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, 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.  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.


The new constitution proposed by Tunisian President Kais Saeed has been approved in a referendum as per preliminary results. 94.6% of voters adopted the constitution, according to numbers released by the Tunisian Independent Higher Election Authority. The referendum was held on July 25 amid widespread calls for boycott by opposition parties and civil society. It took place exactly a year after President Kais Saeed sacked former Prime Minister Hisham Mashiishi. In the subsequent months, he assumed more power. He also sacked a large number of judges after accusing them of inefficiency and harboring terrorists. The draft constitution proposed on June 30 by Saeed faced immense criticism from all sections of the country. Critics said it was a power grab by Saeed. Tunisia s opposition attributed the low voter turnout to illegitimacy of the referendum. Taeed has claimed that low turnout will not affect the prospects of the referendum. The new constitution will replace the existing one that was claimed following the Tunisian Revolution of 2011. It will give the president more powers over legislation and the judiciary. Parties such as Islamist Anhadah, Democratic Progressive Party, Free Disturion Party, and the Workers' Party of Tunisia have been leading various campaigns asking the people to boycott the referendum. They argue that the process through which the new constitution was drafted is not democratic but illegitimate. The proceedings of the committee which drafted the constitution have also been questioned on the grounds that its appointment was arbitrary. The committee was not allowed to consult any political party while drafting the constitution. Despite being invited to Tunisia s largest union, the Tunisian General Labor Union boycotted the committee s consultation asserting that the dialogue was arbitrary and unilateral. The main sort of thrust is that it will be basically doing away with what had been a divided executive between the president and the prime minister. And it will be giving all of the power to the presidency. We'll see that there is still a parliament but it seems that its powers are vastly reduced. There seems to be very few checks on the presidency. Actually, one of the advisors who was involved in the drafting of this draft has given an interview where he's actually condemned the final version. He says that the version he presented to the president, all of the checks that were there have been reduced. And it's no longer just a hyper-presidential system, but he says that it really is a dictatorial system. And that's one of the criticisms we've seen from many political parties. That said, human rights are still protected within the new draft constitution. But many of the institutions and institutional checks that were there to really guarantee them seem to have been taken away. And that seems to be the main concern that we've heard from human rights activists. There are no guarantees of some of the independent state bodies that previously had been assured by the constitution, whether that's the higher electoral board or whether that's an anti-corruption authority. Even there's question marks over the regulatory board for broadcasting in media. And so these are certainly some concerns that people have. You know, it's also at the moment, I can tell you that the debate over the referendum is for them people who are supporting it really are not sort of supporting it on the merits of the constitution. They're supporting it either as a way to sort of say that we don't want to go back to rule under a parliamentary system where the main party, the biggest party, would be the Anasta-Islamist party. Or they're just saying we want to basically give a vote of confidence to the Kai Faiyat, but really not engaging with the substance of the constitution. Lat-scale protests have been going on in Tunisia since the president's power grab last year. Last week, several thousands took to the streets in the capital Tunis against the referendum. The police forces tried to brutally suppress the demonstration and arrested scores of human rights activists and members of the Workers' Party of Tunisia. Alongside the political conditions, Tunisia is witnessing a major economic crisis and the past few years have seen increased frustration amongst labor and social movements. Kai Said has continued neoliberal economic policies and has not been able to manage the crisis. There was an amnesty bill after the revolution in 2011 that took many years. It was finally passed, but activists who had pushed back against it had basically succeeded in limiting it to amnesty only for corrupt officials of the Ben Ali era, but not for corrupt businessmen. And the president basically extended amnesty in the hopes that this would convince the businessmen to bring money back in the form of investment. Now, it's really not clear that that's the case. It's not clear that there's any sort of transparency or accountability in terms of actually tracking that wealth. He's made a few sort of, let's say, at the rhetorical level, pay-ons to tracking down illicit capital flow that has gone out of the country to a debt audit, but he hasn't taken any substantive steps. And in the meantime, he's continuing negotiations with the IMF for a new loan of something like $4 billion, which in many cases we're hearing people say there really isn't an alternative. Even people who are very critical of the IMF are saying that Tunisia is, at the moment, unless there's significant debt forgiveness, unless there's really very strong steps towards economic energy and food sovereignty, there really isn't an alternative at this current moment to a new IMF loan. But it doesn't seem that the president has really tried to offer alternatives that are really substantive to the way that Tunisia was going before economically.


Greetings, my esteemed laddingtons. Shout out to my man, Jean, for gifting me this fine game. Very heartfelt and appreciated. So, first and foremost, I have a question for all English guys, and, well, everyone who's interested in English history. When you meditate and contemplate on the Wars of the Roses, which side are you on? Are you on the Lancaster side or the York side? Let me know in the comment section below. It's of utmost importance that we discuss this interesting topic. Now, I thought to, before I begin to get into the actual gameplay, I would like to give a book recommendation, and that is Corny Guilden's series on the War of the Roses. The Wars of the Roses. Good stuff. I always recommend everything he writes. I have read almost all of his books. I have one book which I bought immediately upon seeing that it was released, but I have saved it. It's set in ancient Greece, so I can't read it now in winter. I'm checking in from Jortunheim, a glorious snow landscape, but I must wait until reading this book until Helios comes back in force so I can sit outside in the sun. With a good cup of coffee and read that book. So I'm savoring it. I'm saving it for a good occasion. Anyway, long story short, if you want a good series on that time, I can definitely recommend Corny Guilden. Another good British author, Bernard Cornwell, recommend his books before. I'll do it again. Now, of course, the Uthred series might be his most famous one, but he has also written about the Hundred Years War, similar to this time as well, so the late Middle Ages. Now, why do I mention all of these things? Now you wonder, yes, because Edward Longshanks is a bit before, so you see there in the description. It's 1264-1307. And also last thing before I begin to get into this the gameplay of our TV series recommendation. It's based on books by Philippa Gregory. It's called The White Princess and The White Queen. Quite good actually, available at HBO. Obviously not a service I recommend per se, but they had it there at least. And worthwhile watching. I can definitely recommend those two seasons. The only thing I need to strongly object to was a sequence where they visit the Spanish court. Now, of course, the Spanish court descended from Visigoths, and the princess they're there to make a match for their son to marry the Spanish princess. She erupts into a flamenco dance. And I can assure you that no Spanish royalty would have erupted into a flamenco dance. It was something that detracted quite a bit from the quality of the series. It was yeah, not good in the least. I am a great friend and admirer of Spanish history and culture, so I need to defend Spain's good name and say that royalty did absolutely not dance flamenco dances. Anyway, that being said, the rest of the series is quite good. The third season is an absolute catastrophe, but I won't blackpill you by talking more about that. Anyway, this particular DLC, you have a new civilization, the Burgundians. I won't get into the grand dukes now, maybe for a coming video, who knows, depending on how popular this one is. Then you have the Hotavilles. I don't know how you pronounce it. To my great shame, I do not speak French, highly barbaric of me. I must redeem myself by praying extra hard in the Temple of Iron. I read German. That was my language in school. You could choose between a few different languages. The most common ones were French, Spanish and German. Anyway, this might be an interesting topic for a podcast talking about the Normans. As you see here, Robert de Hotaville left Normandy in 1047 with a small robber band. Allow me to translate it. He left Normandy with a heroic menerbund of glorious shield brothers, and they sought to mug others. That's the actual story. But in all seriousness, though, the most famous Norman, of course, William the Conqueror, conquered Mary Old England in 1066, but they were quite adventurous, these guys. Vikings imbued with a high level of thumos that made them conquer and explore. I will talk about it in another video, in a podcast, going through the history of the Normans, who can derive inspiration from them. So anyway, that might also be for a coming video if there is any interest in it. Let's get into Edward Longshanks, and you probably recognize this name. He's the villain in Brave Heart when you see this time period from the perspective of the Scots. He is the villain. You can also read about many good books about Robert the Bruce as well. You have a British author, Robin Young, female author, written great books on this time as well, from the perspective of Scots. I can definitely recommend it. Anyway, let's get into the campaign. So we have here Mary Old England, and we'll travel to Wales, Scotland, and to the Holy Land. So let's get into it. Vane Ambition. Boom. Worthy of my father's name. How much longer until these whispers become cries of rebellion? My heart quivers when I visit his tomb in the Abbey. I am wracked by fear. Claws at my soul like the talons of demons. Fear that I will prove them right. That I will disgrace this family's legacy as my grandfather did. Yet I must not give in. I will study my father's reign and hope that it will imbue me with his strength, the fortitude that I so lack. My father Edward spent his youth watching in disgust as his ineffectual father Henry drove England to the brink of ruin. When his subjects could bear his ruinous reign no more, a rebellion led by the Baron Simone de Monfort deposed Henry, leaving his kingdom in tatters. After the disastrous Battle of Lewis, the baronial forces captured and jailed my forever grandfather in Hereford. Some kings begin their rise to the throne from within the palace. Edward began his from the anguish of imprisonment. Yet even when the future seems bleak, fate occasionally extends its hand and offers allies from the most unlikely of places. All right, as always in Age of Empires, epic and atmospheric introductions. So now we're in the game. Edward Longshank's Must Revive and Gilbert de Clair's Must Revive again excuse my French if I mispronounce it. I hope I pronounced it correctly. My man, the extremely aesthetic Guyam, can tell me how my French pronunciation is. Anyway, let's get into it. Let us do so. Edward Longshanks and in no tear, this aesthetics these three golden English lions, Norman lions, absolutely aesthetic need to make some sort of legio gloria garment. On a similar note, do let me know also in the comment section who your favorite English king might be with the most interesting or inspiring or whatever. I always read the comments in Let's Play videos. We must mug these new males. Ha, there were quite many. I need to run around with them a bit so they can mug these guys. I might also edit this video a little bit to make it more exciting I suppose. These guys, they want to question the validity of my claim to the English throne is what Edward says here. The reason I am more responsive on Let's Play videos is that the overall amount of comments is much less so I can have the opportunity to respond to more of them. Otherwise, I usually feel quite bad if I only respond to some comments. Of course, on Gabbmine's Patreon subscribe star, I'm always very responsive. This might be the thumbnail for the video. This aesthetics, the red and then golden lions on it. Absolutely, absolutely, aesthetically pleasing. Ha, we have some other blasphemous enemies here. We need to avoid them if we can. I'm not sure we can actually. Aha, one saw us.  I'll say note on the music. It's very literally speaking music to my ears since I grew up with this soundtrack. I've said it before, I'll say it again. Both H-Vempires 2 and H-Methology are sacred to me. I am extremely thankful for growing up with them. Extremely thankful for the interest in history they supplied me with. And yeah, sacred games, I'm sure many of you feel the same way if you've grown up with it. Good times, good times. We need to fight these guys as well. I don't think I can lose this guy because then we lose the campaign. He needs to survive. He needs to survive the jive. A great channel which I'm sure you're all fully familiar with. If you're not, search for it and enlighten yourself about pagan history and other good stuff. Good stuff, good stuff. So now I will do a little edit here and I will build up and I will get right back when I have progressed further in the campaign. So be right back. All right, so I am back.  I built an additional town center there and one here and now I'm gathering some stone to build a castle since the forces are coming quite soon. So I'll place the castle there I suppose and I don't even know if we have enough time to build it. I suppose not. Oh well, I have created a little army that will withstand the onslaught, the baronial forces. So we'll see what is going on now. Let's say I need to gather some more wood. That is noted. So they didn't actually come to attack us. They expect us to go to them to to mug them. That is fine. I suppose that's what we'll have to do. Here is my favorite unique unit by the way. The English long woman. Highly joyous to create them. Anyway I will attack someone I suppose. The Welsh perhaps they need to be subjugated and we'll see what happens after that. We have reached the Welsh. They're not prepared to face our army or maybe they are actually and I need to make sure that Edward doesn't die here. That would be a bit silly if he does. Maybe I should actually have brought some knights or something.  Otherwise we might have a hard time taking down all of their buildings here. Another edit as you might have noticed. So I subjugated the Welsh. They are now on our side and now I am putting on some pressure on this kennel worth and they're being protected by baronial forces and Simon the Younger. So we need to kill this guy and hopefully get this city over to us. Now we are less than three minutes until the baronial forces attack us and we'll gonna see if we can actually conquer this city in that time. I suppose that will reduce the amount of enemies that will come at us. We'll see how that goes. I have a battering ram in addition to my army here of crossbowmen and longbowmen. So the gate is down. Can we make it in time? I don't really suppose we can. But we'll see. We can at least destroy part of their town. Maybe I shouldn't have left the home base unprotected in case they attack. But it's a gamble. We'll see how it turns out. We have a unit, a sergeant I haven't seen before. Looking quite aesthetic. Okay his knights are coming and we were just a bit too late to destroy this city. Anyway let's look over our defenses. We are at the population limit here so we can't really create more soldiers. But if we can we can train some knights and stuff. I don't really see where this laddington might be though. Maybe he comes with a surprise attack. I will then counter with our surprise flicks. Hopefully that will get him to abandon his hostility to watch us. We'll see. The rest will be compelled to fall before me and beg for mercy. Okay so that was it. We just had to subdue them. Let's continue. Like his Norman ancestors, Gilbert Declair was a shrewd opportunist. Initially a commander in the rebel army. He keenly sensed its divisions and saw that he could further his own ambitions by defecting to Edward. My father moved with alarming speed. Warchester and Glochester, Gilbert's family seat was seized without delay. After a string of further victories, he compelled the leaderless and battered rebel forces to capitulate. Known by many as long shanks, Edward was feared for his imposing stature and foul temper. While Henry was restored to the throne, all were well aware that Edward was now the true power behind it. Good stuff. Good stuff. I thought about ending the episode there, but we're actually going to have a quick look at A Man of God. I think this is the only reasonable course of action, so let's get into it. Ruthless and cunning, Edward did not always adhere to the chivalric codes that the European nobility held so dear. After massacring Simone the Younger's men in a stealthy night attack on Kenilworth, he used the captured banners to disguise his army before smashing Dimonfor at Esham. In an act of alarming brutality, the corpse of Dimonfor, an earl of England, was hacked into carrion and his head mounted on a pike. Edward had won the war and cowed the rebels into submission in just over a year, but the victory came at the cost of his honor. Nobility and clergy alike cringed at the thought of an impious butcher one day ascending to the throne of England to restore his ailing reputation and prove himself worthy of the crown. My father resolved to lead a crusade east to the Holy Land, where the situation grew increasingly dire. The voracious Mamluk Sultanate waged a fierce war of expansion, first against the Mongols and now against the beleaguered crusaders. Several fortresses quickly fell before their unstoppable onslaught. Their leader was Sultan Baibars, a man not unlike my father Edward. Originally a slave of Kumin origin, he rose to prominence through military service and had recently taken power after murdering the reigning Sultan. Having just slaughtered the population of Antioch, his armies now threatened Oka and Tripoli from the east, converging on both cities in a suffocating siege. Edward's army arrived in Oka to find the inhabitants desperate, but not beyond hope. All right, I mainly just wanted to look at that intro. We have a beautiful city here as well we need to defend but I will actually leave it there for this time and I will see you all on Telegram I suppose if you're not already following me there, also follow me on Gabb and Minds. So now thank you for watching, xoxo, boom!


Hi and welcome to the session. I'm Kanika and I'm going to help you to solve the following question. The question says find that 12th term of a geometric progression whose 8th term is 192 and the common ratio is 2. Now before solving this question we should know that geometric progression is of the form a a r a r square a r q and so on. The general term of this progression is given by d n is equal to a into r to the power n minus 1 where a is the first term of the progression r is the common ratio which is obtained by dividing any term by its preceding term or we can say that r is obtained by dividing nth term by n minus 1th term. Let's now begin with this illusion. Now in this question we have to find that 12th term of a GP whose 8th term and common ratio is given to us. Now let a be the first term with the common ratio of a GP. Now the 8th term that is t 8 is equal to a into r to the power 8 minus 1 and this is equal to a into r to the power 7. In the question we are given that 8th term is equal to 192 so we have t 8 is equal to a into r to the power 7 is equal to 192. Let's name this equation as equation number 1. Now the 12th term that is t 12 is equal to a into r to the power 12 minus 1 and this is equal to a into r to the power 11. Now a into r to the power 11 can be written as a into r to the power 7 into r to the power 4 from 1. We know that a into r to the power 7 is equal to 192 and in the question we are given that common ratio that is r is equal to 2. Now substitute the value of a into r to the power 7 and r in this expression. By substituting the values we get 192 into 2 to the power 4 and this is equal to 192 into 16 and this is equal to 3072. Hence the 12th term of a GP that is t 12 is equal to 3072. Hence our required answer is 3072. This completes the session by NJKL.


Welcome to the 24th lecture in the course Engineering Electromagnetics, we continue with our discussion on parallel plane wave guides which we started in the last lecture and the topics we consider today are going to be the transverse electric waves, the field distribution associated with this kind of waves and finally we see how the field of the TE waves can be expressed as a superposition of plane waves. Before we take up these topics let us have a brief and quick recapitulation of the material we discussed last time. We considered a parallel plane wave guide of this kind which consists of conducting planes which are infinite along the z and the y directions. Therefore they impose boundary conditions such that the tangential electric field at these planes is 0 which is the same thing mathematically as saying that the normal component of the magnetic field is 0 at these conducting planes. With this basic configuration and the boundary conditions identified we wrote the Maxwell's equations for the region between the perfectly conducting planes and for the sake of simplicity we assume that this medium is a perfect conductor and we wrote the components of these Maxwell's equations in Cartesian coordinates with these associated wave equations. These equations were simplified by making the following considerations. We assume that the wave propagates in the z direction with the propagation constant gamma bar so that the derivative in the z direction can be replaced by minus gamma bar and since the structure is infinite along the y direction no particular conditions are imposed in that direction. We considered that the derivative with respect to y of all the field components can be replaced by 0. So with these considerations we obtained the simplified versions of the Cartesian components of the Maxwell's equations. These 6 equations and the associated wave equations for the magnetic field and for the electric field. Then we expressed the transverse field components H x and H y and E x and E y in terms of the longitudinal field components E z and H z. Once we reach this stage it is possible to bring in a further order into this major equations. We say that in general either H z or E z must be non-zero for all field components to be so that it does not remain a trivial situation. We will see an important exception to this later on but right now we say that in general one of these two longitudinal field components E z or H z must be non-zero. So depending on which one is non-zero we get a transverse electric field kind of solution which is referred to as either a T e wave or an H wave that is when H z is non-zero and we get T m waves or alternatively E waves when E z is non-zero. From this point onwards we started considering the solution of these equations subject to the boundary conditions identified earlier for the T e type of waves. For the T e waves we were able to see quite simply that the field components which would exist would be H z H x and E y. The other three field components become zero. E z is as it is zero by virtue of the very assumption that we are talking about the T e wave solution and the other two field components also become zero because of that. Therefore we wrote the y component of the wave equation which the electric field satisfies. Essentially what we are doing is we are picking up a field component which is tangential to the conducting plates in the parallel plane waveguide. So that we can in a simple manner apply the boundary conditions namely the tangential electric field is zero at the conductors. So we pick on E y and write the y component of the wave equation in this manner. Recalling that we introduced a symbol H where H squared was omega squared mu of silent plus gamma bar squared. This simplifies to this kind of second order partial differential equation. And then we consider the nature of E y. It has no variation with respect to y. It has this kind of variation along the z direction since it is propagating with the propagation constant gamma bar in the positive z direction and the only variation that is not specified yet is the variation with respect to the x direction. Recognizing this, this partial differential equation simplifies to this ordinary second order differential equation. Once we reach this stage it is quite easy to write the solution for the part of E y which varies with respect to x. That is E y not x is equal to C 1 sin of H x plus C 2 cosine of H x. Now one can look at the structure that we are dealing with once again. This is the structure alright. And we have got a mathematical solution for E y. And E y is tangential to these conducting planes. And therefore at x equal to 0 as also at x equal to a this tangential electric field component must be 0. That is how we have to choose the constants of integration and the parameter H. So, once we apply the condition that it should be 0 x equal to 0. We straight away make out that the constant that we have introduced C 2 must be equal to. Similarly it is also required to be 0 at x equal to a the upper conducting plane. And what would that imply? That would imply that H a should be equal to an integral multiple of pi. Where m may take on values 1, 2, 3 etcetera. As regards the choice m equal to 0 is concerned. We will see that this will lead to all field components becoming 0 a trivial situation. And therefore that is excluded from the possibilities of values that m can take. Which implies that the parameter H must be equal to m pi by a. That is it is related to the spacing between the parallel conducting planes. And also an index m has been introduced. The role of this index will be mentioned. We will have something to comment on this as we proceed. Therefore, now since we have got the x variation specified. We can now write the expression for the y component of the electric field. The complete expression E y is equal to C 1. And then sin of m pi by a x e to the power minus gamma bar z. So, except for this amplitude constant C 1. We have got now the complete solution for E y for the particular structure that we are dealing with. And the boundary conditions that this structure imposes. As far as this amplitude constant is concerned. It will depend upon how much power one has transferred to this particular field solution from the exciting source. And therefore we need not worry about it too much right now. This is as far as the y component of the electric field is concerned. What are the other field components that exist in this case? The other field components that exist are H z and H x. How are we going to get the expressions for these other field components? We go back to our expressions for the transverse field components in terms of the longitudinal field components. And we see that as far as H z and H x are concerned. They are simply related to E y. H z can be obtained from here. And H x can be obtained later on in terms of H z using this expression. So, using these expressions that have been derived earlier. We are now in a position to write down the complete field expressions. We are going to get H x equal to minus gamma bar upon j omega mu c 1 sin of m pi by A x e to the power minus gamma bar z. And H z turns out to be minus m pi by j omega mu A c 1 cosine of m pi by A x e to the power minus gamma bar z. Now since these field expressions are obtained in terms of the index m which governs the order of variation along the x direction. And there is no variation with respect to y direction because of the particular structure that we are dealing with. We can call the solution as the solution for T e m 0 modes. Here new term is being introduced that is the mode. A mode corresponds to a particular choice of m in these field expressions. And it is a very well defined particular kind of field variation with respect to the x and y, x, y and z directions that a particular mode would refer to. A wave may consist of more than one mode. That is the fine distinction between the mode and the wave. One can compare it to the general situation that we could have let us say an electrical signal which in general has any kind of time variation. And the specific types of electrical signals would be once which have specific types of time variation. For example the sinusoidal time variation. And in fact the analogy can be extended further just as an electrical signal with any general time variation can be expressed in terms of components which have sinusoidal time variation. Any general field for a particular wave can be decomposed into the fields corresponding to specific types of modes. So a time variation is now being replaced by a spatial variation. With that distinction modes serve the same purpose as let us say the sinusoidal time variation serves in the case of time variation. So it should be sufficient here to say that a mode refers to a very specific type of field configuration which of course is a solution of the problem at hand. And in a general situation the wave that propagates on the structure could be composed of more than one mode of one kind or perhaps more than one mode of different kinds. So that distinction between a wave and modes can be kept in mind. So depending on the choice of m we are going to get solutions for different types of modes and these will be referred to as let us say T e 1 0 mode, T e 2 0 mode and so on. The first index here the first subscript here refers to the order of variation along the x direction. And the second subscript here refers to the order of variation along the y direction. And you can see that we are choosing a terminology which is going to serve us even when we use waveguides which have confinement in both x and y directions. Now let us consider the parameter gamma bar. Gamma bar we call the propagation constant and it is equal to h squared minus omega squared mu epsilon. If you look up the expression for h squared it will be consistent with that and of course the square root of this entire expression. In general we say that gamma bar is complex and it has a real part say alpha bar and an imaginary part beta bar. Now the case that we have considered in that case also we can note down the fact that h has been obtained as m pi by a. So that on the right hand side both terms are real under this simplifying assumption that the medium that is filling the region between the conducting planes is a perfect dielectric. So that it has a real permittivity. So since both these terms are real gamma bar can either be completely real depending on their relative magnitudes or it can be completely imaginary. It cannot be complex in that sense. By virtue of the simplifying assumption that we have made we are considering a perfect conductor and we are considering a perfect dielectric. So losses are not appearing in this solution and therefore gamma bar can either be real or can only be completely imaginary. When gamma bar is completely imaginary then we will have gamma bar going to j beta bar and correspondingly e to the power minus gamma bar z. The factor that is taking care of the z variation in these field expressions is going to change to e to the power minus j beta bar times z where further beta bar is going to be equal to omega squared mu epsilon minus m pi by a whole squared whole squared root. And one can make out the fact that it is only above a certain frequency where beta bar will be real or alternatively gamma bar will be completely imaginary. Below that frequency the situation will be the other way round and gamma bar will be completely real. Which case is going to correspond to an attenuating situation as far as the positive z direction is concerned. Whereas this situation where gamma bar is completely imaginary is equal to j beta bar is going to correspond to a propagating wave with associated phase shift as it propagates and with no change in the amplitude. Ordinarily this is the range of frequencies where we will be interested in. This is the range above let us say a certain cutoff frequency where the wave would propagate without attenuation under the simplifying assumptions of no losses. And here an important distinction is appearing between the types of waves that are going to be supported on this parallel plane wave guide and uniform plane waves. There there was no such thing as a cutoff frequency. But here we see that depending on the value of frequency the behavior is going to completely change. And only above a certain frequency we will get propagation without any attenuation. This we will comment on a little later once again. So in this propagating frequency range and let us say for one particular mode say m equal to 1 we can write down the field expressions for the TE10 mode that is m is going to be taken to be 1. And in this situation when gamma bar is equal to j beta bar that is above the cutoff frequency that we referred to. We can write down the various field expressions that we have obtained for the TEM0 modes and they will be appearing as follows. We will have E y equal to C 1 sin of pi x by A. And then here we have e to the power minus j beta bar z. The other two field components can also be written. We have H x equal to substituting gamma bar equal to j beta bar in the earlier expression for H x. The expression that we get for H x is as follows. It is minus beta bar by omega mu and then C 1 sin of pi x by A. And then pi x by A e to the power minus j beta bar z. And finally H z expression turns out to be j m pi by omega mu A C 1 cosine of pi x by A e to the power minus j beta bar z. They all correspond to a particular mode which is propagating in the z direction with a certain phase shift constant. So the z variation is identical in all but their amplitudes can be different. Their variations with respect to x direction in this case and with respect to x and y directions in a general case can be different. But since they correspond to a particular mode their z variation must be identical. Now what we need to consider many times is the way the field looks like with reference to the waveguide the field configuration. How if you if you wanted to visualize the field it will look like. It is relatively simple for us to visualize how the y component of the electric field is going to look like. But the magnetic field components we have more than one magnetic field component and therefore they are going to combine in a manner which right now is not easy for us to predict. So for that purpose we consider the corresponding actual time varying expressions for these phasor expressions. Because after all the complex notation is only a mathematical convenience. There is nothing complex about the fields otherwise. So we get the actual time varying expressions corresponding to these field components. The time varying field components can be obtained by using our standard procedure. We multiply by e to the power j omega t and consider the real part. And when that is done we get e y which is going to be c 1 sin of pi x by a and then instead of the exponential factor we get cosine of omega t minus beta bar z. Similarly h x turns out to have an expression which is minus beta bar by omega mu c 1 sin of pi x by a and once again cosine of omega t minus beta bar z. And then we get the it is not difficult to see that h z is going to have an expression which is minus m pi by omega mu a c 1 cosine of pi x by a and then here we have sin of omega t minus beta bar z and then here we have sin of omega t minus beta bar z. And it clearly brings out something to demonstrate our statement earlier that there is nothing complex about the fields in reality. These factors j etcetera that appear they basically refer to a particular phase. Since h x and h z are in phase quadrature therefore they have a phase therefore they have a difference of a factor j. So this one has a cosine kind of variation with respect to time this one has a sin kind of variation if we take the same time reference alright. So with these time varying expressions available to us we can now consider how the fields look like with reference to the parallel plane guide that we have taken up for analysis. The parallel plane guide here for example on the left side is shown in the x y cross section. And therefore we have this as the x direction and this is the y direction x equal to 0 is this plane and x equal to a is this plane. So one can clearly visualize that at some z equal to constant plane we are considering the x y plane cross section of the parallel plane wave kind. Now how would one represent let us say the electric field in this. Electric field has only one field component E y therefore its representation is fairly straight forward. But even here one can choose two different types of representations or notations. For example one kind of notation will be where the magnitude is shown very clearly magnitude is varying only with respect to x. And it is varying in a sinusoidal manner with respect to x at x equal to a it is 0 and at x equal to 0 it is 0. And therefore round the plane x equal to a by 2 we can consider a sinusoidally spatially sinusoidally varying field. And we will have to mark this as E y because otherwise there is no direction included in this representation. And also since E y is a function of z also and so also for omega t one will have to specify the value of this argument omega t minus beta bar z equal to so and so at that particular time and at that particular value of z will be the variation of E y like this with an amplitude corresponding to C 1 and a multiplication by this factor. As we deviate from that value of z or from that value of time the sinusoidal variation nature will be maintained, but the amplitude is going to change. And therefore strictly speaking this representation should be a three dimensional representation with the amplitude E y coming out let us say out of this plane if it is of a positive sign and going into the plane if it is of a negative sign. So that kind of visualization we will have to carry out ourselves. Now this is a magnitude representation of the fields and we see that while it clearly shows the magnitude variation the direction has to be specified by put labeling this curve if it corresponds to E y we have to say that this is for E y. This restriction can be relaxed by going in for a slightly different type of representation by using directed line segments representation where we may have a magnitude and direction kind of representation. Here we draw directed line segments if it is E y that we want to represent that these line segments will be directed in the y direction and using the same axial notation that is this is y and this is x and x equal to 0 and x equal to a as noted in the figure above we may consider this kind of representation. Let us say like this then how do we represent the fact that as we move away from the middle midpoint x equal to a by 2 plane the field is decreasing we increase the spacing between these directed line segments and therefore let say one representation that is possible is like this directed line segments which are spaced closely together where the field strength is more and then the spacing is decreasing where the field strength is less. Obviously it does not very clearly or accurately show the magnitude but it does show the direction and it gives an idea about the manner in which the magnitude is varying. So one can choose either of these two notations to denote field components when we go to a field where more than one component exists then we will find that this notation is somewhat more it is more communicative of the actual overall field. So let us do that and we first use the magnitude representation for the magnetic field magnetic field components and we now since we have the x and z components in the magnetic field we choose the other cross section of the parallel plane guide such that it is the x z plane that we are talking about and of course these planes at x equal to 0 and at x equal to a remain as they were. And now we put down the values of these cross sections at different values of z. We could say that we are taking a snapshot of the field at a particular instant of time so that t is fixed and only z is varying. And the way z appears in this argument as we go to increasing values of z this argument should decrease. And therefore the value of this argument omega t minus beta bar z may be taken to be equal to let us say pi here and then pi by 2 here and then 0 and so on may be minus pi by 2 here. That is the value is a matter of what kind of scale we choose along this z direction. But let these be the planes omega t minus beta bar z equal to pi pi by 2 0 minus pi by 2. Why we have chosen this kind of planes will be clear in a minute. For example at omega t minus beta bar z equal to pi at z is completely 0. So at such a plane we have only h x. And as far as this argument is concerned it has a maximum value and the argument has the value minus 1. Maintaining the same notation that is the positive values are indicated by drawing occur to the right of the reference plane h x can be shown here in this manner h x has a sinusoidal variation. And these two negative signs cancel. So overall it is a positive sign and therefore we are drawing it to the right of this reference plane. And as I mentioned earlier actually it should be considered as something projecting out normal to the plane of the paper. We must clearly understand that this is the variation at omega t minus beta bar z equal to pi. Where we have only h x which exists. If we shift the value of this argument slightly from pi h z will also appear. And the value of this component itself is going to be altered. So these things we have to keep in mind. Similarly when we go on to the value of this argument equal to pi by 2. Then we find that only h z exists at this specific location. Sign omega t minus beta bar z equal to pi by 2. And h z has a negative sign in front. The variation is governed by the cosine function at x equal to a it is cosine of pi. So another negative sign is going to appear there. So keeping these things in mind the h z component is going to have a cosine sinusoidal kind of variation. And maybe it can be represented in this manner. Once we have considered these two curves. And we have to be very careful in going from the phasor notation to the time varying notation. And in transferring these expressions into this kind of curves. We have to be very consistent in our notation. For example positive sign must be represented by in a particular manner consistently. Here we are representing the positive values by drawing the curve to the right of the reference plane. You can choose some other notation also but you have to be consistent in following that notation. Once we have drawn these the curves and the field components that exist at other planes can be drawn just by inspection. For example here once again we will have only h x and it can be represented by a curve which is like this. And at this point we will have only h z which is going to have a variation which is like this. As for the case of the y component of the electric field this is the magnitude representation of the components of the magnetic field. Now actually it is just for the sake of convenience that we talk about these components separately. What exists in the waveguide as the T10 mode is a composite magnetic field with the components being like this at different points. And therefore we will like to combine these in a certain way and see what is the overall magnetic field looking like. And that can be done by going in for the other kind of representation which we call the representation in magnitude and direction. Now please look at this carefully. At this plane where this argument has the value pi we have only h x existing with maximum amplitude anywhere on the guide and it is x directed. So using this kind of notation we can represent this by directed line segments which are oriented along the x direction. And we do not take these up to the conducting planes because at the conducting planes h x also is 0. And as we move away from this reference plane on either side along the z direction the h x component value itself is going to reduce as we noticed earlier. And therefore as we move away from this we draw similarly directed line segments but shorter in length. And with an increased spacing to signify the fact that as we move away from this reference plane the h x value is going to reduce according to the cosine of the argument omega t minus beta bar z. These are all x directed in some manner they represent the magnitude they surely represent the direction very clearly. But whether it is positive x direction or the negative x direction now following our convention since positive sign was represented to the right of this reference plane we can put an arrow sign which is pointing towards the positive x direction. The same procedure can be repeated at this reference plane where this argument has a value 0. The situation is going to be identical with the reversal in the direction. And therefore we can draw closely spaced minus x directed line segments here and with somewhat increased spacing and reduced length lines segments here. These must not extend up to this reference plane because there h x is 0 and only h z has the maximum value. The same representation is now extended to the z component of the magnetic field. And we see that the z component of the magnetic field is maximum at this reference plane. And following our notation of sign consistently we can draw this kind of line segment here and here. And noting the sign the arrow directions must be like this positive z here, negative z here. As we move away from these conducting planes h z is reducing in value. And therefore at a slightly greater distance from these conducting planes we may draw h z in this manner. Once again the directions are going to be like this which representation can be transferred to corresponding planes with a reversal in the sign. And therefore we may draw h z like this. What will be the arrow directions here? It will be negative z here and positive z here and in a similar manner we will have the z component of the magnetic field represented by this kind of arrows properly identified like this. This is in a way this skeleton showing the important features of the overall magnetic field. And now one can very simply visualize that actually the field is of this form giving us the magnetic field in closed loops in this manner. And if we have proceeded correctly using the notation consistently the arrow directions will not clash otherwise they will clash which obviously is going to be an error. And proceeding consistently and correctly we will have the same arrow direction on a particular magnetic field line and this is the representation in magnitude and direction. Depending on the purpose and the patients one can fill it up appropriately taking care that the spacing between the lines or this the number of lines per unit width in a certain direction represents the magnitude of the field line at different points. If you have any clarifications to ask you are welcome. So this is where we like to stop today. Today we have considered the field components for the transverse electric waves and we have introduced the notion of modes. And therefore we have got the expressions for TE M0 modes. We considered in detail the field expressions for the TE 10 mode. We have considered its time variation and we have seen how these fields can be represented to indicate what kind of field lines are there associated with these fields. Thank you.