 In our previous lecture, we were discussing about the derivations of the Navier-Stokes equation. Now what we will do in this lecture, we will work out some exact solutions of the Navier-Stokes equation. As we discussed, the Navier-Stokes equation is not a very simple equation to solve in principle because these are nonlinear coupled partial differential equations but in certain simple cases exact solutions are there that means one may analytically solve the Navier-Stokes equation and it is possible to work out some of those solutions in a very using very simple mathematics and that we will do in this elementary course. We will not go into the details of all possible cases where exact solutions of Navier-Stokes equations exist but certain very simple elementary cases which fall within the domain of an introductory course in fluid mechanics. So if we try to revisit that what was the form that we derived for the Navier-Stokes equation, so we derived this particular form. So this particular form was for a special case. Can you remember what is the special case? Incompressible flow and what are the other assumptions? Okay, let us put a body force maybe to make it a bit more general. So the assumptions are incompressible flow because if it was a compressible flow there is a term which is associated with the divergence of a velocity vector that particular term would appear. Then Newtonian and Stokesian fluid, homogeneous and isotropic. So those were there for the original derivations and with the additional incompressibility constraint this is the form of the equation that we derived. Up to this we did in the previous lecture and why this incompressible form is important is because like whatever exact solutions we will be deriving in this course will be only for incompressible flows. So from now onwards whatever exact solutions we will be deriving we will not explicitly mention time and again that it is incompressible flow but you just keep in mind that that is the assumption that we are following for the remaining derivations. Now it is also not a bad idea to write this in terms of a vector form because this form when you write in terms of the indices i and j this is an elegant form but this we wrote in terms of Cartesian index notation. But if you have a curvilinear coordinate system or the most general case of maybe a non-orthogonal curvilinear coordinate system then for all those cases these types of index notation in the in this form will not work. You require a different form because these are Cartesian index notations but you may have different coordinate systems but vector form is general. So if we write its equivalent vector form then that vector form will be valid independent of the coordinate system. So if you want to write it in a vector form so let us say we are interested to write it for a velocity component say ui. So the next term if you want to write in terms of the vector say this is the velocity vector. So we are just writing one component we will then write the other component. So we have only written one of the terms in vector form. The other terms we have still written in the original index notation. So this is the ith component. So we should remember that i equal to 1 will imply x component, i equal to 2 will imply y component, i equal to 3 will imply the z component. So what we will do that this for i equal to 1 this is x component. So we will take the x component and then multiply that x component with i cap then i equal to 2 that will be y component multiply that with j cap the unit vector corresponding unit vector. Similarly i equal to 3 will multiply with k cap and add those together. So if we add those together then what will be the corresponding form? So rho this will be partial derivative with respect to time u1i plus u2j plus u3k. So that will become the velocity vector u. Similarly this will become minus gradient of p just like taking the individual components and finding the resultant. So this also as we have seen in the previous notation in terms of the total derivative this may be written in terms of d dt of u. So this vector form is important and we will use this vector form for some of the cases and where one coordinate system differs from the other is the del operator is different differently expressed in different coordinate systems and that is how expressions when written in terms of the coordinates of the coordinate system will differ from one system to the other. Somewhere x y z somewhere r theta z somewhere r theta phi depends on the coordinate system that you are using but where all those coordinates appear will be in terms of the del operator because del operator has different expressions for different coordinate systems. But when you write it generically in terms of del operator it is independent of the coordinate system that you are writing and that is one of the good things about this form. Now we will start with the exact solutions for a case where we will come down to a much more simplified version of this particular form and that first exact solution. So exact solution 1 we will consider something called as fully developed flow between 2 parallel plates. This has a technical name it is known as plane poiseuille flow. So we will look into this special case first we will draw a sketch to understand that what is the specific form that we are looking for of the Navier-Stokes equation in this case. Here we can very elegantly use the Cartesian index notation because the coordinate system inherent to this description is a Cartesian one. So the understanding is that you have 2 parallel plates. So let us say that you have a plate top plate and you have a bottom plate. These plates are separated by a gap, gap may be large or small or whatever and these plates are such that the dimensions in this plane are much smaller than the dimension perpendicular to this plane. That means you have a plate like this. So one plate like this, another plate parallel to this where the third dimension is like infinity very large. So the dimension perpendicular to the plane of the board is much much larger in comparison to the dimensions in the plane of the board. So it makes the problem equivalently a 2 dimensional problem. Why? Because if the third dimension is very very large then the gradients in the third dimension are very very small because gradients are basically change in variable divided by the length over which the change takes place. If the length scale in the third dimension is very very large then the characteristic changes in terms of the gradients are very very small and then the effect of the third dimension may be neglected. So if we have, let us say that we give certain dimensions. Let us say this is the center line of the channel. So this height is say h, this height is h. Let us say that the length of the plate is, each plate is l and the width is w such that w is much much greater than 2h. So when you come in terms of the cross section, the effect of the h in terms of creating the gradient is much much more important than the effect of the dimension w that is the width. Now let us try to first understand that physically what happens because we have a keyword fully developed flow which we have not yet explained. So we will try to explain that keyword or collection of keywords through understanding that physically what goes on. So if you have say a fluid flow entering the channel, say with a uniform velocity, somehow let us say that there is a uniform free stream and that enters the channel. So just think that the channel constitutes of 2 parallel plates and for the time being forget about 2 plates. Just consider that you have 1 plate. So if you consider that you have 1 plate then what happens? You will see that because of the effect of the plate the boundary layer starts growing. We have discussed about this earlier. So let us say that from this point the boundary layer starts growing and it grows like this. Why does the boundary layer grow? Let us just recapitulate very briefly that you have the no slip boundary condition at the wall. So the effect of the wall is that it brings the fluid to rest at that point and because of the viscosity of the fluid this effect is propagated towards the outer fluid. So the fluid elements which are very close to the wall feel the effect of the wall the strongest and then the fluid elements which are further and further away from the wall will feel some effect of the wall but not as strong because they are also being pulled by a fluid element on the other side which is moving faster. And further and further away from the wall you go, you feel that the fluid elements which are more and more away from the wall will feel the effect of the wall much much less and we will come to a height beyond which the effect of the wall is not felt at all. That means viscosity is not capable enough of creating further velocity gradient and that means we have come to a state where beyond that the velocity profile is virtually uniform and that location at each and every point at each and every section is given by a height which is some distance from the wall. Depending on how speedy the incoming flow is this distance will be small or large. If it is a high speed flow this distance will be very small. If it is a low speed flow this distance will be large and we will come into the details of the largeness or smallness of estimation of this when we study in details the boundary layer theory in one of our subsequent chapters. But like here we are interested more in terms of qualitative terms that what is happening. So if you draw a velocity profile let us try to draw a velocity profile at 2 different sections. Let us say that we draw it at one of the sections which is relatively closer to the inlet. Let us say this is section 1 and we draw it at a section which is section 2 which is bit further away from the inlet. So at the section 1 if you draw the velocity profile see the velocity will have a gradient till you come to the edge of the boundary layer then the velocity will be uniform. You have to keep in mind that on the other side there is a plate which has perfect geometrical similarity and physical similarity that means boundary condition here is no slip here also it is no slip. So whatever will happen at the bottom plate there is nothing which makes us believe that same thing will not happen at the top plate. So same thing will happen at the top plate and there will be a symmetrical development of the boundary layer. So the understanding of the boundary layer is that within the boundary layer so the red line is the edge of the boundary layer. Below that you are having velocity gradients because of the viscous effects. Outside that velocity gradients are not there it does not mean that the viscosity is 0 but velocity gradients are simply not there that means effect of the wall has not propagated beyond that point. Now if you draw the velocity profile at the other end also it will be very very symmetrical to what it was at the bottom end and then it will be uniform outside the boundary layer. So say this is the velocity profile. Now let us try to draw the velocity profile at a section 2. So when you draw the velocity profile at a section 2 so the effect of the wall will be felt now to a greater distance from the wall because the fluid has now suffered greater effect of the wall by virtue of propagating deeper and deeper into the channel. After drawing the velocity profile of the section 2 if we focus our attention on the velocity profile for the section 1 we see that it is divided into 2 parts. One is the boundary layer region and another is outside the boundary layer region which we call as a core region. So this is what you have for a core region and the boundary layer region. Now similarly at section 2 there will be some core region. Core region is now thinner and boundary layer region is now thicker. So the first question that we will like to ask ourselves is whatever is the velocity at the core region at section 1 say it is uc1 and let us say that the velocity at the core region for the section 2 is uc2. Question is is uc1 greater than uc2 is it less than uc2 or is it equal to uc2? So we will try to answer this logically not by a guess work. So let us try to do that. To do that you have to understand that let us consider that we are assuming that it is a steady flow. So over each cross section the same flow rate is there that means flow rate at section 1 is same as flow rate at section 2. When we say flow rate it is usually mass flow rate but if the density is taken as a constant it is as good as volume flow rate. So volume flow rate how it is obtained by integrating the velocity profile over the cross section that gives the volume flow rate. So if you integrate the velocity profile over the cross section and it remains constant for sections 1 and 2 then what is the feature of section 2? See at section 2 you have now a greater distance from the wall that is suffering the effect of the wall. So the effect of the wall is now felt to a greater distance both top and bottom because they are symmetrical that means if the boundary layer was not growing in terms of its size or thickness then whatever was the extent to which the fluid is slowed down the now the extent to which the fluid is slowed down is more and to compensate for that the fluid in the core region should move faster than the whatever it used to move for the section 1 to keep the integral of the velocity same for both sections 1 and 2. Simply you have a greater distance over which the fluid has suffered greater resistance it has to compensate for that with a higher velocity in the core to have the integral of the velocity over the section conserved that is the flow rate conserved okay. So that means the first understanding is you have uc2 greater than uc1 okay. Now the question is if you have uc2 greater than uc1 then what about the pressures at sections 1 and 2. So when you are having a velocity and when you are trying to relate that with pressure you are very much tempted with use of Bernoulli's equation right. So we have to see that whether the Bernoulli's equation here is applicable or not. So let us say that we consider a streamline along the center line the streamline is connecting say 2 points a and b located on the center line. a is a point on section 1 and b is a point on section 2 both along the center line and that is one of the stream lines because of the symmetry. Now can you apply the Bernoulli's equation between a and b? See there are many assumptions of Bernoulli's equation but the first assumption that you should look for that is it an inviscid flow or a viscous flow then look for all other things because if the first question is not satisfied then you should not look for any other thing because you cannot apply it for a case where viscous effects are there. The interesting thing is that here the flow is what? Flow is viscous flow because it has viscosity but effect of viscosity is not felt here because you do not have a velocity gradient so you have no rate of deformation and the shear stress is the viscosity times the rate of deformation. So if the rate of deformation is 0 it does not matter whether viscosity exists or not. So in terms of filling the shear stress it is there is no shear stress here and that means it is like an inviscid flow. So effect of the wall has not propagated into the center line. So that means you may apply Bernoulli's equation between a and b provided other assumptions of Bernoulli's equation are also justified. So like you may use the steady version by assuming that it is a steady flow and incompressible a special case of incompressible here we are considering a constant density flow. So when you write the Bernoulli's equation between a and b you have Pa by rho plus U a square by 2 let us say that a and b are so close that the difference in height between these 2 is not important. So just it is same as Pb by rho plus Ub square by 2 and we know from this analysis that Ub is greater than Ua therefore Pb is less than Pa. If you are not happy by assuming that a and b are of the same height and you have all the right not to feel happy because you might have a vertical parallel plate like this. So if you have vertical parallel plate and you are considering 2 points say a and b on the center line yes there is every chance that no matter like whatever distance you go it is a vertical change. So it is a total vertical change and it is it may be quite substantial depending on where are the points a and b. So if you still consider that well I am considering a and b which may be located at different heights. So you add even g into we use a coordinate system like h because x, y, z we have preserved for different coordinate so this plus. So we may say that in terms of writing it not just by pressure but if we write it as Pa plus rho gha divided by rho plus Ua square by 2 that we can write Pb plus rho ghb by rho plus Ub square by 2. We have discussed earlier that the equivalent pressure that we have as pressure plus the effect of the gravitational head together is known as the piezometric pressure. So here what we have written is the piezometric pressure. So we call it we will give it a symbol Pa star okay. We give it a symbol Pa star and we give it a symbol Pb star. So when we write it in terms of piezometric pressure the good thing is that it becomes independent of the orientation of the channel that we come to a conclusion that yes in terms of piezometric pressure we should have Pb star less than Pa star. When we are writing this see for a parallel plate channel which is horizontal or maybe having an inclination with the vertical which is slight then this rho gh effect is not important and then it is straight away giving Pb less than Pa but if the height effect the difference in height between 2 points that is important then it is P star or the piezometric pressure effect that is going to be important. So from here we may conclude one important thing that the flow is taking from higher piezometric pressure to lower piezometric pressure right from A to B. So question is that means there is a driving pressure gradient from the point where the fluid is flowing to the point where the fluid is flowing which is giving this pressure gradient. So you must have a driving device which is allowing the fluid to have this pressure gradient so that it can maintain the flow because why it requires a pressure gradient see it has to accelerate along the core its velocity should increase along the core. So there must be a driving force which should make it accelerate and that is a driving piezometric pressure gradient from a higher pressure gradient to lower pressure gradient. Only what may provide that you may have a pump that provides that pressure gradient. So you must have an energizing mechanism it need not necessarily be a pump but you must have something which energizes the fluid so that it may have an acceleration along the core and that is also one of the important technical or technological understanding that what is the consequence of what happens in the core. Then the important thing for our exact solution of the Navier-Stokes equation is that the fluid in the or the boundary layers which are there in the region where the boundary layers are developing the edges of the boundary layer because they are symmetric when they meet they meet at the centre line somewhere. What happens beyond this is something which is interesting so if you draw a velocity profile here from our previous discussions we may conclude that whatever is the velocity of the centre line here that is a maximum of all because the fluid is continuously accelerating. So it comes to a maximum velocity here and at the wall the velocity is 0 and you get a sort of velocity profile. Question is will this velocity profile change any further as you move away from this. See what makes the velocity profile change is as follows in the region which is towards the left of this section the boundary layer is continuously growing. So there is an adjustment at each and every section to keep the flow rate unaltered and that depends on the extent to which the wall effect has propagated into the fluid. Here the wall effect has propagated fully into the fluid that means the centre line also now knows yes there is a wall and I am feeling the effect of the wall because like the edge of boundary layer is now limitingly on the centre line. But beyond this what happens beyond this there is no change because now the entire fluid knows that what is the effect of the wall and it responds to that in the same way there is no further development of the boundary layer. So we say that the flow has become fully developed when there is fully developed there is something which is not fully developed and that is known as a developing flow. So the region which is here is known as a developing flow basically the boundary layers are developing here and it is also known as an entrance region these are just different names. So you have an entrance region and you have a fully developed region and the hallmark of the fully developed region is that now if you draw velocity profile at different sections let us say that you draw velocity profile at a section somewhere else then these velocity profiles would be just identical these are the same. What it means it means that if you are considering that u as the velocity component along x now u becomes function of y only what are our x and y coordinates let us set up x and y coordinates as this as x and this as y or you may write partial derivative of u with respect to x equal to 0 because only variation of u is with respect to y. See fundamentally u could be function of in this coordinate system functions of x y z and time we have considered first assumption is steady flow. So you have any partial derivative with respect to time of any flow property or velocity that is 0. So it could be function of x y and z the gradients with respect to z are approximately 0 because you are considering infinitely large width in the z direction. So you have so width is much much larger than say 2h and that means z direction is along the width. So you have this equal to 0. So when you have use these assumptions to begin with you had only 2 choices possible or rather 2 dependences possible u could be dependent on x and y. Now because of treating it as a fully developed flow so we are now going to our analysis only that is valid in the fully developed region. So u have u not a function of x. So fully developed flow will mean it is only for u for example p varies also this is x. For example p varies with x. So here we have written for any variable but here only for u not for p that you have to keep in mind. So with these assumptions you come to a conclusion that yes u therefore is a function of y only. So the partial derivatives all with respect to u with respect to y will now soon boil down to ordinary derivatives of u with respect to y because it is just a function of y. Now to understand how these equations are simplified we will first come to our objective. Our first objective is to find out the mathematical form of this velocity profile how u varies with y. So that is our first objective. To satisfy that objective what we will do we will solve the fluid flow equations. When you consider the fluid flow equations remember that you have the continuity equation that has to be satisfied always and what is the form of the continuity equation that we are looking for remember we are talking about incompressible flow and 2 dimensional incompressible flow because effectively the flow is 2 dimensional in the xy plane. So 2 dimensional incompressible flow you have the continuity equation partial derivative of u with respect to x plus v with respect to y that is 0. Now you consider the assumptions one after the other come to fully developed flow. See these are the important assumptions that we are going to follow for the problem that we are considering. So when you have fully developed flow then clearly this term will become 0. That means by imposing the constraint of fully developed flow what does the continuity equation give us v is not a function of y right. So v is not a function of y. Since v is not a function of y if we may find out that what is the value of v at some particular y that v should be true for all values of y right. So solution for v is not that difficult if we just have to find out that what is the value of v at some point which is known to us say at the boundary for example. So if you look at the boundary at these points what is the value of v 0 because of no penetration boundary condition. So because of no penetration the v is 0 at the wall. So since v is 0 at y equal to wall so there are 2 walls one is plus h another is minus h that means what we can say that v is 0 for all values of y because v is not a function of y so at some particular y once we have found out the value of v that should be true for all values of y. So this means that v is equal to 0 for all y and when we say that v equal to 0 this is not an approximation that v is approximately equal to 0 this is v is identically equal to 0. Later on we will come to certain situations where we will have v is much less than u that is v is not 0 but we may consider it to be 0 as compared to u those are different cases. This is a case where it is not an approximation it is just an exactness. So when v is equal to 0 then basically you are having to deal with only one velocity component that is u. Now keeping that in mind let us write the momentum equations or the Navier-Stokes equation. So we have the Navier-Stokes equation in terms of the coordinates Cartesian index written. So let us write it for the x and y components for convenience we will write it in terms of first the y component because it will nullify many effects. So let us write the y component y component means i equal to 2 Navier-Stokes equation. So you have rho so u2 is the velocity component along x2 which we give an identical name as v. uvw are the velocity components which we use as simple notations. We are also assuming constant properties and what is the body force along white direction? Let us consider an extreme case as the vertical and extreme case as horizontal the more common case is neither horizontal or vertical but inclined. So I mean arbitrarily inclined. So let us consider that we are talking about a channel which is like this x is the coordinate along the channel y is the transverse coordinate and the acceleration due to gravity is acting along this. So what is the body force that you will have? So you should look for the body force component along y direction what is that? So you have this g you have its own components. Let us say that this angle is theta that is the angle made by the horizontal with the axis of the channel is theta. So then this angle is theta so it is –rho g cos theta okay. Now let us look into the different terms first see look into the assumptions steady flow so this will be 0. You always see this is the systematic way of looking into the problems that we have with the use of the Navier-Stokes equation. First write the full equation then see what are your assumptions accordingly you make a simplification of the equation. So do not remember any equation like a formula. So we will always start with the Navier-Stokes equation and come up with the simplified form based on the assumptions okay. So this is steady flow then next term see v is 0 therefore whatever is involved with v there is no gradient of v as well so these terms are 0 therefore these terms are all 0 and then this gives you is equal to –rho g cos theta. So if you integrate it you have p-rho g cos theta into y so plus some function of x right. Next let us look into the x component of the momentum equation which will be the important equation for getting the velocity profile but the background is important otherwise you cannot simplify that equation. So let us write the x component. So let us substitute different terms first steady flow so the first term is 0 then next term fully developed flow so this is 0 because of fully developed flow v is 0 so the entire left hand side becomes 0 and that gives a lot of advantage because now the non-linear partial differential equation has become a linear equation okay. Now come to the right hand side because u is not a function of x its first order derivative with respect to x is 0 and therefore second order derivative is also 0 so and what is the body force? Just look into the figure and say what should be the body force along x- okay just for bx we will substitute –g sin theta so it is rho g-rho g sin theta. So what you have that mu into this is what we can write. So p has what? p-rho g cos theta y-rho g sin theta-fx-rho g sin theta x-fx that is by substituting p from there which one this one yes this is plus okay then whatever in the bracket it is there so substituting p from the what we got from the y momentum equation plus the rho g sin theta x term. Now if you see that anyway partial derivative of this with respect to x will be 0 because this only contains y. So the remaining term see this is a function of x only and this is a function of x only therefore this term is a function of x only and if you see the effect what is this term actually this is this term is just like a pressure which is varying with x plus the gravitational head because what is x sin theta. So if you go from say this point to this point which is which you traverse along x then this height h is x sin theta if this distance is x. So that is like rho gh okay. So you have some form of pressure which is varying with x on the top of that rho gh so it is giving you a piezometric pressure effect fx plus rho gh. So and the important thing is since this is varying with x we can just write this as dp star dx where p star is fx plus rho gh. Why ddx not partial derivative reason is clear these are only functions of x and this becomes 0 when partially differentiated with respect to x. What do you get in the left hand side? See u is a function of y only therefore the left hand side is mu d2u dy2 because u is a function of y only so partial derivative and ordinary derivative are the same. So now you have an equation where the left hand side is a function of y only right hand side is a function of x only it is possible only when it is a constant. Otherwise you cannot have an effect where some function of x will nullify some function of y. So this implies that each equal to a constant say the constant is equal to c. So once you have each equal to constant the remaining work will be easy because you may integrate it twice to get the velocity as a function of y. So let us quickly do that. So if we integrate it with respect to y you have mu du dy is equal to cy plus c1. Now let us fix up a domain in which we want to solve our equation. See the domain is symmetrical the top half and the bottom half they are symmetrical. So we may just solve it for the top half the bottom half will be just the mirror image of that. So for analytical calculation that is not necessary always but if you want to solve it numerically reducing the size of your effective domain by utilizing the symmetry solves your computational time I mean saves your computational time. So it is something which one should always utilize if you find symmetry. Remember symmetry should be in terms of 3 important things symmetry in geometry symmetry in boundary condition and symmetry in physics. So we have to make sure that all these 3 things are maintained and here all those things are maintained in the top half the bottom half. Now next is using the boundary condition. So our domain is now from y equal to 0 to y equal to h. So what is the boundary condition at y equal to 0? That is the center line du dy is 0 because u is a maximum here or otherwise from symmetry. In some case it could even be a minimum but whatever it is it should have an extremum at the center line. So it is derivative with respect to y is 0. So that means you have c1 equal to 0. Next let us integrate it once more. So you have mu u is equal to c y square by 2 plus c2. What is the boundary condition at the other end? Boundary condition at y equal to h that is the wall u equal to 0 that is the no slip boundary condition. So that means you have c2 is equal to minus c h square by 2. So we have got a form of the velocity profile and let us write that form. So what is the form? Mu u is equal to c y square by 2 minus c h square by 2. So u is equal to c by 2 minus mu into y square minus h square. See it is symmetrical with respect to y equal to 0. So positive y and negative y has no consequence because it is y square. The other important thing is y square minus h square is always less than or equal to 0 because your maximum value of y is h. But you are having u along the positive x direction that means c must be negative and that is what we have seen that you must have a negative gradient of the piezometric pressure. It should be from higher piezometric pressure to lower piezometric pressure and so this dp star dx which is equal to c is negative. So you know it in terms of the pressure gradient and so this c is equal to dp star dx. Sometimes it is not a bad idea to write it in terms of the average velocity. So what is the average velocity? How do you find out what is the average velocity? So average velocity u average is what? Integral of u over the area of cross section divided by the area of cross section. So here the area of cross dA is what you take at a distance y. So let us say that this is the channel. At a distance y we take a small strip of width dy and dA is dy into the width of the channel. So it is integral of u dy into the width of the channel divided by 2h into the width of the channel and the effect of the width of the channel gets cancelled out. And this is actually from –h to –h because of the symmetry this is as good as into integral of 0 to h u dy divided by 2h. So half is good enough and you may substitute you that is c by 2 mu into integral of 0 to h y square –h square dy divided by 2h. So let us just complete this expression. You have u average equal to c by 2 mu into sorry 2 is not there in the denominator. These 2 is not there because it is already cancelled. So c by 2 mu into integral of that so h cube by 3 –h cube that divided by h. So this becomes –2 third so that becomes –ch square by 3 mu. So you can write c as 3 mu u average –3 mu u average by h square and that if you substitute in the expression for u you will get from this expression u by u average is 3 by 2 into 1 – y square by h square and clearly the form shows that is a parabolic velocity profile. So that is the first objective that we have satisfied that we have found out what is the velocity profile. For engineers the next important objective which is quite straight forward and one step from this is to find out what is the wall shear stress. So what is the wall shear stress? So the wall shear stress is basically tau xy at the wall that is for a Newtonian fluid. Remember that is mu into partial derivative of ui with respect to xj plus uj with respect to xi. So just in terms of u and v and x and y these are substituted. At the wall v is 0 and in fact we do not care whether v is 0 at the wall because for fully developed flow v is 0 everywhere. So we have straight away whatever effect of v is there 0. So it is as good as mu du dy because u is a function of y only, mu du dy at the wall. Wall means you have y equal to h. So what is mu du dy at y equal to h? So you find out what is du dy? du dy is cy by mu. So du dy at y equal to h is ch by mu and c u we know is that minus 3 mu u by h square. So we can write this c in terms of tau wall or tau wall in terms of u average from this expression. One important thing is we have not substituted this with a proper sign. Remember that when you write tau as mu du dy then that y is what? A coordinate which is perpendicular to the wall from the wall to the fluid. So we are considering the top wall here. Ideally that y coordinate should have been this one but our y coordinate is opposite to that. So it is better to consider a separate coordinate y1 which we are talking about here which is just like the y1 coordinate which is perpendicular to the wall and towards the fluid. So this is basically at y equal to h or y1 equal to 0 but y1 and y they are just oppositely oriented. So you can write y1 is equal to nothing but h-y. So when you write this du dy at, so this may be written in terms of minus mu du dy at y equal to h. Though it is mu du dy1 at y1 equal to 0 that is as good as, so this plus minus sign is with the convention of the normal of the wall. So that means du dy at y equal to 0 that is cy by mu that is c at y equal to h that is ch by mu. So what is tau wall? Minus mu into ch by mu. So minus mu du dy and c you can write as 3 mu u bar by h square. So 3 mu u bar by h. So that is the wall shear stress. Sometimes we write the wall shear stress in a non-dimensional form. So this is a dimensional form. So the non-dimensional form of the wall shear stress is something what engineers use as friction coefficient. So friction coefficient say cf that is defined as a non-dimensional form. So wall shear stress divided by half rho u bar h square. So see rho into velocity square is a unit of pressure. Stress is also unit of pressure. So this is unit less. Half has a sanctity that is like sort of represents a normalized effect with respect to the kinetic energy. So but like it is for just for non-dimensionalization even one may omit the half. This is just to keep the sanctity of the sort of sense of kinetic energy. So this you have 3 mu u average by h by half rho u average square. So cf is 6 mu 6 by divided by rho into u average by mu. Rho u average into h by mu. And if you want to write it in terms of the width of the channel this is 2h. So this you can write as 12 by rho u average into 2h by mu. Importantly this rho u average into 2h by mu it represents what? It is like if you see rho into velocity into the characteristic width of the channel divided by mu. It is a Reynolds number. So this is 12 divided by a Reynolds number based on the length scale 2h which is the distance between the 2 plates. So the non-dimensional friction coefficient sometimes this is known as Fanning's friction coefficient because Fanning was the person who first introduced this coefficient Fanning's friction coefficient that is 12 by the Reynolds number based on the distance between the based on a length scale which is the distance between the 2 plates. See for a flow taking place internally the velocity that you are taking in the Reynolds number is the average velocity because otherwise velocity varies over the section. So we have to take some reference. The reference is the average velocity rho and mu are properties and the length scale is the characteristic length of the system which is 2h or even you may write it in terms of h but then you have to clearly specify that what is the length scale that you are taking for the Reynolds number. Usually 2h is taken as the length scale. So let us stop here today and we will continue with the exact solutions in the subsequent lectures. Thank you.