 Or we have discussed about various applications of microfluidics and now we shall proceed into the theoretical description of the subject to understand better that how a basic theoretical building block or how a basic theoretical foundation can be used to solve problems in microfluidics. While mentioning that it is also important to appreciate that always continuum hypothesis may not work for applications over small scales. But there is a wide class of problems where continuum hypothesis will still work with certain modified considerations therefore I mean we will try to begin with the continuum considerations. One of the big issues that we would keep in mind is that like these conservation equations are fundamental to even macro scale flows but in micro scale flows there may be certain alterations but before understanding the alterations we will revisit the governing equations and their derivations. Now for the remaining part of this course I will try to follow a teaching methodology which is a somewhat hybrid type of teaching methodology. So what we will do is we will describe the derivations of course in the forms of slides to make this video lecture more effective but at the same time we will work out some of the key steps in details in the board. So we will use the mixed mode of demonstration and I believe it will be interesting and useful to you. It is neither the classical old style of presentation where the entire thing is done in a board nor the seminar type of representation where the entire thing is done through slides but somewhere in between and this is like in line with the spirit of this course that this is neither a very fundamental like core course for a particular department nor it is a fully research course so it is a sort of a bridge between a core course and a full research level course and our teaching style will also be commensurating with that. So to begin with we will discuss about the general philosophy of conservation. So like the general philosophy of conservation when we apply we may use it for conservation of mass we may use it for conservation of momentum we may use it for conservation of energy we may use it for conservation of species we may use it for whatever conservation but generally in fluid mechanics we tend to use an Eulerian approach that means we try to identify a control volume which occupies a fixed region in space across which mass can flow energy can flow momentum can be transported. So this control volume across that something goes in something comes out there is some change within the control volume and the net change within the control volume is nothing but in minus out plus generated. So the total change in the control volume for example let us say if there is a mass that comes in whatever mass comes in if the same mass goes out then there is no change but let us say that because of chemical reaction there is some species which is generated. So whatever has come in minus whatever has gone out plus whatever species that has been generated because of reaction may be the total change in the mass of the species that is under consideration. So that is the basic principle of species conservation. So this kind of conservation principle you can use for any particular scenario. Now what is the important consideration that you have to keep in mind in fluid mechanics. The important consideration is that all our basic laws of fluid mechanics are in general describe I should say basic laws of mechanics classical mechanics I mean forget about fluid mechanics are describe on the basis of a Lagrangian point of view that is you identify fixed mass and write equations maybe equations of motion or whatever for that identified mass which is called as a control mass system. On the other hand in fluid mechanics it is convenient to describe the scenario in terms of a control volume that means an identified region across which mass and energy can flow and the reason is as follows if you have a specified region in space which is your control volume let us say let us say this is your control volume. Now you can sit with a camera and target to the control volume and see whatever changes are taking place across that. So something comes in and something goes out some change is taking place within that you are not bothering yourself about where from that something has come where will that something go. So you are not tracking individual particles and tracking individual particles in fluid mechanics is a very difficult job because fluid elements are continuously deforming during motion. So that it is very difficult to track individual fluid particles. So what instead you do that you write your equations of change with respect to this control volume although the basic laws of mechanics are described for a control system control mass system. So there must be a transformation there must be a transformation that will try to transform from a control mass system to a control volume system approach. So that a control volume based approach so that you can write your equations of motion with respect to control volume rather than with respect to a control mass system it is not just true for equation of motion but maybe even the first law of thermodynamics and so on so different conservation equations. So that is achieved by realizing a transformation through something which is called as Reynolds transport theorem. So in Reynolds transport theorem what we are essentially interested in is that we have a control mass system so we have a Lagrangian approach and we have a control volume concept which is the Eulerian approach. So we want to write a mathematical transformation that transforms the Lagrangian approach to the Eulerian approach. So to understand how this theorem is applied if you look into this diagram which is there in the slide if you look into this diagram let us look into this diagram very carefully. You have this pink colored region I mean this total contour is the system at time t. So the system at time t is the region 1 plus the region 2 together the pink colored region is the system at time t. On the other hand the region with the sky blue color is the configuration of the system at time t plus delta t. If delta t is small the differences in the configuration the differences are not that large but we have sort of magnified it in the diagram we have magnified this in the diagram to sort of make the system configurations distinctly looking in the figure at time t and t plus delta t. So let us say that capital N is an extensive property of the system extensive property means a property that depends on the extent of the system. So you can write N at time t. So if you get back to the slide N at time t is a combination of N at 1 at time t and N at 2 at time t because at time t the zone is zone 1 plus zone 2. Similarly you have N at t plus delta t is equal to N 2 at t plus delta t plus N 3 at t plus delta t. So we are interested in finding out what is dN dt of the system dN dt of the system. So dN dt of the system is what by definition of the derivative it is limit as delta t tends to 0 N t plus delta t minus N t divided by delta t that is the definition of dN dt. Now what is N at t plus delta t? N at t plus delta t. So let us write this in the board maybe to explain it better. So we have a configuration. So we have this is 1, this is 2 and this is 3. So N at t plus delta t is N 2 at t plus delta t plus N 3 at t plus delta t and N at t is N 1 t plus N 2 t. So our calculation is dN dt that is limit as delta t tends to 0 N t plus delta t minus N t by delta t. So we will see that at 2 you have a change in N because of the change in property at 2 with time. So you have limit as delta t tends to 0 N t plus N 2 t plus delta t minus N 2 t divided by delta t plus now you have certain adjusting terms. What are the adjusting terms? This is a term which is a positive plus term and this is a term which is a minus term. So now if you divide this by delta t and take the limit as delta t tends to 0 then what is that? That is the property that is transported across 3. So that is nothing but rate of outflow of the property. Similarly, when you write this one what is this? In the limit as delta t tends to 0 this is inflow of the property inflow outflow with respect to what? With respect to this common region which represents the so called control volume. So because why this region represents the control volume because as delta t tends to 0 this is the region which is that region in space across which if you like focus your attention then you can see that this is inflow and this is outflow. So this is like an identified this is the region in space which you can identify across which you can study the transport. So this is rate of outflow minus inflow. Outflow minus inflow of the property okay. Now let us think about the first term. Let us think about the first term. So what is this? This is the rate of change of n with respect to time within the control volume right. So why we write it as partial derivative of n with respect to time? We write it as partial derivative of n with respect to time because at a fixed location we are studying the change with respect to time that is why partial derivative with respect to time. Now the next question will be that how to express the outflow minus inflow? So let us say that let us identify a region on the control volume and see that how we can express the outflow. Let us say this is a small area dA. Now an area has a orientation. What how do you describe the orientation of an area? You describe the orientation of an area by its outward unit normal vector. So let us say that eta cap is a unit vector in the normal direction of dA. So what is the volume flow rate across dA? The normal component of velocity times the area dA. So if v is the velocity then we have v dot eta dA. So this is the volume flow rate. This multiplied by rho is the mass flow rate and let us say that small n is the property per unit mass that is small n is capital N per unit mass. So if you multiply that with small n this is the total outflow of the property integrated over the control surface. Cs means control surface, surface which is binding, bounding the control volume. So then I mean as an alternative notation many times this can be written as a vector dA. Either you write it as the normal vector times the scalar area or you write it as the area vector itself. I mean these are two alternative ways of looking into it. So this is outflow. Now how do you write inflow? So when you consider inflow let us consider an area here. Now here the normal direction is like this and for the fluid entering the control volume the velocity is like this. So what is the most important demarcating factor between this scenario outflow and the inflow scenario. So if you consider the dot product of v with eta here the dot product itself is positive and by looking into this figure you can understand that here the dot product will be negative. That means when we say outflow- inflow we do not have to use a minus sign. The dot product positive will automatically indicate outflow and the dot product negative will automatically indicate inflow. So for the outflow- inflow we can write together this common term integral of rho v dot dA into n small n where the control surface some part of the control surface may be inflow some part of the control surface may be outflow some other part may be neither inflow nor outflow. Now we have to keep in mind or we have to pay a little bit of attention to what is this v? So this v we have to remember that this v is nothing but the fluid velocity but to be more precise velocity of the fluid relative to the control volume. Let us say that the control volume is a tank just as an example. Let us say that the tank is moving towards the right at a speed of 1 meter per second and water is also moving towards the right at a speed of 1 meter per second. Then what is the rate at which water is coming out of the tank? That rate is 0 because tank is also moving water is also moving in the same way. So relative to the tank no water is moving out. So to get the net flow it is important to understand that the velocity has to be relative velocity. Vr velocity of the flow relative to the control volume this relative velocity is same as the absolute velocity when the control volume is stationary when the control volume is not moving otherwise we have to be careful about this okay. So when you write this vr.da n then it completes the description of this term. Now what about this small n capital N? So if you take a small volume within the control volume of dv because velocity sometimes we use v so we are just using a little bit of different symbol to indicate volume. So what is the mass of this element? Rho dv and small n is the property per unit mass. So this integrated over the control volume is the total n. So we have described this and in the left hand side what is dndt? It is dndt of the system. So with this basic understanding let us now go into the slides to complete the algebra. So you can see here that we have these 2 terms. This is the first term and this is the net outflow-inflow term. So these 2 terms if you write mathematically then you can express the total rate of change with respect to the system in terms of rate of change with respect to the control volume and you can see that there is a correction term. This correction term is because of outflow-inflow. So if there is no flow taking place across the control volume of course the control volume is as good as a control system and there is control mass system and there is no difference or no change taking place across the control volume. Now this particular expression which is highlighted here relates the rate of change with respect to a system with the rate of change within a control volume and that is known as Reynolds transport theorem. So Reynolds transport theorem is a very powerful theorem and we will use this Reynolds transport theorem to derive various governing equations. So to do that we will first go to the case of conservation of mass. So conservation of mass is a very basic requirement that needs to be fulfilled for any problem that we undertake no matter with whether it is microfluidics, nanofluidics, gigafluidics, whatever fluidics you call I mean it has to be satisfying the conservation of mass. So for conservation of mass we write the capital N is equal to m which is the total mass and small n is capital M per unit mass so that is equal to 1 and we assume that the control volume is stationary. So if we assume that the control volume is stationary let us go to the slide. If we assume that the control volume is stationary then we have vr equal to v right. We have the relative velocity same as the absolute velocity then you can write the dm dt of the system. Now by definition of a control mass system a control mass system is a fixed mass. So dm dt of the system is 0 and the right hand side what you can do is that like you first write vr equal to v and then for this term what we do is essentially we consider the volume of the control volume is not a function of time. That is itself violated if you have a deformable control volume. If you have a deformable control volume then volume of the control volume is a function of time but we are considering a non deformable control volume. So if we consider a non deformable control volume then what happens is that this time derivative we can take inside the integral. So if we take the time derivative with inside the integral and then we use the divergence theorem. So the divergence theorem maybe let us go to the board to recapitulate the divergence theorem a little bit. So let us come to the board. If we write if we have a vector function and if we have the integral of that over a surface say over a contour control surface which is the surface that bounds the control volume then this is as good as this is the volume integral. So f dot dA is same as divergence of f dv but you have to be careful that this surface is not just any surface it is a surface that completely bounds the control volume okay. So in this particular example you have instead of f what is the what is there in the example rho v becomes del dot rho v. So if you let us now go back to the slides to see that like what are the consequences that we are having as a simplification of this equation. So you are writing del dot rho v dv equal to 0. So what is happening is you are able to write integral of del rho del t plus del dot rho v this total term dv equal to 0 because your volume choice control volume choice is arbitrary this is possible only if the integrand itself is 0 that means only if the del rho del t plus del dot rho v is equal to 0. This is nothing but the continuity equation. So next what we will do is we will get into the conservation of linear momentum. So let us look into the slides and as we did for the previous equation whenever we find that there are sort of key steps in the slide which we need to highlight more we will go to the board and explain the details of the in between derivations. So in the conservation of mass what we wanted to conserve in the conservation of mass we wanted to conserve the total mass. In conservation of linear momentum we want to conserve the linear momentum. So your capital N is equal to mv and small n is capital N per unit mass. So small n is equal to v. Again we assume stationary control volume so that the relative velocity is same as the absolute velocity. So now we write the Reynolds transport theorem just we replace capital N with mv. Remember one very important consideration that the Reynolds transport theorem the n is not necessarily a scalar. So n can be any scalar or vector property. So here we are using a vector property which is the linear momentum. So n is equal to mv so d dt of mv. So here we have small n small n is v and rho v v dot dA. Now why we write it in this way? What is the motivation? See the left hand side is the rate of change of linear momentum of the system. We can apply the Newton's second law directly to the left hand side because the left hand side talks about a control mass system. So that is the resultant force acting on the system by the Newton's second law. Now the resultant force acting on the system is given by basically we have considered while deriving the Reynolds transport theorem that it is a limiting situation when the delta t tends to 0 so that the system tends to control volume. That means the force on the system is as good as force on the control volume. So next we will see that what kind of forces are acting on the control volume. In continuum mechanics we talk about 2 types of forces. One type of force is called as surface force which is distributed over the surface and the other type of force which is distributed over the volume of the body that is called as a body force. Like for example gravity force is a body force. Pressure is a surface force. I mean force due to pressure is a surface force. So next we will see that how we can characterize the concept of a surface force. So to do that we will try to understand 2 important concepts. One concept is the concept of a traction or a traction vector and the second concept is the concept of stress. So to do that maybe we need a little bit of elaborate discussion. So let us get into the board. Let us say that there is a body and you take away a chunk of material from the body. This may be a fluid body, solid body. See this concept does not differ from a solid to a fluid. I mean it is a general concept of continuum mechanics. So when you take this body chunk out you have to represent. So this is as if you are drawing the free body diagram of this chunk. So you have to represent the force exerted by the remaining part of the body on this one. So if you isolate this then what you do? What you essentially do is you represent a force per unit area by a vector on one elemental surface of the chunk. This is a force exerted by the other part of the material on this small part per unit area, remember. So this is called as traction vector t, traction vector. Let me write it here. Now when you have this traction vector you have to keep in mind that this traction vector is not just an ordinary vector. The reason is that this traction vector that is the force per unit area depends not only on the location of this point where you are calculating the force per unit area but on the orientation of this area. That is depending on the orientation of this area. So how do you describe the orientation of the area? Again by outward unit normal vector. So let us say that is eta. So this traction vector is often designated with a superscript eta to indicate that it is sensitively dependent on eta. Now this because it is a vector the traction vector will have its own component. Just like if you have a vector f then it has its component f i which is the ith component. So i equal to 1 will mean x component, i equal to 2 will mean y component and i equal to 3 means z component. This is Cartesian index notation. So the index will describe the component of the vector. So the ith component of the traction vector may be expressed with this t superscript eta with a subscript i. Now it is very nice to work with this vector but the question is how will you get this vector? You will get this vector not explicitly but by means of using force per unit area. On special surfaces the special surfaces have normal directions either along x or along y or along z. So instead of having an arbitrary surface you consider some special surfaces which have their normal either along x for example the y z plane has normal along x. So similarly you can think of certain surfaces. These special surfaces are such that these special surfaces have their normal directions either along x or along y or along z and you can express the traction vector. So you can write those special surfaces by using an index i for the ith component and a superscript j now where this j also can be 1, 2, 3 because j equal to 1 will mean that the normal direction is along x, j equal to 2 will mean that the normal direction is along y and j equal to 3 will mean that the normal direction is along z. So this kind of representation where i can be 1, 2, 3 and j can be 1, 2, 3 is not for all surfaces. It is only for special surfaces which have normal direction either along x or along y or along z whereas this notation is for any arbitrary surface. So now this in an alternative notation can be expressed like this tau ji where what is this j? j is the direction normal of the surface on which the force is acting and i is the direction along which the force is acting and this is force per unit area. So this is called as component of the stress tensor. So this is something which is a tensor which is I mean of course I mean it is possible to define a tensor in an abstract mathematical way but it is not so important for this course to learn all the mathematical intricacies of a tensor but we will try to see that how does it differ from a vector for example. So clearly you can see that like this requires 2 indices for specification. So this is called as a second order tensor. So that out of the 2 indices one index is giving it a vector like character like the index i but something extended from the vector like character is given by the index j which talks about the direction normal of the area which is considered to represent this tau. So if only one index was necessary then would have been a first order tensor. So a vector is a first order tensor. If no index is necessary to represent that then it is a 0th order tensor that is nothing but a scalar. So scalar vector second order tensor these are all falling in the general category of tensors but here we are focusing our attention to second order tensors and we will see I mean very soon possibly that the second order tensor has a very unique characteristic. What is the unique characteristic? The unique characteristic is that it maps a vector onto a vector. So it creates a transformation that maps a vector onto a vector and we will see that how this tau can map a vector onto a vector. So what we have learnt is basically the traction vector is a general notation but for the special case when the normal is either along x, y or z then that can be represented by an equivalent tau ji type of notation. So with this notation let us now get into the slides back again. So this slide represents some examples that like the proper representation of the tau ij notation. So let us say that this is x1, this is x2 and this is x3 in a right handed coordinate system x1 means x, x2 means y and x3 means z. So let us consider this right hand surface in this surface. Now you can see that let us first consider the normal stress tau the first index is what? The first index is the direction normal. So the direction normal is x2 that is why the first index is 2. The second index is what? Second index is the direction of action of the force itself that is also acting along 2. So this is tau 22. Consider the shear components of the stress. So you have let us say that the force acting the stress tensor component that represent the shear stress along x1. So the first index tau the first index is 2 because the normal to the surface is still 2. The second index is 1 because the force is acting along 1. So I am not repeating all the examples but you can well I mean you can practice this representation by considering by drawing an arbitrary cube and for all the phases of the cube you can try to represent by the corresponding forces corresponding stress tensor components. Now this kind of representation you can see the tau ij or tau ji representation whatever this you can apply only for those surfaces which are having normals along x1, x2, x3 either positive x1, x2, x3 or negative x1, x2, x3. But on an arbitrary surface the direction normal is not along x1, x2 or x3 it is an arbitrary direction normal. So our objective now is to express the traction vector at any point on an arbitrary plane in terms of the stress tensor components at that point. So that we will see in the next slide. So to achieve that what we have done is we have created a volume Oabc. This volume is bounded by how many phases? This volume is bounded by 4 phases Oab, Oac, Obc and Abc. Now why we have chosen such a volume? See whenever we are deriving something the first question that should come to your mind is that why are we going for that particular way of derivation. See we have to relate with what is represented by the tau ij notation with something which is not representable by the tau ij notation. So out of the 4 phases here 3 of the phases have normal along either x1 or along x2 or along x3. So the forces on those can be represented by the tau ij notation. Whereas the Abc is an arbitrary plane the force on Abc cannot be represented by the tau ij notation. So here we consider the surfaces s1, s1 is Oac, this Oac that we call as s1 with surface normal along x1, s2 is Obc with surface normal along x2. You can see when we say surface normal along x1 it is actually surface normal along negative x1. Like if you consider the Oac what is the outward normal to this? Outward normal to this is minus x1 direction. Therefore by sign convention all the forces which are represented on that surface their positive directions will be along the negative direction. So let us come back to it. So what we can see here is that Obc has a direction which is negative of normal direction which is negative of x2. Oab has a normal direction which is negative of x3 and Abc has an arbitrary normal which we call as say eta which is eta1i plus eta2j plus eta3k where eta1, eta2, eta3 are the direction cosines of the vector eta. Now we can apply a force balance on this element. So to apply a force balance let us say that we are writing force along x1. So force along x1 so what will be the types of forces as we have discussed? Type of force will be surface force and body force. So the surface force for the surface s1 for the surface Oac what is the surface force along 1? It is tau11 because it is normal direction is 1 and the force direction we are considering also 1. So tau11 that is force per unit area that multiplied by area s1 is the total force. Why are we putting a negative sign? Because the outward normal of this is along negative x1. So by our sign convention the force on this surface the force along x1 the positive sign convention means that the force along x1 will be in the negative x1 direction because the normal itself is along the negative direction. Similarly the force for s2 that is OBC what is that? That is tau21 why tau21 the first index 2 is because the direction normal of the surface is 2 and the second index 1 is because the direction of the force is along 1. Again the negative sign is because the OBC has outward normal of minus x2. So minus tau21 s2 similarly for OAB minus tau31 s3c for these 3 surfaces we are able to use that tauij type of notation. But for the surface s which is ABC what we do is we use the traction vector t notation because that is an arbitrary surface. So t with superscript eta and subscript 1 because subscript 1 indicates the direction of the force. So the first four terms are the surface force components. Next the body force term. So let us say that B is the body force per unit volume. So what is the volume of this? The volume of this if H is the perpendicular distance from O to ABC then what is the volume? 1 third into the area of the base into height that is the volume of this figure. So 1 third into s into H and B is force per unit volume. So 1 third into s into B 1 third into s into H into B 1. B 1 is the component of B along x1. So this is the body force. So we have represented the resultant force is equal to mass into acceleration. Mass is rho into volume times acceleration along one. Now we can simplify by noting that see what is s1? OAC. OAC is nothing but that can be perceived as a projection of ABC on the x2, x3 plane. It is nothing but the projection of ABC on the x2, x3 plane. So that means it is like a component of the area ABC on the x2, x3 plane. So the component is given by the dot product. So it is basically the dot product of the direction of ABC with the direction of OAC. Direction of OAC has the unit vector i. Direction of ABC has that unit vector eta1i plus eta2j plus eta3k. So if you take a dot product of these 2, so let us go to the board and do this. So let us say that you have, let us go to the board. One vector eta which is eta1i plus eta2j plus eta3k and the other vector is i. These are the 2 normals of the 2 phases. So if you take the dot product of these 2, then you get eta1 because i dot i is 1. So this is the dot product of the component. So this is the dot product of the component. Component of what? Component of the area s. So s1 becomes s into eta1. Similarly, s2 becomes s into eta2 and s3 becomes s into eta3. With this understanding we can simplify the equation. So let us get back to the slides and try to simplify the equation. So what we can do here is that instead of s1 we write s into eta1. For s2 we can write s into eta2. For s3 we can write s into eta3. So s from all sides get cancelled. Then very importantly we take the limit as h tends to 0. Why we are interested to take the limit? What is h? h is the perpendicular distance from O to ABC. If you take the limit as h tends to 0, what will happen? The entire volume will shrink to a point and we are interested to make a representation at a point. So to make a representation at a point what we will do? We will take the limit as h tends to 0. So if h tends to 0 you see the body force term becomes 0 and the acceleration mass into acceleration term becomes 0. So you have t1 as tau11 eta1 plus tau21 eta2 plus tau31 eta3. So you can see we can derive this. Many people derive this by considering no body force and no acceleration. But what we can show here is that the same derivation is true even if there is body force and even if there is acceleration. So this is true for even an accelerating element subjected to body force. Still we can write the same expression relating the traction vector with the stress tensor component. So this is the relationship that relates the traction vector with the stress tensor component. So we can write ti with superscript eta is equal to tau 1i eta1 plus tau 2i eta2 plus tau 3i eta3. So that is actually what? That is summation of tauji etaj where j is equal to 1 2 3. In the notation here we can see that instead of writing the sigma tauji etaj we are simply writing just tauji etaj without with an invisible summation of sigma. This is Einstein's index notation. So Einstein's notation what the notation Einstein introduced is that if you have a repeated index then you must keep in mind that there is an invisible summation over it. So you do not write a sigma on the top of it. So when you write tauji what you essentially keep in mind is that tauji etaj because j is repeated there is a summation over j from 1 2 3. So this theorem which relates the traction vector with the stress tensor component is called as Cauchy's theorem. So using this Cauchy's theorem it is possible to relate the traction vector with the stress tensor component. So we have till now understood that how to represent the surface forces, how to represent the body forces and how to write this the relationship between these forces and the transport of linear momentum across the control volume. So what we will do next? We will try to substitute the expressions of these forces in the Reynolds transport theorem and first try to obtain an expression which relates the change in linear momentum in terms of the surface forces and the body forces and that can be done for any general fluid. And then we will represent the surface forces for special types of fluids which are Newtonian fluids and we will talk about the special types of Newtonian fluids called as homogeneous and isotropic Newtonian fluids and then we will invoke a hypothesis called as Stokes hypothesis by which we will arrive at an equation which is the celebrated Navier-Stokes equation that we use many times in continuum fluid mechanics. So that we will take up in the next lecture. Thank you very much.