 Good morning, till now we had a lot of discussion on basic diffusion equation, the fundamental aspects about Fourier's law and how to solve this differential equation for one dimensional applications. And this is probably one of the most important things that we need to cover what we are going to do now in the form of an application that is heat transfer from an extended surface. All of us have seen or worked with extended surfaces in terms of the geometry as well as we have seen this in motorcycles in various applications we would have seen them, but and probably fins are also taught extensively in an undergraduate program. So, what do we mean by a fin and why should even we think about having an extended surface? Well to our primary equation that is governing heat transfer our main aim actually is to remove heat effectively. So, in most applications you want heat removal rate to be very good and if you want to dissipate heat or remove heat from a system then the one of the primary modes of heat transfer to from the ambient from the heated surface to the ambient is either convection or radiation we have not studied radiation. So, let us look primarily at convection. Coefficient is governed by q is equal to h A T s minus T infinity where h we know is the heat transfer coefficient given by units watt per meter square Kelvin area is of course, meter square which is nothing, but the heat transfer area and this is the temperature difference between the surface and the surrounding to which the heat is going to flow. So, whether it is heating or cooling whatever be the application this is the primary equation that is going to govern. Now, h can be a very complicated function that we do not care, but let us assume for now that h is we can calculate this quantity or we can get this value quite easily that is not the aim of this particular topic. So, h is not a problem T s is specified T infinity is also specified to us and A is also specified. So, given h A T s T infinity this quantity q becomes a finite quantity. So, q can be calculated if I know all these quantities and I will get q as a number. Now, depending on the application how good or how fast the cooling needs to be has to be decided suppose I have to remove there is there is let us take an example of reactor fuel for example, when the fuel when the power is turned off the fuel still has stored energy stored energy means you have basically is long fuel rods in that there is radioactive material which have because of their rho times C p there is stored energy present in them also this stored energy as well as fission products which are there will give out heat. So, this heat has to be removed even though the plant is turned off this heat has to be removed and for effective removal of heat the amount of heat which is being spit out coming out should be equal to the amount of energy which is being carried out by the coolant. So, what we are saying is if the heat comes out at the rate of say 80 watts and the fluid or the coolant is able to carry out heat carry away the heat only at about 60 watts then each time there is going to be a increase in the energy content and that will cause a d t by d t to be positive that means temperature will rise with respect to time. So, we will have a situation where the temperature of the fuel element will keep increasing and when it increases there will be a point when it will reach the melting point of the fuel element and lot of accidents etcetera can happen let us not go into all that, but the point is heat removal removing capability becomes very important and it for effective heat removal we need to be able to optimize these four quantities H A T S T infinity. So, for Q to be good what are the options that we have well options that we have are the following we have T S T infinity A and H we do not have too much choice on T S because T S is specified to us by the application. So, T S is kind of fixed. So, T S let us write down one by one T S is fixed by the application we do not have too much control T infinity the sink temperature or the temperature to which the heat is being removed that is also application driven. So, we have if it is air then it will be ambient air temperature if it is inside a cold room then that temperature if you are talking of the condenser in a power plant T infinity would be the cooling water temperature we do not that is taken from a river or a lake or whatever. So, that temperature we do not have too much control we probably can have a little bit variation, but not a great control H and A are the other two parameters which are going perfect well H in many applications what is H? H is essentially something which depends on the nature of flow. What do you mean by nature of flow? It depends on the geometry it depends on the fluid conditions fluid which is flowing. So, in a given application if you cannot have water as the coolant and you have only air to be able to cool it for example, your refrigerator on the back side of the refrigerator in olden days you would see these coils long serpentine coils which would be and if you touch there it will be slightly warm that is essentially the heat which is being thrown out by the refrigerator into the room. Now, you cannot cool that coil with you know cold water or you cannot circulate cold water to keep that coil cold it is just the ambient air which is going to take away the heat that is why in warm climates in hotter environments the refrigerator load becomes more also. So, I cannot change the fluid in most applications where it is possible I can have a better fluid in case it is the geometry is fixed which is most of the case geometry is fixed. So, the nature or I mean that effect of geometry is also lost the nature of the flow whether the flow is laminar or turbulent that also we do not have too much control say take the example of the refrigerator. I have these coils air in the room is typically not going to be able to go I mean very fast maybe once in a while a fan will be on in the room. So, you will have a better circulation, but more often than not it is a regular natural flow of air which is going to take away the heat. So, the nature of the flow also is not too much under our control even if it is it could be only for a limited period of time. So, the bottom line is T s is fixed by the application T infinity is fixed by again what we have is the medium for cooling h is fixed by the nature of flow geometry fluid over which we do not have too much control. So, what remains I want to increase q I have three parameters delta t these two have to be constant or cannot change too much. So, only thing which I am going to play with is the area available. So, this area which is there if I increase it as much as possible then my heat removing capability will also increase with this objective the concept of fins or extended surfaces has come about. So, what do you mean? So, you just have an extent you increase the area how do I increase the area in many applications space may be a constraint. So, cost may be a constraint you can have other penalties we will see what they are which may affect this increase in area can I increase this area forever infinitely can I increase the area how much is an effective increase in area what are these things what are the controlling parameters what are the boundaries within which I need to work. So, that this concept of extended surface is actually useful and not become a little bit of a problem. So, this concept of extended surfaces or fins is one of the most one of the applications where there is lot of practical relevance and whole topic you know there is there are lot of studies on enhanced heat transfer which are because of this concept of extended surfaces. Now, our common example which most of us give are motor motor bike fins which are there to remove the heat those are conspicuous large you can see them physically very nicely you can associate a dimension with it a thickness associated with the fin, but fins need not necessarily be of that big dimension. If you want to read further there are text books on enhanced heat transfer where these things are covered and I will just give you one thing one example is you can have pipes whose surfaces inside surface can be grooved you know where it looks like you know if you draw it it will look like a spring, but these grooves can be very very carefully machined with small dimensions or if I look at the view they can have this kind of protrusions etcetera. So, something which is going to cause a increase in the area of the surface. So, like this you will see in a blown up view a plane surface versus an extended surface this is the extended extension which has been given to the surface. So, this kind of studies are very very practical application oriented studies and a good branch of heat transfer covers enhanced heat transfer topics and you can a lot of literature available on that. So, the first thing related to this is what we study in an undergraduate curriculum which is fins. Now, examples etcetera we all know. So, fin is essentially an extended surface where you have combined conduction and convection what do I mean by combined conduction and convection I am going to add an extra material to the parent surface to increase the area. So, that extra material will have its own properties will have its own thermal conductivity will have a finite thickness will have a heat transfer area associated with it. So, there is going to be conduction also there is going to be convection on the outside of this element because of the h t infinity which is been there in the problem already. So, it is a combined conduction convection problem and this is what I have told already the q is given by h a t s minus t infinity we have control over only. So, many parameters and therefore, this concept of extended surfaces has come in. So, let me just go ahead here and see this heat transfer rate can be increased by increasing the heat transfer surface area through which convection occurs that is we saw that a then this can be done by providing extended surface or fins that extend from the wall into the surrounding fluid main constraints or main things that you need to keep in mind are thermal conductivity. Now, I have a base material which has to get cooled. So, if the base material is there is heat coming out from this surface in this base material is say some alloy and I want to put extended surfaces or fins on this and I am taking the liberty to draw fins because most of us know what they are. So, if I want to put an extended surface and I cannot put a surface made of plastic because plastic is essentially a insulating material. So, I have to have a material whose thermal conductivity is such that it is not going to behave as a insulation it is something which behaves or which aids the heat flow from the base material to the outside. The heat is going to be conducted from this base along the fin and is going to be dissipated in the environment h comma t infinity refers to the environment. So, for that to be effective the thermal conductivity of the material becomes very important. Ideally fin material should have a very large thermal conductivity to minimize temperature variation from the base to the tip. What does this mean? Let us just go back to our equation that we wrote favorite equation that we have again q is equal to q is equal to h A T s minus T infinity. I have added a fin very nice but what happens in the process? Let us say this is the base surface and this portion is what I have added a fin the base surface is at T s. So, even the shaded portion is at T s. So, if I take this shaded area as A B heat transfer through this base surface or to the shaded area is h A B T s minus T infinity very nice I could get a number for that. Now, I am replacing this base area by a fin I have increased the area. So, the cross section remains the same as that of the base area A B, but I have an additional surface here top surface I have a bottom surface through which additional heat is going to flow and I can write q therefore, nu is equal to h A nu T minus T infinity. Please note that q nu will be definitely different from q this one because I have roughly the same heat transfer coefficient, but I have a larger area A nu is much greater than A base that is ok. But what can I say about T minus T infinity is T minus T infinity the same as T s minus T infinity no because these two are not equal because T s would have been the temperature at this portion at this portion which was actually the original base between this portion and the tip this is called the tip of the fin between the base and the tip there is going to be a temperature drop because of conduction effect and therefore, I will get some temperature which is definitely lower than the base or surface temperature originally. So, this bracket T s minus T infinity is going to be larger in case of the base material alone and it is going to be smaller when I have an extended surface. So, what is the correct thing to do if this product A nu times T minus T infinity becomes smaller than A B times T s minus T infinity then I am actually going to transfer lesser amount of heat then what I did when I had no fins that is not acceptable to us. So, I need to ensure that this product A nu times T minus T infinity is always greater than A base times T s minus T infinity which means the increase in the area should be much more than the drop in temperature of the fin from T s to some value T. So, if I take care of that ideally the drop should be 0. So, that I can have this temperature as T base, but if that is not possible in reality I should have as low a drop as possible that means the fin material should be a highly conducting one. So, that is what this point is in the limit of infinite thermal conductivity entire fin would be at the base temperature thereby providing maximum possible enhancement of heat transfer. Of course, all the applications we know cooling of electric power transformer engine head so on and so forth. And this sketch gives you various types of fins one this is a circular pipe and you have annular fins like this. This is plate over which there are extended surfaces and there is gas flow between the plates and the fins are used to enhance the heat transfer. These are all parts of heat exchangers different fin configurations are there all undergraduate textbooks give this and most of us cover these also. So, this is a straight fin of uniform cross section, this is a straight fin of triangular or non uniform cross section that means that every x location outward from the base surface the cross sectional area is going to decrease. This is a annular fin so this is a circular pipe and fins are attached to it. So, it is an annular fin and the last one is what we call as a pin fin. Pin fin essentially is like what surface and the protrusion or the extended surface and the base surface are completely totally different geometry just protrudes out from the surface. And if you see here these are all occupying a large area this one is occupying much smaller area also. And what fin to choose how to choose all these things are decided by the space, weight, cost consideration, how to manufacture it, how to put it in along with the base surface etcetera. There is one more point here it configuration may depend on the extent to which fins reduce the surface convection coefficient and increase in the pressure drop associated with the flow. Let us just explain these two things pressure drop why should pressure drop increase. Well if I have a base surface I have a flow between two parallel plates or whatever be the geometry. So, I have a finite value of delta p which is given by some f l v square by 2 g d this is a representative equation. So, this need not be d it can be hydraulic diameter does not matter, but this is the pressure drop. Now if I have extension or protrusions which are going to cause enhancement of heat transfer from this what is happening is that I have flow which is happening and it is being obstructed by these so called extended surfaces. Also because of these obstructions there is some kind of a circulation which will happen here first of all there is a reduction in flow area. So, if the flow area reduces m dot is equal to rho a v will give me if the mass flow rate of the coolant is fixed which is going to be the case density is fixed area has reduced. Therefore, velocity has to increase to conserve mass and if velocity increases we can say that you have a much larger pressure drop because delta p is proportional to the square of the velocity. Also you have some kind of you know enhancement or turbulence which will happen because of these protrusions which will cause additional pressure drop. Whatever be the case presence of this extended surfaces in a flow is going to increase the pressure drop. Pressure drop translates directly to pumping power which directly translates to higher cost. So, this is something which I cannot ignore. So, pumping power I cannot ignore I have to have control over what is going to happen to the pumping power when I extend when I increase the surface area by extended surfaces. So, if a given application has a set of pumps which are already in place in having an extended surface should not say that I have to go and redesign the pump or put in new pumps that is not acceptable in real life. So, that pump should have the capability to overcome the additional pressure drop that is imposed. Second thing is the reduction in the heat transfer coefficient reduction in the heat transfer coefficient would occur primarily because of the change in the geometry also of the flow. You would have some surfaces see what heat transfer coefficient we are talking about we give a number h is some 50 watt per meter square Kelvin. It is an average value average means over the entire surface the value of h is uniform. This is obtained by some integration of the local heat transfer coefficient local means at every point you integrate this to get the average value. This average value is what is specified in our applications in our problems and this value would change because the local heat transfer value will also change. What is happening because of an extended surface you do not have flow reaching every nook and corner you are going to have obstructions as we said and imagine this protrusions like this the flow is not going to go around like this very nicely. So, this kind of motion will not be possible and therefore, there would be a reduction in the heat transfer rate also the heat transfer coefficient and this also should not be overlooked. But for our undergraduate course applications problems that we do we say that the heat transfer coefficient that was there given in the problem for base surface essentially we keep it as the same, but this is also a constraint which has to be taken into account in choosing a particular set of fins for an application. Now, let us quickly go to the mathematical aspect here is the general conduction analysis all colleges all universities cover this nevertheless we will do this carefully step by step. So, that this part energy balance it is nothing but energy balance. So, that everybody understands every step of this what is the aim the aim is to perform an energy balance on this extended surface why are we doing that as I said if I go back to the white board where I had written this thing before my parent equation was h a b t s minus t infinity there is no denying that now I have put an extended surface this one is a uniform cross sectional area surface does not matter I am saying that t minus t infinity is not going to be the same as what I had before that means what t is changing with respect to position from the base surface this original surface is called as the base surface t is changing and ideally how do I do the calculation I say if this surface if I have an extended surface here there is actually a figure much later I will go to that first yeah see actually this figure represents the ideal and the actual case this is the base surface and this is the extended surface order fin ideally when the thermal conductivity of the fin is infinite or close to a very very high value we will we want the fin temperature at every point there are 5 dots here 6 dots all of these are points where temperatures have been calculated and ideally we would want the base surface temperature to be 80 every point here also should be 80 that means if I select a region let me just go to the whiteboard here. So, if I divide the fin into say 5 parts each part if it is having the same temperature as the base then I can write q for each area 1 is equal to h a times t base minus t infinity say q 2 for the second surface is going to be h a t base minus t infinity and I can just keep writing and adding summation of q is equal to h a t base minus t infinity. So, I just have to add the areas associated and all surface temperatures are the same. So, q is equal to h times t base minus t infinity summation of area which is the area of the fin this is the ideal case when I can take this out common which means every point is at the same temperature but in real life that is far from the truth you will have a temperature variation which is going to look like this 80, 70, 65 these are just representative numbers and what it means is that this portion the first dot is roughly at the base temperature 80. So, heat transfer from that area surrounding that roughly is h a 80 minus t infinity next surface area if I take 2 that will be h a a 2 which is we can say same as a 1 if you divide it equally but t now is 70 degrees and not 80. So, I have a smaller heat transfer rate through each additional length of fin. So, though I am adding small equal sized elements like this the building blocks are the same you know when children play with these blocks all identical blocks are there each identical block we are adding to make an extended surface but in doing so we are changing the temperature of that surface. So, though the area 1 area 2 is the same the temperature t 1 and t 2 is not the same. So, h a t 1 minus t infinity is greater than h a t 2 minus t infinity is greater than h a t n minus t infinity. So, what this means is that utility of each additional area goes on decreasing and I am not going to ever get this original relationship where t b minus t infinity was the temperature difference for every element. So, in reality this is a major problem. So, for determining the actual heat transfer what do I do I should know the temperature at every point for so that I can write this and keep adding all these h a t minus t infinity local values 1 2 3 4 n elements and get the total heat transfer through the fin and that is my heat transfer calculation. But who is going to give me these individual temperature at every elements that is essentially what we are going to do if I derive a differential equation in for with temperature as the dependent variable and x as the independent variable. I will solve it to get t as a function of x along the fin and that temperature distribution will give me local heat transfer rate and integration of this over the entire length is going to give me the total heat transfer rate. Let us quickly get back to this this is the general conduction analysis we have a base surface a temperature t b and we have taken a general extended surface. So, we will write the differential equation for a general case but solve it for a case with constant surface cross sectional area. So, we look at a general geometry like this and invoke a set of assumptions first one heat transfer is assumed to be one dimensional that is only in the x direction our job is to transfer heat outside from this violet pink surface through this extended surface in the x direction. Yes depending on the geometry of the fin there will be heat transfer in all directions also all other directions also, but as I as we told yesterday the primary direction of heat transfer is important d t by d x is important d t by d x would be much greater than d t by d y d t by d z therefore, we can neglect those two in comparison with d t by d x. So, in reality heat transfer would be in all three dimensions but this is of primary importance to us. The rate at which energy is convective to the fluid from any point on the fin must be balanced by the rate at which energy reaches the point due to conduction of course, that is true. So, conduction heat transfer necessarily must be balanced by convective heat transfer because at steady state you are going to have no heat accumulation in any part of the fin. So, whatever is coming out from the base surface goes to element one that has to be taken out by element two three four and the heat has to go out by convection everywhere. Steady state conditions are assumed the fin is thin and temperature changes in the longitudinal direction are much larger than those in the transverse direction. This is what I said d t by d x is much greater temperature changes in the longitudinal axial direction x are much larger d t by d x is much larger much greater than d t by d y or d t by d z thermal conductivity is assumed to be constant you can take it as a variable and play around with it. But for this general derivation we take it as a constant radiation from the surface is assumed to be negligible why because first of all we do not know how to deal with radiation second thing when I am deriving the governing equation and applying a boundary condition which is having a fourth power of temperature solution becomes complicated. So, for down we deal only with convective boundary condition uniform convective heat transfer coefficient over the surface. As I said this H value can change, but we are saying all through at all points on the surface there is an average value which is uniform which is constant with this background let us look at the conservation equation and again and again I am telling this is nothing but budget. So, my budget starts with taking a control volume every derivation in heat transfer we will deal with a control volume first if you are going to get a differential equation it has to start with a differential control volume. Let us take this as a ring and this ring I am blowing up here this is the primary direction of heat transfer x q x is coming here q x plus d x is this way and heat is going out by convection q convection which is given by h a s a s is the surface area this is the cross sectional area a c. So, we have two areas one is cross sectional area other is the surface area and I will write this energy balance again e dot in minus e dot out plus e dot generated is equal to e dot stored we said steady state. So, this goes to 0 obviously, in the fin there is no heat generation. So, this also goes to 0. So, essentially I will get e dot in minus e dot out is equal to 0. So, I have to say in how many ways energy is coming into the control volume in how many ways energy is going out of the control volume and I have that already done in the figure q x is coming in it is going out as q x plus d x plus q convection. So, one path in two paths out and again professor has told us how to relate q x and q x plus d x by Taylor series expansion. So, q x plus d x would be q x plus d q x by d x times delta x or d x this I can put Fourier's law. So, q x is equal to minus k a c this is where we have to be careful heat is flowing this way. So, the area associated with heat transfer is this conduction heat transfer area is this one cross sectional area associated with convective heat transfer is the surface area. So, a c d t by d x with a minus sign that is my q x. So, I will substitute for e in as q x e out as q x plus d x and there would be additional term q x plus d x plus q convection is equal to 0 and this is q x. So, my equation is this one and now after that it is only algebra. So, I substitute for q x here substitute for q x plus d x q x will cancel off I will have d q x by d x delta x and q convection to balance each other let us that is all done here. So, I am going to go fast on this. So, when I write that I get an equation which is given by this form. So, q x plus d x is this and substitute this back here I would get q x is equal to q x plus d x plus d q convection put all these things and q x will cancel off. So, d q x by d x would be written like this d q convection is nothing but h d a s we are talking of a differential surface area t minus t infinity and when I substitute that and do the maths. So, let me just quickly go through this for us. So, next page I have q convection minus q convection d q convection minus d q x by d x delta x is equal to 0. So, this started off with q x minus q x plus d x plus d q convection equal to 0 and this was q x minus q x minus d q x by d x delta x plus d q convection minus d q convection equal to 0 these two cancels I am left with this part. Now, I will substitute for everything minus sign will go off h d a s t minus t infinity is the convection part plus d q by d x d x that is d by d x of minus k a c d t by d x d x is equal to 0. And surface area is nothing but perimeter times thickness associated with that element. So, I will substitute that I would get h times perimeter t s minus t infinity d x plus d by d x of minus k a c d t by d x of d x is equal to 0. So, essentially this is the differential equation now we have to massage it to get it into a convenient form we said we can cancel off this elemental d x because that is not equal to 0. And then bring out k which is a constant cross sectional area we will have to keep it as a variable because in this geometry that we have chosen the cross sectional area is varying. So, I will get h p t minus t infinity plus or minus k d t by d x apply product rule I will have k comes out u times v and another expansion would give me k a c d square t by d x square equal to 0. So, this is page 14 and per rearrange I will get k a c d square t by d x square all of you please write with me plus k d t by d x d a c by d x minus h p t minus t infinity equal to 0 divide through by k a c I would get a slightly better looking equation k cancels off I would get d t by d x 1 over a c stays here d a c by d x minus h p by k a c t minus t infinity equal to 0. So, this is the differential equation of interest and this is already been derived all all of us know how to do this also I have it here and this in this form equation 3.8 here has d a s by d x it is nothing but the perimeter because d a s is equal to p times d x it has been recast in this way other than that the form is the same what kind of a differential equation this is second order non-linear of course because I I have d t by d x and then there is a product of another derivative sitting here. So, it is a difficult thing to solve I need two boundary conditions associated with temperature and I need to know the variation of area with respect to x a variation of cross sectional area with respect to x d a c by d x to be able to solve this. So, though this is the general equation for one-dimensional conduction in a fin we cannot use this directly at least in case of a class to be able to get a solution of course this can be solved numerically and this is been done for various geometries also. So, what we do how do we then solve this do we leave this as it is no we look at fins of uniform cross sectional area and we say let me put d a c by d x is equal to 0 that means the cross sectional area does not change with respect to x very nice. So, this second term which is the culprit this goes to 0. So, I am left with a nice differential equation d square t by d x square minus h p by k a that is some constant because if I know the geometry I know the perimeter I know the cross sectional area if I know I know the material thermal conductivity I know the heat transfer coefficient h p by k a is a number for us. So, that is also known. So, I get d square t by d x square minus constant times t minus t infinity equal to 0. So, that is the nice differential equation for which solutions are known. So, keeping that in mind with the uniform cross section application the differential equation reduces to this. And just to make life easy you do not want to carry this t minus t infinity throughout you substitute that as theta. So, I d theta by d x is equal to d t by d x d square t by d x square is d theta d square theta by d x square. So, the differential equation becomes d square theta by d x square minus m square theta where m square instead of carrying this h p by k a all the time we call it as m square simply because you do not want to have square root of this parameter in the solution. So, that is why it is m square it has no other significance. Historically it is been called as m square. So, we are also calling this as m square. So, for this differential equation d square d square theta by d x square minus m square theta is equal to 0. We have general solution which is given by theta is equal to c 1 e to the m x plus c 2 e to the minus m x. So, this is a well known solution I am just going to write the two equations and the solution. So, with assumptions that d a c by d x is equal to 0 and h p by k a c is equal to m square theta is equal to t minus t infinity. If I put all this I get a very very nice differential equation which is of this form. And the solution for this if you take any mathematical handbook you will get theta of x is equal to c 1 e to the m x plus c 2 e to the minus m x. What this tells me the temperature distribution therefore, I know this was the aim with which we started right. We wanted to get temperature at every x location. So, that for every x location along the fin I get the heat transfer rate I add up all the heat transfer rate I get the total heat transfer through the fin. So, I got theta which means I have got t which means I know the temperature temperature distribution which means I should be able to calculate the heat transfer rate. So, this I have a solution, but this solution depends on c 1 and c 2 which in turn depends on the boundary condition associated with the problem. So, this is the second order ordinary differential equation a very bad looking equation this one we have converted it to a nice looking second order ODE. And this one we can we need two boundary condition because second order first boundary condition is common for any type of fin that is not going to change that is at x is equal to 0 t is equal to t b or theta is equal to theta b what does this mean it means at the base surface. So, the parent surface on which you are going to put the fin is called as the base surface. So, on the base surface if I am putting this fin at x is equal to 0 the surface temperature essentially whether it is on the fin or the base surface this surface temperature is essentially the base temperature. So, that is a fixed boundary condition for all fins the other boundary condition where do we specify we can specify it at l by 2 we can specify it anywhere, but we say that logically we want to specify it at the fin tip tip means the edge of the fin at the fin tip we are going to specify the boundary condition at x is equal to l. And there are there is a class of four boundary conditions which are specified all of these are known to us through text books. So, four boundary conditions are there one I will just list them out you have done this mathematics I will list them out one by one here yeah this probably is not very clear first one is convective boundary condition at the tip second is adiabatic boundary condition at the tip let me just go here boundary condition at the tip is convective implies d t by d x minus k d t by d x at x is equal to l is equal to h t minus t infinity this is the convective boundary condition you can cast this in terms of theta. So, it will be minus k d theta by d x is equal to h theta adiabatic boundary condition which is the second one which will be minus k d t by d x at x is equal to l is equal to 0 specified tip temperature t at x is equal to l is specified as some t surface. And last one is one very interesting case which is the insulated sorry which is the infinitely long fin what is this infinitely long fin we will just write it mathematically as l tends to infinity it is not a fin which extends 5 kilometers it is an infinitely long fin it refers to a fin for which heat transfer rate is almost equal to 0 beyond a certain length we will see that through an example problem. But what we are saying is the first boundary condition was at x is equal to 0 t is equal to t b or theta is equal to theta b this is common I need two boundary conditions remember this is common whatever be this is a family of second boundary condition which I have second boundary condition is to be chosen from one of this. Now, in real life are we going to have exactly one of these most probably we have a general convective boundary condition you can approximate certain situations as an adiabatic or insulated tip you might come across specified tip temperature infinitely long fin again once we explain that you will be able to appreciate that infinitely long fin also does exist in real life. So, with laying down these boundary conditions we will like to stop here for this part we will solve this and get the solutions later on we are open for questions for 5 more minutes. Javelpur college? Yes. Sir, increase of internal fluid flow problem particularly for transient transfer you mentioned that the heat transfer rate is always increases sorry because of the pressure drop. What what please go please repeat. And the pressure drop increases. Pressure drop increases. And what is the need of to provide the ribs or the roughness particularly in sir you have mentioned that yeah and heat transfer coefficient decreases the question is what is the need to provide such type of roughness particularly in internal fluid flow applications. For the question is one of the participant has asked is in internal flows that is for example a pipe why do we put extended surfaces or why do we put protrusions why do we put protrusions the answer is the heat transfer coefficient does not decrease as you said if you put a protrusion in a pipe if you imagine the ribs what professor was talking about if you imagine in a pipe if you put bangles inside of definite thicknesses if you put multiple bangles various locations spaced proper if you put them what will happen the pressure drop will increase for the same mass flow rate if you have to compare to the smooth pipe case the pressure drop will increase but the heat transfer coefficient is also going to increase we have to locate these enhanced surfaces enhancement enhancers what I would call the enhancers such a way that the heat transfer coefficient is going to increase not decrease enhancers can be anyone it can be a twisted tape it can be ribs it can be springs it can be even tube itself can be twisted. So, the point here is essentially whenever we put enhancer at the cost of the pumping power the heat transfer coefficient is going to increase what professor was saying is that heat transfer rate is what we are interested in increasing either I can increase the heat transfer rate by increasing h or a or both or both. So, here what you the case what you said is increasing h. So, to summarize whenever we put enhancers h is essentially going to increase but at the cost of pumping power ok. Jaipur sir if the heat generation is not constant during the source then what will be effect on the fin surfaces heat radiation. The question asked is if the heat generation is going to be non uniform what will be the effect of the performance of the fin that is the question asked by one of the participants. The answer would be of course it is little premature professor has not yet covered us however professor told that in all of this fin analysis we have to we are making a basic assumption that is the base temperature is constant. But if the variation of heat generation is there with time let us say then my base temperature is also going to vary with time for a given base temperature if I have the solution then essentially fins are going to be effective of course how we how do we make them effective is what we need to study we are going to study and realize that h has to be very high and k has to be very high these are the parameters which we are going to see later on. But as we go along we will realize that the base surface temperature we are going to take it as constant. If the volumetric heat generation is there and that is going to vary my temperature base temperature then my solution at every instant is also going to change that is the answer for this question. Jaipur sir Jaipur sir Jaipur sir Jaipur you when the pressure in the pipeline increases the velocity decreases and how does the mass flow rate varies? Answer is in your question itself you only answer you said that the pressure drop increases velocity decreases. If velocity is decreasing mass flow rate equal to density into area into velocity if I assume that my fluid is incompressible that is density is constant cross sectional area of my pipe is constant then my mass flow rate is bound to decrease that is the answer for your question.