 We have looked at both velocity and temperature boundary layers and we have obtained similarity solutions as well as integral solutions for a variety of conditions of the free stream velocity, the suction and blowing velocity and the pressure gradient which of course is accounted by free stream velocity variations as well as the wall temperature variations with and without effect of viscous dissipation. We now turn to another important class of flows called internal flows and that is the topic of my discussion today, laminar internal flows. I will explain first the relevance of internal flows with important definitions and prediction of developing flow as an example of application. Internal flows are principally of interest in heat exchangers where it is important to have knowledge of the pressure drop for a given length of the tube or what we call the friction factor and also the heat transfer coefficient or the Nusselt number on the tube side of the heat exchanger to facilitate their design. Modern heat exchangers of course employ ducts of both circular and non-circular cross sections. Sometimes even curved ducts are preferred or are necessitated in order to conserve space. Duck passages with internal insertions such as the twisted tape or coils are also popular. Optimally internally structured surfaces such as rib rubbiness, grooves and indentations are also used for augmentation of the heat transfer coefficient or the Nusselt number. Most of these modern applications represent examples of very complex terms as opposed to the very simple case of a flow in a round tube or flow between two infinite parallel plates. So solution of transport equations of mass momentum and energy provide means for obtaining FNNU which is of interest in design. In simple ducts, analytical solutions are possible. In more complex ducts however, one needs CFD solutions and unfortunately that would be a topic outside the scope of the present lectures. But we shall consider a few non-circular ducts and treat them analytically. In the slides to follow, I will show you a few examples of ducts of non-cross sections and structured internal ducts and also curved ducts. So here are a set of examples. The first one on the left here is of course the simple annulus with the inner tube and an outer tube. The flow is between the two tubes. Sometimes the space between the inner and outer tubes is also connected by fins and what we have is an annular sector duct. The fluid flows through this sector and all sectors behave in a similar fashion and therefore, interest is in an annular sector. The plate fin heat exchanger is of course well known to you. This is the top plate, this is the middle plate, this is the bottom plate and there will be several stacks of plates like this. The flow is cross flow in the sense then between the top two plates, the hot fluid meme flow from left to right and here in the bottom two plates the flow may be directed as shown here, the cold fluid. Again the two plates are connected by fins which form ducts of triangular cross section. Here the cold fluid flows through ducts of square or rectangular cross sections. They are very commonly encountered ducts in plate fin geometry of more complexity and in extremely small heat exchangers what are called mini heat exchangers or micro heat exchangers with very high surface area to volume ratios bordering on 1200 to 1500 meter square per cubic meter. Ducts of extremely complex shapes are employed. Here is an example of a duct which is moon shaped and what I have shown here is a half section of the moon x axis being the symmetry axis. This is the inner edge of the moon, this is the outer edge of the moon. Likewise, there are ducts which are of cordoid shape. Again x axis represents the symmetry axis and I mean the symmetry plane and this is the outer boundary of the cordoid duct. In nuclear rod clusters in a circular shell there are large number of rods kept. There is one rod in the center, then there are several rods in the first rod ring, then another set of rods in the second rod ring and so on and so forth. So, it may go like 1 6 and so this is a 19 rod cluster geometry. So, it would go like 1 6 7 and 12 in the outer ring making 19 rods. The flow of course, this is the symmetry plane. Again this is also a symmetry plane and the flow takes place in the interspace between the rods and you can see how complex the flow cross section then is. It is a very good example of a complex shaped duct that is encountered in nuclear reactors. The plate phoenix exchanger that I showed you here, the plates here are flat. In order to save space, many a times the plates are actually wound in a spiral and what you then have are several spirals starting from the center and becoming ever bigger in the radius r and what I have shown here is a typical section of a curved spiral plate heat exchanger deck. So, this is the outer wall, this is the inner wall and the duct itself has a radius of curvature r. The fluid is flowing through this and coming out like that. Another example is that of a plane tube with a circular cross section, but in which a metal strip of width equal to the diameter of the tube is twisted about the axis of the tube. What the tape does then is to divide the circular cross section into two semicircular ones and each semicircular cross section then twists along the axis of the tube as one goes down the tube in the flow direction. So, what you have are two semicircular cross section twisting about the axis of the tube and it forms a very complex curved duct with a non-circular geometry. Internally structured surfaces, the gas turbine blade is a classic example of how surfaces are internally structured. First of all, notice that this is the cross section of the blade and you will see ducts of non-circular cross section. Each duct has ribs in it sometimes straight sometimes at an angle as you can see here. Some ducts are straight whereas others are having a bend. In some part of the blade, you have a duct which has ribs and then one wall is perforated so that the flow goes this way and then also comes out with impinging air on the leading edge of the blade and the remainder air goes out of the blade through the top. Likewise, at the rear end of the trailing edge of the blade, you have flow coming in from here and which passes on through this perforation in this wall as well as some flow goes directly into this trailing edge passage. In this passage, you have number of pin fins, solid pin fins. They are designed to make the trailing edge strong and you have perhaps the most complex of internally structured passages or ducts that you can encounter in engineering practice in a gas turbine blade. A tube of this type which is extensively used in refrigeration and air conditioning industry has spiral grooves etched on both inside surface as well as the outside surface. These grooves are cut out. This is a copper tube and you can see how tiny the near surface structure is. In fact, the height of the grooves is quite small and if you have familiarity with turbulent laminar sublayer in a turbulent flow, then the height of the tube just exceeds the laminar sublayer. The purpose of the tube grooves is to make sure that the sluggish laminar sublayer is continuously disrupted and therefore, enabling enhanced heat transfer. Same is the purpose of internal rib roughnesses and so on and so forth. Our interest here now is on laminar flows. All the many practical applications are in turbulent regime. We shall take up the turbulent flow a little later but let us concentrate first on the laminar flow in a duct. Let us say the flow enters with a uniform velocity u bar let us say. As soon as it encounters the top and the bottom wall, there would be viscous action and therefore, the boundary layer development as shown here. Inside the boundary layer, there will be a reduction in velocity from its initial value and to compensate for that, there would be increase in velocity inside what is called the core region. For all practical purposes, one could imagine the core to be almost of uniform velocity but the velocity itself would be greater than what it was at the inlet. So, we have a region core region in which the velocity continuously accelerates and is greater than that of the inlet velocity but in the boundary layer region, you have a velocity which is lower than the inlet value. Ultimately, the boundary layers from the two walls meet. When they meet, we say the flow is no longer in state of development but has reached a state of fully developedness. By that, we mean that the velocity profile that is now generated would sustain itself without any change in the axial direction. Therefore, the fully developed flow is identified with the du dx equal to 0, where u is the velocity in the axial direction as well as the pressure gradient in the axial direction is constant. What is of interest from practical standpoint is the estimate of development length l v and l v as we can very well imagine would be a function of Reynolds number. The higher the Reynolds number, the higher will be the length l v. In this entrance region of the duct, analytical treatment is usually quite difficult except in very simple cases. For more complex cases of complex cross sectional ducts and so on and so forth or curved ducts, one really has to adopt CFD based procedures, computer procedures. Nonetheless, I would take up a very simple case as shown on the next slide. It is essentially a case in which these are two infinite parallel plates both in the z direction as well as in the x direction. The z direction being and the y direction is measured from the bottom plate going forward. The plate interpolate distance is going to be 2B. So, the distance between one wall and the axial symmetry is simply B as you will see on the next slide. So, we consider laminar flow between infinite parallel plates separated by distance 2B and in the entrance region using boundary layer approximations, the governing equations and the boundary conditions are as given here. First of all, you will have the continuity equation, then the momentum equation in which you have the convective terms and then the viscous term. The velocity at x equal to 0 for all y is the average velocity u bar with which the fluid enters the duct. The v in the inlet plane again is 0 and u bar at any cross section would be 1 over B 0 to B u dy would be the average velocity. At the symmetry plane, d u by d y at all x will be 0 and so will the v at the symmetry plane will be 0. At the bottom wall x 0, u will be 0 and so would be be 0. That means there is no suction or blowing into the duct. All this is quite familiar and understandable by now. We non-dimensionalize this equation first in which velocity is made as u star as u over u bar v star equal to v over u bar p star equal to p over rho u bar square x star the distance x over d h y star y over d h. d h is the hydraulic diameter Reynolds number also is defined based on the mean velocity u bar and the hydraulic diameter and d h for flow between parallel plates separated by distance 2 v is twice the distance between the plate and therefore equal to 4 b is the hydraulic diameter. Equation 5 here shows that the pressure gradient is balanced by both the momentum change terms or the convection terms and also the viscous terms the d 2 u d y square. So, viscous friction as well as momentum change caused by change in velocity profiles along the length the duct balances the pressure gradient which itself varies with x. Now, what makes it difficult to obtain solutions to this type of equations is the fact that remember this first of all these are 2 equations and we have 3 unknowns u v and pressure. So, we really have a problem here that unlike in the boundary layers where we specify the pressure gradient we cannot do so in internal clause and we have to do some tricks. So, the coupling that is that exists between the continuity equation and momentum equation is due to convection terms here and that is what makes this equation set in non-linear equation set because of the coupling and we have to find ways to override this coupling. Now, this is precisely what was done by a man called Langar in a paper published in 1942 titled steady flow in the transitional length of a straight tube and was published in journal applied mechanics volume 9. He said the left hand side or the momentum change terms are functions of x and y, but we shall write them as beta square multiplied by velocity. Beta square is a constant at one cross section its value will change from cross section to cross section, but at a given cross section beta square is a constant and u star is a function of x and y of course, so that the left hand side and the right hand side I have the same dimensions. So, if I substitute for the left hand side beta square into u star then the momentum equation can be written as d 2 u star by d y star square minus beta square u star equal to r e d p star by d x star and you will quite easily derive that the pressure gradient d p star by d x star and I will show you that remember d p star by d x star is essentially rho u infinity rho u bar square into d p divided by d h d x. In other words and we define f l the local friction factor as 1 over 2 d p d x into d h by rho u bar square sorry this should be I beg your pardon this should be d h over rho u bar square. Then you will notice that this is nothing but d p star by d x star is nothing but 2 times f l essentially this is the frictional term this is in a way represents now the convection term and this is the pressure gradient term we will manipulate this a little further. So, we define u dash let me write down this equation the equation that we wrote down is d 2 u star by d y star square minus beta square u star equal to r e d p star by d x star. So, I introduce here u dash equal to u star minus r e by d x star. u by beta square d p star by d x then this equation if I substitute for u star and notice that this is a function of x only and therefore with y this will simply become d 2 u dash by d y star square minus beta square u dash equal to 0. Now, this is with boundary condition u star equal to 0 at the wall y star equal to 0 and at the axis of symmetry where y star is equal to 1 by 4 remember this is the this is the duct of length 2 b this is the axis symmetry we are measuring y this way. So, this is y equal to 0 and this is y equal to b, but d h is equal to 4 b. So, this becomes y star equal to 1 by 4 b divided by 4 4 b and this is also equal to y star. So, at y star equal to 1 by 4 the velocity gradient will be 0 the velocity gradient in y direction would be 0 and that is what I have the equation has 2 boundary condition this equation has 2 boundary condition u star equal to 0 at y star equal to 0 and d u dash by d y star equal to 0 at y star equal to 1 by 4. Now, this is a very familiar fin equation in heat conduction that all of you are familiar with and therefore, the solution is very simple it is a c 1 exponential beta y star plus c 2 exponential of minus beta y star and if I make use of these two conditions I can determine c 1 and c 2 they evaluate in the following way c 1 is Reynolds by beta square d p star by d x star divided by 1 plus exponential of beta by 2 and c 2 equal to c 1 exponential of beta by 2. So, this is quite straight forward algebra to really evaluate c 1 and c 2 which incidentally is a function of c 1 into exponential of beta by 2. Now, to evaluate pressure gradient we make use of the definition of mean velocity we said u bar is equal to 1 over b integral 0 to y u divided sorry 0 to b. If I change y to y star then essentially you get 1 over b into u bar into 4 0 to 1 by 4 u over u bar d y star d y would be equal to 4 b yes. So, this is 4 times b and this would be u bar and therefore, b b gets cancelled. And you will get 1 by 4 will be equal to 0 to 1 by 4 u star d y star. If I now substitute for u star 0 to 1 by 4 u dash minus n o by beta square d p star by d x star d y star. I know u dash as a function of y and therefore, integrate then you will see that you can get r e d p star by d x star equal to f l Reynolds equal to beta times 4 c 1 exponential of beta minus 2. So, knowing the value or assuming a value of beta you can always calculate beta c 1 and therefore, evaluate expression d r e d p star by d x star the local pressure gradient. Another quantity of interest is the central line velocity how does the velocity at the central line change of course, the velocity begins at x equal to 0 u c will be equal to u bar at x equal to 0, but it will go on increasing till the flow become fully developed. So, if we consider this equation again equation 10 this equation again then and write it at the axis of symmetry y star equal to 1 by 4 it will simply become d 2 u d y star square at 1 by 4 minus beta square u star c equal to r e d p star by d x star where it can be shown that the evaluation of d 2 u star d y star 1 by 4 is nothing but 2 c 1 beta square over exponential of beta by 4 and therefore, a little algebra of about 2 3 lines will give you u star c is a function of again beta and c 1. So, we have evaluated the pressure gradient local pressure gradient we have evaluated the local value of the central line velocity for a given beta which incidentally means for a given value of x along the duct. The next question is of course, which value of beta corresponds to which value of x because beta has been arbitrarily chosen numerical value and we still have to work out what is it the connection between beta and x. So, let us go back to equation number 5 this was the over original equation. If I integrate each term of this equation from the lower wall to the axis of symmetry the equation reads as r e times d by d x of u star u star plus d by d y star of v star u star equal to minus r e d p star by d x star plus d 2 u star by d y star square. If I integrate this equation from 0 to y star then I will get d by d x of 0 to 1 by 4 u star u star d y star plus v star u star at 1 by 4 minus v star u star at 0 equal to minus Reynolds by 4 d p star by d x star because v star u star is equal to minus this is not a function of y. So, simply I get at 1 by 4 plus d 2 d u star by d y star at 1 by 4 minus d u star by d y star at y star equal to 0. So, this term remains intact but at the axis of symmetry v star is 0 likewise u star is 0 at 0. So, that term is also 0 this term remains intact d u star by d y star at the axis of symmetry is 0. Therefore, that term goes to 0 and this term of course survives because there is a velocity gradient right at the wall. Therefore, the equation will now read as I have shown here, r e d by d x star 0 to 1 by 4 u star u star d y star equal to minus r e by 4 d p star by d x star plus d u star by d y star y star equal to 0. So, if I now substitute u for u star in terms of u u dash here and evaluate d u star by d y star from velocity profile where u dash is equal to u star. So, I substitute for u star in terms of u dash and then carry out the integration and differentiation. Then after some algebra about one page you derive an equation that r e equal to d f 1 by d x star equal to f 2 where f 1 and f 2 are functions of beta as shown here f 1 would be c 1 star plus i 1 plus i 2 minus i 3 i 1 is equal to that term i 2 is equal to that term and i 3 is equal to that term and f 2 itself would again be a function of beta and c 1. Another way of writing this equation is to say x star equal to r e times integral f 1 value of f 1 at x star equal to 0 to value f 1 at x star equal to x star 1 over f 2 d f 1. It is this that establishes the relationship between x star and beta that we find out because f 1 and f 2 are functions of beta. How is this done? It is done as shown here. We want to evaluate this integral. So, we first assign different numerical values to beta and generate functions f 1 beta and f 2 beta and tabulate them. Then we simply carry out integration by trapezoidal rule. So, essentially you assume certain values of beta f 1 and f 2 a large value let us say I have used here 60. So, you have a value of f 1 60, f 2 60 and so on and so forth. Then 59, f 1 59, f 2 59 and so on and so forth and you go on towards tending to 0. So, we have for each beta value function. So, then what you do is simply in order to carry out you say sum is equal to 0 and then you say sum is equal to sum plus f 1 60 minus f 1 59 divided by half of f 2 60 plus f 2 59. As the integral, this is the integral 1 over f 2 d f 1 and simply go on adding these terms and you carry out the integration. This is the trapezoidal rule which I just wrote. So, now of course, you can make beta arbitrarily large, very large. It can go up to infinity and on the other side will go to 0. I took several values and I found going up to 60 was quite good enough because as beta tends to infinity, x star tends to 0 and as beta tends to 0, x star tends to infinity. So, of course, in lieu of infinity one can choose any value. Now of course, once you have chosen beta values, you can also calculate for these beta values. You can calculate three more quantities. One is c 1, second one is u c and the third one is f l. So, these are the three quantities you can calculate. They are all functions of beta and once the integration relates the value of x star to beta, we know that these are the values that correspond to x star. Although large number of solutions were generated for very tiny steps of beta, I am showing you solutions for select values of beta. So, here are the values. At beta equals 60, c 1 turns out to be extremely small and negative. It corresponds to x equal to 4.60 into 10 to the power minus 6. Here, x is divided by d h, the hydraulic diameter and then divided by Reynolds number. u c star is simply the central line velocity divided by u bar and this is f l multiplied by Reynolds number which is the local friction factor. So, you can see for each value of beta, a value of x has been discovered. As beta goes on reducing, x goes on increasing and finally, for beta equal to 0.1, x is equal to 0.01, u c star becomes equal to 1.4, 998, almost 1.5. So, asymptotically, you can see from about 0.005 onwards, the change in friction factor or essentially the pressure gradient is very small. The change in central line velocity is also very small and essentially a state of fully developedness has been reached asymptotically at infinity. Of course, u c star becomes 1.5 and 24. This incidentally can be shown even from the analytical solution that we already have. So, you can see as the flow develops, the central line velocity increases, friction factor itself, multiple Reynolds number of course decreases with x. Solutions of this type can be generated, here I have generated them for flow between parallel plates, where the resulting equation is a simple fin equation. But you can also do this for flow in a circular tube, where the resulting equation is a Bessel's equation. Also in case of an annulus, entrance rigid of an annulus, the resulting equation is a Bessel's equation. The algebra there is much longer, but here it is a simple algebra and therefore, I chose to do the flow between parallel plates, which in itself is of great interest. This is the variation of friction factor versus Reynolds number, starts from the high value and goes down to a constant 24 at about 0.01. Likewise, the central line velocity goes from 1 and ends up with 1.5. So, fully developed friction factor is 24, fully developed central line velocity is u c by u bar 1.5. We can now treat this distance L v by d h r e approximately equal to 0.01. The behavior near the fully developedness is very asymptotic and therefore, one could not really fix precisely the value of development length, but purely by observation we can say L v by d h Reynolds of 0.01 is a good estimate of the flow development length, which we said was our objective. Now, all these results that I mentioned of the fully developed plow are well known from the u g tex, either from analytical solutions for annulus and circular tube or in the case of ducts of non-circular cross section where CFD analysis is used, you can get values of development lengths and I will show them on the next slide. But here note that instead of the local friction factor that I have plotted here, sometimes people prefer because the local friction factor gives you variation of the local pressure gradient, but people are interested sometimes to measure actually the local value of pressure itself and that means, what they define what is called as an apparent friction factor because this is what you will measure in an experiment. Apparent friction factor is defined as minus 1 over 2 p x minus p x 0 divided by x d h over rho u bar square and it is simply 1 over x 0 to x f l d x. So, our solution for f l can be integrated to get apparent friction factor and it would go something like just slightly above the local friction factor and again predict at long length you will predict that apparent friction factor is also into Reynolds number is again 24. So, let us look at the values of development lengths predicted for different ducts and here I have chosen a few ducts, a ducts of few cross section circular tube its development length is 0.05 l v divided by d h divided by Reynolds for an annulus of different radius ratios r i by r rho. So, if the inner radius is very small it is 0.01944, if it is 0.1, 0.17 and so on and so forth. Note that these values need not be monotonous because the value of d h itself goes on changing for different radius ratios and so does the value of Reynolds number which is a function of hydraulic diameter. So, these need not be monotonously increasing or decreasing the values are simply numbers to be used directly in an analysis. But notice that when r i by r rho tends to 1 this is r i and this is r rho, when r i by r rho tends to 1 which means r i is very very close to r rho then for all practical purposes the flow in annulus is much like the flow in between parallel plates and predictably you are seeing that the development length is 0.01 which we actually calculated through our analysis. Similarly, in rectangular ducts of side ratio b by a or essentially it is 2 v by 2 a then the ratio b by a gives you the aspect ratio. When b by a is 0 again when b is very very tiny you essentially get flow between parallel plates because a then is infinite and the l v by d h ratio is 0.01 as we predicted for 0.125 is 0.227, 0.254 to 7.5 and so forth. And when you tend to 1 that is a perfect square the development length is 0.0752 here is a result for semicircle this has been calculated by numerical analysis and l v by d h Reynolds number is 0.0622. So the development length normalized with hydraulic diameter and Reynolds number has a notionally fixed value for a given duct and is always as you will see when we go on to heat transfer calculations it is very essential that we know what these development lengths are so that we can account for their presence in the heat transfer analysis. So, the integral work of this type is almost now abandoned because most people use CFD based solution procedures, but it would be useful for you to know who were the early contributors to this very interesting aspect of internal flows and therefore I give you a few references. The first one is by Sparrow and Lynn published in 1964 it is called flow development lengths in hydrodynamic entrance region of tubes and ducts physics of fluids volume 17 another one by Hahn hydrodynamic entrance lengths for incompressible laminar flow in rectangular ducts published in 1960 Lundren's parent star pressure drop due to the entrance region in ducts of arbitrary cross section this is in journal of basic engineering 64 and then for annular passages there is the paper by Heaton Reynolds and Case in international journal of heated mass transfer. So as I said the flow in the entrance region is quite complex situation and one needs exact analytical solutions for classroom work can only be worked out for simple cases I gave you example of one case and then concluded with the magnitudes of the development lengths as encountered in different ducts. In the next lecture we will take up the case of fully developed flow inside ducts of circular and non-circular cross section.