 Today's lecture what we will do is we will look at another application where we can possibly get a simplified solution for fluid flow problem okay and the idea is we are not going to be doing stability analysis. We just have to get a solution now to a problem which where we can exploit the presence of different length scales. So this particular problem is going to be called a shallow cavity flow. So as far as shallow cavity flow is concerned, first of all order exactly is a cavity, just think of a rectangular channel okay and let us say there is a top plate which is moving with a particular velocity u0 okay. So there is a fluid which is confined here and this is the rectangular channel and let us say as always it is extending to infinity outside the plane of the board. So we are looking at the story in this 2 dimensional plane okay, so no variations in the direction perpendicular to the board. So what we expect is that the liquid is going to be dragged by the top wall and it is going to circulate, definitely can penetrate the wall. So you have this kind of a situation, so actually this is a very classic problem. This problem is called the lid driven cavity and this is one of the first problems people solve in computational fluid dynamics. So you write down the equation of continuity, equation of momentum in x and y directions and then you solve using some numerical method okay. So this flow field is obtained using computational fluid dynamics okay. So the question of course is whether we can actually make some kind of a simplification and get some idea about the flow field okay and under some conditions. So when you talk about a shallow cavity, we are talking about a cavity whose depth, let us say this is the z direction and this is the x direction and let us say the extent in the z direction is d that is this distance and the extent in the x direction is l okay. So if d by l is very much smaller than 1 okay and we have a shallow cavity. What we would like to do is analyze this problem and see if we can get some idea about the flow field inside okay by exploiting this factor d by l being very much less than 1. Now where is this important? So for example you can have thin films okay or you can have maybe even microfluidic devices. So supposing you have a very thin film of liquid on a solid surface and either the lower surface is moving and the upper surface exposed to atmosphere okay you would have a situation where this is a thin film and that is the direction which is being dragged for example. And this of course instead of a solid solid wall you will have a solid gas wall but again the business of these 2 lens scales is going to be present and so basically the presence of these lens scales d by l being very much less than 1 d by l is a small parameter you can think of already an epsilon coming up here and how I can possibly use this epsilon force simplifying equations and getting some insight that is the idea okay. And microfluidic devices for example we long back spoke about something called a slug flow and the basic idea in a slug flow situation is supposing you have a 2 phase flow and let us say this is oil and this is water you would have things like oil slugs almost occupying the entire channel okay, separated by water. So what is going to happen the oil is going to be flowing okay and this is going to be continuous stream of oil drops or oil slugs flowing and of course there is also water present here which is also flowing. Again if you do what we did yesterday that is work in a moving reference frame that is you are sitting on the oil drop and moving along with it. It would be like the oil is stationary and the wall is moving backwards okay sitting the water occupied portion and moving again the wall is actually moving backwards. So what you can see is when you have this kind of a relative motion between the wall and this you are going to have internal circulations inside the slug okay. In the moving reference frame we have let us say we forget about the film this guy is moving backwards. So this is something like your lid driven cavity problem okay. Only thing is instead of solid walls I have a liquid-liquid interface and it is approximating this to this okay and what I am going to observe is I will have some kind of a flow pattern of that kind and a flow pattern of that kind here. So some vortices are induced okay. So the point I am trying to make here is when you have this kind of a slug flow regime you would have because of the viscosity you would have vortices induced okay and the one way to understand this is to look at this problem in a moving reference frame. So rather than look at the because actually this is an unsteady state problem because at any instant of time you will have one particular portion occupied either by oil or by water okay. So just like yesterday we went to a moving reference frame I go to a moving reference frame and I say look the walls are moving backwards liquid is stationary and what this means is you have these internal circulations. Now what we want to do is we want to understand these internal circulations because that is going to help in mass transfer and heat transfer and things like that. So when you are trying to do some kind of a reaction and let us say there is some species here which has to be transported from the aqueous phase to the organic phase or vice versa. This flow is going to actually help in moving the species okay. So what one wants to do is one wants to understand how these vortices are actually going to be developing and whether we can get some idea about the magnitude of the velocity. Of course one approaches do CFD okay and the other approaches see if we can get some insight without doing CFD. And like I told you at the beginning what you can do is you can get these analytical solutions get some idea and if you are really interested you can do a CFD and get a more rigorous solution and in some limit the CFD can be validated by this analytical solution because at the end of the day CFD is going to always give you some nice pictures nice graphs and results. So the numerical accuracy of these results can be verified only by verifying this and some limit for example that gives you some confidence in your CFD results okay. So the idea is removing reference frame the wall moves back okay and in each slug we have internal circulations okay, internal circulations and this is important to understand mass transfer for example in reactions. So typically what will happen is there will be some species here and another species in this phase both have to come together to one phase for the reaction to occur okay and this convection is going to be deciding the whole thing. So how do you go about analyzing this? So the point I am trying to make here is this problem is very similar to that problem okay. You have a rectangular wall and the only difference is the boundary condition here. Here I have a solid wall but here I have a liquid-liquid interface but if you neglect the fire that is going to get deflected that means again perpendicular component of velocity will be 0 but you will have tangential component of velocity for example okay. So how do we go about, so this is just a motivation for doing this kind of a shallow cavity problem okay and typically in micro channels the slug length can be 5 to 10 times that of the diameter. So maybe we can push our luck and try and get some understanding of the flow field here using the shallow cavity limit. So what we will do is look at the flow in the original problem that we had in the shallow cavity and this is the problem DL maybe I will do what has been done in Gary Lille U0. In Gary Lille he has this business I think this is the x direction and that is the z direction. The lower plate is the one which is moving in the U0 with the velocity U0 in the positive x direction okay. Clearly what is going to happen is liquid is going to get dragged and going to form some kind of a vortex okay. This whole problem can be analyzed by dividing this entire domain into 2 parts okay. The central portion here which is the core region and the portions on the other 2 sides which is the near wall region okay. So I am dividing the entire domain into 2 portions. The central portion which is core and here where the fluid is actually going to be bending backwards. So there is a difference in the physics in both these regions. For example if you focus somewhere in the central portion of the core region you do not have the effect of these walls okay. For all practical purposes you can view the flow as being almost parallel only thing is the flow is the positive x direction in the lower portion and in the negative x direction in the upper portion okay whereas here and therefore for all practical purposes you can imagine that the vertical component of velocity is 0 I mean in some limit of course there is small vertical component of velocity but that is going to be negligible whereas here the vertical component of velocity is going to be comparable to the horizontal component of velocity okay. So in the core region the u and w sorry w is much smaller than u whereas near the wall the w is comparable to u. We are also going to look at macro fluidic applications and very small channels and the flow through the small channels the characteristic Reynolds number is going to be very low okay. So we are talking about very very slow flows. So what we will do is we will try and analyze this situation using the low Reynolds number limit. So rather than write everything and then put Reynolds number equal to 0 I am just going to at the beginning itself put the inertial term is equal to 0 write my equation of continuity and equation of momentum okay. So let us do that. So we focus on lower Reynolds number flows which means we drop the inertial terms on the left hand side of the Navier-Stokes equations okay. And what am I left with the equation of continuity which is du by dx plus dw by dz equal to 0 and then 0 equals minus dp by dx plus mu d square u by dx square plus d square u by dz square okay and 0 equals minus dp by dz plus mu. So what I have done and gravity I am not worried about just imagine the gravity is actually absorbed into one of these pressure terms okay as a gradient. So this is your modified pressure. So that is my equation of continuity the y direction is outside the plane of the board I am neglecting. So this is the regular thing that we have. So what we want to do is we want to simplify this make it dimensionless okay and then exploit the fact that d by l is very much lower than 1 and see what kind of simplification we are going to get. So see this business of the length scale in one direction being much smaller than the other is something which is you have seen in boundary layer flows for example okay. The thickness of the boundary layer is very very small. So what is the argument you make over there? You say that look in order for the equation of continuity to be satisfied for both the velocity component both these terms actually contribute okay. The characteristic scales which are written down there that w is much smaller than u we need to have an estimate clearly what is the characteristic scale in the of the velocity in the x direction it is u0 because that is the lower plate velocity. What about the characteristic velocity in the w direction that is going to be decided by the problem here in the sense that both these terms have to contribute okay and I am going to scale the length in the x direction with l and the length in the y the z direction with d because those are the characteristics scales in the respective directions okay. So u characteristic in the x direction is u0 x characteristic is l and y characteristic is sorry z characteristic is d okay and we need to know what is w characteristic okay. So clearly I am going to substitute this here I am going to get du star by dx star and this is u0 that is the u characteristic okay and l plus w characteristic by d0 r0 okay. So the idea is that these two terms have to balance each other and they can balance each other only if this is of the order of magnitude 1 and so this implies that w characteristic is d by l times u0 or epsilon u0. So it is epsilon times the u characteristic okay and so what I am going to do now is I am going to use this in my momentum equations make them dimensionless okay and get an idea about what is happening. What I want you to understand is the regular flows when you have a pipe flow the pressure drop is something which you impose experimentally okay whereas in this particular problem there is going to be a pressure gradient which is going to be decided by the flow okay. So dp by dx and dp by dz is something which I do not know something which I have to find out earlier the second point is a flow you said because that is a pressure driven flow that is controlled by you experimentally. So you impose dp by dx and then you find out what the velocity field is you get a parabolic profile okay. Here remember the flow is going to be driven by the wall and so dp by dx and dp by dz is something which I need to find out okay. So let us do this I also do not know what is p characteristic because the characteristic pressure is going to be something which is decided by the flow and the fluid properties clearly things like viscosity okay and the dimensions of the channel. So p characteristic is also an unknown. So what I am going to do is I am going to take the x momentum equation make a dimensionless and idea is 2 by dx star times p characteristic by l plus mu times d square u by dx square will give me u0 square okay. I like to get d by l out so I am going to take out d square and the z direction I have d that comes out. So I have d square by l square here okay. Remember this is made dimensionless with l that is coming here this is made dimensionless with respect to d so that comes here I am taking out a d so that d comes there. So clearly this d square by l square is epsilon square. So if I have chosen my characteristic scales properly what this tells me is that the second derivative in the x direction is much smaller than the second derivative in the z direction. See the z direction my distance is very small so that the variations are much sharper. So when I have to compare these 2 I can actually neglect this in comparison to this okay. So basically this is an order of 2 orders of magnitude lower than this depending upon epsilon. This is an epsilon square term but what I need to do is I need to choose my p characteristic so that this pressure gradient in the x direction is going to balance the term in the z direction the second derivative of the z direction okay. So then only this will balance that. So I am saying I am going to neglect this compared to this okay. So this is epsilon square that is epsilon square and I have 0 equals minus dp by dx times p characteristic by l plus mu u0 by d square times d square d z star square. Now p characteristic is chosen as mu u0 times l divided by d square or mu u0 by l times 1 by epsilon square okay. So basically the characteristic pressure which is actually developed inside the gravity is given by this okay. So if you choose this then this becomes equal to that and your simplified equation of momentum in x direction is minus dp star by dx star plus d square mu star by d z square that is your dimensionless equation of momentum okay. So this pressure force basically the 2 important forces are the pressure force and the viscous force okay. Gravity anywhere you are neglecting inertia is gone so that is and there is something very similar to what you have seen in your second partial flow you have pressure and you know your viscous force. Only thing is I need to account for this guy bending back and all that and dp by dx is not known to me okay. So now let us do the other direction 0 that is the z momentum equation okay. That is what we have to do and what I will do is go to the other side of the board. I am going to start with this by d z star. This is p characteristic which I know and this is which I found from that and this is going to be given by d plus mu times w characteristic is u0 times epsilon okay that comes out and I do my usual stuff which is bring a d square outside here right and I am left with so I have the same stuff l coming as the characteristic so that l comes there this gives me d, d comes out, d comes out that d goes there okay. So now remember what is p characteristic? I already found out what p characteristic is I am going to substitute that here and I am going to get using the p characteristic we get 0 equals minus dp star by dz star plus mu u0 epsilon by d square times a d which goes there and p characteristic is coming in the bottom which means I have epsilon square divided by mu u0 and there is an l here okay times between these 2 terms this is much smaller than this d square w star by dz star square because this is epsilon square times that okay. So let us simplify I get d by l is epsilon, d by l is epsilon right yeah that epsilon that goes off and I get 0 equals minus dp star by dz star plus epsilon square times d square w star by dz star square. So in the limit of epsilon tending to 0 or if you actually did a perturbation series solution and if you found out the solution in terms of a power series okay. So what you would get is in the limit of epsilon going to 0 this term is going to be negligibly small compared to this okay. So if you look at the terms of order epsilon to the power 0 this is going to give me minus dp star by dz star equals 0. In other words the pressure variation in the z direction is not there clearly the region is so thin in the z direction for all practical purposes you can actually neglect the pressure gradient in the z direction whereas the pressure gradient in the x direction is given by this momentum equation which we wrote over there okay. So basically what we have done is we have used the fact that epsilon is very small and rather than do a very formal power series expansion what you should do is now you should seek p as p0 plus epsilon p1 w as w0 plus epsilon and then equate coefficients but those equations are independent of epsilon. The epsilon is occurring only here so I am just getting the zeroth order solution directly by putting epsilon equal to 0 okay. I mean your zeroth order solution or the base solution how do you find just put epsilon equal to 0 you get a base solution. So I am going to get the base solution just by putting epsilon equal to 0 okay. You can do a more formal thing like we did earlier and then look at the first term then you will get this because we need to solve and now I am going to drop all the stars okay 0 equals minus dp by dx 0 equals by dz these are the terms at epsilon to the power 0 these are the 3 equations which we have. So let us look at how we can proceed what does this imply dp by dz is 0 implies that the pressure is a function only of x okay is 2 thin in the other direction so neglect the variation that direction this implies pressure is a function of x okay. So if pressure is a function of x alone okay then I can integrate this d square u by dx squared dz squared equals dp by dx that is from this equation okay. For my equation of the momentum in the x direction d square u by dz squared is dp by dx this is a function of x I am going to integrate this twice but since this is a partial derivative in the z direction what I have is these constants we are going to come they can actually be functions of x okay. So integrate this once you get du by dz equals dp by dx dp by dx times z plus c1 of x okay integrate this one more time you get u equals dp by dx times z squared by 2 plus c2 of x these guys are functions of x because I have a partial derivative I have total derivative I have been constant okay. So remember u can be a function of x because the dp by dx can change with x that is what this means dp by dx remember the function b is a function of x so since pressure is a function of x I am allowing for the fight that u can change yeah yeah yeah z that is right. I thought I wrote z but I guess I did not okay. Now you have to put those boundary conditions what are the boundary conditions at z equal to 0 u is 1 because that is how I made a dimension list and that z equals 1 I have u is 0 because the upper wall is stationary right. So when I put z equal to 1 when I put z equal to 0 I have u equal to 1 okay that tells me c2 is equal to 1 c2 of x is 1 u of 0 equals 0 implies c2 of x equals 1 okay and at z equal to 1 u is 0 and 0 equals dp by dx times half plus c1 plus c2. So c2 is 1 and therefore c1 is okay that is c1 so now I actually have the expression for my velocity u is dp by dx times z squared by 2 plus c1 is c1 x is dp by dx times half minus 1 times z plus 1 okay. I am going to group my dp by dx terms together which one minus dp by dx yeah yeah okay thanks yeah as always you guys are right man and this is minus right so I need to put a minus here I think that is it. So that is your velocity field and it is something like it is a parabola of course and remember but have you solved the problem yeah we need to find w but we know u we know we do not know u because we do not know dp by dx dp by dx is something we said I need to find out what is dp by dx remember it is not that I am imposing a pressure gradient it is the wall which is moving. So how do I find out dp by dx I need to use some condition about the flow. So let us look at to give an idea and to prompt you yeah yeah so I am looking at the core region right and the core region how is the flow somewhere on the top is going from the bottom is going from left to right and for all practical purposes you have this kind of a flow. Now what do we expect is there a net flow across this line you have a confined liquid liquid is confined whatever liquid is going to go from left to right we have a steady state situation okay whatever liquid is going from left to right must be the same as the liquid is coming from right to left okay. So what that means is the volumetric flow from left to right must be balanced by because there is nothing the liquid is not leaving your 2 walls you have 2 rigid walls the liquid cannot go out. So whatever liquid is coming from left to right must actually come back from right to left. So the total volumetric flow rate across this line must be 0 okay and that means the fluid going from left to right across in the fluid the flow rate equals the flow rate from right to left okay which means the net flow rate must be 0 or integral u dz from 0 to 1 must be 0 okay because I have confined liquid and then I am just moving this fellow so I am going to use that condition and of course I know u I am going to use that and find out dp by dx. So put this here and if you do the algebra you will get this is used to find dp by dx equals 6 okay. I mean you can just do this integration and you can find out that dp by dx is 6. So what that means is dp by dx is indeed a constant okay and something similar to what you had for your Heggen-Boysel flow when you say that the pressure gradient is a constant. So if dp by dx is 6 now you go back to that equation earlier we said p was a function of x but I know it is linearly varying. So dp by dx is constant which means that u is not a function of x, u is a function only of z okay. This implies u is a function of z alone it is independent of x okay. So u is independent of x therefore what this means is you have something like a fully developed flow situation d u by dx is 0 okay. That means d u by dx is 0 I will just say it is analogous to fully developed flow and from the equation of continuity dw by dz equals 0 from equation of continuity and since you have 2 solid walls w is 0 at the walls therefore w is 0 everywhere implies w equals 0 everywhere. So what we have done so what have I done we basically got an idea about what the flow field is at the 0th order when epsilon is very very small when epsilon is 0 okay. So the 0th order solution tells you that there is no vertical component of velocity in this core region what I am focusing okay. The velocity is almost fully developed so there is no change in the x direction and the exact dependency on z is given by that just put dp by dx equal to 6 you will get that. So you will get something like I mean if you really did the and plot it or something we put z equal to 1 that has to be 0 which is I put z equal to 0 you get some 1 or something okay yeah dp by dx is 6 3 this is z is it yeah thank you yeah this is z okay yeah yeah then I am okay because z equal to 0 should be 1 I was wondering it is not satisfying the boundary condition out of all of this. So yeah now z equal to 0 to 0 and z equal to 1 that is happening that is good if that is happening then everything is fine yeah z square minus z plus 1 minus z yeah. So 0th order solution is like this okay and this is 1 actually bottom the profile you get something like a parabola which is bending back like normally you have a parabola which is 0 and the 2 walls but here you have a parabola which is 1 here drags and then actually goes back to 0 okay. So let me see if I can draw this properly if I draw this properly it is 0 here then it is something like plus 1 is it yeah something like this this is 0 and this is plus 1 of course this is plus 1 that is the value of the velocity okay. So varying from plus 1 to 0 moves to the right and then bends back this is also not right this is also not right I really need to plot this function this is not right why is it not right yeah that is right this is not correct this is not correct. So what is the right thing clearly this is wrong nothing goes beyond 1 that is one reason and the velocity has to be negative and the way I have drawn it everywhere it is positive okay. So you need to draw it right and let us draw it right for a change which is going to be like this 0 and this is plus 1. So that is basically one application for shallow cavity flows and then you can do this also for the near wall region and then you can do it for different boundary conditions and stuff like that that is I think with that will stop.