 So very good morning so last week we were looking at the solution the constant heat flux case which was extended by sparrow and Greg to the basic solution which was done by polehausen and the solution which was obtained by Ostrac later on to the constant heat flux case and here typically we have to define what is called as a modified Grashof number because we do not know up front what the temperature difference is going to be in order to calculate the Grashof number based on ? T therefore we do a simple scaling and convert the ? T in terms of heat flux which is known okay so based on that we have defined a modified Grashof number and if we look at the same similarity variable which polehausen are used okay if you look at the ? equal to X times Grashof number by 4 whole power 1 4th so if you substitute the modified Grashof number into this you find that there is no dependence on X okay which cannot be possible because similarity variable has to be a function of both X and Y and therefore sparrow and Greg modified the definition of ? a little bit they just introduced to the power 1 by 5 instead of 1 by 4 and which is now going to be therefore a function of both X and Y so we will see with this modified definition of similarity variable whether we try to successfully reduce this into a similarity differential equation so before doing that let us also calculate what is the order of magnitude of the reference velocity okay so since we start from the basic assumption in the similarity solution that u by u reference is actually a function of ? so this is the similarity solution assumption that means if you plot a non dimensional velocity profile so it should be only a function of ? correct so for this you need to know what the u references so how do we estimate u reference so this is the same way that we did for the constant wall temperature case so there we equated the order of power of Grashof number and Reynolds number so what did we get there the Grashof number to the power of the same order as Reynolds number correct so when we talk about therefore Reynolds number to the power 1 by 2 this was 1 by 1 by 4 okay so therefore wherever we had similarity variable in the original Blasius equation if you remember this was Y by X REX power half this is your Blasius similarity variable so we replaced REX power half with Grashof number to the power 1 by 4 here that is how Paul Hausen started so now in the present case our with the modified Grashof number we have a modified similarity variable and now therefore how do we do the order of magnitude so in terms of gr star how do we equate it this cannot be the same because now we have modified our similarity variable to be gr start to the power 1 by 5 so only so 1 by 5 should be of the order of magnitude of REX to the power half okay according to the current modification of the definition of similarity variable by Sparrow and Greg so this should be the order of magnitude of your modified Grashof number okay so now can you please substitute our definition of modified Grashof number calculate what is the order of magnitude of you reference huh that is f of f of ? here we started off with okay so you used g of ? okay fine use g of ? then and then we integrated g of ? and call that as f of ? okay fine it is a function of ? that is all it is so now by equating these two you can find out what is the order of your reference tell me in terms of Grashof number because we have Grashof number here okay so you can directly tell me in terms of Grashof number in terms of modified Grashof number so what do you get you get a factor of 5 here outside Grashof number by 5 raise to the power 2 by 5 right into ? by X is that okay right so this is you are you by this is your you reference now we can find out the corresponding transformation of the field from X Y to ? by using ? okay now ? is nothing but integral Udy so which we will substitute from this as U reference into we will write this as integral g ? into dy as dy by d ? into d ? so dy by d ? will take out into d ? and integral g ? d ? is nothing but what you call as f of ? it is another function okay therefore ? is equal to U ref into dy by d ? into f of ? so we know dy by d ? in terms of Grashof number and U reference also we have calculated the substitute and tell me what will be a function ? in terms of Grashof number the modified Grashof what is dy by d ? so X X answers here so this gives 5 into Grashof modified Grashof number by 5 the whole raise to the power 1 by 5 into ? I will put this new here okay so therefore once you find out ? now we know the transformation between therefore ? and so this should be multiplied by f of ? so we know the transformation from X Y the flow field in X Y coordinate to ? through the relation between ? and f okay also we know the basic transformation through the similarity variable okay so therefore now we can substitute this into the governing equation the momentum as well as the energy equation the same way substitute for U V du by dx du by dy d square U by dy square okay and you will be able to reduce this partial differential equations into similarity equation so before doing that also we need to find out the similarity variable for ? okay so let us assume now ? is a function of ? but we have to find the right non dimensionalization here so in the constant wall temperature case we use T – T 8 by – T 8 but the constant heat flux case we do not know the wall temperature and therefore this has to be somehow converted in terms of heat flux again so we can again use the Fourier's heat flux to do a scaling let us say that this is your T wall this is your T 8 at the edge of the boundary layer so what is this distance ? T this is your thermal boundary layer thickness therefore if you apply the Fourier's law this will be K times T wall – T 8 by ? T so therefore we can find a scaling for T wall – T 8 as Q double prime ? T by K that okay so now the order of ? T is the same as ? right so we can just substitute this as T – T 8 by Q double prime ? by K and what is ? now how do you find out ? ? is equal to y by ? correct so therefore we can find this as dy by d ? right so I am going to replace this as dy by d ? okay so substitute this in terms of crash of number so therefore what will be ? of so this will come out to be T – T 8 by Q wall by K x to the power 1 by 5 okay so just check the transformation whether you can write it in terms of dy by d ? tell me rather than writing it out separately like this I will ask you to substitute directly in terms of crash of number so what is dy by d ? in terms of crash of number x into crash of modified crash of number the 5 power – 1 by 5 so this is your transformation is it clear so now you can go ahead substitute this into the momentum and the energy equations so you have u du by dx remember the coordinate this is x is along the plate length and y is perpendicular to that plus v du by dy is equal to nu e square u by dy square plus g beta into okay so where now T – T 8 is equal to therefore T is equal to T 8 plus ? of ? into Q wall ? by K to x into modified crash of number by 5 the whole power – 1 by 5 you can substitute for dT by dx as this you have d ? by so you have basically dT by dx as functional dependence on x directly and also through ? okay similarly when you say dT by dy it is a function of y only through ? so like that you have to substitute for all the terms in the momentum and energy equation so g ? T – T 8 will be again g ? into Q wall by K x correct so you are T – T 8 is P substituted through this so similarly you complete the exercise and I will give you the final solution so you will find that this transforms the equations into a similarity ordinary differential equation as a function of only ? therefore the momentum equation becomes d Q f by d ? Q – 3 times d f by d ? the whole square plus 4 f into d square f by d ? square – ? equal to 0 this is the similarity momentum equation and the energy equation is d square ? by ? square – Prandtl number into 4 f d ? by plus ? into d f okay so this is how you finally transform so you can try this at home so the same way that you did the constant wall temperature case substitute and eliminate the common terms and this is the final set of equation that you get now we also have to make sure the boundary conditions have similarity that means they should not have dependence on x and y how do we do that so we know that the set of flow boundary conditions satisfy that at y equal to 0 that is at ? equal to 0 both you and we are 0 which indicates that d f by d ? is equal to 0 and also f is equal to 0 right and at ? going to infinity our what should happen you should approach you 0 okay so it is not like your Blasius case okay so you should still approach 0 so therefore your d f by d ? should also be 0 at ? going to infinity and apart from that what is the condition on energy equation that ? equal to 0 so let us look at ? going to infinity so there ? will go to 0 okay now we need one more condition at ? equal to 0 but we do not know the temperature at the wall therefore for terms of ? we cannot define but what we know is the heat flux so what can you define in terms of d ? by d ? so you have to once again perform the conversion so at the wall you know – K dT by so write this in terms of ? ? and ? and therefore tell me what will be the condition for d ? by d ? at the wall so what do you get for d ? by d ? – 1 so therefore at ? equal to 0 d ? by ? equal to – 1 right so you have all the boundary conditions required to solve the OD is here and once again you have to use the shooting method you have to start from the wall okay and you need boundary condition at the wall for basically both ? and d ? by d ? but at wall you have only boundary condition for d ? by d ? so you have to therefore guess the value of ? at the wall and then satisfy the value of ? at ? going to infinity same way the iterative method okay so then you will be able to find the complete solution set and let me tabulate the final solution so naturally once again this is a coupled set of OD is you have to solve them simultaneously and therefore the flow solution is also a function of Prandtl number for different values of Prandtl number you will have therefore d square f by d ? square at ? equal to 0 and what do you have for temperature ? at ? equal to 0 so this will be the part of solution that you get correct so you know d ? by d ? ? equal to 0 so you therefore guess ? ? equal to 0 until you iteratively satisfy the condition that ? ? going to infinity equal to 0 right so for different values of Prandtl number 0.1 this is 1.6434 this will be 2.7507 Prandtl number of 1 this is 0.721 1.3574 and Prandtl number of 10 this is 0.306 0.7644 Prandtl number of 100 0.465 1262 these are the different values okay so for a Prandtl number of 1 for example what was the subsequent value of d square f by d ? square for constant wall temperature case if you go back approximately say we had for 0.72 it was 0.676 approximately you can say about 0.68 or 0.69 so therefore this case the velocity gradient is slightly marginally higher than the constant wall temperature case okay so since the flow and the temperature fields are strongly coupled you can see that the velocity gradient is also a function of the boundary condition it should be point yeah okay I have reversed it 0.12 yeah you are right it cannot increase sudden 0.1262 and this should be 0.465 it is correct it should gradually decrease the increasing Prandtl number so now having known ? ? ? equal to 0 how do you calculate the Nusselt number we have to finally derive an expression for Nusselt number as a function of the modified Grashev number and Prandtl number so how do we get the expression for local Nusselt number so you have once again HX by K which is Q wall by T wall minus T infinity into X by K now in this case we know what is Q wall but we do not know T wall minus T infinity okay so therefore put this in terms of ? at ? equal to 0 okay so our definition of ? is I will write it again here so ? equal to T minus T infinity by and this was Grashev number by ? minus 1 by ? and we also had X outside so from this substitute for T wall minus T infinity that is nothing but ? at ? equal to 0 so you simply get this as Grashev number by ? H to the power 1 by ? into 1 by ? at ? equal to 0 correct so corresponding value of you can again fit a correlation as a function of Prandtl number and you can substitute for ? ? equal to 0 as a function of Prandtl number so you will get a expression for Nusselt number as a function of Grashev and Prandtl numbers so now what we will do next is to look at approximate solutions so these are the two exact solutions possible in natural convection when you talk about external natural convection when you talk about internal natural convection it is in a cavity and so on you do not have exact solutions then before going to I mean numerical full set of numerical solutions we will also look at approximate methods or the integral solution yeah this is all modified Grashev number correct so all this is star so is this the same thing I am using this expression and putting this Grashev number this is all modified so whenever you talk about constant wall flux case you have to always use modified Grashev number okay so next what we will do is look at the constant wall temperature case first and try to derive the approximate or the integral solutions similar to the external force convection that we did so this is something less rigorous than the similarity solutions and nevertheless they give you a useful solution which is kind of close to the exact solution so we will start off with the approximate methods so we will only focus right now on constant wall temperature boundary condition you can do a similar exercise for constant heat flux boundary condition also okay so let us assume that now the wall temperature is maintained constant you have gravity acting downward and you have a natural convection boundary layer okay the first point in deriving the approximate method is to derive the integral equations for both momentum and energy so let us try to find the integral equations by integrating the momentum equation and the energy equation across the boundary layer okay so now this solution was done by person called square is called also square solution is a approximate method now when we do the square solution you should understand that once we get the integral equation what is the next step we guess the profiles for velocity and temperature substitute them into the integral equation convert this into a simple OD which we can straight away integrate okay now in doing that we need to know the reference velocity in the blushes case in the flat plate case external force convection you know that this is your u infinity but in the natural convection case we do not know the u ref okay we do not know precisely u ref therefore what are all the unknowns here we do not know u ref and then what is the outcome of solving the momentum integral equation what do we get ? we get an expression for hydrodynamic boundary layer and then by solving the energy integral ? t so in your integral equation applied to external force convection you have only two unknowns ? and ? t your reference velocity is your free stream velocity there but now in natural convection we have three unknowns okay unfortunately we still have only two equations two integral equations so therefore what do we do is in the case of square he made a very simple assumption that since we are talking about natural convection for mostly gases we do not talk about very high prantle numbers are very low prantle numbers but for prantle numbers approximately around one so it is reasonably safe to make an approximation that ? is equal to ? t so finally we therefore reduce the number of unknowns to the number of equations that we have okay so this is an approximation that square makes but we will see whether this will impact the accuracy of the solution much still it will be close to the exact solution since we have to solve two equations we need to only reduce this to two number of unknowns okay and therefore when we integrate this irrespective of whether we are in integrating the momentum or the energy equation will still go from 0 to ? only okay so all of you can now try converting this into an integral equation also we need to write down the continuity equation please do not forget it we need continuity equation in order to relate the V velocity okay so this also we will have to integrate so I have split the second integral here V du by dy dy as d by dy as of u v minus u into dv by dy okay I have split the second integral into these two parts so therefore what do you get when you integrate d by dy of u v from between the limit 0 and ? huh what happens to this integral 0 okay there is no u velocity at y equal to 0 and y equal to ? right that knocks away I can use continuity to write dv by dx as dv by dy as du by dx so now I can therefore combine these two terms right so I have two times 0 to there should be a plus here so I have missed a sign somewhere okay so this is minus here right so this is plus okay therefore two times 0 to ? u du by dx dy this is equal to ? when I say du by dy between the limit 0 and ? what is the value at y equal to ? 0 okay so this will be therefore minus du by dy at y equal to 0 and the last term will be as it is you cannot unless you know the temperature profile okay you cannot integrate it so we just have to retain this as 0 to ? g ? into dy okay so this is therefore the momentum integral equation so let us call this as your momentum integral equation right so similarly if you integrate the energy equation so you have 0 to ? u dt by dx dy and the second term can also be split you can write this as 0 to ? d by dy of vt dy minus 0 to ? t into dv by dy into dy which we can use the continuity equation and write it like this this is equal to ? times once again dt by dy minus ? dt by dy at y equal to 0 so if you integrate this between the limits 0 to ? what do you get at y equal to 0 v equal to 0 but at y equal to ? v is not equal to 0 because we have an entrainment happening at the edge of the boundary layer for the boundary layer to grow okay so that v at ? is obtained from the continuity equation so you can therefore substitute for the continuity here so you have 0 to ? u dt by dx dy plus becomes v at ? times t infinity okay so I am substituting for v of ? as minus integral 0 to ? into du by dx dy into t infinity plus 0 to ? into t du by dx dy this is equal to minus dt by 0 so this can again be reduced to the form which will be d by dx of 0 to ? u of t minus t infinity dy which is equal to minus ? dt by dy at y equal to 0 okay so you can again write this as d by dx of ut minus t into du by dx and that and this will cancel and therefore finally you will have two terms which you can combine it so this is the same energy integral like what you got for the external forced convection okay only the momentum integral is now much simpler the energy integral remains the same and even the momentum integral you can just write it you can take the d by dx term out you can write it as simply d by dx of u square dy which will be the same as 2 u into du by dx okay so d by dx of u square dy so that now once you substitute the approximate profile for velocity now this becomes an ordinary differential equation with respect to x okay similarly once you substitute the approximate profile for temperature and integrate it you get a ordinary differential equation so you get two equations and two unknowns one is your reference velocity the other is your ? okay so you should understand that in this case we are integrating both with between the limits 0 to ? so we do not distinguish ? and ? t right so therefore once we derived the integral equation we have to make a guess for the approximate profiles of velocity and temperature now what will be the proper guess for velocity can we make a linear approximation because I told you that the peak can occur somewhere in between and again drops to 0 so we in order to approximate this kind of a curve what kind of a polynomial should we take at least we should have a cubic polynomial whether we take linear or quadratic cannot predict this maxima here okay so we have to start off with a cubic velocity profile and that is what square did so he assumed a cubic velocity profile however for temperature he used a quadratic profile which is not a bad approximation and so this is where he started off by making an assumption that U by U reference is equal to A plus B into Y by ? C into Y by ? the whole square was D into Y by ? the whole Q you can make ? is equal to Y by ? so this was the cubic assumption similarly for ? okay you can say A1 plus B1 into Y by ? plus C1 into Y by ? square okay so we will stop here tomorrow we will solve the two profiles by making the writing the boundary conditions down substitute them into the integral equations and proceed further okay.