 So, in the previous two lectures, we considered friction factors in fully developed laminar flow in regular section ducts as well as ducts of complex cross sections like any arbitrary cross sections as long as they were singly connected. We now turn our attention to fully developed heat transfer and like we did in the case of friction factor, we shall first of all look at very very simple situations like the circular tube family, annulus and so on and so forth under variety of boundary conditions. To do that, first of all we must define fully developed heat transfer. Consider this duct in which fluid at uniform temperature enters and there is a constant wall heat flux supplied at the wall. Then you can see that the wall temperature will begin to rise and after some distance would rise at a linear rate. The bulk temperature, however, typically from first law of thermodynamic that is m dot Cp into dt bulk by dx mind you heat flux is constant and therefore, m dot Cp into dt bulk by dx into dt bulk will be simply equal to q wall into perimeter into dx and therefore, dt bulk by dx will be simply q wall into perimeter divided by m dot Cp and all these are constants and therefore, you would see that the bulk temperature would rise linearly with x right from the start. We of course, assume that the velocity profile is fully may or may not be fully developed and therefore, this shows the thermal development of the temperature profile. You can see it is started with a uniform temperature, but then as the heat flux comes in it assumes a curved shape. Of course, the gradient at each section would be constant because q wall is constant ultimately the profile would become like that and since heat transfer coefficient is defined as q wall over t wall minus t bulk this is constant. The difference between t wall minus t bulk is small to begin with and therefore, heat transfer coefficient is high near x equal to 0 and it progressively drops as difference between t wall minus t bulk increases and ultimately becomes constant because t wall minus t bulk itself becomes constant. That is depicted on the left figure here. Now, let us consider another boundary condition which is frequently met that of a constant wall temperature. This is like steam heating in which the wall temperature is constant. The fluid enters at a value lower than the wall temperature which is shown here and the fluid bulk temperature would therefore, start rising. You can see here unlike the constant wall heat flux case, the bulk temperature does not rise linearly, but it arises non-linearly with x. To begin with the gradient of temperature is very large and therefore, at the wall and therefore, h which is minus k d t by d y at r or plus here divided by t wall minus t bulk. Although t wall minus t bulk is large, so is k d t d y at r is very large and as a result again you get a variation of heat transfer coefficient h which is very similar to that in a constant heat flux case. That is what I have shown here h goes on changing with x, but decreasing with x. The temperature profile of course, becomes curved because of thermal boundary layer development. Mind you, at each section the t wall will be same. Only thing is the t center line will go on increasing, so that the t bulk goes on increasing and this variation is non-linear as I have indicated here. Of course, if the duct was very very long say going up to infinity, the bulk temperature itself would become equal to the wall temperature. In fact, the temperature at all radii will become equal to the wall temperature, but that would occur only at infinity and it is not a case of importance in practical ducts which have limited length to diameter ratio. Fortunately however, it so happens that the as the temperature profile develops, the gradient at the wall goes on changing as a result the heat transfer rate goes on changing with x. A point is reached beyond which although t bulk is changing and heat transfer the gradient of the temperature is changing at the wall. A point is reached where k dT dr divided by t wall minus t bulk becomes constant or h becomes constant. So, in heat transfer we say fully developed heat transfer is identified with constancy of h with axial distance. To show this let us consider this, what does this imply for the temperature? So, we define for example, phi as a function of x and r in case of a circular tube as t wall x minus t x r divided by t wall x divided by t bulk x. Now, when you t wall and t bulk can only be functions of x where t bulk is of course, rho C p u T dA and rho C p u dA where dA is the area of cross section. In fully developed heat transfer we say that d phi by d x will be 0, the dimensionless temperature profile will go to 0 or essentially dimensionless temperature profile will become constant with x and since phi is a constant with x, we would expect that d phi by dr at the wall will also be constant with x, which of course means that it is equal to minus dT by dr r equal to r divided by t wall minus t bulk x equal to q wall over t wall x minus t bulk x. This definition is used for constant heat flux case, this definition would be would indicate definition of h in case of constant wall temperature and you will see that h becomes constant. So, fully developed heat transfer constant c of h implies that d phi by d x must be equal to 0, it is the dimensionless temperature which must be 0. Remember, we define fully developed flow where d u by d x itself was equal to 0. In heat transfer we say fully developed dimensionless temperature gradient with x is 0 is the condition for fully developed heat transfer and it implies h equal to constant. Although t wall and t bulk and indeed temperatures at every radius may vary with x and r, but as long as t wall, t bulk and the heat flux that ratio remains constant or h is constant, we say the fully developed heat transfer has been reached. So, let us take the first case, the simplest case which you have studied in your undergraduate course and that of a circular tube with constant heat flux at the wall. So, in that case the governing equation as you will recall is simply 1 over r d by dr r dT by dr equal to u divided by alpha, alpha is a thermal diffusivity k by rho Cp into dT d x, but since q wall is constant dT d x would be replaced by dT bulk by d x and it will actually also mean dT wall by d x. From our definition you can see if dT bulk if I were to say d phi by d x equal to 0 and dT x equal to dT bulk x then dT wall by d x will also be 0 or to make it explicit. Let us say d phi by d x equal to 0 would imply that where phi is equal to t wall minus t divided by t wall minus t bulk and this would imply 1 over t wall minus t bulk into dT wall minus t by d x minus 1 over t wall minus t over t wall minus t bulk squared dT by d x t wall minus t bulk equal to 0 which essentially gives me dT w minus t by d x equal to t wall minus t over t wall minus t bulk into dT wall minus t bulk by d x which implies that all dT w by d x is equal to dT by d x and is also equal to dT bulk by d x. The implication is that once the fully developed flow is reached temperatures at all radii increase with constant with x equal to and at all radii dT x is equal to simply dT by d x. So, you will see that dT bulk by d x of course is simply this is m dot into C p and this is q wall into perimeter which of course is a constant and therefore, the equation would simply become 1 over r d by d r r dT by d r and fully developed flow in a circular tube is given as 2 u bar 1 minus r square by r square and therefore, this would be that. So, if I integrate this equation twice with boundary conditions t equal to t wall at r equal to r and dT by d r equal to 0 at the axis of symmetry then I can determine two constants of integration and the result is shown on the next slide here. T is equal to t wall minus 3 by 4 q wall over k r plus q wall over k r into r square minus r 4 r. Now, of course, I evaluate t bulk which is you will recall and we assume constant properties. So, rho C p just cancels and if I were to substitute this for temperature and the velocity profile from the previous slide then I would get t w minus 11 by 24 q wall over k r. Now, remember in all these cases integrations are very important and you have to take a lot of care to make sure that you make no errors in evaluating the temperature, in evaluating the integrals. So, you can now see that Nusselt number which is defined as h d by k it will be equal to 2 r by k q wall over t wall minus t bulk and t wall minus t bulk is 11 by 24 q wall r by a. So, that gives us 48 by 11 equal to very well known result 4.36 36 which you derived in your undergraduate course. A similar analysis for flow between parallel plates which are separated by a distance 2 v say if I have two parallel plates and the distance between the plates is 2 v and if I measure y from the axis symmetry then you fully develop divided by u bar in this particular case is 3 by 2 1 minus y by y square by b square as you can as you recall and if I carry out the similar analysis for constant heat flux at both the walls then the answer I would get is based on hydraulic diameter. Remember the hydraulic diameter for plate distance of 2 v d h is 2 times the plate distance and that is equal to 4 v. So, h 4 v by k will give you 8.235 very simple cases that can be done by pencil and paper integration no difficulty at all you do not need numerical integration or anything like that. Let us now turn our attention to little bit more complicated thing again I am going to consider now a flow in an annulus flow inside an annulus you can get variety of it may be heated from inside or it may be heated from outside q wall o or q wall i. So, we are going to consider this particular case in fact, both the cases in turn. So, let us look at you will recall from our earlier analysis that fully developed velocity profile is given as 2 by m 1 minus r by r o square plus b l n r by r o which has a logarithmic term in it and b itself is this m is this and r star is r i by r o the radius ratio of the annulus. The equation proper of course remains the same it does not change at all only thing is d t dx here will be d t bulk by dx and that would be equal to 2 pi r o q wall o plus r i q wall i divided by rho c p u bar pi r o square minus r i square. We can have two types of boundary conditions in the first case outside wall is heated. So, q wall o is given equal to k d t by d r at r o and the inner at the inner radius r i we have t w i. On the other hand we can also have a case in which the outer inner heat flux is given and the outer wall temperature is t w o annulus solution. So, if I integrate this particular equation with substituting for d t by dx d t bulk by dx and u f d equal to this and integrate this equation twice which is my new quite elaborate integration because logarithms are involved. I would get temperature itself as that where a the multiplier of this square bracket is this and q star is inner wall heat transfer divided by outer wall heat flux. I will consider two cases as I said in the case one where outer wall is heated c 1 turns out to be this and c 2 turns out to be this. In the other case where inner wall is heated it will turn out to be this and c 2 is equal to that. So, these things require very careful algebra in order to avoid any errors. I can express the solution that of the previous slide in a more compact form as t minus t w o q wall o r o by k which has dimensions of temperature q wall r o by k has dimension equal to 1 over m into 1 plus q star r star plus minus q star is q wall i minus q wall o 1 minus r star r star is r i by r o whole square multiplied by a function f 1 minus function f 2 and I have given here the values of f 1 and f 2. We now define let us say in case one where q wall o is heated I will define n u o equal to at the outer wall h o d h by k where the heat transfer is specified and that would be equal to q wall r o by k t wall minus t bulk into 2 into 1 minus r star square. Remember this is nothing but for the annulus d h is equal to 2 times r o minus r i 2 times r o minus r i and that is what is reflected here. So, if I take r i common you will get that. So, with this temperature now you must evaluate the bulk temperature mind you the bulk temperature evaluation becomes extremely difficult because there are logarithmic terms involved and numerical integration is the best way out. It does not require too much effort. For case two where inner wall is heated that is q wall i is given the solution can be expressed compactly in this fashion r star square by m into 1 over q star plus r star or that multiplied by a function f 3 plus all this and f 3 itself is given by a long expression like that. Mind you these expressions take quite a bit of algebra to arrive at and again in this particular case I will define n u i equal to h i d h by k equal to q wall i r o by k divided by t wall i minus t bulk into 2 into 1 minus r star. Again using this temperature profile I must evaluate t bulk which is required here by numerical integration. It is very fortuitous that the n u i n u o for a variety of radius ratios can actually be expressed in the form I have indicated. n u i would be equal to n u i i into 1 minus theta i by q star and n u o can be expressed as equal to n u o o into 1 minus theta naught minus q star where n u o o and theta o n u i and theta i are simply functions of the radius ratio. Now, remember q star was actually equal to q w i divided by q w o is equal to theta i. If suppose that was equal to theta i then this gives a odd result that n u i would be equal to infinity. But that should not worry us because all it implies is that it does not imply infinite heat transfer, but simply that the inner wall temperature turns out to be equal to the bulk temperature. Therefore, n u i goes to infinity. Similarly, if q star is less than theta i then n u i will turn negative which implies negative h i. But again as we said repeatedly that this is not a particularly unacceptable situation all it implies is that since n u i is negative t wall must be greater than t bulk, t wall i must be greater than t bulk, t wall i must be less than t bulk and as a result the n u i has turned out to be negative. So, similar arguments would apply of course, to n u o. So, values of n u i i, n u o o, theta i and theta o are given on the next slide. But and you to write the previous solutions in this manner you again have to do little bit of algebraic manipulation to show that this theta i and n u i i are functions of radius and only q wall minus q wall i over q wall o separates out as a function. So, here are the solutions that I present from r star equals 0 which is the circular tube and you can see n u i i has no meaning, but n u o o is 4.364 and its influence coefficient is 0 and therefore, n u o itself as you can see here influence coefficient is 0 and n u o would be equal to n u o o and therefore, that would be equal to 4.364. As I go on increasing the inner radius then you can see that the coefficients and the n u o o values go on changing. When r star becomes 1 you essentially have flow between parallel plates. In this case both n u i i and n u o o and theta i and theta o are exactly identical and therefore, as they should be and our table confirms that they turn out to be the same. So, usually numbers are not readily available for annulus, but with this method you can see now you can evaluate them for any situation for any q wall i any q wall o that may be prescribed and you can readily recover the solutions for entire family of annulus solutions. I will deliberately consider now a problem in which I am considering flow between parallel plates that is the last entry in that table. Let us say this is q 2 and this is q 1 the velocity is fully developed. So, if you read the problem it is like this the flow between parallel plates which are 5 centimeters apart. Therefore, d h will be 10 centimeters and I have said q 1 is equal to 1 kilowatt per meter square and this is 5 kilowatts per meter square. Of course, since the flow is fully developed the bulk temperature is rising with x and I am simply considering the k x axial position where t bulk is 30 degree centigrade and the conductivity of the fluid is 0.2 watts per meter kelvin. The question is what will be T w 1 and what will be T w 2 that is the question that I asked. So, for example, the solution would run something like this n u 1 will be h 1 d h by k equal to n u 1 1 over 1 minus theta 1 q star and if you see the influence coefficient for parallel plates is 0.346, n u i i is 5.385 and therefore, the n u 1 will become 5.385 1 minus 0.346 divided by q star which is 0.2 and therefore, the Nusselt number turns out to be negative 7.377 minus 7.377 and therefore, the heat transfer coefficient will also turn out to be negative minus 14.753 and since q wall is 1 kilowatt h 1 is minus 14.753 and t bulk is 30, I can evaluate T w 1 as minus 37.8. So, T w 1 evaluates to minus 37.78 degree centigrade. What about the outer wall? You can do the same thing n u 2 equal to h 2 d h by k equal to n u 2 2 minus 1 minus theta 2 into q star and again you will see n u 2 will turn out to be 5.785, h 2 in this case will be 11.57 and T w 2 turns out to be 462.112. You can see what a great temperature difference there is. So, basically you have a situation where a very high temperature on this side and a very low temperature on this side. So, minus 38 here 462 here and the average temperature is 30 degree centigrade. It is for this reason that we are interested usually in finding out the solutions because the wall no wall should become too hot or too cold because it might affect other processes on outside or inside of the annulus. So, in as much as the bulk temperature is only 30 degree centigrade, the wall temperatures can be enormously different for just a very small heat fluxes of 1 kilowatt to 5 kilowatt meter square. So, it is for this reason that we take particular care in evaluating our Nusselt numbers accurately in annulus flowers. I turn my attention to circular tube V transfer in which T w is constant, axially constant and the flow is fully developed. So, you can say that in this particular case N u sub T means T wall is constant. It will be h 2 r divided by k and that would be equal to dT dr at r into 2 r divided by T wall minus T bulk and that should be a constant because h itself we say the fact that it is fully developed means h is constant and therefore, Nusselt number is a constant. Then if you go back to slide 2 and recall that this is the condition for fully developed heat transfer for h equal to constant, then you will see that dT by dx would be simply phi times dT bulk by dx and dT bulk by dx would be from heat balance 2 alpha divided by u bar r equal to q dT by dr. Remember k dT dr at r equal to r is simply the heat flux that is coming in and therefore, dT bulk by dx would be related to dT by dr at the wall. So, if this was the original equation and if I now substitute for dT dx in terms of phi, then this will transform the equation would transform to this form d 1 over r d by dr r star d phi by dr star equal to minus 2 N u T phi into 1 minus r star square. This is nothing but a second order ordinary differential equation with the boundary condition that phi r star at 1 which is the wall it is equal to 0 phi is 0 and d phi by dr star at the axis of symmetry is 0 because it is a symmetry line and r star is equal to r divided by r. So, this kind of a second order ordinary differential equation is really solved by shooting method as I show you here. So, what one does is that let us say this is r star and we want d phi by dr star equal to 0 the actual value of phi would not matter all that we want is d phi by dr star equal to 0 it is a dimensionless quantity and we say d phi by dr star is equal to 0. So, we assume a value of Nusselt number and solve the ordinary differential equation on a computer perhaps and you will come up with a value of phi at r star equal to 1 which is the wall which is the wall where we want phi star to be 0. If our N u guess was incorrect then the chances are that you will come up with this. So, this is N u guess 1 obviously, it is not equal to 0 as I wanted. So, I take another guess and you will see you I may come up with that N u g 2 and the error now is on the positive side the error here was on the negative side and therefore, I can use a bisection method to refine my error and generate a solution ultimately I will come up with a value of N which gives me phi at r star equal to 1 equal to 0 which is what I want. So, this is the N u correct this would be the correct value of N u and if you do that in this equation you have to go on assuming a value of N u t and solve the equation and you will see N u t turns out to be 3.656 or sometimes taken as simply 3.66. It is also possible to solve this equation analytically in a series form and the solution has been given here and again the value of N u t works out to be 3.656. If I were to do the same problem for flow between parallel plates with t wall equal to constant and t wall equal to constant on both sides then the Nusselt number that I would develop is 7.545 but that I leave you as an exercise to be done. We now turn our attention to one more case and that is the case of highly viscous fluids flowing in a tube highly viscous. So, its Prandtl number is much greater than 1 in which case under and there is a constant wall heat flux applied. In that case you will recall from our energy equation that because viscosity is very high viscous dissipation term mu d u by d r whole square becomes as important as the conduction term and therefore it must be retained in the energy equation. U f d of course is 2 u bar 1 minus r square by r square and d t by d x equal to d t bulk by d x is equal to constant d u f d by d r square from this thing will evaluate to 16 u bar square r square by r 4 and therefore substituting these things in here I get 2 u bar by alpha 1 minus r square by r square d t bulk by d x equal to 1 over r d by d r r d t by d r plus 16 times mu by k u bar square r square divided by r 4. This is equation A 2 and the original equation is A 1 and the boundary conditions are again at the axis of symmetry this would be 0 and at the wall it is equal to q wall equal to k d t d r. The main thing is I said d t bulk by d x would be constant but what will be its value that is what we need to evaluate. So, to do that on the next slide what I will do is I will simply integrate this equation. So, you can see the left hand side if I integrate over area of u f d and because d u f d by d x is equal to 0 because it is fully developed u f d d t by d x will actually become d x of u f d into temperature and I integrate this d by d x of 0 to r u f d t d y then you can see this will give me d t bulk by d x because when this is divided by integral 0 to r u f d d y I mean d r r d r rather this is the definition of t bulk and this is a constant u bar r d r is nothing but the mass flow rate through the channel or through the circular tube. So, you will see that to determine d t bulk by d x I would simply integrate this equation from 0 to r of both sides of this equation. The left hand side will give me d t bulk by d x this I will evaluate because I will get r d t d y d r at the wall and r d t by d r at the axis which is 0 and I substitute q wall which is known for the upper boundary condition and likewise you integrate this from 0 to r. The result is d t bulk by d x would turn out to be 2 q wall alpha k u bar by r which is the case when there is no viscous heating is included but now you see that the bulk temperature rise is also influenced by the amount of viscous heating that has taken place and that is the equation A 3. So, if I substitute d t bulk by d x in equation A 2 which is this equation. So, if I substitute for d t bulk by d x I will have an equation which looks like this I must integrate this equation twice and determine two constants of integration to determine the temperature profile. The term of the profile looks something like this remember again if mu was 0 that is if viscous dissipation was neglected then that term would be 0 and here that term and we will recover our original case of solution without viscous heating and the t bulk then evaluates this temperature must be integrated with u f d and t bulk minus t bulk would evaluate to 11 by 48 q wall d by k plus mu u bar square by k. Therefore, now it is very easy to define the if I divide both sides by q wall d by k then the Nusselt number would be obtained as here 11 by 48 plus mu u bar square divided by q wall d raised to minus 1 and this quantity is called the Brinkman number after the scientist who first solved this kind of a problem and this can be written as 192 divided by 44 by 192 multiplied by Brinkman number. Of course, if Brinkman number is 0 then you would readily recover 192 by 44 equal to 4.364. So, the effect of Brinkman number for high Prandtl number fully developed heat transfer becomes very high the Brinkman number depressed would be the heat transfer coefficient. I now consider the case of liquid metals when Prandtl number is very small as you will recall for liquid metal the Prandtl number is less than 0.01 usually Prandtl number is less than 0.01 and therefore, the actual conduction term which we have been neglecting so far becomes important particularly when the wall temperature is constant wall temperature is constant and you have liquid metal heat transfer. Then the governing equation would look as I have shown here there will be the radial conduction term and the axial conduction term equal to u f d by alpha d t by d x. Now, of course this is now a two dimensional equation it involves dependent variables r and x it can be solved analytically or nowadays much more simply by finite difference method and I give below the finite difference method solutions for different values of Peclet number. Now, I want you to appreciate how Peclet number comes about so the governing equation is 1 over r d by d r r d t by d r plus d 2 t by d x square equal to u f d divided by alpha d t by d x. Now, if I define x star is equal to x by r and r star is equal to r by r then you will notice that this will become simply 1 over r star d by d r star into r star d t by d r star plus d 2 t by d x square r star square and equal to u f d by alpha d t by d x star into r and you will see that this becomes sorry and there would be 1 over r star here also. So, if I multiply through by r star r square then you will see this becomes d by d r star into r star over d t by d r star plus d 2 t by d x star whole square is equal to u f d r divided by alpha d t by d x star and what is that u f d into r divided by nu into nu divided by alpha I do that then you will see that this is Reynolds number divided by 2 and this is Prandtl number nu divided by alpha. The product of Reynolds and Prandtl is called the Peclet number. So, this is essentially Peclet number divided by 2 d t by d x star and therefore, the equation can be written as Peclet number into 1 over r star d by d r star r star d t by d r star and so on and so forth. So, the Peclet number essentially determines the behavior of the solution. I have obtained this remember is because Prandtl number is so low the product of Reynolds and Prandtl can be very low in laminar flow. So, I have taken values of 0.1, 0.51, 1.5, 2, 3, 5, 7, 0.5 and 10 and obtained solutions by finite difference method and here are the solutions for very low Peclet number then a set number is 4.057, but as I increase the Peclet number which means allowing for more and more axial conduction then you will see that this becomes even more the Nusselt number goes on reducing and when Peclet number is about 10 and the wall temperature is constant mind you. So, in large Peclet numbers the axial conduction effect goes on almost becoming negligible and you arrive at 3.85. In effect as Peclet number tends to 0 you get only constant heat flux solutions or nu equal to 0.364 and as Peclet number tends to infinity you get the constant wall temperature solution 3.667. There is yet another case of considerable interest which you might like to know and this is the case of a tube which may be receiving let us say radiant heating then clearly around the circumference it will have a variable heat flux q wall theta although actually it will be constant at each cross section q bar will be simply 0 to 2 pi 1 over 2 pi q wall d theta would be a constant with x. So, this is a case which will occur for radiant heating sometimes it will also occur for a tube of uneven thickness let us say and it is being electrically heated electrically heated. So, there is internal heat generation within the tube and therefore it will receive circumferentially varying heat transfer. So, in that case the equation would become again because of the constant heat flux condition dT bulk by dx would be constant given by that and but now you must allow for conduction both radially as well as in the circumferential direction. Again this 2D equation can be solved both analytically or by a finite difference method and for q wall theta as this a simple circumferential variation of heat flux the solution turns out to be this 1 plus b cos theta 11 by 48.5 b cos theta where b is simply a parameter. So, you can see in this case a nu theta along the periphery of the duct can assume both positive as well as negative values, but that should not disturb you because that is expected all it tells you is whether the T wall bulk is greater than T bulk or less than T bulk, but if b is equal to 0 which implies q wall theta is uniform along the then you readily recover 4.364 and that is what we had. So, this particular type of problems are very important because hot spots have to be avoided on walls when you have uneven heating on all the side. In the next lecture we shall consider heat transfer in non-circular and arbitrary section ducts.