 We are numbering them sequentially, 1 to 10 was given yesterday, today we will start with number 11. So, there is a long cylindrical QL element in a nuclear reactor, there is energy generation which occurs uniformly, the diameter of the rod has been given to us. So, diameter has been given to us as 25 mm and there is a thin aluminum cladding. So, there is a thorium rod and then there is a cladding made of aluminum thorium. This diameter has been given to us, it is proposed that under steady state condition system operates with a volumetric heat generation rate of 7 times 10 to the power 8 watt per meter cube and h comma T infinity are also specified. So, h is 7000 watt per meter square Kelvin and T infinity is 95 degree Celsius. So, we are asked to find for safe operation the clad outside surface temperature should not exceed 600 degree centigrade. So, this is the limit on the outside surface temperature. So, we are asked to find whether this cooling condition is acceptable is satisfactory. That means, with this given amount of cooling h and T infinity will the amount of heat generated be taken out. So, that the temperature on the outer surface of the clad is not going to be greater than 600 degree centigrade. So, where is the clad? This is the cladding aluminum. So, on the outer surface of the clad we should not have temperature greater than 600 degree centigrade. So, it is a simple problem where we are going to deal with energy balance only. So, list of assumptions we will write we are dealing with steady state known constant properties one dimensional conduction in radial direction outward radial direction real conduction uniform. Now, you will appreciate what is why this assumption is very important. There is flow of air from this side let us say we know for a fact that h is going to vary, but we are going to assume uniform h and T infinity known constant uniform volumetric heat generation. Then you have this is a no contact resistance between the fuel element and the cladding. So, if I apply energy balance E dot in minus E dot out plus E dot generated equal to E dot stored this goes to 0 because of assumption 1 which is steady state. You do not have energy coming in you just have energy generation and that has to be transported out. So, there is no energy coming in therefore, E dot G is equal to E dot out. E dot G is been given to us in the form of a volumetric heat generation. So, let me write that Q dot is given to me as 7 times 10 to the power 8 watt per meter cube volume of the cylinder I do not know the length. So, we can do the calculation per unit length basis per unit length basis we will do. So, I will not get total Q instead I will get Q prime which is Q by L that is nothing but Q dot times pi r square. So, pi r square L is the volume I bring the length here. So, I get the linear heat rate this r is the fuel element radius not the cladding. So, this comes out to be Q prime comes out to be 7 times 10 to the power 8 times pi into 25 by 2 times 10 to the minus 3 whole square 12.5 millimeter and this is nothing but 3.4361 10 to the power 5 watt per meter. So, this is the linear heat rate this has to be removed what happens if there is heat accumulation temperature is going to rise. So, we want to see at steady state with this convective heat transfer coefficient on the outside we do not know what is the outside temperature of the fuel element this is the clad we do not care about this we are just worried about this being P naught this. Obviously, calculations would have come from safety aspects of the fuel we are not concerned with that, but we know for a fact whatever heat is generated has to go out and this is the clad. So, Q prime therefore, is equal to h pi D clad out times T out minus T infinity y D clad out y pi D it is actually pi D L, but that length has been taken here because this is Q prime that is Q by L. So, pi D clad out times T naught minus T infinity this value is given to us as 600. So, h pi D clad out T infinity minus 600. So, we will substitute for that substitute we can do one of two things we can substitute for h pi D clad out 600 minus T infinity we will get a value of linear heat rate that linear heat rate if it comes to be if Q prime now comes out to be less than Q prime calculated that means this cooling arrangement is not sufficient. Conversely, we can put the value that of Q prime calculated earlier and estimate that T naught or the outside surface temperature if T naught comes less than 600 we are safe if it comes greater than 600 we are unsafe. So, let us do we have done we have used this value. So, that is 3.4631 times 10 to the power 5 watt per meter is equal to h 7000 into pi into 0.025. Why we are using the same dimension essentially because thickness it has been told is negligible thickness or very thin aluminum cladding. So, we are using this as same as D naught of the fuel element T out minus T infinity. So, from this if I calculate I get T surface comes out to be 725 degree centigrade which is much greater than 725 degree centigrade this is T naught which is greater than 600 degree centigrade which is permitted. Therefore, the cooling arrangement is poor or not satisfactory how do we increase this only way logically probably is to increase the value of the heat transfer coefficient. So, that effective heat removal is going to take place. So, some of you can try the other thing put 600 see what is the Q it has to come out to be lower than 3.463 times 10 to the power minus 5 watt per meter. So, we are doing these kind of simple problems in the beginning just to emphasize the concept of energy balance. Now, we will shift to two questions for some time maybe for next 10 minutes. So, what we would encourage you to do is you be solving the problems we will hand pick any random problem out of 10 to sorry 11 to 20 and then solve, but please be solving while we are taking questions from various centers. So, yesterday we had requested the coordinators to take a look at the solved tutorials from all participants today. So, I hope that was done and almost all of them have solved all the problems that was assigned yesterday. So, if that is not the case please let us know there could be multiple reasons that either they are having difficulty in which case we need to address that issue or if people are having other problems we need to address that also. So, Ms. College will do. Could you please explain how the Heisler charts are constructed? Heisler charts I think. So, that is very good question one of the participants question is how are Heisler charts constructed and how are they different from the analytical solutions which have been shown. So, that means we have not connected both of them. So, the analytical solution what was shown let me reshow that. So, this is the analytical solution. So, using this analytical solutions only we are plotting what is there on the x axis and what is there on the y axis of my charts in the y axis I have t minus t infinity upon let me show that yeah what is that what is that we are showing. So, here what is there this is Fourier number and this is 1 by biot number and this is theta number what is my analytical solution telling analytical solution is telling that this theta is a function of biot number how directly I cannot see this lambda 1 is a function of biot number this is the solution of a transcendental equation. So, for a given biot number I will get lambda 1 that is the first truth itself if I take only the first truth or even multiple roots I get for a given biot number I will get lambda 1 and a 1 if I plug in that lambda 1 and a 1 I can construct temperature as a function of x and t minus t infinity minus t i minus t upon t i minus t infinity. So, answer your question the chart is nothing, but the graphical representation of this equation. So, in fact historically why charts were generated because at that time when charts were generated as you have seen in the historical perspective that is in 1952 or 1960s where there was no not even calculator not even computers computers and calculators were not there. So, at that time charts were very convenient. So, that is the reason they had generated charts that means they had one time sat down and calculated may be using logarithmic tables and generated those charts. So, that every time I can use those charts. So, to answer your question the charts are nothing, but the charts are nothing, but the graphical representation of these equations. And if you look at the graph also it is the center line theta naught. So, what is done is this term has been pushed to 1. So, that you only have a 1 e to the minus lambda square tau in fact it is shown here this is the center line thing. So, this has been plotted as a function of two parameters one is tau and other is 1 by biot number or biot number and then what they have done this fraction goes from 0 to 1 all the time. So, the second plot which is there is this one. So, theta by theta naught it is essentially going from 0 to 1 again as a function of 1 by biot number, but the dependency of the local position x by l which you see here x by l that is there in that. And biot number comes in there because biot number decides how at a given location what would be the temperature at relative to the center line temperature ok. We will take one more question, Jabalpur. Over to Jabalpur college. When we were talking about the transient heat conduction we obtained a exact solution from where we can obtain that series from which method we obtained that series. It is the solution of the different the question asked by one of the participants is that we have given a series solution where did this series solution come from of course, yes one thing we have to admit we have not provided the mathematics how we have solved this differential equation which is in the non dimensional form we have only given the solution. So, it is we have not represented the mathematics in this solution may be why do not you post this question in the forum we will write down the complete derivation and put it across no problem, but here in this because usually why we have done this because usually for u g we do not give this solution. So, that is why for under graduates we are not given however as a teacher you would like to know you yeah it is there in many books actually if we say heat conduction by ozizic or any conduction books if we see it is there but not in u g text books it is there ok. Professor Gaitande says that it is there in professor Sukathme's book also. So, basically it is there in any text books, but if you are not able to get it you post it in the forum we will give you the solution not an issue ok. We will move on to tutorial problem 16 problem number 16 what is the question asked the question asked is stainless steel ball of diameter 3 centimeters let us write in known stainless steel ball diameter equal to 3 centimeters is uniformly heated to a temperature of 800 degree Celsius that is T i equal to 800 degree Celsius it is to be hardened by first cooling in a oil bath to a temperature of 100 degree Celsius that is T we can take this as T final if you want T final equal to is hardened by first cooling in an oil bath to a temperature of 100 degree Celsius to a temperature T final we can call T final is 100 degree Celsius and the heat transfer coefficient and the oil bath temperature are heat transfer coefficient is 700 watts per meter square Kelvin note this the heat transfer coefficient value is very high why because the fluid used is oil that is the reason although even natural convection is taking place the h is quite high and oil bath temperature is 40 degree Celsius that is T infinity is 40 degree Celsius what is the time required for this process let us answer this question there is a second part first let us take this part that is let us before we move on let us state the assumptions the first question see we have so many techniques which we are there so first thing is we have not moved beyond of course we have done multi dimensional but let us go ahead and make the assumption that it is one dimensional and we know the properties all properties are constant and remember all the properties are going to vary with temperature nevertheless we are going to make an assumption that properties are constant and next thing is I am neglecting radiation why I am making the statement that I am neglecting radiation because the temperatures are high so that is why I am making a statement that I am neglecting radiation another important thing is that now I think we will appreciate better I am saying that the heat transfer coefficient is uniform heat transfer coefficient is uniform it need not be on the top it might be better on the bottom it might be better so it depends how the natural convection is taking place nevertheless I am making the statement going the assumption that heat transfer coefficient is uniform the first check in a transient conduction problem the first check in a transient conduction problem is to check whether my Bayard number is less than 0.1 or not that is the first check I need to make so if I have to do that check what should I calculate h l by k h l c by k and this is a sphere ball so the characteristic length would be volume by area that is pi d cube by 6 upon pi d square so I get d by 6 so that is characteristic length is d by 6 we have done in the morning so h into that is 700 into 700 into Bayard number equal to 700 into here diameter is 0.03 upon 6 that is equal to k is 61 so please go ahead and calculate this number the number is going to be d is 0.03 divided by 61 that is 0.6 0.057 0.0574 0.0574 so which is very much less than 0.1 so that means what our life is easy now that means we are going to use lumped capacitance lumped on a lumped methodology so that is lumped capacitance is valid so for lumped we know the temperature distribution theta by theta by theta i equal to e to the power of minus t by tau minus t by tau what is theta here tau is yes h as t by rho v ct so just we have to plug in now it is a plug in problem we say plug in problem because we have to just plug in the known thing unknown thing unknown what is unknown here what is the unknown thing I need to find out what is the time required so time will come out that is 100 minus 40 upon 800 minus 40 is equal to exponential of minus 700 into t into into actually 6 by d as by v equal to 6 by d that is 200 it is but nevertheless let us put it 6 by d as 0.00 0.03 it turns out that time equal to 65.61 seconds 65.61 seconds now the next part of the question is next part of the question is determine the heat removal rate from the oil bath per minute so that its temperature remains constant at 40 degree Celsius so now I need to calculate the heat removal rate we have already done this we have already done this we can go to the document back and see that relation so if you see that we have got that as q equal to rho v cp theta i into 1 minus of e to the power of minus t by tau yes let us do that that is so that is equal to heat removal rate equal to rho v cp theta i into 1 minus of e to the power of minus t by tau I think it might be a good idea to calculate tau so that we get the feel of tau actually let us calculate tau for the heck of it tau equal to rho v cp upon h a so rho is 7865 7865 v by a is d by 6 that is 0.03 by 6 and cp is 460 upon h is 700 okay so if I calculate this 7865 into 0.03 into 460 upon 6 upon 700 I get 25.84 seconds you can see that it is taking too much time to respond because you see it is 3 centimeters number 1 for the thermocouple you remember in the morning class we solved it was just hardly 4.52 seconds so here in this problem because that diameter was hardly 0.07 mm 0.7 mm 0.7 mm now you see the diameter it is 30 mm because of which the response time the time constant has gone up that is why it is 25.8 seconds now let us put this back in our earlier relation q equal to rho is again 7865 into v is pi d cube by 6 d is 0.03 cube into cp is 460 into 800 minus 40 into 1 minus of e to the power of minus t we have just now found 65.61 upon tau we have just now found 25.84 that means you see how I should look at 65.61 divided by 25.84 that means it is almost I have taken 2.6 times the time constant okay so why because I want to reach higher and higher I want to cool it to I mean here I am I yeah I want to cool it to lower and lower temperature so the power comes down to 1326 joules q equal to 1326 joules this is for one ball but the question asked is for question asked is for 100 balls so for 100 means all that I need to do is multiply by 100 so that means 1326 into 100 that means 132.6 kilo joules that is the heat removal rate I need to have in a minute this is the heat I need to remove so that bath temperature does not increase yes that is the if this much heat is not removed then this 40 degree temperature bath would be increasing in its temperature so you have to provide appropriate cooling for the bath so that this much amount of heat has to be removed okay so then we will move on to the next problem or maybe we can take some questions for a few minutes over to Anna University Chennai. Okay see actually can we postpone this question tomorrow in fact in a great the question asked by one of the participants is very good question actually for a pipe why do we take the characteristic length as diameter and for a flat plate why do I take the length of the pipe length of the plate length of the plate and so these characteristic lengths vary for different configuration why do we take differently for different configuration I think we will postpone this question to convection in convection we are going to deal with this in great detail so actually characteristic length to put it in a very just to clear your suspense so that you do not feel worried we have to take the characteristic length actually the boundary layer thickness but we cannot take boundary layer thickness because we do not know the boundary layer thickness so it is the engineering necessity which makes us use the characteristic length which is measurable that is either the diameter or the length but Professor Bajani in his book shows that if we take the Reynolds number defined an appropriate boundary layer thickness no matter what is the class of the problem that is whether it is pipe flow that is internal flow or external flow or impinging flow or natural convection Reynolds number of the order of hundred is the transitional Reynolds number in fact for different for different flow situations you will get different critical Reynolds numbers that is for flow over a pipe flow in a pipe you saw the critical Reynolds number as 2300 for flow over a flat plate you will see it is to 5 into 10 to the power of 5 for jets you will see that the critical Reynolds number is of the order of 100 why is this difference because it is because we are taking different characteristic lengths if we take the appropriate boundary layer thickness all the critical Reynolds numbers for all class of flows will collapse to around 100 so the point is we will if you are not understood this answer do not worry at all we will address this issue in great lengths in convection. Baramati Over to Baramati any questions please sir in case of unsteady state conduction process Biot number we are considering less than 0.1 so in certain text books they are considering if Biot number is also greater than 0.1 they are considering it as an lumped capacitance method is it appropriate or it is not appropriate if it is appropriate in what are the exemptions for this Biot number to be taken as less than 0.1. I have not the question asked by one of the participants is that few text books are saying that Biot number greater than 0.1 also we can take it as lumped but to the best of my knowledge we have not come across any text book where in which Biot number greater than 0.1 can be taken as lumped is stated so to the best of my knowledge or our knowledge we will just stick to our value Biot number less than 0.1 only might be taken as lumped anything greater than 0.1 might not be taken as lumped. We know the other solution techniques so we can solve it and see with the correct technique and see how much error is also being introduced if you assume lumped capacitance so that is a check for you to do also. So that is we usually do this see we have a closed form solution for our plane wall case you take the plane wall problem whichever we have and you the plane wall problem we have a solved problem in our in our notes so you take the same plane wall problem for and now calculate the using the lumped assumption and see how much the temperature difference between the plane wall that is the closed form solution and the lumped capacitance method solution will differ you yourself will realize that Biot number around 0.1 is the is a good approximation. Hello. Yeah. Sir one more question Biot number less than 0.1 as well as Fourier number point greater than 0.2 so both together are applicable for this lumped capacitance method or only one condition is sufficient. See. Four to you sir. Let me answer this question in the question ask this for lumped whether I should take both Biot number and Fourier number together or is Biot number enough or Fourier number enough see if we see in the notes we have given only Biot number I will answer it this way when I make an assumption that Fourier number it is lumped then what is that I am taking Biot number equal to alpha t by sorry h l by k less than 0.1 but what is there in Fourier number Fourier number is alpha t by l squared that means it is dependent on the characteristic length that is Fourier number but let me answer it in a simple manner that is for lumped analysis the only non-dimensional number which has to be looked for is Biot number and Biot number has to be less than 0.1 for lumpedness we are not going to be worried about Fourier number at all. So, we will move on to tutorial. So, this time we will solve a problem. So, we are choosing the problem because we are not done any problem on fins. So, let us take up one problem on fin that is problem number 18. So, let me read out the problem a brass rod in the form of a fin this is fin fin actually 100 mm long that is l equal to 100 mm and 5 mm in diameter. So, diameter equal to 5 mm extends horizontally from a casting which is a 200 degree Celsius that is T b base temperature is 200 degree Celsius. The air temperature is 20 degree Celsius and provides a heat transfer coefficient of 30 watts per meter square Kelvin that is h equal to 30 watts per meter square Kelvin and T infinity equal to 20 degree Celsius. What is the heat transfer from the rod? That is the question there is a second half let us not worry about the second half what is the heat transfer from the rod? Evaluate the temperature of the rod at 50 mm from the base of the at the free tip. So, I think we will first approach the first part of the question. So, let us take up our circular fin. So, let us just brush up what we have done for fins just go to fin. So, if you see the fin because you see the first thing in this case is that we need to calculate the efficiency of the fin. Let us see for this configuration how does the efficiency of the fin look like why because what are the x axis what are the y axis I should be looking this is the pin fin. So, this is the pin fin case. So, for that pin fin what is that I need to calculate that is zeta I need to calculate zeta and I need to calculate the another parameter which is zeta and area of the fin if I get the area of the fin and zeta from this curve I can get the efficiency of the fin. So, first thing is to do is I need to sit down and calculate the area of the fin and this zeta. So, let us write area of the fin equal to pi D L into pi D L plus pi D squared by 4. So, if I substitute diameter and length diameter as 5 mm and length as 100 mm I get the area in meters if I substitute that I will get that as 1.5904 into 10 to the power of minus 3 meter square meter square. Now, zeta equal to L plus D by 4 within the brackets into square root of 2 h by k D actually here capital D and small d are same sorry we have just used replacing like it is ok. So, if I substitute for L D and L is 100 mm that is 100 into 10 to the power of minus 3, D is 5 into 10 to the power of minus 3, h is 30, D is 5 into 10 to the power of minus 3, k is 380. You note here the thermal conductivity is quite high thermal conductivity is quite high because it is made of brass if I substitute that I get zeta equal to 0.569. If you go back to the graph see this is the graph zeta is 0.569 the curve for that is this the below one. So, I get an efficiency of 0.8 I get an efficiency of 0.8 that is for 0.569 0.569 I get an efficiency of 0.8. In fact, we do not even need area of the fin for calculating efficiency we need area of the fin for calculating the heat law heat transfer rate ok. So, Q maximum equal to h a f theta b T b minus T infinity h is 30 area we just now found 1.5904 into 10 to the power of minus 3, T b is 200 and T infinity is 20. So, I get 8.58 watts. So, Q actually equal to Q maximum into efficiency or efficiency of the fin into Q maximum we have just now found efficiency as 0.8, 0.8 into 8.588 I get 6.87 watt per meter. So, this is the actual heat transfer rate no sorry this is watts it is not watt per meter sorry it is watts it is watts it is just the watts. Now, that answers the first part of the question that is what is the heat transfer from the rod. Now, the second part of the question is evaluate the temperature of the rod at 50 mm from the base ok. So, what is the temperature distribution given by we will go back and flash the table for various boundary conditions we have solved in this. So, which is the boundary condition I am solving here. Convective boundary condition. It is the convective boundary condition that is H comma T infinity that is the first case which is a very lengthy equation ok, but we we cannot take any shortcut we have to use that that is which is involving cos hyperbolic sin hyperbolic control let us do that. So, that is theta y theta b equal to cos hyperbolic m l by 2 m l by 2 that is right plus h sin h by m k sin hyperbolic m l minus l by 2 upon cos hyperbolic m l plus h by m k sin hyperbolic m l. First let us find out m and then we will substitute that m equal to square root of h p by k ac h is 30 h is heat transfer coefficient is given to be 30 and perimeter is pi d cross sectional area is pi d square by 4. If I substitute for perimeter and thermal conductivity being 380 4 by d 4 by d that is right 4 by d pi d by pi d square by 4 gives me 4 by d. If I substitute for diameter as 5 mm that is 5 into 10 to the power 0.005 all under square root yeah I get m of 7.947 what is the unit of m per meter meter to the power of minus 1 right. So, if I substitute m and l by l in the above equation h and k I get theta by theta b at 0.807 theta b is 200 minus 20 minus 20 equal to. So, I get t at x equal to l by 2 as 165.37. Now, although the problem is not specified here if we ask ourselves a question if we decrease the length of the fin also see we have shown demonstrated how to calculate the fin temperature at x equal to l by 2. Following the same procedure one can calculate at x equal to l. So, you will get t at l equal to 155 does that make sense or not the base temperature is sitting at 200 and at the half of the fin length it is at 165 and at the fin tip it is at 155. So, continuously the fin temperature is decreasing. So, that is what we had mentioned yesterday. So, that the fin tip when we calculated for maximum heat transfer we assumed that throughout the fin surface throughout the fin length we took it temperature as equal to t b for calculating q fin maximum. So, but actual temperature there is a decrease in the temperature as we move along the length. The material being what it is bronze with 380 watt per meter Kelvin as a thermal conductivity the drop is not very high it is there about 45 degree drop, but it is not very high. Yeah, it is essentially because it is also high it is 30 ok. Now, I will give you a historic perspective for this problem historic in the sense that this problem came from a students question saying that if I have a fin of length l and I replace it by 2 fins of length l by 2 will I have better heat transfer so on and so forth. These were the questions. So, we cooked up a problem in the tutorial. So, that they could understand this mathematically also looking at the numbers. So, the second part of the problem essentially is that you make the fin of 100 mm into 2 fins of 50 mm length and then you are asked to calculate the total heat transfer rate for 1 fin multiply by 2 will give you the total heat transfer rate for this 2 fins and you are asked to calculate the tip. So, it is going to be I will just show you the diagram here. This was the first case where this was the length of 100 mm. Now, I am going to have 2 fins like this instead of 100 mm each one is 50 mm ok. This is 100 mm I am going to have 2 separate fins this is fin 2 and this is fin 1. So, what was asked is what is whether there is an increase in the heat transfer or decrease in the heat transfer and we just added one more part. So, here 165 was the temperature at l by 2 let us say what is the temperature when it is equal to 50. So, it is a fin tip temperature for the 50 mm fin whether it is more or less we urge you to do the calculation for this it is exactly on the same lines only thing that you have to take care in part b is that l becomes 50 mm everywhere other than that the entire problem remains the same. So, you do the calculation for q. Yeah. The answer is if we have 2 fins q per fin turns out to be 4.13 watts that means for 2 fin it will become 8.26 as opposed to 2 what we had got as 6.87 watts for the 100 mm fin. So, that means having 2 fins which are shorter which are half of the length is more effective than the and reason is not far to see because in the second half of the fin or when the fin is long the temperature drops also meaning you have a larger t minus t infinity the temperature drop is more. So, the efficiency of the fin is going to become less. So, that is the reason and the tip temperatures that is t at l is equal to 50 mm which corresponds to this one is 186 degrees centigrade which is much more than 165.37 which we calculated when it was a long fin of 100 mm length. So, combination of this is going to give you a higher heat transfer therefore, 2 fins appear to be better than a single fin. So, this is again coming back to our tens of questions which came through yesterday since yesterday ML equal to 2.65 that is all it is. So, you can relate this again to ML equal to 2.65 of course, this is not an insulated infinite length k it is a convective boundary condition k, but it is more or less like that. So, that is as you increase the length of the fin its effectiveness is going to decrease that is what this message from this problem is going to come up. Thanks to the student who asked us this question and we made a who made us realize this little point. So, I think we will stop solving problems for today. So, but we strongly argue to solve all problems which we have not solved that is all other problems of course, there are few three problem number 12 I would strongly recommend you to solve because it is not difficult, but it is too much of book keeping, but I but you will be able to appreciate when the area is changing how to handle the temperature distribution. So, that is important to see because otherwise in all the classroom we have start we have solved only the constant cross sectional area. So, this is the problem which has been cooked so that the cross sectional area of the fin is also changing. So, just please I urge you to solve a problem number 12 it is lengthy do not lose your heart it is just simple book keeping. So, it is just once you formulate the problem you have to just apply the boundary conditions 13 and 14 are trivial you can solve them and of course, 15 is again a fin problem, but I would strongly urge you to solve so that you get to know how to use and read the charge for different fin configurations 16 we have solved 17 is also I would urge you to solve because it is again how to use the charge you will get to know 18 has been solved today of course, partially, but you please solve and till 18. So, then 19 and 20 is not there. So, that is it for tutorials for today we will take for 10 or 15 more minutes couple of questions before we sign off. Erode any questions please. Sir, I have the two questions. The questions are application oriented. My question number one in case of submarine engines in case of submarine engines the heat rejection is taking place in the sea water. Whereas, in case of on-road vehicles the heat rejection is that in the ambient. Submarine engines engines which are used for submarines heat rejection is taking place that is heat is rejected to the sea water. How it happens? Number one. Number two in case if it is rejected in the sea water the heat is rejected in the sea water because as you know the HP horsepower of the marine engine is extremely high. So, the heat rejection is also going to be extremely high. So, in such case can we apply this transient heat conduction problem to find out the heat affected zone in the sea. Can you get the question? Yeah, yeah let me rephrase the question. The question asked is the little the question asked is in submarine engines the heat is rejected to sea water. So, can I apply the fundamentals what are taught in transient conduction for for determining the heat rejection rate to the sea water. But the question is little bit vague to where are we looking at? The first question is we are not directly releasing the heat to the sea water. There are I think to the best of my knowledge in submarines or in various ships they do not reject the heat directly to sea water because we have to take care of aquatic animals. So, the rejection has to the whatever heat is generated in submarine engines or in ships that has to be utilized internally and that is what is called as waste water waste heat recovery. Waste heat recovery so that whatever heat is being rejected it does not increase the ambient that is the sea water temperature significantly why because we need to keep the aquatic animals safely. To answer your second question that is can I apply transient conduction whatever applied whatever has been taught now the question would be where am I looking for the heat transfer. The question has to be little more specific then only I can answer the question where to apply. Are you saying that it is for the region just outside of the ship into the water can I use transient conduction is that what you are saying? You cannot apply transient conduction. If the heat is not rejected directly to the sea water then how it happens how the regeneration happens the heat recovery as you said in case of marine engines because some heat should go out now. Some small amount does go out. Small amount yeah the question has this some heat has to be rejected where does it go. Yes some heat has to be rejected but that heat how is that heat rejected handled ok I can answer. Physically how that heat is. So the rejection is I will pump in let us say I will pump in large amount of sea water but large mass flow rates of sea water if I take large amounts of mass flow rates of sea water the temperature of the sea water what is going out will not be large. The delta T I will keep same maybe the load is high but C P is fixed. So if I increase the mass flow rate my delta T that is the T in which has been taken into the submarine and sent out T out that is T in minus T out is kept as minimal as possible not exceeding 1 or 2 degree Celsius then in that case my aquatic animals are safe. I think that is the only way I can think of answering this question. This is MCP delta T basically it is m dot C P delta T I will have to have huge pumps to pump in the water inside my regenerators to pump in at a large mass flow rate so that the delta T in my regenerator is almost constant. I mean the delta T is very very small and the temperature in one side that is the sea water side is almost constant because my mass flow rate is very high over to you professor. If the mass flow rate of sea water which is used for heat rejection is extremely high then that may add more weight to the weight to the submarine. Then what will be the effect of? No you need see adding weight does not mean that I have to put pumps now if without putting pumps how will I move my ships? I have in my engine rooms I have huge pumps I need pumps I cannot do away with pumps just because that I cannot add the weight even that is true in plane. We have a compressor sitting in a plane under the wing so here also I need a pump I cannot say that I do not have a compressor or I do not have a pump I have to have a pump there is no other option I need to have pumps. The question is the question is professor can we apply the transient heat conduction problem and solution and equations to this application. Professor I need to rephrase the question and ask you back which part of the ship you are asking me to look ship or water which part of the submarine engine you are looking for the transient conduction. This is too generic I would request you to put this question in forum properly phrasing it. So, let us continue this discussion because this is this is becoming a too much of an open-ended problem. Ok, ok professor M.A.N.I.T. Bhopal any questions please. In conduction 4 slide number 45 sir problem in area where the air temperature this one. Yes sir this is the only for the one dimensional condition this is the not how to apply the two-dimensional condition. The question asked is in this problem is it one-dimensional can I apply this semi-infinite medium solution for two-dimensional. The very fact that in the way we set up this equation set up the semi-infinite medium itself semi-infinite medium itself the temperature is varying only with X and that is 1 D and time. So, semi-infinite medium solution cannot be applied for 2 D at all. So, that is the answer for your question because here for semi-infinite we have taken temperature as a function of X and T. Over to Anna University any questions please Anna University. Neutr's law of cooling which is given by Q equal to HA T S minus T infinity. At the same time we have also seen that the convection is due to both the diffusion and adduction that is bulk fluid motion. My question is the equation Q equal to HA delta T takes into account of both random molecular motion that is diffusion and bulk fluid motion. The question raised here is Q equal to HA delta T and one of the participants says that this in this equation the H is taken care of by both advection that is the bulk motion and also the conduction within the boundary layer. I just should implore upon all participants that all these issues will get cleared as we go along in convection. It is a little premature to answer this question. I would just suggest that you please be patient with us for at least one more day. We can take these type of questions maybe tomorrow evening. Why because by that time we would have answered through our lectures all the questions regarding the heat transfer coefficient. We have just introduced the definition today. Thank you sir. Over to Jane to you please ask your question. In the multi-directional conduction problem we are trying to superimpose the individual conduction rates of the heat in order to find out the overall heat transfer. Now what are the kind of assumptions we have to make in case of multi-directional conduction problem? The one of the major assumptions in multi-dimensional conduction problem is or for that matter for all solutions what we have ok. Let me rephrase the question. The question asked is in multi-dimensional conduction problem which we have taken as a product of the individual solutions that is for let us say short cylinder we have taken temperature as a product of radial direction and the temperature as a product of axial direction that is x and r. So, what are the assumptions involved? The same assumptions whatever we took for individual cases. In the individual cases whatever the assumptions made one question one assumption was temperature is a function of phase in 1 d that is x and also it is a function of time. Another thing is that properties are constant that is thermal conductivity in the x direction is constant. Otherwise I cannot pull out that k from my if you remember the heat diffusion equation del by del x into del by del x of k del t by del x. I have pulled k from this that means k into del squared t by del x squared. If I have to do that I have to make that assumption that k is constant in the direction of x. So, this is another assumption and of course I have to assume that rho and C p are both constant in terms of time and space. Sorry there is no space here. It is constant it is constant with temperature. Temperature. It is constant with temperature that way yes it is space also homogeneous. So, these are the assumptions involved in multidimensional conduction problems. Are you okay? Okay. Over to Jabalpur college. During fixing the criteria for lumped system analysis we discussed that heat transfer coefficient the convection coefficient should be low. But if the convection coefficient is low then do not you think that the heat transfer will be less? The question asked is if for the biot number for the lumped capacitance validity we said that biot number has to be less than 0.1. The question asked is do not we think that for this for this criterion to be valid my h also has to be small. So, is it not an anomaly if the h is less then the heat transfer rate is less that is the question. H being less and less cooling is less. But what we are trying to say is spatial uniformity in temperature can be better achieved if h is small. That is what we are trying to say. We are not commenting on how fast or how slow the heat is going to be removed. We are saying that spatial uniformity of temperature can be a good approximation for situations where h is small. That is what we are trying to say. In that situation you are going to probably have a poor value of heat removal rate and mind you that biot number also depends on the conductivity of the material. So, if h is small and k is also small proportionately then still you might not be able to achieve lumped capacitance. So, another way of looking at it is if you go to the transparency if you go to the slide we have put that as resistance. So, if you see that we have put h l by k as conductive resistance upon convective resistance. So, what is that we are trying to say? We are saying trying to say that the convective resistance has to be the convective here conductive resistance upon convective resistance. When will be biot number less when the convective resistance that is conductive resistance compared to convective resistance is less. So, even if h goes up accordingly if I increase my thermal conductivity the overall biot number is less. So, it is just a interplay between conductive and convective resistances. We cannot just say that it should be low or it should be high. Yes, in general it should be low for a given thermal conductivity that is all we are trying to say. To be valid over to Anna University. We are using a brass rule that you took the value of brass conductivity as 380 watts per meter Kelvin. But in data books it is written as at 20 degree Celsius it is around 60 to 80 80 watts per meter Kelvin. As the temperature increases slight difference is there but this much difference how it is possible. The question asked is if I understand correctly the thermal conductivity of brass rod we have taken as we have taken in the problem as 380. But actually yes see I can always get away with the answer see there are various alloys various types of brass alloys whose thermal conductivity is closer to 380. But if you are unhappy with the value 380 of brass take it as copper that does not matter. Copper is around 400 so 380 is little lower than copper. But to be honest honestly answering your question I think there can be a brass rod which can be made brass is made of different compositions. So if I play with the compositions I can still get 380 I do not think 380 is out of the world number at all. But if you are very uncomfortable with that number you can imagine that this fins what we are considering here are made of copper or fully copper. So I do not think there is any conceptual problem here. Any other questions please. JN2 Hyderabad appears to have a question over to JN2 Hyderabad. When a space vehicle is carrying a satellite to orbit what are the modes of heat transfer that takes place and which of the one is more predominant. All to you sir. Before leaving the earth it is having it is it is convection but once it leaves the earth it is in space it will be predominantly by radiation. So that is all one can answer but of course before it leaves the space one thing is for sure it is high speed that is Mach numbers are very high. It is supersonic or even hypersonic I think it would be hypersonic not even supersonic. What I mean by hypersonic is that Mach numbers are much greater than 4 or 5. So Mach numbers are high means the flow is going to be compressible it is not incompressible. Of course we are not taught now that you have asked the question to the extent possible we will try to cover that is we cannot make the assumption that the density variations are negligible. Density variations are important until we get out of the earth. So that is it is going to be highly compressible flow where in which the density variations are to be considered. Now when I get out and I get into space of course radiation is important but if I have to consider convection it is rarefied gas dynamic. What is rarefied gas dynamic? Whatever we are studying or whatever we are going to study tomorrow it is all continuum that is continuity sorry continuum mechanics. That is the molecular mean free path is much much smaller but in rarefied atmosphere the distance between the molecules is comparable to my characteristic length. For example rocket diameter itself or rocket length itself the molecular mean free path becomes comparable. That means there is one more non dimensional number which comes into picture that is what is called as Knudsen number. But for continuum mechanics we consider that Knudsen number is very less but for rarefied gas dynamics that is in space Knudsen number is very high. Then my Nusselt number that is the heat transfer coefficient will not only be dependent on Reynolds and Prandtl but also Knudsen number. But all sudden done the heat transfer coefficient are going to be smaller in space. Radiation will dominate over convection. If one wants to nitpick and wants to find the heat transfer coefficient the Nusselt number is going to be function of not only Reynolds and Prandtl but also Knudsen number. Knudsen number is the ratio of molecular mean free path to the characteristic length that is in case of rocket we can take perhaps the Nose diameter. I think that is the answer for your question professor. Thank you. Thank you for the day.