 In the previous lecture we were discussing about the exact solution or analytical solution of a pressure driven fully developed flow through a rectangular channel. Now we will continue with that. So just to recapitulate this is the cross section of the rectangular channel we set up our y axis and z axis like this and the dimensions of the faces are a and b. We basically split this problem into 2 problems. In one case the problem is like a flow through a parallel plate channel problem which we had already solved and in the later case and in the second problem we are basically interested to solve for this problem this is the second problem. We will keep a note of the boundary conditions on one side of the board because we have to use the boundary conditions for solving the problem. So at y is equal to 0 del u2 del y is equal to 0 at y is equal to a by 2 u2 is equal to 0 at z is equal to 0 del u2 del z equal to 0 z is equal to b by 2 at z is equal to b by 2 u2 is equal to minus u1 which is a function of y. This function of y is obtained by some solving the parallel plate problem. Now this is a problem where the governing equation is homogeneous and 3 of the 4 boundary conditions are homogeneous so we can go for the separation of variables. So we sort a solution of the form u is equal to a function of y into a function of z. Remember that method of separation of variables is one possible way by which you can convert a PDE to a ODE that is a partial differential equation to an ordinary differential equation for which maybe more convenient analytical solutions can be obtained as compared to the original partial differential equation. So that means you will get from this equation f double dash g plus g double dash f is equal to 0 and f double dash by f is equal to minus g double dash by g is equal to yesterday in the previous lecture we discussed that this has to be a negative constant to satisfy the boundary condition so that is equal to minus lambda square same. So if that be the case then you have one equation which is f double dash is plus lambda square f equal to 0. So f double dash plus lambda square f is equal to 0. What it means? What are the solutions of this? Sin and cos so f is equal to say c1 sin lambda y plus c2 cos lambda y. Similarly for g you have g double dash minus lambda square g is equal to 0. So the solution is e to the power plus lambda z and e to the power minus lambda z the linear combination of that. So that we can equivalently write as a combination of cos hyperbolic and sin hyperbolic because cos hyperbolic and sin hyperbolic is again a linear combination of exponential plus and exponential minus. So g is equal to say c3 sin h lambda z plus c4 cos h lambda z so the solution is the product fg but we have to see whether that is the general solution or not. We will take up this issue in a moment but first let us apply the boundary conditions to in an attempt to get the 4 constants. So first let us appeal to the boundary condition that at y equal to 0 del u to del y equal to 0. At y equal to 0 del u to del y equal to 0 means basically df dy equal to 0 because in u u to f is the function of y. So df dy equal to 0 means when you differentiate this with respect to y sin will become cos and cos will become sin with plus minus sin I am not bringing that into the picture. So then this coefficient of c1 when it is df dy this will be cos and coefficient of c2 will be sin. So at y equal to 0 the sin will be automatically 0. So to make this 0 at y equal to 0 you must have c1 equal to 0. So this boundary condition gives let us write this in a box c1 equal to 0. Similarly let us write let us consider the third boundary condition at z equal to 0 del u to del z equal to 0 that means dg dz equal to 0. So when you make dg dz sin h will become cos h and cos h will become sin h and by the same logic you will have c3 equal to 0. So that means you get f into g now you have these 2 terms which are underlined here are 0. So you have f into g is equal to c2 c4 cos lambda y cos h lambda z this c2 c4 let us call it a a new constant. Now you have at y equal to a by 2 u2 equal to 0. So we apply the third boundary condition second boundary condition at y equal to a by 2 u2 equal to 0 that means cos lambda a by 2 is equal to 0. What does it mean? It means lambda a by 2 is equal to what or multiple of pi by 2 2n plus 1 pi by 2. So let us write it in a bit of more elaborate way here. So you will get lambda is equal to 2n plus 1 into pi by a. So how many possible values of n lambda are there? If you have infinite number of possible values of n then for each value of n there will be a value of lambda and see because it is a linear differential equation for each value of lambda if this is the solution then the general solution is the summation of the solution for all possible values of lambda. So instead of writing the solution in this way we will write u2 is equal to summation of a n cos lambda n y cos h lambda n z. The reason is pretty clear that for each value of n you have a corresponding solution the resultant solution is a linear superposition of all these solutions because the governing differential equation is linear. The only task that remains is the determination of a n and the only condition that remains to be used is the fourth boundary condition. So we will use the try to make use of the fourth boundary condition for determining a n. To determine a n let us consider that equation f double dash plus lambda square f is equal to 0. So this is true for all values of n so we can write with f equal to f n lambda will be equal to lambda n. So d2 fn dy2 plus lambda n square fn is equal to 0. For all possible lambda n see lambda n is not I mean here you cannot substitute all possible values of lambda n but the lambda n which satisfy this condition lambda n is equal to 2n plus 1 into pi by a only that value of lambda n not all possible values of lambda n. So there are only these possible values of lambda n for which you have a non-trivial solution because you could otherwise argue that if I do not take cos lambda n y as 0 then also I could get a solution then I would set a n equal to 0. Instead of setting cos lambda n y equal to 0 I could have set a n equal to 0 to get the solution but then if a n becomes 0 u2 becomes trivially 0. So a n equal to 0 or a equal to 0 is ruled out that is why for using the boundary condition at y equal to a by 2 u2 equal to 0 we did not set a capital A equal to 0 but cos lambda a by 2 equal to 0. So if we had set capital A equal to 0 that could have satisfied this boundary condition but that could have given rise to a trivial solution u2 equal to 0. u2 equal to 0 means it is a trivial solution we are interested for a non-trivial solution. So these possible values of lambda n are the values which are giving rise to non-trivial solution of u2. So these are called as eigenvalues okay. So you are familiar with eigenvalue problems in linear algebra like this is a sort of an eigenvalue problem in differential equations. So meaning is something very similar that you are getting a non-trivial solution for these values. Now these corresponding functions fn which correspond to this non-trivial values of lambda n which are obtained from this. So if you substitute lambda n from lambda n here you will get some fn that fn is called as eigenfunction just the corresponding concept in linear algebra is eigenvector. So here instead of eigenvector it is a eigenfunction. So let us proceed with finding out a condition which the eigenfunctions will satisfy for this problem. So what we do is let us use a different color we multiply with fm, m is not equal to n in general but a special case is there where m is equal to n. So we multiply this with fm and integrate, integrate over y that is the domain minus 0 to a by 2 here because our domain is one fourth of the domain. So I am not writing the limits here because the limits are not important at this stage but I mean if you are very particular about it your limits are 0 to a by 2. Now what we will do is we will integrate it by parts. So one of these functions we have to consider the first function and the other function we have to consider the second function. Typically the rule is first function into integral of the second. So we typically take the higher order derivative as the second function because we can reduce the order of the derivative by integrating through integration by parts. So first function this is second function. So first function into integral of the second minus integral of derivative of first into integral of the second. In the next step what we do we just do the same thing but swap m and n. So if we swap n and n that means if we interchange m and n what we get. So fn just interchange n and m in the above expression, m and n are arbitrary. So you could swap I mean these are just indices okay. Now let us look into the boundary term. The term in the square bracket is the boundary term at y equal to a by 2 u2 equal to 0 right, u2 equal to 0 means f equal to 0 and at y equal to 0 del u2 del y equal to 0 that means df dy is equal to 0. So that means this boundary term becomes 0. See this is where we require the homogeneity of the boundary condition. Had the boundary conditions not been homogenous this boundary term would not be 0 okay and that would give rise to some sort of complication. You may avoid the complication in a way which I will discuss in a moment but here this boundary term is 0, this boundary term is 0. Now you subtract these 2 equations. This one say the second one from the first one. So what you get? If you subtract these 2 equations you will see that you will get so I am writing it at the top subtracting. So this second term gets cancelled. So lambda n square-lambda m square integral fn fm dy is equal to 0 from 0 to a by 2. So what can we conclude from here? What can we conclude? If this is 0 then to generalize integral fn fm dy is equal to 0 if m is not equal to n because if m is not equal to n this term is non-zero in general. So this has to be 0 to make it 0 but if m is equal to n then that is not necessary. So this is not equal to 0 if m equal to n. This is called as orthogonality condition. So remember see I will give you a little bit of qualitative perspective that why out of so many names it is called as orthogonality condition. See usually we are familiar from our school level studies. We are familiar with the orthogonality condition in terms of orthogonalities of vectors. Two directions if they are orthogonal or perpendicular we say that that is a situation with orthogonality of vectors. Now here we are not talking about orthogonality of vectors. We are talking about orthogonality of functions. So what is basically happening here that in the function space if you see when you write orthogonality that means what? Orthogonality means you have if you have two vectors which are orthogonal to each other forget about functions just take an example of a vector. If two vectors are perpendicular to each other then what is their dot product? Dot product is 0 right. Dot product means what? Dot product means one component of this vector multiplied with the corresponding component of the other vector and that is summed up for all components. So here instead of the vector components being multiplied it is a functions being multiplied. So it is a special case of something called as inner product of the functions. So let us not get into more details of the functional theory here because our objective is to solve this fluid mechanics problem and we will like proceed with this but now let us look into C. Many times we say that for method of separation of variable to work we require the some of the boundary conditions to be homogeneous and you can see that while deriving this orthogonality condition it is clear that why we require the boundary conditions to be homogeneous. So that these boundary terms are absent. Now in one way even if one of the boundary conditions was inhomogeneous it could be possible that this is cancelled by this while subtracting by some way or the other then also it will work. So somehow this boundary term has to be cancelled when you subtract. So whether you require or what kind of boundary condition you require it is up to you to decide but mathematically these two terms should cancel. I mean either individual terms are trivially 0 or if they are not trivially 0 when you subtract one from the other that should be cancelled. So that will give rise to this orthogonality condition. Now why this orthogonality condition is important for solving the problem using the separation of variables method. Let us look here. So in this expression you have a summation. So it is possible that to calculate this you require to evaluate a large number of terms but if you now multiply this with am sorry fm this is fn. If you now multiply this with fm and integrate both sides at z is equal to b by 2 then out of all the terms in the integral only the case of m equal to n will remain. Other terms will be 0. So instead of a summation you will get just a single term a with a coefficient of an that is why we require this orthogonality condition. I am telling these things in details because if you refer to books you will see that I mean up to this is fine suddenly the books will come up with an orthogonality condition and then I mean there will be a lot of step jump and then that will give you the solution. So I want to give you the details of the solution maybe for one or two cases and I mean that is how I want to design this particular course initial few derivations I want to be as detailed as possible but because this is a sort of a bridge course between like basic course and research level course. So I would expect that in the later derivations if I skip a few steps it will be possible for you to follow that and work it out by yourself and that will give you a good training of how to do how to solve these problems by yourself. Of course I will give you a complete picture of how to give the solution for almost all the problems that we discussed in the class but at least I will leave something on you so that you can complete the exercise. Here I will live on you only some trivial integration and algebra but I am trying to work out the in between concepts to whatever depth possible. So now so we apply at z is equal to b by 2 u2 is equal to minus u1. So basically this equation u substitute minus u1 y is equal to summation of an fn cos h lambda n b by 2. Next step is multiply by fm and integrate with respect to y okay. So I am repeating the step it is multiplying by fm and integrating with respect to y. So when you do that if you use this condition only one case that remains is m equal to n other cases are absent. So instead of a summation it becomes just an and it becomes integral of fn square dy. So you will get minus integral of u1 y fn, fn is what? fn is cos lambda ny dy from 0 to a by 2 is equal to an cos h lambda n b by 2 integral of 0 to a by 2 fn square means cos square lambda ny dy right. This fm becomes fn so it becomes fn square fn square is cos square lambda n where the value of lambda n is given by this equation. So from here you can find out the value of an the only thing that I am leaving on you to complete this exercise evaluation of this integral and evaluation of this integral at this level this is very straight forward okay. So you can find out an that completes the solution of u2 okay. So u2 is equal to summation of an cos lambda ny cos h lambda nz where an is given by this expression and lambda n is given by this expression and u is equal to u1 plus u2. So it is clear how do we get the solution to the problem this is the exact analytical solution to the problem. Now I give you some homework based on this solution which will give you a good lot of physical insight. So if you have a rectangular cross section like this of the channel this is a this is b. So based on this in method one we had discussed about one approximate method one we have obtained an approximate velocity profile. So from that velocity profile you can find out a non-dimensional pressure drop. If you recall the velocity profile was something like I cannot remember it exactly but like something into y square-s square by 4 into z square-b square by 4 something into dp dx I mean some constant a square b square some type of constant may be there. So then what you do is you find out u average okay and then express u average as a function of dp dx. u average you know how to find out integral u dy by integral dy integral u da by integral da here not dy dy dz integral from y-a-a-2 to a-2 and z-b-2 to b-2. So you can get u bar as a function of dp dx and dp dx you can write as-hf rho g by l. So from that you can find out an expression for the friction factor f just in the same way as we did for the parallel plate channel. So find out f the friction factor the Darcy's friction factor okay. Now from the analytical solution also plot the friction factor I mean obtain the friction factor. So u is equal to u1 plus u2 from here you find out f it involves a little bit more mathematics to calculate f from here because the velocity profile is little bit elaborate it is not as simple as a polynomial like this but the principle is the same. Now when you use the general solution I mean there are certain insights you know when you if you work the problem by yourself then only you will understand. See in the classroom we cannot give you always a complete picture something we should live on you as an exercise because like one important question that should come to your mind even as an implementer that this is an infinite series how many we cannot calculate infinite series how many terms of the series should be good enough to give you a solution which is correct. So that I ask you to work out right everything should not be spoon fed in the class. So you work out that how many terms in the series. So you can easily check the convergence by considering a particular term up to a particular term and then one additional term what is the relative change. If the relative change started starts to be progressively almost ending to 0 if you go on adding additional terms that means the series has almost converts practically from numerical calculation point of view and you will see that not many terms will require I mean from will be required like from the perception of an infinite series it is often an elusive idea that possibly very large number of terms will be required but not very number of not very large number of terms will be required but that you practice by yourself and figure out that how many terms will be required to get a convert solution for you in the series. So the homework is on one side you have a velocity profile given by this on another side you have a velocity profile given by this this is also function of dpdx where is whereas the dpdx come I mean it is I mean again it is elusive you cannot see anywhere dpdx you get u1 is a function of dpdx not only that in u2 also you have this integral of u1 y cos lambda and ydy this contains dpdx okay because u1 itself contains dpdx. So dpdx is very much there so on one hand you have this solution on one hand you have this solution you make a plot in whatever plotting software you use let us say matlab you make a plot of u as a you make a plot of u as a function of y and z may be a 3 dimensional plot if you want u as a function of y and z one from this solution another from the exact solution for what values for different ratios of a by b or b by a for different a by b or b by a ratio whatever either you choose a by b or b by a this is the first exercise second exercise is plot the value of the Darcy's friction factor for different values of a by b and b by a for obtain from using this solution and using the analytical solution you will see that there are scenarios when the velocity profile which is given by this differs considerably from the velocity approximate velocity profile and that is what you expect because this is quite heuristic this was just made to satisfy the boundary conditions nothing more than that and you can see that here is a summation of cos h and all and this is just a simple polynomial so you expect certain deviations but interestingly the deviations that that you figure out are not that much a prominent when you talk about the friction factor the friction one of the reasons is that in calculating the friction factor like here you use u average instead of u so some see you have a u between these two limits you are satisfying the boundary conditions this is one boundary condition this is other boundary condition so in one case you have a velocity profile which is a very accurate like this in another case it is not so accurate like this okay so this difference is clear but the integral requires area under the curve for which this little bit of error is not so bad okay so we get rid of this error in mathematics because we are going for description through the friction factor requires really the information for average velocity not the local variation of velocity this is very analogous to analysis in the boundary layer theory in the boundary layer momentum integral method we use approximate velocity profiles and many approximate velocity profiles are widely different I mean some are linear some are quadratic some are cubic some are sinusoidal you have worked out these problems in fluid mechanics basic fluid mechanics but you will see that as an engineer our objective of interest many time is the velocity profile but many time more commonly to calculate the frictional loss or the wall shear stress so that is given by some integral of the velocity profiles not the velocity profiles directly so errors in the velocity profiles do not show up that much provided that errors in the integrals of the velocity profile are not very high so here also you will see that there will definitely be some error and the error will be perhaps most severe when the aspect ratio is perfect that is a is equal to b because then it deviates most prominently from the parallel plate channel case but otherwise if you take other limits that if you take either a by b tending to 0 or b by a tending to 0 that makes a parallel plate channel case then this friction factor and this friction factor calculated from here should match I mean that is something that what I am saying is from an intuitive viewpoint I am not imposing this on you but I would leave it on you as an homework that please make a plot of this and compare the analytical solutions with the approximate analytical solutions and then comment on the deviation between the two that will be the homework corresponding to this particular problem okay. So we have worked out certain cases of fully developed flows steady fully developed laminar flows for cases where it is a unidirectional I mean basically one dimensional variation that is u as a function of y only that is the parallel plate channel and rectangular channel where u is a function of both y and z. So these two cases we have solved and we have also solved for the circular geometry. So far so good but we have discussed about fully developed flow. Now the question is that in microfluidics we are not always interested about fully developed flow I mean the flow may be fully developed may not be fully developed but the condition demands something which is a little bit different from fully developed flow which is a low Reynolds number flow. So how do we compare the implications of a fully developed flow with the low Reynolds number flow. So we will do that in a moment and we will start with the discussion on low Reynolds number hydrodynamics remember it need not be fully developed flow low Reynolds number hydrodynamics. This is the heart and soul of microfluidic flows low Reynolds number hydrodynamics. So we will to begin with we will make certain simplifications see we are interested to get the concept rather than the mathematics. So we will make the mathematics as simple as possible and get the physics or the physical concept out of that. So we will assume a 2 dimensional flow I mean 3 dimensional flow we could have easily done that would add terms but we would not give a new physics. So 2 dimensional flow and constant density and let us assume that you have parallel plate channel with a height of h this is the center line x and y axis are laid like this the z direction in z direction the channel v is very large in in dimension. So that does not come into the picture in the analysis it makes it a 2 dimensional problem. So what we will do is basically we will first figure out what are the scales of length velocity time etc. length scales velocity scales time scales in describing the very basic hydrodynamic problem and through the continuity and the Navier-Stokes equation. So you we start with the continuity equation. So you have del u del x plus del v del y equal to 0 right. So order of magnitude wise what is the order of magnitude of this basically maximum value of u by maximum value of x. Let us write this let us first write and then we will discuss about this. Let us say u reference is the maximum velocity along x and let us say that l is the axial length of the channel. This is let us say v reference is the velocity scale. See it is not fully developed flow so v is not identically 0 right. We are not committing see we are now trying to generalize things beyond fully developed flow but low Reynolds number flow. The fully developed concept it does not matter whether it is low Reynolds number or high Reynolds number or whatever because the inertia term gets dropped out identically. Now here we are bothered about low Reynolds number flow because in micro channel the characteristic length scale is small that makes the Reynolds number based on the characteristic length scale small but v is not identically 0. This v by h it does not matter whether you write h by 2 or h scale wise it does not matter order of magnitude wise it does not matter okay. So now if you compare these terms if you now compare these terms see in a governing equation when there are 2 terms and they are some of them is 0 and this is not identically 0 then what it means that order of magnitude of these 2 terms must be equal one may be of the opposite sign as compared to the other because some total has to be 0. So if this is of a particular order this if this is non-zero that also has to be of the same order okay. So you have you can write from here v reference is of the order of u reference into h by l. So one inference from here is that if h is much much less than l that is very common in some microfluidic and in many microfluidic experiments let us say h is of the order of micron and l is of the order of millimeters right. If you see typical micro channels we will show you in our experiment demonstration and you will see that how these micro channels look and all. You will see that most of the times they will look like a scratch on a substrate the substrate may be a polymer substrate or silicon or whatever. So you will see that it is like a scratch the width of the scratch is not visible. So it is like human here like if you just take a small I mean tear away a small piece of here and just stretch it it is like a micro channel. So if you it is like the dimension of a micro channel so width is of the order of micron but length is still a few millimeters if not centimeter. So h by l that ratio is typically small that means v reference is much much less than u reference but there may be exceptions and in science or technology it is very interesting to discuss about the exceptions rather than the rules because the rules are quite obvious exceptions are not so obvious. I will talk about one such exception. Let us say so we have considered the reference length scale along x as l right. So x reference typically I should have written it as x reference but I mean just straight away I have written it as l. Question is will always x reference be l? There is a very interesting answer to it and for that I give you a technological perspective. See this whole discussion of microfluidics very often I will discuss about technology and applications and very often I will discuss about deep mathematics and we will try to merge it these two together to solve practical problems that should be the objective of learning this course. So in you have learnt by this time that typically micro channel flows will be low Reynolds number flows there are exceptions but typically very low Reynolds number flows. So in low Reynolds number regime it is very difficult to achieve good mixing because it is difficult to it is difficult to realize turbulence. So what instead you can do to get so mixing requires vortices in the flow. So you can create vortices in the flow by creating patterns on a microfluidic substrate. So for example you can create patterns like this patterns of variable weightability that means some of these patches may be hydrophobic some of these patches may be hydrophilic. So you can do chemical treatment or physical treatment to do I will show you in a later part of this course that how to design the weightability of a substrate. What is the technology that you need to design the weightability of a substrate but for the time being you assume that instead of having a constant weightability you could design the weightability of the substrate by patterning the substrate with says some patches. So then if you pattern the surface with such patches you can create local vortices here. Now forget about that if you pattern the surface with these patches then the length scales of your concern are the span of these patches not the length of your channel. So if your length of the channel is l and then this is actually your x ref so x ref is not equal to l. What is the characteristic length you must remember the characteristic length is a length over which characteristic changes take place. So if you have a pattern substrate like this characteristic changes do not take place over the entire length of the channel the characteristic changes take place over the patches and many times this length of the patch is compared to the height of the channel comparable with the height of the channel. So then you could get a v reference of the order of u reference even. So if you get a v reference of the order of u reference you can see that that can give rise to a nice vertical structure in the flow. So I am trying to impress upon you that you should not have a prejudice you should not have a prejudice that v reference is always less than u reference it depends on what is x reference. So for the standard scenario when the surface of the channel is not patched or pattern so that you have a uniform weightability you can consider the length of the channel as the characteristic length scale along x but that is not obvious that we should keep in mind. So let us summarize what we have learnt through the basics of low Reynolds number hydrodynamics we have just touched upon the continuity equation. So we had made the assumptions that it is a 2 dimensional flow and constant density flow. So we have referred to the continuity equation we have done an order of magnitude analysis of the continuity equation and from that we have related the reference velocity scale along y with the reference velocity scale along x in terms of the reference length scales along x and y. So we have done this much up to this much today we will start with the order of magnitude of the momentum equations and we will discuss about their consequences for low Reynolds number flows that we will take up in the next lecture. Thank you.