 In the previous lecture we were discussing about the concept of a developing flow and a fully developed flow and we start now with the exercise of finding out the velocity profile in a fully developed flow through a parallel plate channel. So as we discussed in the previous lecture we considered a channel like this y axis is important here and basically our objective is to find out how velocity varies with y. Let us start with the basic equations before doing that let us put some dimensions to this. So let us write the basic equations continuity. This is the continuity equation in vector form or if you want to write this in index form del del xj of rho uj whatever I mean either forms are equivalent. So now let us assume that we are dealing with incompressible flow right I will like to state here is that in microfluidics in pressure driven microfluidics there are extensive studies on compressible micro flow and incompressible micro flow. So because compressible micro flow requires a prerequisite which is compressible fluid mechanics that is not commonly covered in undergraduate text and this is just like sort of a preliminary course in microfluidics or a beginners course in microfluidics and we do not have the scope of bringing the discussions of basic compressible flows into the picture. We will avoid the discussion and mainly consider incompressible pressure driven micro flows although over time to time we will discuss about some aspects of compressible micro flows and we will discuss about that but primarily our focus will be incompressible micro flows. What is the definition of an incompressible flow? See we have discussed about this or we have hinted about this earlier but I will again reiterate here that incompressible micro flows or incompressible flows rather should not be confused with incompressible fluid. So incompressible flow is just a kinematic concept which says that there is no volumetric strain of the fluid element. So if that is there now let us simplify this equation. So del rho del t plus so if you have u v w as the velocity components so you can write this equal to 0 right that is by using the product rule of derivative. Now for incompressible flow which one is 0? This term is 0 for incompressible flow, incompressible flow. If that term is 0 for incompressible flow then what can we conclude? We can conclude that this is also 0 for incompressible flow, this is what? This is the total derivative of rho. So the total derivative of rho has to be 0 for incompressible flow. Does it mean that rho is a constant does not necessarily mean that for example let us say that we can have such a situation may be a bit hypothetical and mathematical but we can have such a situation when we have this del rho del t term as of some magnitude and of some sign whereas these terms are of the same magnitude and opposite sign. So that some total of these 2 terms is 0 but rho could still vary with position and time. Such flows are called as variable density incompressible flows okay. Now a very common wrong concept is propagated in many text books which says that assuming incompressible flow rho is equal to constant. So assuming incompressible flow does not give you a guarantee that rho has to be a constant but if you take rho as a constant then that will surely satisfy the condition of incompressible flow okay. So basically we normally deal with that example which is a special case of incompressible flow which is a constant density flow. But for incompressible flow the right constant is either the total derivative of density equal to 0 or basically the sum total of del u del x plus del v del y plus del w del z equal to 0. Yes, you have a, see I will give you an example when the density can vary with time. Let me give you a physical example. So let us say that you have a control volume okay, a control volume. There is some liquid here and you are heating this with a burner. This liquid is getting converted into vapor okay. So at a given location, let us say this is the location, when liquid gets converted into vapor the density is decreasing. So at a given location the density is changing with time. So there is a del rho del t okay. So it is possible that there is a del rho del t because of change of phase. It is not just true from liquid to vapor phase change. It can be true for any phase change, solid to liquid, liquid to solid. So in all those examples and it is possible that you have that term. Now that term will be sort of, so if for example if there is an expansion then where will that extra volume of fluid go? That will only pass through transfer across the phases of the control volume. So that will give rise to these terms okay and some total of that if that is 0 then that flow is incompressible otherwise it is compressible okay. Now let us come back to our specific example of this flow through parallel plate channel. So let us write the continuity equation incompressible flow. See here we are considering incompressible flow. It may be constant density or variable density because we are using this as the condition for incompressibility of the flow. Absolutely perfect. No problem. When we say that assuming incompressible flow density is a constant that is not always a fair statement. But when we say assuming incompressible flow this is right that there is no problem. So now for fully developed flow this term is 0 because u is not a function of x. For flow through a parallel plate channel this term is not important because variations along z direction is infinitely large as compared to the smallest dimension. So that can be neglected. So you basically have del v del y equal to 0. When you say del v del y is equal to 0 that means what? That means v is not a function of y. When we say that v is not a function of y that means that whatever is v at the wall y equal to h or y equal to minus h the same v is there for all values of y because v does not vary with y. So if you find out v at some specific y that value of v should be same for all values of y. That specific convenient choice of y is y equal to plus h or y equal to minus h. At those locations v is 0. Why is v 0 at the wall? It is 0 at the wall because of no penetration boundary condition not no slip boundary condition. It has nothing to do with slip or no slip. Right? Like these are very common mistakes. Why I am repeatedly telling this is that again whether it is classical fluid mechanics or microfluidics please keep your fundamental basis very strong. That will help you to solve problems no matter whether it is microfluidics or classical fluid mechanics. See v means what? v means the normal component of velocity slip talks about the tangential component of velocity. It has nothing to do with the normal component of velocity. Normal component of velocity is 0 because the fluid cannot penetrate through the wall. If there were holes at the wall then that v would not be 0 at the wall. So because there are no holes at the wall that is no penetration boundary condition v is equal to 0 at the wall. So v is equal to 0 at y is equal to plus minus h. So because v is equal to 0 at y is equal to plus minus h that means v is equal to 0 for all y okay v is equal to 0 for all y because v is not a function of y. So if you have found out v for a particular y that should be true for all values of y okay. So our summary from the continuity equation is that for fully developed flow v is equal to 0 for all y. Then for a fully developed flow this would have not been true if there are holes at the wall okay. Now let us write the Navier-Stokes equation. So we have made use of the continuity equation and the summary of the findings from the continuity equation is that v is equal to 0. Next we write the Navier-Stokes equation. When we have written this we have already made some simplifications. What are those simplifications? Incompressible flow because we have not included that mu by 3 that term that is there because of mu by 3 del uk del xk if you remember del del xi of mu by 3 del uk del xk that term we have not considered. Body force we have not considered because we as I told in the beginning that for our discussions in the chapter we will include the gravity as the body force in the pressure itself. So this p is like p star. I am not writing going to write it as p star because it is inconvenient to write it always like that but we assume that that p is p star this is ui. Now we assume steady flow we will begin with steady flow but we will discuss about several unsteady flows which are relevant to microfluidic applications in this particular chapter so but we will start with steady flows. So for steady flow this is 0. So let us write the various components and the other thing is mu is a constant. So it is essentially does not matter whether you keep it out of the derivative or inside the derivative mu is a constant. We also assume that rho is a constant. So now we are boiling down to the special case of incompressible flow which is a constant density incompressible flow. So we are assuming that all the properties are constants. Rho is constant, mu is constant like that. So steady flow and constant properties are 2 additional assumptions that we are making at this stage. Now if you write this equation in the index notation see we started deriving the equations with the index notation. So I will first write it with the index notation and see that how we can convert that notation to the well known x, y, z notation that you commonly use. So rho u1 del u1 del x1 plus u2 del u2 del x2 sorry del u1 del x2 so i equal to 1 plus u3 del u1 del x3 is equal to minus del p del x1 plus mu del 2 u1 del x1 plus del 2 u1 del x2 plus del 2 u1 del x3 squares. Now you just write x1 equal to x equivalent to x, x2 is equivalent to y and x3 is equivalent to z. So rho u del u del x plus v del u del y plus w del u del z this is the x momentum equation. Now let us simplify this equation for fully developed flow. So for fully developed flow u is not a function of x that means del u del x0 so this term becomes 0. We have just now derived that v is also 0. So this term becomes 0, this term anyway is not important because it is essentially a flow in the xy plane, gradient in the z direction is not important. Then right hand side because u is not a function of x del u del x is equal to 0 its second derivative is also 0. So this is 0, this term is not there. So you can see that the left hand side has become identically 0. The non-linearity of the Navier-Stokes equation which gives rise to certain complications in the solution is because of its left hand side and you can see that by invoking the assumption of fully developed flow the left hand side is becoming 0 and that means the acceleration is 0. That is what we have intuitively discussed earlier that because the velocity is not changing with along x the acceleration along x must be 0 and that we have shown by Rigor's mathematical analysis. So what you are left with is this okay. So you can see that the viscous term is balanced by the pressure gradient term okay. Now let us come to the y momentum equation. i equal to 2 is the y momentum equation just replace u with v okay. Just we have replaced u with v here. Now v is 0 so because v is 0 all these terms are 0. So what we get from here this is equal to 0 that means p is not a function of y. If p is not a function of y then p can be a function of what? See what are the 4 variables x, y, z and time it is steady flow so p is not a function of time it is the flow is in the xy plane so p is not a function of z. So potentially p could be function of x and y but because p is not a function of y p is a function of x only because p is a function of x only you can write the partial derivative of p with respect to x same as dp dx remember this is actually dp star dx. So you can write not only that u is a function of y only. So you can write this equation as mu or yes mu d2u dy2 is equal to dp dx or d2u dy2 is equal to 1 by mu dp dx. Now this is a function of y only this is a function of x only so a general function of y this could be any general arbitrary function so a general function of y is equal to a general function of x that is true only if it is a constant otherwise that is not possible let that constant equal to c. You can integrate this twice du dy is equal to c1 u is equal to okay before that we can use a boundary condition so du dy cy plus c1 not c1 because we took that as c. So we can find out c1 by noting one of the boundary conditions maybe we will use the boundary conditions later on. So u is equal to cy square by 2 plus c1 y plus c2 now let us apply the boundary conditions what are the boundary conditions see for analytical solution it may not be that important but for numerical solution it is important that if you have the problem which is symmetrical then you can solve for one symmetrical half that reduces the size of your domain for analytical work it is of no great consequence but for numerical work it saves a lot of computational time. So let us say that we are interested to solve half of the domain in that spirit. So you can see that at the centre line which is y equal to 0 the velocity is what the velocity is maximum that means du dy must be 0. So one of the boundary conditions is at y equal to 0 du dy equal to 0 and the other boundary condition at y equal to h u equal to 0 no slip. So the boundary condition 1 at y equal to 0 du dy equal to 0 means c1 is equal to 0 boundary condition 2 at y equal to h u equal to 0 is equal to ch square by 2 plus c2 c2 is equal to minus ch square by 2. So what is the velocity profile u is equal to c by 2 mu into y square minus h square. Again this may give you an illusion that because y is always less than h in magnitude this term is negative this may give you an illusion about negativity of u but that is not true because to have positive u you have negative c c is actually dp dx. So that negativity combined with this negativity makes a positive u for a negative dp dx it depends on c because mu we have kept with the other term that is why c into y square minus h square because mu we have absorbed with the dp dx term. Then see many times instead of this we are interested to express the velocity in terms of the average velocity right to express the velocity in terms of the average velocity. Why we do that because average velocity is very easy to measure if you have a flow meter then the flow meter can calculate or can measure the volumetric flow rate you just divide it by the cross sectional area to get the average velocity. So how do you define the average velocity u average is equal to integral u dy from minus h to h divided by 2h assuming width of the channel as 1 or even if the width is w it gets cancelled from or b it gets cancelled from both numerator and denominator. So this is basically integral 0 to h u dy by h u is c into y square minus h square c by 2 right yes c by 2 not c c by 2 right c by 2 into y square minus h square by h. So u bar is equal to c by 2 into h cube by 3 minus h cube divided by h. So that is what minus c h square by what is this c? c is equal to minus 3 u average by h square we will use this expression for calculating the pressure drop in the fully developed section but we can put that in the velocity profile. So u is equal to u by u bar is equal to 3 by 2 into 1 minus y square by h square by substituting in this expression okay. So you can clearly see that u varies quadratically with y that means it is a parabolic velocity profile. So the fully developed pressure driven flow velocity profile is parabolic. Now we will find out 2 important engineering parameters of interest and these 2 parameters one is related to the dimensionless wall shear stress another is related to the dimensionless pressure gradient because the wall shear stress and pressure gradient are sort of related by a simple equation for a fully developed flow these 2 dimensionless measures will also be related. So first we will find out a dimensionless wall shear stress what is tau wall for a 2 dimensional flow mu into the velocity gradient at the wall. So you can see that this term is not important because v is 0 at the wall. Now by tau wall when we write tau wall we are always interested about the magnitude. So you can see that if you draw the velocity profile now if you draw the velocity profile in this figure a parabolic velocity profile like this then if you go at so wall is what? Where is wall? Wall is y equal to h. At y equal to h what is du dy? Positive or negative? Negative because u is decreasing with y as per this coordinate system. So but tau wall we always consider as positive we are writing tau wall means magnitude of tau wall. So we put a mod here just to indicate that we are interested about the magnitude of this. So what is this? This is 3 by 2 mu u average 2 by h 3 mu u average by h. What is dimensionless tau wall? We had mentioned this but we are not defined that is Cf. This is nothing but this is called as Fanning's friction coefficient. So Cf is equal to what? Tau wall by half rho u average square. So 3 which means let us write Cf here 3 mu u average by h into 2 by rho u average square. So 6 by rho u average h by mu. So you can see that the denominator is nothing but Reynolds number based on the height h. So 6 by here comes an important consideration. Many times Reynolds number is mentioned as or interpreted as the ratio of inertia force by viscous force. Now we have to keep in mind that that interpretation should be very carefully and restrictively used for a fully developed flow. Inertia force is what? Inertia force is mass into acceleration. For a fully developed flow we have just now seen that acceleration is 0. Does it mean that the Reynolds number is 0? No. The thing is that inertia force by viscous force is one possible interpretation of Reynolds number but it could be given several interpretations. So one of the possible interpretations is that it is a dimensionless velocity or it can be thought of as inertia force by viscous force provided that whatever is the energy of the flow if all that energy could have been used to accelerate the flow then that equivalent inertia force by viscous force could have been the interpretation of Reynolds number. But it is not always as straight forward as inertia force by viscous force. So that has to be understood. So 6 by Reynolds number based on H. The next thing is that we will come to this interpretation that C is equal to minus 3 u average by H square. So this C is what? C is 1 by mu dp dx. dp dx means basically dp star dx. Now there is a, so this is basically this is a constant. So this can be written as minus delta p by L. Right? dp star or dp dx because it is a constant it is some delta p by delta x. What is the delta p? Let us say that delta p is this one. That delta x is the length of the channel that you are considering for your analysis. Let us say that you have put 2 pressure tapings. One pressure trapping here, another pressure tapping here. Let us say you have a pressure sensor by which you are measuring the pressure drop. So you have put 2 tapings at 2 sections 1 and 2. Then your L is the axial distance between the sections 1 and 2. Not necessarily the length of the total channel. It is the distance between the sections across which you are measuring the pressure drop. So you have this as minus delta p over L. Why minus? Because there is a delta p is always a magnitude. So there is a drop in delta p. So delta p is equal to 3 mu u average by h square. Now we can write delta p as equivalent some head into rho into g because simply because the unit of pressure is h rho g. Unit of h into unit of rho into unit of g. So instead of delta p you can write some equivalent h hf into rho into g. This hf is called as head loss due to fluid friction. They were divided by L is there. Delta p by L hf rho g by L. Now you can write hf is equal to 3 mu u average by h square into L by rho g. Next step is that we have a general expression of hf which is given by f L by hydraulic diameter into u average square by u g. This is called as Darcy-Wesbach equation. So this f is called as Darcy friction factor. This f is called as Darcy friction. This is the definition of Darcy friction factor. So this is a non-dimensional essentially a non-dimensional pressure drop. Just like cf is a non-dimensional wall shear stress. This is a non-dimensional pressure drop. Because the wall shear stress and pressure drop are related, cf and f must be related. And it is very easy to find out the relationship. So to do that, I mean for this special case, so you can find out see hf is equal to 3, okay hydraulic diameter. First what is the hydraulic diameter? It is 4 area by weighted perimeter. So it is sort of an equivalent diameter for a section which is not circular. Let us say that the width of the channel is b. So what is the cross sectional area? 4 into 2h into b, so 4ab by what is the perimeter into 2h plus b. You divide both numerator and denominator by b and take the limit as b tends to infinity. Because that is what is the definition of a parallel plate channel. So as h by b tends to 0, when b tends to infinity. So you will have hydraulic diameter is what? 4h. So hf is 3 mu u average by h square l by rho g is equal to f l by 4h into u average square by 2g, okay. So we cancel g from both sides, cancel l from both sides, cancel one of the h, then f is equal to 24, 3 into 4 into 2 that is 24 by rho u average h by mu. Do not confuse it with your well marked up and well known formula 64 by Reynolds number that is for circular pipe. Here we are not discussing about circular pipe, okay. So this is 24 by Reh. So you can clearly see that f is equal to 4cf. So from engineering relevance it does not matter whether you are talking about f or cf because equivalence can be drawn, okay. What are the important inferences or what are the important conclusions that we can draw from this analysis? First conclusion into Re is equal to constant for fully developed laminar flow, right. This Re may be on the basis of h, 2h, 4h whatever. Friction factor into Reynolds number is a constant. That constant depends on what? That constant depends on the geometry of the cross section. Like it is 64 for a circular pipe, right. So it depends on the geometry of the cross section. Now we will put a question to this, question mark to this for microfluidic applications. One of the reasons is that see for classical flows see what we are saying? When we are saying that this is a constant we are saying that it is not dependent on the roughness of the wall, right. So if you recall the Moody's diagram, the famous Moody's diagram that graphically demonstrates the friction factor in a flow through a circular pipe, you will see that in the vertical axis if you have f, in the horizontal axis you have Reynolds number in the logarithmic scale and on the right hand side vertical axis you have epsilon by d which is the relative roughness of the wall. The wall roughness length scale divided by the diameter of the pipe and in the laminar fully developed region the f versus Reynolds number is f into Reynolds number equal to constant. So in the log plot it will look like a straight line and that constant does not depend on the roughness of the wall. But typically those data have been considered to be important when you are talking about macro scale systems where the surface roughness elements do not compare with the characteristic system length scale, right. But if you are having a micro channel you may turn up with surface roughness length scales which compare with the characteristic length scale of the system. So how will that influence the friction characteristics? Whether this still the same thing works or not is not a resolved issue. So that has to be carefully understood. So the difference at least we have understood one difference is that this in this case we have not considered the effect of the length scale of the roughness and length scale of the roughness may become comparable with the system length scale for micro channels. The second is that hf let us write the expression of hf 3 mu u average by h square l by rho g. Now you can replace u average by what? In terms of volume flow rate q by a, right. So what is q by a? q by 2h into b, right that is the area. So you can write this as hf is equal to 3 by 2 mu q l by rho g h cube b. One important thing that we can gather from here is that the head loss is inversely proportional to a power of the dimension h. Therefore, this is for small h as you reduce your h from say meter to micron scale you will see that the head loss goes up tremendously. So you require a huge pressure gradient to drive the flow through micro channels. So that is one of the issues associated with pressure driven microfluidics that is because of this kind of relationship between head loss and the length scale characteristic length scale h. We have not yet discussed about the circular geometry but if you consider the circular geometry the head loss will be 128 mu q l by rho g pi d to the power 4 that is the Hegan Poiseuille's law. So for a circular geometry the head loss is inversely proportional to the fourth power of the hydraulic diameter. So no matter whether it is third power or fourth power you see if you reduce the height or the hydraulic diameter you will see that there is a tremendous rise in the head loss and you have to supply the necessary pumping power to drive the flow. So this is one of the issues associated with pressure driven microfluidics. So these are the 2 points that we have to keep in mind while we are proceeding further in our analysis. So in the next lecture what we will do is we have till now discussed about steady flow but a flow where the variation of velocity is one dimensional that means u varies with y only. So that has been the case when it is parallel plate channel that is the width is infinity but if the width of the channel is finite or width is comparable with the height how will the solution change? How will we find out the velocity profile? How we will find out the friction coefficient? All these things we will take up in the next lecture. Thank you.