1
00:00:14,210 --> 00:00:28,530
so let us look at a specific example so there
is a composite slab and one end of this slab
2
00:00:28,530 --> 00:00:34,920
is maintained at no flux condition that is
adiabatic conditions and the length is specified
3
00:00:34,920 --> 00:00:45,470
as fifty millimeters and twenty millimeters
so thats the thickness of the slab and lets
4
00:00:45,470 --> 00:00:58,550
say that the q dot a and q dot b are the volumetric
generation of heat in these two slabs and
5
00:00:58,550 --> 00:01:09,080
if k a and k b are the corresponding conductivities
and if water is flowing outside ok water is
6
00:01:09,080 --> 00:01:16,620
flowing outside with at a temperature of thirty
degree c and the heat transport coefficient
7
00:01:16,620 --> 00:01:24,680
is thousand watt per meter squared kelvin
ok thats the heat transport coefficient and
8
00:01:24,680 --> 00:01:30,090
you can throw in some number
so if the temperatures are specified as t
9
00:01:30,090 --> 00:01:40,890
naught t one and t two so if q dot a is one
point five into ten power six watt per meter
10
00:01:40,890 --> 00:01:49,750
cube and k a thermal conductivity is given
by seventy five watt per meter kelvin and
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00:01:49,750 --> 00:01:55,930
if q dot b is zero which is which means that
there is no heat generation and the second
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00:01:55,930 --> 00:02:06,359
slab and if conductivity is given by one fifty
watt per meter kelvin ok so we need to find
13
00:02:06,359 --> 00:02:19,260
the temperatures so the question is we need
to find the temperatures of these three locations
14
00:02:19,260 --> 00:02:26,201
t naught is at the location where the adiabatic
condition is maintained and t one is the interface
15
00:02:26,201 --> 00:02:30,880
between the two slabs and t two is the exterior
ah exterior surface
16
00:02:30,880 --> 00:02:33,340
so how should we proceed
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00:02:33,340 --> 00:02:41,820
yeah
18
00:02:41,820 --> 00:02:44,750
right balance equation for both can we do
better than that
19
00:02:44,750 --> 00:02:50,340
hm
20
00:02:50,340 --> 00:02:54,280
we know the solution separately of course
we can write the balances and we can find
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00:02:54,280 --> 00:03:02,950
the solutions ok can we do it better than
that so look at the system and intuit what
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00:03:02,950 --> 00:03:09,010
whats the happening here what is happening
here so you should not blindly go and write
23
00:03:09,010 --> 00:03:14,120
equations all the time you have to see what
what the system is doing can we do better
24
00:03:14,120 --> 00:03:23,480
than that can we do in a simpler fashion can
we solve the problem
25
00:03:23,480 --> 00:03:25,150
so what is happening at this interface
26
00:03:25,150 --> 00:03:30,970
yeah
27
00:03:30,970 --> 00:03:37,020
heat flux is same because whatever heat that
is being transported from slab a is being
28
00:03:37,020 --> 00:03:51,040
transported to slab to at that interface ok
what is the flux at that interface
29
00:03:51,040 --> 00:03:56,290
so note that this is adiabatic right so there
is nothing that is leaving on this end so
30
00:03:56,290 --> 00:04:02,280
whatever heat that is being generated should
actually be the flux at which the heat is
31
00:04:02,280 --> 00:04:08,010
being the total amount of heat that is generated
because of the heat generation term must be
32
00:04:08,010 --> 00:04:14,680
transported only in one direction right so
thats sort of obvious from the way the problem
33
00:04:14,680 --> 00:04:19,690
has been defined but you must intuit these
kinds of things so so the total amount of
34
00:04:19,690 --> 00:04:32,620
heat that is generated
so q dot a into the volume of this flat a
35
00:04:32,620 --> 00:04:38,610
which is lets say if this is l one length
l one will put the numbers later
36
00:04:38,610 --> 00:04:47,720
so this is length l one l one into a so thats
the total amount of heat that is generated
37
00:04:47,720 --> 00:04:52,150
and if we specify that its a steady state
condition thats what we will be looking at
38
00:04:52,150 --> 00:04:59,300
that should be equal to the rate at which
heat is being transported in slab b because
39
00:04:59,300 --> 00:05:04,560
there is no heat generation there so whatever
comes in here has to go out if the steady
40
00:05:04,560 --> 00:05:15,580
state condition has to be maintained and that
should be equal to h into t two minus t infinity
41
00:05:15,580 --> 00:05:23,800
right very simple so because there is constant
heat transfer rate in slab b because q b dot
42
00:05:23,800 --> 00:05:29,590
is zero heat transfer rate is constant therefore
whatever heat that is actually flowing at
43
00:05:29,590 --> 00:05:34,749
this interface whatever rate at which heat
is being flowing from at this interface excuse
44
00:05:34,749 --> 00:05:40,610
me should be transported at this interface
if you have to maintain steady state conditions
45
00:05:40,610 --> 00:05:47,639
therefore q dot into l one into a should be
equal to h into a right so that is rate we
46
00:05:47,639 --> 00:05:56,969
are looking at rate not flux right so we know
what q dot a is so this means q dot a into
47
00:05:56,969 --> 00:06:07,069
l one should be equal to h into t two minus
t infinity so we know q dot a we know t infinity
48
00:06:07,069 --> 00:06:14,029
we know the heat transfer rate so we can find
out what t two is right and so that will be
49
00:06:14,029 --> 00:06:24,889
q dot a into l one by h plus t infinity if
i throw in the numbers it will be thirty plus
50
00:06:24,889 --> 00:06:33,199
two seventy three watch the units ok i am
using s i units into ten power six multiplied
51
00:06:33,199 --> 00:06:44,430
by fifty into ten power minus three divided
by thousand so thats approximately hundred
52
00:06:44,430 --> 00:06:50,639
and five plus two seventy three so that was
one hundred and five degrees
53
00:06:50,639 --> 00:07:01,900
so thats the temperature t two ok is it clear
to everyone how do we find t one
54
00:07:01,900 --> 00:07:10,310
how do we find t one we know that the heat
transfer rate is constant and we can simply
55
00:07:10,310 --> 00:07:16,360
write resistance network right if we know
how to write resistance network so the simply
56
00:07:16,360 --> 00:07:30,259
based on resistance network t one t two and
that is given by l l two by k two into a right
57
00:07:30,259 --> 00:07:39,860
so thats the resistance network and this is
q right so can we find t two from here
58
00:07:39,860 --> 00:07:50,650
no q b is zero there is no heat generation
in the second slab if there is no heat generation
59
00:07:50,650 --> 00:07:58,430
in the seconds slab you would actually write
the resistance network right so how do we
60
00:07:58,430 --> 00:08:10,509
solve this you dont know the area right so
what do we do do we really need the area so
61
00:08:10,509 --> 00:08:16,719
how do we solve how do we draw the resistance
network correct resistance network to find
62
00:08:16,719 --> 00:08:17,719
temperature t one
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00:08:17,719 --> 00:08:23,759
area
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00:08:23,759 --> 00:08:30,379
we know q equal to h into a right so so there
are two ways of doing it you know what q is
65
00:08:30,379 --> 00:08:39,419
you could use this resistance network or you
can draw overall resistance network also t
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00:08:39,419 --> 00:08:45,970
infinity one call it t infinity and t one
so you can use either of overall network or
67
00:08:45,970 --> 00:08:51,180
you know what q is and you can find out what
what the temperature t one is ok so t one
68
00:08:51,180 --> 00:09:00,721
will be i will just put the number its very
easy to calculate one fifteen degree c what
69
00:09:00,721 --> 00:09:17,680
about t zero how do we calculate t zero how
do we calculate t zero yeah
70
00:09:17,680 --> 00:09:22,330
so we have to use the solution here we have
to write the model if the model is not given
71
00:09:22,330 --> 00:09:27,810
to you obviously you have to write the model
and find the solution because the resistance
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00:09:27,810 --> 00:09:33,840
concept is not valid for that slab and therefore
we have to write the model equation and find
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00:09:33,840 --> 00:09:39,450
the solution and so i am not going to rewrite
the equations here so we solve the equation
74
00:09:39,450 --> 00:09:47,240
just now so you can find out that ah q dot
will be in sorry t naught will be t one plus
75
00:09:47,240 --> 00:09:56,520
q dot l square by two times k is equal to
one hundred and forty degrees so remember
76
00:09:56,520 --> 00:10:07,490
that slab a is nothing but if you have if
you have a slab which is twice the length
77
00:10:07,490 --> 00:10:13,570
of l one right so thats the problem that we
just solved right so where the at the midpoint
78
00:10:13,570 --> 00:10:18,120
you have ah adiabatic condition right so thats
the same problem so if you find what is the
79
00:10:18,120 --> 00:10:19,180
yes
80
00:10:19,180 --> 00:10:29,460
yeah sure so so remember that we solve this
problem right at x equal to zero minus l n
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00:10:29,460 --> 00:10:34,161
plus l and if you have a heat generation we
solve this problem and we found the temperature
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00:10:34,161 --> 00:10:39,740
profile correct and we know what is the temperature
distribution what we need to know is what
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00:10:39,740 --> 00:10:45,010
is the temperature at this location right
thats the midpoint so remember that we also
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00:10:45,010 --> 00:10:53,790
said solving this problem is equivalent to
solving half slab starting from x equal to
85
00:10:53,790 --> 00:10:59,790
zero with the adiabatic condition why is it
adiabatic because if you impose a symmetry
86
00:10:59,790 --> 00:11:05,410
condition on this where the temperatures on
both sides are same then you have a maxima
87
00:11:05,410 --> 00:11:10,930
at the center so the profile looks like this
so you have a maxima at the center which means
88
00:11:10,930 --> 00:11:16,110
that d t by d x is zero which is exactly the
flux condition is exactly the flux boundary
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00:11:16,110 --> 00:11:22,860
condition so this problem here is as good
as solving a problem which is twice the size
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00:11:22,860 --> 00:11:26,930
and if you get the temperature profile you
can find out what is the temperature at the
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00:11:26,930 --> 00:11:32,100
midpoint and thats what we solved a few moments
ago
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00:11:32,100 --> 00:11:36,380
so once you know the temperature this is simply
given by this expression here and you will
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00:11:36,380 --> 00:11:43,720
be able to find out so remember that the solution
is given by i can give you what the solution
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00:11:43,720 --> 00:11:47,400
is
sir
95
00:11:47,400 --> 00:11:49,240
yes
96
00:11:49,240 --> 00:11:51,080
where
97
00:11:51,080 --> 00:11:52,390
yeah conduction is included in q b
98
00:11:52,390 --> 00:12:01,340
this is a balance at the interface the wall
that is generated so not that the conduction
99
00:12:01,340 --> 00:12:08,080
is happening inside the slab right so at the
boundary what comes out of the boundary is
100
00:12:08,080 --> 00:12:12,380
the total amount of heat that is generated
inside the slab of course it comes through
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00:12:12,380 --> 00:12:18,900
conduction how does heat come here because
of conduction inside the slab but we are looking
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00:12:18,900 --> 00:12:23,370
at what happens with that interface because
there is no heat that is leaving on the other
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00:12:23,370 --> 00:12:27,680
side everything has to leave through this
side and so whatever comes here should be
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00:12:27,680 --> 00:12:32,860
the total amount of heat that is transported
the second slab because there is no heat generation
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00:12:32,860 --> 00:12:46,700
in slab b clear so the solution of this equation
would simply be t x is q dot l square by two
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00:12:46,700 --> 00:12:54,381
k into one minus x square by l square plus
t s so you put x equal to zero thats what
107
00:12:54,381 --> 00:13:01,590
you would get
so t s is the temperature of the two surfaces
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00:13:01,590 --> 00:13:06,420
which is same and that is nothing but p one
and our problem here and so if you put x equal
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00:13:06,420 --> 00:13:11,880
to zero here what you get is the solution
here yes
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00:13:11,880 --> 00:13:13,650
where
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00:13:13,650 --> 00:13:20,190
you dont need to know you know what q is q
is h into a into t two minus t infinity so
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00:13:20,190 --> 00:13:25,320
the area will cancel out the area will cancel
out when you actually write it its easy so
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00:13:25,320 --> 00:13:33,460
we can write so t one maybe i will write it
here for the benefit of everyone i will write
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00:13:33,460 --> 00:13:44,510
it here so what is this so this is t one minus
t two divided l two by k a k two a h into
115
00:13:44,510 --> 00:13:52,860
t two minus t infinity right so area will
cancel out and you know what t two is already
116
00:13:52,860 --> 00:14:05,410
you know what t infinity so you should be
able to find t one ok any other questions
117
00:14:05,410 --> 00:14:10,790
so i mean the take home message of this particular
example the reason why i showed this is we
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00:14:10,790 --> 00:14:15,950
should not blindly just go and write equations
for a given problem you should first intuit
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00:14:15,950 --> 00:14:22,180
what is happening in the problem can i make
some estimate about certain quantities without
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00:14:22,180 --> 00:14:26,850
solving the equations of course you can solve
the equations and verify them but can i make
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00:14:26,850 --> 00:14:32,240
some entity for example the intuitive judgment
we made about heat transport a second slab
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00:14:32,240 --> 00:14:37,020
such kind of judgments is very very important
when you are handling these kinds of engineering
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00:14:37,020 --> 00:14:40,830
problems
so you must always pay attention and give
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00:14:40,830 --> 00:14:47,740
utmost importance to your intuition ok so
the next topic that we are going to see is
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00:14:47,740 --> 00:14:53,010
extended surfaces so i will go to briefly
give introduction to that in the next five
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00:14:53,010 --> 00:15:06,050
minutes or so and then we will go deep into
it in the next few lectures so so far we said
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00:15:06,050 --> 00:15:15,220
by newtons law of cooling we said that q equal
to h a into lets say the surface temperature
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00:15:15,220 --> 00:15:21,290
minus t temperature of the fluid to which
the heat is being transported so that k lets
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00:15:21,290 --> 00:15:26,990
say it could be a slab like this where the
fluid is actually flowing past the slab and
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00:15:26,990 --> 00:15:32,910
this is the surface temperature and this is
the temperature of the fluid now the question
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00:15:32,910 --> 00:15:38,650
comes in most of the engineering problems
is how can i increase the total amount of
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00:15:38,650 --> 00:15:44,810
heat transport
what are the ways by which how to increase
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00:15:44,810 --> 00:15:58,420
its a very very important question
how to increase q so the purpose of heat transport
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00:15:58,420 --> 00:16:05,050
is basically to transport heat from one location
to another so the better i can do the better
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00:16:05,050 --> 00:16:11,660
placed i am right so if i can design a system
which can transport heat better and obviously
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00:16:11,660 --> 00:16:20,180
i am much better place so if you look at these
three quantities here so if i call this as
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00:16:20,180 --> 00:16:27,710
delta t ok some temperature difference so
i can rewrite it as h into a into delta t
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00:16:27,710 --> 00:16:35,450
so the net heat transfer rate depends upon
three quantities right one is the heat transport
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00:16:35,450 --> 00:16:42,070
coefficient the other one is the area of heat
transport and the third one obviously is the
140
00:16:42,070 --> 00:16:46,530
temperature difference so these are the three
quantities that needs to be tweaked so if
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00:16:46,530 --> 00:16:51,220
you want to increase heat transfer rate you
can increase either of them individually or
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00:16:51,220 --> 00:16:55,190
you can increase multiple of them right
many of them simultaneously if you are able
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00:16:55,190 --> 00:16:59,490
to increase then you are going to increase
the total amount of heat that is transported
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00:16:59,490 --> 00:17:06,380
ok so when we discuss convection we are going
to look at how to play with the heat transport
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00:17:06,380 --> 00:17:11,390
coefficient what are the methods by which
heat transport coefficient can be estimated
146
00:17:11,390 --> 00:17:16,010
and what are the factors on which the heat
transfer coefficient depends upon and how
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00:17:16,010 --> 00:17:22,179
to increase them is what we will see for most
of the convection chapter ok so now what we
148
00:17:22,179 --> 00:17:28,529
are going to see is how to increase the area
heat transport and thereby increase the net
149
00:17:28,529 --> 00:17:35,630
heat transfer rate so one way to do that is
what is called the extended surfaces or in
150
00:17:35,630 --> 00:17:42,999
classical literature its also called as fins
ok so there are several equipments where fins
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00:17:42,999 --> 00:17:47,389
are being placed in order to increase the
heat transport
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00:17:47,389 --> 00:17:52,990
for example in the radiators in car where
you want to dissipate the heat as soon as
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00:17:52,990 --> 00:17:58,950
possible which means you need to have a very
high heat transport so in order to increase
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00:17:58,950 --> 00:18:05,490
the net heat transport you design a system
which has very high surface area i in fact
155
00:18:05,490 --> 00:18:18,100
the way its done is so the fins sort of looks
like this
156
00:18:18,100 --> 00:18:19,179
so they sort of look like this
157
00:18:19,179 --> 00:18:51,929
so so you have a base here ok so there is
a base system which is maintained at a certain
158
00:18:51,929 --> 00:19:00,519
temperature so supposing if the temperature
of the base surface is t s ok now you want
159
00:19:00,519 --> 00:19:06,340
to increase the heat transport from the base
to the fluid which is circulating around ok
160
00:19:06,340 --> 00:19:12,280
so there is a fluid which is flowing here
it could be any fluid in case of car radiators
161
00:19:12,280 --> 00:19:17,720
is basically air which is being circulated
so you want to maximize the amount of heat
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00:19:17,720 --> 00:19:22,530
that is transported from this surface to the
fluid so the way to increase the heat transfer
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00:19:22,530 --> 00:19:32,389
area is you create a thin protruding ah surfaces
like this and so you have ah heat transfer
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00:19:32,389 --> 00:19:37,840
occurring from the upper and the lower part
of this protrusion that comes out of the surface
165
00:19:37,840 --> 00:19:43,379
and so you can have many such protrusions
and each of these protrusion is what is called
166
00:19:43,379 --> 00:19:48,360
as a fin
so you have now added some extra area for
167
00:19:48,360 --> 00:19:54,399
heat transport and thats how you enhance the
net amount of heat that is being transported
168
00:19:54,399 --> 00:20:00,789
now if you look at it this is a slightly different
problem than what we had dealt with so far
169
00:20:00,789 --> 00:20:08,600
so compare this kind of a structure with the
structure that we had dealt with so far where
170
00:20:08,600 --> 00:20:15,080
the fluid which is being which is flowing
is actually at the outer end of the system
171
00:20:15,080 --> 00:20:21,240
that you are dealing with and the flow of
heat is actually in the direction perpendicular
172
00:20:21,240 --> 00:20:29,320
to that of the fluid flow right unlike here
in this case the conduction is occurring in
173
00:20:29,320 --> 00:20:39,299
this direction while simultaneously there
is heat loss from the from these surface right
174
00:20:39,299 --> 00:20:47,649
so here there is no simultaneous conduction
of heat and heat loss from the surface while
175
00:20:47,649 --> 00:20:53,799
in the extended surface you have now introduced
the system where there is simultaneous conduction
176
00:20:53,799 --> 00:20:59,690
of heat and also loss of heat to the fluid
which is surrounding it so in this sense its
177
00:20:59,690 --> 00:21:03,950
a slightly a different problem and we are
going to see how to account for these kinds
178
00:21:03,950 --> 00:21:09,350
of systems in fact that answers one of the
questions we asked a short while ago what
179
00:21:09,350 --> 00:21:14,539
about heat lost from the system so here we
are going to account for simultaneous loss
180
00:21:14,539 --> 00:21:21,210
of heat from the solid which is actually conducting
heat in one direction so this is what we are
181
00:21:21,210 --> 00:21:26,470
going to see for the next couple of lectures
and how to quantify heat transfer process
182
00:21:26,470 --> 00:21:31,480
and how to find out the net amount of heat
that is transferred from these kinds of fins
183
00:21:31,480 --> 00:21:31,679
ok