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the question is where you want to incorporate
the assumption and when you want to incorporate
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the assumption you could incorporate the assumption
after you write the generalized model now
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this generalized model does not include the
assumption that the cross sectional gradient
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is negligible and this is the way to do that
you cannot arbitrarily neglect the ah second
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order derivative in the laplacian because
the boundary condition is now going to be
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different and the way to see that made to
appreciate that is ah i would probably urge
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you all to take a look at the solution of
two different types of problems one is you
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take a diffusion operator and you put zero
boundary conditions ok dirichlet or zero boundary
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conditions and you take the same problem put
zero flux boundary condition and you take
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the same differential operator and put a ah
two flux boundary conditions ok like something
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like this and you have to convince yourself
that the nature of the solution you get in
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all these three are completely different
particularly in the third case it will be
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completely different ah if you want me to
state it i would like you all to convince
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yourself by solving this problem you solve
d square t by d x square equal to zero with
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three types of boundary conditions you can
say t at x equal to zero equal to t at x equal
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to one equal to zero ok thats one problem
and then you say d t by d x at x equal to
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zero equal to d t by d x at x equal to l equal
to zero thats the second problem and the third
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one is d t by d x at x equal to zero equal
to some h into h by k into t at x equal to
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zero minus t infinity and d t by d x h by
l into t at x equal to l minus t infinity
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thats the third problem one could in principle
have mixed boundary conditions you should
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actually even try to solve that also and you
must appreciate that the these three classes
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of problems have completely different solution
and in fact if you understand that thats the
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reason why you cannot incorporate this assumption
directly into the generalized model and this
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is the correct way to incorporate the assumption
that the gradients in the radial direction
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is zero particularly when you have this kind
of a a convection or flux boundary condition
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if you have these two types of boundary condition
then you would assume you can directly ignore
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the gradients and you would not make a mistake
because of the nature of the solution that
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00:03:02,430 --> 00:03:07,449
we will get for these three types of problem
i its a very good exercise all of you should
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actually try to solve these three types of
problems and you will see these three classes
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of problems in almost all the engineering
systems that you will see in this program
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and also probably elsewhere its a very very
common type of differential equation problem
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that you will see in many many engineering
systems and there are some very good reasons
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for that i will not get into the but there
are very good reasons why these classes of
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problems will appear in almost all the natural
systems that you will try to model assuming
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cross sectional area temperature to be uniform
is not an appropriate and not a realistic
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thing if the aspect ratio is reasonably good
if the aspect ratio is very small then you
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can assume that the temperature is almost
uniform
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now of course you dont have to assume that
to be uniform it just makes your life easy
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to solve the problem thats all now you could
in principle take the full problem and get
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00:03:59,430 --> 00:04:03,709
the full solution and you will see that the
solution that you will get from this equation
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00:04:03,709 --> 00:04:08,219
will not be very different if you impose the
condition that the temperature is nearly uniform
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in the cross section it doesnt matter you
can go either way it doesnt matter you can
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00:04:13,049 --> 00:04:16,690
start from the generalized equation and then
you can integrate you will get exactly the
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00:04:16,690 --> 00:04:22,150
same and you can start from you can you incorporate
the assumptions while building the model you
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00:04:22,150 --> 00:04:27,720
will get exactly the same two errors it wouldnt
matter and in fact some of your things about
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00:04:27,720 --> 00:04:32,039
the integral balance is already incorporated
here this is exactly what you do when you
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00:04:32,039 --> 00:04:38,410
write an integral balance ok so lets take
one of the questions that he asked what happens
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there a x is constant ah before that we should
actually go to the boundary conditions of
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00:04:49,640 --> 00:04:59,289
this problem
so so in principle there are four different
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00:04:59,289 --> 00:05:07,159
possible boundary conditions that could have
one is obviously you can have a a flux boundary
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condition d t by d x at x equal to l will
be h t at x equal to l minus t infinity so
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00:05:16,939 --> 00:05:21,310
thats one possible boundary condition that
you would obviously guess that just like how
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00:05:21,310 --> 00:05:25,759
heat transfer is occurring from this cross
sectional area you would expect that the heat
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transfer would occur from the other edge of
the fin well now the second possible boundary
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00:05:34,010 --> 00:05:38,830
condition is that there is no flux may be
that section is insulated so if the boundary
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00:05:38,830 --> 00:05:44,610
is insulated with some insulator then there
is no heat exchange that is you impose a an
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adiabatic condition at the other end of the
fin and then the third one is you could impose
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a constant boundary condition so this is very
unlikely but just for hypothetical case you
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00:05:56,680 --> 00:06:02,710
can assume that t at x equal to l is some
constant temperature ah there are some example
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00:06:02,710 --> 00:06:10,550
but they are not that many and the fourth
one which is generally used for some ah standardization
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00:06:10,550 --> 00:06:14,139
purposes
so supposing if the fin is infinitely long
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infinitely long fin what will be the boundary
condition is fin is infinitely long want to
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00:06:28,340 --> 00:06:33,719
try if this fin is infinitely long and if
the temperature of the fluid is t infinity
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00:06:33,719 --> 00:06:34,719
you what would you expect
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00:06:34,719 --> 00:06:43,210
t equal to equal to t infinity right so you
expect that t at x equal to l l tending to
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00:06:43,210 --> 00:06:48,960
infinity will be equal to t infinity so that
just for standardization purposes you will
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00:06:48,960 --> 00:06:54,280
never achieve such an ideal situation however
long the fin is going to be but its just for
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00:06:54,280 --> 00:06:59,941
standardization purposes such kind of boundary
condition has been prescribed ok all right
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00:06:59,941 --> 00:07:09,370
so next we go to a specific case where so
supposing we assume that the cross sectional
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00:07:09,370 --> 00:07:11,129
area
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00:07:11,129 --> 00:07:19,960
is constant which means that the cross sectional
area is same and therefore the curved surface
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area is actually a curved surface area of
a cylinder so this equation would simply which
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00:07:26,900 --> 00:07:37,639
means that d a c x by d x is zero which means
that the gradient of the cross sectional area
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00:07:37,639 --> 00:07:42,479
with respect to x is zero because its a constant
and so we can simply reduce this equation
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00:07:42,479 --> 00:07:54,879
at v square d by d x square thats equal to
h by a a c t minus t infinity what is d a
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00:07:54,879 --> 00:08:05,389
s by d x for constant cross sectional area
its the its two pi r perimeter right its the
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00:08:05,389 --> 00:08:16,749
perimeter so if p is the perimeter p is the
perimeter of the fin that you are considering
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00:08:16,749 --> 00:08:27,240
and that should be the that should be equal
to d a s by d x the change in the cross sectional
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00:08:27,240 --> 00:08:31,590
surface surface area for heat transport is
simply the perimeter of that particular fin
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how do i solve this equation is it complete
do i need to know something else to solve
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00:08:42,402 --> 00:08:48,970
the equations i need to know the boundary
conditions right so if i specify the boundary
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condition that x equal to zero is the temperature
of the base ok so note that the fin is attached
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00:08:56,079 --> 00:09:01,529
to a base surface and so if the temperature
of the base surface is specified so thats
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00:09:01,529 --> 00:09:07,110
going to be the boundary condition at x equal
to zero and i could say that t my if i assume
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00:09:07,110 --> 00:09:15,769
that there is a convection boundary condition
so if i specify this boundary condition then
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00:09:15,769 --> 00:09:31,829
we can solve the equation how do we solve
this equation how do we solve this yeah its
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00:09:31,829 --> 00:09:40,279
not that hard some exponential so you say
theta is t minus t infinity you make a substitution
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00:09:40,279 --> 00:09:48,470
and then this will become d square theta by
d x square equal to if i call this m square
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00:09:48,470 --> 00:09:59,280
m square theta and this will be theta at x
equal to zero is t b minus t infinity which
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00:09:59,280 --> 00:10:03,600
i can call it as theta b right so i get it
right
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00:10:03,600 --> 00:10:15,410
so t is theta is t minus t infinity right
so ok and then minus k d theta by d x at x
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00:10:15,410 --> 00:10:25,410
equal to l thats equal to h into theta with
a pretty easy and you can solve this is an
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00:10:25,410 --> 00:10:31,250
exponential solution its like an eigen value
problem what are the eigen values of this
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00:10:31,250 --> 00:10:43,300
differential equation eigen functions eigen
functions are e power yeah plus m x and minus
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00:10:43,300 --> 00:10:48,440
m x so you can solve the eigen value problem
and substitute the boundary condition i will
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00:10:48,440 --> 00:10:57,259
just give you the final form so theta x pi
theta b this is the base temperature is given
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00:10:57,259 --> 00:11:12,720
by the ratio cos m l minus x plus h by m into
k into sin hyperbolic m l minus x divided
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00:11:12,720 --> 00:11:26,560
by cos hyperbolic m l plus h by m k sin hyperbolic
m l ok
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00:11:26,560 --> 00:11:39,480
so thats the solution and the profile will
be so if theta b thats the profile that you
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00:11:39,480 --> 00:11:51,940
will expect so thats the profile that you
can expect from this solution and in fact
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00:11:51,940 --> 00:11:56,829
you should go and plot this and see its sort
of obvious to read it out if you know how
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00:11:56,829 --> 00:12:01,430
cosine hyperbolic and sin hyperbolic functions
look like if you dont know i think its a good
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00:12:01,430 --> 00:12:07,850
exercise to go on plot cosine and sin hyperbolic
and convince yourself that this is the solution
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00:12:07,850 --> 00:12:17,269
for this kind of a problem ok so now ah we
said that when we started this extended surfaces
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problem we said that the purpose of extended
surface is essentially to increase the heat
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transfer rate right so therefore we need to
find out what is the total heat that is transferred
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00:12:30,100 --> 00:12:35,980
by the fin ok so thats what thats the quantity
of interest which is going to quantify the
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00:12:35,980 --> 00:12:43,399
problem that you are looking at so we need
to find out what is the total heat transfer
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00:12:43,399 --> 00:12:58,649
rate
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00:12:58,649 --> 00:13:04,970
how do we find this how do we find the total
heat transfer rate that is the total amount
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00:13:04,970 --> 00:13:11,430
of heat that is transported by the fin so
note that there are two processes one is it
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00:13:11,430 --> 00:13:17,839
is transferring by conduction and the heat
is being lost from the other end of the fin
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00:13:17,839 --> 00:13:22,630
and while doing that it is also simultaneously
losing heat way of convection so how do we
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00:13:22,630 --> 00:13:24,269
find the total heat transfer rate
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00:13:24,269 --> 00:13:26,420
yeah
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00:13:26,420 --> 00:13:31,130
they are supposed to be
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00:13:31,130 --> 00:13:43,160
added how so one could integrate h into d
a s into t minus t infinity ok across the
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00:13:43,160 --> 00:13:48,959
whole fin so that will give you the total
amount of heat that is transferred so thats
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00:13:48,959 --> 00:13:55,520
one way to do that but so note that i always
told you that you should use your intuition
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00:13:55,520 --> 00:14:02,220
so there is an intuitively simple way to do
this ok so we assume that this system is under
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00:14:02,220 --> 00:14:11,980
steady state right is that the system is under
steady state conditions so if it is under
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00:14:11,980 --> 00:14:17,610
steady state conditions it means that the
temperature profile remains constant right
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00:14:17,610 --> 00:14:27,279
it doesnt change the time so whatever heat
so supposing if this is my fin this is my
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00:14:27,279 --> 00:14:34,670
fin because i we impose a steady state condition
and there is no heat generation inside so
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00:14:34,670 --> 00:14:41,980
whatever heat that is actually transferred
to the fin at the base should be equal to
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00:14:41,980 --> 00:14:48,310
whatever total heat that is being transferred
right its a very simple simple thing there
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00:14:48,310 --> 00:14:52,949
are is it clear to everyone
whatever is the total amount of heat that
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00:14:52,949 --> 00:14:57,690
is transferred from the base to the fin should
be the total amount of heat that the fin is
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00:14:57,690 --> 00:15:02,829
able to transfer because i said its a steady
state condition the temperature profile has
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00:15:02,829 --> 00:15:07,709
to be maintained so whatever heat that comes
in should actually go out otherwise the temperature
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00:15:07,709 --> 00:15:14,251
profile is going to change our assumption
is not valid so in order to satisfy the assumption
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00:15:14,251 --> 00:15:18,990
that we made that its under steady state condition
an intuitive way to find the total amount
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00:15:18,990 --> 00:15:23,910
of heat transfer is what is the amount of
heat that is actually transferred from the
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00:15:23,910 --> 00:15:28,269
base to the fin at x equal to zero
in fact if you integrate you will find that
135
00:15:28,269 --> 00:15:37,259
it will be exactly the same ok and that will
be give will give you the expression so that
136
00:15:37,259 --> 00:15:44,589
will be minus k so the total amount of heat
that is transferred by the fin q f is minus
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00:15:44,589 --> 00:15:52,750
k into cross sectional area into d t by d
x at x equal to zero so thats the amount of
138
00:15:52,750 --> 00:15:58,290
heat that is transferred to the fin at x equal
to zero and that should be equal to the total
139
00:15:58,290 --> 00:16:08,000
amount of heat that is transferred so that
will be equal to
140
00:16:08,000 --> 00:16:22,519
and that should be the theta b square root
of h p k c k a c into sin hyperbolic m l plus
141
00:16:22,519 --> 00:16:41,620
h by m k into cos hyperbolic m l divided by
cos hyperbolic m l plus h by m k sin hyperbolic
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00:16:41,620 --> 00:16:50,370
m l ok so thats the total amount so note that
its the total amount of heat that is transferred
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00:16:50,370 --> 00:16:56,060
the local heat transfer rate is not constant
so remember the resis when we discussed about
144
00:16:56,060 --> 00:17:02,779
the resistances that the whenever there is
heat generation or heat loss at local location
145
00:17:02,779 --> 00:17:08,380
from this from the solid that you are looking
at the total heat there the local heat transfer
146
00:17:08,380 --> 00:17:13,890
rate is now going to be a function of position
and in fact that is part of alluded to in
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00:17:13,890 --> 00:17:19,300
this expression so note that the total heat
transport rate is now a function of the length
148
00:17:19,300 --> 00:17:24,420
of the fin that you are considering
if it well to be constant then irrespective
149
00:17:24,420 --> 00:17:28,660
of the length the heat transfer rate should
be exactly the same thats not true so its
150
00:17:28,660 --> 00:17:33,120
a function of the length i mean sort of it
is reflected in the total heat transport rate
151
00:17:33,120 --> 00:17:38,320
so if you know the length of the fin then
you should be able to calculate what is the
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00:17:38,320 --> 00:17:43,110
total amount of heat that that particular
fin can actually transfer from the base and
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00:17:43,110 --> 00:17:48,830
this is an important design parameter so you
have if you want if you know what is the total
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00:17:48,830 --> 00:17:54,260
amount of heat that you want to transfer then
you could use this expression in order to
155
00:17:54,260 --> 00:17:59,620
design the length of the fin so this is often
an important parameter that you have to find
156
00:17:59,620 --> 00:18:04,050
what is the length of the fin that you have
to use in order to achieve a certain amount
157
00:18:04,050 --> 00:18:08,030
of heat transport rate is that clear to everyone
ok
158
00:18:08,030 --> 00:18:14,440
so ah i will just give you the expressions
for the other type of boundary conditions
159
00:18:14,440 --> 00:18:19,840
but i would ah strongly encourage you to derive
them and convince yourself so supposing if
160
00:18:19,840 --> 00:18:27,280
you use no flux boundary condition so thats
the second type of boundary condition d t
161
00:18:27,280 --> 00:18:40,380
by d x at x equal to l is zero and the solution
will be theta by theta b is cos h m l minus
162
00:18:40,380 --> 00:18:51,550
x divided by cos hyperbolic m l and the total
heat transfer rate q f is given by square
163
00:18:51,550 --> 00:19:04,100
root of h p k times a c into theta b into
tan hyperbolic m l and for the third boundary
164
00:19:04,100 --> 00:19:10,680
condition where you have a fixed temperature
that is theta at x equal to l is theta l the
165
00:19:10,680 --> 00:19:21,340
solution will be theta by theta b equal to
theta l by theta b multiplied by sin hyperbolic
166
00:19:21,340 --> 00:19:39,740
m x plus sin hyperbolic m l minus x divided
by sin hyperbolic m l and for the fourth type
167
00:19:39,740 --> 00:19:44,620
of boundary condition where you have long
fin infinitely long fin
168
00:19:44,620 --> 00:19:53,620
so theta at x tending to t infinity thats
equal to zero that is the temperature and
169
00:19:53,620 --> 00:19:58,460
the other end is equal to the temperature
of the fluid itself and so that case theta
170
00:19:58,460 --> 00:20:10,290
by theta b is simply given by e power minus
m x and q f is given by square root of h p
171
00:20:10,290 --> 00:20:18,400
k a c into d theta b so those are the four
different solutions and i would encourage
172
00:20:18,400 --> 00:20:25,090
all of you to actually derive these expressions
and convince yourself that this is the solution
173
00:20:25,090 --> 00:20:33,500
ok so there is another definition and some
expression for the definition with that we
174
00:20:33,500 --> 00:20:45,750
will finish todays lecture what is called
the efficiency so note that when we said the
175
00:20:45,750 --> 00:20:52,050
purpose of a fin is basically to enhance the
heat transfer rate so therefore from design
176
00:20:52,050 --> 00:20:58,410
point of view it is useful to define a quantity
called efficiency how efficient is the fin
177
00:20:58,410 --> 00:21:04,580
that i have designed can i calculate the efficiency
and find out what so if i want to compare
178
00:21:04,580 --> 00:21:10,810
two different fins in instead of comparing
what is the actual total heat transfer rate
179
00:21:10,810 --> 00:21:16,420
it is useful to compare the efficiency of
the ah fin that has been designed
180
00:21:16,420 --> 00:21:25,570
so therefore the efficiency of a fin is basically
defined as the total heat transport rate that
181
00:21:25,570 --> 00:21:34,320
is the total amount of heat that is transferred
by the fin divided by the maximum possible
182
00:21:34,320 --> 00:21:41,800
heat that can be transferred divided by the
maximum possible heat that can be transferred
183
00:21:41,800 --> 00:21:55,150
by the same fin what is q max how can we find
the maximum possible so lets pose a slightly
184
00:21:55,150 --> 00:22:01,420
different question when can we achieve maximum
heat transfer from a fin of a given geometry
185
00:22:01,420 --> 00:22:18,040
when can we assume when can when can we get
maximum heat transport rate under what conditions
186
00:22:18,040 --> 00:22:21,930
temperature
187
00:22:21,930 --> 00:22:28,610
but thats thats like infinitily long fin you
never going to achieve it in fact if the temperature
188
00:22:28,610 --> 00:22:33,980
the right end is t infinity then the temperature
gradient or driving force for convection is
189
00:22:33,980 --> 00:22:40,260
almost zero at the end of the fin so which
means that the ah the the end of the fin are
190
00:22:40,260 --> 00:22:47,092
almost of no use the any heat transport rate
is governed by the temperature gradient which
191
00:22:47,092 --> 00:22:51,810
is the driving force so if the temperature
at the end of the fin is equal to that of
192
00:22:51,810 --> 00:23:00,730
the fluid temperature its no use so you dont
want to design a system where some parts of
193
00:23:00,730 --> 00:23:07,730
your fin is going to have a zero driving force
thats useless its a waste of money ok so how
194
00:23:07,730 --> 00:23:16,090
do we find maximum when do you achieve maximum
heat transfer rate its hypothetical you are
195
00:23:16,090 --> 00:23:19,559
not going to achieve it temperature is
constant
196
00:23:19,559 --> 00:23:22,930
constant where
197
00:23:22,930 --> 00:23:27,140
throughout the length ok what is that constant
thats equal to the
198
00:23:27,140 --> 00:23:31,280
thats equal to the base temperature ok so
whats the purpose of a fin the purpose of
199
00:23:31,280 --> 00:23:37,860
a fin is to transfer heat from the base to
the fluid around when can you achieve maximum
200
00:23:37,860 --> 00:23:44,020
heat transfer rate when the temperature in
the fin is constant and not just at any constant
201
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it should be equal to the temperature of the
base itself because thats the maximum possible
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heat that you can transfer so therefore q
max is simply given by h into the overall
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conductive heat transport area multiplied
by t b is the base temperature so the maximum
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possible rate so note that this is maximum
possible rate but what you will actually achieve
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in reality will obviously be lesser than this
so this is the maximum possible and therefore
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one can define efficiency as q f divided by
h a s into t b minus t infinity
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so thats the efficiency of a fin and in fact
for this problem the efficiency would be ah
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so for the second type of boundary condition
where its an adiabatic case so eta f simply
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given by square root of h k perimeter into
a c into theta b tan hyperbolic m l divided
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by h into a s into theta b theta b is nothing
but t b minus t infinity so thats the substitution
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that we made when we solved the differential
equation and so thats nothing but tan hyperbolic
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m l divided by m l ok what happens when l
equal to zero yeah
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what does tan m l by m l when l equal to zero
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ok what is tan hyperbolic when l is zero
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yeah
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zero so it is zero by zero so what is it oh
its not that simple how do you know that its
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one
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when you have an indeterminate system what
do you do
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you do an l hospitals rule l hopitals rule
and so you will find out that eta f is equal
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to one thats sort of obvious to intuit and
the reason why you can intuit is you said
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that the maximum temperature in the fin is
achievable only when the fin size is zero
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you can never achieve a constant temperature
which is equal to the base temperature so
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therefore the efficiency if at all you have
the base cross sectional area which is sufficient
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to transport heat as much as whatever is required
to be transported then the efficiency is going
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to be one but this is not a realistic case
its just hypothetical yes
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number you designed ah
will be zero
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thats right so which means that as l goes
to infinity eta f goes to zero thats right
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so its not that you design a long fin and
you are going to have very high heat transport
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so its completely counter intuitive so although
intuitively you would guess that i design
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a very very long fin and i will be able to
achieve as much as heat transport that i want
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thats not really true and the reason why the
efficiency is very poor is that the as you
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as the length increases the driving force
is constantly decreasing because you have
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two processes which are occurring simultaneously
one is the conduction and the other one is
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the convection so the driving force which
is actually forcing the efficiency to be zero
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ok