 Today, we will look at fully developed laminar flow in a duct. We will begin with definition calculate friction factor for a simple circular cross section duct as well as annular cross section duct and then somewhat more complex ducts like rectangular duct or an annular sector duct. You will recall from lecture 14 that fully developed flow region occupies greater part of the tube length in ducts in which length to diameter ratio is very large. This is particularly obtained in micro tubes where diameter is so small that any practical length gives you very high length to diameter ratio. Fully developed flow friction factor F sub f d provides the lower bound to the apparent friction factor as well as the local friction factor. These two were evaluated for flow between parallel plates during our last lecture. In laminar flows as a rule, fully developed flow friction factor multiplied by Reynolds number is an absolute constant and that constant depends on the duct geometry for the given duct. How is fully developed flow evaluated? It is simply from force balance over an actual distance delta x of the tube. So, the pressure drop over length delta x multiplied by the area of cross section of the duct equals the shear stress at the periphery of a cross section multiplied by the perimeter multiplied by delta x tau wall bar is the averaged shear stress. Now, you know that the friction factor is the ratio of shear stress divided by kinetic energy which we in this case define as rho u bar square by 2 where u bar is the mean actual velocity. From this relationship, it will be obvious that it would become 1 by 2 modulus of d p d x d h by rho u bar square. I have put a modulus to remind that d p d x is negative in any ducted flow. The hydraulic diameter d h is 4 times area of cross section divided by perimeter of the duct. In this definition, friction factor is called the Fanning's friction factor named after Fanning. Let us take the case of a circular tube which you all very well know. This is simply to refresh your memory. When the flow is fully developed, then the radial velocity as well as the circumferential velocity components are 0 and the axial velocity gradient is also 0 because the boundary layers have now met which results in the axial pressure gradient to become constant. Hence, the axial momentum equation reduces to 1 over r d by d r r d u by d r equal to 1 by mu d p by d x equal to a constant with boundary conditions u equal to 0 at the tube wall r equal to r and the gradient d u by d r equal to 0 at the axis of symmetry. Integrating this equation will give you once or twice with respect to r and using these two boundary conditions to evaluate two constants of integration will give you u equal to minus r square by 4 mu d p by d x 1 minus r square by r square. Hence, evaluating the mean velocity as 0 to r u r d r divided by 0 to r r d r gives you minus r square by 8 mu d p by d x. Notice the negative sign here as well as here which simply allows counts for the fact that d p d x now is of course, negative which gives us u over u bar equal to 2 into 1 minus r square by r square, a very well known result that you will recall from your undergraduate days. What about the wall shear stress? Well, wall shear stress is tau equal to minus mu d u by d r at the wall r equal to r and that becomes equal to r by 2 d p by d x with a negative sign giving you 4 mu u bar by r that would be circumferentially constant. Hence, the friction factor which is tau wall over rho u bar square by 2 as we saw in the previous slide as well as this definition gives you 16 by r a. Note that f f d into Reynolds number therefore, is 16 and constant as we said before and also for a circular tube the hydraulic diameter itself becomes equal to the diameter. Often you may have encountered the fact that f f d into r e is taken as 64 rather than 16 that simply follows from the definition that f f d is 2 times d p by d x d by rho u bar square. However, as we showed on this slide that the correct definition of f f d is according to the fanning friction factor and which results from the force balance. So, sometimes although the friction factor is taken as 4 times the fanning friction factor, we would continue in our lectures with the fanning friction factor. Let us now turn to the annulus for the fully developed flow in an annulus equation 1 will again remain as it is that is no difference there because again v r v theta r 0 and axial derivative of velocity is 0, but the boundary conditions now will be u equal to 0 at r equal to the inner radius and u equal to 0 at r equal to outer radius and that is what is mentioned here the no slip boundary conditions at r equal to r i and r equal to r o. If I integrate the equation 1 twice I would get two constants of integration c 1 and c 2 and applying these boundary conditions would give me c 1 equal to that c 2 equal to that integrating this equation to evaluate u bar the average velocity gives me that expression in all these cases r m is really the velocity is the radius where d u by d r is 0. Remember if I have axis symmetry like this is the inner radius r i this is the outer radius r o then the plane of the velocity profile will look like that and with the 0 velocity both then the point at which d u by d r is 0 is the radius r m. Remember r m will not equal r i plus r o by 2 in general in fact as this expression shows r m is very much a function of the radius ratio r i by r o. So, we use locator r m and therefore u over u bar becomes 2 times r o square minus r i r square plus 2 r m square l n r by r o divided by r o square plus r i square minus 2 r m square. So, as I said where r m is the radius of maximum axial velocity of the location of d u by d r equal to 0. Mind you this integration to evaluate u bar is not very straight forward because l n r is involved and one needs to take special care to evaluate l n r r d r properly. Further based on hydraulic diameter now the hydraulic diameter for a duct annulus section will be 4 times a cross sectional area will be pi by pi into r o square minus r i square and the perimeter will be 2 pi r i plus r o which gives us pi and pi gets cancelled 2 gives me 2 2 times r o minus r i. So, twice the flow width r o minus r i is the hydraulic diameter. So, if I use that definition of hydraulic diameter it is very easy to show that f into Reynolds number would be 16 times 1 minus r star square r star is the radius ratio of the duct r m star is r m by r o and d h is 2 into r o minus r i is the hydraulic diameter. Now, of course, as you can imagine when r i by r o tends to 1 that means the 2 radii are very close to each other then we practically have the case of flow between 2 parallel plates and for which f into Reynolds number is 24 as you derived in your undergraduate class. Let us look at the numbers. So, for different values of r i by r o you can evaluate the friction factor. In fact, as r star tends to 0 you will see f into Reynolds number tends to 16 which is the lower bound indicated by the circular tube. The r i by r o standing to 1 is the upper bound and that is the flow between parallel plates for all other radii product of friction factor Reynolds number would lie between 16 and 24. Let us now consider more complicated case and that is flow in a rectangular duct. Now, in this case you will appreciate if I have a rectangular duct like so with this as the symmetry line in x direction and let us say this is this direction is z this direction is y then the boundary layers will develop on the y max and y 0 wall as well as on z equal to 0 and z equal to z max walls. Therefore, I would have a structure of boundary layers growing like so and ultimately the boundary layers will meet at some point. After the boundary layers, you will see components w and v. So, w will be 0, v will be 0 and d u by dx will be 0 and therefore, you are left with an equation of the fully developed flow in a rectangular duct is simply d u d 2 u by square plus d 2 u by d y square equal to minus 1 over mu d p by dx equal to constant and that is what I have shown here sorry beg your pardon this is this should be plus d p and that is equal to constant. So, if you divide u divide by minus 1 over mu d p dx then you get d 2 u star by d z square plus d 2 u star by d y square equal to minus 1. This is a Poisson equation with a constant right hand side with a constant right hand side and what are the boundary conditions? If we say that the duct dimension is B on the narrower side and A on the longer side then u star is equal to z at plus and minus a by 2 and it is also equal to 0 at plus and minus b by 2. Now, a Poisson equation of this type can be solved by employing double Fourier series with the method of undetermined coefficient and I will briefly explain what that method is. In the most general case both sides of the Poisson equations are multiplied by f 1 into f 2 where f 1 is a function of z only and f 2 is a function of y and they are taken as Fourier series a m cos m pi z plus a by a plus b m sin m pi z by a. So, this is a function of z and likewise this is a function of y again a Fourier series. In the present case the boundary conditions require that the term considering sin functions must vanish on the wall the velocity being 0 you will see that sin m pi by 2 would be a finite and that would make f 1 a finite at the wall which we do not want and therefore, the sin must be sin terms must be 0 only the cos term survive and therefore, our equation would be u star equal to c m and f y z m equal to 1 3 5 to infinity n equal to 1 3 5 to infinity where f y z is equal to cos m pi z a which is the product of that and that and c m n is really a m into c m as you can see. So, we need to determine the constant c m n so that we can get value of u star this is the double Fourier series. You must remember what the function is f y z is cos m pi z by a multiplied by cos n pi y by b. So, the solution procedure is like that the given Poisson equation is multiplied by f y z on both sides and integrated from limits on z and y. So, we have integrated it from minus a by 2 to plus a by 2 minus b by 2 to plus b by 2 d 2 u d z square d 2 pi square f y z d y d z is equal to the right hand side which is minus 1 into f y z d y d z. Integration by parts would give the left hand side equal to minus pi square m square by a square plus n square by b square minus a by 2 plus a by 2 minus b by 2 u star f y z d y d z simply these terms are reduced to that multiplied by u star into f y z. The right hand side on the other hand would give you minus 4 a b m n pi square minus 1 raise to m plus 1 n by 2 minus 1. So, if you substitute for u star and equate the left and right hand sides what is u star? u star is simply this. So, if we substitute for u star in this expression equate left hand side with right hand side we get c m n equal to the expression given here and again here substituting for f star square y z would give me that expression for c m n. Therefore, the velocity is now given by c m n that expression multiplied by cos m pi z by a plus sin n pi z y by b is the f y z function. Of course, you must now integrate this over the cross section of the duct to obtain u bar star which comes out to be 64 by pi raise to 6 into b square into all this and I have now used gamma as the aspect ratio the ratio of the shorter side to the longer side gamma is the aspect ratio of the duct. So, the final solution then is this where f y z again as I said is product of cos m pi z by a and cos n pi y by b. Therefore, the friction factor into Reynolds number would be given by 1 by 2 d h square by u bar equal to that. d h of course, for a hydraulic diameter in this particular case in the rectangular case would be b and a. So, 4 times area of cross section a b divided by 2 times a plus b which gives me 2 a b divided by a plus b or if I were to divide this by a then I get this is equal to hydraulic diameter. So, I get this as 2 b divided by 1 plus a b by a which is d h by b will be equal to 2 divided by 1 plus gamma that is what I have shown here d h by b would be 1 plus gamma and gamma is b by a the aspect ratio. If we evaluate friction factor Reynolds number by substituting different values of gamma we obtain this results that are shown here. So, when gamma is equal to 1 that is the aspect ratio is exactly equal to 1 then you get a square duct this both shorter and longer side are equal length and we get a square duct in which case the maximum velocity would be 2.08 u max divided by u bar would be 2.08 and the fully developed friction factor into Reynolds number will be 14.26 and that is the case of a square duct. As you go on reducing aspect ratio you see that the maximum to u bar ratio in general goes on falling and the friction factor into Reynolds number goes on increasing. When we come to 0 that is when the shorter side is much smaller than a then of course you approach more less the flow between parallel plates and you can see u max over u bar as you recall would be 1.5 and f r e is 24 which is very well known result. In all these cases you have to choose the values of m and n in this series and I have chosen m and n equal to 101 of course it varies with duct shape but once you evaluate these things on a computer you can give any value and once the series has converged it is good enough to take any value of m and n beyond 101 the results will not change. Now, let us look at even more complicated duct and that is a sector of an annulus or a circle whichever way typically you get ducts which are of this type. This is the inside radius r i and this is the outside radius r o and this is the included angle which I have called here as theta naught. So, we have two parameters here r i by r o and theta naught. So, we expect from friction factor f d into Reynolds number to be a function of both these parameters. In the fully developed state the radial velocity v r as well as the tangential velocity v theta will be 0 this v r equal to v theta will be equal to 0 and the d u by d x in the axial direction will now be 0 which renders the momentum equation in the cylindrical polar coordinates to the form shown here 1 over r d by d r r d u by d r plus 1 over r square d 2 u by d theta square equal to 1 over mu d p d x equal to a constant and here is a little trick which you must bear in mind z equal to ln r by r o and u star equal to u divided by minus r o square by mu d p d x. Then you will notice that d 2 u star by d z square plus d 2 u star by d theta square equal to minus e raise to 2 z again we get a Poisson equation, but unlike the case of a rectangular dirt the right hand side is now a function of z or the function of radius r z is ln r by r o and therefore, the right hand side is function only of the radius because u star is equal to 0 at z i ln r i by r o and 0 is equal to 0 at theta plus minus theta naught by 2 yes at the two walls at the two walls. This particular case is the one in which the inner radius is 0 here the inner radius is finite. Second sectoral ducts as I said are formed in slots of electrical stampings or the smallest salamistre sector of a internally fined annular duct. Now, the method of solution for this case is exactly same as in the case of a rectangular duct that I showed you earlier. So, we multiply both sides by an appropriate Fourier series function cos m pi theta by theta naught sin n pi z by z i where these functions are obtained by noting the boundary conditions and f m n is the Fourier coefficient. We can evaluate that in the manner I explained earlier to f m n would be equal to f m f 1 divided by f 2 f 1 of course, is given by that z i as you know is ln r i by r o and f 2 would be given by this which also includes the included angle theta naught. Now, we can evaluate u bar u bar star would be simply u bar star would be u r d r d theta r i to r o and r i by r i by r i by r i 0 to theta naught or if you like minus theta naught by 2 to plus theta naught by 2 divided by minus theta naught by 2 to plus theta naught by 2 r i by r o r d r d theta and if you carry out that evaluation you get this equal to f 3 by 4 the sum given by that. Remember m varies 1, 3, 5 to infinity n on the other hand varies as n equal to 1, 2, 3 to infinity and f 3 is given by that and f 4 is given by that and therefore, the friction factor multiplied by Reynolds number is simply d h by r o square by 2 u bar star square where u bar is taken from there. The hydraulic diameter in this case would be as you can see it looks a very complicated evaluation but very simple idea is d h is equal to 4 times the cross sectional area divided by perimeter. So, 4 times theta naught into r o square minus r i square divided by theta naught into r o plus r i plus 2 times r o minus r i that is what it is all about and if I were to divide this through by r o then note that z i is simply l n of r i by r o and therefore, r i by r o will be equal to exponential of z i and that is what I have shown here. So, it is possible to evaluate friction factor Reynolds number as a function of these two parameters r i by r o and theta naught. Here are the evaluations for a few values of radius r star equal to 0.75 r i by r o, 0.5, 0.25 and 0.001 which is a very small value of course, theta naught equal to 180 degrees will give you semicircular duct. As you can see here if theta naught was 180 degrees that would give you a semicircular duct and then 90 degrees would give you that kind of a duct and so on and so forth coming down to very narrow angle of 5 degrees and you can see the friction factor at 0.75 for example, is 25 for a semicircular duct going down to 17.6. Here at r star equal to 0.5 it goes down to 9 goes down and then increases to 19. This is 0.25 and this of course, would correspond to r star equal to 0 again for a semicircular duct 16 down to 12. There is a very monotonic decline in this. So, these two parameters ducts the annular sector family as well as the rectangular duct family are amenable to relatively easy solutions using Fourier series. However, nowadays with extremely compact heat exchangers and extremely in electronics for example, where the flow passages are so miniaturized that the compactness requires that the ducts are often highly squashed or curved. As I said in lecture 14 they can be moon shape, they can be sinusoidal shapes and so on and so forth and these the method that I described is as long as the boundary shape is nice and regular which is describable by a nice function. But in order to deal with ducts which have very complex shapes, we need to consider certain special methods and that is what I will turn to in my next lecture on complex duct cross sections. The fully developed laminar flow in complex ducts of complex cross sections.