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alright so we looked at ah we stopped in the
last lecture by observing that ah there are
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ways to quantify the conduction process so
we are going to go ah deep into that today
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and look at what are the ways to quantify
and look at certain examples and certain types
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where we can look at the quantification process
and try to understand what is the temperature
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00:00:33,890 --> 00:00:41,460
distribution so we stop by observing that
we need to find the temperature distribution
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so in order to quantify the system we need
to find the temperature distribution so if
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00:00:54,590 --> 00:00:59,710
we know the distribution obviously we will
be able to find out the gradient that is the
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00:00:59,710 --> 00:01:05,100
d t by d x the temperature gradient at any
location in a system and if we know that we
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know the transport rate we know the flux and
we multiplied by the transport area we will
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00:01:08,280 --> 00:01:14,930
know the heat transfer rate ok
so we need to write energy balance so which
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00:01:14,930 --> 00:01:27,069
means that we need to write a energy balance
ok so lets take a a general differential element
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so x y to z what comes out is q z plus to
delta z q x x q y so those are the rate at
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00:02:28,879 --> 00:02:35,060
which heat is entering different surfaces
of the element and we could add a we can say
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00:02:35,060 --> 00:02:54,879
that q dot is the rate of heat generation
per unit volume so in principle something
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00:02:54,879 --> 00:02:59,780
could be happening inside for example there
could be electrical heating which leads to
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00:02:59,780 --> 00:03:05,430
generation of heat inside the system that
you are considering so q dot could be the
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00:03:05,430 --> 00:03:10,390
volume at rate of heat generation that is
the amount of heat that is generated per unit
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00:03:10,390 --> 00:03:15,040
volume of that system
so the volume of this small element v is nothing
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00:03:15,040 --> 00:03:22,790
delta x multiplied by delta y multiplied by
delta z ok so we know that so lets write a
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simple balance so the general thumb rule ok
for writing any balance whether it is heat
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00:03:28,159 --> 00:03:44,760
or mass balance is that we say input to that
element minus output plus generation term
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that should be equal to the amount that is
accumulated that will be accumulation so you
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could call this i we call this o we could
call this g and you could call this a ok so
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this is the manthra of all the balances that
you would be writing in this course and perhaps
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in many other courses till you finish your
b tech and in fact this is the ah manthra
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in almost all the balances that you would
write in any engineering discipline irrespective
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of whether it is chemical mechanical etcetera
so this is the general this is the general
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manthra and we wont go dwell more into this
so we will use this manthra here today and
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then we are going to write a general balance
for ah conservation of energy for this system
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and we will use that balance and try to simplify
for different kinds of systems that we are
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going to look into in the future alright so
what is the input to the system the amount
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of heat that is input is q x right plus q
y plus q z so thats the total amount of heat
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rate rate at which heat is being input to
that system that we are considering and what
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00:05:04,080 --> 00:05:16,569
leads is q x plus d x q y plus d y minus q
z plus d z thats the amount that is leading
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00:05:16,569 --> 00:05:21,650
the element that we have considered plus whatever
is being generated so you have to be very
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00:05:21,650 --> 00:05:24,970
careful
so when you have a heat loss term when you
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00:05:24,970 --> 00:05:29,500
have a sink term or heat loss term then you
have to use a negative sign so you have to
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very careful about the sign convention that
you use for any balance that you generally
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00:05:35,020 --> 00:05:42,009
know ok so that could be q dot so this is
defined as rate of heat generation per unit
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volume so therefore you have to multiply by
the volume of the system that you are considering
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00:05:50,330 --> 00:05:55,910
and thats equal to the accumulation which
is given by rho c p which is the density of
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00:05:55,910 --> 00:06:02,100
the material that you are using multiplied
by the specific heat capacity into heat temperature
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gradient with respect to time multiplied by
the volume so thats the heat balance it is
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00:06:11,460 --> 00:06:18,280
very simple input what goes out we subtract
that and whatever is generated you add to
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00:06:18,280 --> 00:06:21,379
the system that should be equal to whatever
is being accumulated
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00:06:21,379 --> 00:06:26,520
so this term tells you what is the amount
of heat that is actually being stored what
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00:06:26,520 --> 00:06:31,360
is the amount of energy that is being stored
by the system because of the heat transport
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process so this tells you what is the capacity
of the system that you are looking at and
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c p is the parameter intrinsic parameter which
quantifies that capability of that system
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has to how much heat that it can store ok
so now we can simply divide these equations
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00:06:49,090 --> 00:07:00,470
by the volume and so it will be for the units
matching here c p is for unit volume ok are
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00:07:00,470 --> 00:07:11,379
the units matching look at the units i want
you to pay attention to the units
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00:07:11,379 --> 00:07:15,780
the units matching for all the terms here
you have to convince yourself when you are
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00:07:15,780 --> 00:07:24,590
actually in the class everything thats happening
is right is it right no what is wrong
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00:07:24,590 --> 00:07:25,590
which one
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00:07:25,590 --> 00:07:27,449
where this one here
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00:07:27,449 --> 00:07:37,639
yeah why is that a problem here
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00:07:37,639 --> 00:07:43,000
but its already taken right watts per meter
cube is the unit of q dot we multiplied by
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00:07:43,000 --> 00:07:54,659
volume it becomes watts what is what is the
units of q x is watts it is watts its hot
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00:07:54,659 --> 00:08:00,550
flux its transfer rate is rate of heat transfer
keep in mind the convention that we use ok
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00:08:00,550 --> 00:08:07,349
you should always remember this when i say
q it is rate when i say q single prime its
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00:08:07,349 --> 00:08:13,539
rate per unit distance and when i say q double
prime its p flux that is the rate of heat
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transfer per transport area ok alright so
now i divide equation by so q x plus d x divided
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by delta x into one by
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00:08:27,780 --> 00:08:41,479
excuse me joule per kilogram kelvin and rho
is k g per meter cube so thats why you have
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00:08:41,479 --> 00:09:12,890
multiply by volume ok plus q y minus q y plus
delta y divided by into one by ok so moment
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00:09:12,890 --> 00:09:30,330
we formulate it in this form all i have done
is i have just divided the equation by the
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00:09:30,330 --> 00:09:35,510
volume and i have just collected the terms
which corresponds to x y and z together so
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00:09:35,510 --> 00:09:44,350
the important thing to observe here is that
this delta y delta z which appears as a coefficient
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00:09:44,350 --> 00:09:51,130
for the rate terms in x direction that is
nothing but the area of heat transport in
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the x direction so that tells you what is
the cross sectional area at which the heat
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transport is occurring in the x direction
that is basically this plane here you see
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this plane here delta x its the plane in the
z and the y direction similarly delta x delta
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00:10:10,130 --> 00:10:17,030
z is the cross sectional area of heat transport
in the y direction and similarly for the z
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direction ok is that clear to everyone ok
alright so so we can rewrite this as so when
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i say that limit delta x goes to zero delta
y goes to zero and delta z goes to zero
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so we can write this as d q by d q by d x
one by delta y delta z plus its a minus sign
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00:10:51,540 --> 00:11:03,750
minus d q by d y delta z z o t ok so with
this can i get the temperature distribution
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00:11:03,750 --> 00:11:32,810
yes or no can i get the temperature distribution
with this so if we know what yes
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00:11:32,810 --> 00:11:38,670
is some constitutive relationship so we dont
know what q is so if there is a constitutive
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00:11:38,670 --> 00:11:48,080
relationship if you have not heard this word
called constitutive relationship if we know
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00:11:48,080 --> 00:11:59,350
the constitutive relationship between q and
temperature we are done so if we know the
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00:11:59,350 --> 00:12:06,120
constitutive relationship between the transfer
rate and the temperature then we are done
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00:12:06,120 --> 00:12:10,490
we have completely described the temperature
distribution we dont know the distribution
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00:12:10,490 --> 00:12:15,690
because we have to solve the equations but
we have described it completely ok and this
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00:12:15,690 --> 00:12:24,140
is what is given by fouriers law that we saw
in the last lecture so q is given by the area
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00:12:24,140 --> 00:12:30,351
of heat transport in whichever direction you
are considering supposing if it is x its the
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area of heat transport in x direction multiplied
by the corresponding flux in the x direction
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00:12:37,760 --> 00:12:46,580
right and so this is given by minus k so supposing
i assume the system is isotropic if i assume
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that its a isotropic system there is no reason
that you should not assume it but lets say
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00:12:54,180 --> 00:12:59,810
that for simplicity we assume that its a isotropic
system although all the the framework does
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00:12:59,810 --> 00:13:04,610
not change whether the system is isotropic
or non isotropic but lets say for simplicity
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purposes we assume that its isotropic
so there will be k into what s the area of
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00:13:09,630 --> 00:13:16,690
heat transport in x direction delta y delta
z so we observe that so thats the coefficient
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that comes out in your energy balance into
d t by d x so thats the representation of
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fouriers law in the x direction similarly
we can write for y
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minus k
y and q z is k z so we can substitute all
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those into the model equation now so this
is what is called the energy balance or model
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00:14:08,720 --> 00:14:22,380
equation thats what is called the model equation
so we can substitute these fluxes from fouriers
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law into the model equation d by d x into
k into delta y delta z plus d by d y k into
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plus q dot equal to rho c p into tau t by
ok so supposing if the area does not change
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with the cross section area is constant supposing
for a system in case of cartesian coordinates
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we will see that the cross sectional area
of heat transport remains constant
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so if we assume that the for a system where
area of heat transport is constant we will
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see in future we will see few examples of
area is not constant and how these model equations
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will change but lets say to start with we
assume that the area of heat transfer is constant
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which means we can plot delta x delta y etcetera
from the derivatives inside and so this will
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simply become k into d square t by d x square
plus d square t d y square plus d square t
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00:16:25,880 --> 00:16:45,330
by d z square plus q dot equal to rho c t
tau t so this is the energy balance ok what
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00:16:45,330 --> 00:16:53,350
is this term in the bracket called as classically
called the laplacian so one could rewrite
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00:16:53,350 --> 00:17:06,380
this expression as del square t plus q dot
divided by k equal to rho c p divided by k
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00:17:06,380 --> 00:17:16,780
into so this is the ah a cute way of writing
the energy balance what is this term does
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anyone know
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00:17:17,780 --> 00:17:30,460
yeah number is non dimension its a dimensionless
number this has a dimension rho is k g per
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meter cube c p is joule per kilogram kelvin
and k is watt per meter kelvin so what are
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the units of this quantity so the units are
its k by rho c p the unit is meter square
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per second yes
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because in principle the system need not be
isotropic right so if it is not isotropic
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then you would expect that this k s will be
different and of course the gradients can
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obviously be different so therefore you have
to distinguish the fluxes in all three directions
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and the transfer rate in all three direction
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its oh oh it should be q x q y sorry thanks
thanks for pointing that out please correct
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your notes which should be q x q y its a typo
so so q x q y and q z are the transfer rates
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in the corresponding direction ok alright
so the units of this k by rho c p is meter
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square per second can you guess what it is
what this quantity is from the units
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its diffusivity so its called so this quantity
is called thermal diffusivity thermal diffusivity
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so so what it signifies is k is the thermal
conductivity and rho c p is the capacity of
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the system right so this is the ratio between
the ability of the system to conduct it versus
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its ability to store the energy and so rho
c p remember what i told you a few moments
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ago rho c p is characterizes its in the intrinsic
property that quantifies the ability of the
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00:19:30,720 --> 00:19:38,330
system to store heat within itself store energy
within itself and k is the intrinsic property
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of the system which characterizes the ability
of the system to transfer heat from one location
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to the other via conduction so this is the
ratio between the ability of the system to
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conduct heat versus the ability of the system
to store heat within itself and that is what
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is called thermal diffusion yes
divided by
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so so this is the so what happens so there
are so there are two processes which are occurring
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simultaneously one is the energy is being
transferred because there is collisions between
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the electrons or there is lattice waves which
is happening and so there is transfer of heat
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from one electron or one molecule to the other
molecule now when the temperature is increased
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the energy state of that molecule also is
going to be increased right so the rho c p
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00:20:39,149 --> 00:20:45,850
it signifies the ability of the molecules
to store heat because of higher temperature
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00:20:45,850 --> 00:20:50,760
and the interaction between them is the one
which is going to signify the ability to transport
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00:20:50,760 --> 00:20:58,130
the heat so this ratio is basically the thermal
diffusion and the reason why it is ratio is
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thermal diffusion is supposing if the material
has a very capability to store heat then the
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00:21:03,759 --> 00:21:08,320
amount of heat that it can transfer because
of the interaction is going to be very small
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so its a competition between the ability to
transfer heat because of the interaction versus
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00:21:13,809 --> 00:21:19,490
the ability of that system to store the heat
in the same location with the same molecule
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00:21:19,490 --> 00:21:26,429
and thats why its called diffusion in fact
thats the definition of diffusion any other
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00:21:26,429 --> 00:21:36,059
questions ok so is it enough to just write
the balance have we completely described can
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00:21:36,059 --> 00:21:44,799
we solve this equation now is it possible
to solve what do you need this is a partial
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00:21:44,799 --> 00:21:52,039
differential equation what makes a p d e complete
you need boundary conditions right so let
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00:21:52,039 --> 00:21:57,470
us now discuss for the next few moments as
to what are the different types of boundary
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conditions what are the general classes of
boundary condition ok ok so there are three
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00:22:19,159 --> 00:22:26,169
basic classes of boundary condition or three
types of boundary condition one is the constant
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00:22:26,169 --> 00:22:33,059
temperature boundary condition constant temperature
boundary condition
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00:22:33,059 --> 00:22:40,519
so for example if you take a system ok so
lets say this is x equal to zero ok one possible
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00:22:40,519 --> 00:22:46,259
boundary condition is you could fix the temperature
of that particular boundary ok we can say
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00:22:46,259 --> 00:22:52,299
that the temperature is fixed at some constant
so typically i use subscript s for surface
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00:22:52,299 --> 00:23:00,139
so it could be surface temperature so we can
say that the temperature at x equal to zero
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00:23:00,139 --> 00:23:06,289
is maintained at a certain constitutive temperature
t x ok so thats one type of boundary condition
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00:23:06,289 --> 00:23:15,540
so there are several examples of this so supposing
if you want to lets say you have you have
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00:23:15,540 --> 00:23:20,639
water geyser right so we have geysers at our
residence in our bathroom where we get hot
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00:23:20,639 --> 00:23:28,610
water right so the temperature of the surface
so you see there will be a heating coil so
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00:23:28,610 --> 00:23:33,110
the basic geyser works is there is a heating
which is electrically heated and there is
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00:23:33,110 --> 00:23:38,519
a thermostat which maintains a certain temperature
of the heating coil so note that its not the
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00:23:38,519 --> 00:23:42,610
temperature of the water that it maintains
its actually the temperature of the heating
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00:23:42,610 --> 00:23:46,210
coil that it maintains
so if you are trying to write a model of this
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00:23:46,210 --> 00:23:52,919
system to find the temperature distribution
of water then one of the boundaries to which
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00:23:52,919 --> 00:23:57,890
the coil is exposed to to which the water
is exposed to a coil will actually be maintained
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00:23:57,890 --> 00:24:02,700
at a certain constant temperature so its a
constant boundary condition constant temperature
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00:24:02,700 --> 00:24:09,269
boundary condition is what you should use
at that term ok another type of boundary condition
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00:24:09,269 --> 00:24:21,580
is called the constant flux boundary condition
ok so this is the second type there are two
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00:24:21,580 --> 00:24:32,690
different sub types in this case one is zero
flux so note that zero is also a constant
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00:24:32,690 --> 00:24:45,799
so zero flux and then non zero flux non zero
constant flux so the way we look at it is
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00:24:45,799 --> 00:24:52,340
supposing if you have a zero flux ok at x
equal to zero if you have zero flux then you
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00:24:52,340 --> 00:25:00,240
would expect that the so if i look at the
temperature profile inside near x equal to
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00:25:00,240 --> 00:25:05,240
zero because the flux is zero at that location
you will see that the temperature is going
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00:25:05,240 --> 00:25:11,249
to be flat its going to be parallel to the
x axis so we could describe this by ah an
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00:25:11,249 --> 00:25:30,850
expression we can say d t by d x at x equal
to zero so we can describe this boundary condition
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00:25:30,850 --> 00:25:40,960
by the following expression d t by d x at
x equal to zero equal to zero so thats the
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00:25:40,960 --> 00:25:47,580
no flux boundary condition or zero flux boundary
condition and then you can have a constant
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00:25:47,580 --> 00:25:56,809
flux where you say d t by d x at x equal to
zero is some constant oh there should be a
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00:25:56,809 --> 00:26:02,190
k
so the flux is k into d t by d x and with
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00:26:02,190 --> 00:26:08,000
the a minus sign so keep in mind that you
should not forget the minus sign why is there
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00:26:08,000 --> 00:26:15,419
a minus sign because the energy transfer is
in the negative temperature gradient so minus
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00:26:15,419 --> 00:26:21,820
k d t by d x at x equal to zero is equal to
constant flux ok and then the third type of
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00:26:21,820 --> 00:26:31,929
boundary condition is the variable flux boundary
condition or it sometime simply called as
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00:26:31,929 --> 00:26:37,799
flux boundary condition ok but really its
variable flux boundary condition just to distinguish
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00:26:37,799 --> 00:26:46,840
between the second type ok and so that is
simply given by minus k d t by d x at x equal
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00:26:46,840 --> 00:26:59,600
to zero equal to some flux which is f function
of the local temperature ok so how can we
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00:26:59,600 --> 00:27:15,019
define this flux is there a way to quantify
this variable flux is there a way to quantify
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00:27:15,019 --> 00:27:26,840
this flux remember that for conduction we
said the quantification is done by fouriers
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00:27:26,840 --> 00:27:37,190
law what about this any suggestions i will
give you a hand its also called as convective
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00:27:37,190 --> 00:27:42,749
boundary condition or convection boundary
condition