 So, welcome back. So, here is a situation which is what we call a fully developed flow down an inclined plane. So, what we have is as my highlighter is tracing right now and inclined plane inclined at an angle of theta with respect to the horizontal gravity is acting vertically downwards. The coordinates are chosen such that the x direction is along the plate and y direction is along the normal. And what is happening here is that a liquid film is flowing down the inclined plane and the liquid film is such that it has a thickness of h. So, y equal to 0 is the plane y is equal to h is the free surface of the liquid film. It is at the thickness of h and the free surface is exposed to a constant value of atmospheric pressure. So, that is the setting. Again here we are also going to look for a velocity field determination and under the assumptions that are listed here you will see that the velocity field will be only a velocity profile. In other words, there will be only one component of the velocity left and that one component will be a function of one coordinate. So, that rather than generating any general u as a function of x, y, z sort of a situation we will again come up with a solution that will say that u is only a function of y. So, let us see the specific solution though. We assume that it is a steady constant density flow. We assume that it is a 2D flow. Again it is a Cartesian system now. So, 2 dimensional will mean that there is no z component of the velocity and since it is an infinitely long z dimension. So, to say we will say that the z differentiation of any quantity will result to 0. So, there are no variations with respect to the z direction in any quantity. That is the idea. We will still stick to this fully developed flow situation which means that we will assume the flow to be such that the axial velocity will be independent of the axial direction which means that it is not going to be a function of x u velocity. That same procedure we start with continuity, simplified and immediately we realize that there is only v as a function of x the y velocity. Employ the boundary condition that at y equal to 0 which is our plate. Since it is assumed to be a solid impervious plate, the normal velocity component will be 0 which then forces us to conclude that this function of x is exactly equal to 0 and therefore, we again generate the same conclusion that there is no y component of velocity in this situation. So, it is again coming from the use of fully developed condition. Now we get on to the y momentum equation. If you go back to the problem statement or the problem setting I would say, y direction is along the normal direction with respect to the plane. So, what we need to realize here is that when we want to include the body forces. Now remember that this is a force, so this is a flow that is occurring under the action of gravity and the gravity is acting vertically downwards. However, with respect to the given coordinate system, there will be a component of gravity in the x direction and there will be a component of the gravity in the y direction. So, given the geometry, we will be in a position to include these two components appropriately in the y direction and x direction momentum. So, let us see how we can do that. So, we go to y momentum equation and we add this last term which my highlighter is now showing which is minus rho times the appropriate component of the body force which is g times cosine theta. So, if you go back here, let me try to draw the sketch. So, this is the inclined plane at an angle theta with the horizontal and if we are looking at the gravity, we resolve it parallely and perpendicularly. So, this is g and our coordinate system is x along the plate y normal to the plate. So, this is theta and therefore, from geometry you will realize that this is also theta here and therefore, the y component here is going to be g times cosine theta and the x component will come out as g sin g times sin theta and this y component is acting in the negative y direction and that is the reason we choose in our y momentum equation minus rho times g cosine theta. Everything else is as usual, so far we had not considered this body force terms in the problems that we have seen. This is the first problem that we are seeing it and that is why I wanted to point that out. Again, our assumptions are steady, constant density flow etc. So, writing the y momentum equation in the non-conservation form and then explicitly writing the convective derivative and the local derivative which has not been written because it is identically equal to 0 since the flow is steady. If you see the convective derivative because v function is identically equal to 0 everywhere that is the v velocity is identically equal to 0. The derivatives will simply go away and nothing is left on the left hand side. Similarly, the viscous terms will go away because v is identically 0. What you are left with is minus the pressure gradient in the y direction and minus this rho times g times cosine theta equal to 0. So, you rearrange this, integrate the resultant dp dy equation partially with respect to y. So, we obtain p as minus rho g cosine theta times y after you integrate it with respect to y partially and because it is a partial integration with respect to y, we do not have a constant of integration, but we have a function of integration which would be a function of x. So, the question is what should be this function of x? To do that, we go and employ a boundary condition for the pressure which is what? Which is at this free surface pressure is equal to a constant value p atmosphere. Where is the free surface? It is at y equal to h this precisely what has been written here pressure at y equal to h is equal to p atmosphere. So, substitute this in our general expression for pressure and you will realize that the function can be determined. It turns out to be p atmosphere whatever it is let us do it in fact and then we plug this back in our general expression for pressure sorry here and what we have is an expression for pressure. It turns out that the pressure is then simply p atmosphere plus rho g times cosine theta whole thing multiplied by h minus y, h is obviously the constant height of the film. So, as you can see from the determined pressure expressions that this is not a function of x at all. So, therefore, we realize that the x pressure gradient is equal to 0. So, this is a useful intermediate result. Finally, we go to the x momentum equation and simplify it by using our previous results including this x pressure gradient 0 v equal to 0 fully developed flow assumption as always as we have been doing and if you simplify it completely what you will realize is that the only two terms that will be left will be mu multiplied by d 2 u d y squared which now by the way this is going to be a total derivative second that is with respect to y of u. The reason is because right at the beginning as always when we said fully developed flow we concluded that u is not a function of x it can be a function of only y and the appropriate body force component in the momentum equation in the x direction is plus rho times g times sine theta and therefore, what we are left with is mu times d 2 u d y squared plus rho g sine theta equal to 0 you integrate this twice with respect to y and you will generate a profile for the velocity u as a function of y with the two integration constants in bit. So, we need to determine these two integration constants using appropriate boundary conditions. So, what are those one of them is very obvious which we have used before and that is at the solid inclined plane that is at y equal to 0 there will be no relative tangential velocity that is in the x direction because of the so called no slip condition which we have used several times now. So, that is what I am mentioning here u at y equal to 0 is 0 because of no slip on the other hand in this case the other boundary condition turns out to be at y equal to h and this boundary condition is essentially comes from a physical force balance if the interface between two fluids is flat we will say that the shear stress across them is continuous. So, what we have in this case is a y direction going from a liquid film to an air mass. So, there is an interface between the liquid mass and air mass. So, therefore, this is a free surface between a liquid and an air mass and it is essentially a one dimensional flow in the sense that we have already seen u being only a function of y therefore, the shear stress is simply using our one dimensional expression for the shear stress mu multiplied by du dy the continuity of shear stress across the liquid and air interface which is a physical boundary condition forces us to say that mu times du dy on the liquid side at y equal to h is equal to mu times du dy on the air side at y equal to h and let me go back to whiteboard and show you what we are going to argue. So, this is our boundary condition at y equal to h what we will do is we will come up with a simplified version of this boundary condition to do that we divide by the density of liquid throughout the equation and we actually transpose it to the right hand side and then look at this ratio the ratio of the density sorry the viscosity of air to the viscosity of liquid typically viscosity of any liquid will be much larger than the viscosity of air may be even 1000 times higher or and or so therefore, what we can say about this ratio is that this ratio is going to be very very small a small number divided by a large number and as a simplification we will simply say that this is so small remember this is just a simplification that this is identically equal to 0 and thereby we will come up with a simplified boundary condition which will be du dy that is the y derivative of the velocity u velocity on the liquid side as we are interested in finding the liquid side velocity profile is equal to 0 at y equal to h. So, this is our simplified boundary condition such simplifications are routinely done when the kinds of analytical solutions that we are looking at are evaluated again the idea is not to any time obtain an exact result clearly already there are so many assumptions built in that we are simplifying the governing equations to something very simple. So, clearly we are not really doing anything exact, but at least some idea of what sort of a flow field to be expected if you have roughly a situation where a liquid film is sliding down an inclined plane with the free surface exposed to constant atmospheric air can be obtained through an analysis such as this. So, that is precisely what I am writing here mu liquid will be much higher than mu air and therefore, we obtain a simplified boundary condition at y equal to h that the u velocity differentiated with respect to y will be equal to 0. So, we have the required two boundary conditions the no slip at y equal to 0 and a 0 gradient as we say at y equal to h plug it back in our profile and you obtain the solution here. So, as you can see here y for u velocity as a function of y will turn out to be again parabolic type profile. So, if you want to now look at the way it looks at y equal to 0 which is our plate it is going to be 0 because that is the boundary condition anyway. At y equal to h what can you see it is not 0 it is non-zero, but it will attain a maximum value equal to rho g sin theta over mu you just substitute y equal to h. So, you will see h square minus h squared over 2 which is simply h squared over 2. So, if you want to plot it let me go to the wide world. So, this is our inclined plane and this was our film thickness. So, this is y equal to h and usually the velocity profiles will be shown in this fashion. So, this is u as a function of y. So, it starts at 0 here and reaches a maximum value with a functionality of y square. Obviously, this is not going to be how it is in real life if you see a liquid film sliding down or a flat plate, but roughly speaking at least it can be close to a y square. If you want you can say that one can perform a far more detailed CFD analysis and I will find out what should be the profile exactly and that is perfectly fine. But the value of the analytical solution is that within a few minutes or may be half an hour you can actually get a reasonably good idea of what to expect in a real life situation. Many times in engineering most of the times in engineering I should say we are really not after any exact solution per se. We just need a reasonably good idea on how the flow field is behaving with that guideline normally we go ahead with our systems design that is at least as far as practical engineering is concerned. The idea of obtaining more and more correct solutions for the flow field is still restricted to research where we want to try to obtain more and more accurate solutions using our CFD type analysis. But many times you will see that in practice when based on some CFD type analysis a systems design is to be considered. People will go ahead with a scoping idea I mean roughly a reasonable idea of how the flow field is behaving is what they are looking at and anyway there will be several factors of safety built into the system with which they can proceed. So, this quest in some sense of finding more and more accurate solutions for problems and the eventual practical applications in some sense will go hand in hand one will drive the other and so on. So, please keep in mind that all these analytical solutions are to be interpreted from that perspective that they are giving us some reasonable idea as to how in a given situation the flow field is behaving. So, this was about this falling film of liquid on an inclined plane. This is the problem that I was talking to you about earlier. This is a problem which we are going to utilize as our benchmark problem for the lab that we will perform tomorrow afternoon. So, this is the transient flow between infinite parallel plates. So, it is the same setting as what we had for the plane Poiseuille flow and the Couette flow. What we have is two infinitely large parallel plates at y equal to 0 and y equal to h separated by a distance h filled with a viscous fluid in between x is along the plate y is normal to the plate. Now, this is not a steady flow situation. This will be treated as an unsteady flow situation will be still treated as a constant density flow and will neglect body forces. So, an unsteady problem will need in addition to the boundary conditions and initial conditions. So, the way this problem is set physically is that at time equal to 0 there are these two plates and the space between the two plates is filled with viscous fluid and everything is stationary at time equal to 0. So, that is what we will call an initial condition the third line here that everything I am writing only u velocity because we are going to realize that there is only u velocity present in this situation as we simplify our equations. So, the u velocity for all y locations would be 0 at time equal to 0. So, the second number 0 here denotes the time equal to 0. So, that is the initial condition. So, this is the initial condition with which we will begin the problem setting is such that at time equal to 0 then other side of 0 if you want the bottom plate is kept stationary all the time. However, the top plate is suddenly set into motion with a velocity of capital U. So, for all times greater than 0 the top plate will be considered to be moving in its own plane with a velocity of capital U. So, the idea is that we start from everything stationary let the top plate in motion with a velocity of capital U at time equal to 0 plus if you want other side of 0 and then see how the flow field develops as a function of time. So, as you can imagine many times it is worthwhile thinking about the problem a little bit and then carrying on with the analysis you can perhaps imagine that in this case we will obtain an axial velocity which will be not only a function of the y coordinate, but will also be a function of time because what is going to happen is that as you set the top plate in motion at time equal to 0 for small time ahead of time equal to 0. The bottom plate will really not know what has happened or the fluid near the bottom plate will not really know what has happened at the top. Only the fluid which is very close to the top plate will be set in motion because of the action of viscosity the viscosity will transmit this axial momentum from the top plate to the fluid that is close to the top plate and as time progresses more and more region in the fluid as you go in the negative y direction will be set in motion. So, therefore this u velocity the axial velocity will be a function of both y and time in this situation. We will still maintain our 2D assumption and the fully developed assumption which we have talked about now several times, but other than that the setting of the problem has been now explained. So, now see what we can do. As always we begin with the mass conservation equation there is no difference here it is a constant density flow transient or otherwise it does not matter it is still divergence of velocity equal to 0. Simplify that as usual this is exactly the same as what we have done before we will realize there is no y component of velocity v is 0. Then we go to the y momentum equation simplify again exactly the same manner realize that there is no y pressure gradient. So, this is exactly the same as before. So, I am not spending any more time on that right now. Now here when you go to the x momentum equation you will realize that we need to maintain this du dt term which is our unsteady momentum term because we said that in general we will u is going to be in this case a function of y and time we cannot get rid of this local acceleration term. On the other hand the convective acceleration is 0 because it is assumed to be a fully developed flow and v is equal to 0. So, what is left then after the simplification in the x momentum equation is that after dividing through by the constant density that is partial derivative of u with respect to time equal to minus 1 over rho times pressure gradient in the x direction. Now I am choosing to write it as total derivative the reason is because y momentum showed that dp dy is 0 p is not a function of y plus mu divided by rho will give me the kinematic viscosity and what is left is the second derivative of u with respect to y. But remember now I need to maintain this as a partial derivative because u is a function of y as well as t. Again what we will assume is that the imposed axial pressure gradient which by the way I did not perhaps explain the problem setting was that at time equal to 0 when you started this motion of the plate you also activate some axial pressure gradient you can imagine that. So, at time equal to 0 two things happened one was you set the top plate in motion with a velocity of capital U and maintain that velocity at all times after that. At time equal to 0 you also imposed an axial pressure gradient equal to dp dx and you maintain that axial pressure gradient. So, we assume that that known axial pressure gradient dp dx is maintained. So, that minus 1 over rho times dp dx will be considered as some known constant and I am simply calling that as s for the purpose of illustration. So, therefore what we are left with is a governing partial differential equation with a pair of boundary conditions and one initial condition. So, the governing partial differential equation that is left out to solve is at the bottom of the screen. Partial derivative with respect to u with respect to time of u equal to the kinematic viscosity multiplied by the second derivative of u with respect to y all partial plus this constant s subjected to u equal to 0 at y equal to 0 for all times which is our lower plate boundary condition u equal to capital U at all times at y equal to h which is our top plate boundary condition. And the initial condition is that u for all y and time equal to 0 is equal to 0 that means we start with a stationary fluid at time equal to 0 and let the flow develop in time. So far this was all setting of the problem in some sense in the sense that based on some problem setting physics that was explained and given we have simplified the governing equations to finally a reduced x momentum equation in the form that is at the bottom of the screen subjected to or to be solved subjected to these boundary conditions and initial conditions. So, from here onwards fortunately or unfortunately depending on how you want to look at it everything is mathematics. So, far it was all in a way of physics because we wanted to simplify terms based on some physical arguments. As many of you perhaps are aware of this is what we call a diffusion type equation and a diffusion equation which has a source term in the form of this constant s. It is going to now sort using the technique of one what is called as superposition and two what is called as a separation variable which is a very standard technique in the solution of partial differential. So, here I think from now onwards I presume that people with applied mathematics would be somewhat happy because this is right in their realm. We have a governing PDE as we say partial differential equation which is a diffusion type of equation with a source term and then there are boundary conditions and initial conditions. So, let me take a minute and then before we actually get on to the details say that I do not necessarily expect people to really know these mathematical techniques. I am just going to describe how the problem is solved and I will give you a solution which will come in the form of an infinite series because typically these separations of variable type techniques will result in infinite series solutions. We are going to utilize this solution as a benchmark problem when we solve in our lab tomorrow the same governing equation that is sitting on your screen right now using a finite difference approach as we call it. So, in that sense you do not have to get bothered too much if you are unable to really follow the technique of this superposition and the separations of variable. If you know it nothing like it is great if you do not know it do not worry about it right now you can always go back to some of those applied mathematics texts which I had which I had mentioned on day one and if you go through the partial differential equations chapter carefully of course you will be able to understand the techniques quite reasonably. So, having said that let me move on and the way this solution is obtained is that we are looking at a transient solution starting from time equal to 0, but eventually we know that in this situation as time goes very very large what is going to happen? We expect that the solution will reach a steady state solution and the steady state solution is something that we already know we have worked out what is the steady state solution? Here is the steady state solution here is the steady state solution which is essentially a steady problem where we have been assuming that this top plate has been moving steadily everything is steady and then finally we had obtained as our first example in this exact solutions techniques the solution for a steady problem. So, the steady problem solution is already known to us going back to how we are going to approach the present problem we then simply say that we expect the solution to be composed of a steady part and a so called unsteady part such that the unsteady part will simply go to 0 as time tends to infinity and then in the mathematical sense in reality the time dependent part can go to 0 very fast, but at least in terms of an argument to say that we are expecting a superposition of two solutions which will be a steady solution superposed with a transient solution with the expectation that the transient solution as time tends to a large value will simply tend to 0 there by automatically giving us the steady solution. So, this is actually a very standard technique which is called as the superposition technique if you want in applied mathematics many of the people I am sure those who have the applied mathematics background I should not be telling them this all this they will know this, but those who are perhaps unfamiliar can simply note it down that this is a very standard superposition type solution that we obtain. So, the idea is that the steady part which I am going to call as unsteady meaning that time dependence is not there the only dependence that the steady solution will have will be a u dependence. So, un will be a function of only y on the other hand the transient part will actually have both y and time dependence. So, that the transient part is what I am going to call u 2 as a function of both y and time. So, what is the steady part going to satisfy? Steady part will have its own governing equation and a set of boundary conditions. Let us just go back if you see if you are talking about a steady solution there will not be any time derivative of u in the steady governing equation at all. So, what we will have is 0 here minus 1 over rho dp dx will be there which we are calling as anyway and nu times d 2 u which will be now actually a total derivative, but not just u it is u 1 the steady solution. So, the steady solution will actually satisfy the governing equation 0 on the left hand side equal to s plus nu times d 2 u d y square. And that is precisely what has been pointed out here the steady part will satisfy the governing equation nu times d 2 u and d y square plus s equal to 0 steady solution will carry the boundary conditions along with it which are at y equal to 0 u equal to 0 which is u 1 in this case because it is a no slip condition at the top we have the top plate moving at capital U. So, that is the second boundary condition solve it we already have the solution we already had the solution before. So, it is the same thing that I am repeating here from the earlier slides this is nothing different initially I just wrote it in terms of s, but then s we know is 1 minus 1 over rho dp dx etcetera. So, that is how we have obtained this. So, it is exactly the same solution that we had obtained for our first problem of the infinite parallel plates in a steady setting. Now, let us see how the transient part is going to satisfy. So, the transient part will satisfy a governing equation