 these are the figures which we have discussed yesterday and this is the equation, okay. So if you have ideal packed bed reactor meaning it is a following plug flow then these are the two equations material and energy balances what we have to use to design the reactor. So what I have told yesterday was that if it is an adiabatic system then you draw the straight lines using this equation 8 and then we know how to get the number of stages if it is required otherwise a single stage simply drawing the line like this if it is exothermic or like this if it is endothermic and take corresponding x a and these are the rate curves, okay. This is the rate curves, yeah. So the intersection between x and then rate curve you will take and x verses 1 by minus r a take area under the curve and you will get the weight of the catalyst or volume of the reactor, okay. This procedure is very very simple straight forward one. So still if you have adiabatic system and then theoretically real optimization if you want to do that means you would like to get the minimum reactor volume or minimum weight of the catalyst for a given conversion so you have to be there theoretically on this line. That means it is isothermal condition which will give you the maximum rate and you will have the minimum reactor volume. So how do you operate that if you want to be on isothermal condition? So because most of the time we have been telling that you know isothermal isothermal conditions but any reaction will be either exothermic or little bit exothermic or endothermic unless otherwise you dilute the reactant to such a level where delta h r is almost zero. Then you will have isothermal that no one does. Why? Because you have so much waste material, inert material where the reaction is not allowed, the inert material is not allowing the reaction then you have to separate all these inert material from the products yeah. So then it is again very very costly for us. So that is why no one will do. The other possibility also is for packed bed reactors take inert catalyst and mix with active catalyst and then again conduct the reaction. There again you are deliberately limiting the rate of reaction. So rate of reaction limiting means delta h r you know the amount heat released will be very very minimal. Again you will have isothermal condition but you can imagine that if you have one gram of catalyst so probably you have to use hundred grams of inert material. So that means almost hundred times the volume will be larger when compared to the only active material correct no. So that is these are the other methods but if it is absolutely necessary people will go for that otherwise no. So that is how you make system isothermal otherwise you have to go for multi stage theoretically speaking we have to be here that mean it is an exothermic let us say it is an exothermic reaction easy to understand and when I have exothermic reaction adiabatic case no heat removal there. So adiabatic case and we start somewhere here and it is adiabatic reactor exothermic. So you will move that means I am just drawing this line okay you will move in this direction because this adiabatic line as the conversion as the temperature increases conversion increases. So then you cannot cross beyond this why it may be dangerous for the reaction for the reactor and also for the catalyst all these. So then you have to cool again again heat again cool again heat so like that you have to be almost be touching here you know this border you should not cross but always you are on this line but just imagine this is only three stages if I want to be completely on this line how many stages are required infinite. So that means I will start here go like this go like this again go like that you know you theoretically you will be on this line which is not possible that means so many heat exchangers so many sections you have to make right. So that is why multi stage always will be limited to a certain number may be five six depending on your thing but not infinity infinite is the correct one right so infinite means I will get the minimum reactor volume or minimum reactor I mean minimum catalyst weight. So similarly if I want to do the same thing it is optimal for endothermic reaction it is optimal that means you know the lowest amount of catalyst we should get but it is an endothermic reaction in this case reversible how do we do it we start from these and so we know that definitely we are going to have you know it is adiabatic system so that means when you are conducting endothermic reaction adiabatic system temperature falls as conversion increases so you move in this direction okay so of course this length will be depending on your slope so you may go something like this again heat now not cooling heat the system because if you go further you are going to get less and less conversion so that is why you again heat it come here again go heat so like this you can continue till the desired conversion if this is the desired conversion similarly here also if this is the desired conversion so till then you have to do that right. So this is how we use multi stage adiabatic reactors if you want more conversion and if you are very near to optimum right so these are the things what we have to do here and here what we do yesterday I have already explained this theoretically speaking we should have we should be on this line like this this is the optimum line that means that the maximum rate line where if I am there on this line because it is a packet bed reactor every place you will have a rate of reaction so you have between this and this you have infinite rates between this and this you have infinite rates between any two rates also you have infinite number of rates so some of those rates will be in your reactor existing okay all that so that is the reason why theoretically speaking you should be on this line but it is not possible that is why again you go here start and come here go here and if it is three reactors what you are using if this is the desired conversion you have three stages but you see we are almost on the line there right this is optimum line you are somewhere on the average you are only staying on that particular line but if you want to I mean really minimize that means optimize you will get the minimum reactor volume possible then you have to be only go like that right so that needs infinite number of stages infinite number of stages is nice word infinity mathematical word but I think no one has ever seen infinity right so if they have seen they would never come back to tell us I have seen infinity okay so both are possible so that is the reason why we never go to infinity and then we will choose some finite number of stages and finite number of stages depend on of course the outlet conversion and in fact there is a trial and error procedure you know graphical procedure gives you lot of information as I told you yesterday here this is anyway this is fine these two but here we start at this point let us say I have T zero that is what your starting point okay this is what is starting point but how do you know that this starting point is right we do not know so that is why again here we will have trial and error okay no kinetics cannot tell you what is the starting you know temperature the reactor inlet temperature reactor temperature yeah see it is exothermic adiabatic system if you already know that the dangerous temperature is around let us say 500 degrees centigrade if you start at 500 it is exothermic system definitely you will have higher temperature then you will have catalyst spoiled already okay so that is why it must be something less how much less okay so that we do not know that is why again we should have a trial and error business here I will just guess okay let me start at this point T not okay so may be 350 I am starting so then I draw the line because this information I know beta one by beta values are nothing but you have f total f and f a not delta h r and one more thing c p okay all these things I know that means I know the slope so I will draw with that slope right so I will move on this line till a point where my criteria the other criteria yesterday what you have written x a in yeah x a out equal to dou by dou t of 1 by minus r a dx a equal to 0 right what I do is every time I draw this line take all the rates I can draw any number of rates here so then take all the rates take the slope of that if I plot with temperature because I know also know the temperatures that is the beauty with this you will know the temperature you will know the conversion you will know the you will know the rate right so all this I can have there that information and I extend this line till this integral becomes right and from the dynamic programming it automatically falls that this same rate if it is stopping somewhere here let us say on this rate so I should have my cooling system till this temperature where because I have to be on the same rate logical know because it is a heat exchanger there is no catalyst there is no rate going on okay so that is why I will be on this rate only and through dynamic programming also it can be proved r a out equal to r a in after I mean okay r a in for heat exchanger is also equal to r a out for heat exchanger that means heat exchanger does not have any it is not it is not a reactor it is only just removing the heat so that is why you will be there on the same line you cannot draw beyond this and then have because there is another line which is going like this you cannot go there this is automatic so that is why you stop here from again you draw the line and of course again on this line this criteria must be tested that is only the optimum and like that you also go to the third one cooling third one like that again this criteria is tested finally when it is zero then you will have some conversion here now you have to check whether this conversion is the conversion which you wanted or not because it is for a given cat for a given conversion only you are trying to minimize the catalyst so that is the reason why you go back again if it is not telling you go back again a move either this side or this side depending on you will get some idea whether it should be you will be you will be having you know a temperature this side or this side higher or lower so like that with few trial and errors you can easily locate that temperature which will end up with the exact conversion what you wanted that is how it is done for exothermic reversible reactions okay yeah which condition yeah you know this condition tells you that always you are on this okay theoretically you should be on this line okay so on this line is this is the maximum rate line maximum rate locus I have not written earlier maximum rate locus right okay I know I have to be on this line but it is not possible for me exactly to be on that line so that is why I have to be around that line giving me this you know the average rate supposed to be the maximum and that rate will be maximum average rate that is important average route only you will get average 1 by minus r a is the one which you get for a plug flow reactor or packed bed right so to get that I have you know this condition will tell me that I am now when I am moving in this direction if I am able to follow till here yesterday I have also drawn that graph you know you remember you have this kind of thing okay so if I plot this is the one which I have to plot here dou by dou t of 1 by minus r a yeah and this side I have x a so this condition will automatically tell me that I am on the optimal path okay so but mathematically we can prove that this is you know it is simple optimization where your first derivative equal to 0 okay that is the one only but the same derivatives are used but the objective function and all that will be a slightly different function if you go to actual dynamic programming if I am not able to cover that but if someone is interested a little bit of that in a most understandable way given in from ant and bishop yesterday I have written that book name chemical reactor analysis and design it is in the packed bed reactor there is a chapter on packed bed reactors highly complicated you know really taken very very complicated level but I think I cannot cover all that but only I can give you the information if someone is interested you can see there and as far as dynamic programming there is concern the mathematics are not that difficult you get this condition very easily it is not at all complicated and you also get the first condition as r 1 equal to r 2 that means this this condition you should be on the same rate Levenspiel also mentioned without mathematics the same conditions okay I think in third chapter in third edition it must be 21st chapter or 22nd chapter packed beds okay catalytic packed bed reactor design is given but I think analytically only he just explained okay so that is why to to satisfy this then only I have combination of increasing rates combination of decreasing rates which will give me the maximum point that means around this I will be there that point is achieved if I go beyond this okay if I do not go beyond that let us say I have drawn till there let us say that means I am not crossing this so when I am not crossing that so all these rates that means I have to move necessarily this way when I am going this way the all these rates are lower rates because this is r a equal to 0.1 r a equal to 1 r a equal to 10 like that so that is why when I am doing that definitely I am not doing the correct job because I am now only taking only the lower rates you can also say that I can go this way right when I am going in this way then I am going to again lower final conversions equilibrium conversion itself is you know this if I draw a line here it is so small a conversion for me so that is why that combination will give me this kind of equation where I am on optimal part so that is the reason why we have to satisfy that condition and then by trial and error also you can fix this t not otherwise which t not is the better t not we do not know this is the reason why you know flow chart is automatically developed if I am using this kind of reactor and my t not says that okay your feed stock that is coming from some other place may be 100 degree centigrade but t not will be let us say 250 degree centigrade okay yeah you have to preheat so that means you have to add another extra heat exchanger so that is an additional equipment in the plant that is how the flow chart is automatically developed so at the end of the reactor if you want to separate the gases then you have to use some other technique probably you have to cool it and then it becomes liquid and then use distillation to separate them so now you see there is a condenser for the gases for condensing the gases to in the liquid form and then afterwards you again distill to separate the products and reactants you have another equipment so that is how automatically the equipment in the flow chart increases and in between anyway when you are pumping from one place to the other place you need the pumps so that is why in every plant you have the pumps heat exchangers and you know condensers distillation columns or absorption columns reactor reactor is only one that is why I said if I have my dream reactor you do not need all that everything is happening at room temperature 100 percent conversion 100 percent pure you know the other day last Monday I was there in BHU when I told that people are also really have happy when I said that okay because SAR I think it is very good I also told that you do not have to read any chemical technology because there is no flow charts it is only reactor you learn other than that you do not need anything okay unless otherwise you have that kind of reactor you have to necessarily have all these possibilities that is why really thrilling things are there in the reactor design okay so that is as far as graphical procedures concerned now using these okay if you hate graphs because some people may not like art I like drawing graphs for me on the board is excellent right so but some people you know always like writing equations so for them we can use these two equations to design in fact what you are doing is we are only trying these two by solving here so this line is energy line okay so adiabatic you know that is energy balance and these things are yeah middle balance line so the solution between every point when it is crossing then that intersection is a solution okay so that is what what we are indirectly doing but now let us see those people who hate this and can use these equations but you can see that here I have three unknowns like d z I do not know length of the reactor right and also I do not know corresponding d x a in that d z and I also do not know what is delta t the increase in temperature so temperature I do not know conversion I do not know corresponding length I do not know how many equations I have I have only two equations if three and there is no problem because there are only two equations now I have to go and guess what is happening so what we do normally here is to solve equation 2 and 5 and please remember I have changed this a little bit I do not know whether you observed this or not yesterday I have written d a dashed thus that d a dashed is nothing but you know that is the pie yeah pie d into that length and this length because I am taking about you know these are called the finite difference methods so I take a finite difference that means finite difference in conversion you can also take finite difference in temperature finite difference method okay so I will take for example delta x okay let me write this is pie d dz if I take the entire reactor then of course I have pie d l that is no problem okay so that is why I have substituted here then I have here yeah this is delta t t minus t t not I know okay t not I know and then dz yeah t I do not know length z I do not know starting with z equal to 0 and dz I do not know dx I also I do not know that is why what we do is to solve equation 5 and 2 first guess not guess assume step 1 assume delta x for example point 1 okay if you use point not 1 you will get more accurate good yeah so then for this delta x assumption that means you are now telling that let me assume that I have 10 percent conversion now guess what do you guess what is the temperature in that 10 percent conversion you know initial conversion initial temperature is may be 250 degrees a degree so for 10 percent conversion probably you may expect 270 270 degrees so t not I know so definitely delta t I can now calculate okay so that much increase is there in the temperature so then yeah using this temperature and using delta x equal to point 1 I know now this I know also this okay the temperature because I am now guessing the temperature and now you can calculate what is dz okay so number 3 is calculate delta z because finite difference method I have written so that is why I am writing delta z calculate delta z from equation 5 I gave yesterday's number so that is why in between you have that okay so yeah then what do you do because I have to check whether my guess is right or wrong so then what I have to go I have to go to equation 2 equation 2 has dz and also dx and also I have temperature because in this minus r a I have k into you know ca for example if it is a first order reaction simple otherwise there will be k so that k value has temperature now I know the temperature I guessed I know x I assumed and now I calculated delta z all these three values substitute in equation 2 okay so 4 is substitute t z and x x a in equation that means of course equation 2 has left hand side right hand side if this equal to this by substituting all that temperature is this side and also length is this side conversion is this side so if LHS equal to RHS or otherwise the difference between those two is 0 then you will have correct that means you guess at the correct temperature if you have not guessed at correct temperature again you have go to step 2 okay substitute okay substitute equation 10 to check LHS equal to RHS okay in equation 2 if number 5 if LHS is not equal to RHS in equation 2 in equation 2 then what you do again guess so go to step 2 and then repeat okay go to step 2 and step 2 and repeat and repeat so that is how what we do of course I do not know whether your memory is still working or not we have done the same thing in the last semester okay yeah we have solved also I think one problem or not solved exam I have given I do not know whether I have given in the exam or not okay yeah so this is the one but now the same procedure is also extendable generally trial and error I have not given there in the examination but this time I can give because this is high level course now so that is LKG this is now PhD so so that is why that is why I can give that one now and the same procedure used by Smith chemical engineering kinetics in 13th chapter he has solved a problem okay I think this is some styrene production also styrene production wonderful problem do not neglect that try to understand but notation will be different and equations may be different in the sense rate will be expressed per volume of the catalyst or weight of the catalyst or surface area of the catalyst but you should not get confused because we know that in heterogeneous system you have to be very careful with the units okay and another thing CP sometimes CP can be expressed in terms of mass or moles lot of difference between those two lots of difference between those two okay yeah so that is the reason why you have to be careful and then please see that one I think example 10.3 or so if I am right or 10.2 or 10.3 styrene production real industrial type problem so how to calculate conversion and then corresponding weight of the catalyst both are I think he has solved that problem and even though it is solved example but understand the solid example first then I am going to give the assignment tomorrow I will give the assignment okay so this is the one and these are all simple because still we are in the dream world in the sense that assuming that only we have plug flow right and the dream world is most of the time true in chemical industry because we do not want to complicate things so when I calculate with this procedure the catalyst let us say 2 tons I will happily take 2.2 tons okay so then all radial non uniformities are axial non uniformities all that will be taken care of okay so that is why we do not solve those equations but as academicians we have to know how to solve those equations so that is why now we will develop equations for the you know general equations for design of packed bed reactor okay so general equations for the design of packed bed reactors assuming again pseudo homogenous model assuming pseudo homogenous model yeah so here let me say that we have a packed bed is normally very packed packed bed so L by D should be little bit better so this is yeah we have the catalyst particles throughout touching one other they are not hanging okay so you may have input this is the output and I may take this is the centre of the bed so I will take a small volume here so this is the volume element and through which I have material entering that means you know the reactant or reaction mixture at this point entering let me say that this is is delta z this thickness okay this is I have to also show that this is z direction and this is r direction from the centre okay radius so now this will be this will be delta r yeah so here z equal to L so now we have to this is r and then this may be total length is capital R good so now we have to write our universal expression that is input equal to output all that stuff yeah so we do write maybe I think I will use this and then remove that later okay so here material balance or energy balance also same thing we have a balance over this over this where we have output material balance is always for a component so material balance of A if I take A as the reactant okay yeah so output or not input equal to output plus yeah accumulation plus or minus generation or disappearance if it is generation it will come this side okay maybe I think I better put minus plus okay generation is equivalent input no so it is happening there so this is the equation so now for mass we have to over this element and we never said that it is plug flow right but how the material is moving material is moving it is entering and we never said it is ideal plug flow that means there may be axial dispersion okay and also there may be radial non uniformity in the sense that if you have concentration gradients in the radial direction you also have transport right so that is why you have at this point just entering same thing also here at this point also yeah I have material entering by in this direction axial direction entering by plug flow plug flow means convection I do not say plug flow by convection and also by axial diffusion both are in the axial direction because convection is in this direction because flow is in the direction mainly flow is in the axial direction so you have the convective term and also axial dispersion as I told you axial dispersion can be easily imagined when you are walking on a conveyor belt conveyor belt speed alone is the convection okay and there are all molecules standing there and some molecules cannot move and some molecules will move because of the concentration gradient okay so that is why the base molecules which are not moving they are moving only with the speed of conveyor belt is the convection and over that when you are walking then that is the diffusion in plug flow what we normally do is we neglect that diffusion that means almost there is no axial dispersion okay or even if it is there it is negligible when compared to total convection so that is the reason why we neglect that okay so those two terms will be there in the axial direction and mass transfer also in the radial direction that is coming to radially entering here and leaving here because we are writing material balance over this okay so that means if I turn like this assuming that I have the thickness we are now writing what is entering there and how it is moving and in this thickness what is coming out right so temperature also same thing and concentration also same thing material balance also same thing but now we are first writing the material balance that means in the input itself I have three terms in the input itself that means what are three terms convection itself diffusion and radial diffusion okay so all three will be there so that is why we will write here please take that okay let me write here afterwards I will tell you input equal to bulk flow or convection shall write convection or bulk flow convection plus I have axial dispersion plus we have radial dispersion okay in this element that means here this is the one which I have taken from here inside right here when I say actually what is entering by convection it is something entering in that element and also by diffusion something is entering and in the radial direction because of the concentration gradient something is coming out okay good so that is the one and if I write the equations let us say this term as one two third term if I write I think these three terms separately then you will understand so the first term is it is C E concentration C U is the convection term C U what are the units of C U? C into moles for flux it is flux C U is flux moles for meter square second or moles per centimeter square second okay it is a flux right yeah so that is why it is a flux multiplied by I have to write the cross sectional area that is why I asked that okay cross sectional area so when I write that cross sectional area what is the cross sectional area I am talking how where it is entering it is entering this annular space that is what my control volume yeah so like this know this is the thickness where it is entering so what is that area 2 pi r into d r okay yeah so that is all there so you will have here 2 pi r okay delta r okay all yeah and of course I can also write if I want yeah what are the units of now this moles per second so if I want to write only mole balance that I can also put there into delta t time units so that means moles entering not rate moles entering moles leaving moles reacting moles accumulating because unsteady state equation where I want to write that is why delta t also I am taking so that delta t tending to t 0 delta r tending to 0 delta z tending to 0 will give me a differential equation okay taking all these things into account so number 2 that second term will be this is axial dispersion we will write now this one as d e l axial direction d e is effective diffusivity l is in the axial direction so d e l this is now dou c a by dou z because the time is varying r direction it is varying so that is why this dou yeah and this multiplied by 2 pi r delta z of course this is minus this entire thing multiplied by delta t okay that is axial direction good so this is the one and third term will be the radial dispersion that is the one so that term will be minus d e r okay effective diffusivity in the radial direction into yeah this is dou c a by dou r yeah into delta z no no 2 pi r 2 pi r delta z where you are telling this one yeah right correct correct correct thank you so this again multiplied by delta t okay similarly all this is at r equal to at z equal to z at this point this is z this is z plus delta z so all those 3 terms again will be there for at delta okay now let me say that this is at z equal to z in that in this figure all these 3 terms as output will be there at z equal to z plus delta z so I am not going to write that same terms so finally I will give the equation so but what are the other terms we have accumulation term and generation term right so accumulation term will be okay output I think shall I write here okay this is this is input I have here output same as above but at z equal to z plus delta z that is the output exactly same thing and time checker you do not have class no class no class today so today is what so only Thursday and Friday here so now I have to write the accumulation term yeah accumulation term will be what I have minus r minus r a o b in fact okay minus r a I will just write yeah accumulation it is not generation yeah so accumulation yeah accumulation term will be the volume and you know delta c a okay yeah so I have delta c a 2 pi r d r into d z okay or otherwise I think I should not write d r is I think there delta or so I have to write yeah this is delta I am making mistakes because time whether I can complete this I do not want to write whole thing again next class so that is why yeah so this is the term accumulation and now we have the generation or disappearance because here we are writing we are writing for a no a is disappearance yeah disappearance term will be minus r a into the volume 2 pi r delta r and delta z that means this minus r a is based on what volume and you have to be really careful with this because it is heterogeneous system so if I base it on let us say weight of the catalyst yeah then I have to multiply by bulk density bulk density because entire volume we are taking you know not individual particle density so all that I think I have to tell you that okay so now substituting no accumulation delta t will not come you check the units see every time we are talking about moles you know by the by what are the units of this moles only and here also moles right so here also simply moles concentration verses volume yeah dosy by dot e if I am not taking time here if I remove time here time here time here then that will be automatically accumulation okay if I remove time here time here time here time there okay then I have to write dosy by dot e because now I am balancing moles per time that is the reason so now if you if I am substituting let us say this equation 1 okay and yeah all this is 2 3 4 yeah substituting equation 2 3 4 okay I think here also there is an imaginary equation this is 3 out later so this is 4 this is 5 right 3 is the same equation but only at z equal to z plus delta z so if I substitute 2 3 4 5 in 1 and then take substituting equations 2 3 4 and 5 in 1 and taking delta z yeah delta r and delta t tending to 0 all 3 because normally many people will not do this balance for every term so that is why I thought at least one time I will do it so that you will know how to write this balance but only thing is you have to substitute them patiently and then write the entire equation and you yourself see in terms of only finite differences like delta t delta z delta r and then take the limits on your own because for every class yesterday there was some people from Milano, Politecnico Milano, Politecnico di Milano or something you know from Italy they told that the courses will have credits of how many they said I think 20 to 30 credits one course I said what is this I think here we have 1 hour equal to 1 credit so if I have 3 classes in a week then we call 3 credit courses but how can you call 30 credits it seems they are going to calculate that credit for the students work for every hour the student is supposed to work it seems 6 hours or so depending on the subject okay so all that also counted as hours if you are working let us say 6 hours in a particular course in room plus 1 hour here right then sometimes you know parallel you have the laboratories right in laboratory also you work some hours for that also you have I think there they said laboratory they are working before us so if they work 4 hours 4 credits so 4 plus 7 plus 1 how many 12 that is 12 credit course so like that why I am telling is then I told that we do not know whether our students are working or not so that is why we never take that is why we never take how many hours they are working we take that how many hours we are working yeah good okay so substituting all this and then taking what you get for material balance is I think I have here this equation which you would have seen many times this is D e L dough square C a dough square C a by dough z square plus D e R dough square C a by dough R square plus 1 by R dough C a by dough R minus dough of C a u by dough z minus R a okay minus of minus R a minus R a is an indication minus means we are writing for do not take literally that is a negative term okay the value what you substitute is only positive K C a only not minus K C a right yeah so this is equal to dough C a by dough T so this is equation number 6 similarly you write the heat balance equations that means now you have to write only in terms of calories alone okay calories alone so that means you can write row C p u all the terms I am not writing there so what you get exactly same thing that is what you know transport phenomenon approach so I do not have to tell the other thing you will get exactly similar equation for heat balance now this is K e L thermal conductivity effective thermal conductivity in axial direction that multiplied by dough square T by dough z square plus K e R radial direction effective thermal conductivity dough square T by dough R square 1 by R dough T by dough R we have minus dough by dough z of equivalent yeah row u C p T right exactly equivalent to your this u this entire thing okay so then you have minus delta H R minus delta H R minus of minus delta H R and minus R a yeah so there is another term here which I have not shown at the if you really take at this outlet that comes as a boundary condition is a boundary condition if I expand this entire thing here then you will get a term for heat removal from the from the walls heat removal from the walls time okay heat removal from the walls okay anyway so this is equal to correct no yeah this is equal to dough T B by dough T so this is equation 7 and for equation 7 you have to add I think last one we have to add to equation 7 the heat removal term heat removal term for D D T diameter of the tube H W T minus T W okay otherwise that also will be 4 pie and not 4 pie 2 pie D Z 2 pie D Z okay 2 pie D D D Z you know I do not know whether you have really those people who have observed this sometimes we use 4 by D T and sometimes we use pie D pie into D into D Z you know what is the reason yeah cross section but I think that also you know heat is removed only through the this outer walls this way okay no please remember this once we use directly the actual area that is removed that is pie D into D Z other times what you do is surface area per unit volume of that element surface area that means pie D by your pie D D Z divided by you have unit volume of that element what is the unit volume of that element pie D square by not pie D square pie by 4 D square into D Z that is why you get here 4 by D T many people would have not noticed that even in heat transfer even in transport phenomena in real heat transfer you would have got that see there are 2 ways but you are talking about same thing but how are you writing the balance that is what only most most important there this is based on per unit volume of the element where you are writing the balance otherwise you simply take like the other one what you have done was simply taking the area okay so these 2 methods I think please make a note somewhere otherwise I am sure that you are going to forget but in the books some books use this equation some books use pie D into D Z so then you should not get confused you told wrong it is not wrong it is only based on on what basis you are writing the heat transfer okay good so that is why anyway add this one later to this term and this is purely now adiabatic system okay adiabatic system anyway here I do not have to add anything because material is not coming out of the system and I can tell you variation of this is your membrane reactors you know membrane reactors where through the walls also you have some amount coming out there you have to add similar term in this basic these 2 are the basic equations in almost all chemical engineering okay so if there is no reaction these terms will not be there all other terms will be there to be solved so this is what the general model unsteady state with all diseases like radial non-uniformity okay and axial non-uniformity all kinds of things now we have to see that you have to make it more and more healthy that means best healthy person is only plug flow so remove this term remove this term remove this term only these 2 terms very happy okay I think I will stop here next class we will discuss a little bit about this