 So, we will get started with the final chapter of convection that is natural convection. So, hopefully within one hour we should be covering this convection. So, the idea here is in the natural convection I had lot of questions in the moodle just few minutes back also I have answered again for natural convection to take place what are the things which are required. Two conditions have to be satisfied number one there has to be temperature gradient because of which density differences are generated. Number two there has to be acceleration due to gravity because of this only because of these two only body forces created. If one of the two are not there let us say G is there, but there is no temperature difference no natural convection let us say there is temperature difference because of which density gradient is there, but there is no density there is no acceleration due to gravity like in space then there is no question of natural convection. So, the point is both the conditions that is temperature difference and also the acceleration due to gravity have to be there otherwise natural convection cannot occur. Now, coming back so that is what is being told in this transparency. Now we need to characterize this temperature difference to density difference how do we do that? For that we take recourse of what is called as coefficient of volume expansion. So, coefficient of volume expansion is a measure of change in volume of a substance with temperature at constant pressure. I am sure most of you you have studied this in thermodynamics that is coefficient of volume expansion is a measure of change in volume of a substance with temperature at constant pressure that is 1 upon V dV by dt at constant pressure. This can be replaced volume can be replaced by density that is equal to minus 1 by rho d rho by dt at constant pressure why minus sign is there? Why minus sign is there? Increase in density will decrease in volume so that is why these two are inversely proportional volume and density are inversely proportional density equal to mass by volume as is same. So, you can see here if I take larger a substance with large beta will undergo large change in dV by dt and similarly for a substance with small beta will undergo small change of volume that means large change in density. So, not large change in density same change in density but only thing is that it takes a negative sign so that is about beta. So, now this we will be using little later little later for now you remember that beta equal to minus 1 by rho d rho by dt at constant pressure. Now, what will happen to my boundary layer when there is a vertical flat plate? I have a vertical flat plate and the boundary layer both the hydrodynamic and the thermal boundary layer is growing I have taken Prandtl number equal to 7 because of which both the thermal and the hydrodynamic boundary layer thicknesses are same. But if the Prandtl numbers are different than 1 the same whatever principles we studied for forced convection are valid here also the Prandtl numbers are less than 1 delta is going to be lesser than delta t and other way round if Prandtl number is greater than 1 delta is going to be greater than delta t. But here in this case I have taken Prandtl number equal to 1 which is almost like air whose Prandtl number is 0.71. So now what are the assumptions I am going to take natural convection flow is laminar I am taking that boundary layer is laminar flow is two dimensional that is I am having only V and U flow is steady that is there are no variations with time and fluid is Newtonian that is shear stress is proportional to shear strain and properties are constant all thermo physical properties are constant except that the density difference between the fluid inside the outside the boundary layer that gives rise to bioenzy force is the only thing except in the body force everywhere the property variations are considered to be constant. With this let us write the equation same equations they are not going to be different at all that is del u by del x plus del v by del y equal to 0 I think I will write this so that everyone would be feeling sleepy let us write this del u by del x plus del v by del y equal to 0. Now similarly what is the x momentum equation rho u that is rho into u del u by del x plus v into del u by del y equal to minus del p by del x plus mu into del squared u by del x squared plus del squared u by del y square. Similarly, rho into u del v by del x for y momentum equation plus v del v by del y equal to minus del p by del y plus mu del squared v by del x squared plus del squared v by del y squared minus rho g what is this minus rho g that is the body force that is the density of the fluid added upon multiplied by gravity. So, that is what we have put here minus rho g y minus because it is acting downwards y upwards is positive that is what we have taken now what is the energy equation looks like u into del t by del x plus v into del t by del y equal to one of alpha into let me just check it out alpha into del squared t by del x squared plus del squared t by del y square. Now, which of the two momentum equations are not so important compared to the other one like flat plate here in this case yes which one will not be important the you see here we have x is u and y is v u is not so important as v unlike floor a flat plate where u was dominant over v. So, here v is more important than u so that is what we need to note down that is what we need to note down you see here u is very much smaller compared to v so complete x momentum equation vanishes. So, that means this complete x momentum equation vanishes that means pressure is going to be a function of only which implies that pressure is going to be a function of only y it is going to be a function of only y not a function of x. So, now what do we do now let us take out of these two terms out of these two term which term is important sorry out of these two term which term is important what is the scale of this term del squared v by del x squared it will be some velocity let us say v upon x squared is delta squared and v upon this scale is v upon del squared that means this will be very much smaller compared to del squared v by del x squared. So, that is why this can be neglected now what about this del p by del y which is going to be d p by d y what is this d p by d y now going to become equal to what is this d p by d y yesterday only we told that the viscid portion talks with the inviscid portion. Now in natural convection is there any flow no so it is still that is it is hydrostatic pressure. So, that is del p by del y is nothing but d p infinity by d y which is nothing but the hydrostatic pressure gradient who is dictating it the density that is minus rho infinity into g that is d p infinity by d y please correct this this is not d x this is d p infinity by d y that is d p infinity by d y equal to minus rho infinity into g that means my x moment my y momentum equation is going to reduce to rho into u del v by del x plus v del v by del y equal to rho infinity minus rho into g plus mu into del squared v by del x squared del squared v by del x squared. So, that is what we have written here that is what is here before we move on we have to similar arguments similar arguments bit in the temperature also which of these two terms I can neglect this is delta t upon delta t squared that is thermal boundary layer thickness squared delta t upon l squared. So, this term can be thrown out. So, that is this term can be neglected in the same arguments the way we argued it for this term. So, now let me write the refined equations let me write the refined equations that is that is del u by del x del x plus del v by del y equal to 0 continuity equation has remained positive and this is the momentum equation. Now, we need to do we need to express this rho infinity minus rho in terms of temperature how do we do that that is beta equal to minus of 1 by rho d rho by d t at constant pressure. So, beta equal to minus of 1 by rho rho infinity minus rho upon t infinity minus t. So, I can write rho infinity minus rho g beta minus has been absorbed inside sorry not g beta rho beta into t minus t infinity if I substitute in this equation that is why momentum equation I get rho g beta t minus t infinity. So, now if I divide throughout by rho g beta t minus t infinity here I want to stop and tell you something see always students will come back and ask me the question you are taking you are assuming constant properties you are assuming constant properties, but you are taking variation of density in the body force. What we are saying is that the density in this term is assumed to be constant only the and viscosity here is assumed to be constant but the density variation only in the body force is assumed is taken as varying why if I do not take that as varying what will happen then there is no bio ency then there is no body force then there is no movement. So, this is essentially the driving force my rho g my rho g this my rho g is the driving force this is my driving force if this is not there is no driver there is no natural convection. So, we cannot have any more free convection at all. So, that is precisely what we what we say when we tell that the rho g is I mean density variation is taking into account only in the body force in all other locations we are taking that constant why are we taking in all other locations as constant because we have to get the closed form solution otherwise it become difficult to get the closed form solution. So, this is my momentum equation now let me rewrite or recast the same equations del u by del x plus del v by del y equal to 0 u del v by del x plus v del v by del y plus nu del squared v by del x squared plus g beta t minus t infinity u into del t by del x plus v del t by del y equal to alpha del squared t by del x squared what are the boundary conditions no slip boundary condition u equal to 0 v equal to 0 at x equal to 0 and v equal to 0 as extends to infinity here because it is still air and t equal to t s at x equal to 0 and t equal to t infinity as extends to infinity one thing I want you to note in these equations is that you see temperature in momentum equation. You never saw temperature in momentum equations when we were solving forced convection in forced convection the momentum equation and the energy equation are uncoupled let me write this in the momentum equation in forced convection the momentum equation and energy equation are separate or uncoupled or decoupled they are not coupled. So, first you solve the momentum equation and what do you get when you solve the momentum equation you get the velocity profile taking this velocity profile we solve the energy equation to get what temperature profile taking this temperature profile we computed nusselt number unlike this in case of natural convection what is happening in case of natural convection in case of natural convection we have momentum and energy equation coupled when I say momentum and energy equation are coupled mean temperature is sitting in my momentum equation. So, I need to solve both of them simultaneously I cannot go one after another I have to get simultaneously I have to solve momentum and energy equation in order to get velocity and temperature profile that is V and that is what essentially the summary this is what we need to emphasis whenever we are teaching natural convection. So, now after this we if we non dimensionalize if we non dimensionalize I think let us do that otherwise we will be going too fast let us do that or so what are we saying is these are the equations we will non dimensionalize we will non dimensionalize again using x star equal to x by L y star equal to y by L u star equal to u by infinity v star equal to v by u infinity and t star equal to t minus t infinity upon t s minus t infinity for a minute let us not worry about what is u infinity there is no forced convection velocity this is some representative velocity and we do not know what is that velocity let us take that velocity u infinity if we take that continuity equation is what we will get the same thing there is no confusion there and these two terms are also same on the left hand side I do not have to do again on the right hand side first term also we got last time 1 upon r e l I do not think I need to focus on that but what is the term I need to focus on I have g beta g beta t minus t infinity I have so what is that I should be doing for g beta t minus t infinity when I non dimensionalize I get g beta t s minus t infinity into l cube upon nu square that is I am going to get I think I will not I will just state that that is yeah t s minus t infinity upon nu square upon nu square into l cube into t star upon r e square r e l square so what is this number this number is nothing but Grashof number Grashof number is g beta delta t l cube upon nu square so that is what has been written here so we get now the nu non dimensional number is u star del v star by del x star plus v star del v star by del y star equal to 1 upon r e l del squared v star upon del x star squared plus Gr upon r e l into r e l square t star this is what is that so what is the nu not again if you non dimensionalize the energy equation you are going to get r e p r let us not do that but what is the nu non dimensional number which is come g r so Grashof number few people would have been relevant I mean very comfortable with Rayleigh number Rayleigh number is nothing but Grashof number into Prandtl number which comes down to g beta t g beta delta t l cube divided by alpha nu because P r is nothing but nu by alpha okay so you get g beta delta t l cube upon alpha nu for Rayleigh number what is the importance of this non dimensional number gr by r e l square let us stop and take a look at this the point is lesson learned is we have got a nu non dimensional number what is called as Grashof number which is relevant only when only in case of natural so that is because of this body force t s minus that is the density gradient which is because of the g so now you can understand why both g and t s minus t infinity should be there if g is not there and only t s minus t infinity is there let us say in space there is no Grashof number Grashof number is 0 then there is no natural convection okay so now if g r by r e squared is very much less than 1 that means what g r is very less compared to Reynolds then it is forced if g r is very much greater than r e squared then natural convection dominant if it is g r is of the order of Reynolds number then both natural convection and forced convection are important and again we will have if you remember in all of this we made one basic assumption that is my natural convection flow is laminar now the question is what is the critical Grashof number at which my flow is going to transit from laminar to turbulent that is given by critical Grashof number is around 10 to the power of 9 for Grashof number 10 to the power of 9 we get the flow transiting from so this is that is the transitional 10 to the power of 9 so now professor Arun wants me to emphasize it little different way that is if g r by r e l squared is very much less than 1 means what we said forced convection is dominant if it is very much less means what will happen to this term this term will vanish if I make this term vanish what is my equation essentially it is going to be forced convection that is very beautiful way of reducing the natural convection to forced convection so that is how we can understand the physical significance of this Gr by r e l squared so now and of course now you can expect the correlations we are not going to give you the solution in fact similar way as we did for blacious similarity solution is possible you can see this in either incorpora or devet or chengal does not give but incorpora and devet does give the similarity solution so nusselt number you can expect it to be a function of Grashof number and Prandtl number so you may ask where is this Reynolds number gone Reynolds number I do not know the velocity is being generated because of Grashof number only but then why Rayleigh number because in the energy equation I have Prandtl number so I have in the momentum equation I have Grashof number and in the energy equation I have Prandtl number that is why nusselt number is a function of Grashof and Prandtl so nusselt number is a function of Rayleigh number it is typically of the form C r a l to the power of n where r a l is G beta delta t l cube upon alpha nu this is alpha nu that is nu squared into Prandtl number which is what you get as alpha nu film temperature the properties are evaluated at mean film temperature like flow over a flat plate that is T s plus T infinity by 2 so there are plethora plenty of correlations already I have answered several questions actually in the moodle there are several questions what should be the characteristic length if my plate is oriented characteristic length is continues to be l only all that I am going to do is replace this G by G cos theta in case of Rayleigh number definition that is this is the theta that is all that is all the different otherwise you see here I have for a flow over a vertical plate nu equal to 0.59 r a l to the power of 1 by 4 and for 10 to the power of 9 to 10 to the power of 13.1 but r a l to the power of 1 by 3 so that is how for entire range this is the correlation which is there so likewise you can get for hot plate again hot plate which is heated on the top and heated at the bottom you can see that the Rayleigh number the Nusselt numbers for this case is higher than the Nusselt numbers for this case because here the natural convection has to be generated and move up so the plate itself is creating resistance for the flow to take place that is it has to come from here and move out upward but here it will move out if you are not getting it because during coordinators workshop large discussion went into this so if I have a plate which is hot on this side and cold on this side what will happen is the flow would go something like this if it is hot here and cold here then it is no way it can come down it will just move up like this so natural convection currents are generated in different way based on the boundary condition what you have so that is about various correlation for cylinder you have for sphere also you have I do not think we will solve a problem also because it is pretty straight forward okay let me take it on a fast track and finish it off what is this problem telling the problem is telling that I have a 1 pipe of 6 meter long and it is horizontal and the room temperature is at 20 degree Celsius but my outer surface of the term pipe is 70 degree Celsius this is for constant heat constant wall temperature whatever correlations I have given they are all for constant wall temperature for heat flux we will define Rayleigh number little differently and it is a different constant heat flux boundary condition we will handle that little differently so do not have to worry about constant heat flux let us bother about only constant wall temperature if someone is very much interested in constant heat flux put it across in the moodle we will answer that okay so in fact if you can reform Rayleigh number equal to g beta delta t l cube divided by alpha nu delta t has to be replaced from Fourier slough conduction that is q double dash equal to k del t by l that is all for del t you replace as q double dash l by k that is your Rayleigh number new Rayleigh number based on heat flux so there are correlations for Rayleigh number for new Rayleigh number that is Rayleigh number based on constant heat flux if I have gone very fast see Rayleigh number equal to g beta delta t l cube upon alpha nu so q double dash equal to k del t by l from Fourier slough I can write that is g beta l cube upon alpha nu into for del t this is sorry del t so del t is q double dash l upon k so r a h star r a h star for constant heat flux that is the notation Professor Bajan uses r a h star equal to g beta l cube q double dash upon alpha nu k so this is the Rayleigh number based on constant heat flux boundary condition there are standard correlations for this as well yeah this is l to the power of okay so coming back to this problem I have to take the properties at 70 plus 20 by 2 at 45 these are the property beta is 1 by t f that is 1 upon t 18 Kelvin even if you do not make this assumption the properties are available in the charge so if you get to calculate Rayleigh number by putting g beta delta t alpha nu you get it as 1.869 into 10 to the power of 6 that means it is laminar for laminar or turbulent we have identified this correlation which we can use that is this correlation for the entire range this correlation is valid if you go ahead and use that correlation I get a h of 5.869 watts per meter square degree centigrade and apply pi d l that is the surface area and multiply by delta t I am going to get q dot as 443 watts and here remember we have neglected radiation now what if I take radiation into account and taking epsilon as 1 that is a black body epsilon a sigma t s to the power of minus 4 and t infinity to the power of 4 if I take I get 553 watts what was the natural convection heat transfer rate it was 443 the point by taking radiation is that in most of the natural convection problem radiation is significant or dominant if there is natural convection we cannot neglect radiation most of the most of the times so this is what is coming out from this problem so there is another problem I am going to definitely skip this problem I want you to solve it yourself and there is mixed convection here there was one question on the very first day or second day where how do you handle mixed flow we were told we were asked so all that we would say is that mixed flow is taken by the same equation but we have to combine we have to we have to take like almost like of course we are not put the I think I do not yeah it is we will take like it is not going to be 1 plus 1 equal to 2 it is going to be I will handle as if it is forced convection problem separately natural convection problem separately that is for those cases where g r is of the order of r e square so if that is the case I will separately handle force separately handle natural and this appropriate n's are put so that combined heat transfer coefficient I get this is how we handle natural convection coefficient so professor wants to go to the problem and he wants to tell something to us so yeah in this problem if you see the value of h it is around 6 it is a extremely low value so typical numbers that you get in natural convection heat transfer coefficients of this r of this order 5, 10, 15 so on and so forth and if you go back to the thing on radiation see many times we say the temperature difference is very small so we neglect radiation most of the situations here also the temperature difference is hardly 50 degree centigrade but you see in fact radiation is almost equal or if not equal little bit more than natural convection so most of these are coupled radiation and natural convection problems so many times we neglect radiation that means we are severely underestimating the heat removal rate or heat transfer behavior. So I think with this we have come to a big full stop for convection so of course honestly we have gone reasonably fast for external convection and also for internal convection wherever correlation based approach was there essentially because we were told in the co-ordinators workshop that correlation are all well chosen well understood and well applied but what is not understood was boundary layers that is why we have dealt lot of time on boundary layers energy equation momentum equation and continuity now I think for next 15, 20 minutes we will be taking questions and later on we will start off with radiation ok Kolhapur Institute of Technology Kolhapur. So is there any standard procedure to generate the correlations ok I have had at least 5 questions in the model for which I have even sent the regression procedure of my measurements course notes PPT I have uploaded see the point is how do we how do we get the regression procedure ok let me do let me just do a thought experiment now let us say I take a flow in a pipe and let us say for some configuration I have put some ribs or Y I have put some twisted tapes let us say I have put some twisted tape and now I want to get the Nusselt number how do I do the experiment I do I will keep the Reynolds number fixed this is for should convection experiment and I have maintained in the laboratory constant heat flux this is constant and constant heat flux I will maintain and this is the Reynolds number this is the Reynolds number when I do the experiment I will do the experiment for different velocities V 1, V 2, V 3, V 4, V 5 ok Reynolds number is constant that is 0.71 there is no problem now what do we do V 1, V 2, V 3, V 4, V 5 I have got now I have put in each of these cases I would have got Q double dash 1, Q double dash 2, Q double dash 3 I am going very slow intentionally because plenty of questions on plenty of guys have asked the same question so Q double dash 4, Q double dash 5 now at the same time let us say I get average temperature T average 1, T average 2 and T average 3 and T average 4 and T average 5 now I will take bulk temperature of course bulk temperature I will take and this is wall temperature similarly I take bulk mean temperature let us say I compute H 1, H 2, H 3, H 4, H 5 so what do I get I get N U 1, N U 2, N U 3, N U 4, N U 5 for R E 1, R E 2, R E 3, R E 4, R E 5 now let us say this is turbulent flow if it is turbulent flow what will happen is that I know that N U is a function of R E to the power of M and P R because it is what is that turbulent flow we will take it as let us say 0.3 P R is not a function because P R we have not varied so I cannot take that as a function so what will happen I will get N U by P R, P R to the power of 0.3 is equal to C R E to the power of M. Now what will happen let me linearize this equation otherwise it becomes very difficult so let me call this as Y and let me call this Reynolds number as X so what will happen log Y I am taking log Y equal to log C plus M log X so if I this is like my Y and this is like my X and this is like my C Y equal to M X plus C if you do the regression you will get this constant C and M that is how I can now put N U that is how I can give C and M if you put it in excel sheet also it will give you but we should know how the excel sheet works I request you to please see the regression procedure for both polynomial fit and linear fit which I have posted yesterday midnight ok is that ok ma'am. Yes, yes. S G S I T S indoor any questions please. My question is how to measure shear stress practically in a pipe flow at any radius smaller. See the question asked by one of the participants is that in a pipe flow how do you measure the shear stress at any given location what is shear stress near wall it is laminar sub layer we have been telling laminar sub layer and above that it is buffer layer and above that is turbulent boundary layer by that time we would have reached the center of the pipe. Now this is turbulent boundary layer this is buffer layer in the laminar sub layer in the laminar sub layer what is shear stress mu into del U del V by del R in this case mu into del V by del R if I take a pitot tube and measure near the wall I should be able to get the velocity or I by now I have told the hot wire also actually we do not use hot wire we use hot film for what I have here it is ok if you use hot wire or hot film we will be able to get the velocity profile from that velocity profile the shear stress in the laminar sub layer is measured in case of turbulent boundary layer what is shear stress it is minus rho U prime V prime bar. So here U prime is the velocity this is U and this is V usually if you plot the U prime that will be significant U prime squared bar is greater than V prime squared bar which is greater than W prime squared bar similarly U prime V prime bar is greater than U prime W prime bar which is indeed greater than V sorry it is U prime V prime bar which is greater than V prime W prime bar which is greater than U prime W prime ok. So, the point is you have to using the hot wire anemometer you separately measure fluctuating component u prime and v prime and multiply that at every instant and take the average of that you are going to get the turbulent shear stress in the turbulent boundary layer. That is how you can handle both the shear stress in the laminar boundary layer and the turbulent boundary layer that is how we measure any more questions. One more question here my question is related to energy equation my question is can we use this energy equation for stationary fluid also and if yes then what will be the modified energy equation. So, what is the energy equation let us write energy equation for the heck of it. What is energy equation looking like? What is that? I have rho C p u del t by del x plus v del t by v del t by del y equal to k into del square t by del x square plus del square t by del y square. Let us not worry about viscous dissipation pressure term or not. What is u and v? You said stationary what is u and v? u is 0 and this term also gets vanished because v is 0 what am I left out with? I am left out with simple conduction no nothing is there steady state term is 0 if it is assumed that it is steady unsteady term is 0 if it is assumed that steady. So, what is that I am left out with simple free air slope conduction or the heat diffusion equation del square t by del x square plus del square t by del y square equal to 0 if there is no heat generation otherwise energy equation reduces to heat diffusion equation in the absence of no flow that is all it is that ok SVNIT Surat. Hello sir in the case of natural convection in enclosed body most of the correlation involves equivalent thermal conductivity. Yeah most of the correlations in case of natural convection in enclosures involve what is called as equivalent thermal conductivity yeah there is something called equivalent thermal conductivity actually we have not dealt with that if you post that problem whatever specific problem you have we will give, but it is like saying that the equivalent thermal conductivity again comes from the correlation only. So, I have not answered your question. So, what I would request you is please put up this question in the moodle we will answer this question. Hello sir you told me that you told that grace of number upon natural square equal to 1 then both natural and force convection exist is there any physical phenomena exist then tell me ok. See physical phenomena the question is g r by r e squared less than 1 or sorry greater than 1 is natural convection or g r by r e squared of the order of 1 is mixed convection any physical phenomena which you come across now what are the physical phenomena which we come across if it is pure natural convection problem there is no let us say I have I am in an enclosure that is in my room and I have not put on the fan and my room has got exposed to sunlight for and it is a very high temperature very worst day in a in Mumbai and it is in third floor or fourth floor and especially my top floor is like this that is yeah and they are putting black coated in in fact this is what exactly is happening on my house to that paint is put to avoid the seepage, but unfortunately it is black and emissivity is high. So, what is happening the radiative heat transfer is taking place and what is happening because of the natural convection current it is just hot air is just there. So, when I go in the night and put on my fan which is just there somewhere here all the hot air comes down. So, till I put on the fan the mode of the heat transfer is pure natural convection that is one physical. When I put on fan for some time there is natural convection plus horseback convection, but after fan has rotated for let us say 15 20 minutes then thoroughly the heat transfer is modeled by natural convection. So, that sorry by forced convection. So, that is how I look at it that is one of the examples which I can think of Amrita Koyambattur. Sir, we have two doubts sir. Yeah. One is that in the morning you have explained about the temperature variation I mean of the fluid for flow through fluid for a constant heat flux condition and the heat transfer and we have seen that the non-dimensional temperature ratio is a constant. Now, for your constant wall temperature condition what will be the temperature variation once it becomes fully developed. The question is in the morning Professor Arun has taught us for constant wall temperature constant heat flux case the wall temperature increases as we move downstream, but for constant wall temperature what will happen to the profile that is it. Fully you are asking about fully developed right madam yeah when dealing with this definition I had chosen this diagram right this diagram and this definition when I had chosen this one I had put constant wall constant wall heat flux case just to tell you that the fluid is trying to get hot and you can have a bulging of the temperature profile. When I put constant wall temperature this these two lines this one here and the corresponding location for the next profile will be the same. In that case what is going to happen still this expression is going to be valid nothing will change with this expression except that this is going to be a constant T s is a constant T s minus T local divided by T s minus bulk mean temperature ok, but let us go back to our set of nodes where we said I remember this we had told that the slope of the curve becomes equal for both the profiles this is what we had written yeah these slopes became equal in constant wall heat flux case why because minus k dT by d r at r equal to r which is the slope represents the heat flux that is constant. Now if I have constant wall temperature case at this location the slope will be like this at the next location I will just draw so that you are not confused constant wall temperature case. If I take this is the width T s next location I will put the same width because that is my constraint T s, but the fluid has gained energy because it has gotten heated and what will happen this center portion which is there that would have bulged out a little bit more. So, I will get a little bit more fatter profile. So, if I look at the slopes here this slope is different m 2 is different from m 1. So, m 1 is great sorry is m 2 is greater than m 1 indicating d T by d r at wall for 2 is greater than d T by d r at 1 which means heat transfer q 2 is more than q 1. In fact, constant wall temperature is easy to visualize non dimensional temperature d by d it will be like this yeah what is the question yes that is how I am showing it has become fatter here. So, energy content of the fluid has increased as it has moved from here to here it has taken more energy. So, it is bulk mean temperature has increased. So, T s is fixed in both numerator denominator local temperature has adjusted itself to satisfy this relationship while T m x has increased. Any other question Amrita? My question is for constant heat flux h is independent of x. Yes. And we have said we have found out that h is not varying with x I mean x also, but this bulk mean temperature T m is it independent of x. No see we have this graph see this d T m by show pointer show point sorry sorry. Yeah you see this transparency in this what is happening the wall temperature is continuously increasing in the top one and at the same time the bulk fluid temperature is also increasing. Only thing what we are saying is that the slope is constant that means the temperature difference is constant it so happens that T s minus T m is constant. So, what is wrong in this? So, now h how can it be a constant because T m is varying with x is not it sir? No. So, in fully developed flow h equal to q double dash upon T s minus T m h equal to q double dash upon T s minus T m are you with me? Yes sir. Yeah. So, q double dash is constant q double dash is constant T s minus T m is also constant. So, my heat transfer coefficient has to be constant in a fully developed flow. No you are not understood you are asking why is h a constant is that the question your is is why h a constant is that the question? Sir check the definition of h that you have given for a radial system. Here see this is this correct? No in terms of that k into dou T by dou r that equation sir. Yeah this one. That equation. This one. Then there this bulk mean temperature we take it is varying with x is not it? Yes madam what we are saying is d T by d r at r equal to h yes this is the definition of h you are asking this is changing with respect to x the how is h a constant is that correct? Everything is changing with respect to x. So, for that the answer is the what is this? What is k d T by d r at r equal to r? k d T by d r at r equal to r is nothing but the heat flux correct? If my constant heat flux boundary condition is applied the numerator is a constant is that right? Heat flux yeah understand now. And because heat flux is a constant and I am talking of thermally I think you have got it. That is it. Turba college indoor. My question is that for cryogenic fluids for cryogenic fluids how the thermal conductivity and coefficient of h vary? The question is how does the heat transfer coefficient and thermal conductivity for cryogenic fluids vary? Cryogenic fluid is also either a liquid or a gas based on the vaporization temperature or what is the vaporization temperature or the saturation temperature. So, it is there is nothing different except that it is handling low temperature instead of heat transfer most of the times there we are handling cold transfer. So, there is nothing specific or sacred or different for cryogenic fluids except for the fact that the temperatures are low. We have to keep track again on parental numbers and keep track on thermal conductivity that is all. Sir another thing is that suppose this cryogenic fluid having in atmosphere in atmosphere it will operate suddenly. In that case the convective is more as compared to conduction. The question is if the carbon dioxide or liquid nitrogen all of a sudden if I open it to atmosphere the liquid this liquid state fluid will evaporate itself into vapor. So what is happening when it was liquid if it is moving yes convection would have been there if it is not moving then conduction would be there. If it were to be moving or if it is still when open to atmosphere it is getting vaporized. When it is getting vaporized it is moving out so when it is moving out then it is almost like natural convection is not it. So, because I am not pumping it out it is coming out or leaking out by itself by virtue of temperature difference. So, I see here I am seeing the thermal conductivity of carbon dioxide I see the parental number is 0.74 the thermal conductivity of carbon dioxide is 0.74 that means carbon dioxide is going to behave almost like air only. So, there is no difference anything as such except that specific heat is again around that air only thousands thermal conductivity is 0.01 but only thing what I am seeing is that temperature with temperature the thermal conductivity is varying significantly. But the point is carbon dioxide or liquid helium or nitrogen or any other cryogenic fluid is also going to behave like any other Newtonian fluid that is all the answer is. I think this is the fourth question on cryogenic liquids I am answering today. So, let me restate this for any cryogenic liquid all that it depends is whether it is liquid or gas we have to just check out. If it is liquid it will take the liquid properties if it is gas it will take the gas properties otherwise convective heat transfer wise there is no special preference or there is no special antagonism against cryogenic fluids that is it. Irma University any question? The convective heat transfer coefficient decreases in case of laminar and turbulent zone and it increases sharp plane transition zone can you elaborate this? The question asked is from laminar to turbulent there is a jump in heat transfer coefficient in case of external flows or in case of internal flows. So, here why is it that there is a sudden jump in heat transfer coefficient when I am transiting from laminar to turbulent. It is because of turbulence only turbulent heat transfer that is rho C p v prime t prime contribution is significantly larger compared to that of simple del T by del y. So, the turbulent heat flux is very very large compared to that of the heat flux because of the temperature gradient that is the reason why turbulence heat transfer is higher than that of laminar heat transfer. Why it is dropping? It is dropping because the boundary layer thickness is increasing in case of laminar flow boundary layer thickness is increasing you can imagine that like L by k a is that is delta by k a is increasing because delta is increasing because of which the heat transfer coefficient is decreasing. This is true both for laminar and turbulent. There is one more question sir. We proved that that skin friction relation between skin friction coefficient and friction factor, but what is the physical significance of these two terms? Both mean the same. See we showed in the morning that C f equal to tau all upon rho u infinity squared. How did I get f? f is also shear stress only which is computed on the basis of the measured pressure drop. Both mean same C f means shear stress tau all also means shear stress that is what we are telling here. C f is represented as tau all and f is also represented as tau all, but only thing is that in f I do not see tau all directly because I am expressing that in terms of delta p, but what is it? Both are shear stress only both are non dimensionalized shear stress that is all it is. If you want to if you want to nitpick then friction factor represents the non dimensional pressure gradient whereas coefficient of friction represents a non dimensional wall shear stress. If you want to remember it like that you can put it that way. But still both are shear stress both are the resistances offered by my pipe for the flow to upper. Wall shear stress I mean delta p is a manifestation or a measurable manifestation of the wall shear stress that is all it is that is all it is ok. Government call it Salem. How can we ascertain the flow is fully developed thermal flow? What are all the basic conditions? You have stated that d t by d r at wall condition is constant. Is it the only condition or in what are the other ways by which we can say that? So what we usually do the question is how do we ascertain the thermally fully developed condition while we are doing experiments. Typically we do constant heat flux boundary condition why because that is easy to generate constant heat flux boundary condition in the lab. So when I take constant heat flux boundary condition what do I plot I will generate constant heat flux boundary condition and I will plot wall temperature versus wall temperature versus wall temperature I will plot. Whenever my wall temperature variation becomes whenever the wall temperature variation there is a slope change you see there is a slope change this is developing flow and this is fully developing flow. Whenever there is a slope change I know ok here my developed fully developed region is starting this is for constant heat flux boundary condition. For constant wall temperature it is little difficult to I have to take the temperature distribution and measure for constant wall temperature it is little difficult to experimentally figure out but for constant heat flux boundary condition it is straight forward to figure out this is that ok. One more question can you show me the profiles of temperature variation in case of laminar and turbulent. In case of hydrodynamic boundary layers velocity profile is almost it is parabolic in case of hydrodynamic boundary layer velocity profile is parabolic in laminar and almost flattened in turbulent. Basically the question asked is what is the temperature profile in fully developed laminar and fully developed turbulent it will just show the slope. Now we are going to draw the temperature profile for fully developed laminar and fully developed turbulent. See the temperature profile is going to be looking like bulged only all that is going to be different in laminar and turbulent is that the slope in case of laminar will be lower than the slope in case of turbulent. While we are saying turbulent heat flux is higher in case of turbulent compared to that of laminar that is all it is. If this is the slope for laminar and this is for turbulent so that means the slope is very high this is for laminar first one on the left on the right it is turbulent. So that is how the temperature distribution that means turbulent temperature profile will look more bulged or more full or fuller than what you see for laminar flow that is all the difference. Hello sir. Yeah. Sir in the case of flow through the rectangle that one surface is heating all other surfaces are insulated. So how we can find out velocity and temperature profile? Same the question asked is in a rectangular channel or in a square duct if three surfaces are insulated and one surface is either constant heat flux or constant wall temperature boundary condition. How does one get the temperature distribution in the velocity profile? For velocity profile there is no problem it is the same as what we did earlier for constant for the temperature profile we will have to go ahead the way we went ahead for the other cases it need not in fact in most of the real life cases it is not going to be all walls heating or all walls cooling. So it is going to be few walls not being heated and few walls being insulated. If you take a gas turbine blade cooling passage you top and bottom wall are heated and the two side walls are insulated actually that would be the simulation. So the fundamentals are not going to change at all you are still going to apply the bulk mean temperature and you are still going to apply the heat transfer coefficient the way we have defined so far. So there is nothing which is going to be different if the boundary conditions of the any of the walls are changing. In fact we can get the closed form solutions for all of these cases. If you take any of the data handbooks Rosenhoes heat transfer data handbook for all of these cases he has derived and kept. Why forced convection is discussed earlier than natural convection always is there any specific reason? Okay one of the question is why forced convection is discussed always earlier and then dealt with the natural convection. See I think only simple reason what I would think for is that we understand boundary layers thermal and hydrodynamic boundary layer very easily if it is forced convection. In fact we have enough headaches when even when we are handling forced convection itself. Natural convection is little difficult why because I have one more non-dimensional number coming into picture that is Grashof number. Here we have only Reynolds and Prandtl there I have Grashof number which is coming into picture. So it is like going from simple to complex first forced convection is simpler than natural convection that is the reason we go from forced to natural. Another thing yeah it is very obvious why because simple to complex when I say in the forced convection the momentum equation and the energy equations are not coupled. First I can solve the momentum equation get the velocity profile substitute that in the energy equation get the temperature profile. So it is going linearly that is one step at a time. But in natural convection my momentum equation and the energy equation are coupled. So my mathematics becomes that much more difficult. In fact even today we have not solved for natural convection why because coupled equation solving is difficult we have to take similarity approach it is quite difficult that is the reason in the in the keeping the principle that we go from simple to complex we have gone from forced convection to natural convection that is how in all textbooks also it is being dealt with. Hello I am Srinivas speaking I would like to ask one question on this one fully developed flow and fully developed for constant heat flux the Nusselt number is somewhat higher than the constant wall temperature method. So how will you quantify this in terms of constant heat flux as well as for constant wall temperature see this drying point is almost constant inside wall temperature is almost constant it is as good as constant wall temperature how will you quantify in terms of physical phenomena or something any other thing is there. Yeah the question is for constant heat flux boundary condition the heat transfer coefficient that the Nusselt number is the 4.36 and for constant wall temperature it is 3.66 I have answered this already twice the main reason we attribute for these two is that the temperature difference between the fluid and the wall as long as the temperature difference between the fluid and the wall is larger I would get the higher heat transfer that is the higher heat transfer coefficient as we can see in the constant heat flux boundary condition case here we can see that for the constant heat flux case I can see that T s minus T m is constant on the other hand for constant wall temperature the temperature difference is continuously decreasing because of which my overall delta T l m T d if I were to compute the overall delta T l m T d for the constant wall temperature will be lower than that what I would get for constant heat flux. So that is the reason why the heat transfer coefficient for constant heat flux case is higher than that of constant wall temperature over T or put it in other words if I have to remove the same amount of heat in a constant wall temperature case my driving delta T has to be larger. In this clarification is that inside wall temperature see when the fluid is flowing through a cube inside wall temperature is constant is not it is as good as constant wall temperature is it right. No, no, no. See this is. See maintain outside wall temperature you mean okay the question is in constant heat flux boundary condition case inside wall temperature is constant only the outside wall temperature is varying what is that we are talking about we are talking about inside wall temperature only we are not talking about outside wall temperature maybe we are getting carried away because we measure outside wall temperature that is why I said in the morning my thickness of the tube has to be as thin as possible why because even if I measure the outer wall temperature it should be same as the inner wall temperature. So what means conception number one is that inner wall temperature for constant heat flux boundary condition case is constant with x is not correct it is going to increase we have shown that T s is going to increase with T s is increasing with distance what is this T s? T s is inner wall temperature but I am measuring outer wall temperature. So there is in fact between inner wall and outer wall temperature this is conduction which is playing if the conductivity is less and my temperature my thickness is less than inner wall temperature and outer wall temperature will be just the same. Yeah that is what I have drawn here T s o could be higher than T s i but it is not going never going to be a constant these two will be having the same slope as T bulk mean ok. Now our aim is to bring these two curves together so that the resistance associated with the wall is neglected. Yes log R 2 by R 1 upon 2 pi L k is as minimal as possible. So R 2 should come closer to R 1 number 1 and k should be lower higher as possible sorry k should be as lower as possible this should not be conductive. PSG Koyam Bhuturi any questions please. Sir in the free convention also we have a flow of air and up to what velocity we can consider that as a free convection and beyond that we can consider that as a force convection problems. So the classification of free and forced convection is in question now up to what velocity I can consider the convection as free and above which I can consider it as forced. See it is in non-dimensional numbers Grashof by R e squared. So Grashof should be as higher as possible compared to R e squared. Who is generating this R e? R e do I know? R e I do not know R e is being generated because of Grashof that is because of delta T velocity is being generated. If this G R e is of the order of R e then only both forced and natural convection are important. So we cannot talk velocity alone or delta T alone we should be talking this in terms of Grashof with reference to Reynolds number. I think for today we are going to wind up the session.