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alright so we stopped by looking at the resistance
network qualified by strength thermal resistance
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network ah for composite walls so thats where
we stopped in the last lecture
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so supposing there are multiple walls which
are associated so if the length is l one l
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two and l three and we have fluid which is
flowing on either side of the composite wall
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so thats the composite wall ok and so if the
conductivities are k a k b and k c so we said
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that we could construct the network which
has essentially five five resistances
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thats this x direction and if the heat transport
coefficient is h one on this side and the
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temperature is t infinity one and this is
h two and t infinity two so we said that its
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t infinity one and and t infinity two and
the resistances are one by h one a l by l
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one by k one by a l two by k two a and l three
by k three a and one by h two a ok so one
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could define we know what are the total resistances
so the total resistance r total it will simply
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be sum of all the resistances sum of all the
individual resistances which is essentially
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sum of these
now what we are going to start within today
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lecture is we are going to look at what is
called the overall heat transport coefficient
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ok so one could define ah overall heat transfer
coefficient and the reason for doing that
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is ultimately from [measure/measurement] measurement
point of view in fact is what i would measure
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is the temperature here and temperature here
so i need to know what is the total amount
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of heat that is transferred from lets say
hot fluid on this side ah i use one ok so
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i can put one two one three ok and cold fluid
on this side ok so i want to know what is
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the effective heat transport that is occurred
from the hot fluid to the cold fluid and that
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is simply because i may not be able to measure
the temperatures in between so one could define
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the total amount of heat that is transported
similar to the way we defined newtons law
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of cooling as the constitutive equations so
we could write we could define a overall heat
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transport coefficient u multiplied by a multiplied
by the net temperature difference of the observable
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temperatures or measure ah measurable properties
ok and that should obviously be equal to that
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should be equal to what
yeah by r total so thats right so that should
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be equal to i infinity one minus t infinity
two divided by r total simply by the definition
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of the resistances ok so from here we can
clearly read out that r total equal to one
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by u a ok and that should be equal to this
sum of the individual resistances sum of all
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the resistances that is involved in the system
that we are considering so this is a a ubiquitous
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property of ah a definition of any heat transport
system so in principle one could define a
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overall heat transport coefficient for any
heat transport system and what we are doing
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essentially is we are lumping all the properties
all the transport processes which are occurring
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inside everything is lumped into these one
quantity called universal or overall heat
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coefficient and we will see many variations
across the overall heat transport coefficient
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that depending upon the system that we are
considering that we are going to see in todays
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lecture and several lectures in future so
its very important to understand the definition
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of overall heat transport coefficient and
note that its a fictitious quantity it can
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be detected based on the individual properties
of the system that we are considering however
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over all heat transport coefficient is mainly
for convenience purposes and for calculation
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purposes it really helps in ah defining such
kind of a quantity
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so there is one small aspect about resistances
there is something called contact resistance
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so so far and all the ah that couple of examples
that we described in the last lecture and
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start of todays lecture the we assume that
the contact between the slabs is supposed
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to be a smooth contact however in reality
that need not necessarily be the case for
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example if you have ah when you say smooth
its smooth all the way upto the microscopic
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level right so its not this smoothness that
you observe in your eyes its the smoothness
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that you would observe when you see under
a microscope so that you will not expect you
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there is no reason why you should expect that
the wall that you are having has a a smooth
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surface all the way upto the microscopic level
so therefore there will always be certain
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resistance which is offered by the non smooth
contact between the two surfaces of the composite
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wall so as a result one could define something
called a contact resistance and it depends
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upon the smoothness properties so there is
no clean way to estimate what should be the
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value of this contact there are some correlations
we will not go into those correlations in
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this course but as when it is required particularly
from the problem solving or exam point of
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view these kinds of numbers will be provided
to you ok so give me a sec
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so supposing if you have a another wall here
ok now if i want to construct resistance network
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including the contact resistance then what
i would do is so supposing i have fluid which
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is flowing here ok then you will have so in
if there is no contact resistance then there
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total of five resistances but because the
contact position is not smooth its going to
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offer a certain resistance to heat transport
and therefore you will have an additional
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network which is basically the contact resistance
which will come in series so the the other
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five are the same thing what we saw a short
while ago so we need to include a contact
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resistance which is present in between this
supposing we want to include a contact resistance
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between the second and the third wall we could
do that we could include an another resistance
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here any questions
why should ah because of there is a lack of
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correct
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yeah so supposing if they are not in contact
with each other ok so a a very simple example
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is supposing you are holding a coffee cup
ok so there are two ways of holding it i could
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hold it like this with all my fingers on the
curved surface of the tumbler or i could hold
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it just at the top so if you look at the workers
who are like ah drinking coffee and tea the
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hot tea they usually hold the tip or they
hold a cup like this and the reason is that
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you do not want to have a direct contact with
the surface which is hot now taking that parallel
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here if this surface is not smooth then the
contact the transport of heat from one slab
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to the other depends upon the overall effective
surface area which is available for transport
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now if the contact is not very smooth then
the total surface area which is available
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for heat transport from one slab to the other
slab is not as much as the overall surface
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area which is available and therefore therefore
the therefore the total amount of heat that
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is transport is not exactly the amount of
heat that comes at this end and therefore
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this offers a resistance yes you had a question
ah yes there will be a dissipation because
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of that see what is the resistance note that
the resistance is basically characterizes
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the total amount the ability of the system
to transport heat what is the resistance that
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it offers to transportation of heat from one
location to the other because the contact
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point is not significantly good there is going
to be some dissipation and therefore that
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that offers a certain resistances and thats
what is captured by the contact resistance
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ok any other question ok
alright so next what we are going to see is
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we assumed so far in all the cases that we
considered that the area of heat transport
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is constant so today we are going to look
at the varying area systems so how to characterize
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and quantify conduction process through a
a system where the surface area or cross sectional
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area for heat transport is constantly changing
a very simple example would be that supposing
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i have a truncated cone ok i have a truncated
cone ok and lets say i am looking at so this
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is lets say at equal to zero ok so i want
to know what is the heat that is being transported
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from lets say let me call this as x direction
x equal to x one x two i wanted to call it
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radius so thats the centers of radius is zero
so i want to know what is the amount of heat
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that is transported from x one to x two and
i am going to make an assumption because i
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am looking at one d systems i am making an
assumption that the cross sectional temperature
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that means that every at every cross section
i assume that the temperature is uniform in
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the cross section and the gradients are zero
ok
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so now so this is my radius at any location
and so i could write my ah balance and if
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i continue to assume that its a steady state
system and heat generation is zero same assumptions
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as what we made before so i could simply write
the total amount of heat that heat transfer
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rate q x is given by minus k which is the
conductivity of that material multiplied by
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the cross sectional area of heat transport
so note that now the cross sectional area
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is a position of the function position ah
is a function of the position excuse me ok
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multiplied by d t by d x ok so what is the
objective we need to find the temperature
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profile that is the objective of the problem
right so we said that if we know the temperature
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distribution if we know the temperature profile
we are done we have quantified the system
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so thats what we need to find so lets say
we integrate this equation ok so we say that
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q x is minus k what is the area its pi cross
sectional area is pi into radius square ok
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pi r square right multiplied by d t by d x
right so supposing i say that r goes as a
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a linear function of the position if i say
that the radius of the local radius goes as
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the linear function of the axial position
then i could simply write this as minus k
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pi a square x square into d t by d s into
d t by d x ok and now what will be q x will
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it be constant or it will change rate of heat
transfer that will be constant why will it
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be constant because of the energy balance
you see that there is no heat that is being
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generated so whatever comes in it has to go
out here at steady state condition note that
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steady state is very important yes right so
the question is if you dissipation of heat
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how will it be constant but when we say that
there is no generation or less of heat which
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means that the dissipation is zero we will
will come to that so there are ways to consider
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that we are actually consider when we are
doing a two dimensional system you can actually
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look at dissipation from the outside walls
and we will actually see it in one of the
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examples in the future lectures alright so
because q x is constant we should be able
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to integrate this expression to find the temperature
profile
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so supposing i integrate between t one and
t one is the temperature at the boundaries
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of this system ok t one is the temperature
at the boundaries of this system i integrate
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between t one and t thats equal to q x minus
q x by k pi r square of k pi a square into
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d x by x square going from x one to x ok its
a pretty simple integration so its q x by
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00:16:01,029 --> 00:16:11,500
k pi a square into one by x minus one by x
one ok so note that because its one by x square
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the minus sign will go away because of the
integration and this is t minus t one ok so
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00:16:22,190 --> 00:16:36,639
therefore t is t one plus q x by k pi a square
into one by x minus one by x one is it a complete
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00:16:36,639 --> 00:16:45,220
description we dont know the q x value right
so its not a complete description yet so how
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do we find q x so we know that the temperature
on the other boundaries is t two so we can
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use that property to find out what is q x
so we dont know what q x is so how we are
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00:16:57,920 --> 00:17:09,380
going to do that we are going say t two is
t one plus q x by k pi a square multiplied
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00:17:09,380 --> 00:17:22,579
by one by x two minus one by x one ok
so from here q x is given by t two minus t
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00:17:22,579 --> 00:17:38,660
one divided by one by x two minus one by x
one multiplied by k pi a square multiplied
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00:17:38,660 --> 00:17:51,179
by k pi a square and so now we can plug this
into our solution we can plug this into our
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00:17:51,179 --> 00:18:04,270
solution so t equal to t one plus k pi plus
t two minus t one divided by one by x two
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00:18:04,270 --> 00:18:13,580
minus one by x one multiplied by one by x
minus one by x one so thats the distribution
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00:18:13,580 --> 00:18:21,580
temperature distribution in the system with
the varying cross sectional area so an important
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00:18:21,580 --> 00:18:29,140
message of this example is that what you need
what is preserved here or what remains constant
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00:18:29,140 --> 00:18:35,770
in this system because there is no heat generation
or dissipation is the heat transfer rate and
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not the flux ok so you have to make a distinction
here so what we said is the heat transfer
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rate
so this remains a constant however the flux
145
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which is given by q prime which is minus k
d t by d x this is not a constant so this
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00:19:05,880 --> 00:19:11,549
is an important observation which you would
not have made in the simple one d system where
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00:19:11,549 --> 00:19:16,410
the cross sectional area is constant the cross
sectional area of heat transport is constant
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00:19:16,410 --> 00:19:21,750
where you would not be able to make such a
distinction between the two so when you have
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00:19:21,750 --> 00:19:26,000
varying cross sectional area what is really
conserved and what is really preserved is
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00:19:26,000 --> 00:19:31,590
the heat transfer rate and in fact thats the
reason why you write a a rate balance and
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00:19:31,590 --> 00:19:37,130
not a flux balance so this is very important
to understand this distinction its important
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to write ah transfer rate balance because
thats the final quantity that is importance
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and the flux need not necessarily remain constant
even in a small element so its very important
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00:19:48,820 --> 00:19:53,539
to understand these distinction and we are
going to next see how these things are going
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00:19:53,539 --> 00:19:57,580
to play a role when you are looking at radial
systems where this becomes extremely important
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00:19:57,580 --> 00:19:57,850
ok