1 00:00:14,500 --> 00:00:29,270 alright so we stopped by looking at the resistance network qualified by strength thermal resistance 2 00:00:29,270 --> 00:00:39,199 network ah for composite walls so thats where we stopped in the last lecture 3 00:00:39,199 --> 00:00:44,950 so supposing there are multiple walls which are associated so if the length is l one l 4 00:00:44,950 --> 00:00:55,420 two and l three and we have fluid which is flowing on either side of the composite wall 5 00:00:55,420 --> 00:01:11,380 so thats the composite wall ok and so if the conductivities are k a k b and k c so we said 6 00:01:11,380 --> 00:01:27,720 that we could construct the network which has essentially five five resistances 7 00:01:27,720 --> 00:01:34,380 thats this x direction and if the heat transport coefficient is h one on this side and the 8 00:01:34,380 --> 00:01:40,840 temperature is t infinity one and this is h two and t infinity two so we said that its 9 00:01:40,840 --> 00:01:52,270 t infinity one and and t infinity two and the resistances are one by h one a l by l 10 00:01:52,270 --> 00:02:04,490 one by k one by a l two by k two a and l three by k three a and one by h two a ok so one 11 00:02:04,490 --> 00:02:12,700 could define we know what are the total resistances so the total resistance r total it will simply 12 00:02:12,700 --> 00:02:18,170 be sum of all the resistances sum of all the individual resistances which is essentially 13 00:02:18,170 --> 00:02:22,610 sum of these now what we are going to start within today 14 00:02:22,610 --> 00:02:27,720 lecture is we are going to look at what is called the overall heat transport coefficient 15 00:02:27,720 --> 00:02:44,930 ok so one could define ah overall heat transfer coefficient and the reason for doing that 16 00:02:44,930 --> 00:02:50,590 is ultimately from [measure/measurement] measurement point of view in fact is what i would measure 17 00:02:50,590 --> 00:02:55,890 is the temperature here and temperature here so i need to know what is the total amount 18 00:02:55,890 --> 00:03:03,770 of heat that is transferred from lets say hot fluid on this side ah i use one ok so 19 00:03:03,770 --> 00:03:12,620 i can put one two one three ok and cold fluid on this side ok so i want to know what is 20 00:03:12,620 --> 00:03:18,340 the effective heat transport that is occurred from the hot fluid to the cold fluid and that 21 00:03:18,340 --> 00:03:26,030 is simply because i may not be able to measure the temperatures in between so one could define 22 00:03:26,030 --> 00:03:32,110 the total amount of heat that is transported similar to the way we defined newtons law 23 00:03:32,110 --> 00:03:43,440 of cooling as the constitutive equations so we could write we could define a overall heat 24 00:03:43,440 --> 00:03:50,850 transport coefficient u multiplied by a multiplied by the net temperature difference of the observable 25 00:03:50,850 --> 00:03:59,069 temperatures or measure ah measurable properties ok and that should obviously be equal to that 26 00:03:59,069 --> 00:04:07,129 should be equal to what yeah by r total so thats right so that should 27 00:04:07,129 --> 00:04:15,019 be equal to i infinity one minus t infinity two divided by r total simply by the definition 28 00:04:15,019 --> 00:04:25,250 of the resistances ok so from here we can clearly read out that r total equal to one 29 00:04:25,250 --> 00:04:33,740 by u a ok and that should be equal to this sum of the individual resistances sum of all 30 00:04:33,740 --> 00:04:40,000 the resistances that is involved in the system that we are considering so this is a a ubiquitous 31 00:04:40,000 --> 00:04:47,010 property of ah a definition of any heat transport system so in principle one could define a 32 00:04:47,010 --> 00:04:52,310 overall heat transport coefficient for any heat transport system and what we are doing 33 00:04:52,310 --> 00:04:57,990 essentially is we are lumping all the properties all the transport processes which are occurring 34 00:04:57,990 --> 00:05:03,840 inside everything is lumped into these one quantity called universal or overall heat 35 00:05:03,840 --> 00:05:10,160 coefficient and we will see many variations across the overall heat transport coefficient 36 00:05:10,160 --> 00:05:13,880 that depending upon the system that we are considering that we are going to see in todays 37 00:05:13,880 --> 00:05:19,210 lecture and several lectures in future so its very important to understand the definition 38 00:05:19,210 --> 00:05:24,190 of overall heat transport coefficient and note that its a fictitious quantity it can 39 00:05:24,190 --> 00:05:30,310 be detected based on the individual properties of the system that we are considering however 40 00:05:30,310 --> 00:05:36,110 over all heat transport coefficient is mainly for convenience purposes and for calculation 41 00:05:36,110 --> 00:05:40,230 purposes it really helps in ah defining such kind of a quantity 42 00:05:40,230 --> 00:05:53,530 so there is one small aspect about resistances there is something called contact resistance 43 00:05:53,530 --> 00:06:03,280 so so far and all the ah that couple of examples that we described in the last lecture and 44 00:06:03,280 --> 00:06:11,960 start of todays lecture the we assume that the contact between the slabs is supposed 45 00:06:11,960 --> 00:06:19,780 to be a smooth contact however in reality that need not necessarily be the case for 46 00:06:19,780 --> 00:06:25,889 example if you have ah when you say smooth its smooth all the way upto the microscopic 47 00:06:25,889 --> 00:06:31,430 level right so its not this smoothness that you observe in your eyes its the smoothness 48 00:06:31,430 --> 00:06:36,310 that you would observe when you see under a microscope so that you will not expect you 49 00:06:36,310 --> 00:06:42,819 there is no reason why you should expect that the wall that you are having has a a smooth 50 00:06:42,819 --> 00:06:48,490 surface all the way upto the microscopic level so therefore there will always be certain 51 00:06:48,490 --> 00:06:56,550 resistance which is offered by the non smooth contact between the two surfaces of the composite 52 00:06:56,550 --> 00:07:04,580 wall so as a result one could define something called a contact resistance and it depends 53 00:07:04,580 --> 00:07:09,660 upon the smoothness properties so there is no clean way to estimate what should be the 54 00:07:09,660 --> 00:07:14,449 value of this contact there are some correlations we will not go into those correlations in 55 00:07:14,449 --> 00:07:20,569 this course but as when it is required particularly from the problem solving or exam point of 56 00:07:20,569 --> 00:07:26,090 view these kinds of numbers will be provided to you ok so give me a sec 57 00:07:26,090 --> 00:07:34,130 so supposing if you have a another wall here ok now if i want to construct resistance network 58 00:07:34,130 --> 00:07:39,550 including the contact resistance then what i would do is so supposing i have fluid which 59 00:07:39,550 --> 00:07:49,710 is flowing here ok then you will have so in if there is no contact resistance then there 60 00:07:49,710 --> 00:07:55,860 total of five resistances but because the contact position is not smooth its going to 61 00:07:55,860 --> 00:08:01,490 offer a certain resistance to heat transport and therefore you will have an additional 62 00:08:01,490 --> 00:08:12,699 network which is basically the contact resistance which will come in series so the the other 63 00:08:12,699 --> 00:08:17,520 five are the same thing what we saw a short while ago so we need to include a contact 64 00:08:17,520 --> 00:08:21,879 resistance which is present in between this supposing we want to include a contact resistance 65 00:08:21,879 --> 00:08:26,470 between the second and the third wall we could do that we could include an another resistance 66 00:08:26,470 --> 00:08:34,150 here any questions why should ah because of there is a lack of 67 00:08:34,150 --> 00:08:35,150 correct 68 00:08:35,150 --> 00:08:41,700 yeah so supposing if they are not in contact with each other ok so a a very simple example 69 00:08:41,700 --> 00:08:46,440 is supposing you are holding a coffee cup ok so there are two ways of holding it i could 70 00:08:46,440 --> 00:08:53,260 hold it like this with all my fingers on the curved surface of the tumbler or i could hold 71 00:08:53,260 --> 00:08:58,910 it just at the top so if you look at the workers who are like ah drinking coffee and tea the 72 00:08:58,910 --> 00:09:04,860 hot tea they usually hold the tip or they hold a cup like this and the reason is that 73 00:09:04,860 --> 00:09:10,330 you do not want to have a direct contact with the surface which is hot now taking that parallel 74 00:09:10,330 --> 00:09:16,690 here if this surface is not smooth then the contact the transport of heat from one slab 75 00:09:16,690 --> 00:09:22,500 to the other depends upon the overall effective surface area which is available for transport 76 00:09:22,500 --> 00:09:27,130 now if the contact is not very smooth then the total surface area which is available 77 00:09:27,130 --> 00:09:32,790 for heat transport from one slab to the other slab is not as much as the overall surface 78 00:09:32,790 --> 00:09:38,760 area which is available and therefore therefore the therefore the total amount of heat that 79 00:09:38,760 --> 00:09:43,860 is transport is not exactly the amount of heat that comes at this end and therefore 80 00:09:43,860 --> 00:09:49,130 this offers a resistance yes you had a question ah yes there will be a dissipation because 81 00:09:49,130 --> 00:09:54,610 of that see what is the resistance note that the resistance is basically characterizes 82 00:09:54,610 --> 00:10:00,550 the total amount the ability of the system to transport heat what is the resistance that 83 00:10:00,550 --> 00:10:06,060 it offers to transportation of heat from one location to the other because the contact 84 00:10:06,060 --> 00:10:11,220 point is not significantly good there is going to be some dissipation and therefore that 85 00:10:11,220 --> 00:10:15,830 that offers a certain resistances and thats what is captured by the contact resistance 86 00:10:15,830 --> 00:10:25,490 ok any other question ok alright so next what we are going to see is 87 00:10:25,490 --> 00:10:32,329 we assumed so far in all the cases that we considered that the area of heat transport 88 00:10:32,329 --> 00:10:46,310 is constant so today we are going to look at the varying area systems so how to characterize 89 00:10:46,310 --> 00:10:54,279 and quantify conduction process through a a system where the surface area or cross sectional 90 00:10:54,279 --> 00:11:03,380 area for heat transport is constantly changing a very simple example would be that supposing 91 00:11:03,380 --> 00:11:19,140 i have a truncated cone ok i have a truncated cone ok and lets say i am looking at so this 92 00:11:19,140 --> 00:11:28,579 is lets say at equal to zero ok so i want to know what is the heat that is being transported 93 00:11:28,579 --> 00:11:38,110 from lets say let me call this as x direction x equal to x one x two i wanted to call it 94 00:11:38,110 --> 00:11:43,900 radius so thats the centers of radius is zero so i want to know what is the amount of heat 95 00:11:43,900 --> 00:11:50,541 that is transported from x one to x two and i am going to make an assumption because i 96 00:11:50,541 --> 00:12:07,389 am looking at one d systems i am making an assumption that the cross sectional temperature 97 00:12:07,389 --> 00:12:12,710 that means that every at every cross section i assume that the temperature is uniform in 98 00:12:12,710 --> 00:12:15,780 the cross section and the gradients are zero ok 99 00:12:15,780 --> 00:12:27,630 so now so this is my radius at any location and so i could write my ah balance and if 100 00:12:27,630 --> 00:12:34,560 i continue to assume that its a steady state system and heat generation is zero same assumptions 101 00:12:34,560 --> 00:12:42,250 as what we made before so i could simply write the total amount of heat that heat transfer 102 00:12:42,250 --> 00:12:49,120 rate q x is given by minus k which is the conductivity of that material multiplied by 103 00:12:49,120 --> 00:12:56,230 the cross sectional area of heat transport so note that now the cross sectional area 104 00:12:56,230 --> 00:13:01,950 is a position of the function position ah is a function of the position excuse me ok 105 00:13:01,950 --> 00:13:10,240 multiplied by d t by d x ok so what is the objective we need to find the temperature 106 00:13:10,240 --> 00:13:15,110 profile that is the objective of the problem right so we said that if we know the temperature 107 00:13:15,110 --> 00:13:19,779 distribution if we know the temperature profile we are done we have quantified the system 108 00:13:19,779 --> 00:13:27,620 so thats what we need to find so lets say we integrate this equation ok so we say that 109 00:13:27,620 --> 00:13:38,490 q x is minus k what is the area its pi cross sectional area is pi into radius square ok 110 00:13:38,490 --> 00:13:51,860 pi r square right multiplied by d t by d x right so supposing i say that r goes as a 111 00:13:51,860 --> 00:13:58,950 a linear function of the position if i say that the radius of the local radius goes as 112 00:13:58,950 --> 00:14:05,790 the linear function of the axial position then i could simply write this as minus k 113 00:14:05,790 --> 00:14:20,630 pi a square x square into d t by d s into d t by d x ok and now what will be q x will 114 00:14:20,630 --> 00:14:29,630 it be constant or it will change rate of heat transfer that will be constant why will it 115 00:14:29,630 --> 00:14:37,580 be constant because of the energy balance you see that there is no heat that is being 116 00:14:37,580 --> 00:14:42,910 generated so whatever comes in it has to go out here at steady state condition note that 117 00:14:42,910 --> 00:14:52,920 steady state is very important yes right so the question is if you dissipation of heat 118 00:14:52,920 --> 00:14:57,310 how will it be constant but when we say that there is no generation or less of heat which 119 00:14:57,310 --> 00:15:03,730 means that the dissipation is zero we will will come to that so there are ways to consider 120 00:15:03,730 --> 00:15:08,250 that we are actually consider when we are doing a two dimensional system you can actually 121 00:15:08,250 --> 00:15:12,960 look at dissipation from the outside walls and we will actually see it in one of the 122 00:15:12,960 --> 00:15:19,209 examples in the future lectures alright so because q x is constant we should be able 123 00:15:19,209 --> 00:15:24,220 to integrate this expression to find the temperature profile 124 00:15:24,220 --> 00:15:32,980 so supposing i integrate between t one and t one is the temperature at the boundaries 125 00:15:32,980 --> 00:15:38,459 of this system ok t one is the temperature at the boundaries of this system i integrate 126 00:15:38,459 --> 00:15:51,860 between t one and t thats equal to q x minus q x by k pi r square of k pi a square into 127 00:15:51,860 --> 00:16:01,029 d x by x square going from x one to x ok its a pretty simple integration so its q x by 128 00:16:01,029 --> 00:16:11,500 k pi a square into one by x minus one by x one ok so note that because its one by x square 129 00:16:11,500 --> 00:16:22,190 the minus sign will go away because of the integration and this is t minus t one ok so 130 00:16:22,190 --> 00:16:36,639 therefore t is t one plus q x by k pi a square into one by x minus one by x one is it a complete 131 00:16:36,639 --> 00:16:45,220 description we dont know the q x value right so its not a complete description yet so how 132 00:16:45,220 --> 00:16:51,680 do we find q x so we know that the temperature on the other boundaries is t two so we can 133 00:16:51,680 --> 00:16:57,920 use that property to find out what is q x so we dont know what q x is so how we are 134 00:16:57,920 --> 00:17:09,380 going to do that we are going say t two is t one plus q x by k pi a square multiplied 135 00:17:09,380 --> 00:17:22,579 by one by x two minus one by x one ok so from here q x is given by t two minus t 136 00:17:22,579 --> 00:17:38,660 one divided by one by x two minus one by x one multiplied by k pi a square multiplied 137 00:17:38,660 --> 00:17:51,179 by k pi a square and so now we can plug this into our solution we can plug this into our 138 00:17:51,179 --> 00:18:04,270 solution so t equal to t one plus k pi plus t two minus t one divided by one by x two 139 00:18:04,270 --> 00:18:13,580 minus one by x one multiplied by one by x minus one by x one so thats the distribution 140 00:18:13,580 --> 00:18:21,580 temperature distribution in the system with the varying cross sectional area so an important 141 00:18:21,580 --> 00:18:29,140 message of this example is that what you need what is preserved here or what remains constant 142 00:18:29,140 --> 00:18:35,770 in this system because there is no heat generation or dissipation is the heat transfer rate and 143 00:18:35,770 --> 00:18:42,840 not the flux ok so you have to make a distinction here so what we said is the heat transfer 144 00:18:42,840 --> 00:18:54,490 rate so this remains a constant however the flux 145 00:18:54,490 --> 00:19:05,880 which is given by q prime which is minus k d t by d x this is not a constant so this 146 00:19:05,880 --> 00:19:11,549 is an important observation which you would not have made in the simple one d system where 147 00:19:11,549 --> 00:19:16,410 the cross sectional area is constant the cross sectional area of heat transport is constant 148 00:19:16,410 --> 00:19:21,750 where you would not be able to make such a distinction between the two so when you have 149 00:19:21,750 --> 00:19:26,000 varying cross sectional area what is really conserved and what is really preserved is 150 00:19:26,000 --> 00:19:31,590 the heat transfer rate and in fact thats the reason why you write a a rate balance and 151 00:19:31,590 --> 00:19:37,130 not a flux balance so this is very important to understand this distinction its important 152 00:19:37,130 --> 00:19:42,770 to write ah transfer rate balance because thats the final quantity that is importance 153 00:19:42,770 --> 00:19:48,820 and the flux need not necessarily remain constant even in a small element so its very important 154 00:19:48,820 --> 00:19:53,539 to understand these distinction and we are going to next see how these things are going 155 00:19:53,539 --> 00:19:57,580 to play a role when you are looking at radial systems where this becomes extremely important 156 00:19:57,580 --> 00:19:57,850 ok