WEBVTT
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welcome to the next lecture in the course
effective engineering teaching in practice
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this lecture will be on something called active
learning let us see what that is all about
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so what is active learning the understanding
is rather broad right so there are very many
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different ways of addressing this the understanding
is something like this let me give you two
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statements for understanding active learning
anything that involves students in doing things
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and thinking about the things they are doing
is called active learning anything that involves
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students and doing things and thinking about
the things that they are doing whats the definition
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[give/given] given by or the understanding
given by bonwell and eison in their paper
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in nineteen ninety one published in nineteen
ninety one thats one way to understand it
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let me give you another statement that will
be helpful in understanding anything course
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related that all students in a class session
are called upon to do other than simply watching
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listening and taking notes is called active
learning this understanding was given in the
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paper by felder and brent in two thousand
nine so that is what it is as we go along
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it will become clear what active learning
is all about so although i dont state the
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learning outcomes explicitly it is all there
and if you look at the various chapters or
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various topics it would be there in the background
and you could you may want to pick it up its
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the teacher must be aware of the learning
outcomes and its good that these students
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are also made aware of the learning outcomes
in a certain case different people have different
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styles of doing things ok lets move forward
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problem based done learning that we introduced
in one of the earlier chapters earlier
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topics as a means of an extension of the tradition
lecture can also be done in an active learning
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mode ok remember i mentioned that problem
based learning can be done in various different
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modes and i am going to present will be one
of the modes of relevance to extending the
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effectiveness of a lecture
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so in this in this chapter we are going
to look at problem based learning in that
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can promote an active learning mode or
that that is implemented in an active learning
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mode for example first pose the problem to
the class this is what we did earlier also
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then give them the needed time or give
them the needed background and principles
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ok first pose the problem then then give the
principles the same thing that we mentioned
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earlier as an extension to the traditional
lecture after that guide them to solve the
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same problem in class itself ok and the learning
would be that much better earlier the teacher
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worked out the problem the teacher gave
the problem as a context to introducing the
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principles then probably the teacher worked
out the problem himself or herself to give
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to present the principles in that context
instead of that if the teacher can guide the
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students to work they work it out themselves
the learning would be that much better and
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we are all after learning by the students
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since the above would need student engagement
much beyond passive receipt of information
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and some excitement which is generated with
the initial problem statement that extension
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itself would have generated some excitement
or how do i solve this particular aspect it
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seems interesting and so on so forth in the
students and then they are participating in
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the solution itself and this is this is
active learning i am going to demonstrate
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this with a problem that i mentioned earlier
that will all be done at various levels
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if you look at it that way that way the appreciation
is much better so the same problem that was
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mentioned earlier very briefly i said you
could pose this problem and then work it out
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or give the principles and then work it out
here we slightly change things we are going
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to pose the problem and then what comes next
after posing is going to be slightly different
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let me read this problem out in the context
of this particular topic the humidifier is
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fed with dry air no water vapour it is removed
during the processing of air to avoid contamination
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of the bioreactor and clean liquid water in
other words a dry air and clean liquid water
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are fed to the humidifier the liquid water
flow rate is eighteen centimeter cubed per
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minute five mole percent of oxygen are needed
in the in output stream of the humidifier
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for supply to the bioreactor
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let us determine the molar rate at which or
determine the molar rate at which air should
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be supplied to the humidifier when it operates
at steady state a diagram is given here the
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humidifier here this is the air flow this
is the water flow the molar flow rate of water
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the molar flow rate of air this is the product
flow the molar flow rate of product and the
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mole fraction of oxygen in the product has
to be point naught five this is the condition
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of the problem this is a humidifier
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so if we apply our problem solving strategy
here what is known many different things that
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there is a water stream here and air stream
here the water stream flow rate is known eighteen
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cc per minute and the product stream is here
the product stream of composition the oxygen
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composition has to be five mole percent or
mole fraction needs to be point naught five
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all this is known and there are many other
things that are known which we reduce which
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will become clear as we go through what do
we need to find the molar flow rate at which
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air should be supplied to the humidifier this
particular value m dot of air and the stream
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has to be formed this is the molar flow rate
of air m air dot
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how do we relate what is known to what we
need to find ok this becomes a solution to
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the problem we will have to first consider
the system which is the humidifier this comes
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from material balance princi applying material
balance principles so let us consider the
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system as a humidifier here indicated by the
dotted line let us work with moles because
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the requirements of the problem mole as we
know is mass divided by the molecular mass
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and if there is no change in the species say
due to a reaction during the process the mole
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balances when individual species are as good
as the mass balances however when there is
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a reaction be a little careful when in doubt
always do mass balances ok in this case we
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know that there is no chemical reaction and
therefore mole balances are as good as mass
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balances remember this mass by molecular mass
the mass would be the same and so on
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what is the composition of air you ask this
question to the class ok give them time and
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then they would come up with hopefully its
made up of twenty one percent oxygen and seventy
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nine percent of nitrogen by volume or mole
if we ignore the minor components such as
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carbon dioxide and other things in the air
for the purposes of this problem let us consider
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air to consist of twenty one percent oxygen
and seventy nine percent nitrogen by mole
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or volume they are the same
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now this is the question to pose to class
now can you express the molar flow rates of
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oxygen and nitrogen in the air stream in terms
of the molar flow rate of air right remember
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we had earlier shown or the problem statement
had the molar flow rate of air that is what
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we need to find ok so we need to work around
that and therefore the need here is to express
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the molar flow rates of oxygen and nitrogen
in the air stream in terms of the molar flow
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rate of air ok give them time ok couple of
minutes and probably walk around the class
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to figure out whats happening you have you
will have very good idea as to who is struggling
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who is very enthusiastic who is very comfortable
with this and so on so forth just by this
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exercise ok
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so during this active learning process there
is so much information about the students
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learning their abilities and so on so forth
that can be gleaned by just observing so molar
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flow rate of oxygen and air is point two one
the molar flow rate of air twenty one percent
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by mole is oxygen and by extension the molar
flow rate of nitrogen and air must be point
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seven nine one minus point two one point seven
nine molar flow rate of air ok this would
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come from the students mostly ok wait till
the average student gets it there will be
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one or two would be a very bright and come
up with the answer immediately acknowledge
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them definitely dont put them down but give
time for the average student to get to it
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ok and then when you walking around you
could ask the students to think in various
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different terms to get into get to this answer
provide them with clues and so on so forth
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now ask this question now what is the molar
flow rate of water its already given in the
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problem but in a form that needs to be converted
the form given in the problem is eighteen
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cc per minute is the molar flow rate of water
the density of or it is eighteen gram per
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minute you can write this down and ask the
students why some would give the answer
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and then you need to you might need to explain
it to some masses volume times density and
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therefore the density of water is one gram
per cc eighteen cc per minute would be eighteen
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grams per minute eighteen grams per minute
mass by molecular masses moles the molecular
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mass of water is eighteen therefore mass one
gram divided by moles eighteen divided
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by molar mass which is eighteen you would
get one mole per minute thats what is given
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there mole is mass by molecular mass molecular
mass of water is eighteen
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now let us perform the mass balance is to
find out the molar flow rate of air present
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this and then say can you do the balance on
oxygen and tell me what you get see you are
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directing them to do a balance on oxygen ok
doing a balance on oxygen is straightforward
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thing that you are asking them to do in
the problem they will have to figure it out
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themselves here you are directing them ok
so give them some hints and they could do
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this is the first time they are being exposed
to the problem
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so tell me what you get or expose to the aspect
of applying material balances to a closed
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ended problem so can you do the balance on
oxygen tell me what you get then they would
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do the balance they would they would focus
on the system they would write the material
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balance equation and figure out how the mass
balance on oxygen works out the details could
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be something like this steady state process
therefore for all species the time derivative
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can be taken to be zero thats input rate plus
generation rate minus the output rate minus
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the consumption rate equals accumulation rate
the useful form of the material balance equation
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for processes
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since this is a steady state process the accumulation
term goes to zero and therefore the oxygen
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balance would be the input rate the molar
flow rate of oxygen in the input generated
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out minus output minus consumption there is
no generation of oxygen in the system therefore
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that term goes to zero there is no output
of oxygen in this sorry there is no consumption
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of oxygen in the system therefore that term
goes to zero and therefore what remains is
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molar flow rate of oxygen in the input stream
minus the molar flow rate of oxygen in the
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output stream equals zero that is your material
balance
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so point two one molar flow rate of oxygen
molar flow rate of air is the molar flow rate
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of oxygen in the input stream point naught
five mp dot is the molar flow rate of oxygen
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in the output stream in terms of the variables
that we have defined for ourselves therefore
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input minus output equals zero rates that
is and therefore we have mp dot in terms of
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the molar flow rate of air and then its straightforward
as tell them see here there are two variables
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molar flow rate of the product which is unknown
and the molar flow rate of air that is unknown
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we need the molar flow rate of air therefore
we need to do and a we need to have another
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equation in terms of these variables or related
variables and we need to solve this
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so to provide that equation tell them can
you now do a balance on the total moles and
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tell me what you get total moles is very straightforward
then they will start doing it you go around
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first time you might need to give them more
time give them hints as you go along since
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there is no generation or consumption of moles
in this process this is the solution to it
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the molar flow rate of air plus the molar
flow rate of water this is the input minus
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the molar flow rate of the product that is
the output with the system in focus is zero
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right therefore you substitute the terms
in terms of whatever we have found out
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in the previous parts of the problem you will
get molar flow rate of air plus one minus
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point two one divided by point naught five
the molar flow rate of flow rate of air which
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is the molar molar flow rate of water from
the previous problem i am sorry this is
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point two one minus point naught five molar
flow rate of air is the one that we got
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from the previous problem one as the molar
flow rate that comes in there ok
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molar flow rate of air is m dot air molar
flow rate of water is one and mp dot is point
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two one divided by point naught five molar
flow rate of air ok that is what it is and
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if you solve this it is all in terms of molar
flow rate of air we get point three one mole
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per minute and this is our answer ok so this
is active learning we approach the problem
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then we give them the principles and then
we guide them through solving the problem
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the learning is far far better for an average
student compare to we just telling them what
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it is ok they have processed this and therefore
they learning is that much better
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the second way by which i am going to give
you a few examples by which active learning
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can be done this is by no means exhaustive
ok there are so many other different methods
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by which one can implement active learning
i am just giving you some examples so that
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you would know the second examples you could
ask them ask the class to do simple calculations
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and give them time and wait ok waiting is
something that is very difficult for inexperienced
15:19.459 --> 15:25.250
people in this particular strategy but
wait go around the class help the class and
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then it will be far far better learning
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simple calculations such as these can be given
for example can you estimate the mass of air
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in this room ok this i normally do when i
do when i teach principles of
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biology to engineers and they they need numbers
and so on to work with therefore i give them
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this problem but with an idea of getting them
to appreciate that microbes are present in
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the air all the time at reasonable concentrations
ok so i asked them to find out what is how
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can you estimate the mass of air in this room
and then wait for the answers walk around
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so how do you find the the mass of air is
what is needed what is known you can figure
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out the dimensions of the room and you know
the density of air is one point two nine kilogram
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per meter cubed and therefore volume times
density is going to be mass as simple as that
16:23.049 --> 16:28.600
so if you look at this particular room the
quick thing to fix is the height which would
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be typically about ten feet here yeah about
ten feet here and then the width this pretty
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much a square room the width if i go by the
benches the benches would at least be five
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feet three of them with about three feet in
between so fifteen plus six thats twenty one
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feet so about seven meters so seven meters
by seven meters forty nine forty nine times
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three or lets say fifty times three one fifty
meter cubed is the volume of this room
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so the volume is known the density is one
point two nine so one fifty times one point
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two nine is the mass of air that is present
in this room in kilograms ok so you can thats
17:13.870 --> 17:19.019
a little startling to people at times to figure
out there are hundreds of kilograms of air
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in a typical room ok so this causes the
students to participate in the process of
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learning and give them simple calculations
in the context of whatever you are doing to
17:32.210 --> 17:39.500
come up with better learning you could
even ask them to do derivations in part ok
17:39.500 --> 17:45.289
and then they are involved in the process
of the derivation itself the basal principle
17:45.289 --> 17:50.201
and how to use that principle to get maybe
a useful form of an equation they can be a
17:50.201 --> 17:51.201
part of
17:51.201 --> 17:55.360
let me illustrate it by this example this
in the context of transport i am going to
17:55.360 --> 18:02.470
derive one of the fundamental useful equations
and transport let us consider a fixed volume
18:02.470 --> 18:10.540
in space the volume delta v is delta x delta
y delta z this is a cuboid through which the
18:10.540 --> 18:17.870
fluid flows let us consider the right handed
coordinate system for analysis you know when
18:17.870 --> 18:24.029
if x is this and y is this when you go from
x to y the direction of the right handed screw
18:24.029 --> 18:30.019
should be z you go from x you turn x to y
z is like this so if you write x like this
18:30.019 --> 18:35.649
and y like this x to y is like this your z
will be like this ok
18:35.649 --> 18:41.279
so let us choose a right handed system in
this case i have given our x is like this
18:41.279 --> 18:47.090
our y is into the board or into the space
here and therefore x cross y your right handed
18:47.090 --> 18:53.130
screw comes up like this therefore your z
axis should be like this here we have drawn
18:53.130 --> 19:01.850
a cuboid the length here is delta x the length
here is delta y the length here is delta z
19:01.850 --> 19:06.519
the coordinates of this point therefore i
mean are xyz and therefore the coordinates
19:06.519 --> 19:12.430
of this point are x plus delta x y plus delta
y and z plus delta z so this is a very general
19:12.430 --> 19:17.269
kind of a thing which we use to derive some
very fundamental equations which can be applied
19:17.269 --> 19:21.779
to anything here i am going to do something
about material balance for in a flow situation
19:21.779 --> 19:28.220
so let us begin the general material balance
equation that we already know the accumulation
19:28.220 --> 19:32.510
rate is input rate plus generation rate minus
output rate minus consumption rate ok this
19:32.510 --> 19:37.480
has to be valid over any system in this case
we are going to concentrate on the cuboid
19:37.480 --> 19:44.809
that we have defined and let us apply this
equation there we are considering total mass
19:44.809 --> 19:49.519
here we are going to do a total mass balance
therefore there is no question of any generation
19:49.519 --> 19:55.070
or consumption and therefore the equation
will reduce to the accumulation rate is input
19:55.070 --> 20:00.389
rate minus output rate and since this is a
three dimensional flow we need to consider
20:00.389 --> 20:05.149
the contributions from all the directions
x y and z
20:05.149 --> 20:09.380
and let us consider them one by one all this
you can state right in the beginning because
20:09.380 --> 20:13.970
this is too early to bring them in you need
to bring them in in a guided fashion ok especially
20:13.970 --> 20:18.580
this is something that they havent seen at
all before we go forward let us note that
20:18.580 --> 20:26.019
density times velocity would be the mass flux
you know rho times v rho is kilogram per meter
20:26.019 --> 20:31.000
cubed velocity is meter per second even in
terms of units kilogram per meter square per
20:31.000 --> 20:38.419
second which is nothing but amount of mass
transferred per unit area proportional
20:38.419 --> 20:43.559
perpendicular to the direction of motion
per unit time or the rate of mass transferred
20:43.559 --> 20:50.000
per unit area perpendicular to the direction
of transfer therefore which is nothing but
20:50.000 --> 20:54.970
flux of mass flux of that quantity flux of
mass and therefore rho times v density times
20:54.970 --> 20:57.759
velocity is going to give us mass flux
20:57.759 --> 21:05.940
so what is going to be the rate of mass through
the phase at x in other words this particular
21:05.940 --> 21:17.830
thing so this is the this is the x axis this
is the phase at x so what is going to be the
21:17.830 --> 21:24.470
flux of mass or the rate of mass inorder to
do balance we need rates the rate of mass
21:24.470 --> 21:30.510
through the face at x ok this is the question
that i am posing to class might be difficult
21:30.510 --> 21:34.980
for anybody in class to answer in the beginning
and therefore let us give the answer after
21:34.980 --> 21:43.630
waiting maybe for a short while that would
be rho times vx this is the flux at x ok at
21:43.630 --> 21:50.860
that x times the area of that phase which
would be delta y delta z you point out the
21:50.860 --> 21:57.070
area at that phase as but this is the area
that we are interested in what is that area
21:57.070 --> 22:03.690
this is the area of the rectangle here this
is delta y times delta z ok
22:03.690 --> 22:09.010
so the flux times the area is going to give
us mass and therefore this is flux rho vx
22:09.010 --> 22:14.880
is flux at x times relevant area perpendicular
to the direction of transfer delta y delta
22:14.880 --> 22:21.220
z now what is the rate of mass with the face
at x plus delta x in this particular thing
22:21.220 --> 22:30.299
yeah x plus delta x this is the this is x
this is the x plus delta x face right this
22:30.299 --> 22:36.110
particular face here through which is going
out so what is the rate through that face
22:36.110 --> 22:42.570
its going to be rho vx at x plus delta x times
delta y delta z so once we have given them
22:42.570 --> 22:49.179
this then students are able to relate to what
we need for this particular derivation then
22:49.179 --> 22:54.740
pose the question now can you tell me what
the rate of mass n through the face at y is
22:54.740 --> 23:00.640
y is this direction therefore what what is
the rate of mass through n and then they would
23:00.640 --> 23:06.039
work it out when he would get it with this
some might need help and then after waiting
23:06.039 --> 23:12.419
for some time write this out on the board
rho v y at y and the relevant phase area is
23:12.419 --> 23:17.789
delta x delta z and we can play this game
till we finish up all these terms
23:17.789 --> 23:23.841
so rate of mass in and rate of mass out through
the various relevant phases are given the
23:23.841 --> 23:30.350
only term that remains in the material balance
is the rate of mass accumulation within the
23:30.350 --> 23:35.980
volume element and that can be represented
as this again might the students might need
23:35.980 --> 23:42.900
help based on experience dou dou t of rho
v ok rho into v density into volume is mass
23:42.900 --> 23:48.520
therefore dou dou t of mass is dou rho the
delta v can be written as or v can be written
23:48.520 --> 23:54.070
as delta x delta y delta z thats a small volume
that we are considering there the volume of
23:54.070 --> 24:01.350
the system of the control volume dou rho
delta x delta y delta z times dou t and since
24:01.350 --> 24:07.809
delta x delta y delta z are not functions
of t we can take them out and say dou x dou
24:07.809 --> 24:12.080
y delta x delta y delta z dou rho dou t ok
24:12.080 --> 24:17.660
now that we i set up all these terms now
can you substitute the above terms in the
24:17.660 --> 24:23.019
relevant terms in the mass balance equation
tell me what you get its a question that can
24:23.019 --> 24:27.700
be posed so they will work it out give them
time they will work it out so you are essentially
24:27.700 --> 24:32.250
bringing them into the derivation process
itself they put this all together wait
24:32.250 --> 24:38.190
for some time and then you could write this
down on the board and then you tell them now
24:38.190 --> 24:42.610
divide throughout by delta x delta y delta
z and tell me what you get or you could if
24:42.610 --> 24:46.390
you if you feel that its a little too much
for first timers then you write down the earlier
24:46.390 --> 24:50.509
expression and then say now divide by delta
x delta y delta z and tell me what you get
24:50.509 --> 24:51.970
thats straight forward
24:51.970 --> 24:59.000
so the dou rho dou t is one by delta x plus
something plus one by delta y and something
24:59.000 --> 25:05.830
and so on and then you ask them can you see
something familiar there some definition of
25:05.830 --> 25:10.980
something math relevant when especially when
we impose a limit of delta x tending to zero
25:10.980 --> 25:15.470
delta y tending to zero and delta x tend z
tending to zero many would be able to see
25:15.470 --> 25:21.659
that they are all definitions of the derivative
some might need help but this is only you
25:21.659 --> 25:25.370
know directing them therefore you dont have
to worry about the entire class in this particular
25:25.370 --> 25:31.509
case entire class getting it in this particular
case ultimately theyll get it and therefore
25:31.509 --> 25:39.820
when you apply the limits you get the differential
equation dou rho dou t is minus in according
25:39.820 --> 25:43.720
to the definition of derivative its the other
way around and therefore minus dou dou x of
25:43.720 --> 25:50.120
rho vx plus dou dou y of rho vy plus dou dou
z of rho vz and in vector notation it can
25:50.120 --> 25:58.880
be written as dou rho dou t is minus del dot
rho v del is nothing but dou by dou x plus
25:58.880 --> 26:04.070
dou by dou y plus dou by dou z and dot rho
v is going to give you that you could explain
26:04.070 --> 26:07.529
the vector notation little more i am not going
to do it here
26:07.529 --> 26:15.690
and this process we have brought the students
into the actual derivation they are participating
26:15.690 --> 26:21.760
in the derivation and thereby their you know
involvement with the learning process is from
26:21.760 --> 26:27.740
ground zero ok and that is something very
powerful also did you realize that we have
26:27.740 --> 26:32.730
told a story here remember earlier i had mentioned
that you know even math related things can
26:32.730 --> 26:39.690
be presented in a story form and this is a
story here a logical development of the process
26:39.690 --> 26:44.380
a derivation in this case with the backgrounds
given with the details given and so on so
26:44.380 --> 26:51.100
forth to help students is what would engage
the students much better and it is a story
26:51.100 --> 26:53.509
right thats what i meant ok
26:53.509 --> 26:57.299
another i think this is the last example that
i am going to give here there are very many
26:57.299 --> 27:02.990
examples you could read books on these papers
on these and employ your own means this is
27:02.990 --> 27:09.090
a popular thing think pair share that was
put forward by lyman and its one of the popular
27:09.090 --> 27:14.830
things therefore i thought i will include
this as one of the examples of active learning
27:14.830 --> 27:19.909
think pair share work something like this
give a short question it could be can you
27:19.909 --> 27:24.019
estimate the mass of air in this room right
give five minutes for students to work it
27:24.019 --> 27:29.480
out ok individually individual students to
work it out the regular classroom then after
27:29.480 --> 27:36.240
five minutes ask neighbors to discuss the
solutions with each other right if you
27:36.240 --> 27:42.070
have clicker possibilities and so on you
could monitor the progress as things go along
27:42.070 --> 27:48.610
but this itself is good enough ask neighbors
to discuss their solutions with each other
27:48.610 --> 27:53.710
and then ask them to revise solutions based
on the discussion ok
27:53.710 --> 27:59.360
that itself is good enough the neighbors can
since they are in the process this very relevant
27:59.360 --> 28:02.970
the conversations begin quite quickly they
are already tuned to the conversation they
28:02.970 --> 28:07.970
begin quite quickly they know where they are
confused so they could ask their neighbors
28:07.970 --> 28:13.179
and that discussion can go on ask them to
revise the solution based on the discussions
28:13.179 --> 28:18.590
if there is a need and finally you could either
provide the solution or call someone to work
28:18.590 --> 28:23.559
it out on the board or discuss the solution
to the class within the entire class and so
28:23.559 --> 28:24.559
on
28:24.559 --> 28:29.360
there are variants of this that people have
done which are more effective than just this
28:29.360 --> 28:35.559
you can look it up later so this is the other
example so the above techniques are some means
28:35.559 --> 28:41.570
by which active learning can be implemented
there are many many more i think thats all
28:41.570 --> 28:46.279
i have in terms of active learning for you
active learning is essentially getting
28:46.279 --> 28:52.159
their students engaged in the learning process
and making them do activities to improve their
28:52.159 --> 28:58.669
own learning and this is a significant improvement
from the traditional lecture ok initially
28:58.669 --> 29:04.279
problem based learning and then now active
learning along with learning outcomes and
29:04.279 --> 29:10.749
so on so forth we are slowly going to more
and more effective means of learning ok different
29:10.749 --> 29:18.740
strategies to improve that ok lets stop here
and lets meet again later