 We are looking at practice problems in energy balance in stirred vessel in this lecture. Now, let me explain what the problem we want to look at today. We want to look at a semi batch operation of a stirred vessel. This is the exercise that we want to look at. What we have is a vessel which is well stirred. Reaction that takes place is 4NH3 plus 6HCHO giving you N4CH26 plus 6H2O. So, this is hexamethylene tetramine which is made from ammonia and formaldehyde. What we have is you have ammonia coming in and you have here formaldehyde. So, you have an initial charge of formaldehyde in the equipment in the vessel into which you add ammonia. The reaction is instantaneous which means as soon as ammonia contacts formaldehyde the reaction and the ammonia is completely consumed. The initial composition of the vessel is all the data is given. Let me just put down all the numbers. I call this as A, I call this as B. So, what is in the equipment is that let me write here B I equal to 2000 liters. It is given C B I which is the initial formaldehyde present here is 15 g mol per liter. Now, the concentration this is component A C A 0 C A 0 is given as 15 gram moles per liter per liter. Now, this temperature T A is T A 0 is 0 and then T I this this temperature T I this T I is given as 0. C P specific heat C P is given as volumetric specific heat is 1000 calories per liter C for the mixture for mixture it does not change assumption is it does not change as heat of reaction is given as minus 74 kilo cal per mole H M T for every mole H M T formed. So, much of reaction is taking place. Now, the question that we have to address is the following. He says now that the temperature initially is 0 calculate the time required for complete consumption of first part is calculate the time required for complete consumption of formaldehyde that is 0.1. And second part is the time required for the temperature to reach 60 is it starts at 0 how long does it take to go to 60. Third part is how much heat is added or removed to maintain the temperature at 60 hence forth. So, let me state the problem once again we have one first part is how long how long does it take does it take for H C H O to be completely consumed. This is the first question that we want to answer second question we want to answer is if process is adiabatic that means we do not allow any heat loss adiabatic. How long does it take does it take to reach 60 C that is second part. Third part is if reactor reaction mixture mixture is to be maintained maintained at 60. See how much heat heat is to be added or removed. So, how much heat is to be maintained. So, these are the three questions that we need to answer how long does it take for H C H O to completely get consumed. If you operate adiabatically how long does it take to reach 60. And then if it want to be maintained at 60 how much heat you must add or remove this is the three points that we want to address. Let us look at this one by one first is reaction is instantaneous. Therefore, what shall we say that if volume of the equipment is V and r a is the rate at which component a is getting consumed. So, reaction is n h 3 plus n h 3 plus 4 n h 3 plus 6 H C H O giving you n 4 C H 2 6 plus 6 H 2 O. So, this is hexamethylene tetramine. Now, V times r a now what is V times r a equal to by definition. We say that it is minus of r 1 times V times minus 4 that is equal to which is equal to F A 0 equal to 10 liters per minute multiplied by 15 gram more. What is 10 liters per minute that 10 liters per minute is the manner to mention this here this V naught this V naught is 10 liters per minute. So, what are we saying here what we are saying here is that this the ammonia is entering at 10 liters per minute multiplied by 15 that is 150 moles per minute is entering. And that must be equal to what the rate at which if V is the volume of the equipment and then r a is the rate of chemical reaction we know r a is minus of r 1 multiplied by 4 and all that. Therefore, this gives us r 1 times V equal to 150 divided by 4 equal to 37.5 gram moles per minute. Now, we write a balance for B what is balance for B input output plus generation which is r B times V equal to d by d t of N B what is B N B is the moles of B which is now what is now we know this there is no continuous input there is no continuous output sorry output of a therefore, r B is what r 1 with a minus sign into 6 times V equal to d by d t of N B this is clear. And this is constant because this is clear therefore, what is the time shall we say time required equal to. So, this whole thing becomes total amount of N B is N B I divided by r 1 r 1 is r 1 V 37.5 multiplied by 6 r N B I is 15 multiplied by 6 r N B I is 15 multiplied by 2000 divided by 6 multiplied by 37.5. So, that gives you the time that is taken to completely consume the formaldehyde which is present in the equipment. Please let us understand this once again what are we saying what we saying is that if you write a material balance for component B there is no input of component B there is no output of component B whatever is the material is present inside it is undergoing reaction as per this rate function r B r B is what by definition is r 1 multiplied by the stoichiometric coefficient that is what I have written here. So, it is r 1 is stoichiometric I have to put this minus sign here. So, it is minus sign should have been put here minus 6. So, this is r 1 is 37.5 multiplied by 6 15 into 2000 2000 initial volume. So, 133 minutes is what it takes for the material to be completely consumed. Now, the next part of the question is how long does it take for it to reach 60 degree C what are we saying now. Now, as this reaction proceeds heat is generated and therefore, this whole thing gets heated up from initial temperature of T is 0 degree C and then we want to know how long it takes to reach 60 C for which what we have to do is to write the energy balance. What is your energy balance let me write the energy balance in the most general form n i times d h i by d t we have we have derived all this at an earlier stage. So, I am not doing it again sigma f i naught h i naught minus of h i we have derived this also plus sigma r i times v i equal to 1 to p reactions. We have done this also minus of delta h i star time plus q minus of W s that we have removed. So, now n i d h i by d t now what is n i adiabatic operation therefore, v this is not there. So, we have the left hand side is written as v times C p volumetric specific heat d t d t and then this this term i equal to 1 to n species. So, this term I write as v naught times C p some another volumetric specific heat times t naught minus of t and this term there is this reaction I write as r 1 v times minus of delta h 1 star is this clear. So, the left hand side is volumets. So, notice that this volume v is actually v i plus v naught times t and d t d t equal to the left hand side is 10 multiplied by C p is 1.0 times 0 minus of t plus r 1 is 37.5 r 1 v multiplied by this delta is minus of 74.6. So, this notice here this v naught is 2000 this is 10 t times d t d t equal to 10 with a minus minus 10 t plus 37.5 times. So, this is minus minus minus minus minus minus plus because exothermic this is exothermic. So, this plus only multiplied by 74.6. So, I can also write this is minus 10 plus 2797.5. So, this is we have to only integrate this to find out how temperature changes with time. Let us do that let us do that. So, we get please notice here this is the equation we have. So, 2000 plus t d t d t is minus of 10 t plus 2797. So, we can integrate this we can integrate this I have done this like this. So, d t let me see whether I can do it like this plus t divided by 2000 plus 10 t equal to 2797 divided by plus t. This is let me see how it how it shapes up I will put it here. So, d t d t I am dividing t divided by 2000 plus 10 t 200 plus this is not 200. Now, this integrated form the solution looks like this. So, after integration after you can you can do all this not very difficult to do integration. We get 200 plus t multiplied by 200 plus t multiplied by 60 equal to 2797 2.7 times t. This is the solution this is the solution therefore, plus t t is here turns out that t equal to 1200 12000 divided by 219.7 that is about 54 minutes. Now, what is it that we know is that what is the time that is needed when this becomes 60 that is what is. So, you have to put t equal to 60 here and find time here that is what I have done. Now, the next part of the question is I hope this is clear you are just integrating it this is the solution to the integral and then we want to find out what is the time corresponding to t equal to 60 what is the time that you put t equal to 60 a fine time and this what I find to be about 54 minutes. Third part of the question is maintain t equal to 60. Here what we have done for t equal to 60 c what is the time what is the what is t this is what we have found out. Now, you want to maintain t at 60 how much heat is to be added or removed. So, for that we write the energy balance what is the energy balance in our form we write the energy balance as v naught c p t naught minus of t plus r 1 v times minus of delta h 1 star this is plus q minus w s that is c equal to v c p minus d t d t this is our energy balance and we want this to be and we. So, we want q this is what we want. So, we can put all the numbers let me see therefore, I am putting all the numbers 10 c p is 1 kilo cal per minute and then temperature is going from 0 to 60 plus r 1 v is 37.5 multiplied by 74.6 plus q this is clear therefore, q equal to it turns out to be 297 kilo cal per minute. So, what we are saying is that in order to maintain the temperature at 60 we have to so much of heat has to be. So, what is this t naught t naught is the temperature at which material is coming in and this is 60 is the temperature to which it must be heated. Therefore, this much heat has to be continuously supplied this is the heated which heat is generated. Therefore, this q and then q is positive and negative q is negative showing that so much of heat has to be removed heat has to be removed. So, what is it that we have tried to learn from this exercise is that when we have dealing with very explosive reactions we find it more convenient to do a semi batch operation. So, that the concentration of ammonia in solution is kept 0. So, it is very safe number 1 number 2 since the heat is generated heat is generated we were able to use that heat to heat up the fluid. So, that we are able to achieve some amount of energy conservation and then subsequent to reaching the appropriate temperature we are able to remove the heat or in other words we are able to adjust the rate of ammonia addition. So, that the heat removal that we are able to achieve without causing any explosion. So, this kind of semi batch operation might be useful when you are dealing with very explosive reactions. Now, let us go on to the next exercise second exercise of interest to us now that you finish one we go to the next one. The second exercise that we want to look at is the following here we have a stirred tank reaction A goes to B and then reaction A goes to C and this R 1 equal to k 1 C A and R 2 equal to k 2 C A. On other words we are conducting a reaction and then what as the call the data is given here let me write down all the data temperature volumetric flow density heat capacity heat concentration coolant heating medium all the data is given and all the data for the reactions lot of data is given let me explain all that in a minute. So, the important point that we want to convey in this is that when you have a reaction where you have let us say this is desired and this is undesired undesired. So, our interest is of course to operate the equipment under conditions where the desired reaction is favored or the product B is the preferred product that we would like to see and therefore, we should like keep product C as small as possible under conditions of reaction. How do you achieve this what kind of process operation you should choose etcetera is the point of interest in this exercise. Now, let us put all the data because this we need all the data to be able to solve this problem let me put all the data. So, we have temperature which is given as 54.5 volumetric flow which is given as 066 cubic meters per second density of reaction mixture is given as kg per cubic meter heat capacity C p is given as 0.8 kilo cal per kg degree C. So, density is given and therefore, volumetric specific heat can be found out feed concentration is given as 15 mole per mole per litre reactor volume is given as 1 cubic meter heat transfer coefficient is given as 0.55 kilo cal per square meter degree C second and heating medium is given heating medium heating medium is saturated steam. At 100 C this is some data lot more data is given regarding the reaction also if you look at the reaction. So, let me just put a line here. So, that we have some idea what is what so the reaction. So, we have a 1 is 3 10 raise to 8 per second a 2 equal to 2 10 raise to 14 per second e 1 is given as 15,000 calories per mole e 2 is given as 25,000 calories per mole delta h 1 is given as minus 9,000 calories per mole delta h 2 is given as minus 9,000 is given as 13,880 calories per mole. These are some data regarding the reaction. What is the reaction? Reaction as you have said is A goes to B and A goes to C. A goes to B A goes to C rate functions are given K 1 C A K 2 C A desired and undesired all the data etcetera is given and our question in front of us is the following. What is the temperature which maximizes the production of B? This question 1 let me just write down just in to provide the context. So, we want to know. So, first part is first part is show that temperature which maximizes maximizes production production of B is given as K 2 tau equal to e 1 divided by e 2 minus of e 1. So, that means the choice of temperature should be such that this equality satisfied. Then only we get maximum rate of production of B that is first thing we have to show. Then find now that find conversion conversion reaction, reactor temperature for the conditions above 3. What is the area of cooling heating coil required? Now the context is frequently we encounter problems like this where there are the desired reaction, there is an undesired reaction and therefore, we need to choose conditions under which our objective of making the desired product. Therefore, first thing is to show what that the temperature which maximizes the production rate of B is given by this expression. Once you have found that then for those conditions what is the conversion the reactor temperature all those conditions have to be fully specified. So, this is the question that is of interest to us. Let us try to go through this now, you have this problem here and there is a coil. I will show this coil like this, not coil like this which provides the heating or cooling. So, let us write the stichometry, stichometric table, what does it say? It tells us what happens to A B. B C because of reaction. So, the reaction takes away. So, this is and B is F A 0 X 1 and C is F A 0 X 2 correct plus F B 0 which is taken as 0 plus F C 0 is taken as 0. So, what are we saying here? That because stichometry if the reaction extends or X 1 and X 2 this is what it is. Therefore, material balance if you do a material balance for B for B we get what input minus of output which is F B F B F B which is plus R B times B equal to 0. So, 0 minus of F B is what F A 0 times X 1 what is R B is K plus K 1 C A 0. C A 0 times 1 minus of X 1 minus of X 2. Notice here that C A is F A by V therefore, it is F A 0 times 1 minus of X 1 minus of X 2 V is V 0. So, that is equal to C A 0 times 1 minus of X 1 minus of X 2. So, that is what is written here multiplied by V. Similarly, we can write a material balance for component C. Let me write material balance for component C material balance for component C. What is it say? 0 minus of F A 0 input output and then what is K 2 C A 0 times 1 minus of X 1 minus of X 2 times V equal to 0 is metal balance. So, if I call this equation 2 if I call this as equation 1 simply by 1 divided by 2 gives you from 1 and 2 1 and 2 what do we get? X 1 divided by X 2 equal to K 1 divided by K 2. Straight forward if you just look at this, this is X just look at these 2 it is quite obvious that you know X 1 X 1 divided by it is very obvious you can see here. So, X 1 divided by X 2 is equal to K 1 by K 2. So, you can call this equation 3. Now, you can substitute for this relationship in equation 1. So, you can put putting I can putting equation 3 in equation 1. We get what F A 0 minus F A 0 X 1 plus K 1 C A 0 times 1 minus of X 1 minus of K 2 by K 1 X 1. Is this clear? Minus of X 1 minus of X 1 go through this now. So, this gives us simply for rearranging X 1 multiplied by 1 plus K 1 tau plus K 2 tau equal to K 1 tau. Please you can divides this V is here. So, if you divide through what V 0 this becomes C A 0 cancels of V 0 this becomes K 1 tau 1 plus K 1 tau and so on. So, this is what I have done. So, this gives you X 1 equal to K 1 tau divided by 1 plus K 1 tau plus K 2 tau. Is this clear? The rest is fairly straight forward. Let us just go through what we have done once again because it is important that we understand what we have done. So, we have let me run through this once again. We have a stirred tank. This is a CSTR and then this reaction A goes to B, A goes to C. Rate functions R 1 and R 2 are given. Now, what we are saying is that when we write the material balance when we write the material balance we have input minus of output plus generation equal to accumulation. So, what is the rate at which this gets generated? The rate functions are given here. So, you can see here R B is generated from reaction where A goes to B. Therefore, R B is generated. It is a positive quantity. This is what we have written here. Similarly, R C can see here R C. It is generated. A goes to C. A goes to C also. So, that is what we have to do plus sign here. Therefore, we have put it in the same form and dividing throughout by V 0 and then cancelling out C A 0 and so on. We get this relationship. What is it that we would like to maximize maximize? See, we would like to maximize maximize X 1 with respect to with respect to T, which means that we have to differentiate d X 1 by d T. So, we have d X 1 by d T. We have to find out what is it equal to? We can differentiate this. Now, I am just writing down the differentiation. Differentiation of K 1. See, we know this. This is not new to you. Derivative of d by d K, d by d T of K is K 1 even by R T square. So, I will write here K 1 E 1 by R T square times tau divided by 1 plus K 1 tau plus K 2 tau. The second term is K 1 tau and then we have to differentiate the denominator. So, K 1 E 1 by R T square tau and then K 2 E 2 by R T square tau divided by 1 plus K 1 tau plus K 2 tau. So, this is what we have to set it equal to 0. Now, we can do some algebraic manipulations and so on. So, we will not spend too much time on that. When you do that and go through the algebra, you will find that the result you will get is E 1 by 1 plus K 2 tau equal to K 2 E 2 tau or K 2 tau equal to E 1 divided by 1 plus K 2 E 2 tau. This is what we will get. So, the result that we have wanted to prove we have been able to show. Now, it is important to understand what we have done. So, what is being said is the following is that if you have a stirred tank and then this reaction A goes to B, A goes to C and this rate constants are given as it is given. So, then the best choice to get the maximum production of B is simply to choose your K 2 such that this equality satisfied. Now that we know what is the best way to do this reaction. Now, we have put some numbers to see what it has to say. In the data E 1 is given as E 1 is given as 15000 in units of calories per mole. E 2 is given as 25000 in the units of calories per mole. Therefore, if you put these conditions into this one, therefore you should have K 2 tau for max production of B, max B equal to 15000 divided by 25000 minus of 15000. That is equal to what 1.5 minus of 15000. That is equal to what 1.5. So, we should have and what is tau? Tau by definition is reactor volume by V naught. That is given reactor volume is given and then the V naught is given as 0.06. So, that is equal to 15.1 second. So, tau is known 15.1 seconds. Therefore, K 2 tau which is equal to 1.5. Therefore, K 2 equal to 1.5 divided by tau. That is 1.5 divided by 15.1. That is equal to 0.1 second. So, what have we found out here? What we have found is that the temperature, the value of K 2 that we should choose should be such that K 2 should be 0.1 per second. So, that comes out of R. We know that from our this one K 2. So, we know that K 2 equal to A 2 e to the power of minus of E 2 by R T. This we know. This A 2 is given as 2 in 10 raise power of 14 e to the power of this is given as 25000 R is 1.987 and T. So, essentially this equation defines the temperature at which we must operate the reactor. So, we can solve for T and that finds out to be about 83.5 C. Please note that accuracy of calculation is important. Now, if you know the temperature to be 83.5. So, notice here that you get this temperature T in Kelvin and it is converted to centigrade. So, what is K 1 at 84, 83.5? To find out we can say that is equal to 3 times 10 raise to power of 8. It is all given a exponential of minus 15000 divided by 1.987 and then temperature of is 357 which is 83. This is what is given. So, that turns out to be 0.197 per second. So, we have got K 2 from there we have got temperature from there K 1 we have got 0.197. Therefore, what is the value of? Now, we want to know what is the composition? What is X 1? We want to find out what is X 1 which is K 1 tau plus K 2 tau. So, K 1 tau is 0.197, tau is 15.1 seconds divided by 1 plus K 1 is 0.197 multiplied by 15.1 and then K 2 is 0.1 multiplied by 15.1. So, that gives you X 1 as 0.542 and then X 2 by definition we have already got that is K 2 by K 1 times X 1 that is 0.54 to multiplied by 0.1 divided by 0.197. So, that comes out to be 0.273. So, what we have said is that if you have chosen the temperature which maximizes the production of B and that temperature is given by this equation which you already derived. The temperature is comes from this tau is specified in the problem. Therefore, you can find K 2 once you find K 2 we can solve for K 2 from here where T is a temperature in Kelvin which is converted to centigrade here. Once you know value of temperature we can find K 1 also the Arrhenius dependence is given therefore, K 1 value is known. Once K 1 value is known we have already derived this X 1 is given by K 1 tau by 1 plus K 1 tau K 2 tau therefore, X 1 is 0.4 and X 2 is K 2 by K 1 you found 0.273. So, what we saying is that in this multiple reaction where A goes to B A goes to C X 1 about 54 percent is X 1 and about half of it is component X 2 which is component C. You notice here that in the rate functions are such that a significant part of the reactants go towards undesired product which is not very desirable. This problem itself is chosen to illustrate to you that a huge quantity of the reactant material is often going towards products which are wasteful and this is where choice of a good catalyst. So, that the specificity towards our desired product that is required and therefore, requirement development of suitable catalyst to target towards your required product is very very important in reaction engineering. The next question is how much energy see we have already said now that we have we have how much heat or heat is to be added or removed to be able to achieve what we want. So, for that we look at our energy balance which says d n i d h i by d t equal to sigma f i naught h i naught divided by h i plus sigma i equal to 1 to p reactions r i times minus of delta h i star times v plus q minus of W s. What we are saying is to be able to achieve what we have achieved here what we have achieved here we need to be able to ensure that the heat that is required for the process is a appropriately added or removed. So, that we need to find out how do you find out now d h n s let me go through this calculations. So, what is h i h i equal to h i at 0 at some reference temperature plus c p i times t minus of t r correct. Now, if these effects are not important then it is you know we can simplify the whole thing simplifies. Let us put all the numbers to see how it works out. So, we have here putting all the numbers we have v naught. So, the energy balance simplifies like this for our case plus r 1 v minus of delta h 1 star plus r 2 v minus of delta h 2 star plus q. So, I will put all the numbers here which is 0.006 into 720 into 0.8 into 54.5 minus of 83.5. So, this is the feed temperature to reactant temperature plus the next is r 1 v we notice here that this is equal to f a 0 x 1 is very elementary this is equal to f a 0 x 2. So, this r 1 v equal to f a 0 x 1. So, I am putting that f a 0 x 1 is f a 0 is 0.006 multiplied by 15 multiplied by 0.454 into 9060 this is the and then the next one is 0.006 15 into 0.27 to minus of 13 880 notice here please notice here our heat of reaction the data that is given you can see here delta h 1 is exothermic delta h 2 is endothermic. So, this effect I have taken into account. So, minus delta h is positive minus delta h is negative here I can see plus q. You put all the numbers you will find that q equal to 13 it is positive 13 kilo cal per second q equal to if you put all the numbers here you get q equal to 13 kilo cal per second. Now, how do you ensure that 13 kilo cal per second is actually achieved that is by looking at the heat transfer surface you know that q equal to h times a heat transfer area. So, I will put it like this h a times t c minus of t h is given as how much is h heat transfer medium is given as 0.55 kilo cal per second. So, it is 0.55 in the units of seconds a you do not know coolant temperature is given as 100 and then reactor temperature is given as 84 and q is 13. So, a equal to 6 about 8 is about 1.5 meters. So, what have we said that this problem a going to be a going to c we need a heat transfer surface is about 1.5 square meters to be able to achieve the maximum production rate of component b is that clear. Let me just summarize what we have said without losing the thought. So, we have this multiple reaction in which you have a heat transfer found out the best conditions for achieving the desired. Now, there is a related question that might that might be of importance to us is the related question is we just run through this once again. So, you have this reactor here we have this heating medium or cooling medium whatever in this case it is the heating medium equal to t c feed is coming in feed is going out. Now, what we would like to know is that whether this reactor operation that we have chosen whether it is stable or unstable. Now, to answer questions like this are best answered by looking at our stability criteria which we already done and as per that stability criteria if you look at the values of l m and n will be able to tell whether the process is stable or unstable. So, what we have tried to do in this particular problem is that we have taken two exercises case one in which it is a semi batch operation where we have tried to understand how we can deal with very explosive reactions. So, that you know and then ensure that the temperature at which reaction takes place is kept under control and also the rate at which the heat is released because the reaction is also taken into account and we operate the process. So, that there is no explosion at the same time your products are formed at the rate at which you can handle that was the first exercise we have done. The second exercise we have done is an exercise in which there are two reactions a goes to b a goes to c in a stirred tank and then we try to see how best we can maximize the production of the desired product which is component b. What we have done is that we set up the equations for production of component b and then we said all right if this is the rate at which this component b is produced what is the conditions under which the maximum rate of production can be obtained which we did by differentiating the expression for x 1 with respect to temperature. And we found out that the best temperature at which we can operate this process is defined by this relationship which is k 2 tau equal to e 1 divided by e 1 minus a v 2 that is something that we did correct sorry k 2 divided by e 2 minus v 1. Now, once we did that what we said is that then clearly the temperature at which we are going to operate the process can come by looking at the value of k 2 and hence finding out the value of temperature corresponding to the given value of k 2 because all the arenas you know parameters are given. So, once k 2 is known we find out k 1 because temperature is known and therefore, we find that the x 1 which is the extent of reaction and therefore, what is the composition of the outlet stream and also what is the amount of product C which is undesired product is formed. So, you know the desired product you know an undesired product the conditions under which this problem must been problem statement has been given we find that the desired product if it is 0.54 the undesired product is 0.27 showing that the undesired product is not small at all it is very significant indicating that our design must look at this problem of wasteful production of undesired product which means that you know we do not have the catalyst which is good enough to specifically target your reactance towards the desired product. In fact, that is one of the most important decisions and developmental work that we must do always to see that our catalyst always ensures that the reactance substantially goes towards the desired product and the production of undesired products kept as low as possible. Now, it is not just that the undesired it may be a I mean your reactance may not be very expensive and therefore, we might be able to afford to throw away a significant part of our reactance as an undesired product. But the question is that whatever is the undesired product that is produced this must will go into our environment and create problems of pollution. So, since the cost of polluting the environment has become quite serious as you all will understand we cannot afford to have a product which is producing a lot of undesired product a process cannot afford to produce undesired product. So, that is something that you and I must recognize and remember in our design making sure so that our not only we have an economically good process you also have a process which is environmentally sound environmentally safe apart from being economically attractive. Thank you.