 Good morning. We will quickly go through our epsilon NTU method. Most of us again are comfortable with this topic. So, we are not going to spend too much time or too much into the details of it, but some physical aspects are going to be given nevertheless first and foremost. See, we have LMTD and epsilon NTU method. Why do we have these two methods? Well, first LMTD method I am sure professor Prabhu has told is essentially a sizing method, where I want to get the dimensions meaning if I know the heat load which heat exchanger or what is the size of the heat exchanger which is going to give me this performance or this Q or heat transfer ability. So, it is what we call as a sizing problem. So, when I have temperatures known mass flow rates known or in other words if I have the heat load known one fluid inlet outlet temperature mass flow rate, another fluid inlet or exit temperature whatever it is I will know this so called LMTD by some energy balance then I will know my overall heat transfer coefficient because I know H of each of these fluids and if I know the conduction resistance I can get U once I get U I know LMTD I know the Q I can apply this Q is equal to U a LMTD and I can get the area or which is the size of the heat exchanger. So, knowing a diameter I can get the length or knowing the length I can get the diameter and appropriately I can size I can give a physical size to the heat exchanger well now in real life do we know all these things do we know three temperatures that is cold fluid inlet hot fluid inlet cold fluid exit or hot fluid exit. So, I need to know three of these four to be able to apply energy balance because what is energy balance energy balance essentially is we are going to say heat loss by hot fluid is equal to m dot h C p h T h i minus T h o. So, if I just take a simple counter flow heat exchanger T with respect to x and this is hot fluid T h o T c i T c o here T h i here. So, what am I saying I need to know three out of these four temperature T h i T h o T c i T c o if I know any three by energy balance I can get m dot C C p c T c o minus T c i. So, knowing three I can get the fourth one knowing any of these two combination either this one or this one I can get the heat load. So, I know knowing all three I know the fourth one going three temperatures get q get fourth temperature and then get area is equal to q by q times delta T log. So, this is the standard procedure that we have followed very nice, but in real life we most occasions we will know only this and this temperature. We will know the mass flow rate m dot h we will know the properties hopefully depending on the working fluid. We know mass flow rate m dot h m dot c we know T h i T c i because these are under our control for example, oil is coming out from a lubricating oil from some application is coming out you know the exit temperature from that application. So, that is the inlet to the heat exchanger. So, it is being cooled by if you are in a ship it has to be cooled by sea water. So, you know the sea water inlet temperature apart from that I do not know anything. So, for me to use this delta T log mean approach is going to be very difficult you can do it we are not saying you cannot do it, but it will be an iterative approach. It will not be like you know how you have taught how you are done in class we cannot do it so simply. So, what do we do therefore, we say when I am going when I do not know 3 out of the 4 temperatures my problem essentially is intractable by this method. So, what do I do I do not solve the problem that is not an option. So, we go to what we call as this so called epsilon NTU method what is how and means why did people come up with this epsilon NTU method or how did somebody know that this method is going to be useful for solving heat exchange related problem. What is this epsilon what is this NTU how did they come up well this has not come from you know magic or something this is come from very very fundamental concept. What is this epsilon we will see NTU we will see and this ratio CR that we are going to see before that I want to tell you one other thing which I suddenly remember if you are talking of class classroom type of problem you can use this kind of solution you know where you have everything known inlet exit temperatures are known you can do this close form solution like this, but in real life if you are trying to do in application or if you are in a design environment where you want to calculate the length accurately where you will want to take the property variation etcetera. What is typically done is you split the you march in every incremental length. So, you choose a particular length DX you let me let me just put it. So, the provost says we will tell this at the end after we have finished both epsilon NTU method it is correct because in it can be valid for both methods. So, it is not specific to LMTD approach anyway. So, let us get back to this epsilon NTU method. So, what is this epsilon what is this NTU what is this specific heat ratio and how did somebody know that this is going to give me some kind of a dimension of heat exchanger meaning after you see it becomes obvious, but how well it has come from a very very fundamental thing by case in London. So, this is done mainly to avoid iterations. So, what is known is the type and size of a heat exchanger and whether I am going to be able to achieve this particular performance is what is going to be understood by this method. So, it is the inlet temperature of the two fluids are known mass flow rate specific heats type and size of the heat exchanger of course is known our idea therefore is to predict the outlet of outlet temperature of the hot and cold fluid also to see whether this thing is going to satisfy our application. This problem of course can be solved by LMTD, but it is going to be a iterative approach. So, what do we do? So, we define two or three quantities and then we will proceed further with the derivation. So, first quantity of interest is this so called effectiveness. What is this effectiveness? It is actual heat transfer rate that is what you calculate. So, let me take a counter flow heat exchanger just for simplicity. I will put this diagram T function of X, T H I, T H O, T C I, T C O, this is the direction of flow I know T H I and T C I, M dot H, C P H, M dot C, C P C. I know the area or the size this is known. So, what are we saying? We are going to say that there is some heat transfer application for which this heat exchanger is there that application basically will tell you that means you know the heat load also. Is this going to work and how good is it? Is it a very good heat exchanger or a very bad heat exchanger? So, first definition the easiest ones we will take M dot C P of cold you calculate, M dot C P of hot fluid we will calculate. This quantity we are calling as capital C C capital C H. Specific heat this is now basically specific means any specific quantity specific enthalpy specific internal energy would be mass basis per unit mass. This is basically you are multiplied by the mass flow rate. So, this is the actual heat capacity. So, that is this defined here already. So, we are going to stick with those definitions this C is C min by C max. What is C min by C max? I have calculated M dot C P for hot fluid M dot C P for cold fluid one of them is going to be smaller than the other. So, take the smaller one divided by the larger of the two that is called as a C or certain textbooks will call this as C R. C or C R will be nothing but C min capital C min by capital C max. So, this is one quantity we define. What is this? What is the C min C max? It is not any magic again. It is just temperature difference ratio. How did I get that? Q is equal to C H capital C H. This is I hope the capitals are clear. T H I have inlet minus T H outlet heat loss by hot fluid is correct. This is M dot C P. This is another M dot C P. Therefore, let us say C C by C H that is M dot C P cold divided by M dot C P hot is equal to T H I minus T H O divided by T C O minus T C I. So, in case C C is smaller than C H this is the definition in case C H is smaller than C H the reciprocal is the definition. But whatever it is we need to appreciate that this is just the temperature difference ratio. That is the first thing. Second definition that we have is this so called NTU. We have seen this NTU in when we looked at internal flow constant wall temperature case. We remember we came up with this delta T local divided by delta T inlet is equal to E X P minus H P X by M dot C P. We came up with this. This is nothing but your surface area and we came up with this very useful term and we call this as NTU. Remember the diagram I am just putting this again for those might have forgotten. This is the profile correct constant wall temperature and this is your fluid temperature which is increasing T surface T X. Now, this derivation when we did we did it for what for a pipe where a fluid was flowing in and the inside wall temperature was at T S and the fluid temperature varied with respect to the length direction. Well I can do the same derivation when I have not just U sorry not just H at the inlet this H was H at inlet H on the convective heat transfer coefficient for the fluid on the inside which is what we are doing. Now, I can do the same thing for the same fluid which is flowing here. If I am given this surface temperature T S O which is what we measure many times this is T S I and there is a conduction resistance here and this is T M of X heat transfer rate is the same right. So, there I could write Q is equal to H A S T S minus T M local remember we wrote this and that is how this H A S came into the problem. Now, I can write this as Q is equal to instead of H A S I will define overall heat transfer coefficient times A which is nothing but 1 over summation of all the thermal resistance where R thermal is 1 by H I A I internal flow convective resistance plus log R out by R in divided by 2 pi L k. I can do this per unit length so on and so forth I am not going into the details. So, essentially this H A S which was T S minus T M X divided by R convection can now be replaced by T T S out minus T M X divided by summation of R correct summation of R thermal which is these two or that is equivalent of saying U times A T S O minus T M X correct. So, if I do this my whole derivation I can do and still say delta T X which is nothing but T S out I am the driving temperature difference for this resistance is T S out minus T M local divided by T S out minus T M inlet is equal to exponential d k of minus U times P X divided by M dot C P that is all. So, this is U A this is your surface area local till that location from inlet to that location. So, U A by M dot C P and this quantity we called there without going into so much detail H A by M dot C P we decided to call this as N T U remember our we called that quantity as N T U and that N T U is not come from anywhere it is just come from this concept of reciprocal of thermal resistance. So, I can write U A S by M dot C P is nothing but N T U number of transfer units what is this transfer units are there some 15 units no what we are saying is this is a representation of the physical size of the heat exchanger this are related to the flow which is given to us this is the again a flow related quantity this quantity is a size related quantity size gets embedded in this A term. So, larger the number of transfer unit larger the N T U larger is the dimension or larger is the dimension larger is the N T U I will just quickly go to this slide in internal flow which has this concept H A S by M dot C P this is dimensionless parameter now instead of H A S I am calling U A S this H A S was what 1 over convective resistance. So, in fact it is N T U is nothing in when we did internal flows we define N T U as H A S by M dot C P this H A S was 1 over convective resistance. So, it is M dot C P 1 over M dot C P times 1 over R convection now we are saying this N T U is U A S by M dot C P which is nothing but 1 M dot C P R total. And we for recap for doing a recap I am going to say this thing again N T U of 5 indicates the limit has been reached for heat transfer that means what if I fix this inlet temperature and keep changing my surface area here and keep changing the surface area H is fixed M dot is fixed C P is fixed I will get a different set of exit temperatures that exit temperature increases with increase in length because diameter is fixed H M dot C P are fixed as my length increases N T U increases exit temperature increases what are we saying N T U of 5 use me exit temperature of 99.5 N T of 10 is giving me exit temperature of 100. So, it is hardly again we are coming back to the driving temperature difference decreases with further and further increase in length in this case. So, same thing here we are saying N T U is a representation of the dimension and again I am going to show you this is nothing but the ratio of temperature differences what is that how is that going to come will see in a minute. So, this N T U has been defined like this Q again let me say is U A S delta T log mean L M T D this Q is also equal to M dot C P let us say cold fluid is this is a minimum specific heat fluid. So, T C O minus T C I so dimensionally this U A S by M dot C P is nothing but T C O minus T C I by delta T log mean this has units of temperature this has units of temperature N T U is a dimensionless quantity, but again it is directly related to the temperature differences. So, this also is a temperature difference ratio wow. So, C R or C is a temperature difference ratio N T U is also a temperature difference ratio very nice. So, what are we where are we going from here one last thing that we have to define and this is the more difficult thing is the so called effectiveness. Once we define this there is hardly anything more to be done effectiveness is defined as actual heat transfer rate divided by maximum heat transfer rate actual everybody can write heat load M dot C P cold times T C O minus T C I divided by Q max or this is M dot C P H T H I minus T H O by Q max what is this Q max this is what we do not know this is where most questions occur. Let us take let us try to understand Q max Q max refers to the maximum possible heat transfer that could occur when can this heat transfer be maximum heat transfer can be maximum when my length tends to infinity. So, if my length were infinite if my heat exchanger was infinitely long then I am going to have maximum heat transfer. So, what is that parallel flow or counter flow what is it going to give me if I take for example, parallel flow I am going to do here after a particular point in parallel flow we can very easily see that the driving temperature difference is going to be almost equal to 0. Whereas in counter flow if I see let us take this purpose I have to draw it like this big and with a different slope T H I T H O T C I T C O temperature versus X M dot max is given by we are going to say the formula first and then we will say why that is the correct way of writing C min times T H I minus T C I this is all of us do this all of us know this also you want to say why this is true. So, in the limiting condition maximum heat transfer is going to be given by the maximum temperature difference that is possible in this device it cannot be this minus this cannot be this minus this it has to be the maximum temperature difference. Now, maximum temperature difference times some M dot times C P what is that M dot times C P why is it C minimum that is what we will see now think of this fluid here I will use the same diagram think of this fluid which fluid hot or cold has a smaller heat capacity how do we know that we can get that by the slope of this curve how the slope is what M dot C P is essentially Q divided by the delta T whatever hot or cold fluid is Q divided by delta T. So, delta T is large Q is fixed hot fluid exchange heat with cold fluid so this is going to be fixed. So, if my M dot C P is small delta T has to be large M dot C P is small delta T has to be large that means in this case this is the lower M dot C P because I have drawn this with a larger slope larger temperature change this one is going to be a smaller that means lower sorry higher M dot C P or a lower temperature drop. Then why should C min come here why not C max well it is because this fluid if length were infinite this fluid eventually the exit temperature of the cold fluid because it is having a lower M dot C P lower heat capacity is going to reach somewhere at infinite length asymptotically it will reach T h i correct why not why not the higher M dot C P fluid we will argue for that also T h i it started off if length went to infinite this thing is never going to be able to reach this T c i because of its inherent capability that it cannot take that much of temperature difference because it has a larger M dot C P associated with it. So, the colder fluid will asymptotically or in the limiting condition reach T h i. So, this C min essentially is going to be the one which controls the maximum heat. So, the cold fluid in the limiting condition will undergo a temperature difference of T h i minus T c i where T h i was the exit temperature T c o became equal to T h i in the limit of L tending to infinity that is the maximum heat transfer. Now, just for clarification I will do the other thing also because our students will have infinite confusion on this. So, let us say my M dot C P of hot is smaller than M dot C P of cold still my formula for Q max will come out to be the same C min T h i minus T c i it will come out to be the same you will see that. So, hot fluid is going to show a larger temperature change let us say this is the cold fluid T c i T c o hot fluid is showing a very large temperature change T h i this is T this is X what am I saying this will in the limit this is actually quite easy to understand in the limit this fluid will eventually reach the inlet temperature of the cold fluid. I do not have to even explain this too much just by looking at the diagram if I increase the length the inlet temperature is there quickly the hot fluid will reach the exit temperature of the cold fluid in the limiting case. So, this T h o will become equal to T c i therefore, Q max will therefore, be M dot C P of hot times T h i minus T h o which in turn is T c i. So, whatever be it whichever be the situation my Q max is nothing, but C min because that fluid because of its smaller heat capacity can take in larger temperature change. So, T h i minus T c i now I will just write epsilon and then we will close for T is Q by Q max again I am going to show you this is the temperature difference ratio if I write this as let us say hot fluid because we have this diagram here M dot C P hot T h i minus T h o divided by M dot C P hot T h i minus T c i this cancels of I am left with a temperature. So, effectiveness N T u and C r or C are all delta T something by another delta T. So, temperature difference ratios now we will understand this. So, I have three non-dimensional numbers just bear with me for two minutes dimensional analysis hot and fluid mechanics tells me I have three non-dimensional numbers which are all related to the same quantity some temperatures one of them has to be a function of two others. So, I can have C r as a function of epsilon N T u I can have epsilon as a function of C r and N T u I can have N T u as a function of C r and epsilon C r is known to us because that is a flow M dot and C P. So, that we will not take as an unknown quantity that is a known quantity. So, depending on my problem situation whether N T u is known or epsilon is known I can in general write the following I can in general write epsilon is a function of N T u and C r or N T u is a function of epsilon and C r why I could write that because these non-dimensional numbers we are not giving them any special name these are all non-dimensional quantities only these non-dimensional quantities are all functions of temperature difference ratios only nothing else. And since that is the case by our logic that we have learnt in dimensional analysis one of them is going to be a function of the other two I can even write for sake of completeness C r is a function of epsilon N T u, but we will not find that useful because C r is something that we know already. So, this we will not see we will see only these two cases from now I will go very fast because it is all things that we have studied this explanation sometimes students get bold. So, we have to spend some time on it we will take questions for about 5 minutes and then quickly get started any questions N i T T T. For condensation we are saying that the temperature of the steam remains constant. So, for example when we are allowing a very low temperature cold fluid. So, it will take the latent heat as well as the sensible heat when it takes the sensible heat the temperature of that horizontal line will decline, but why we are saying it is horizontal the same case for boiler. Okay professor the question is if I take a super heated vapor or sub cooled liquid then I will have some sensible portion and I will have some and I will have a portion in which my temperature is constant. For the sensible portion you design the heat exchanger as if you are designing for single phase flow and then use the delta T L M T D approach for constant temperature. So, basically the heat exchanger design will be subdivided into three parts, sub cooled latent heat and the super heated portion. So, basically it will become three portions that is what essentially we do in design of condensers or evaporator. In this part the line is horizontal that is all we never said that the line is horizontal all through any other question. You have used the hydraulic diameter as D2 minus D1, but in a double pipe heat exchanger the double the hydraulic diameter is different for heat transfer cases and pressure drop measurement. So, which hydraulic diameter we have to use? See that is what I emphasized while solving the question asked is we have something called for heat transfer application we use one hydraulic diameter for flow rate applications I use another hydraulic diameter. See there is no confusion here actually. See how is Reynolds number defined? Reynolds number is defined as rho V D by mu. Velocity it will come by velocity equal to m dot upon rho into area. Area is annular area no confusion I get velocity, but which diameter I should use? I should be using the hydraulic diameter. I should be using the hydraulic diameter which is D0 minus Di, but where is the heat transfer taking place? Where is the heat transfer taking place? It is taking place only on the outer wall. So, I have to take D0 as the hydraulic diameter for heat transfer and D0 minus Di for Reynolds number, but few people argue and say no for both of the things I will take D0 only. It is okay actually. So, D0 few people argue and say that D0 only I will take for both actually it does not make much difference, but only thing is that Reynolds number will be slightly bloated. So, what if you have if you are nitpicking and if you have to do things correctly for Reynolds number I have to take hydraulic diameter which is 4 into wetted area upon perimeter and for heat transfer that is Nu equal to HD by K D has to be D0 outside diameter of the inner tube for heat transfer he mentioned that D2 square minus D1 square by D1 for pressure drop he used D2 minus D1 for heat transfer the outer side for annular side he used D2 minus D2 square minus D1 square by D1 for pressure drop he used D2 minus D1. I do not think that is right see D2 square minus D1 square there is no question to get it solved okay. So, what is that we are saying is for fluid dynamics we are going to use the hydraulic diameter as D2 minus D1. Yeah, but for heat transfer we are taking the outer diameter of D1 that is all I am saying D1 plus thickness did we answer your question yeah please repeat your question for example pressure drop estimation the hydraulic diameter is 4 times of hydraulic radius hydraulic radius is cross sectional area by wetted perimeter. So, for annular flow pi by 4 D2 square minus pi by 4 D1 square D1 plus D2 this case we will get D2 minus D1 correct okay other case for heat transfer for heat transfer we will get the same numerator by denominator is pi D1. So, we will get D2 square minus D1 square by D1 4 times area by what is the what is the numerator again please area same area is saying 4 into what is saying is 4 into yeah D2 squared minus D1 squared upon pi into D1 into that is it D1 into L D1 yeah only D1 that is all by D1 this is what is given in Karn vetted perimeter that is what is given in Karn yeah and Karn this is given yeah yeah yes sir yeah yeah people define like this people define like this, but there I said there are two schools of thoughts there are one can take the hydraulic diameter like this also people take like this also, but you can go back and check even if you take pi see basically what is happening where is the boundary layer where is the boundary layer growing the bond the thermal boundary layer is growing only on the outer wall. So, why should I take pi into D2 in pi by 4 into D2 minus D1 square that has no meaning. So, what where is the boundary layer growing is what I should be taking. So, that is why I said there are two schools of thought one school of thought is what professor Karn has given as DH equal to D2 squared minus D1 squared by D1 that is one school of thought, but the other school of thought is what we have stated in our problem that is simply pi D D1 into L that is all is that ok. So, having defined these three quantities epsilon NTU and AS now it is just a matter of putting things together and coming up with a formulation. So, what are we saying we want to relate these three quantities. So, that we are able to get a formulation which is based on these three non-dimensional variables. So, in broadly if you are going to write we will say NTU is UA by C min, C min is this and effectiveness is given by the function of C and NTU. So, this derivation essentially comes from fundamentals. So, Q by Q max is actual heat transfer rate by maximum heat transfer rate and that is the effectiveness. So, what I am doing now is just manipulation I am going to put all these things all this Q, Q max everything I told you in the beginning itself all these are ratios of temperature differences what I am doing I am writing this in this form. So, Q is equal to effectiveness times Q max which is nothing but effectiveness times C min T hi minus T ci definition of Q max this Q is nothing but what do I get UA by delta t log mean right UA by delta t log mean is used and I am going to substitute for these quantities T hi minus T ci how did I get that let me just write this step. So, that you do not get confused Q is equal to UA delta t log mean what is delta t log mean UA times counter flow we are going to use T hi T hi T ci T ci. So, T hi minus T ci T ci. So, T hi minus T ci minus T hi minus T ci divided by log of T hi minus T ci divided by T hi minus T ci. So, this is my LMTD business. So, what I do Q is equal to effectiveness C min T hi minus T ci. So, I am going to substitute now this relationship here and substitute for T hi minus T ci. So, what I what do I get. So, this is going to be log part remains there it is. So, I am just going to take it to the other side is equal to UA T hi minus T ho can be replaced. So, if I go back to this whiteboard T hi minus T ho I will club together T ci minus T co I will club together. So, this difference T hi minus T ho will be related to m dot h T th plus T co minus T ci will be m dot c T t that is all I am putting. So, once I put this what is the what is the aim we do not want any of these temperatures in picture. We want only this F a C r and N T u somewhere in picture. So, let us say I bring all these things together in form of C c and C h this remains as it is the log term and C c by C h we keep carrying this all the time C c by C h plus 1. In this thing I have told I am taking that C min comes out to be the C c C min is C c. If C min is not equal to C c then I will use the appropriate the lower one for the division otherwise it does not matter the way of doing is the same. So, Q is C c T c out minus T c in T h C h times T c in T h in minus T h. So, we have assumed in this derivation that the C min is this quantity. So, then it is just algebra I am not spending time on this this we will do some manipulation here because I want to get these in terms of all the temperatures. So, that I can write these as temperature difference ratios. So, if I all the steps have been given. So, I will right hand side remains as it is this is T h i minus T c i plus T c i minus C c by C h blah blah it comes out from here in terms of. So, this looks like an N T u to you this looks like a C r plus 1. So, we are going to have something like that and everything else here is going to be related to the other non dimensional parameter. So, I am going to substitute for all these using these two steps. So, this is a manipulation of this term T h o minus T c o. So, that I am going to write in this way from here where where did this come from this came from energy balance. So, I want to eliminate T h o. So, I write it what do I why do I write it like this I write it like this because T h i and T c i are the only known temperatures everything else is unknown. So, I do not like keeping all these in picture. So, I am writing T h o T c o I am going to write in terms of T h i first. So, that manipulation was for this and the other one will be having the T c i here. So, if I substitute all these and do the match I am just going to get some all the steps are there please do it. We are not going to do this I am going to get log of 1 minus 1 plus C c by C h epsilon C min by C c that is going to be equal to u a s. So, on and so forth. So, rearrangement will give me this expression this is what you see in the texts epsilon is a function of N T u times specific heat ratio. Now, this was derived for counter flow you can have a family of such derivations for various other geometries and all the text books that are there will have these formulae given. This formulae and this formulae are essentially the same let me just. So, this is N T u is an unknown effectiveness and C c are the parameters whereas there effectiveness was the unknown N T u and C c were the parameters. So, this and that are essentially one and the same you use what you want depending on what is known. So, if you know N T u and C c use this to get effectiveness if you know the effectiveness which is possible you use the other formulae to get the N T u however whatever be it these are essentially the same and these are tabulated in the form of graphs for us. So, this I will leave it to professor Prabhu to quickly go through these and then there are there is one problem which I think we are not going to thank you.