 So, another thing let us get back to what I had postponed I have been answer I am saying that I will I will answer it later I will answer it later now I think one of you should answer me why for a flat plate dp by dx is 0, why for a flat plate case we are now into external flows now you can shoot any question why is dp by dx equal to 0 in all textbooks it takes of course few guys try to take planes to make us understand and few guys do not why do you think it is 0 see let us let us write the momentum equation once again u del u by del x plus v del u by del y is equal to minus 1 by rho del p by del x plus nu del squared u by del y squared I am not writing del squared u by del x squared because I know that it is not important. Now if I write this equation for the outside of the boundary layer I am getting out of the boundary layer now I am getting out of the boundary layer what will be my u? u infinity square u infinity u infinity into du infinity by dx why because u infinity what about next term what about next term why why is it 0 v is not there v is not there this is 0 because v is not there what happens to my next term minus 1 by rho dp infinity by dx I am going to call it as p infinity because I am outside what about the next term what will happen to next term will it become nu d squared u infinity squared divided by dy squared yes it is 0 why it is 0 yes there is no viscosity there is no viscosity what is this equation by the way what is this equation what is this equation what is this equation if you integrate this what will you get it is a very familiar equation Bernoulli's equation yesterday I told no from Navier's talks you can reach Bernoulli provided if I throw my viscous terms but what is this equation telling me what was the question I asked why is p in p is 0 but what is this equation telling me flow over a flat plate u infinity is u infinity is constant so what will happen to my dp infinity by dx so what does it tell what does this tell this is the pressure which is being imposed on my boundary layer the pressure which is there outside has to impose itself onto the boundary layer that is how the inviscid we are what did Prandtl do us he differentiated he differentiated into two layers one was the viscous portion another was the inviscid portion so how will inviscid portion talk to viscid portion how will the interaction happen through this pressure term through this pressure term if I know that is how inviscid solution is also important in the inviscid from the inviscid solution what will you get from your inviscid equation which you have done doublet and all I do not know whether you remember or not you get the u infinity variation for flow around a cylinder for example u infinity sin x you get for various profiles you get u infinity variation what am I going to use that it will give me the see this equation if I know the velocity distribution around my bluff body inviscid portion which I get from inviscid solution I will get my pressure variation that pressure variation is what I am going to impose in my impose in my boundary layer equation that is how inviscid portion is going to talk to with my viscid portion so it is not inviscid solutions are useless that is not right statement that is what Prandtl did it to us that is the greatness of Prandtl okay so now coming back what was the question we asked ourselves now I can answer u infinity is 0 sorry u infinity is constant for flow over a flat plate so dp please note this down this is what we need to tell our students otherwise you cannot answer my student why is dp by dx equal to 0 for a flow over a flat plate most of the textbooks are not going to tell us this okay so that is how you get so let me repeat pressure around says that we need to repeat this see what did I do I took the momentum equation applied it for so I have I have flow over a flat plate if I take I have boundary layer which is viscid portion I am calling why am I calling viscid portion because that is where the viscosity is important I cannot neglect but now all this portion all this portion is inviscid portion I can get away without taking viscosity see if I solve inviscid equations I am going to get my u infinity flow around a cylinder if you might be remembering when you correct me flow around a cylinder is it right u infinity sin x u equal to u infinity sin x it is u infinity sin x with my x varying from radially okay I may be wrong little bit but yeah something like that so point is I get this flow distribution u infinity distribution as a function of x from my from my inviscid solution inviscid equation that I will use to get my inviscid equation of course I am not doing that but I will have to tell that that is it inviscid equation gives me u infinity as a function of x that u infinity of x I am going to get use and get my get my dp infinity by dx this is what is being imposed imposed on my on my boundary layer very beautiful it is actually okay we cannot study inviscid now but it is very nice if you can demonstrate let me see before our general workshop if I can cook a small case and show that this is how it is so that I will make a transparency whatever I am writing now only see when I am teaching when we are teaching we teach this on the board but now only I am realizing that it has to come on the power point yeah yes it is the beginning I have to come from there only I cannot go this way I have to come from there from inviscid portion to visit portion here we are lucky for a flow that is precisely why we take flow over a flat plate why because we can throw my dp by dx so when I say imposed on the boundary layer I mean that del p by del x is equal to dp infinity by dx or I should be writing dp by dx equal to dp infinity by dx that is what I mean when I say it is imposed on the boundary layer if you want better treatment of what I have told either this I have learned from professor andreson's textbook I have not learned this from anywhere else but touch and go is there in boundary layer theory by Schlichting ok but proper better treatment is there only by aerodynamics by andreson I would recommend strongly that all of you read as a story you can read aerodynamics by andreson ok so ok now coming back now coming back let us now there are two points now let us get back to this whichever derivation we were doing earlier what were we doing this is this is what we did ok now let me apply the scale to x momentum equation what is my x momentum equation reduced x momentum equation I am writing same thing again and again but that is ok because it should get registered in your mind so much so that even while traveling back when you are manning this you should be getting it in your mind you del u by del x plus v del u by del y equal to minus 1 by rho dp by dx is not there now because we have appreciated that that is 0 nu into del squared u by del y squared what is the scale of the right hand side left hand side just see u infinity squared by this should be of the same order of what is there on the right hand side nu u infinity by delta square ok now very beautiful answer I am going to get now ok let me write this as delta squared on the left hand side they are of the same order they are not equal delta squared is of the order of what do I get can anyone help me with the algebra prevails our delta squared is of the order of nu by nu l l by u infinity let me divide both sides by l squared let me divide both sides by l squared what do I get delta squared by l squared which is of the order of nu by u infinity l what do I get now what do I get this implies delta is of the order of l what is this u infinity l by nu rl to the power of to the power of minus 1 by so if you derive from similarity solution what is it you are going to get you are going to get a constant delta by x you use this correlation very left and right delta by x equal to some constant into r e to the power of minus of r e x you see without I have done no mathematics I have done no mathematics but I have reached the solution I can even reach up to cf I can even reach up to cf but why I did this because I am going to take a similarity variable eta equal to I am going to use a similarity variable eta equal to y into u infinity by nu x because what am I trying to do in semi-infinite medium what did we do in Nithin Gulhane please tell me what did we do in semi-infinite medium I am able to see your name that is why I am calling you no other intention ok you better remove that if you do not want me to you are getting caught too many times yeah may be marry up on where one what did we do in semi-infinite medium we reduce the t equation which was the function of x and y x and del t to eta so pde was converted to od that is precisely what we are going to do this pde I want to convert it into od so I am going to write now can you link these two equations and tell me why did I write eta equal to y into square root of u infinity by nu x see for a minute and you can do it only thing is that you have to take delta as y and x as L as x is that right is it obvious should I be doing that no I guess a month is blank a month could not see no problem so what it is what is it I wanted to go fast but that is ok that is y by x is of the order of rex to the power of minus half so what am I doing what am I doing so it is just there y into square root of yeah yeah that is ok ok ok you push it now y is of the order of x into what is rex u infinity x by nu square root u should be helping me here but I am not getting any help from is that right y by x am I am no no haha sorry this is wrong this is wrong y is of the order of x into square root of nu upon u infinity x so what do I get y is of the order of square root of nu x upon u infinity now you should be seeing y eta is like that I have just taken eta equal to y into square root of u infinity by nu x ok so now professor will take over and so that is how eta choice is important but I think two points I want to summarize at the cost of being dubbed as repetitive one thing is dimensional similarity we should emphasize students number one whether even if you are teaching integral approach and second thing is that how inviscid portion talks with the inviscid portion these two points if we emphasize them they will not be scared of boundary layer theory anymore in their life so now what is that I am doing now let us just mundane I say mundane because it is all algebra correlations the point is I have understood that friction factor I am going to see correlation friction factor in friction factor correlation what is that I am going to see Reynolds number and Nusselt number correlation I am going to see Reynolds and and now we can see that Reynolds for flow over a flat plate I am going to see friction factor Reynolds number to the power of minus point okay so you now know that because we have derived that already yeah he wrote Nusselt number you know we have not defined Nusselt number as yet you know there was a transparency yesterday which we said we will come back later yeah what is the physical meaning of Nusselt number that is right he wrote the partial derivative of t star with yeah let me open that again because it is good to guess that is the definition of this one and then there was a transparency which had a Nusselt number definition introduction to convection no I have written now h equal to I will h equal to minus k del t by that is right h equal to minus k del t by del y at y equal to 0 upon t s minus t infinity morning also we reminded ourselves about this equation so this was h but now what professor is asking is Nusselt number we consciously skipped that because we thought that once we get enlightened with all the background we will come back to this now someone said convective resistance and conductive resistance professor wants the elaboration on that because that is that is very important because textbooks tell us q convection upon q conduction equal to h delta t upon k delta t by L question is L is what you rightly said that L is boundary layer thickness it is not the pipe diameter or the flat plate because conduction where is the conduction happening just within the boundary layer so that we have to because even chungal writes this as hl by k but I guess he knows that we have understood that it is boundary layer thickness that is what he is assuming this has been taken from chungal only but we need to emphasize that this L is not the characteristic length it is delta then only this Nusselt number equal to 1 has a what is that meaning any u equal to 1 means conduction equal to convection that is conduction within the boundary layer that has to be emphasized is that okay. No convection means we always attribute that conduction within the boundary layer no slip means what you are coming back yeah coming back as your your your puristic thing is minus k delta t by delta y at y equal to 0 who is deciding my delta t by delta y at y equal to 0 who is deciding me that my temperature profile within the boundary layer both answers are right both answers are right delta t what is that you are telling is it is delta t by delta y at y equal to 0 why should I care about y equal to little above but I am saying I should care about that because that is what is going to decide my delta t by delta y at y equal to 0. Yeah, yeah. So, temperature gradient is the function of the heat transfer coefficient heat transfer coefficient is dictated by the temperature gradient again temperature gradient is dictated by the temperature gradient. Yes, yes, yes. How do I get the orientation effect we will post on this question. That is flat plate no orientation effect means what I have to do I have to do experiments if it is laminar I can perhaps get the solution yes we will decide okay you should be able to answer now how will how will I get for different orientations how will I get the answer for nacelle number how will I get who is giving me for flow over a flat plate nacelle number that is the velocity distribution for orientation also for orientation also I will have to get my velocity distribution but then in that case dp by dx I cannot take it as did I answer your question or not I do not know now what was your question. Here for nacelle number no this is because okay okay okay I understood your question now see why I do not define nacelle number as h delta by k why do not I define it as h delta by k. Why I do not define nacelle number as h delta by k I take it as h l by k and take characteristic length ultimately what is that I am trying to do if you get back I want what heat transfer coefficient right I have to non dimensionalized heat transfer coefficient and I have non dimensionalized nacelle number in terms of characteristic length but not delta why do I know delta up to any case do I know delta so I should be representing that in terms of characteristic length it is an engineering necessity that is all. Yes correct no no here what we mean by saying that there are two things here we are we are mixing up two things it is this is not particularly for internal or external or general h delta by k should be the nacelle number definition okay why if I have to understand it as convective resistance to conductive resistance but I define nacelle number in terms of characteristic length because I do not know delta because why we alerted you because nacelle number 1 if I take l has no it has to be delta it has to be delta the point here is we give nacelle number correlation in terms of characteristic length because I know characteristic length and I know velocities and I know fluid properties that is why we define nacelle number in all correlations in terms of characteristic length but that does not tell if that characteristic length based nacelle number is less let us say equal to 1 you cannot say that conductive resistance equal to convective resistance that is all I am telling for our engineering necessity I have to take correlation I have to use it no length okay I guess now I have that is how I think you wanted to get fine no problem please bear with me if I am impatient okay even Reynolds number for that matter we take rho v l by mu okay it is actually inertia and viscous force within the boundary layer I should be taking there also delta isn't it but why I cannot take boundary layer thickness because I do not know so that is why I take characteristic that is why in fact that is a good point I got the point why I get transitional Reynolds numbers different for different class of problems for flow over a flat plate my Reynolds number very good point now I will make you understand Reynolds number critical Reynolds number transitional critical Reynolds number for flow over a flat plate is 5 into 10 to the power of 5 RE pipe RED defined on the characteristic length of the diameter of the pipe is 2300 for jet impingement RE is around 100 why they should be different now let me define on the basis of boundary layer thickness it will be of the order of 10. You take the boundary layer thickness relevant to flat plate or to flow through pipe or for jet impingement or for natural convection then RE delta is of the order of not 10 sorry 100 anything exceeding 100 will be Reynolds anything less than 100 it will be I have reached you know so RE defined on the basis of delta is unique is unique but RE I always define on the basis of characteristic length why because that is my engineering necessity I have to use it ultimately I have to use it I do not know delta no for every case my delta is different I cannot generalize it that is how now I have reached it now I have reached it okay okay that was a good question that was a good question really good question although I was impatient please bear with me my impatient okay because I have to have another objective of I have to complete things in time okay so I have to two rough edges I have to corner I mean smoothen them fine no problem so now what is that yeah flow over a flat plate now I will tend to go fast so now that velocity profile I know and temperature distributed this is the continuity equation momentum equation and the energy equation so now these are the boundary conditions now I know eta if I take this and non dimensionalize sorry transform these equations in terms of eta I get my equation as f triple prime plus f f double prime by 2 where f is u by u infinity is f prime okay I am going to go very fast I know but if you do that all that algebra and you get this equation with the appropriate boundary conditions if you solve that is these are the boundary conditions you are going to get that re equal to sorry delta equal to 5 delta by x we saw no delta by x is of the order of rex to the power of minus of that is what it is all that I am now getting is 5 by solution you see the power of scale analysis you see the power of scale analysis so then you get from this delta you get tau wall equal tau wall equal to mu into del u by del y you get c of x relation okay when you go back home you see this algebra I do not want to spend time on this algebra because I want to focus on insides only so c of x equal to what is that you get 0.664 into rex to the power of minus is that okay so that is how you get the shear stress now similarly you can do the energy equation also by similarity solution you get theta double prime plus Prandtl into f theta prime by 2 and if you solve that equation you are going to get delta t equal to 5x upon pr to the power of 1 by 3 upon rex okay in fact we can do this from scale analysis also okay but I am not going to do that okay so you can show that the boundary layer thermal boundary layer thickness is this so all that you are seeing is what delta t by x is of the order of re and pr so rex to the power of half and pr to the power of 1 by 3 so the solution for turbulent laminar flow is c of x is this and for turbulent flow c of x is this we can derive also but this has been corroborated through experiments also okay so sorry that brings us to the last yesterday I made a mistake I said for flow over a flat plate professor Arun was right I was wrong c of x was professor Arun said boundary layer thickness turbulent boundary layer thickness you asked a question yesterday re boundary layer thickness grows faster is thicker in turbulent boundary layer compared to laminar we showed but I told re to the power of 0.8 it was wrong professor was right re to the power of 0.2 but still nevertheless it does not matter why because re is very high that makes it thicker you can go back and check it out okay but I was wrong because I had told re to the power of 0.8 so now we have to be worried because I have recorded that so that I have to correct it myself that is what it occurred to me in the night yesterday fine so I thought anyway I will correct it here so this is these are the c of x is okay now if we integrate the way we integrated yesterday for local heat transfer coefficient if I integrate that I will be getting the average friction factor okay so now rest is straight forward what is that we what is that we will get correlations for rough surface smooth surface all those cases and similarly you will get nusselt number nusselt number you will get 0.332 re x to the power of 0.5 pr to the power of 1 by 3 for Prandtl numbers this is valid only for Prandtl numbers greater than 0.6 this is not valid for which fluid it is not valid for liquid metal because liquid metal is lead tin sodium okay these are all used in nuclear reactors so the Prandtl number is 0.002 0.03 lead is around that okay so very low Prandtl number this equation is not valid for that we have a different equation and this is again we can derive it also and we can do through experiments also for a flow or a flat plate but in general I want to make a statement in general irrespective of the class of the problem what we are handling for laminar cases only you can what is the statement for laminar only you can derive you can derive for turbulent it is not possible to derive for few cases for flow over a flat plate you can derive but flow through pipe you cannot derive you have to do through that is how detus bolter correlation comes it is coming through lot of experiments okay that is how experiments are important okay. So nowadays if it is laminar you can do through if it is a complicated case you can use CFD and generate but for turbulence it is little sticky again because which turbulence model to use it is a problem the point is for laminar you can generally give the closed form solutions they are mathematically tractable but for turbulent generally it is not possible okay that is what we need to understand why it is not possible no no in the background of what was taught yesterday you should be able to tell whatever the governing equations when we did Reynolds stress averaging yesterday we got 9 6 additional terms you remember minus rho u prime squared minus rho u prime v prime minus rho u prime w prime minus rho v prime square bar minus rho w prime square bar they are all unknowns to us so that is precisely what makes life difficult for turbulent okay. So this is in fact this was plotted already yes laminar flow how do I differentiate a flow is laminar or turbulent Reynolds number for all Reynolds numbers greater than 5 into 10 to the power of 5 less than 5 into 10 to the power of 5 it is laminar I did not get professor what is your question no I think I did not reach you for liquid metals I have a different correlation that is all it means that is all actually for liquid metals you will get rex to the power of half and pr to the power of half by scale analysis also we can do it okay so then then comes then you get the average heat transfer coefficient here it is here it is for liquid metal you have 0.565 rex to the power of half and pr to the power of half for liquid metal this is again laminar this is again laminar that is true this is okay okay okay I am answering that question I am answering that question Churchill's correlation is independent of the range of the Prandtl numbers I am handling it is valid for both liquid metals and so for all Prandtl numbers less than 0.05 only this is valid but your question is what should I do between 0.05 and 0.6 but generally we do not come across a fluid liquid metals are very less okay so generally we handle air so anything between 0.05 and 0.6 0.7 is not there that is the reason but if we come across we can use Churchill's correlation okay so that is that is a good question okay so now there are various correlations if I do not start heating if I do not start because there we started heating right from the tip of the boundary sorry the flat plate that is the leading edge but here I am starting heating later on so velocity boundary layer overtakes the thermal boundary layer so for that case actually you can derive this from fundamental from fundamental so but then this derivation definitely for UG will be beyond the scope but definitely we will not derive I think you have how do you teach this these correlations you give them the list or they are supposed to mug them you give the handbook great so then they are supposed to use them. In exams how do they get the? Allowed? Allowed? Great. Achha Kodan Raman we have to see. Mumbai University it is not allowed. Good. In our days we had to mug up these equations. It is good if students are not allowed to mug it is really nice. You tell me okay just looking at this graph I think we will appreciate that the heat transfer coefficient is higher for turbulent flow so many times in applications you know if your boundary layer is grown sufficiently it is what is broken and then allowed to redevelop for the simple reason that you know if the boundary layer is continuously growing there the heat transfer coefficient will drop but if it is broken and allowed to reform what will happen is there will be again a spike in the heat transfer coefficient so the so called average heat transfer coefficient will be larger in case of stripping of the boundary layer and reattachment. That is coming back to yesterday's thing laminar sub layer proof. See what do we do in a pipe what do we do in a pipe I think I have actually worked out the problem which you had asked I have just worked out I will when we get into internal flows I will solve what professor is saying is that what do you do if you have to roughen this what do you do for this flat plate you will put small small sticks protruding from this what professor is saying is that you place them so close that so close that it will not grow too thick too thick so that your heat transfer coefficient because as the boundary layer thickness is increasing that is what yesterday your point was but that does not mean that turbulent boundary layer thickness is smaller what we are saying is that our boundary layer thickness we continue to try to keep turbulent but that turbulent boundary layer thickness we try to keep it as small as possible because we want to be in this domain I think we have no I am going to talk about that PhDs are done ok so very very useful example is there in reactors especially practically it is done you know reactor will have several fuel assemblies fuel rods and they are held together by what you call a spacer grid spacer grid is nothing like your you know in olden days milk used to come in bottles right so the bottles would be kept in a crate which has the circular locations for the bottles to be held egg crate is one example so such a such a thing is there made of a metal and the design is confidential so it is not revealed so what happens is that is placed and designed in such a way that at specific locations where it is placed the boundary layer is stripped off so mixing of the flow also happens because of the of the way the grid is designed and stripping of the boundary layer happens so very good enhancement in heat transfer at for inner reactor you need better cooling so whatever it is heat transfer coefficient should be as high as possible and you are talking of flow areas or flow dimensions which are very very small you know the pitch between two fuel rods is of the order of a few millimeter 30 millimeters or something like that so you are talking of sub channel dimensions which are very small in that if the boundary layer is growing from all four if I am imagining area of four rods it is going to grow and completely occupy the channel area so the boundary layer is going to cause tremendous reduction in heat transfer I should cut it off correct and allow it to reattach and in the process the design of the spacer grid makes life means is so complicated that you know how well it is mixing what is it going to do to the flow to enhance heat transfer that is what is the design characteristic because otherwise it is nothing but I create geometry or you know honeycomb geometry or whatever so enhancement due to good mixing and stripping of the boundary layer both are there and it is very useful. Let me complete see there is another application in gas turbine blade you want to cool the gas turbine blade these are the cooling passages because gas turbine gas temperatures are quite high so metal temperature the only we know that gas temperature has to be increased to increase the overall efficiency so but there is a metallurgical limit so the only way is to cool it this cooling passage if I take it like this you imagine a math stick put like this because we are discussing I thought I will discuss this in internal flows but now that so much discussion is happening let me finish it off here these are math sticks you imagine these are a math stick being placed what we are saying is that one math stick and the next math stick is placed so close that I am not allowing my boundary layer two things it is doing the placement of the math stick ensures that my laminar my boundary layer is turbulent laminar sub layer is broken that is what decides me the thickness or the height it is actually called as E and who decides the pitch which pitch is the optimal pitch that is the boundary layer I will not I should not be allowing my boundary layer to substantially grow of course all this comes at a cost pumping power will go up this is this is the this is the blindly one doesn't increase the height and all 1 mm 2 mm why 1 mm and all when do we break the brine laminar sub layer I will come back to that when we take up that example yes transfer so is it possible to break it in laminar flow so we can get the higher heat transfer coefficient if you break it it will be turbulent because you are broken okay so maybe what you are trying to say is if I break in such a way that it is laminar sub layer if I am still in the laminar sub layer that will not serve my purpose at all. The second question is sir what as I go along the boundary layer I am having the negative slope of this heat transfer coefficient if I not allow the boundary layer to grow over the surface should I not get the constant heat transfer from right from the beginning that is what we are attempting to do see if I take it back here yeah if I take it back here what will happen what will happen between these two let us try to plot nusselt versus distance what will happen here what will happen here actually it will decrease and then again increase it will again shoot up from here decrease it will come back to this and again it will shoot up so periodically but point is on an average I have increased but if I were not to put this ribs my heat transfer coefficient would have been somewhere here this is the number versus x this is okay this is the done by because your boundary layer is growing and you are breaking up by putting the ribs over the surface that is okay but my question is instead of placing the ribs over the surface suppose I am make the change in the surface itself so as the boundary layer does not grow like if I am using the porous media yeah I would like to suck the boundary that is what aerodynamic people are doing they are not trying to stop the boundary layer with suction that is possible to enhance the heat transfer coefficient but you see it is quite complicated in any application you see most of the applications that have you come across any application like that honey come you rightly pointed out that thing usually as far as I know the applications wise for increasing the heat transfer we usually in heat exchangers or in gas turbine blade cooling or the examples what he took in nuclear the easiest thing is to rough on the surface blowing means I have to I have to I have to locate them appropriately see I will give you an opposite example you told suction I am going to give you blowing there are several ways of cooling my surface same gas turbine blade how do I how do I cool it in another way there are holes throughout flow comes out flow comes out how do you cool your potato you take the potato from your cooker how do you how do you how does your wife or mother peel it out have you any time peeled your potato how do you peel it out when you take it out of the cooker you will put it under tap you will keep it very low flow rate and you will peel what what exactly you are doing there you are creating a small film such that the my hand does not know that my potato is hot I am having a small film around it with cold water so what exactly I am doing I am blowing I am not sucking your example was suction I am saying here blowing same thing for gas turbine blade also I can blow it and still make it cool so there are various ways in which you can operate. That may be in terms of house the distance between the blade tip and the casing so if I am having the boundary layer grows over the surface of the casing no but the no no it is not like that I have to now place my holes such that I have made a proper film it is not going to be one hole it is going to be several holes and optimization of that holes is involved I am talking about I think we will stop here because we will we will discuss we will go for a tea and then come back.