 We will get started. So, we had just understood something related to internal flow. And we said that this concept of thermally fully developed is something which is a little bit tricky to understand and hopefully we have got a feel for it. All we are saying is that a non-dimensional temperature which we are defining as the ratio of the difference between the surface temperature and the local value divided by the surface temperature and the mean temperature, that difference that ratio is invariant in X. And I think that definition is easier to understand especially if I take a constant wall temperature case, that way when I do that these surface temperatures become constant. So, all I am saying is that when I do the, you will see in one of the exercises you will see how to take the derivative of that. It is basically u by v form. So, v du by dx minus u dv by dx by v square that whole thing is equal to 0. And what it tells me essentially, if again just for ease of understanding I am keeping wall temperature constant, I am saying that the variation, the nature of variation of the mean temperature is similar to the nature of variation of the local temperature. And I think that is logical because one feeds into the other. So, with this background let us just go further. So, in a thermally fully developed region, this derivative with respect to X is 0 by definition. And therefore, if the derivative is 0 that means it is a constant with respect to X that means it is invariant with respect to X independent of X. So, we are saying if that function this T s minus T, T s minus T divided by T s minus T m, if this temperature difference ratio is not a function of X which we said in fully developed, then its derivative also should be independent of X. So, if this implies derivative of this function with respect to R must also be independent of X, am I right? See what I am saying is again T s minus T divided by T s minus T m with respect to R, but T s is some quantity, T and T m are both varying with R and X, is that right? Is that a fair statement? T s, T and T m are both functions of R and X, T is a function of R and X, T m is also a function of R and X, essentially what it means is not, function of R and X that is not the correct way to write, this is essentially a function of X because it has been integrated over the radial direction. So, what we are saying is if this ratio is not a function of X that means the derivatives of derivative of this term with respect to R should be independent of X also, meaning T definitely is a function of R and X, nobody is denying that, T m is only a function of X which that means T m is independent of R, so the denominator is independent of R. So, all I am claiming is that this T s minus T derivative with respect to R is also independent of X, it is a function of R and X, but in fully developed, thermally fully developed flow, this gradient is also going to be independent of X, I know you are, I see you are not convinced, that is okay, del, del T by, okay, this is theta, okay, del theta by del R, no it is not a function of X, I am not saying it is a constant, it is not a function of X, if I fix the radius at the same radial location at varying X locations that is constant, so I think now I have answered your question, I am not varying the radius, I fix the radius, that is why I am saying D by, at a given, that is independent of X, okay. No, no, no, again we are coming back to temperature profile, I drew this, superposition, see I had drawn this here, okay, it is a crude diagram, but I think what I am saying is the temperature profile will bulge in or collapse or whatever you want to call, but if you are looking at the temperature profile alone, that is the problem, you do not look at the temperature profile, you look at the ratio, you are talking of this some local temperature difference divided by some pseudo temperature difference based on an average temperature, okay, this ratio we have claimed to be independent of X. Now what I am saying is, if this is, this function is independent of X, okay, that means it is derivative with respect to R, it is a, I am not saying anything about its dependency with respect to R, in fact it is a function of R, all I am saying is it is independent of X, therefore its gradient also should be independent of X, mathematically also if you take a function which is not, which is independent of X, you can take any mathematical function, its derivative also obviously will be independent of X, do not look at temperature in isolation, we are looking at the non-dimensional temperature, which is essentially a difference in temperature divided by some reference, I do not know what that reference is, so I choose the best possible thing which is Tm, which incidentally is obtained by massaging this temperature profile to come to an average value, okay, then massage has a scientific basis because it is an energy balance, okay, Ts minus Tm, Tm is not a function of R because it has been integrated with respect to R, so Ts is not a, Ts is not a function of X, sorry Tm, Ts is not a function of R, Tm is not a function of R, only local temperature is a function of R, all I am saying is if this condition, if this condition is what is my definition of thermal, thermally fully developed, then this is straight from improper, then the derivative of this ratio with respect to R should also be independent of X, okay, so I do not know how else to, it is fair, right, because I am convinced means I am happy, so what we are saying is this quantity when I evaluate, why am I doing it at R equal to R, what is my aim, I need heat transfer coefficient, heat transfer coefficient is defined as minus k dT by dy at the wall divided by some reference temperature difference, correct, so therefore I will take the derivative with respect to R, again this is our favorite coordinate system, in fluid mechanics we have always converted R to Y, if R equal to 0, Y equal to capital R, R equal to capital R, Y equal to 0, so it is a downward, radially downward coordinate system starting from the pipe form, so this is R equal to R, I will differentiate this, the denominator is not a function of R, so it stays constant, Ts is a constant, so the derivative is 0, so this whole thing is not a function of X, and then all I am doing is now saying q heat transfer, heat flux is essentially local heat transfer coefficient, which is there, which I do not know what it is right now, times a reference, a temperature difference, what is the temperature difference which I am dealing with, surface minus the fluid, now for me to take a fluid temperature here does not make sense, because fluid temperature is continuously varying, therefore the whole idea of this bulk fluid temperature to replace T infinity, that is what is the purpose, so instead of that we have a Tm, Ts minus Tm has come, that is equal to k Dt by Dr, where is the minus sign gone, minus k Dt by dy, y is equal to R minus R, so I think that I will lose you if I do this, so the coordinate system is like this, this is y, this is the center line, this is R, R equal to 0, R equal to R, y equal to 0, y equal to capital R, and we will say y is equal to R minus R, so dy is equal to minus Dr, that is where the minus sign gets absorbed, so all I am saying is now, if I take this derivative, why am I doing this, I am doing this essentially to prove a very, very, very important relationship, this thing is not a function of x, that means in thermally fully developed region, this some ratio, I do not know what it is, some this ratio is independent of x, it is a constant with respect to x, now I will play with this ratio and say this Dt by Dr at R equal to R, if I multiply by k and I have this division already here, this quantity is nothing but your heat transfer coefficient. And therefore, in thermally fully developed portion of a tube, local heat transfer coefficient is a constant independent of the x location, this is clear, fully developed, that is why if you go back now in your flow over a flat plate, we had a local Reynolds number, local heat transfer coefficient which were all functions of x location, am I right, what is the characteristic dimension for flow through a pipe diameter, do I care whether it is 15 meters or 5 meters long, I do not care, as long as the flow is fully developed, there is a representative characteristic Reynolds number, there is a characteristic heat transfer coefficient associated only with the boundary condition of the flow at the surface, not it is independent of the length, fully developed, I have lost you completely, which part, clear, getting confused, which part? I check see if it is constant, no I got confused. It is okay, you might be expressing, there will be 5 others who do not want to tell it, it is okay, please ask. No, I did not understand this properly, because of how Hx comes to be constant, see. What professor is saying is that, do you understand this part, now I am just recasting this thing in the form of heat transfer coefficient, what is heat transfer coefficient, my definition for heat transfer coefficient, let me write it for your benefit, it is nothing but H is equal to minus k dt by dy at y equal to 0 divided by Ts minus Tm, I do not know T infinity, I do not like T infinity, so I will use Tm here, this I am saying is nothing but minus k dt by dr, this becomes a plus, because of the coordinate transformation and r equal to capital R divided by Ts minus Tm, and now if you look at that expression which was there on that slide, this quantity is essentially like that only k, only k is missing, that is all, this whole thing times k is like your heat transfer coefficient, that is what is given here, and the minus sign has got an absorb because of the coordinate transform, so all I am saying is this thing is your heat transfer coefficient, k is the property of the fluid, it does not depend on the distance, it does not depend on x, so if A times B, A is the constant, B is the independent of x, A times B is also independent of x, therefore heat transfer coefficient is independent of x, is that clear, so this is a typical variation of laminar flow friction factor curve and the heat transfer coefficient, and this entrance length which we call as hydrodynamic entry length is approximately correlated as. I want all of you to draw one thing, draw the boundary layer growth which is what we ask always, I want we drew for flat plate case, boundary layer growth of thermal and hydrodynamic for Prandtl number less than 1, greater than 1 and equal to 1, I want all of you to draw for pipe, pipe flow. You understood, I will just show you what sir is asking, pipe you draw, three pipes you draw, center line you draw, draw one hydrodynamic boundary layer, so obviously it has to meet somewhere, I am showing only delta, now draw delta T with respect to this for PR equal to or of the order of magnitude 1, PR very much greater than 1, PR very much less than 1. Please draw it, I think you will be able to understand much better if you draw that. PR equal to 1 profiles are same, so roughly the same length, so PR greater than 1 what happens, PR greater than 1 thermal boundary layer is inside, that means what, that means what, that means what, so PR greater than 1, so Mehta is telling me this is what it will be, means is this what you are telling me sir, okay, I like this, this is delta of T of x for PR greater than 1, okay, what does this mean? Yeah, that is what we learnt yesterday, from this point of view of fully developed, what does it mean? When? At this point. At this point. So here it was only hydrodynamically fully developed, here it is thermally fully developed, what does it mean? In this part I am in for trouble because flow is not thermally fully developed, but it is hydrodynamically fully developed, it is difficult for me to estimate heat transfer coefficient, meaning not difficult, what I am saying is it will vary, it cannot be treated as constant. Beyond this point I can read heat transfer coefficient and friction factor both as constant, here h will vary with respect to x, f will be constant, in this part both will change with respect to, so what does it tell me, why did sir ask you to draw this? If you know to draw for a plate then you should know to draw for this, from a application point of view or practical utility point of view, what does this tell you? Portion of the pipe, so as the flow gets fully developed, when it passes through in tubes. Is that why you ask them to draw? Yeah, because I just wanted to tell that, see in the next case if we take, what will happen for Prandtl number less than 1? For Prandtl number less than 1 you just draw, so I will get delta t has reached hydrodynamically developing length is very very long, much larger than the thermally developing region, so what does that mean? Does that simplify here unlike flat plate case? Correct, the point is now only temperature has become invariant, but the velocity has not still become invariant, so it becomes much more complicated unlike flat plate case, in flat plate case for liquid metals was most sought after, why because it is in the thermal boundary layer, what was my velocity? Did I know my velocity or not? Thermal boundary layer, for a flat plate case how will it look like? For a flat plate, yesterday professor had drawn. Liquid metal. Liquid metal, for Prandtl number less than 1, delta is very much smaller than delta t, so there the life was easy because within the thermal boundary layer my velocities were all equal to approximately infinity, but here it is not so, here it is not so, that is what I wanted to bring out that is all, it is not going to simplify here, but in flat plate it is a very win-win situation, because you can get, you do not have to solve momentum equation, you will get you infinity to get the Nusselt number distribution, you only have to solve the energy equation. All I have done here is plotted the friction factor and heat transfer coefficient variation with respect to angle, this is for flow through a pipe you have the Nusselt number variation for x by d, so with varying location axial location you have the Nusselt number variation and I think now we will quickly go through this, this is the good part of the internal flow. General thermal analysis is essentially energy balance, nothing new we are doing, only thing we are saying is we want to now write things in terms of the heat transfer coefficient, so if a fluid is coming in with mass flow rate m dot and T i is the inlet temperature going out at some temperature T e, we want to do a analysis for the fluid element. So, q is heat coming in, this heat if you write e dot in minus e dot out there is no energy generation it is a steady state m C p m dot C p T i plus q dot is equal to m dot C p T e which essentially boils down to m dot C p delta T is equal to q dot, so this can be approximated, see we for us to deal with practical problems it is good to have at least one of the two things constant, if either the heat flux should be constant or the wall temperature is constant and life becomes a little bit easier otherwise life becomes difficult to deal experimentally, so constant wall temperature essentially will be available to you when you have one fluid either condensing or boiling on the surface of another. So, condensation if there is negligible pressure drop or boiling if there is negligible pressure drop occurs at constant temperature, so we can assume constant temperature boundary condition when you have a fluid condensing on the other side of the tube. I was telling 1 by bio number why we are plotting, why 1 by bio number was coming to constant wall temperature we can link why boiling and condensation could give me should give me constant wall temperature very high heat transfer coefficient that heat transfer coefficient is achieved in boiling and condensation it is of the order of lags, so that is why I can reach I can tend towards constant wall temperature because convective resistance is negligibly small. And constant wall heat flux of course when you have electric resistance heating or radiation I always. So, if I do an energy balance here, so q double prime instead of that we are calling it surface heat flux times a is equal to m dot c p delta t which I want to use I want to bring in heat transfer coefficient. So, this is h times t surface minus local fluid temperature, so this is a differential control element control volume m c p t m m dot c p t m is the local mean temperature of the fluid t m plus d t m is the local fluid temperature at the exit of the control volume after the fluid has gained some amount of heat d q that d q is given by the look the heat transfer coefficient times the temperature difference times the which area surface area perimeter times d x d a s perimeter times d x. So, this relationship here gives me surface temperature is equal to mean fluid temperature plus q double prime by h we will have to use this a little later to generate this. And this energy balance here e in two paths for energy to come in one path for energy to go out e dot in is equal to e dot out will give me the same thing m dot c p delta t is equal to q double prime p d x which gives me a differential equation d t m by d x is equal to q dot s p by m dot c p q dot s is nothing but q double prime heat flux. So, if this is integrated for constant heat flux condition q double prime is constant m dot c p of course are constant perimeter unless the geometry is varying it is not going to be changing. So, whole right hand side is constant d t m by d x is constant what does this tell me temperature of the fluid varies linearly with respect to x magically I have already drawn this for you. So, linear variation in the bulk fluid temperature is obtained when I have constant wall heat flux condition am I clear on this now we will use this part again this is nothing new it is just telling you q double prime is equal to h delta t T s minus T m. So, now if I overlap what I had studied a little earlier I like to draw these three together that is why I am drawing it again h fully developed developing region fully developed region right. Bulk fluid temperature is varying with respect to x correct surface temperature locally is nothing but bulk fluid temperature plus q double prime divided by h, h is also varying locally ok. For constant wall heat flux condition this is constant right if this becomes constant denominator h also becomes constant this quantity is a constant number that means what the difference between the fluid and the surface temperature T s minus T m is constant when h is become constant when is h constant h is constant beyond fully developed region. So, for fully developed region I will have a line which is parallel to the bulk fluid temperature line surface heat temperature curve has the same slope as the bulk fluid temperature what happens in the other part in the other part heat transfer coefficient is increasing this way as I move from here to the inlet or if I go from inlet to the location of fully developed region h is decreasing. So, this quantity is decreasing therefore, this difference will have to increase. So, this is the variation of surface temperature it reaches a maximum value and stays there what is the maximum value dependent on it is dependent only on the value of the heat transfer coefficient which is constant for constant wall heat flux fully developed condition. So, if this is kept constant and this is constant your delta T is constant ok. So, only equation to remember is this there is nothing to memorize and of course, our dear m dot C p delta T is equal to that is all. That is coming from E dot in that is also I do not even have to remember that E dot in is equal to E dot out is this clear. So, your students should be able to draw this we should emphasize on how this comes why these lines are parallel why this is diverging here all these things you have to clearly emphasize. There is another experimental implication of this most of the time experimentalist wants to do constant heat flux because of the simple reason why because the bulk fluid temperature is going to vary linearly. And that is one of the basic test what we do when we do experiments is that whether these two lines are I get the what temperature measured and bulk fluid temperature if I assume linear are those two parallel if they are not parallel there is some problem in my experiment. That can be checked even for what is that in the extra days x 1 1 3 4. Sir, these two when we have where are we handling two phase 2? No, here in this analysis where are we handling two phase 2 we have localized ourselves only for single phase 2. And I am saying convictive heat transfer I am saying only for single phase 2 I am not saying until and unless it is specified that I am getting into burning sensation. I am not saying that it is two phase 2. Sir, when is your having the non boiling hydrofluid? I said single phase when I say single phase period no two ways about it there is no other phase. There is a second phase. Yeah. This is universally accepted for single phase. Single phase it is not applicable for air water. I have not told. It is not applicable for air steam it is not applicable for any other condition other than single phase single fluid flow. So far so far we have been handling since yesterday when we started off with convection we have not gotten into two phase flow ok we are still in single phase flow. So, no confusion. Sir. TM is linear provided Cp is constant. That is there. It is not a function of temperature though it is a constant heat flux. See here in all of this we are assuming that the properties are constant. Because otherwise I cannot take the Cp out no. If Cp is a function of x then definitely my dT by dx also will be different. See. Yes. No no it cannot be infinite it cannot be infinite. So, you were telling about next problem it will be typically 2 to 3 times more than the full detail. But I cannot say for sure always 2 to 3 times. I am just saying someone asked while coffee how much how many orders it will be high. So, I just said 2 to 3 times roughly roughly ok. So, therefore what happens because of this because of this equation T s is equal to T m plus a constant. So, that means what in thermally fully developed region derivative of this with respect to x will go to 0. So, q dot q double prime is constant h is constant. So, this is the number. So, dT by dx dT s by dx is equal to dT m by dx essentially that is what we have written. And this relationship of course we have for the definition of thermally fully developed d by dx of T minus T s minus T divided by T s minus T m is equal to 0. So, this implies T s T m this is like this times dT s by dx minus dT by dx set equal to 0. So, it gives you the same thing ok. So, and that derivative is essentially is equal to q double prime perimeter by m dot C p which is a constant ok. So, it is a fully developed flow in a tube subject to constant surface heat flux. The temperature gradient is independent of x and thus the shape of the temperature profile does not change along the tube. Now, is where we can relate it directly to how you did velocity profiles shape because constant wall heat flux. Still I am not saying profiles are the same profile shape is the same. There profiles were same I could cut one profile place it on the other there would be no difference. Here I am saying one would be thinner than the other or thicker, but it would essentially be like this. So, if I had a temperature profile one of them like this other at some location would have the same dT by dx the slopes would be the same it probably is become thinner or thicker according to the heat coming in or coming out ok. It is not the same the shape is the same similar not the same that is closes that we can get to velocity ok. So, for circular pipe of course you can substitute for perimeter and write it in a convenient form that is all ok. Now, the easy part was constant wall heat flux condition. Here the other boundary condition which is possible is constant surface temperature which we say is a little bit more difficult to maintain also, but it is also an engineering application based. So, we are doing it. So, Q again is h A delta T which is T s minus T m some average value. What is the average value? Why are we using an average value? Because if T s is constant this T m is going to change. So, I have to take a difference at the inlet a difference at the exit and then take some kind of an average value. That is all I can give a number for this value of Q. All of us will appreciate this especially if you have done heat exchange all of you are teachers. So, you know in heat exchanger all of us like to give this special case problem you know when you have this I think why we like to do this special case I do not know, but all of us like to do this this case evaporator or condenser ok. So, one will be horizontal other will be changing. This is one delta T and this is one delta T. So, which is the delta T that I should use for writing the Q? What does this figure tell me? Is Q constant everywhere? No. No. If I take this strip the driving temperature difference is given by this T s minus this bulk fluid temperature local. So, Q is directly proportional to T s minus T m of x. So, largest driving temperature head is in this part smallest is at the exit. So, heat transfer rate is going to decrease from this left hand side to right hand side correct. So, to give an average value we have to do this kind of an average delta T. So, what is done is essentially delta T i plus delta T e by 2 which is equal to T i T s minus T i plus T e by 2 or this is some so called T b. This arithmetic mean temperature difference is nothing, but an average of the temperature difference between the surface and the fluid temperature and the inlet and the exit of the test section. What in this definition what is implied is that the mean fluid temperature varies linearly along the tube this we know is not the case that means what instead of this variation we are taking this variation that is an assumption, but again we are engineers. So, amongst the many assumptions we use this is one of the assumptions. We expect the right answer to the third decimal place from our students that is the paradox. So, this simple approximation often gives acceptable results not always so, we have to we need a better way to evaluate and that better way to evaluate all of us know what is it? LMTD. LMTD. So, we can go fast. So, once it is familiar home ground LMTD we can go fast. So, why is LMTD needed? You have answered the question because the delta T is going to change at every x location arithmetic average is not the correct way to do it. So, a logarithmic average has to be incorporated. So, essentially we are doing that here. So, consider heating of a fluid inside the tube of constant cross sectional area whose inside surface is maintained at constant temperature mean temperature of the fluid will increase in the direction of the flow because of heat transfer. So, m dot Cp dTm is equal to h a delta T and all I am saying is writing this in this familiar form dTm is minus d of T s minus T m and separation of variables. So, h p m dot Cp everything is constant dx is the variable along the length. I am going fast purposely because this is similar to LMTD analysis. Anybody who is feeling left out then please let me know if I integrate this I will get T s minus T bulk fluid at the exit we will call it T m at exit or T e minus log T s minus T mean at inlet which is T i. This will be log of these two quantities is equal to minus h p x by m dot Cp p x or p l is nothing but surface area perimeter times length is the surface area. So, if this is the case let me just spend a second here. My temperature distribution is going to be like this and I will get T surface minus T m of x divided by T surface minus T m at inlet is equal to exponential of minus h p x by m dot Cp. When this x becomes equal to l this will become the exit location and I will write this as T e. If it is any local location all I have to replace put this the appropriate x value I will get the appropriate local fluid temperature. The driving temperature difference continuously changes is not a constant. The driving temperature difference is constant in case of fully developed constant wall heat flux condition. This difference has to be clearly brought up. Because the driving temperature difference is not constant heat transfer rate is not constant and a typical temperature profile is this exponential decay. This temperature difference decays exponentially from inlet to the exit and the magnitude of that exponent is minus h a by m dot Cp and we are very familiar with this from our heat exchanger business. This dimensionless quantity is called as NTU. I think heat exchanger everybody is happy. It is a measure to day after tomorrow even if you have one hour we will finish heat exchange. But for Friday to come Thursday has to go. So that is radiation. So NTU of about 5 indicates that limit is reached for heat transfer. Here sir has made a nice table of NTU which is nothing but h a by m dot Cp and has evaluated the exit temperature for sir given inlet condition. Inlet is 20 m dot Cp everything has been chosen and NTU essentially has been taken and he has calculated the exit temperature. NTU of about 5 indicates that the limit is reached for heat transfer. Look at the difference 99.5 and 100. What is the NTU tell you physically? Everybody has used I have also used NTU number of transfer. What is transfer? It is transfer units, dimensions, size, size of the heat exchanger. Larger the NTU larger is the size available heat transfer area and which of these terms here A s. Larger the length if you fix the diameter is the larger length if you fix the length is the larger diameter. So no matter what we do beyond 5 there is hardly any utility in the exit temperature. A large NTU and thus a large heat transfer surface area may be desirable from a heat transfer point of view but may be uneconomical. This is common sense. And this of course is your LMTD I am not going to spend time on this log mean temperature difference. LMTD is obtained by tracing the actual temperature profile. It is an exact representation than an average temperature difference between the fluid. It reflects this exponential decay function when delta T e differs from delta T i by no more than 40 percent. This is for analysis the error in using arithmetic temperature difference is less than 1 percent. So you can use that. We should always use logarithmic mean temperature difference when determining convective heat transfer coefficient for constant wall temperature case. So now we can I am going to go at a breakneck speed. I am not going to go slow now because now we know what we are studying. So now next thing is just to take recap of fluid mechanics again steady flow in pipes. So I am just taking Heggen Poiseuille flow and here I am taking the continuity equation and the Navier-Stokes equation written in cylindrical coordinates. So I will write them for viscous incompressible steady flow both the continuity and the Navier-Stokes equation. Now I will consider fully developed flow. Now I will consider fully developed flow. What is happening for fully developed flow? What was happening in Cartesian coordinates for fully developed flow? U is only a function of that is how we define. The velocity is only a function of y. It was not function of y and z, x and z. Similarly here velocity is only a function of radius but so v theta and vr are going to be 0. So if I make that not assumption, if I take that condition which is applicable for fully developed flow I am going to get that del P by what is that I get? So if I reduce these equations accordingly, I am not going to spend time. So I get the velocity profile something like this. I would like you to spend time with this nodes because this is fluid mechanics. I do not want to spend time on this. So consciously because if I go on going step by step I will be losing at least half an hour. I am not going to do that. So I will get the paraboloid of revolution. That is what you were in the morning professor was talking about paraboloid. So that is this velocity profile. So now I can get the maximum velocity, sorry velocity profile in terms of maximum velocity and I know the average velocity. If I have to get the average velocity, if I put in this velocity profile and integrate that, if I integrate that I should be getting the average velocity. What do you see here? How many times my average velocity lower than the maximum velocity? Yes? Or maximum velocity how many times it is more than the average velocity? Two times. For turbulent flow, for turbulent flow, this is laminar flow. I am going to go this way, that way because now we know where we are. So what will be the turbulent flow? How will it vary? Yes? It was sharp. So it is actually power law. We all have studied for turbulent flow. 17th power law. But u by, if I am wrong, u please correct me, u by u maximum, is it there? u by u maximum is equal to 1 minus r by r to the power of 1 by n. So always 7 is not going to come. 7 is going to be a function of, that n is going to be a function of, function of what? Remember we are trying to represent mathematically my turbulent velocity profile. So what is this n going to be a function of? Reynolds number and epsilon by d. If I take hydraulically smooth pipe, I am using the word hydraulically smooth pipe. What does that mean? Hydraulically smooth pipe means as long as my geometric roughness of the pipe is lower than the viscous sublayer or the laminar sublayer. That means it is smooth. I am not breaking the laminar sublayer. If it is smaller than that, see geometric roughness does not mean anything for fluid mechanics. As long as my geometric surface roughness height is lower than the laminar sublayer, it is called hydraulically smooth pipe. Geometrically it may not be smooth. If you touch by hand, it might be rough. But if my thickness or the surface roughness is lower than the viscous sublayer thickness, it is hydraulically smooth surface. It is obvious. But you see the words how they are so nicely put. So that is, this is the velocity profile for turbulent flow. I am just going this way, that way because I wanted to remind you about the turbulent flow. So now the shear stress, you can, now you see I was telling in the morning, see because there was yesterday also taught I know pressure drop, I know shear stress. So you see from the pressure drop, I can get the shear stress. So shear stress is going to be 4 mu vz average by R. So now from this shear stress, I can show that Cf equal to 16 by Re and F equal to 4 Cf that is 64 by Re. Most of the textbooks do not use the notation Cf. They use F replacing. So that is why we will have fights. No, no, my F is 16. No, no, my F is 64. It is not my F or your F. It is the question of whether it is skin friction factor or Darcy's friction factor. Is that okay? Yeah. What is the difference between the no-sleep boundary condition and the boundary condition which has shear stress equal to 0? Shear stress equal to 0 is not a boundary condition. Why do you say shear stress equal to 0 is a boundary condition? Actually, professor has covered in the morning shear stress equal to 0. First thing, I have to rephrase my question. No-sleep is a boundary condition. Shear stress is not a boundary condition. Yes, I use that boundary condition at one point. What is that point? At the point of morning, morning professor touched that at the point and he told also, at the point of separation, at the point of separation only my velocity gradient is 0. So that is not a boundary condition, I would say. But if I have to answer your question, when will my shear stress be 0 at the point of separation? Sir, when we take an interface, then... It is interface. Again I am... No. Please, you are not going to ask me questions on micro channel and you are not going to ask me questions on two-phase flow. Because we will keep them for cup of coffee but not here. Because we are in one track, we will be sidetracking. Let us not do that, please. Okay. So now, similarly on the same note, so on the same note, if I take, now let me go little slowly. Of course, I think I have not put complete derivation but I have. So this is my energy equation in cylindrical coordinates. Now you see Q dot and phi V. Q dot and phi V, what are these terms? Phi V is what? Viscous dissipation term and of course pressure term is not there and this Q dot I have just put because this is some energy is being released within my control volume because of some combustion. Some energy has been added because of some combustion within my control volume. That I did not take up in my energy equation derivation in the morning. Some energy has been generated because of some chemical reaction has occurred, some combustion has occurred, some endothermic or exothermic reaction has occurred. So that is this Q dot, that is also not going to be there. So that is not there. So what will happen to my equation here? What will happen to my equation? Is del T del T there? Del T del T, steady I am taking, steady that will not be there and now V theta is not there and I am only having and V r is not there. I am only having V z because I have taken hydro dynamically fully developed. So my equation got reduced to rho Cp V del T by del x, x being the direction of the flow and now what about this? What about these terms? These three terms. Temperature is not a function of theta because we have been taking always with rings. So it is only a function of r and del square T by del x square is, we have harped so much since morning. Del T by del x is T minus, then T s upon T s minus T m is invariant with x means, means my T is del T by del x is not a function of x. So that means del square T by del x square also should not be a function. So that means it is not there. So that is for a fully developed flow this is not there. So this is my equation. So if I solve this, how do I solve this equation? What is the boundary condition? I have imposed myself constant heat flux and this is what couple of minutes back professor has derived. So dT s by dx equal to dT m by dx. I think I should write this as ordinary derivative. That is a mistake. dT s by dx equal to dT m by dx equal to 2 q s dot by rho C p V m r equal to constant and this is my velocity profile. So if I substitute this velocity profile here and substitute for dT by dx and solve this equation. Now it has become a function of only r. So if we integrate this and find these constants, if I integrate this and find these constants I am going to get the temperature distribution. One interruption please spend half an hour to do all the mathematical steps in this derivation. Because this is probably the only derivation or only solution of this energy equation which is possible analytical solution. I am sure you are all asking this in your answer. For 0.36. Because this is one thing which we do not have to remember anything. You can come from first principle. So we need to insist. Of course our professor Vedula has very complicated questions. In one of his found question is you can think of the last time question was top portion is constant heat flux, bottom portion is insulated. I have a pipe. Half pipe is constant heat flux and half pipe is insulated. Now how to compute my bulk fluid temperature? I have to break my head. So what I am trying to say is you can cook problems. I just took this example because you can cook problems such that they are derivable. For that we have to be innovative. This type of problem I have not seen in any of the exercise problems in any of the textbooks. It is quite innovative in generating questions. It is an art. And all students you ask any of our students they are scared of professor RPVs question papers. That is why they attend all these lectures. So that was the digression. What I am trying to say is if you keep your questions tight they will be serious. So again same thing I can come back for the bulk fluid temperature. Now you see bulk fluid temperature required temperature distribution and velocity distribution. I know that. So if I plug in that I can get the bulk fluid temperature. Now if I compute Nusselt number that is h equal to Q s dot upon T s minus T m. If I have got Q s dot in terms of T s minus T m if I plug in that I get 4 point. This is for constant heat flux. Similarly one can do constant wall temperature but it is not easy. So please do not give that as assignment. Please do not give this as assignment for constant wall temperature. If it is a PG course I would recommend but not for UG because it is iterative. One can do this. It is not that it is impossible. One can do this but you can only for UG you can only state. In convective heat constant mass transfer course we do not derive we give that as assignment again but PG students are supposed to do it. If it is advanced heat transfer course the way you are teaching you can give that as assignment provided you have solved it beforehand. So that I get Nusselt number equal to 3.66. I am going to ask you one question. For constant wall temperature and constant heat flux Nusselt numbers are different. For constant heat flux you got 4.36 and for constant wall temperature you are getting 3.61. Then okay so now it is you can go on deriving this. We derived it for circular. We can derive it for square. Square you can ask. You have derived it for circular no? You can ask for square. They should be able to derive. All that will changes perimeter but the concepts are going to be same. In fact it is easier for them because they are handling Cartesian coordinates. That you can ask even for constant heat flux boundary condition you can ask for square. So that is what we should ask. Whatever we have derived in the class should not be asked for the exam. You make it open book. Let them see these steps and do it for square fair enough but if they can follow these steps and do it for a new configuration. Isn't it? So that is what we need to try when we set our question papers. So now developing laminar flow if you see for developing laminar flow there are various correlations. These are all correlations. These can be derived. I will not get into each of these correlations. There is a problem. I am skipping this. One important concept between laminar and turbulent we need to understand when it comes to heat transfer. In laminar flow we said the heat transfer coefficient or the Nusselt numbers are dependent on boundary condition. Now when I put Dittus-Bolter now we will take up your question whatever it is on Dittus-Bolter. Dittus and Bolter collected way back in 1954. He collected around 200 papers data and collated all that information and came up with this correlation. So now is this correlation, is this correlation valid for constant heat flux or constant wall temperature? Both. Both. Why? Or which other cases it is valid for? That is true. That is true. As long as it is single phase, as long as it is single phase and Newtonian fluid we are handling. So this is going to be valid. Why? Why in turbulent flow my Nusselt numbers are independent of boundary condition? This is very important to tell them because this until we tell them our students are not going to notice. In our data handbooks also we do not specify. Because we just give turbulent flow the moment it is turbulent flow I will use Dittus-Bolter. But I can confuse him by asking him half of my pipe is constant heat flux, half of my pipe is constant wall temperature. He has to use the same correlation. I can confuse him. Those are the tricks we need to use to confuse them in the exam so that we make sure that their principles are well understood. That is when Manan in the exam gets started. Now coming back. Why is it independent? It is independent of boundary condition. Why is it independent of boundary condition? Because of the nature of the flow, how is my turbulent flow? It is random. You guys said yesterday it is random. It does not know what is the boundary condition. It is a boon for an experimentalist. Why is it a boon for an experimentalist? Why is it a boon? You do your experiment based on your convenience because convenient thing is wrap, heater wire and do with the constant heat flux. So it is valid for any boundary condition. It is valid for any boundary condition. Half of my pipe is constant heat flux, half of my pipe is constant wall temperature. You give this as an exam problem. Half is constant heat flux, half is constant wall temperature. Now you calculate the heat transfer coefficient. So then there are various correlations. Pettucco proposed correlation, Nielinski's correlation, Dittus and Boltzner correlation usually works only well for gases. For liquids, Dittus Boltzner correlation is erroneous as high as 20%. So that is why Nielinski around 1970s, 1973, I have these papers. What I will do is in Moodle I will upload Dittus and Boltzner paper and also Nielinski's paper. So that you can see how much effort they have taken in collecting that data. So Nielinski's data is usually recommended for validation. Why these correlations are important? If you are building a setup or you are written in numerical code, you want to check whether they are first, before you start doing, Sandeep Sonwane has done with nanoparticles. Before doing nanoparticles, he has to first check his setup whether it works well for zero nanoparticles and Dittus Boltzner correlation has to be valid. So Nielinski's works well for liquids rather than gases. So gases usually Dittus Boltzner, but people usually with whichever correlation it matches, they say it matches and they will simply go ahead. So that is a human tendency. But all experimentalists appreciate the problem, so we do not need to pick too much about the validation as long as it is getting validated with one or the other correlation. So this is for liquid metals. Here you can see that for liquid metals, there is a slight dependence on, but not much slight dependence on boundary conditions, very slight I would say, very slight. Anyway liquid metals are always special, they are different from our general fluid. So now coming back to rough surfaces, I think I have that much time to cater to your question yesterday. I will just, so for rough surfaces, this is the Kohlbrug's correlation where in which I have friction factor as a function of, what is this? Epsilon by D, surface roughness non-dimensionalized with the pipe diameter and this is again Re by square root of F. This is an implicit equation, I have to do it iteratively. Now present day calculators I guess can do. This is for flow through annulus, there are correlations again for this and these are all various correlations, I am not going to touch upon that. There is a problem also, I am not going to solve this problem, but I am going to answer only one question that is, now let us say, now let us say you are asking how do I know whether I have put a surface roughness exceeding my laminar sublayer or not? That question I told I will cook up a problem, I have cooked it. So if, let us, I want all of you to solve along with me. See let me choose my Reynolds number as 10 to the power of 5, come on, let us take Re equal to 10 to the power of 5, I am not going to show you anything. For epsilon by D, for epsilon by D equal to 0.05, I get a friction factor of 0.071, how did I get this? How did I, either from Moody's chart or Colebrook's correlation which I showed a little while ago. If I plug in Reynolds number and epsilon by D I am supposed to get F. You will get near about 0.071. Now what is the question being asked? What is the question? Whether this epsilon by D or epsilon is breaking my viscous sublayer or not? That is the question asked. What should be for the laminar sublayer? What should be the y plus? y plus should be around 5. What is y plus? What is y plus? I know I cannot expect you to remember this. y u tau by y plus equal to, y plus equal to, come right along with me, y plus equal to y u tau by nu. What is u tau? Square root of tau wall by rho. What is tau wall? How will I get my tau wall now? Quite confusing, isn't it? How do I get tau wall? We have defined friction factor. What is kinfraction factor? Cf equal to tau wall upon half rho u average square, right? F equal to, what is F equal to? 4 Cf, 4 Cf equal to 4 Cf. 4 tau wall upon half rho u average square. What am I trying to do? What am I trying to do? I am trying to get tau wall. Then I can get my y plus is given. Y plus is given to be 5 because I know for laminar sublayer it has to exceed 5 and u tau consists of tau wall by rho. How do I get my tau wall now? F is 0, no? F is 0.071 and another thing I have not given. Let us take the diameter of the pipe as 25 mm. That is 1 inch pipe I have taken. So, if I take 25 mm as my diameter, Reynolds number is 10 to the power of 5 equal to rho v average d by mu. What is the density of the water? Let me take water as my fluid. 1000, very good. 1000 into v average diameter is 25 into mu. What is the density of the water? 10 to the power of minus 3 to convert it to meters. What is viscosity of water? At room temperature it is 0.07975 Pascal. So, if I plug in this, I get v average equal to 3.19 meters per second. Quite a huge velocity if it is water. In your labs or in our tap waters, the velocities are not so high. They are quite less. Now, plug in this average velocity here and get Cf. F is 0.071, 4 tau wall upon half rho into 1000, u average squared is 3.19 squared. I get a tau wall of 90.3 Pascal. That is Newton per meter squared. Do not worry. I do not remember this. I have worked it out. I am just trying to, you got it so far? Any questions on this? No, right? Now, what next? What next? y plus. y plus is known. Why I have to find? 5 equal to, I have to first find u tau. u tau equal to square root of, square root of tau wall is 90.3 divided by rho is 1000. So, I get 0.3 meters per second. 0.3 meters per second. That is why it is called u tau because it is having a unit of velocity. So, now if I substitute, what is 5 equal to? y plus equal to y u tau by nu. y is what I need to find. u tau is 0.3 and by nu, nu is the kinematic viscosity. Dynamic viscosity I have told 0.0007975 upon 1000. I get y of 1.329 into 10 to the power of minus 5 meters which implies that it is 0.013 mm. So, what do I comment on my answer? What do I comment on my answer? What is my comment? I have broken my sublayer or not? What is my actual surface roughness? No, epsilon. You will have to calculate epsilon. What I have given is epsilon by d. Epsilon by d equal to 0.05, epsilon equal to 0.05 into 25 that is 1.25 mm. Actual surface roughness is 1.25, but corresponding viscous sublayer for this kind of velocity and with this kind of surface roughness you have only 0.013. 0.013 mm is so small is enough to break the viscous sublayer. So, but then the trick here is you will come to know only how did you come to know this because you knew that means apriery beforehand you should know you can make only after measuring only you will come to know whether you have if you put ribs of your own beard shapes and sizes and put them at different pitches you will come to know whether you have broken your laminar sublayer or not only after doing the measurements. So, this is the physical interpretation of this laminar sublayer is that okay? So, twisted tape what will happen there is a tape which is twisted if I take this paper and I twist it I hold it on one side and twist it on the other side I usually we generate it by putting fixing it on lathe on one side and twisting it on the other side. So, you get a twisted tape and you would have seen few leaves twisted all around actually that twisted tape if I insert in a pipe what it is going to create my flow is laminar my flow is laminar I am not going to turbulent I will go to turbulent little later my flow is laminar what is it going to do it is going to twist so that is going to create swirl that means flow is now not to go through laminae now I have it is generating V it is in addition to U it is generating that is how Nusselt number increases that is how the heat transfer coefficient increases on the same note perhaps that is why I say that Nusselt number is independent of may be we can this is convincing Nusselt number is constant and independent of Reynolds number may be because it one layer does not talk with the other layer may be that is a convincing answer perhaps because in twisted tube only I am able to create the swirl and that swirl is dependent on Reynolds number then only Reynolds number comes into picture in twisted tape there are n number of enhancement devices every other day every other journal you open you are you are going to get one more paper Chinese are minting papers on this okay so you get hundreds of papers if not thousands on enhancement devices but still I think there is enough to do because there is so much optimization there are so many new shapes new augmentation devices coming out so sky is the limit but why should before you tell anything why should in a single phase flow the diameter affect as long as my non dimensional numbers are kept constant but I would expect that the Nusselt number variations will be within the experimental uncertainty I do not expect any variations in your answer before you tell me the answer what you got this is my first perception based on the anyone would comment based on the theory what we have studied yes with the with the increase of there are plethora of issues here when you are doing experiment surface roughness can vary from one pipe to another pipe number one number two experimental uncertainty is anyway going to be there no sir this is only theoretical theory turbulent how did you do theory no this was assumed problem that all parameters were kept same theoretical problem yeah what do I do in theoretical problem correct theory means you are you going to use the detest voltage yeah then what is theory there there is no theory H turns out to be a function of diameter in terms of inversely proportional to d to the power of minus 0.2 if you take dimensionary but point is I never want to think things in dimensional format at any point of time point is see I do not want to think and I should not think also I will say it why whether in a Reynolds number I change the whole idea of non dimensionalization why did it come why did Nicaragic came up with the non dimensionalization how on earth he knew that he had to plot the Moody's chart F verses are you only how did he do that he didn't change the diameter of the pipe he changed only the velocity he didn't change the fluid that is the strength of the non dimensionalization why should I again come back to dimensional things I should not think in terms of dimensional parameters at all if I have to talk about diameter H if I reduce it H from the dimensional non dimensional relationship yes H is going to come out as d to the power of 0.2 e to the power of minus 0.2 but I don't see any need for doing experiments to get myself convinced no no no you are telling exactly opposite of what it is see what we are trying to say please don't get confused with this question this is this is for everyone as long as I keep my Reynolds number same okay for friction factor let us say friction factor is dependent on what Reynolds number for a minute let me say epsilon by d is constant or I am handling hydraulically smooth pipe so Reynolds number is constant mean I can either be changing rho or I can be changing mu that means I can be changing the fluid or I can be changing the diameter or I can be changing the velocity it should not matter which one I am changing as long as my Reynolds number is same that is the strength of dimensional analysis no two ways about it you are telling exactly opposite professor so it is dimensional analysis cannot go wrong okay so there should not be any confusion.