1
00:00:25,680 --> 00:00:32,180
so we ah initiated discussion on extended
surface in the last lecture so what it really
2
00:00:32,180 --> 00:00:37,730
means is that the on account of convective
mode of heat transport is essentially given
3
00:00:37,730 --> 00:00:44,590
by the product of heat transport coefficient
multiplied by the area of heat transport so
4
00:00:44,590 --> 00:00:52,610
if i put a ah subscript s ah its basically
the surface area of heat transport because
5
00:00:52,610 --> 00:00:59,640
its multi phase system where the heat transport
is from the solid to the fluid which is circulating
6
00:00:59,640 --> 00:01:07,000
around so there has to be ah its the surface
area is responsible for the heat transport
7
00:01:07,000 --> 00:01:13,250
via convection from the this fluid around
so if h is the heat transport coefficient
8
00:01:13,250 --> 00:01:19,240
a is the surface area of convective mode of
heat transport here and delta t is the corresponding
9
00:01:19,240 --> 00:01:23,210
temperature
so this is essentially newton's law of rule
10
00:01:23,210 --> 00:01:33,350
so essentially newtons law of rule is and
we said that the purpose of any heat transport
11
00:01:33,350 --> 00:01:38,960
equipment or heat transport system is essentially
to either increase the heat transfer rate
12
00:01:38,960 --> 00:01:43,490
if you want to dissipate the heat or if you
dont want to dissipate the heat you want to
13
00:01:43,490 --> 00:01:49,350
cut down you want to maintain the to a certain
certain chamber with a with a certain temperature
14
00:01:49,350 --> 00:01:54,740
then you want to cut down the amount of heat
that is lost so so essentially it boils down
15
00:01:54,740 --> 00:01:58,830
the total amount of heat that is transferred
from the system to its surroundings so one
16
00:01:58,830 --> 00:02:06,570
way to increase as we observed in the last
lecture is to increase the surface area of
17
00:02:06,570 --> 00:02:12,220
heat transport and that is achieved by ah
a set of systems one class of system called
18
00:02:12,220 --> 00:02:17,980
the fins so what we are going to see in todays
lecture is we are going to look at what are
19
00:02:17,980 --> 00:02:24,980
the different ways of fins and how to understand
heat transport process in these kinds of surfaces
20
00:02:24,980 --> 00:02:33,930
so lets take a little general example ok so
so we said that there is a base supposing
21
00:02:33,930 --> 00:02:52,650
if there is a
base solid ok so note that we we purpose is
22
00:02:52,650 --> 00:02:57,560
to increase the heat transfer rate into the
fluid which is surrounding so one way to do
23
00:02:57,560 --> 00:03:06,580
that is to have some sort of a fins which
are actually protruding out of these surfaces
24
00:03:06,580 --> 00:03:17,910
and it could have many such fins many such
fins which are protruding out of these base
25
00:03:17,910 --> 00:03:23,060
surfaces and there would be heat transfer
from the top and the bottom surface of this
26
00:03:23,060 --> 00:03:36,060
fin and of course in the sides also so let
us consider our generalized an arbitrary shape
27
00:03:36,060 --> 00:03:44,260
ok when a general arbitrary shape so let us
assume that this is the varying cross sectional
28
00:03:44,260 --> 00:03:51,560
system and we are going to take specific examples
and specific cases of this generalized a generalized
29
00:03:51,560 --> 00:04:07,590
pin to extended surface ok so if we want to
quantify heat transport process in such a
30
00:04:07,590 --> 00:04:12,680
generalized system we need to write a we need
to find the temperature distribution and in
31
00:04:12,680 --> 00:04:16,470
order to find the temperature distribution
we need to write the energy balance thats
32
00:04:16,470 --> 00:04:25,570
what we have been doing
so lets say we take a small element so this
33
00:04:25,570 --> 00:04:37,460
is d x and lets this b d x direction going
to l ok and let us assume that there is fluid
34
00:04:37,460 --> 00:04:44,360
which is flowing past this object everywhere
at temperature t infinity ok so there is some
35
00:04:44,360 --> 00:05:03,580
fluid which is flowing and if i assume
the steady state process so let me may at
36
00:05:03,580 --> 00:05:11,169
we make a few assumptions saying that at the
steady state and let us assume that the heat
37
00:05:11,169 --> 00:05:17,490
transfer is primarily one dimensional which
means that the heat transfer is only in the
38
00:05:17,490 --> 00:05:30,900
x direction and i assume that cross section
temperature is uniform ok so i assume that
39
00:05:30,900 --> 00:05:36,229
every cross section the temperature is uniform
which means that the gradients are zero in
40
00:05:36,229 --> 00:05:45,539
this cross section ok but if the aspect ratio
that is if the if the radius of the cross
41
00:05:45,539 --> 00:05:50,129
section that we are considering if that is
substantially small compared to the length
42
00:05:50,129 --> 00:05:55,250
of the ah of the object that we are considering
then its a reasonable assumption to make that
43
00:05:55,250 --> 00:06:01,050
the gradients in the cross section is negligible
if it if the aspect ratio is significant so
44
00:06:01,050 --> 00:06:06,250
if it is close to one certainly its not correct
to assume that the cross sectional temperature
45
00:06:06,250 --> 00:06:10,469
is uniform there will be gradients in the
cross section and you will have to consider
46
00:06:10,469 --> 00:06:15,680
that and that would re combinational energy
balance which we will see a few lectures down
47
00:06:15,680 --> 00:06:17,850
the line
thirty five forty three
48
00:06:17,850 --> 00:06:23,590
ok so what is the energy balance so remember
the mantra i told you in one of the lectures
49
00:06:23,590 --> 00:06:29,610
whatever comes in in this element whatever
it leaves plus whatever is generated inside
50
00:06:29,610 --> 00:06:35,639
the system or whatever is lost from that location
should be equal to accumulation right so thats
51
00:06:35,639 --> 00:06:46,270
the mantra that we wrote so input minus output
plus generation equal to accumulation this
52
00:06:46,270 --> 00:06:51,360
term could be accounted in the generation
by putting appropriate sign so i will not
53
00:06:51,360 --> 00:07:01,520
write it as a separate term here so supposing
i draw the cross section ok so this is dx
54
00:07:01,520 --> 00:07:13,289
thats dx and a subscript c is the cross sectional
area of our conduction and note that the cross
55
00:07:13,289 --> 00:07:17,860
sectional area could in principle be a ah
function of the axial position so this is
56
00:07:17,860 --> 00:07:26,949
in principle a function of x ok and the convective
mode of heat transport is actually occurring
57
00:07:26,949 --> 00:07:33,930
from the curved surface area of this object
so along the curved surface areas area we
58
00:07:33,930 --> 00:07:38,909
see that there is a fluid which is flowing
and therefore there is constant heat exchange
59
00:07:38,909 --> 00:07:45,340
from the solid to the fluid along the curved
surface area and of course there will be heat
60
00:07:45,340 --> 00:07:50,939
transport from the end also
so note that the x equal to zero basically
61
00:07:50,939 --> 00:08:02,699
corresponds to the it is attached to the base
it is attached to the base equipment so its
62
00:08:02,699 --> 00:08:08,129
not exposed to the fluid and there will be
heat exchange from the curves surface and
63
00:08:08,129 --> 00:08:16,900
also from the lagging edge of the ah of the
fin that we are considering ok so the supposing
64
00:08:16,900 --> 00:08:26,729
i call d q convection is the differential
amount of heat that is exchanged or that that
65
00:08:26,729 --> 00:08:33,940
leaves this surface and is sent into the fluid
stream ok so thats the amount of rate differential
66
00:08:33,940 --> 00:08:41,289
rate at which heat is being exchanged and
if the amount of heat that the rate at which
67
00:08:41,289 --> 00:08:48,510
heat is transferred to that element input
is qx and if the rate at which it leaves this
68
00:08:48,510 --> 00:08:57,050
q of x plus d x so at steady state accumulation
is zero so therefore the balance is simply
69
00:08:57,050 --> 00:09:08,250
q x which is the input minus qx plus dx which
is the rate at which heat leaves this element
70
00:09:08,250 --> 00:09:14,050
minus the amount of heat that is lost because
of exchange of heat from the solid to the
71
00:09:14,050 --> 00:09:22,290
fluid which is flowing past it ok
so minus dq convection thats equal to zero
72
00:09:22,290 --> 00:09:29,820
very simple so whatever comes into that element
qx minus whatever it leaves that is q of x
73
00:09:29,820 --> 00:09:37,230
plus dx minus whatever is lost because of
convection from the heat transport from the
74
00:09:37,230 --> 00:09:48,570
solid to the fluid which is flowing past that
object ok so q from fouriers law we know that
75
00:09:48,570 --> 00:10:00,899
qx is minus k ac of x into dt by dx so note
that in principle the cross sectional area
76
00:10:00,899 --> 00:10:06,089
could be a function of the position and so
its a a variable cross sectional area problem
77
00:10:06,089 --> 00:10:16,080
and similarly from newtons law of cooling
dq convection is heat transport coefficient
78
00:10:16,080 --> 00:10:28,430
into das and if the local temperature is t
multiplied by t minus t infinity so thats
79
00:10:28,430 --> 00:10:35,030
the d q convection ok so this comes from fouriers
law and this comes from newtons law of cooling
80
00:10:35,030 --> 00:11:11,880
so when we substitute all this minus k minus
k s into t minus t equal to zero ok any questions
81
00:11:11,880 --> 00:11:32,930
on this so far so now i could rewrite this
as d by dx k a of x dt by dx equal to d by
82
00:11:32,930 --> 00:11:48,980
dx h s into t minus t infinity so if i assume
that the thermal conductivity and the heat
83
00:11:48,980 --> 00:11:55,700
transport coefficient are not function of
position then one could pull it out and say
84
00:11:55,700 --> 00:12:17,190
k d by dx a of x equal to h is t minus t infinity
a function of position in this term here yes
85
00:12:17,190 --> 00:12:21,889
or no
86
00:12:21,889 --> 00:12:32,899
did you watch out so this is the d by dx rate
of change of convection convective term right
87
00:12:32,899 --> 00:12:38,569
so here what is changing is the area changing
with dx or t minus t infinity what should
88
00:12:38,569 --> 00:12:50,949
you consider
so note that this is the differential balance
89
00:12:50,949 --> 00:12:57,459
right so you are looking at what is changing
in this small element so when we say newtons
90
00:12:57,459 --> 00:13:03,459
law of cooling we assume that the cross sectional
temperature or the ah ten local temperature
91
00:13:03,459 --> 00:13:09,870
of that element is constant and therefore
what is changing here with respect to convection
92
00:13:09,870 --> 00:13:16,810
from the solid to the fluid is the cross sectional
area so therefore we can rewrite this as da
93
00:13:16,810 --> 00:13:27,180
s by d x into t minus t infinity ok so keep
in mind that keep in mind keep in mind that
94
00:13:27,180 --> 00:13:35,540
the what is changing here the ah so in this
small element whenever you look at convective
95
00:13:35,540 --> 00:13:42,360
heat transport from the solid to the fluid
along the cross sectional area so in the differential
96
00:13:42,360 --> 00:13:47,110
balance so note that this is very very important
assumption that you make in differential balance
97
00:13:47,110 --> 00:13:52,470
that the local quantity that you are considering
in the control volume you assume that the
98
00:13:52,470 --> 00:13:57,742
local quantity the element is small enough
such that the local quantity remains constant
99
00:13:57,742 --> 00:14:06,410
in that element and therefore therefore the
t minus t infinity in this small element must
100
00:14:06,410 --> 00:14:10,290
remain constant its not a function of dx
101
00:14:10,290 --> 00:14:13,850
excuse me
102
00:14:13,850 --> 00:14:19,220
thats already a counter here in the gradient
so the temperature change that you are seeing
103
00:14:19,220 --> 00:14:24,790
here is inside the solid that variation is
already accounted with the diffusion term
104
00:14:24,790 --> 00:14:32,009
so the how what what causes the temperature
variation inside is the conduction process
105
00:14:32,009 --> 00:14:38,149
so the variation of temperature inside the
solid is already accounted for here but when
106
00:14:38,149 --> 00:14:43,230
you are considering convection what is causing
convection the driving force of convection
107
00:14:43,230 --> 00:14:47,889
is the local temperature inside the solid
and the local temperature in the fluid outside
108
00:14:47,889 --> 00:14:53,020
hm
mathematically correct
109
00:14:53,020 --> 00:15:08,600
they are correct
because my ah input is ah input is
110
00:15:08,600 --> 00:15:14,379
that is correct if we use the chain rule so
note that look at dq here this is the differential
111
00:15:14,379 --> 00:15:19,600
amount of heat that went in this differential
amount is because of the differential area
112
00:15:19,600 --> 00:15:24,769
and not because of the differential temperature
gradient so if you use the chain rule the
113
00:15:24,769 --> 00:15:29,449
differential temperature gradient is zero
because you assume that the temperature in
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00:15:29,449 --> 00:15:35,010
this small element is same and thats the basic
assumption of any differential balance in
115
00:15:35,010 --> 00:15:40,740
a control volume you always assume that the
temperature is constant and therefore the
116
00:15:40,740 --> 00:15:45,600
differential conduct amount of heat transfer
that occurs because of connections because
117
00:15:45,600 --> 00:15:50,889
of the differential amount of a cross sectional
area or surface area which is present and
118
00:15:50,889 --> 00:15:55,949
therefore even if we use chain rule you could
use even if you use chain rule in this term
119
00:15:55,949 --> 00:16:02,310
the differential temperature difference will
be zero because you by by nature of the ah
120
00:16:02,310 --> 00:16:08,129
ah the the the base is the fundamental premise
of differential balance is that the ah local
121
00:16:08,129 --> 00:16:13,580
temperature gradient is assumed to be constant
so if you use that property then even if you
122
00:16:13,580 --> 00:16:20,290
use chain rule that temperature gradient will
be zero ok so we can simply write this as
123
00:16:20,290 --> 00:16:31,000
k into a of ac of x i should have put a subscript
here so ac of x into t square t by dx square
124
00:16:31,000 --> 00:16:53,470
plus a into d ac by dx that should be equal
to h into t minus t infinity into da s by
125
00:16:53,470 --> 00:16:59,380
d x ok
so note that in the conduction term the gradient
126
00:16:59,380 --> 00:17:04,970
in the or the net heat transport rate is now
going to be a function of both the temperature
127
00:17:04,970 --> 00:17:11,030
gradient and the cross sectional area gradient
unlike what you see in convection term because
128
00:17:11,030 --> 00:17:16,350
the convection here occurs along the cross
sectional area and you assume that the the
129
00:17:16,350 --> 00:17:22,069
driving force for convection is the local
temperature gradient so thats the key difference
130
00:17:22,069 --> 00:17:28,270
why this this kind of why if you use the chain
rule the temperature gradient will be zero
131
00:17:28,270 --> 00:17:37,220
ok any other questions ok alright so this
is the general balance so we can simply rewrite
132
00:17:37,220 --> 00:17:54,850
this as v square t by dx square plus one by
ac into dt by d x thats equal to h into t
133
00:17:54,850 --> 00:18:07,730
minus t infinity divided by k into a c by
d a s write d x so this is the ah general
134
00:18:07,730 --> 00:18:17,120
balance for the arbitrary geometry the ph
is considered ok
135
00:18:17,120 --> 00:18:21,000
any questions on this so far yes
136
00:18:21,000 --> 00:18:29,480
s is the curved surface area curved surface
area of the object maybe you came late you
137
00:18:29,480 --> 00:18:34,520
didnt see the exact picture yes
138
00:18:34,520 --> 00:18:48,300
ok so so lets draw the picture again supposing
if this is the arbitrary fin ok so the conduction
139
00:18:48,300 --> 00:18:54,920
occurs in this direction so this is the heat
transfer rate because of conduction ok so
140
00:18:54,920 --> 00:19:01,010
conduction occurs from so this is the base
of the object so the fin is actually protruding
141
00:19:01,010 --> 00:19:08,223
from the base and so the heat is now carried
from the base to the fin and there is conduction
142
00:19:08,223 --> 00:19:15,540
mode of heat transport which is carrying heat
across now we have fluid which is actually
143
00:19:15,540 --> 00:19:26,870
flowing past this object on all directions
ok so now so the heat is lost from every location
144
00:19:26,870 --> 00:19:36,680
in the curved surface ok so at any location
x ax is essentially the supposing if the radius
145
00:19:36,680 --> 00:19:45,010
is given by supposing if the radius is given
by r ok so at any differential a s any location
146
00:19:45,010 --> 00:19:51,580
x the differential area of conductive mode
of heat transport it is simply given by two
147
00:19:51,580 --> 00:19:54,390
pi r time dx if the thickness is d a
148
00:19:54,390 --> 00:20:04,740
so you cannot define because the you need
a see for any any heat transport locker you
149
00:20:04,740 --> 00:20:10,320
need a cross sectional area correct now you
need some cross sectional area at which the
150
00:20:10,320 --> 00:20:16,150
heat transport is going to occur and so at
any given location you need a definite thickness
151
00:20:16,150 --> 00:20:22,250
in order to define heat transport process
so as supposing if i look at the so this is
152
00:20:22,250 --> 00:20:26,040
the cross sectional area supposing if i know
what is the total length that i know what
153
00:20:26,040 --> 00:20:32,020
is the total surface area of heat transport
but if i want to know what is the heat transport
154
00:20:32,020 --> 00:20:36,570
at a given location where the cross sectional
area is changing then you have to define a
155
00:20:36,570 --> 00:20:42,300
specific thickness otherwise area is not defined
so if you if you define a specific thickness
156
00:20:42,300 --> 00:20:48,260
then thats the area that you have to consider
so definitely when you want to define an area
157
00:20:48,260 --> 00:20:52,560
of heat transport you need two dimensions
you cannot define with just one dimension
158
00:20:52,560 --> 00:21:00,010
area is not defined so you need two dimensions
in order to define a area so you need to define
159
00:21:00,010 --> 00:21:05,080
if it is a varying cross section then you
need to have at least a small thickness at
160
00:21:05,080 --> 00:21:10,180
which the conductive heat transport is going
to occur now if it is non varying cross section
161
00:21:10,180 --> 00:21:15,030
that is if the cross section is same then
you could actually consider what is the overall
162
00:21:15,030 --> 00:21:23,910
area of heat transport is that clear to everyone
any
163
00:21:23,910 --> 00:21:29,200
boundary condition we want to see that so
note that conduction is occurring in the x
164
00:21:29,200 --> 00:21:34,750
direction so the whatever heat transport that
is occurring in this boundary is accounted
165
00:21:34,750 --> 00:21:39,090
as the boundary condition thats what we are
going to see it does not matter whether you
166
00:21:39,090 --> 00:21:42,670
do integral or the differential balance in
fact we are going to see in a short while
167
00:21:42,670 --> 00:21:47,580
ah in order to estimate some of the properties
there are many different ways to estimate
168
00:21:47,580 --> 00:21:53,050
properties whether you use the integral method
or whether you use the overall maximal heat
169
00:21:53,050 --> 00:21:58,050
transport method you are going to get exactly
the same answer so the differential and the
170
00:21:58,050 --> 00:22:02,220
integral balance can actually be merged together
in some sense there are formal ways to do
171
00:22:02,220 --> 00:22:08,140
that so there is no need to worry about whether
if i use integral balance will i get a different
172
00:22:08,140 --> 00:22:13,840
answer you will not you get exactly the same
answer whether you use an integral or differential
173
00:22:13,840 --> 00:22:14,860
balance
174
00:22:14,860 --> 00:22:20,980
you want to see that in a short while that's
the next case we are going to consider and
175
00:22:20,980 --> 00:22:24,850
that's that's rightly pointed out that's the
obvious case that you would want to consider
176
00:22:24,850 --> 00:22:29,131
right that's the immediate thing you want
to check whether if we say cross sectional
177
00:22:29,131 --> 00:22:32,360
area is independent of the x direction then
what will be the case
178
00:22:32,360 --> 00:22:46,900
so in the original energy balance equation
supposing you write multi dimensional energy
179
00:22:46,900 --> 00:22:59,960
balance equation ok so lets take that for
a moment let's digress for a a couple of minutes
180
00:22:59,960 --> 00:23:03,960
so in principle you could derive the same
equation from the generalized energy balance
181
00:23:03,960 --> 00:23:12,160
also so the generalized energy balance is
k del square t so this is the generalized
182
00:23:12,160 --> 00:23:22,770
energy balance right generalized energy balance
is k del square t plus whatever is the amount
183
00:23:22,770 --> 00:23:28,410
of heat that is generated inside the solid
ok and that should be equal to the accumulation
184
00:23:28,410 --> 00:23:33,370
term so if you assume steady state accumulation
is zero and if we assume that there is no
185
00:23:33,370 --> 00:23:42,800
heat generation inside thats also zero ok
so k del square p equal to zero is your model
186
00:23:42,800 --> 00:23:47,900
but keep in mind that the heat transport that
is occurring from the curved surface is now
187
00:23:47,900 --> 00:23:53,990
going to appear as a a boundary condition
in the radial direction so if i if i write
188
00:23:53,990 --> 00:24:00,470
the boundary condition at r equal to lets
say capital r which is a function of position
189
00:24:00,470 --> 00:24:10,870
x then i will be minus k d t by d r thats
equal to h into t minus t infinity so that
190
00:24:10,870 --> 00:24:16,330
will appear now as a a boundary condition
now the way to get this model from here is
191
00:24:16,330 --> 00:24:21,960
all you have to do is you say that so note
that we did not include one of the assumption
192
00:24:21,960 --> 00:24:27,580
we said that the cross sectional temperature
is uniform so how do we incorporate that so
193
00:24:27,580 --> 00:24:46,250
you integrate model equation with respect
to r if you integrate the schema second if
194
00:24:46,250 --> 00:24:50,350
you integrate with respect to r
so keep in mind that you have to use the correct
195
00:24:50,350 --> 00:24:55,510
weights ok so note that you are now dealing
with these radial coordinates so you have
196
00:24:55,510 --> 00:24:59,810
to account for when you integrate the radial
coordinates you will have to account for change
197
00:24:59,810 --> 00:25:07,200
in the cross sectional area so the correct
integral is essentially rdr t times rdr ok
198
00:25:07,200 --> 00:25:16,020
so what you will get is a a cross sectional
averaged temperature ok so this t is essentially
199
00:25:16,020 --> 00:25:21,330
we are assuming that the radial distribution
is uniform right we said that there is no
200
00:25:21,330 --> 00:25:26,760
gradient in the radial direction which is
equivalent to saying that its a a cross sectionally
201
00:25:26,760 --> 00:25:32,520
average temperature quantity so now if you
integrate this equation in the r direction
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and substitute the corresponding boundary
condition you should get that and in fact
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its a good exercise to check so account for
the varying cross sectional area and you incorporate
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this integration into your model equation
and you will be able to get that so this is
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the way to go from the generalized equation
and in fact we will not discuss this but this
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sort of assumption is called lumping lumping
or averaging in engineering literature so
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its a very very common technique used when
you model different engineering systems so
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00:26:06,520 --> 00:26:12,450
you can use a lumping method if you know that
the aspect ratio of a certain problem is very
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very small that is i define aspect ratio as
the radius divided by the length if that is
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very very small then you could actually do
what is called a lumping assumption and then
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you get a one dimensional model