 as plant liqueur by 2 y by l into 1 minus y by l that is what we have. Now, if you plot plot y by l as the coordinate t minus t naught by t 1 minus t naught as the abscissa plant liqueur becomes a parameter. So, if you plant liqueur equal to 0 means there is no velocity of the plate. So, it is a static thing and then the heat transfer process is purely conduction and please at this point assume t 1 greater than t 0 for our discussion purposes t 1 greater than t 0. So, you will find when you plot this when plant liqueur equal to plant liqueur is the fluid. Accord equal to 0 there is heat transfer from top plate to the bottom plate because t 1 is higher than t naught that you can easily make out that profile will tell you. Now, as you now put in some value for plant liqueur starting greater than 0.5 1 1.5 2 3 4 5 6 something interesting is happening that is why I take up this problem I will just do my bit and go at plant liqueur equal to 2 rather I would say plant liqueur greater than 2 there is something very interesting happening to the temperature profile. Now, when you draw this temperature profile you will take the abscissa as theta from 0 to 1 and of course y by 0 to 1. So, one would like to think that all the profiles would be between this 0 to 1 0 to 1 thing, but when you go beyond plant liqueur equal to 2 suddenly you will see the temperature profile extending outside theta equal to 1. Number 1 number 2 not only that when we say the top plate is at t 1 you know what it means it is held at t 1 it is artificially deliberately held. So, if there is some heat transfer into that you are taking away that heat to hold it that is what it means it would not go beyond that. So, t 1 is held constant the temperature profile is extended beyond 1 that means the temperature there is now higher than the temperature of the plate t 1 what it means is although you are holding the temperature t 1 greater than t 0 you are holding it at certain plant liqueur number meaning at certain u 1 u 1 value the profile extends temperature profile extends beyond the theta 1, but you have to bring that profile back to that one point. Now, you see the gradient has changed its direction originally the gradient was like this for plant liqueur equal to 0 gradually when you increase the plant liqueur it become at 2 it become vertical at y equal to l and when you went beyond plant liqueur equal to 2 the profile is changing that means the direction of the heat flow is changing. So, what does it mean although you are holding t 1 and t 2 constant deliberately t 1 bring greater than t 0 at a value of plant liqueur equal to just beyond 2 heat has now started flowing into the top plate rather than from the top plate to the fluid that means there is a heat reversal has happened heat flow reversal has happened this is something now which is coming out because coming out of our flow because we took the viscous dissipation term into consideration du by dy whole square we did not neglect it we said here du by du by du by may be important du by dx we neglected du by dz we all those we neglected when we have taken a situation the problem where du by dy was is finite and it should be considered you find now a beyond certain value of u 1 heat is started flowing into the upper plate question or not this only if you plot you will understand then you think what is happening to the physics of the problem now the answer is that somehow this increasing velocity increasing a cut number increasing velocity has created has generated heat in the system that heat has caused the fluid temperature to rise that rise in fluid temperature beyond the t 1 value is what is the cause of heat transfer into the t 1 therefore increasing velocities are finally the root of at this this phenomenon of heat generation within the fluid and therefore heat transfer to the top plate this is happening this phenomenon of heat generation within the flow because of u 1 is viscous dissipation actually it is viscous heat dissipation viscous energy dissipation now this is viscous dissipation not simply kinetic energy it is not simply u 1 square it is actually mu if you look at the energy term energy equation mu du by dy whole square mu is always present so it is a combination of the viscosity and the velocity gradient so it depends upon what are the values of these two mu can be extremely high du by dy can be very small viscous dissipation is important when is mu very high for oils so when there is a oil involved even small du by dy can generate enough viscous dissipation on the other hand if you go to low viscosity fluids typically air is extremely small but you are under those conditions if du by dy is very large mu into du by dy whole square will be considerable so for low viscosity fluids high velocity gradients will cause viscous I mean there is heat generation and therefore heat transfer for high viscosity fluids even small gradients will lead to the same effect I am trying to make an important point here it is not simply viscosity that counts it is not simply the velocity gradient it is a combination of these two which brings in a very important physical phenomenon into this system where heat is being generated when heat is generated it has to escape somewhere it tries to go to towards the plates and that can happen only when the temperature of fluid itself is very high higher than these two plates that is what you see when you plot these temperature profiles you can go up to 4, 6 and 8 even time so viscous heat generation has now caused heat flow reversal at the top one number one if there is a heat the temperature profile reversal there should be a maximum for that profile please find out what the maximum temperature is this is all a case T1 greater than T0 only we are talking about what is the maximum temperature and at what location it occurs please find out now this is purely by plotting the profiles you can make out how else can you find out that there is a dT by dy equal to 0 at the top wall please look at the file can you find out by an analysis what is the value of plant lecate equal to 2 if you plot it you will know that but derive an expression for the value of plant lecate equal to 2 when dT by dy is 0 at the wall that is when the change of direction takes place here the temperature profile temperature profile is there Ty differentiate that dT by dy at y equal to L you put 0 and obtain plant lecate equal to 2 that is what I am telling you to do you can get it through the profile if you plot which you have not plotted but at least analytically find out from the T equation at what value of plant lecate is dT by dy equal to 0 3 or 4 steps please obtain plant lecate equal to 2 as this critical number beyond which heat gets transferred into the top wall although T1 is greater than T0 we emphasis this point simply because there is a mindset amongst all of us generally that as long as T1 is greater than T0 it should always flow from T1 to T0 but what this problem tells you is no that is not right your mindset is not right there is what is called as a viscous dissipation that is possible it is possible for low viscosity fluids at high velocity gradients and high viscosity fluids at low velocity gradients heat is generated because of this viscous shear that shear force which is depends upon the kinetic energy is converted into heat that heat is now trying to escape the if the temperature of the fluid goes beyond the T1 temperature of the top plate there is heat transfer into the T1 do not always look at the physical model and say T1 I mean heat transfer will be always from T1 to T0 that is what it says this is the effect of viscous dissipation importance of mu and du by dy always look at mu du by dy not mu alone not du by dy alone please obtain that as partly equal to 2 partly equal to 2 also where does the y maximum occur T maximum occur this is again dt by dy at t0 you have to find out where dt by dy become 0 and please obtain this I am really not going to write anything on the board I am a little unhappy with nothing being done by you you are all intelligent people please do this yourself I will give you the final y max this is where the T max occurs in this case of T1 greater than T0 so take the temperature equation differentiated dt by dy and find out where this is equal to 0 that is the maximum point please obtain this following expression y maximum is equal to I am going to dictate wrote down if you are interested y maximum is equal to k by mu u1 squared again and again this mu u1 squared k combination comes in all this problem mu u1 squared by 2k k 2kl mu u1 squared you can look at the term mu viscosity u1 squared kinetic energy y max equal to k by mu u1 squared into T1 minus T0 this is the imposed temperature difference plus why is this half plus half not plus half it should be plus l by 2 y max equal to k by mu u1 squared into T1 minus T0 plus l by 2 I think l you should get and one thing you can check here if T1 equal to T0 equal temperatures where will the maximum occur so check that here from this expression if T1 equal to T0 y maximum will be l by 2 and find out what that temperature is do that now I will ask a question if T1 equal to T0 equal temperatures occurred equal to 0 there is no heat transfer at all it is entire fluid body is at T0 whatever equal temperature it is a T0 T0 the moment you introduce the plate velocity and depending upon that now heat transfer comes into picture a non heat transfer problem in adiabatic problem you are converting into it it is converting itself into heat transfer problem simply because of the imposed flow that you have created you have not use any pump an imposed flow because of the movement of the plate that you have imposed action. So now there is a viscous generation that heats up the fluid only thing is it is doing it symmetrically because both the plates are at the same temperature the heat that is generated has to go in these two directions you will find a nice symmetric temperature profile y equal to 0 T equal to T0 y equal to l also T equal to T0 you can call it a TL but both of them are equal and there is a parabolic temperature profile with a maximum at l by 2 find out what that maximum in I have asked you to do this also for oils you will find this temperatures could be very high for air it does not matter but this is taken for our lubrication problems T max is the most important physically significant thing coming towards from quiet flow in the case of lubrication films what would be the temperature of the oil film that will attain at various u 1 why is it important higher the temperature for liquids what happens to viscosity for liquids will it increase or decrease for higher temperature decrease that will destroy the lubrication characteristic of that particular so we want to keep the maximum temperature within a certain limit therefore this this comes in so you will find by looking at physical situations you should be able to formulate your own problem when you look at a heat transfer problem you should be able to identify its conduction convection radiation what are therefore the important parameters coming into picture what is the physical model all that you have to do it is not simply that this is given as something very specific what it tells you what you should understand is you look at the heat transfer situation think about it write down the physical model and find out what are the parameters find out how heat is transferred and what variables control the heat transfer now I have also asked you to look at calculate heat transfer at the lower wall both lower wall and upper wall you have the temperature profile okay how do you write down a small expression for heat transfer at the lower wall how do you rather calculate it from the temperature profile qw equal to minus k at have you done that y equal to 0 now can you tell me therefore what is the final expression for q at y equal to 0 is very important after all temperature differences are there thermal conductivity is there L is there this I will give you 3 minutes to do this please I will just give you a hint qw or whatever is equal to k by L into give me what is there in the brackets k by L it should be of the order of k dt by dx so k by L you take give me what you get in the brackets in terms of pranthal number q at the bottom wall equal to k by L multiply by something in the brackets in terms of pranthal number yeah tell me k by L I have taken it outside the bracket in the brackets you tell me what they simplify it here there is no more L if I you have taken L outside the bracket there will be no more L k by L has come up give me the thing in the bracket anybody else whatever is it okay okay there are several ways of how you can put it can you please do this see whether you can put it in this form can you tell me here now what I can write then this is simple conduction equation as you can see k by L t0 minus t1 please see whether you can get this into this form slight manipulation k by L you got it anybody has got it k by L now I hope you have got it now look at this expression this is the q at bottom wall so there is usually one would think k by L into t0 minus t1 but now we know that is possible only when cut number equal to 0 so fine and if t1 is greater than t0 you will get a that directions you will get you know you minus and all that but you see from this expression the driving temperature difference when viscous dissipation exist is not simply t0 minus t1 but it is t0 minus you take this together t1 plus Prandtl u1 square by 2 Cp look at Prandtl u1 square by 2 Cp obviously the units of that term must be temperature it has to be temperature so this is the heat flow at the bottom one mark it leave it we will go to the next step we will come back to this equation later so when t1 is greater than t0 both the plates are at some temperature control the heat flow at the bottom wall equal is equal to k by L into t0 minus bracket t1 plus Prandtl u1 square leave it at that now take up the next problem now you have a quiet flow you have done this hopefully you have done this in your exam you have a quiet flow the bottom plate is insulated get me get me the temperature of the bottom the insulated wall it has some temperature there is some temperature attained by the insulated wall what is the temperature again go to the same temperature equation is the same that this now another thing you will learn here is temperature equation is the same different boundary conditions t1 greater than t0 t1 equal to t0 t0 greater than t1 dt by dy x0 at wall all this give will give you different information and of course answers but more than that information that you can call out of it one we have now seen that viscous dissipation is important beyond a certain Prandtl cut and it has the tendency to make the heat reverse even though the temperature is high we have said that now heat transferred the bottom wall you get you got it as k by L into t0 minus of t1 plus Prandtl u1 square by 2 cp keep at the side take a new case where the same quiet flow bottom flow bottom plate is insulated get me a simple expression for the temperature of the bottom wall y equal to 0 dt by dy is 0 y equal to L t equal to t1 same 2 boundary conditions 2 constants get me the temperature for the bottom wall which is actually dt by dy at y equal to 0 very simple expression you got it tell me t at y equal to 0 is what can you give it in terms of Prandtl number yes little matter whenever you find mu and k you put a cp there make it into Prandtl divide the other thing by multiply divide by cp this is okay no problem it is permitted that is why it should be easier for them the temperature profile y equal to 0 dt by dy is 0 y equal to L t equal to t1 tell me if you have completed if you have made a mistake it does not matter tell me this is okay tell me now any one of you know start with the temperature term here t1 plus Prandtl not rho now what is the problem you have taken quiet flow with y equal to 0 dt by dy equal to 0 when I said dt by dy equal to 0 what is can you use a word to describe that surface it is a dt by dy equal to 0 means what it is a adiabatic it is an insulated surface it is an adiabatic surface that means the temperature t y equal to what you have got is actually t adiabatic wall or t adi whatever so in an insulated condition you have now the temperature now please see it is always higher than t1 we did not specify the temperature as t0 we specified an insulated boundary condition then when you got it actually in terms of the other variables you found the at y equal to 0 which we now call the adiabatic wall temperature not the adiabatic temperature adiabatic wall temperature or insulated wall temperature you found t1 plus something which depends upon u1 of course the fluid also very important leave it now go to the previous case where you found the heat transfer at the bottom wall look at that heat transfer at the bottom wall qw equal to k by l multiplied by t0 minus you are getting this but that was not adiabatic wall that was t y equal to 0 t equal to t0 so for the case of a non adiabatic situation here so actually what now I can write I can actually write it as adiabatic for the non adiabatic case I can still invoke the adiabatic wall temperature concept now although it is not adiabatic I am communicating here the t1 not equal to t0 k is not adiabatic but when you calculate the rate of heat transfer at the bottom wall you are getting a combination of parameters which happen to be give you the adiabatic wall temperature how do I know it is adiabatic wall temperature that is what you did just now putting dT by dy at y equal to 0 equal to 0 you said the temperature that is attained by the insulated wall is this but this happens to be same as this that means ladies and gentlemen even in a non insulating or non adiabatic wall condition now what is this this is the equation for heat transfer the driving force driving temperature difference is not simply the imposed t0 minus t1 but it is a imaginary temperature difference the driving potential which has now adiabatic wall temperature input if you say sir there is no adiabatic wall yes there is no adiabatic wall so this is the concept of the adiabatic wall is coming into the non adiabatic thing meaning the first thing that now when you when we got this expression we did not use the term anyway that is a high velocity flow we have not used it we did not set high velocity flow we said we went on saying Prandtl occurred what is the value of Prandtl occurred the effect of the Prandtl number and the velocity is coming into picture which simply did not say high velocity flow because you have done high velocity flow in force conversion high velocity flow is a du by dy occurred number you said now I did not say high velocity I simply said we will take mu by k du by dy whole square into consideration let us see what happens now in that situation the actual analysis is telling me the heat transfer at the lower wall for a certain Prandtl number for a certain du by dy is not simply k by l into t not minus t1 but it is k by l into t not minus a new temperature which is t1 plus Prandtl u1 square by 2 cp then you found it is adiabatic so what it means is in all such situations and it is true for normal convective situations the there is always a velocity of the flow either the flow is there is a velocity of the flow or in this case there is a velocity of the plane because convection is relative velocity wherever there is relative velocity therefore that is in convection the correct driving temperature potential is this and not simply the temperature difference that you see on paper therefore for all convection cases the first thing that you have to do is calculate taw even if it is not adiabatic case that is the right temperature the moment there is u1 there the question is u1 square by 2 cp is there it may be small large very large that is a different matter but it is there u1 is not 0 the moment u1 has come into the picture where it is a quiet flow and then you go to the boundary layer flow there is a kinetic energy there is a viscosity therefore there is a viscous dissipation the question is is it important how important now when you do not know what it is what you should do is take this expression for calculation if that you have understood the next question is how do I get the ta adiabatic wall temperature go one step ahead now look at this ta adiabatic wall is equal to what is it currently u1 t1 plus t1 plus we will make a small again manipulation here bring t1 to the left hand side so taw minus t1 divided by u1 square by 2 cp can you tell me what it was now that you have done the high velocity flow you have done the high velocity plate Arvind what is this term in high velocity flow now what is it called I will I will put a symbol for this tell me in high velocity flow you got a very similar expression you already knew about ta adiabatic wall temperature earlier actually taw minus t1 divided by u1 square by 2 cp I will give you a hint before I left in the last class I said can you compare the quid flow to something else which you are which you know that is the boundary air flow is not it now what is t1 corresponding to then in boundary air flow what does t1 correspond to and u1 corresponds to so suppose you put taw minus t infinity here u infinity by 2 cp now do you recollect what it was no that is h by rho infinity cp look into the notes if you have I want you to give me the name of this term maybe you are more familiar with that flat plate problem therefore in this I am replacing t1 by t infinity 1 by u infinity that is a boundary air problem it is that the way of looking at it is right but the what you get on the right hand side will be different but what is this term called taw minus t infinity divided by u infinity square by 2 cp please find out u infinity by 2 cp is a temperature term therefore it is non-dimensional you want me to give you a hint beyond that you do not have the notes earlier notes high velocity boundary air flow with air what is it called now you thought recovery factor was only in high velocity boundary air flow now I am giving a recovery factor in a simple quit flow which you have solved by hand complete set of navier stokes equation is not a boundary air flow it is a parallel flow you took the governing energy equation including a term which you did not you did not know what it was but d o by d o whole square you it did not cancel out so you took that into consideration then when you take the temperature profile and put various boundary conditions you are getting so much of information first thing you got was a cut number then we know now there is a heat reversal at the wall even though t1 is greater than the bottom temperature that is because of viscous generation and there is therefore some critical value where viscous heat comes into picture that you found out you should find out plant leg cut equal to 2 by dt by dy at that equal to 0 now you go one step beyond you find that the non-adiabatic case the bottom wall temperature is something which is very similar for which you need a temperature from a insulated boundary condition which is the insulated or adiabatic wall temperature so you take the adiabatic wall temperature term put it into the non-adiabatic thing that is the right driving equation for convection is what I am now trying to tell you and now go if you go further now I asked you t infinity now I will come back to taw minus t1 because it is a quit flow situation there is no infinity here this is called the recovery factor and what is the recovery factor in this case this is equal to from this expression so much comes out of this problem you know why is it called recovery factor recovery factor it is recurring something something is being recovered what is being recovered you can guess do not say heat loss the heat that is generated by viscous dissipation that is being recovered in the case of the adiabatic wall where is it being recovered if you see it is going to the by the way what is the max I will just go back what is the maximum temperature in the adiabatic case and where does it occur I think I have asked you this question where is the adiabatic temperature maximum temperature where does it occur you know the t1 t0 profile you have drawn so max t1 t0 is held constant so nothing can happen the maximum temperature occurred in the fluid t1 equal to t0 same thing tmax occurs in the fluid at y equal to l by 2 in the adiabatic case which is the maximum temperature where yeah yeah at the wall itself yeah so draw the profile at y equal to say that is the maximum temperature how it has come about some of the heat that is generated is being recovered at the plate this may or may not be the very right physical interpretation but this gives you an idea the heat that was generated by friction has been recovered by the system itself it has not let it go as a loss if it is a loss still the insulated surface would have some temperature maybe it is slightly lower but when it has recovered taken the heat and supplied part of it in insulated case there is no heat loss from the plate dT by dY is there but at that point whatever it has absorbed it has raised its temperature so in an insulated boundary conditions case insulated plate temperature is the highest and therefore you can draw the profile I want all of you to draw the quet flow draw the profiles for t1 greater than t0 case t1 equal to t0 case and the adiabatic case see the temperature profiles how they change okay now you go beyond that so we are saying here the frictional heat is being recovered therefore it is a recovery factor now we go to the next step so what is recovery factor equal to can you tell me from your notes what was the recovery factor for that particular case that you so what is the value of recovery factor for the high velocity laminar boundary layer flow whatever you have done laminar or turbulent laminar I think that r value what is it you got the r somewhere there in your notes so r there should be a relationship r equal to what suppose I want to calculate you got tw minus t infinity divided by infinity by 2cp equal to r that r equal to what for that particular situation I am sure you have that value there your notes what will in terms of what you will find the value of r at least no you must have you must have you got the definition of r in your class notes cover I should not be asking you so many questions here you said recovery factor fine so there is an expression for recovery factor I am asking you what is the value of that in that particular case the laminar flow flat plate case find out it should be there we will go further I wanted to bring the point here that r is a function of Prandtl number only in terms of number but it changes from situation to situation in terms of the geometry and the flow so a flat plate laminar problem has a certain recovery factor is a function of Prandtl number a laminar sorry a flat plate turbulent flow has its own recovery factor which is a function of Prandtl number a quiet flow we did not specify the fluid here but fluid is the Prandtl number so for the quiet flow the recovery factor it does not matter whether it is adiabatic or non that is not the point at all recovery factor value is equal to Prandtl so if you know the Prandtl number half the solution is not now I do not know whether you have missed for laminar boundary layer it should be a Prandtl power half or something laminar high velocity no no you have missed it you should be there somewhere there should be square root of Prandtl number this is very important for us to know that the recovery factor has a value which is some factor I do not want to say function yeah okay function of Prandtl number Prandtl power one half root one third or whatever in the case of quiet flow it is equal to Prandtl number that is all for high velocity laminar boundary flow with air it is I think Prandtl power half you please check for turbulent fluid Prandtl one third I think please check it out this is very important two aspects one you should know that recovery factor is a function of Prandtl number and also how much it is because one is a physical signal query now so tell me when you want to solve a problem for your how does this help you this expression what is it that you know what is it that you do not know in this expression this is expression to calculate the adiabatic wall temperature because you always know the fluid is that air for us you know air oil it could be water it does not matter at all so the moment you know what is the fluid of course you know the plate velocity this is given plate velocity plate temperature Taw is known the moment Taw is known to go to the heat transfer equation do not simply put T not minus T1 that is what I am trying to convey to you same thing in the case of laminar boundary air flow usually what is the Newton's law of cooling q is equal to H A delta T delta T is Tw minus T infinity I am saying do not do that take Tw minus T adiabatic you do not have to worry now T adiabatic has u1 square all it simply means is if u1 is small this is negligible if u1 is considerable that will be important it could be a rise of 0.001 degree centigrade when u1 is small but it is there you may neglect it but it can never be 0 viscous dissipation is 0 only when u1 equal to 0 but then we do not talk about convection at all in this case so the moment there is a u1 that is coming into picture fluid velocity there is kinetic energy there is a viscosity there is a velocity gradient therefore mu du by dy whole square becomes important that comes from Taw w equal to mu du by dy so viscous shear leads to heat generation heat generation heats of the fluid and the temperatures in the fluid will be slightly higher than what actually the T infinity but it may be negligible if T infinity is 150 degree centigrade and the viscous dissipation has increased at 0.1 degree you do not care but if T1 is 150 degree centigrade and viscous dissipation has increased by 40 degree centigrade now on what does that depend either 0.1 or 40 two things Prandtl number the fluid and the velocity gradient how a cut number Prandtl a cut number so when you want to calculate the rate of heat transfer in convection in any situation this is an example the same thing you can do in the laminar here it very clearly says do not worry about what u1 and T1 are there get the adiabatic wall temperature use this expression T0 is actually Tw in a way our our old thing and immediately you have the heat transfer coming into question so we do this squared flow generally in the beginning of the laminar flows we never said laminar we did not say turbulent because it is all for laminar we wrote the expressions for laminar we do this therefore to con to first of all understand the procedure of solution of a convective heat transfer problem starting from the physical model I will never I say this very often you have to start with the physical model ad nausea as I say physical model mathematical model simplification of the equations if they are very complex and neighbors equations are complex then take the simplified version so in a boundary air flow it can be Prandtl equations but in this case there is no name given to those 4 equations you know D w d square u by d y what name there is no name but you please look at them it is like in conduction what is the simplest expression for conduction d square d by d y square is equal to 0 you cannot get anything simpler than that with two boundary conditions similarly you have here d square u by d y square is equal to 0 d w by d y square is equal to 0 why momentum has taken care of taken care energy equation we wrote everything we said there is no heat generation so we did not say there is no viscous dissipation purposely so that we wanted to study because it has d u by d y now in this after that you have to solve you do not have to rush to the computer big fluent code you can do it by hand this is the only problem and the corresponding Pozolet-Phozo problem which you can do by hand with three different boundary conditions T 1 greater than 0 T 1 equal to T 0 I want you to do that and a boundary condition at the insulated boundary condition you get different temperature profiles but importantly you find lot of physics the effect of viscous dissipation the possibility that the heat transfer can go to a temperature which is higher than the other temperature in the system simply because the fluid has increased temperature a practical application in lubrications or anywhere where oil is involved as a lubricant it loses its viscosity at this higher temperature that is important thing so you have to find out what is it that is why the gaps are given as 3 mm 4 mm you know otherwise you know it is not it is not a duct kind of a thing and then you come the new non-dimensional term which is a cut number which you never thought you do that you did not say it is a high velocity problem but it came in a cut number then when you temperature profiles were drawn you got the heat reversal then you go to the adiabatic wall temperature there you find you can get the adiabatic wall temperature expression but that is exactly is what appears in the non-adiabatic case therefore in the non-adiabatic case even for you to calculate heat transfer first calculate the adiabatic wall temperature that can be obtained through the definition of a recurring factor which is nothing but Tiw minus T1 by Q1 square by 2 Cp which is a recurring factor in this particular case it has a definite value which is equal to the Prandtl number if it is oil it is 1000 if it is there it is 0.72 and immediately you know the effects will not be very high in the case of air but it will be extremely high in the case of oil there itself the Prandtl number effect is simply hitting you in the I actually and therefore when you want to find out the rate of heat transfer in any convection problem you say right thing to do fundamentally calculate the adiabatic wall temperature which is non-existent which is imaginary thing it is as if there is a adiabatic wall temperature there is no adiabatic wall temperature in your normal our cases this is very important for you to understand because students will always say sir why do we calculate the adiabatic wall temperature there is no it simply so happens physics tells you when you have this non-adiabatic equation there is a term which is nothing but the adiabatic wall temperature so you say there is a imaginary adiabatic wall temperature imaginary temperature which is equally adiabatic wall once you put that in that is the right equation for heat transfer and that is a function of Prandtl number which comes to a recovery factor recovery simply means the frictional heat that is being generated is being required as which is possible within the system that is all it means so for you to solve this problem for today heat transfer in all convection system get the adiabatic wall temperature through the recovery factor which is simply a function of Prandtl number we can actually plot all this you know as a function of Prandtl and R for various flows and go back and calculate the adiabatic wall all these new ideas physics of the problem how to solve the convection equation how to put in the boundary conditions actually solve now if you take a complex Navier stokes equation you will not do this you will go to a code that you will not know how it is done you have now integrated every equation the temperature equation twice every time to get the two constants depending upon the two boundary conditions so the whole thing about convection is encapsulated in the squared problem if you know this problem completely you can play with any equation even if you go to the code you at least you know what is happening how did this come out how did the rate of heat transfer calculated q is equal to minus k dT by dy that is why in the beginning stage I said physics wise I want for my heat transfer dT by dy at y equal to 0 for me to get that I have to have the entire temperature profile for you to have the temperature profile you have to have the energy equation all with the boundary conditions and that comes from nature which is the first of thermodynamics so from nature to rate of heat transfer in an adiabatic or a non adiabatic situation you can understand provided you actually do the calculations by hand not by a computer put plant record various values plot it earlier we used to have these graph sheets on which we wish to plot and as the plot we are plotting the whole thing comes visually though you do not have to accept what I say about the computer says you do it by hand this is the only problem which you can do by hand and see the solution as I say coming out because of you you are churning all the information mathematical information physics information you integrate these together to calculate the rate of heat