 in the last class arrived at the simplified forms of the governing equations. So let us continue with the governing equations which are written here. First let us consider the r momentum equation. So in the r momentum equation first of all let us see that what is this Br. Br is the body force component in the radial direction. There is a body force component in the radial direction that is true but until and unless this extent of the pipe is very large it is like a very high radius pipe then that effect of that body force is not going to be important along the r direction. So this therefore may be neglected. So you come up with pressure gradient along r equal to 0 or approximately equal to 0 that means p is not a function of r only not a function of r that means p is a function of z only. So you can write when you come to the z momentum equation this as dp dz. You can combine that term with the body force term to have ddz of p plus rho gz sin theta is equal to mu into 1 by r d dr of r d vz dr because vz is a function of r only that we showed in the last class therefore it boils down to an ordinary derivative. So this p plus rho gz sin theta is like p plus rho gh so this is like p star the piezometric pressure. We again are having an equation where you have dp star dz is equal to mu into 1 by r d dr of r d vz dr where the left hand side is a function of g only the right hand side is a function of r only. So this is function of g only this is function of r only and again this may be valid if each is a constant. So each equal to constant say equal to c. So the first thing that we may do is to find out the velocity profile from this so we consider the right hand side and integrate it so you have d of r d vz dr is equal to c by mu r dr. So this you may integrate so once you integrate you have r d vz dr is equal to c r square by 2 mu plus c 1 which means d vz dr is equal to c r by 2 mu plus c 1 by r. So you may integrate it once more so if you integrate it once more you will get vz as a function of r. So let us do that so you have vz equal to c r square by 2 mu plus c 1 ln r plus c 2 sorry 4 mu yes 4 mu plus c 1 ln r plus c 2. Now c 1 and c 2 are the constants of integration that we have to find out from the boundary conditions. So what are the boundary conditions here boundary conditions so one of the boundary conditions that is at the wall is straight forward at small r equal to capital R you have vz equal to 0 no slip boundary condition. What happens at small r equal to 0 that is a point of singularity so to say and but physical problem is that vz has to be defined at small r equal to 0 I mean you cannot have a situation where there is vz undefined at small r equal to 0 because it is a physical problem the pipe is entirely filled up with a fluid with some velocity at r equal to 0 there it should be a well defined velocity and that is possible so vz is at r equal to 0 vz must be finite it cannot be infinite or it cannot be undefined in other words therefore from the second condition you must have c 1 equal to 0 this is from the boundary condition number 2 and from the boundary condition number 1 so this is boundary condition number 1 and this is boundary condition number 2. So from boundary condition number 1 you have 0 equal to c r square by 4 mu plus c 2 that means c 2 is minus c r square by 4 mu so the velocity profile becomes vz equal to c by 4 mu into or if we write minus c by 4 mu into capital R square minus small r square and as usual you see that the c which is like dp star dz that has to be negative to drive the flow along the positive z direction which means that vz has to be positive if c is negative. Now you may express c in terms of the average velocity what we did for the plane Poiseuille flow for the Heggen Poiseuille flow also the same thing is valid. So what is the average velocity v average it is integral of vz over the area of cross section divided by the area of cross section. So vz what is the area of cross what is the elemental area that you can choose say at a distance r you take a strip of width dr. So if you if you just draw its other view so you have taken at a radius r some strip so you are talking about such a strip of width dr and this strip is 2 pi r dr so vz into 2 pi r dr from 0 to capital R divided by pi capital R square is the area. So let us do this integration quickly so 2 into integral of 0 to r you have the minus c by 4 mu so what is v average in terms of c so minus c by 2 mu then you integrate it r square so r 4 by 2 minus r 4 by 4 by pi r square sorry pi is not there pi has got cancelled. So this is equal to minus c r square by 8 mu so this is one important result which will use subsequently but at least let us write the expression for the velocity profile by writing c in terms of this. So vz becomes in place of c you write minus 8 mu v average by r square that into 1 by 4 mu again there is a minus sign into r square minus small r square that means vz by v average is 2 into 1 minus small r square by capital R square this is the velocity profile in terms of the average velocity. Recall the flow between 2 parallel plates it was like 3 by 2 into 1 minus y square by 8 square so it is quite similar the coefficient is different just because of the different geometry and it is also a parabolic type of velocity profile because it is square with the local radius. Now you can find out what is the wall shear stress so let us find out what is the wall shear stress so it will be like mu we will adjust the sign positive or negative later on mu into the rate of deformation this is like tau rz proper subscript with the index now one of the important things is see positive tau with like mu du dy so this is like just like mu du dy first of all vr is 0 so this term is not there so it is just like mu du dy but the thing is see the r direction is towards the solid boundary it is not away from the solid boundary so that has to be compensated with the minus sign here. So if you have a solid boundary like this if you have this as y then you have like mu du dy so y is away from the solid boundary towards the fluid the reason is you want to write it as a positive shear stress so you want u to be increasing in the y direction so y equal to 0 is like the no slip and then u increases but here the r direction is opposite to that so you have to make it up with the minus sign so the wall shear stress therefore if you just differentiate it with respect to r so dvz dr is like 2v average so this entire thing evaluated at small r equal to capital R that is at the wall wall is located at small r equal to capital R so 2v average into minus 2 so minus 4v average by r is the dvz dr right therefore the wall shear stress is 4 mu v average by r what is the coefficient of friction or the Fanning's friction coefficient that is tau wall by half rho into v average square this we introduced in one of our previous classes this friction coefficient it is a non dimensional way of writing the shear stress so it is 4 mu v average by r basically it becomes 8 mu v average by r rho v average square so it is 8 divided by rho v average r by mu the denominator is the Reynolds number based on radius r but usually for this type of geometry the length scale that is taken is 2r or the diameter of the pipe so if you just consider that then the cf becomes 8 into 2 divided by rho v average into the diameter by mu where diameter is 2 into r so this is 16 by the Reynolds number based on the diameter of the pipe as engineers what we want to do with this friction friction coefficient so what importance or implication it has to us so if it is large it is small we will understand that if it is large maybe the frictional effect is large but how do we take care of that in a design and that may be visualized more effectively if you consider now the dp star dz term which is equal to c so to do that let us come back to this equation v average written in terms of c so what is c? c is equal to dp star dz so let us write dp star dz in terms of the other parameters so it becomes minus 8 mu v average by r square so minus 8 mu v average by r square and what is this dp star dz this is if you have pressure at the inlet as say p in if you have pressure at the outlet as say p out and the distance over which this pressure difference is there is l then first of all we could derive that dp star dz is a constant that means p star versus z is a linear profile so this is as good as p star out minus p star in by l so this we just write as p star out minus p star in by l so in other words it is minus of p star in minus p star out by l I mean why we are writing it in this way is see the p star in minus p star out is the driving delta p star for the flow the driving difference in piezometric pressure that should drive the flow see why this driving pressure difference is necessary it is necessary because you have viscous resistance in the flow so this is a manifestation of the consequence of the effect of viscous resistance that is being exerted in the flow so you have to have a driving pressure gradient which is a favorable pressure gradient that is dp star dz negative to make this flow occur so when you have this one this delta p star so what it in effect does in effect it overcomes the viscous resistance so the viscous resistance also may be expressed in form of a pressure unit so the viscous resistance therefore you may say that delta p star is nothing but the delta p star because of the viscous resistance and we may write it in terms of a head some unit of length so we remember that the pressure is nothing but some length unit into rho into g and the length unit is the equivalent unit of pressure expressed in terms of length with which we call as head so that is what we discussed also earlier so if we write it in terms of a head length unit as head with a subscript h f for friction f for friction so we can write it h f rho g where this h f as engineers we understand that this is a loss of head due to fluid friction therefore it is possible to write this expression dp star dz if we come back here in terms of this delta p star as h f rho g so you have h f rho g by l is equal to 8 mu v average by r square of course it is possible to write it in terms of in terms of the diameter so it is so 8 becomes 32 here okay so it is also possible to write the average velocity in terms of the flow rate that is sometimes important because if the flow rate is q it is just q by pi d square by 4 because in the experiment you usually measure the flow rate so it is we are trying to write the equation in terms of experimentally measurable parameters so from this what you get is h f is equal to 128 mu q l by rho g pi d to the power 4 okay and this is what we are trying to and this equation is known as Hagen Poiseuille's equation so what it gives? It gives a direct indication of the loss of head due to viscous effects in the flow if the flow had viscosity 0 there would have been no loss of head and we can see that the loss of head is directly proportional to the flow rate it is proportional to the length over which the fluid is flowing which is understandable and inversely proportional to the fourth power of the diameter so if you make the diameter very very small the loss of head will be very large that means to drive flow through a very small tube say a tube of micron size you require a huge pumping power because there is a huge head loss because of the frictional effects that is one of the challenges in having a flow in a micro or a nano channel in a very small channel so the whole understanding is that like those are advanced topics but you see that the basic of fluid mechanics gives us a clue that what are the challenges as you make the sides smaller and smaller what are the engineering challenges in terms of having the flow now it is also possible to express it in terms of a non-dimensional form by writing this in this way by writing this as some factor f into l by d into v average square by 2g so this is just a non-dimensional way of writing it see l by d is a sort of aspect ratio of the pipe it is a non-dimensional parameter so and v square by 2g is the kinetic energy head so the head loss expressed as fraction of the kinetic energy head this f is that sort of fraction so what is that so if you just equate these two so you have hf as rho g rho sorry f l by d v square by 2g and that is equal to whatever expression of hf what we had so 32 mu v bar by d square so from here what we can conclude what is f 64 divided by rho v bar into okay by just by equating these two so that means this f becomes 64 by the Reynolds number so this sort of non-dimensionalization was first introduced by an engineer known as Darcy. So this friction factor is also known as Darcy's friction factor so it is just a different way of writing the friction or frictional coefficient you can see clearly that f and cf are related by f is equal to 4 into cf so just is the same thing expressed in terms of in different ways so one looks into it in the view point of a pressure drop another looks into it in the form of a wall shear stress and here they are related because you have to overcome the wall shear stress and therefore you have a pressure drop so that is how they are related and this 4 coefficient is just because the difference in which they are defined but physically they imply the same so if it is Manning's friction factor that is cf and if it is Darcy's friction factor it is f so now so we have a clear idea that if you have flow through a pipe there is a head loss the head loss is because of the viscous effects in the flow and you require a driving pressure gradient the driving pressure gradient is to overcome an energy loss the energy loss may be expressed in terms of a head and that head is given by this expression so that is what we learn and as engineer it is important because say you are given a length of the pipe see I mean let us think about how this helps in design say you are given a length of a pipe say 10 centimeter length of a pipe you are given a flow rate that you expect out of it say some meter cube per second you know what is the viscosity of the fluid and density of the fluid like that so your problem is that what should be the power of that pump that is necessary to drive the flow this is one of the very basic elementary problem so to know that you have to know that what is the pressure drop that is taking place and first of all if you find out what is the hf what is the head loss over that length then that into rho into g is the pressure drop and the pump should have enough power to overcome that so the pump should have a power that is developed that is given by rho g into q into the head loss it has to overcome that loss so this is like a unit of power so if you know that what is the flow rate if you know what is the head that the pump has to supply then you know that what is the power of the pump that is necessary and so it comes into so a very basic elemental mathematical work also is necessary even for a very crude design work which is there in the day to day life of fluid mechanics like flow through pipes okay now we come to another example example 5 and that will be like the final example before we will be moving to the next chapter so the final example is about flow between 2 rotating cylinders long rotating cylinders this is the problem that we are going to revisit because this problem we introduced when we were discussing about viscosity and its measurement so you have 2 cylinders concentric cylinders of radii r1 and r2 let us say that the inner cylinder is rotating with an angular velocity omega 1 and the outer cylinder is rotating with an angular velocity omega 2 and these angular velocities are like these are not functions of anything these are constant angular velocities and we are considering a steady flow of a Newtonian fluid of constant properties which is which are kept the fluid is kept in the gap between the 2 cylinders this type of problem is important because we have discussed that it gives a principle of the measurement of viscosity of an unknown fluid so principle of a device known as viscometer so we have earlier qualitatively used the one dimensional form of the Newton's law of viscosity to like get an estimation of what happens here but now what we will do we will try to do it in a more careful way by using the proper equations of motion so first of all we have to understand that what are the important variables which are involved here so again by the nature of the problem the geometry of the problem the cylindrical polar coordinate system seems to be a proper choice so the r theta g coordinate system now the corresponding velocity components are there vr v theta vz the first assumption that we will make is that the length of the cylinder which is perpendicular to the plane of the figure is very large what is the consequence of making the length large large means large in comparison to the radius so if l is large then what is the consequence in terms of your analysis it effectively boils down to a 2 dimensional problem in the r theta plane so the z gradient is not important so when you come to the r theta component you have vr and v theta these things are there next is what about the partial derivative with respect to theta see when you are solving a problem it is not that the assumptions are given to you it is important to come up with the assumptions based on the physical description of the problem see the inner cylinder rotates with an angular velocity which does not does not have any preference over theta right it is not that at different theta it is different the outer cylinder also rotates with an angular velocity that does not have any preference over any theta and therefore in between we expect that it whatever is the behavior should not have any preference over theta so that means the derivative with respect to theta for whatever the flow parameters that should be 0 that should automatically follow from the physical description of the problem next what we will do we will now go to the basic equations again we will go to the cylindrical coordinate system equations navier stokes equation but first we will go to the continuity equation so if we go to the continuity equation you see we are talking about the continuity equation in the cylindrical polar coordinate system first term because we are considering the steady flow the density gradient derivative that is 0 with respect to time the second term the second term is like if you want you should keep it the third term and the fourth term see because there is no derivative with respect to theta the second term is 0 no gradient with respect to z so that term is 0 so only one term is r into vr the derivative of that that means d dr of r vr that is equal to 0 just like what we had for a Heggen Poiseuille flow and what will be the obvious conclusion that vr is not a r into vr or vr is not a function of r so we know that at 2 different radii vr is 0 r equal to r1 and r equal to r2 and that is good enough for us to say that vr is equal to 0 for all r because vr is not a function of r so the problem has boiled down to only one velocity component that is v theta okay so with this understanding let us now go back to the momentum equations let us look into the momentum equations first the r momentum equation that is the first equation written in the slide so you have vr equal to 0 so all the terms in the left hand side are 0 except the term involving v theta so in the left hand side you have minus rho v theta square by r okay and it is the centripetal effect so we may understand physically what it is so when it is a rotating thing it is a centripetal acceleration because of the tangential component of the velocity right hand side you have first term minus dp dr that term is there the remaining terms first term has vr so that is 0 second term has again vr that is 0 third term again has vr and fourth term has the derivative with respect to theta so that is 0 there is no body force along r okay so you have the r momentum equation what it gives dp dr is rho v theta square by r no that is what I said that is always a special case so you also have to satisfy the boundary conditions and consider it for all r or for all possible cases vr proportional to 1 by r is only a special case but you have to consider that it should be true for all possible velocity fields it is not just for one particular velocity field not only that you have to satisfy vr say 0 at r equal to r1 and r equal to r2 so if you have say r into vr as some constant c so that is vr equal to c by r then at r equal to r1 how do you ensure that vr will be 0 you have to ensure that the boundaries you have normal component of velocity 0 that you have to keep in mind so the r momentum equation you have dp dr as rho v theta square by r let us look into the other momentum equation see the z momentum equation is useless here because it involves all the terms which have vz and there is the nobody force along z and only it gives that dp dz equal to 0 that is the z momentum equation only term that remains important is dp dz equal to 0 the most important momentum equation will be that theta momentum equation so theta momentum equation if you see that is the second equation written in the slide first you go to the left hand side you have first the flow is steady so the first term is 0 next you have vr is there so vr into that is 0 third term derivative with respect to theta so that term is 0 fourth term because vr is 0 that is equal to 0 fifth term vz is 0 so or even if you do not consider that that gradient with respect to z is 0 because z is large so in either way the left hand side totally becomes 0 right hand side first term dp d theta is 0 because there is no theta variation second term is very much there because you have v theta as a function of r which is which is what we are going to solve next term it is a gradient with respect to theta so that is 0 the term after that you have derivative with respect to z that is 0 the next term you have vr that is 0 and you have no body force no body force along theta therefore it boils down to mu sorry theta momentum let us write the terms 1 by r mu into 1 by r sorry ddr of so ddr of 1 by r ddr of rv theta that is equal to 0 so see now v theta could be function of what v theta could be function of r theta and z it is not a function of z because there is no gradient with respect to z the length is large v theta is not a function of theta because there is a there is no variation with respect to theta for anything so v theta is a function of r only and therefore we may write it as an ordinary differential equation so we may just write it equivalently as ddr of 1 by r ddr rv theta is equal to 0 so if you integrate it you will get d of 1 by r ddr of rv theta that means 1 by r ddr of rv theta is equal to say some c1 so rv theta is equal to c1 or d of rv theta is c1 rdr right so if you integrate it you will get d of 1 by r so if you integrate it rv theta is equal to c1 r square by 2 plus c2 and v theta therefore becomes c1 r by 2 plus c2 by r so it is like of the form ar plus b by r right so if you look into this equation can you recall that the corresponding components of this velocities are related to the vortex flows that we have studied earlier so what is the first component that is like a forced vortex and this is free vortex or irrotational vortex so it is like a combination of free and forced vortex that is going to be the resultant velocity field now and if you want to get the pressure distribution you have to go back to the r momentum equation to get the radial pressure distribution so you know now v theta as a function of r you may substitute and integrate it with respect to r to get the pressure as a function of r we are not going into that but what we will do is we will find out what is the expression for the wall shear stress so we are interested in tau r theta or just shear stress at different locations so if you want to write so there is an expression for tau r theta and I am just writing it down here and I will tell you that what is the origin or rationale behind this expression so this is for a Newtonian fluid just written just express in terms of the r theta coordinate system so it is just like just think of 2 coordinates like if it was tau xy that was what mu so basically you are having a cross gradient x component of velocity with y gradient and y component of velocity with x gradient so similarly you have a theta component of velocity with r gradient and r component of velocity with theta gradient and this 1 by r these adjustments are there because of the use of the cylindrical polar coordinate system so these adjustments are also quite obvious that if you have say this term so you have vr with respect to theta see you want a gradient that means it should be the velocity divided by a length so this is d theta so r d theta is like an elemental length in a cylindrical polar coordinate system so that is why 1 by r term has to come outside so in this way you may relate this term with the general understanding of the curvilinear systems so again it is not very important to go into the derivation of this term but we will just utilize this to calculate the stress so if you now write this is mu r d dr of v theta by r so v theta by r is we have not yet completed the expression for v theta because a and b are 2 constants of integrations to be evaluated but that we can straight forward do from the boundary condition but at least let us look into the form first so v theta by r is a plus b by r square so v theta by r is a plus b by r square and therefore d dr of v theta by r is minus b by r cube sorry minus 2 b by r cube so you have this as minus 2 b by r cube theta derivative is not there so minus 2 mu b by r square what is the so one important thing is independent of theta but dependent on r the next thing is that what is the elemental shear force because of this so if you consider an element at a radius r you have a line element like this because it is independent of theta otherwise you could have taken a small line element with r d theta but because it is independent of theta we are taking a total peripheral line element so what is the elemental shear force on that so the shear force is f at a distance r is tau r theta into 2 pi r into l so 2 pi r is the perimeter of this times the length is the outer surface over which this tau r theta is tangentially acting so that becomes minus 2 mu b by r square so into 2 pi r l and what is the moment of this force with respect to the axis of the cylinders at a radius r that is f r into r and you can clearly see that is equal to minus 2 pi mu b l which is independent of r so if you remember that while dealing with the understanding of the principle of viscometer we said that same moment or torque is transmitted at all radius and this is what it shows that at any arbitrary radius the torque which is there is the shear torque the torque due to the shear force is independent of the local radius this is one of the very important understandings now to find out the constants a and b you have to use the boundary conditions so what are the boundary conditions let us just write the boundary conditions and try to find out a and b so the boundary conditions are at r equal to r1 p theta is omega 1 into r1 no slip boundary condition so velocity of the fluid same as velocity of the solid and at r is equal to r2 v theta is omega 2 into r2 so you substitute that so omega 1 r1 is equal to a r1 plus b by r1 and omega 2 r2 is equal to a r2 plus b by r2 so it is possible to find out a and b from these two right these are like capital as per our notations so this is the boundary conditions so a r1 plus b by r1 a r2 plus b by r2 now let us say that we want to solve a problem which is a bit of a modified version of this one what is that modified problem so the modified problem is you have one single cylinder solid cylinder rotating with an angular velocity omega in a fluid of large extent so one cylinder of radius r and of large length rotating with omega so there is no inner or outer concept so how can you solve this problem by considering the solution of this one so omega 1 so r1 is equal to capital R omega 1 equal to omega r2 tends to infinity and omega 2 equal to 0 so if you do that let us see that what forms the constants of integration state for that particular case so you have r2 tends to infinity and omega 2 equal to 0 so when r2 tends to infinity this term is 0 omega 2 is 0 that means a equal to 0 so this will straight away give you a equal to 0 that means b is equal to omega into r square so tau r theta in that case becomes minus 2 mu omega into r square by r square and what is the v theta there v theta is just b by r because a is 0 so omega r square by r so it has become like a free vortex we have seen earlier that a free vortex flow is an irrotational flow right so that much we know that means the vorticity vector if you calculate that will be a null vector that we showed by calculating the circulation and the vorticity now let us try to find out what is the shear force on the fluid and that is something which is very very interesting so we have calculated the shear force locally at a radius and may be they are taught due to that but over the volume what is the total shear force that is acting to do that what we will do is we will write the momentum equation just in a vector form so the momentum equation is rho okay so this is the vector form we have derived this vector form and the advantage with the vector form is whatever vector operation that we will be doing on it so long as it is in a vector operation form we do not explicitly mention which coordinate system it is so it remains valid for all types of coordinate systems so the problem or the situation that we were looking for in that particular situation the left hand side was 0 always by the consideration of fully developed flow and steady flow so this term was 0 and this term was 0 because of the sort of either fully developed or may be 0 gradients in certain direction say theta or z direction for whatever reason the left hand side always came to be 0 for the example that we were talking about so in the vector form also it is like that so you have the gradient of p is equal to mu into this now let us try to use this particular term and see this is what this is basically a force per unit volume right so if we want to find out that what is the force per unit volume because of viscous effects this is the term that is coming into the picture okay so this you can say that is the f per unit volume mu into this now what we will do is let us say we want to take the curl of both sides or just by looking into another way let us try to find out this term I mean we may take a curl of this side this and do but let us just look into this so we want to take the curl of the curl of the velocity vector so that is given by the vector identity so in this vector identity let us now try to use this vector identity so what we will do is we will consider an incompressible flow so incompressible flow when you are considering an incompressible flow what is the consequence of that you have the divergence of the velocity vector equal to 0 so when you have divergence of the velocity vector 0 then you have that del square v is equal to minus of curl of curl of v right and curl of v is the vorticity vector right therefore you can write that del square v is equal to minus curl of the vorticity vector but the zeta is the vorticity this force that is mu del square v therefore we can write this as mu minus mu into curl of the vorticity vector so this is what this is the viscous force per unit volume right so this is shear force per unit volume it does not matter whether it is the equation is simplified to this form or not even if the left hand side all the terms are there still this happens to be the viscous force per unit volume so this particular term so only these terms were simplified to this extent for the problems that we discussed but for all problems it will not be like that but for all problems for a Newtonian fluid with a constant viscosity this is going to be the viscous force per unit volume now you look into this special case so this for this free vortex we know that it is an irrotational flow so that means curl of the velocity vector is 0 that means the vorticity vector is a null vector so that means this term is 0 okay so you see an example if you recall one of our very introductory lectures in fluid mechanics we mentioned one thing that fluid mechanics is a beautiful subject because it gives something which is non-intuitive and this is one example you have shear stress not equal to 0 but shear force equal to 0 if you recall we mentioned this as a non-intuitive example that you have shear stress not equal to 0 right this is b is not equal to 0 it is varying with the radius but you now find that because of the irrotationality the shear force is 0 with this example one of the very non-intuitive things we learn out of fluid mechanics that there may be cases where the shear stress is not 0 but the shear force is 0 and that we can get from very elementary mathematical consideration of this type of flow. So to summarize we have looked into examples of exact solutions of the Navier-Stokes equations mostly for steady flows we have not done it for unsteady flows that is not there in the purview of this elementary course but at least for steady flows and fully developed types of flows we have worked out the solutions and we have come to conclusion that simple closed form solutions of Navier-Stokes equations are possible for such types of flows and these give rise to certain important insights from engineering and scientific principles. So we will conclude this chapter here and from the next lecture we will start with another important facet of the equations of motion for viscous flows and that is the turbulent flows. So we will start with the introduction of turbulence from the next lecture. Thank you.