Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati

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1 Heat Transfer Prof. Dr. Aloke Kumar Ghosal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module No. # 02 One Dimensional Steady State Heat Transfer Lecture No. # 05 Extended Surface Heat Transfer 2 Welcome to lecture 5 of module 2 on heat transfer. This lecture will continue our previous discussions on extended surface heat transfer, and the previous discussions what we have seen that why we need that extended surface heat transfer. And then for different case we have developed a generalized equation for temperature profile for a field with constant area of heat transfer, conduction heat transfer area. And then we seen that how the generalized temperature equation can be reduced to a specific form depending upon a see different boundary conditions that we have sorry people defined cases and in both the in all the 3 cases we have different boundary conditions that have been applied, and also you have seen how to find out the efficiency for the fins. Now, we will carry on that discussion and we will try to see that how are the efficiency expressions for the different cases that we have discussed in case of fins, and then later on we will take up the problem on this. (Refer Slide Time: 01:34)

2 So, to carry on the discussion then in the first case, case 1 we had as we have. So, Case 1 we have already discussed and this is the case 1 the field is in find at a long, and for that situation we have found out that temperature profile happens to be like this T naught T bar by T naught bar equals to e to the power minus m x, and for this kind of situations we know that if we also we have seen in case of find heat transfer that efficiency expression is like this it is minus k into A into dt by dx at x is equal to 0 and this is dt bar by dx at x equal to 0 divided by 8 into A T into T 0 bar. And this here. So, if we try to see that from this expression if we find out what is d T by d x. So, d T bar by d x at x equal to 0, this is equal to minus in T bar 0 and if we put this here, then it becomes minus minus cancels. So, k into a m into A into T bar 0 divided by 8 into. Now what is this A T, if we go back to our previous discussions, we will see that A T is that b this area plus the surface area of this. So, if this is L into this is the L into say B you can say this is the surface area and in the bottom also we will be having similar surface area and here also this. So, we can say that if this is the perimeter p into this length p into L will be the surface area of this side and the bottom side and this will be the surface area extra that is b into h, then this will be h into that 8 A T will be equal to the total perimeter total perimeter is P into L length plus B into H, that will be the total area into T 0 bar and we know that m is equal to we have seen we have already decided that m is equal to root over h P by k into A. So, then it becomes actually, if we substitute this value of n over here we will get root over of h P k into A divided by what we will do we will assume that and also it is very practical in case of long a field that B H this area will be much much less compared to P L. So, therefore, we will write at we will simplify it to be 8 into P L. So, if we write like that we can also write it as k A root over of k A by root over of P L into 1 by L sorry. So, this is equal to root over of k A by root over of h into p into 1 by L and you know that root over A 3 by k l is equal to m. So, we can write as one by m into l; that means, what for a long field the efficiency is 1 by m into L and is we understand as we understand that if l tends to infinity that indicates that eta efficiency is tending to 0. So, that is what I was telling that because of this reason infinitely long finishes got no practical relevance in the thing. In the point of view there is no efficiency if it is getting 0

3 and second point of this from the application of point of view for a long very long fins what are you going to fit it what are you going to accommodate it. So, that is not a practical solution. So, any way, but for case one that efficiency expression is becoming 1 by m L. (Refer Slide Time: 06:43) Now, if we go for case 2, in case of case 2 T bar already we have seen that T as we have seen in previously T bar by T 0 bar and that is equal to cos hyperbolic m into L minus x plus L by m k into sin hyperbolic m into L minus x divided by cos hyperbolic m L plus h by m k sin hyperbolic m into L. Now, d T by d x bar by d x at x is equal to 0 in this case is equal to minus m T 0 bar into sin hyperbolic m L plus h by m k cos hyperbolic m L divided by cos hyperbolic m L plus h by m k sin hyperbolic m L. Then eta efficiency is again as we have seen our previous case it will become root over of h p k A by h of 8 into p L plus B H into this sin hyperbolic m L plus h by m k cos hyperbolic m L divided by cos hyperbolic m L plus h by m k sin hyperbolic m L. Now, if we neglect again in certain situations we may neglect this and if we neglect B H if B H is less than p L then we can write it again as 1 by m L into as we have seen in the previous cases that this one is equal to 1 by m L h p L is equal to this equal to 1 by m L. So, it will be 1 by m L into this part. So, that is a factor that is coming for this case, which is sin hyperbolic m L plus 8 by m k cos hyperbolic m L divided by cos hyperbolic

4 m L plus h by m k sin hyperbolic m L. So, this we can see that with respect to the previous case here there is a factor that is come into picture when you have case of that a finite length of the field is considered. (Refer Slide Time: 10:11) Now coming to Case 3 when the tip is insulated in that case we have already find out a temperature profile as T bar by T 0 bar and that is equal to cos hyperbolic sorry m into L minus x divided by cos hyperbolic m into L. That is what we have got now if we again find out that eta if eta to find out. So, then we have to find out d T bar by d x at x is equal to 0 that for finding out the actual heat transfer at x equal to 0 then it is becoming minus n sin hyperbolic m L by cos hyperbolic m L into T bar. So, that says that if you put again that eta f this is equal to again root over of h p k A by h into p L plus B H and then tan hyperbolic m L and again in the same lump sum if you take that if B into h the product is less than p L, then we can write eta f is equal to tan hyperbolic m L by m L. So, that becomes the efficiency for the case 3 and in all the cases you can say that that the area of heat transfer by the field at the tip where which is equal to the breath of the fin into the thickness of the fin and in most of the situations, the thickness of the fins are in mille meters and then the total area with respect to the total area the p into L that main area of heat exchange by the fin happens to be much much higher compared to this area B into h. So it becomes a practical assumption engineering

5 assumption that B H is much much less compare to p L. So, B H is taken to B negligible or taken to be assumed to be 0. Under that situation, we have found out how what happens to the efficiency terms. Now whatever we have seen as of now is that 3 efficiency terms you have found out 3 efficiency expressions we have found out for t3 different cases and whatever we have found out in all the cases we have found, we have seen that heat conduction area for the fin that was constant. It is never changing, but as we have shown you in previously, in previous lecture there are different types of field we have shown and in many a situation we have seen there can be a variable area for the fins and when there is a variable area of the fin is involved then this derivation does not work out. Then we have because while deriving this we have taken area to be constant, but that area is a function of x. So, we have to take into consideration the function area as a function of x and then you have to go for the a derivation and then it looks to be a little bit more complicated. So, for the time being we are not discussing about that complications and not going for that derivations students interested in doing this they can he can always find out in that case he has to do that area is a function of x he has to find out it can be a linear function, but particularly when there is a trapezoidal kind of things when there is a conical end it can be linear function. But when there is a parabolic profile is there, then it need not be a linear function the area is need not be a linearly decreasing with respect to x and then resistant has to be little bit cost as above that. So, therefore, we just wanted to restrict ourselves with the simplest situation by that area of heat conduction through the fin is remaining more or less constant it is not changing as we go along the length of the length of the fin.

6 (Refer Slide Time: 14:51) Now, we will take of a problem on this fin heat transfer and we will see that how it work out. So, let us start the problem, problem 1 it says a carbon steel carbon steel pipe the outside diameter o d is equal to 60 mille meter has 8 longitudinal fins of thickness 1.6 mille meter the fin the fins extends sorry the fins extend the fins extend 4 centimeter from the pipe wall thermal conductivity of fin material of the fin material is 50 watt per meter per Kelvin the pipe surface temperature is 200 degree centigrade ambient temperature is 30 degree centigrade and heat transfer co efficient is and the heat transfer is 100. We have to calculate the percentage increase calculate the percentage increase, in the rate of heat transfer for the as well as the percentage fin effectiveness for the finned tube over the plain tube. So, this is the problem description it states it says that there is carbon steel and a carbon steel pipe and this pipe has got 8 longitudinal fins and they are of thickness 1.6 mille meter and the fin e extends 4 centimeter and that necessary information is like thermal conductivity the temperatures heat transfer coefficients are given. So, you have to find out what would be the speed efficiency and what would be the heat transfer and all this.

7 (Refer Slide Time: 19:12) So, if we just try to see that it is like this. So, we have a pipe and say this is the longitudinal fin that is being there and say this is the center line. So, there are like this way there are 8. So, if I take front sorry a top view of this it will look like this. So, there will be like this there will be some something like that. So, there will be sorry it will be like this it will be and 8. So, These e 8 fins are there at different sideways and this value has been given as diameter is this. So, 0.3 centimeter and this portion has been given as 4 centimeter. So, like this way there are 8 fins that is there here longitudinal fins are there fine. Now if you try to see that in how to tackle the problem or to find out the values of efficient first and to find out the efficiency, we need to know first which conditions we have to use this is a case straight forward inside case 2; that means, fin is of definite length fin is of definite length and that is too though tube is not insulated under that situation, if you try to calculate the efficiency of the fin the efficiency of the fin is eta is equal to 1 by m L sin hyperbolic m L plus h by m k cos hyperbolic m L then cos hyperbolic m L plus h by m k sin hyperbolic m L. So; that means, we have to calculate the value of m L and we have to calculate the value of m k h by m k now h by m k is nothing but we know h is equal to 100 and m is that we have to calculate. So, if you see that m is equal to root over of h p by k into area h p by k into area. So, this is that equal to h into root over of h into the p is equal to 2 into p is

8 equal actually b plus h as we have seen in the previous diagram we have seen that the weight blast thickness that is this is the p perimeter is this is the perimeter and this is the sorry this is B this is B this is H and this is H. So, 2 into B plus H is the perimeter and divided by k into area now if we say that h is relatively as you know is smaller than d, because we know that h is equal to 1.6 millemeter been given h is equal to. So, if we consider that perimeter like this is say B this is B this is H and this is H, B is that that breath like the length of the pipe and H is the thickness. So, we know that B plus B plus H plus H is the perimeter and in general we know that h is the thickness and which is much much less compared to the B values and therefore, we can write that 2 into B plus H we can write nearly equal to 2 into B. And therefore, what we can get is from here H into 2 into B divided by we have k into and area is equal to B into H. So, so this B B it cancels. So, we can get this is equal to root over of twice h by k into h now this is equal to m. So, m L is equal to sorry I will write here. So, m L is equal to then root over of twice h by k h into L and in this problem, this value is all this value has given h is equal to 100 k is 50 and H we know So, this is root over of 2 into 100 by k is equal to 50 into h is equal to into L is equal to we have 0.04 this is length this is L. So, then we can we will get the value of m L is equal to 2, you get the value of m L is equal to 2 and then and also we get the value of m is equal to value of m also we get if we put this square we get the value of m is equal to 50. So, then h by m k is becoming 50 into m is equal to 50 into k is equal to another 50. Then this value is becoming again So, h by m k is becoming Now if you put the value of all these here m L is equal to 2 h by m k is equal to So, eta value is becoming 1 by 2 into sin hyperbolic 2 plus 0.04 cos hyperbolic 2 by cos hyperbolic 2 plus 0.04 sin hyperbolic 2. And then this value is coming to be So, efficiency when we consider in these 2 situations the efficiency is becoming Incidentally also that if we consider even the case 3 here case 3 remains that T p is in; that means, there is no heat transfer from tube which is merely equal to assuming the heat transfer area negligible, because if we assume that B into H is very small if we assume that this heat transfer again is very small for the heat transfer which is; that means, the ratio heat transferred through the tube will be almost 0 which is equivalent to the case assuming that T is insulated.

9 (Refer Slide Time: 27:43) So, that way if we apply case 3 to find out efficiency, we will find that efficiency expression,if we apply that is eta is equal to tan hyperbolic m L by m L, and we have seen the values that it is equal to tan hyperbolic m L is equal to 2 by 2 and this value is So, we say that previous case the efficiency is marginally more we have seen that in the previous case the efficiency was and now in present situation it is 0.48 only. So, there is a very small difference in the efficiency and which is only this much difference in the efficiencies obtained. So, this is almost negligible therefore, either of case 2 or of case 3 can be used for this kind of analysis, because that tube area this it depends upon the thickness of the fin where the tube area fin area is almost negligible and. So, heat transferred through this is almost negligible. So, case 2 and case 3 are almost coinciding with each other.

10 (Refer Slide Time: 29:04) Now, after we find out the efficiency then we have to find out the heat transfer we have to find out. So, for tube without fin. So, heat transfer rate if we try to calculate for that we have to find out that area of bear tube or according to the problem this is the pipe plus pipe. So, it area of the bear tube or pipe and that is equal to you know pi d 0 into in the into the L of the small l length of the here pipe. And if we this will become now this will become now pi into 0.06 is the outer diameter into say one is the per unit meter length. So, that we are assuming the length is equal to 1 meter this much meter square we are assuming L is equal to 1 meter length of the pipe. So, then it is becoming meter square and then heat transfer rate p is equal to h into area of the bear pipe into delta t and this is equal to h is 100 and area of the bear pipe is in is equal to into delta T is equal to 170. So, this becoming kilo watt.

11 (Refer Slide Time: 31:32) So, this is the heat transfer rate by the heat transfer rate by the pipe when there is no fin associated to it. Now if we try to find out the fin area of the fin fins transferring heat if we try to see here we will get that 2 into 1 into 0.04 this is the if we go back to our this thing we will see that we have taken this one to be 1 meter. You have taken this one to be 1 meter. So, 1 into 0.4 that is the area and it is in both sides heat is transferred from both the sides. So, then it becomes 2 into 1 into 0.04 this mille meter square this plus the area of tube and if we neglect the area of the tube. So, this is becoming actually 0.08 meter. And if we try to see that area of tube for each case is equal to into 1; that means, meter square. So, we can see that that area of the tube is very small compared to the total area therefore, we can simply neglect that area and then for 8 fins the area is 8 into 0.0 sorry 0.08 meter square and that is equal to 0.64 meter square this is the area of the fins that is transferring heat and then we have to find out the area which is not occupied by the fin and that area of their area of the pipe which is not occupied by the fin that is also responsible for heat transfer. So, that amount will be. So, they are area of pipe of finned pipe. This will be equal to 0 point I am sorry 0. Yeah. We have seen this is the area of the total pipe minus this is the area occupied by 1 fin there will be 8 fins. So, minus 8 into meter square and then it becomes meter square.

12 So, total area heat exchange area of the finned pipe and that is equal to 0.64 plus and that is equal to meter square. Now we have found out the area of this finned pipe. (Refer Slide Time: 35:10) So, heat transferred by the fin pipe heat transferred rate by v finned pipe p is now efficiency of the fin we have found out. So, 0.48 into the area of the fin is 0.64 into 170 is the driving force delta T into h is 100. So, this is actually we should write this will be heat transfer co efficient into area of the fin into driving force into eta A. So, this eta A efficiency of the fin is 0.48 then heat transferred area is equal to 0.64 then driving force is delta T is this and h is equal to 100 this plus this is the from the fin part and from the unoccupied part of the pipe and that is equal to say 8 again into a pi into delta T and that is plus will have h is equal to again 100 into area of the pipe here is just we have seen into 170 this area is the area of the pipe surface area of the pipe minus the portions which are occupied by the fins. So, then it becomes actually kilo watts. So, we have seen when the pipe was not finned, then that heat transfer was kilo watt and now we have found that heat transfer rate is kilo watt. Therefore. So, change percentage increase equals to minus by into 100. So, this is equal to percent. So, we can see that it is 156 times. This many percentage 156 percentage increase in the heat transfer rate is there, when the this kind of fin is there and

13 effectiveness is percentage effectiveness equals to that heat transfer just with fin divided by into 100 and this is may be called to percentage. So, effectiveness is 256 percent and percentage increase in heat transfer rate is percentage. So, this gives an idea that how fin helps us to increase the rate of heat transfer this problem will help us to understand that. Now one thing I should clear here, in case of circumferential fins or radial fins the area of conductive heat exchange that changes and that varies and in that situation, this kind of analysis will not be working to find out that values of m L and efficiency. So, what is being done in that case is there are certain charts are available to find out the values of efficiency for circumferential fins and or radial fins some charts are available. (Refer Slide Time: 39:27) And this charts are basically with respect to fin efficiency what says an expression that is h by k into A m whole to the power half into l to the power of 3 by 2 where A m is called profile area h is heat transfer coefficient k is thermal conductivity of the fin material and A m is called profile area this is equal to L into H the length of the fin into H thickness of the fin and L is the length of the fin. In some situations it may happen that that L is replaced by L c that is corrected length fin length. Then in that case the efficiency, if the plot is efficiency one says H by k into Am whole to the power half into L c to power 3 by 2 and L c is equal to L plus H by 2 and L

14 c is equal to L plus d by 4 this is for rectangular profile and this is for fin thin fins thins names this circular rod and d is the diameter of the rod fin rod. So, this way we find out the from the chart we can find out the values of efficiency for fins which has got a variable idea and then once we know the efficiency then rest of the part is the actually, heat transfer we can calculate that all that we have seen through this problem also actual heat transfer rate that is equal to that efficiency of the fin into a heat h A into T 0 bar. So, A is the total area of heat transfer h is the conductive heat transfer coefficient and T bar is the driving force delta T. From this, we can find out that actual heat transfer rate by the fins by the fins that has got variable conductive surface area. Now this way we try to explain to some extend on the extended surface area and how a heat transfer takes place through the fins and how the temperature profiles are there how are the efficiencies what is the effectiveness of the fin let us say that we have discussed. (Refer Slide Time: 43:12) Now, one specific topic I just wanted to tell here is that is called Thermal Conduct Resistances thermal conduct resistance now from the name itself we can understand like here.

15 (Refer Slide Time: 43:39) We can see that when 2 bodies come in physical contact A and B they are coming in physical contact, then what may happen there can be some extra resistance encountered due at this contact point. If you see that this is a just magnification of this is been shown here and once you see this that there is basically, some kind of roughness always they are in any surface there is some roughness when this 2 surface joins with each other then there will be some empty space that is been developed. And these empty spaces are being occupied by some gases it may be here it may be some other gases the empty spaces are occupied. So, what happens when the heat is being transferred say from A to B when the heat is being transferred in this direction from A to B what is going to happen is, that part fins is going through the direct contact of the solids in some places there is no contact directly by the solid the contact is by the gas phase inside these between the solids. So, part of this is by the direct contact between solids A and B and part of the transport is taking place through the gas or air which is entrapped in the wide space. Now, we understand or we know that thermal conductivity of the gas or air is very low compared to the solids and then therefore, what happens as because the as because it is expected that the whole things would have been the solid contact, but as because it is not the case then what happens there is a an extra resistance that is being developed and therefore, they we will find that there is a drop in temperature in the contact zone

16 between these 2 bodies and this is called this drop in temperature is happening due to the resistance being offered by the fluid entered into this place and therefore, it is called thermal contact resistant. And if you see that that if we draw a profile of the temperature we will find the suppose this is as we have seen that this is A and this is B and if we draw the profile here we will see heat transfer is taking place from A to B. So, here the heat transfer is taking place. So, now, this is these are the 2 temperature profile this is the temperature profile. So, they say this is T 1 say this is T 2 A this is T 2 B and this is say T 3. So, what is happening from T 1 to T 2 A there is a profile change there is a profile this is this is depending upon the thermal conductivity of a between T 2 B 2 T 3 again there is a another profile this is depending upon normal conductivity of the B, but at this junction there is a drop from T 2 A to T 2 B and this drop is basically, due to the resistance being offered due to the pressures of some gases entered in the wide space that is being created when they are in contact and that is being created due to the non smoothness of the surfaces or rather roughness of the surfaces. Now, this can be much easily understood if we increase the external pressure here, or in the other words other around, if we decrease the pressure here then what will happen, if we just decrease the pressure of the surrounding then the pressure of the fluid interrupt in this point then in this thermal contact region will be reduced. So, the number of molecules of the interrupt molecules will be reduced and when the numbers of molecules interrupt is reduced then what will happen thermal conductivity will be further reduced and that situation the drop will be further more. So, resistance will increase. So, resistance will increase with decrease in, we decrease in surrounding pressure there is one case another case is that, if we apply if we apply some kind of pressure at this junction at this junction. So, that some kind of deformation takes place here, at this junction this is the application of some kind of pressure then what will happen that give, if the deformations have to deformation takes place then thus if we can bring about more smoothness on this surface then that wideness will be reduced. So, amount of gas entered into the gas material will be reduced and then thermal contact resistance will be reduced.

17 That means what you say is that if we increase the joint pressure by some means resistances will we decrease with increased joint pressure this is what if we put some kind of pressure at the joint between this 2 body then what will happen the there will there will be more contact and, because of more contact less area for the air or gas has to be entered. So, less conduction by the airy material more conduction by this solid material and therefore, the thermal contradiction will decrease and in some in many situations these thermal contact resistance are reduced by putting some greasy material at this junction point if you put some greasy material we can reduce the thermal contact resistance. So, this way we have tried to see that in case of steady state heat conduction situation particularly in one dimension what happens point there is a cylindrical circulation when there is a plain wall and there is a spherical wall and there is a spherical body when there is an extension of the surface areas by extended surface resistant in the wire applying fin and then also we tried to see that what is the physical significance of thermal contact resistances in case of heat transfer. Now, in the next lecture onwards we will discuss on unsteady state heat transfer we will see that how or what will be the situation how are the performances when temperature changes with time what happens to the heat transfer rate if it changes if temperature changes with time all these things we will discuss from next lecture onwards. Thank you very much.

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