 So let us today continue on the discussion with respect to the fully developed both hydrodynamically and thermally fully developed region so I denote this as region 3 we solved for the velocity profile in the hydrodynamically fully developed region we got a parabolic velocity profile and we also started the solution to the heat transfer problem the first case that we took was a constant wall flux boundary condition so Q wall double prime is constant and for this particular case we have also shown that if your Q wall is constant that means your DT wall by DX should be equal to DTM by DX this comes because of the fact that this is equal to H into you all minus Tm okay also we have expanded the fact for thermally fully developed flows you have the non-dimensional temperature gradient variation axially is zero so from which we had obtained an expression for DT by DX in terms of DT wall by DX and DTM by DX that was DT wall by DX minus T minus T wall by Tm minus T wall into DT wall by DX plus T minus T wall by Tm minus T wall DTM DT mean by DX correct so you can note the partial and the standard derivative that I am using here so this is a function of 1 DX this is a function of only X whereas temperature is a function of both R and X therefore I use a partial derivative for DT by DX okay so now if you substitute this result okay if you call this as number 2 and this is number 1 substituting the consequence 2 x 1 you cancel of this terms because they are identical and therefore you come to the conclusion that my DT by DX should be equal to DT wall by DX which in turn is equal to DTM by DX and we have seen that this can be possible only if these are equal to a constant because this is a function of both R and X but these are only a function of X so this can be holding true the slopes can be the same if this is equal to a constant and therefore if you plot the variation of temperature axially so you can visualize for example a temperature profile so where you have T wall X here you can calculate T mean X and of course this is your T X, R so if you plot your temperature somewhere here local temperature along the axial direction so you can say that this follows a profile linear profile like this may be somewhere at this radial location okay so this is the center and the mean profile will be slightly lower than the temperature of the at this particular radial location so you can say that this is your so this is at some particular radial location we can say R equal to R1 for example okay that is this particular location where I am plotting this R equal to R and at the wall of course that will be the highest temperature amongst all the three and that will also vary linearly like this okay so this is the characteristic of the temperature variation as far as the constant heat flux boundary condition is concerned so all the three variations are identical they are just parallel to each other but they keep varying with the axial position so that is the characteristic nature of the fully thermally fully developed of course they have to be hydrodynamically fully developed also now having seen this we wanted to calculate actually the temperature profile itself so if you substitute the fact that your DT by DX is actually a constant that is equal to some DT wall or DT wall by DX or DTM by DX into the energy equation okay so therefore it becomes much easier to integrate the energy equation directly so you have U by ? now the DT by DX can be replaced as some DTM by DX now which is a constant equal to T square T by DX square plus okay so we have neglected the axial conduction in comparison to the magnitude of the radial conduction so therefore will retain only the radial conduction terms are DT by DR okay so now since the left hand side is a constant we can just integrate it out the same way we integrated the velocity profile where your DP by DX was a constant so this is a much simpler way of looking at the solution so if you integrate it twice I am not going to spend time integrating I will give you the final expression 1 by a DT by DX which is actually a constant so I can either use DT by DX or DTM by DX they just mean the same and they are they are just constants so you have UC R square by 4-R power 4 by 16 R0 to the power 2 plus C1 ln of R C2 so this is what I get somewhat similar to the velocity profile where you had a pressure graded in term in the place of this DT by DX you had DP by DX there and you had a similar variation okay so now we have to find the constants C1 and C2 and in order to do that we have to apply boundary conditions once again two boundary conditions are required okay so what are the two boundary condition okay so at R equal to R0 let us say the temperature is equal to some T wall of course which is a function of X okay so if you want to apply at a particular axial location so that there is a particular wall temperature which is a boundary condition and at R equal to 0 the temperature has to be finite now if you look at this directly similar to the velocity profile if for a finite temperature at the center your C1 should directly be 0 because this term otherwise goes to infinity okay so therefore we directly eliminated C1 so we can use the other boundary condition R equal to R0 T is equal to T wall to calculate the second constant C2 so if you substitute you can determine the constant it will come out as so at R equal to 0 R equal to R0 T will become T wall this will be T wall of X minus this entire thing goes to the left hand side this is UC by A into so now this is evaluated at R0 so this is R0 square by 4 minus R0 square by 16 okay so if you simplify it it comes out as so you can take one R0 square into DTM by DX okay I can maintain this is DTM by DX into 3 R0 square by 16 I think can you just check if this is correct this is T wall minus UC by A R0 square okay this R0 square cannot come here I think this R0 square should not come here right so it is just 3 R0 square by 16 okay so this is your C2 and therefore you can substitute for C1 and C2 and write your final expression if C1 is anyway 0 therefore T of R, X so you can substitute for C2 and you see UC by A DTM by DX so this is a common term here and here so that can be just taken out as a common term so of course you have your T wall of X which can be written as this minus UC into we can also pull out R0 square as common by A so UC you can take out one R0 square so that everything can be expressed as non dimensional form and R0 square also is here into DTM by DX that is also common so this will be left with 3 by 16 which is this plus you have 1 by 16 so you have already R0 square taken out so this will be 1 by 16 R by R0 the whole power 4 okay and this remaining term will be R by R0 1 by 4 this will be minus 1 by 4 R by R0 the whole power 2 okay so this will be a final expression for this let us call this as equation number 3 so therefore you can see at this state it appeared like T is a function of only R because your DTM by DX was constant but once you calculated this constant C2 now it is a function of X through T wall therefore the final expression for T is a function of both R and X all right okay so once you got the expression for temperature we will go ahead and calculate the expression for the mean temperature which we defined or the bulk temperature or the mixing of temperature so how was the mean temperature defined so we now know the variation of T with respect to both X and R now we should get an expression for the mean temperature varying with X okay and of course the wall temperature or T minus Tm minus T wall okay the difference between them so how do we define the mean temperature yeah so it is basically a mass weighted average of temperature and how do you mass weight 0 to R0 T into so what is the mass ? x A x U okay in the numerator and denominator the row can cancel off so you have U and of course U is a function of only R okay and what is the differential area 2 ? RDR 2 ? cancels in numerator and denominator you have RDR in the denominator it has to be divided by the mass flow rate okay so 0 to R0 you have U of R into RDR okay so therefore you have the velocity profile I will just write down the velocity profile for you again so you are U of R is 1 by ? okay – 1 by 4 ? into DP by DX R0 square into 1 – R by R0 a whole square so this is your velocity profile and this is your temperature profile right here so you have to plug both these inside and you know so this DTM by DX is a constant DP by DX is a constant okay so therefore you will have to integrate this with respect to R you have to multiply all the terms with respect to R and then you have to integrate it it is a little bit lengthy integration which I am not doing you can also if you find it too difficult you can also use the Mathematica and try to do the integral but finally if you do that the expression for TM comes out as T wall of X – 11 by 48 into UC R0 square by 2 a into DTM by DX okay so this let us call this as number 4 this is the variation of mean temperature with respect to X okay so therefore you can see your TM – T wall is going to be a constant because DTM by DX is a constant okay so the difference between the two TM – T wall has to be a constant alright so once you got your TM now we can take calculate the non-dimensional temperature so your non-dimensional temperature is defined as T – T wall of X by TM of X – T wall of X okay so now you have of course T – T wall an expression for T – T wall here from the temperature profile and also an expression for TM – T wall here okay so if you just substitute those expressions so T – T wall will be so you have your UC R 0 square by A into DTM by DX into 3 by 16 plus 1 by 16 R by R 0 to the power 4 – 1 by 4 R by R 0 square that is basically T – TW there is a minus sign but in the numerator and denominator the minus sign cancels off similarly TM – T wall will be minus of that I cancel the minus sign this will be 11 by 48 into UC R 0 square by 2 alpha into DTM by DX so DTM by DX cancels your UC R 0 square cancels alpha cancels out okay so this will give you an expression as 96 by 11 because you have 48 into 2 so that goes up 96 by 11 into this entire factor here 3 by 16 plus 1 by 16 R by R 0 to the power 4 – 1 by 4 R by R 0 square so this is your final expression for ? now once you calculated your non-dimensional profile now you can see the non-dimensional profile is not a function of X now okay the way we defined your ? in the thermally fully developed region ? is supposed to be independent of X so and that is what comes out correctly okay so this proves the fact that the assumption what we did of course we use that assumption also but that correlates with what we find finally for ? so this is only a function of R now you can use this definition of ? and calculate your heat transfer coefficient we had shown that in the case of thermally and hydrodynamically fully developed flows the heat transfer coefficient has to be a constant now we have to determine what is that constant value so for the constant heat flux case we have arrived at the particular profile for non-dimensional temperature so I will probably give you about 5 minutes you can from this step calculate what is the heat transfer coefficient I suggest all of you to do it yourself and check okay so therefore finally you get your heat transfer coefficient as a constant value for a given diameter K by D into 48 by 11 therefore you can define a Nusselt number okay for internal flows based on the diameter of the duct so it is not a local Nusselt number anymore and if you define like that you get a constant value 4.36 so for internal laminar flows both hydrodynamically and fully developed and for a constant heat flux boundary condition this is your constant value of Nusselt number 4.36 okay so that therefore you see that internal flows are quite different there is no local variation once you reach a completely fully developed condition and after that the heat transfer coefficient is becoming a constant value so even in your external flows one of the quiz problems I had given you the porous flat plate where your suction is continuously happening on the surface and you are asked to prove at large values of X your boundary layer thickness is becoming a constant and therefore as a consequence of this the heat transfer coefficient also becomes a constant so even in external flows if you maintain suction boundary condition you can show that your heat transfer coefficient can become a constant somewhere down the line in the case of internal flows that is true if you have a fully developed flow region okay and now so this is a straight forward case as far as relatively straight forward when you look at in terms of the simplification that we have done for the constant heat flux now let us look at the other boundary condition which is your what constant wall temperature so only these two boundary conditions are the primary boundary condition that we will see one is a Dirichlet boundary condition the other is an Hyman boundary condition okay now when you look at constant wall temperature what this means your DT wall by DX is 0 correct so now let us go back to our equation for DT by DX and substitute this expression what do you get so if you substitute DT wall by DX is 0 in that expression yeah correct this will be partial derivative you are right so I want all of you to also participate you have to tell me if you apply this condition DT by DT wall by DX is 0 what will be the resulting expression for DT by DX yeah T minus T wall by T M minus T wall into DT M by DX okay so the other first two terms get cancelled off you have only this last term so now this is your approximation for DT by DX right so of course your mean temperature is only a function of X but this T here this is a problem this is a function of both X and R so now if you substitute this into the energy equation now what happens so you in your energy equation on the left hand side you have your DT by DX so if you substitute in place of that you have now U by alpha into DT by DX I am going to substitute this particular expression here T minus T wall by T minus T wall into DT M by DX so this is equal to of course your 1 by R D by DR into R DT so now this is what you need to solve let me let me call this as another equation number one okay so this is the energy equation which you need to solve to find the temperature profile now the problem with this now earlier when you had constant heat flux the left hand side was completely a constant okay you had only DT M by DX which again was a constant therefore you could directly integrate it out now you have T minus T wall by T M minus T wall now T is actually not a constant now so this entire expression now has cannot be solved analytically the way that we did last case so therefore we have to do it numerically and also iteratively so one way of doing it is you can guess some value of temperature profile preferably from the earlier case the constant heat flux case and substitute as a first guess and then you can integrate it along the radial direction okay and again get a new temperature profile and again keep putting it on the left hand side and keep doing it until you reach a converged solution okay so the other smarter way of doing it is again go back to our shooting technique so we should try to reduce this equation to another simpler form where we can apply shooting technique and solve that equation so that is what we are going to do now okay what I am going to do now first let me before going into the shooting technique we have to prepare we have to get appropriate expressions for DTM by DX so to do that we will do a small energy balance so let us look at the profiles of temperature T as a function of X of course your wall temperature is now a constant okay now we want to know how the mean temperature is varying earlier wall temperature mean temperature and temperature at any location where just straight lines with equal slopes so now it is not the case right so there from the Newton's law of cooling you directly showed that the slopes have to be the same but here it is not true therefore in this case the wall temperature is of course a constant but how the local temperature how the mean temperature vary is not clear so to determine the variation what we will do is just take a simple energy balance so you take a duct so you take a control volume so this is your heat transfer ? Q differential amount of heat transfer for this differential control volume and you have some enthalpy coming in an enthalpy which is leaving of course there is a mass flow associated with this enthalpy so therefore you can apply energy balance and say my del Q dot is equal to M dot CP into DTM okay so that is del Q is equal to M dot into ? H DH and DH corresponds to CP into DTM because this enthalpy is the mean enthalpy or bulk mean or mean enthalpy that I am talking about so far the given heat transfer this is the corresponding change in the enthalpy of the fluid and I can relate my differential amount of heat input to the change in the mean temperature corresponding to the change in the enthalpy so this is the starting point of the energy balance so I can just divide everywhere by DX so this will be now DTM by DX will be equal to now this is your heat transfer rate okay so this can be expressed in terms of heat flux you can write this as Q double prime which is in watt per meter square into if you assume that the control volume has a differential length DX into DX into perimeter right so this will be your surface area where you are adding heat alright so this will be divided by M dot CP and of course you are dividing by DX so DX DX cancels and now your Q double prime you can apply Newton's law of cooling and write this as H into T wall minus T mean okay therefore you have DTM by DX is equal to P by M dot CP P is your perimeter into H into T wall minus T mean so this is this is the expression coming out of the simple energy balance that you are doing so we can what we can now do since your T wall is a constant we can write this as D by DX of T wall minus TM correct T wall is a constant so you can just introduce D by DX of T wall minus TM okay now there will be a minus sign so we can put a minus sign on this side also P by M dot CP into H into T wall minus TM is already here okay so let us call T wall minus TM as some delta T okay so with this you can integrate this expression so this will be D ? T by ? T you can next integrate this from ? T1 to ? T2 so what it means if you have a duck long duck at the inlet that will be your ? T1 that that is the difference between your wall temperature and the mean temperature and at the exit somewhere that is your ? T2 okay so once I have an equation for this temperature difference T wall minus T mean I can just simply integrate it from the inlet to the outlet so that this should be equal to minus P by M dot CP into HDX so this I integrate from 0 to your entire length or wherever whichever location there you want to find the appropriate temperature difference okay so this will give me Lawn of ? T2 by ? T1 equal to minus P by M dot CP H into X okay so or my T wall minus T mean at some location 2 by T wall minus T mean 1 will be equal to exponential of minus P HX by M dot CP so this is my expression which tells me how the mean temperature varies because my wall temperature is constant if you know the mean temperature at some inlet from that you can calculate what is the mean temperature at some location X using this expression so if you plot this expression okay if you start with some mean temperature at the inlet you will find that the mean temperature keeps varying in a logarithmic fashion like this and it will asymptotically go and meet this T wall when your X goes to infinity so this becomes 0 and therefore your T mean will become T wall where your X goes to infinity okay so this is the variation of your TM okay the mean mean temperature of the fluid okay now having known this and we have this particular expression here let the let us call this as expression number 4 okay I started with 1 okay let me call this as 2 so we will use this expression now simplify for DTM by DX because why we are doing this is we have a DTM by DX term in this equation so we have to simplify this a little bit so for that we are using the energy balance now we will simplify it also we should know how the mean temperature profile varies so for that we can use that equation and integrate it out so now from 2 so you can write your DTM by DX as okay for the case of circular duct your perimeter is 2 pi R so pi D divided by your mass flow rate will be rho into pi by 4 D square rho AV so the velocity is mean velocity into CP into H T M so H T wall minus T mean I can replace this by your heat flux Q double prime okay so this cancels pi cancels here D cancels so this will give me your Q double prime will be K DT by DR at R equal to R0 this will be divided by your row into U M into D will be R0 by 2 and I can write this as R0 by 2 into CP this will be 2 R0 yes so that is correct so now I can club this as K by rho CP K by rho CP is a so this can be written as DTM by DX is actually a times DT by DR at R equal to R0 divided by this will be 2 times so U M R0 so this is my expression for DTM by DX in terms of DT by DR at R equal to R0 this is a little bit of mathematical manipulation nothing more so as far as circular duct is concerned you are just writing in terms of your diameter and of course your mass flow rate little bit simplification and you get a relationship between DTM by DX and DT by DR at R equal to this is the gradient of temperature at the what so this can be substituted into equation one for DTM by DX so substituting into one so of course on the right hand side you have the same on the left hand side you have so I am going to write my I am going to substitute my velocity profile fully developed velocity profile for you so that will be twice U M into 1 minus R by R0 the whole square by A and DTM by DX is substituted from here so which will be now this is T minus T wall by TM minus T wall DTM by DX will be 2 alpha by U M R0 into DT by DR at R equal to R0 so this is equal to the right hand side which is 1 by R D by DR of R ET by DR okay so here U M U M cancels alpha cancels here okay so therefore your final expression which you have to solve will be 1 by R D by DR of R DT by DR that should be equal to 4 by R0 2 into 2 4 divided by R0 into 1 minus R by R0 the whole square 1 by R by R0 the whole square into T minus T wall by TM minus T wall into DT by DR at R equal to R0 okay so let us call this as my equation number 3 so this is the equation finally which comes to the form which I want to express as a ODE in order to apply my shooting technique okay so you please stop and ask me if you have any questions any doubts anywhere so these are all just mathematical manipulations I am going a little bit fast here so now I am going to introduce a non-dimensional temperature ? or ? I am using the ? here to represent the non-dimensional temperature I can write it as T minus T wall by T mean minus T wall and a non-dimensional radial coordinate R star which is nothing but R by R0 so this I am going to substitute into my equation number 3 and rewrite the entire equation in terms of non-dimensional variables P and R star okay so if I do that so from this expression I get that my DT by DR is equal to TM minus T wall into D5 by DR so I can substitute for D by DR as TM minus T wall into D5 by DR TM minus T wall cancels here okay and I can also define my Nusselt number as DT by DR at R equal to R0 divided by T wall minus TM into diameter so diameter is nothing but twice R0 so if I use these expressions and substitute into this governing equation 3 I request all of you to do that and tell me what will be the final non-dimensional non-dimensional equation for ? in terms of ? and R star so with that we will stop for today so what do I get so I have now I can write this was 1 by R star D by DR star R star now for DT by DR I can substitute as TM minus T wall TM minus T wall is a function of only X so that can be taken outside the derivative right so I have TM minus TR into D5 by DR star this should be equal to 4 by R0 into 1 minus this is R by R0 is R star so this I can write as R star square T minus T wall by TM minus T wall this is ? so this will be times ? and DT by DR is nothing but again TM minus T wall into D5 by DR so this I can write as TM minus TR T wall into D5 by DR at R equal to R0 so now so I have RR this cancels here so I have R0 into R0 R0 square so I have divided by R0 square here and here I can have again R0 and I can write this as R star so R0 square R0 square will cancel TM minus TR TM minus TR cancels so therefore I am left with the final non-dimensional form D by DR star into R star D5 by DR star which is equal to now I have got the expression for Nusselt number as minus 2 R0 into D5 by DR so I can replace D5 by DR R by R0 as Nusselt number by 2 R0 so this will be a minus 4 into 1 minus R star square into ? into this can be written as Nusselt number by 2 R0 so there is a factor of R0 which is somewhere I have to cancel so this is this is R0 R0 here cancels this is R0 square and on this side oh okay I think yeah so this is anyway I can write this as D5 by DR star you are right so this is already D5 by DR star so therefore this will be in terms of Nusselt number okay so this is my final expression you see this has reduced to an ODE now I will call this as number 4 so this entire expression is now a function of ? now remember ? is a function of only R so earlier I had a partial differential equation now I had reduced that somehow by any means of manipulation and non-dimensionalization to an ODE which is a function of R and this is of course an higher order ODE I can break it up into two first order ODE is and you shooting method so we will see that in the next class okay I will just give a summary of the shooting method which you are already used to to solve this equation okay.