 Okay so let us quickly go through what we covered last time so we were looking at using integral methods for the thermal entry length region and here we confine this integral method for internal flows to only Cartesian coordinate system so since it is difficult to integrate your energy equation in cylindrical coordinate system and apply the integral technique so most of the discussion related to integral method for internal flows is confined to plane ducts so that is basically flow between two parallel plates so we will take up one such example where we have a thermally developing region that is basically region to where we have the fully developed velocity profile the parabolic velocity profile and we have applied the constant wall flux boundary condition so when you write the velocity profile basically you use a coordinate system which is aligned to the center of the channel okay so that we have to transform to a coordinate system which is starting from the plate because the integral method is basically integrated from the plate till the edge of the boundary layer so that is where we transform the coordinates and then this is the velocity profile in the transformed coordinate okay and of course if you integrate the energy equation very straightforward integration you directly get the energy integral which is the equation number one so all you need to do is know the velocity profile and the guess some value guess some profile for the temperature the velocity profile here is coming from the exact solution because you cannot again use an integral method solution in the fully developed region okay so once the boundary layers merge then the integral method cannot be applied because these are all boundary layer equations so therefore we use the fully developed velocity profile from the analytical solution but for the temperature profile in the developing region we just make a guess for a cubic polynomial as usual and these are the boundary conditions is specified wall flux at y equal to 0 and at y equal to ? T the gradient is 0 and the temperature at the edge of the boundary layer is equal to the inlet temperature okay so from here we get if you substitute all the boundary conditions you can end up calculating all the coefficients and this is the resulting cubic profile for temperature that you get so the temperature profile is a function of your wall flux your thermal boundary layer thickness and of course the non-dimensional y coordinate so we will introduce certain non-dimensional variables here your ? is your non-dimensional y coordinate same way this is the same nomenclature I am using similar to your external boundary layer flows so y by ? T and similar to your external flows where you used non-dimensional Zeta for ? T by ? here we do not have any boundary layer the momentum boundary layer thickness because we are considering already fully developed flows therefore instead of ? we use D here which is the the half separation distance okay and we substitute this profile into your momentum integral you integrate it out and we neglect all the higher order terms of Zeta since you are looking at the thermal entry length you are supposed to be looking at region where your ? T is much smaller than ? okay so therefore since your ? is less than your ? is your Zeta is less than 1 we neglect all the higher order powers of Zeta of the order of quadratic powers and higher so if you do that you get a direct expression for the Zeta which is nothing but your non-dimensional thermal boundary layer thickness okay so from here we have to go on and get the expression for the heat transfer coefficient so we define our heat transfer coefficient H of X which is varying locally in the thermally developing region as Q X divided by T wall minus T M so this is our definition in internal flows so now we need to get the denominator which is T wall minus T M okay so we can get T wall minus T I from the profile that we have this is our profile let me call this as equation number 2 so T wall minus T I here is at Y equal to 0 okay and we also need to know what is my T wall what is my T mean minus T I so once I know T wall minus T I T mean minus T I take the difference I get T wall minus T L okay and that will give me the final expression for heat transfer coefficient so now we have to determine what is the mean temperature so to do that how do we determine mean temperature we integrate our temperature profile this is a mass weighted average okay so that is T minus T I and weight it with the velocity and here for the channel case we have to consider only the integration along this why now the velocity profile that we have now we are looking at boundary layer growth from the top plate and bottom plate okay now we are focusing on only one plate at a time so for this the velocity profile we have to consider is from 0 to D okay where you have your velocity profile that is a maximum velocity there all right so therefore we will do this integration from 0 to D because this is a mass weighted average we are taking and divided by 0 to D U D Y okay so we know that I can calculate my mean velocity based on 1 by D integral 0 to D U D Y so I can define a mean velocity from 0 till D because the boundary layer growth at the bottom plate is affected by the velocity profile from 0 to D and for the top profile it is exactly symmetric okay so I am looking at the mean velocity which is basically affecting the thermal boundary layer growth on the bottom plate okay so that therefore I am concerned only with the region from 0 to D here okay not the complete 0 to 2 okay so if you integrate it out so you get your relation between your 0 to D U D Y as U M into so therefore you can write this as 0 to D D minus TI into U by U M D Y into 1 over D okay so now I can also express the relationship between D and the hydraulic diameter so the hydraulic diameter for the case of plain duct we have seen that this is basically four times D so therefore we can replace D by DH by four okay so this will be DH by four so how we got this I think all of you know if you consider a plain duct in the three dimensional sense and this is your separation between the plates which is 2D you consider width in the third direction as W so the equivalent hydraulic diameter here will be 4A by P which will be four times 2D into W by perimeter will be twice W so therefore DH will be four times D okay so finally this is my relationship for calculating the mean temperature so I simply have to substitute my temperature profile into this I know my velocity profile U by U M okay which is so you are in the case of channel flows again your center line profile is 1.5 times your mean velocity okay so this can be substituted and you can integrate and therefore determine the expression for T mean minus TI so we will quickly do that so I am substituting my velocity profile as well as the temperature profile so this becomes 3 by 2 into 4 3 K DH over non-dimensional so I am converting this into eta which is Y by delta T okay in place of Y so therefore I am substituting all of this it becomes 0 to 1 essentially and this is 2 eta zeta minus eta square zeta square into 2 minus 3 eta plus eta cube into D okay so if you plug in for your temperature profile which you already have it here in terms of eta so your Y by delta T here can be expressed in terms of eta and in case of your velocity profile your Y by D can be expressed as eta into zeta right so we plug it in plug it in in terms of eta and zeta and you integrate over the non-dimensional Y coordinate that is basically your eta between 0 and 1 so this gives my expression for Tm minus TI as 2 cube double prime delta T square by K DH and I do not now eliminate any a higher order terms of zeta okay so unlike the case where I substitute into the momentum integral and there the moment the expression for zeta zeta square is under the derivative D by DX so there if I make the approximation that the higher order terms for zeta can be eliminated so the differential equation becomes easy to solve whereas here I do not have any difference this is an algebraic equation so I retain all the higher order terms as much as positive okay so this is my expression for Tm minus TI let me call this as equation number 4 and this is the expression for T wall minus TI this is equation number 3 so therefore T wall minus Tm can be written as T wall minus TI minus Tm minus TI okay so you see there are common terms here your Q2 Q double prime delta T by K is common okay so we will just take that out so that I can write my expression for T wall minus Tm as 2 times Q double prime DH by K into zeta by 12 minus zeta cube by 40 plus zeta power 4 by 192 so in fact the Tm minus TI can also be substituted you can simplify this a little bit and write it in terms of zeta cube by 10 minus zeta power 4 by 48 so here I am using the relation that my delta T is nothing but zeta times D so I am substituting for delta T in terms of zeta times T so this becomes zeta square so this multiplies this becomes zeta cube here zeta power 4 and this is my D so this is my expression for Tm minus TI and this is my expression for T wall minus TI when I subtract both this is what I get for T wall minus Tm let me call this as the equation number 5 so therefore so once you have the denominator here so directly I can get the expression for HX so HX will be nothing but Q all double prime that is your constant heat flux applied to the blades divided by this particular expression here for T wall minus Tm so my Q all double prime cancels here now I can directly get the expression for Nusselt number you can see that this can be written as HD by K okay so my Nusselt number in the thermally developing region which is based on the local heat transfer coefficient times the hydraulic diameter divided by K comes out to be of this form which is I can also take some terms common from here I can take write this as 2 by 4 and I can make it 1 by 2 here okay so that this becomes 3 10 and 48 okay so therefore this becomes 1 by 2 and goes to the numerator this becomes 2 by zeta by 3 minus zeta cube by 10 plus zeta power 4 by 48 okay where your zeta is nothing but we have already derived the expression for zeta by solving the energy integral that is 80 times X by DH divided by Peclet number whole power 13 so this non dimensional ratio here Peclet number by X by D is nothing but grades number okay so you can also call this as grades number one this is inverse of grades number so this is Peclet number by X by DH okay so once you basically know your non dimensional zeta so this gives you the variation of your local Nusselt number with respect to the non dimensional axial location okay so this is the relation for the case of flat plate now the question is whether if you use the asymptotic value whether the asymptotic solution leads to the case of thermally fully developed flows that is the question so therefore if you look at the asymptotic case where your zeta goes to very large value okay whether this particular solution becomes equivalent to your solution for fully developed thermally fully developed flows okay so in that case if I substitute since my zeta is in terms of delta T by D what happens if the flow is thermally fully developed what should be the value of asymptotic value of zeta one okay so if for large values of zeta so that should asymptotically go to one so your Nusselt number for zeta going to one from this expression comes out to be 7.86 now I think in the last class I have given you a table for flow past channel and what is the value if you apply a constant heat flux in the fully developed region 8.8.23 so that is the actual analytical solution so compare that with what you are getting from the approximate method so it is not leading to the correct solution you know it is there is a quite a bit of an error between these two values this is because the integral method is not valid for the asymptotic limit okay why the boundary layer is much so therefore the boundary layer approximation is not valid anymore and the other thing we have in the energy integral we have neglected the higher order terms of zeta correct so in fact for we are looking at the asymptotic case which is for the large values of zeta and that we have neglected here so even if the boundary layer approximations where assumed to be valid since we have made an approximation here so together put together finally the asymptotic values are not that close to the exact values okay so therefore you should consider the integral method when applied to the internal flow case this is a very approximate method and it is valid only in the region where your boundary layer approximation is valid that is the initial part of the thermal entry length if you go down downstream to the place where it merges and finally becomes fully developed then you cannot apply the approximate methods okay so this is to just give you an overview one example how we can use the approximate method in internal flows and that is also restricted to the Cartesian system Cartesian coordinates and again you use the exact solution for the velocity profile and you work out the case for thermal entry length so you can do this for also a similar analysis can be done with the constant wall temperature case so I am not going to do this in fact I have posted the assignment number 4 on the moodle you can just go and check that and I left that as an exercise in the assignment 4 where you are supposed to sit and work out for the constant wall temperature case a very similar exercise very straight forward the same similar kind of an exercise which you did for the external flows okay so I will just give you the final solution for that before we stop the discussion on laminar internal flows so if you consider that instead of constant wall flux you have constant wall temperature boundary condition okay so for the case again you can assume a cubic polynomial for temperature and you should of course use the appropriate boundary conditions in this case so rather than saying the gradient is equal to minus q wall by K you say that at y equal to 0 your T is equal to T wall which is fixed boundary condition and the other boundary conditions remain the same you get profile which is something like this T minus T wall by TI minus T wall 3 by 2 y by delta T minus half of okay so this is this is if you recollect this is this is the exact cubic profile that you got for external flows okay the same profile only the non-dimensionalization is a little bit different you have T minus T wall by TI instead of T infinity minus T wall here your free stream temperature is replaced by your inlet temperature at the edge of the boundary layer okay so you can assume this to be a non-dimensional temperature theta for the thermal entry length problem and your energy integral now so this is your energy integral this can be rewritten so your q wall can be written as minus K dT by dy at y equal to 0 okay and that is equal to d by dx you have 0 to delta T now I am going to use the non-dimensional coordinate for y which I define as Eta which is equal to y by delta T so therefore I can integrate across the non-dimensional coordinate so that limits of integration becomes 0 to 1 okay so row CP into u into T minus TI I can write in terms of theta so how do I express T minus TI in terms of theta 1 minus theta so that becomes 1 minus theta will be T minus TI by T wall minus TI okay so therefore T wall minus TI is constant because my wall temperature is constant inlet temperature is constant so I can introduce T wall minus TI and both sides it will get cancelled off so this can be written as u into 1 minus theta into dy so instead of dy I substitute in terms of Eta okay and if I convert y to Eta I have to multiply by delta T okay so there is a delta T also which is coming here so once again this K can be written as alpha row CP right so row CP cancels on both the sides and I can convert my y in terms of Eta so I have to also multiply by delta T in the denominator so this becomes Eta equal to 0 okay so this is my modified energy integral slightly modified form for the case of constant wall temperature so now I have the cubic profile and of course the velocity profile is again no the fully developed velocity proof okay parabolic profile so I substitute and then I proceed I had the same way I did for the constant heat flux calculate my mean temperature and then finally I get an expression for the local variation of the Nusselt number I will just give you the final expression okay so the final expression comes out to be NUX is equal to 2 by ? by 3 minus ? Q by 10 plus ? 4 by 48 okay so you can compare the expressions that you got from the constant heat flux they are coming to be the same here I think you have to do it and check probably I made a mistake here okay you please check in the assignment when you do that in the I have given this as an assignment problem so you please check what is the expression so I am not going to give you this right now okay so I myself have not worked it out so you please do that and check it has to be of course different so this is to conclude our discussion on approximate methods in the last class I have also given you a tabulation where I have compiled all the fully developed Nusselt number values for different geometries we start from plain duct with parabolic velocity profile then circular duct for slug flow profile plug flow profile parabolic flow profile and also for other cross sections like triangular cross section and the order of decreasing Nusselt number starts from your channel flow the highest value comes for the plug flow case with the constant wall flux then constant wall temperature then your parabolic flow velocity flow so it goes in that order starts from 2D duct flow and then your 3D circular ducts and then finally your triangular cross section so this is the hierarchy in which you are decreasing Nusselt number follows that pattern and before before I conclude I just want to talk a little bit about the project that I was thinking in mind I also posted another document on Moodle which you will see it is called project natural convection in a square cavity I will just briefly describe the problem and also how you are going to solve it and I since you have now more than a month's time with you I think you should probably take it up a little bit seriously and try to do the project you know so far whatever you learnt in the class they are all based on some theoretical discussion and deriving some analytical expressions okay and that is not probably sufficient for you to gain appreciation of subject like convective heat transfer okay a few of you were coming and asking me in this subject only dealing with the deriving expressions okay there are so many equations and the mathematics is very rigorous and is it limited to that okay so in order to satisfy those kind of people it is the better that you also do a practical hands-on project take up a case where you can really feel that you can apply the concept that you learnt of course you know I am not asking you to do analytical solutions for everything so you can also try some numerical solutions okay which are nowadays gaining immense popularity and replacing the analytical solutions you should try that and see for yourself as a you know fundamental research problem how you are able to gain your you know better your understanding of what you learnt in the class so for that I want to you know give this small problem of course it takes some coding you know you have to write a small code you know based on finite difference method it is a very straightforward technique once you know that and the document which I posted as all the details it has details including how to write the finite difference expression for the governing equations how to apply the boundary conditions and how are how are you going to solve iteratively and it also gives the solution so I want you to just simply read the document thoroughly and understand and try to implement it it may take some time that is why I am announcing this one month before your final exams okay so I will just give you a brief overview any questions on what we have covered I hope everything is clear okay so anyway when you do the assignment problems I think things will get much more clearer all right okay so coming to your project of course I am being a little bit hasty here before the topic of natural convection is taught to you I am giving you the problem but I think most of you have a basic heat transfer background and you will be able to quickly appreciate what I am explaining here so you can consider a square cavity just for geometrical as well as computational simplicity nothing else it can have any aspect ratio and you can consider that the left hand wall of this cavity is given some temperature you know you can call this as the temperature which is higher than the right side temperature okay so this is a heated cavity and the top and the bottom walls are insulated okay so in this case you do not have basically any flow to start with okay so that is no inflow outflow here it is just a cavity which is enclosing a space and you it is filled with fluid and now you already know some of natural convection so because of the temperature difference there is a boundary layer growth which happens due to the density difference and this will start a convection process okay so this convection process is called the natural convection okay so it is also called as a free convection so now for small temperature differences so in natural convection we characterize the non-dimensional number which is called the Grashof number also written sometimes as Rayleigh number this is your G beta into TH minus TC into if you look at the dimension if it is a square cavity dimension is HQ by kinematic viscosity into thermal diffusivity okay so this is a non-dimensional number which characterizes the ratio of buoyancy force to the viscous force and this will give you the strength of the convection which is happening so now you can see that that is linked the buoyancy force is linked to a temperature difference the higher the temperature difference the greater is the density difference and that creates the convection patterns to be stronger and stronger so initially for small delta T is you will find the convection pattern is very mild and it will be mostly conduction so if you look at the isotherms you will find that the isotherms go like this almost and there is a linear profile from the left to the right that is indicating that it is conductive mode of heat transport and as the Rayleigh number increases due to the temperature increasing temperature difference so the convection pattern becomes dominant and you will see nonlinearity coming in the temperature profiles do not look so good and then you start seeing streamline patterns so initially your streamline patterns will show a circulation like this later on it may become more and more convection dominated okay so you can start from Rayleigh number of say 10 power 3 and go up to 10 power 5 okay so from something which is close to conduction to something which where you can see the dominant effects of convection but still it is laminar okay so this is what I want you to do with the project you take a small cavity square cavity and you vary the Rayleigh number from 10 to 10 to the power 3 to 10 to the power 5 and you look at the modes of heat transfer when it when it starts from being from conduction then transition to convection and then becomes pure natural convection and to solve this you do not have to solve the Navier stokes equations in the true sense you do not solve the momentum equations but you solve it in the stream function vorticity method which I have derived in the very beginning okay so we introduce a stream function and vorticity and solve the equation so just to give you an overview how the equations look for the case of natural convection okay so anyway if you introduce your stream function the advantage is that your continuity is automatically satisfied so you do not have to solve for the continuity equation and now you can eliminate the two momentum equations by taking for example you differentiate the x momentum with respect to y and the y momentum with respect to x and you subtract the two therefore you can eliminate the pressure gradients and finally you can get a single equation divide of pressure okay so you have one variable less to solve now okay and therefore the number of equations also come down so in that case you are vorticity equation finally if you do that can be written as you d omega so I am writing this under 2d incompressible and steady state approximation so you do not have to look at the unsteady or transient patterns you just directly go to the steady state and want to get the steady state solution okay therefore the steady state equations will be minus g beta it will be dt by dy okay so now you introduce stream function in terms of you to replace the velocity in terms of stream function so I can say my u is equal to d ? by dy if I assume a stream function and v is equal to minus d ? by dx so I can write this in terms of the stream functions so d ? by dy into d omega by dx minus d ? by dx into d omega by dy is equal to the rest of the things on the right hand side is the same okay so this term is a body force term which comes in actual convection and you take the derivative of the y momentum equation with respect to y so therefore you have a dt by dy and the gravity is acting downward here right so therefore you it acts only on the y momentum equation okay if it was inclined cavity then you have gravity acting in both the momentum equation so therefore when you take the derivative of the y momentum with respect to y and subtract it so therefore you get this minus g ? into dt by dy okay so now this is your equation for what is it so what is it transport equation so this is the first equation that you have to solve and obviously you see to solve this you also need to solve the energy equation along with it because you need the information of temperature gradient so therefore we solve the standard 2d incompressible energy equation we do not use the viscous dissipation we neglect the viscous dissipation so and that will be u dt by dx plus v dt by dy this is the convection term on the right hand side you have a d2t by dx2 plus d2t by dy2 so once again for you and we you can write in terms of the stream function this becomes d ? by dy minus d ? by so this is the second equation this is your energy equation so you have your vorticity transport you have your energy and what else is required do you have enough equations to solve for all the variables so this is a governing equation for vorticity right and this is the governing equation for solving temperature but what is the equation for getting the stream function so there we use the definition of vorticity okay you define your vorticity how do you define vorticity in two dimensions so your dv by dx minus du by dy so there we substitute in terms of stream function for you and we here okay so your v is minus d ? by dx okay so therefore this becomes d2 ? by dx2 plus d2 for you it will be d ? by dy so this is equal to minus ? so this is my third equation for stream function so I call this as equation number 3 so therefore you have three equations three unknowns okay so one for stream function one for vorticity and one for temperature three partial differential equations and three unknowns which you can solve by using finite difference method and that is very clearly explained I think all of you have some basic understanding of finite difference in the heat transfer course in the undergraduate course already I think some of you we have done the finite difference expressions okay for the 2D conduction problem and I think all the M tech students also have taken numerical methods in thermal engineering this semester so it should be fairly straightforward to express all the derivatives in terms of the finite difference expressions so you can have a look at the document I have posted and that clearly gives you how to write the finite difference expression for each of the derivatives the first order derivative the second order derivative and so on and you solve these equations together iteratively okay so solution for omega requires of course the knowledge of stream function as well as a temperature solution for temperature requires a knowledge of stream function solution for stream function requires a knowledge of vorticity so all these three equations have to be solved simultaneously but it can be solved iteratively okay that means you do not have to invert a matrix together so first you start with the solution to stream function equation where you guess a value for vorticity field to begin with that is the initial guess value you solve this and get the field for the stream function once you have the stream function field you come to your vorticity equation make use of the guest value for so that is your iterated value for the stream function the field that comes out so that field you make use of you also guess the field for temperature and use this equation to solve for the vorticity okay once you have the vorticity field and your stream function field then directly your temperature field also can be solved so like this you keep solving take the newer values the latest values use that in the next iteration and so on till the difference between two consecutive iterations is very small you can use that as some 1 e power minus 5 or 1 e power minus the same way that you are using shooting methods okay you have a convergence criteria and then finally stop and then when you plot these contours in MATLAB or wherever tech plot or MATLAB and you will find this nice convective patterns in terms of no stream function if you plot you can directly find out the convective patterns and isotherms also can be plotted so and the coming to the boundary conditions to solve this you need boundary conditions okay so the boundary conditions here as far as the temperature is concerned these two left and the right walls are fixed temperature so you know the temperature at those points you do not have to solve them the top and bottom have adiabatic conditions so in the simplest case will be extrapolation from the inner point so you divide the domain into of course your grid which you all know and you have your locations where you are solving for the governing equations and then at the boundary at the top and bottom you simply extrapolate from the interior point okay and for as far as the boundary condition for stream function is concerned you have to apply the no slip condition at the walls u equal to 0 and v equal to 0 what does it mean in terms of stream function it means my d psi by dy equal to 0 and d psi by dx is equal to 0 so if I take the value of stream function say at this bottom left corner as some 0 I can take any value so the thing is the boundary condition for stream function is in terms of gradient so it does not depend on the value as long as the gradient condition is satisfied so if you take for example the value here as 0 then you can look at this isothermal wall you can apply d psi by dy 0 so therefore the stream function has to be 0 everywhere okay so if it is 0 here again for this wall you can apply d psi by dx 0 right so therefore it has to be 0 here even here so therefore in all the walls you can directly put stream function as 0 okay and for vorticity you can use the remaining condition for example for the left wall you have used d psi by dy as 0 but you have not used d psi by dx is equal to 0 so this can be used in the vorticity condition so you have you have the expression for vorticity here so for the left boundary your d psi by dy is equal to 0 therefore your d square psi by dy square is 0 so this will be 0 for the left bound okay so therefore your vorticity is d square psi by dx square you can write as simple Taylor series expression for the boundary in terms of the interior point and you can express this in terms of the vorticity at the the stream function at the boundary so therefore you can directly calculate your vorticity at the boundaries so all this is very clearly explained in the document okay so please read through the document if you have any questions you can ask me okay but it is very straightforward and just you have to construct the finite difference expressions and you will be therefore converting the partial differential equation into an set of algebraic equations so that is the method how you have to numerically solve differential equations finally you get a resulting set of algebraic equations which you solve by any method whether whether you have to consider this as a matrix and use inversion methods are do iteratively okay so here you can do a very iterative solution which is going to take only a few lines of code do loops where you just iteratively keep doing it till the convergence criteria satisfied and then you can plot it it plot the results for stream lines already in the document for 10 power 3 I think has given the stream lines and also for 10 power 4 and you should also plot the Nusselt number that you get okay as a function of the Rayleigh number I think in the document he has also given that you have for you start with say 10 power 3 and then say 10 power 4 10 power 5 and then so you can stop with 10 power 5 you know just just a representative and you can you can see the computed values which he has given there and your computations you compare both of them on the same plot for Nusselt number apart from the Nusselt number you also have your stream function which you have to plot and compare with his stream function and you also have your isotherms which you can plot okay so this gives you a very good idea practically how you solve the Navier Stokes equation in two dimensional case so there you have the gift of basically reducing that in terms of stream function vorticity and you can get a direct solution through differences and I think that will hopefully give you a better appreciation of subject when you go to the research level okay in research level we need not do everything by similarity solution so that gives you also an exposure to using some numerical methods okay so so that the deadline for the submission of the project will be on the last day of the class or maybe if required if required you feel some difficulty then we will have it on the day of the final exam okay and my suggestion is that if you feel that individually you have problem working with that you can also discuss with your colleagues you know there is no harm in discussion but whatever code that you write should be individual it should not be copied and you have to submit your code along with your report.