 In our previous lecture, we were discussing about the some of the important aspects of load Reynolds number hydrodynamics and we started with analyzing the continuity equation. So to summarize what we observed during our analysis, let us look into the slides. So we made an assumption that the flow is 2 dimensional and constant density flow and the continuity equation revealed a relationship between the velocity scales with the corresponding length scales. So we made an important comment that the length scale if the vertical direction is the y direction then the length scale along y is h but the length scale along x, it may be the total length of the channel or it may be even less than that depending on whether there are characteristic changes taking place over shorter distances. Like for example, if the channel surface is patterned with a particular weightability then it is possible that x reference, the reference length scale along x is much much less than the total axial span of the channel. With this background, let us now move on to the board to discuss about the corresponding implications for the momentum equation. So let us write, so let us consider the physical problem in hand. So there is a channel like this. You have x axis lead in this direction, y axis lead in this direction. So x momentum equation del u del t plus u del u del x plus v del u del y with a rho is equal to minus del p del x plus mu. As obvious I am not repeating again that we are making an assumption that the flow is incompressible. So that you know the additional term, source term due to the compressibility effect is not appearing here. Now let us make an assumption that it is a steady flow. We will discuss in details about some unsteady flows. So first let us consider that it is a steady flow. So example, steady flow. So the steady flow will have del u del t equal to 0. Then we consider the orders of magnitude of various other terms. So this term is not there for steady flow but let us consider the other 2 terms. So rho u reference square by x reference that is the order of magnitude of this term. That is the order of magnitude of this term rho v reference into u reference by y reference. Now v reference is related to u reference through the continuity equation. So let us write it once more the continuity equation. Our analysis revealed that this is of the order of u reference by x reference. I am not committing whether x reference is equal to the length of the channel or not because we discussed that it depends on the physical situation on hand. And this is of the order of v reference by y reference. So v reference is of the order of u reference into y reference by x reference. So that means this term is also of the order of rho u reference square by x reference. So if this term is important, this term is also equally important until and unless v is identically 0 and this is something which is not very intuitive because intuition says that v is much much less than u if y reference is much much less than x reference. Despite that v del u del y is of comparable order as that of u del u del x because although v is much much less than u, if x y reference is much much less than x reference, still del u del y may be much much greater than del u del x. The gradient in the y direction is likely to be much sharper or steeper as compared to the gradient in the x direction. That makes up for the fact that v may be much much less than u. So the product u del u del x and v del u del y, they are coming out to be of comparable order. Let us now look into the viscous term, the pressure gradient term we will consider separately. Let us consider the viscous term. So the viscous term, what is the order of this term? Mu u reference by x reference square and this term is of the order of mu u reference by y reference square. If y reference is much much less than x reference, again I am telling that this is not always true but this is true in many cases when there is no pattern of slip or pattern of weightability along the axial direction of the channel, then you can say safely that y reference is much much less than x reference. So if y reference is much much less than x reference, then this term will dominate over the other. So you can see then then this term is going to be the deciding factor as compared to the other term. Look into the similarity between this and the boundary layer theory. In boundary layer theory, you have the boundary layer thickness delta much much less than L. So there the delta is a naturally evolving length scale. Here the y reference may be governed by the confinement height of the channel. The micro channel confinement can decide or can dictate what is y reference. But although scale wise the treatments are very similar, you have to understand very carefully that the boundary layer theory is valid for high Reynolds number whereas here we are doing a low Reynolds number hydrodynamics. So physics is completely different. Although scale wise you can see that there is some analogy. This has to be remembered very carefully that this analysis is not what we intend to do for boundary layer theory. Although scale wise they are happening to be similar. So now let us write the ratio of inertia by viscous. So let us take an example with y reference equal to h, x reference equal to L which is much much greater than h. Let us say this is an example that we take. So let us write what is the ratio of inertia force by viscous force. Inertia force by viscous force. So what is the inertia force? The left hand side is the inertia force. So inertia force is of the order of rho u reference square by L. Now we have taken an example with x reference equal to L and the viscous force mu u reference by h square. I am sorry that this is not visible in the board. I will write it at a separate place. So write it here. Inertia force by viscous force is of the order of rho u reference square by L by mu u reference by h square. So this is what? Rho u reference h by mu into h by L. So if h is the characteristic length scale of the channel, this is the Reynolds number based on h. So you can see that the ratio of inertia force by viscous force is the Reynolds number into h by L. So one very important and interesting observation is that if the Reynolds number is small, h by L we have already committed as small, then the inertia force can be neglected as compared to the viscous force. So the left hand side of the Navier-Stokes equation becomes irrelevant. So then that equation is called as Stokes equation. Stokes equation is Navier-Stokes equation with the inertial terms equal to 0. I mean approximately 0, not identically equal to 0 but approximately equal to 0 or neglected that is called as Stokes equation. So then when you have the Reynolds number relatively high, you can see that then also it may be possible that inertia force is negligible as compared to viscous force. Because the Reynolds number may be relatively high not too high but h by L is small. So the product of Reynolds number and h by L is important. This is the very important take home message that it is not the Reynolds number that we are bothered just about but it is the product of Reynolds number and h by L. So even if the Reynolds number is not too small but the product of Reynolds number and h by L that is small then this effect is that the inertia force may be negligible as compared to the viscous forces. So then what is our equation that we are left with? We are left with the Stokes equation. Let me write that. Of course it can include other body forces but here we are discussing mainly pressure driven flow so I have not included any other body force. So you have 0 approximately equal to, I have purposefully written both of the viscous terms because it might so happen that they are equally strong but still the left hand side can be omitted if the inertial force is negligible as compared to the viscous force. So for the example that we have talked the special form of the Stokes equation boils down to because this term is negligible for that example. What about the y momentum equation? What about the y momentum equation? For the y momentum equation you can make similar analysis and write, okay. Similar just replace u with v. Now the question is out of these 2 terms which term is more dominating? So to do that let us write what is the order of del square v del y square? What is that? This is of the order of v reference by h square and order of del square u del y square is of the order of u reference by h square, v reference is much much less than u reference which is true if y reference is much much less than x reference then this term is much much less than del square u del y square this is u. So now apply a logic you have these 2 equations. Now in this equation the pressure gradient term is being balanced by the viscous term. In this equation also the pressure gradient term is being balanced by the viscous term but the viscous term in the second equation is much much less than the viscous term in the first equation therefore the pressure gradient term in the second equation must be much much less than pressure gradient term in the first equation. So this analysis implicates that del p del y is much much less than del p del x if that is so if that is so then you can write del p del x approximately as dp dx okay I am writing it again because this part of the board is not visible. So del p del x is approximately equal to dp dx. So then we are left with this equation where this can be replaced by dp dx approximately okay. So in low Reynolds number hydrodynamics we get an equation if y reference is much much less than x reference we get an equation which remarkably resembles with fully developed flow that is why we discussed with fully developed flow to begin with. So now the question is where is the similarity and where is the dissimilarity. So let us write the equation. So we will divide the board into 2 parts in one part we will write the low Reynolds number equation on another side we will write the fully developed flow equation that will help us to compare. So we assume that the physical problem on hand is flow through a channel 2 dimensional situation. So fully developed flow we had del u del x is equal to 0 and 0 is equal to minus dp dx plus mu d2u dy2 right low Reynolds number this is fully developed flow this is low Reynolds number flow low Reynolds number flow does not require del u del x equal to 0 right. So low Reynolds number still you will have del u del x plus del v del y equal to 0. So in fully developed flow if there are no holes in the walls then that will mean v equal to 0 identically but low Reynolds number flow does not require v is identically equal to 0 right. So v may be small as compared to u if length scale along y is much small as compared to length scale along x but if the length scales are comparable u and v may be comparable. On the other hand when you have no holes on the wall of the channel then for fully developed flow v is identically equal to 0 no question of any approximation much much less or close to this kind of approximation nothing is required. Not only that the x momentum equation is approximately equal to right although while solving problems we will not write always approximately equal to we will assume that is equal to just in the casual spirit of solving the problem. But fundamentally it is approximately equal to because the inertial term is negligible as compared to the viscous term but the inertial term is not identically equal to 0 for fully developed flow the inertial term is identically equal to 0. Not only that you see that this in place of del p del x we are written dp dx here for fully developed flow that is an exact behaviour that is an exact condition. On the other hand here in place of del p del x we are written dp dx that is an approximate consideration by comparing the terms in the x momentum with the terms in the y momentum equation. Finally here you can see that these are partial derivatives whereas this is ordinary derivative why these are partial derivatives because now u is a function of both x and y because u del u del x is not equal to 0 identically not identically equal to 0. So u is a function of both x and y so this is a partial derivative whereas this is an ordinary derivative. Many times however see these are small intricate tidbits but these tidbits are important in the conceptual paradigm but for somebody who is just interested for solving a problem maybe there is no great difference between this equation and this equation. So that is why we have worked out some problems related to fully developed flow that can give you a glimpse of how to solve related problems on low Reynolds number flow need not be necessarily fully developed. So we had made this discussion which is sort of like a comparison between low Reynolds number flow and fully developed flow. Let us now summarize these discussions as we go to the slides. Yes this is dpdx right this is dpdx yes now let us go to the slides. So I mean because of some version problem in the computer what you are seeing here as square box is basically of the order of tilde. So that has not come properly in the version of the power point used in this computer so that but assume that like well tilde has come, tilde has come properly what has not come is much much less. So much much less has not come so y reference much much less than x reference. So this much much less thing has come as a box in the slides so just take that as much much less. So h is much much less than l so v reference is much much less as compared to u reference. So we have performed an order of magnitude analysis of the momentum equation x momentum equation and have found out that the inertial term by viscous term is of the order of Reynolds number into h by l. So there are 2 cases for which the inertial term can be neglected as compared to the viscous term. We have discussed all this but every time what I am trying to do is that after discussing something we are trying to pause for a moment and summarize what we have discussed. So that it becomes easier for you to digest the material. So if Reynolds number is small or h by l is very small, if h by l is very small in that case even if Reynolds number is large small h by l makes the product small. So I will tell you like when you say large small I mean what are the typical scenarios. Let us say h is of the order of micron and l is of the order of millimeter okay. So 10 to the power – 6 by 10 to the power – 3 okay. So 10 to the power – 3 is the ratio of h by l. So even if the Reynolds number is of the order of 100 you can see that the inertial term by viscous term can be 100 into 10 to the power – 3. So inertial term may be 1 order less as compared to the viscous term. So even with 100 Reynolds number say 100 is not a very small Reynolds number but still you can have inertial term much negligible as compared to the viscous term. So to generalize here in the slides we have written the forms of the final forms of the x momentum equation with a body force term Bx which in the board when we were working it out we neglected that body force term. And then in the y momentum equation sorry in the x momentum equation we can we neglected mu del 2u del x2 term as compared to mu del 2u del y2 term because mu u reference by l square is much much less than mu u reference by h square. The similar analysis we did for the y momentum equation and then the y momentum equation has also illed the same form as that of the x momentum equation we have detailed in the board how that is possible and then by comparing the corresponding terms we figured out that del p del y is much much less than del p del x. So that we can write in place of del p del x we can write dp dx. So if we do that then the system of equations for low Reynolds number flow comes out to be del u del x plus del v del y equal to 0 and 0 is approximately equal to minus del p del x which is again equal to dp dx approximately equal to dp dx plus mu del 2u del y2. Finally we discussed about the difference between low Reynolds number hydrodynamics and hydrodynamically fully developed flow. So for low Reynolds number flows the left hand side the inertial terms are approximately equal to 0 that is negligible on the other hand for hydrodynamically fully developed flow the inertial terms are identically equal to 0 no question of any approximation is there. So that not only that if there are no holes in the walls of the channel then for fully developed flow that means that v is identically equal to 0 whereas for low Reynolds number flow v is not identically equal to 0 it may be it may be lower than lower in order as compared to u depending on what is length scale along x and length scale along y. Now let us come to the board again and discuss a little bit about the time scales. So far for order of magnitude analysis we have considered steady flows but let us consider examples of unsteady flows. So when you have unsteady flows then you write the equation let us write the left hand side anyway we begin with the full equation first because that will explain the context. If it is an unsteady problem there is one term in this equation which must be important what is that term the del u del t term right because unsteadyness means the velocity field at a given point is expected to change with time then only the unsteadyness comes. So obviously this term is important now this term is associated with a time scale. So we can write the order of magnitude of this term as rho u reference by t reference question is what is this t reference physically what is that characteristic time scale it is the characteristic time over which characteristic changes in velocity occur within the flow field okay. So now the question is that what is that characteristic time what is that characteristic time over which characteristic changes will take place. So now it depends on this term is always important but out of the remaining terms some of the terms may be dominating over the other. So based on that so you may have a time scale which is based on this scaling of this with this you may have a time scale based on a scaling of this with this that is if these two are comparable that means this is the this is dictating the important physics or you can have a time scale based on this. Let me discuss it in a more elaborate manner. So if you look into the forcing parameters here you have an advection term you have a pressure gradient term and you have a diffusion term let us say that the advection effect is in creating the unsteadiness is much more as compared to other effects. So if the advection effect is much more important then what happens then this term is of the same order of magnitude as this term because other terms are of less order of magnitude. Then the time scale that you get is called as advection based time scale. So what is the order of magnitude of this term rho u reference square by x reference. So advection based time scale when you have rho u reference by t reference is of the order of rho u reference square by x reference. So what is t reference x reference by u reference. So what is this t physically? Physically it is the time required for the perturbation to get advected by a distance x reference. If v is the velocity then the time required by the perturbation to traverse the distance x by advection is simply x by u but that is due to advection only. We next consider so advection time scale. So this is what it is we will discuss about three different time scales one which highlights the contribution of this term the other which we will do now is that highlights the contribution of this term and the third which highlights the contribution of this term assuming that this term is always important. Now what is the correct time scale of a problem depends on the physics of the problem that out of this forcing parameters which one is governing the physics. There may be situations which are little bit tricky situations when you cannot rule out the effect of one as compared to the other and then you have to really like critically discuss that whether the time scale would be advection dominated or viscosity dominated or so to say diffusion dominated time scale. So what is the diffusion dominated time scale or diffusion based time scale? So diffusion based time scale is that the situation when the unsteady term scales with the diffusion term. So what is the order of magnitude of the diffusion term assuming x reference much much greater than y reference so this is of the order of mu u reference by y reference square. So diffusion time scale is governed by rho u reference by t reference is of the order of mu u reference by y reference square. So t reference is of the order of y reference square by mu. Mu is mu by rho the kinematic viscosity. So you can see that the kinematic viscosity for a physical problem is very very important and we should discuss very carefully about its physical implication. See it is a ratio of viscosity by density. So viscosity is what viscosity talks about that there is a disturbance in momentum how fast does it propagate within a fluid. So if there is a wall if there is a solid boundary that creates a disturbance in momentum. So that disturbance in momentum is diffused within the fluid by virtue of the fluid property viscosity. On the other hand fluid has a ability to sustain its momentum. See the fluid which is flowing further away from the wall that is disturbed because of the viscous effect but that also tends to retain its own momentum because of what? Because of its inertia and that inertia is related to the density because it is related to the mass. So the denominator rho represents its ability to sustain its momentum whereas the numerator mu represents the ability to create a disturbance in the momentum. So it is basically the sustenance of the disturbance in the momentum relative to the sustenance of the momentum that is important not just the disturbance in the momentum itself that is where the kinematic viscosity comes into the picture. So you can see that this T reference so for example if y reference is equal to h then this is called as the diffusion based time scale h square by nu. So what it is this time this T reference is the time required by the momentum disturbance to diffuse by a distance h that is the physical meaning of this characteristic time. That is if you allow the disturbance to propagate then the characteristic time that it takes to traverse to diffuse is h square by nu. Now there may be physical problems where the time scale is not governed by either the advection term or the diffusion term but the time scale is governed by the pressure gradient term which is itself pulsating in nature. So let us consider an example del p del x is equal to some a0 plus a1 sin omega t okay. So we are applying a we will work out a problem of this type and see that what is the velocity field as a consequence of this kind of pulsating flow that is very important for many microfluidic devices and this kind of devices can be used as a flow meter as a pulsating flow meter that means you can measure the flow rate as a function of time using many such devices. So these have important applications in microfluidics. So when you have a pulsating pressure gradient when you have a pulsating pressure gradient then you must have this term always important because otherwise there is no use of giving the pulsating pressure gradient right. If you are giving a pulsating pulsatile pressure gradient and you want that to be physically effective in a problem then that means the pulsating term has to be important. So the pulsating term if it is important that means we can say that the corresponding time scale is nothing but 1 by omega right. You have to keep in mind that the scale is then decided by the frequency of pulsation okay. So to summarize the scale may be decided by an advection based phenomenon, a diffusion based phenomenon or a time dependent forcing parameter that is what is summarized in this slide. So time scales you have advection based time scale if x reference is of the order of l it is t reference is l by u reference, diffusion based time scale it is h square by nu or depending on the forcing parameter t reference is equal to 1 by omega. So you have to decide that which scale is governing the physics okay. Now we will consider some examples of Lore-Enoz number flows, we will consider only one example of steady flow and we will establish that the solution of the problem is quite similar to fully developed flow and then we will consider some examples of unsteady flows which are not normally covered in the undergraduate level fluid mechanics course. So we will first start with the steady Lore-Enoz number flow as an example. An example of steady Lore-Enoz number flow. So let us consider that you have x axis like this, y axis like this, h is the half height of the channel. So let us write the governing differential equation assuming x reference equal to l and y reference equal to h or 2h whatever I mean scale wise there is no difference between h and 2h. So now let us call it dp dx although you may keep it generically as del p del x but let us call it as dp dx. So now you integrate this equation. When you integrate this equation what you will get del u del y integrate means integrate with respect to y plus a function of x this is no more a constant. For fully developed flow this was a constant right. For a general Lore-Enoz number flow whether it is fully developed or not we do not care this is because with respect to integration with respect to y or differentiation with respect to y a function of x is like a constant okay. So if you integrate it once more right you have c2 as another function of x. Now you have to evaluate c1 and c2. So what are the boundary conditions? One of the boundary conditions y is equal to 0, y is equal to 0 is the center line del u del y equal to 0. Can you tell or talk about a condition when this may not be true? Just give an example give a practical example when at y equal to 0 del u del y equal to 0 is not true. Let us say both the plates are stationary. Both the plates are stationary it is a channel where both the plates are stationary. Let us say that the weightability of the 2 channels the weightability of the 2 channels they have different characteristics. So that the hydrodynamic boundary condition we will discuss about slip boundary condition later on but just as a qualitative remark the hydrodynamic boundary condition is not symmetrical at the upper plate and the lower plate. If the hydrodynamic boundary conditions are not symmetrical at the upper plate and the lower plate then at the center line the situation will not be symmetrical okay. And then in fact you cannot solve half the problem here we are able to solve for half of the domain with an understanding that there is symmetry always remember in a physical problem the symmetry demands 2 things. One is symmetry in geometry another symmetry in physics. So these 2 must be there in tandem even if the hydrodynamic boundary conditions are different you can still see that there is a symmetry in geometry but that does not ensure that there is a symmetry in the problem. Symmetry in geometry is just a necessary condition but it is not a sufficient condition. So you also have to see that there is a symmetry in hydrodynamics at the 2 walls. So if that is not the case then this is not true but if that be the case which is more often the case that we consider then at y equal to 0 you have del u del y is equal to 0. So what I am trying to do is that I am trying to give you a specific picture but trying to also create a broad picture which may deviate from the specific example. See we can work out only 1 or 2 specific examples but when you discuss about a theory your understanding of the theory should be broad based so that it can address problems beyond what we are discussing. So at y equal to 0 you have del u del y equal to 0. Plates if plates are of different weightability or weightability variations along the plate are different then the hydrodynamics which is described by the slip length we will discuss what is the slip length later on. I am not trying to that is why I told that we are trying to give just a qualitative picture at the moment. I will discuss in details what is slip length and how does it come into the analysis of solving equations and so on but for the time being we are assuming that at least it will it can break the symmetry of the problem. Different weightabilities or different variations in weightability of the 2 plates can be a factor of symmetry breaking in the problem despite the problem being geometrically similar. So then you have at y is equal to 0 you have del u del y is equal to 0. So at y is equal to 0 if you have del u del y equal to 0 then c1x is equal to 0 right may be in general it is a function of x but in this case not right. So it will literally be a function of x if you have a slip length which is varying along x. So in this case you have c1x equal to 0 what is the other boundary condition at y is equal to h u is equal to 0 right at y is equal to 0 du dy equal to 0 that means c1 equal to 0. So at y is equal to h u equal to 0 so if at y equal to h u equal to 0 then you will have 0 is equal to 1 by mu dp dx h square by 2 plus c2 okay. But it could so happen that at y equal to h u is a function of x for example at y equal to h let us say that the channel is patterned with some substrates whereas where on this shaded region there is slip and there is a velocity of slip which is different from 0. I am just giving an arbitrary example it is not absolutely arbitrary you can make engineering devices using this principle but arbitrary at least at this stage when we have not introduced the slip. So let us say that there is a slip velocity over these patches okay. So let us say these patches are slip regions and no patches are no slip regions that means at the wall you will have u as a function of x then you will have the c2 as a function of x which has really come to a constant if this does not vary along the wall okay. So in general c1 c2 are functions of x now for the specific common problem that we address for that particular case you can see that this equation is coming exactly same as the fully developed flow equation. So sometimes knowingly or unknowingly we mix up the low Reynolds number hydrodynamics with the fully developed flow and still get the same answer and same solution there is nothing wrong with it provided those considerations are still valid in this example those considerations are still valid but if you make now this functions of the velocity at the wall as a function of x alternate slip and no slip as an example then you will get this as a function of x. Not only that once you get u as a function of x you will get del u del x if you once you have u as a function of x you can calculate del u del x that means from that you can calculate del v del y because del u del x plus del v del y is equal to 0 then you can integrate that with respect to y to get the variation of v and if c1 or c2 or both are functions of x then there will be a v as a function of y otherwise also v is a function of y that we cannot bring out from this equation simply because of the fact that we have neglected the inertial terms otherwise we could have brought that out from this equation itself but if c1 c2 are functions of x that becomes more explicit that brings that comes out from the analysis otherwise that comes out from the concept rather than the mathematical analysis. So to summarize what we have discussed as an example of a steady flow let us say that like if we consider the velocity profile after evaluation of c1 and c2 we could get the velocity as a function of y and the wall shear stress. So that means that if you are working with a case of a low Reynolds number flow where you do not have a breaking in symmetry or you do not have patterned slip at the wall then you can use equations which are almost or virtually same as the fully developed flow equations and then you can solve for those equations. Now this is so far as our discussion goes for steady flows but there is an interesting paradigm that remains to be discussed for pressure driven flows which is unsteady flows. So we will start with the discussion of unsteady flows in the context of first the classical fluid dynamics and then specifically in the context of microfluidics that we will do starting from the next lecture. Thank you very much.