 So, good morning again. Yeah, so I hope you are all getting used to the number of derivations that we are doing so far I think this entire course you will be doing such derivations and finding similarity solutions. So, it is not so difficult except that you should get used to the notations and things like that. So, tensor notation is one of them which we introduced yesterday I think if you have time you please read up a little bit on the tensor real notation it is not that difficult to understand also you should have some basic knowledge of solution to ordinary differential equations because in all the similarity solution problems we will convert the partial differential equations into an equivalent ordinary differential equation find the similarity solution so that has to be done numerically and I expect I will be giving some assignments where you have to solve those OD is numerically by some techniques okay so otherwise if we just stop there and give you the solution you may not appreciate exactly how the solution is come and it will all look to a mathematical okay unless you do it yourself you will not appreciate the solution process. So we will look into the second law of for a close system so we were till now deriving the conservation equations we were looking at two different approaches we started from a Cartesian coordinate control volume and derived the other one we had a coordinate free representation applied Reynolds transport theorem to a closed system and converted the closed system rate of change of properties to rate of change with respect to an open system and that is another way of deriving in a coordinate free representation. So after we have looked at the first law it is also very important to look at the second law because as we all know most of the times we stop with the first law without too much of botheration to the second law but if you look at any practical heat exchanger device entropy generation is very important you know there are lot of irreversibilities within the system and we have to quantify to some extent the extent of these irreversibilities and we should also identify what are the sources of the irreversibilities okay so that is where we are going to apply the second law and we are going to derive a conservation of what is called as conservation of entropy okay just like you have conservation of mass momentum and energy so there is also a possibility of deriving conservation of entropy and that is what we are looking at here. So we start with applying a Clausius inequality for a closed system okay and if you express this in the form of a rate equation so you know that the total derivative dds by dt should be greater than or equal to 1 by t into the differential amount of heat transferred okay so I will just represent this as ? q dot okay. Now let us apply the Reynolds transport theorem to express this total derivative for a control mass with respect to the partial derivatives in a control volume so you can define a property called a specific entropy okay and in the Reynolds transport theorem the a is a property per unit volume right so then that that can be expressed as ? times the specific entropy so we can expand this using the Reynolds transport theorem like this now the right hand side applies for an open system okay so now for an open system therefore if you substitute this Reynolds transport theorem expression into one so this is your equation which defines the conservation of entropy for a open system okay so this is nothing but conservation of entropy for open system okay or you can also if you do not want to look at this as a conservation of entropy you can also look think this as an entropy balance okay entropy balance equation okay so now what we are going to do all the surface integrals can be converted to volume integrals by Gauss divergence theorem and therefore we can further put this I can take the integral over V out so d by dt ? s d V plus this will be again a divergence operation right here this will be ? dot ? s V okay and I am going to bring this to the left hand side here so this will be again minus ? Q okay so this is actually a vector right here so I am just going to give this vector notation okay so this is all integrated over the differential volume because I have already applied the Gauss divergence theorem so on the right hand side that that should be equal to the rate of generation of entropy okay so therefore I can I can also say that if I express this in okay or maybe I can also look at this way I want to look at the entropy generation for the entire volume so I can also bring this inside minus a s dot gen okay d V equal to 0 right so that I can separate out the integrand okay and write a finally a partial differential equation okay so strictly speaking this entropy generation is per unit volume here rate of rate of entropy generation per unit volume so if I can multiply by ? V okay so I can say this can be multiplied by d V okay so then I can put this on the left hand side group all the terms together and that can be equal to 0 therefore the integrand has to be 0 so then this will give me my s dot gen should be equal to d by dT of ? is dV plus this which I am going to write in terms of the total derivative notation which is ? dS by dT okay minus del dot ? Q I think this should be ? Q by T okay I think I omitted by T because that that is there in the Clausius inequality okay so this should be del so this divergence operator is on this entire term okay so so this is how I finally can calculate the entropy rate of entropy generation now provided I can also expand this particular divergence operator on del Q by T so I can rewrite this a little bit so I can express this as s dot gen is equal to ? dS by dT plus 1 by T so I can expand this I can split it up into I can take 1 by T out del dot I can write this as del Q that is one the other term is 1 by T square del Q dot del T okay right so because this is a gradient operator so gradient this is a divergence operator on a vector so this has to be a scalar so therefore you should make sure that you only get scalar operators out of so this is again a vector dotted with vector you you basically get a scalar here okay so you make sure that the final resulting operation also is consistent the original operator is that clear this splitting up is clear okay so I am just so if you have a gradient the same way that you write you have to write the same way for a divergence operator also only thing you should make sure the resulting operator is also scale giving you a scalar okay so therefore now what I can do here so now I am going to evoke the Gibbs theorems one of the Gibbs theorems if you can remember so for a close system again you know that du is equal to Tds-pdv okay so I am going to write as pd x 1 by ? okay so this is coming from straight from your first law right where you substitute Tds for ? Q and you have pdv work okay so this I can write in terms of the rate equation and you can tell me how this should look if I want to write ds by dt okay so if I if I express ds by dt so I can write this as du by dt so I can take 1 by T on the other side minus so this should be if you can express this as d ? by dt so what should be the remaining term ?2 and T right so I am I am taking this T dividing it all the side so I am now going to multiply this by ? okay throughout so that this we can be written as oh so this that should be a P here P by ? T right so this is a little bit of manipulation to suit my convenience here because now I have to find some relation for ? ds by dt so I am just expressing this from the Gibbs equation okay I am connecting that to change in the internal energy and a change in density with respect to time now if you look at incompressible flows okay for that this has to be 0 right so there is no change in density with respect to time therefore the change in the entropy has to be directly the change in the internal energy of the system that is directly linked there so let us also express the first law now if you write the first law so you write d ? du by dt should be equal to minus ? dot q okay plus u ? okay so I am writing in the final coordinate free representation okay so instead of saying del dot q you can also say del dot del q in fact if you if you want to maybe rewrite to just avoid some confusion you can also write this in terms of yeah okay just to avoid some confusion you can express this as q instead of del q here so that finally it is consistent with that maybe you can do this small change okay so both are consistent all right so therefore what I am going to do now is to substitute for ? du by dt from this expression into this and this I am going to substitute into this okay so the resulting expression will be s. Gen will be so this will be basically ? ds by dt is nothing but ? by t du by dt which will be minus ? dot q by t plus ? ? by t okay so that is this term plus you have these two terms right plus del dot q by t minus 1 by t2 q dot del t okay so these two terms cancel off therefore your final expression for s. Gen turns out to be if you you can also now substitute the Fourier's law for q as k ? t and express this as minus okay so this is minus of q dot which is again minus k ? t so that will become k ? t2 divided by t2 okay plus ? by t ? okay so therefore the components of entropy generation or two so one is coming from the entropy generation due to the conduction part okay so within the system the conduction of heat the other is due to the viscous dissipation okay so this part till here I think straight forward only after this we apply the Gibbs law and just manipulate a little bit so that we can eliminate some of these total derivative terms okay so finally we write everything in terms of the heat transfer by conduction and the viscous dissipation so now there was this person called Adrienne Bayjan I think there is one textbook also which you are referring convective heat transfer okay so he came up with a very ingenious method that he wanted to give a non-dimensional number which is actually referred to as the Bayjan number okay so this is called Bayjan number so notation is given as BE you wanted to look at the contribution of the entropy generation by means of conduction okay as a fraction of the total entropy which is generated okay so that he has taken this term on the numerator divided by this entire thing on the denominator okay so this is called as Bayjan number in fact the idea of entropy generation due to heat transfer versus idea so this derivation actually is done by him and there is also a paper in 1990s you know with this student where he has derived his expression for heat transfer irreversibility due to heat transfer okay and I think the Bayjan number was credited to I mean it was basically given due to his contribution for this work particularly and this measures if your Bayjan number is some value say about to it tells you that the majority of entropy generation is through by conduction okay and if it is much much lesser than to it tells the majority of entropy generation is through viscous dissipation okay so basically this can be plotted just like you plot your isotherms and heat flux lines you can plot the entropy generation due to each of these and you can visualize and see how it looks okay so very useful tool I mean and you can see where the in which location in a particular system your entropy generation is by the conduction part and where it is by viscous dissipation and if you can probably try to reduce the viscous dissipation by some method so that will also reduce the entropy generation yeah both are dominant kind of both are equally important okay so any questions I think I wanted to just finish the governing equations therefore I wanted to touch upon the second law also apart from the first law which you are already familiar okay so I think many of the textbooks do not talk about the second law conservation and things like entropy generation much okay so any if any other questions on this any doubts any anything that requires some clarity or is that okay okay fine so then we will proceed to a new topic now still we are in the introductory portion of this course where we are deriving the conservation equations and so on so you have seen that the Navier-Stokes equations are quite complicated so we cannot solve them analytically except when we are making some approximations now we can also bypass this and we can introduce another set of equations which can be solved in place of Navier-Stokes equations especially for two dimensional flows and two dimensional incompressible flows okay so for 2d incompressible you know we have to solve if you solve the Navier-Stokes we need equations for u v and p okay so u v p and if you are solving energy again temperature okay now we do not have a separate equation for pressure okay we have equation for u and v and continuity equation which is like a default equation so therefore numerically there are some techniques to overcome this hurdle where we construct an artificial pressure equation and things like that if you want to overcome that we can rewrite the Navier-Stokes equation into what is called as a stream function vorticity formulation okay and that is very useful when you are looking at 2d incompressible flows okay there you have only two variables to equations you can solve that straight away and the equation solution is also slightly simpler than solving the Navier-Stokes okay so we will just quickly derive those formulation today and in fact in the project that you are supposed to do you will be using those 2d stream function vorticity equations with the energy you will be solving them numerically okay I will send you some reference papers and you will apply that to a problem of natural convection okay so now as in a natural convection will be covered you can because this is a nice rectangular cavity so the taking you know finite differences will be much easier alright okay so we will do that now so the next topic will be for 2d so this 2d is the approximation that is required okay we cannot write this in 3d again okay 2d incompressible flows okay so let us write down the Navier-Stokes equations first and from there we will try to derive the stream function vorticity equations so from the continuity equation so you have to tell me for 2d incompressible flows what is a continuity equation let us assume Cartesian coordinate system okay so du dx plus dv dy equal to 0 so now the complicated equations slowly will get simplified now as in when we go through the solutions okay and the x momentum du by dt plus would be equal to minus what about density has to be divided right plus dynamic or kinematic kinematic this is your Laplacian operator right so and similarly your y momentum which will be dv so we will also write your energy equation after we derive the stream function vorticity formulation will quickly go back to the energy equation and simplify that okay so what is your 2d incompressible energy equation we can directly write for temperature okay plus is equal to alpha because that is k by ? Cp which is nothing but the thermal diffusivity okay the thermal diffusivity alpha plus okay so now I am going to neglect the viscous dissipation now okay I do not want to put too many terms into this so I can safely neglect the viscous dissipation okay so this is neglecting okay so now this is more familiar to you have been working with these equations in your earlier courses now what we have to do we have to think a little bit see there is a pressure term in the momentum equations and we want to somehow eliminate it because as you know that we do not have a separate equation for pressure and when we want numerically solve that we want to simplify this problem so how can we probably do that take derivative with respect to y here respect to x subtract these two right so that will eliminate this pressure derivative term so exactly that is what we are going to do okay so we will differentiate x momentum with respect to y okay and so I am going to say differentiate y momentum with respect to x minus differentiate x momentum with respect to y okay so if you do that so you can group these two terms the temporal term you can say d by dt of so you can write this as dv by dx minus du by dy is that right okay so I am taking d by dt common so I am differentiating with respect to x minus differentiating this with respect to y all right so the other terms you get lot of terms here because now if I say I am going to differentiate this with respect to y so I have to split it up again right so this will be minus you have to tell me now so this is d by dy of you du by dx okay so I can write this as u into d square u by dy dx minus yeah du by dy into du by dx okay similarly this term can also be expanded that will give you minus v into d square u by dy square right minus du by dx into dv by dy this should be dv by dy into du by dy I am sorry okay let me check all the terms again yeah okay so the same thing now I am going to do differentiate the y momentum with respect to x okay so this minus this right so these terms will be positive here so plus you have du by dy into dv by dx plus you have u into d square v by dx square plus I am going to differentiate with respect to x into dv by dy plus v into d square v by dx dy okay so when I differentiate this with respect to x this and subtract they are going to cancel okay on the right hand side I will have nu into minus d cube u by dy dx square minus d cube u by dy cube plus so I will have these two terms right here d cube v by dx cube plus d cube v by dx dy square yes 1 2 3 4 5 6 this one do you by dou x yes you are right because we are differentiating with respect to x so all the other terms are correct please do this and check once so I am now going to group these terms together in some particular fashion and we will see that that grouping will help in reducing we can actually define a new function which can be substituted for those grouped terms so I am just going to group like this so you take common and what are all the terms common to you so you have this term you have this term so I can write this d square v by dx square minus d square u by dy dx so these are the terms common to you okay so again plus v now you have to tell me which terms I can group dou square v by dou x dou y minus all right so we have somehow 1 2 3 4 okay so four terms we have taken care and before that okay I am going to add the temporal derivative term so that is d by dt of d v by dx minus du by dy plus these terms right so this term is also taken care now we have 1 2 3 4 4 terms which are still on the left hand side that have to be grouped together so what I am going to do is I am going to take d v by dx common okay so if I take d v by dx so that means this is common so du by dx plus d v by dy which is nothing but two dimensional continuity which is automatically satisfied okay this is a nice trick to eliminate all the unnecessary terms okay plus next I am going to take du so this is actually minus du by dy out so du by dy so you have du by dx and d v by dy so that will also be satisfying continuity so this goes this also goes on the right hand side I can just group them as d cube v by dx cube minus d cube u by dx square dy okay so I am grouping these two terms these two terms okay plus d cube v by dy square dx okay minus d cube u by dy cube alright so I am just grouping them now what I am going to do is to introduce a function called the vorticity which you are all familiar right so how do we define vorticity many of you can recollect from your incompressible flows in fact you can get a clue from the terms that we have grouped d v by dx minus yeah exactly okay now you can see why we have grouped those terms because they are going to be directly in terms of the vorticity that we have defined and now we are going to take derivatives and check it will come exactly to those terms okay so for example if you now take the derivative of vorticity with respect to x now you please tell me what the term should be that should be exactly the first term on the spatial derivative term right so that is del square v by del x square minus del square u by del y del x okay so d omega dy should be what d square v by dx dy minus d square u by dy square okay that is this term right here okay so now if you take second derivative d square omega by dx square can you tell me what the second derivative will be okay so this is this term right here okay so regarding this no Q v by okay and this is no Q u by no y Q okay that is this term right here okay so therefore you find that are grouping all these terms make sense you can write them in terms of the vorticity and its derivatives therefore I am going to substitute in terms of vorticity so this will be d omega by dt plus u d omega by dx plus v d omega by dy should be equal to nu into so this is my governing equation for vorticity okay or you can also say that this is the vorticity conservation equation so have you derived this before earlier incompressible flows vorticity okay so this is a this is an important equation and it simplifies because now your pressure terms are not appearing anymore okay now still we are not done because we have simplified the momentum equations okay but the thing is still you have your u and v velocities appearing here okay and we have to somehow eliminate them huh stream function so how are we going to do we have to define a function which automatically satisfies the continuity equation and therefore you do not have to again solve for continuity separately right so therefore what is that function which does it that is a stream function okay so from the stream function you can calculate your velocities for example u is related to your stream function as d psi by dy and v is minus d psi dx so naturally you can see that du by dx plus dv by dy will be 0 so the stream function is automatically satisfying continuity so you can plug in for you and v in terms of the stream function and the other thing the stream function should also satisfy this equation because this is the this is how the vorticity is defined and vorticity is related to the velocity terms the velocity derivatives so therefore the stream function should make sure that it satisfies this particular definition of vorticity so if you substitute for stream function into that so what do you get so your omega will be so if I take dv by dx this is minus d square psi by dx square and du by dy minus so minus of d square psi by right so you have you can now substitute this in terms of the stream function therefore you have eliminated the velocities okay so this is your equation number one this is the conservation of vorticity equation number two is now since you have introduced stream function you have to solve for stream function okay and that is done by means of second equation which relates your stream function to vorticity okay so this is your second equation so now you have two equations two unknowns right one for psi and the other for omega so this is much better to solve rather than the Navier stokes correct okay so where you have to solve for three equations and you do not have an equation for pressure so that is why that is why it is very popular technique for numerical solution to two-dimensional incompressible flows okay most of the journal papers that you take for 2d incompressible flows still until recently okay where you do not did have a powerful computing facility to solve the full Navier stokes equations they were employing the stream function vorticity technique okay so and this is what you are going to try out also now the same thing in energy equation if you substitute okay so apart from the flow field you can also calculate for temperature so the energy equation becomes you can substitute for you again in terms of stream function so this is your energy conservation okay so you solve for three equations for three unknowns omega psi and temperature okay so these are the equation that you exactly have to solve in your particular project and I will be giving the papers you have to apply the corresponding boundary conditions to solve this problem so for a particular problem you have to do it and then you can you can probably look at steady state solution where you do not have to consider the time derivative okay only the spatial derivative and you can do this iteratively so there are couple of techniques like Gauss Heidel or Gauss Jacobi iterative techniques which you can do it and you can use finite difference methods simple finite differences to basically discretize these derivative terms alright okay so with that I think most of the conservation equations we have seen and also different variants of the conservation equations what I am going to do now is to slowly get into the theme of this course which we are going to apply for both external and the internal flows okay now we have to make some approximations to the Navier Stokes equations when we apply that to external flows we cannot solve the Navier Stokes as it is okay so these approximations are called the boundary layer approximations as far as the external flows are concerned and before doing that first we will try to non-dimensionalize the Navier Stokes equation we will try to define some non-dimensional numbers we will see what non-dimensional numbers are governing the flow and heat transfer parameters and we will also identify the regimes based on these non-dimensional numbers and finally when we go to boundary layers we have to use these non-dimensional numbers to make certain approximations okay so once the boundary layer equations are derived from there we will start our process of solving for different configurations okay so we will I will just introduce you little bit we have some more time about 7 8 minutes so I am going to talk about the different parameters different variables and how we are going to non-dimensionalize them okay so I think this is probably familiar to you because you have done this earlier in your incompressible flows and also advanced heat and mass so very quickly we will go over it and you can yourself try to non-dimensionalize I am going to write down only the final non-dimensional equations okay so here after we will be dealing only with incompressible flows and that also in two dimensions okay so all the complicated terms will be dropped off one only will retain terms in these two dimensions so 2D incompressible in fact I should not have it as the Navier stroke so the same thing applies here and anyway I will just do it write it quickly again I am going to also include the pressure term okay so here I am what I am doing is 2D also incompressible and steady so this is the approximation that I am going to bring here because after we non-dimensionalize and apply this to boundary layer flows we are pretty much going to do this for steady state solutions okay so we will stick on to this particular form throughout so all your most of your solutions will be for 2D incompressible and steady state and also we are including the viscous dissipation here to non-dimensionalize and see what kind of non-dimensional numbers come out okay suppose you take any flow say past an airfoil or past your flat plate okay so you have your free stream which is described by your free stream velocity and temperature and you can also maintain the surface of the airfoil okay at an isothermal condition such that your T wall is constant and let us say the characteristic dimension of this particular airfoil is given by the chord length which is L all right so this is these are some of the variables that you have fixed you know you have your U infinity T infinity defined okay the geometry is well defined and also you are fixing the boundary temperature to an isothermal condition so with these parameters how are we going to non-dimensionalize so the as far as the coordinate system is concerned okay let us take a coordinate system X and Y okay like this which are along the chord length and normal to the chord so I am going to define a non-dimensional coordinate system capital X which is X by L similarly capital Y will be Y by L and as far as the velocities are concerned I am going to introduce capital U okay which is what U by U infinity similarly capital V is equal to V by U infinity okay now coming to pressure okay so velocities coordinates are done so I am just going to introduce a capital P which is equal to the small p dimensional p divided by so how do we non-dimensionalize the pressure okay so I am going to cut the half term because it does not make much difference O infinity U infinity square okay so the free stream density is row infinity all right and finally coming to the temperature I will define a non-dimensional temperature ? such that at the wall the value of the non-dimensional temperature equal to 1 and in the free stream away so the value should be 0 so how do I non-dimensionalize that T- T infinity by T wall- T infinity okay so that at the location where at boundary T equal to T wall this becomes 1 and at the free stream somewhere that ? goes to 0 all right so with this I just write the final equation you can in fact derive this and check for yourself if you substitute these non-dimensional variables into the dimensional Navier Stokes equations so you are now going to tell me so we will write down only the final version so what will happen to the continuity this is right and we just substituting for the non-dimensional in terms of the non-dimensional variables there all right they cancel out everywhere okay as far as the momentum equation is concerned you du by dx plus V du by dy I am writing all in non-dimensional terms is equal to minus D P by dx plus something which you have to tell me what it is d square u by dx square plus d square V by dy square okay the same thing for the this is capital U dV dx plus V dV dy minus so these are all in non-dimensional variables so you have to check and tell me what terms should be there and similarly finally you have d ? dx plus I am going to introduce a non-dimensional viscous dissipation V star so you have to fill in the blank how what are the terms that are going to appear here okay so what should be the term right here ha Reynolds number so if your Reynolds number is very large so what happens this term is more important than the convective term is that correct it should be 1 by Reynolds number right 1 by RE based on the length so where your REL is nothing but ? U 8 L by ? correct similarly the same thing comes here 1 by REL now what should be the term here this half minute we are done okay so I am just going to give it you please check it 1 by RE Prandtl number okay where your Prandtl number is equal to u Cp by K which is also ? by ? okay and finally here I am just going to introduce a non-dimensional number which probably you have not encountered so this is going to be what is called as Eckert number by Reynolds number here where your Eckert number is defined as U 8 square by Cp into T wall minus T 8 okay so what I give you as a homework is to please substitute all that carefully and check whether we arrive at this non-dimensional formulation okay I think you can do that directly you can see that this comes directly straight away only this combination and this combination you have to check again all right so we will stop and we will meet on Tuesday okay we will start with the boundary layer approximation.