 Welcome you to the second lecture of multiphase flows okay and what we are going to do today is basically look at a very specific problem in microfluidics and that is the problem of stratified flow in microfluidics. So like I had mentioned we will have one consider a system which is has a t shape okay and we have 2 inlets through these 2 limbs okay. What I want to be clear about is that the view I am drawing is the view of the channel as we see from the top. So the t junction is placed horizontally on the board on the table and we are viewing it from the top and so the 2 fluids are flowing side by side and you would have 2 different phases an aqueous phase and an organic phase which are flowing. Now the objective today is going to be only to understand the hydrodynamics okay. What we are interested in is microfluidics which means 1 of the dimensions is less than 1 millimeter. So one of the characteristic features of any flow in a microfluidic channel is that the Reynolds number is very low because the Reynolds number as you know is going to be defined as the density multiplied by the velocity average velocity multiplied by the characteristic dimension divided by the viscosity okay. So now we are looking at channels which have a very low diameter so we are talking about very low Reynolds number flows. So what does this mean? This means the inertial forces are negligible and viscous forces are dominant okay. So we are looking at dominant viscous forces and very low inertial forces. So more specifically since we are looking at lower Reynolds numbers we are looking at the laminar regime okay and so when you talk about laminar regime flows are very well defined and so that helps us go towards getting an analytical solution. So whereas if you had a large Reynolds number you would possibly go towards the turbulent regime and then you would possibly have to go to a numerical solution. So our objective in this course is to try and get understanding of flow problems transport processes analytically and so to keep things simple we are basically going to exploit the fact that the Reynolds number is low. So we exploit the low Reynolds numbers seek to obtain analytical solution. So let us come back to this figure here. What do I expect? I have 2 liquids now which are actually immiscible okay. The aqueous phase and the organic phase they do not mix with each other and let us say they are completely immiscible. So we are going to be looking at stratified flow. The aqueous phase is going to flow through this inlet channel it is going to come and flow along this main channel here the organic phase comes and turns around. So this is going to be the path of the aqueous liquid this is the path of the organic liquid. Now if your channel is sufficiently long one of the flow regimes that is possible is what we have seen is the stratified flow which means that there is a very well defined interface between the 2 liquids and that is the aqueous phase here and you have the organic phase here. So what we want to focus on is how is the flow field in this particular portion which is far away from the inlet here okay. This particular portion close to the inlet is where you have what are called entrance effects because the fluid has to actually negotiate the bend okay and so you could have some mixing you could have some lateral interchange of mass and momentum. But we are not focusing on this because in a typical situation what happens is the entrance length is just going to be a small fraction of the entire length. So whatever you are really going to be actually visualizing which you are actually going to be seeing is going to occur over this part of the channel which is going to occupy a significant portion of the length. So as engineers you may be interested in understanding what is happening in this region which we will call the fully developed region okay. So typically the fully developed region occupies a significant portion of the channel okay. So if you are interested in how a chemical reaction is taking place between these 2 liquids or how mass transfer is taking place between these 2 liquids it is more sensible for you to focus on the fully developed region okay because that is where the actual process is going to take place. So what we are going to do today is mainly look at the flow because the flow is going to decide what the mass transfer behavior is going to be okay. So today we will just focus on how to determine the flow behavior how to determine the velocity profile when you have 2 liquids flowing in a fully developed manner in a micro channel okay. Now since we are interested in developing analytical solutions our focus is on getting not only mathematical solutions but also getting physical insight. So we want to make sure that we simplify the problem in such a way that we retain the important physics. So we want to make simplifications which will make it mathematically easy for us to solve a problem. So the solution is going to be mathematically tractable but then we do not want to lose information about what is actually happening inside the system okay. So our objective is make simplifications so to our strategy is as follows we make simplifications such that the important physics is retained okay but we can and still we can solve the problem analytically okay and so the simplifications are mainly mathematical. So now coming back to the fully developed flow now what I am going to assume is all of you have done a course in fluid mechanics okay this is a PG level course. So we will just go with that assumption later on we will actually derive some of the fundamental equations again. So let us look at just the fully developed region now which is x let us say this is the flow direction which is the x direction and this is the y direction and have the z direction which is coming out of the board okay. Now I talked about making simplifications so one of the simplifications I am going to make now is I am going to assume that instead of having a channel which has a finite cross section which is rectangular I am going to assume that these plates are actually going to extend to infinity in the z direction okay. So what that basically tells me is that things are basically not going to be changing in the z direction okay I am also going to assume that things are going to be sufficiently long in the x direction. Remember I am going to have to keep it small and finite because I am looking at a micro channel so the simplification that we are looking at is we assume the x and z directions to go to infinity to be infinitely long okay and in the y direction the width between the plates is capital H okay. In order for me to specify the problem what I also need to know is the location of this interface. So I need to know so this gap between the 2 plates is capital H the fluid is occupying a the lower fluid is occupying h small h region of the entire space between the walls okay. So this is liquid 1 this is liquid 1 which has properties rho 1 and mu 1 this is the density of the first fluid viscosity of the first fluid this is the second liquid which has properties rho 2 and mu 2 okay. Now clearly the first fluid is occupying the first fluid occupies the distance 0 less than y less than h and the second fluid occupies h less than y less than capital H. So one of the things that we are interested in is trying to understand how the velocity is going to change inside this because the velocity profile is going to decide you know for example the time spent by the fluids inside the channel and also going to decide any other transport processes which are going to possibly occur. Now my simplifying the z direction going to infinity basically make sure that I can neglect the velocity component in the z direction and I can also neglect changes in the z direction okay. So that is basically what whereas the flow direction is also going to infinity but what I am going to do now is only focus on the fully developed profile because that is what I am interested in okay. So z direction going to infinity allows me to neglect the flow in the z direction as well as changes in the z direction because the z direction I do not have any confinement or walls. So if the walls were there then the walls would actually affect the flow in the z direction okay. Now what I want to do is go back to understanding which velocity component is going to be important okay. So what we have done is we have simplified things by just saying that the z direction things do not change. So we are going to focus only on the x and y directions now. So in order to focus on the x and y directions we are going to look at first the equation of continuity which is since we are looking at liquids we are going to say that the liquids are incompressible no changes in the density and remember I am going to write this equation of continuity for each phase. So I am going to write the equation of continuity for this liquid I am going to write the equation of continuity for this liquid okay. So I am going to just say that the equation of continuity for phase 1 the divergence of rho 1 u1 equal to 0 okay density is a constant. So I can actually take out the density and it basically boils down to divergence of u1 equal to 0 since the liquid is incompressible and this essentially means if I am going to write it in the scalar form where u, v and w are the velocity components in the x, y and z directions okay, u, v, w are the velocity components in the x, y and z directions. What I want to do is I want to look at simplifying this equation okay. So the first thing we do is we observe that we have assumed things to be infinitely long in the z direction okay. So there is going to be no changes in the z direction besides the velocity is also 0 in the z direction. So I can essentially drop off this term. So this term essentially goes to 0 okay. We have also assumed that the flow is fully developed. So when I say things are fully developed what I mean is that there is no variation in the direction of the flow. The direction of the flow is x okay. So when I am saying that the flow is fully developed I basically mean du1 by dx is 0. So because of my fully developed flow assumption and that is what I am focusing on du1 by dx is 0 because it is fully developed okay. So what that leaves me with is that I have dv1 by dy is 0. So my equation of continuity for one of the liquids basically simplifying to this. I could have written the same thing for the second liquid and I would have gotten a similar expression dv2 by dy equal to 0 okay. Now from this what do I conclude? I conclude that v1 does not change with y that is what this means. Now what I have here at y equal to 0 is a solid wall okay. So the liquid cannot penetrate the solid wall. So the normal component of velocity which is the y component of velocity is going to be 0 at this wall and there is going to be 0 everywhere okay. So what we know is since at y equal to 0 we have a solid impermeable wall v1 equal to 0 okay at this boundary and we also know that v1 does not change with y. So from this I conclude that v1 is 0 everywhere inside the first fluid okay. So this implies that v1 equal to 0 everywhere inside the first fluid. I could do a similar argument for the second fluid. I would use the fact that the top wall is a flat plate impermeable the velocity here is 0. I would eventually come with dv2 by dy being 0 use the boundary condition here and come to the same conclusion that v2 is 0 everywhere inside the second fluid okay. Similarly we can conclude that v2 equal to 0 everywhere inside the second fluid. So I want to emphasize here I have not assumed that the velocity in the y direction is 0. I have only assumed that the flow is fully developed okay. I have only assumed that the flow was fully developed and that told me basically that there is no velocity in the y direction. Now the fact that there is no velocity in the y direction the velocity is 0 here of this liquid velocity is 0 here in this liquid basically tells me that the interface is going to remain flat okay. We will take a look at what the pressure is we will also find that there is no change in the pressure across this interface. So basically the interface is going to be flat. What this means is that the stratified flow is a possible solution that is we are being internally consistent. We assume that a stratified flow can possibly exist. What I am showing here is a stratified flow can exist okay under fully developed conditions. So basically what this means is we can have a stratified flow okay. Sometime later on we will show you some videos wherein we have examples of stratified flow being seen. So experimentally you can see this but that is also consistent with the mathematical theory. So in particular what I am saying is you can have fully developed stratified flow where the interface remains flat interface does not get deformed okay. So everything is coupled fully developed implies interface remains flat. Interface flat in turn implies there is no vertical component of velocity because if the interface had deflected that means there was a vertical component of velocity which actually caused the deflection okay. The fact that V1 is 0 follows from fully developed follows from the fact that the interface is flat okay. And you know I am doing this in rectangular Cartesian coordinates because that is really simple and easy for you to possibly visualize. Later on we will possibly look at curve coordinate systems. So having taken care of the equation of continuity and what that has told me is that the velocity component in the y direction is 0 and need to now understand our objective is what is to find the velocity profile. So the only component of velocity we need to worry about is the x component of velocity okay. So what you want to do is write down the Navier-Stokes equations in the x direction because that is what is going to decide what the x component of velocity is okay and simplify that with whatever we have already found out okay. So that is what we are going to do. So let us write out the Navier-Stokes equations and I am just doing it for one of the fluids I am not putting the subscript 1 it is the same thing for both the liquids. So basically what I have done is written down the Navier-Stokes equations or the equation of momentum this is the equation of momentum in the x direction okay. I am writing this equation here and I want to use the information that I already have to simplify this. So the fact that it is fully developed tells me that u does not change with x and that allows me to drop off this term because it is a fully developed flow. Remember that is what I have assumed fully developed okay. I am also assuming that the flow is at steady state. If I assume that the flow is at steady state this goes off because at steady state this goes to 0. What about this term? This term is going to drop off because we have already established that v is 0 okay everywhere inside the liquid. So this is 0 because v equals 0 everywhere inside the liquid and we have assumed things are not changing in z. We have assumed that things are not the velocity component in the z direction is 0. So this drops off. Let us come to the right hand side. What about dp by dx? The flow is going to be driven by a pressure gradient okay. So dp by dx has to exist without the only thing which is driving the flow is the pressure gradient. So that is something which I am going to be externally imposing on the system by means of a pump okay. So that is going to be there. There is something which I am controlling as an experimentalist. What about this term here? This term is going to be 0 because of fully developed flow okay and this term is 0 because things are not changing in the z direction because it is infinitely long in the z direction. This is u is infinitely long in the z direction. So things do not change in the z direction. I am having a flow only which is one dimensional. I have the u component of velocity the x component of velocity which is flowing in the x direction. So basically what this means is the equation of motion in the x direction simplifies to the following equation minus dp by dx plus mu d square u by dy square equals 0. I want you to pause for a minute and think about what I said earlier about flow being fully developed and low Reynolds numbers. Now if you remember these terms on the left basically represent inertial terms okay. This is your pressure forces. These are your viscous forces. Now low Reynolds number flows basically correspond to very low inertial terms. So basically what I have done when I am simplifying I could have done this earlier. I could have just told you look we are looking at lower Reynolds number flows. We are just going to drop off this left hand side and then we could have just retain the right hand side and done the analysis okay. That could have been one approach which you can take because these are the inertial terms which are negligible in the regime of microfluidics okay. So basically these are inertial terms so which we have neglected. The left hand side represents inertial terms okay and these negligibly small okay and this is consistent with our assumption of low Reynolds number. So I just want to tell you that what we have done now is basically neglecting inertial terms and remember I started off with low Reynolds numbers which means low inertial forces, large viscous forces they are fine. So basically the momentum equation tells me that the viscous forces are balanced by the pressure forces okay. That is basically what it is. So I need to also look at the momentum equation in the y direction okay. In the z direction I am not particularly worried about the z direction because nothing is changing in the z direction but I want to simplify the momentum equation in the y direction okay. So let us do that. So look at the momentum equation in the y direction. What do we get? I have just written the Laplacian as it is. So now I like to know because clearly there is some information which is present in the momentum equation and I like to extract that okay. So let us see what information is present here. You know that we have already established the vertical component of velocity. The v component in this y direction that is v is 0 everywhere inside the domain. So essentially what this simplifies to is that the pressure is independent of y okay. So what this basically means is that the pressure does not change in the y direction. So that is the information which you are getting from the momentum balance in the y direction okay. And so whenever you are doing a problem you need to make sure you get a result which is actually consistent. So remember okay if the pressure does not change in the y direction there is going to be no pressure discontinuity across this interface. There is no velocity component. As a result the interface does not deform because the interface can deform also from a pressure change okay. Now we will look at what the regular boundary condition is at the interface okay the normal stress boundary condition. But I just want to tell you here that the pressure is independent of y okay. And this means the pressure can only be a function of x okay. This implies that pressure can only be a function of x. And clearly that is what we expect. We expect the pressure to decrease as you go along the flow okay. So pressure will change with x. But what you have found out is that there is going to be no change in the y direction. So the pressure is uniform in the y direction. So if you had a particular pressure at the inlet if you have a particular pressure this is going to vary along x but there is going to be no variation induced in the y direction okay. So the pressure there is going to be no pressure change across the interface. So there is no velocity component across the interface there is no pressure change in the interface remains flat. So far things are consistent okay. Now what we do is we need to find out how velocity is changing in the y direction okay and that tells me mu d square u by dy square equals dp by dx. This is my differential equation which I am going to work with okay and now I am going to go back to the fact that remember what about this, what about dp by dx? It is a function of only x okay. This is a function of x. What about d square u by dy square? It is a function only of y. It does not depend on x because it is fully developed, velocity is fully developed okay. This is a function of y. So the only way a function of x can be equal to a function of y is if both of them are equal to a constant okay and so what this means is that the dp by dx that you have is actually a constant okay. So you have a constant pressure gradient okay along the flow which is driving your flow. So this implies that mu d square u by dy square equals dp by dx which is equal to a constant okay. Now what do you want to do is we want to get the velocity profile. Now I am going to go back and put my subscripts 1 and 2 okay because I need to find out what the velocity is for liquid 1, what the velocity is for liquid 2. What I have done is gotten rid of subscripts now because I just wanted to tell you that this is generic for both the fluids. So in the domain 0 less than y less than h we have mu 1 d square u 1 by dy square equals dp by dx okay. In the domain I just got mu 2. Now see I have the same pressure gradient driving both the fluids. I have a particular pressure at the inlet okay and I have already told you that I have pressure only varying in x direction. So I do not use a different constant for this pressure gradient and a different value for this. Both of them are the same. The pressure gradient in both the liquids are the same okay and that is one of the simplifications we have because our flow is parallel okay and straight. So if the interface have been bent then there would be a pressure difference between the 2 liquids which I would have to incorporate. We will worry about that later right now we will do the simple problem. So the fact that this is a constant allows you to actually integrate this out and u 1 depends only on y. So you can actually integrate this out and this would on integrating twice would give you u 1 okay maybe I should do this step by step. I do not want to make a mistake here. Let us just integrate the first guy du 1 by dy integrating once du 1 by dy gives me 1 by mu 1 dp by dx plus c 1 and u 1 gives me 1 by mu 1 dp by dx y square by 2 plus c 1 y plus c 2 okay. So this is how my u 1 is going to be varying with y c 1 and c 2 are arbitrary constants which I need to determine. So all I have done is integrated this differential equation. Similarly I can conclude that u 2 is going to be 1 by mu 2 dp by dx y square by 2 okay because I will have mu 2 here plus c 3 y plus c 4 okay. I need to now find out these constants and for that I need boundary conditions okay and what are the boundary conditions I am going to use? So we have 4 constants and we need 4 boundary conditions okay. So the boundary conditions are going to be clearly applied at the walls and at the interface. So at y equal to 0 what are the 4 boundary conditions? y equal to 0 this is the lower wall we have the no slip boundary condition okay. The no slip condition implies u 1 is 0 now at the upper wall and y equals capital H again the no slip boundary condition because the wall is flat stationary the wall is not moving okay the flow is being driven by pressure. This implies u 2 is 0 it is the second liquid okay because the second liquid is going from small h to capital H. So I got my 2 conditions from my no slip boundary condition and I need 2 more. Now what I am going to do is these 2 conditions are going to be at the interface. So at the interface when y is equal to small h we have the continuity of velocities I want a physically admissible profile I expect the velocity field to be continuous okay. I do not expect that to be discontinuity in the velocity field. So what this means is u 1 must be equal to u 2 okay and the other thing which I expect is that the stresses the shear stress the tangential stress exerted by the first fluid on the second must be equal to that exerted by the second fluid on the first okay. So we want this is coming from the continuity of the velocity okay and we also have continuity of the shear stress along the interface. So we will derive the formal equation of how to go about calculating the shear stress but right now what I will do is okay the this is the x direction and this is the y direction. The shear stress at this interface is going to be given by tau yx okay the first subscript tells you the direction of the normal and the second subscript tells you the direction along which you are trying to find the stress okay. So I am interested in finding the stress on a surface which is perpendicular to the y direction and along the x direction because that is where the flow is okay. I am interested in finding so I want continuity of shear stress. So I want tau yx in the first fluid must be equal to tau yx of the second fluid and if you are going to assume your liquid is going to be Newtonian the two liquids are Newtonian. So the shear stress tau yx in the first fluid is given by mu 1 du1 by dx sorry du1 by dy plus dv1 by dx. This is the shear stress at the interface. The shear stress at the interface given by the second fluid is mu 2 times du2 by dy plus dv2 by dx. Remember these are evaluated at the interface y equals small h. Now since the interface is not deflecting v1 is 0 everywhere at the interface as a result of which dv1 by dx is 0 and we have similarly dv2 by dx is also 0 because of the fact that the interface is not deflecting and the vertical component of velocity is 0 for all x. What this means is the equality of the shear stresses gives me for a Newtonian fluid remember this has been written for a Newtonian fluid mu1 du1 by dy equals mu2 times du2 by dy at y equals small h the interface. So these are the four boundary conditions. This is the fourth one where we have equality of the shear stresses at the interface then we have the continuity of the velocity at the interface and we have the no slip boundary conditions at the two walls. These four boundary conditions are used to find out the four constants of integration which arise when you integrate the second order differential equation of the velocity in each phase and that gives you the velocity profile okay. So these are the four conditions which I have to use to get my velocity field that is to determine my constant c1 c2 okay and they are just algebraic equations which you will use and you will find out what the constants are c1 and c2 and then for particular values once you have found out c1 and c2 you can actually plot these velocity fields okay. So what I am going to do is I am going to ask you a couple of questions for you to think about and then we will stop. So now the two velocity profiles have a parabolic dependency on y in the y direction because it is quadratic in y okay. So I am going to ask you a question which of these flow fields can be seen? One this is my interface. So what I want to emphasize is I have drawn these three flow fields keeping in mind two things. The velocity is continuous remember what I have is that the derivatives do not have to be continuous. The derivatives do not have to be equal okay. There can be a difference in the slopes at the interface that is how this tells you the slope in the one liquid is going to be the ratio of the viscosity is multiplied by the slope in the other liquid okay. So that is a discontinuity in the slopes that you are going to observe. So this is a possible flow field that is the one maxima in this liquid and there is no maxima here. There could be a maxima in this liquid that is no maxima here or there could be a maxima in both the liquids okay. So what I want you to do is think about whether these are possible. If they are possible under what conditions will you actually see this kind of a behavior or this kind of a behavior or that kind of a behavior or if you are not going to be able to see any of them you have to know we will discuss this the next time. Thanks.