 So, we look at practice problem in energy balance in this lecture. Let me first describe the problem to you, so that we understand what the situation is. We have here a stirred tank and it has a cooling coil. The reaction A going to B and it is quite reversible, so this is the kind of reaction taking place. Feed V naught and then composition C A naught coolant temperature T c then feed temperature T naught. These are some data that is given is as follows C A naught equal to 1.66 k mole per cubic meter T naught is 21 c V naught 0.6 cubic meter per hour. Activation energy for reaction 1 this is 1 and this is 2 is 25 thousand k cal per k mole. Coolant temperature T c is 10 c and heat transfer coefficient this is between coolant and reactance that is 1000 k cal per square meter per hour c. So, this is the data that we have. Now, some more data we would require because the reaction is reversible therefore, we need some equilibrium data which is also available which I will. So, we have T k 1 and capital K equilibrium constant 293, 303, 315, 323, 1.06, 4.37, 21.357, 57.2, 21.6, 6.96, 1.97, 0.89. So, this data on reaction velocity constant and equilibrium constant are given. The question in front of us is the following. Now, that we this is where is heat of reaction heat of reaction I have forgotten and just heat of reaction this delta H heat of reaction is minus 20,000 a going to be this is for a going to be heat of reaction calories k cal per k mole. So, this is a reversible reaction and plus it is exothermic. So, we know that if you have a reversible reaction which is exothermic it has some interesting features showing that it as a maximum reaction rate and so on which we have already you know discussed. So, let us look at first part of the exercises one is derive locus of max rates. What is the locus of max rates? How do you do this? So, suppose I say r b is the rate of formation this is our reaction is a going to be a goes to be and this way this one and this is 2. So, r b is what k 1 c a minus k 2 c b. Now, since there is no volume change I write this is c a naught times 1 minus of x minus k 2 c b naught plus c a naught x and we take this as 0. So, essentially is k 1 c a 0 minus of x k 2 c a 0 x. Now, locus of maximum rate is understood like this del x del r b del x at constant t this is what we want to find out. That means del r del x at constant t. So, del r del r del t at constant sorry sorry this is or we want to know how the rate of formation of component b changes with temperature. Let us let us do that we just go through that very elementary manipulations. So, we have r b equal to k 1 c a 0 times 1 minus of x minus of k 2 c a 0 x we want to find out del r b del t at constant x. So, we have to differentiate k 1 with respect to t that is k 1 e 1 by r t square at constant x. Therefore, we do not change this similarly, we differentiate k 2 with respect to temperature we get k 2 e 2 divided by r t square c a 0 times x. Now, we have shown that an exothermic reversible reaction r b goes through a maxima we have shown that earlier. And therefore, at the point of maxima we should have this I will call this x as x m. So, 1 minus of x m divided by 1 minus of x m multiplied by k 1 e 1 by r t square c a 0 equal to k 2 e 2 by r t square c a 0 times x m or x m divided by 1 minus of x m equal to x m divided by x is equal to k 1 by k 2 e 1 by e 2 or some equilibrium constant times e 1 by e 2. This is something that we all know based on the elementary principles of differentiation. Now, suppose you look at the same reaction at equilibrium same reaction at equilibrium which is you have r b equal to k 1 c a 0 1 minus of x minus of k 2 c a 0 times x at equilibrium you have at equilibrium equilibrium we have r b equal to 0. Therefore, we should have k 1 c a 0 1 minus of x e equal to k 2 c a 0 x e or x e divided by 1 minus of x e equal to k 1 divided by k 2 or also means x e equal to if I call this as k you know equal to k x e equal to k by k plus 1. At equilibrium we have the equilibrium conversions given by k by k plus 1. Now, we also said that x m by 1 minus of x is equal to therefore, this also means x m equal to if I call this term as delta if I call this as delta this becomes that means this is delta. So, becomes k delta divided by k delta plus 1 is that clear. So, if we want to make a plot of if you want to make a plot of what happens what happens to x what happens to x for different t. On other words x e equal to k by k plus 1 or it is also equal to 1 by 1 by 1 plus k. Now, for a reaction which is exothermic for an exothermic reaction we know that k decreases as t increases we know that this is a familiar leach act as principle. Therefore, we have as k decreases as t increases. So, when k decreases what happens to this denominator this increases correct. So, as t increases you know that x keeps on decreasing. Therefore, the equilibrium if this is the equilibrium line I hope you understand this it will go through this once again as t increases k decreases for an exothermic reaction we are talking about exothermic reaction exothermic reactions for exothermic reactions as t increases k increases. When k sorry as t increases k decreases therefore, this quantity keeps on increasing. So, 1 plus the denominator keeps on increasing. Therefore, x e keeps on decreasing as t increases that is what I have drawn. Similarly, you have this x m which is given by k delta divided by 1 plus k delta. And this also can be written as 1 by 1 plus 1 by k delta. Therefore, the behavior of x m also will look something like this this is x m what is x m x m by definition is the locus this is the locus of locus of max rates. And what max are we talking about del r b by del t at constant x. So, this is this is the locus we are talking about. Therefore, in our design by and large we would be interested in ensuring that the reaction rates that we attain in the equipment is as high as possible. What is the rational for this I mean if the reaction rates are very large then clearly the equipment that is required to do that processing would be the smallest. And therefore, it sort of makes sense to have your process operating at the highest reaction rate. So, what we are saying now is that this is the first part of the question which says the question let me state the question once again question is this we call the first page where we did all this. We said that we have here we want this x say specify best conditions required to get a conversion of x equal to 0.52. So, the problem that we would like to address now is what are the best conditions that is required to get an outlet conversion x from reactor 1 equal to 0.52. So, this is the question now how do we address this we address this by recognizing. So, let us just state that question once again we have we have a stirred tank we have a coil this is a coil coming in and going out fluids coming in fluid is going out. So, we want this x to be 0.52 number 1 number 2 we also want that the reaction rates chosen here be the highest or the size of the equipment for this choice should be the smallest. That means, we want to get the highest reaction rate or the optimum conditions optimum conditions to achieve x equal to 0.52. Now, we said x m divided by 1 minus of x m equal to 1 by 1 plus k delta. That means, the choice of the conditions should be such that the this condition be satisfied. Now, if I put x m as 0.52 by 1 minus of 0.4 0.52 equal to 1 by 1 plus 1 by k delta essentially this equation defines what should be the value of k capital K or equilibrium constant at which I should operate the process. Now, you can solve this and then find the value of capital K that gives you the that satisfies this inequality. Now, to be able to do that to be able to do that what is it that we require to be able to do that delta the way we have defined delta is even by e 2 which means that if this is the reactance these are reactance these are products. Now, this is given as heat of reaction heat of reaction is given as how much is the heat of reaction heat of reaction is given as 20,000. So, this is 20,000 or 20,000 calories sorry 20,000 calories per mole it is given. What is this? This is e 1 and this is this is e 2. Therefore, e 1 minus of e 2 equal to delta h this is given as this is equal to 20,000 this is what is given I am sorry what I wrote not write e 1 minus e 2 and e 1 is given as e 1 in the problem statement is given as 25,000 minus of e 2 equal to this exothermic heat of reaction minus 20,000. Therefore, e 2 equal to 45,000 calories per mole. What are we saying? What are we saying is that in the problem that is specified in front of us the activation energy for the forward reaction is given as 20,000 and the heat of reaction is also given as 20,000 but exothermic. So, I put all those conditions here and we find e 2 which turns out to be 45,000. So, we have in this reaction that e 1 is 20,000 e 2 is 45,000 therefore, delta. So, what is delta equal to 20 25,000 divided by 45,000. So, this is the heat of reaction which is 5 by 9 correct that is equal to 0.5 0.55. So, that is delta value. So, we recognize that to be able to. So, what we are saying is what we are saying is the following that our process runs in such should run in such a way that the exit conversion should be 0.52 for which you have to appropriately take care of the heat load using this device the heat removal system. So, what we are saying is that that if k delta. So, we have x m which is equal to 0.52 equal to equal to 1 k delta divided by 1 plus k delta. Therefore, this equation defines what is the value of k delta what is the value of k delta can we calculate. I have calculated that and found k delta values to be. So, if you put all the numbers you get 0.52 divided by 1 minus of 0.52 equal to k e we do not know this and then 25,000 e 1 45,000 e 2. So, this gives you a k value of 1.97 this is what we get. So, that means if you choose a temperature choose. So, choose temp T of reactor reactor so that equilibrium constant k becomes equal to 1.97. Now, how do we do this? So, we have to look at the data that we have after all this data available to us it is given to us. It is given our data for this reaction is already given corresponding to 1.97 we find that the temperature of the reactor must be chosen as 315 degree k this is degree k. So, what is it that we have done? So, when implies T equal to 315 k. So, what are we saying? What we are saying is that we must operate the process in such a way we must operate this process which means to be able to operate the process corresponding to the maximum reaction rate. That means you want this 0.5 this x equal to 0.52 you want yes, but you want this in such a way that this process should run here at the highest reaction rate. That means the conditions to be chosen should be on the locus of maximum reaction rate which means what we are saying is the following. For x m equal to 0.52 we have shown that this capital K must take a value of 1.97 which means the temperature at which this equipment this reactor must be operated the temperature must at which this must be operated is T equal to 315. So, it is corresponding to if this is 315 then we can get 0.52 which means if 0.52 is the conversion at which you must operate then you have to choose temperature to be 315. So, as to satisfy the criteria that x m equal to k delta by 1 plus k delta ok. So, if that means if you want to choose a temperature of T equal to 315 how do we get this we get this by recognizing that this you can get 315 if we satisfy the energy balance which means what the whole procedure says is that the maximum reaction rate locus defines temperature and the energy balance defines the heat load to be handled. So, as to achieve this kind of operation. So, having said that let us see how we can get forward. So, now the question is that we have this reactor now to which feed is coming in it is it is you know and this x is given as 0.52 and we have said for this x to be then the T should be 315. So, what are the unknown quantities now we know we not, but we do not know the residence time here. So, residence time tau is unknown that is 1 and similarly the amount of heat to be removed q is also unknown. Therefore, once we specify tau and q the problem is fully solved how do we do that to get this we let us look at the material balance what is material balance tell us material balance tells us that input minus of output plus generation equal to accumulation and at steady state accumulation and at steady state accumulation is nil. So, what is f a 0 and f a. So, we can put all these numbers here f a 0 minus of f a is simply f a 0 times 1 minus of x and what is r a r a by definition is k 1 c a minus of k 2 c b times v equal to 0. So, we can simplify this recognizing that c a equal to c a 0 times 1 minus of x and c b equal to c b 0 plus c a 0 x and then c b 0 is 0. So, putting all the simplifications we can get this is. So, f a 0 this becomes f a 0 x first term minus of k 1 c a 0 times 1 minus of x minus of k 2 c a 0 x times v equal to 0 we divide throughout becomes x minus 1 minus of x times k 1 tau minus of k 2 equal to 0. So, we can simplify this as it be right here we get x equal to x multiplied by 1 plus k 1 tau by 1 plus k 2 tau equal to k 1 tau it comes directly from here I am this collecting the coefficients of x. So, x multiplied by 1 plus k 1 tau plus is it right it is 1 minus of x 1 minus of x is it all right x just a minute minus is plus. So, it is. So, we get x multiplied by 1 k 1 tau is equal to k 1 tau. So, this is taken to as a plus k 1 tau it is fine. So, this gives you x equal to k 1 tau divided by 1 plus k 1 tau plus k 2 tau now k 1 at 3 1 5 and k 2 at 3 1 5 k we can get from the data that is already given. Therefore, on the in this equation only unknown is tau because k 1 at this temperature 3 1 5 and k 2 at this temperature 3 1 5 is given in the table. Therefore, you can calculate in the if you given x you can calculate tau. So, putting numbers x is 0.52 and k 1 and k 2 at k 1 at 3 1 5 and k 2 at 3 1 5 these numbers I have calculated and then I will tell you this numbers k 1 at 3 1 5 and k 2 at 3 1 5 k 1 at 3 1 5 is 2.34 2.34. k 2 at 3 1 5 is 10.08. So, I have calculated this numbers units are per hour all these are per hour all the data is per hour I hope let us just see k 1 k 1 is in per hour. So, all per hour everything is per hour. So, you can put this numbers. So, what we get is that k 1 is 2.34 tau divided by 1 plus 2.34 tau plus 10.8 tau. So, if you solve this we get tau equal to 0.11 per hour. So, this is material balance what we have done see what we have done is we have this is our problem. Our problem is we have a start tank where this exit conversion is specified. It is also said we must operate along the locus of maximum reaction rates which means that we should have x m by 1 minus x m satisfying this relationship. So, putting the value of x m which is specified as 0.52 we can get k delta delta is also known because delta is even and e 2 are known. Therefore, k 1 therefore, all the numbers are known and on that basis from the material balance which is given by this equation we can find out the residence time provided others are known in this case others are known that gives you residence time of 0.11 hour. So, our problem now is that the residence time tau is known in this tau is tau is equal to this is known. So, flow rates are known all these are known x is known here. So, only unknown quantity now is how much heat to be added or removed to maintain the temperature here as 315 degree c 315 degree k. So, how do you do this now we look at the energy balance and see what the energy balance says as far as maintaining this kind of temperature. So, what is the energy balance saying for a start energy balance for a start tank because this is a start tank. So, we write the equation that we know d t d t equal to v naught c p times t naught minus of t plus r 1 minus of r 2 times minus of delta h 1 star times v plus q minus of w s. So, in this case steady state therefore, steady state therefore, this goes to 0. So, r 1 minus of r 2. So, everything is known here accepting q. So, if you put all our numbers let us do those numbers. So, it is 0 equal to 0.6 this is cubic meters per hour c p is 1000 and this is 40 minus of 42 t naught is given as 40 is that right t naught is given as 40 and then r 1 minus of r 2 r 1 minus of r 2 this from our material balance we know it is simply f a 0 times x I will not show this now. So, it is simply f a 0 f a 0 is what c 0 which is 0.6 c 0 is 0. So, this is simply f a 0 times x times minus delta h 1 star times v plus q. Now, what is v reactor volume reactor volume is equal to volumetric flow rate times residence time volumetric flow rate is given as 0.6 and then residence time we calculated as 0.1. So, this becomes 0.6 cubic meters per hour is that clear. So, we have got volume reactor volume to be 0.6 cubic meters not to not per hour 0.6 cubic meters is 0.6 0.06 this is equal to 0.06 cubic meters. So, we can put this in here and then the only unknown here is q we can put all the numbers I have put most of the numbers. So, let us say this is 0.6 times 1.66 times 0.52 times plus 20,000 plus 20,000 and then 0.52 0.52 is also to be put here. So, essentially what we are saying is that r 1 minus r 2 times v is f a 0 x therefore, this is not there it is plus q. So, this gives us value of q equal to minus of 3158 k cal per hour and what is minus sign mean in the first law convention minus sign means the heat is going out of the system. Heat input is taken as positive in first law the heat output is therefore, the negative sign means so much of heat will have to be removed. So, as to be able to maintain the temperature of 315 degree C. So, the most important thing here is that we have been able to specify what is the temperature at which we must operate. So, that we get conditions corresponding to the maximum reaction rates having done this there are few simple calculations we can do after all you know the design will require you to specify things properly. So, if q is equal to minus of 3158 kilo cal per hour. So, what is the surface area heat transfer area we have to give. So, heat transfer area is what h a times delta t equal to q q is given therefore, a equal to q divided by h delta t correct q is 3158 h is 1000 what is delta t our delta t is temperature is 315 which is 270 and then 315 means 42. So, it is delta t is 42 minus of 10 that is equal to 0.26 square meters this is given this we have found out as 315 k and this temperature is given as 10 C. So, that is how it becomes what it is. So, we have been able to calculate what is the heat load and what is the heat transfer area. Now, the second related question in all this is to be able to see whether the process that we are running will be stable or unstable. On other words we must check for the stability of our process for various kinds of disturbances. So, that is what we would like to do now check how stable is our steady state. So, to be able to do this what is called as steady state stability what we try to do is we try to plot we try to plot what is called as u g and u r what is u g and u r u g is in generation. And u r is heat removal and our steady state our steady state is about the equality of heat generation heat removal when they are equal we know that the process it is at steady state. We can plot this curves u r just write down what is u r and u s something we have done before, but it is not a very complicated thing anyway. So, u r u r s u g and s is r 1 s minus of r 2 s multiplied by tau and j 1. And then it is we can show this is no point in doing it once again we know that r 1 s minus of r 2 s is equal to c a 0 x s divided by j. I will not show this because you already done this before therefore, u g s equal to c a 0 x s times j which is equal to c a 0 k 1 s tau tau divided by j 1 1 plus k 1 s tau plus k 2 s tau. See these things we have derived before therefore, I am not doing it again notice here that if you want to plot u g s all that you need to know is what is the temperature corresponding to this x s or in other words you want to plot u g s if on the right hand side k 1 s is the function of temperature. And k 1 s we know k 1 is what some k 1 0 e to the power of minus of e 1 by r t and therefore, since the right hand side is fully known k 1 is known therefore, you can plot u g as a function of temperature. Similarly, what is u r or u r by the f u r s equal to 1 plus beta we have done all this thing therefore, I am not doing it again t s minus of t c star where t c star equal to t naught plus beta t c divided by 1 plus beta. So, all these we have done where beta is equal to h a by v naught c p all this is known. So, what we are saying is that we can now plot u g versus temperature we can now plot u r versus temperature therefore, the points at which the intersect are the points at which steady state occurs. So, this is what we are trying to say and you can plot this how do you plot this and this show you how to plot this they are not difficult to do, but I just do some calculations to illustrate how this plotting is to be done see let us say t and k 1 say k 2. So, k 1 tau k 2 tau x s u g s and u r s what is u g s c s 0 x s j 1 what is u r s. u r s 1 plus beta times t s minus of t c star and all these are known. So, I will just put some say 293 I will just 2 1 or 2 293 3 naught 3 3 1 5 and 3 2 3 I am just calculating some numbers. See we put all these calculations you can find at 3 naught 2 9 3 this turns out to be 3.32 minus 15.6 and then that is 293 3 naught 3 is 9.96 minus 1.29 then this is 17.2 comes out to be 15.9 and then 26.1 and 41.6. In other words what we are saying is that you can now plot see this data here you can see here this data got this data here u g s you can plot u g s versus temperature and then u r s you can plot u g s versus temperature this curve looks something like this and then u r s versus temperature like this and this is the point of intersection which is 3 1 5. So, what we have try to say here is the following if you have a stirred tank in which an exothermic reaction is taking place and then you have to operate this process at the maximum reaction rate which means it specifies the temperature at which you should operate and so on. Now you can also understand this steady state by plotting u g s this is u g s or u r s this is u r s or in other words the point at which the intersection takes place is the point of steady state. So, we can understand steady state in various ways this is another way of understanding steady state our interest now is now that we know it is there is one point of intersection how stable is this steady state is it stable is it unstable. Now to understand stability of steady states we have done this already I will just quickly run through this procedure for understanding stability of steady state. We have shown in an earlier discussion that an exothermic stirred tank stability can be understood in terms of 3 numbers l m and n where l is defined as 1 minus tau by c a 0 del by del x of r 1 minus of r 2 m is equal to 1 plus beta where beta is h a by v naught c p and then n is equal to j 1 tau within brackets del r 1 minus of r 2 divided by del t at steady state. And stability is l plus m greater than n l m greater than n. So, steady state is stable steady state is stable if these 2 criteria are satisfied. So, what is l m and n these are all numbers which is determined by what is given here where residence time here is known c a 0 is known r 1 minus of r 2 at s all these are known. So, for a given reaction kinetics l m and n is fully specified. So, what we want to do now is to illustrate to you how we can actually calculate l m and n for the specific problem that we have in front of us I want to calculate this in front of you. So, let me go through this whole thing quickly. So, r 1 minus of r 2 for example, for us to calculate l m and n we should know r 1 minus of r 2. So, that we can differentiate with respect to x and t and find out the value of these numbers at steady state that is what I am trying to do now r 1 minus of r 2 is k 1 c a 0 1 minus of x minus of k 2 c a 0 x. So, del r 1 minus of del x equal to k 1 n s with the minus sign minus k 2 k 2 s multiplied by c a 0 it is obvious from here. So, that is equal to minus of k 1 s plus k 2 s times c a 0. Therefore, what is l l we have to calculate l here we have to calculate l because tau is known everything else is known. So, let us calculate l. So, l is equal to 1 plus tau by c a 0 within brackets of k 1 s plus k 2 s or this is equal to multiple c a 0 is there equal to 1 plus k 1 s tau plus k 2 s tau. So, k 1 s is known k 2 s is known tau is known the l can be calculated. So, will I put numbers if I put these numbers this becomes 1 plus 21.3 times 0.11 plus 10.08 into 0.11. So, that is equal to 4.45 the value of l for this particular problem where tau is 0.11 per hour k 1 and all that we have calculated l is 4.45. Now, we have to calculate m what is m? m we said m is equal to 1 plus beta which is 1 plus h a by v naught c p that is equal to 1 plus h is let me just put this numbers here calculated here somewhere else h is 1000 and we have calculated this area just now we calculated area is 0.24 square meters divided by v naught c p v naught is 0.6 and c p is 1000 kilo calories per cubic meter particular c and so on that comes out to be 0.43. So, l please notice here m is this now we have to calculate n this is j 1 how to calculate n n equal to j 1 tau del by del t of r 1 minus of r 2 at s. So, when you do this differentiation this becomes n equal to j 1 tau del by del t becomes k 1 s e 1 c a 0 times 1 minus of x divided by r t s square that is 1 and the other side is k 2 e 2 c a 0 x s divided by r t s square. Now, once again all these are known so we can substitute calculate this we can put all the numbers I will just put all the numbers here j 1 notice here the j 1 by definition is minus of delta h divided by c p. Volumetric specific heat I am putting all the numbers here and then when you do that it turns out to be 0.03 I put all the numbers here this whole calculation turns out to be 0.03. Now, our steady state stability criteria a criteria for stability is l plus m greater than n value of n is 0.03 please guess n is 0.03 and l value we calculated as 0.03 l is 4.45 and so let me let me put all the numbers together in front of you we get l plus m. So, l if you summarize here l equal to 4.45 m equal to m is 0.43 n equal to 0.03. So, l plus m greater than n l m greater than n we can see here both the criteria l plus m both the criteria are satisfied are satisfied which means what which means that the steady state the steady state that we have chosen steady state stable by stability what we mean is what we mean is that to a disturbance the steady state would return to its original state because after the disturbance by by by this criteria what we really mean is that that after disturbance it will take time to reach the steady state, but it will reach that old steady state that is the point this is meant by this satisfying of these two criteria. Now, the question is that this whole this whole analysis is based on what is called as small deviations from steady state which means that we our disturbance should not be so large that the linear approximation that we have assumed in this steady state analysis that should not be violated. So, small disturbances these are all correct, but if the disturbance are very large then clearly our our approximations in the analysis are not satisfactory there were alternative techniques would be required I will stop there. Thank you very much.