 So, what we are doing is this the Navier-Stokes equation right. So, we derive the Navier-Stokes equation last class which was del v del t plus p dot del v is equal to minus 1 by rho gradient of the pressure plus eta by rho plus plus v plus whatever body force is there right like gravity or whatever. So, today what we look at is we will try to solve this in one or two simple cases. We leave the complicated cases for later, but before I do that I just want to say about a couple of limits of this equation. So, this is the full Navier-Stokes. So, this is Navier-Stokes remember for an incompressible fluid for an incompressible fluid otherwise we will have another second derivative term and Newtonian fluid. So, one of the limits that is one of the limits of this equation is when you have a fluid with no viscosity. So, eta is equal to 0 which is the inviscid limit this is called the inviscid limit. So, this term sort of drops out right and things like super fluid helium and so on come under this limit. The other limit which is what we will be focusing on is this limit of low Reynolds number this. So, once we have taken out the eta from this equation this is called the Euler equation I think yeah this is called the Euler equation. The other limit is low Reynolds number or Stokes flow where I say that I neglect the inertial terms which means I neglect this v dot del v and I work in the stationary limit which means I neglect this del v del t right. So, this is low Reynolds number plus stationary quasi stationary and then the only terms that survive are these terms on the right hand side which means that this then gives me the equation that gradient of p is equal to eta Laplacian of v plus whatever body forces that you have. So, if I can neglect the inertial terms in this equation remember my Reynolds number was a measure of how strong my inertial forces were compared to my viscous forces inertial forces compared to my viscous forces and if this Reynolds number is very small which means that the inertial forces are negligible compared to the viscous forces I neglect these inertial terms in the equation and then what I get is these terms on the right hand side which means that my Navier-Stokes equation now reduces to this equation which is called the Stokes equation and the nice thing about this equation is that unlike this full Navier-Stokes this is a linear equation. I have gotten rid of the non-linear term this v dot del v right which was the main problem in solving them which is the main problem in solving the Navier-Stokes. So, once you have neglected that this is now a linear equation and you can often write down solutions depending on the geometry or at least similarly. So, when we talk of lower Reynolds number physics the hydrodynamics is sort of governed effectively not by the full Navier-Stokes, but by this limiting case of the Stokes equation. So, this is the equation that is more relevant for biological flows mostly. Now, there is a different there is a different way I can arrive at the Stokes equation. So, here I arrive by neglecting the neglecting these inertial terms, but I can arrive at it at a different way which is this case of one dimensional flows. This is case of one dimension which means that let us say I have a velocity vector which has my three components v x v y v z. For a one dimensional flow only one of these velocity components will be there right. So, if it is along a pipe or something the fluid is just flowing in one dimension which means that this v y is equal to v z is equal to 0. So, if you have a flow like this only v x is there then we can see what that says for the full Navier-Stokes equation. So, if I still say that my fluid is incompressible which means that divergence of v is equal to 0 right. That means is that del v x divergence is del v x del x plus del v y del y plus del v z del z equal to 0 right. Now, because this is a one dimensional flow v y and v z are not there right. So, therefore, the v x is not a function of x del v x del x is equal to 0 which means that v x is some function of y and z only. Now, if you work out this term this v dot del v. So, if I now work out v dot del. So, this is my inertial term v dot del v v dot del is v x del del x plus v y del del y plus v z del del z operating on the full vector v which is this v x of y z x cap right. What will this give me? 0 ok. So, for this case for this case for one dimensional flow explicitly this inertial term drops out. So, therefore, if you are looking in the stationary limit where this del v del t also you can neglect this one dimensional flow reduces back to the Stokes equation where you just have this gradient of p plus eta Laplacian of v plus body force is equal to 0. So, this is true therefore, for general flows in the low Reynolds number. So, a general low Reynolds number flow, low RE flow or a strictly 1D flow. In both cases the Navier Stokes equation reduces to the Stokes equation where you have got in rid of the non-linear term. To do this the canonical example is sort of the Spiesoli flow. So, we can just quickly work that out. So, the idea is this that we will look at fluid flows inside a tube ok with some velocity. And one of the cases where you might imagine this is relevant in biology flows like this through tubes is for example, blood flow through capillaries. So, blood generally flows through all these arteries, veins, capillaries and so on. If you look at all these sort of vessels the diameters of these would be in the ranges of a few microns, 2, 3 microns ranging up to for the major blood vessels up to few centimeters. So, these of course, these are branched networks and so on. But if you look at a small segment of this network that is like blood which that is blood which is flowing through this parallel tube of some diameter microns or whatever ok. So, that is the sort of biological picture that we have in mind. So, this we can solve using the Stokes equation and we can see at least what. So, this has many sort of approximations, but even within those approximations we can try and see what it tells us a little bit. So, here is the idea. So, this is I will assume that so, blood is a Newtonian fluid I will assume that blood is a Newtonian fluid with viscosity eta and density rho. This is not true blood is not Newtonian, one can do stress strain measurements and show that blood is not Newtonian fluid, but let us make this assumption since that is what we can do little bit analytically. It is not too bad at least it will give us a sense of how bad it how bad this approximations. So, this is my assumption one that I will assume that blood is a Newtonian fluid and secondly, I will assume that this blood vessel is a pipe blood vessel is a straight pipe of diameter D. So, in principle of course, capillaries or whatever will be will be flexible. So, they can be some shape I will just assume for the time being that these are straight pipes. So, what I want to solve is this Navier-Stokes equation and I need a boundary condition. So, I will use the no slip boundary condition no slip boundary condition which means that the velocity at the walls of the pipe the velocity of the fluid or blood at these walls the walls of the pipe is going to be 0. So, this is fluid velocity is 0. So, that is that and so, let me draw this a little better. So, here is my blood vessel it is a cylinder like this of some length let us say L ok and some diameter D. Let us say there exists some pressure difference of delta P between the 2 ends of the pipe. So, this is the pressure difference between the 2 ends of the pipe this pressure difference competing with viscosity is what causes the flow. So, this is basically P 0 minus P L and my blood is flowing let us say in this direction and let us say this is my z axis. So, my velocity the fluid velocity vector that has only a component along the z axis and if you think about the symmetry of the problem the cylindrical symmetry the magnitude of the velocity can only depend on the distance from these walls. So, it can only depend on this radial distance from the walls. So, this is r equal to 0 this is r equal to d by 2. So, this is the form of my velocity vector V of r in this travelling in the z gap ok. And let us say that I only I want to look again at steady state solutions. So, I want to look at steady state solutions. So, del V del d in the Navier stokes is 0. I know that again because this is a one dimensional flow my inertial term will also drop out. So, V dot del V is equal to 0. So, what I will solve there are no other external forces. So, the equation that I need to solve is gradient of P is equal to eta Laplacian of P ok. So, this is the setup of the problem. It is a standard problem in fluid mechanics pipe flow we will just adapt it or not even adapt it. We will do it with the background of this sort of blood flow through capillaries at the back of our mind. So, when we put in some numbers we will put in numbers that are for this viscosity diameter whatever we will put in numbers that are that correspond to this sort of capillary blood viscosity and so on and then see what that tells us ok. So, the obvious choice is to work in cylindrical coordinates right because the problem has cylindrical symmetry. So, the gradient of P in cylindrical coordinates is r hat del P del r plus theta hat 1 by r del P del theta plus z z hat del P del z and the Laplacian the z component of the Laplacian is del 2 v z del r 2 plus 1 by r square del 2 v z del theta 2 plus del 2 v z del z 2 plus 1 by r del v z del r. So, this if you go back to your curvilinear coordinates the cylindrical coordinates and check this should be the z component of the Laplacian. This term does not is not there because v is a function of r only this term is also not there these terms are not there which means that if I substitute this gradient of P and eta Laplacian of P the equation sort of reduces to del P del z is equal to eta into. So, let me bring the eta this side del 2 v z del r 2 plus 1 by r del v z del r. So, this is the equation that I need to solve this two terms I can just combine and write as 1 by r del del r r v z you can integrate. So, this is the equation that I need to this is what the Navier stokes equation reduces to for this pipe flow within this assumptions. You can solve this differential equation. So, you can integrate directly once with respect to z and twice with respect to r right there are two derivatives of r. So, if you for example, integrate with respect to z over the length of the pipe this will just give you delta P right the difference in pressure between the two ends over here and on the right hand side you will get a L basically which is the length of the pipe and then you can integrate twice with respect to r. So, just do this I will write down the solution ok. So, you can solve for this v z of r or v of r. So, you can check. So, this is minus delta P the difference in pressure divided by the viscosity into the length of the pipe r square by 4 minus c 1 log of r plus c 2. So, you can check this and substitute back and check that this is the correct solution. So, now I need to find out this constants which means I need to apply the boundary conditions. So, I have one boundary condition here what other thing can I use right remember this is my radial axis this is 0 this is d by 2 at the surface so what can I say about this constants c 1 and c 2. So, that is 1 because v at d by 2 is equal to 0 that is 1, but I have two constants. So, I need to use something else what other thing can I use maximum velocity maximum velocity at the center ok. So, what would I write for the mathematically v at r equal to 0 is v max something like that, but then this you do not know right. So, you are just introducing another constant into the system as such it instead of c 1 c 2 1 you can get rid of by using this. The other instead of writing c you would write in terms of v max it is true, but you have not you have to evaluate that. What else can you use? c 1 is equal to 0, why c 1 equal to 0 right. So, if you put r equal to 0 which is the center of the pipe this term blows up right which is not physically possible you need to have some velocity. So, the fact that this thing is some finite number implies that c 1 is equal to 0 and now you can put this other condition v is equal to d by 2. So, let me get rid of that term. So, this term is not there anymore. So, now we can put r equal to d by 2 and at that the velocity goes to 0 which gives you c 2. So, what is c 2? Yes. So, this comes out to be delta p by 4. So, 4 eta l d square by 4. So, c 2 comes out to be d square by 16. So, if you put that in this is the velocity profile which as you say does indeed have a maximum when you put r equal to 0 the magnitude of that maximum velocity is delta p by 4 pi eta l sorry 4 eta l d square by 4 and of course, by construction when you put r equal to d by 2 this velocity goes to 0. So, the profile looks like that of this pipe flow at the walls it is lowest. So, there is no flow at the walls it is maximum along the axis of the cylinder this is called persuading flow ok. So, this is my velocity profile profile like this along this is alright. So, now, if I say that let us calculate a few now that I have the velocity I can try to characterize it in different ways. So, for example, I could ask that what is the average fluid velocity across the cross section. So, what is the average fluid velocity across the cross section which means that I need to take the average. So, v average across this cross section is this v of r into 2 pi r d r right r going from 0 to d by 2 divided by pi r square. So, pi d square by 4. So, this is right this is the average fluid velocity over the cross section v r 2 pi r d r divided by the cross section over here. So, if you put it in this v of r and you do this integration this is an easy enough integration you put in the limits again I will just write down the answer this comes out to be the pressure difference times d square by 32 beta l this is the average fluid velocity. You can also ask what is called as the flow rate or the volumetric flow rate. What is the volumetric flow rate which is the amount of fluid flowing through a cross section of the area per unit time volumetric flow rate and that is just that is called q and that is just this let us say this average velocity across the cross section into the cross sectional area. So, pi d square by 4 which therefore, comes out to be pi delta p d to the power of 4 by 32 into 4. This is the amount of fluid that flows past a cross sectional area per unit time. The thing to sort of note is that this is proportional to the fourth power of the diameter or the radius of this pipe that you are flowing through. So, this is Poisson's result actually very famous result in flow flows, but if you think about in the context of blood flows for example, what it says is that very small changes in your diameter or the radius of the capillary that you are flowing blood through can impact this flow rate by a large amount right. This goes proportional to the fourth power of the so, if you have a blood capillary and where your blood is flowing and you have some plaque deposition or something which reduces the effective radius of these tubes that effect even if it is a very small change in the radius. So, if you have a capillary and you have some deposition in this even if this amount of change is not a lot sorry this reduction in the radius is not a lot that affects the flow rate sort of disproportionately because of this decreases this flow rate significantly ok. So, what I wanted to do was to take this result which is the standard result and just put in some numbers and see what sort of velocities I get in the context of blood flow. So, let me say I take a typical capillary diameter of some 5 microns it is roughly 2 to 10 microns a capillary. So, I take 5 microns let us say I take a capillary length of around a centimeter and a pressure difference which is close to the experimental value 3000 Pascals and I need so, I have a delta P I have a D. So, I need a viscosity. So, let me just assume that I take the viscosity of water it is slightly different it is somewhat different from water, but whatever let me take the viscosity of water which if you remember it is 10 to the power of minus 3 Pascals seconds. If you put in all of these numbers into this into let us say the velocity expression the average velocity expression you can get what is the average velocity of this blood flow in a capillary of in a pipe like capillary of 5 micron diameter. So, if you put in these numbers what you will get is roughly around. So, you can check what you will get is roughly around 0.02 centimeters per second and measurements of blood flow in actual capillaries gives you a number. So, let us so, this is my theoretical estimate the experimental range is something like 0.05 centimeters per second which is actually pretty good. We have done an extremely sort of simple naive modelling we have thrown away any complication we have taken a pipe straight pipe and so on and so forth, but still we get an estimate which is pretty close at least order of magnitude it is correct to whatever is the actual flow. So, even though we have made these simplifications the analysis is not too bad. If I wanted to go from this sort of a value to this if I am wanted to increase the velocity and I say that this is so, I have taken that eta blood is equal to eta water. Let us say I say that ok, this is not correct ok. I need to take the correct viscosity of blood. Should I take a larger viscosity or a smaller viscosity to get the correct average velocity? Smaller viscosity right. So, if I wanted to get this number closer to the experimental estimates it would say that I need to take a smaller viscosity. Do you know what is the viscosity of blood? Anyone any idea? Is it more viscous than water? Is it less viscous than water? More viscous than water. So, actually blood is around if I it is around three times the viscosity of water which means that this estimate that I did by doing this approximation is actually better is actually slightly misleading the actual estimate is worse than this because I have taken the viscosity of blood to be the same as water. And you cannot really do a correct estimate unless you so, now if you want to get the numbers right. So, this is actually the theoretical estimate should be something like 0.002 by 3 if I take the correct viscosity of blood.