 We were discussing about the exact solutions of the Navier-Stokes equation and we will consider now a second example which is known as a Kuwait flow. So what is the difference between this and the plane Poiseuille flow that we looked into in the yesterday's class. If you have say 2 plates which are parallel to each other and if you have a pressure gradient which is acting on the fluid, then the flow which is being created the fully developed part of that that we call as plane Poiseuille flow. Now in the Kuwait flow there is a modification to that. What is the modification? Now the plates are not stationary but one plate is moving relative to the other. That means you now have say as an example the top plate is moving with a velocity ut relative to the bottom plate. So it is not always necessary that the top plate will move with a velocity or bottom plate will move with a velocity. Important thing is there should be a relative velocity between the 2 plates in whatever sense. So that relative velocity is being created by creating say a design motion of the top plate. There are cases when this design motion may be oscillatory or time dependent in nature and that type of problem is not within the scope of this particular course but in an advanced fluid mechanics course which is there in the subsequent years that is discussed. So here we are talking about that there is a uniform time independent velocity with which the plate is being moved. So with that understanding we have to see that what extra effect does it create beyond the plane Poiseuille flow that we have discussed. See if you forget about the effect of the plane Poiseuille flow just think that this is there this relative velocity is there. What it creates in a fluid is a velocity gradient. So it is an automatic inducer of velocity gradient because the bottom plate has 0 velocity, top plate has velocity ut and let us say that the gap between the 2 plates is h. Then roughly of the order of ut by h is the rate of change of velocity or the velocity gradient. So it creates a rate of deformation by itself by the boundary condition. So this is the way in which you create a shear by the boundary condition itself. So this is sometimes known as a shear driven flow. So this is the mechanism of inducing shear or in other ways if there is a shear in the fluid it is a one way by which you may simulate it artificially by having a velocity gradient which is roughly like ut by h. If you want to simulate a rate of deformation this is by giving a velocity to the plate and keeping a gap it is possible to impart the order of magnitude of the rate of deformation that you want. On the top of that you might be having a pressure driven flow because of by virtue of a pressure gradient some dp dx is there some negative dp dx is there. We have seen that that is what that can drive the flow through pressure gradient. So you have a pressure gradient driven flow, you have a shear driven flow and the resultant is a combination. What kind of combination that we will like to see here. So let us set up a coordinate system. Say this is y axis and this is x axis and let us write the equations of motion here. So we will keep all the assumptions which were there in the plane Poiseuille flow valid for this problem also. So steady flow, constant properties and fully developed flow incompressible of course all these things. So all when you say constant property by default it is incompressible that means density is constant. So with all these considerations into account we have seen that you can write the equation of motion as mu d2u dy2 is equal to dp star dx, right. See that was there for the Poiseuille flow problem, plane Poiseuille flow problem. For this problem it does not change because what has changed is the boundary condition. But in terms of like whatever is there within the domain of the problem there is no special effect that has come into the picture or that has gone away. So the governing equation remains the same. Now of course it is very easy and trivial to integrate it twice to get the velocity by using these boundary conditions. But what we will do is we will just do it in a different way. Why the different way? We want to isolate the effect of this additional thing which has come into the picture. So we will consider that say u1 be the velocity field due to the pressure gradient. Let us say u2 is the velocity field due to this plate motion that is induced shear. So let us write a governing equation and boundary condition for u1 and u2. On u if you want to write the boundary condition, original boundary conditions at y equal to 0, u equal to 0 by no slip at y equal to h, u equal to ut, right. That is also by no slip boundary condition. Now let us say that we are writing this problem in this way, mu d2 u1 dy2 is equal to dp star dx with the boundary condition y equal to 0, u1 equal to 0 and at y equal to h, u1 equal to 0 and mu d2 u2 dy2 is equal to 0. boundary condition at y equal to 0, u2 equal to 0 and at y equal to h, u2 equal to ut. If you just add these 2 equations, you will get mu d2 dy2 u1 plus u2 is equal to dp star dx and if you add these 2 effects, it makes u equal to 0. If you add these 2 effects, it makes u equal to ut. So what u, u is nothing but u1 plus u2. Why we are able to do it so easily? Because of the linearity of the governing differential equation. Since this is a linear differential equation that is the governing equation, u equal to u1 is a solution and u equal to u2 is a solution then u1 plus u2 is also a solution. So that is why you may use this linearity or exploit this linearity to decouple it into 2 different problems. See the first problem with u1 is what we actually solved yesterday that is the Poiseuille flow problem. Only the coordinate axis we have shifted from the center line, we have taken the y equal to 0 as the bottom plate. So that is just a cosmetic change, in principle it changes nothing. So that problem is there as one part and the second part with u2, it totally isolates the effect of the motion of the plate. So you are now in a position to adjudge or pinpoint the effect of the plate from the solution of the u2 problem. So let us quickly solve the u1 and u2 problems. So the u1 problem if you want to solve, so basically you have to integrate it once. So mu or let us write du1 dy is 1 by mu dp star dx into y plus c1 and u1 is equal to, so use the boundary conditions for u1. So at y equal to 0, u1 equal to 0 that means c2 equal to 0. So the boundary conditions are c2 equal to 0 and at y equal to h, u1 equal to 0. So c1 is minus h square by 2 mu dp star dx. So the u1 profile is 1 by 2 mu dp star dx, maybe we put a minus sign outside into h square minus y square, sorry h square minus yh. Now so this is for u1, for u2, for u2 it is even simpler. So d2 u2 dy2 equal to 0 which means du2 dy is c1 and u2 is c1y plus c2, maybe another c1 plus c1 say c1 prime, c1 prime plus c2 prime. So use the boundary conditions at y equal to 0, u2 equal to 0 that means c2 prime equal to 0 and at y equal to h, u2 is equal to ut that means c1 prime is ut by h. So u2 becomes ut into y by h. It is possible to write the expressions, the velocity profiles for u1 and u2 in a non-dimensional form. For example, you can write u2 by ut is equal to y by h. This is a non-dimensional way of writing it and let us give a new variable name y bar equal to y by h just for writing convenience. Which one? Last line yes, yes y square minus sorry y square minus yh minus y square, yh minus y square because minus sign is absorbed already. Now if you non-dimensionalize u1, let us also non-dimensionalize this with respect to ut. So we may write this minus 1 by 2 mu ut dp star dx, maybe multiply both numerator and denominator by h square. So it will become y bar minus y bar square and see if you have a careful look into these terms, this term in the bracket is non-dimensional, the left hand side is non-dimensional. So whatever is here is also non-dimensional okay. So let us just give a name to this. Let us say that this is alpha which is a non-dimensional parameter. It depends on, see this alpha is a physically a measure of what? Physically a measure of the relative strength of the pressure gradient in driving the flow with respect to the plate velocity in driving the flow right. So although it is a mathematically non-dimensional parameter but physically it represents the relative driving effects of the pressure gradient and the shear imposed. Now with this understanding, let us try to write the resultant velocity which is just simply u1 plus u2. So you have u equal to u1 plus u2 which is equal to or maybe u by ut as u1 by ut plus u2 by ut that is equal to alpha into y bar minus y bar square plus y bar. So one important thing is that this alpha maybe positive or negative. See think of a case when ut is equal to 0. If ut is equal to 0 and if you want to drive a flow along the x direction, only way positive x direction, only way is you must have dp dx as negative that we have seen or dp star dx as negative if you consider the piezometric head as the total driving head. Now if you come to a situation when ut is positive then there might be a case when dp dx is positive but still the flow is maintained along the positive direction because positive dp dx will try to have a sort of resistive effect on the flow along x but ut is trying to drive the flow along the x direction. The resultant may be such that ut is successful and then the flow may be along the positive direction. So let us try to see that what are the typical velocity gradients. So for different values of alpha let us first find out what is du dy and we will get a important picture. So du dy is ut or du dy bar say is equal to ut into alpha into 1 minus 2y bar plus 1. So let us try to find out what is du dy bar at y bar equal to 0. So that is equal to ut into 1 plus alpha. So if you want to draw the velocity profile, so this is the sketch where we want to draw the velocity profile. So along the x axis we will plot velocity along the y axis we will plot the y coordinate. So if you see let us consider a value of alpha which is negative. So value of alpha negative means what? Value of alpha negative means what happens to this. So alpha is negative means dp dx is positive. So dp dx positive is adverse pressure gradient. So we have discussed it earlier. So always remember that positive dp dx or dp star dx implies adverse because we have seen that a negative dp star dx is something which helps in maintaining the flow along the positive x direction. So this is called as favorable pressure gradient. So adverse means it tries to have an opposition to the flow. So when you have positive dp dx that is if you have say if you have a positive dp dx that means you have a negative alpha. Now the situation is the du dy at y equal to 0 depends on the extent of negativity of alpha. So there may be one range of alpha where alpha is less than minus 1 and another range is alpha is greater than minus 1. If alpha is greater than minus 1 although it is negative but sum of this will be positive. So you will have a positive slope of the velocity profile but if you have alpha less than minus 1 let us take an example say alpha equal to minus 2. So if you take an example of alpha equal to minus 2 this is minus u t in a non-dimensional form of course. So what it means is that if you have a negative slope of the velocity profile that means the velocity see at the wall it is 0 because of no slip. Then the velocity actually becomes negative because the slope is negative and if you want to consider the upper plate see let us say here is the upper plate here you have to have the velocity as u t which is like this that is by the boundary condition. So let us say that is equal to u t. So somehow this started with a negativity but it has to match with here. So it must somehow cross this axis. So in the process it will have a sort of minima at a point which you can easily find out by differentiating u with respect to y and setting it to 0. So you are having a case where you may have a sort of back flow say this is an example with alpha equal to minus 2 as an example. So the important understanding is that you may get different sorts of velocity profiles depending on the value of alpha. Say you have alpha equal to 1 as an example. So if you have alpha equal to 1 you see that the situation is different. For alpha equal to 1 you have du dy as something which is positive at y equal to 0. And so the when you want to sketch the velocity profile what are the things you should look for and this you should practice at home I am not going to do it for all possible ranges of values of alpha but that you should do yourself. But important characteristics of the plot you should look for what is du dy at y equal to 0 that is how it starts off. And where is the location of du dy equal to 0 that is a sort of location of maxima or minima in the velocity profile. If it starts off with a negative slope then it will be a minima if it starts off with a positive slope it will be a maxima somewhere. So that you have to find out whether it is a maxima or a minima. Where is it located is it because it has to be eventually located between 0 to h and why we have chosen this example for demonstration say alpha as a negative number which is in magnitude greater than 1 because it gives a sort of interesting competition between the pressure gradient and the driving shear. If you take alpha as something positive then this dp star dx negative is a trivial case. Then what happens the pressure gradient also drives the flow in the positive direction and the motion of the top plate also drives the flow along the positive direction. So that is like they just aid each other but here is a case where the pressure gradient is adverse it tries to oppose the flow whereas the motion of the top plate tries to aid the flow where the opposition effect is strong. Opposition effect is strong close to the bottom plate which is the furthest away from the location where the external induction of the motion along positive x is there. So here you see the locally back flow and as you go further and further away towards the top the effect of the top plate will dominate. Now there may be a situation when let us say that whatever is the effect of the integrated effect of this velocity is same as the integrated effect of this negative velocity. Then no matter whether you have some velocity distribution the flow rate over each section may be 0. So that means you may come up with a situation where integral of u dy from y equal to 0 to y equal to h that is equal to 0 that is q. So you find out what is that condition what should be the value of alpha for that that you may easily do by integrating this and finding the value of alpha. You will see that it will come for a value of alpha corresponding to dp dx positive because if dp dx is negative that will never occur it will both effects will help each other. This is the case when it will occur when both effects will oppose each other so that the resultant flow rate is 0. So although you have velocity is locally but the integrated effect of the velocity is 0. So the combination of different pressure gradient and shear that you should try to look into by trying for different values of alpha. What values typically you should try say alpha between minus 1 to 1 and alpha greater than 1 example say alpha equal to 2 like that and alpha less than minus 1 that example that we have taken as 1 alpha equal to minus 2. So for these ranges you try to assess the nature of the characteristics and try to make a sketch of the velocity profile. This type of thing is very important. So you make a sketch of the velocity profile does not mean that you actually plot it. You assess the nature of the variation by the slope and so on and try to make an assessment of qualitatively how the velocity will vary from bottom plate to the top plate. Now a very simple situation corresponding to a special case of this is alpha equal to 0. So when you have alpha equal to 0 that means there is no pressure gradient. It is neither favorable nor adverse and then just the solution is very simple u equal to ut into y by h. This is known as the simple quiet flow that is a quiet flow without any pressure gradient. So this is a case where only the shear is driving the flow and the velocity profile for this is linear. So you can see that if you want to superimpose to velocity profiles to get the resultant velocity profile. So you have the velocity profile as the sum of the velocity profile due to Poiseuille flow which is like this for a dp dx which is negative this is for negative dp dx plus a linear velocity profile because of the velocity profile motion of the plate. So this is u1, this is u2. So if negative dp dx the resultant is the algebraic sum of these two just that is what we got from the differential equation u1 plus u2. But if the dp dx is in the other way you might have the this is say positive dp dx. You might have the pressure driven velocity profile directed in the opposite sense and then the resultant of these two may be either positive or negative that is very obvious but that is one of the important observations from the individual velocity profiles. So alpha equal to 0 is a special case of the just the simple quiet flow. Let us take a third example. Let us say that you have a inclined plane over which you have a thin film of liquid of film thickness h uniform film thickness h. So here there is some liquid and here there is air. So this line which is at the top is the interface between the liquid and the air. Gravity is acting like this and the inclined plane makes an angle theta with the horizontal and pressure at air at all locations is P atmosphere. So you have even like the left of the film P atmosphere, the right of the film P atmosphere like that and this is just a liquid film. So if h is very thin it is known as a thin film flow and we will like to see that what kind of velocity profile in the film you get in this case. And again let us assume steady constant property and fully developed flow. These are the three assumptions that we are considering here also valid. So let us write the momentum equations. What is the consequence of the continuity equation? So let us set up the coordinates. Most of the things are very similar to what the two cases that we have already done. So we will summarize that. Let us say that this is like x axis and this is like y axis. So what if you try to use the continuity equation then what is the consequence of the continuity equation? v is identically equal to 0 for fully developed flow. So the situation does not change here y. You have the continuity equation for incompressible flow equal to this. For fully developed flow u is not a function of x. So that means v is not a function of y. So here you are not maybe sure about the interface between the liquid and the air. But the bottom wall at least you know that at y equal to 0, v must be equal to 0. No penetration boundary condition. So at y equal to 0, v equal to 0. This means v equal to 0 for all y. So you are having to deal with a situation where you have only u as the velocity component. Again we are modeling it as a two dimensional flow. So the third dimension perpendicular to the plane of the board is quite large in comparison to the x and y dimensions. So if you write the y momentum equation, so again it will look like the same as what we did for the Poiseuille flow. So we are just summarizing it. Left hand side you have all the terms involving the gradients of v. So that those are 0 because v is 0. Then minus this plus again all the terms involving v plus the body force along y. So what is the body force along y? So minus rho g cos theta. So you have p equal to minus rho g cos theta into y plus some function of x. Then our more important concern is the x momentum equation. So let us write the x momentum equation. So if you write the x momentum equation for steady flow, so you have rho. For the steady flow the unsteady term goes away. So anyway let us write all the terms and see which term goes away. So this is y momentum. Now we are looking for x momentum. Then what is the body force? What is the body force here? Rho g sin theta. It is along positive x. g sin theta is along positive x. So now let us see which terms are there and which terms are not. First of all because of steady flow this term is 0. Then fully developed flow means this term is 0 and v is identically equal to 0. So left hand side again has become equal to 0. Since fully developed flow you have the second order derivative of u also with respect to x0 and u is a function of y only. So this becomes the ordinary derivative d2u dy2. So you have mu d2u dy2 is equal to, now what is dp dx? dp dx is df dx basically. One important assumption is that you have this gap h is very small. So that we call as a thin film. So if h is small what is an important consequence? The effect of the body force along the y direction is not important because of the very thin gap the effect of gravity within a thin gap is not important. So because h is small this particular effect is small. That means fx is approximately equal to p or for a thin film it is as good as p. There is no difference. So you have basically this is equal to dp dx – rho g sin theta. So you can write it a bit differently. You can write mu d2u dy2 – rho g sin theta is equal to dp dx. Again you see that this left hand side you can write as function of y only. Of course this is like a constant but it constant is a special case of function of y. So this is a function of y only. This is a function of x only and this implies each must be a constant. So when each is a constant then dp dx is a constant means what? Let us say that l is the total length of this. dp dx is a constant means the variation of pressure with x is linear. So if you say that the left end is 1 and the right end is 2 then basically you have dp dx is what? p2 – p1 by l. Now from the boundary condition you see that both p1 and p2 are p atmosphere right because it is exposed to a uniform atmosphere. So this is p atmosphere and this is p atmosphere. Therefore c is equal to 0. So this is a case where c is equal to 0 but the way in which the flow is driven is because here the gravity acts like a driving pressure gradient. So we have seen it is a piezometric head that is important and the gravity part of the head is giving the piezometric head. So it is from a higher elevation to lower elevation and that is what is driving the flow. That is what is like a equivalent – equivalent negative dp star dx. If you include this in the p star although p change is not there but the gravity effect is there and then like it is very easy and straightforward to integrate it twice to get the velocity profile. We are not repeating anymore but what we will now try to pinpoint is what will be the boundary conditions. See many problems have same governing equation but the solution of one problem differs from the other through the boundary condition. So let us try to see that what are the boundary conditions here. So let us write the boundary conditions. So boundary conditions. What are the boundary conditions at y equal to 0? What is the boundary condition? Yes? What is the boundary condition at y equal to 0? u equal u equal to 0, that is fine. What is the boundary condition at y equal to h, yes? y du dy equal to 0, yes? Who says that there will be no shear stress here? God has said that there will be no shear stress here, where from you get that principle? Which fundamental principle tells that there will be no shear stress here? Which fundamental principle you tell? There is no such fundamental principle. The fundamental principle is like this here, you have 2 fluids. Air is a fluid and this liquid is a fluid. So fundamental is like a problem where you have say fluid 1 and fluid 2. You are thinking about the interface condition. At the interface what should be the condition? You have a continuity in certain things. What are the things which are continuous? First of all velocity is continuous because at a point you cannot have different velocities when you consider 2 different fluids. So velocity at fluid 1 is equal to velocity at fluid 2 at the interface. But that is something what we cannot use here because we do not know what is the velocity. The other thing is that there should be a continuity in shear stress because whatever is the shear stress at the interface in the fluid 1 side, same shear stress should be there in the fluid 2 side. So it is a continuity in the shear stress because at a particular interface there cannot be different momentum transport at the 2 sides. So the shear stress must be continuous. So the continuity in shear stress means here it is just a one-dimensional type of flow. That means mu du dy must be continuous. So at y equal to h you must have mu of the liquid into du dy measured from the liquid side at the interface is equal to mu of the air into du dy at the air okay. Because the viscosity of air is much much less than that of the liquid, mu a is much much less than mu l. This term is much much less that is mu a by mu l if you write. This term is so small that it multiplies with whatever that smallness remains. So this becomes as good as du dy is equal to 0. So whatever you have said du dy equal to 0 here as a boundary condition or equivalently 0 shear is correct but in an approximate sense. If I now replace air with another liquid then you will be in trouble because then if you do not know what is the fundamental principle from which it comes out still you will write a wrong thing. So important is the fundamental principle is continuity of the shear stress. Here the special case is just because air it is a coincidence that air has much much less viscosity than the liquid water say this is water. So roughly like 1 by 1000. So then in that case this right hand side is so small that we can say that it is as good as 0. So now the two boundary conditions are at y equal to 0 u equal to 0 and at y equal to h du dy equal to 0. And then based on these boundary conditions you can integrate this to get the velocity profile that I am not going to do very simple exercise okay. Now what we will do is we will see the consequence of having two fluids which are such that the viscosity of 1 is not negligible in comparison to the other. So we have seen one such movie earlier and what we will do is we will look into the one of those movies again. So let us look into one of those movies where we were basically looking into the effect of viscosity is of a two layered fluid. So just try to see that you have a two layered fluid you can just perceive from the difference in color that you have two different layers and just color dies are there two different color dies are there in the two fluids to indicate the velocity profile. And you see that the velocity profiles are such that at the interface there is no continuity in the slope that is du dy is discontinuous at the interface that you can clearly see. The reason is it is not the continuity of du dy that is important continuity of mu into du dy because mu of the two fluids are different du dy at the interface has to be adjusted. So when you have such two different fluids with two different viscosities you should not look for a continuity in du dy it should look for a continuity in mu into du dy that is what is the important thing okay. Now let us come back to what we were discussing and we will now try to look into one example problem where we will use a different coordinate system. So till now we have used the Cartesian coordinate system to solve certain problems. Now we will come to an example 4 and then subsequently one example 5 where we will be using a different coordinate system and the choice of the coordinate system is often driven by the geometry of the flow. So here what we are going to do is we are going to discuss something which is very important from engineering applications fully developed flow through circular pipes. So that is known as Hagen-Poiseuille flow that is basically fully developed flow through circular pipes or tubes or whatever. So let us say that we have a circular tube or pipe like this when we draw it it looks like a parallel plate channel because of course we are drawing the projection of it in the plane of the board but keep in mind that we are talking now about a problem with cylindrical symmetry where because the pipe surface here we are talking about is a circular cross section. So you have a pipe of say maybe we are talking about an axial length of L and fully developed flow. So the coordinate systems what we use for this case are the cylindrical polar coordinate systems which are more common. So you have small r as the radial coordinate, z as the axial coordinate and if you take the cross section of this, so if you draw the separately the sectional view it is a circle. So the cross section wise the coordinate system is a polar coordinate system. So at a point if you consider it is r, theta. So this is r and this angle is theta. So the coordinate system requires r, theta, z and these are again mutually orthogonal coordinate system because theta direction is perpendicular to r and r is perpendicular to z and that means you have now r, theta, z all these are basically mutually orthogonal to each other. Now we will be using therefore now the Navier-Stokes equation in the r, theta, z or the cylindrical polar coordinate system. So already we have communicated the corresponding forms of the equations through the course website. You must have those equations with you. Important thing is it is not necessary that these equations you have to remember. We will provide all these equations to you I mean during exams and so on because our test is not whether you can remember these equations. So whatever is there with you is something what you will have if there is a question related to that it comes in the exam. Our important objective will be how to simplify those equations and come to the solution of this problem. So let us look into some of those equations. I will project those equations in the screen so that which are already there and it will be possible for you to follow it easily. So let us look into the cylindrical coordinate form of the Navier-Stokes equation and we will try to solve the Navier-Stokes equation with the cylindrical coordinate form. To do that we will always start first with the continuity equation and see that what the continuity equation gives to us. So let us look into the continuity equation form in the cylindrical coordinate. So if you look into the continuity equation if you look into say the slide the second equation. So the first term is that the density variation with respect to time. So we are considering a constant density fluid so that term goes away. So let us see what the continuity equation gives to us okay. Next let us come back to the slide again. See the second term in the continuity equation. The second term it remains because we are not very sure that what is going to be there. So 1 by r d dr of r into rho into vr, rho r vr. So that is the term and what about the third term? So you have a third term. What about this term? First of all rho is a constant. So rho is a constant and hence it comes out of the derivatives. So here also rho comes out of the derivatives. Then this term boils down to what? It boils down to the derivative of the gradient of vz with respect to z okay. What is the fully developed flow? See here also the concept of parallel plate channel does not change. That is you have the boundary layer which are merging and beyond that the velocity profile whatever is going to be there that does not change further with the axial coordinate. The axial coordinate is the z coordinate. So that means vz, vz is the velocity component along the axial direction. That does not change with the z coordinate any further. So fully developed flow means this is equal to 0. Just like du dx was 0 for parallel plates, here just replace u with vz and x with z axial coordinate. So then you are left with d dr of r vr that is equal to 0 right. We are not considering that singular case. See these are the singularity at r equal to 0. So we are not considering that singular case but for a general case. So r that means r into vr is not a function of r right. Only way it is possible there are 2 ways. So the first one way is if you do not look into the boundary condition. Say you do not look into the boundary condition but you just want to look into this. So then if you have vr equal to c by r of that form. Then that will be a conclusion made without looking into the boundary condition because this has to be satisfied within the constraints of the boundary condition not just as a general case. So what is the constraint of the boundary condition that at the wall at small r equal to capital R you must have vr equal to 0. That is the no penetration boundary condition. That is the fluid cannot penetrate to the wall in the radial direction. So which means that you have since at r equal to capital R vr equal to 0 which means vr equal to 0 for all r from this one except r equal to 0 that is a singularity we are not talking about that. So whenever we are coming up with a solution we are cleverly avoiding the singularity point and then for the remaining that from this since this is not a function of r basically you forget about the r equal to 0 case that means it is as good as vr not a function of r. So if you have found out that vr equal to 0 at r equal to capital R it should be 0 for all r. So from the continuity equation our conclusion is that vr equal to 0 just like v equal to 0 for parallel plate channels. Next we will go into the momentum equation. Let us say so we have how many versions of the momentum equation. So if you see the cylindrical coordinates you have the r, theta and z. So let us look into the case with theta. See first of all what about the derivatives with respect to theta. So if you look into these equations you will see that there are lots of terms which are having derivatives with respect to theta. So if you want to figure out that what happens for the derivatives with respect to theta you look into the equations very carefully. Now when you have derivatives with respect to theta in all the momentum equations such terms are there. So terms of the form derivatives with respect to theta. Question is that when will it be 0? See theta is what? Theta is the polar coordinate. So if you have a symmetry with respect to the axis that means that is called an axis symmetric flow. Then the symmetry with respect to axis means that there will be no variation with respect to theta. It is only symmetric with respect to axis at which theta it is located is immaterial. So axis symmetric problem will always mean this equal to 0. When you have axis symmetric problem which means that this equal to 0 you have to keep another thing in mind that axis symmetric problem does not ensure that v theta equal to 0. You could still have v theta but that not a function of theta. So if you have a v theta component it is something called as a swirl component because it tries to have a rotation or swirl in the flow. But here we are considering that there is no v theta. So if there is no v theta I mean if you look into the continuity equation see the continuity equation is satisfied let us go back to the continuity equation. You see that in the continuity equation there is a term dd theta of rho v theta. So that term we omitted because of the axis symmetric condition. So in the continuity equation we did not consider v theta that does not mean that v theta is not there simply the theta gradient of v theta was 0 because of the axis symmetric condition. So till now we have never utilized the condition that v theta is 0 but now we will utilize it by noting that it is not a swirling type of flow. There may be swirling type of flow in a pipe where there is a rotationality that is important to the flow but here we are considering that such rotationality is not important to the flow and that means you have for this problem so no swirl is considered that means you have v theta equal to 0. So you see certain conclusions are there conclusions at the end are important but where from you arrive at the conclusions I feel is even more important. So now we go back to the momentum equations again. So in the momentum equation let us look into the theta component of the momentum equation. So theta component of the momentum equation has all terms involving the derivatives with respect to theta right except in fact all. So the theta component of the momentum equation will have only one term which does not involve derivative with respect to theta. What is that term? So if you look into the theta component. So in the theta component the first term is dv theta dt is equal to 0. Second term vr dv theta dr equal to 0 because v theta is 0 that term is 0. Third term has vr and v theta so when v theta is 0 that is equal to 0. Fourth term also has v theta. So left hand side is 0. Right hand side no p variation with theta so that term is 0. Next term you have a v theta so although the r derivative is there but because v theta is 0 that is 0. So if it is a swirl flow that term is not 0 but when there is no swirl that term is 0. Remaining terms have either v theta or derivatives with respect to theta either of those are 0 so it is like 0 equal to 0. So theta momentum equation does not give us anything for this problem. So then the r momentum equation. So let us look into the r momentum equation. So for the r momentum equation you see the first term. First term is dv r dt that is because of steady flow the first term is 0. Second term vr is equal to 0 identically that you have to keep in mind because of fully developed flow. So that means the second term is 0 and the third term is 0 either way either vr is 0 or the theta gradient is 0 either way it is 0. Fourth term you have a v theta square by r. v theta square by r is like the centripetal acceleration term and because the swirl component or the rotational component is not there the fourth term is 0. Then the fifth term vr is 0 so that term is also 0. Right hand side minus dp dr now this term is there so at least we have found out one term which is there. So minus dp dr then next term. Next term is again 0 because vr is 0. Then the remaining term so d2 vr d theta 2 that is 0 d2 vr dz2 equal to 0 dv theta d theta equal to 0. You have a rho vr. So rho vr is like if you have a body force along the r direction. So if you have a body force along the r direction then that is what is very much possible. So you see that there is only one component which is vz and there is no vz term in this. So you have basically the r momentum equation giving what? 0 is equal to minus dp dr plus rho into vr. Now we come to the z momentum equation which is going to be the most important equation for our velocity profile solution because our velocity profile is vz. So look into the terms first term so we are talking about the last equation which is there in the slide. So first term the unsteady term it is 0 then next term is vr is there so that is 0. Third term v theta is there that is 0. Fourth term dvz dz is there which is 0 for fully developed flow. So the left hand side has become 0. Right hand side you have minus dp dz which is there. Next term the first term is very much there because vz as a function of r is what is reflected in that term. Second term in the square bracket vz is not a function of theta. So that is 0 and the third term is 0 by fully developed flow. So we are left with the z momentum as 0 equal to minus this 1 plus mu into 1 by r this into r plus where what will be bz here. So if you have the acceleration due to gravity like this so let us say that this axis of the pipe is inclined at an angle theta with the horizontal so you have this angle as theta and this angle as 90 minus theta. So you have a brs so br as minus g cos theta and bz as minus g sin theta. So this is minus g sin theta okay. So the forms of the equations that where we have arrived are something which are very much similar to what we have arrived for parallel plate channels only because of the cylindrical symmetry you have this 1 by r d dr this type of term. So but if you write it in the divergence form in the vector form or here the Laplacian operator form that is basically identical. So vector form is identical just because of the shift of the coordinate system that is what the second derivative of vz with respect to r looks like whatever was there in the Cartesian system equivalent in this cylindrical system okay. So we stop here now and we will continue with this in the next class. Thank you.