 So good evening and we will look at the last part of natural convection today so before we move on to turbulent convection so far we have focused on the methods to obtain analytical solutions through similarity methods or approximate methods like such as the square solution and all these have been focused on a very simple vertical flat plate configuration okay. So now what happens if we deviate from this plane configuration we will see that it is not possible to get analytical solutions for more complex geometry so we will quickly go over some of the other configurations and I will give you the empirical correlations which have been obtained through experiments and maybe you can just use them when you are whenever you solve natural convection problems involving complex shapes. So one deviation from the simple vertical flat plate is that if you tilt the plate so let us say that this is your plate here okay and your gravity is of course acting downward vertically and let us say you have an angle ? with respect to the vertical direction so this is the tilt of the plate facing downward so if you heat this plate such that T ball is greater than T infinity still the same natural convective boundary layer will form but the only difference is the strength of the gravity will be slightly reduced in this case okay so the same governing equations will be applied for example if you denote this coordinate as X along the plate and perpendicular to the plate as Y you can still write down the momentum equation this is the momentum equation along the plate length this is what we call as the X moment and we have ? D square U by dy square plus the only difference here is that we have cos ? multiplied to G okay and the energy equation is still the same okay so in therefore in such a case we can have the same solution whatever we did for the vertical flat plate and we will only redefine the definition of Grashof number okay so we will just include G cos ? into ? T wall – T infinity into X cube by U square so we will just include the inclination angle cos ? directly into the definition of Grashof number so so that whatever solution that you already encountered whether it is the similarity solution or the approximate solution is still valid for this case with the slight modification in the Grashof number okay so but now we have to be careful that if you have a limiting case where you tilt it facing upward that is your ? in the minus direction and you make it – ? by 2 okay so what happens so when you have a configuration where you tilt it this way so that is the plate is facing upward instead of downward what happens when you tilt it such that ? becomes – ? by 2 in this direction becomes a horizontal plate okay so you will have a heated plate position horizontal gravity acting downward okay so can we use then the same equations here so you have X and this is your Y the limiting case where your ? becomes – ? by 2 so you have a horizontal surface now can we still apply the same equation to this configuration we cannot because as you see the buoyancy is now acting in the Y direction there is no buoyancy at all in the X direction so therefore we have to be careful okay for reasonably small inclinations the earlier approach is okay but if you have completely horizontal plate then we have to redefine the governing equations so in that case what will be the X momentum okay so in this case you cannot include the business type approximation in the X momentum because there is no gravity in the X direction so you just write your boundary layer equations so this will be your X momentum now the effect of buoyancy will be felt in the Y momentum equation okay so therefore although you neglect the Y momentum in the advection part there will be a pressure gradient in the vertical direction which will be balanced by the buoyancy force so this will be equal to therefore G ? x T – T 8 so this is the modified set of momentum equations for the horizontal heated plate configuration okay so for this also we can find some similarity solutions which I am not going to go through in detail but I will only give you the final expression for the Nusselt number this was done by Pern and Gebhardt PRN so this is Pern and Gebhardt and for the isothermal case they have given the following so this is coming from the similarity solution 0.5013 ? of number the power 1 by 5 Prandtl number or 1 by 4 okay so to derive this they used a similarity variable ? which was Y by X ? of number by 5 the whole raise to the power 1 by 5 so this is the kind of similarity variable that they used okay so you see that the advection component the momentum is neglected but in this case you do not have DP by DY is 0 so you have the pressure gradient in the vertical direction balanced by the buoyancy force whereas in the axial direction you do not have any buoyancy force but the most dominant momentum will be in the axial direction so this is the kind of you know configuration which is different from what we have seen so far. So now all these are assuming that the plate lengths are infinity right so what happens if you talk about finite plate lengths so if you for example make this finite so then you have edge effects coming into picture okay so let us discuss this configuration which is also kind of common so this is the case of horizontal finite plate in the horizontal direction so we have a case in which we keep the plate in the horizontal configuration and this is finite in length okay so one case case a we heat the plate okay and we can have case B where we cool the plate that means T wall is less than T infinity yes but in the the most the most important component which is governing the variation of Nusselt number along the x direction is your momentum in this direction as you can see the Nusselt number here is calculated as a variation along the plate length so therefore it is actually the axial momentum which is the reason why the Nusselt number is itself varying in that direction so it has been made an approximation that we can neglect the y momentum in the advection part and we can still solve this to get the profile variation the variation of the u momentum in the x direction which actually influences this but this is an approximation of course you can you have to do a finite difference or finite volume a method based solution in order to solve this exactly you cannot neglect also the vertical momentum there will be some patterns streamline patterns which will rise due to the buoyancy in the vertical direction and that will definitely cause some vertical momentum but while deriving the similarity solution it has been assumed that that is not going to impact the local heat transfer okay so of course we have to make some assumption in that case if you also include the vertical momentum we cannot find a similarity solution right so let us come back to the finite plate configuration where T wall greater than T infinity now what do you think will happen when I have a heated plate and surrounding is T infinity and this is finite length so naturally the fluid in contact with the plate which is getting heated up as a lower density and its tendency will be to rise up so therefore since now you have a finite length so this will try to sneak in through the edges of the plate and this will be symmetric from both the edges so if you draw a vertical line so you will have a symmetric patterns in the flow about this line so you will generally see that so this is the kind of patterns that you will end up okay so that is that is the hot fluid which is in contact with the hot the plate at the bottom surface will now have a tendency to escape through the edges if this was infinite so this hot fluid will be just staying there it can ever rise up but due to the finite length so the edge effects now cause the flow of the hot fluid to rise up and also the hot fluid on the top surface what will happen so this will also have a tendency to rise up correct so this is a very complex flow pattern that you will have okay so in this case you cannot neglect the y moment so everything will become important so therefore you cannot find an analytical solution now what is the case where you cool the plate so in this case again the density of the air surrounding the plate will be heavier compared to the ambient density so that therefore what is the tendency it should be the reverse it should try to come down sneak downwards therefore if you draw a vertical line so about that you have a pattern which is something like this and at the same time the heavier fluid at the bottom surface here will also tend to move downwards so in one way the flow pattern at the top surface of the heated plate and the bottom surface of the cold plate are identical and vice versa in the flow patterns in the bottom surface of the heated plate and the top surface of the cold plate are also identical okay so therefore the empirical correlations have been developed for one pattern which is this one the other pattern which is this one and the same pattern for this can be used for this also and similarly the Nusselt number for this pattern can be used for this also so an obvious view should understand that the Nusselt number for the top and bottom surfaces will be different because the flow patterns are locally different okay so I will just give the correlation these are obtained from some experimental observations okay so let us identify this region to be so I will call this as region 1 region 2 here so that is region 1 is the upper surface of the heated plate region 2 is the lower surface of the heated plate and 3 is the upper surface of the cold plate 4 is the bottom surface so for regions 1 and 3 so these 1 and 4 so these are identical flow patterns so the average Nusselt number based on the length of the plate so you can call this plate length to be some capital L this is calculated as 0.54 times Rayleigh number the power 1 by 4 okay so how do you now so far we have not defined Rayleigh number right we have used everything in terms of Grashof number so we will define Rayleigh number now Rayleigh number is nothing but another non-dimensional combination of Grashof number into Prandtl number so many a times you see that Grashof number Prandtl number come together so many times you can kind of say see that in some complex configurations the power of Grashof number and Prandtl number might be identical such as this case okay both Grashof number and Prandtl number raise to the power 1 by 4 so in that case they can be grouped and you can have one single non-dimensional number call the Rayleigh number so this will be nothing but G beta ? t so now if you define this based on the plate length this will be L cube by ? into ? right there is the product of Grashof and Prandtl number so similarly for regions 2 and 3 so region 1 and 4 so region 1 is the upper surface of the heated plate 4 is the cold surf for the bottom surface of the cold plate now for regions 2 and 3 that is the bottom surface of the heated plate and upper surface of the cold plate we have again identical flow patterns so therefore we have another correlation for that this is given by 0.27 times L A number raise to the power 1 by 4 okay so this region 2 and 3 so they seem to have a smaller value of heat transfer rate compared to region 1 and 4 okay so this these are some correlations for a more complex configuration like this we can also look at other configurations for example the case of a heated cylinder okay heated cylinder which is the suspended in the ambient where T wall is greater than T infinity okay so in this case how does the flow pattern look so again you will have something like this okay so you have rounded edges now instead of the sharp corners here so therefore the heated fluid will tend to rise up due to buoyancy and you will get nice flow patterns looking like this okay so from the upper surface you also have some flow patterns which are rising up so similarly for cylinder also for similarly for sphere this is this is cylinder so that means you do not have any variation in the depth it will be uniform within the perpendicular to the board but if you look at sphere there will be variation in the third direction as well okay so but nevertheless if you take a cross section of that you will find a similar onion bulb kind of a flow pattern so for these there are some correlations available the most comprehensive correlation was developed by Churchill and Chu this is the most commonly used correlation and this is valid for rally number now in the case of cylinder and sphere the rally number is based on the diameter okay so this is valid for a rally number greater than 0.1 and less than 10 power 12 that means a very very wide range the 13 orders of magnitude variation so this has been correlated over a wide range of experimental data so this is one of the most popular correlations so very complex looking dependence on rally number and Prandtl number but this is the same kind of formulation which is also done for the vertical flat plate because as you can see the vertical flat plate whatever we have done so far is for laminar flows now if you want to extend it to turbulent flows again we cannot use a similarity solution so the same set same people Churchill and Chu they have developed an empirical correlation for wide range of rally numbers so I will also give you their correlation for vertical flat plate which is of the same form so for vertical plates so this is valid for rally number from 10 power minus 1 to 10 power 12 and this is the correlation for the average nusselt number based on the length of the plate not the local nusselt number this is something similar form instead of the 0.6 you have 0.825 but this also is the same this becomes 1 by 6 here divided by 1 plus 0.492 Prandtl number 9 by 16 8 by 27 very similar form to the cylinder case some coefficients are slightly different and RAL is raised to the power 1 by 6 whereas there it is raised to the power 1 by 2 so again for general heat transfer calculations this is a very accurate for a wide range of rally numbers all the way from laminar to turbulent right and in the laminar regime this will be recovering the correlation by Ostrich okay so this will agree very well with Ostrich solution now for sphere for the case of sphere we have a simpler correlation which is fairly accurate 2.2 plus 0.43 again rally number based on the diameter of the sphere but this correlation is valid only in the laminar regime okay so let me just quickly check its validity because this seems to be quite small rally number range so looks like this package does not have yeah there is yeah so looks like this is valid for a small rally number regime that is correct less than 10 power 5 okay we do not have a very accurate correlation for sphere for a wide rally number range okay so these are some common complex configurations so you might encounter anything close to a cylinder you can use the cylinder and three dimensional object you can use a sphere and so on so kind of with these kind of correlations you can approximate slightly more complex surfaces and you can apply these correlations so next what we will do is quickly look at the presence of natural convection in internal flows okay so all these have been on external flows there where you place a plate in a free stream and you observe the buoyancy forces which aid the development of the boundary layer now what happens when you confine this completely okay so that means you can look at simple objects like cavities and where you do not have per say anything like a external free stream temperature and density so what happens to these natural convection boundary layers within enclosed spaces therefore there is a separate discussion on natural convection inside enclosures and in fact these are far more interesting than the external natural convection because external natural convection we are doing in detail because of the availability of solutions but the more interesting phenomena occur when you have natural convection in cavities so one of them is that you can have two parallel plates in 2d configuration okay you can for example heat the bottom plate and keep the top plate cold okay and what happens now again due to the buoyancy forces so this will be an unstable configuration the bottom plate will heat the air and the top plate will cool the air and therefore the tendency will be to rise up and this will be to descend downwards in the end you will be getting some kind of structures vertical structures like this where the hot fluid tends to rise up cold fluid will come down and this can happen symmetrically okay so depending on therefore the length to the height ratio so the aspect ratio of this configuration you can have several number of these vertical structures appearing so this kind of vertical structures are attributed to what is called as Rayleigh-Bernard convection okay so essentially this is happening due to an instability that means if you say this is your why so what should be your d rho by dy should be should it be positive or negative it should be positive so then only this can be a unstable configuration now what happens if you heat the top plate and cool the bottom plate so in this case what will be d rho by dy negative so what do you think will happen in this case this will be a stable configuration okay the heated fluid will be stratified here and the cold fluid will be here okay the stratification will happen the hot fluid will stick to the top surface cold fluid to the bottom surface but you won't have any of this convection patterns so this is a stable configuration and no convection cells will be formed right so in order to produce the Rayleigh-Bernard instabilities you need to have unstable configuration where d rho by dy is greater than 0 so this you can very nicely see the vertical structures happening and you can also see this in three dimensions in three dimensions you will see a honey comb kind of a structure from T H to T C there must be correct but T H the lighter fluid will be only can rise up it cannot go down so therefore you cannot have a convection from the hot to the cold in that case okay so it will always this is always buoyancy aided convection okay so you cannot have a convection which is opposing the buoyancy and going only the buoyancy has to aid it so the natural direction for buoyancy to aid is d rho by dy greater than 0 right so for this case these are very complex instabilities and you can to some extent do some linear stability analysis and predict the onset where the instabilities can become Rayleigh-Bernard convection cells but beyond that you cannot use any analytical methods okay so these are usually predicted by numerical methods okay you have to solve the entire set of Navier-Stokes equation with the business approximation and you will be able to capture this kind of roles convection roles so I will not have a very detailed discussion on this these are topics beyond the regular classroom thing but the other interesting configuration in 2D is to have an enclosure that is a cavity which is closed on all sides okay and you can heat the left wall of this cavity and you can have a cooler right wall the top and bottom being insulated okay so this is a another interesting configuration okay so now what happens when you have a differential heating between the left and the right walls again so you will have a boundary layer which can actually grow okay and again a boundary layer which will descend in this direction so what will happen together in an enclosure will be the net effect will be a vorticity which will be driving a convection cell in this direction the clockwise vortex okay from the hot side it can only rise up and from the cold side it can discern down so you will be ending up with the clockwise vortex pattern okay now very small temperature differences if you define for example Rayleigh number based on this temperature difference between the left and the right walls and let us say the cavity is a square cavity with sides dimension H so for small Rayleigh numbers less than say 10 power 3 in this case the temperature difference will be too small to produce a convection pattern so it will be only conduction which is happened okay so therefore when you look at the isotherms you will find that there will be a simply a linear variation from the hot temperature to the cold temperature so once you increase the Rayleigh number about 10 power 3 that is when you find the nonlinearities in the isotherms so the isotherms will now start deviating like this correspondingly you will find this vortex pattern appearing so as you keep increasing your Rayleigh number you will find the vortex pattern becomes more complex you might get more than one vortex there okay and what will happen to your Nusselt number that is Nusselt number on these two walls they will increase with your Rayleigh number as your Rayleigh number increases your Nusselt number also consequently increases it increases the convective motion and therefore higher heat transfer rate okay now this is again we can find the solution only by numerically solving the equations Navier stokes equations okay so what I suggest is that this is a very interesting problem for you to solve the equations as such so I I would suggest that you follow the solution procedure as given in this book by Ust-Uisen and Naylor convective heat transfer okay so I will give you also the section number where he has explained the complete numerical solution to this cavity problem so he has given a wonderful explanation section 8.9 that is chapter 8 8.9 in chapter 8 natural convection so the topic is natural convection heat transfer across a rectangular inclusion so there he has listed down the equations the boundary conditions and also how he is solving them with finite difference methods so this is introduction to convective heat transfer by whose two is an Patrick whose two is an and David Naylor actually I think this is an e-book now which is available for free I think most of you can all of you can access this is an open access e-book okay so this is available on the if I remember right this is part of the open access thermal fluids the thermofluids they have an open access of journals and books so you thermo thermofluid central it is called okay so thermofluid central if you go to their website they have some free open access e-books can although originally it was a published as a hard copy now they have a free e-book as a soft copy available which you can download okay so especially you look at this section he has very nicely shown the way he is solving these equations is by what is called the stream function what is it a method so rather than directly solving the continuity and the momentum equation because if you use stream function it makes the continuity equation redundant so therefore rather than having so many equations you have only two equations one for the stream function the other for vorticity right so you introduce the definition of your stream function d ? by dy v is equal to minus d ? by dx and similarly vorticity this is dv by dx minus du by so this is then cast the equations the Navier Stokes equations both the x and y momentum are then rewritten okay we do some manipulation there like for example we take the derivative of the x momentum with respect to y and y momentum with respect to x and then we subtract so we can actually recast these equations in terms of a vorticity equation and since the stream function satisfies continuity okay so we will have a simple stream function vorticity equations which can be solved and the appropriate boundary condition and the procedures are all very nicely explained so you can just go through and he is using only simple finite differences I am sure most of you would have already gone through some numerical methods course so you should be able to quickly understand how to solve the finite difference equations okay so you can use an iterative scheme for example like Gauss Seidel okay and try to find the solution so it is a very interesting problem because once you code it so you can change the rally numbers and you can look at the different streamline patterns so I suggest that you please take this up as part of your final project okay look at rally numbers in the range from 10 power 2 okay till 10 power 5 that is 3 orders of magnitude variation in the rally number and you can actually use the following correlations to compare your numerical solution for the Nusselt number you can assume that your Prandtl number is equal to 1 okay and so for the rally number between 10 power 3 and 4 this is the correlation and for rally number between 10 power 4 and 6 the correlation becomes 0.065 rally number or one-third so you can actually try up to 10 power 6 if you can run the code if you cannot run up to 10 power 6 you can stop it till 10 power 5 okay so you run the code get the results for Nusselt number and compare it plot them as a function of rally number so your average Nusselt number as a function of rally number so you plot your numerical results and also the correlation should be able to match very close and also you can look at the streamline patterns you can plot the streamlines because already you are solving for the stream function so in matlab you should be able to plot the stream function isotherms all this for the different rally numbers okay so yeah so this is a very interesting exercise of all the things that we have done in force convection and so on because you see flow patterns which vary drastically with the changing rally number okay so so please try to submit this project when you come for the final exam okay so that that will be mostly I think 3rd May 3rd May is the D slot exam so we will have the submission of the project also on the same day along with your final assignment the assignment final assignment I have posted it has both natural convection and turbulence turbulence will be covering in another 4 classes so meanwhile so you will have time to work on the natural convection part so all this together the assignment which I have posted assignment 5 and a project together on the day of your entrance exam okay so these are this is a very standard problem if you go through some literature survey you should be able to find many papers to validate and programming is given very nicely explained in this book okay so any questions so in that case we will conclude one last topic which I would talk very briefly a few minutes is the case of mixed convection okay I have mentioned about this in the beginning of the natural convection part but I want to just emphasize because most of the practical heat transfer problems are you know having the combination of both natural and forced convection you cannot ignore either of them so in that case what happens to the flow patterns so for example if you talk about gravity acting downward and you have a forced motion bulk motion of fluid and along with that a natural convective boundary layer okay so this is one case the other case where the buoyancy is acting upward but the forced motion pushes the fluid downward okay so in this case you will have a boundary layer growth okay you should understand that since there is a forced convection motion in the upward direction so this will kind of tend to compress the boundary layer compared to the case where you push the bulk motion downwards so in this case the boundary layer will look more stretched out okay so when we do these kind of flows now what is the way to calculate the similarity solution is there a similarity solution for this problem well the thing is what is the correct similarity variable to use because when you talk about forced convection boundary layer we use the Blasius variable but natural convection involves Russia of number so which similarity variable can we use to describe this problem so therefore if you look at the cases where the ratio of Richard's a number that is your grass of by RE square is of the order of one we cannot find any similarity solutions because both of them are very dominant but for the limiting cases for very small Richard's a numbers okay so there the forced convection similarity variable might be the correct one and similarly if you go to the other limiting case where Richard's a number is basically very very large so there the rash of number using rash of number in the similarity variable might give you the similarity solution but these are only in the limiting cases okay but most of the regimes where both are dominant we cannot find any similarity solution to this again we have to go for numerical solutions but as I told you so the simpler way of doing this is to just blend the Nusselt number from forced and natural convection by using a simple power law expression like this so wherever you have a forced convection which is assisted by buoyancy so you use a plus sign and wherever you have a forced convection opposing the buoyancy you use a minus sign so naturally the case where you assist the buoyancy will be the one with a higher Nusselt number because you compress the boundary layer and therefore the ball temperature gradient will be higher correct so this is a very simple hands-on approach if you want a very detailed flow patterns then you have to go for again numerical solution so I will try to quickly cover the turbulent convection also okay so I think it should be sufficient okay so thank you very much and please take note that you have to also complete the project along with the assignment when you come for the N semest right.