1
00:00:24,519 --> 00:00:31,800
so similar to furriers law which is the constitutive
relationship between the flux due to conduction
2
00:00:31,800 --> 00:00:37,829
and the local temperature gradient newtons
law of cooling is the constitutive relationship
3
00:00:37,829 --> 00:00:45,699
for heat transport from a solid system where
conduction is the mode of heat transport inside
4
00:00:45,699 --> 00:00:51,969
and there is a a conduction outside the solvent
ok so the way to describe it supposing if
5
00:00:51,969 --> 00:01:00,579
we have a solid here ok and there is some
fluid which is flowing past this solid so
6
00:01:00,579 --> 00:01:06,550
this is x is equal to zero will be flowing
at a certain velocity v ok we will see a lot
7
00:01:06,550 --> 00:01:11,090
more about it when we discuss ah convection
ah topic may be about ten lectures ten fifteen
8
00:01:11,090 --> 00:01:20,340
lectures from now ok and so there could be
a a transport coefficient h ok so we have
9
00:01:20,340 --> 00:01:36,060
to now define a heat transport coefficient
and the reason for defining such a coefficient
10
00:01:36,060 --> 00:01:41,909
is that note that here is heat transport across
different phases so here this is a solid and
11
00:01:41,909 --> 00:01:46,640
this is a liquid so you have transport across
different phases and so we need to define
12
00:01:46,640 --> 00:01:52,509
some part of a coefficient thats a fictitious
coefficient and we will actually see how to
13
00:01:52,509 --> 00:01:58,899
estimate these coefficients ok its for convenience
purposes such a coefficient has been described
14
00:01:58,899 --> 00:02:08,459
and so newtons law cooling simply says that
q prime at x equal to zero that the flux at
15
00:02:08,459 --> 00:02:16,150
x equal to zero should be equal to the heat
transport coefficient h and supposing i assume
16
00:02:16,150 --> 00:02:21,770
that the temperature here is t infinity and
if the temperature inside
17
00:02:21,770 --> 00:02:26,870
so if this temperature is higher than the
temperature at the boundary ok so this is
18
00:02:26,870 --> 00:02:35,360
the surface temperature so if i assume that
the fluid temperature is higher than that
19
00:02:35,360 --> 00:02:39,610
of the solid temperature which means that
the heat is transported from the fluid to
20
00:02:39,610 --> 00:02:46,670
the solid then one could define the flux as
t infinity minus t s which is basically at
21
00:02:46,670 --> 00:02:53,160
x is equal to zero so thats the newtons law
of cooling and so if i flux this into my third
22
00:02:53,160 --> 00:03:02,270
boundary condition and get minus k d t by
d x at x equal to zero equal to heat transfer
23
00:03:02,270 --> 00:03:12,370
coefficient into t infinity minus t s or you
can say t at x is equal to zero ok this gradient
24
00:03:12,370 --> 00:03:19,990
will take care of that so note that when t
infinity is less than t s then you are going
25
00:03:19,990 --> 00:03:24,430
to have a profile which is different ok
supposing if t infinity is greater than t
26
00:03:24,430 --> 00:03:30,410
x you are going to have a profile which looks
like this alright because there is heat that
27
00:03:30,410 --> 00:03:35,490
is going from the fluid into the solid now
supposing if t infinity is less than t s we
28
00:03:35,490 --> 00:03:41,240
are going to have a a different profile so
the gradient will care of ah the ah sign whether
29
00:03:41,240 --> 00:03:47,270
it is t infinity is greater or lesser ah t
s so this is a general form irrespective of
30
00:03:47,270 --> 00:03:53,460
what is the ah sign of t ah t infinity minus
t s because the gradient is what is going
31
00:03:53,460 --> 00:03:58,770
to dictate what is going to be the sign and
so also you should note its very important
32
00:03:58,770 --> 00:04:07,740
to note here that this gradient describes
the flux inside the solid and this formula
33
00:04:07,740 --> 00:04:15,740
here represents the flux just outside the
solid so this is a balance between flux of
34
00:04:15,740 --> 00:04:22,370
the heat transport just inside the solid and
flux of heat transport just outside the solid
35
00:04:22,370 --> 00:04:34,040
so this balance really comes about because
we assume that ah the temperature is continuous
36
00:04:34,040 --> 00:04:43,159
is that continuous variable
because its a continuous variable there has
37
00:04:43,159 --> 00:04:49,069
to be a a gradual change in the temperature
and if the if the way to account it mathematically
38
00:04:49,069 --> 00:04:55,710
is that the gradients have to be equal thats
how you ensure that the temperature is a continuous
39
00:04:55,710 --> 00:05:02,090
variable that is it changes continuously without
any jump at the boundary point ok so this
40
00:05:02,090 --> 00:05:07,230
is very important to understand how this balance
comes about you will see very often not just
41
00:05:07,230 --> 00:05:11,840
in this course in several other courses in
reaction generating etcetera in the future
42
00:05:11,840 --> 00:05:17,509
you will see that these kinds of ah flux balances
at the boundary will come into picture ok
43
00:05:17,509 --> 00:05:28,580
so let us take a a specific example so lets
say we start with ah ah one dimensional steady
44
00:05:28,580 --> 00:05:41,910
state conduction process ok so we start with
a simple one dimensional conduction so lets
45
00:05:41,910 --> 00:05:51,710
say we have a slab so thats a slab and we
assume that its one dimensional so let us
46
00:05:51,710 --> 00:06:02,139
say that the conduction process is in the
x direction ok and if you assume that the
47
00:06:02,139 --> 00:06:07,559
so this is the temperature axis so if the
temperature here is t two and lets say temperature
48
00:06:07,559 --> 00:06:15,931
here is say t one and t two ok but x equal
to zero the temperature is t one and x is
49
00:06:15,931 --> 00:06:25,780
equal to l the temperature is t two ok so
what is the energy balance for this so keep
50
00:06:25,780 --> 00:06:34,110
in mind keep in mind that the general energy
balance we should always try to sort of look
51
00:06:34,110 --> 00:06:48,870
at the general framework with del square t
plus q dot by k equal to rho c p by k into
52
00:06:48,870 --> 00:06:56,389
d t by so you can always start with the general
energy balance so we said that its a a steady
53
00:06:56,389 --> 00:07:03,789
state system so with time derivative goes
to zero to d t by d t is zero because its
54
00:07:03,789 --> 00:07:09,780
a a steady state system we are considering
a a steady state problem and then we said
55
00:07:09,780 --> 00:07:17,309
it is one dimensional system so which means
that the laplacian is simply in the x direction
56
00:07:17,309 --> 00:07:24,189
d square t by d x square ok and supposing
if you also assume that q dot is zero that
57
00:07:24,189 --> 00:07:31,360
is the there is no energy generation or loss
inside the system no matter whatsoever and
58
00:07:31,360 --> 00:07:36,781
so q dot is also equal to zero so these are
the immediate translation of some of the assumptions
59
00:07:36,781 --> 00:07:44,830
that we have made and so based on this we
can write a simple model it will be k into
60
00:07:44,830 --> 00:07:52,699
d square t by d x square equal zero thats
the simple model equation for the system that
61
00:07:52,699 --> 00:07:58,219
we have considered where it is a one d steady
state conduction with no heat generation slash
62
00:07:58,219 --> 00:08:04,199
heat loss
what are the boundary conditions so at x equal
63
00:08:04,199 --> 00:08:16,509
to zero equal to t one and t of x equal to
l equal to t two ok can i solve this equation
64
00:08:16,509 --> 00:08:26,270
of course yes we have done this its not a
big deal so temperature is given by t two
65
00:08:26,270 --> 00:08:37,110
minus t one t two minus t into x by l plus
t one ok so thats the temperature oops t one
66
00:08:37,110 --> 00:08:44,120
t two minus t one into x by l plus t one thats
the solution very trivial if you dint understand
67
00:08:44,120 --> 00:08:48,810
how this comes about you should go home and
try to work it out and get the solution very
68
00:08:48,810 --> 00:08:57,010
easy to get this ok alright
so now we want to find out what is the heat
69
00:08:57,010 --> 00:09:02,660
transfer rate ok we want to find out what
is the rate at which heat is being transferred
70
00:09:02,660 --> 00:09:09,010
for this system so remember in the last lecture
i am mentioned that heat transfer rate is
71
00:09:09,010 --> 00:09:16,820
the quantifying ah description of the transport
of heat via conduction and in general it is
72
00:09:16,820 --> 00:09:22,780
the quantifying description for any heat transport
process so you need to find out what is the
73
00:09:22,780 --> 00:09:27,410
heat transfer rate and also you should keep
in mind that its very very important from
74
00:09:27,410 --> 00:09:32,500
application point of view so when you are
looking at ah certain system what somebody
75
00:09:32,500 --> 00:09:37,470
wants to know is how much heat is being transported
what is the amount of heat that the system
76
00:09:37,470 --> 00:09:42,180
would generate at one wall of or one end then
what is the amount of heat that i am supposed
77
00:09:42,180 --> 00:09:46,380
to take from that thats that end of the wall
so this is an important question you will
78
00:09:46,380 --> 00:09:51,600
always have when you from the application
point of view but finding the energy distribution
79
00:09:51,600 --> 00:09:56,640
is really leading towards finding the heat
transfer rate and the way to find out is very
80
00:09:56,640 --> 00:10:07,970
simple we know that minus k area of heat transport
multiplied by the corresponding gradient so
81
00:10:07,970 --> 00:10:15,710
its very easy what is the gradient d t by
d x its t two minus t one by l right so there
82
00:10:15,710 --> 00:10:29,600
will be minus k a into t two minus t one by
l ok so with this framework we could introduce
83
00:10:29,600 --> 00:10:34,690
what has called the resistance that is offered
by the system so we should reflect that one
84
00:10:34,690 --> 00:10:39,600
of the points that we discussed a short while
ago what is meant by thermal diffusion is
85
00:10:39,600 --> 00:10:43,700
that there are two competing process these
which are occurring one is the ability of
86
00:10:43,700 --> 00:10:48,130
the system to store heat and the other one
is the ability of the system to transport
87
00:10:48,130 --> 00:10:57,590
heat because of conduction so one way to look
at is the resistance that is often a clean
88
00:10:57,590 --> 00:11:06,750
and elegant way to capture the one of the
competing processing is to calculate the resistance
89
00:11:06,750 --> 00:11:10,980
that is offered by that particular system
for that process its similar to what you would
90
00:11:10,980 --> 00:11:15,640
have seen in your electrical engineering or
some of the electricity things if you have
91
00:11:15,640 --> 00:11:19,670
studied ok
so we can find out what is the resistance
92
00:11:19,670 --> 00:11:36,390
offered to towards conduction towards conduction
and the way it is defined is so r is generally
93
00:11:36,390 --> 00:11:43,290
is a symbol i will use for resistance so r
for conduction sometimes i use a subscript
94
00:11:43,290 --> 00:11:48,690
sometimes i dont but generally r means the
resistance which is involved ok so resistance
95
00:11:48,690 --> 00:11:58,690
offered is simply given by delta t divided
by the rate at which heat is transported in
96
00:11:58,690 --> 00:12:10,080
that direction ok and that is given by t one
minus t two divided by k into a divided by
97
00:12:10,080 --> 00:12:21,960
l ok so i have just taken the minus sign inside
and so that is nothing but l by k into k so
98
00:12:21,960 --> 00:12:31,170
the resistance that is offered by the system
for thermal diffusion is given by l by k a
99
00:12:31,170 --> 00:12:36,500
so note that it is not just conductivity but
also the length and the area of heat transport
100
00:12:36,500 --> 00:12:41,510
plays an important role here and that you
would sort of intuitively guess if the length
101
00:12:41,510 --> 00:12:46,350
is longer then you have ah a longer span for
diffusion to occur if the length is smaller
102
00:12:46,350 --> 00:12:50,350
you have a shorter span so obviously you would
intuitive that length and the cross sectional
103
00:12:50,350 --> 00:12:55,170
area is going to play a role and that sort
of clearly comes out from the analysis of
104
00:12:55,170 --> 00:13:01,130
the moral that we have written so far ok so
therefore we can easily translate this into
105
00:13:01,130 --> 00:13:15,950
a a simple resistance network so we can easily
translate it into a a resistance network and
106
00:13:15,950 --> 00:13:20,860
this sort of network you are familiar with
from your p j e d days where we can say that
107
00:13:20,860 --> 00:13:26,190
there is a network and it is connected by
the temperature of the two surfaces of your
108
00:13:26,190 --> 00:13:31,890
system and the resistance of the network is
given by l by k f ok
109
00:13:31,890 --> 00:13:39,540
alright so we can now add a small complication
to it or rather small addition to it so lets
110
00:13:39,540 --> 00:13:50,710
say that there is some fluid which is flowing
on both side lets say i call this as the hot
111
00:13:50,710 --> 00:14:00,060
fluid ok and i call this is the cold fluid
and if the temperature of the hot fluid is
112
00:14:00,060 --> 00:14:13,340
t infinity one temperature of the cold fluid
is t infinity two ok what will be the resistance
113
00:14:13,340 --> 00:14:23,760
offered for heat transport from the hot fluid
to the surface how do we find that we need
114
00:14:23,760 --> 00:14:30,630
to find the resistance which is offered for
heat transport from the hot fluid to the lets
115
00:14:30,630 --> 00:14:37,760
say surface at x is equal to zero how do we
find that is described by newtons law of cooling
116
00:14:37,760 --> 00:14:47,260
ok so q which is the rate of heat transport
at this location is given by h into a which
117
00:14:47,260 --> 00:14:52,930
is the heat transport coefficient multiplied
by the cross sectional area which is outside
118
00:14:52,930 --> 00:15:00,790
the board into t infinity comma one minus
t at x equal to zero which is t one right
119
00:15:00,790 --> 00:15:08,040
so thats the rate at which the heat is being
transported from the hot fluid to the slab
120
00:15:08,040 --> 00:15:14,780
and we can use the resistance network on step
and we can say this is the convection ah resistance
121
00:15:14,780 --> 00:15:22,030
and thats simply given by one h a its very
simple to see this this is nothing but t infinity
122
00:15:22,030 --> 00:15:33,100
comma one minus t one divided by q very easy
to see this its a same formula i have used
123
00:15:33,100 --> 00:15:42,030
delta t divided by q ok h depends on many
things like flow velocity properties of the
124
00:15:42,030 --> 00:15:47,130
fluid we will see all that we will see how
to find h ah its not very easy to explain
125
00:15:47,130 --> 00:15:52,130
in the next five minutes so there are about
almost like twelve lectures we are going to
126
00:15:52,130 --> 00:15:57,110
have in this course which describes what is
the method by which you can estimate h so
127
00:15:57,110 --> 00:16:03,130
it really requires that much time so we will
actually look at it ah some time around the
128
00:16:03,130 --> 00:16:09,320
mid sem and after that ok
alright so so the so now therefore we can
129
00:16:09,320 --> 00:16:16,230
now extend this [reti/ resistance] resistance
network into three resistances ok so supposing
130
00:16:16,230 --> 00:16:23,430
if the heat transport coefficient from the
hot fluid to the slab is lets say h one and
131
00:16:23,430 --> 00:16:30,360
from the ah slab to the cold fluid that is
h two ok if thats the heat transport coefficient
132
00:16:30,360 --> 00:16:37,160
then we can say that the temperature of the
hot fluid is t infinity one and the resistance
133
00:16:37,160 --> 00:16:45,670
offered for heat transport is one by h one
a ok and the temperature at the boundary is
134
00:16:45,670 --> 00:16:52,710
t one and the resistance offered by the slab
for heat transport inside the slab is given
135
00:16:52,710 --> 00:17:01,529
by l by k a and the temperature on the boundary
is t two ok and similarly the the resistance
136
00:17:01,529 --> 00:17:09,280
here is given by h two into a so we can almost
read out what the resistances are so thats
137
00:17:09,280 --> 00:17:13,870
the power of resistance method that you have
learnt in your electricity things so you can
138
00:17:13,870 --> 00:17:18,689
always read out the properties and present
it in terms of the network on there is something
139
00:17:18,689 --> 00:17:24,839
even more cute you can do so if this is the
total amount of rate at which the heat is
140
00:17:24,839 --> 00:17:31,730
transport from the overall system from the
hot fluid to the cold fluid then one could
141
00:17:31,730 --> 00:17:43,590
represent q x as the total resistance or may
be we should write so one could say that the
142
00:17:43,590 --> 00:17:54,059
overall resistance overall resistance is simply
given by the overall temperature difference
143
00:17:54,059 --> 00:18:00,650
divided by the total amount of heat that is
being transferred right and very easy
144
00:18:00,650 --> 00:18:08,059
what is the overall resistance its in series
so just some of the resistances very trivial
145
00:18:08,059 --> 00:18:22,490
ok so its one by h one a plus l by k a plus
one by x two into a alright so thats the overall
146
00:18:22,490 --> 00:18:29,070
resistance that is offered by the whole system
together so this gives you an elegant opportunity
147
00:18:29,070 --> 00:18:34,670
so if you know what are the measurable quantity
so this is the first time i am using measurable
148
00:18:34,670 --> 00:18:39,430
quantity so when you are doing any modeling
work when you are representing any system
149
00:18:39,430 --> 00:18:44,660
using mathematical representation or model
equations you must always understand what
150
00:18:44,660 --> 00:18:49,710
are the observable quantities what is it that
that you can measure experimentally what is
151
00:18:49,710 --> 00:18:55,350
it that you can measure in physical system
and actual system so whenever you have if
152
00:18:55,350 --> 00:19:00,340
you take the example of the geyser you know
so all that you can measure is the temperature
153
00:19:00,340 --> 00:19:06,720
of the fluid that comes in and goes out so
can use can we write our model equations and
154
00:19:06,720 --> 00:19:12,290
can we write our mathematical expression in
such a way that we are able to use the observable
155
00:19:12,290 --> 00:19:19,540
parameters for observable quantities in order
to estimate the properties of the system for
156
00:19:19,540 --> 00:19:24,650
example it could be i want to know what is
the temperature t two but i all i do all i
157
00:19:24,650 --> 00:19:30,809
know is i can only measure the ah temperatures
of the outside fluid so can i use the resistance
158
00:19:30,809 --> 00:19:36,400
network so thats what these kind of resistance
network method provides you an elegant opportunity
159
00:19:36,400 --> 00:19:41,770
to use the measurable quantity and find some
of the quantities that is not known to you
160
00:19:41,770 --> 00:19:47,800
via normal experimental measurement ok so
we are going to see some examples of these
161
00:19:47,800 --> 00:19:53,740
mostly in the next lecture ah so we will just
take a little bit more complicated case of
162
00:19:53,740 --> 00:19:59,270
this resistance network and we will finish
with that for todays lecture
163
00:19:59,270 --> 00:20:09,610
so let us take the next example of composite
wall
164
00:20:09,610 --> 00:20:14,840
composite wall so supposing instead of one
wall i have multiple walls which are stacked
165
00:20:14,840 --> 00:20:19,150
with next to each other and this is the very
very common thing that you would observe in
166
00:20:19,150 --> 00:20:24,820
a insulating system so supposing if you have
an insulator usually it is stacked with different
167
00:20:24,820 --> 00:20:31,700
walls and each of them have different conductivities
its very very often very common to design
168
00:20:31,700 --> 00:20:40,740
system with multiple wall which actually have
different properties ok so you can have k
169
00:20:40,740 --> 00:20:45,050
one k two and k three each of these could
have different conductivities so you could
170
00:20:45,050 --> 00:20:49,950
imagine that you might want a certain material
which is exposed to the hot fluid but you
171
00:20:49,950 --> 00:20:55,429
dont want the second material which has better
conductivity to be exposed to the hot fluid
172
00:20:55,429 --> 00:20:59,830
because there could be some reactivity with
the fluid so in order to avoid that its a
173
00:20:59,830 --> 00:21:05,580
very common practice to design these kinds
of composite walls and so if i assume that
174
00:21:05,580 --> 00:21:13,380
the length of this first wall is l one length
of the second wall is l two length of the
175
00:21:13,380 --> 00:21:20,429
third wall is l three and if i have lets say
hot fluid which is flowing before the first
176
00:21:20,429 --> 00:21:30,970
wall and there is a cold fluid which is flowing
after the third wall and if the temperature
177
00:21:30,970 --> 00:21:36,679
of the hot fluid is seen t infinity one and
if the heat transport coefficient for newtons
178
00:21:36,679 --> 00:21:44,030
law of cooling is h one here and if it is
t infinity two and h two and i can now specify
179
00:21:44,030 --> 00:21:54,800
temperatures t one t two t three and t four
so i can easily read out and draw the resistance
180
00:21:54,800 --> 00:22:04,270
network with minimal effort so how many resistances
are there how many five right so we have five
181
00:22:04,270 --> 00:22:12,240
resistances the first one being the resistance
offered for transport from a transport of
182
00:22:12,240 --> 00:22:21,300
energy from the hot fluid to the first slab
so that is one by h one into a comma one t
183
00:22:21,300 --> 00:22:45,180
one t two and this is l one by k one a t three
l two by k two a t four l three by k three
184
00:22:45,180 --> 00:22:56,720
a and this is t infinity comma two and thats
the amount of heat that is being transported
185
00:22:56,720 --> 00:23:02,880
from the system from hot fluid to the cold
fluid and thats q x ok i think the good point
186
00:23:02,880 --> 00:23:06,010
to stop
so in the next lecture ah what we are going
187
00:23:06,010 --> 00:23:11,630
to see is we said we assume that the cross
sectional area is constant so we are going
188
00:23:11,630 --> 00:23:17,170
to start with a a specific case where the
varying cross sectional area of system and
189
00:23:17,170 --> 00:23:20,720
we are going to see how the equations are
going to change and how the temperature distribution
190
00:23:20,720 --> 00:23:25,179
is going to change by the way one important
point before we close what will be the temperature
191
00:23:25,179 --> 00:23:31,310
distribution in this slab is linear so very
obvious from the expression so its linear
192
00:23:31,310 --> 00:23:37,170
ah temperature profile and what will be the
temperature profile if there is aa varying
193
00:23:37,170 --> 00:23:42,300
cross sectional area and we look at other
geometries we looked only at cartesian coordinates
194
00:23:42,300 --> 00:23:45,660
here we will start looking at radial systems
and other coordinates ok