 So, we are coming to the last leg of the day. So, this is the last session which is the tutorial session. So, these are the problems I am sure you have all these problems with you. No, I do not intend them. So, I do not intend to flash the problems. So, I am sure all the coordinators have given the problem sets which was uploaded in the morning one set with the each of the participants I am sure about that. So, we had a question between 3 to 330 whether we should be collecting the tutorials and correcting and all. So, we had internal discussion see the way we are planning to conduct the tutorial is that we will give first 15, 20 minutes to solve may be let us say today we have total 10 problems. So, we have 1 and a half hours. So, 1 and a half hours means we can take may be 3 problems in every 15 minutes roughly, but first 15 minutes we will give you time. So, that you can sit down and solve each of these problems. So, first 15 minutes you can solve and then we will perhaps solve problem 1, 2 and 3. So, let us see this way how many we solve. We do not plan to solve all 10 problems today. Yeah and another thing today being the first day we are given 10 problems because most of them you will see are very very straight forward direct application of formulae, but later on tomorrow onwards we will give about 5 to 6 depending on the difficulty level and the time involvement for each of the problems. So, as we go further and further in the course especially problems on heat exchangers you will have 2 or 3 in the given 1 and a half hours time. So, but today things being very straight forward we will try to I mean we would expect you to solve all 10 of them in the allotted time. So, for the next 15 minutes I would request all the coordinators to insist all the participants to solve the problems. So, for next 15 minutes now may be 10 minutes we will take questions from various centers if they have any. So, may be we are going to Bharamati. Yeah, if I remember some centers participants started the quiz that around 340 they will be doing it till about 4 10, 4 15 those guys can join the tutorial session after that, but beyond that let us not discuss anything related to the quiz today. We will have a discussion on the quiz and the solution tomorrow sometime no discussion on the quiz no questions on the quiz today ok. So, over to Bharamati VP COE Bharamati. Ok sir my question is regarding the fins part how to decide the pitch of the fins or design point of fins. So, that is a good question actually see in the previous the question is how to decide the pitch of the fins. So, if we take our problem which we had solved earlier. So, here the pitch was given. So, what we are saying is that see in the previous problem what we said what is the pitch given the pitch given is given to be 3 mm. So, if you ask me how does this pitch affect in the pitch on the efficiency of the fin if you see in the problem in the efficiency the pitch is not coming anywhere, but the pitch how will it decide pitch is decided on the base of on the basis of how can I place. See here if we see the pitch is around 3 mm the pitch is around 3 mm, but this is a very difficult question to answer directly. So, why I say very difficult to answer directly because if I decrease the pitch very close to each other then I am impeding the natural convection within this space. So, my space has to be reasonably large so that the natural convection is aided, but how will I know that whether the heat transfer coefficient is going to get affected or not. So, it is quite difficult to answer that question whether I can just by simple calculations what is the optimal pitch, but if you see as far as the efficiency is concerned it is not dependent on the pitch at all as you can see here in any of these parameters the efficiency is not dependent on the pitch, but the pitch will affect only the heat transfer coefficient. So, we can say if we assume for a minute that heat transfer coefficient is not going to be dependent on the spacing we can it will be decided on the basis of the fabricational difficulty. Here we can see it is 3 mm I cannot get closer than 3 mm because I need to have access either for soldering or welding or any other fabrication procedure. So, that would be the answer for your question. One other thing probably is the pressure drop associated if your pitch becomes very small then your calculated delta P would be much larger and remember velocity square is there and actually I think professor Prabhu can correct or add typically what is done is when you put these enhance surfaces or these geometries for improving the heat transfer you take the ratio of the new heat transfer by the old heat transfer divided by new pressure drop by old pressure drop and that ratio should come out to be some number greater than 2 or something like that which will justify the use of fins or extended surfaces even though there is a pressure drop related penalty. So, basically what we are saying is that the heat transfer coefficient has to be taken along with the pumping power when you decrease the pitch as professor said the pumping power is also going to increase. So, we have to take the pressure drop and the heat transfer coefficient together that will actually that is an indirect way from these calculations it is not going to get decided that has to come independently. I hope you have answered your question. Yes, sir. Thank you. Okay. Over to MK Triple Spune. Good afternoon. In case of two dimensional heat conduction which you had explained yesterday if all the sides of the plate are maintained at constant temperatures which are in general different what can we say about the corners because at the corner I suspect that there might be some sort of mismatch between the two temperatures coming from both sides. Okay, that is a mathematical issue, but physically it is not possible to maintain at the interface all of a sudden temperature jump it has to gradually change. If I have an interface one wall is at 40 degree Celsius and the other one at 150 how can I have all of a sudden 40 become 150 that is not possible. So, it has to gradually change it can be mathematically handled, but not physically. So, that is all I would say definitely as you said at the corners there has to be tapering down. You will have a tapering. Tapering down. Definitely the answer is definitely there will be tapering at the interfaces that is at the corners. Yes. Okay. Thank you. Over. Bhopal. May 90 Bhopal. Over to you. Good evening sir. This is Dr. Suresh from NIT Bhopal. So, sir actually I have one application question regarding TGA and TGTA. Once you are going for find out the temperature exothermic or endothermic reaction. Okay. In that normally when TGA, TDA we will use to advise that minimum amount of material minimum amount of material to minimize the heat transfer effect. We have to take minimum quantity of material to analysis that TGA, TDA the process. So, what is the minimum amount of material to be used? Can you suggest? What is TGA and what is TDA? Over to you. Sir, it is in thermogravimetric analysis. DTA is differential thermogravimetric analysis in which temperature we use a material to analyze its characteristics. So, when we use at various heating rate and with different material quantity the curve is differing. So, it is advised to use minimum amount of material and minimum heating rate to avoid the heat transfer problems. So, that the material can transfer better. There will not be any difference in temperature. Okay. So, can we find what is the method we have to put it over there to solve this problem? Okay. Now, the question asked is for all participants for the benefit of all participants there is something called thermogravity, gramimetric analysis. In thermogravimetric analysis one would try to measure the constants that is the Arrhenius constants in a first order reaction in any exothermic or endothermic reaction. So, the question asked is how much amount of the mass has to be put in a thermogravimetric analyzer. So, that my results are independent of the mass. Honestly, I have no answer for this because I do not know I have never used thermogravimetric analysis. However, here we have a professor, Professor Kananayar. We will get back to him. We will ask him this question and get back with the answer. I would request you to please put this question in the forum with proper formulation. So, that we will definitely answer this question. This being very specific only you would be interested in this. So, we will get back to you with the answer in the forum. Okay. Surely, we will put the in moodle, sir. Over. Question over. Thank you. Okay. We will get started with the, okay. So, let me start off with tutorial problem one. The question asked is tutorial one this is that is first problem. So, let us take what is given here is find. Let us write find the problem number one. So, what is given? T1 is given to be 400. I am saying find in each of the case what we have to find is the heat flux. We need to find heat flux and its direction. In the coordinate system what is being shown and also in addition we need to find dt by dx change over of 10. So, that is what we are saying is what is to be found. We need to find the heat flux okay and temperature gradient dt by dx dt by dx we need to find for each case. So, what are the assumptions involved? One is steady state, second is known properties. All properties are constant that is thermal conductivity all are constant. In this case properties means only thermal conductivity and it is 1D conduction it is 1D conduction and there is no volumetric heat generation. It is not an assumption because it is not given in the problem but still nevertheless we will state that. So, schematic here is given in each case. So, let me take the first schematic that is let me draw that T1, T2. So, what is known? Known is given to be thermal conductivity of the material is given to be 100 watt per meter Kelvin and thickness is the thickness of this is given to be thickness is given to be 100 m and I have T1 that is T1 is equal to 400 Kelvin and T2 is equal to 600 Kelvin. So, these are these are the things which I know. So, I am switching over now to another notebook. Let me rewrite this dt by dx. So, what is known? Let me rewrite this quickly known is T1 is 400, T2 is 400 Kelvin and T2 is 600 Kelvin and thickness is 100 mm that is 10 centimeters. It is little smaller than smaller scale and the figure is given what is to be found out is heat flux and dt by dx. So, let me first take there are three situations. Let me take the first situation this is temperature the y axis and the x axis is x and given temperature is T1 and T2. So, the temperature is increasing with the increase of x. So, T1 is 400 T2 is 600 this is x equal to 0 and this is x equal to L. So, first let us find in each case dt by dx. So, dt by dx equal to T1 minus T2 upon x1 minus x2 that is x1 minus x2 that is 0 minus L or you can write this as T2 minus T1 upon L minus 0 either ways. So, T1 is we have already found T1 is we have been given 400 minus 600 divided by L is given to be 100 mm that is 0.1 minus 0 minus 0.1. So, numerator happens to be minus 200 minus 200 upon minus 0.1 that means 2000 Kelvin per meter 2000 Kelvin per meter that is dt by dx is positive number 1 and number 2 it is 2000 Kelvin per meter. So, that answers dt by dx, but now what is heat flux? Heat flux is given by minus k dt by dx straight forward there is no question here. So, k is minus is there as it is k is 100 Watt per it is a good idea actually in few textbooks they put units even while substituting it is a good idea if we can substitute nothing like it Watt per meter Kelvin into dt by dx is 2000 Kelvin per meter. So, Kelvin Kelvin gets cancelled out. So, I end up with minus 2 into 10 to the power of 5 Watt per meter square. What does this mean? Note here it is 200 kilo Watt per meter square number 1 quite a huge heat flux and if you have to give the feel of the heat flux generally the heat flux is what we come across in laptops and all are of the order of 10000 Watt per meter square. So, this is quite high compared to that now negative sign negative sign means what? It is moving in the negative x direction. So, heat transfer is in heat transfer is in negative x direction. If you are not careful anywhere we are most likely to go wrong. So, what is that we need to do is systematically do dt by dx and then take minus k dt by dx according to coordinate system that is all. So, there is nothing difficult here. Just to add if you look at the sketch itself in the positive direction t 2 is higher than t 1. So, the gradient it has to be positive. So, that sign 2000 with a positive sign itself is a correct answer looking at this low. So, you have an acute angle in the positive x direction. So, mathematically also without knowing heat transfer you can say the slope has to be positive here. That is temperature is increasing with the increase of x. Yeah. So, and then k minus k dt by dx has a negative sign associated with it. So, there will be a minus sign in the answer for heat flux and obviously, heat is going to flow from right to left and that is the physics will tell us because higher temperature to lower temperature. So, if we correlate physics mathematics everything I do not think these things are any difficult. Let us take up next part that is next part we are most likely to get confused. So, let us take that. So, I will do only the second one third one I will skip. So, that is this is temperature t of x this is t 1 t 2, but my x direction unlike the last problem it is in this direction. So, again dt by dx equal to 0 and L. So, dt by dx equal to t 2 minus t 1 upon 0 minus L there is no confusion on this or t 1 minus t 2 upon L minus 0. So, what is t 2 we have 600 minus 400 upon minus 0.1. So, that means I get minus 2000 watt per meter. That means temperature is decreasing with the increase of x I think it is right that is what we have got dt by it is indeed physically that is what is happening temperature is decreasing with the increase. So, I have made a mistake here here Sameer and others are correcting me. So, it is Kelvin per meter you are right. So, that is 2000 Kelvin per meter. So, that is temperature is decreasing with the increase of length x. So, that is physically realizable. So, q double dash is equal to minus k dt by dx. So, what is minus k minus 100 into dt by dx is minus 2000. So, the unit of k is watt per meter Kelvin and this is Kelvin per meter. So, Kelvin Kelvin gets cancelled out. So, I get plus 2 into 10 to the power of 5 watts per meter square. So, you notice here positive sign naturally the heat flux is moving in the positive x direction because it is moving from higher temperature to lower temperature. So, no matter however it is twisted and given now I think you can handle this all that we are saying is find out dt by dx plug in q double dash equal to minus k dt by dx. So, now whichever way it is twisted and given to us we can now simply sit down and calculate it. So, that is why I am skipping the third part ok over to problem 2. What does that problem 2 say? An electrical resistance heater is embedded in a long cylinder of diameter 30 mm when water at a temperature of 25 degree Celsius and velocity 1 meter per second flows crosswise over the cylinder. The power per unit length required to maintain the surface at a uniform temperature of 90 degree Celsius is 28 kilowatt per meter. So, what is given? The cylinder is there let me start sketching. So, let me just say schematic. So, I have a small cylinder that is electrical resistance wire heater I am imagining as a cylinder and there is a flow taking place across the cylinder. And the velocity of the flow is 1 meter per second velocity is equal to 1 meter per second that is known is diameter of the cylinder is 30 mm and T infinity is 25 and velocity is given to be 1 meter per second. And the wire is being maintained at a temperature of wire temperature is being maintained at let me call that is T wall at 90 degree Celsius. And the power generated is per unit length that is I would call that as Q dash is 28 kilowatts per meter. So, when air also at 25 degree Celsius, but with a velocity of 10 meters per second the power this is at 1 meter per second. But when air also at 25 degree Celsius, but with a velocity of 10 meters per second is flowing then the power per unit length this is water temperature is at 25 degree Celsius. So, but in case of in the second case I have air in second case I have air whose velocity is 10 meters per second, but the power I can give is quite high as you can see no quite less in case of air all the velocity is 10 you see in the earlier case it was 28 kilowatt that is 28000, but in this case we are able to give only 400 watt per meter. So, why we can give only 400 watt per meter because my wire will burn out. So, because it is air so air convective heat transfer coefficient is going to be less compared to the tough water let us see whether that does that happen or not. So, here essentially what we are checking is we are trying to check in this problem whether we have understood how to apply Newton's law of cooling. So, Q equal to h A into Q equal to h A into T wall minus T infinity. So, h equal to which implies h equal to Q upon area into T wall minus T infinity. What I have here is this is equal to Q dash L upon Q dash L upon pi into D into L note I am not taking pi D squared by 4 I am taking pi D L this is because it is surface area. Professor Bajan calls this as bathing area. So, it is getting bathed my cylinder is getting bathed. So, pi D L into T wall minus T infinity. So, L gets cancelled out I do not need to know L. So, rest is substitution now for the case of water let me calculate that that is h equal to Q dash is 400 sorry 28000 upon pi into diameter is 30 mm 30 into 10 to the power of minus 3 into T wall is 90 minus T infinity is 25. If you just press calculators we are going to get this as 4.57 kilo watts per meter square Kelvin or 4570 watts per meter square Kelvin remember that this is in thousands. So, now let us do the same problem for air that is which is at 10 meters per second and Q dash is at 400 watts per meter. So, that is we have h equal to Q dash upon pi D into T wall minus T infinity. In this case my Q dash is given to be let me just take Q dash is given to be 400. So, 400 upon pi into my diameter of the wire is 30 mm 30 into 10 to the power of minus 3 and T wall is 90 minus 25. So, if I substitute that I get h of air as 65.3 watt per meter square Kelvin. You see h of water we got 4570 watts per meter square Kelvin, but h of air is found to be 65.3. So, as we go along in convection we will see that this is very natural. So, for water it has to heat carrying capacity of water is very high. So, we should be able to take away more heat in case of water as opposed to that of air. So, we will see that it is actually dependent on Prandtl number higher the Prandtl number higher the heat transfer coefficient. So, for now it would suffice to say that h of heat transfer coefficient of water is always greater than heat transfer coefficient of air. In fact, this is how one measures the heat transfer coefficient also. So, the next question was calculate and compare the convective. So, we have answered all questions. So, all that we are saying is h of water is greater than h of air. So, now let us move on to the third problem. So, what does the third problem say? Problem number 3 that is we have a surface of an area what is known we have a surface of an area 0.5 meter square emissivity that is it is a perfect black body. So, we do not have to worry emissivity is 1. Let us go ahead and write it as 1 and the temperature of the body is 150 degree Celsius. So, one important thing whenever we handle radiation is that we need to convert degree Celsius into Kelvin. So, that is the safest way because in radiation we handle in our Stefan Boltzmann law sigma t to the power of 4 t is in Kelvin. So, I have to convert this temperature to Kelvin. So, 150 plus 273 we get 423 Kelvin and evacuated chamber whose walls are maintained at 25 degree Celsius surface area and the temperature is placed in a large evacuated chamber. Chamber wall temperature is found to be 25 degree Celsius this is the body temperature T s. So, I would say chamber I would call this a surrounding temperature that is 25 degree Celsius that is 25 plus 273 equal to 298 Kelvin. What is the rate at which the radiation is emitted by the surface? So, what is the radiation emitted by the surface? We will just take the find we will take first that is not coming out find that is net energy emitted q emitted that is equal to sigma area of the body into T s to the power of 4 emissivity emissivity is there. So, but here emissivity is 1. So, emissivity is 1 sigma is 5.67 into 10 to the power of minus 8 into area is 0.5 into T s is 423 to the power of. So, I get q emitted equal to 907.64 watts. So, now q net what is q net that is the next part of the question is what is the net rate at which the radiation is exchanged between the surface and chamber walls that is q net is equal to q emitted minus q received. So, that will be equal to sigma A into T s to the power of 4 minus T surrounding to the power of 4. So, if I substitute for sigma 5.67 into 10 to the power of minus 8 into 0.5 into T s is 423 point to the power of 4 minus 298 to the power of 4 I am going to get 684.07 watts naturally this number has to be lower than that of emitted. So, I do not think any comment we can add here just for the simple reason all that I want to carry the message by solving this problem is that temperature should be in Kelvin for radiation this is the point we want to emphasize. So, that we do not commit mistake because in Stefan Boltzmann law T s is in Kelvin T s is in Kelvin. So, what we will do is next 10 minutes we will take questions either on these problems what we have solved or any other questions which we have on the topics which we have covered so far that is now it is 10 minutes to 5 till 5 we will take questions before we start taking the solutions from problem number 4. NIT Trichy any questions? Sir, I have one doubt related to nanoscale and microscale heat transfer. Okay, you ask it? Sir, is it applicable this law? Sir, 4 year law is applicable for nanoscale and microscale. I have no answer for this question. The question asked is is Fourier law of conduction applicable for nanoscale and microscale on the face of it? Yes. In fact, if you see incorpora and David fourth edition not fourth edition sorry sixth edition which is come out the recent edition which has come out there is a problem in which the nanoscale problem has been solved. So, actually there also again parallel circuits are used and again resistance analogy is used and Fourier law of conduction is used. In principle yes even for microscale and nanoscale one can use Fourier law of conduction. So, there is no doubt about that but if you ask me whether thermal conductivity is going to be constant or is it going to be varying I have no answer for that but for usually they are all made of silicon, silicon carbide. So, silicon carbide thermal conductivity is very well documented. The answer is in simple terms even for microscale and nanoscale Fourier law of conduction is still valid with may be some exceptions but I am not aware of those exceptions but in general yes it can be applied. What is the advantage of this micro fins over the conventional fins? The question asked is by one of the participant what is the advantage of micro fins over conventional fins? Micro fins are indeed conventional, there is nothing like very unconventional about micro fins. Since ages micro fins have been used in refrigerating circuits for evaporators and condensers. So, micro fins is indeed conventional but micro fins are used on the inner side of the tube not on the outer side usually. Why because micro fins essentially now it is little early now that you have asked the question I have to answer it. Micro fins actually break the laminar sub layer and make the boundary layer completely turbulent which will enhance the heat transfer coefficient and decrease the convective resistance on the inner side of the wall. So, micro fin tubes are very much used in refrigeration circuits. Sir good evening sir another question sir. So, yesterday in refrigeration for suction line we have to provide insulation and the answer given is that for the compressor the entry to be liquid so that it should be insulated but normally it is a vapour sir I do not know whether sir that point please explain sir. See one of the participant is asking the question that yesterday she is referring back to one of the questions yesterday what we had answered that is the compressor inlet is usually he is claiming that it is vapour it is inside yes the inlet is vapour and by the time it gets out it is going to be superheated. So, yes it is vapour did we say that it is going to be liquid. So, no if we have told that it is liquid we stand corrected we are wrong yesterday if you have told that it is liquid we are wrong yes at the inlet it is going to be little lower than the saturated but it will be saturated at least it will be saturated liquid and as it is coming out of the compressor it is going to be vapour. Now, the answer given to that question was why is the insulating medium given only on the suction side why not on the pressure side honestly I know answer for that may be professor Arun can chip in and answer that. Probably because suction line is the difference between the temperature T infinity or the room temperature and the suction line temperature is substantial meaning 30 40 degrees but for the discharge line the temperature of the fluid is almost close to the surrounding temperature usually because it is at the higher temperature compressor inlet typically is at lower than atmospheric temperature. So, the chances of heat coming in from the atmosphere are high when the delta T is large when the delta T is small the amount of heat transfer is negligible is small therefore you do not really care about insulating the discharge side probably that is the better explanation. Sir, another question shall I ask sir. Go ahead. Sir, another question from the refrigerator itself in cold climate in cold climates the condenser is exposed to the atmosphere whether there is any heat load more heat load given to the compressor sir. The question asked is by one of the participants in cold climate condenser is exposed to atmosphere is there any additional heat load from the ambient to the condenser. So, condenser condenser is generally sitting at a lower temperature at a higher temperature but then if it so the condenser usually is sitting at it is operating it is taking the fluid the refrigerant to the condenser is coming at a higher temperature. Higher temperature and pressure. Higher temperature and pressure ok we can perhaps understand like this. If I imagine a tube in tube condenser inside the tube let us say we have the refrigerant going at a higher temperature but that has to be cooled by chilled fluid which is which is to be at a lower temperature. So, which is already sitting at a lower temperature that is why perhaps it is exposed to the atmosphere that is all I can say but I do not think we can generalize this because it depends on refrigerant to refrigerant if it is R 12 or R 22 I do not think we can afford to expose. But if it is R 134A yes my chiller temperatures are going to be sufficiently low they are almost negative side then perhaps I can expose it to atmosphere provided my atmosphere is also sitting at negative but if my atmosphere is again sitting at higher temperature then there will be heat loss. So, I would not like to expose the condenser heat gain. So, I would not like to have the heat gain from the ambient to the condenser. So, I think the condenser is exposed as you said in the cold condition only for those refrigerants where on the chiller temperature, chiller fluid temperature and the ambient temperature are of the same order that is all I can think of as the answer ok. Thank you. Kolhapur KIT Kolhapur any questions please. Yes sir there is no question when we are required to use the concept of corrected length of fin that was not touched in that session. The question asked is let me let me rephrase the question asked is by one of the participants what is the corrected when do we use the corrected length of the fin. So, when do we use the corrected length of the fin ok ok. So, we understood your question see corrected length is taken here already for example, of course that figure is not there let me see. Yeah, yeah here it is there you see this zeta is already the corrected length L plus t by 2 into square root of h by k t is already the corrected length. In fact, I do not think it can be called as corrected length it is. It is opening the fin it is cutting the fin and opening it into two flat surfaces so that you take care of the rim that is what it is. That is all it is ok. Thank you. Sir one more question. Please go ahead. We have solved problem number 3 in that we have taken that areas surface area A is located placed in an evacuated chamber. Does the heat transfer rate or net heat transfer rate heat transfer will depend upon how that area is placed in the chamber. One of the major assumptions in that problem is that the object is very very small compared to the chamber. So, wherever you place it however you place it all the energy from the chamber walls are coming to the surface. So, till we do radiation heat transfer we do not we do not calculate anything called as view factors at present. We just use the actually it is heat given by the object minus heat received by the object. So, if I have to write heat received by the object it has to be sigma epsilon t surrounding to the power 4 times a view factor associated between the room and the object. So, that factor and all the area multiplication etcetera there is something called as reciprocity rule which ultimately gives you this formula. So, inherent assumption is that the object is very very small compared to the overall dimensions. So, no matter how the object is oriented it is not going to affect the final heat transfer rate. Thank you. Thank you. Over to J N T U Hyderabad any questions? J N T U Hyderabad. The problem is all no there the velocity of water and velocity of air is given. The problem can be we can solve by external flows no. Bulk mean temperature T F is equal to wall temperature average of T W plus T infinity. Sorry to interrupt. So, meanwhile so in order to solve the second problem Reynolds number and Prandtl number we have to calculate no. I would just like to rephrase one participant is asking that in problem number 2 that was taken up. We can calculate Reynolds number Prandtl number etcetera and try to calculate the heat transfer coefficient. Well that we will do once we cover convection. Right now we are at a stage where we just know Newton's law of cooling. We do not know convective heat transfer we are just applying this equation. Later on when we are going to do convective heat transfer such kind of problems obviously will involve computation of H naught by in that case Q will not be given to you. In fact, Q will have to be calculated by you for the wire basically what you would be asked is how much is the heat removal rate if air at 25 meter or 10 meter per second and 25 degree centigrade flows and how much is the heat removal rate when water at 25 degree centigrade and 10 meter per second flows. So, you would not be given Q in that case you would be asked to calculate Q for which the first step would be to calculate the heat transfer coefficient. This is a beautiful problem I would say why beautiful because this is a method of measuring convective heat transfer coefficient. If one is trying to measure the convective heat transfer coefficient this is the method I take a wire put it in water go on increasing the or for a given fluid velocity I will go on increasing the input power that is voltage and current what I am dumping into the wire until it burns out. So, this is this is a very elegant problem in the sense that this is a this is showing me the method to measure the heat transfer coefficient cylinder in case of external flows. Yes, we are going to do that in convection when we do the questions what you asked. Thank you. One more question sir. Sir, in heat exchanger by inserting pin pins we can increase the heat transfer coefficient so the pin pin analysis can be done either analysis can be done with the help of effectiveness or efficiency sir. See question one of the questions asked by the participants is pin pins can be used in heat exchangers whether efficiency or effectiveness is to be used for evaluating the pin pin performance. Pin pins generally pin pins generally in the they are not used under convective conditions. So, if I have to answer this question little more detail I will have to use the draw something. So, generally pin pin cooling where it is being used. So, it is not being used in convective condition if I take a gas turbine blade this is my gas turbine blade which is there in all aircraft gas turbines or fighter aircraft so here in this portion you use convective cooling that is the fluid goes and comes back it is almost like my hand this is the size also will be roughly around 6 centimeters. So, we put fins throughout here the fins are put throughout the pin fins are put here why because the space here is very less the space here is very less. So, instead of convective cooling I cannot make a passage and have convective cooling. So, in order to have effective heat transfer I put fins here vertically sorry I drew it wrong. So, vertically from top to bottom there will be pin fins if I redraw this portion again. So, there will be fins throughout here one fin the next one fin. So, like that pin fins are going to be there. So, pin fins why are we using this here pin fins because number one there is no space for convective cooling passage. Number two here by simple conduction itself I can effectively remove the heat transfer because the thickness is very small. So, pin fins generally are not used in convective cooling passages that would be the answer for your question. We will take up problem four onwards for now ok. Yeah, I think there is one more question. In the morning we had one problem relating to finding out the length of the fin. We took the condition adiabatic fin tip is equal to infinitely long fin heat transfer rate and we have equated it to find length. So, why did we do that? We considered these boundary conditions for this and we found the length of the fin. Yeah, the question asked is in one of the problems in the morning we took here estimate how long the rods must be there. Of course, in the problem itself it is stated for the assumption of infinite length. Even if it is not stated we would go ahead with this infinite length. See why this is taken as infinite length? By common sense one would think that as I go on increasing the fin length the heat transfer rate will increase that is heat transfer rate that is number of wattage is the amount the wattage will increase with the increase the length, but then that in real life it is not so we solve the problem. In the problem what did we see? Again we have shown this table several times nevertheless I would like to show that table again that is we have plotted M L versus tan hyperbolic length that is infinite length is infinite only for mathematician. How infinite is infinite for an engineer is decided by tan hyperbolic l? I can decide only when I take the infinite boundary condition. So, that is the reason why we have taken infinite boundary condition. I think that is the only logical answer I can look at I can give for this question. Another thing was I mentioned in the morning as the fin length keeps on increasing the delta t available for heat transfer keeps on decreasing. So, ultimately an infinitely long fin would essentially have a tip which is equivalent to a location where t b is almost equal to t infinity. So, that there is no heat transfer there and that is why we are saying that in the limiting condition we can say an infinitely long fin at the tip has a boundary condition which is equivalent to the adiabatic boundary condition that was the reason for taking this equality infinite length. Because we need to get a length. So, to get a length we need to know what is the heat transfer associated if in the limiting case of this infinite fin it behaves as an adiabatic tip condition and that is logical. Thank you. We are going to problem number 4. Problem number 4 says it is radioactive waste are packed in a long thin walled cylindrical container and the waste generate some thermal energy non uniformly according to some equation which is given it is a parabolic equation in r and q dot or that is a volumetric heat generation rate. This is one of the cases where definitely you will have volumetric heat generation and that is a non uniform volumetric heat generation. All through when we did various derivations we said volumetric heat generation is uniform. This is one of the situations where it is non uniform and the equation is given as a function of the radius of the cylinder. Steady state conditions are maintained by submerging the container in a liquid that is a T infinity and provides a uniform heat transfer coefficient h. We are asked to find the total rate at which energy is generated per unit length and from this to get an expression for the surface temperature at the container wall. So, assumptions in this actually what is there in this problem if I say given we are given a cylindrical container which has some radioactive waste. So, they are being transported let us say. So, there is q which is a function of r volumetric heat generation that is given to us. So, q dot of r is given to you as q dot 1 minus r by r naught whole square. So, this is given to us and q naught is a known positive constant. This to keep things cool this is being surrounded by a fluid which is providing your heat transfer coefficient h and the fluid is at T infinity. So, you are asked to find expression for rate at which energy is generated per unit length. Please note q is the heat transfer rate q dot we call as volumetric heat generation watt per meter cube. In nuclear reactor parlance this will be called as q triple prime triple prime corresponding to per unit volume, but here heat transfer textbooks most of them we use q dot q double prime logically is heat flux watt per meter square q prime is the linear heat rate watt per meter. So, logically this should be q triple prime instead of q dot, but anyway we stick with this nomenclature. So, we are asked to find q prime linear heat rate and also obtain an expression for p surface. So, bunch of assumptions will be there. So, first assumption is of course, steady state this we have not dealt any unsteady state related problem. First is steady state given uniform heat transfer coefficient third thing T infinity remains constant. So, you are adding heat means nuclear reactor waste is giving away heat to the surrounding fluid, but it is a sink in thermodynamics we have studied sink temperature will not change. So, T infinity remains constant otherwise we have a problem that is one thing then yeah. So, if I look at the solution it is just a mathematical exercise we want to calculate the volumetric heat generation per unit length. So, let us first do it per unit volume we can write q dot r d v this is going to give me watt per meter cube times meter cube which is going to be watts and d v for a cylinder volume of a cylinder is pi r square l. So, d v is going to be 2 pi r d r times l. So, this l is what is going to come down. So, total q dot is nothing but integral q naught 1 minus r by r naught whole squared times d v which is this one I bring this l down. So, q by l which is q by l which is given by volume q naught 1 minus r by r naught whole squared 2 pi r d r this l has already been brought down and the limits of this integration I will replace this by r is equal to 0 to r is equal to r naught that is where the problem ends we can pull out q naught 2 pi out of the integral you have 1 minus r squared by r naught squared multiplied by r d r integrated from 0 to r naught this is going to give me 2 pi q naught integral of r d r would be r squared by 2 integral of r cube this is to be multiplied r cubed is r to the power 4 by 4 r naught squared applying limits 0 to r naught and that is going to just give me q dot by l equal to 2 pi q naught if I apply the limits this will be r naught square by 2 minus r naught so this is going to give me r naught squared by 4. So, this will be r naught squared by 2 minus r naught squared by 4 would be r naught squared by 4 this 2 cancels with this 4 will get me 2 the answer is pi q naught r naught squared by 2. Let us check whether it is correct in terms of units q naught would have been in watt per meter cube times radius square which would be meter squared answer has to be watt per meter which is what we require. So, this is the first part of the problem second part is very straight forward you are asked to find the surface temperature. So, I have to do an energy balance. So, if I take the cylinder heat is being generated e dot g this has to go out by convection e dot out and my favorite equation is there e dot in minus e dot out plus e dot generated is equal to e dot stored steady state this goes to 0 nothing is coming to the control volume e dot in is 0. So, e dot out is equal to e dot g e dot g we already had. So, let us write e dot out minus h a surface area p surface minus t infinity is equal to minus e dot g minus is cancel of e dot this is going to be h times pi r naught 2 pi r naught l is the surface area p s minus t infinity is equal to q this l I will bring here I get q by l which we have calculated. So, I will get h times pi r naught 2 times h pi r naught t s minus t infinity is nothing but q by l which I substitute from the first part as pi by 2 q naught r naught square pi pi r naught r naught square. So, therefore, I get t s is equal to t infinity plus q naught r naught by 4 h that is the answer this is h. So, it is a very good exercise in book keeping and of course, mathematics simple integration and more importantly application of this conservation equation e dot in minus e dot out plus e dot generated equal to e dot stored. So, this equation we are applying after we have obtained an expression for e dot generated and just things can go wrong if you do not take care of the length properly. So, this is already q by l that is what we need to keep in mind. So, once that is done there is no issue about the problem. We are skipping problem 5, 6, 8, 9 and 10. 8, 9 and 10 are very very straight forward application of thermal resistance concept for insulation properties have been given one case you find k other other case find the heat transfer rate. So, I think all of us know these things. So, we will take up problem number 7 which is slightly lengthy it is not difficult it is slightly lengthy. So, that we try to appreciate these kind of problems also. So, what is given is in problem 7 I am doing problem number 7 of this tutorial. It is one dimensional steady state conduction with uniform internal heat generation occurs in a plane wall of thickness 50 mm. So, as I am reading I am going to write plane wall is given to us 1 d steady state conduction. So, this is one of the assumptions let me as I am writing the problem I am writing assumptions also in view of time one dimensional steady state it is a plane wall dimension is given thickness is 50 mm constant thermal conductivity k is constant k is equal to 5 watt per meter kelvin for these conditions temperature distribution is given to me in the form t x is equal to a plus b x plus c x square. And we are also given couple of other boundary conditions surface at x is equal to 0 t is equal to t naught is 120 degree centigrade and at x is equal to l we have insulated boundary condition which I want to translate mathematically as d t by d x at x is equal to l is equal to 0. We are also given another condition for x is equal to 0 that is the left phase that is a convective boundary conditions. So, at x is equal to 0 let me let me draw a schematic. So, this is the coordinate direction x 0 l there is k which is given to me h comma t infinity on this side. And this surface is at t naught which is specified to be as 120 degree centigrade and this is insulated wall. So, d t by d x is equal to 0 at x is equal to 0 we have h is equal to 500 watt per meter square kelvin and t infinity is 20 degree centigrade. So, this is the plane wall which is given to us whatever we supposed to find applying overall energy balance calculate the internal heat generation. So, find internal heat generation 1 and determine coefficients a b and c by applying the boundary conditions to the prescribed temperature distribution. So, for this problem you are supposed to find the temperature distribution and it is expected that it is a polynomial distribution as we have studied in the class. We want to find the coefficients a b and c. So, as we started we wrote couple of assumptions let us see we have any more 1 d steady state constant properties uniform h and t infinity so constant and uniform values and volumetric heat generation rate is uniform. So, that is also an assumption we make we do not have any direction or location specific volumetric heat generation. So, I have tried to fill as much as possible here anyway. So, now our we have to find the heat generation rate. So, overall energy balance is what we have to do overall energy balance for the entire wall. So, I take the entire wall as the region of interest and I do e dot in minus e dot out plus e dot generated equal to e dot stored steady state this goes to 0 e dot in is e dot in is 0 how do I know that I know that because t infinity is 20 t surface at x is equal to 0 at x is equal to 0 t surface is 120 degree centigrade. So, heat is not coming from the outside to the inside heat is going from the wall to the outside. So, in fact physically if I look at the problem the left this phase is insulated. So, nothing can go out from the right hand side phase heat is being generated. So, obviously heat has to go out if steady state conditions have to be maintained from right hand side to left hand side from the object to the fluid. So, this surface temperature obviously has to be higher than the free stream temperature. Therefore, if I take this control volume which is the entire available unit of the plane wall there is no e in it is only e dot generated equal to e dot out that is the balance. So, e dot generated is what I have to find that is equal to e dot out and that comes out to be q dot times volume which is a times l is equal to e dot out which is h times a t naught minus t infinity we use t naught because that is the notation we have used. So, areas get cancelled what is this area that I am talking about this area is the area into the plane of the paper. So, this area is what we are cancelling of this thickness is l which is given to us that we have. So, q dot times l is equal to h times t naught minus t infinity therefore, q dot volumetric heat generation rate comes out to be 500 times 120 minus 20 divided by 0.05 meters which is 10 to the power 6 watt per meter cube. So, this is the energy balance and this given as a volumetric heat generation. Now, you are asked to find what is the is a b and c associated we can I will just outline the steps for this you can solve this at home. So, we have to solve the differential equation with volumetric heat generation. So, 1 d steady state equation with volumetric heat generation is d square t by d x square plus q dot by k is equal to 0 and therefore, d t by d x comes out to be minus q dot x by k plus c 1 and t of x comes out to be minus q dot x square by k plus 2 k plus c 1 x plus c 2 this is known to us from class. Now, I will apply the boundary conditions which are there for the problem. So, t of x is a plus b x plus c x square. So, d t by d x is nothing but b plus 2 c x. So, d t by d x at the right hand wall at x is equal to 0 is 0 which comes out to be b plus 2 c l equal to 0 therefore, I get b is equal to minus 2 c l 1 relationship. Second, convective boundary condition at x is equal to 0 we get h times t naught minus t infinity is equal to minus k d t by d x at x is equal to 0. So, now here we have to be careful about the sign. So, d t by d x at x is equal to 0 let me evaluate at x is equal to 0 comes out to be b from here I substitute x is equal to 0 I get b h t naught minus t infinity everything is known k is known I get b is there a problem with this equation sign related. So, please take a look at this equation in the form of the sign that we are doing because x is equal to 0 is a left hand boundary it is going this way. So, we have to take care of the sign appropriately. So, left hand side is right. So, this one has to be taken care correctly. So, that d t by d x on the right hand side would be a plus value and I will get b appropriately. Next, I can substitute the overall into the temperature distribution I can get a from substituting for the temperature at the surface at t naught is equal to 120 degrees you get from here b you have got c from here t naught at x is equal to 0 substitute in that equation you will get a and that should explain the solution. So, a comes out to be 120 and b is correct b is equal to 10000 and c is equal to minus 1 lakh. So, this is what you would get. So, I know we went fast in the last problem, but straight forward energy balance related concepts only that nothing new or nothing difficult here, but see here we have to take care as I said about the sign because the direction is important. So, if you to the bookkeeping correctly I do not think there is going to be any problem. We urge all of you to do the remaining problems each of them should not take more than 5 or 10 minutes to do. Tomorrow morning before you come at 9 o clock before we start the session I request all coordinators to collect the tutorials and during the day before lunch just check if all the problems have been attempted the solutions of this tutorial will be available from tomorrow morning. So, we will we will upload the solutions the day after the tutorial is over. So, we urge all participants to spend the evening trying to finish this and we will meet tomorrow morning at 9 o clock where we will start transient conduction. Thank you. Any questions if any we can take yeah. Vijay T. I am Mumbai are you with us. Sir I have we have few questions here my first question is that while solving one problem in the morning it was found out that the product M into L was 2.65. Okay. Whereas when you showed one table in that table for for the infinite infinitely long fin you showed that M into L has to be 5. How is this discrepancy? There is. Okay. The question asked is in the problem solved in the morning ML in the problem it has been stated as 2.65 but in the table it is shown that ML of 5 is required. So, let us get back to the table I I think we have flashed the table N number of times nevertheless that shows the importance of the table. So, ML if we see ML variation with tan hyperbolic ML see what we are saying is that see for 2.65 for an ML of 2.65 what is the efficiency? Efficiency is around 0.99 already it is almost 100 percent for an engineer I think 0.99 is equal to 1. So, but then ML of 2.65 and ML of 5 the length of the fin for a given M that is HP by KAC for a given M length of the fin is almost double but I have not gained much in terms of heat transfer rate. So, that is the reason why we are saying in the problem that ML of 2.65 is enough for us I think that makes sense because ML of 2.65 is already reaching 0.995 times over to you. Okay. Have we clarified? Yeah. Sir, one more thing is that while you were solving a problem in the morning you said that K is equal to K 0 plus 80 that problem I am referring to what I understand from this discussion is that K is never then a constant in a real life problem is that correct? The question asked is K is never constant in a real life problem I do not think this statement is right see our operating temperature ranges are not going to be always from 0 to 1000 degree Celsius. So, they are limited for example if I am taking air let us say if I am taking the temperature between 30 to 50 I do not think temperature the thermal conductivity is going to vary significantly between 30 to 50 but let us say I take water and it is entering at 30 degree Celsius and getting out at 100 degree Celsius yes thermal conductivity is going to significantly vary. So, case by case basis the thermal conductivity variation is important but we cannot make a statement that thermal conductivity is always varying or thermal conductivity is always constant it depends on situation to situation and temperature range ok thank you we are ending this session for the day