 So, we are going to be looking at energy balance today. What we are trying to do here is look at energy balance for a stirred vessel. Now, stirred vessel in which there is stirrer where the composition to the system is uniform at different positions or in say the intensive properties of the system is the same everywhere because of the mixing that you have introduced. You have F j moles coming in, F j 0 moles coming in, F j moles going out. You have so much of heat coming in, so much of work coming out. The system itself has N i moles of component i each of energy E i and the energy of the system we denote as E system. And the statement of energy balance is that input minus of output plus generation is accumulation. What we recognize here is that in energy balance there is no generation of energy or destruction of energy. Therefore, generation term should not appear unlike in the material balance where we accept that there is generation of material because it can form another product. Therefore, this term is removed. So, the statement of energy balance is simply input minus of output equal to accumulation. Stating this in mathematical representation is d by d t of E system equal to sigma 1 to N E i F i input and then E i F i output summed over all species plus Q minus W s. Where Q is heat input, W is work output, this is the energy that is coming in with all the species energy that is going out with all the species. We are neglecting heat of mixing that is neglected here. We will include if required at a late stage. E i is the energy content of each of the species and this consists of a kinetic amine. This is internal energy, kinetic energy, potential energy. Notice that the units of this is Joules per kg or in SI units can be also called as meter square per second square. Now this term W dot can be understood in some ways by looking at a term called flow work. What is flow work? Flow work is the work that is required to get material in and out of a system in the absence of friction. Something that has to be understood by actually making a calculation of the term F i P t omega i. Now F i P t omega i in the absence of chemical reaction does not change. But P t omega i can change depending upon the conditions under which your system is working, high pressure, low pressure, whatever. So, in a sense P t F i P t omega i which is units of Joules per second or watts, this really depends upon the reacting system. Therefore, this difference between F i P t omega i out minus F i P t omega i in, this can be positive, it can be negative. If it is negative, it means what? This amount of work has to be put into the system, so that the material can be put in and taken out. So, energy that is required to put material in and out of the system is what is flow work in the absence of friction. So, we can understand of the total amount of work that comes out of the system W in terms of shaft work which we can measure. Shaft work we can measure by measuring the amount of generator output and so on. While flow work we cannot measure, but we can calculate because F i P t omega i is a known quantity. You can measure and therefore, these two terms can be, this can be calculated, this can be measured and therefore, you can calculate W dot. So, keeping this in mind, what we are saying is that, if you look at as an example, we make a calculation of how much is the flow work required to put material in and out through a sulphur dioxide plant for example, say 3000 tons per day sulphur dioxide plant, what is the flow work we are talking about. So, we can calculate a sigma F i P t omega i out F i P t omega i in because we know what is P t typically one atmospheres, omega i we can calculate or we can simply say F i omega i is simply volumetric flow. So, we know how much is the change in volumetric flow or the inlet to outlet, why does it change because temperature has changed. So, we can actually calculate this whole term and estimate of this is what I have got here, the flow work for a plant of this size turns out to be about 125 kilowatt, it is a minus sign indicating that you have to put that amount of work. So, the material can be put in and brought out, but we have done calculation in an earlier class as to the amount of energy that is required in terms of pressure drop, we have calculated that for a 3000 tons per day plant and we found out this close to about 2.5 megawatts. So, what we are saying here is that even in the absence of friction, you would have to spend 125. In other words, the minimum energy that is required to put material in and out is not 0, this is the point that we have to appreciate when you have to deal with all these numbers. We can also calculate what is the energy that might be associated with expansion of steam through a turbine going from 0.1 to 0.2, this is a steam turbine cycle producing power, this is producing power. What does it do? We have a boiler which provides energy into steam and then we have a condenser, the q 1 minus of q 2 is the amount of work that would come out of the system, first law energy balance and that comes out partly through this, partly used up here the network is W t plus W p, this is what comes out q 1 minus of q 2, this we know, but what is the interest here is to find out what is the flow work from 1 to 2. So, I have got here what is the pressure, what is the temperature, what is the specific volume, so we can calculate p t omega i 2 minus p t omega i 1. So, this if you put all the numbers down, you will find that is about minus 2.63 kilo calories per kg, showing that there is a small amount of work that you have to do to be able to put material in and out, even for a steam turbine. So, what are we saying is that our statement of energy balance, which is d by d t of e system, which is energy in minus of energy out plus q which is heat input and this work have separated the work in terms of shaft work and flow work. This term phi p t omega i out phi p t omega in can be appropriately clubbed with the functions in and out. So, when you do that this equation looks something like this, where e i plus p t omega i in and then e i plus p t omega i out multiplies the appropriate terms. So, e i consists of which is called as internal energy, kinetic energy and potential energy. So, this e i substitute here the whole term looks something like this, that the total amount of energy which is in the system, which is e system for example, consists of n i moles. So, I put n i here multiplied by what is the energy, energy is u i which is internal energy, kinetic energy, potential energy. You notice here that p t omega i occurs only in the flow term and not in the accumulation term, therefore that term is not included there. So, this I have clubbed u i plus p t omega i can be clubbed we recognize that p t omega i is p v u plus p v is h. So, it becomes enthalpy plus the kinetic energy and potential energy. So, the energy balance term when you club all these terms together it looks something like this d by d t of n i the number of moles in the system multiplied by h i u i square g z i minus p t omega i I put it like that means, I put it in terms of h i even on the left hand side even though h i u i was there I have called u i is h i minus p t omega i. In the right hand side term I have clubbed the u i plus p t omega i is h i in h i out. So, leaving these two terms which is kinetic energy and potential energy plus q minus w s. So, what seems to happen in reacting systems is that these terms are not very important generally these terms are not very important. On other words by and large we are dealing with the enthalpy of the materials which we are dealing with. So, a simplified form of the energy balance can be looked up like this you have d by d t of n i h i and then you have f i h i in f i h i out plus q minus w s. So, this something we could have written even without having to go through all this something that we know for a long time, but the point of writing this is to draw your attention to two important features. When we write d by d t of n i h i we are actually neglecting the effect of p t omega i we are neglecting the effect of u i square and g z i. So, this is important to recognize because there could be instances where these are not negligible. So, you must recognize that these assumptions are involved in the simplified representation of the energy balance that is written here. So, that means on the left hand side n i h i is actually that h i was not there u i was there we replaced it by h i. When can we do this we can do this if u i and h i are not very different which is often the case as far as reacting systems of concern in our domain of interest, but it may not be always. So, we have simplified on the left hand side here on the right hand side we have neglected kinetic and potential terms which may not by and large may be right in most cases, but there are situations where they may not be correct. So, just to summarize what we are saying is that a simplified representation which involves only n i h i on the left hand side and f i h i on the right hand side is a very simplified representation which will apply for most cases, but may not be for all cases. Just to understand this let us just look at a hydral turbine. So, what is the hydral turbine what is the hydral turbine you have a reservoir which has water a huge quantity of water it is available at a certain height h. So, that you can use that potential energy to run a turbine how do you apply our energy balance what is our energy balance d by d t of system. So, what we are trying to say here is that when there is a huge amount of energy here and we are drawing a small amount on a daily basis or hourly basis the rate at which the energy of the system changes we can neglect under the quasi steady state approximation. That means if the time scale for which we are observing the system is small in relation to the time scale over which the reservoir gets consumed the left hand side can be seen as quasi steady therefore, can be deleted. On the right hand side what is this term what is the right hand side term we said if you look at here the right hand side terms are h i u i square by 2 g z i and p t omega i we have neglected on the here h i u i square by g z i, but in a hydral power station this term is important. In a hydral power station this term is important therefore, this term we should take into account that is why I said whenever we delete a term we should be very careful. So, what I have done is that I have again looking back at this I have put all the terms that are appropriate to a hydral power station. So, what is it look what is h i u i square by 2 and then g z i all that I have put in here. So, it looks like this. So, you have the effect the left hand side is 0 the right hand side has the effect of pressure and at position 0 and position 2 both are open to atmosphere I have taken it as p 0 both p 2 and p 0 are at the same pressure. So, I have knocked it out I am doing a energy balance between position 0 and position 2 and what is h i I have taken it as c v t naught minus of t r I could put c v or c p both are not very different and then you have I have neglected this term u naught term is not as very important. Therefore, you have g z naught as the term here and similarly you have a g z 2 at this here. So, what have we done we have written the energy balance where the left hand side is 0 we have knocked out the pressure terms we simply have what is called as a simplified form of the energy balance which looks like this. Which means the amount of energy that we produce per unit mass of flow is g z g time z naught minus z 2 and c v times t naught minus of t 2. Now, if t naught equal to t 2 then this term goes away then the amount of power that you produce if w if w if I have taken f as 4000 it is 2 megawatt is that clear. So, I have taken c v this term as 0. Suppose, we find there is a small increase in temperature small I have taken as 0.005 or 0.005 it really depends upon how well you design the conduit leading from the reservoir to the high delta. So, you notice here if you allow an increase of 0.05 you have lost more than 50 percent of the power this is what I want you to appreciate. Our fluid mechanics which designs this conduit from the reservoir to the to the hydro turbine which might be several kilometers away it may not be nearby. And the velocities for part of the reason why I have reason I have knocked out this term u naught because velocity would be the same in both the places that is why I have knocked it out. So, the point to remember here is that we can lose a large amount of energy, but not taking care of the design of the conduits that fluid mechanics is an important part of design. See we all think that mechanical energy can be transmitted without loss of efficiency we always think of Kano inefficiency, but here is an instance of mechanical inefficiency which is as high as 50 percent. So, this care we should take in the design. So, let us see what happens in a steam turbine steam turbine we have done anyway we already calculate a steam turbine we have calculated here if you do the same calculation for a steam turbine it is about minus 0.2.63. So, we can calculate the flow work required for a steam turbine or for a for a sulphur dioxide and so on. Now, we also said that this is not common, but this is something that we would like to do that we can apply our energy balance also to run what we call as a reacting system which will produce power. How do you make a reacting system produce power? That means a reaction like this 2 N O 2 times N 2 O 4 this reaction of course reactions known to be instantaneous. Therefore, depending upon the free energy change of this reaction there are 2 reaction one is N 2 O 4 going to N O 2 or N O 2 going to N 2 O 4. So, you can see here. So, you can here this picks up energy and then delivers energy picks up energy delivers energy. So, here it picks up energy at Q 1 may be at 80 degree centigrade and then at Q 2 at may be at 25 degree centigrade it delivers. So, it is Q 1 minus of Q 2 is the work produced if you can make produce work. So, in some cases you are not able to produce work or you are able to use it for a space heating application. Since, a lot of low temperature heat is available in many places you can pick it up using a chemical reaction and deliver it to a for a space heating. So, the important point here is that there is tremendous scope for designing working fluids which is able to pick up heat at low temperature for space heating application. Otherwise, you will have to use more expensive materials here is an instance was. So, much of heat is available at this low temperature which is required for space heating applications. So, reacting fluids and our energy balance is able to make use of all these because if the energy balance is written keeping in mind that the energy what is the energy balance? Our energy balance only saying that d by d t of n i h i is f i h i and f i h i out f i h i in. Basically it is saying how much energy is coming in how much energy is going out how much heat you put in how much work you take out. So, just to summarize what we are saying is that you have an energy balance where the system has amount of energy equal to amount of energy coming in amount of energy going out plus Q minus W s. Just to put it in the context of a stirred vessel what we have said is that h i system therefore, is also equal to h i out our nomenclature is that the intensive properties of the system is also equal to the intensive property of the exit stream. Therefore, I have denoted as h i and then I put a 0 to take care of input streams. So, in this equation I have written it slightly differently taking into account the fact that h i in the system is same as h i in the exit stream while h i and f i in the inlet stream has got a 0 here to indicate that it is a inlet stream. So, this is the statement of energy balance d by d t of n i h i summed over all species equal to summed over all species f i h i at at inlet minus sigma f i h i at the outlet no subscript plus Q minus W s. Now, our interest is to be able to put this whole equation in a form that we can use for our reacting systems. Let us see how to do this to do this what we have done here is that this term this term sigma n i h i d by d t I am just separating it as two terms n i d h i d t h i d n i d t. The first term here n i d h i d by d t is written as two terms. Now, the advantage of doing this is that when you look at a single reaction alpha 1 1 a 1 plus alpha 1 2 equal to alpha 1 n a n equal to 0. We recognize that the rated which species a 1 reacts divided by the stoichiometric coefficient rated species 2 reacts divided by stoichiometric coefficients equal to rated which species n reacts divided by stoichiometric coefficient. So, all these equality equal to r 1 which we denote as the intensive rate of that reaction. So, if I ask you what is the rate at which component a 1 is getting formed you will say that it is r 1 multiplied by alpha 1 1 what is the rate at which component a n is reacting you will say it is r 1 multiplied by alpha 1 n. So, that is the nomenclature which you are doing. Therefore, our material balance in this third vessel for component i will look as input output generation what is the generation term for component i r 1 times alpha 1 i because multiplied equal to d by d t of n i. So, our statement of material balance for component i now our interest is what our interest is to see whether we can generate this term h i d n i by d t because you can generate it from this by multiplying by h i and summing over all species. What is meant by multiplying by h i summing over all species what is meant is we write one equation f 1 0 f 1 minus f 1 r 1 alpha 1 1 v equal to d n 1 d t that is one equation. Similarly, there are n equations. So, each of those equations you multiply by the appropriate species enthalpy and sum over all species. So, that is what when you multiply throughout and sum over all species what you find is this. That means, if you multiply throughout by h i and sum over all species what you get is sigma h i f i 0 please note here. I am just multiplying by h i and summing over all species correct multiplying by h i summing over all species. Then what I will get h i f i 0 h i f i and then the last term. So, what I have done here is I have multiplied this by h i therefore, it becomes h i f i 0 h i f i and then I multiply this term by h i and summed over all species therefore, I get r 1 alpha 1 i h i v equal to h i d n i by d t and summed over all species. So, what I have done now. So, now we can simplify what do you do now now we have this equation here. So, we have this equation where h i d n i by d t we have generated here we have generated. So, this we have generated using material balance. So, I can substitute for this term from this equation 4 is that clear. So, we have generated this h i d n i by d t by looking at the material balance. Therefore, in this equation 3 you can replace h i d n i by d t by the left hand side of this equation. So, when you do that when you do that what you find is this. So, I put all the terms are written here you find left hand side right hand side all the terms you find it sort of some terms knock off. And we recognize that this term r 1 times alpha 1 i h i v is essentially r 1 times alpha 1 1 h 1 alpha 1 2 h 2 times alpha h n h n which means what this is the enthalpy change for reaction 1. So, this term becomes enthalpy change for reaction 1 therefore, this whole term becomes n 1. So, how the energy balance looks like now is it is n i d h i by d t the first term second term h i f i 0 h i f i 0 h i not f i. So, you can combine these two terms you get h i. So, what I get here is you can see here the common thing here is what I have a mister term here f i 0 should be sorry. So, it simplifies as it simplifies as n i d h i by d t and f i not f i not from here can I see both h i minus h i not. And this term becomes r 1 times delta h 1 star which is the enthalpy change for reaction plus q minus w s. So, now it is in a form which is a little easier for us to make use of even for example, suppose there is a reactor in which many streams are entering many streams are entering. And those streams might be entering at different temperatures different quantities all those effects can be appropriately accounted because each stream coming at different temperatures coming at different enthalpies. So, the form in which this equation is written helps you to manage various inputs into a reactor various temperature all those kinds of things can be handled. For a special case we can do lot of simpler things. So, this term r 1 times v times delta h 1 star this is what this is for a single reaction. You can write this r 1 in terms of what is called as r 1 times alpha 1 k and divide this by alpha 1 k what does this mean? This is rate of formation of component k multiplied by the enthalpy change express with respect to component k this how data is often available. So, what is important is to recognize this term and put it in forms that you and I can understand r 1 times alpha 1 k refers to rate of formation of component k. This is often readily available because we measure in our experiments we will be measuring this term. So, then in that case you should multiplied by the rate of the heat of formation sorry enthalpy change for the reaction express with respect to that component that is important then only it will be consistent. So, in this form the way in which this equation is written it is probably the most general form in which you can write a chemical reactor energy balance. It takes into account all sorts of variations that might occur. Having said that having said that we recognize that this term h i can be put in terms of temperatures because the enthalpy are available to us in terms of a standard heat of formation plus various c p delta t which takes into account the effect of the specific heats. So, what we have said is that if there is no phase change then clearly we do not have to account for phase change solid to liquid liquid to gas gas all those things is not required. We simply can look at enthalpy at temperature t is standard heat of formation plus an integral which goes which is the average specific heat in that interval between t and t r. So, we can express h i at t like this h i at t 0 also like this. So, this is an assumption what is the c p this c p is the average in this range this is the average. So, every time we look at look at this we look at the average specific heat in the range t to t r in t to t 0. So, expressing our looking back if our equation here where is our equation. So, this is our equation n i d h i. So, now I can replace h i in terms of a temperature and specific heat therefore, this whole equation n i d h i by d t simplifies as something like this. So, first term is n i d h i with this what we have written here let me just go through this is n i which is written as sorry v i sorry this is v i this is v i. So, n i is v i time c p i correct is this clear please understand please understand n i d h i by d t n i is what v i v i time c i c i is missing correct. I have written it here I have written it properly here. So, the first term let it stay like that let it stay like that this n i let us know harm in that. So, n i c p this is written in this form v time c p i c p means specific heat d t d t similarly, the right hand side. So, what have we done now? So, we have what we have done is that this each of this terms h i and h i naught and h i we have put in terms of specific heats. Therefore, the right hand side now looks in terms of v i c p i this term c i c p i the only term that has to be summed over all species is v i c p i on the right hand side the only term you have to sum is c i 0 c p i. On other words what we have done is that by expressing the enthalpies in terms of specific heat even those if you look at this h i naught and h i h i naught is known h i changes depending upon the extent of reaction. Similarly, h i changes because of extent of reaction, but by this formulation of writing in terms of volumetric specific heat what we have done is that c i c p i c i changes, but c i c p i for a mixture does not change very much. On other words a rather difficult problem we have converted into a simpler problem by combining c i c p i. So, that the left hand side now simply looks like a volumetric specific heat multiplied by the reactor volume d t d t. Why it has been possible because we have written this sigma c i c p i as a volumetric specific heat which does not change during reaction this is the assumption and it is not a bad assumption. You can do calculations for several systems you will find that sigma c i c p i for all the species taken together for the reaction may not have changed very much. Similarly, sigma c i 0 c p i summed over all species also would not have changed very much. So, it simplifies otherwise you know messy kind of system the whole equation looks in this form v c p d t d t is v 0 c p t 0 minus t. The assumption here is that all the feed stream is coming at t 0. See in the previous formulation that assumption was not there, but now we have said all the streams are coming at t 0. So, this formulation is not useful when you have many streams coming in different temperatures, but this formulation is very useful. So, what is important for you is to recognize that we must use the appropriate formulation for our application. So, we have this statement for a single reaction saying that r 1 v this is heat generation if it is exothermic reaction or heat absorption if it is endothermic reaction heat added work output. Typically, this may not be important in many cases because we are not trying to you know most cases, but it may be important in some and appropriately you will have to delete terms that are relevant to your application. Now, if we go from a single reaction to a multiple reaction. So, what do we do? We had only one reaction here correct. Now, we have many reactions and other words how do you go from a single reaction to multiple reaction. So, what we have done for that case is that the left hand side we are there was one term on the left hand side of the sigma n i d h i by d t h i d n i by d t which we did some manipulation we can do the same here. How do we do the same here? We recognize that we have n we have p reactions alpha 1 1 a 1 to alpha 1 and similarly alpha p 1 a 1 to alpha p 2. There are p such reactions which are taking place in our stirred vessels. So, what we want to do is that we want to generate what is this term h i d n i by d t for the case of multiple reactions. So, how do you do that? We do that by writing the energy by the material balance for the i th species input output the generation is because of r 1 alpha 1 i r 2 alpha 2 i r p alpha p i. Notice that all the rate processes appear in the generation term multiplied by v equal to 0. So, this is the statement of energy by material balance and we want this term h i d n i by d t which means what we have to multiply this by h i and some over all species. So, let me say it once again what we are saying here is that when we have a multiple reaction this term h i d n i by d t we want to replace from energy of material balance and we do that by writing the stoichiometry writing the energy material balance for species i which is what f i naught minus f i plus the generation from all the p species r 1 alpha 1 i r 2 alpha 2 i r p alpha p i. So, now we can multiply by h i and some over all species. Once you do that we generate this term that is what. So, when you do that and it sort of simplifies nicely. So, we need not go through this, but when you do that the whole thing simplifies like this that this equation this that means this which means this term for a multiple reaction when a substitute it looks like this. What we have done is that the second term h i d n i by d t which is substituted from material balance and then the whole equation looks something like this. So, only change that has taken place is that wherever this reaction term occurs in the reaction term previously there was one reaction now there are p reactions. So, this term now becomes j equal to 1 to p r j delta h j star v. So, this term the heat generation heat absorption previously it was because of only one reaction here only one reaction correct and now we have p reactions. Therefore, this term becomes r j alpha j that is what it becomes. So, it becomes r j multiplied by the heat of enthalpy formation enthalpy change for formation of the j th reaction j going from 1 to p. So, the energy balance for a single reaction a multiple reaction it does not really matter because the form is identical the form is identical. So, now we can replace everything in terms of temperature when you do that if all the feed stream if all the feed stream is at t naught this is how the equation looks like. Why is the summation over i equal to 1 to n this is now this term alpha 1 i h i we have replaced it as delta h i star h 1 star. So, this is our single reaction this term this is important thing is that this term is actually delta h 1 star for single reaction. Now, when there are many reactions then it will become for j th reaction it will become delta h j star r j is the the intensive rate of formation of intensive rate of that reaction j. So, r j j going from 1 to p there are p processes rate processes is that clear. So, all the feed is coming at same temperature this is the assumption, but in this form it is not so bad because that effect is there. So, here in this form you can deal with different components coming at different temperatures, but from this point this point all the components are coming at same temperature. So, which whether use this or use this depends upon what is the application you are looking at that you should take care whether it is single reaction multiple reaction when there is one reaction then you j going from 1 to 1. So, this this form equation 4 is a very general statement of energy balance for a stirred vessel where v c p which is volumetric specific heat rate of change of temperature of the system equal to v naught c p t naught minus of t is a flow term is the reaction term heat addition and work. So, this is how it looks if you if I ask you for example, suppose we have an instance of a multiple reaction taking place in a let us say a stirred vessel where the temperature is maintained constant. It is still it is still as in the sense what I am saying is that there is an instance let us say you have an equipment which is into which fields are coming, but nothing is going out what are we saying v naught is finite, but there is nothing going out. Now, this the way this equation is return the v naught is finite, but nothing is going out. So, people tend to think that you know this second term should not be there, but second term is coming because of h i what is h i it is a enthalpy of the stream in the equipment you understand. So, we have typically we think the second term should be should be removed, but no this is the important point and I have seen many instances where people knock out the second term this is this is within the h i is what h i is the enthalpy of stream in the equipment. So, see it is multiplied by v naught v naught is input not output is that clear. So, you should be very careful you know in a in a given exercise the equations how they play out in your application you have to be very careful. Otherwise you might knock out terms which are actually not appropriate to be knocked out and I thought we did some exercises where I pointed this out to you also in one exercise some of you said that no no no we should knock it out. And then we look back in the equation I realize that no we cannot knock it out that is in semi batch you will find this kind of problems. Now, let us look at the same problem for the case of a plug flow vessel we looked at stirred vessel we look at plug flow vessels. What is a plug flow vessel? A plug flow vessel is one in which there is no mechanical stirring. So, we have here a vessel through into which fluids are coming in fluid is coming in fluid is going out and our material balance is for this differential element. So, what is the statement of energy as input of energy minus the output of energy plus q which is the heat input per unit volume. Therefore, I have multiplied small q by delta v and then W s times this is the heat work output per unit volume. So, input output generation there is no generation this is heat input work out equal to 0 for steady state we are writing the material balance for steady state we have not had time to look at the steady state here. So, in the limit as delta v tends to 0 the general material energy balance for a plug flow vessel looks like this. This is the so d by d v of f i h i becomes q minus W s. Now, d by d v of h i f i is what we have to deal with and put it in appropriate forms. So, that we can see what happens to temperature of the equipment for that what we have done we know that we can expand this is f i d h i d v and h i d f i d v and our material balance equation tells us what is d f i d v. So, if you look at d f i d v how does it look d f i d v due to is is equal to rate of generation of component i what is rate of generation of component i r 1 alpha 1 i up to r p alpha p i. So, d f i d v in terms of all the species therefore, when you multiply by h i and sum over all species h i d f i d v becomes sigma r 1 alpha 1 i h i what is this we know this is heat of enthalpy change for reaction 1 alpha 1 i h i alpha 2 i h i is enthalpy change for reaction 2 alpha r p i h i is enthalpy change for reaction p. Therefore, this term this whole term h h i d f i by d v what we have done is that this if you recall here here you have this is f i d h i d v h i d f i d v therefore, the term h i d f i d v you can now replace from here correct. So, what I have done is. So, this equation here this this equation I have written it as f i d h i d v plus h i d f i d v this is what I have written I am written it again here. So, our equation 1 I have written it by expanding the term f i d h i d v. So, I have written it in this form two terms are appearing this h i d f i d v can be substituted from here which is what r 1 delta h 1 star r 2 delta h 1 star r p delta h p star. Therefore, this term becomes r 1 delta h 1 star r 2 delta h 2 star similarly, up to r p delta h p star. Therefore, our energy balance now becomes f i d h i d v plus all this r 1 delta h 1 r 2 r p equal to q minus w s. So, so this if you take into the other side it becomes this f i d h i d v putting in terms of our volumetric specific heat becomes volumetric flow times volumetric specific heat that sigma c i c p i d t d v this term is simply summation over j equal to 1 to p there are p rate processes here plus I have taken it to the other side. Therefore, I have put a negative sign this is taken to the other side this is negative sign plus q minus w s. So, for a plug flow s l where there is no stirring we find that v c p d t d v which is the accumulation per unit volume this is v c p d t d v this is rate of change of temperature of the equipment and this is the rate of heat generation if it is an exothermic reaction this is heat input this work output. The only thing the difference between a stirred vessel and a plug flow vessel here q is heat input per unit volume it is per unit volume that is the difference it is not heat input per unit time. In the previous case it was heat input per unit time. So, that is the difference that we should take care all right. Now, let us see what simplifications we can do this term q heat input per unit volume assuming that our equipment is a pipe or a tube. So, tube surface area see the volume of a I have just calculated here our heat transfer is h heat transfer coefficient surface area multiplied by the driving force. Therefore, if you have a unit volume here suppose this is how the heat is coming in heat is coming in like this coming like like this this heat per unit volume I have written in where a h is heat transfer surface per unit volume. What is a h it is heat transfer surface per unit volume for a pipe what is it equal to surface area divided by volume of the pipe which is 4 by d or in other words this a h for a pipe is 4 by d. Therefore, small q for a pipe is actually 4 h by d t c minus of c understand. Therefore, a energy balance for a pipe looks v c p d t d v summation over all p rate processes plus 4 h by d times minus w s this may turn out to be typically 0 it may not be I am says it may turn out to be may not be. But, what are we saying now see we are saying that in a plug flow vessel the rate of change of temperature with volume is heat if it is a heat generation and this is per unit volume per unit time you can see this per unit volume per unit time plus what is the heat that is removed by see what is this t c t c minus of t this is the heat that is added to the system typically we will find this term is negative because you will remove heat. Therefore, t c minus t will be negative so you will find this will be positive. So, this is a positive term typically for an exothermic reaction this is turns out to be a negative term therefore, this positive and this negative when they balance each other then it had to be the hot spot or the point of maxima in the profile temperature profile. The point of maxima in the temperature profile occurs at a point where the heat removal becomes equal to the heat generation then this becomes 0. So, this is negative this is positive therefore, this becomes 0 the point at which where it becomes 0 is the point of maxima. Now, the interesting thing that is of relevance to us is generally we like to see that the temperature at which the maxima occurs is within the allowable limits of the material of construction in more importantly allowable limits of the catalyst that you may be using. Most of the cases it is that catalyst which is crucial to ensuring that this temperature never goes beyond what is permissible. So, what you would do is that you would like to see that this when it goes to 0 what is the temperature at which this equality take place and what is the temperature for a given choice of pipe size for a given choice of size of the pipe you will see what is this temperature whether that temperature is good enough for the catalyst to be acceptable. Now, what you will do is that frequently when this when this pipe is large you will find that this temperature becomes unacceptable. So, you will find that the choice of pipe size or the size that you will choose will be determined by the hotspot temperature your system will permit. So, this calculation you will do to find out what is the best temperature best choice of pipe size that you can use when you use a small pipe because it occurs in the denominator the heat transfer is very large. So, it is able to dissipate heat very fast when you use a small pipe when you use a large pipe it is not able to dissipate heat very fast and therefore, the hotspot temperature goes up you understand. So, in this when you solve this equation these are the points that is a greatly important to us what is the choice of D. So, that this temperature is what you would like to see or in other words when you do your numerical calculation what you will do you will choose different pipe sizes and you will generate this hotspot profile and see what is the profile that is acceptable to you and you will choose that profile which is acceptable to you or that pipe size which is acceptable to you this is clear. So, this is the most important point as far as the design of tubular reactor is a concern you will find in many cases we would like to operate it adiabatically. That means this q is this is 0 there are many instances particularly if you go to ammonia plant or sulphur dioxide plant it is an adiabatic operation. Why is adiabatic operation? That means we set this equal to 0 that means you allow this temperature to rise you allow this temperature to rise, but you do not allow it to rise beyond a point. On other words you have very short bed the bed is very short allow the temperature to rise and you cool you understand. So, that is the idea of you allow the temperature to rise you cool allow the temperature to rise you cool allow the temperature to rise you cool. So, that you are able to reach the extent to which you want to reach. So, all those strategies are built into this equation now, because you will write this equation for every reactor. There are 7 reactors, we will write it for every reactor. So, all those are integrated into this statement of energy balance. Now, there are people who talk about what is called as radial flow reactors. What is radial flow? You can have flow outwards or inwards. So, now you will find the way this is formulated, the way this is formulated, you can take into account radial flow quite comfortably, but here you should change this, because this is for a pipe, this we have written for a pipe. So, for a radial flow reactor you will change this, this A H you will change for a radial flow. So, what I am trying to point across is that, see whatever simplification that we make, it is under the assumption that we are using a pipe, but moment it is not a pipe, then the left hand side is formulated properly, but the right hand side is not satisfactory. So, you should formulate this 4 H by D, which is for a pipe for a radial flow reactor, then it becomes appropriate for a radial flow reactor. So, what in other words, what you are trying to point out here is that, the equations are written in a fairly general form, so that you can use it for the design that is appropriate to your application. What is important is to replace those terms, which are not appropriate, so that it applies to your situation appropriately, that clear what we are saying. So, whether it is a stirred vessel, we had in a stirred vessel, we had generation of energy term, we had heat transfer term, we had work term and there we did not have this kind of problems, but here in tubular reactor, because there is no mixing, we have these new features, for which you have to appropriate choice of pipe sizes are required. And much of tubular reactor design that you will see in the process industry, a great deal of effort is spent in trying to make a good choice of the size of the pipe, it is important.