 Welcome to this lecture number 23 on the NPTEL course on fluid mechanics for chemical engineering undergraduate students. The topic of our discussion in the last lecture was differential momentum balance and we were proceeding towards the derivation of the complete differential momentum balance and in that context we introduce the notion of the stress tensor. The differential momentum balance is essentially derived the way we did it is to take an infinitesimal control volume of a simple geometry let us say cube and of sides delta x delta y delta z in the three quarter directions that is the volume of the cube delta x times delta y times delta z and essentially we wrote down the integral momentum balance as applied to a control volume in the limit as the size of the control volume shrinks to a point. Then we obtained the following simplified relation we obtained the fact that sum of all forces acting on this control volume is equal to rho times the substantial derivative of velocity volume times del v where del v is the volume of the control infinitesimal control volume. Now there are two contributions to forces acting on the control volume one are called body forces the other types of forces are called surface forces body forces a common example is gravity. So, if you have a fluid with density rho in present in the control volume and the acceleration due to gravity vector is g then the force acting on this control volume is mass times g m g m mass is rho times the volume. So, g is acceleration due to gravity these this is the body force if the body forces gravity is due to gravity there could be other body forces such as electrical body forces present in ionic liquids or magnetic body forces those will not be of concern to us in this introductory course, but the most common body force that is encountered in many applications is of course gravity. So, that will be the force that we will include in this discussion in our course. Now surface forces are little more tricky now the surface forces are due to the fact that suppose you have a surface of that cube which we just drew and let us draw it the full cube there are six phases to the cube x y and z. So, each phase is denoted by a unit outward normal. So, this phase has a unit outward normal n is in the plus i direction whereas, this phase as a unit of normal n is in the minus i direction likewise this phase as unit outward normal n is in the plus j direction and this phase has unit off normal in the minus J direction likewise, you can do it with the plus k minus k direction I will refrained from doing that because it will reduce the clarity of the picture because one cannot draw three dimensional pictures very well in a paper. So, I will just leave it like this. Now, on each phase in general the force exerted by the fluid that is present outside on this phase can in general be in some direction. That force is since we are all we are considering only infinitesimal forces I will denote it with the delta f. The force can act in general on a given surface in an arbitrary direction. This force acting on a surface whose unit normal is in the direction n can have two components in general. Suppose I take an arbitrary surface of area delta a and unit normal n. There can be two contributions to the forces. One is the direction along the normal those are called the normal forces and then there are two tangential directions to this surface. So, there will be two components along the direction of the tangent. So, in general you will have delta f t 1 delta f t 2 along the two tangential directions on the surface. So, the normal forces are predominantly due to pressure because pressure acts purely normally plus also due to the fact that there could be normal viscous stresses. The tangential forces are purely due to or also called as shearing forces or shear stresses. If you define it in terms of per unit area these are only due to viscous stresses in a fluid. So, it is normally convenient to write the force per unit area as the normal force per unit area sigma n as limit delta a tending to 0 delta f n by delta a. Likewise the shear stress is denoted as tau n 1 is limit delta a tending to 0 delta f t 1. So, tau t 1 let me explain the notation. This denotes the fact that you are considering a surface whose unit normal is in the n direction and this is the tangent vector in the one direction and that is similarly there is also a tangential force in the other direction. All we are saying is that for any given plane any surface with an arbitrary orientation n the force that acts on that surface due to the fact that fluid is the neighboring fluid the force exerted by the neighboring fluid on this surface could have three components one along the normal and two in the two tangential directions along the surface. That is all we are saying, but now we have considered a cube with six faces in our control volume x y and z. So, each face will have three components this face will have the unit normal is in the plus i direction and in general the force exerted will be the force exerted will have three components. So, the three components of the force are we can write this as one along the normal direction which is actually the x direction here. Because that is the direction of the unit normal then the two tangential directions are y and z. So, delta f y and delta f z. So, we can write three components of the stresses the force per unit area as follows sigma x x. So, let me write the two x's in two colors sigma x and I will explain you the reason write it slightly bigger sigma x x is equal to limit delta a tending to 0 delta a tending to 0 where a x is the let us say this is the area of this face with unit outward normal in the plus x direction delta f x by delta a x. This is the normal component of the stress acting on that surface that shear stresses are usually denoted by tau the greeks symbol tau. So, now I am going to use two colors to the two subscripts tau x y is equal to limit delta a x tending to 0 because that is the face we are considering. But the force is now in the y direction the area on which the force is acting is in the x direction similarly can have tau x z is equal to limit delta a x tending to 0 delta f z by delta a x. Now, let me explain the colors for the two subscripts that explain the reason for the two subscripts in the orange color refers to the direction of the unit normal to the surface the unit outward normal to the surface on which we are considering the forces. So, this tells you the orientation of the surface orientation. Now, that is the purpose of the first subscript the second subscript the green subscript tell you the direction of the forces the forces can act in either x y or z direction on a surface whose unit normal is in the plus x direction which is i. So, this on this surface you could have forces with three components. So, therefore you need two subscripts to denote the two issues that is one is the direction of the force the direction other is the direction of the unit outward normal to the surface on which we are considering the force likewise you could also write tau y x sigma y y tau y z this is you consider a phase whose unit normal is in the y direction j direction and you are considering the forces in the x y and z. So, the first subscript denotes the direction of the unit outward normal and the second subscript tells you the direction of the force. So, you need two subscripts and likewise you can have in the on a plane whose unit normal is in the z direction that is plus k tau z x tau z y sigma z z. So, as the control volume shrinks to a small point this the forces on this tiny cubic volume element can be summarized as an array of numbers sigma x x tau x y tau x z sigma sorry tau y x sigma y y tau y z tau z x tau z y sigma z z. So, this is an array of numbers that describe the state of stress on a cubic volume element with nicely defined surfaces along the three coordinate directions x y and z namely. Now, what is the this is this collection of the state of stress present in a less in a fluid is called the stress tensor y is it called a tensor it is because of the following reason if you have a vector like velocity you can write it as v x times i plus v y times j plus v z times k. You need only one subscript to denote the direction of the components in which the velocity is pointing velocity has is a vector has only one direction associated with it one physical direction associated with it namely the direction of the velocity itself. But if you consider something like a stress which is denoted with two underlines two underlines to denote that it is something more higher than a vector it is actually a tensor because it has two physical directions associated with it associated with it. So, velocity is a vector stress is a tensor. Now, what are those two physical directions the two physical directions are the direction of the unit normal on which the force is acting and the direction of the force itself. Stress is force divided by unit area force as a vector has a direction and the area itself has a direction because of its orientation. Now, so at each and every point in the fluid if one has a knowledge of this matrix or the tensor sigma x x tau x y tau x z tau y x sigma y y tau y z tau z x tau z y. If one has the collection of this one has an information about this nine elements then there is a result due to Cauchy which states that if I have at a given point and a surface with arbitrary orientation that is you have put a coordinate system x y z. But this orientation of the surface is not along any of the three directions x y and z, but it is an arbitrary orientation n x times i plus n y times j plus n z times k. It is it is oriented in an arbitrary fashion not necessarily along one of the three coordinate directions then and the area of the surface is delta a let us see. Then what is the force the force on this surface will in general have three components in the x y and z direction. So question one can ask is what is the force acting on such a surface what is the force per unit area what is the stress acting on such a surface. The answer due to Cauchy which I will not derive is that simply take the local value of the stress on this surface take the stress tensor acting at that point. So, remember this is an infinitesimal it is a point and you have an infinitesimal area surrounding that point. So, at that point you should know what is the stress tensor sigma x x tau x y tau x z tau y x sigma y y tau y z tau z x tau z y sigma z z and according to Cauchy's result you simply have to do a matrix multiplication of the three components of the unit vectors with the stress tensor. This is like a matrix multiplication of a matrix three by three matrix with a three by one vector. So, the answer will be a three by one vector which is this. So, by having the knowledge of the stress tensor at a given point which the stress tensor is a given point in the Cartesian coordinate system tells you what are the forces per unit area acting on directions which are perpendicular to the three unit vectors in the Cartesian frame of reference. But if you have an arbitrary phase arbitrary area then you can get the force per unit area acting on that surface with an arbitrary orientation in this fashion. But by merely multiplying the elements of the stress tensor or the stress matrix with the column vector containing the components of the unit vector. So, that is the power of doing the Cauchy construction. Now, having understood what are the stresses acting on the surface also the sign convention for the stress which we mentioned last lecture. Suppose you have a surface that is pointing in the plus j direction and suppose we say that tau y x is 10 Newton per meter square this is the x direction this is the y direction. What this means is that you have a force. So, y represents in our convention the direction of the unit normal. So, that is in the plus j direction x tells you the direction of the force. So, if it is positive tau y x is positive then the force of 10 Newton per meter square force per unit area that is it acts on the plus x direction on a surface use unit normal is in the plus y direction. So, if you have both the direction of the unit normal and so if you have the answer of the stress tensor to be positive. That means that both the direction of the unit normal and the direction of the force are positive or it could also be the following that on the same surface at the same point you could consider a unit vector in the minus j direction. You could consider and each plane has two unit normal one is the normal pointing in outward direction other is normal pointing in the inward direction. So, if you consider the unit vector to be pointing in the minus j direction then the force acting on this surface is in the minus x direction this is plus x the magnitude of the force is 10 meters 10 Newton per meter square. So, if somebody says that you have a force or stress tau y axis plus 10 Newton per meter square it could either mean that in this could either mean that a force of plus 10 is a Newton per meter square is acting in the plus x direction in a surface on a surface whose unit normal is in the plus y direction or a force of 10 Newton per meter square is acting in the minus x direction on a surface whose unit normal is in the minus y direction. So, that is the meaning of positive terms positive stress values. Suppose you have negative stress values if somebody says the element of a stress tensor is negative that means the following if you have a unit normal in the plus j direction the force is acting in the minus x direction and the magnitude of the force is plus 10 magnitude of the force is 10 or vice versa if you consider the unit normal in the minus j direction then the magnitude of the force is in the plus x direction the force acts in the direction of the force is in the plus x direction while the magnitude remains 10. So, if the stress value is negative then it means that the direction of the force and the direction of the unit normal or opposite to each other. If the stress value is positive then the direction of force in the direction of unit normal or along the same direction. They both could be positive here or both could be negative. Here, if one is negative, the other is positive and vice versa, one is positive, the other is negative. So, that is the sign convention for the stress tensor. Having done all this, we went back to the differential momentum balance, which simply said that is rho g times delta v plus the surface forces acting on the cube. Now, the way we did was to consider each phase of the cube, take each direction x, y, z. So, we first wrote down the x momentum balance. How do we do that? We did that by saying that take the x component of this equation delta v is rho g x delta v plus all the surface forces acting along the x direction on this control volume. What are the various surface force acting on the x direction? On this phase, there is a force sigma x x. On this phase, there is a force tau y x. On the front phase, there is a force tau z x. All these are pointing, the direction of the force is remember the second subscript. So, all these are pointing in the x direction, but there are not just three phases. There are also these opposite three phases, which for which the unit normal is in the opposite direction. We have to consider those forces as well and then use the Taylor series expansion for tau at x plus delta x in terms of tau at x to arrive at this simplified relation. We finally ended up with this equation rho du dt delta v is delta v times rho g x plus partial sigma x x by partial x plus partial tau y x by partial y plus partial tau z x by partial z. So, the volume of the control volume drops out of the equation, because both terms are multiplied by the same quantity. Now, we also said that in a static fluid, the only force that is acting are the normal pressures. So, the stress tensor becomes and this pressure acts inward. So, you will have a minus p. In a static fluid, the stress tensor is this. In a flowing fluid under flow, we want to write the stress tensor as this, which would be the case if the fluid were to be static plus stresses that come up because of the viscous resistance of the flow of the fluid. So, this is written as x y. So, sigma x x becomes minus p plus tau x x sigma y y becomes minus p plus tau y y and sigma z z becomes minus p plus tau z z. Under flow therefore, these are the terms that will happen only these are non-zero only under flow non-zero only if flow exists. This is these are the viscous stresses. The stresses that come about because a fluid resist motion continuous deformation. So, these are the viscous stresses that are generated by the fluid to resist the deformation. So, then we wrote rho d u d t is minus partial p partial x plus rho g x plus partial tau x x by partial x plus partial tau y x by partial y plus partial tau z x by partial z. This is the first term the x component the y component of the momentum balance this is x y and the z component. So, these are the three momentum equations the differential form the momentum equations these are also sometimes referred to as the Cauchy momentum equations. These are the differential form of the momentum balance these are also called the Cauchy momentum equations. And we also have the mass balance if you remember for an incompressible fluid the continuity equation which in Cartesian coordinate means. So, we have four equations three components of the momentum equation and one continuity equation. So, we have number of equations is equal to four. So, far and the number of unknown variables that we have we can count as follows three velocities the three components of the velocities one pressure and the nine components of the stress. So, the problem is highly under specified because the number of equations. So, far is less than the number of unknowns. So, we cannot solve this problem. So, we cannot solve this problem as such right now. So, that is the idea that we cannot solve the problem because you have far two equations than unknowns. Now, we have to look at the problem why we have this situation that we are not able to solve the problem completely. And that is because we have not said anything about whether the material is fluid or a solid even now. Because all that we have done is to apply Newton second law to a control volume and then took the limit as the control volume shrunk to 0 size 0. But nothing has been told. So, far whether the material that resist deformation is a fluid or a solid that is the constitutive nature of the material has not yet figured in our framework. So, we need to specify something about the nature of the fluid that will relate the stresses to the velocities and thereby we will be able to achieve some closure. But before that we can also write down equation for angular momentum that will tell us that tau y x is tau x y tau z x is tau x z tau y z is tau z y. That is the stress tensor is a symmetric tensor that is the matrix of elements is a symmetric matrix. That is you have whether you write it like this tau x x tau x y tau y x whenever you write it like this these two terms are the same and these two terms are the same the green terms and likewise these two pink terms are the same. They are the same because it turns out that when you do the angular momentum balance and if there are no internal couples that generate angular momentum within the fluid which is true for almost all fluids. Then one can show that the stress tensor is identically symmetric always. So, we are left with only a symmetric matrix has only six elements because there are three equalities that relate three that give you three additional equations. Therefore, a symmetric matrix has only six independent components as opposed to the nine independent components for a second order tensor six independent components. But nonetheless we are still left with more unknowns than the number of equations that we have to solve them. So, we have to specify what is called the constitutive relation. In this context one way is to do one way to approach the problem of constitutive relations is to do experiments on known fluids and try to correlate the stresses with velocities and so on. That is one way and that is often done in labs to characterize fluids. The other way is to guess the form of constitutive relation based on some general principles. Now, let us try to guess the form of constitutive relation from an intuitive or physical point of view. Now, firstly let us take a very simple system that is flow is happening between two plates x and y bottom plate is stationary static top plate moves with the velocity v. So, you will have a velocity that looks like this in between the two plates. Now, the question that one can ask is the relevant component of the stress is you have a plate whose unit normal I mean I am just drawing across section actually it is a plate it is extends whose unit normal is in the plus j direction and the direction of the stress would be in the plus x direction. So, the component of stress will that will be of interest is tau y x. Now, what should tau y x be a function of what can tau y x depend on? We can say the tau y x will depend on the velocity velocity of the flow, but that is not all because that does not tell you the whole story. Because if it just depends only on the velocity then you can say that well what about the distance between the two plates will it be the same if I have the same velocity, but a very large distance will it be the same if I have very small distance. So clearly the velocity tau y x cannot be a function only of the velocity somehow the distance between the two plates have to come in. Also from a physical point of view the stress cannot be in a fluid if the entire fluid moves with the constant velocity then there cannot be any internals viscous stresses generated because that is merely rigid body motion. Stresses are generated in a fluid only if there is relative deformation between two points. Two points move past each other relative to each other. Now that so only then stresses will be viscous stresses will be generated. Now that part is accurately captured by what is called the gradient of velocity. So stresses in a fluid cannot be generated purely by a rigid body motion of the entire fluid. There has to be relative deformation between two points in a fluid and that is captured by what is called the gradient of velocity. That is if you have flow in the x direction the velocity component of interest is v x the x component of the velocity. If the x component of the velocity is different at two different points then we say that there is a gradient. So this is denoted by d v x y d y. So in this particular case there is indeed a variation in velocity in the y direction. So there will be stresses that are generated because there is a difference in velocity at between two different points in the fluid. If you say both the plates are moving with the constant velocity v then of course there would not be velocity gradients and there would not be any viscous stresses because this amounts to merely moving the entire fluid as a rigid body. So only gradients of velocity can generate stresses not absolute velocities. So in this simple context which we just discussed let me just show the picture here itself. You have two plates one at y equal to 0 other at y equals h and the top plate is moving with the velocity v then the relevant component of the stresses tau y x because you are considering the force acting on this part of the fluid that is the top surface on in the x direction because that is the relevant component of the stress. So tau y x must be proportional to p y. The constant of proportionality is called viscosity of the fluid this is called the viscosity. Now the dimensions of viscosity area can be estimated as follows. Stress is force per unit area this is force divided by area. The dimensions of velocity gradients are simply one over time. So the dimension of viscosity stress can be written as m is times t to the minus 1. If we take to the other side then the dimensions of viscosity is m l to the minus 1 t to the minus 1 k g per the units in s i units are k g per meter second. So this is a material property of the fluid it is a material property of the fluid of the fluid. So this is the simplest expression because we consider a relatively simple flow between two parallel plates the flow is only in one direction and but the in general the flow can be much more complex. But before I go to the complex situation this is often called the Newton's law of viscosity. This is merely an expectation from our side intuitive expectation from our side. It is not guaranteed that all fluids should obey this expression this relation. In fact many fluids do not. For example if you consider fluids like toothpaste or shampoo or ketchup they do not obey this relation they are much more complex. But many simple fluids like air, water, glycerin and so on they do obey this simple relation. Such fluids are called Newtonian fluids the fluids which obey the Newton's law of viscosity which says that the stress the shear stress is directly proportional to the velocity gradient and the constant of proportionality which is a true constant it is independent of the flow parameters like velocity or velocity gradient is called the viscosity of the fluid. Now this is for the simplest case but Newton's law must also be written for the most general case. So the Newton's law of viscosity for the most general case where the stress the flow can be in all the three directions tau x y tau. Remember the stress tensor is symmetric from angular momentum balance. So let us hope that I have used mu use mu is the symbol used for viscosity normally tau x z tau z x is mu tau x sorry tau y z is mu and then you have the normal stresses tau x normal viscous stresses is 2 mu is 2 mu tau z z is 2 mu partial w y partial z. And this is the constitutive relation these are called constitutive relations because they tell they have some information about the constitutive and material that make up the fluid these are called constitutive relations. And the ones we have written are for a simple Newtonian fluid such as air and water. Now we have written all the stresses in terms of the velocities or the gradients of velocity. So now we have exactly the same number of equations as you have unknowns because you have three momentum equations and one continuity equation and we have three components of velocity and one pressure as the unknown. So that completely specifies the problem. So now let us substitute all this in the momentum equation. So let us go step by step for the x momentum equation plus rho x plus we had partial tau y x by partial x plus partial tau y x by partial y plus partial tau z x by partial z. Now let us substitute all the three components. So let us I am going to focus only on this term now and then we can substitute this back in this expression. I am going to substitute each terms partial tau x x is 2 mu partial u by partial x plus partial tau y x is mu plus partial tau z x is mu plus partial u partial z. Now I can write the first term as since it is 2 I can write it as sum of two terms partial partial x mu partial u partial x plus again the same thing. Now in all these equations I am going to take out I am going to write the first term separately all the other two sorry all the other two terms. I am going to write separate of the first term plus partial partial y of mu partial v partial x plus partial partial z of mu partial w partial x. But still we have those two terms plus partial partial y of mu partial u partial y plus partial partial z of mu partial u partial z. Now I am the order of these derivatives partial derivatives are interchangeable. So let me write this as partial partial x and we will assume that mu is constant independent of special portions. So we can pull out mu of partial u partial x that is from this term. We also have if you interchange these two partial derivatives then you will have partial partial x of partial v partial y and likewise from this term if I interchange the partial derivatives partial partial x of partial w partial z plus mu then you can write this equation this sorry this term as partial squared u partial x squared this term as partial squared u partial y squared plus this term as partial squared u partial z squared. Now this term is identically 0 from continuity equation the mass conservation equation says this is identically partial u partial x plus partial v partial y plus partial w partial z is nothing but del dot v is 0. So if you look at this expression the only thing that survives is that this becomes mu partial squared u partial x squared plus partial squared u partial y squared plus partial squared u partial z squared. So I am going to write the x component of the momentum equation after using the constitutive relation for the Newtonian fluid as rho the substantial derivative of the x component of the velocity is partial p partial x partial gradient term rho g x the gravity term plus mu partial squared u by partial x squared plus partial squared u by partial y squared plus partial squared u by partial z squared. This is nothing but the Laplacian del squared of u where del squared is the operator partial squared by partial x squared plus partial squared by partial y squared plus partial squared by partial z squared. So the x component of the momentum equation can be simply written as rho partial substantial derivative of the velocity is minus partial p partial x pressure gradient term plus the gravity term plus the viscous stress term which becomes like this. So I can write similar equations for the y component of the momentum equation just by analogy one can of course go through the algebra in the same manner and you will end up the same thing this is x component y component this is the z component mu del squared w and of course one has continuity equation mass conservation equation the mass balance differential form of mass balance which is partial u partial x plus partial v partial y plus partial w partial z is 0. Now I can write this in a coordinate free form just as the mass conservation equation in a coordinate free form became del dot v 0. I can multiply these three components by the unit vectors and add them up and I will get the following equation d v d t is minus gradient of pressure plus rho g vector now plus mu del squared velocity vector. These two equations are called the Navier Stokes equation for an incompressible for an incompressible Newtonian. So this is the mass balance this is the differential momentum balance. So the next task for us is to understand these equations by providing some solutions of these equations in simple physical context. We will continue this in the next lecture.