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.  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.  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.


And this seems to be constant in the news anymore and I know you guys cover it because it's a cultural phenomenon now We got a we got to acknowledge and that is biological men dominating women's sports yet again Yeah, so a news source in Canada actually got a tip from presumably another volleyball player There was a game on January 24th where five biological men were competing in a women's college volleyball tournament our game Michael I don't I don't know when this breaks Maybe it doesn't or we're gonna talk about this for the rest of our lives, but do you think there is a cultural swing back to this? Madness no longer happening or it's just kind of the new cultural norm Well, I think You know, I think the movement is toward More female empowerment Biological female empowerment, you know Riley Gaines the swimmer from the University of Kentucky has been on the forefront of this And last year she just flat-out said we have to have girls who when the whistle blows They don't run. They don't swim they stand up on the block and they don't go


Oh, it is a hot one, but we've got a celebration. I'm reviewing one of the movies that I worked on Bitter kind of like the movie That's a hot mug guys I've worked in the film industry for about four years now But I've actually never funny enough watched anything that I've worked on I've worked on cop and a half worked on TV shows Like you me her I've worked on several hallmark garbage movies I've worked on some big shows too like man high castle away, which is coming on Netflix soon but the thing that I really enjoy doing in this industry aside from doing FX is Location scouting for those of you who don't know what that is Essentially, you're the guy who is tasked with trying to find the place that the directors and the producers want to shoot their movie They'll give you a description of what they're looking for whether it's a business a house a building Whichever and then they'll kind of give you an area of where they want to shoot for bigger productions They kind of have the willy-nilly to shoot wherever but with Vancouver BC film especially smaller projects like this They have you look for something in a particular area according to budget and scheduling conflictions So I got to scout for sniper assassin's end, which was originally called Lady Death Originally wasn't even going to be a sniper movie this long running line of Tom Bairinger movies This film follows Tom Bairinger's son as he's being falsely accused of a murder of a big high official While there is this woman this Yakuza sniper who is taking out targets according with a big business big Pharmaceutical blah blah blah Tom Bairinger gets kind of wrapped in it to try and help out his son.  It's a VOD was it fun to find the stuff for oh you bet you was it even fun or seeing the things that I found Oh hell yes considering some of the things that I found were not originally intended to be Used the in the way they were for instance the beginning of the film has Lady Death setting up a sniper shot in a hotel room They had me looking for a hotel that was within the vicinity of the Vancouver art gallery that had Floor to ceiling high windows now there was one building, but it cost $30,000 to shoot in there a day and obviously that wasn't within the budget So I went and found the next best option and so on and so forth I found them different areas of the film quarries houses businesses and Honestly, that was so fun to see these things that I found I'm literally two big portions of the film take place in a forest and in this really nice mansion by the water Which I didn't have any part of that was not me for finding that but a lot of other places I found for instance the road where they have the the trap set up and they have to shoot out That was on a road that I knew of and found in Abbotsford There's this big fancy meeting in this restaurant called the chameleon in Maple Ridge the house that they were Supposed to kick the door down But instead they did fucking blow it in half that I found in Fort Langley the quarry I secured while I didn't get to secure the big locations like the forest the main house and the safe house I still had a great time finding these things because you get to meet different and interesting people you get to learn about their homes you get to learn about the businesses and Sometimes you get to eat at them got to eat at Rocco's for free that was in mission That was when they were trying to have me find a gas station. I went as far as Agassi, which for those of you don't know it was about 87 kilometers away from the office, so that's like about 40 miles So it's a bit of a drive when you're trying to make everything Titan concise I was kind of surprised that they were even having me go further than Abbotsford the film is okay It's it's a VOD. It's not the greatest the sound is oddly lacking And I know the guys who worked on the film are good at their job So I don't know if it was post-work the post-work is a little weird in this film music's kind of non-existent Yeah, it wasn't that hot. It's not as hot as it is right now like my god I'm sweating doing this review right now.  It was just kind of like I'm very nervous like oh hey Mr. Bairinger can I get a photo with you loved you in platoon, but then later on in the day I saw him standing by himself and I just went up to him and I said I don't know if you get talked to about this much But I really loved you in Gettysburg and that totally changed the day around actually because he's a very big Civil War Enthusiast and I have actually found over the years a very big fascination with the Civil War especially with Ken Burns's Series, which we talked about that was something really cool to connect with about just the love of history about this Crucial moment in American history and it was really really fun to talk about with that with him and just kind of talk about his Experience in the film like he talked about the reenactors He talked about just the scale of the film. It was really really cool And unfortunately I got cold in a way and he kind of looked a little bit It was like oh that was a good conversation And I'm sorry that I had to leave because we never got to talk again And that that's a that's a missed opportunity But I'm still happy that I took that chance to do that So you guys want to check it out the film is called snipers assassins and it was fun to scout for this I have since scouted for a few projects since then all of them being hallmarks But you know what it's still fun to do It's a really great experience is definitely the most fun I've had in the film industry Aside from getting to do a blizzard on a car for a snow show That was fun. That's a story. I guess maybe I'll tell another day. Anyways guys. I hope you enjoyed this video It's more of a personal thing for me, but if you guys want to check out something I worked on I'm not in the credits at all, but neither is any of the locations team aside from the ALM and ALM and Neither is the special effects team except for the head coordinator and the lead guy If you want to check out my IMDb though it exists. I'll put that link in the description below Anyways, that's all for me guys I hope you enjoyed this video if you did leave a like and if you're interested in more subscribe Otherwise stay safe out there. Try and stay cool. I'm failing obviously. Oh, Jesus Anyways, that's all for me. See you guys next time Thanks for watching the video My name is Knitz and you might remember me from the animated cult classic TV show undergrads It's been a while, but I'm happy to say the click is finally getting back together in an all-new movie Thanks to a successful Kickstarter campaign But we are still asking for your support to see any and all updates about the upcoming undergrads movie Be sure to check out and like the bring back undergrads Facebook page And with any luck, we'll see you guys soon


Fiorella LaGuardia is kind of the villain of the piece. Why did LaGuardia ban Pinball? Yeah, LaGuardia came in definitely on a crusade to stop organized crime, which I think was a lot of a goal. But it was very hard to do, and it's very hard to show people that he was succeeding. In 1933, Fiorella LaGuardia ran for mayor, promising to rid New York City of corruption and organized crime. Pinball, he said, was gamble, ruled by the mob, a racket to steal lunch money. So he banned it, you know, for the children. And it wasn't just a ban. It was a public relations crusade. Major raids throughout the city squads of police swarming into bowling alleys, bars, anywhere they could find them.


Barn destroyed by fire, missing teacher ID, Evers proposed new water quality funds. These stories and much more coming up on Community News Review. This is Community News Review, a service WSES TV news content provided by WHBL. Hello, I'm Maddie Fister and welcome to Community News Review for Monday, February 11th, 2019. A fire destroyed a barn in the town of Lima Saturday afternoon. The Sheboygan County Sheriff's Department received a 911 call reporting the fire on Bridge Road. When deputies and firefighters arrived, the barn was fully engulfed. There were no animals inside, but the building was used to store some lawn equipment. Nearly a dozen departments responded to that scene. A Wisconsin gymnast now says she was sexually assaulted by Larry Nasser for the former USA gymnastics doctor accused of groping hundreds of athletes. WAOWTV reports Nicole Cassidy, a Kenosha native who went on to compete at Michigan State, says Nasser assaulted her more than 60 times beginning when she was just 14 years old. The first incident came during a camp at Michigan State where Cassidy said she injured her hamstring the first time she had suffered a major injury in the sport.  She later accepted an offer to compete at Michigan State where she says Nasser would continue to assault her. I mean, I knew that that was the treatment that had been done to me but I didn't realize that it was actually sexual assault until the other woman had come forward, Cassidy said. Cassidy is part of a second wave of women coming forward in hopes of receiving healing assistance from Michigan State. Nasser was sentenced to between 40 and 175 years in prison last January, essentially a life sentence. He has since been appealed for a new sentence. Autonomous technology startup Neuro said on Monday it raised $940 million from Softbank Group Corporation which valued the Silicon Valley based company at $2.7 billion. Neuro was co-founded in 2016 by two former engineers of Google's self-driving car project, Dave Ferguson and Zhuzhan Zhu. The funding by Softbank came through its $100 billion vision fund which usually picks up big stakes in fast growing technology companies. Last year Neuro launched a self-driving delivery vehicle to run errands from picking up groceries to delivering dinner and partner with U.S. supermarket chain Kroger Company to extend its services. The firm which largely operates in the San Francisco Bay area said it will use the funds to expand its delivery service to new areas, add new partners, expand its fleet and grow its team. Neuro has raised more than $1 billion from investors including Greylock Partners and Gorogong Capital which had participated in its series A funding round. It's the news that Kathleen Ryan's family did not want to get. Police in Illinois confirmed on Thursday that the 56-year-olds remains were found outside her burned out car last week. Dental records were used to confirm her identity. Ryan went missing after she was supposed to be meeting her son at a basketball game in Northern Illinois. She was a teacher at St. Augustine Preparatory Academy in Milwaukee. Investigators are saying how she died and they are not saying if they have any suspects in her death. Anyone with information about the cases asked to call the Lake County Sheriff's Department at<PHONE_NUMBER>. Governor Tony Evers wants to borrow millions of dollars to replace lead pipes in older neighborhoods across the state. The Department of Natural Resources and the State of Agricultural Department would borrow up to $70 million. The proposal will be part of Evers's budget to be presented to state lawmakers on February 28. Evers in an Associated Press interview described that portion of his budget as an investment to improve water quality around the state. He did not say where the money to repay the loans would come from. The Department of Natural Resources would be the lead agency on lead pipe replacement. They would bond up to $65 million and create a fund that municipalities could draw upon. Local communities would have to fund half of the cost of pipe replacement. Evers said another $5 million would be available to mitigate agricultural impacts on water quality. Evers suggested that the DNR may draw up a new regulation for manure spread for farms to prevent groundwater contamination. Evers also said some of the water quality initiative would speed up stricter water quality standards for the Wisconsin, Milwaukee, St. Croix, and Rock Rivers. Public health experts are closely watching measles outbreaks in Oregon and Washington state. It could happen here, Tracy Saldar, an expert on pediatric primary care at UW Madison says the measles virus is easily transmissible. It is spread by respiratory droplets and the thing about measles is it can remain in the air for up to two hours. The measles vaccine, when administered twice in a child's first year, results in 95% immunity to measles. So in states like Wisconsin where there is a waiver and parents can opt out of the vaccines that decrease immunity, it puts others at risk. Wisconsin is among 18 states that allows parents to opt out of the childhood vaccines if they have personal, moral, or philosophical beliefs against immunizations. And finally, the funeral for a slain Milwaukee officer is set for Wednesday. Matthew Rittner was a 17-year veteran of the department. He was shot to death last Wednesday as a team of officers attempted to serve a search warrant at a Southside duplex. The shooting suspect, Jordan Frick, was under investigation for illegal gun and drug sales. He is accused of firing an AK-47 semi-automatic pistol through the door after officers opened using a battery ram. A forensic investigation says four shots were fired, one struck and killed Officer Rittner, piercing his lungs and his aorta. A judge set bond at $1 billion for Frick. And that is all we have for today. Join me again next time for another recap of our local news on Community News Review. This is content for this program provided by WHBL in cooperation with WSCSTV.