 We are going to be looking at time dependent operations in this lecture. Of course, the need for time dependence of course, is quite obvious. We are looking at batch operations you know time is the essence you know how long you take to complete the process and so on. And there are several other cases in which your time dependence becomes important. For example, if you are starting up a process you know how long does it take to reach steady state. And what are the conditions that you could make use of to see that this time that is required to which steady state is as quickly as possible you know or there could be a semi batch operations in which you know you have to remove a material from a process and so on. And how long it takes you know there were time is a crucial element in several operations where we are talking about batch or semi batch kind of operations. So, here we look at some examples to illustrate how we can deal with such situations. And come to a way by which we can formulate our equations to take care of this kind of situations. So, first exercise we would like to look at is for example, start up we will just take the example of start up. So, start up of a CSTR. So, what do we have we have a CSTR let us say to which material is coming in and material is going out it is well stirred. So, it is C A 0 coming in at V 0 as an example we will say that there is a reaction A going to be taking place. And this rate functions R A let us say is K time C A as an example. Let us say that initially this is the material here the initial C A in reactor equal to C A i let us say this is the initial condition which we want to. So, we want to see how long it takes for this process to reach steady state number 1. And what else that we can do to see that you know the steady state is attained as quickly as possible. It is something that is of great interest to us if you are starting up a CSTR on a daily basis for some process. So, how do we deal with this let us say we write our material balance which is what input output plus generation equal to accumulation. This is our statement of material balance input I will say input this is output this is generation this is the accumulation. Let us see how what we can do with this I will write this as V 0 C A 0 that is input. And then say V C A that is output let us say our reaction is K time C A equal to d by d t of what is N A which is V time C A. So, we can say that V 0 equal to V which means that the flow or the inlet. So, I will draw it once again here what we are saying here is that if this is V 0 we are saying this is also V 0 this typically what we might expect V equal to V 0. And then we assume that capital V is constant which means that the amount of fluid in the equipment remains constant that is what we are saying. So, for this can come out of the derivative and so on. Therefore, we can write this as C A 0. So, if I say tau the residence time as V by V 0 therefore, this becomes minus tau and then C A by tau minus of K time C A equal to d by d t of C A or I will write this as C A 0 minus of C A minus of K tau C A equal to d by d t of C A. So, I just forgot to multiply by tau. Therefore, let me write this as tau times d by d t of C A plus 1 plus K tau times C A equal to C A 0. So, this is the equation that describes the stirred tank in the doing the unsteady state. So, what is the integrating factor integrating factor is e raised to the power of 1 plus K tau multiplied by. So, I will put this tau here. So, that I put this tau here. So, it is tau integral d t. So, this is the integrating factor. So, we can integrate this and I will write the solution. So, that we do not spend too much time in trying to do what we know quite well. So, I will write the solution as the solution I mean we can solve that by the integration factor and all that. So, solution looks like this solution is C A equal to C A 0 by 1 plus K tau plus constant of integration multiplied by constant integration multiplied by e raised to the power of minus 1 plus K tau d by tau. So, solution to the C A what we are saying once again let us just not forget the physics of the situation. We have a stirred tank materials coming in and going out the reaction is taking place is V naught and C A naught and C A and so on. And so, this is the solution we want to find the constant of integration. So, at t equal to 0 we have said that C A equal to C A i this is the assumption. So, if you put this here you get the constant of integration becomes quite obvious constant of integration you can just see very easily that C A i minus of C A naught divided by 1 plus K tau. So, this is the constant of integration you can substitute here therefore, solution becomes C A equal to C A naught divided by 1 plus K tau plus C A i minus of C A naught by 1 plus K tau this is the constant of integration multiplied by e raised to the power of 1 plus K tau multiplied by t by tau. So, this is the solution to the case of a start up of a stirred tank. So, what is it that we are saying that if you have a stirred tank and you start it up with C equal to C this is initially C A i this is initial C A i. So, this is the unsteady state behavior of the stirred tank. Now, let us see what happens at t equal to infinity this becomes 0. So, what happens at t equal to infinity C A equal to what shall we say at t equal to infinity this whole terms appear therefore, C A becomes C A naught by 1 plus K tau this is something that we know from our understanding of stirred tanks. But what is more important is if I ask you how long does it take if we appropriately choose C A i the question is if we chose C A i equal to C A naught divided by 1 plus K tau then the time to reach steady state is nil or in other words the process operates at steady state from time t equal to 0. On other words by an appropriate choice of the initial state in the C S T R in the equipment you can actually avoid the time that you will lose in reaching steady state. This is the point that is to be remembered that this time required to reach steady state. So, let me just put it on this if C A i equal to C A naught divided by 1 plus K tau if C A i is so chosen that it is equal to C A naught then you find that C A at any time t of becomes equal to 1 plus K tau. On other words the output is independent of time. So, the C S T R is able to reach steady state immediately so that you are able to avoid the time that is lost in reaching steady state. So, cut the long story short what we trying to say is that start up of a C S T R you can actually achieve practically the final end point by appropriate choice of C A i. If you choose C A i properly your steady state is obtained very quickly and therefore, your process is able to run at steady state without losing too much time. So, this is the most important aspect of this unsteady state that we should recognize by an appropriate choice of the initial state. We are looking at an illustrative exercise to understand how we can apply the design equations we have derived. The problem is we have a stirred tank C S T R to which component A enters and goes out. A reaction 2 A going to B is taking place a dimerization reaction the rate function r A is given as minus of K times C A square the rate constant K is 60 liters per gram mole per minute. The reactor works at 3 atmospheres and temperature of 400 C. The problem is in two parts first part we have to find out what is the volume of this equipment for a conversion of 0.5. We will do that first and then we look at the second part a little later. Now to be able to handle a problem like this where there is a change in volume because of chemical reaction we have to write the stoichiometric table and take into account how the volume change can be appropriately taken into account. So, you have component A and component B component A comes in at F A 0 there is no component B at the inlet at the outlet. If conversion is defined with respect to component A we know that the outlet flow will be F A 0 times 1 minus of x statement of material balance and component B nothing is coming in at the inlet and it reacts as per this twice A going to B or A going to half B. Therefore, the amount of F B form will be F A 0 x by 2. So, if you add the inputs and the outputs input if you add it is F T 0 it is the total moles of component A coming in which is same as F A 0 and the total moles of what goes out as you can add A and B together you find that it is F A 0 times 1 minus of x by 2. So, that you know that F T 0 which is F A 0 and it becomes F A 0 times 1 minus of x by 2 and you can see there is a change in the number of moles because of chemical reaction. Now to be able to account for the effect of this what we do is that we apply what is called as gas law and because this reaction twice A going to B is a gas phase reaction. So, we know from our gas law that V by V 0 is F T by F T 0 T by T 0 P 0 by P and so on and since gas is taken as ideal we have simply it is F T by F T 0. Why is that because there is no change in pressure there is no change in temperature and therefore, those effects are removed therefore, you have V by V 0 is simply F T by F T 0. What is V? V is the volumetric flow at the outlet and V 0 is the volumetric flow at the inlet. So, V by V 0 tells us the change in volume because of chemical reaction F T by F T 0 we can see from here F T divided by F T 0 is simply 1 minus of x by 2 therefore, we get V by V 0 is 1 minus of x by 2. So, essentially we have now taken into account the fact that there is a change in volume and that effect of change in volume is now expressed in terms of extent of the action on conversion with respect to component A as reference. Now if you have to calculate what is C A and C B because we require C A in our rate expression we now want to calculate what is C A. Now C A by definition F A divided by V where F A is the molar flow rate at the exit and V is the volumetric flow rate at the exit. Now F A from here we find it is F A 0 times 1 minus of x and volumetric flow is V is V 0 times 1 minus of x by 2 therefore, we find C A at the exit is F A 0 times 1 minus of x divided by V 0 times 1 minus of x by 2. So, that we get C A as C A 0 times 1 minus of x divided by 1 minus of x by 2. So, we are now able to substitute for this C A in the rate expression. So, that now we can use that result effectively. Now what is C B? C B by definition is F B by V F B by V and that becomes C A 0 x by 2 divided by 1 minus of x by 2. So, what we have done is that using the fact that there is volume change, we have been able to express concentrations in terms of conversion. We have done all this. Now what is important is we have to calculate what is the values of these numbers. For example, what is C A 0? C A 0 by definition is P by R P. This is gas law. What is pressure is given as 3 atmospheres. The gas constant is 0.082 litre atmosphere per gram mole degree k and the temperature is 673 which is 400 C A. So, that you find that C A 0 is 0.0544. Now what is C A? We have just now said C A is C A 0 1 minus of x divided by 1 minus of x by 2. Therefore, we can now put all the numbers and find out that C A at the exit is equal to 0.0 C A at the exit. This is the exit, exit C A. C A at the exit is 0.036 moles per litre and what is C B? It is F 0 x by 2 divided by B 0 1 minus of x by 2 which I have already said and if you put all the number it becomes 0.018 mole per litre. So, we have calculated what is C A and what is C B as well as we also calculated what is C A 0. Now we will have a position to put these things in our design equation to find out what is the volume of the equipment that is required to obtain a conversion of 0.5. Let us do that now. So, what have we done now? We said V we said V equal to F A 0 x divided by minus of r A and I am just substituting for minus of r A which is what is minus of r A minus of r A is if k times C A squared and therefore, I am putting k is here and what is C A which is C A 0 squared multiplied by 1 minus of x divided by 1 minus of x by 2 which have derived all this. Therefore, now we know the volume of the equipment is given by the right hand side. Now we know what is F A 0? We know what is x? We know all the numbers here. If you put all the numbers you can get the volume of the equipment to be 6.35 litre putting all the numbers k is 60, C A 0 is known. So, everything is known here and therefore, we can calculate what is the volume of the equipment which turns out to be 6.35 litre. Now the second part of the exercise is quite interesting. The second part of the exercise what is this is the second part of the exercise? What it says is now as this C S T R which is running at steady state at an instant of time when the steady state is achieved and the process is running at x equal to 0.5, we close the outlet valve. All the outlet valves are closed and the inlet flow F A 0 is adjusted. So, that the pressure remains constant, temperature remains constant, temperature remains constant because we maintain temperature constant. So, question is how long would it take for the outlet mole fraction here, sorry how long the mole fraction inside the equipment to become 0.9. Let me repeat the question is when the process is running at steady state, we close the outlet valves and we adjust the flow entering the equipment so that pressure remains constant. And we want to find out how long it would take for the equipment for the composition inside the equipment to reach y b or mole fraction of y component b is 0.9. So, this is what we would like to do now. Now clearly this is an instance of an unsteady process. Therefore, we have to set up our balances to take into account the fact that it is an unsteady state process. Now let us see how to address this. Now what we said what we said is that we are adjusting we just draw it once again we are adjusting we have this feed is coming in, but there is no output this is all F A 0 and it is a function of time. So, if you write a material balance for A for A and B. So, you have material balance for A and material balance for B let us see what it is input F A 0 of t which is a function of time output there is no output plus generation equal to accumulation input of B output of B generation of B equal to accumulation of B. We know that there is no output of A because the valves are closed therefore F A disappears. Similarly, if you look at component B there is no input of component B there is no output of component B and therefore rate of generation of component B equal to accumulation of component B. So, the material balance for A and B gives us these equations 1 and equations 2. Now let us recognize let us recognize that this is a gas phase reaction where pressure is constant temperature is constant volume is constant volume is constant because the equipment volume does not change. Now here is an instance of a gas phase reaction where temperature is constant pressure is constant volume is constant therefore we should expect that the number of moles inside this equipment from the time we start this process would not have changed. On other words if you add equation 1 and 2 d by d t of n A plus n B the right hand side should go to 0 that is the requirement of problem statement because pressure is constant temperature is constant volume is constant therefore we have the right hand side is 0 therefore we have if we add the left hand side F A of it is a function of time plus R A plus R B multiplied by V become 0 therefore F A of time F A at any F A 0 at any time t equal to minus of R A plus R B times V. On other words the material balance taking into account the fact that gas law gives us the condition that d by d t of n A plus n B is 0 gives us a result which tells us how we should regulate this process to be able to achieve constancy of pressure which says we must regulate the flow of F A 0 so that it becomes always equal to minus of R A plus R B times V. So this is the condition that we will impose on the control system that will regulate the flow of component A into the system. Once we know this the rest seems fairly straight forward let us let us look at all that once again. Now what we have said we have said that the material balance tells us that we must regulate F A 0 coming into the equipment as minus of R A plus R B times V. We also know from the statement of a stoichiometry that the rate at which chemical reaction occurs R A by minus 2 must be equal to R B by plus 1 because A is reacting B is getting formed and as a result we should have the rate of formation of B equal to minus of R A by 2 follows by from the fact that stoichiometry tells us that A equal to half of B it follows from that. So that means if you can substitute this result that R B is minus of R A by 2 in our equation 3 here so that now we get F A 0 of t equal to minus of R A and then R B is minus R A by 2 I will just call this as equation 4 so you can get R B as minus of R A by 2 so that F A 0 of t becomes minus of R A by 2 multiplied by V. This is clear notice here that R A is minus of K C A square is already given in the problem statement which is what is mentioned here where R A is minus of K C A square. Now we can substitute for R A here so that now we get the rate at which the material is entering the equipment F A 0 of t how it must change with time is now given as which is R A which is K times C A square V divided by 2 minus there are 2 minus signs here so that now we get F A 0 which is a function of time must be K times C A square by 2 times V. On other words what is being said is the problem statement is that we must adjust F A 0 of t so that it is always equal to K times C A square by V C A square by 2 V so we must adjust F A 0 of t so that it satisfies this equality always this is the control system that we must implement. Having said this having said this now the rest is even fairly straight forward now what does it mean it means that if you look at the material balance for component A what is the material balance for component A input output generation equal to accumulation there is no output because the equipment is closed there is nothing coming out of the equipment now so this valve is closed so whatever comes in we can only accumulate therefore we have output is 0 so input F A 0 of t is already shown in the previous description as V times K C A square by 2 and R A is what K C A square this also we know because R A is given as K C A square in the problem statement so with the minus sign so we have F A 0 of t which is V times K C A square by 2 and R A is minus of K C A square by 2 times V so that we get d by d t of N A is minus of K C A square by 2 multiplied by V so what we have been able to do by utilizing the material balance in the problem statement is to show that the d by d t of N A is now equal to K times C A square V by 2. Now having said this we can go forward and carry out the integration now left hand side is minus K C A square V by 2 the right hand side N A is by definition N A you know this N A is always V times C A and V is constant therefore we can come out of the derivative which cancels off so that we get now that the differential equation which governs the variation of C A with time inside the equipment is given by minus of K C A square by 2 equal to d by d t of C A which on integration gives us the result 1 by C A minus of 1 by C A I equal to K T by 2. So what we are saying now is that we are now in a position to tell how long we must run the process so that we get concentration of C A as might be specified in the problem. What is specified in the problem? What is specified in the problem is the following problem statement says problem statement says that let me draw it once again draw it once again it says this is closed we want this y b here to be equal to 0.1 this is what it says problem statement says that how long does it take for the contents of this equipment to reach which y b equal to y b equal to 0.9 y b equal to 0.9 or y a equal to let me write this here you want y b equal to 0.9. Now this one point we must remember is that at the instant of time when we close this valve the process is running at steady state at that time composition of A the composition of A and B I have written at C A I and C B I the composition of A and B at the instant when we close the valve was the values corresponding to the steady state achieved corresponding to x equal to 0.5 in the first part of the problem. On other words at the time when we close these valves so that it ran as a semi batch operation C A I and C B I was the steady state values that we had achieved in part one of the problem and we already found that values to be 0.036 gram moles per liter and C B as 0.018 gram moles per liter. So that C A plus C B I at the time we close the valve was 0.0544 gram moles per liter. Now we have also said when we looked at the material balance that since P equal to constant T equal to constant volume equal to constant and therefore the total number of moles cannot change. Therefore the concentration inside this equipment C A I plus C B I must always be equal to C A plus C B because that is the statement of the problem. On other words what we are saying is that C A I plus C B I which is 0.0544 when we close this valve that total cannot change as this reaction proceeds because that is the problem statement. So what we are saying now therefore is that if you want Y B equal to 0.9 or Y A equal to 0.1 C A by C A plus C B is 0.1 where C A plus C B is specified as 0.0544. On other words what we are saying is that the value of C A at which we want to stop the process is given by this number. C A must stop at 0.005004 we started at C A I as 0.03006 we want to stop at 0.005004 or in other words C A we want the time required to reach Y B equal to 0.9 or Y A equal to 0.1 we must substitute C A value of 0.005004 and C A I value of 0.03006 that will be the time that is required to reach Y B equal to 0.9. If you put all these numbers you will find that the time required to reach Y B equal to 0.9 is 5.2 minutes. On other words this whole exercise that we have tried to illustrate the use of design equations is essentially to tell how we can use the design equations for unsteady state process like a stirred tank to understand how a semi batch process can be understood, can be modeled, it can be regulated depending upon the statements of the process or depending on the requirements of the process we are dealing with. The third exercise we want to look at is the following that you have a chemical reactor. Now, the output goes to a condenser and then you have the reaction taking place is A plus B going to C plus D. D is volatile and the reaction is minus of K C A and C B. You have a batch starting with N A 0 and N B 0. So, this is the exercise in front of us is you have a reaction A plus B going to C plus D and the product D is volatile. Therefore, we are able to boil it off and then condensate. Now, the question in front of us is how the volume of fluid in this equipment will change with time. Of course, this is relevant even in a commercial practical situation because if you are having this in a practical situation you will have to find out I mean what is the volume that is left unreacted. So, that you know you can start the next batch you must know whether it is to what extent the fluids have been consumed. So, this is got a very practical significance the data given is N A 0 equal to 1 kilo mole N B 0 is given as 1 kilo mole. Then density is given as 20 kilo mole per cubic meter this for the fluid C A 0 is given as 10 kilo mole per cubic meter and C B 0 is also given as 10 kilo mole per cubic meter. The problem statement is that you have a batch equipment in which you have N A 0 moles of A and B 0 moles of B and then the fluid density mixture is given as 20 kilo moles per cubic meter C A 0 and C B 0 are given and this reaction A plus B going to C plus D and the rate function is minus of K C A C B second order reaction. Question is what is the volume of fluid left in the equipment after a given extent of time. So, let us see how to understand this our stoichiometry is what N A equal to N A 0 times 1 minus of x we know this N B equal to N B 0 times N A 0 x we know this N C is N C 0 which is plus N A 0 x and N D is N D 0 plus N A 0 x I have cancelled this off because it is not there initially. Now if you write a material balance for A balance for A what do we get input and sorry I am writing a material balance for D sorry not D input there is no input of D here output you notice here notice here is that this D this D is going out here this is D is coming out here. So, F D plus R D times V equal to 0 why is it 0 is 0 because N D does not accumulate what is the question we are having in front of us is that as the reaction occurs the D form being volatile it is rapidly removed therefore, there is no accumulation of D in the equipment that is why we have this input this is input this is output this is generation of component D this is accumulation of component D. So, there is no input F D R D what is R D R D by definition what we have said is that R D is K C A C B times B on other words R D is same as I mean we can even say this R D is same as R A with the minus sign is it all right because whatever is the what shall I say whatever is with the minus sign whatever is the D A form A consumed is the D form all right. So, essentially what we are saying is suppose you write a material balance material balance for A we get D and A by D T there is no there is no input no input there must be equal to R A V now N A is what N A 0 times D X D T that we know equal to minus of R A V this also we know N A 0 D X D T equal to what is what is R D what is R this we know from our stoichiometry R A by minus 1 equal to R D by plus 1. So, R A R D this is minus R A V is equal to plus R D. So, this is simply equal to R D V is that clear what we are saying what we are saying is that F D you can see here it is no need to write this F F D equal to R D R D V. So, this actually gives you let me just write it here F D equal to R D V this we come from the material balance for D now R A equal to minus of R D therefore, this is equal to F D that is what we are saying. So, what we are saying is that this is equal to F D now we also know the rate at which volume must be equal to volume must be equal to F D whatever is the change in mass of the fluid in the equipment must be equal to the rate of change of F D. So, is it clear where rho L refers to molar density of the mixture and so on. So, we can look at these two equations you can see this equation and this equation we can see here since they are both equal to F D therefore, N A 0 D X D T must be equal to D by D T of V L let me write this for you and then let us look at it once again. So, what we are saying is that N A 0 D X D T please note here N A 0 N A 0 D X D T equal to F D D by D T of V L is also equal to F D. So, you notice here D by D T of V L here V L is this is decreasing therefore, there must be a minus sign here. So, that is why I put a minus sign here. So, what we get? So, what we get here is that N A 0 D X D T must be equal to minus of D by D T of rho L times V that is equal to minus of rho L times D V D T equal to N A 0 D X D T or N A 0 D X we can write N A 0 sorry D V D X D V D X equal to N A 0 N A 0 divided by rho L with a minus sign is that what we are saying that means rated which volume changes with respect to extent of reaction is N A 0 by rho L. So, I have written this as we can integrate this and so on. So, let me integrate this I will get V equal to V 0 minus of N A 0 X divided by rho L. I will write this as V 0 minus of C A 0 X by rho L is it alright C A 0 X by into V is it N A 0 is C A 0 times V X by rho L. So, I can write V 0. So, I just write this here. So, V 0 multiplied by 1 minus of epsilon L times X what is epsilon L epsilon L equal to C A 0 by rho L this is what we are saying. So, what we have achieved what we have achieved is that we have an instance of a chemical reaction let me just go back to this chemical reaction that we have we have this chemical reaction A goes A plus B goes C plus D D is volatile. So, that as this as the reaction proceeds D is being volatile it can be boiled off. Therefore, the volume keeps on decreasing because D is going away. So, that is the question that we have in front of us and we have shown from our material balance from our material balance that N A 0 D X D T is F D and similarly D by D T is also minus of F D. Therefore, we can equate these two that is what we have done and found that the rated the volume change is related to conversion like this. So, V 0 and conversion are related like this. So, this is the relationship we will keep with us. So, we have let me write that once again we say that V equal to V 0 minus sorry V 0 1 minus epsilon L by X. Now, so let us go back to our equations N A 0 times D X D T. This is our equation which is minus of K C A times C B times V. Now, this V we can substitute from here that was the idea. So, we have N A 0 D X D T equal to minus of K C A what is C A which is C A is N A by V and what is C B is N B by V and this is V or equal to minus of K N A 0 1 minus of X they are both equal. Therefore, I will put like this it is V which is V 0 times 1 minus of epsilon L X. So, it is equal to so 1 N A 0 gets cancelled off. So, it is K C A 0 1 minus of X holds square divided by 1 minus of epsilon L X equal to D X and D T. So, we can solve this we can solve this now it is a fairly simple thing to solve this. So, we have now our equation is D X D T D X D T equal to minus sorry plus plus K both sides sorry N A 0 this will be a minus sign here will be a minus sign here I forgot the minus sign here the minus sign here. So, D X D T becomes K C A 0 1 minus of X holds squared divided by 1 minus of epsilon L X. So, you can solve this and then the solution is fairly simple K 0 T equal to 1 plus X divided by 1 minus of X plus epsilon L L under 1 minus of X alright. Now, we can see basically what it means is that once you are given what is the extent to which you want to remove once once X is given you can find T or when T is given X can be found out basically this is the equation which determines what happens to the process and V equal to V 0 into 1 minus of epsilon L times X. So, all the data is given therefore, you can calculate what is the I mean basically X T V there are three things here. So, if you want to calculate if you are if two of two are given the third can be found out. So, essentially depending upon what is specified in the exercise in this particular exercise what is specified is 80 percent X is given as. So, if you find put X equal to 0.8 you can find T once you put X equal to 0.8 you can find V. So, here for X equal to 0.8 you will find T equal to I have found T equal to 3.09 hours and V equal to 0.04 cubic meters. So, what we have tried to do in this exercise is the following that we have taken three examples of unsteady processes first exercise was to find out how long it will take for a steady state to be achieved in a CSTR. Second exercise we took is look at a chemical reaction in which we want to hold volume pressure constant over a period of time we said how it can be done. Third exercise is that we are looking at process in which you are continuously removing the liquid and how long it takes for you to reach a certain extent of reaction. So, on all three cases essentially we set up the equations and found out how the different things are related. Thank you.