Reactor Sizing and Conversion Principles

Here we are in chapter 2 where we're going to talk about conversion and reactor sizing. We are in lecture 6 segment 1. In this segment we're going to introduce chapter 2, we're going to define conversion and we're going to develop the design equations in terms of conversion. Okay, before we start, I have some reflections to share with you which actually coming from the authors Davis and Davis and fundamentals of chemical reaction engineering they state the subject of chemical reaction engineering was initiated and evolved primarily to accomplish the task of describing for a given reactive system How to. So CRE is good or developed to help you describe how to, how to what. Choose the proper reactor. You want to know which reactor is the best for you. Is it batch reactor? Is it plug flow reactor? Pack bed reactor? Semi-batch reactor? Fluidized bed? And so on. CSTR probably. what else calculate its size not only we should know what is the proper reactor what what type of reactors is the most suitable for my application should also be able to calculate its size what else determine its optimal operating condition that's also very important so if we have we know what's our the best type of reactor to use and we know we calculated the required size but we also need to determine what's the optimum operating condition what's the best operating condition that i should run the reactor at so that i get maximum conversion or maximum conversion and selectivity and so on okay so let's go into chapter two In the first chapter, the general mole balance equation was derived and then applied to the four most common types of industrial reactors, the ideal reactors, of course, that we have discussed. In this chapter, okay, so this is what we did, basically, okay, derive the general mole balance equation and apply it to the four ideal reactors. In this chapter, we're going to define conversion. We're going to develop the design equations. In terms of x, we're going to size CSTR and size black flow reactors given these kinetic data, minus RA versus x, and then we're going to compare conversions for reactors in series. Definition of conversion. In defining conversion, we choose one of the reactants, preferably the limiting reactant, as the basis of calculation. So if we have this reaction, we have A moles of A react with B moles of B to give you C moles of C and D moles of D. We have to choose the limiting reactant. Okay, so we check the feed and from the feed and the stoichiometry, we determine which reactant is the limiting reactant, which compound is the limiting reactant, and we make it the basis of our calculation. Therefore, we're going to divide the whole equation, our stoichiometric equation, on the stoichiometric number of, stoichiometric coefficient, I should say, of A, for example, if A was the limiting reaction. So, we rewrite the reaction, stoichiometric equation, this way. Okay, now, every quantity is on a per mole of A basis, meaning if I say, B over A, and this is a number, maybe it is 2, maybe it's 1, whatever it is, okay, B over A, moles of B is reacted with every 1 mole of A reacting. And we say we have C over A, that's again a number, okay, maybe 3, maybe 4, maybe 1, maybe half, whatever, okay. C over A moles of C. is produced per every mole of A reacted. Okay, so now we chose our basis of calculation. The conversion X is defined as, okay, so here we have XA, okay, moles of A reacted divided by moles of A fed, right? And this is something you have learned in previous courses. The conversion is moles of A reacted. How many moles of A reactant? Pair mole of A fed. And usually, Shabab, we don't write the subscript A. It's understood that if we say X, X is based on the limiting reactant. For irreversible reactions, the maximum conversion is 1, of course. For reversible reactions, the maximum conversion is the, and yes? What's the maximum conversion for a reversible reaction? Correct, it's the equilibrium conversion xe. You cannot go beyond equilibrium conversion. If A is the limiting reactant and B is provided in excess, can xb reach one, any hundred percent conversion of B? So, for this reaction if A is my limiting reactant but I decided to define conversion with respect to B not with respect to A which is we don't do this okay so can XB reach 100% of course not because A will finish before B correct so we'll have some B left over there did not unreacted B therefore do not use the excess species as our basis of calculation because it doesn't make sense okay it's much better to choose the limiting reactant as basis of calculation so i know that the conversion would be somewhere between zero and one minimum zero conversion no reaction maximum one for our reverse for a irreversible reaction Let's talk about batch reactor design equation. In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to products until until what? Well, until either equilibrium is reached or the limiting reactant is exhausted, meaning you get 100% conversion. Okay, let's write x equation symbolically. So, we know that xA is mole of A reacted divided by mole of A-fet. So, how do you write moles of A reacted? Well, you write it as NA0 minus NA, right? We know that number of mole is not conserved, so it can be destroyed or created, meaning it can be consumed or produced. Okay, so let's see. We have NA0, which is the initial number of mole. Okay, we have NA, which is the number of mole left. And the difference, where did it go? The difference basically reacted. Correct? Reacted. Okay, divided by moles of a fit, which is NA0. This is how we calculate X. Okay, let's do some. manipulation. So let's do cross multiplication. We have NA0 multiplied by X equals NA0 minus NA. Let's take NA to the other side, or let's take this guy to the other side and bring NA here. So NA equals NA0 minus NA0 times X. We could take NA0 times X. common factor, we have 1 minus x, so now we have NA equals NA0 minus x. Okay, very good. So, here we go. That's the equation that we have now. Okay, let's see, what does this equation mean? So, every equation tells you a story. This equation actually is telling you the following story. that the remaining number of moles of A The remaining number of moles of A basically equals what? Let's see. We know that x is the fractional conversion, right? The fraction of the material converted. Okay, so what is 1 minus x? 1 minus x is the fraction unconverted, right? It's the fraction unconverted, meaning it's the fraction remaining. Remaining out of what? Remaining out of Na0. So if we multiply the fraction unconverted into Na0, we get the quantity unconverted, Na. Okay, so starting from the general mole balance equation, derive the batch reactor design equation in terms of X. Let's do that. Okay, so this is the differential form of batch mole balance equation used for interpretation of reaction rate data. Okay, so this is the general mole balance equation that we derived in chapter 1. So, let's now try to introduce x so that we have the design equation in terms of x. Okay, so we just said that we have an A. correct equals NA0 1 minus x so and we said this can be written as NA0 minus NA0 times x so this is NA0 now let's take the derivative of both sides okay let's take the derivative of both sides so we have here d NA equals what's the derivative of NA0 and what's the derivative of Na0x, well the derivative of Na0 is 0 because Na0 is constant. So its derivative is 0 and since Na0 is constant, it goes outside the derivative and we have Na0 dx. Okay, great. So we come to this equation and instead of writing dNa, which is this, It's not the DNA of biology. Okay. So instead of writing DNA, we write minus NA0 DX. Okay. And then we have DT equals RA times V. We can rearrange it if we want to write DX by DT equals minus RA times V. That's what we always like to write. R A as minus R A divided by an A naught. So now we have what is known as the design equation for batch reactor written in terms of x. So here we go. Now we have the differential form of batch design equation usually used to find V. or can also be integrated as you can see here so now we have the integral form the integral form of batch design equation usually used to find t okay and therefore voila we have a design equation for a batch reactor written in terms of conversion and remember the longer the reactants are left in the reactor the greater will be the conversion okay let's go to design equations for flow reactors for a battery reactor conversion increases with time spent in the reactor so the more time the molecules spend in the Reactor, the more chances of reaction, so more reaction will take place because more collision would take place, and therefore you get more conversion. For continuous flow systems, this time usually increases with increasing, increasing what, Shabab? Increasing the reactor volume, correct. So the larger the volume, the larger the time. the molecules spent inside the reactor especially we're talking about the reactants okay and the more chances of collision therefore more reaction and more conversion the bigger or the longer the reactor the more time it will take the reactants to flow completely through the reactor and thus the more time to react therefore we say conversion is function of volume the larger the volume Probably the larger the higher the conversion. For continuous flow systems, X can be defined mathematically as, okay, can you guess how we write it? Yes, it will be written as F A naught minus F A. So we have F A naught, which is introduced to the reactor, whether it is CSTR or plug flow reactor. This is F A naught. and then you can compare it to F A. Where did the difference go? Well, if we take F A naught minus F A, the difference is reacted. So that's why this is mole A reacted divided by mole of A fit. And similarly, we can write the following equation. Therefore, F A equals F A naught times 1 minus X, where X is the fraction converted. 1 minus x is the fraction unconverted we multiply it by how much Fid so how much Fid multiply by the unconverted fraction gives you the unconverted number of moles gives you the remaining number of moles okay recall that if a0 equals Ca0 times epsilon 0 you know Ca0 is the entering concentration of A and O is the entering volumetric flow rate and this will give you if A0 you're not sure you write the units so let's see this is mol of A right divided by let's say liter okay multiplied by volumetric flow rate is liter let's say per second so you have mol of A per second which is If A0, this is how much is the molar flow rate of A at the entrance. For liquid systems, C A0 is given in terms of molarity, correct? So whether it is specified when you buy the feed, or you could always take a sample from the feed and take it to the lab and measure its concentration. the report will be given in terms of molarity okay for example one molar half molar and so on okay for gas systems for gas systems CA0 can be calculated from how do we calculate CA0 so how do we calculate CA0 for a gas flow system if you want to define of course our issue is that we need to know epsilon naught and we need to know C A naught here in order for us to be able to calculate F A naught we know that this is done through the volumetric flow meters so now we'll talk about this for liquid we said we take a sample we to go to the lab and do this analysis what about gases how do you calculate C A naught how do you determine C A naught well you will determine it through an applicable equation of a state. Correct? Okay, how is that? Well, you know already that P, if you want, we can do, well let's start with a gen, let's say ideal gas. Okay, so make it simple. So P times V, or in this case because we have a continuous flow reactor, so we have times epsilon, okay, equals the molar flow rate so that would be Ft okay multiply by RT and we talked about the feed so this will be not this will be not this will be not and this will be not okay so what about if we concentrating on a We're concentrating on A, so that would be P A naught, the partial pressure of A, correct, times the epsilon naught, and here this will be F A naught times A times T naught, and let's see what happens if we do the following. If we take here F A naught divided by epsilon naught, okay, and this side we have P A naught, over RT0, then we know that this guy, this guy is simply CA0, right? So CA0 equals PA0 over RT0. Okay, great. So let's continue talking about this. Okay, so for gas systems, at high temperature and low pressure. CA0 can be calculated from this equation that we have just derived together. You can always use the mole fraction at the entrance and the total pressure and of course at the temperature as well. So you can do either this or you can do this. Okay, so why we are saying this is only applicable at high temperature and low pressure? Well, because at high temperature and low pressure, most gases behave ideally, right? Therefore, we can use the ideal gas equation of state. Why is that? Why do they behave ideally at high temperature and low pressure? Well, basically, at high temperature, you have high volume, right? And low pressure, you have even larger volume. In other words, you have small density. correct which means if the volume is so large and the number of moles is that given number of moles that means the molecules are far from each other so they do not interact with each other okay because they are far from each other okay and having them not interacting that will take you to the ideality right because if the now if the moles interact with each other either attracting or repulsing from each other then That's not an ideal gas. Okay. And the other thing is, remember, we say the volume of the gas-U-S system is coming from the, basically, the void between the molecules, not because of the volume of the molecules itself. It's not like a liquid. For a liquid, you have a volume for a specific number of mole because the volume is coming from the volume of the molecules. and that's even more true for solid but for gases the volume is based on that void the total space without the volume of the molecules however if you have very high pressure and very low temperature and the molecules are very close to each other so the total volume or let's say the volume of the molecules will become appreciable compared to the volume of the total system. But if you have high temperature and low pressure, the volume is so large and therefore the volume of the molecules are negligible compared to the total volume. So we don't have to take into consideration the volume of the molecules, which again, this takes us to the ideal gas low. And therefore the ideal gas low here is correct regardless to the type of the gas regard therefore regards to their volume. Okay, design equation for CSTR starting from the general mole balance equation derive the CSTR design equation terms of x. So this is the general mole balance equation and we said let's say for A for instance okay this can be written as FA0 minus FA, and for a CCR, the reaction mixture is well mixed, so the rate of reaction is not function of volume, right? The rate of reaction is not function of volume because simply the reaction mixture is well mixed, so the temperature is the same everywhere, the concentration is the same everywhere inside the reactor, therefore it's not function of volume. volume so it goes outside and upon integration we have this guy and we said c scale is operated usually under a steady state so that's what we have it's called zero accumulation equals to zero What else? Well, let's see, we said we can take this to the other side, so we have fA0 minus fA equals minus star A times volume. Let's look at this guy, and remember the objective is to introduce x into this equation. What is the definition of x? Well, the definition of x is fA0 minus fA divided by F A naught. Okay, so that means, that means F A naught minus F A equals F A naught times x. So here we can write F A naught times x equals minus R A times V, which takes us to this equation, V equals F A naught times x divided by minus RA. So now we have a design equation for CSTR written in terms of X. Excellent. Taib, so here we go. We said the general mol-basic equation can be written this way, and then we say this is how we introduce X. Okay, let's look at the rate of reaction. Okay. Let's look at the rate of reaction. In fact, this rate of reaction is the rate of reaction inside the reactor, right? It's inside the reactor, okay, because we say the volume is a function of the rate of reaction. The larger the rate of reaction, the smaller the required volume to achieve a given conversion. However, why are we evaluating the rate of reaction at the exit? Well, simply because the CSTR is perfectly mixed. Therefore, the exit composition and temperature, of course, is identical to that inside the reactor. So, the temperature and the concentration at the exit is exactly the same as those inside the reactor. Okay, so they are the same. So, you can evaluate the rate of reaction, whether inside the reactor or right at the exit. For simplicity, we evaluate it at the exit. Therefore, the rate of reaction is evaluated at the exit. Exit conditions. Let's go now to plug flow reactor. And remember what we're trying to do is we take the design equations that we derived for the four ideal reactors in chapter one. We're trying to take these equations and introduce x into them so that we have design equation written in terms of x. Okay, it's the turn of plug flow reactor. We model the tubular reactor as having the fluid flowing in plug flow, i.e., no radial gradients in concentration, temperature, or rate of reaction. So, for a given plug flow reactor, we know that if we take a plug, one of the plugs here, one of the plugs, we know the concentration in the middle, same concentration here next to the wall, same concentration here, the same thing for temperature, and the same thing for rate of reaction, right? We said that. just as a revision however as the reactants enter and flow axially okay so it goes through the axis so it flows axially down the reactor they are consumed and the conversion increases along the length of the reactor so the conversion here is zero at the beginning and then continues increasing the months starting from the general mole balance equation Derive the plug flow reactor design equation in terms of x. Okay, so starting from here, we learned that in order to convert this to a suitable equation for plug flow reactor, we said you take the derivative of both sides, correct, derivative of both sides, and we said simply we, this is of course zero, right, this the derivative of this is zero, so we'll end up minus d fi by dv and this function is first integrated with respect to v then derived with respect to v so it remains as is so we get this equation when we do the balance for a. Okay now how can we introduce x here in this equation? Well this is the differential form of black flow reactor more balanced equation or the design equation and we remember that a equals if a0 1 minus X therefore therefore can be written as fA0 minus fA0 times x and now we can take the derivative of both sides so we have dF A equals of course dF A0 is 0 because fA0 is constant minus fA0 A x so basically dF A equals minus f A naught dx. So we come here, instead of d f A, we can write minus f A naught dx by dv equals r A. Of course, we can always take the negative to the other side, okay, or we can also do this, equals minus r A divided by f A naught. Okay, now we have a design equation for plug flow reactor return terms of x. Okay, so here we go. Differential form of plug flow reactor design equation. And then this is the integral form of plug flow reactor design equation. Great. What is needed in order to integrate the above equation? We need to integrate here. part of the equation. How do we, what do we need? Well, basically, we need minus rA as a function of x, correct? So that we substitute for minus rA with an equation where you have x, and then integration would be possible. Okay, so we need to know minus rA as a function of x, and this is something we're going to learn in chapter 3. Okay, the last ideal reactor. We need a design equation for this reactor, for packed bed reactor, written in terms of x. Packed bed reactors are tubular reactors filled with catalyst particles. The packed bed reactor design equation in terms of x will be something similar to the design equation of plug flow reactor, except that we are using W instead of V, and we're using minus RA prime instead of minus RA. Okay, the differential form must be used when analyzing reactors that have pressure drop along the length of the reactor. So we're talking about pressure drop. Okay, and this is something we're not discuss in chapter 4, not now. Okay, if pressure drop is zero, the integral form can be used straightforward. And now we have the integral form for packed bed reactor. design equation. Now in this segment we developed together the design equations for the four ideal reactors written in terms of x. In the following lecture we're gonna learn how to use them. See you soon.

Choose the proper reactor. You want to know which reactor is the best for you. Is it batch reactor?

Is it plug flow reactor? Pack bed reactor? Semi-batch reactor?

Fluidized bed? And so on. CSTR probably.

what else calculate its size not only we should know what is the proper reactor what what type of reactors is the most suitable for my application should also be able to calculate its size what else determine its optimal operating condition that's also very important so if we have we know what's our the best type of reactor to use and we know we calculated the required size but we also need to determine what's the optimum operating condition what's the best operating condition that i should run the reactor at so that i get maximum conversion or maximum conversion and selectivity and so on okay so let's go into chapter two In the first chapter, the general mole balance equation was derived and then applied to the four most common types of industrial reactors, the ideal reactors, of course, that we have discussed. In this chapter, okay, so this is what we did, basically, okay, derive the general mole balance equation and apply it to the four ideal reactors. In this chapter, we're going to define conversion. We're going to develop the design equations.

In terms of x, we're going to size CSTR and size black flow reactors given these kinetic data, minus RA versus x, and then we're going to compare conversions for reactors in series. Definition of conversion. In defining conversion, we choose one of the reactants, preferably the limiting reactant, as the basis of calculation.

So if we have this reaction, we have A moles of A react with B moles of B to give you C moles of C and D moles of D. We have to choose the limiting reactant. Okay, so we check the feed and from the feed and the stoichiometry, we determine which reactant is the limiting reactant, which compound is the limiting reactant, and we make it the basis of our calculation. Therefore, we're going to divide the whole equation, our stoichiometric equation, on the stoichiometric number of, stoichiometric coefficient, I should say, of A, for example, if A was the limiting reaction. So, we rewrite the reaction, stoichiometric equation, this way.

Okay, now, every quantity is on a per mole of A basis, meaning if I say, B over A, and this is a number, maybe it is 2, maybe it's 1, whatever it is, okay, B over A, moles of B is reacted with every 1 mole of A reacting. And we say we have C over A, that's again a number, okay, maybe 3, maybe 4, maybe 1, maybe half, whatever, okay. C over A moles of C.

is produced per every mole of A reacted. Okay, so now we chose our basis of calculation. The conversion X is defined as, okay, so here we have XA, okay, moles of A reacted divided by moles of A fed, right? And this is something you have learned in previous courses. The conversion is moles of A reacted.

How many moles of A reactant? Pair mole of A fed. And usually, Shabab, we don't write the subscript A.

It's understood that if we say X, X is based on the limiting reactant. For irreversible reactions, the maximum conversion is 1, of course. For reversible reactions, the maximum conversion is the, and yes? What's the maximum conversion for a reversible reaction?

Correct, it's the equilibrium conversion xe. You cannot go beyond equilibrium conversion. If A is the limiting reactant and B is provided in excess, can xb reach one, any hundred percent conversion of B? So, for this reaction if A is my limiting reactant but I decided to define conversion with respect to B not with respect to A which is we don't do this okay so can XB reach 100% of course not because A will finish before B correct so we'll have some B left over there did not unreacted B therefore do not use the excess species as our basis of calculation because it doesn't make sense okay it's much better to choose the limiting reactant as basis of calculation so i know that the conversion would be somewhere between zero and one minimum zero conversion no reaction maximum one for our reverse for a irreversible reaction Let's talk about batch reactor design equation. In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to products until until what?

Well, until either equilibrium is reached or the limiting reactant is exhausted, meaning you get 100% conversion. Okay, let's write x equation symbolically. So, we know that xA is mole of A reacted divided by mole of A-fet. So, how do you write moles of A reacted? Well, you write it as NA0 minus NA, right?

We know that number of mole is not conserved, so it can be destroyed or created, meaning it can be consumed or produced. Okay, so let's see. We have NA0, which is the initial number of mole. Okay, we have NA, which is the number of mole left.

And the difference, where did it go? The difference basically reacted. Correct?

Reacted. Okay, divided by moles of a fit, which is NA0. This is how we calculate X.

Okay, let's do some. manipulation. So let's do cross multiplication. We have NA0 multiplied by X equals NA0 minus NA. Let's take NA to the other side, or let's take this guy to the other side and bring NA here.

So NA equals NA0 minus NA0 times X. We could take NA0 times X. common factor, we have 1 minus x, so now we have NA equals NA0 minus x. Okay, very good. So, here we go.

That's the equation that we have now. Okay, let's see, what does this equation mean? So, every equation tells you a story.

This equation actually is telling you the following story. that the remaining number of moles of A The remaining number of moles of A basically equals what? Let's see. We know that x is the fractional conversion, right? The fraction of the material converted.

Okay, so what is 1 minus x? 1 minus x is the fraction unconverted, right? It's the fraction unconverted, meaning it's the fraction remaining. Remaining out of what? Remaining out of Na0.

So if we multiply the fraction unconverted into Na0, we get the quantity unconverted, Na. Okay, so starting from the general mole balance equation, derive the batch reactor design equation in terms of X. Let's do that. Okay, so this is the differential form of batch mole balance equation used for interpretation of reaction rate data. Okay, so this is the general mole balance equation that we derived in chapter 1. So, let's now try to introduce x so that we have the design equation in terms of x.

Okay, so we just said that we have an A. correct equals NA0 1 minus x so and we said this can be written as NA0 minus NA0 times x so this is NA0 now let's take the derivative of both sides okay let's take the derivative of both sides so we have here d NA equals what's the derivative of NA0 and what's the derivative of Na0x, well the derivative of Na0 is 0 because Na0 is constant. So its derivative is 0 and since Na0 is constant, it goes outside the derivative and we have Na0 dx. Okay, great. So we come to this equation and instead of writing dNa, which is this, It's not the DNA of biology.

Okay. So instead of writing DNA, we write minus NA0 DX. Okay. And then we have DT equals RA times V. We can rearrange it if we want to write DX by DT equals minus RA times V. That's what we always like to write.

R A as minus R A divided by an A naught. So now we have what is known as the design equation for batch reactor written in terms of x. So here we go. Now we have the differential form of batch design equation usually used to find V. or can also be integrated as you can see here so now we have the integral form the integral form of batch design equation usually used to find t okay and therefore voila we have a design equation for a batch reactor written in terms of conversion and remember the longer the reactants are left in the reactor the greater will be the conversion okay let's go to design equations for flow reactors for a battery reactor conversion increases with time spent in the reactor so the more time the molecules spend in the Reactor, the more chances of reaction, so more reaction will take place because more collision would take place, and therefore you get more conversion.

For continuous flow systems, this time usually increases with increasing, increasing what, Shabab? Increasing the reactor volume, correct. So the larger the volume, the larger the time. the molecules spent inside the reactor especially we're talking about the reactants okay and the more chances of collision therefore more reaction and more conversion the bigger or the longer the reactor the more time it will take the reactants to flow completely through the reactor and thus the more time to react therefore we say conversion is function of volume the larger the volume Probably the larger the higher the conversion.

For continuous flow systems, X can be defined mathematically as, okay, can you guess how we write it? Yes, it will be written as F A naught minus F A. So we have F A naught, which is introduced to the reactor, whether it is CSTR or plug flow reactor. This is F A naught. and then you can compare it to F A. Where did the difference go?

Well, if we take F A naught minus F A, the difference is reacted. So that's why this is mole A reacted divided by mole of A fit. And similarly, we can write the following equation. Therefore, F A equals F A naught times 1 minus X, where X is the fraction converted. 1 minus x is the fraction unconverted we multiply it by how much Fid so how much Fid multiply by the unconverted fraction gives you the unconverted number of moles gives you the remaining number of moles okay recall that if a0 equals Ca0 times epsilon 0 you know Ca0 is the entering concentration of A and O is the entering volumetric flow rate and this will give you if A0 you're not sure you write the units so let's see this is mol of A right divided by let's say liter okay multiplied by volumetric flow rate is liter let's say per second so you have mol of A per second which is If A0, this is how much is the molar flow rate of A at the entrance.

For liquid systems, C A0 is given in terms of molarity, correct? So whether it is specified when you buy the feed, or you could always take a sample from the feed and take it to the lab and measure its concentration. the report will be given in terms of molarity okay for example one molar half molar and so on okay for gas systems for gas systems CA0 can be calculated from how do we calculate CA0 so how do we calculate CA0 for a gas flow system if you want to define of course our issue is that we need to know epsilon naught and we need to know C A naught here in order for us to be able to calculate F A naught we know that this is done through the volumetric flow meters so now we'll talk about this for liquid we said we take a sample we to go to the lab and do this analysis what about gases how do you calculate C A naught how do you determine C A naught well you will determine it through an applicable equation of a state.

Correct? Okay, how is that? Well, you know already that P, if you want, we can do, well let's start with a gen, let's say ideal gas. Okay, so make it simple.

So P times V, or in this case because we have a continuous flow reactor, so we have times epsilon, okay, equals the molar flow rate so that would be Ft okay multiply by RT and we talked about the feed so this will be not this will be not this will be not and this will be not okay so what about if we concentrating on a We're concentrating on A, so that would be P A naught, the partial pressure of A, correct, times the epsilon naught, and here this will be F A naught times A times T naught, and let's see what happens if we do the following. If we take here F A naught divided by epsilon naught, okay, and this side we have P A naught, over RT0, then we know that this guy, this guy is simply CA0, right? So CA0 equals PA0 over RT0. Okay, great.

So let's continue talking about this. Okay, so for gas systems, at high temperature and low pressure. CA0 can be calculated from this equation that we have just derived together. You can always use the mole fraction at the entrance and the total pressure and of course at the temperature as well. So you can do either this or you can do this.

Okay, so why we are saying this is only applicable at high temperature and low pressure? Well, because at high temperature and low pressure, most gases behave ideally, right? Therefore, we can use the ideal gas equation of state.

Why is that? Why do they behave ideally at high temperature and low pressure? Well, basically, at high temperature, you have high volume, right?

And low pressure, you have even larger volume. In other words, you have small density. correct which means if the volume is so large and the number of moles is that given number of moles that means the molecules are far from each other so they do not interact with each other okay because they are far from each other okay and having them not interacting that will take you to the ideality right because if the now if the moles interact with each other either attracting or repulsing from each other then That's not an ideal gas. Okay.

And the other thing is, remember, we say the volume of the gas-U-S system is coming from the, basically, the void between the molecules, not because of the volume of the molecules itself. It's not like a liquid. For a liquid, you have a volume for a specific number of mole because the volume is coming from the volume of the molecules. and that's even more true for solid but for gases the volume is based on that void the total space without the volume of the molecules however if you have very high pressure and very low temperature and the molecules are very close to each other so the total volume or let's say the volume of the molecules will become appreciable compared to the volume of the total system.

But if you have high temperature and low pressure, the volume is so large and therefore the volume of the molecules are negligible compared to the total volume. So we don't have to take into consideration the volume of the molecules, which again, this takes us to the ideal gas low. And therefore the ideal gas low here is correct regardless to the type of the gas regard therefore regards to their volume.

Okay, design equation for CSTR starting from the general mole balance equation derive the CSTR design equation terms of x. So this is the general mole balance equation and we said let's say for A for instance okay this can be written as FA0 minus FA, and for a CCR, the reaction mixture is well mixed, so the rate of reaction is not function of volume, right? The rate of reaction is not function of volume because simply the reaction mixture is well mixed, so the temperature is the same everywhere, the concentration is the same everywhere inside the reactor, therefore it's not function of volume.

volume so it goes outside and upon integration we have this guy and we said c scale is operated usually under a steady state so that's what we have it's called zero accumulation equals to zero What else? Well, let's see, we said we can take this to the other side, so we have fA0 minus fA equals minus star A times volume. Let's look at this guy, and remember the objective is to introduce x into this equation. What is the definition of x?

Well, the definition of x is fA0 minus fA divided by F A naught. Okay, so that means, that means F A naught minus F A equals F A naught times x. So here we can write F A naught times x equals minus R A times V, which takes us to this equation, V equals F A naught times x divided by minus RA. So now we have a design equation for CSTR written in terms of X.

Excellent. Taib, so here we go. We said the general mol-basic equation can be written this way, and then we say this is how we introduce X. Okay, let's look at the rate of reaction. Okay.

Let's look at the rate of reaction. In fact, this rate of reaction is the rate of reaction inside the reactor, right? It's inside the reactor, okay, because we say the volume is a function of the rate of reaction. The larger the rate of reaction, the smaller the required volume to achieve a given conversion. However, why are we evaluating the rate of reaction at the exit?

Well, simply because the CSTR is perfectly mixed. Therefore, the exit composition and temperature, of course, is identical to that inside the reactor. So, the temperature and the concentration at the exit is exactly the same as those inside the reactor.

Okay, so they are the same. So, you can evaluate the rate of reaction, whether inside the reactor or right at the exit. For simplicity, we evaluate it at the exit.

Therefore, the rate of reaction is evaluated at the exit. Exit conditions. Let's go now to plug flow reactor.

And remember what we're trying to do is we take the design equations that we derived for the four ideal reactors in chapter one. We're trying to take these equations and introduce x into them so that we have design equation written in terms of x. Okay, it's the turn of plug flow reactor. We model the tubular reactor as having the fluid flowing in plug flow, i.e., no radial gradients in concentration, temperature, or rate of reaction.

So, for a given plug flow reactor, we know that if we take a plug, one of the plugs here, one of the plugs, we know the concentration in the middle, same concentration here next to the wall, same concentration here, the same thing for temperature, and the same thing for rate of reaction, right? We said that. just as a revision however as the reactants enter and flow axially okay so it goes through the axis so it flows axially down the reactor they are consumed and the conversion increases along the length of the reactor so the conversion here is zero at the beginning and then continues increasing the months starting from the general mole balance equation Derive the plug flow reactor design equation in terms of x.

Okay, so starting from here, we learned that in order to convert this to a suitable equation for plug flow reactor, we said you take the derivative of both sides, correct, derivative of both sides, and we said simply we, this is of course zero, right, this the derivative of this is zero, so we'll end up minus d fi by dv and this function is first integrated with respect to v then derived with respect to v so it remains as is so we get this equation when we do the balance for a. Okay now how can we introduce x here in this equation? Well this is the differential form of black flow reactor more balanced equation or the design equation and we remember that a equals if a0 1 minus X therefore therefore can be written as fA0 minus fA0 times x and now we can take the derivative of both sides so we have dF A equals of course dF A0 is 0 because fA0 is constant minus fA0 A x so basically dF A equals minus f A naught dx.

So we come here, instead of d f A, we can write minus f A naught dx by dv equals r A. Of course, we can always take the negative to the other side, okay, or we can also do this, equals minus r A divided by f A naught. Okay, now we have a design equation for plug flow reactor return terms of x.

Okay, so here we go. Differential form of plug flow reactor design equation. And then this is the integral form of plug flow reactor design equation. Great. What is needed in order to integrate the above equation?

We need to integrate here. part of the equation. How do we, what do we need?

Well, basically, we need minus rA as a function of x, correct? So that we substitute for minus rA with an equation where you have x, and then integration would be possible. Okay, so we need to know minus rA as a function of x, and this is something we're going to learn in chapter 3. Okay, the last ideal reactor.

We need a design equation for this reactor, for packed bed reactor, written in terms of x. Packed bed reactors are tubular reactors filled with catalyst particles. The packed bed reactor design equation in terms of x will be something similar to the design equation of plug flow reactor, except that we are using W instead of V, and we're using minus RA prime instead of minus RA.

Okay, the differential form must be used when analyzing reactors that have pressure drop along the length of the reactor. So we're talking about pressure drop. Okay, and this is something we're not discuss in chapter 4, not now. Okay, if pressure drop is zero, the integral form can be used straightforward.

And now we have the integral form for packed bed reactor. design equation. Now in this segment we developed together the design equations for the four ideal reactors written in terms of x.

In the following lecture we're gonna learn how to use them. See you soon.

Transcript for:Reactor Sizing and Conversion Principles

Transcript for:
Reactor Sizing and Conversion Principles