1
00:00:13,630 --> 00:00:20,020
so i would still look at the one d system
we will look at two d systems later but lets
2
00:00:20,020 --> 00:00:29,050
look at one d system in the radial coordinates
so the general equation as we derived in the
3
00:00:29,050 --> 00:00:40,700
second lecture is that the its k into del
square t plus q dot equal to rho c p into
4
00:00:40,700 --> 00:00:48,230
partial derivative rho t by tau t right so
now we need to write the laplacian in the
5
00:00:48,230 --> 00:00:54,220
radial coordinates and thats not a difficult
task to do people have worked it out so del
6
00:00:54,220 --> 00:01:01,910
square t in radial coordinates so when i use
subscript r in laplacian it means radial coordinate
7
00:01:01,910 --> 00:01:14,860
system so that will be one by r d by d r r
into d t by d r plus one by r square d square
8
00:01:14,860 --> 00:01:20,329
t by d pi square plus d square t by d z square
9
00:01:20,329 --> 00:01:31,609
well in a moment i will draw the coordinate
so if this is a cylinder so thats the z direction
10
00:01:31,609 --> 00:01:40,240
and this is the radial direction and the curve
around it so there is an angle around every
11
00:01:40,240 --> 00:01:46,899
radial point and so that angle is what is
called as pi ok so these are the three coordinates
12
00:01:46,899 --> 00:01:52,850
of a cylindrical system and this is the general
laplacian in the cylindrical coordinate system
13
00:01:52,850 --> 00:02:00,810
ok so now what we are going to do is we are
going to look at the coaxial cylindrical system
14
00:02:00,810 --> 00:02:31,530
ok so the problem we are going to take is
coaxial cylinder ok ok so the centre is here
15
00:02:31,530 --> 00:02:43,990
this is r equal to zero ok and if the radius
of the inner cylinder is r one ok and the
16
00:02:43,990 --> 00:02:52,950
radius of the outer cylinder is r two ok now
let us assume that its a solid volume wall
17
00:02:52,950 --> 00:03:04,940
here ok so its a solid wall so we are looking
at heat transfer between the coaxial cylinders
18
00:03:04,940 --> 00:03:10,180
ok it could even be filled with a liquid it
could even be filled with other kinds of fluids
19
00:03:10,180 --> 00:03:18,400
but lets consider that its a a solid wall
for now and let us assume that the k is the
20
00:03:18,400 --> 00:03:29,900
thermal conductivity of the material of the
solid wall and so the what is the model equation
21
00:03:29,900 --> 00:03:38,310
so supposing we assume that q dot is zero
and steady state condition what is the model
22
00:03:38,310 --> 00:03:47,050
equation so its del square t is zero so if
we assume that its one d system so that is
23
00:03:47,050 --> 00:04:02,900
purely radial radial conduction purely radial
conduction then your model is one by r d by
24
00:04:02,900 --> 00:04:13,480
d r equal to zero thats the model equation
and if i specify that the temperature here
25
00:04:13,480 --> 00:04:19,400
is t one and temperature here is t two and
i could easily read out the boundary conditions
26
00:04:19,400 --> 00:04:27,720
so the boundary conditions are t at r equal
to r one its a t one and t at r equal to r
27
00:04:27,720 --> 00:04:38,760
two is t two so thats the boundary condition
whats the general solution of this equation
28
00:04:38,760 --> 00:04:46,840
you want to try yeah
29
00:04:46,840 --> 00:05:01,680
yeah yeah it is log what is it you want to
try c one lawn r plus c two right so the general
30
00:05:01,680 --> 00:05:16,960
solution so the general solution is c one
lawn r plus c two l n stands for the natural
31
00:05:16,960 --> 00:05:23,740
logarithm and so now we can substitute the
boundary conditions so t one will be c one
32
00:05:23,740 --> 00:05:33,510
lawn r one plus c two and t two will be c
one lawn r two plus c two and so from here
33
00:05:33,510 --> 00:05:44,740
we can find out c one is t one minus t two
by lawn r one by r two and c two is t two
34
00:05:44,740 --> 00:05:55,570
minus t one minus t two lawn r two divided
by lawn r one by r two so its quite easy you
35
00:05:55,570 --> 00:06:02,110
can almost read out by solving these simultaneous
equations linear equations and so when we
36
00:06:02,110 --> 00:06:05,690
substitute we can find that the temperature
distribution
37
00:06:05,690 --> 00:06:21,460
so the temperature distribution t is given
by t one minus t two by lawn r one by r two
38
00:06:21,460 --> 00:06:31,449
multiplied by lawn r by r two plus t two so
thats the so all i have done is i have just
39
00:06:31,449 --> 00:06:36,840
substituted these constants into the general
solution of that equation and this is the
40
00:06:36,840 --> 00:06:47,460
temperature profile ok so what should we do
next we know the temperature profile whats
41
00:06:47,460 --> 00:07:07,400
the next step in characterizing to find the
heat transfer rate ok what is q minus k into
42
00:07:07,400 --> 00:07:15,650
a into d t by d r but the patch here is that
area is now a function of the radial position
43
00:07:15,650 --> 00:07:21,420
line right so as you go along the radius the
cross sectional area of heat transfer is going
44
00:07:21,420 --> 00:07:27,040
to be different so you have to factor that
into the calculation of the heat transfer
45
00:07:27,040 --> 00:07:37,050
rate ok so d t by d r ok what is a r minus
k two pi r l
46
00:07:37,050 --> 00:07:44,780
so if the length of the cylinder is l length
of the cylinder is l then it is minus k two
47
00:07:44,780 --> 00:07:59,490
pi r l into d t by d r ok so d t by d r is
t one minus t two divided by lawn r one by
48
00:07:59,490 --> 00:08:09,650
r two into r two by r into one by r one by
r two right so thats the and that is nothing
49
00:08:09,650 --> 00:08:20,610
but one by r t one minus t two into lawn r
one by r two so i can substitute the gradient
50
00:08:20,610 --> 00:08:34,610
and so that will be q r equal to minus k two
pi r l into t one minus t two divided by r
51
00:08:34,610 --> 00:08:52,180
into lawn r one by r two ok so that will be
q r equal to minus two pi k l into t one minus
52
00:08:52,180 --> 00:09:02,400
t two divided by lawn r one by r two so why
is there is a negative sign here is the heat
53
00:09:02,400 --> 00:09:10,210
transfer rate negative
54
00:09:10,210 --> 00:09:15,810
ok supposing if i say t one is greater than
t two ok
55
00:09:15,810 --> 00:09:22,130
r one is smaller so lawn r one by r two is
negative so there is now issue with sign here
56
00:09:22,130 --> 00:09:28,450
as long as you follow correct convention and
you write your model equations properly there
57
00:09:28,450 --> 00:09:32,830
will be no sign mistake that you will find
in your solution so this is very important
58
00:09:32,830 --> 00:09:37,370
you must be able to intuit what should be
the correct sign of all the terms that you
59
00:09:37,370 --> 00:09:42,680
write in your model equations its very important
to understand this and in fact i would urge
60
00:09:42,680 --> 00:09:48,390
all of you to practice all these model equations
as many number of times and convince yourself
61
00:09:48,390 --> 00:09:53,480
that this is the correct sign that you have
to put at each and every term in the model
62
00:09:53,480 --> 00:10:01,850
equation ok alright so so i can simply write
it as two pi k l into t two minus t t one
63
00:10:01,850 --> 00:10:08,680
divided by lawn r one by r two and once again
i could define the resistances the way i have
64
00:10:08,680 --> 00:10:19,490
done before right so r resistance is given
by lets say t two minus t one by q and so
65
00:10:19,490 --> 00:10:31,070
that will be lawn r one by r two divided by
two pi k l so thats the resistance between
66
00:10:31,070 --> 00:10:37,790
the that is offered by the solid wall for
heat conduction in the cylindrical system
67
00:10:37,790 --> 00:10:42,410
so note that the form that you get for the
resistance is quite different than from what
68
00:10:42,410 --> 00:10:49,140
you got in a one d system and that is this
two pi comes because of the change in the
69
00:10:49,140 --> 00:10:55,010
cross sectional area and also its important
to note that you got only the length scale
70
00:10:55,010 --> 00:11:00,240
which is appearing here unlike in the other
case where you have a by l where you have
71
00:11:00,240 --> 00:11:05,870
per unit length right so here we should note
that because of the cross sectional area you
72
00:11:05,870 --> 00:11:12,839
have a slightly different form than what you
got in a a one d system ok what is the temperature
73
00:11:12,839 --> 00:11:24,510
profile
i want to now draw the temperature profile
74
00:11:24,510 --> 00:11:28,540
between r one and r two what will be the temperature
profile
75
00:11:28,540 --> 00:11:46,170
its not negative but t two so when i say t
one is greater than t two right so i have
76
00:11:46,170 --> 00:11:54,399
to define yeah so then you have r two by r
one so depending on how you define so be very
77
00:11:54,399 --> 00:11:58,670
careful when you define your quantities you
have to follow the correct sign convention
78
00:11:58,670 --> 00:12:07,900
ok ok so now what will be the temperature
profile if i say t one is greater than t two
79
00:12:07,900 --> 00:12:16,480
so its t one and t two i said t one is greater
than t two so what will be the temperature
80
00:12:16,480 --> 00:12:29,800
profile it will be a logarithmically decreasing
function there will be a so its quite easy
81
00:12:29,800 --> 00:12:35,860
to read it out so unlike what you got in cartesian
coordinates where the cross sectional area
82
00:12:35,860 --> 00:12:43,480
of heat transport was constant you had a linear
profile here you have a a logarithmic profile
83
00:12:43,480 --> 00:12:49,920
so remember that you must be able to intuit
what the profile is going to be even before
84
00:12:49,920 --> 00:12:54,900
you write the model equations so right now
we wrote the model equations and we found
85
00:12:54,900 --> 00:13:00,510
out what the profile is but its also important
to do vice versa otherwise the solution that
86
00:13:00,510 --> 00:13:05,459
you get you may not be able to figure out
whether the solution you got is right or wrong
87
00:13:05,459 --> 00:13:10,310
so its very important to intuit what is going
to be the natural temperature profile that
88
00:13:10,310 --> 00:13:16,300
you would get even without writing the model
so this is very very important in any modeling
89
00:13:16,300 --> 00:13:21,890
exercise you should be able to intuit what
should be the solution or what will be the
90
00:13:21,890 --> 00:13:27,360
approximate solution that you would get even
before writing the equation solving them so
91
00:13:27,360 --> 00:13:31,560
this is a very important insight that you
should gain while doing various courses in
92
00:13:31,560 --> 00:13:40,779
the department ok alright so lets look at
composite wall and radial systems so as you
93
00:13:40,779 --> 00:13:50,820
would immediately intuit that these are just
extensions of what we saw in the last lecture
94
00:13:50,820 --> 00:14:00,910
so
essentially you have many concentric cylinders
95
00:14:00,910 --> 00:14:06,300
and what should be the model equation and
what should be the resistance catalog ok so
96
00:14:06,300 --> 00:14:19,010
i am going to draw only the cross section
here ok
97
00:14:19,010 --> 00:14:34,700
so now if radius is r one r two and r three
and lets say we have hot fluid which is flowing
98
00:14:34,700 --> 00:14:42,970
here and the cold fluid which is flowing outside
and there is lets say the heat transport coefficient
99
00:14:42,970 --> 00:14:52,340
is h two and t infinity two are the temperatures
and the temperature here is t infinity one
100
00:14:52,340 --> 00:15:00,269
and h one and if i know the interface temperatures
t one t two and t three so now i need to draw
101
00:15:00,269 --> 00:15:06,870
the resistance network how many resistances
are there if i assume that the walls are smooth
102
00:15:06,870 --> 00:15:12,600
there is no contact resistance how many resistances
are they are five right so they are totally
103
00:15:12,600 --> 00:15:38,110
five resistances ok and what are they what
are the five resistances how many are there
104
00:15:38,110 --> 00:15:55,190
four or five four i am glad you are all awake
its four very good ok so what are the resistances
105
00:15:55,190 --> 00:16:03,920
whats the first one t infinity one and t one
whats the first resistance
106
00:16:03,920 --> 00:16:16,040
one by one by h one a now what is the area
so i should say one by h one a one so note
107
00:16:16,040 --> 00:16:23,920
that the area is now different in the inside
and the outside two pi r one l right so this
108
00:16:23,920 --> 00:16:32,050
area is two pi r one l so its not the cross
sectional area the heat transport is actually
109
00:16:32,050 --> 00:16:41,139
occurring along the curved surface
so if this is the length so the heat transport
110
00:16:41,139 --> 00:16:46,420
is actually occurring along the curved surface
of the cylinder and so the area is two pi
111
00:16:46,420 --> 00:16:57,149
r one into l ok what about the resistance
here at lawn r two by r one divided by two
112
00:16:57,149 --> 00:17:06,409
pi k one if i call that k one is the thermal
conductivity here and k two are the thermal
113
00:17:06,409 --> 00:17:14,600
conductivity there so its lawn r three by
r two divided by two pi k two into l and this
114
00:17:14,600 --> 00:17:23,850
will be one by h two into a two so this is
two pi r two into l ok so thats the heat transport
115
00:17:23,850 --> 00:17:31,850
area is that clear to everyone and so r total
which is the total resistance is given by
116
00:17:31,850 --> 00:17:48,500
one by h one two pi r one l plus lawn r two
by r one divided by two pi k one l plus lawn
117
00:17:48,500 --> 00:18:06,760
r three by r two divided by two pi k two l
plus one by h two ah two l r three sorry should
118
00:18:06,760 --> 00:18:17,280
be r three thanks
once again the way we did before we could
119
00:18:17,280 --> 00:18:25,580
define a a overall heat transport coefficient
so q in the radial direction we can write
120
00:18:25,580 --> 00:18:41,490
as u times a into t infinity one minus t infinity
two ok and so one by u a so resistance total
121
00:18:41,490 --> 00:18:50,870
is given by t infinity comma one minus t infinity
comma two divided by q r right and so that
122
00:18:50,870 --> 00:18:55,500
is one by u into a
123
00:18:55,500 --> 00:19:02,470
right so thats the question so here what should
be the area that you should choose for defining
124
00:19:02,470 --> 00:19:07,770
the overall heat transport coefficient in
a one d system in cartesian coordinates it
125
00:19:07,770 --> 00:19:17,090
didnt matter because the area was same so
here you can now define two different heat
126
00:19:17,090 --> 00:19:31,150
transport coefficient u one and u two and
so this u one is based on the inside area
127
00:19:31,150 --> 00:19:43,610
that is two pi r one l and this is based on
based on two pi r two l so now if you have
128
00:19:43,610 --> 00:19:51,390
varying cross sectional area system then the
overall or the universal heat transport coefficient
129
00:19:51,390 --> 00:19:58,510
is defined based on the area of interest so
the question comes in which one should i choose
130
00:19:58,510 --> 00:20:05,440
i have two options now which one should i
choose so it completely depends upon the problem
131
00:20:05,440 --> 00:20:10,510
that you are handling and it depends upon
what are the quantities that you want to estimate
132
00:20:10,510 --> 00:20:19,330
for example if i want to estimate the ah temperature
of the hot fluid ok lets say i know i know
133
00:20:19,330 --> 00:20:23,540
what the temperature of the hot fluid is but
lets say for some particular problem i want
134
00:20:23,540 --> 00:20:28,690
to estimate what should be the temperature
of the hot fluid in order for certain amount
135
00:20:28,690 --> 00:20:32,230
of heat to be transferred across a given composite
system
136
00:20:32,230 --> 00:20:38,549
so in that case i should use the inside area
and define find out the overall heat transport
137
00:20:38,549 --> 00:20:43,860
coefficient based on the inside area so if
i want to find a hot fluid properties then
138
00:20:43,860 --> 00:20:49,990
i need to use the outside ah area and find
out the ah overall heat transport coefficient
139
00:20:49,990 --> 00:20:56,490
defined based on the outside area when we
140
00:20:56,490 --> 00:21:05,140
yes it includes everything so we are going
to see that in a short while it includes everything
141
00:21:05,140 --> 00:21:09,850
the only advantage you will have is that it
just has some convenience in terms of calculations
142
00:21:09,850 --> 00:21:14,790
thats all it doesnt matter whichever area
you choose but if you want to have things
143
00:21:14,790 --> 00:21:19,100
to be little bit more convenient to do your
calculations then you define your heat transport
144
00:21:19,100 --> 00:21:27,419
area based on that ok so supposing if i define
so one by u a so if i define heat transport
145
00:21:27,419 --> 00:21:33,270
overall heat transport coefficient based on
the inside area and that will simply be one
146
00:21:33,270 --> 00:21:50,110
by h one plus lawn r two by r one into r one
by k one plus lawn r three by r two into r
147
00:21:50,110 --> 00:22:02,490
one by k two plus one by h two into r one
by r three so that will be the oops sorry
148
00:22:02,490 --> 00:22:08,059
one by u one thanks so one by u one all i
have done is i have just multiplied the equation
149
00:22:08,059 --> 00:22:16,500
by the corresponding area and similarly one
by u two will be one by h one into r three
150
00:22:16,500 --> 00:22:30,750
by r one plus lawn r two by r one into r three
by k one plus lawn r three by r two into r
151
00:22:30,750 --> 00:22:48,850
three by k two plus one by h two so which
clearly means that one by u one a one should
152
00:22:48,850 --> 00:22:54,780
be equal to one by u two a two so that sort
of obviously you can intuit from the way it
153
00:22:54,780 --> 00:22:58,600
has been defined ok any questions on this
so far
154
00:22:58,600 --> 00:23:18,549
yes ok so so i am going to quickly summarize
in the next couple of minutes what we the
155
00:23:18,549 --> 00:23:35,960
first topic that we have finished so we looked
at the general energy balance to quantify
156
00:23:35,960 --> 00:23:52,640
conduction process so its key purpose is to
quantify conduction process mode of heat transfer
157
00:23:52,640 --> 00:24:04,010
ok then we looked at a very simple case of
one d conduction
158
00:24:04,010 --> 00:24:16,090
without heat generation and here we looked
at cartesian coordinates cartesian coordinate
159
00:24:16,090 --> 00:24:23,100
system we looked at radial coordinates we
looked at radial system so these are the things
160
00:24:23,100 --> 00:24:28,740
that we have seen so this is the sort of the
first aspect of conduction process so what
161
00:24:28,740 --> 00:24:34,340
we are going to see the next topic is going
to be the so the third topic in that list
162
00:24:34,340 --> 00:24:48,550
which is what we are going to see now is what
happens when we include heat generation and
163
00:24:48,550 --> 00:24:58,470
i should also mention that we included resistance
concepts along with this the
164
00:24:58,470 --> 00:25:03,320
systems network concepts along with these
so what we are going to see now in the next
165
00:25:03,320 --> 00:25:08,320
five minutes and in the next lecture which
starts ten minutes from now will be what happens
166
00:25:08,320 --> 00:25:13,120
when you include heat generation off we so
far assume that heat generation is zero what
167
00:25:13,120 --> 00:25:18,890
happens if we include that and how these analysis
that we have we have done so far would look
168
00:25:18,890 --> 00:25:34,280
slightly different from what we saw so far
169
00:25:34,280 --> 00:25:47,610
so supposing we take a slab ok supposing we
take a slab and lets say that q dot is the
170
00:25:47,610 --> 00:25:53,340
volumetric heat that is being generated inside
the slab it could be be many different ways
171
00:25:53,340 --> 00:25:58,500
for example it could be like there is an electrical
heating system inside and so that is generating
172
00:25:58,500 --> 00:26:03,970
heat inside the slab or it could be some reaction
which is inducing heat inside there are solid
173
00:26:03,970 --> 00:26:09,120
reactions so it could be that solid reaction
is inducing some heat some exothermic reaction
174
00:26:09,120 --> 00:26:13,490
which is inducing some heat inside the slab
so there could be many different mechanisms
175
00:26:13,490 --> 00:26:19,539
by which heat is actually being introduced
into a system and so there is a a simultaneous
176
00:26:19,539 --> 00:26:26,890
heat generation and heat transport
so supposing just for sake of convenience
177
00:26:26,890 --> 00:26:33,360
purposes i define midpoint as x equal to zero
in fact you will see in a short while why
178
00:26:33,360 --> 00:26:43,190
i did that ok and lets say that plus l and
minus l are the coordinate systems for this
179
00:26:43,190 --> 00:26:50,640
geometry and if i assume that temperature
on one end is t one and the other end is t
180
00:26:50,640 --> 00:26:58,150
two so by representation i have assumed that
t one is greater than t two ok and suppose
181
00:26:58,150 --> 00:27:08,640
i proceed that its a steady state problem
ok so its a a steady state problem then what
182
00:27:08,640 --> 00:27:16,370
is the model equation
183
00:27:16,370 --> 00:27:26,480
its the k del square t laplacian plus q dot
equal to zero so thats the model equation
184
00:27:26,480 --> 00:27:34,870
and so we can write it as k d square t by
d x square plus q dot equal to zero and the
185
00:27:34,870 --> 00:27:45,020
boundary conditions are quite trivial minus
l is one and t at x equal to plus l is t two
186
00:27:45,020 --> 00:27:52,780
ok so i can solve this equation without much
effort so i am not going to derive the solution
187
00:27:52,780 --> 00:28:06,940
here so i will give you what the solution
is ok so the solution is t of x equal to q
188
00:28:06,940 --> 00:28:22,559
dot into l square by two k into one minus
x square by l square plus t two minus t one
189
00:28:22,559 --> 00:28:33,520
by two into x by l plus t one plus t two by
two ok so thats the solution for the model
190
00:28:33,520 --> 00:28:39,210
ah just a second order differential equation
so its not a very difficult exercise to solve
191
00:28:39,210 --> 00:28:44,290
them so you should actually solve the equation
and convince yourself that this is the correct
192
00:28:44,290 --> 00:28:45,150
solution ok