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ok lets get started
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ok so we start by ah making this depiction
and we said that the solution of or the temperature
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distribution is given by q dot l square by
two k into one minus x square by l square
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plus t two minus t one divided by two into
x by l plus t one plus t two by two ok so
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thats the solution what will be the temperature
profile
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parabolic
parabolic ok
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will it be symmetric will it be non symmetric
why is it non symmetric because so it will
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be non symmetric where will be the maxima
can you guess
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closer to t one
closer to t one right so look I mean without
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actually going and plotting there are several
intuitive things which you will be able to
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see simply by looking at the system it is
very very important to capture these intuitive
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behaviors of the system ok so we will have
a maxima so thats the kind of profile that
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you would expect for this this kind of a solution
this is the temperature distribution now supposing
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if t one is equal to t two equal to lets say
a surface temperature ok so then the profile
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will be symmetric right so we should have
a a symmetric profile we will have a a symmetric
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temperature profile and so the solution would
be t two is equal to t one so the second term
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will disappear and this will simply be t s
so the solution will be t x equal to q dot
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l square by two k into one minus x square
by l square plus t s so thats the solution
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when the temperatures of the two surfaces
either surfaces are equal to t s
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now this is an important observation and in
fact now i will explain why i chose this kind
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of a coordinate system right so this solution
is also the solution of the problem where
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you have a an adiabatic boundary ok so lets
say you have a slab where you have an adiabatic
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boundary at x equal to zero that is there
is no heat loss or there is no heat transfer
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complete heat heat transfer at x equal to
zero is zero so remember the second boundary
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condition we discussed constant flux where
here the constant is zero so there is no flux
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boundary condition so this is also the solution
of this problem and the reason why it is is
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this a solution is because if you look at
the solution of because a solution is symmetric
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its symmetric at x equal to zero the flux
is the flux is zero right d t by d x is zero
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at that location and therefore the solution
of this problem is also the solution of this
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problem so you got the solution of another
problem for free simply by redefining the
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coordinate system and looking at the problem
so this is an important piece that you have
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to check so any system that you are looking
at you will also have to see how to define
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your coordinate system so that your solution
is going to give you much more insight than
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what you would normally get so this is very
very important when you are solving engineering
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problems how to define a coordinate system
correctly so that you would get very good
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much better insight than what you would normally
get otherwise ok alright so can we use resistance
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concept here can we define resistance for
this problem
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what do you think if the temperatures are
not same then there is a overall temperature
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difference whats the answer yes no why
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thats right so when you have a generation
term so this is very important when you have
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a heat generation term the whole concept of
resistance network hinges in the fact that
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the total amount of heat that is generated
inside the system is zero there is no heat
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generation or heat loss from the system which
means that the net heat transfer rate from
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the system is constant when you have heat
generation the net heat transfer rate at different
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locations in the system is not constant so
this is very important to realize this distinction
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that you cannot use a resistance network concept
whenever there is a heat generation or a heat
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sink term which is present in the system so
we cannot use resistance concepts I should
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qualify by saying thermalresistance thermal
resistance network if the net heat transfer
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rate is not constant and in fact its not difficult
to realize this the net heat transfer rate
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depends upon the first derivative of the temperature
right so q x is minus k a into d t by d x
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right
so now thats going to be a function of position
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as you can clearly see its a parabolic system
and so the first derivative is going to be
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a a function of position which means that
the q x is now going to be a a positionally
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dependent quantity so the heat transfer rate
is no more constant at any position inside
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the slab so this is a big difference between
what you have with a system where there is
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no heat generation or heat sink versus the
system where there is heat generation or a
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heat sink this is very important to understand
this distinction so the resistance concept
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is not valid for the system where the net
heat transfer rate is not constant if its
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a function of position you cannot use the
resistance network concept alright so what
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do we do so if we cannot use resistance network
concept what do we do we have to deal with
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the original distribution
so note that when we use the resistance network
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concept we didnt care about the distribution
of the temperature as long as we know the
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resistances simply based on the resistances
we will be able to represent the whole ah
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heat transport process in the form of a network
and we are able to detect some properties
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but if we cannot represent using network then
we have to deal with the original equation
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itself so is this is a complete description
can we find the temperature profile sure why
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not if we know q dot we can find it right
so if if q dot is known then we can find the
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we can describe or find the temperature profile
and all the properties in terms of rate etcetera
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can be simply found by taking the first derivative
now supposing we dont know what the supposing
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we dont know what q dot is what should we
do supposing we dont know what is the net
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amount of heat that is generated per unit
volume what should we do any suggestions
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is it still is it is it still a solvable problem
can we still quantify what do you think do
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we need some additional information so the
observation is at x equal to zero when you
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said x equal to zero then there is some simplification
that occurs on this equation right so if q
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dot is not known then we should be able to
measure some other temperature right so if
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q dot is not known then we need to at least
know the temperature at some other location
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so lets say we know how to do that ok so the
rate is done and practice is that so there
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is something called a thermocouple ok so thermocouple
is basically a a copper wire ah which has
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a high conductivity so that the amount of
heat that is transferred the time it takes
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for transferring the heat to measure the total
amount of heat that is transferred and find
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the temperature is going to be extremely small
so its a fast response system and so you put
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a thermocouple in some location and then you
can measure what is the temperature
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so you will see lots of thermocouples in the
lab course that you will do in this semester
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most of the experiments will have some measurement
of temperature and so you will see lots of
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thermocouples going in different directions
into the ah thermometer ah reader ok so you
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will see there is a a digital thermometer
reader and you will see thermocouple going
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from different locations so so you could use
a thermocouple so basically stick in some
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thermocouple and the appropriate point so
this is a designed question supposing i have
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to stick in what is the appropriate point
at x equal to zero so we can stick in at x
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equal to zero and measure the temperature
and lets say if that is call t zero so t zero
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is the temperature at the centre of the system
centre of the slab that we are looking at
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so t at x equal to zero is nothing but q dot
l square by two k plus t s so its obviously
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not a function of the position and so from
here we can find out that q dot is t zero
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minus t s divided by two k by l square ok
so thats q dot and so substituting that in
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the original equation you will find t x that
s equal to ah t zero minus t s divided by
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two k by l square into l square by two k into
one minus x square by l square plus t s
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so therefore so if it is unevenly distributed
then you will have to know what is the distribution
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of q dot in the slab and you could consider
that in our equation for example so the question
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is what happens if it is if q dot is unevenly
distributed so you could simply rewrite your
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model so if q dot itself is a function of
position what if q dot is a function of position
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so you could simply rewrite your model q dot
as a function of position and solve so it
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wouldnt matter you can easily take into account
so when i assume q dot as constant it is completely
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without loss of generalization so if you know
what is the function of q dot you could put
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that and you could solve the equation if it
is solvable if it is non linear then there
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are some restrictions it it is linear you
should be able to solve by enlarge most of
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the linear system so t x minus t s so we can
rewrite the expression as minus t s thats
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equal to one minus x square by l square
so this kind of a reformulation of the solution
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is sometimes called as similarity solution
we will actually see some of the similarity
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solutions when we talk about convection in
a tube laminar flow in a tube but these kinds
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of representation by using the ah maximum
value or the symmetric temperature and scaling
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the solution is sometimes called as the similarity
solution ok so we will look at that sometime
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but just for historical purposes i want to
mention that these type of representation
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is called similarity solution so this ratio
is now going to be a a parabolic function
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and this is a a non dimensional temperature
so note that the quantity on the left hand
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side is non dimensional and this kind of non
dimensionalization is often done in many engineering
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systems when we will actually use these kinds
of non dimensionalization when we actually
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deal with convection problem ok any questions
so far
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alright so lets move into the next problem
so lets look at radial systems or cylinder
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with heat generation ok so lets represent
the cylinder so lets say that we have a solid
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cylinder and there is heat that is being generated
at a volumetric rate of q dot as t watt per
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meter cube of heat that is being generated
and if r naught is the radius of that cylinder
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and lets say that t s is the surface temperature
ok so one could write a model and its very
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simple the way we have done before k and if
we assume that its a one d system one d system
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steady state one d steady state system then
we can say k into one by r d by d r plus q
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dot equal to zero so thats the model what
are the boundary conditions you want to try
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t at r equal to r equal to r naught thats
equal to t s and what about the other boundary
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condition its a two point boundary value problem
right
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ok so d t by d r at r equal to zero is zero
you can see no flux why is it no flux at r
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equal to zero so its because there is a symmetry
which is present in a natural system so this
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symmetry imposes the condition that d t by
d r equal to zero at r equal to zero ok so
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again i am not going to solve this problem
here solve this equations here so the solution
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is t r is minus q dot by four k into r square
plus t s q dot r naught square by four k and
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so that we can simply rewrite as q dot by
four k into r naught square one minus r by
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r naught square plus t s so once again if
we introduce a similar scaling as what we
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did before like looking at the midpoint temperature
etcetera so supposing we know what is the
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midpoint temperature so t naught is given
by q dot r naught square by four k plus t
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s so thats the temperature at the midpoint
and so q dot is given by t naught minus t
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s divided by four k by r naught square ok
so its exactly what we did before and so we
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can substitute that into the temperature distribution
and we will find that t r equal to t naught
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minus t s divided by four k by r naught square
into r naught square by four k oops did i
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make a mistake
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yeah it should be the other way into four
k by r naught square so t naught minus t s
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into four k by r naught square multiplied
by r naught square by four k into one minus
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r by r naught square plus t s so
we can rewrite this as t r minus t s divided
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by equal to one minus r by r naught the whole
square so that s the temperature profile and
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once again we can clearly see that the temperature
the heat transfer rate is a function of the
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radial position so its not constant anymore
unlike what we saw in the radial system with
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the ah with no heat generation we saw that
the heat transfer rate is constant however
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when there is generation heat transfer rate
is not constant so whatever you would intuitively
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guess or whatever you would intuitively identify
is something that will fall out from the equation
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if it does not fall out then what you intuited
is most likely wrong or the model equation
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is wrong either of these two
so you have to look at these two things whenever
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you look at any problem both in the problem
solving things for this course and for any
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other engineering system so you have to intuit
first and then you write the model equation
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ok