 We are looking at advanced reaction engineering design equations. We have talked about design equations and we derived the design equations for various situations. Let me quickly run through all that. So, we said for a batch the reaction time is n is 0, 0 to x a d x a minus of r a times v. We have derived this. Now, we can plot this if for constant volume batch for example, then v you can take it out of the integral. So, that this becomes c a 0 by minus of r a. This is what I have plotted here c a 0 by minus of r a versus x a. This integral will be in 0 and x a whatever is the area under the curve becomes the reaction time. On other words if you have some data on your reaction you plotted in the form of c a 0 by minus of r a versus x a simply area under the integral from 0 to x a gives you the reaction time. That is for a batch that is only for a constant volume batch. Now, if the volume does change due to various reasons then appropriately we have to take into account the effect of volume change which also we have done and we have done some exercises to illustrate this as well. Now, the second situation the in fact I have shown this here the for the case of constant volume batch is what you are getting. Now, if it is for a c s t r we have written the design equations also we have shown that the residence time for a c s t r is given as c a 0 x divided by minus of r a. Once again we can make a plot of c a 0 by minus of r a versus x a. This data comes from our experiments I mean we have shown how to do these experiments and so on. So, this data comes from our experiments. So, that you can plot this and then if you want to take the reaction from 0 to x a then the area under this rectangle gives you the residence time for a c s t r. This what is written here residence time for a c s t r is c a 0 x by minus of r a which is the rectangle that you can see here. Once again the same data c a 0 by minus of r a versus x a is used to find out residence time for a c s t r. Now, we can do the same thing with this with a p f r once again we have derived a little earlier that the residence time for a p f r is c a 0 integral of minus d x a by minus of r a. So, when you plot c a 0 by minus of r a versus x a once again area under the integral 0 to x a this area gives you the residence time for a p f r. Notice here that the residence time for a p f r the integral the reaction time for a constant volume batch which is given as t r the area the integrals are the same. In other words we can look upon reaction time in a batch as equivalent to the residence time in a p f r they are they are the same because it talks about the time of time that the fluid element spent in the reaction environment. This is something that we have understood based on whatever we have done. We can continue this and perhaps look at a more interesting situation is something that could happen or it could have various other kinds of uses. Let us for example, look at a sequence of stirred tanks. What is the sequence of stirred tanks? Sequence of stirred tanks is reactor 1 reactor 2 reactor 3 I will put 1 2 3 here just to just to indicate that there are so many tanks. Now, what is it that we have? We have reaction taking place in reactor 1 reactor 2 reactor n that means fluid elements entering reactor 1 and the moving to reactor 2 then moving on up to reactor n and coming out. Now, we have written the design equations for a CSTR I mean something that we have written for quite some time. So, accordingly I have written what is the reactor volume? Reactor volume v 1 is f a 0 x 1 by minus of r a 1. What is the reactor volume 2? f a 0 x 2 minus of x 1 divided by minus of r a 2. Notice here in a CSTR reaction takes place at the exit conditions and the reaction rates at the exit condition has to be taken. f a 0 x 2 minus of x 1 is the moles of component a that is under gone chemical reaction divided by the rate of reaction gives you the volume of the equipment. Now, we can put it in the form of residence time by dividing by volumetric flow at the inlet that gives you the residence time in the stirred tank 1 residence time in stirred tank 2. Similarly, react volume of stirred tank n is f a 0 x n minus of x n minus of 1 this is x n minus of 1 what is entering here is x n minus of 1. Therefore, x n minus of x n minus of 1 multiplied by v a 0 this is the moles of component a that has been react reacted divided by the reaction rate gives you the volume. And residence time is simply divided by the volumetric flow. So, you get c a 0 x n minus of x n minus of r a n or in other words tau 1 is c a 0 x 1 minus of r a 1 tau 2 c a 0 x 2 minus x 1 divided by minus of r a 2. So, if you make a plot of c a 0 by minus of r a something that we already said versus x. So, you get this curve which is the performance of the of the reaction now if you look at this area this area going from 0 to x 1. So, this area you can see is tau 1 and you can see here this area this area is tau 2 c a 0 x 2 minus of x 1. You can say x this is x 2 x 2 minus of x 1 divided by minus of r a 2 this is the residence time for reaction 2 for reactor 2. And similarly, for residence time for reaction 3 and reaction n and reactor 3 and reactor n and so on. So, on other words what we are trying to say here is that if you have a stirred tank if you have a stirred tank sequence of stirred tanks simply you have to construct these rectangles. And then area of those rectangles gives you the residence time plotted in the form of c a 0 minus of r a versus x. So, what we are trying to say here is that we could have situations in which the reaction kinetics are not very easy to determine. But what is not so difficult to determine are this kind of data c a 0 by minus of r a versus x this kind of data is relatively easy to determine. And therefore, it is relatively easy to determine what is the reaction time that is required for a given extent of reaction. And that way you can you can use may be a number of stirred tanks or may be 1 p f r and so on to drive the reaction to the end point of your interest. So, this is the point that I was trying to clarify. So, our job now to determine what is the relationship between concentration and conversion for a single reaction that is what we are considering right now. If it is a gas phase reaction the reaction once again let me write down is a a plus b b equal to c c plus d d c as an example. So, what you have got here is that if it is a gas phase reaction this is a gas law which is p v equal to z and r t z is a compressibility factor. Now, at the initial state or at the entrance it is p 0 v 0 z 0 n t 0 r t it is a batch system. So, this ratio gives you v by v 0 z by z 0 n t by n t 0 r cancels gas constant t by t 0. Now, essentially what we are saying now is that since we already derived the relationship between n t and n t 0 from our stoichiometry. We know that v by z 0 can be written in terms of this ratio which is already derived in the last class 1 plus y a 0 is mole fraction of the inlet x a is the extent of reaction delta a is the change in number of moles per mole of the reference component a. So, in t by t 0 p 0 by p z by z 0 on other words now we are able to tell what is the volume of the equipment how it depends on the variables of the system which is change in number of moles arising from the number of moles changing because of reaction and change in temperature if any change in pressure if any change in the compressibility factor if any. Typically if concentration is what we are interested in because we want to express the reaction rate function in terms of concentration we need concentration which is simply n a divided by v. Now, n a by definition n a 0 times 1 minus x a because that is how we define conversion and v comes from the above equation as v 0 times 1 plus y a 0 x a delta a multiplied by time temperature t by t 0 z by z 0 p 0 p on other words using this relationship we are able to express concentration in terms of conversion x a. Similarly, we can do the same thing for components b components c component d and so on therefore, we can express the reaction rate function in terms of conversion this is as far as the batch system is concerned. Suppose, we have a flow system flow system means our previous relationship p v equal to n r z n r t is replace by p small v where small v is the flow rate through the system. So, we have p small v times z n z f t r t f t is the flow rate v is smaller volumetric flow. So, p 0 v 0 z 0 f t 0 r t once again we get relationship very similar to what you already talked about v by v 0 in terms of f t by f t 0 p 0 by p t 0 by and z 0. So, that now we can express concentration c a as f a by v f a is smaller flow of a at any position and v is volumetric flow therefore, by definition we know that f a is f a 0 times 1 minus x a because that is how we have defined conversion and v which is now coming from this equation as v 0 times 1 plus y a 0 x plus delta. This relationship v 0 in this ratio f t by f t 0 we already shown is equal to 1 plus y a 0 x a delta. So, we are able to express concentration in terms of conversion. Similarly, we can do for components c components c d and i. So, what we have done is that you have use a stoichiometry of reaction and then using the stoichiometry of reaction and the gas law we have been able to convert concentration in terms of conversion. So, that now we can substitute this c a function in the reaction rate function and carry out the necessary integration or solution of the algebraic equations. Now, just like we have done for component a we can do for component b if it is a batch system it becomes n b by v. Once again we get a nice relationship in terms of initial composition of component b and conversion x a and c a 0 and so on. Similarly, just like we have got for c a and c b we can do for c c and c d. So, what we want to now do is we just quickly show you a problem sheet from which we want to look at one of these problems. The problem we want to solve is that we have a reaction which is acetaldehyde undergoing decomposition thermally giving you methane and carbon monoxide. This reaction takes place at 520 degree c it is at one atmospheric pressure. We want to conduct this reaction to the extent of x a equal to 0.5 which is we want to convert 50 percent of it. We can conduct this reaction in a CSTR continuous input of component a continuous output is in a CSTR. We can conduct this reaction in a PFR where component a comes in and reacts and goes out. We can conduct this reaction in a constant volume batch reactor in the sense we put all the reagents inside close all the valves and conduct the reaction at 520 degree c or alternatively we can also conduct this reaction in a constant volume batch reactor. How do you get constant volume you have a piston which is able to move up and down depending upon the change in pressure inside. Therefore, if you charge this acetaldehyde in this volume and as the reaction proceeds there will be an increasing volume and it will keep changing the volume. So, that the pressure will keep changing, but volume will change pressure will keep constant will co volume will change. On other words you can conduct this reaction in CSTR at constant pressure in a PFR at constant pressure in constant volume batch reactor or constant pressure batch reactor. What we would like to do is to do perform a calculation to see what is the size of the equipment or the reaction time that you would require for each of these reactor operation choices. This is the problem we would like to solve. So, let us how do you go about doing carrying out a solution to a problem like this. Of course, we first have to write the stoichiometry to understand what happens component A component B component C. What we start with we are starting with pure component A what goes in is pure component A. Therefore, B C there is no inert these 2 are 0. What comes out by definition we have said whatever is the x A is the extent or conversion then what comes out is given by F A 0 times 1 minus x A. How much B is produced by stoichiometry we can say so much is produced and similarly, so much of C is produced. Therefore, the total number of moles that is coming out of the equipment is F T is F A 0 times 1 plus x A total number of moles you have put in is simply F A 0. So, what do we have you are putting in F A 0 moles which is coming in and what goes out is F A 0 times 1 plus x A which means there is an increase in the moles because of chemical reaction. It is a gas phase reaction. So, I said here does the gas these are all gases at the temperature of reaction. So, we will have to apply gas law to understand how volume changes as the reaction proceeds. So, we have already shown little earlier that V by V 0 is F T by F T 0, T by T 0, Z 0 Z by Z 0 and P 0 by P. It is a gas law statement we know this for from our early understanding of gas laws. Now, F T by F T 0 you can see from here F T by F T 0 is simply 1 plus x A. Therefore, if the temperature of the reactor does not change this we maintain the temperature at constant at 520 C and at the conditions the pressure also does not change much and therefore, this is 1, this is 1, this is 1. Therefore, simply V by V 0 is F T by F T 0 from this stoichiometric relationships we get V by V 0 as 1 plus x A. So, what we have been able to do is that we are able to tell how volume volumetric flow changes as reaction proceeds. Now, to be able to tell what is the concentration we know concentration is F A by V if it is a flow system F A is F A 0 times 1 minus of x A V just now we have shown is V 0 times 1 plus x A. Therefore, we are now able to tell that concentration at any conversion is C A 0 multiplied by 1 minus x A by 1 plus x A. So, in essence what we have been able to do now is that you are able to express concentration in terms of conversion. Now, in this problem it is given that it is a second order reaction that means R A is given as minus k times C A square it is given it is given in the problem statement. So, that now we have that if design equation for C S T have already done. So, that volume of the equipment is F A 0 times x A times minus of R A and what is minus of R A is simply from what we got here minus of is k times C A square. So, that is what is done here C A is C A 0 1 minus x 1 plus x A. So, I have put here. So, you get minus of R A you are taking the science into account. So, k times C A 0 square 1 minus x A whole square 1 plus x A whole square. So, the volume of this equipment is F A 0 times all this. So, when you put all these numbers first you calculate concentration which is P by R T calculated as 0.0514 moles per liter at the temperature of 520 C. So, what do we get? We get reactor volume you put all the numbers here F A 0 is given as 1 mole per hour we want x A of 0.5 or concentration at C A 0 is 0.0154 you can put all these numbers. We get C S T R volume as 16 liters and C S T R residence time is 878 seconds. We are looking at case 1 where we have a C S T R we want to get 50 percent conversion or x A equal to 0.5. We were not saying that to get 0.5 conversion we need residence time of 878 seconds the reactor volume should be 16 liters for this 1 mole per hour. Second exercise now we want to do the same thing we want to do the same thing in a in a P F R at constant volume. Which means we want to do this problem in in a constant pressure P F R where the molar flow is 1 gamma mole per hour pressure is maintained at 1 atmosphere temperature is maintained at 520 C. How do we do this? We recall that our design equation for a P F R is simply F A 0 integral 0 to x A d x A by minus of r we have done this earlier. What is r A is k times C A 0 square 1 minus of x A 1 plus this also we have just now done because we have derived C A in terms of conversion. So, you simply what is the volume of the equipment F A 0 times 0 to you have to go from 0 to x x A and we notice that we want x A of 0.5 in this case is what we want. So, we have put the rate function here. So, we simply have to integrate this to find the value of the size of the equipment that is required to do this. Now, this is a very simple integration I have done this integration. So, this how the answers look like that volume of the equipment is given by this relationship and you can put all your numbers F A 0 is known C A 0 everything is known. So, you can calculate the volume of the equipment to be 6.3 liters. Please recognize that in the previous case when you used a C S T R we got 16.0 liters. Now, the volume required is only 6.3 liters and the residence time defined as V by V 0 is 346 seconds we got 878 second last time. Now, what are we saying that means when you use a C S T R as the reaction equipment we require much bigger equipment and when you use P F R then you have much smaller equipment. So, we will come back to this in a minute. Now, one issue that we would always like to know because there is a change in volume and we are calculating our residence time as tau as V divided by V 0 where V 0 is the volumetric flow at the inlet. But the fact is that because of volume change the volumetric flow is increasing in this case therefore, actual time of residence inside the equipment is different from what we have found out based on V by V 0. So, it is a interest to us to actually find out what is the residence time that is experienced by the fluid elements. How do we do this then we derive what we call as the actual residence time that is required for this equipment. How do you define actual residence time d tau which is the actual residence time is simply d tau is d V by V that means if you take an elemental volume d V at divide by the volumetric flow at that position that gives the residence time for that differential element. Therefore, we can use the design equation for P F R which is d F A d V equal to R A and then simplify this and get what we call as actual residence time in terms of the rate function and so on. We already derive all these things in our area class therefore, this basically gives us what is the actual residence time now in terms of other measure properties like X A and so on. This can be integrated and we get a result like this it says that the actual residence time in a P F R where there is increase in volume can be integrated to get what this function where X is the extent of reaction or conversion. You can put your numbers X is 0.5 we know that K is known C A 0 is known I put all the numbers here you get actual time of residence is 261 seconds. What are we saying? What we are saying is that if we have a plug flow reactor entering at 1 mole per hour and then temperature is 520 because of this reaction even though the residence time based on inlet molar flow is 346 seconds actual time of residence is only 261 seconds because of the fact that there is an increase in volume or increase in volumetric flow rate as the reaction proceeds. The next issue that we would like to look at is what happens suppose we conduct the reaction in a constant volume batch reactor. How does it look? Let us see how it looks now our constant volume batch reactor we have looked at the design equation for batch input output generation accumulation. There is no input there is no output therefore, this is DNA by D T this is the statement of material balance for a batch reactor something that we have done. Now, once again C A is what N A by V and since this constant volume batch therefore, V is V 0 therefore, C A becomes C A 0 1 minus of X A. Please notice that the function C A if for a constant volume batch is C A 0 1 minus of X A, but for the case where there is constant pressure we have the effect that we have already shown. Now, you can substitute N A in terms of N A 0 and so on finally, the kind of equation that you have to solve is this is the kind of equation that you have solved. Notice here this equation looks very different from what we have done for a constant pressure batch. So, we can put all the numbers and then integrate and so on. So, we will get for a constant volume batch the reaction time is given by X A by 1 minus of X A times 1 by K C A 0. So, this is the integrated form for the case when there is a constant volume batch. Please notice that the final result we get for a constant volume batch is quite different from what we got for the other two cases we have considered. If you put your numbers you find that the reaction time for a constant volume batch is 196 seconds. What are we saying? What we are saying is that when we do this in the case when this is for a constant pressure CSTR, this is for a constant pressure PFR then we also looked at what is the actual residence time here. Now, we are looking at a constant volume batch reactor for this case we find that the reaction time for a conversion of 0.5 is about 196 seconds much lower than these two cases we have considered so far. So, the reaction time required for a constant volume batch is only 196 seconds. Let us see what happens if we conduct this the same reaction in a constant pressure batch. How do you get constant pressure? We get constant pressure by providing a piston which is able to move up and down. Depending on changes in pressure inside it keeps the volume it is volume changes keeping the pressure constant. So, what do we do? You allow the volume to change because the piston is able to move. So, we are looking at a case of constant pressure variable volume. This is the case of variable volume. So, let us quickly run through this how it looks. So, once again we are writing the material balance there is input output generation accumulation there is no input because the batch process there is no output because the batch process therefore, generation equal to accumulation therefore, d by d t f equal to r a v. Once again v by v this gas law will hold temperature does not change compressibility changes are not important there is no change in pressure because we are keeping the pressure constant therefore, v by v 0 that means change in volume of the equipment is n t by n t 0. We have shown from our stoichiometry that n t by n t 0 1 plus x a therefore, volume of the reaction equipment divided by the volume at 0 time is 1 plus x a. Now, what is concentration? We said concentration is number of moles divided by volume number of moles is n a 0 times 1 minus of x a therefore, we get we get that c a at any conversion is c a 0 1 minus of x a by 1 plus x a something similar to what we already got for constant pressure case earlier. So, we get the same kind of result c a is c a 0 1 minus of x a by 1 plus x a. Now, we know what is the rate function rate function we can therefore, substitute from our we can substitute for r a our rate functions are known therefore, we can substitute and integrate this to find our solution that is what has been done. So, left hand side d n a by d t is replace like this the right hand side r a v is replace like this. So, that you get d by d t of x a is given by this relationship. So, our result will come from the solution of this equation for the case of constant pressure variable volume batch when you integrate this is the result you get. On other words for the case of constant pressure variable volume batch reactor the answer we get is reaction time equal to this result. And you will notice that this is exactly the same result we got for the actual residence time enough p f r. On other words the result that we are getting for actual residence time in a p f r and for constant pressure variable volume batch is identical. Therefore, when you put all the numbers you get the actual time that is required t r is equal to 260 second that is what we got for the case of actual residence time in a p f r. So, what are we saying what we are trying to say is the following this is important we have the problem we are trying to solve is this that you have a reaction which is undergoing decomposition to give you methane carbon monoxide we are conducted the reaction in a constant pressure CSTR you conducted the reaction in constant pressure p f r you conducted the reaction constant volume batch and then you conducted this one in the case in a variable volume constant pressure batch. So, there are 4 cases notice here the initial pressure is one atmosphere pressure would change because of the reaction. Now, to understand this what I have done is I have just looking at how the different equipments would perform constant volume batch constant pressure batch constant pressure p f r constant pressure CSTR. I have just set out here what is the rate expression that we are getting for each case for each case what is the rate expression that we are getting. If you look carefully you find that constant volume batch the rate expression is k z 0 1 minus x whole square while in the other cases there is a denominator 1 plus x a whole square or in other words in a constant volume batch the rate at which chemical reaction occurs is always much much higher than the rate at which chemical reaction occurs if it is a constant pressure batch. And because the fact that the reaction rates are much higher in a constant volume batch the time of residence or time of reaction is much lower compared to the constant pressure batch. Similarly, if you look at constant pressure p f r once again we find that the reaction time is higher than constant volume batch. The reason is the rate expression itself shows you very clearly that the rate of chemical reaction in constant pressure batch constant pressure p f r is much lower than the rate of chemical reaction in constant volume batch that is why you get a higher time of reaction. Now, if you look at a constant pressure c s t r our rate expression if you see carefully is such that the it is much much lower than the others and therefore, you find that the reaction times are very large. So, what are we trying to tell what we are trying to say is that in our reactor design we must take care to see that the rate expression that we would choose or the equipment that you would choose should have very high reaction rates at every point of the equipment that is important at every point of the equipment. And that is why because it is very favorable in constant volume batch we get very small reaction time it is not so favorable in a c s t r therefore, we get very high reaction time. Therefore, choice of equipment really depends upon how well we understand the reaction kinetics. So, that we can maximize or minimize maximize the benefit or minimize size of the equipment that we use for our application. Let us just quickly recall what we have said what we have said is that we have a reaction and this in this reaction we are trying to look at what is the performance of various kinds of devices for a given reaction the reaction of our interest is we wrote this reaction as acetaldehyde C H 3 C H O giving you C H 4 plus C O correct C H 4 plus C O. Now, we looked at various devices are just summarize this just to bring to your attention some important features. What we have shown through our calculations is that constant volume batch reactor has the rate expression if you see the rate expression for a constant volume batch and compare this with constant pressure batch this is also a constant pressure C S T R. The rate expression shows very clearly that the reaction rates that we can achieve in a constant volume batch is much much higher than the reaction rates that you will get in constant pressure and then and in a constant pressure C S T R. In both these cases the reaction rates are less because the rate expressions are like that therefore, if we can actually utilize a constant pressure constant volume kind of batch equipment will get higher reaction rates that is one important point that we trying to get across. The second point we try to get across to you is that even if you are having a constant pressure process for example, a constant pressure P F R and a constant pressure C S T R. Now, both cases the reaction rate expressions are the same, but notice that the residence time is 261 for a P F R and 878 for a C S T R. How do we explain this? We explain this by recognizing that there is a fundamental difference between C S T S C S T R and a P F R because the residence times are very different. In other words C S T R because it operates at exit conditions of conversion conversion exit conditions of conversion the reaction rates that is achieved in a C S T R is that of the reaction rates at the exit. While in a P F R the reaction rates are at the instant positions at which the reaction is taking place. In other words on even though the reaction rate expressions are the same P F R has a much higher average reaction rates compared to C S T R. This point must be borne in mind in the choice of reaction equipment. So, this is one example that you would like to recognize. Now, there is another example we would like to take to illustrate how our system would perform when there are certain important changes taking place in the equipment. For example, the problem we would like to now address is there is a plug flow reactor. There is a plug flow reactor in which there is a reaction A going to be reaction A going to be and this plug flow reactor contains a catalyst. Now, what comes in the feed is coming in at 20 atmospheres and because of pressure drop the pressure that goes out is less than what is coming in and then the pressure gradient that we are expecting in this process is given as d p d w is minus of 0.2. On other words as fluid moves the pressure drops at this rate d p d w is minus of 0.2 pressure keeps on decreasing and because of decrease in pressure we expect some effect on the process and that is what we would like to quantitatively understand and evaluate. What are we saying that rate of chemical reaction is given as minus of k times C A. It is a first order reaction and k is rate constant rate constant is given everything is given. Now, what is being said in this problem is that you in an experiment in which you have feed coming in feed going out and so on you are experiencing as 86.5 percent conversion. Now, what is asked of you is what do you expect to see if there was no pressure drop you understand that this there was no pressure drop what do we expect to see that is one. Second point is that if instead of using a P F R if we had used a C S T R what do we expect to see. That means you have got some data based on in your experiments where you find 86.5 percent is the conversion because of this pressure gradient and you would like to know suppose first by some design of this catalyst you could have substantially eliminated this pressure drop due to this catalyst. What is the likely benefit you will see in this equipment P F R or alternatively what is the likely benefit you will see if you had a C S T R. So, this is the question that we are trying to answer. Let us see how to understand this. Now, our data says that it is 86.5 percent conversion using the existing equipment where there is pressure drop in this D P D R will use 0.2. So, we have to first model this equipment to understand what are the features of this equipment that we are using. What does it say D F A D V say by this P F R equation K A which is given as minus of K C a first order reaction. Now, it is a gas phase reaction therefore our gas law will hold V by V 0 F T by F T 0 T by T 0 P 0 by P Z by Z 0. This is gas law we know from our problem statement that pressure changes because it is of the pressure gradient D P D minus of 0.2. So, that when you integrate this you get that the pressure at any point is pressure at the inlet multiplied by 0.2 times W that means pressure keeps on decreasing by this relationship. So, that if you want P by P 0 we get this relationship. So, why are we writing this relationship we are writing this relationship because we know that our volumetric flow depends on pressure. So, how the pressure changes accordingly or volumetric flow is affected and if volumetric flow is affected we know concentrations are affected. So, you will find the pressure will have an effect on concentration and therefore it will have an effect on the rate of chemical reaction. So, it is that effect that we want to quantitate in this problem. So, what are we saying what are we saying is that D F D A D V is R A which is minus of K C A. Now, we have been given catalyst W. So, it is not volume that is given what is given is weight of catalyst. So, we would express this in terms of the data that is given that is what is done here volume is expressed as W divided by rho is volume. So, I put the rho on the numerator. So, we have this equation which tells you the first our what is called as our design equation. So, C A is what we said C A is F A by V, but F A is what F A 0 1 minus of X A from stoichiometry and what is V we have said just now we have said just now from here V by V 0 is what P 0 by P what is. So, that means this relationship appears here. So, when you substitute this into this relationship we find V by V 0 becomes our V becomes V 0 times P 0 by P. So, this effect this effect of volume V by V 0 what is V equal to V 0 times P 0 by P because T by T 0 is 1 Z by Z 0 is 1 F T by F T 0 is 1 because there is no change in volume due to reaction. So, but there is change in volumetric flow because of pressure variation that is the effect we are trying to quanticate. So, what we have got here concentration at any position is F A 0 times 1 minus of X A it comes stoichiometry when volumetric flow changes because of pressure variation is V 0 times P 0 by P. So, that now we see concentration C A is written as C A 0 1 minus of X A multiplied by this term which takes into account the effect of pressure drop. Now, we can substitute all these things into our design equation here. So, that you get D by D V or D by D W of X A rated which conversion changes with weight of the catalyst is given by K times rho times V 0 where are we this how it comes you have substituting for C A here. So, it comes relationships like this where K by V 0 rho which we do not know X A is given as 0.85 for this particular case of P F R. So, if you integrate this if you integrate this this equation essentially tells you how the P F R that we have got in our experiment is performing. That is what has been done that is what has been done and we find that when we integrate this we get a relationship like this. This is the final result we get we integrate the final result we get is for this P F R containing 60 K G of catalyst catalyst 1 minus of X A is multiplied alpha times W minus beta and so on. Where beta is what beta is W by beta is beta equal to 0.2 I may have mentioned beta is 0.2 by p. So, that is beta. So, X A is given what is the alpha alpha is this which is unknown p 0 is known. So, this quantity is known beta by p 0 is known. Therefore, we can substitute the value of X A coming from our experiment that is what I have done. What I have done here is that in this solution I have substituted the data with X A is given X A not X A X A is 0.865 W is given. So, many K G beta is given p 0 is 0.01. So, we can put all the numbers and find out what is the value of alpha that is appropriate to our experiment. So, essentially what we have done so far is that from the data that is given to us we have calculated what is the value of we have calculated what is the value of parameters that describes this data. Our parameters are alpha by is the parameter of the process. Therefore, alpha is equal to 0.0476. So, what have we done what we have done is that we have taken a problem in which data is given X A is 0.865 then the weight of the catalyst is given in that pressures are given. So, we have characterized this in terms of what is the value of alpha that is responsible for this kind of chemical reaction. Now, the question is now that the alpha is known we can answer the question that we want to answer what happens if there is no pressure drop. What happens if you instead of using P F R if you use a stirred tank let us try to answer the question. So, suppose we have a plug flow reactor in which there is no pressure drop for which this value of alpha is known this is alpha this is alpha which is known in this case given as 0.0476. Therefore, what we simply have to find out integrate this when you integrate this you find that our P F R in the absence of pressure drop is given by this relationship this is the solution our solution is this. And what is the value of X A if I ask you alpha is known because alpha is 0.4076 W is known which is 60 kg. Therefore, you can find out what is the conversion that we would expect if there was no pressure drop. Now, the context to this question is the important why are we looking at the case of no pressure drop it tells us that what was 86.5 when there is pressure drop actually becomes 94 when there is no pressure drop. On other words there is considerable benefit in a process to significantly reduce pressure drop by a suitable design of your catalyst that is the point that is being made. The second question that is that we are trying to answer is suppose instead of using a P F R we had used a stirred tank that means what is the stirred tank you have a catalyst 60 kg of catalyst you want to support in on a basket that means you have a spinning basket into which you have to support the 60 kg of catalyst. It is not a very difficult thing to do appropriately already design this stir and so on. And once you do that you are able to put your feed your feed comes in and your feed goes out product go out. And your reactor design equation is V by F V equal to F A 0 X A by minus of R A. So, that you can put all your rate functions here rate function R A is we have just now been shown that R A is minus of K times C A the first order process. Therefore, you can find out what is the weight of the catalyst weight of the catalyst in terms of the parameters of the system what are the parameters of the system here alpha alpha is given. Therefore, you can find out what is the extent to which you can drive the reaction what is the value of X A what is the value of X A. So, let me do this calculation for you W 60 and then V 0 alpha is alpha is K by K by V 0 that is 0.0076 0.0476 equal to X A by 1 minus X A. So, you can find out the value of X A. So, what we are saying is that we can find out the value of X A since we know the alpha values. So, the problem that we are try to solve is in the absence of pressure drop there are significant benefits to be achieved if you have a plug flow reactor without pressure drop. Now, the idea of doing what is called as a CSTR spinning basket is something very important that you should recognize. Now, generally to be able to get good data on kinetics on a catalyst it is important to have temperatures around the catalyst to be uniform and spinning baskets give you very good temperature uniformity and this is what makes spinning basket reactors extremely valuable for catalyst evaluation. How well the catalyst performs under various conditions get good kinetic data. So, that you can model the kinetics properly which you can use for design. So, you will find that CSTR data is what is generally useful on the point of view of trying to get good kinetic data for design. Having said this about pressure drop there are few things that I would like to draw your attention from the point of your pressure drop in a process. I have taken an example here the example is that suppose we look at sulfur dioxide reaction sulfur dioxide giving you sulfur trioxide SO 2 plus half O 2 to the high SO 3. The very old process running may be for last 150 years vanadium pentoxide catalyst is well known. I have just taken some data from a 3000 tons per day SO 2 plant. The plant itself it is in 4 stages I mean for at after every stage the sulfur dioxide produced is taken for absorption and then it is put back. So, what is called as double contact double absorption process, but our concern here is something else typically the beds are about 0.25 to 0.75 meters thick and about 70 to 80 tons of catalyst what is used and the pressure drop that is observed is about 0.05 atmosphere. This typical pressure drops that we see in the sulfur dioxide plant. Now a 3000 tons per day sulfur dioxide plant because of nitrogen coming in along with oxygen and so on the actual flow at the reaction conditions is around 317 cubic meters per second. So, fairly large flow you have a pressure drop of 0.05 atmosphere. I put it in terms of newtons and then if you take the efficiency of blower to be about 70 percent which is about typical you find that the energy that is required to actually pump these gases into the system requires 2.5 megawatts. The point I am trying to put across to you is that the energy that is required to overcome pressure drop is a very significant part of our cost. And therefore, any work that we can do to take care of this pressure drop is would be extremely valuable. So, it usually shows in a sulfur dioxide plant we are spending about 2.5 megawatts of energy to be able to push the gases through the system and solve because of pressure drop. So, just to sort of emphasize the point that anything we can do to design catalyst better it will be great advantage from the point of view of energy consumption of course, from the point of view of cost as well that is the point I am trying to across. So, let us just summarize what we have tried to say in this exercise is that we have looked at design equations for idealized cases. And then we have said that the design equations various choices exist. And these choices give you various kinds of reactor designs and these reactor designs you know give you various kinds of performance. And those performances have implications in terms of cost and so on. And we took the example of pressure drop to illustrate how our designs must look for catalyst which is not very high in terms of pressure drop. We will go to the next item now we will is suppose instead of looking at what is called as a plug flow reactor we put a recycle. Let us say we are considering a reaction a a plus b b going to c c plus d d this is the reaction we are considering. Now, our interest in trying to understand a recycle are many why do we recycle we generally the recycle that you will up you will find in process industry is mostly because we want to make better use of this of the of the energy here. Because it is hot we want to make use of the heat that is one reason that of course, there are other reasons for looking at recycles. But before we do that let us just look at recycle from a more fundamental point of view. We want to set up design equation therefore, we want to understand how to handle recycle. So, to illustrate this what I have done I have just put down suppose there was no catalyst that means you have a recycle ratio I have taken an example here. Let us say the recycle ratio is 4 how do you define recycle ratio moles in in stream 4 to moles in stream 3 it is defined that way recycle ratio is defined as moles in stream 4 divided by moles in stream 3. There are some people who define this volume volumetric flow in stream 4 divided by volumetric flow in stream 3 instead of moles people might define in terms of volume. But you should be careful here because temperatures at 3 and temperature at 4 may be different therefore, those ratios are be carefully looked at therefore, I have taken recycle ratio in terms of moles if a moles at 3 to moles moles at 4 to moles at 3. Now, I ask you suppose there is no chemical reaction there is no there is no catalyst here therefore, you are putting a recycle ratio of 4 what are the flow rates at various points. Now, it stands to reason F A 0 if it is 1 what is F A 3 it has to be 1 because you know whatever course it must come out. Now, what will be F A 4 4 times this I have put 4 here what is F A 2 and 3 I have put the numbers on other words what we are trying to say here is when there is no chemical reaction or in other words when there is no catalyst in the reactor the molar flow rates through the system at position 1 and position 2 position 1 is 5 and position 2 is 5. Why is position 2 5 because there is no reaction why is position 1 5 because there is no reaction. But the important point we should bear in mind is that the fluid that is entering at position 1 is 5 units 5 units are entering. Now, let us let us say what happens to flow rates at all these positions when the extent of reaction is about 40 percent of x A is 0.4. So, if I say that we are entering at 1 mole per hour 1 mole per whatever the number may be per hour per second whatever is 1 if I say the conversion is 0.4 and if I ask you what is coming out at position 3 you will say it is 0.6. Why because 40 percent is the conversion. So, if I say conversion is 40 percent as I said mentioned here how much is coming out at position 3 clearly it will be 0.6 because 40 percent is converted. Now, if I say that recycle ratio is 4 how much is F A 4 you will say it is 2.4 4 times 0.6 2. How much is F A 2 what you will say it is F A 3 plus F A 4 therefore, it must be 3. What is the F A 1 what you will say F A 1 is F A 0 plus F A 4 F A 0 is 1 F A F A 4 is 2.4 therefore, F A 1 is 3.4 this comes from our material balance. So, on the basis of the numbers that I have given you will find that the flow rate at position 1 is 3.4 flow rate at position 3 is 0.6 position 2 is 3.0. The question that I want to ask you is what is the conversion it at position 1 what is the conversion at position 2. How would you answer this question what is the conversion at position 1 what is the conversion at position 2. A simple answer you would give me is between position 2 and position 3 there is no chemical reactor. Therefore, we should expect that the conversion at position 2 and position 3 must be the same. So, that means at position 2 conversion should be 0.4 position 3 also conversion should be 0.4 here we know that the conversion is here we know that conversion is 0.4 anyway based on our physical understanding. We want conversion here also to be 0.4 how can we make conversion here 0.4 by an appropriate choice of our reference. So, if we chose 5 as our reference that means if we chose 5 units as the reference on the basis of which we will define conversion inside the recycle loop then clearly we find that conversion here at position 2 is 0.4. What I am trying to get across to you is that whenever you are looking at a recycle device the choice of reference inside the recycle device is important. And that choice should be the moles of component a entering at position 1 in the absence of reaction because our reference is always on the basis of absence of reaction. How much is going into the equipment in the absence of reaction we know that at position 1 5 units are entering in the absence of reaction you can see here 5 units are entering in the absence of reaction. Moment reaction took place the amount of coming here is 3.4. Therefore, what I want to ask you is that what is the conversion that is appropriately defined in the presence of reaction at position 1. Now, you will tell me that in the absence of reaction 5 units are entering in the presence of reaction it is only 3.4. Therefore, conversion at position 1 is 3.4 divided by 5 1 minus. So, conversion here it would be 1 minus 3.4 divided by 5 this would be the conversion. Why divided by 5 because 5 was the units that was entering in the absence of reaction 3.4 is in the presence of reaction and 1 minus this is the conversion. So, what is it 5 3.4 divided by 5. So, this becomes 0.32. So, this becomes 0.32. So, what are we saying what we are saying is when we have a recycle device we need a proper reference to be able to work inside the recycle device. To work inside the recycle device we must have a proper conversion. As we can see in the slide that inside the recycle device if you define conversion on the basis of 5 units you find that conversion at 0.1 is 0.32, conversion at 0.2 is 0.4, conversion at 0.3 is also 0.4. Showing that as you go from 0.2 to 0.3 the conversion remains consistent with our understanding. So, this is the most important point in the recycle device we must choose conversion which is consistent across 0.2 and 0.3. Similarly, conversion at 0.1 should be consistent with respect to what enters at position 1 in the absence of reaction which in this case is 5 units. But, in expressed in terms of our variables what we are saying is that if F A 0 is entering the process what enters recycle system is R plus 1 F A 0. What we are saying is that what enters the reactor what enters the reactor in the absence of chemical reaction is R plus 1 F A 0. When R is 0 then what is entering is F A 0 when R is finite what enters the reactor in the absence of chemical reaction is R plus 1 F A 0. Therefore, inside the recycle loop our reference for defining conversion should be R plus 1 F A 0. We take up more of this when we meet next time. Thank you.