 Last time we were discussing about some examples of unsteady flows and in one example we were considering and we have just started with that example is that there is a oscillating boundary if you look into this slide it summarizes the problem that you have a solid boundary the solid boundary oscillates with a velocity u0 sin omega t. So it is a sinusoidal oscillation in this case and we have to figure out that how the velocity varies with position and time in this particular scenario. So to understand that we started with our governing differential equation and the governing differential equation for this particular scenario summarizes to this one this del u del t is equal to nu del square u del y square and we started identifying various scales pertinent to this problem. So we will just have a brief recapitulation to the various scales maybe let us go to the board and try to revisit the various scales for this problem. So you have the oscillating boundary this one and you have your governing differential equation this is your y axis and you have the governing differential equation. So let us write the various scales so this scale this is of the order of u0 by the reference time scale and because here there is a frequency of oscillation the choice of the reference time scale is 1 by omega for this particular term we are considering y ref as the reference scale for the scale along y we are considering that as y ref. We have to keep in mind that there is no natural length scale along y for this particular problem because there is no confinement but the length scale will come from the physical parameters. So if you equate the scales of these 2 terms then you are left with y reference is of the order of square root of nu by omega. So this becomes a physical length scale along y so what does it indicate? It is that when the wall is oscillating now how does the effect of oscillation of the wall propagate to the fluid the effect of oscillation of the wall propagates to the fluid by virtue of the fluid viscosity. So and to be precise we have seen that in these particular cases kinematic viscosity is the governing parameter rather than the viscosity itself. So the penetration depth up to which this effect of oscillation is failed scales with square root of nu by omega. So when the plate oscillates with a velocity angular velocity omega then or a frequency omega then the effect of that frequency is failed over a distance which scales as square root of nu by omega. So with this particular scale so you can write y bar is equal to y by y ref u bar is equal to u by u ref. So if you keep these scales then you are left with when t bar is omega t because of the harmonic nature of the solution we can assume a solution of this form. So this will have a sine and cosine component of time because e to the power i t is cos t plus i sine t. So because it will have a sine or cosine it will be a in effect it will be a total net sine or cosine with the same frequency but with a phase shift. So there will be a shift in phase and there also is likely to be an attenuation in the amplitude. So the net solution the resultant velocity will be a somewhat representative of the signature of this type of variation but with certain modifications. Modifications there will be some decay in the amplitude expectedly because of the dampening effect of the fluid there should be some decay in the oscillation as you go along y and there should be a phase shift from this sinusoidal variable from this sine omega t there should be a phase shift. So all those are nicely reflected if you take the solution of this form e to the power i t f y the f y will take care of the amplitude as a function of y and e to the power i t will take care of the phase and imaginary the reason is pretty clear that at the boundary we want to satisfy the boundary condition and at the boundary you have sine omega t is the imaginary part of e to the power i omega t. So that is why the imaginary here. So this sums up what we have discussed in the previous lecture and we will take it up from there. Now so let us substitute now the solutions. So just to make the writing easy we will not write imaginary all the time but we will assume that we are taking the imaginary component of the solution. So del u del t is i e to the power i t f y. So now if you substitute these in this equation what you eventually get f double dash minus i f is equal to 0 right from this equation you substitute these 2. So the solution is f is equal to c 1 e to the power root over i y bar plus c 2 e to the power minus square root of i y bar. To get c 1 and c 2 you require the 2 boundary conditions. Boundary condition number 1 at y is equal to 0 what is f. So at y is equal to 0 u is imaginary to the power i t non-dimensional u right. So f is equal to 1 and at y tends to infinity far away from the plate there is no velocity because far away from the plate the fluid does not feel the effect of the plate. So at y tends to infinity f will tend to 0. So if you consider that at y tends to infinity f tends to 0 then what will be the value of c 1. So boundary condition 2 will tell that c 1 is equal to 0 because the solution has to be bounded and boundary condition number 1 at y is equal to 0 f is equal to 1. So c 2 is equal to 1. So that means your solution is f is equal to e to the power minus square root of i y bar and u bar is equal to imaginary e to the power i t e to the power minus square root of i y bar. See so far as fluid mechanics is concerned this solution is good enough but to get a little bit of more insight we can do some algebraic manipulation over this expression that is pure algebra and it has nothing to do with great details of fluid mechanics. From now onwards we will not be doing all the details all the time because we have covered a few examples which tell you that how to take care of those manipulations but maybe for one or two more occasions including this one we will try to see that what is the final form of the solution simplified form because that keeps certain physical insight. So let us try to go for one or two more steps of simplification again I am repeating those simplifications are not necessary for plotting the solution but that can give you certain insight. You see that e to the power i theta you can write cos theta plus i sin theta you can substitute theta equal to pi by 2. So e to the power i pi by 2 is equal to cos pi by 2 plus i sin pi by 2 that is nothing but equal to i. So square root of i is equal to e to the power i pi by 4 right. So that is cos pi by 4 plus i sin pi by 4 both cos pi by 4 and sin pi by 4 are 1 by root 2. So this is 1 plus i by root 2. So you can write this one as u bar is equal to imaginary e to the power i t into e to the power minus 1 plus i by root 2 y bar. So this is imaginary e to the power. So first take the real component minus y by root 2 right into e to the power i t minus y by root 2 right. So now out of these 2 terms see this has no role with the imaginary or the real component of the complex number e to the power i theta. So this you can just write as an amplitude of the solution. So u bar e to the power minus y by root 2 into imaginary of basically e to the power i t minus y by root 2. So sin. So you can see here this is the final form of the solution which gives a good amount of physical insight. See as you have this oscillation this oscillation forces the fluid to oscillate with what frequency? With the same frequency that you can see that here this t is basically omega t. So it is the same omega but with a phase shift that phase shift is this okay. And what is the amplitude of oscillation? If there was no damping effect then the amplitude of oscillation would have been the same that is u0. So in that case u this would have been 1 but because of the damping effect which varies exponentially with y you can see that the amplitude now becomes this is y bar basically e to the power minus y bar by root 2. So it is a situation like a damped oscillation. So the fluid in this case behaves like a damped oscillator okay. Now having done this particular case we will now like to ask again a question which is very much pertinent to microfluidics. Well we have a solid boundary that oscillates but you have an infinite dimension along the y direction. So that the fluid is unbounded but what happens if the fluid is bounded? That is what is important for microfluidics considerations. So instead of a channel like this instead of a single plate like this if you now consider a channel where the fluid is bounded and the fluid is being oscillated then what is the response of the fluid against that oscillation that is what we are going to study next. So the motivation of studying the Stokes second problem in the context of microfluidics is not that it is really a microfluidics problem but it gives a basis which is a case of an unbounded flow so that we can consider the case of a confinement or a bounded flow with this particular case to see that what is the explicit role of confinement towards driving the solution for the velocity profile. With that motivation we will consider the next problem. So let us say that we have a channel like this. Now to make the problem a little bit more realistic and a little bit more close to a biophysical scenario which is important for microfluidics applications we will not consider the plate is oscillating but we will rather consider that there is a oscillatory pressure gradient that a pressure gradient that oscillates sinusoidally with time. So let us say that this is a channel we have minus dp dx is equal to a0 plus a1 the previous one we took as sin just for a change we will make it cos we could keep this as also as sin as well without any problem okay. So we have a steady part of the pressure gradient and we have a time oscillatory part of the pressure gradient. Let us set up some axis coordinate axis so let this be the y axis let this be the x axis and let half height of the channel be h. So the physical setup of the problem is pretty clear that you have a driving pressure gradient which is oscillatory with time you want to find out the velocity field within the channel as a function of position and time okay. Assuming that other assumptions that we made for the previous problems previous series of problems that we have been solving all those assumptions are valid. So we will not get into the details of those assumptions anymore we will try to see that what are the different aspects of the velocity profile that you can get from this. Now you can see that you have pressure gradient which has decoupled steady and time dependent part they are not really coupled. So that means we can exploit the linearity of the governing differential equation and obtain solutions for the velocity profile once steady solution based on this and one unsteady solution based on this and the net solution is a linear superposition of the 2 solutions okay. So let us say that u equal to u1 corresponds to the steady solution corresponding to a0 that is you apply a steady dpdx and you get the corresponding solution as a steady velocity profile okay. So that is a steady solution corresponding to a0 and this we have already worked out quite some time back when we first started with flow through a parallel plate channel fully developed flow through parallel plate channel. Now what we will do is we will consider the unsteady part and let us say u equal to u2 is a unsteady solution corresponding to a1 cos omega t. So now what we will do is we will write the governing equation for u2 or first let us write the governing equation for u1 at least we have done the solution but just to recapitulate. So what is the governing equation for u1? So this will become right. So what are the boundary conditions? What is the boundary condition at y equal to 0? At y equal to 0 you have maximum velocity so you have du dy equal to 0 du1 dy in this case and at y equal to h you have u1 equal to 0 that is the no slip boundary condition. So with these 2 boundary conditions you can straight forward get the solution of this equation okay. Now so that is the solution for u1. We will pay more attention on how to get the solution for u2 and the net solution is a linear superposition of the 2 solutions. So solution for u2 okay and boundary conditions same as the previous. So the resultant solution which is u1 plus u2 if you consider u equal to u1 plus u2 that will be governed by the equation rho del u del t is equal to minus del p del x which is a0 plus a1 cos omega t plus mu del 2 u del y2 okay. So now we will devote our attention to analyze the various scales of this problem. See understanding the scales is very important because it gives us a physical insight of the phenomenon that is taking place as we are going for the solution. Otherwise it is just a mathematical exercise but understanding the scales gives a physical picture that what is the range in which the velocity varies, what is the depth of penetration up to which this variation of velocity is failed all these things we can decide a priori. So to understand that we divide first both sides by rho. So this becomes a1 by rho and this becomes the kinematic viscosity we just give a new name this as capital A which is a1 by rho. So that now we have 3 terms in the governing equation. What are these 3 terms? You have an unsteady term, you have a forcing term and you have a term which is the viscous term. So let us study the various scales. What is t reference? The time scale. See for this problem to be important that means for this oscillatory component of the pressure gradient to be important the time scale should be 1 by omega. Otherwise you can solve this problem but this component is not important. If this is important it must have emphasis of its own role that means this unsteady part of the pressure gradient should influence the time scale of the problem. So that means t reference is 1 by omega. What is y reference? Now you have a confinement. If you want your confinement to be important if you want it to be important now that is our wish that the confinement effect is important. We will see which parameter and then decides that whether the confinement effect is important or not. But if our wish is fulfilled that means if the confinement effect is the governing parameter then y reference has to be h. What is u reference? You really do not know because you are forcing the system to oscillate not with a natural velocity scale but with a natural pressure gradient scale. So you have to figure out the corresponding velocity scale. So to do that now you can non-dimensionalize it. So this is u reference omega, this is a cos t. The bars are indicative of non-dimensional quantities okay. Now let us divide both sides by capital A. Now tell me out of these 3 terms what are the terms which you expect to be always important in terms of priority? In terms of priority if you are interested to see what is the influence of an oscillatory pressure gradient then the oscillatory component of the pressure gradient has to be important. So this is priority number 1 right. Why because there are several terms we have to prioritize their relative importance. This is priority number 1 because if this is of no priority then the problem that we are solving is a different problem where the oscillation is not important. So if we are interested about solving a problem where the oscillation is important then this has to be a priority. So this is like priority number 1. What is the second priority? The second priority is the unsteady part because if you do not have an unsteady component of the solution then the unsteadiness of the problem is not important but here one of the focus is to figure out the unsteadiness of the solution. So this is like priority 2 towards dictating this unsteadiness whether the viscous term will be whether viscous effects are important or not that will be dictated by this term. That may be important even may not be important we do not know we have to figure out. Now what is the order of this term? This is of the order of 1 right. So if this term has to compete with this term then this term also has to be of the order of 1 right. If this term has to compete with this term this is of the order of 1. So if this term has to be equally competitive then this also should be of the order of 1. Now this particular term is constrained to be of the order of 1 by virtue of non-dimensionalization because you have non-dimensionalized U2 with respect to its reference value so that that is U2 by U reference is of the order of 1. You have non-dimensionalized the time by t by t reference that is also of the order of 1. So this derivative is constrained to be of the order of 1. Therefore you have omega into U reference by A is of the order of 1. The derivative is of the order of 1 because you have chosen non-dimensional U as U by U reference which is of the order of 1 because U is of the order of U reference and non-dimensional t by t reference t is of the order of t reference so t by t reference is of the order of 1. So it is if you write delta del U2 by del t is of the order of some delta U2 by some delta t then this is of the order of 1 and this is of the order of 1 okay. So omega U reference by A is of the order of 1 that means U reference is of the order of A by omega. So this gives us a nice physics that if you have a driving pressure gradient which has an oscillatory component how can you get the corresponding velocity scale? Even for engineering this is important because say you may have a pump that pumps the flow through an oscillatory manner by by an oscillatory manner. Now as an engineer you may be interested that well this is the oscillatory pumping power. Now what is the velocity that I will typically get from that? So from the amplitude of the velocity amplitude of the pressure gradient which is A1 by rho which is A so you can get the corresponding U reference. This is the order of magnitude of the velocity. And at least without getting into very deep and involved calculations you can get an order of magnitude of the velocity that you can get. Now what about the viscous term? That is very interesting so this particular term. So in this term so U reference is A by omega. So nu A by omega by AH square. So A gets cancelled. So you can see a very interesting thing. The relative influence of the viscous term does not depend on the amplitude of the pressure oscillation. Does not matter what is the amplitude of the pressure oscillation? It is large, it is small, medium whatever that does not influence the relative influence of this term. So what are the parameters on which it depends? So let us call this as 1 by lambda square. Just where lambda is the parameter for this problem. So what is lambda? So lambda square is equal to omega H square by nu. So H square by nu by omega. See first of all because all the coefficients are non-dimensional. This also has to be non-dimensional. So nu by omega has to be square of some length scale. What is that length scale? That is the length scale that we got from the Stokes' second problem. This is if you just look into your notes for the Stokes' second problem you will see that this is y reference square for Stokes' second problem. So you can see that the relative importance of the viscous term depends on the relative extent of the confinement with respect to the Stokes' penetration depth. That is this lambda is an indicative of this is H by y reference Stokes 2 square. H is much much less than y reference Stokes 2. Then what happens? y reference Stokes 2 is the penetration depth and if H is much much less than y reference Stokes 2 then lambda square is very small. This term is 1 by lambda square. Then the effect of this term is amplified. So if you are coming into a confinement where the confinement effect is strong that is H is small. If H is small with respect to the Stokes' of course see this all these things are important try to analyze critically. Sometimes we say that the confinement is small. Question is in the world there is nothing which is absolutely large or absolutely small. So if we say that the confinement is small we have to compare. We have to say small with respect to what? Like 1 millimeter is small with respect to 1 kilometer but is very large as compared to atomic dimensions. So smallness or largeness has to be compared with respect to a basis. So the confinement whether it is a very narrow confinement or the confinement effect is not important what is the parameter that will decide? The parameter that will decide is the length scale of the Stokes' second problem and that is nu by omega. So that depends on the driving frequency and the kinematic viscosity. Even if you take the kinematic viscosity let us say that the fluid is water. So say that that property is fixed. So mu is roughly 10 to the power minus 3 Pascal's second and rho is 10 to the power 3 kg per meter cube. So kinematic viscosity is roughly 10 to the power minus 6 meter square per second. There is a like rough calculation for the kinematic viscosity of water. Even if you keep that as fixed but omega whether this is hertz kilohertz megahertz depending on that you have to figure out that what is this length scale and take the square root of that and then compare that length scale with the confinement. So that will decide if that is small then the resultant term here is 1 by lambda square. So then that will be amplified. So that governs the physics of the problem. See I will work out the solution and we can do that pretty quickly but always it is important to gain a physical insight while working out the solution rather than taking it just as a mathematical exercise. So let us write the equation. This is our governing differential equation and the important parameter is lambda. So just like the previous case what will be the nature of the solution for getting a solution a separable solution of time and y. So because the forcing component is cos omega t so we will take the real of e to the power i t not imaginary this time because e to the power i theta real component is cos theta. So real e to the power i t into f y. You have to be little bit careful in writing. I mean I understand it is not so easy to write a consistent symbol in the board always but important to give it a distinctive symbol because re in fluid mechanics is also Reynolds number. So just to make sure that you do not jumble up with the Reynolds number because sometimes for working out equations non-dimensionally Reynolds number will automatically come. So you have to be careful that you should not mix up the Reynolds number with the real. So this is not Reynolds number. This is the real component of this one. So I mean if you are very confused just write re al instead of Reynolds number to make sure that you are meaning the real component. So again we are not going to write the real all the time. So e to the power i e to the power i t f of course it will be real but I am not writing again and again real and so this equation gives i e to the power i t f is equal to what we will write for this e to the power i t because it is real of both sides. Real of e to the power i t is cos. So this equation is obtained by taking by doing all these things and taking real of both sides. Real components of both sides. So plus 1 by lambda square e to the power i t so that means you have i f is equal to 1 i f is equal to 1 plus f double dash by lambda square. Just to ease the algebra we consider a new variable f bar which is f plus i. So that these 2 terms will combine together give one single term. So you can write i f is equal to i f bar minus i square plus i square. No we have taken it on the other side so it has become minus okay. i f is equal to i f bar minus i square and i square is minus 1. So this is i i f is equal to or i f is equal to i f bar plus 1. So that means i f minus 1 is equal to i f bar right. Not only that because it is just an addition of constant f double dash is same as f bar double dash. So you can write i f minus 1 as i f bar is equal to f bar double dash by lambda square. So f bar double dash minus lambda square i f bar is equal to 0. So you can write the solution in terms of exponential or just for little bit more algebraic convenience we will write it in terms of sin h and cos h all the same. So f bar is equal to c 1 cos h lambda square root of i y bar plus c 2 sin h lambda square root of i y bar. Only one thing remains that is to get the values of c 1 and c 2. So for that we will be using the boundary conditions. So at y is equal to 0 the boundary condition. At y bar equal to 0 what is the boundary condition? del u to del y equal to 0 that means f dash is equal to 0 right. f is velocity so f dash is velocity gradient and that means f bar dash is 0. So if f bar dash is 0 which constant is 0? See when you differentiate this cos h will become sin h and sin h will become cos h this term will be automatically 0. So c 2 will be 0 and at y bar equal to 1 y bar equal to 1 means at the wall what is f? f is equal to 0 no sleep boundary condition u is 0 this is u 2 is equal to 0 that means f equal to 0. Our solution does not involve f bar it involves f bar. So f bar is equal to what? f bar is equal to i. So you have i is equal to c 1 cos h lambda square root of i. So that will give you what is c 1 right and that completes the solution of this problem. So what I will do is that I will utilize the slides to sum up the solutions for the stokes second problem and this particular problem and I will advise you to plot the stokes second problem solution as compared to this problem for different values of lambda and see that for which values of lambda this problem essentially boils down to a problem which has solutions similar to the solution of the stokes second problem but for what values of lambda it does not okay and because these are non-dimensional parameters you do not have to work with a large number of physical parameters non-dimensional parameters will do. So let us go to the slides. So this is the stokes second problem I will straight away go to the solution I will not like go through the algebra of the solution but I will just show you a graphical representation of the solution of the stokes second problem. So you can see this graph dimensionless u versus dimensionless y for different capital t is the capital t which is plotted here is omega into t or small t bar okay. So you can see that for different values of capital t you are having different u versus y solution. Okay and like these are very important and this eventually decides that like this solutions we could obtain at other times but these are just certain representative values of time dimensionless time okay. So the dimensionless times typical dimensionless times which are plotted are say pi by 4 pi by 2 3 pi by 4 and all this I will advise you to plot at other intermediate dimensionless times and figure out that how the solution is progressing as a combined spatiotemporal characteristic of position and time confined flow with an oscillatory pressure gradient. So just the recapitulation that you have u equal to u1 as the solution for the first problem and which is the steady part of the problem and for the second problem you have various terms and relative importance of these various terms we have discussed and the final solution you will get in terms of the cos hyperbolic and sin hyperbolic and then the final solution you can get in this form. So you can see the cos hyperbolic form along with this e to the power it that is there. So basically I mean you can do one more step of or a few more steps of algebraic simplification to extract the real component out of this expression. But you know if you are just trying to make a plot of this using some plotting software then you can directly use the real or imaginary component like if you are doing the using the matlab for plotting this solution then you can directly use this form to plot. So always the final most simplified form of the algebraic expression may not be necessary but sometimes it is important to derive that to gain some physical insight into the problem as we did for the Stokes second problem. So to summarize what we have discussed today and over the previous few lectures we have mainly discussed about some solutions to the Stokes equation I mean Navier Stokes equation but with negligible advection terms which makes it boil down to the Stokes equation and some solutions to that for both steady and unsteady flow problems. In the next lecture we will take it up further forward from that thank you very much.