 In the case of the rotational flows, we have defined something called as potential function. Sir, can you give you a brief physical intuition of what this potential function is, just like what we say for stream function? The question asked by one of the participants is that, what is the physical significance of potential flow and can you give the brief introduction about the potential function? No, the first point is what I want to tell you is that, as I said, when I have a flow over a flat plate, professor Arun has already taught us that, we can differentiate the flow into two domains, one is the viscid domain, another one is the inviscid domain. In the viscid domain, that is within the boundary layer, there is effect of viscosity and in the inviscid domain, there is no viscosity. So, if I do not take velocity, if I do not take viscosity, if I write the momentum equation, what will I have, what will I have u del u by del x plus v del u by del y equal to minus del p by del x. So, but now, to solve this problem, what people used to do is to write potential equation, that is I write u equal to del phi by del x v equal to del phi by del y, why because the beauty of that is because it will reduce to del square del square phi by del x square plus del square phi by del y square equal to 0, if you put that in continuity equation. So, del square phi by del x square plus del square phi by del y square for various boundary conditions, mathematically has been solved and kept Laplace, that is Laplace equation. So, the solution of the Laplace equation is available or was available, those solutions were adapted for various configurations, that is what is called as irrotational. You take appropriate boundary conditions, actually people do that it is a very interesting topic actually, but it is a topic where in which one can cover all of that in 5 to 10 classes, you need complex variables concept and you have to take the Laplace's equation and solve that. So, if you take complex variables and take the Laplace's equation and appropriately put the boundary there, may be it is flat plate, may be it is aerofoil, one can get the u infinity variation, you see for a flat plate we took u infinity equal to constant. If I do the same thing for flow around a cylinder, I will get u by u infinity as a function of theta or if I take an aerofoil, I will get u infinity as a function of location that comes essentially from potential flow solution, that is which is the solution of the Laplace's equation. So, why what is so great about Laplace's equation, because Laplace's equation solutions are available. So, Laplace's equation has been attacked by so many mathematicians, that is why potential flow solutions are so relevant for fluid mechanics, but then how is this inviscid region solution is useful for my viscid region, I have already told. If I know the u infinity, if I if I get u infinity u infinity d u infinity by dx, use me dp by dx, this dp by dx is sitting in my boundary layer equation. So, I get the pressure impressed upon the boundary layer by dp by dx. So, this is how the boundary layer and the inviscid region get connected with each other, that is the significance of irrotational flow or the inviscid solution, okay. While we define the potential function, the x component of velocity u equals we write minus dou phi by dou x and v equals minus dou phi by dou y. So, what is the significance of the negative? No, I do not think we write for phi minus, that is for stream function, for that is that is for continuity, see for stream for potential function, we do not write minus professor, we write u equal to del phi by del x and v equal to del phi by del y. So, we for stream function, we write like that. So, I think I have answered beyond beyond what we should be, but if you think that I have not answered your question comprehensively, please put this up in the moodle, we will answer this question much more comprehensively. This is much to do with the fluid mechanics rather than the heat transfer, but we do not want to wish it away, you if you are not happy with this answer, you please put it in moodle. Even for stream function, see professor Arun has written here. So, it is actually basically whatever function I choose, it has to satisfy my continuity equation, that is del u by del x plus del v by del y. So, to satisfy that we are defining it this way, either the stream function or the potential function. I think now we have time left for solving one more problem, we will take up the problem. Let us take up problem 30, it is a simple problem, a thin flat plate that is 0.2 meter by 0.2 meter, 0.2 meter by 0.2 meter, that is length is 0.2 meter and width is 0.2 meter, is oriented parallel to the atmospheric air stream having a velocity of 40 meters per second, u infinity is 40 meters per second. The air is at a temperature 20 degree Celsius, T infinity is 20, while the plate is maintained at 120, T s is 120 degree Celsius. The air flows over the top and bottom surfaces. This is where we always give it to students. See, we always write boundary layer grows only on the top, it is not so. If the flow is there on the bottom, boundary layer grows on the bottom also. So, that means the drag force is there on the top surface also and the drag force is there on the bottom surface. And both are identical, whatever I do on the top, I have to just multiply by 2 to get the drag force. So, and measurement of the drag force reveals that a value of 0.075. So, net drag force F D equal to 0.075 force means it has to have a unit Newton. What is the rate of heat transfer from both the sides of the plate? Properties of air are rho equal to 1.125 kg per meter cube, mu equal to 18.4 into 10 to the power of minus 6 Pascal second, Prandtl number equal to 0.71, k equal to 0.02635 watt per meter Kelvin. So, the first thing is we need to calculate to get q, q equal to what? H A s into T s minus T infinity. It is a constant wall temperature case, I need to find out H. So, how do I get H? So, how do I get H? H is to be obtained from, yeah the point here is the very fact that drag force is given, we need to capitalize this. So, I do not have to calculate again for floor or a flat plate, again R e to the power of half or P r to the power of 1 by 3, that is not required here. Why because I can invoke Reynolds analogy? Why because drag force has been measured? So, measured is always the most sacred thing. So, now what does Reynolds analogy say? Stanton equal to C f x by 2, Stanton number equal to C f x by 2. So, Stanton equal to N u by R e P r, C f x by 2, Stanton is defined as N u by R e P r. So, what is C f x? We have to find C f x from drag force. What is C f x definition? C f x equal to drag force upon F drag force upon F d upon half rho u infinity square into area. So, F d is given to be 0.075 into 2 divided by 0.075 because it is not very clear. That is right. So, it is not divided by 2, we have to take drag force net C f x, both sides we have to take. So, no divided by 2 because this is the net drag force both on the top and the bottom one. So, upon half into rho is 1.225, 1.125 and u infinity square is 40 square into area is 0.2 into 0.2. I get C f x as that is all I will keep it like that. I have not calculated yet I am substituting for C f x by 2 later on. So, this is C f x by C f x by 2 equal to C f x equal to. Now, I need to calculate R e, let it be like that R e equal to R e equal to let us find that R e equal to rho v infinity l by mu rho is 1.125 v infinity is 0.075. So, this is 40 l is 0.2 mu is 18.4 into 10 to the power of minus 6 Reynolds number comes to 489130. Now, PR is 0.71 is given. So, if I substitute all these 3 terms in the top, I will be getting Nusselt number line. If I substitute all of this in the top equation not in that C f x equal to mu by T R. So, if I substitute all of that in that. So, I get I get a nu equal to 180.87 which is equal to h l by k f h l by k f h is to be found out l is 0.2 k f is 0.02635. So, I get an h r which is 0.02635. So, I get h of 23.83 watts per meter square kelvin. So, q equal to h a into T s minus T infinity h is 23.83 area is 2 into y 2 into both the sides of the plate 0.2 into 0.2 120 minus 20. So, I get 190.64 watts. So, this is how one can compute the heat transfer rate in case of conversion it is divided by 2 drag force is divided by 2. So, here I have my students here are saying that drag force has been divided by 2. Yeah, when I am measuring in the problem if I really understand the air flow flows over the top and bottom surfaces of the plate and the measurement of the drag force reveals the value of 0.077. So, I have to divide this for both sides because C f x is written only for one side of the plate. So, C f x both sides has no meaning actually it is a non dimensional thing. So, with this I think we will move back to questions again. So, I think we will not solve the problems anymore. So, I would request you to solve all the other problems. In fact, this type of problems we have to still cover in our subject material after that we will I think you will be much more confident in solving these problems. S V N i t is Surat. Is irrotationality of fluid element is an ideal condition for laminar flow? The question is irrotationality an ideal condition for laminar flow. See the point is irrotationality has nothing to do with laminar or turbulent flow. Generally inviscid flows are irrotational. Generally I am using this word generally that means irrotational flow generally should not generate workplaces. I think it is very obvious irrotational the fluid particle cannot rotate. If it cannot rotate it cannot generate vorticity. However there is an exception there are exceptions. If I take flow over an aerofoil on the other day also to my student I have answered this question. If I take flow over an aerofoil the pressure on the top surface and the pressure on the bottom surface are different. So, if I have an aerofoil something like this because of the discontinuity there is a vorticity form or rotationality generated. Flow the rotationality can be generated even in inviscid flow only in case of flow over an aerofoil. This drag it is called separate it is what is called as induced drag. It is neither called as pressure drag or form drag. This is called induced because it is induced because of the pressure difference in the top and the bottom surface. So, point is generally all irrotational flows are inviscid with few exceptions. That is the answer him. J N T U Hyderabad. Some first question is thermal conductivity of a ice is 4 times that of water. How is it? The second question is if the odd fluid is made to flow parallel to the flat plate how the velocity boundary layer and thermodynamic boundary layer varies. And the third question is if the fluid is flowing perpendicular to the flat plate how the formation of hydrodynamic boundary layer and thermal boundary layer are formed. One of the professors has asked this three questions. I could not recollect the second question, but the first question is thermal conductivity of ice thermal conductivity of ice is 4 times the thermal conductivity of water. Thermal water is liquid and ice is solid. The only way to look at it is the thermal conductivity of solid is greater than the thermal conductivity of liquid. So, that is how I think one can understand this number one. Number two, what is the hydrodynamic boundary layer formation if my plate is 4 times normal to the flow. Normal to the flow not parallel it is perpendicular to the flow that was my question. Anyway, if I have phrased the question wrongly you will correct me because second question for second question I will have to come back to you. So, how does that happen? The boundary layer it first what happens here it first comes and hits. It is actually flow over an impinging plate. If my main flow is laminar actually what happens I will tell you because this is the stagnation region. So, that is very close where it has hit in the stagnation region I do not know whether you will be able to catch me or not it is laminar boundary layer with a pressure gradient actually. It is a laminar boundary layer with a pressure gradient it is a laminar boundary layer with a pressure gradient subsequent to that there will be transitional boundary layer and subsequent to that it will be turbulent boundary layer. If you do not it is not required to make a differentiation between various boundary layer all that we can say is that boundary layer will again start forming like this that is all it is boundary layer will start again forming like this and that is how it is going to be. And definitely nusselt numbers in this case are much larger than the nusselt numbers compared to that of the parallel flow because it is going and getting stagnated that is why you know locally if I have to cool something I use jet impinging that is what essentially this is this is what is called as stagnant flow. Professor will you repeat your second question and please correct me if your third question I have interpreted it wrongly. My third question is if the hot air is flowing over the cold plate how does the hydrodynamic boundary layer and thermal boundary layers are formed are they similar to that of cold air flowing over the hot plate. The question is second question asked by one of the participants is that whether the question asked is does it matter since morning we have been saying that T s is greater than T infinity T s is greater than T infinity does it matter if T infinity is lesser than sorry T infinity is greater than T s that is hot air is blowing on the cold plate. See one thing I have I have realized is whether it is heat transfer or cold transfer fundamentals are same it is only the difference is direction of the heat transfer is changing boundary layer is boundary layer boundary layer is because of the temperature gradient. So, a professor Arun is drawing for us so that is in the top case the T infinity is lower than T s in the bottom case T s is lower than T infinity but the boundary layer is there you see the boundary layer nature is just the same and the heat transfer coefficient is also going to be the same. So, it does not matter whether I heat the plate or cool the plate heat transfer coefficient just depends only on the boundary layer as long as the boundary layer is formed whether I am heating or cooling heat transfer coefficient is independent of the direction of the heat transfer all that gets changed is the direction of the heat transfer is that ok professor. Next question sir in energy equation we are using viscous dissipation so in momentum equation also there is viscosity effect now why we are not naming as or why we are not specially focusing on viscous dissipation first question and next question is why we are especially going for non dimensional x star u star we star like that we are. So, why cannot you solve the same continuity equation momentum equation and energy equation without adapting dimensionless over to you sir. The second question is that why is that I am always non dimensionalizing first question I remember now why is we are non dimensionalization why not take continuity equation momentum equation and energy equation as they are in terms of u v and t and x and y rather than taking u star v star t star and p star y see the point is as follows if let us take only the momentum equation if I take momentum equation what is that I got momentum equation what is that I am getting I am getting friction factor is equal to for example, for flow over a flat plate what I am getting friction factor equal to some constant some constant into r e to the power of half p r e to the power of half this is what we get you remember the constant ok. So, I get constant into r e to the power of half now you see r e to the power of half what is r e rho u infinity l upon mu what am I trying to say if I have to generate a function for f for an unknown situation some constant into r e to the power of m let us say I have to find if I find this m for one fluid that is for air it is enough because for momentum equation it is independent of Prandtl number all that it is dependent is it is dependent on Reynolds number as long as you keep the Reynolds number same no matter whether you use for air water oil or any other Newtonian fluid this f equal to constant into r e to the power of m is valid. Now, for the other portion that is for if I take the temperature equation that is the energy equation what is there temperature here Nusselt number is a function of r e to the power of m p r to the power of m. So, let us say r e we have understood p r p r let us say my oil is there whose viscosity is 6 ok. Now, I can take I can generate same Prandtl number by taking water at room temperature or the oils viscosity is lower than this 6 let us say 4 then I will take water only instead of oil, but I will keep the Prandtl number same what I am trying to say is you take fluid a or fluid b as long as you keep the Prandtl numbers the same then my Nusselt number distributions are going to be same. So, that is the beauty of non dimensionalization if I had non dimensionalized let me get you through the history. Now, all of us know Moody's chart you all know Moody's chart what is Moody's chart friction factor versus r e is plotted in the Moody's chart both for laminar and turbulent for turbulent you see as a function of epsilon by d also that is surface reflash. Why f is plotted only against r e is that Moody's chart applicable for water only or air only no it is applicable for any Newtonian fluid, but how do you know that f is dependent on only r e because of this non dimensional analysis had I not done this non dimensional analysis f is a function of r e only we would now never arrived at ok. Similarly, in fact professor Nusselt has got has become professor because of doing this work this is what is called as dimensional similarity for the work what he has done what we studied or what professor Arun taught us through non dimensionalization that is what is called as dimensional similarity that is that is done by professor Nusselt whatever he did that time is all in textbooks and that is what we are studying the point is we are non dimensionalizing when we non dimensionalize it does not matter which fluid we take as long as the Reynolds number and Prandtl number are same Nusselt number is going to be same provided my fluid is Newtonian. So, that is the strength of that is the strength of non dimensional analysis coming to your first question why do we not worry about viscous dissipation in momentum equation and why do we worry about viscous dissipation only in the energy equation. Now where have we neglected viscous viscosity when I say viscous dissipation it only means viscosity where have I neglected viscosity in momentum equation. If I neglect viscosity in my momentum equation then it reduces to inviscid flow that is Euler's equation that is Bernoulli's equation I have not neglected if you remember mu upon rho into del squared u by del x squared and del squared v by del y squared del squared u by del y squared we have taken that is vicinity if you want to call it you can call it as what is that viscous dissipation you can call that as viscous dissipation in case of momentum equation there is nothing wrong there is nothing wrong, but in the energy equation there is no other place where viscosity is there that is the only that is phi is the only place where viscosity is there that is why it is called as viscous dissipation. Viscous dissipation we said that that is the energy lost because of wasteful frictional losses here also if you look at the momentum equation this is the term which has come because of viscosity. So, this is also something like frictional loss only this is not something which can be recovered. So, this is a when you do the balance of forces this is the frictional effects or this is come because of the shear stress. So, the role is the same the role of this is analogous to the role of this except that if we throw this out then we are totally throwing the baby out with the bathtub that is it. One more question. So, in that energy equation the pressure is we are writing substantial d rho by d t. So, we are writing energy equation in x direction now. So, it is must be I am thinking that small d. So, why we are putting as a substantial derivative the question asked by the participant is that d p by d t in the energy equation is the total derivative why not a ordinary derivative we have not written it while deriving it has come. Remember energy equation is not having x y direction x direction y direction z direction energy equation we have only one equation for momentum only for momentum transport only we have x direction y direction and z direction and for a for that matter p does not have direction p is scalar, but y is d p by d t is the it is the pressure being convicted or the work done by the pressure force by virtue of velocity u v and w and at the same time if it is unsteady pressure is going to vary with time. So, we have not written or cooked up or generated d p by d t it has come out from fundamentals when we derived this. So, pressure indeed is varying with pace and also with time because of convection convective velocity u. So, it is total derivative and it is not about just only x direction or y direction or z direction it is the work done in all directions because of the velocity and pressure force is that ok. We had stopped with flow over parallel plate flat plate flow over flat plate and we had realized that now I need not have to spend too much of time on this. We had realized that initially it is laminar and then followed by turbulent of course, there is transition, but we usually do not know how to handle transition. So, we just say it is all of a sudden that is from critical Reynolds number once it reaches a critical Reynolds number of 5 into 10 to the power of 5 that is for a given velocity I can figure out at what location the boundary layer transits from laminar to turbulent. It is very nicely shown here it is all mixing flow is there and here it is laminated. So, this is how we can visualize parallel flow over flat plate and these are the correlations are summarized in one location. Laminar boundary layer you have delta by x equal to 5 r e x to the power of minus 5 and turbulent boundary layer you can see that it is delta v by x equal to 0.382 r e x to the power of minus 1 by 5. So, here it increases because the Reynolds number is very high. So, it increases rapidly as opposed to you can see here the turbulent boundary layer in the turbulent the turbulent boundary layer rate of increase with the increase the distance is much larger compared that of the laminar that is coming out from this and shear stresses of course, if I know the delta I can get the shear stress. The shear stress for laminar is 0.664 r e x to the power of half and C f x for turbulent is 0.0592 r e x to the power of minus 1 by 5 and this is valid for 5 into 10 to the power of 5 to 10 to the power of 7 because it has been experimentally demonstrated. In the absence of anything else even for higher Reynolds number than this one can go ahead with this correlation itself. Now, what is this? So, this is the local friction factor now if I have to get the average friction factor what should I do? I need to integrate this that is 1 upon L C f x d x between 0 to L. So, if I substitute for C f x 0.664 r e x to the power of half d x. So, I will get I will if I pull out this L here that is 1 by L is already outside I need not to pull out v x by nu to the power of minus half. So, if I differentiate this x to the power of minus half not differentiate integrate x to the power of minus half I am going to get 1.328 r e L to the power of half. So, average skin friction factor coefficient for a length of L it turns out to be 1.328 upon r e L to the power of half. Here L in r e L has to be taken as characteristic length L because it is averaged over the length L similarly for turbulent flow if you do you end up getting 0.074 upon r e L to the power of 1 by 5. So, that is essentially what is being shown here. If I have to cover net skin friction for laminar and turbulent complete that is right from the leading edge to the trailing edge I have to integrate over laminar that is 1 upon L 0 to x c r c f x laminar d x plus x c r p L c f x turbulent d x. So, that is what turns out to be like this if you do that this is what you get if it is for rough surface it is found that this can be derived from fundamentals it is not coming from experiment it can be derived that for rough surface for turbulent 1.89 minus 1.62 log of epsilon by L to the power of minus 2.5. What is this epsilon? Epsilon is nothing but my geometric roughness. So, this relation for turbulent delta v x for turbulent 0.382 x upon r e x to the power of 1 by 5 can be derived from fundamentals we are not doing that because in a new G class it is quite difficult to do that. Now, for turbulent laminar heat transfer coefficient we have derived this yesterday 0.332 r e x to the power of 0.5 p r to the power of 1 by 3 you can see that the heat transfer coefficient and the skin friction factor decrease with the increase of the length as long as it is laminar and in the transition all of a sudden it shoots up. And again in the turbulent flow the boundary layer thickness is growing because of which the heat transfer coefficient and the skin friction coefficient in the turbulent boundary layer decreases. Remember that the heat transfer coefficient and the skin friction factor in turbulent flow is very large compared to that of laminar. If you cannot believe me what I would request you is to take the laminar flow and the turbulent take a Reynolds number and take a velocity let us say 30 meters per second take air and start from leading edge to the trailing edge and put these equations in excel sheet and calculate this and plot this graph the graph will look more or less like this. Of course in your graph there will be nothing like transition there will be a sudden jerk there will be a sudden jump in the heat transfer coefficient. So, this is just a connector you can see here this connector is linear because I do not know how the distribution in between looks like. So, I have just joined this point with this point with a straight line. So, this is for turbulent flow 0.0296 rex to the power of 0.8 pr to the power of 1 by 3 note the difference pr to the power of 1 by 3 is same both for laminar and turbulent, but for rex the rex power you see here it is 0.5 here and here it is 0.8 that means the rate at which the heat transfer coefficient is increasing with the increase of the Reynolds number in turbulent flow is much large compared to that of laminar flow. So, after that similar way the way we did for CFX we can do the heat transfer coefficient averaging for laminar and turbulent that is h equal to 1 upon L 0 to xcr h laminar dx plus xcr to L h turbulent dx. So, this is what is the average heat transfer coefficient relation. So, that average heat transfer coefficient turns out to be something like this if you integrate that is 0.037 reL to the power of 0.8 pr sorry minus 871 pr to the power of 1 by 3 this is another correlation for covering all Prandtl numbers that is from 0.6 to 60 and this is for turbulent that is 5 into 10 to the power of 5 to 10 to the power of 7 you do not have to get confused between this correlation and this correlation both of them reasonably yield the same numbers they are not very different they are suggested by two different people that is why it has been represented if you see it is not very different here it is 0.0296 re x to the power of 0.8 pr to the power of 1 by 3 here it is 0.037 reL to the power of 0.8, but there is a deductive term minus 871 that will take care of the differences. So, for liquid metals yesterday we have seen that for liquid metals we have derived that pr to the power of half, but for flat plate for higher Prandtl numbers it is going to be pr to the power of 1 by 3 for liquid metals always liquid metal domain is different from what we handled for general fluid. So, Prandtl number less than 0.05 we have nu x equal to 0.565 re x pr to the power of half although Prandtl number is small hx by k k of thermal conductivity of lead tin are all very high because of which the heat transfer coefficient will be very high. So, this is the Churchill's correlation which will cover for all Prandtl numbers no matter whether I am handling air or water or liquid metal this covers all correlation you do not have to bother about which Prandtl number you are handling in that case Churchill's correlation can be used. So, now if I have a slight unheated starting length that means I will start heating my plate from here these were all these correlations were all for constant heat flux I remember that I have not mentioned, but I am stating now these all correlations whichever I stated they are dependent they are for constant heat flux for laminar. We will see later on for turbulent flow boundary the boundary condition does not matter for laminar flow only the boundary condition matters this we will elaborate when we do flow through pipes. So, if I have an unheated starting length that means if L is the length of the plate and zeta is the length of the plate over which I have not heated and I have started heating only after zeta then the Nusselt number this can be derived if someone is interested in this derivation he can go to go and see this book that is convective heat transfer by case and Crawford case and Crawford by convective heat transfer by case and Crawford. If you are not able to get this book you please Moodle I have derived this in my hand written notes I can scan that and upload it for you. So, that is the Nusselt number so all that you can see here is nothing much different from the previous correlation the top that is the numerator is just the same the denominator it has to be decreased little bit because the boundary layer is growing little later that is what is taken care by 1 minus zeta x to the power of 3 by 4 to the power of 1 by 3 this is for laminar and this is for turbulent that is for unheated starting length. As I said for uniform heat flux boundary condition this is the relation in fact for turbulent if you see only for laminar this was what 0.332 and here this was for 0.02916 not much of difference between 0.02916 and 0.038 that is what I was saying 0.0296 so and 0.03. So, for constant heat flux it does not for turbulent flow the boundary condition does not matter much for heat transfer coefficient why I will answer this when I go to flow through pipes please do not ask me that question now. So, heat transfer rate is equal to heat flux equal to heat transfer coefficient into T s minus T infinity so I can get the T s equal to T infinity plus q s dot upon h s after getting the heat transfer coefficient. Now, let us quickly take up a problem I think we have solved this problem I do not have to spend too much of time on this problem there is a flat plate and T infinity is 60 degree Celsius and my plate is at maintained at 20 degree Celsius one of the questions yesterday was what about hot air blowing conditions cold plate same correlation I am going to use it does not matter whether I am heating the plate or cooling the plate heat transfer coefficient is heat transfer coefficient heat transfer coefficient only temperature knows only the temperature gradient it does not bother about the direction of the heat transfer. So, here we need to remember that the properties of the temperature have to be taken at film temperature all these correlations are suggested for the film temperature properties that is T s wall temperature plus T infinity by 240 degree Celsius there was a question in the model which I have answered yesterday night why we have to take average film temperature not other temperature. Why because within the boundary layer the temperature is varying from wall temperature to the fluid temperature, but I have to take continually the variation it is very difficult. So, the only way to do is to take average when I take average means what I am assuming that the property variations within the boundary layer are linear they may not be linear, but this is the best thing what I can do. So, that is why I do it this way another thing is that I know only these two temperatures I know the temperature I do not know the temperature gradient if I know the temperature distribution then I can take the property variations and then integrate it that is not easy all the time for engineering problems. So, that is the reason we take the average film temperature. So, if I take these temperatures and at this temperature that is 20 plus 60 by 240 and I take the properties and plug in in my correlation first thing in any problem in convective heat transfer is to check the Reynolds number this is what I insist I have seen most of the time I myself making mistake while designing heat exchangers not calculating Reynolds number and directly taking an appropriate correlation for turbulent flow assuming it is turbulent I make this mistake most of the times myself. So, the first thing to check in convective heat transfer is Reynolds number whether it is internal flow or external flow same thing I am doing here v l by nu if I substitute that I get 41000 it suggests that it is laminar and I take the appropriate correlation remember I have taken average skin friction factor I get c f equal to 6.533 into 10 to the power of minus 3. I need to know skin friction factor is usually not above 1 it is always less than 1. So, if some number I get c f 100s or 1000s I have to be careful somewhere something has gone wrong and drag force equal to c f a s rho v square root by 2 c f I know area is length of the meter 5 and width of the plate is 1 meter I have taken and if I substitute all I get 57 Newton nusselt number equal to 0.664 re to the power of 0.5 pr to the power of 1 by 3 oil you see Prandtl number is 2870 Prandtl number is so high. So, if I put that I get a nusselt number of 1918 and if I multiply it with the thermal conductivity and divide it by the length I get h s 55.2 watt per meter square degree Celsius and h a s t infinity minus t s if I do I get 11 kilo watts. So, this is how we handle convective heat transfer problems. So, first thing is I need to take the properties of the average film temperature second thing is that I need to compute the Reynolds number to check whether I am in laminar domain or in turbulent domain so that I choose an appropriate correlation. Now, coming to flow around the cylinders now you can go on we have derived it for flow or flat plate now you can go on for cylinder for square that is what you can see the correlation there is a correlation for cylinder there is a correlation for square cylinder there is a correlation for square cylinder replaced diagonally all our functions of REPR a nu equal to CRE to the power of m PR to the power of n C m n are going to change for laminar flow for all of these configurations one can sit down and derive that has been done by that has been done by professor Shah and London. If you see this book Shah and London by laminar heat transfer you will be getting all sorts of correlation for all of these derived both internal and external. So, next is various shapes are there and they are all presented here I will come to flow around cylinder little later on. So, now let us take up flow around a cylinder I do not intend to take up the complete fluid dynamics of flow around a cylinder first thing is here I am plotting nusselt number versus theta theta is starting from here 0 and this is 90 and this is 180 I am only traversing from 0 to 180 because the heat transfer distribution on both sides it is going to be symmetric same. So, now you can see here for if it is 70800. So, here first thing is nusselt number is increasing with the increase of Reynolds number that is the first thing I think that is very obvious I do not think I need to say this because with the increase the velocity my heat transfer coefficients have to increase I think everyone appreciates this I do not have to explain that. But now if you take 70800 nusselt number is very high at the stagnation point that is this point is the stagnation point that is where the nusselt number is very high and subsequently as the boundary layer is growing then the nusselt number decreases why does it again pick up from here why does it pick up again from here because here the boundary layer was laminar then it may the boundary layer may transit it to turbulent. So, that is what happens for subsequent on once also only thing is that there is a shift of laminar to turbulent location in angle it is shifting towards right with the increase of Reynolds number. So, if you have to see the complete catalog of flow around cylinder you please see shifting boundary layer theory new addition in the introduction chapter itself is having he has put up a table where in which he shows that how the boundary layer separates how the boundary layer separates actually you actually flow downstream a cylinder you see vertices being formed. So, these vertices also are responsible for increase of the heat transfer. So, I am not going to spend too much of time because there is lot of local data is required to explain all of this, but one thing I can definitely tell two three points I can definitely tell that is this is the stagnation point the flow is hitting directly. So, at this stagnation point one would expect the heat transfer to be high and subsequently the heat transfer is decreasing because the way it decreased for flow over a flat plate because as the boundary layer thickness is increasing the heat transfer coefficient decreases, but here the heat transfer coefficient starts increasing because the flow the boundary layer may turn into turbulent boundary layer. So, that is what is happening there is a sharp increase at about 90 degrees is due to the transition from laminar to turbulent flow the later decrease is again due to the thickening of the boundary layer. Nusselt number reaches second maxima at 140 which is the flow separation point in turbulent flow and increases with the with as a result of intense mixing in the turbulent wake region. We need to understand what is called as flow separation. So, before we move on let me just say what is flow separation see if I take a flow separation can be understood in a flow in a backward facing step. This is what is called as backward facing step backward B A C K backward facing step. If I take a backward facing step and if there is a flow like this the flow goes and it separates here the boundary layer has to separate here. The boundary layer is growing and then all of a sudden the boundary layer separate and this is the dead zone means there is no velocity it is not like that it is just that there is a negative velocity generated then later on it grows up. So, this is what is called as reattachment point typically at the reattachment point shear stresses are very high at the reattachment point that is the reason in fact if you take ribs if you take flow around ribs flow over ribs if you put because this is the question which has been asked time and again the boundary layer separates and reattaches and again grows separates reattaches and again grows. So, at the reattachment point the heat transfer coefficient is very high and here in this zone the heat transfer coefficient is decreasing why because it is dead zone because this is there is negative velocity. So, the point is this is the separation point this is the separation point this happens even in case of let us say we are all of a sudden let us say there is an expansion there is a sudden expansion in a pipe. So, same thing flow reattaches it is like having two backward facing steps forward facing steps not backward facing steps. So, the flow separates and you have the negative velocities here in this zone. So, that is similar to that is happening here. So, there is a flow separation at 140 degree Celsius. So, at this separation point there is the flow is separating and as a result of which there is intense all that we say is that it is very easy to say that boundary layer becomes turbulent and therefore the heat transfer increases. So, with this we will move on there is another problem which I do not intend to solve this is again flow around a cylinder it is just that the correlation is different I have just used the Churchill's correlation I do not intend to solve this problem it is more or less like flat plate problem this is also same thing, but only thing is that I have taken a sphere instead of cylinder. So, I do not intend to solve this problem I request you to go back and solve this problem.