml. So, this is one thing and now, after using
the dilution water, so, volume of sample plus volume of diluted water, so, I am just writing
diluted seeded water. And this will always be equal to 300 ml as
earlier told. Now initial DO in the bottle is determined
to be 8.5 milligram per liter. So, as soon as water and both the things are
mixed together, so, the initial demand was only 8.5 milligrams per liter DO was this. Now, k value at 20 degrees centigrade is given
already given 0.23 per day, now, DO you after 5 days at 20 degrees centigrade is found to
be 5 milligram per liter. So, we have to find out the BOD5. Now, in this case it is not given that whether
it is seeded water or not. So, that means, here there is no seeded water
only dilution water without any oxygen demand was there. So, that means the, it is simple water otherwise
the data may have been given for seeded dilution water also that for seeded dilution waters
bottle the initial value was this and final value was this. So, if it is not given that this here we are
assuming that the seeded dilution water had no oxygen demand by itself. So, in this case, the BOD5 will be the formula
that earlier was written the BOD5 is equal to DO initial whatever is that DO initial
minus DO final. So, like i and f divided by P. And that P can further be written. So, we have this formula DO 0 or DO initial
minus DO 5 upon Vw into this. So, all the values are kept and we can see
here it is 175 milligram per liter for 5 days. So, this is there. And if it is also many times required to find
out that what is, what will be the ultimate BOD, what will be the maximum BOD that will
be required. So, ultimate BOD remember ultimate BODu it
is written or L 0. Both are same. So, BODu, we can find out the formula with
respect to BODt was BODt is equal to L0, 1 minus e raise to minus kt. So, the same thing is written here. And we just find out the BODu value by dividing
and it is 256 milligram per liter. Then, there is another question which is written
here, the BOD6 of a wastewater is determined to be 400 milligram per liter at 20 degrees
centigrade. The k value is 20 degrees at 20 degrees centigrade
is 0.23, per day, what will be the BOD8 value if the test runs were conducted at 15 degrees
centigrade. So, remember, we use the generic formula is
written like this, the BOD at for t days is equal to L0, 1 minus e raise to minus kt. But we can modify this formula, L0 is always
equal to BODu, this was highlighted. So, there is no change here. But we can use a trick at what temperature
we are writing. So, this formula can be rewritten like this. So, BOD for time t, at what temperature we
are finding, then L0 and BODu they do not have any effect with respect to temperature
because ultimate demand will be always be the same. Only rate by which we are reaching that ultimate
demand will be different. And similarly for the k term because k term
is temperature dependent, we can write k capital T, where T represents the temperature multiplied
by time. So, this is there. So, and this is the same formula that is written
here. Now, already for k, we know that k at any
temperature T is equal to k20 and theta raise to t minus 20. Now, for this case, we are assuming that theta
value to be 1.047. So, we can find out at any temperature, so
at 15 degrees centigrade, we can find out what is the value of k. So, from here using this formula we can find
out the value at 15 degrees centigrade. So, this will be the first step, we can use,
so k15 can be found out using the k20 value. And so, this already this is 047 raise to
15 minus 20. So, this we can solve and this the value we
can find out here. And now BOD 6, we have to find out the BOD
6 at 20 degrees centigrade is given to be 400 milligram per liter, k20 is given. Now, by using this value and implying this
formula we can find out in the first step what is BODu. So, in the first step this BODu or ultimate
BOD has been found out and all the, we are using the same formula which is given by equation
1 and we can find out okay ultimate oxygen demand is this. Now, later on, since we have to find out the
BOD at 15 degrees centigrade, we are using this same equation 1 only we have to use the
k15. So, k15, we can find out from here and we
can solve, the answer will be this. So, we can see that, we can find out the BOD
at any temperature or any number of days if we know the value at some specific temperature
and day. And we can use this modified formula always
along with the second formula. So, both in combination can be used for finding
out the BOD for at any number of days for any number of days at any temperature. So, with this we end today’s lecture and
will continue in the next lecture. Thank you very much.