 So, let us get started with our second session. So, what we were up to was that we are up to conservation of momentum that is famously called as Navier Stokes equation. So, the Navier Stokes equation what it is essentially doing is the conservation of momentum. So, we said that there is rate of increase of the x momentum minus rate at which the x momentum enters and rate at which the x momentum leaves and that is equal to the sum of the x component forces applied to the fluid in the control volume. So, various forces acting are normal surface forces and body forces. Surface forces are per unit area and body forces are per unit volume. Surface forces are normal stresses, shear stresses, pressure. So, we have surface forces which are contributed basically because of normal stresses that is because of stretching as I said del u by del x del v by del y or del w by del z. Shear stresses because of angular deformation as we saw in the morning del delta alpha or delta beta that is essentially because of velocity gradients in the other direction other than the velocity itself that is del v by del x that is v not in y, but in x or z direction would contribute to the shear stress. And similarly u not in x, but in y and z that is del u by del del y and del u by del z would create angular deformation. So, not create would be responsible for angular deformation and next is the pressure. So, what is pressure? Pressure is as I said it is because basically because of intermolecular collisions. So, we should not get confused both normal stresses and pressure are normal to the surface, but the cause for each one of them are different as professor around told just before we went for lunch. And body forces are gravity forces, Coriolis forces, centrifugal forces, electromagnetic forces, electrostatic forces you can go on adding like that various body forces, but usually we do not consider all those influence of the body forces in the classroom problems. But in the derivation at least in our case we would keep them as f x, f y, f z f x being the body force in the x direction, f y is the body force in the y direction, f z is the body force in the z direction. So, now this is what we did just before we went this is the mass flow rate in the x direction rho u delta y delta z into rho u that is delta y delta z is the area u is the velocity rho is the density. So, this gives the mass flow rate in the x direction the momentum in the x direction entering is u into this m dot in the x direction. Similarly, mass flow rate mass momentum because the mass flow rate in the x direction is which is getting out is given by the same term that is m dot x plus del m dot x upon del x into delta x. If I do that I get the same term here again plus del of rho u square by delta x del x into delta x plus delta y into delta z. So, x direction there is no problem in the y direction I have rho v delta x delta z is the mass flow rate in the y direction, but that is this mass flow rate is going to carry an x momentum because of the velocity u. So, that is at the entrance that is what is the entering what is getting out is the same term plus del of rho u rho u v upon del y delta x delta y delta z. Similarly, one can go ahead and write for the z direction. So, there is no problem. So, now if I do the book keeping that is I will let me handle only the left hand side left hand side terms of this that is rate of x momentum increase minus rate momentum entering plus rate at which the x momentum leaving. So, entering is negative getting out is positive. So, what will happen the first terms of the getting out will all get cancelled out this will get cancelled out with this term this term will get cancelled out with this term. So, I will be left out with rho u squared let me write that what am I going to be left out with I am going to be left out with del of rho u squared upon del x plus del of rho u v upon del y plus del of rho u v upon del z plus del of rho u by del t in all of these terms I have taken out delta x delta y delta z that is what I am writing is LHS upon delta x into delta y into delta z this is what I am writing. So, what is this term del of rho u by del t that is the rate of that is this term that is rate of increase of the x momentum that is with respect to time. So, that is the x momentum is del of rho u delta x delta y delta z this is of course the volume x momentum is rho u. So, del of rho u upon del t is the rate of change of x momentum. So, the left hand side is these are the terms what we got. So, if I expand this if I expand this let me go on expanding the each of these terms first let me keep let me expand this that is u into that is del of the first term that is from this del rho by del t plus the second term is rho of del u by del t whatever I done I have written I have expanded this term first is rho del u by del t that is this plus u del rho by del t that is similarly plus of what will be coming out from this term del of yeah del of no rho u square I have taken out one u. So, I should be left out with I am writing this as rho u into u this term rho u into u that means once I am keeping rho u constant and then differentiating u next I am keeping u constant and differentiating rho u. So, that means u I have taken out that means I have kept constant. So, I should be left out with del rho u upon del x plus plus of rho u rho u means rho I have kept outside. So, it should become u del u by del x similarly plus of yes here again I am writing u into rho v. So, what will happen del rho v by del y plus plus v del u by right plus del of rho w by del z that is u into rho v by del z that is u into rho w plus w rho del sorry w del u by del z is that right. So, what is this this is equal to LHS upon delta x into delta y into delta z. So, what is this term what is this by the way can anyone see what is this yes yeah why this is nothing, but my conservation of mass can you see that or you need to expand that. So, this is nothing, but if you expand this this is nothing, but d rho by d t plus rho of del dot v equal to 0 this is that this is my continuity equation. So, I can cancel this. So, what am I left out with what is this by the way del u by del t plus u del u by del x plus v del u by del y plus w del u by del z this is a total derivative of u. So, rho d u by d t is equal to LHS upon delta x by delta x delta y delta z. So, this is my left hand side let us take stock what we have done see what have I done. So, we took this is the left hand side which we are budgeting now. So, we took the control volume and we took the budgeting of x momentum only and when we substituted all of that here we ended up for left hand side of course, I have taken per unit volume if I take per unit volume I get rho d u by d t. Now, we have to take care of right hand side. So, what is there on the right hand side some of the forces. So, we need to take care of surface forces and body forces. Let us take care of first normal and shear stresses. Let us see how do we do that. So, what are the normal stresses and shear stresses which are acting you see the notation here please watch this carefully. This is sigma x x this is sigma x y sigma x z first thing sigma x x are what is sigma x x those are normal. Sigma x x and sigma y y are normal stresses and sigma x y sigma x y sigma x z sigma y x sigma y z they are shear stresses. Now, what is the notation what do I mean by x and x the first subscript represents the direction of the normal to the plane on which the stress acts that is this is the x direction it is normal to the plane on which this is the direction this is the x direction of the normal to which this to the plane on which this stress is acting. Second is the direction that is x now if you do not understand this here we will understand in sigma x y y represents the direction in which my stress is acting x represents the plane the that is the direction of the normal to the plane that is this is the plane in the x plane I have taken. So, that is why this is sigma x x sigma x y and this is sigma x z similarly this is this plane is in the y plane. So, that means all these stresses here would be having first subscript as y and this is x direction so sigma y x this is y direction so sigma y y this is z direction that is why sigma y z. So, this is the notation of normal stresses and shear stresses now let us see what are these. So, now what is that I have done if I take a normal stress sigma x x if I take a normal stress if I do the bookkeeping let me do that on the white board so that we do not get lost that is if I take the control volume same control volume which we have been taking which is delta x delta y delta z the normal stress in the x direction is sigma x x. So, what would be the force in the x direction this is delta x delta y delta z. So, stress into area should give me the force so sigma x x into delta y into delta z now what is this stress which is acting here now you might be asking me all this while we used to write in the inward and then outward here why am I writing both the stresses outward that is the sign convention that is the sign convention. So, usually normal stresses let me write here normal stresses are considered as outward positive normal stresses are considered outward positive this is the notation we use positive and for pressure inward is considered as positive this is the notation by convention. So, what is getting out now sigma x x delta y delta z plus del of sigma x x by delta del x into delta x delta y delta z is that ok. So, similarly I can write for pressure what is the pressure acting here p into delta y delta z what is the pressure acting here what will be this pressure p into delta y delta z plus del p by del x into delta x delta y delta z which is the other stress I should be taking in the x direction means let us go back and see which all are contributing in the x direction this sigma x x we have taken pressure of course we have taken in the x direction which is the other stress which is contributing sigma y x sigma y x. So, let me take that so that is sigma y x will be acting on this this is the shear stress which is acting on this surface. So, this will be this shear stress is here at the bottom is sigma y x delta into first y represents the y plane x represents the direction sigma y x is acting over area which area delta x delta z delta z and what is getting out the shear stresses should be you see please note this arrow mark which I have drawn the arrow mark here is towards left and this is towards right why why the directions have taken opposite for normal stresses I took both outward, but for shear stresses I am taking 1 towards left and 1 towards right why shearing means what it has to shear shearing means it has to occur only when it is occurring in the opposite direction one stress has to be in the left to right and the another stress has to be from right to left then only it can shear shear means what it has to get cut it has to have the feel of cut. So, it has to shear means these stresses should be in the opposite direction. So, this stress should be sigma y x into delta x delta z plus del of sigma y x upon upon delta del y into delta y delta x delta z. Same thing I have written here ok. So, same thing I have written here. So, if we do the book keeping what will happen what we what is that we are saying we are saying that some of the forces acting some of the x component of the forces acting. So, that means positive force means it should be in the positive x direction. So, this force will get cancelled out with the this force and this force will get cancelled out with this one and this force will get cancelled out with this one. So, what am I left out with some of the forces means del p by del x, but what is the sign it takes del p by del x it is acting you see it is acting in words this is in the negative x direction. So, that means this is minus del p by del x. So, similarly what about this what about this term I am I am taking away delta x delta y delta z per unit volume I am writing plus del of sigma x x upon del x. Next plus del of sigma y x upon del y and plus del of sigma z x I have not shown z because my figure becomes clumsy I have not shown z, but z x by del z plus f x plus f x is equal to r h s upon delta x delta y delta z. So, what was there on the left hand side on the left hand side I had rho d u by d t. So, that means I have this is equal to rho d u by d t this is the x momentum equation. But have you solved what are we writing equations for what are we writing equations for what are we doing we are writing momentum equation, but why are we writing momentum equations and conservation of mass for solving velocity profile and pressure. But do I see velocities everywhere no that means still these stresses are to be expressed in terms of velocity gradients. You see we said in the morning normal stresses should be stretching it has to be velocity gradients either del u by del x or del v by del y or del w by del z. So, similarly sigma y x means it should be del v by del x plus del u by del y. So, it has to be represented in terms of velocities we have not done that let us do that. So, before we do that this equation is called as Cauchy's equation which is expressing in terms of stresses. So, I think now I will stop using white board because I will be landing up into huge equations. So, this is what is that we have done. So, this was the right hand side and this is the left hand side which was there rho du dt and this is the pressure force this is the force because the normal stress in the x direction this is the shear stress in y and x plane and then this is the shear stress in z x. So, this is the body force in the x direction. So, what is that we are getting this is the Cauchy's equation. Now our attempt is to express the sigma x x sigma y x sigma z x in terms of velocity gradient. So, if I do that sigma x x is equal to 2 mu del u by del x minus 2 by 3 mu del dot v I have not derived this for you I just like that shot here. There is a lengthy derivation I am going to present that I am not going to explain that to you I have at the end of this presentation those derivations for now you are going to take my words and take that sigma x x equal to 2 mu del u by del x minus 2 by 3 mu del dot v. So, sigma x y is the shear stress which is mu into del u by del y plus del v by del x this is quite easy to see why quite easy to see I say viscosity what is the definition of viscosity shear stress to shear strain. So, shear stress is sigma x y and shear strain is what we had derived in the morning delta gamma dot. So, that is the shear strain. So, shear stress to shear strain rate this is not shear strain this is actually shear strain rate because velocity is sitting per unit time. So, that is viscosity so similarly sigma x z I can seeing x z itself I should be able to write now why because the velocities involved are u and w, but not in the same direction that is del u by del z and del w by del x. So, it is quite straight forward once you understand stretching and strain. So, if I substitute this sigma x z sigma x y and sigma x z in my previous equation what is that previous equation we just now derived right this is del p by del x del sigma x x by del x plus del sigma y x by del y. So, if I substitute these term in this equation if I substitute those in the above equation which I showed earlier that is the Cauchy's equation I am going to end up with a huge equation something like this nothing to get bogged down it is very systematic actually. So, if I expand this in order to expand this I am making an assumption what is the assumption making from this step to this step what is the assumption I am pulling out this viscosity that means what I am making an assumption that viscosity is independent of direction I am pulling out here I am pulling out here that means viscosity is independent of direction x y and z. So, if that assumption if I make and pull that out the first term in this is del squared u 2 del squared u by del x squared, but I am keeping only one del squared u by del x squared and the next one I am writing it here that is mu into del squared u by del x squared that is the next one of this is written here one of this is written here similarly I have del squared u by del y squared which is what I have written here and del squared u by del z squared I have written there. So, I have consumed first term here first term here and first term here and what is left out is the second terms of each of these brackets that is del by del x of minus 2 by 3 mu del dot v that is here minus 2 by 3 mu del dot v what is del dot v by the way for those people who are not with me for del dot v what is del dot v del dot v is we are just derived in the morning del dot v is del u by del x plus del v by del y plus del w by so no confusion on that score. So, that is what if we substitute we are not substituted we have just kept it as it is del dot v and the second terms that is del by del x of mu del u by del x I have finished and second term del v by del y you see here it actually if I involve this what do I get mu del squared v by del y del x in which order you differentiate it does not matter this we have studied in mathematics. So, del by del x of del v by del y next term that is del squared w by del z del x I am writing it as del by del x of del w by del z. So, if I do this and if I expand this that is rho del u by d u by d t minus del p by del x plus this is del squared u all of this and what is this by the way what is this del dot v. So, 1 minus 2 by 3 1 minus 2 by 3 I will be getting 1. So, that is what is this del by del x of mu by 3 del dot v plus f of x similarly this is the x momentum equation similar arguments one can make and derive and get the y momentum equation and derive again z momentum equation that is what is being written here. This is x momentum equation this is y momentum equation this is z momentum equation. Let us take stock what is each term you know each of these equations what is this term rho d u by d t let us not forget this is the net momentum that is momentum increase minus momentum entering plus momentum getting out that is what is this rho d u d t that is net rate of change of momentum is rho d u by d t. Next is this happens to be total derivative this happens to be total derivative of u and this is coming out because of the pressure force and this is this two terms are coming out because of the viscous forces this is pressure force this is these two are viscous forces and this is body force. So, we should never forget these terms. So, this is inertia force so this is pressure force this is viscous force this is also viscous force this is body force. So, now if there is no body force f x becomes equal to 0 or f y becomes equal to 0. So, if I have to this is same thing I have written in vectorial notation that is rho d v d t minus del p plus mu del squared v plus mu by 3 del of del dot v plus m. So, you if you are not comfortable with vectorial notation do not worry at all you can continue to go ahead with this non vectorial notation except this del dot v that del dot v we have used because otherwise if you are uncomfortable with this del dot v you write del u by del x plus del v by del y plus del w by del z period there is no problem at all. So, this is the navier stokes equation which we have derived. Now what should happen if viscosity is equal to 0? What should happen if viscosity is equal to 0? It has to reduce to Euler's equation that is rho d v d t equal to minus del p plus f we have all derived this in fluid mechanics not in this way it looks for us I have written that in the next term the same thing that is rho d v d t I have expressed as rho del u by del t plus u del u by del x plus v del u by del y plus w del u del u by del z is equal to minus del p by del x what is this term? This is the body force minus rho g if I take along a streamline let us say s. So, what will happen and I take steady flow this is not there and if I go along streamline these two terms would not be there and I am going streamline. So, rho u del u by del s minus del p by del s minus rho g del z by del s this is what I think most of us are very convergent with when we are deriving Bernoulli's equation is that right? If I integrate this I am going to get Bernoulli's equation. Bernoulli's equation what are the assumptions stated in Bernoulli's equation? One is steady flow second is it is for incompressible that is density variations are not there and the other thing is it is for inviscid flow when I say inviscid viscosity is 0 that is essentially what I did in the Navier-Stokes equation. In fact so this viscosity if I put 0 only I am getting this Euler equation when I integrate that Euler equation I am going to end up with this equation. So, that is that is the greatness or actually historically it did not come like this. Historically inviscid flow was understood first that is Bernoulli derived Bernoulli and Euler together represented in fact Bernoulli lever wrote this equation in this form. Actually it is this is a very debatable issue actually, but it is said that Euler could see this equation in this form in the Bernoulli's text book this is not written in this form actually. So, we can say that conveniently Euler and Bernoulli together gave us this equation, but this was derived before the Navier-Stokes equations were derived. Navier-Stokes equations came later on where viscosity was there. So where viscosity was equated to 0 and then reduced to Bernoulli's equation. Historically it is reverse first Bernoulli Euler and then we have got the Navier-Stokes equation. So coming back so these are the let us write all equations together. So this is the continuity equation d rho by dt plus rho of del u by del x plus del v by del y plus del w by del z equal to 0. This is the x momentum equation this is the y momentum equation and this is the z momentum. So is there how many unknowns are there here? There are 4 unknowns u v w and p. You may ask how do I get density? Density I will be getting by p by r t that is the gas equation if I take I should be getting by density equal to p by r t that way I will have 5 equations and 5 unknowns. So here it is 4 equations and 4 unknowns. Mathematically it sounds that okay I can get a closed form solution that means number of unknowns are equal to number of equations, but it is not it is it is not mathematically tractable that is I cannot get closed form solutions or I cannot get analytical solutions for there is no general solution for Navier-Stokes equation. If you were to have general solutions for Navier-Stokes equations our life would have become easy we could have built bridges we could have designed bridges we could have designed all fluid mechanic machines much easily unfortunately it is not so. So it is there is no closed form solution the point is there is no closed form solution for these 4 equations to get what for to get velocity u v w and pressure why do I need these velocities and pressures to measure the to calculate the shear stresses and from there to calculate the friction factor and then to calculate the pumping power pumping power. So we to calculate the pumping power I need these velocity solutions. So this is mathematically well posed, but there is no general solution because it is non-linear second order differential equation. So why it is non-linear because u into del u by del x that product is there. So because of that product we get into non-linear. So that is non-linear second order partial differential equations. So these are Navier-Stokes why are they called Navier because they were independently derived by Navier and Stokes almost simultaneously. So Navier happened to be a French mathematician and Stokes happened to be English machinician. So both of them derived so that is why we call them as instead of conservation of momentum equation it is better to call them as Navier-Stokes equation keeping in the back of the mind we are conserving momentum. So with this we have derived the continuity equation and conservation of momentum equation that is the Navier-Stokes equation. But one question I postponed I said I will come back to that question that is sigma xx equal to 2 mu del u by del x minus 2 by 3 mu del dot v to get this I need to do at least 10 page derivation. But however we have done that thanks to professor Yuan this derivation is taken from foundations of fluid mechanics by Yuan. Yuan is none other than student of Von Karman. So that is why his book is so nice actually it is if someone wants to study fluid mechanics in mathematical way he can read this book Yuan. It is all language is very less English is less it is full of mathematics that is full of equations. So but then with one sentence to understand you have to spend even few days sometimes because it is so cryptic it is so cryptic. So what was I telling I was telling how to relate stresses with rate of strain I am not going to derive all of this I am just going to tell you the salient point it is actually this derivation runs to what is the number 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 10 transparencies I am not going to spend time on these 10 transparencies definitely no but if I would suggest that you go through this derivation if not write it out at least take a print out and go through this derivation if you do not understand something in this derivation please moodle or please put that question on the moodle we will get back to you but there is nothing great in this except there are three points I will just stress on those three points and the rest all is mathematics what are we saying in elasticity stress is proportional to strain that is Hooke's law that is what we are applying here also. Second thing is that there are the generalized Hooke's law states that each of the six stress components may be expressed as a linear function of six components of the strength that is again Hooke's law the validity of this assumption how should I believe this this has been validated by experiments for continuous homogeneous isotropic material I am taking my fluid particle also having these properties that is continuous homogeneous and isotropic. Now what am I trying to do is I am making three more assumptions that is the same thing I am stating as three assumptions one is stress component can be expressed as linear functions of rate of strains number one number two the relation between stress and strain is invariant of coordinate that means if I take x y z like this and now if I transform that to other coordinate system my relations between stress and strain do not change that is what it means the relations between stress components and the rate rates of strain component is independent of or do not change to coordinate transformation consisting of either rotation or mirror image. So that is what mirror reflection of axis and the stress components must reduce to hydrostatic pressure when all the gradients of velocities are 0 this professor Arun has already told us if you put all the velocity gradients 0 then I have I am supposed to end up with simple hydrostatic pressure. So these are the under these three assumptions if I write sigma x x as proportional to sigma epsilon x x epsilon is rate of strain in x direction epsilon y y gamma x y this is normal strains these two and this is shear strain similarly if I set it up for x x y y and x y actually you can do in 3d but here actually we are doing in only 2d 3d becomes tensor notation I will have to use so in order to avoid that professor Ivan has given only in 2d I think it is quite nice because we can see how it goes. So now I am transforming this transforming to new transformation that is this is x and y my new coordinates are x prime and y prime. So I write again stresses in new coordinates and relate stresses new stresses to the old stresses and then relate these stresses again in terms of no problem there is some distortion but when you download it is not appearing so basically if I go on doing the algebra and substituting various coordinate transformation if I do I am going to end up with I have to skip all this I cannot help it I will get this equation sigma xx equal to 2 mu epsilon xx plus b blah blah blah you see here I am getting sigma xx equal to 2 mu epsilon xx b we had taken minus 2 by 3 how it is coming also it is stated here b equal to minus 2 by 3 I had written this as del dot v and here we are writing minus p because in the normal stresses we have added the in the in his approach he has put the pressure also few textbooks take in the normal stresses pressure also keeping in the mind outward stresses are positive and inward pressure is positive so that is why this becomes minus okay. So I am not I am not here trying to prove all this or explain all this I would implore all of you to go through this derivation carefully so that you understand otherwise this becomes a black box where from this relation I have written you will not be able to appreciate just to explain that I had to use 10 transparencies. So anyway that is the that is how we have derived navier stokes equation and continuity equation before we go to turbulence maybe we can take two three questions and then professor Arun will take over for turbulence so I am going for institute of road and transport technology road you have any questions I have two question process yes question number one the concept the concept of thermal and boundary layer thickness place a dominant role in separating the viscous region and invasive region right frozen correct my question is what are what are the effects of variation of thermo physical properties of new boots passing over the date on the thickness of thermal boundary layer and velocity boundary layer hydrodynamic boundary layer yeah okay see I think last time also you had asked this question let me rephrase this question how are the thermo physical properties going to affect the velocity boundary layer thickness and the hydrodynamic boundary layer thickness let me answer this actually it is a little premature but still I will go ahead and answer delta and delta t is a function of Reynolds and Prandtl delta is a function of Reynolds number alone delta t that is the thermal boundary layer thickness is a function of Reynolds and Prandtl so we will be deriving this in tomorrow's lectures we will be deriving not tomorrow Monday's lectures we will be deriving this so if I have Reynolds number and Prandtl number Reynolds number is rho V D by mu so the two thermo physical properties are rho and V sorry rho and mu and Prandtl number all are thermo physical properties mu Cp and K so that way the velocity boundary layer thickness and the hydrodynamic boundary layer thickness are affected by the thermo physical properties okay professor thank you very much professor thank you very much thank you Javelpur college any questions please explain the as we move from the leading edge the question asked is why does the boundary layer thickness increase with the increase of the location from the leading edge see one way of looking at it is actually the velocity gradient as I move away from the plate that is the leading edge of the plate the velocity variation is felt for longer distances extended distances why because initially it cannot extend itself directly to the farthest end it takes certain distance for itself to extend to the farthest distance so that is why it has to the in fact professor Arun had taken the example in the morning that is the farthest person will not be able to feel what is happening very much near the plate so similarly here also as I move away from the plate that is the leading edge of the plate only the velocity gradients are felt for larger distances from the flat plate tip so that is the reason why the boundary layer thickness increases with the increase of the location as I move from the leading edge of the flat plate ok thank you sir one more question please sir yeah my question regarding with the conservation of momentum ok the momentum is balanced by the forces why do not we consider the viscosity in surface forces surface forces we are taking why do you say that it is not there yeah the question asked by one of the professors is that why do not we consider viscosity in the surface forces surface where is viscosity coming into picture in the normal stresses and the shear stresses and what are these stresses going to contribute they are going to contribute surface forces so viscosity is indeed taken into account in surface forces through normal stresses and shear stresses so that answers your question professor ok thank you