 for the design of the acrobat reactors. This is the one, I think you will be sufficient here, okay, I will try. This is reactor conditions, reactor conditions, then this is design equation to be used, good. So first reactor conditions if it is isothermal, okay. See we are referring to the same equations what we have written now. You have dispersion term, you have dispersion axial, dispersion radial, so then you have convective term, right, then you have unsteady state term and yeah, reaction term, okay, this is fine. And conservation equations for the design of reactors at steady state, so straightaway dose A by d2 will go, right. Now you have isothermal, that means you have completely energy balance equation out. Now you have only mass balance equation, dose A by dhoti out and it is isothermal, okay. So most of the time we are neglecting, not radial axial, L by dp is very large, okay, most of the time industrial reactors, yeah, that comes later, yeah. So radial, I mean axial we anyway neglect because most of the time L by dp will be greater than that 150, 130 like that. So then only one term now radial dispersion term is there because there is no heat transfer, no isothermal and all that. So you can also assume that we have only radial uniformity. So finally you will have this equation, yeah, if I simplify that you will get minus u dca by dz equal to that is the equation to be used. This is just nothing but I was telling you know w by f not, depending on the units of minus rA kg per catalyst means w by f, this you can convert this as w by f not equal to integral 0 to xA dxA by minus rA, you can convert that. So for isothermal case these are the conservative equations, we do not have to worry about all other terms, yeah. Most of the time when you have isothermal we can use only this is the convective term and that is the rate of reaction term, okay, all other terms you can neglect there. So now if you have adiabatic case, yeah this is one then you have adiabatic case. Adiabatic case means straight away the other term will go that 4 by dt or uA da that delta t, delta t, okay. So that will not be there. So then we have different conditions. In adiabatic case, yeah in adiabatic case if I have, yeah adiabatic case means, first of all adiabatic case when do you use? Isothermal is delta hr approximately 0. Adiabatic means you normally use adiabatic reactions when you have the delta hr somewhat intermediate, may be around 20 to 30 kilocalories per mole, okay. So approximate values, they are not universal values that everyone must follow that, okay. How serious temperature again anyway you have to simulate and then try to find out. But definitely you cannot go to adiabatic if you have heat of reaction around 60, 70 kilocalories per 60, 70 is very high, for example ammonia reaction and all that. So that is why you have to go for only thalic anhydride, right. So you have to only go for non isothermal, non adiabatic that means heat must be removed, okay. So these are the general thumb rules. So that is why when I write adiabatic here you can also write delta hr moderate and we have a feeling that around it may be 20 to 30 kilocalories per mole, okay. So under those conditions if you have this still reactor conditions, okay if I have l by dp this is one group, udp by da if it is greater than 300, you got the meaning of this, what is udp by da, what is udp by da, what is the number, is there any name for that, that group. So come on you told something, it is peculate number but peculate number based on particle diameter, what is that we are now multiplying by, what is this called, l by dp called aspect ratio, okay. So when you multiply aspect ratio with peculate number based on particle diameter, you know what do you get, cancel out dp dp, yeah. Then also it is a peculate number but based on reactor length. So peculate number sometimes they are expressed as length of the reactor and sometimes particle diameter. So if this column peculate number, okay reactor peculate number if it is greater than 300 then you have to use because it is adiabatic, so definitely energy balance also will come, due temperature also, dou ca by dou z equals to minus rub and other equation is rho cp dt dz equal to minus delta hr minus rub. In fact this is what we have already done, when we are drawing those graphs and all that, okay. This one if I solve, I will get t as a function of xa, t as a function of xa, may be you know these symbols are different but finally you will get the same thing, fn r and all that. Savitha is blank, totally blank. Yeah this equation is nothing but your mass balance, energy balance equation for adiabatic system, what we have done last semester and also sometime back also I have shown you that, okay. So that is why this is very simple to solve because this is now only ideal plug flow, that means again we are neglecting radial because when you are not removing heat, so there is no temperature gradients across the radius, okay. Only temperature gradients in the axial direction, that is what is this, dt by dz. So I mean spend some time you know to understand the physics of the equations, that is why I am trying to explain. You could not write because that physics still has not gone into your blood, okay and in chemical engineering similar equations will come in almost all subjects, okay and I do not know whether you notice it or not, maximum order of differential equation what we use in chemical engineering is second order, beyond that we do not go, okay. Only civil engineering people will go, mechanically also same, aeolonautic people also same, all, most of the engineering problems the second order differential equations, enough except civil engineers where they use for bending, bending what that beams and all that, okay cantilever beams. So they only go to to the 4 of 4 and to the 4 of 5, all others second order differential equations that means only two boundary conditions we require for every system and all that, okay. So that is why this is again already you are familiar with and why we are neglecting all other terms like radial dispersion because you are not removing heat, temperature is uniform that will not create any concentration gradient that is so rate effect. If you have concentration temperature change along the radius then that will create automatically concentration difference. So that is why we have to take radial dispersion term but here it is adiabatic you do not have to take, good, yeah. So then if it is less than this, this number if you have LDP into UDP by DEA axial dispersion number less than 300 then it is only necessary you have to check yeah and sorry I think I have to also write here another condition this condition plus Reynolds number condition, Reynolds number greater than yeah I think here I can write this is common to both Reynolds number greater than 10. If you have again Reynolds number greater than 10 and this condition less you have the same equation dou C A by dou Z equal to minus R A O B and you can use dou C P dou T by dou Z minus K E A dou square T by dou Z square that is the axial heat transfer term may be necessary that means minus R O B if necessary, okay. So if I give number this is 1 this is again of course 1 this is 2 this is again 1 this is 3 if necessary if not then you have yeah same thing equation 2 necessary means you have to see you have to take this term and then solve the equations axial term temperature term, okay and if your conversion is almost same even if you take this or do not take this then this is enough yeah equation 2 this is enough, okay. So with computers and all that now one has to do the simulations, good yeah so that is why this is only just caution whether it is really required or not one has to check depending on the problem good so that is the one then you have non isothermal non adiabatic case same similar table this is the end of this so here we have non isothermal that is the reactor conditions non isothermal non adiabatic so sometimes we call this NENA reactor non isothermal non adiabatic right some books will give NENA yeah so under these conditions if I have R E P R by D P greater than 4 greater than 4 then you have to use this equation, okay under this it will come this is radial dispersion dou square C A by dou R square plus 1 by R dou C by dou R I think I will extend beyond this boundary U dou C A by dou Z equal to minus R E O P so this is equation number 3 this is equation number 4 in this yeah we have to also use K E R dou square dou square 1 by R dou T by plus dou Z plus I have 4 by D T this term also will come H W T W minus T equal to minus minus delta H R minus R E O P yeah so this is equation number 5 yeah thank you thank you thank you right right good still alive one kicking, okay yeah so dou T by dou Z so that means this is the only serious case why you know R by D P what is the meaning of R by D P radius by D P yeah yeah so how many particles you can put across up to it across the cross section okay so if you are putting more than that heat transfer effects will come because I told you packet bed is a low g conductor okay so that is the reason why you have to take those things but you see here axial dispersion neglected correct no axial dispersion neglected in fact axial heat conduction can always be neglected axial one except here it is only necessary just you know if you do not see much difference this also you do not have to take K E A is the axial heat dispersion term okay that heat conduction term okay good and I hope you know what is E and what is D E this E and this E effective diffusivity where we are clubbing all the mechanisms like for example effective diffusivity is a combination of configurational diffusion bulk diffusion Knudsen diffusion all three diffusions together we do not know how to separate them mathematically we know each expression we can if you know we can find out otherwise it is a fitted parameter where through experiments D E A is evaluated through experiments and then fit in terms of correlations right in terms of velocities in terms of particle sizes and all that good so this is the another expression and now if I have row R by next one DP less than 4 so R by DP if it is less than 4 which are the terms that may go out there you cannot see he is blocking you what are the terms you have to take what are the terms you have to neglect use some physics use some brain you know I think what are you do not close eyes huh huh why in both the places that means mass balance and heat balance in both the equations both yeah but then why why not mass if heat is neglected automatically you may not have concentration gradients know but what is the meaning of R E P R by DP less than 4 smaller diameter of the column so radial uniformities can be expected so both will dissipate okay that is the logic you know that is the reason why I gave this test I thought you know you will use that kind of logic and then quickly finish it off okay yeah but where is the logic every logic is only in the examination hall afterwards it will not work nothing will work huh oh my god every time it is boring every time we have to talk about only exam exam exam exam yeah okay that way I like this our domes department you know the management the guys who do management they have passion for doing management in fact I told the research even research scholars are so passionate then I told them then why do not you come to engineering departments and then walk across every day some 2 hours around that so that your some of your passion will diffuse from you and then go to these departments because there is zero diffusion gradient is there so that is why easily it can diffuse so chemical engineering department 2 hours mechanical engineering department another 2 hours okay yeah building wise really they have tremendous passion engineering absolutely no passion only examinations okay so that is why I think you know always they sit in coffee day and also discuss management students huh their system is quarterly system what you are studying in 4 months is they are studying in 2 months all 2 great courses okay so that is why they have passion otherwise they cannot survive in management because it is usually so boring subject I mean what is management common sense and logic okay that is all but I think most of us wont use common sense most of us do not use logic in engineering so that is why we do not have that passion but they try to use and you are not answering me well sir they are getting lot of salary we are not getting that much salary you have to defend yourself I say I have to put the question I have to answer also okay yeah because they are starting salary may be 20 lakhs and you are starting salary is 20,000 okay so that is why you do not have passion and they have passion this is what I have fought once with one big company man and then afterwards he never look at me and then I was member and I was removed okay so why I why the I just asked why I think you know in IT industry people should be paid more what is that they are doing extra when compared to an engineer okay yeah because engineer is doing his duty in the plant he is doing the IT person is doing whatever project is given in the IT industry okay why they should be given lakhs and lakhs and where these people are given only 1000s there was no answer except anger okay that is all no answer if that answer and anger should have been better but anyway good so yeah this is also we have another condition here Reynolds number Reynolds number also is based on particle diameter any packet bed I think all the time we only base the based on the particle diameter okay so that I do not have to repeat again yeah so in this case we have equation dou C A by dou Z equal to minus R A O B and rho C P U dou C A dou T dou T by dou Z equal to minus delta H R and minus R A O B so this is again equation 1 other equation is yeah equation 2 see you have to really appreciate that engineering mind you know simplifications even though I have non isothermal non adiabatic and you know fairly large amount of delta H R and all that still I use only these two terms sorry you have heat transfer term that that will come now yeah yeah that one is plus 4 by dt H W T W okay T minus T W it depends on okay depends on you are adding or you know endothermic or exothermic term okay so this is equal to minus delta H R minus R A O B this one then this equation will be 3 5 6 yeah so these are the equations that is all what we have to use okay please remember them I think you know even though you are not in chemical engineering sometimes I think it is better to know this okay yeah so I think there are so many beautiful things are in chemical engineering because there are so many people are waxed up and then anyway I do not want to do anything because sometimes it happens if you have too much work you do not do any work because you do not have a way to start okay so maybe because of there are too many concepts and too many beautiful things here people too many things are there what do I do so leave it maybe that is the reason why you are not interested otherwise you know in other countries USA, Germany and all that chemical engineering is in fact chemical engineering chemical engineers who are getting the maximum salary in US when we were studying I think that went on till may be 80 85 but of course no one can beat them in medical doctors in the US I think they get maximum the next one out of all engineers who are chemical engineers I am talking about 70s and 80s okay after IT came and all things have gone down only they came up too much buoyancy so came out okay yeah so that is why so many beautiful things are there I think you know I think you see you have taken so many courses you know at least in one course have you solved these equations at least in one course and you are supposed to solve this in transport phenomena because transport phenomena here you do not have time to solve here is in reaction engineering course my idea is only to give you the concepts not the mathematical techniques mathematical techniques you are learning from mathematical courses plus transport phenomena transport phenomena is 30 to 40 percent concepts the remaining 60 to 70 percent solving solutions there you have to solve that okay then you will know then you will really enjoy if you solve those equations then you know how to plot you know in all these things what you get at the end you always get temperature conversion and length of the reactor that is all what you get because T X and you have what is the other one C A so what you have to generate here is the graphs this is X A and T okay X A will increase like this this is X A versus this is Z X A versus Z and the other one is temperature so depending on whether you know adiabatic non adiabatic cases and all that adiabatic means simply this also will be another line like this right so non adiabatic case means depending on heat removal or heat addition you may get like this is T versus Z these are the things what you do so you are getting here if someone asks you okay what is my reactor length for 90 percent conversion so then you know you go to 90 percent here that is the reactor length where you will get even the outlet temperature so that that is why here you cannot have the analytical solution you have to go by most of the time the technique used is finite difference method you take small delta X and then calculates small delta Z and also temperature change there like we had know so you have know I think the second order differential term how you have to expand in terms of finite differences I think definitely someone would have taught you in numerical technique courses and all that okay so those are the technique what we use and try to solve them convert them into algebraic equations and then try to find out okay I do not have time to do that I mean if I have only packet bed reactors alone then we could have tried this we have to I have to do so many things because in this course idea is only to give the concepts okay and of course to make you comfortable a little bit of you know some problems so that you are also don't feel that you have done only concepts and then know actual solving the problems that the reason why we give assignments and all that okay so that is why that is why is the surprising to me I have also never solved all those things together when we were studying every time this is going on like that that is why I was telling that at least give that in B Tech project or in M Tech project M Tech project also is not that simple to solve all these equation then and design the packet bed if you are able to take all the parameters how to design the packet bed and also use these mathematics to get the solution then I think you know your confidence level will shoot up and if you are able to solve this any problem you can solve because you don't go beyond this order of differential equation so many boundary conditions heat transfer coming mass transfer coming and fluid flow coming reaction kinetics coming diffusion coming you know in the particle thing and all that okay this R O B is nothing but A tie into R B okay so all those things everything will come in reaction engineering so that is why at least even if you don't solve you remember that you are not solved anything okay that is also very important first of all you should know what you don't know okay so then I that is the first point for first starting point for any learning yeah I think so credit is there is a quotation you know the stepping stone for knowledge is ignorance but you should know you are ignorant that is most important otherwise if you sit here and then like this if you do I think as if you know everything you can't learn anything because you think that you know everything okay so that is why so credit is that quotation is beautiful okay the first step to learning is ignorance which you know that you are ignorant right not I should know but only you should know that you are ignorant no point in me for me knowing that you are ignorant I know that already okay so no point in knowing again further and you also will know that I am ignorant okay that is also true it is not only reversible reaction it is not only one way okay good so tomorrow I will start fluidized beds and just to make you comfortable with units I will solve one simple problem because in heterogeneous systems units you will make lot of mistakes indirectly I am also trying to coach you for examination so that you will not make any examination mistakes okay that one simple problem I will solve how to find out you know weight of the catalyst or conversion if the catalyst weight is given so afterwards I think we can