 we have been discussing on various aspects of low Reynolds number hydrodynamics and so far we have considered only internal flows that is flows taking place through channels or ducts these kinds of examples we have primarily considered. Although we have considered some cases of external flows with regard to unsteady flows we have considered some cases of flows over flat plates and so on but mostly we have considered internal flows. Now external flows are important in the microfluidics context because of several reasons right. Let us consider an example of let us say that there is a charge species this charge species is moving say this species has a negative charge this charge species is moving under the action of an electric field. So let us say there is a fluid medium there is an electric field that is applied on this let us say that Q is the charge. So if Q be the charge and if E be the electric field then Q into E is the force that is acting on this charge species. Now if the charge species is not accelerating then what is happening that this driving force should be balanced by the viscous drag force this must be equal to F drag and this drag force is a function of the relative velocity between the fluid and the charge species okay. So what is the velocity at which this charge species will move relative to the fluid that will be decided by what is the drag force on this particular charge species. And to know what is the drag force on the charge species we may consider several situations. Now one idealized situation is an idealized model of the charge species where the idealized geometry may be for example spherical geometry. Now you may argue that spherical geometry may not represent all the charge species and that is true right while spherical geometry may represent some of the ionic species a spherical geometry may not represent charge species like DNA for example. So if you want to go for a very accurate model you have to consider the shape the geometrical shape of the charge species. However for simplicity for many complex shaped objects you may consider them to be of some equivalent radius where the radius defines the radius of an equivalent sphere and that radius can be thought of as the radius of gyration of the system. So if you are trying to model the situation by an equivalent sphere or sometimes maybe equivalent cylinder we will take that example of a equivalent sphere then for the equivalent sphere it may be possible that we can calculate the drag force by using simple analytical tools and that is a very classical problem. Of course for more complicated shaped objects we can follow the same governing equations but the same kind of analytical solution will not hold and numerical solutions may need to be obtained for complex shaped bodies. So the idea is that we are interested to calculate the drag force on a body of idealized geometry that is there moving relative to the fluid. Now the question is how is it moving? Now we are typically considering low Reynolds number hydrodynamics so we are considering that the Reynolds number based on the diameter of the sphere is pretty low. So I mean typically the Reynolds number much less than 1 so that viscous forces are dominant as compared to the inertial effects. So our objective for today's lecture is that we have let us say a sphere in a low Reynolds number background flow so it does not matter whether the fluid is stationary and the sphere is moving or the sphere is stationary and the fluid is moving because it is the relative velocity that matters. So let us say that the sphere is stationary and the fluid is moving past it and the typical Reynolds number based on the diameter of the sphere is low. So for that particular purpose what we can do is we can refer to a very classical problem which was solved by Stokes long back and the idea of the classical problem is to solve the velocity profile, the pressure profile and hence the drag force that is acting on this spherical object as you have a low Reynolds number fluid flow past the same. The Reynolds number of course is classically calculated based on the diameter of the sphere. We will try to solving this problem at this stage because this problem will require the use of a coordinate system with which you may not be that familiar with. So we will start with an equivalent problem description which we will represent in terms of a vector notation for any arbitrary coordinate system and then we will specify the particular case for this type of coordinate system. So to do that let us say that we are interested to write the governing equations of motion in terms of the Cartesian coordinate system okay. So now in terms of the Cartesian coordinate system let us say we write the Navier-Stokes equation although it is low Reynolds number I am still writing the full form just to give you the form of the Navier-Stokes equation in vector notation. We are considering incompressible flow I am not adding any body force here I mean you can add the body force in a way in which you like. So now you can write this term v dot del u because this v dot del is what this v is ui plus vj plus wk and what is this del I del del x plus j del del y plus k del del z. So if you take v dot del it becomes u del del x plus v del del y plus w del del z and the operator is operating on u okay. So this is the x component of the momentum equation x momentum similarly y momentum and z momentum let us quickly write y momentum it will be v dot del v and the z momentum okay. Now what we do is we consider this into i cap where i cap is the unit vector along x this into j cap and this into k cap and add. So that gives the vector sum of the 3 and if you do that then what you get rho del del t of ui cap plus vj cap plus wk cap so capital V vector plus v dot del ui cap plus vj cap plus wk cap so that is v is equal to minus i del p del x j del p del y k del p del z so minus grad of p plus mu del square of so see here the operator is Laplacian operator del square del x square plus del square del y square plus del square del z square of ui plus vj plus wk so this one right. So this is the compact way of writing the Navier-Stokes equation in vector form we have seen how to write it in index notation this is the compact way of writing that in vector form and writing that in vector form like index form is very much specific to Cartesian nature of the indices Cartesian tensor notation I mean the vector form the advantage is that it does not require the description of any coordinate system. So you can use it for a very general case and if you have a body force just add a plus b vector for indicating the body force. Now for low Reynolds number you can drop the LHS we will see later on whether the same is justified or not for the case that we are investigating but for the timing it is intuitive that if it is a low Reynolds number flow then the left hand side can be dropped this is what we have learnt from our previous experiences. Now so far as what we are going to predict whether that particular inference is still working or not we will discuss about that as we come to the end of the solution of this problem. So then you have null vector basically we will use a vector identity to simplify this problem but first of all before using the vector identity we have to find out a strategy of solving this problem. So the strategy of solving this problem see this equation has 3 components if we use some other coordinate system we may reduce the number of components by using some symmetry or that kind of thing but in a Cartesian coordinate system see we are not discussing this specific problem we are discussing about a general problem that can be described in terms of a suitable coordinate system we have not yet fixed up the coordinate system. So this equation involves how many unknowns in general 3 velocity components and pressure. So you have 4 unknowns and for an incompressible flow you have 4 governing equations also so that closes the system but analytically it is very difficult to couple the continuity equation with what with the momentum equation with the Navier-Stokes equation analytically difficult to couple numerically it is relatively more convenient. So what becomes the deterring factor the deterring factor is this pressure these governing equations are equations governing velocity but in that extra unknowns in form of pressure appear but we do not have an explicit governing equation for pressure. So one possible strategy one possible strategy may be so our origin of headache is the pressure. So the solution is I mean this kind of solution is a very radical solution the solution is if you have headache you cut your head. So if you do not have head you do not have headache right. So you have pressure in this equation so you somehow eliminate pressure then there will be no problem with the pressure term. So how can you eliminate pressure from this equation? So one simple strategy is if you take curl of both sides you can use the vector identity that curl of grad of a scalar is a null vector. So this will go away. So take curl of both sides. So if you take the curl of both sides this becomes 0. Now we will use a vector identity this is a very commonly used vector identity and the interesting part of the vector identity is that you have a divergence of velocity in one of the terms. So if it is incompressible flow as we are considering for this case then this term will go away for incompressible flow not for all cases only for incompressible flow. So instead of del square v you can write minus del cross del cross v right using this identity and what is del cross v what is the curl of the velocity that is nothing but the vorticity. So let us this is vorticity. Let us say this zeta we use for vorticity okay. So we can write this is curl of vorticity and we have additionally taken a curl to make this term 0. So that is why curl of curl of this. Now we can now use the same identity here just replace v with zeta right. So we can write just I mean here instead of v we are using zeta. Now what is this? This is del dot del cross v and by a vector identity this is 0 divergence of curl of a vector this is 0. So we are left with del square of zeta is equal to 0. Now let us consider an example of a 2 dimensional flow in Cartesian system described in Cartesian system. We are starting with the Cartesian system not because that we will use the Cartesian system to solve the problem of the sphere but because of familiarity you can understand certain concepts which we will borrow for solving the problem of the sphere. Now for this particular problem now for 2D flow zeta this is the scalar component of vorticity. So for a 2D flow the vorticity vector is having an effect which is perpendicular the unit vector is perpendicular to the plane of the flow and the magnitude of the vorticity is this one. Now so this is zeta with respect to z axis the vorticity. Now for a 2 dimensional flow see we had said that for eliminating this term by eliminating this term pressure we are not having an explicit linkage of the continuity and the momentum equations. But there has to be some reference to the continuity equation otherwise how do you ensure that you are solving the velocity field from the momentum equation that satisfies mass conservation that has to be ensured. So somewhere continuity equation has to be brought into the picture. How we bring into the picture here is by defining this velocity in terms of stream function. So we define let us say u is equal to del psi del y and v is equal to – del psi del x. So what are these u and v? These u and v are the velocity components and psi is the stream function. So you can see that this definition of psi automatically satisfies the continuity equation del u del x plus del v del y equal to 0 for 2 dimensional incompressible flow. So you can write zeta z now for defining the stream function we have to be careful that right u and v have to be carrying opposite algebraic sign. So it does not matter I mean here we have taken this as plus this as minus but we could have easily taken this as minus and this as plus that would have made no difference. So this is in 2D this is – del square psi. So del square zeta z is equal to 0 that means you have del square of del square psi equal to 0 right. This is what we learn from our 2D experience 2D Cartesian. Now let us come to the problem of the sphere flow past the sphere. So if you now use a coordinate system which is radial r and angular coordinate theta then instead of u and v you could use the components vr and vtheta but because of the change in the coordinate system then this del square operator will not work. So instead of del square we will use a different operator which is called as e square operator. We will come into that. So just try to understand the analogy of the concept. The operator is simply different because we have to adapt to a different coordinate system but concept wise it is very similar. So e square of e square psi will be 0 and this psi is not the normal stream function in the Cartesian system but the stream function defined for this coordinate system which is also called as stoke stream function. With this background the definition of stream function will be different but that will satisfy why it has to be different because that has to satisfy the version of the continuity equation for incompressible flow in the rtheta coordinate system not the Cartesian system. Now next what we will do is we will try to assess that what will be the velocity field far away from the sphere. See we intuitively expect that far away from the sphere the effect of the sphere will not be important. So the velocity field if this is u infinity that will again become u infinity far away from the sphere in the downstream. So if you have this as vr then this will be the vtheta this is the resultant which is u infinity. This angle is theta and this angle is 90 degree. So technically here at r tends to infinity where vr is equal to u infinity cos theta and vtheta is equal to minus u infinity sin theta. Why minus? Because positive theta direction is opposite to this vtheta direction right it will see this vector triangle. So now the next step is we will write this vr and vtheta in terms of the stroke stream function. And then we will solve the stream function we will find out the stream function at r tends to infinity. So we need to now express vr and vtheta in terms of the psi. These expressions what we are going to write for the rtheta coordinate system are quite involved. I mean it is very difficult to remember this I never try to remember this. I mean it is up to your wish you may try to remember this but that will not give you any credit at least. If you are thinking of getting Nobel prize you will not get Nobel prize by remembering rtheta coordinate system equations. So I mean we will just write the standard form of the equation vr and vtheta as a function of or in terms of psi the stroke stream function and theta and then we will proceed. So I am not writing from my mind I am writing whatever is there given in the slide I am just copying from there. So do not think that this is to be written in written by memorizing it is not necessary. If these things are to be written or these things are required in some place some examination on somewhere assume that these things will be provided rather you apply your mind for analyzing the problem. So now vr is equal to 1 by r square sin theta del psi del theta. So which is equal to u infinity cos theta at r tends to infinity. So if you integrate this so psi u is equal to psi is equal to u infinity r square it is sin theta cos theta d theta. So let us say that sin theta equal to some z. So cos theta d theta is dz. So it is integral of z dz that is z square by 2. So that means u infinity r square by 2 sin square theta plus what? See we are partially integrated with respect to theta. So function of r is like a constant. So plus a function of r remember this is at r tends to infinity okay. We can write a similar expression for v theta is equal to 1 by r sin theta del psi del r which is equal to minus u infinity. So this is minus sin is there minus 1 by r sin theta del psi del r this is equal to minus u infinity sin theta at r tends to infinity. So if you integrate this now with respect to r this is u infinity integral r dr is r square by 2. So r square by 2 sin square theta plus some function of theta at r tends to infinity. Now look at these 2 expressions. If you compare now that those must be the same whether you have got it from the expression of vr or the expression for v theta psi should not be different. So that means you can assign f1r or equal to f2 theta equal to some arbitrary constant and that constant for a reference case you can choose as 0. So then you can write psi is equal to u infinity r square by 2 sin square theta at r tends to infinity okay. Now this is at r tends to infinity when r is not tending to infinity but r is a location which is somewhat closer to the surface of the sphere then what will what is expected to change r dependence or theta dependence. R dependence is expected to change as you bring the situation from a far field to the near field because that has not created a change in theta that has created only a change in r. So for a general case therefore we can write psi is equal to some function of r into sin square theta. This is the solution that we are this is the form of the solution that we are expecting this function of r we have to figure out what it is okay. So to get this function of r we will now use the governing equation e square of e square psi equal to 0. So what is e square the expression for e square this is the expression for e square. So what is e square psi first let us calculate the theta dependent term and then we will calculate e square psi we are calculating this term. So you can write this as sin theta by r square into del del theta into 1 by sin theta del del theta of sin square theta is 2 sin theta cos theta right this becomes del del theta of cos theta becomes minus sin theta. So this becomes minus 2 f sin theta sin square theta by r square. So what is e square f double dash see del square del r square of this particular term will become f double dash sin square theta. So sin square theta is common f double dash minus 2 f by r square sin square theta e square psi yes. So e square psi we can write f double dash minus 2 f by r square sin square theta this is say g of r into sin square theta then we have e square of e square psi is equal to 0. So another e square of this that means g double dash minus 2 g by r square that must be equal to 0 because sin theta is not equal to 0 in general. This is from analogy if this was f f has become f double dash minus 2 f by r square g will become g double dash minus 2 g by r square with the same operation. Now one way of getting a solution of this equation is assuming that f is equal to some c r to the power n this is the trial solution for equations of this particular form that is a standard way of getting the solution that is you assume f is equal to c r to the power n. So f dash is equal to c n r to the power n minus 1 f double dash is equal to c n into n minus 1 r to the power n minus 2. So f double dash minus 2 f by r square what it is f double dash minus 2 f by r square what it is so this is c n into n minus 1 minus c n into n minus 1 minus 2. So this is n square minus n minus 2 so that is what n minus 1 sorry n plus 1 into n minus 2 n square minus n minus 2 right. So for f double dash minus 2 f by r square we have got this particular term. So now what is g? g is this one so this so g is equal to let us write it here maybe somewhere here. So g r sorry yes g r is equal to what f double dash minus 2 f by r square is equal to n plus 1 into n minus 2 into r to the power n minus 2 there is a c then you have g double dash minus 2 g by r square that is equal to 0. What is the difference between the form of the f and g? Now c is replaced by c into n plus 1 into n minus 2 and r to the power n is replaced by r to the power n minus 2. So the same thing will come another 2 terms with n replaced by n minus 2 right. So eventually this will lead to n plus 1 into n minus 2 replace n with n minus 2 so n minus 1 into n minus 4 equal to 0. So you will get 4 roots for n so n is equal to 1, 2, 4 minus 1. So if that be the case then what is your solution? Let us write the solution here is equal to c 1 r plus c 2 r square plus c 3 r to the power 4 plus c 4 by r. Each n is see the solution that we assumed is c r to the power n. So there are 4 n's therefore 4 c's. So now to get the stream function what we require? We essentially require to obtain the values of c 1 c 2 c 3 c 4 so we require boundary conditions. Now what are the boundary conditions? So boundary one boundary two boundary conditions you are obtaining on the surface of the sphere. What are the boundary conditions on the surface of the sphere? No slip and no penetration. So the boundary conditions number 1 r is equal to capital R let us say d is equal to 2 r so capital R is the radius of the sphere. So at small r is equal to capital R no penetration how will you write? vr must be 0. So instead of the rth theta system if it was xy system how would you express that? You would express that as v equal to 0. v is equal to what? So the question is how do you express this in terms of the stream function? Let us discuss in a little bit different way so that you understand it better. Let us say you have a flat plate forget about a sphere. So what will be a streamline which is representing the solid surface? Is it a streamline? First of all is the solid surface a streamline? It is always a streamline because the definition of a streamline is there is no flow across the streamline. So the solid boundary there is no flow across the solid boundary so by definition it is a streamline. So streamlines are lines with constant stream function. Therefore a solid boundary the no penetration boundary condition in terms of the stream function is equivalent to a boundary condition with psi equal to constant. What is the value of the constant? Value of the constant is arbitrary because see your definition u and v depends on the gradients of psi. So basically difference of psi is important not the absolute value of psi. So this constant can be arbitrary but for calculation convenience we take that constant to be 0 for the solid boundary. So at r is equal to capital R you must have psi equal to 0 and psi equal to 0 means f equal to 0. Then this is no penetration. What about no slip? No slip means what? Which component of velocity is 0? v theta. v theta how do you write? Del psi del r just like u is equal to del psi del y instead of y it is r. So del psi del r is equal to 0 that means what? df dr equal to 0 at r tends to infinity. What is f? This we have already calculated. What was f? u infinity r square by 2 that means what can we say from here? One of the constants we can. So c2 is what is c2? u infinity by 2 because this form should be satisfied even at r tends to infinity. Not only that we can make an additional observation. The additional observation is c3 must be equal to 0 because the highest order dependence is r square dependence. You cannot have a dependence with r of order beyond that. The infinity condition gives the highest order dependence. So you cannot have any order dependence beyond that. So you have c3 equal to 0. So see the situation is a little bit tricky. You have 3 boundary conditions and using that you are calculating 4 constants. Appears to be something which is not that straight forward. So that needs to be critically discussed. So with these conditions now it is possible that you can substitute these boundary conditions and calculate c1, c2, c3, c4. That algebra is quite trivial. I will not work out that instead we will go to the slides to see that what are the constants that you will get from there. So if you look into this slide, no penetration, no slip will give 2 conditions. Then at r tends to infinity, so the problem is here different notations are used for c1, c2, c3, c4 I think possibly. Let me see. No penetration, it is same. Okay. Same notations are there. Then it is fine. So whatever notations I have used in the board, same c1, c2, c3, c4. So we can use these slides directly. So at r tends to infinity, fr is u infinity r square by 2 that will give you c2 is equal to u infinity by 2 and c3 is equal to 0 that we have already discussed why c3 has to be 0. So if you calculate now c1, c2, c3, c4, so basically from the condition at r tends to infinity you are straight away getting 2 constants. So the remaining 2 constants you can calculate from the no slip and the no penetration boundary condition that will give you c1 is equal to minus 3 by 4 u infinity r and c4 is equal to u infinity r cube by 4. So you get psi equal to fr into sin square theta. So once you get psi, you get everything because once you get psi, you get the velocity field. Velocity field is expressed in terms of psi. Now you can also calculate the pressure field. So you have cut your head. Now you somehow put your head back on the top of your body. So earlier pressure was eliminated. Now what you do is that you refer to the momentum equation where now the velocity field is known. You substitute the velocity field and integrate the momentum equation to get the pressure field. Not only that you can use the velocity field to get the shear stress and the normal stress. So getting the shear stress, normal stress and the hydrostatic stress in terms of pressure you can get the distribution of stress on the surface of the sphere and that will help us to calculate the drag force. So we will take up that in the next lecture. Thank you very much.