 In our previous lecture, we were discussing about the equation of motion for a fluid flow without any specific emphasis on what type of fluid it is to begin with and then we gave some particular attention to some special type of fluid which is called as Newtonian fluid that is a fluid that will obey the Newton's law of viscosity. So we found that for a Newtonian fluid if you look here you will see that for a Newtonian fluid we have come up to this relationship that the deviatoric component of the stress tensor tau ij deviatoric is nothing but cijkl ekl. So the cijkl is a fourth order tensor that maps a second order tensor onto a second order tensor. So there are 81 components and symmetry will reduce the number of independent components further but those are still too many number to characterize a fluid in a practical manner. So we will go for some special considerations which deal with homogeneous and isotropic fluids. So to get further into this issue let us go to the board and try to understand that cijkl this is a fourth order tensor. Now let us say that if it is isotropic it is possible to convert this into an isotropic scalar to convert this into an isotropic scalar we can take aid of some vectors. Why we can take aid of some vectors is because see to make it a scalar we should make it index free. This has now 4 indices to make it a scalar we should make it index free. So to make it index free we should have another repeated index i, another repeated index j, another repeated index k and another repeated index l. So each index can be donated by one vector because a vector can be specified by one index. So if you have a vector a that can I mean so to say donate the index i a vector b that can donate the index j a vector c that can donate the index k and a vector d that can donate the index l. So if you have cijkl into a i b j ck dl then this i is repeated j is repeated k is repeated and l is repeated that means there is an invisible summation of this over this and it becomes a scalar because it becomes index free. Now what makes it an isotropic scalar if it is an isotropic scalar then it should not be dependent on the absolute orientation of the vectors that means if you have the vector a and b let us say this is vector a this is vector b. Now if you rotate this system so that the vector a comes to this new location and vector b comes to this new location then the scalar should not depend on this rotation that means it would be same for the original orientation of a b and final orientation of a b there should not be any change. So that what it means is that the result should depend on only the angle between a and b their relative orientation but not their absolute orientation because during rotation the angle is preserved right the angle between a and b that is preserved after it is rotated. So this is like reminiscent of rotation of coordinate axis. So if you have 2 axis say a and b are like say for example x and y angle between them is 90 degree. So if you rotate them still the angle between the new x and new y is 90 degree right that does not change. So what we can say is that it will depend on the angle between a and b but not the absolute orientation of a and b and angle between a and b may be specified by the dot product of a and b. So this can be written as if s is an isotropic scalar, isotropic scalar becomes a function of the dot product of vectors taken 2 at a time. So a dot b, a dot b can be written as a i b i right because a dot b means what the multiplication of one element of the vector a with the corresponding element of the vector b and summing it up for i equal to 1, 2, 3. So this is a dot b actually. So if we have a considered with b then we have c considered with d this multiplied by a factor plus. So how many such combinations are there? You can combine a with b then c only, c gets combined with d. You can combine a with c then b gets combined with d and you can combine a with c and b gets, you can combine a with d and then b gets combined with c. So there are 3 such possible combinations. Each combination can be associated with a scalar. Let us say this is alpha, this is beta, this is gamma. These are just coefficients. So see just by considering isotropy we can bring down the number of coefficient necessary from 81 to 3 okay. Now the thing is the next step is a little bit of algebraic manipulation. Here in the left hand side we have here you see you have bj but here you have bi to make left hand side comparable with the right hand side we can use the Kronecker delta. So that you can write bi is equal to bj into delta ij because delta ij is equal to 1 only when j equal to i then this j becomes i so it becomes bi. So this is a trick by which you can switch indices. So if you want to switch from i to j, j to i like that you can use this Kronecker delta. So now so in place of bi we can write bj delta ij. So ai bj ck in place of dl, a dk if you write dl then what would be the next thing delta kl. Next term ai bj then ci if you want to make it ck delta ik and dj delta jl. Similarly for the last term ai bj then cj to be converted to ck. So ck delta jk and di to be converted to dl so dl delta il. So you can see the right hand side also all the terms are of the form something into ai bj ck dl. So you can write ci jkl is equal to what ci jkl is equal to alpha delta ij delta kl plus beta delta ik delta jl plus gamma delta il delta jk. Now let us look into the conditions under which we have derived it we have used the isotropy. What does the homogeneity come into the picture we the homogeneity is implicit that means this alpha beta gamma are position independent that is the homogeneity I mean these are constants not depending on position these are 3 constants. We can bring the number of constants further down by noting that you have tau ij is equal to tau ji that means ci jkl is equal to ci jkl because tau ij is equal to tau ij is what ci jkl ekl tau ji is ci jkl ekl therefore ci jkl and ci jkl must be the same. So that means in this equation if you swap i and j that should be the same so that means this is equal to you swap i and j. So alpha delta ji delta kl plus beta delta jk delta il plus gamma delta jl delta ik. Now delta ij and delta ji are the same it is a symmetric tensor so what it follows that if these 2 are equal then what it follows beta is equal to gamma right so we have brought down the number of constants to 2 constants now alpha beta and gamma is also same as beta so alpha and beta. Now let us write an expression for tau ij tau ij deviatoric that is ci jkl ekl. So tau ij deviatoric is equal to ci jkl into ekl that is equal to alpha delta ij delta kl plus beta delta ik delta jl plus gamma is also equal to beta so plus beta delta il delta jk this multiplied by ekl. Now let us simplify this remember that ij these are sort of indices which are there appearing in the left hand side so we are not able to manipulate with that because ij are already there in the left hand side so kl these are sort of indices with which we can manipulate. So delta kl when is it 1 it is 1 only when l is equal to k so when l is equal to k this becomes what ekk so alpha into ekk right that is the first term second term i and j we cannot disturb so ekl now you see delta ik will become 1 when k is equal to i and delta jl will become 1 when l is equal to j so this will become eij k will become i and l will become j so beta into eij beta now you tell what should be this l is equal to i so and k equal to j so eji so plus beta eji so what is eij you recall we defined it in the previous lecture half del ui del xj plus del uj del xi and this is same as eji so alpha ekk plus this is basically 2 beta eij so beta del ui del xj plus del uj del xi what is ekk ekk is e11 plus e22 plus e33 this is what del u1 del x1 plus del u2 del x2 plus del u3 del x3 this is nothing but the divergence of the velocity vector v so this in the shorthand index notation you can write del uk del xk right so ekk is del uk del xk so we can write we can relate the deviatoric stress with the term which is the first term and which is the term which is the second term now how do they relate with the physics of flow that we should understand carefully now the first thing that you must appreciate that this as well as this these are kinematic parameters right these are pure kinematic parameters now what do these parameters represent the first kinematic parameter that is del uk del xk this represents the volumetric strain within the fluid so what is the rate of change of volume per unit volume of the if there is a fluid element then what is the rate of change of volume of the fluid element per unit volume that volumetric strain rate is represented by this so if there is a fluid element that does not change its volume then such a flow is called as incompressible flow so this term is 0 for incompressible flow remember incompressible flow is a pure kinematic consideration there is a distinction between incompressible flow and incompressible fluid that must be understood very carefully incompressible fluid is a fluid where the density is weakly sensitive to pressure the density is not changing with pressure substantially so the change in density due to change in pressure can be neglected and it can be shown that like for Mach number typically less than about 0.3 that is the reality so that that is the definition of an incompressible fluid on the other hand the concept of incompressible flow is a pure kinematic concept and that follows from the consideration that if a fluid is not suffering from any volumetric strain then the flow can be considered as incompressible so this is incompressible flow should not be confused with incompressible fluid so this relates to the compressibility of the flow or the volumetric dilation volumetric change within the flow and this relates to what this relates to the deformation linear or angular if i is equal to j this is linear deformation and if i is not equal to j that is shear deformation or angular deformation so this is related to volumetric deformation this is related to linear or angular deformation so then these constants have special meaning physically so what is this constant this is nothing but the viscosity of the fluid because this is the rate of deformation and the deviatoric component of the stress one part of that is linearly proportional to the rate of deformation and that proportionality constant is known as the viscosity of the fluid that is what is Newton's law of viscosity okay but you know this we have arrived at by considering a special scenario a homogeneous isotropic fluid so that is why this is still not the most generalized form of the Newton's law of viscosity but it is generalized enough considering the homogeneity and isotropy of the fluid and this is a constant typically given a symbol lambda in text books this is known as volumetric dilation coefficient so this relates to volumetric deformation there is a del ij here right there is a delta ij here yes we miss this delta ij so there is a delta ij here right so this delta ij is very important because this shows that this is only the normal component like and this is quite clear the volumetric strain should not give rise to any shear deformation it can give rise to any only normal component of deformation normal component of stress not deformation so now we can write tau ij the total tau ij is the hydrostatic plus the deviatoric component so minus p delta ij plus lambda del uk del xk plus mu del ui del xj plus del uj del xi where lambda and mu are some constants 2 material constants so this is true for what type of fluid homogenous isotropic and Newtonian fluid okay now so far so good let us examine this tau ij for the normal component so this is tau ij in general we will have normal components and shear components so let us investigate the normal components of tau ij so you have tau 11 this is one normal component tau 11 is what minus p plus lambda del uk del xk plus 2 mu del ui del x1 this is also delta ij right delta ij so this is tau 11 right so very important and interesting thing is that see when we start studying fluid mechanics in the undergraduate level because complicated things cannot be discussed to begin with sometimes simplifications are made while deliberating the concepts and sometimes that is dangerous enough to create misconcepts like for example I mean in many scenarios one of the preliminary chapters in undergraduate fluid mechanics is fluid statics and there when we talk about the normal component of stress through pressure we start believing that the normal component of stress can be attributed only to pressure but here you can see clearly that the normal component of stress can be attributed even to viscosity we have an intuition and not a very correct intuition that viscosity cannot be associated with linear deformation see here viscosity is associated with linear deformation so not that viscosity always has to be associated with angular deformation it is associated with linear deformation and the volumetric dilation coefficient this is related to volumetric deformation so volumetric deformation and linear deformation can give rise to normal component of stress it is not just pressure that can give rise to normal component of stress okay so this is tau 11 tau 22 let us write it as it is so del uk del xk plus 2 mu del 2 del x2 tau 33 is minus p now if we add these 3 1 1 plus tau 22 plus tau 33 this is minus 3p plus 3 lambda into del uk del xk plus if you take 2 mu common that also becomes del uk del xk because that is del u1 del x1 plus del u2 del x2 plus del u3 del x3 so 3 lambda plus 2 mu del uk del xk so yesterday in the previous lecture we made a passing remark that there is a distinction between mechanical pressure and thermodynamic pressure so how do we define mechanical pressure we define mechanical pressure as the arithmetic mean of this tau 11 tau 22 tau 33 with the negative sign the reason of the negative sign we have already discussed so that is that makes this as p by 3 yes that is what divide by 3 so now we are encountering one term mechanical pressure another term thermodynamic pressure so question is are they equal and intuitively in our exercises with fluid mechanics and thermodynamics and heat transfer we take these 2 as equivalent we never distinguish commonly between mechanical and thermodynamic pressure see the pressure when we write as an equation of state when we write pb equal to rt nrt let us say for an ideal gas that pressure how do we measure we measure it in the form of some normal component of force that is being exerted per unit area so essentially we are measuring the mechanical pressure but we are sort of using it as an equivalent to thermodynamic pressure so the basis of measurement is this and the thermodynamic basis is this so are they equal we see that first answer is in general they are not equal because you can see that there is a correction term now stokes you know stokes has been a very famous scientist who has contributed greatly in the area of fluid mechanics and typically stokes contribution to Lorentz number fluid mechanics which is one of the heart and soul of microfluidics has been tremendous so in this particular course we will be discussing about stokes hypothesis stokes law stokes equation sometimes you will be confused everything is in the almost everything is in the in the name of stokes so we are beginning with something which is stokes hypothesis this is not stokes equation or stokes law this is stokes hypothesis and stokes hypothesis is that this term is 0 so stokes hypothesis so that term is 0 in general when when lambda is equal to – 2 third of mu okay lambda is because this is not in general equal to 0 this is 0 only for incompressible flow so to generalize it you must have lambda is equal to – 2 third mu now what is the sign of mu sign of mu is positive because you can see that viscous dissipation we will see later on when we discuss about energy equation that viscous dissipation gives rise to a positive term a positive heat source and that gives rise to entropy generation that is possible because viscosity is positive if viscosity is negative that can give rise to negative entropy generation which will violate the second law of thermodynamics so mu is positive because mu is positive lambda is trivially negative so when is so I mean why is lambda negative I mean the physical implication is that see look at the term that accompanies lambda that is the volumetric deformation so you can see that if the volumetric deformation is positive then the corresponding normal stress is negative that means if a fluid element is already expanding the incremental amount of stress to stretch it further is negative that is what is the physical implication of negative lambda if it is already expanding then the incremental stress that is necessary to stretch it further is actually negative that is what is the physical meaning of the negativity of lambda okay so now let us discuss about the stokes hypothesis so we have discussed about the y lambda negative the second issue is that stokes hypothesis implies that mechanical pressure is equal to thermodynamic pressure when is it really true is it always true now to understand that we have to consider that the mechanical pressure what it considers it considers only the translational degrees of freedom of the molecules on the other hand the thermodynamic pressure considers all possible degrees of freedom translational rotational vibrational all possible degrees of freedom of the molecules the question is that when they are equivalent when they they are equivalent when all the degrees of freedom can be eventually converted into translational degrees of freedom and that takes some time so this time is called as relaxation time of the fluid typically thus this relaxation time is very short that means if you create a change in the system then the fluid thermodynamically response to that change by coming to a new equilibrium thermodynamic state local equilibrium thermodynamic state almost instantaneously however if the change is very rapid for example if you have a bubble which is expanding and contracting expanding and contracting very fast then what happens is that the time scale over which you create the change on the bubble by making it expand and contract expand and contract that time scale may be comparable as compared to the relaxation time scale that is the time scale necessary for the relaxation time scale is simply what the time that the fluid needs to adjust to the change in very simple words it is like this it is not a formal definition but the time that it takes to adjust to the change so like for example like if you wake up early in the morning and you find it is very cold so what you immediately do after coming out of your bed is you wear a warm cloth to adjust to the change the time that you take to do that is the is your relaxation time so you just require some time to adapt to the change so what it means is that the relaxation time is so typically what we are considering is that the time scale of change is significantly larger as compared to the relaxation time but if the time scale of change is comparable with the relaxation time like a bubble continuously expanding and contracting then for such a case stokes hypothesis will not be valid if it is doing it very fast so that is why see this is not called as a stokes law this is a hypothesis just like the avogadro's hypothesis this has some conceptual logic logical evidence logical proof and experiments have supported this but there is no direct theoretical proof of this explicit proof of this that is why this is hypothesis. So lambda is equal to minus two third mu leads to mechanical pressure equal to thermodynamic pressure but just for a hypothetical discussion there are situations when lambda is not equal lambda may not be minus two third mu or you do not care what is lambda but still mechanical pressure is equal to thermodynamic pressure that situation is a trivial situation when the flow is incompressible so when the flow is incompressible you see that this term is 0 so it really does not matter what it is this correction term becomes 0 and mechanical pressure is equal to thermodynamic pressure what are the other situations when you have mechanical pressure equal to thermodynamic pressure you have if you have a dilute monoatomic gas so you have the translational degree of freedom only so that you have the mechanical and thermodynamic pressure they are identically the same so this is some discussion on the stokes hypothesis and basically what we will do is that now we will get back to the navieres equation in the navieres equation we will substitute this tau ij for homogeneous isotropic and Newtonian fluid on the top of that we will substitute lambda is equal to minus two third mu which is valid for fluids that obeys stokes hypothesis those fluids are called as stokesian fluid the fluids which obey the stokes hypothesis so homogeneous isotropic Newtonian and stokesian fluid will give rise to a special form of the equation of motion and that is known as the navieres stokes equation so let us get back to the slides to summarize what we have discussed so far we will use the slides mostly to summarize the discussion and then again we will come into the board to complete the derivation so we had derived this form of tau ij is minus p delta ij plus lambda del u k del x k delta ij plus mu del u i del x j plus del u j del x i then we have derived the stokes hypothesis and these are important remarks that the stokes hypothesis is automatically satisfied for incompressible flow and also for monoatomic gas and the time scale of change needs to be larger as compared to the molecular relaxation time next we will use this for deriving the navieres stokes equation so let us come to the board the final result is given in the slide but we will come into the board to derive the final result so let us get back let us get started with the navieres equation what motivated us to go for further analysis beyond the navieres equation the motivation was that you are getting too many number of unknowns as compared to the number of independent equations to close the system of equations so let us say so this is the navieres equation so then what we basically require is to calculate delta yj del xj so that we will calculate from this expression which was already written in the board so we will use this to calculate delta yj del xj so what will be this so del del xj of this so this delta ij is equal to 1 only when j is equal to i so that will become minus del p del xi lambda del del xi of what del uk del xk plus you tell now because I just want to see that you are familiar with the notations so del del xj of mu del ui del xj this is very straight forward the first term if mu is constant you can as well take it out then del del xj of mu del uj del xi now we have already considered it as a homogeneous fluid so mu is a constant so this mu is actually there out of the derivative now consider the continuity of partial derivatives so the continuity of partial derivatives demands that it does not matter whether you differentiate with respect to xj first or with respect to xi first so this as well becomes mu del del xi of del uj del xj right and del uj del xj is as good as del uk del xk this is just a repeated dummy index does not matter you can write jklm whatever so now we can combine this term in the box with this term in the box so the next step is minus del p del xi plus lambda plus mu into del del xi of del uk del xk right plus the remaining term and if you use the stokes hypothesis then this lambda is minus 2 third mu so this term total becomes mu i3 so the right hand side so let us write it once more del del t of rho ui plus del del xj of rho ui uj is equal to minus del p del xi plus del del xj of mu del ui del xj plus mu by 3 del del xi of del uk del xk this is nothing but the celebrated Navier stokes equation so what are the assumptions under which is this what are the assumptions under which this equation is valid homogeneous isotropic Newtonian and stokesian fluid normally whenever we are dealing with liquids we are not bothered about this term because this term is 0 for incompressible flow okay so we will simplify this equation further and we will make a note of that now this form of the Navier stokes equation is known as conservative form why it is called as conservative form because this form has been derived starting from the basic conservation of momentum principle but the conservative form may be very well suited for CFD calculations but not so well suited for analytical calculations so to make analytical calculations this conservative form may be reduced to non-conservative form by taking help of the continuity equation so how we do that let us simplify this term so you can write just the product rule of derivative we have used okay so now we can combine these 2 terms and write this as ui del rho del t plus del del xj of rho uj plus rho del ui del t plus uj del ui del xj this is what 0 by continuity equation this is just the continuity equation so we are left with this so what is this basically the term in the square bracket can you relate it with a kinematic parameter this is nothing but the acceleration of flow so you have 2 parameters this is what this is the temporal component of acceleration or unsteady component of acceleration that means at a given location if the fluid velocity is changing with time then that can give rise to a component of acceleration this is what this is the advective component of acceleration so this advective component of acceleration is because of what the fluid has come to a new position at the new position it has a new velocity and that has given rise to an acceleration so the total change so this is basically given by the total derivative capital DDT of ui so the total change or the substantial change whatever I mean there are many terms which are used to specify the same thing the total change is the sum total of the change had it been at the same location had the fluid particle been at the same location but the velocity at the same location would have changed with time but in addition to that there is a component of acceleration which is because of the fact that the within the time interval the fluid particle has migrated to a new location there it has encountered a new velocity field and that has given rise to acceleration one very important remark so the non-conservative form is that the left hand side is just written as rho d ui dt now this is a very illusive form see this form gives an illusion that rho has come out of the derivative so somebody who does not know the derivation will assume that because rho is a constant it has come out of the derivative but that is wrong it does not matter whether rho is a constant or rho is a variable it has come out of the derivative because of simplification with the aid of continuity equation and general form of the continuity equation not a special form so this rho out of the derivative does not imply that rho is a constant even for a variable rho this rho is out of the derivative so now what you can see is that the left hand side is what mass into acceleration by volume because density is mass per unit volume and right hand side is nothing but the force on the control volume per unit volume so we have missed one term which is a body force and these are surface forces so you have the right hand side force per unit volume on the control volume and right hand side left hand side is mass into acceleration per unit volume so this is nothing but representation of the Newton's second law just in the Eulerian framework nothing more than that so Navier-Stokes equation is nothing but Newton's second law represented in the Eulerian framework nothing more than that now to conclude discussion on the Navier-Stokes equation we must account for the number of equations and number of unknowns that motivated us to come to this simplified stage now what are the unknowns so you have rho u1 u2 u3 remember this equation has 3 components i equal to 1 will be the x component i equal to 2 will be the y component and i equal to 3 will be the z component so you have u1 u2 u3 for the 3 velocity components then pressure and rho so how many are there 5 how many equations are there the continuity plus 3 components of the Navier-Stokes so total 4 again this does not match this matches exactly if rho is not a variable if rho is a constant or rho is specified by some other means that is it is not a variable then you have 4 equations and 4 unknowns that closes the system but if rho is a variable if rho is a variable then what you do rho will depend in general on what rho will depend on pressure and temperature so you will have another equation rho as a function of pressure and temperature this is what equation of state but in that process you incorporate another unknown which is temperature right so you have invoked another equation but you have invoked a new unknown t so for solving for that t we require the energy equation and that we will discuss in the next lecture we will take a short break and then we will discuss about the energy equation thank you.