Attainable Region IIT B notes

Lecture Notes on

Attainable Region by Prof. Arun S Moharir Production Support by Satyanjay Sahoo Anshu Shukla

Chemical Engineering Department Indian Institute of Technology Bombay

Introduction Attainable Region or AR is an important concept in design of reactor systems. In any process, it is the reactor or a reactor network where value is created by converting relatively low priced reactants into high priced products. The value thus created is partly consumed in preparing the feed for the reactor, operating the reactor itself, and processing the product from the reactor. All these consumptions must be minimized while the value creation in the reactor maximized to make the overall process optimally profitable. This, in a nutshell, is the task of chemical process design. Chemical engineering curricula world over, especially the undergraduate ones, aim at designing or creating the process designers for future who will churn out newer process designs. Process design thus must take a philosophical and holistic view rather than getting constrained by individual design tasks such as designing a heat exchanger, a distillation column, a reactor etc. Individual designs of unit operations and reactors are covered in dedicated courses such as heat transfer, mass transfer, reactor design, fluid mechanics etc. It is the process design course which must integrate all these concepts to illuminate the path to that one final all important activity of process flowsheet development and design. The course on process design is thus the culmination of all the courses to finally build something whole and complete. No wonder it is one of the final year, final semester courses in most chemical engineering curricula. IITB, for whose final year undergraduate students this note is prepared, is no exception. The note deals with Attainable Region, AR, an extraordinary concept developed by Prof. Glasser [1, 2] for reactor system design. It is also one of the more recent philosophical ideas which has immensely enriched chemical process design task. While the concept is well understood and further researched upon by teachers, its teaching has been a challenge because it has been difficult to separate complex mathematics from its basically simple philosophical content. It would not be wrong to say that AR entered the research arena without entering the textbooks of basic chemical engineering curricula. This note is an attempt to

present AR to undergraduate students of chemical engineering in palatable and less mathematical form. A small percentage of these budding engineers, who end up being researchers, can always build on this and appreciate and contribute to the concept later and comprehend the vast literature available now on AR. The idea is to keep aside the brute mathematical force used on this concept and convey the simple structural beauty of the profound concept of AR to the beginners. Attainable Region Let us call it AR from now on. Stated simply, AR is a set of all product compositions that are attainable or achievable from a given feed composition by employing any conceivable combination of reactors in any series-parallel combination. A single reactor is a minimal embodiment of a reactor network. If composition of a stream is stated in terms of concentrations of all components in the stream, AR is a set of all concentration vectors that can be achieved from the given feed concentration vector. To begin with, we use a simplifying assumption that all the reactors in the network operate isothermally and at the same temperature. This is the temperature at which the feed is assumed to be available. So, generation or absorption of heat in the reaction which could be exothermic or endothermic is assumed to be handled by suitable heat exchange to maintain the reactors at designated temperature. This is also the temperature at which reaction kinetics should be available. As can be appreciated, AR will depend on reactor kinetics apart from the feed composition. Let us start with some simple examples to develop appreciation for the concept of AR. Example 1: For reaction A  B, construct AR if the feed is pure A with concentration 1 kmol/m3. The rate constant is 1 hr-1.

Let us agree to some conventions for the rest of the note. An arrow like the one used in representing the reaction as above indicates irreversible reaction. Kinetics of the reaction can be either explicitly given or is easily inferable from units of rate constant. For example, above reaction is first order as indicated by the units of rate constant. Rate constant value will be temperature dependent. The given value will be valid only at that temperature. Unless otherwise stated, it is assumed that the isothermal reactor operates at this temperature. With only two chemical species involved in the above case, the concentration space is 2-Dimensional. Using a dominant component of the feed on X axis is a preferred practice, although not necessary. In the present case, the other component is on the Y axis.

For the given stoichiometry, the maximum concentration of A or B is obviously 1 kmol/m3. The feed point can thus be represented by the vertex (1, 0) of the 2-D concentration space as shown in Figure 1.

Feed composition is always a part of the AR. This statement needs no qualification. This is the concentration already available and hence achievable if one chooses to have no reactor at all. Although trivial, let us emphasize this by marking this fact as Rule 1 in constructing any AR in future.

Rule 1: A point in concentration space representing feed composition is a part of the AR. Irrespective of the type of reactor we use, if it provides infinite space time, entire A in feed will be converted to B and the product will have B at concentration 1 kmol/m3 in the present case. This is so because of the stoichiometry and the irreversible nature of the reaction. This infinite time composition of the product is also obviously attainable. Rule 1 says that zero-time composition, i.e. feed composition, is a part of AR. Rule 2 could thus be:

Rule 2: A point in concentration space representing infinite space time composition of product is a part of AR. For the above reaction, the stoichiometry indicates that the sum of concentrations of A and B in the reaction mixture during reaction must be the same as the corresponding sum in the feed. This is so because one mole of A produces 1 mole of B by reaction. In the present case, it means: CA + CB = 1 or CB = 1 – CA Any composition achievable through reaction carried out for any time should thus lie on a straight line with intercept 1 and slope -1 in our 2-Dimensional concentration space. The two ends of the line were developed earlier through Rule 1 and Rule 2. Note that the feed point (1, 0) and infinite time product point (0, 1) both satisfy above equation of the line. Any concentration attainable through reaction is thus on a straight line joining these two points.

CB (Kmol/m3)

1

Figure 1: Attainable region for A  B; CA 0=1, CB 0= 0, k=1hr-1 0.5

0 0

0.5

1

CA (Kmol/m3)

Figure 2: Attainable region for A  B; CA 0=1, CB 0= 0, k=1hr-1 Figure 1 shows the feed and infinite time product points as well as the above line. You can verify that if the feed is passed through a reactor of some size and some type (say CSTR, PFR) which gives some conversion, the resultant product composition has to lie on the line in this figure. No other composition is possible

to achieve starting with this feed. The straight line is thus itself the desired AR in this case. Being a straight line, the AR in this case can be said to be 1-Dimensional. Please note that the concentration space was 2-Dimensional, while the AR is 1Dimensional. Also worth noting is that we did not use any particular reactor type in developing the AR. Also, we made no use of the reaction kinetics or rate constant explicitly. All we used in arriving at the AR is stoichiometry, nature of reaction (reversible/irreversible) to arrive at infinite time composition and the feed composition. With this first acquaintance with the concept of AR, let us look at some variations of the above example to consolidate and expand our appreciation of AR. Example 1A: Same as Example 1 as far as the kinetics and stoichiometry go, but with the only difference that the feed is an equimolar mixture of A and B with each at concentration 0.5 kmol/m3. Figure 3: Attainable region for

CB (Kmol/m3)

1

A  B; CA 0= CB0 =0.5, k=1hr-1 0.5

0 0

0.5

1

CA (Kmol/m3)

Figure 2 shows the AR. The feed point is now (0.5, 0.5), infinite time product point is (0,1) and the straight line joining these two points is the AR. Slope and intercept of the line are the same as in Example 1.

Example 1B: Same as Example 1 as far as the stoichiometry and feed composition go. The only difference is that the reaction is second order, with say some given rate constant value. Figure 3 shows the representation of AR for the mentioned conditions.

All arguments put forth for Example 1 as regards feed composition, infinite time product composition and the linear relation between concentrations of the two components for any conversion are valid. The AR is thus the same as in Figure 1. The kinetics does not seem to play any role and the AR is same for Example 1 and 1B although the kinetics was different. Neither the reaction order nor the rate constant value had any bearing on AR which is a straight line, i.e. AR is 1Dimensional.

CB (Kmol/m3)

1

Figure 4: Attainable region for A  B; CA 0=1, CB0 = 0, k=1 m3/kmol/hr

0.5

0 0

0.5

1

CA (Kmol/m3)

Example 1C: Same as Example 1 as far as the stoichiometry and feed composition go. The difference is in the nature of reaction which is now a reversible reaction with equilibrium constant given as 1. Let is agree to use = sign to represent reversible reactions and say that the reaction is A = B.

The feed composition is a point (1, 0) in the 2-D concentration space. The Infinite time product concentration (i.e. product concentration achievable by employing a reactor providing infinite space time) will now be (0.5, 0.5) because A and B will

split as per equilibrium constant value (CB/CA = 1 as given). The linear relationship between concentrations of A and B will be valid as per the stoichiometry as earlier. The AR will thus still be a straight line as shown in Figure 4. Figure 5: Attainable region for A = B; CA0 =1, CB0 = 0, k=1hr-1

CB (Kmol/m3)

1

0.5

0 0

0.5

1

CA (Kmol/m3)

If you compare the ARs developed by us so far, they are all part or whole of the same straight line. We are not budging from this line in spite of changes in reaction nature, reaction order, and feed composition. We still have a 1-D AR and a 2-D concentration space. It is tempting to make a generalization and proclaim Rule 3 stating that reaction kinetics and feed composition have no bearing on the geometric nature of the AR, at least qualitatively speaking. Or may be Rule 4 stating that dimensionality of AR is one less than the dimensionality of the concentration space. Let is desist from doing that and explore further. AR cannot be, after all, such a trivial concept!

So, let us cook up some more variations of the same example. Example 1D: Same as Example 1C except that the feed is an equimolar mixture of A and B with both concentrations as 0.5 kmol/m3.

The feed point is now (0.5, 0.5). The infinite time product point is also (0.5, 0.5) given the equilibrium constant as 1.0. The AR is thus just a point in a 2Dimensional space. The AR is Zero-Dimensional so to say. Stated simply, the composition remains same and cannot be changed. Whatever be the reactor

type and size, the feed emerges unchanged as a product. That is because the feed is at equilibrium itself. The AR for the above system is shown in Figure 5.

CB (Kmol/m3)

1

Figure 6: Attainable region for A = B; CA0 = CB0 = 0.5, k=1hr-1

0.5

0 0

0.5

1

CA (Kmol/m3)

This example at least gave us some deviation from something we were on the verge of formulating as a rule. The dimensionality of AR is 1 in Examples 1, 1A, 1B and 1C and is 0 (Zero) for example 1D. The concentration space was 2-D in all cases.

The 0-Dimensional AR, say the point AR, is still on the same lines which emerged as ARs for the earlier cases.

All reactions so far were having same stoichiometry. Let us make some deviation from the stoichiometry in the next example. Example 1E: It is similar to Example 1 except that the reaction is 2A  B + C. There are now three species and the concentration space is 3-Dimensional. What about the AR in this 3-D space?

The feed concentration in terms of (A, B, C) is represented by point (1, 0, 0) in this 3-D rectilinear space. The infinite time product composition, i.e. product composition for complete conversion of A is (0, 0.5, 0.5). Both these points must be on AR. What is the shape of the region which includes all other possible

concentrations of A, B and C by employing any reactor of any size? A little thought will indicate that it is still a straight line joining the feed and infinite time product points. The AR is 1-Dimensional in that sense in this case also. It is not a surface but a line. It can be represented in the 3-D space as in Figure 6.

In engineering, one invariably presents 3-D objects in their various 2-D views, such as plan, elevation or profile views. The 3-D perspective is created by the viewer in his mind from such views. We can try doing so in this case also and show the AR by its projection on the A-B, A-C and B-C planes. These representations are also shown in Figure 7. The AR is a straight line in all such representations.

CC( Kmol/m3)

1 0.75 0.5

0.25 0 0

0.25

0.5 3

CA ( Kmol/m )

0.75

1 0

0.75 1 0.5 0.25 CB ( Kmol/m3)

Figure 7: Attainable region for 2A  B+C; CA0 =1, CB0 = CC0 =0, k=1hr-1

The two dimensional views give the attainable concentrations of any two of the 3 components. The concentration of the remaining component can be deducted from the overall mass balance which is CA + CB + CC = CA0 + CB0 + CC0. The feed is pure A with concentration 1 kmol/m3 in this case. Therefore, CA + CB + CC = 1 irrespective of the conversion of A. The 2-D representations are thus adequate and provide an idea of all attainable compositions.

You can now try constructing 2-D or 3-D representations of AR as shown in Figure 8 for the following examples which are similar variations of basic reaction as we tried earlier. 1

0.5

0

1

CC (Kmol/m3)

CC (Kmol/m3)

CB (Kmol/m3)

1

0.5

0.5

0

0

0.5 CA

1

(Kmol/m3)

0

0

0.5 CA

1

0

(Kmol/m3)

0.5 CB

1

(Kmol/m3)

Figure 8: 2D Projection of attainable region for 2A  B+C; CA0 =1, CB0 = CC0 =0, k=1hr-1 Example 1F: Same as Example 1E but the feed is (0.5, 0.25, 0.25) Example 1G: Same as Example 1E but the reaction is second order in A and reversible with equilibrium constant 0.25 Figure 9 shows the graph between CB versus CA with feed as CA0=1 and CB0=CC0=0. Example 1H: Same as Example 1G but with the feed composition as (0.5, 0.25, 0.25): The representation in Figure 10 shows AR for Example 1H. You get 2 ARs which are 1-D in whatever the representation you choose. In one case, the AR is a point or is 1-D. We thus got 0 and 1-D ARs in this 3-D concentration space. You can increase the dimensionality of the concentration space by making the reaction have 4 or more components. For example, try A  B + C + D or A + B  C + D etc. Irreversible nature could then be converted to reversible nature of reaction. If you play around a bit with feed concentration and equilibrium

constant, you will realize that the AR is still 0-D or 1-D. Representation of AR in 4-D concentration space is not conceivable and the best choice is to use a 2-D space with selected pair of components. AR will be a straight line or a point in such a representation.

What is stopping us from getting higher dimensional ARs, say a 2-D or 3-D AR, such as a surface or a volume? We have played with almost everything except that we have been considering only one reaction. Let us try multiple reactions for a change and see what happens. Example 2: Reaction A  B  C is carried out with pure A as feed. Feed is pure A with concentration of A (CA0) is 1 kmol/m3. The rate constants are 1 hr-1 for both the reactions. Construct an AR.

We have a series of two reactions, both irreversible, first order and with given rate constant values. This is a reaction handled as the first encounter with multiple reactions in any worthwhile book on CRE or Chemical Reaction Engineering. CRE deals with two ideal reactor types, a CSTR and a PFR. Let us see what product compositions one can achieve by employing a CSTR to begin with. It can be of any size offering space time τ of any magnitude (0 to ∞). Mass balance on species A and some jugglery gives the following relation between concentration of A in CSTR product stream as a function of reactor space time.

C 0A CA = 1+ k 1 τ Similarly, mass balance on B gives:

CB =

c 0B + k 1 τ C A 1+ k 2 τ

CB (Kmol/m3)

0.25

1 3 0.125

Figure 9: Performance curve for A 

2 6

5

B  C ,CA0 =1, CB0 = CC0 =0, k1=k2=1 4

hr-1, in CSTR

0 0

0.5

1

CA (Kmol/m3)

If concentrations of A and B are calculated at different values of τ from τ = 0 to τ = ∞, and plotted in the CA-CB space, one gets a curve as shown in Figure 8. The line indicates product compositions obtainable or attainable from a CSTR if all feed is pushed through a single reactor of this type (CSTR). So, is this the AR we are looking for?

Not really! AR is the set of all compositions attainable through any combination of any type of reactors, and not just a single reactor of a particular type. So, let us say, we are ready to have two CSTRs in parallel, one processing mass fraction x of the feed and the other the remaining fraction of feed (1-x). Depending on the size of individual reactors, the product from each reactor will be some point on the CSTR performance curve we have generated. Consider points 1 and 2 on the curve as one such pair. When a product of composition as represented by point 1 and with mass fraction x is mixed with the product of composition as represented by point 2 and mass fraction (1-x), we will get a combined product of composition as shown on the line joining points 1 and 2. The exact location of the point will split the line in two parts, one of fractional length x towards point 2 and one of fractional length (1-x) towards point 1. This can be appreciated by simple mass balance. That means, by changing x from 0 to 1, any point on this line is attainable. This is true also of any other pair of points on the reactor performance line, such as points 3 and 4, or 5 and 6 etc. In fact, this is true of a line joining the two ends of the performance line (1, 0) (feed point) and (0, 0) (the infinite time product composition). The same logic holds if one point is the feed point and

another any point on the performance curve. The line joining these two points is also attainable. In practice, one would achieve any composition on any such line by passing a part of the feed through the reactor and mixing the remaining with the product of the reactor by bypassing the reactor.

If all such possible lines are drawn and accepted as representing attainable compositions, the overall AR that is attainable using none, one or two CSTRs with or without bypass is as shown in Figure 9. The AR now is surface and not a straight line. We got an AR which is two dimensional for the first time. It is a surface in a 2-D plane. The composition of the third species (i.e. C) can be obtained by overall balance as seen earlier.

CB (Kmol/m3)

0.25 Figure 10: AR for A  B  C ,CA0 =1, CB0 = CC0 =0 k1=k2=1 hr-1, in

0.125

CSTR

0 0

0.5

1

CA (Kmol/m3) But is this then the AR for the present case?

Actually, we do not know. Whereas we have allowed more than 1 CSTRs while arriving at attainable product compositions, we have not yet considered using PFR instead of, or in addition to, the CSTR. Let us do it now.

Similar to the above treatment, one can develop a PFR performance curve in the 2-D concentration space. Mass balance on A and B for a PFR gives the following relationships. C A = C 0A e − k1 τ

CB = C e 0 B

− k 2τ

[

k1 C 0A −k1τ −k 2τ + e −e k 2 − k1

]

If concentrations of A and B are calculated for varying τ values (0 to ∞) and plotted, we get this curve. It is shown in Figure 10 along with the attainable region with CSTR constructed earlier. We thus see more product compositions attainable outside the region we developed using CSTR.

CB (Kmol/m3)

0.375

Figure 11: Performance curve for A 0.25

 B  C ,CA0 =1, CB0 = CC0 =0,

0.125

k1=k2=1 hr-1, in both CSTR and PFR

CSTR PFR

0 0

0.5

1

CA (Kmol/m3)

Figure 12: Performance curve for A  B  C ,CA0 =1, CB0 = CC0 =0, k1=k2=1 hr-1, in both CSTR and PFR Extending the logic used earlier (if any two compositions are attainable, then the compositions lying on the line joining these two points are also attainable), we can see the AR as the entire surface enclosed by the PFR line. This region includes the region attainable using only CSTRs. For the given feed and the reactions (stoichiometry, nature and kinetics), the AR is thus possible to construct. For the series reaction considered above, the AR is as shown in Figure 11. Composition outside this AR cannot be attained by any series-parallel combination of PFR/CSTR with or without bypass from the given feed.

At this stage, we can formulate a more comprehensive definition of AR as follows.

AR or Attainable Region is a set of all compositions possible to achieve from a given feed composition by employing one or more reactors of basic ideal types such as CSTR and PFR in a network comprising of any seriesparallel combination of them with or without a bypass stream and for given reaction

scheme

in

terms

of

its

stoichiometry,

kinetics

and

thermodynamics. CB (Kmol/m3)

0.375

Figure 13: AR for A  B  C ,CA0 =1, CB0 = CC0 =0, k1=k2=1 hr-1, in both

0.25

CSTR and PFR

0.125

CSTR PFR

0 0

0.5 CA

1

(Kmol/m3)

Let us restrict at this time to all reactors operating isothermally and at same temperature. Also, we do not consider restrictions on reactor sizes or constraints on feed availability in terms of its quantity.

All the ARs we developed so far fit into this definition.

We have argued during above development that if any two points representing two compositions are a part of AR, then the line joining these points must also be a part of AR. Mathematically, this means that the geometrical shape of the AR must be convex In the above example, we constructed on a 2-D concentration space an AR for a reaction scheme involving 3 species. The same could have been depicted in the actual 3-D concentration space. Verify that it would still have been a region marked on a plane surface. What we saw on the 2-D space was only a projection of this actual AR in 3-D space on the chosen 2-D view. The AR could have been similarly depicted on a CA-CC or CB-CC space. The looks could be quite different and this is worth trying as an exercise. The expected AR looks in these

alternative choices of depiction are shown in Figure 12. Also shown are the corresponding reactor lines. Whatever may be the preferred representation style, the information content of an AR is the same. This content is very profound and can be used to design the best suited reactor network for a given task as will be seen later.

1

CSTR PFR 0.5

CSTR PFR

CC (Kmol/m3)

CC (Kmol/m3)

1

0.5

0

0

0

0.25 CB (Kmol/m3)

0.5

0

(a)

0.5 CA (Kmol/m3)

1

(b)

Figure 14: AR for A  B  C ,CA0 =1, CB0 = CC0 =0, k1=k2=1 hr-1, in both CSTR and PFR in (a) CB = CC and (b) CA = CC plane

As of now, you could construct AR for the following variations of the above example.

Example 2A: Same as Example 2 except that equimolar mixture of A and B is used as feed. Feed concentration of A (CA0) is 0.5 kmol/m3. Example 2B: Same as Example 2 except that pure B is used as feed. Feed concentration of B (CB0) is 1 kmol/m3. Example 2C: Reaction A = B = C is carried out with pure A as feed. Feed concentration of A (CA0) is 1 kmol/m3. The forward rate constants are 1 hr-1 for both the reactions. Equilibrium constants for both the reactions are 1.0.

Example 2D: Same as Example 2C except that feed is equimolar mixture of A, B and C with feed concentration of A (CA0) as 1.0 kmol/m3. The forward rate constants are 1 hr-1 for both the reactions. Equilibrium constants for both the reactions are 1.0. Example 2E: Reaction A  B  C is carried out with pure A as feed. Feed concentration of A (CA0) is 1 kmol/m3. The rate constants are 1 m3/kmol/hr for both the reactions.

And so on. You can cook up your own examples and practice more.

Some of these examples required you to go through a lengthy derivation, some called for numerical techniques to solve equations to generate data to be plotted to construct AR. If you have been efficient and correct, the examples took you through these steps to handle reaction kinetics. The end result was you got good looking ARs of various dimensions. You can try more examples with permutations and combinations of reaction types (irreversible, reversible), kinetics, feed compositions etc. You could redo all examples by changing the reaction to parallel instead of series (A  B, A C) for variety. This will give you a lot of feel for concentration space, dimensionality of AR, convexity of AR etc. The mathematics involved even with simple looking and few reactions is often formidable. We will now try something which will keep this reaction-related mathematics at the lowest ebb and allow AR related things to be emphasized. We will consider zero order reactions, without bothering as to how realistic such kinetics is. This will help us appreciate AR and its geometrical content even more. Example 3: Reaction A  B  C is carried out with pure A as feed. Feed concentration of A (CA0) is 1 kmol/m3. The rate constants are 1 and 0.5 kmol/m3/hr for the first and second reaction respectively. Construct an AR.

Let us create a CSTR performance curve. Mass balance on A will give the change in concentration of A as follows. CA = CA0 – k1τ This is valid only for reactors providing space time τ from 0 to τ* where τ* is CA0/k1. For larger reactors providing τ > τ*, CA will be zero. Similarly, mass balance on B will give expression of CB in terms of τ. Till such time A is available, concentration of B will build up in the reactor because consumption rate of B is less than formation rate of B as can be seen from the two rate constants. Once A is completely consumed, concentration of B will also start dropping and eventually reach zero for reactor space time CA0/k2. C is formed continuously at the rate of k2. The reactor performance curve is comprised of two straight line sections in CA-CB space. The feed point is (1, 0). At τ* = 1 hr, CA = 0, CB = 0.5 and CC = 0.5. The straight line joining (1, 0) to (0, 0.5) is valid till τ*. Then onwards, B drops from 0.5 to 0.0. By time τ = 2 hr, all B is consumed. This is point (0, 0) on the CA-CB space. The straight line joining (0, 0.5) and (0,0) is the second part of reactor line. In the actual 3-D concentration space for (A, B, C), the corresponding points are (1, 0, 0), (0, 0.5, 0.5) and (0, 0, 1.0).

Using the earlier logic to construct a convex AR from reactor lines, the AR in A-B, A-C, B-C and A-B-C space are as shown in Figure 13. The triangle is a convex region. The third line of the triangle in each case was arrived at by joining the feed point and the point corresponding to 2 hr space time when both A and B deplete to zero. This was done assuming CSTR as a reactor. PFR should also have been included. But is PFR different from CSTR for zero order reactions? Think and get convinced that the ARs in Figure 13 are the ARs in totality and in compliance with the definition given earlier.

The zero order reactions simplified reaction mathematics and one could visualize ARs with much less computation. Consider two small variations of this example and construct ARs in various views suitably 1

0.5

0

CC (Kmol/m3)

1 CC (Kmol/m3)

CB (Kmol/m3)

1

0.5

0

0

0

0.5 CA

1

0

(Kmol/m3)

0.5

0

1

0.5 CA

CB (Kmol/m3)

(a)

(b)

1

(Kmol/m3)

(C)

Figure 15:

1 CC (Kmol/m3)

0.5

AR for

reaction A  B  C; CA0=1 CB0 =CC0=0, in

0.5

(a) CA-CB , (b) CA-Cc , 0

(c) CB-Cc plane and (d) 0 0.5 CA (Kmol/m3)

CA-CB-CC space 1 1 0

0.5 CB (Kmol/m3)

(d)

Example 3A: Reaction A  B  C is carried out with pure A as feed. Feed concentration of A (CA0) is 1 kmol/m3. The rate constants are 1 kmol/m3/hr for both the reactions. Construct an AR. Example 3B: Reaction A  B  C is carried out with equimolar mixture of A and B as feed. Feed concentration of A (CA0) is 0.5 kmol/m3. The rate constants are 0.5 kmol/m3/hr for both the reactions. Construct an AR.

8 views, without label, are given in Figure 14. These include one 3-D view and three 2-D views for each of the above two examples. Do your own development and label these eight views of AR suitably with selections from following list. (1) Example 3A, 3-D

(2) Example 3A, CA-CB

(3) Example 3A, CA-CC

(4) Example 3A, CB-CC

(5) Example 3B, 3-D

(6) Example 3B, CA-CB

(7) Example 3B, CA-CC

(8) Example 3B, CB-CC

Appreciate how the look and feel of the AR changes with the choice of 2-D concentration space. Information content is of course the same. All ARs have been represented with coordinates of vertices of the space clearly given. The borders or edges of the AR were anyway straight lines and needed no equations to be specifically mentioned. However, if the edges of AR are curves, it is good to write the equations of all distinct curves. ARs with coordinates of vertices and equations of edges mentioned become that much more readable and also usable for reactor design. We will deal with reactor design using AR later. You can try similar cases with parallel reactions: A  B and A  C and learn more. Do not expect the ARs to be even remotely similar to those for series reaction. The similarity between the problems is deceptive. Did you observe the same thing whey you did similar thing earlier with reactions being first order?

ARs with zero order reactions are interesting. ARs with some reactions as zero order and some as non-zero order are even more interesting. Let us try some.

1

0.5

CC (Kmol/m3)

1

CC(Kmol/m3)

CB (Kmol/m3)

1

0.5

0

0

0 0

0.5

0

1

0.5 CA

CA (Kmol/m3)

(a)

1

0

(c) 1 CC (Kmol/m3)

CC (Kmol/m3)

CB

0.5

0.5

0

0 0.5 1 3 CA (Kmol/m )

(d)

1

CB(Kmol/m3)

1

0

0.5

(Kmol/m3)

(b)

1 (Kmol/m3)

0.5

0.5

0

0

0.5

1

0

0.5 CA (Kmol/m3)

CB (Kmol/m3)

(e)

(f)

3

CC (Kmol/m )

1

0.5

(g) 0

1 0

0.5

0.5 3 CA (Kmol/m )

1 0

3

CB (Kmol/m )

3

CC (Kmol/m )

1

0.5

(h) 0 0

0 0.5 0.5 3 CB(Kmol/m )

Figure 16: 8 views of example 3B

1 1

3

CA(Kmol/m )

1

Example 4: Construct AR for the series reaction A  B  C. The first reaction is zero order while the second reaction is a first order reaction. The rate constants are 1 kmol/m3/hr and 5 hr-1 respectively. The feed is an equimolar mixture of A and B with CA0 = CB0 = 1 kmol/m3. The procedure should be the same as earlier. We should construct CSTR and PFR performance curves to begin with. The starting point of both the curves will be the feed point i.e. point (1, 1) on the CA-CB space. For either reactor, the infinite space time product concentration is (0, 0) in this plane. By overall mass balance, the concentration of C for any combination of CA and CB can be found as CC = 2 - CA – CB. Thus CC is 0 at feed point and 2 at the infinite space time point.

For a PFR or CSTR, the mass balances can be solved to get CA and CB as functions of τ and pairs of CA and CB for same τ values can be plotted to get the reactor (PFR, CSTR) curves.

For a CSTR, the mass balance on A gives: CA = CA0 – k1 τ

(k1 is the rate constant for the zero order reaction)

This is true till A reacts out completely. This happens for τ ≤ CA0/ k1 After this CA0 stays 0 and only second reaction is active. For the given values, this time is 1 hr. Concentration of A thus drops linearly with τ for one hour and then stays 0. Mass balance on B can be written separately for τ ≤ 1 and > 1. For τ ≤ 1, concentration of B varies with space time as follows.

CB =

C 0B + k 1 τ 1+ k 2 τ

At τ = 1 hr, A has completely depleted (CA = 0), B has a concentration of 2/6 or 1/3 as per above equation (CB = 1/3) and concentration of C is found from overall balance (CC = 5/3). Beyond this time, B depletes as per first order kinetics of the second reaction giving the following.

CB =

1/ 3 1 + k 2 ( τ -1)

If we choose to represent AR on a 2-D CA-CB space, it is better to develop an expression for CB in terms of CA. This can be easily done if above expression for CA is used to express τ in terms of CA and this is substituted in the corresponding expression for CB. For example, from the expression for CA derived above, one can write τ = (CA0 – CA)/k1 This when substituted in the expression for CB valid up to τ ≤ 1 hr, one gets

C 0B + C 0A - C A CB = 1 + k 2 (C 0A - C A ) / k 1 This is the equation of the line joining feed point (1, 1) to the point (0, 1/3). Beyond this point, CB also drops to zero at infinite space time. That line is along the ordinate in our 2-D representation. It is just a vertical line joining points (0, 1/3) to (0, 0). In any other representation (say CB-CC space) one would have to get the equation of that line too by similar procedure. The CSTR performance line is shown in Figure 15. The procedure can be repeated for the PFR also.

Balance on A yields the same expression for concentration of A as a function of space time. This is so, because the type of reactor has no bearing on reactor equation for a zero order reaction. A will thus deplete from feed concentration to zero concentration for a space time of 1 hr as earlier. The balance on B till such time is a first order ODE and its solution gives the following expression for concentration of B.

C B = C 0B e −k 2 τ +

[

k1 1− e − k 2 τ k2

]

At τ = 1 hr, A vanishes completely and concentration of B is obtained by substituting τ = 1, k1 = 1 k2 = 5 and CB0 =1 in the above expression. This value is CB(τ = 1) = 0.8 e-5 + 0.2 After this space time, B depletes solely by the second reaction as per first order kinetics. The expression is given as follows. CB = [0.8 e-5 + 0.2] e-5(τ-1) The equation for CB in terms of CA can be obtained by substituting for τ in terms of CA as was done for a CSTR. This will allow plotting the PFR line till A vanishes. Beyond this, B drops to zero as per above expression. The PFR and the CSTR curves are as shown in Figure 15. The PFR line is below the CSTR line in the A-B space. What is then the AR?

CB(Kmol/m3)

1

Figure 17: Performance curve for A  B  C, CA0 = CB0 = 1

0.5

kmol/m3, k1= 1 kmol/m3/hr, k2= 5 CSTR PFR

hr-1, both in CSTR and PFR

0 0

0.5

CA

(Kmol/m3)

1

AR has to be convex. Therefore, feed point (1, 1) must be joined with CSTR point (1/3, 5/3) by a straight line. The concavity of the CSTR curve is bridged by this line which is attainable by a CSTR with bypass stream.

There is also a concave region below the PFR curve. All points below the PFR curve which can be attained by 2 PFRs in parallel or a PFR in parallel with CSTR should be enclosed into the AR. This region is retrieved by drawing a tangent to the PFR line from point (0, 0). Since the point at which this line meets the PFR line is attainable and (0, 0) is also attainable. All points on this line are attainable. Now, one has a convex AR which is shown in Figure 16. The equation of the tangent and coordinates of the point where it touches the PFR line must be evaluated. That is left to you as an exercise.

This is the first time we encountered reactor lines which have non-convexity on both sides. It was bridged on one side by joining two prominent vertices while on the other side, one had to draw a tangent. This is something new we saw by conjuring up a series reaction with combination of zero and first order reactions. It is suggested that you construct AR for this case in the other two choices of views (in A-C and B-C space). The looks will be different but the information content the same.

CB(Kmol/m3)

1

Figure 18: AR for A  B  C, CA0 = CB0 = 1 kmol/m3, k1= 1 kmol/m3/hr, 0.5

k2= 5 hr-1, both in CSTR and PFR CSTR PFR

0 0

0.5

CA(Kmol/m3)

1

For zero order reactions, one has to realize that one or more reactants can be completely consumed with concentration(s) dropping to zero. Once that happens, that particular reaction is as good as not there. This should be kept in mind while developing the reactor curves on way to development of ARs.

Since we are with reaction schemes with a sprinkle of zero order and first order reactions, the following 2 examples would be interesting to attempt. If you do these yourself, you would have revised all concepts we developed so far, without much mathematical complications. .

Example 5: 10 different cases are listed below. These are different in the reaction scheme and/or the feed specification. Identify for each case whether the AR is 1-D, 2-D, 3-D or of higher dimension. Construct the AR only for cases where AR is 1-D or 2-D. Select the x-y axes for your AR judiciously. Indicate on the AR itself the equation of the edges and coordinates of the corners of the AR. 0 1 0 1 0 Reaction Scheme 1: A  B  C  D  E  F 1 0 1 0 1 Reaction Scheme 2: A  B  C  D  E  F The integers above the reaction arrow are the orders of those reactions. For example, the 5 reactions in scheme 1 are of order 0, 1, 0, 1, 0 respectively. For reaction scheme 2, the orders of the 5 reactions involved are 1, 0, 1, 0, 1 respectively. Case 1: Reaction Scheme:1, Feed: pure A with concentration 10 kmol/m3 Case 2: Reaction Scheme:1, Feed: pure B with concentration 10 kmol/m3 Case 3: Reaction Scheme:1, Feed: pure C with concentration 10 kmol/m3 Case 4: Reaction Scheme:1, Feed: pure D with concentration 10 kmol/m3 Case 5: Reaction Scheme:1, Feed: pure E with concentration 10 kmol/m3 Case 6: Reaction Scheme:2, Feed: pure A with concentration 10 kmol/m3

Case 7: Reaction Scheme:2, Feed: pure B with concentration 10 kmol/m3 Case 8: Reaction Scheme:2, Feed: pure C with concentration 10 kmol/m3 Case 9: Reaction Scheme:2, Feed: pure D with concentration 10 kmol/m3 Case 10: Reaction Scheme:2, Feed: pure E with concentration 10 kmol/m3 The rate constant values for the zero order reactions are 10 kmol/m3/hr. The rate constant values for the first order reactions are 1 hr-1.

At first sight, the cases look formidable. With 6 species involved, the concentration space is of a much larger dimension than we discussed so far. ARs could thus also be of higher dimensions than we are comfortable with. However, you are asked to only state the dimension of AR in each case and construct only ARs of dimension 2 or less. As you think more, these cases are not beyond your reach.

We will discuss the cases a bit more and then leave the rest to you once the solution seems attainable.

The series reactions are alternately of zero and first order. Let us consider Case 1.

The first reaction is of zero order and feed is pure A. The rate of consumption of A to form B is very high due to high value of the rate constant of the zero order reaction. B will thus be formed rapidly and till A is present. B will simultaneously react to C as per first order kinetics. B will not be consumed entirely at any time, except in a reactor with infinite space time. Concentration of any species cannot exceed the feed concentration of A for the given stoichiometry. Concentration of C that is formed is thus always less than 10 kmol/m3. Rate of consumption of C to form D by zero order reaction is more than rate of formation of C from B for the given rate constant values. Therefore, C will not be there in the reaction mixture at any time. The reaction C  D is thus instantaneous for all practical purposes

and C will be converted to D as soon as it is formed. The reaction C  D could thus be dropped from the reaction scheme without any effect on the end result. D converts to E as per first order reaction and will be there in the reaction mixture all the time. E converts to F by zero order reaction. By arguments as put forth above, E will immediately get converted to F for given feed concentration and rate constants. The effective reaction in this case is thus A  B  D  F. The first reaction is zero order and the following two first order. The concentration space is 4-D and the AR is 3-D (check this statement. It could be misleading). So, we do not have to construct AR in this case.

Case 2 is similar but feed is pure B. There is no A in feed and A cannot be formed because the reactions are irreversible. We are thus basically starting with a curtailed reaction scheme B  C  D  E  F. If you argue on the same lines as above, you would realize that C and E will be consumed as soon as they are formed and will not be present in the reaction mixture any time. The effective reaction is thus B  D  F with both reactions as first order. We have handled this case earlier. The AR will be 2-D and can be constructed as we did earlier.

Case 3 has pure C as feed. The first two reactions and species A and B are thus irrelevant. The curtailed reaction is C  D  E  F which further reduces to C  D  F. The first reaction is zero order and the second first order. We have handled AR for this series reaction previously, although for different values of rate constant and different feed composition. You have to only do it for given rate constant values and feed composition.

Case 4 has pure D as feed and the first three reactions and species A, B, C are thus redundant. The effective reaction scheme is a single zero order reaction D  F. AR is 1-D and a straight line joining points (10,0) and (0,10) in D-F concentration space.

Case 5 is anyways a single zero order reaction E  F with similar 1-D AR in E-F concentration space.

You thus essentially develop ARs for cases 2 and 3. Both are series reactions and have been handled by us earlier. See if you can explore case 1 more.

Cases 6-10 can now be tackled successfully by you. You will surmise that the ARs are respectively 3-D, 3-D, 2-D, 2-D and 1-D (Be careful). The 2-D and 3-D ARs are similar to corresponding ARs for the earlier 5 cases.

The definition of AR given earlier is the most commonly used. One can however extend it further by removing some restrictions one by one. For example, can we develop AR for a case where more than one different feed compositions are available? Similarly, can we develop AR for the case where a reactor of given size is available and must be the only reactor to be used? We are basically trying to find out what happens to the AR if there are constraints on resources such as feed and reactor vessels. This is important because we ultimately want to use AR for reactor design. There is no practical situation where there are no constraints on resources.

Let us get back to our first example and extend it a further. Example 6: For reaction A  B, construct AR. Two feed compositions are available. One feed has only A and inert with concentration of A as 1 kmol/m3. The second feed has A, B and inert with concentration of A and B as 1 and 0.2 kmol/m3 respectively. The reaction rate constant is 1 hr-1.

On the A-B concentration space, the compositions attainable from the first feed is a straight line joining (1,0) and (0,1) points. From the second feed, any composition on line joining (1,0.2) and (0, 1.2) is attainable. For the individual feeds, the ARs are 1-D as expected. But if both the feeds are available, the AR

will be a parallelogram bounded by these two straight lines (Figure 17). The AR is thus 2-D. Any composition within the parallelogram is attainable by suitably mixing the two feeds and processing the combined feed to a certain conversion in a suitable reactor. It is also possible to process two feeds in two separate reactors and then mix the products to achieve the desired composition.

If a restriction is put that the feeds should not be mixed at all prior to reaction, but a feed could mix with product formed in reactor using another feed, what will be AR? If an additional restriction is put that you can employ only one reactor, what will be the AR? Think hard on this.

What is important to note is that the availability of alternative feeds has increased the dimensionality of AR. 1-D AR for a single feed and for given reaction became a 2-D AR if two feeds are available. So, what happens if we have 3, 4 or more feeds available? What will happen if in the above case, we had another feed also available with concentrations of A and B as 0.6 and 0.2 kmol/m3 respectively? Do we get a 3-D AR?

What is AR in this 3-feeds case if restrictions such as no mixing of feeds prior to reaction and employing only one reactor are applied? Again, think hard and develop this AR. What if we relax the restriction a bit and say that you could use up to 2 reactors, but not 3? Think carefully and develop this AR.

CF (Kmol/m3)

1.2 Feed 1 Feed 2 0.8

Figure 19: AR for A  B with two feed composition

0.4

0 0

0.5 CE (Kmol/m3)

1

That was of course a stupid question. The AR is 2-D in this case as you can verify. It is not a parallelogram, but is a polygon with three vertices as three feed compositions and other points obtained by infinite time composition with individual feeds. The ARs with other restrictions on mixing and on number of reactors could be different. Example 7: Feed stream of pure A with concentration 8 kmol/m3 is available. A CSTR of volume 25 m3 is also available and must be the only reactor used for carrying out the following reaction. A  B  C. Both the reactions are zero order with rate constants 16 and 8 kmol/m3/hr for the first and the second reaction respectively. Construct an AR.

We have handled series zero order reactions earlier. The only new thing here is that the reactor has to be a CSTR of given size. In any case, reactor type (PFR or CSTR) makes no difference for zero order reactions. So, the only effective restriction is the reactor size. However, by adjusting the feed rate through the reactor, any space time can be achieved. Therefore, as far as AR is concerned, it is no restriction. Therefore, AR can be constructed as earlier.

Rate of formation of A is more than rate of consumption. B will thus build up in the reactor till A is available. After that, B will start depleting and eventually reach a zero concentration level. A will take 30 minutes or 0.5 hr to react out completely. In this time, A will react to form C at the rate of 8 kmol/m3/hr. At 0.5 hr, the concentration of A, B and C in the reactor product will be 0, 4 and 4 kmol/m3. After this, B will deplete to zero over the next 30 minutes. At 60 minute space time, the concentrations of A, B, C in product stream will be 0, 0, 8 kmol/m3. The AR is thus as shown in Figure 18.

CB (Kmol/m3)

8

Figure 20: AR for reaction A  B  C, k1=16 kmol/m3/hr, k2= 8 4

kmol/m3/hr, with a fixed reactor volume.

0 0

4 CA

8

(Kmol/m3)

Now let us consider a little variation in this example as follows and construct an AR.

Example 7A: Same as Example 7 except that availability of feed is restricted to a maximum of 50 m3/hr.

We are seeing a restriction on feed availability for the first time. This is not as trivial a change as it appears. There is a devil in it. It would be nice if you mull over this example for a day and then read ahead. The AR without any restriction on feed availability was given in Figure 18. With the given restriction on feed availability, the AR will be a part of this AR.

The convexity of AR was logically arrived at by arguing that if two parallel reactors operating at different space times give two different product compositions, then any point on the line joining these two attainable points on the reactor line can be achieved by adjusting flow rates processed by the two reactors and mixing the product streams. Another justification was that if any composition on the reactor line is possible, the reactor product can always be mixed with feed stream which bypasses the reactor and any composition on the straight line joining any point on the reactor line to the feed point is also attainable. AR became convex due to these logical arguments.

In the present case, there is only one reactor vessel. Therefore there is no possibility of reactors in parallel. What is possible is only a bypass reactor. But, there is a restriction on how much feed can be mixed with the reactor product due to limitation on the total feed availability. Both the basic arguments used in constructing AR are thus challenged in this example.

Let us see what part of the reactor line is attainable in this case out of the reactor line comprising of a straight line joining (8,0) with (0,4) and another straight line joining (0, 4) with (0,0) in Figure 36.

The minimum reactor space time that is achievable is when the entire feed passes through the available reactor. With maximum feed availability as 50 m3/hr and reactor size fixed at 25 m3/hr, the minimum space time that can be achieved is 30 minutes. Reactor operation at space times from 0 to 30 minutes are thus not feasible. This range was represented by the straight line portion joining (8,0) to (0,4) in the previous case. The only portion of the reactor line that is attainable by processing part or full feed through the available reactor is thus the line (0,4) to (0,0) in Figure 19.

Now consider the reactor operating different flow rates to achieve different space times from 30 minutes (all feed through reactor) to higher space times.

With 30 minutes, the only point attainable is (0,4). Let us operate the reactor at 40 minutes space time. A feed rate of 37.5 m3/hr will thus pass through the reactor. The product will be a point (0, 8/3) in the A-B concentration space of Figure 19.

This reactor product can be mixed with a bypass flow rate not exceeding the balance available feed (12.5 m3/hr). If the entire balance (12.5 m3/hr with A, B, C concentrations of 8, 0, 0 kmol/m3 is mixed with a reactor product of 37.5 m3/hr with concentrations of A, B, C as 0, 8/3, 16/3 kmol/m3. the resultant stream will

be a mixed cup concentration. It can be checked that the resultant concentrations of A, B and C will be 2, 2, 4 kmol/m3. The bypass reactor can attain any point on the line joining point (0, 8/3) on the reactor line with this point (2,2) by choosing to bypass suitable feed quantities between 0-12.5 m3/hr.

CB (Kmol/m3)

8

Figure 21: AR for reaction A  B  C, k1=16 kmol/m3/hr, k2= 8

4

kmol/m3/hr, with a restricted feed rate.

0 0

4 CA(Kmol/m3)

8

If we now consider product from a reactor operated at space time of 60 minutes, it will have the concentrations 0, 0, 8 kmol/m3. It is point (0,0) in Figure 19. The volume processed through the reactor to achieve this space time is 25 m3/hr. That leaves up to 25 m3/hr feed available for bypass and mixing with reactor product. A product composition of 4, 0, 4 kmol/m3 or a point (4,0) in the concentration space of Figure 19 is thus attainable by sending 25 m3/hr through the reactor and mixing the rest with the reactor product. If bypass stream flow is reduced appropriately, any point on the line joining (0,0) to (4,0) is attainable.

If these arguments are used for other reactor space time possibilities, the attainable region for a reactor with bypass seems to be the triangle joining vertices (0,4), (0,0) and (4,0) on the A-B concentration space. Is this then the attainable region? It is convex alright. But can AR exclude the feed point which is always attainable?

Let us look at vertex (0,0) of the concentration space. It is attainable with a reactor of space time 60 minutes. However, it should be noted that for the zero order reactions of concern here, this composition is attainable also by reactors

operating at space times larger than 60 minutes (i.e. flow rates smaller than 25 m3/hr). If smaller flow rates are processed through the reactor, larger quantities will be available for bypass. This will mean that we can attain points on the CA axis to the right of (4,0) point also. For example, if we process only 12.5 m3/hr in the reactor, we get a space time of 2 hr. All A and B would have reacted to give C. The product will have concentrations of A, B and C as 0, 0, 8 in that case. This will then be mixable with 37.5 m3/hr of feed to get a combined product of concentrations (6, 0, 2). If lesser and lesser feed is processed through reactor leaving larger and larger feed flow rates for bypass and mixing, we can attain concentrations all the way up to the feed point on the X axis in Figure 19.

The Attainable compositions are thus captured by the AR as shown in Figure 19. The AR is thus a triangle with a line attached to it. This region is not convex. And still the triangle and the remaining line contain the only attainable product compositions by employing a given CSTR with feed bypass.

For the first time, we are seeing an AR which is not convex (or did you get any non-convex AR earlier?). This can happen when there are simultaneous constraints on reactor sizes and feed availability.

The example has placed a question mark on our confident statement earlier that AR must be convex. True, but only if the premises which led to the need for convexity are satisfied. And these premises were the possibility of using more than one reactors and also using a reactor with feed bypass with no limits on feed availability.

Non-convex ARs are rarely talked about in the literature on AR. But keep these restrictions in mind as the restrictions have a practicality about them.

A more challenging case will be the following variation of the above example.

Example 7B: Same as 7A except that both the reactions are first order with rate constant 1 hr-1 for both the reactions.

The AR will not be as easy to develop as for zero order reactions. Resorting to computer calculations will be called for. However, one would expect similar observations. Develop this AR.

We have dealt with a case of two feeds earlier. We also dealt with a case of AR with limitation on feed availability. Let us combine the two in the following example, which is amenable to simpler calculations but will require excellent analytical skills.

Example 8: For the following reversible reactions, develop AR for given feed compositions, feed availability and desired production rates. The reactions are all reversible and reach equilibrium as soon as appropriate conditions (temperature, presence of catalyst etc.) are created.

A = B (equilibrium constant K1) B = C (equilibrium constant K2) C = A (equilibrium constant K3) It is given that 2 K1 = K2 = 2. Four feed streams are available with following compositions (concentrations in kmol/m3).

Feed I:

CA0 = 0, CB0 = 0, CC0 = 4

Feed II:

CA0 = 1, CB0 = 0, CC0 = 3

Feed III:

CA0 = 1, CB0 = 1, CC0 = 2

Feed IV:

CA0 = 0, CB0 = 1, CC0 = 3

In a concentration space of A-B, construct AR to produce 48 m3/hr of product. All feeds are available at 12 m3/hr each.

It is actually a trivial problem. It may be noted that all feeds have the same total concentration (4 kmol/m3). The given reactions do not change the number of moles. Therefore, any product obtained by mixing the available feeds before and/or after reaction will also have the total concentration of 4 kmol/m3. Given the concentration of any two components in any feed or product stream, the third concentration is thus easily calculated. Representation on a 2-D concentration space (such as given A-B space) is thus adequate.

For this choice, the four feeds are at 4 points (0,0), (1,0), (1,1) and (0,1) as shown in Figure 20. All are available at a maximum rate of 12 m3/hr.

Equilibrium constants are given for the first two reactions. The same for the third reaction can be deducted easily as 0.5.

Any feed or any mixture in any proportion of any number of feeds, if pushed through a reactor reaches equilibrium instantaneously. The equilibrium composition can also be deducted from the equilibrium constants as CA0 = 1, CB0 = 1, CC0 = 2. It is the same as Feed III corner in Figure 40, i.e. point (1,1) in the A-B space. Any feed stream can thus be converted to a composition same as that of Feed III by reaction. Since, we must have a production rate of 48 m3/hr, we need to us all the four feeds at their maximum availability. AR will be the region, any concentration in which is attainable by reaction and or mixing.

The kinetics is not very important here as the equilibrium is reached instantaneously as soon as we decide to carry out reaction and create suitable conditions. Also, type of reactor (CSTR, PFR) is irrelevant because of the same

reason. To generate AR, we thus do not need much calculations, but some geometrical interpretation of the feeds occupying four corners of our concentration space. Think along the following lines to get the AR.

If we decide not to employ any reactor and simply mix all the feeds available, we will get a stream at 48 m3/hr with concentration represented by point (0.5,0.5). That is, concentrations of A, B and C as 0.5, 0.5 and 3 kmol/m3. This is thus an attainable point. Similarly, if we react Feed I and II and mix the product with Feed III, we have a stream represented by (1,1) at 36 m3/hr. If this is mixed with Feed IV at 12 m3/hr, we get 48 m3/hr of a stream represented by point (0.75, 1.0) on the A-B concentration space. Similarly, composition (1, 0.75) is attainable. If we react Feed II to equilibrium, we have 24 m3/hr of stream with composition represented by (1,1). If we mix Feed I and IV, we get 24 m3/hr of stream with composition represented by point (0,0.5). If we mix these two streams, we get 48 m3/hr of stream with composition (0.5, 0.75). Similarly point (0.75, 0.5) is attainable. Thinking on these lines, you can arrive at the AR as shown in Figure 20.

You can check that no composition outside this AR is attainable by any scheme.

Using the above as a way to construct AR in this case, let us now look at some variations of the problem. We only change the production rate that we want to achieve.

CB (Kmol/m3)

1

Figure 22: AR for system of reaction A = B, B=C, C=A, with 4 0.5

different feed to 48 m3/hr of product.

0 0

0.5 CA (Kmol/m3)

1

Example 8A: Same as Example 8 except that the desired production rate is 36 m3/hr.

We thus have now surplus feed available. This will allow us more flexibility in reaction and mixing and we expect the AR to be a larger area as compared to AR for Example 8. Develop the AR and verify that it is as shown in Figure 21.

CB(Kmol/m3)

1

Figure 23: AR for system of reaction A = B, B=C, C=A, with 4 0.5

different feed to 36 m3/hr of product.

0 0

0.5

1

CA(Kmol/m3)

Example 8B: Same as Example 8 except that the desired production rate is 24 m3/hr.

Example 8C: Same as Example 8 except that the desired production rate is 12 m3/hr.

Example 8D: Same as Example 8 except that the desired production rate is 6 m3/hr.

The AR keeps on expanding as we reduce the product requirement. When the production rate reaches 12 m3/hr, the AR fills the entire concentration space. For any further reduction in production, the AR remains the same and occupies the entire concentration space. The AR for Example 8B is shown in Figure 22. AR for Examples 8C and D are squares joining the four feed points and are not shown explicitly.

CB (Kmol/m3)

1

Figure 24: AR for system of reaction A = B, B=C, C=A, with 4 0.5

different feed to 24 m3/hr of product.

0 0

0.5

1

CA (Kmol/m3)

Now let us make one small change to the above cases. So far, we have developed AR for some reaction scheme or the other. What if the reactive transformations are not possible. For example, what are the attainable compositions if only mixing of streams is allowed and reactions are not possible. In short what is the AR in the above 5 examples if reactive change is not possible or disallowed? The ARs for the case where production rates are 48, 36, 24, 12 and 6 m3/hr are jumbled up in Figure 23. Identify which one belongs to which case. 1 CB(Kmol/m3)

Figure 25: AR for only mixing of 0.75

feed streams for a product of 48, 36, 24, 12 m3/hr .

0.5 0.25 0 0

0.25

0.5

0.75

1

CA(Kmol/m3)

If we carve out respective ARs which indicated compositions which can be attained by simple mixing as developed above from the ARs which were developed earlier for the case where reaction and mixing both were allowed, we

are left with AR for the case where reaction must be resorted to or a reactor must be employed. Note that these ARs are not convex. We thus got another example of non-convex ARs.

We have so far played with almost all possibilities: single/multiple reactions (in fact reaction and no reaction), irreversible/reversible reactions, different orders of reaction, single/multiple feeds,

constrained/unconstrained availability of feed,

variable/fixed reactor sizes etc. This must give confidence that AR can be developed although at times calculations may be involved. The examples also took us through zero, first, second and higher dimensional ARs. With this feel for ARs, one can now appreciate reaction schemes and ARs that have caught attention of researchers. Most of these reactions are series-parallel, a combination we have not considered explicitly. To create challenging problems for research, authors have played with the structure of reaction schemes, orders, types of reactions, feed compositions and rate constant values. This has given rise to extraordinary examples, some of which inherit the names of the developers. These reaction schemes are called model reactions, the famous ones being van de Vusse, Denbigh, Tramboze reactions etc.

It is not the purpose of this primer to dwell on these examples. The objective of this compilation was to prepare you to appreciate these higher complexity examples, should you choose to work in reactor network domain. In any case, can anyone do better than Prof. Glasser’s creation of van de Vusse reaction and its AR? That must be read in original and appreciated as such.

We would now revisit the dimensionality of AR to understand it better. Then we will see how AR can be useful to design a reactor system for a given task.

We have seen geometrically ARs of dimensionality 0, 1, 2 and 3. Dimensionality of AR for a given feed composition is primarily decided by the reaction network. There is a simple way of deciding the dimensionality of AR before even one

launches into calculations. Let us understand this with some of the reaction schemes we used in the examples discussed so far. Example 1 had a simple irreversible reaction A  B. To appreciate the dimensionality of AR, one should simply write the net rate of formation of each individual species from all the reactions. In the present case, we have two species A and B. The rates of formation of A and B per unit volume are as follows.

dC A = − k CA dt dC B = k CA dt where k is the reaction rate constant. The dimensionality of the AR for a single feed is the number of species whose concentrations are involved in some or all the rate expressions for all the species together. In the above case, only concentration of A is involved in the expressions and hence the AR is one dimensional, i.e. a straight line. This is what we observed earlier also.

This holds true for Example 1A and 1B also. In Example 1C where the reaction was second order, only the concentration of A will be raised to power 2 in both the expressions. However, number of concentrations involved is still 1 and AR is 1-Dimensional.

Example 1C had a reversible reaction A = B and the rates of formation of A and B should take the forms as follows.

dC A = − k1 C A + k 2 C B dt dC B = k1 C A − k 2 C B dt

Two concentrations are involved on the right hand side of the rate expressions. The AR should thus be 2-D. However, it should be noted that the stoichiometry demands that the total concentration of A and B at any time must be the same as total concentration of A and B in the feed. Therefore, concentration of B can be expressed in terms of the feed concentration and the concentration of A. Therefore, the rate expressions have only concentration of A on the right hand side and the AR is 1-D as we found out earlier.

Same is true for other variations of Examples 1. In a couple of variations, although the AR should be 1-D as per above analysis based on rate expressions, it was actually zero order. Both these examples had reversible reactions and a feed already at equilibrium. Further reaction was thus not possible and the AR shrunk to the feed point and became 0-D.

We also considered AR for the cases when there were more than one feeds. In such cases, the dimensionality of AR could increase by one as was seen in some examples. Except for these special cases where reversible reactions are involved and feed is at equilibrium itself or where there are multiple feeds, the dimensionality of AR can be deducted from the rate expressions for all the species.

You can check this logic with so many other examples we considered.

Now let us see how the AR can be used for designing the reactor networks for a given task. This is the main practical reason why one would develop AR.

[1] Glasser, D.;Hildebarndt, D.; Ind. Eng. Chem. Res.1987,26,1803. [2] Hildebarndt, D.; Glasser, D.;Crowe, M. C.; Ind. Eng. Chem. Res.1990,29,49.