 Okay so where we stopped yesterday we were looking at the solution for heat transfer similarity solution and this is also called the pole house and similarity solution okay after named after pole house and who continued the velocity boundary layer similarity solution of Blasius and he extended that to thermal boundary layers and we saw the derivation of this particular expression subject to these boundary conditions and of course you know from the from the intuition that for Pantel number equal to 1 the if you replace the temperature variable non-dimensional temperature ? with the non-dimensional velocity the momentum and the temperature equations are identical exactly so pole house and got the idea that if for Pantel number equal to 1 they have the same solution so for other Pantel numbers there should be some kind of similarity solution possible so then he intuitively substituted the same similarity variable which was used for velocity boundary layer and finally he finds that this falls into the category of similarity solutions okay and of course we have seen by direct integration you can reach to this particular step where this gives you the solution of course you need the F from the solution to the velocity similarity similarity and from there you plug it in and you can numerically integrate it or we have also seen the the standard way of doing all the solving all the OD is will be here after by using the shooting technique so you solve the OD directly numerically so that also will give you the same solution for ? but for all this the solution for F is required okay so you have to solve both of them simultaneously solve the momentum similarity solution once you get the value of F you substitute into the energy similarity solution find the solution for ? so all this also depends upon the Pantel number therefore your ? is a function of your location as well as your Pantel number so far a given Pantel number you substitute and you integrate it out and you find a solution so for different values of Pantel number you will get different values of ? as a function of ? so therefore if you plot the similarity solution coming out of this if I plot 1- ? that is I just convert it as T- T 8 by T wall- T 8 as a function of ? you see for each Pantel number the slope of the curve changes higher the Pantel number greater is the slope so that which means now now you all know that the slope of the temperature profile at the wall directly governs your near wall heat transfer rate okay so we are now going to calculate the heat flux which is carried away from the wall and we all know that the heat flux is directly related to the slope of the temperature profile at the wall so greater the slope greater will be the heat flux carried away and therefore your heat transfer coefficient which is defined from this also will become higher so higher the Pantel number the higher the heat transfer coefficient okay so now you can probably once you know the solution you can convert this in terms of ? and derivative with respect to ? and then plug in those values okay can you have you know we have defined ? in this manner T- T wall by T 8 so this can be written as – K so DT by DY will be D ? so T 8- T wall into D ? by D ? I am transforming the variables here into D ? by DY at Y equal to 0 or in fact I should apply this condition to D ? by D ? at ? equal to 0 okay now I also know my similarity variable ? is Y times square root of U 8 by ? X okay so I can just substitute this should give – K T 8- T wall into what is D ? by DY square root of U 8 by U X into D ? by D ? at ? equal to 0 now D ? by D ? ? equal to 0 so this is my D ? by D ? if you integrate this once you end up with this solution for the slope D ? by D ? okay so you want to evaluate this at ? equal to 0 so we will say that my D ? by D ? at equal to 0 should be the value of this F double prime okay at ? equal to 0 okay the F double prime to the power PR at ? equal to 0 by 0 to infinity F double prime at ? equal to 0 to the power PR D ? right since your F is a function of ? so the value of slope from that equation will be evaluated exactly at F double prime ? equal to 0 okay so I mean when you are doing this numerically so all you have to do is you have to substitute the value of F double prime from the Blasius solution that you got right so you know the value of F double prime at ? equal to 0 that is nothing but the curvature okay you substitute and then you integrate it okay so then that should give you the value of the temperature profile temperature slope a slope of the temperature at ? equal to 0 okay so when you do this numerically once you have the complete Blasius solution it is just a matter of substituting the curvature for in terms of F and then calculating the slope in terms of ? okay so once you do this you will find out in fact this was done numerically by Paul Hausson and he has actually fitted nice curve so he gets different sets of values of D ? by D ? for different values of Prandtl number so depending on the Prandtl number you get different values of this okay so now he fitted a nice curve for different ranges of Prandtl number for each range you know he found the curve fit which is sufficiently good enough to describe it and this is what Paul Hausson did it so Paul Hausson's solution was expressed as D ? by D ? at ? equal to 0 for the case where your Prandtl number was extremely small okay so I say Prandtl number approaching 0 okay so for such a case he found the curve fit like 0.564 times Prandtl number raised to the power half this was the best fit which can describe the approximation for low extremely low Prandtl numbers okay for moderate Prandtl numbers between 0.6 and less than 15 he found 0.332 into PR power 1 by 3 okay and for Prandtl number which is very large 0.339 PR power 1 by 3 in fact it is very close you know even this could have approximated all the way from Prandtl number 0.6 to infinity it is only the third decimal point which is just changing but for very small Prandtl numbers the functional dependence on Prandtl number itself is different okay so in fact this is a very useful information for us he has completely covered the entire Prandtl number regimes okay so in fact it is a good exercise you can when you are solving this by for example shooting method you can you will be directly solving for D ? by D ? at ? equal to 0 right you have seen this just like the case of Blas's solution where you do not know the curvature you guess it and try to match the other boundary condition okay f prime of infinity should be equal to 1 the same way here you guess D ? by D ? ? ? equal to 0 you guess this and you finally match this solution okay so ultimately when you converge finally you will directly converge to the correct value of the slope at ? equal to 0 okay and if you probably check with the Paul Haussens correlation you can get a good idea how good this correlation works okay so you can repeat this for different Prandtl numbers right okay for different Prandtl numbers you solve this equation you get the slope and then you substitute and check whether it satisfies the Paul Haussens fit okay so that is a good way of for you to learn and understand whether this fit is very accurate or not okay so this is what Paul Haussens got we let us accept it for the time being so most of the times we are interested in fluids in the kind of intermediate Prandtl numbers okay we are not going into liquid metals or fluids with extremely high viscosity but with intermediate Prandtl numbers between 0.6 and 15 so we will take that value and then substitute for d ? by d ? for calculating the heat flux okay so therefore my heat flux will become now 0.332 x k times T wall – T 8 and just flipping the signs here x Prandtl number power 1 by third square root of u 8 by ? x so this is the equation for the wall flux I want to now go one step forward and calculate what is the heat transfer coefficient okay so my heat transfer coefficient as a function of x will be your wall flux divided by how do I define it wall temperature – reference temperature is my free stream temperature okay so this will be 0.332 k here power half square root of u 8 now I will non-dimensionalize the heat transfer coefficient in terms of Nusselt number local Nusselt number which is defined this way okay so this will be 0.332 PR power 1 by 3 so I multiply by x here so I will be getting square root of u 8 x by ? right so which is nothing but the Reynolds number okay where your local Reynolds number is u 8 x by the kinematic viscosity okay so this is the final expression that you probably find in the textbooks if you have taken a basic heat transfer course you are all straight away given the final expression or flat plate Nusselt number okay so this is how it is coming out so this is a very straightforward correlation and probably most of you know by heart and this is valid for the intermediate Prandtl number range that we are talking about that is Prandtl number greater than 0.6 and less than 15 okay so we can also do one more thing rather than being interested in the local variation of heat transfer coefficient and Nusselt number we can kind of calculate an average value of H for the entire plate so therefore we can also define an average value H over bar which is 1 over L 0 to x HD x okay so this is the way I define my average over the entire plate length okay so if I substitute you can probably do that as a nice exercise you can substitute and integrate it with respect to x you will find that the resulting expression will be 0.664 into K PR 1 by 3 square root of U 8 by NU L okay which is exactly twice of H of x but here x will be equal to L okay so this is exactly twice of the heat transfer coefficient corresponding to x is equal to L okay now same way if you define an average Nusselt number for the entire plate okay depending on the average heat transfer coefficient and based on the length instead of the local coordinate so you will be getting 0.664 into RE based on the length and Prandtl number one third okay so this is this is exactly twice that of NU at x is equal to L alright okay so these are some I think correlations which are familiar to you so I am not going to spend too much of time and so in all these cases you know that the Prandtl number is a very important parameter which is the ratio of NU by A so which gives you the ratio of momentum diffusivity by thermal diffusivity okay and you also know from the scaling analysis which we did when we derived the boundary layer equations the relationship between Prandtl number and ? we showed that my ? T by ? is approximately Prandtl number the power minus 1 by 2 for one case for higher Prandtl numbers this relationship becomes Prandtl number to the power minus 1 by 3 but nevertheless you can see that the thermal boundary layer thickness is inversely proportional to the Prandtl number okay so therefore if you are saying that my Prandtl number is much lesser than one so your thermal boundary layer thickness is much greater than your momentum boundary layer thickness and for Prandtl number equal to 1 both are equal and vice versa okay so this is the kind of observation for higher Prandtl numbers for Prandtl numbers greater than one of the order of one and greater than one you can clearly show that ? T by ? scales as Prandtl number to the power minus 1 by 3 okay so this something which was concluded once you calculate your thermal boundary layer thickness from here and the momentum boundary layer thickness from the Blasch's solution you can simply take the ratio and you find that it exactly scales by this Prandtl number power minus 1 by 3 this is something that which you can observe yourself okay for higher Prandtl number for the low Prandtl number regime it scales as Prandtl number power minus 1 by 2 so in fact you can see here itself okay the slope for the low Prandtl number regime is of the order of PR power half here for higher Prandtl number it is PR power 1 by 3 okay so these are some of the observation that you can make and most of the times if you are looking at oils which are very viscous you are looking at high Prandtl numbers or if you are looking at liquid metals you are looking at very very small Prandtl numbers most of the practical fluids air water and so on they fall in this intermediate Prandtl number regimes so that is why we have specially derived the expression for Nusselt number for those majority of the fluids which fall within that Prandtl number range okay so in fact this will be also very nice exercise which I can give you probably in your homework you can calculate the thermal boundary layer thickness once you get your Blasius profile right so that is that is the value of Y where your ? corresponding to the ? where your U by U 8 goes to 1 or 0.99 or F prime that you calculate should be approaching 0.99 the corresponding value of Y will be ? the same way thermal boundary layer thickness is defined as the point where your ? goes to 0.99 okay the corresponding value of Y so you can calculate both you can take the ratio of that and you can see how it scales you check you check the scaling with respect to Prandtl number okay so you can keep doing this for different Prandtl numbers and you will find that it exactly scales in that manner alright so so with this we will move on to the next topic but actually what could have been done I could have actually started off from the other the next topic that I am going to talk about now rather than doing the flat plate solution because a flat plate solution is a special case of that particular type of flows okay so this type of flows are called as Falkner's can flows okay so let me just give you an introduction to the Falkner can scan type of flows before we go into the similarity solution so any questions so far on the flat plate flat plate is a very basic case that is why I had to start so that you can understand how the similarity solution is obtained how we can solve the ordinary differential equations and so on the other cases are little bit more not that difficult but it is just one order of approximation higher and the flat plate case is going to be a special case of those kind of flows I did not want to directly start off from there and show you the flat plate I could have done that to save the time but nevertheless it would have not been a very good learning experience for you okay okay one more thing which I probably will just mention casually before signing off from this I think all of you have heard about Reynolds analogy okay I think Reynolds analogy is something which is taught in any heat transfer course so you can also show I mean for the case of Prandtl number equal to 1 okay there is a very good relationship between the Nusselt number and the skin friction coefficient okay so it is expressed in terms of standard number is equal to exactly CF by 2 okay so this was called the Reynolds analogy because it was discovered by Osborne Reynolds and this relationship is very useful because once you know the skin friction coefficient you can directly calculate the corresponding Nusselt number also okay your standard number is another non-dimensional number which is a non-dimensional group which is defined based on your Nusselt number Reynolds number and Prandtl number all these are grouped and just another name is given to that now for Prandtl number not equal to 1 people have found that still you can use an analogy now that is called as the Reynolds-Colburn analogy because it is an extension of the Reynolds analogy for Prandtl number 1 according to that your standard number into PR power 2 by 3 is exactly equal to CF by 2 so wherever your Prandtl number is not equal to 1 you can still use the Reynolds-Colburn analogy to calculate the Nusselt number from the skin friction coefficient so in any way I mean already when Paulhausen did it he understood the relationship between the Nusselt number and skin friction directly for Prandtl number 1 okay for the other Prandtl numbers it was extended based on the Reynolds-Colburn analogy and therefore these are very useful expressions for flat plate is concerned you do not have to really be bothered about the heat transfer solution once you get the fluid flow solution you can apply the analogy and calculate the heat transfer solution especially if we are interested in the heat transfer rate at the wall okay so therefore we are more interested in the Nusselt number and things like that okay so with that I think you can show and derive this yourself it is not that difficult I think you can do it I am not going to spend time and therefore with this we will move on to the Falconer scan type of flows okay so so these are flows with pressure gradient okay so so far as you all know in the flat plate case we have seen that the pressure gradient is neglected and therefore it becomes very simple to solve now what happens when you have flows with pressure gradient okay can we derive a closed form analytical solution or maybe can we find at least a similarity solution which we can solve numerically okay so this was not addressed till the 1930s so when Falconer and scan the extended they studied the similarity solutions derived for flat plate and they try to extend that for flows with pressure gradient term also okay and that is why they have been called as Falconer scan solutions at least the flow part okay so heat transfer part was added to the flow part because the heat transfer equation has nothing to do with the pressure gradient heat transfer already has been solved okay so once you solve the flow that f that comes out of the flow goes into the similarity solution for temperature okay so that is going to be much simpler so the flow is the major complication here in fact Falconer and scan they discovered that for flows with adverse pressure gradient they have in fact you can do this from the potential flow theory that is from the inviscid solution that you can derive the velocity profile the free stream velocity which is a function of X now it is not it is not a constant anymore so can be shown to be relationship with respect to X a C X power M okay where C is a constant and M is also a constant but it is related to what is called the wedge angle okay so beta by 2 minus beta and beta pi is your wedge angle okay now what does it mean by the wedge angle is if you draw a very a general kind of a problem with describing this kind of a free stream velocity profile okay it will be something like this okay so you have a wedge okay this is a wedge profile okay so you stick this wedge into the free stream and so initially here your free stream which is approaching will be constant and now once it encounters this wedge so there is of course a pressure gradient okay so the flow will try to for example here accelerate okay so therefore here the local free stream velocity will be a function of X okay correct okay and this profile is described by this relation C X power M for this kind of a case what is the wedge angle so this is your wedge angle beta pi okay so this is the so this is a general figure configuration for which Faulkner and scan has shown the similarity solution and you can derive special cases of this particular solution okay now you can see if my beta equal to 0 what happens what happens to the wedge it becomes a flat plate okay so these two collapse and this is just simply flow past a flat plate and if my beta equal to 0 what what is the value of M M is also 0 right so therefore my U infinity will be constant so which becomes basically the flat plate solution so flat plate solution is a special case of the wedge solutions are the Faulkner scan solutions okay now if my beta equal to 1 for example how does this flow look okay so let me draw the different cases beta equal to 0 as you can see that this will be a flat plate okay now if my beta equal to 1 what will happen to M 1 okay so therefore my U infinity of X will be C X okay now how does how does the configuration look if my if my beta equal to 1 the wedge angle will be what pi so it will be a vertical plate instead of a horizontal plate but the flow will be still coming along the horizontal direction so this will come an impinge the flow comes like this it impinges and it goes this way okay so this is called a stagnation point solution because you can see that the flow comes hits the velocity has to be 0 so this is a stagnation point okay so this is a stagnation solution stagnation point okay so this is I mean since it is 2D this is 2D stagnation flow or stagnation point flow okay so these are some special cases of the Faulkner scan now what happens if my beta is negative so far I have been discussing the case where beta is 0 and beta is greater than 0 so beta going up to 1 now what what can have can also you can also visualize another configuration or beta is negative okay so how could how could beta be negative so you can for example take this you can wrap it around inside it becomes 0 when it touches here and again you turn it inside so that becomes negative and when you do that if you do completely wrap around this way you will find this flow cannot flow past this surface okay but instead this flow will have to come and then deviate this way okay so therefore I am just tilting it upside down if you if you tilt it upside down you can say for beta less than 0 for example where your M also will be negative correct if my beta is less than 0 M also will be negative so I can have a flow which is something like this okay this is my wedge this is how my wedge it becomes okay so I take this this will be 0 and again I put it this way and make this horizontal okay so this will be horizontal and like this the flow will pass like this okay the same thing I am drawing upside down okay so the flow will be like this so what kind of a pressure gradient here it is is it favorable pressure gradient or adverse pressure gradient this is an adverse pressure gradient okay because you are because why your M is less than 0 so your velocity has to decelerate correct so this is an adverse pressure gradient flow all right so whereas the for the case where your beta is greater than 0 what should happen the flow should accelerate your M is positive okay so wherever your wedge angles are positive like this it is an accelerating flow where your wedge angle is negative like this it is a decelerating flow or it is an adverse pressure gradient flow in this case it is a favorable pressure gradient flow okay so therefore you can see there is a family of solutions that can be derived once you have the basic solution for the free stream this is coming from the inviscid this is coming from the potential flow or the inviscid solution okay so for this kind of class of problems there are different configurations that are possible depending on the values of M which depends on the wedge angle okay beta pi and the limiting cases some of the limiting cases that we can have a look at is the case of beta equal to 0 which gives you the flat plate solution beta equal to 1 which gives the stagnation point solution and negative values of beta which is an adverse pressure gradient solution now if I reduce the value of beta so small in fact if I make it too negative what will happen is beyond a certain point the flow will separate the adverse pressure gradient will be so strong that flow separation will take place and once the flow separation takes place the boundary layer theory is not valid anymore okay the boundary layer theory is valid only within the boundary layer once the flow separates there is no boundary layer there okay therefore you have to be cautious there is the lowest low there is a lower limit for beta till which you can find solutions okay below that flow separation takes place and you cannot find a similarity solution to those problems okay so now what we are going to do is for these kind of class of problems we will write down the try to reduce the partial differential equation to a similarity equation so can you all try to write down the governing equations for this kind of problems okay so for flows with pressure gradient so how does the boundary layer equations look okay the continuity is still the same how about the momentum equation everything is the same except only one term which is the pressure gradient term and we are now neglecting the okay now when we are not still written the energy equation first we will take the flow solution and then we will apply that to the energy okay so this is your governing equation now we all know how do we calculate the pressure gradient okay I probably mentioned this in the very third or fourth class okay when we derived the governing equations for boundary layer so how do we evaluate the pressure gradient inside the boundary layer upstream so since we have shown that the pressure is invariant of why okay so we can evaluate this outside the boundary layer and the same value should be valid inside also okay so therefore if you apply the governing equations outside the boundary layer so you will get U 8 D U 8 by DX okay only your free steam velocity is there your V velocity 0 so that should be equal to – 1 by ? DP by DX and there are no viscous effects also there okay so you can simply replace your pressure gradient with your equivalent velocity gradient therefore this term will become U 8 by DX all right so now for the class of problems we are looking at the wedge flow problems okay this is the form of the velocity profile C X power M okay so therefore we know the functional dependence of U 8 on X we can just simply substitute it can you tell me how does it look since my U 8 is equal to C X power M my D U 8 by DX will be C M X power M – 1 okay so therefore this becomes U D U DX V D U DY is equal to I can write again U 8 I can say that this C X power M is again nothing but U 8 okay so I have 1 U 8 here U 8 here which I club together as U 8 square and apart from that I have M and I have X power – 1 okay so I can write this as M by X plus new okay so this is the form of the momentum equation that I will be working with for the Falconer scan flows okay now what are the boundary conditions at Y equal to 0 U and V will be 0 no slip boundary condition and Y going to infinity my U approaches my free stream velocity now be careful your free stream velocity is function of X it is not a constant anymore okay so which is actually of the form C X power M okay so now how do we reduce this to an ODE the same approach that we did with the Blasius flow you assume a similarity variable let us assume the same similarity variable that we got for the flat plate flows okay because functional dependence of boundary layer thickness on X is going to be the same whether you have a pressure gradient term or not therefore your eta which is a function of Y by delta that functional dependence of delta is still the same therefore you can assume the same similarity variable holds good okay only thing here U 8 is a function of X okay so therefore you can write this as Y square root of C by new into this is X power M this is minus 1 X power M minus 1 half okay M minus 1 by 2 and also we can define the stream function and show that stream function is related it is a function of eta the same way that it was there for the flat plate also so those things remain unchanged where your stream function is defined such that U is equal to V ? by dy V is equal to minus okay so till now this is nothing new okay this is the same thing we did for the Blasius solution so all you need to do is substitute for the derivatives now okay in terms of the stream function of stream function is a function of F so therefore you can calculate U V and you can substitute all of them here now only thing is you have this M and U 8 is a function of X okay so once you substitute into this you will find that that I will leave as a nice exercise to you it will reduce to a ODE in terms of F and eta without any terms from Y and X appearing okay so therefore this confirms that a similarity solution to this class of problems is possible it is not too difficult to show once you plug in all the velocity gradients and your velocities into this expression let me call this as number 2 your stream function already satisfies the continuity okay so you have to just plug it in here and substituting into 2 you get your final ODE which is free of X and Y terms it is only function of F and eta okay so therefore this shows that for this kind of flows similarity solution is perfectly possible okay and this is the Falkner scan similarity solution so now you can see now here it is also a function of M okay now if you put M equal to 0 okay what happens this term completely vanishes you have 1 by 2 F d square F by d eta square equal to 0 so this is nothing but the Blasher solution okay so in fact without even touching the Blasher solution we can straight away started you could have started from Falkner scan and showed Blasher solution as a limiting case okay so all kinds of problems can be approached in this manner you can put any value of M that you are interested in for that particular configuration and get the similarity profiles for that particular configuration so this is a function of M right here so what are the boundary conditions the same boundary conditions that have that we have used for Blasher solution apply here ? equal to 0 you are F equal to 0 why you are F equal to 0 which boundary condition does it satisfy no slip correct but exactly which whether does it correspond to U equal to 0 or V equal to 0 U equal to 0 what is U in terms of F what is the relation between U and F U by U 8 therefore F equal to 0 does it correspond to this so then U equal to 0 should be what F prime should be 0 so then what what does it correspond to V equal to 0 so go back to your the way that we derived the Blasher solution look at the expressions for U and V okay so if you put F equal to 0 then you then only your V become 0 because already from the condition that U equal to 0 we know that F prime equal to 0 so for we have V to be 0 then this has to be 0 okay and what is the remaining boundary condition ? going to infinity the same same thing that we have done for the flat plate okay so what is the condition at ? going to infinity U by U infinity should be 1 or DF by D ? F prime should be equal to 1 okay so so this is this is called the Falkner scan similarity solution so now how do we solve it shooting method okay exactly the same same technique that you did for Blasher solution okay you have boundary condition for F you have boundary condition for F prime but you do not have a boundary condition for F double prime at ? equal to 0 so you have to guess something and finally match this satisfy this condition okay you do not worry about because even if you have this equation I will just write down if you apply the shooting method it will simply reduce to three ODEs which I will just write and stop there or maybe I can write it here so the first OD will be what F prime equal to G right second OD will be G prime equal to H and what is the third OD you substitute in terms of F G and H here okay so that will be H H prime will be equal to minus you have M plus 1 by 2 into F into G G prime or H D square F by D ? square H okay this has to be F into H plus M into 1 minus D F by D ? the whole this is yeah so this has to be F prime square so this has to be G square okay so this is my third OD so these are the three first order ODEs okay all I have to do is March from ? equal to 0 till ? some value of 8 or 10 okay and the same set of boundary conditions you know you guess by Newton's method successively you reach to the better guess and then check the condition that at ? equal to large value some ? equal to 8 or 10 your F prime which is nothing but your G should be equal to one okay so that equation has to satisfy till then you keep on iterating by Newton's method you keep guessing better values and you have to solve the set of ODEs for all the points in the domain okay the same exactly the same technique all you need to know is the value of M so which value of which configuration that you are looking okay so once you know the configuration the solution procedure is identical for different for each configuration you get a class of solutions so we will stop here so tomorrow we will complete the heat transfer part of the Falkner's can solution also.