 So now let us move on to 1D conduction. So this is what we are doing is as we said as professor said we are going to do 1D conduction. So just to take recap of what we had derived the heat diffusion equation that is del square T by del x square plus del square T by del y square plus del square T by del z square plus q dot by k q dot being the volumetric heat generation is equal to rho C p by k del T by del t. So professor had told few minutes back around 215. So if it is 1D let us say temperature is only a function of x. So this term would be 0, this term would be 0 and let us say there is no volumetric heat generation and I am taking steady state then this term is also 0. So then I am having only d square T by dx square equal to 0. So he also showed that if I have two boundary conditions we can get the solution. So that is what precisely what we are going to do in the next couple of minutes. So this is if you take the steady state condition and if I take the plane wall, how does the plane wall look like? So the plane wall looks like here as shown here. So you have x equal to 0, x equal to L. So if I draw the same thing, if I draw the same thing this is my plane wall and I have T equal to T s 1 and T equal to T s 2 here at x equal to L that is this is x equal to 0 and this is x equal to L that is this distance is L. So now you may ask me a question how does one generate a constant wall temperature boundary condition? So yes one can easily in fact the most easily a generatable boundary condition in the lab is constant wall temperature that is just by making the hot water flow, hot water or cold water flow for long time. So it will reach steady state. So that is how in fact this constant wall temperature boundary condition T s 1 is being generated, this constant wall temperature boundary condition is being generated by this convective boundary condition that is h 1, T infinity 1. So you have T infinity 1 here and T infinity 2 flowing here. So that is what if I have this is h 1, T infinity 1 and this is h 2, T infinity 2. So for us the boundary condition is going to be T s 1 and this is boundary condition T s 2. So we need to generate any equation for this. What is the governing equation? Just now we found d square T by dx square equal to 0. So what are the boundary conditions I have x equal to 0, T equal to T s 1 and x equal to L I have T equal to T s 2. So now if I put this back if I put this back with these boundary conditions. So as we had seen if I differentiate this I get T of x equal to c 1 x plus c 2 and I have x equal to 0 T s 1 and x equal to L T s 2. So if I substitute this for x equal to 0 I get c 2 equal to T s 1 and c 1 equal to T s 2 minus T s 1 by L. So I get the equation as I get the equation as T of x equal to T of x equal to T s 1 plus T s 2 minus T s 1 into x by L. So basically as you can see here this is like y equal to m x plus c m is my this slope and of course you can see that it is negative actually T s 1 minus T s 2 by L it is negative slope and this is my constant. So that is precisely what we have done here. So you have q x equal to this is the temperature distribution now taking this temperature distribution I can find the heat loss or the conduction heat transfer rate that is q x equal to minus k A d T by d x, k A d T by d x is T s 1 minus T s 2 by L because I am differentiating this and minus sign I have absorbed here. So that is why T s 2 minus T s 1 has become T s 1 minus T s 2. So heat loss rate is equal to k A into T s 1 minus T s 2. So this is what I think we are all very convergent with this so I do not want to spend too much of time k k by yeah this is no k L by A k by L it is yeah k by L it is it is in this equation 2.26 please note that when I divide this by A I get k by L into T s 1 minus T s 2. So now we always like to take analogies so here we take thermal resistance analogy so but then before we do this let us write down see we have studied fluid mechanics we have studied electricity we are studying now heat transfer what is common in all of this is the question. So we all know that when there is a pressure drop when there is a pressure drop there is flow rate of the fluid this all of us can feel very easily without any difficulty. So similarly when there is a voltage drop then only current can flow this also very easily understandable. Similarly for heat transfer whenever there is a temperature difference there would be heat flux that is the delta T is the one which is causing the heat transfer for the heat transfer to take place there has to be temperature gradient like when there is pressure drop only there will be flow rate when there is pressure drop equal to 0 then the flow rate would not be there at all then the flow would not the fluid would not flow similarly when there is no electrical electric potential difference there would not be current. So it is best the best way to understand the heat flux and the temperature difference is through this analogy I always think through flow rate and pressure drop because this is the most this is the closest one to my mind so this is how I think so this is way this is how people also have done and represent this in electrical analogy that is what we are going to do now. So here let me skip this and move on to the next one so I have here TS1 TS2 the temperature difference is TS1 minus TS2 so this is Q double dash sorry Q heat transfer rate equal to L by KA whatever I have done I have put this this is nothing but my Fourier law of conduction. So what is L by KA? L by KA is my resistance that is Q equal to KA T1 minus T2 by T1 minus L. So if I put electrical resistance this is T1 this is T2 and if I take Q as my flow rate current so the resistance would be L by KA right so yeah so that is this is equal to T1 minus T2 upon L by KA this is my current this is my voltage difference delta V and this is my resistance so that is how we are visualizing this electrical resistance analogy. So that is what I have written here so similarly you can say L by KA instead of L by KA you have L by sigma A where sigma is the electrical conductivity. So with this analogy we can apply for various situations let us say I have convective boundary condition. So if I have convective boundary condition by Newton's law of cooling from by Newton's law of cooling I know that Q equal to HA into TS minus T infinity so TS minus T infinity upon Q that is voltage difference upon current should be equal to resistance that is equal to 1 upon HA so that means for convection for convection my resistance would be 1 upon HA. So if heat transfer coefficient is high convective resistance is going to be less that is what we try to do in case of heat exchangers it is too early to reach there but still this is the significance of the convective resistance. So now likewise we can combine all of this I just put here simple Fourier's law of conduction in a solid in terms of electrical resistance and this was convection in terms of electrical resistance and then I am combining all three that is conduction, convection and convection, conduction and convection. So if I put that so this is my solid and this is my TS 1 temperature this is TS 2 and this is let us say H 1 comma T infinity 1 and this is let us say H 2 T infinity 2 so I have TS 1 minus TS 2 upon this is my length L and this is the solid whose thermal conductivity is K and this is area is A so this is L upon KA this should be equal to T infinity 1 minus TS 1 upon 1 upon HA this also should be equal to TS 2 minus T infinity 2 assuming that this is the hot side and this is the cold side TS 2 minus T infinity 2 upon this is H 1 upon 1 upon H 2 A. So this is what I am doing here this is what I have done here so this is putting electrical resistance. So if I write in terms of T infinity 1 minus T infinity 2 then I have to combine all these resistances that is heat loss or the heat transfer is equal to T infinity 1 minus T infinity 2 upon 1 upon H 1 A plus L by KA plus 1 upon H 2 A that is what I am doing in the next slide. So you have T infinity total resistance is this so this is what we are all very convergent with this so I do not I do not want to spend too much of time on this and we have three materials here material A material B material C then all these resistances are in series I can go on combining them for different materials. So next that is you have convective resistance conductive resistance LA by KAA conductive resistance LB by KBA and LC by KCA and 1 upon H 1 A so this is what I am doing what is done for two different materials but these are in series. Now let us take up in a case in which you have materials in parallel materials in parallel that is here I have a material E and I have F and G parallel but here again this H is in series. So accordingly my resistance circuit would become so I would have T 1 and this is the interface T 2 so this interfacial temperature is assumed to be same so T 1 I have LE by KEA and this resistance would be LF by KF please note here I am taking A by 2 because the cross sectional area through which the heat transfer is taking place is only half here so that is why I am taking A by 2 and here again I have LG that is this length LG upon KG into A by 2 of course here LF equal to LG and of course this all this has to be in series with LH by KA so this is how I can rewrite the same thing for the first half and the second half and these two circuits essentially are same they are not any different from each other. So these are all same so here there is only one thing that is this is also called as overall heat transfer coefficient I think it is little too early for Newton's law of cooling I am skipping this overall heat transfer coefficient thing we will reintroduce this may be in heat exchangers and in convection for now let us not worry too much about this UA so this is U means it is sort of heat transfer coefficient let us keep that in mind. So now the next important thing is that see in the previous problem we just took that in this problem we just took that at the interface there is perfect contact that is there is no gap between material A and material B but in the real life it is not going to be so there is going to be a gap so and that gap is generally filled with air or water as one of your one of the participant had asked there is an interface there so if there is an interface this is the perfect interface that is T1 is equal to T2 that is this is the T1 at the layer 1 end at the layer 2 beginning it is T2 but if there is no interface T1 is equal to T2 that is this can be imagined that the layer 1 and layer 2 are held together by applying maximum pressure but now let us say we are not applied any pressure and they are just held by hand if you hold by hand there is going to be an interface like this that means by naked eye we will not be able to see this interface but if you put that under the microscope definitely you can see those voids or the air gaps when you see those air gaps there is definitely the thermal conductivity is there in that void and the temperature is going to be different that is there is a gradient that is there is a heat transfer taking place within that interface so that is what is the thermal conductivity that interface is essentially because of the thermal conductivity of the air and we know that the thermal conductivity of air is very less as I said in the morning it is 0.02635 as opposed to thermal conductivity of the solid which runs into tens or hundreds so convective that is the thermal contact resistance is going to be T1 minus T2 upon qx double dash and that is the interface resistance this interface resistance is going to be very important if I have to quote examples for this interface resistance for example all of us have seen on a mother board that is a copper plate copper thing which is there to cool the mother board in the computer that copper piece is glued on to the mother board with a glue so the interface is created by that glue usually that glue is called as thermal paste the thermal paste is chosen such a way that the thermal paste thermal conductivity is reasonably high however high it is it cannot match that of the solid material that is the copper and the silicon which is the mother board material generally made of so the interface is a serious issue and in those in those locations the interfacial resistance has to be quantified the only way to quantify is to measure them you cannot model them you have to measure why because the porosity is the way the interface is there we cannot model that easy so with this intro to interfacial resistance hello we are just taking few questions for next 10 minutes this is NIT th richinopoli any questions hello hello one is this NIT I understand that normally going from the normal thing that there are only two modes of heat transfer convection also is a form of conduction only that is what many people talked about that was one thing which came in the morning the second thing is you are talking about the moving boundary that question also came in that in you said one side is convection one side is conduction the latent heat also will come into picture I think when it is using latent heat of fusion and when it is melting vaporization also will come into picture right yes so how do you solve such problems very complicated how do you solve such problems yeah actually very complicated issues because it is complicated in fact in fact we cannot handle this kind of problems with 1D conduction in fact incidentally one of my PhD student is working on melting and solidification problems here there we cannot solve using 1D conduction or just by taking latent heat we have to write whole set of equations Navier-Stokes equation and the energy equation and then put the appropriate boundary conditions and solve it so it is not going to be straightforward the way we had discussed in the morning that was just a simplified presentation but we cannot solve it thoroughly see for example I said that there is natural convection in the liquid region how can I capture natural convection until I solve the full momentum equation all the momentum equations that is the Navier-Stokes equations and the energy equations and people generally what they do is that for this type of problem they assume in fact the solid liquid interface is also not going to be clear solid and clear liquid what they call in between these two is what is called as mushi zone so in this mushi zone they handle it in it is almost like porous medium so they take mushi constants and using those mushi constants they solve these equations so that so it to answer your question it is not going to be a straightforward simple 1D conduction either using Fourier's law or something we have to solve whole lot of Navier-Stokes and energy equation with proper assumptions by taking porous medium so that we can take the mushi zone into account over to you what about the first question so that the only two modes of heat transfer basically you see this is conduction and radiation that many people are still talking on that here see this is just a matter of opinion although we say science is objective although we say science is objective science is not objective science is also subjective I would still go by three modes of heat transfer that is conduction convection radiation although we say convection is conduction within the boundary layer but I need to handle fluid dynamics so let us keep it little different from conduction I know it is conduction within the boundary layer but that is boundary layer has to be identified so that is why if you see almost all books almost all books which we refer they say that there are three modes of heat transfer conduction convection and radiation if one wants to nitpick and here split and argue yes there are only two modes of heat transfer over to you thank you sir thank you we can take one question impurities added will essentially decrease conductivity of the material as alloys and adding impurities of the better conducting material molecules will also decrease the conductivity of alloy yeah see I have no answer but this much I know that if I take two best conductors if I if I take copper and nickel copper thermal conductivity is very high nickel thermal conductivity conductivity is very less the alloys thermal conductivity will be lower than the both the both nickel and copper but if you ask me why I do not know okay I do not know the answer but I definitely know as you said the thermal conductivity of two pure metals are higher but when I join them or combine them the alloys thermal conductivity will be lower than the lowest among the between the two we will think about this question and come back we will put this on hold again over to you okay sir I will put this on the model please try to answer because I have not yet found the answer for this definitely okay so we will get started with cylinder cylindrical coordinates we are still in one-dimensional heat transfer and we have a cylinder that is we have r theta z coordinates so in the r theta z coordinates what we are saying is that the temperature is varying only in the r direction but not varying in theta and z direction so if that is the case of course we are not derived this and I would request all the participants overnight today go back home and derive let me write this I want all of you to derive the heat diffusion equation derive the heat diffusion equation in cylindrical coordinates and spherical coordinates let us come back so when you derive the heat diffusion equation in the cylindrical coordinates you get 1 by r d by dr k r dt by dr equal to 0 so the heat transfer rate equal to minus k a dt by dr equal to minus k a is 2 pi r l that is the surface area so you have 2 pi r l into dt by dr dt by dr yeah so professor Arun wants me to write the cylindrical coordinates full heat diffusion equation he wants me to write I definitely do not remember that equation so I am going to write it for the benefit of everyone many many questions are asked of the time of the kind please derive this equation in an exam it is very nice to ask such questions but the problem is it is so mathematically intensive and students often memorize these things for lack of any other way of doing it so these are not questions which I think we should be focusing our attention in actual examination you can give the equation and ask them to simplify or apply so but please I think for the full derivation in cylindrical or spherical coordinates is not exactly the best okay so here if we see that this is the heat conduction equation one dimension heat conduction equation and this is the heat transfer rate so as you can see one important point what we need to note here is that the heat transfer the heat flux earlier in a plane wall heat flux is independent of x but here it is not so this is very important point the conduction heat transfer rate that is q r is not independent of r is not independent of r but k r dt by dr is independent but not k dt dr so we have to be very careful about this heat flux in case of cylindrical coordinates okay so what is that we know what we did for the plane wall similarly we should do for cylindrical coordinates that is we have to integrate this if any integrate this I will be getting the equation t of r equal to c 1 log of r plus c 2 so I have two boundary conditions that is at r equal to r 1 t equal to ts 1 and at r equal to r 2 I have t equal to ts 2 so if I substitute these boundary conditions I will have ts 1 equal to c 1 log r 1 plus c 2 ts 2 equal to c 1 log r 2 plus c 2 so I have t of r equal to ts 1 minus ts 2 upon log r 1 by r 2 log of r by r 2 plus ts 2 so the point to be noted here is that the temperature distribution is no longer linear unlike plane wall it is logarithmic it is not linear if you see the heat transfer rate so we have 2 pi l k into ts 1 minus ts 2 upon log of r 2 by r 1 so the resistance would be the resistance would be q r equal to ts 1 minus ts 2 upon log of r 2 by r 1 upon 2 pi l k so let me get back and check whether I have written no yeah right so this is this is my resistance this is my resistance this is my conductive resistance which is for plane wall which it was l by k a so it was l by k a so now coming back so this is the heat transfer rate so now we have total heat transfer rate similarly with the way we did if we get back to our problem we have convective boundary conditions on the outer wall and convective boundary conditions on the inner wall this is a pipe actually what I have taken is a pipe which is having inner radius r 1 and the outer radius r 2 so we have h 1 comma t infinity 1 on the outer wall and h 2 comma t infinity 2 on the inner wall as you can see this temperature distribution is logarithmic but not linear so if I combine the same way as I did for the plane wall I have t infinity 1 minus t infinity 2 this is you can see this is the convective resistance this is also convective resistance these are the multiple components if I have multiple pipes of different materials of a b and c so this is same as extension of plane wall let me reemphasize reemphasize that here it is k r d t by d r which is independent of r not the conduction heat transfer rate so this is to be very very very carefully noted we should not be confusing this with the plane wall so this is what I had shown you this is the three materials I have a b and c so you have h 1 comma t infinity 1 h 2 h 4 comma t infinity 4 so these are various conductive resistances these are the convective resistances you see here one thing you can notice that these are logarithmic temperature dips which are there is a temperature gradient but at the wall there is a sudden jump this is essentially this temperature gradient is within the thermal boundary layer you can see that here also and you can see that here also this is the general question which students ask us why is this temperature gradient is different is it linear no it is not linear it is that is the temperature gradient within the thermal boundary layer it depends on how I am making the flow take place so definitely it is not linear that is the point we need to note there so that is about the composite cylindrical wall and next is sphere I know I am going very much fast because this is 1 d conduction so generally I am told in the coordinators workshop that these are the things which usually teachers spend more time so they wanted us to spend more time on heat generation and fins and transient conduction so that is why I am going little faster here in case of cylinder and sphere which is an extension of plane wall but the concepts are same so if I take a sphere again 1 d conduction if I take steady state 1 d conduction and conduction being only in the radial so I have what is the conduction equation I have 1 by 1 by r square del by del r of what is that k r square d t d r k r square d t by d r this is total equal to 0 okay so this is equal to 0 so we have got the equation for sphere let me write the full equation for sphere so that you know how this equation came of course we are not deriving this I want all of you to write along with me so that we do not miss out things 1 upon r square del by del r of k r square del t by del r plus 1 by r square sin square theta del by del phi k del t by del phi for spherical coordinates it is going to be r theta phi so you have plus 1 upon r square sin theta del by del theta k sin theta del t by del theta plus q dot generated is equal to rho c p del t del t so what we are saying is that we are saying that the variation of temperature in the theta direction and variation of temperature in the sorry in the phi direction and in the variation of temperature in the theta direction are not there and there is no energy generation term and we are talking about steady state so the heat transfer or the heat diffusion equation is going to be essentially containing only this term that is what is written here so if I get back so this is the equation which I wrote so and the heat flux or the heat transfer rate equal to minus k a dt dr a here is 4 pi r square dt upon dr so now we have again I have a second order differential equation second order differential equation I need two boundary conditions I need to solve them so if I integrate this equation which I am not doing step by step but that is fine I would like you to go back and derive that so you get t of r equal to c 1 by r plus c 2 so if you substitute the boundary conditions at r equal to r 1 ts 1 and r equal to r 2 ts 2 so you are going to get two equations if you solve these two equations this is my temperature distribution definitely here again it is not linear so this is my temperature distribution and if I take this temperature distribution and substitute that in this heat transfer rate that is minus k a dt dr and find out dt dr using this equation so I get 4 pi k ts 1 minus ts 2 upon 1 upon r 1 minus 1 upon r 2 so this 1 upon r 1 minus 1 upon r 2 upon 4 pi k is the conductive resistance in spherical coordinates this is the this is the resistance in the spherical coordinates so before we go to this problem I just want to take a recap we will definitely take up that problem I just want to take a recap of everything because 1D conduction so this is a very good transparency because everything is there here so if you take the heat diffusion equation 1D see we are saying that we are handling one dimensional steady state no heat generation so these are the governing equations d square t by dx square for plane wall this is for cylindrical wall this is for spherical wall and if I take constant wall temperature remember we have not solved these problems for any other boundary condition we are solving only for constant wall temperature boundary conditions so for constant wall temperature boundary conditions these are the temperature distributions this is linear and these two are not linear this is the heat flux you see here this heat flux is independent of x but these two heat fluxes are not independent of r this is the important point to note and this is the heat transfer rate in case of plane wall cylindrical wall spherical wall and of course these are the resistances as usual conductivity will be sitting in the denominator that means larger the conductivity smaller the resistance that answers why in all heat exchangers the heat exchanger pipes are made of copper pipes because the thermal conductivity of copper pipe is very high so the conductive resistance between the two fluids is very less anyway we will touch upon this again when we study heat exchangers and let us come back to this problem spherical there is a spherical thin walled container which is used to store liquid nitrogen at 80 Kelvin note this temperature this is much lower than the ambient temperature it is 80 Kelvin it is 80 Kelvin and the container has a diameter of 0.5 meters and is covered with an evacuated reflective insulation composed of silica powder see usually liquid nitrogen is very important and because this is what is used in most of the sensors for cooling so liquid nitrogen has to be carried in a container from one lab to another lab that happens even in IIT most of the institutions liquid nitrogen plant will be there and liquid nitrogen will be carried so that liquid nitrogen container is this and this liquid nitrogen container because you see why is this insulation important here because there is temperature of the liquid nitrogen which is sitting in the container is at 80 Kelvin which is much much lower than the ambient temperature which is at 300 Kelvin let us say so here the ambient ambient temperature is given to be 310 Kelvin not 300 it is 310 the container has a diameter of 0.5 meter and is covered with an evacuated reflective insulation composed of silica powder the insulation is 25 centimeters thick sorry 25 mm thick it is just 2.5 centimeters not big and its outer surface is exposed to ambient at which is at 310 Kelvin so at the ambient means one would expect natural convection so one would expect to have natural convective heat transfer coefficient and that heat transfer coefficient should not be very high it should have the order of tens that is what is being said in the next sentence that is the convective heat transfer coefficient is around 20 watts per meter squared Kelvin and the latent heat of evaporation and the density of the liquid nitrogen is 2 into 10 to the power of 5 joule per kg and 804 kg per meter and the thermal conductivity of silica powder you see the thermal conductivity as I said in the morning please note these numbers I keep saying this very many times numbers are very important for engineers we need to know the order of magnitudes of each numbers I keep telling this in the morning I told thermal conductivity of copper is of the order of 400 thermal conductivity of water is of the order of 0.6 thermal conductivity of air is of the order of 0.026 now you see thermal conductivity of silica powder is 0.0017 you see it is 2 orders of magnitudes lower than the thermal conductivity of air itself such a nice insulating material but in spite of that there will be leakage that is what we are going to see now so question asked is what is the rate of heat transfer to the liquid nitrogen that is how much is the so now where is the mode which is the mode of the heat transfer that is by convection and conduction the heat is going to transfer from ambient that is the atmosphere to the liquid nitrogen because ambient is sitting at 310 and the liquid nitrogen is sitting at 80 Kelvin now I have what is the rate of heat transfer to the liquid nitrogen and what is the rate of liquid boil off this that is to calculate the this liquid boil off only latent heat is required till that time do not worry about latent heat so let us see how do we go about this problem this figure is self explanatory I have a cylinder and ambient is given sorry sphere it is not a cylinder sorry this is a sphere the h is 20 watts per meter square Kelvin T infinity is 310 and this is the insulating material whose thermal conductivity is 0.0017. Now what are the what are the known parameters I have listed what is that we have to find also we have listed what are the assumptions definitely it has to be steady state we have not handled so far unsteady and it has to be one dimensional in which direction of course radial and there is negligible resistance to the heat transfer to the container wall and the container you see here I am neglecting the interfacial resistance between the container wall and the insulating medium no matter how well I wrap the insulating material there is going to be interfacial resistance however I am neglecting that in this problem for the sake of simplicity this assumption actually this assumption that is negligible container wall and the container inside the nitrogen that is there is no natural convection taking place inside the container no natural convection and of course the interfacial resistance is also not there that is not stated here and properties are assumed constant how does it affect if you see here you see the thermal conductivity of silica powder is given to be at 300 Kelvin but actually the insulating material temperature will be quite different it will not going to be at 300 only but still I am going to assume that the properties are going to be constant that is the thermal conductivity is going to be constant and radiation we are excluding radiation we are excluding so now with this so if I if I write the thermal circuit if I draw the thermal circuit I have T infinity 1 which is which is at 80 Kelvin that is inside liquid nitrogen which is sitting at T infinity 1 at 80 Kelvin and T infinity 2 which is at 310 Kelvin so I have two resistances here I have two resistances one is the conductive resistance of the container and the convective resistance because of the ambient the conductive resistance we have just derived that is 1 upon r1 minus 1 upon r2 upon 4 pi k so r1 and r2 if I substitute that is 1 upon let us keep it let us keep it like that and substitute at the end the conductive resistance is 1 upon r1 minus 1 upon r2 upon 4 pi k and convective resistance is 1 upon h a a being 4 pi r2 square so I have T infinity 2 and T infinity 1 that is 310 and 80 and these are the two resistances which are in series this is what these two conductive and convective resistances are in series so the heat transfer rate or the rate of the heat transfer is just by plugging in all the known parameters if I just plug in all the known parameters it turns out that the heat transfer rate is 13.47 watts now that is only first part of the problem rate of heat transfer to the nitrogen from the ambient to the liquid nitrogen now the question is at what rate the nitrogen is boiling off that is if there were to be a small hole here if there were to be a small hole and definitely with this heat input it is going to leak out how what is the rate at which it is going to leak out is the question so how will I do that I will have to do the energy balance e.in minus e.out plus e.g equal to e.st here e.g and e.st are not there because there is no energy generation and e.st is not there because this is a steady state problem e.in and e.out what is e.in e.in is the heat which is coming from the ambient to the liquid nitrogen and what is e.out which is leaking out because of the latent heat that is m hfg so q equal to m hfg if I do so I have m dot equal to q upon hfg q we had just found 13.47 and latent heat in the problem is given to be 2 into 10 to the power of 5 joule per kg so if I plug in that 13.47 upon 2 into 10 to the power of 5 I get 2 6.74 into 10 to the power of minus 5 kg per second it sounds innocuous that means it sounds harmless this leakage we can live with but if you see if you convert that into volumetric basis so mass loss per day if I compute that for per day you see here if I multiply this kg per second and for one day how many seconds 3600 seconds for 1 hour and 24 hours in a day it turns out that 5.82 kilos of liquid nitrogen I am going to use which is substantial so although 13.47 that was sounding very minimal the leakage loss is very high so this problem is only suggesting that the insulation is so much important and there needs to be no leakage so that is the point which is going to be which is to be understood through this problem. So that is about cylindrical and spherical coordinates now let us move on to what is called as critical radius of insulation that is liquid nitrogen problem there are only two resistances given so just keep that in mind because actually you would have to deal with four resistances if you are going to take the conduction through the wall and any kind of convection through in the liquid nitrogen we are making a big assumption that temperature of the liquid nitrogen which is 80 Kelvin is equal to the temperature at the surface of inner surface of the insulating material that is why we are able to draw the circuit diagram with only two resistances as opposed to four that would be there if I took the conduction resistance through the wall and any convective resistance in the fluid. Professor Arun will take critical radius of insulation for next may be 15 minutes and after that we will take enough questions till the end of the session.