on the control of competing chemical reactions

Chemical Engineering Science, 197 1, Vol. 26, pp. 853465. Pergamon Press. Printed in Great Britain.

On the control of competing chemical reactions R. JACKSON and R. OBANDO

Rice University, Houston, Texas 77001, U.S.A.

and

M. G. SENIORt

University of Edinburgh, Edinburgh, Scotland

(Received 15 August 1970)

Ah&act-The control of competing chemical reactions which are of different order with respect to one of the reactants is considered. Both temperature and the rate of addition of reactants are considered as control variables, and it is shown that the nature of the optimal control policy depends on the relation between the ratio of the activation energies of the competing reactions and the ratio of their orders with respect to the distributed reactant. There is a discontinuous change in the nature of the optimum policy when these two quantities are equal

INTRODUCTION

IN ATTEMPTING to generate a desired substance by chemical reaction, one is often troubled by competing side reactions which degrade the reactants into other, undesired substances. If the reaction is to be carried out batchwise in a well- stirred reactor, one may attempt to influence its course by varying the temperature or by adding one or other of the reactants continually as the reaction proceeds. Exhortations to employ either or both these techniques can be found in most texts on preparative organic chemistry, couched in rather general terms. The same methods are also available in continuous flow tubular reac- tors, where distance along the reactor replaces time as an independent variable. Our object in this paper is to give precision to these ideas and to see what physical features of the competing reactions determine the best method of control.

Control by temperature regulation and by distributed addition of reactants have each been considered previously in a number of publica- tions. Among these we might mention Horn’s [l] original treatment of optimum temperature control for parallel competing reactions, van de Vusse and Voetter’s [2] discussion of the advan- tages to be gained from reactant distribution in

the case of reactions of different orders, and Jackson and Senior’s[3] work on the optimum reactant distribution policy for parallel reactions of the first and second order with respect to the distributed reactant.

Here we will consider, once again, competing parallel reactions of different orders with respect to one of the reactants, but will permit both the temperature and the rate of reactant addition to be used as control variables. In spite of the appa- rent complexity of the resulting optimization problem, it turns out to be possible to find the form of the optimum control policy in general, and without detailed calculation, by application of the necessary conditions for optimality provided by Pontryagin’s Maximum Principle. The detailed reasoning is quite lengthy and is best presented formally as a series of mathematical lemmas and theorems. These will be found in the Appendix, which actually forms the bulk of this paper. Stated briefly, and without certain necessary qualifications, the principal result is that a combination of control by simultaneous variation of temperature and reactant addition rate is never optimal. It is always best to use either pure batch operation, controlling by temperature variation, or purely isothermal operation, controlling by the

tPresent address: Imperial Chemical Industries, Ltd., Heavy Organic Chemicals Division, Teesside, England.

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rate of reactant addition, and the change from one method of control to the other is related in a remarkably simple way to the relative orders and relative activation energies of the competing reactions. Apart from this striking result, the problem is of unusual technical interest because of the reasoning presented in the Appendix, which permits the form of the optimal policy to be deduced generally.

Statement of the problem

The reactions considered are of the form

A+Bkl-C

A+nBkl-D

where C is the desired product and D is waste. The kinetics are assumed to be of the mass action form and to correspond to the stoichiometry of the reactions as written. Thus, the rate of the first reaction is proportional to the concentration of substance B while that of the second reaction is proportional to the nth power of this concentration.

To be definite we suppose that the reactions are carried out in a well-stirred vessel immersed in a thermostatically controlled bath, as indicated in Fig. 1. Heat transfer delays in the thermostat, and between the thermostat and the reactor are negligible on the time scale of the reaction, so the temperature of the reaction mixture is at all times uniform and its instantaneous value can be controlled as desired between certain limits Pin and Pax. a, moles of substance A are intro- duced into the reaction vessel at time zero, and substance B can be run into the vessel from a container, as shown. The instantaneous rate of addition of B can be controlled at all times between specified limits zero and rmr where the upper bound rm is determined, for example, by the maximum gravity flow through the tube from the reactant reservoir to the reactor. Frequently, r,,, will be very large, but it is important that r be bounded in connection with the mathematical reasoning in the Appendix. The total amount of B to be added is also specified, and is denoted by

R. JACKSON, R. OBANDO and M. G. SENIOR

Fig. 1. Illustration of the problem.

Q. It is physically reasonable to assume that all the available substance A should be added at the beginning of the batch, and this point has also been discussed mathematically [3].

To avoid unnecessary algebraic complication it is assumed that the concentration of each substance in the reaction vessel is proportional to the number of moles of that substance present, and is unaffected by the quantities of the other components of the reaction mixture in the reactor. This is the case, to a very good approximation, when the reaction is carried out in dilute solution in an inert solvent, for example. It is also exactly true for a reaction in a gaseous mixture held at constant volume.

The problem is then to determine the temperature T and rate of addition r of B, as functions of time, so that the concentration of substance C at some specified terminal time 7 is maximized.

Discussion of results

Write p = E,/E,, the ratio of the activation energies of the two reactions. Then for sufficiently large r,,, and sufficiently small Tmin it is proved in the Appendix that the nature of the optimum control policy is determined entirely by the relative values of p and n, and changes completely as their ratio passes through unity.

854

On the control of competing chemical reactions

When p > n it is shown to be optimal to add all the reactant B as &.rickly as possible at the beginning of the batch, and to control the reactions by variation of the temperature during the batch time. (In the limit r,,, + m this corresponds to adding all of the substance B, as well as all the substance A, at t = 0). It is, of course, well known[l] how to find the optimum temperature policy in this case, and examples of such optimum temperature policies for p = 2.1 and p = 3.0 are shown in Figs. 6(i) and 7(i), both of which correspond to n = 2, i.e., a side reaction of the second order with respect to B, and values of other problem parameters indicated in the cap- tions. Note that the value of kl, rather than the temperature itself, is plotted as a function of time. The optimum policy is composed of an interval in which the temperature increases monotonically to its maximum value, followed by an interval of isothermal operation at the maximum temperature.

When p < n it is no longer optimal to add all the substance B as rapidly as possible at the beginning of the batch; instead, after adding part of the available B as rapidly as possible, the remainder is retained to be added gradually in the course of the reactions. Furthermore, the optimum temperature policy keeps T at its upper bound Tmax at all times after the initial brief interval in which B is added as fast as possible. In the limit as r,,, + cQ, the length of this interval tends to zero, and the temperature remains on its upper bound at all times. The optimum rate of addition policy r(t) is shown in Fig. 3, which corresponds to the problem for which optimum temperature policies are shown in Figs. 6 and 7. Note that part of the substance B is added instantaneously at t = 0, as indicated on the graph, and the function r(t) plotted shows how the remaining B should be added.

To summarize these results, when p > n all the available B is added at the beginning of the batch and control is exercised by temperature variation, while when p < n, the system is operated isothermally at Tmax and control is exercised by regulating the rate of addition of substance B. This switch in the nature of the

optimal control policy can be illustrated very clearly by considering what happens as p is varied, keeping constant all other parameters of the problem, including kimax and kzmax, the maximum values of the velocity constants. In Fig. 2, the horizontal straight line (a) shows the maximum value of c(7) which can be obtained by isothermal operation at the maximum temperature, using the rate of addition of B for control purposes. Since k,““” and kzmax are taken to have the same values (1-O and 2.0, respectively) for all p, the operating policy is the same for all p, and is the one shown in Fig. 3. It necessarily follows that c(7) takes the same value for all p. The curve (b) shows the maximum value of C(T) which can be obtained by pure batch operation, with all the B added at t = 0 and control exercised by temperature variation. The value of c(7) depends on p in this case, and so does the optimum temperature policy; indeed the optimum temperature policies for p = 1.0, 1.9, 2.1 and 3.0 are shown in Figs. 4(i), 5(i), 6(i), and 7(i) respectively. The two curves in Fig. 2 cross at p = n = 2, curve (a) being higher when p < n and curve (b) when p > n, and this corresponds to the switch from an isothermal optimal control policy to a pure batch optimal policy with temperature control.

It is interesting also to consider this example

I.0

09-

OS- (b)

o-7 -

06- (0)

05-

04-

03-

“:: ” I c 3

P

Fig. 2. Optimum C(T). n = 2, a0 = 1.0, Q = 2.5, kImax = 1.0, &In”” = 2.0 /( min = &I%‘” = 3 I 0, T= 4.0, rm = m. Curve (a) Isothermal operation at pax, optimized with respect to distributed addition of B. Curve fb) Pure batch ooeration with all B added at t = 0, optimized with respect to tempera-

ture variation.

855

S-

4-

L 3-

b,=0.1867

2

t

‘i 0 t/r

Fig. 3. Optimum rate of addition of B for isothermal operation at T= Tm”“. n=2, a(O) = a, = l-0, b(0) = 0.1857, k, = @‘- = 1.0, kz = &- = 2-0, Q = 2.5, r = 4.0, r, = -.

The form of r(t) is independent of p.

P’l.0

R. JACKSON. R. OBANDO and M. G. SENIOR

(i)

0.6b

Fig. 5. (i) Optimum temperature policy with all B added at t = 0. (ii) Corresponding adjoint variables: n = 2, o(O) = a, = 1.0, b(0) = Q = 2-5, klm” = 1.0, hm” = 2.0, k,“‘” =

kzm’n = 0,~ = 4.0,~ = l-9.

I (ii)

Fig. 4. (i) Optimum temperature policy with all B added at t = 0. (ii) Corresponding adjoint variables: n = 2, a(O) = a, = l-0, b(0) = Q = 25, klmu = 1.0, kpmax = 2.0, k,* =

4W”=O,r=4.O,p= 1.0.

in relation to the simple necessary condition for optimality of pure batch operation given by Jackson and Senior[3]. According to this, a necessary condition for optimality of a policy in which all of a substance is added at the beginning of the batch, is that the adjoint variable (see Appendix) corresponding to this substance should take its largest value at t = 0. The adjoint variables A, and hb corresponding to substances A and B are shown in Figs. 4(ii) to 7(ii) for pure batch operation with p = l-0, l-9, 2.1, and 3.0, respectively. For p = 2-l and p = 3.0, hb takes its greatest value at t = 0, so the necessary condition for optimality is satisfied. For p = 1-O and p = l-9, on the other hand, &, has a stationary maximum value, and the condition is violated, as might be expected, since we know that pure batch operation is not optimal in these cases.

The simple form of the optimum addition rate

856

(i)

0.6

0 t /T I.0

(ii)

Fig. 6. (i) Optimum temperature policy with all B added at t = 0. (ii) Corresponding adjoint variables: n = 2, a(0) = a0 = l-0, b(0) = Q = 25 k,“” = 1.0, km= = 2.0, klm’” = kp*

= 0,T = 4*o,p = 2-l.

policy shown in Fig. 3 is not peculiar to the particular problem parameters considered. For sufficiently large r,,, and for p # n it is shown in the Appendix that the optimal r(t) can consist, at most, of

(a) an initial interval in which r = r,, (b) a subsequent interval in which r increases

monotonically with increasing t, and (c) a final “coasting” interval in which r = 0.

In certain circumstances, for example if Q is small, the second of these may be omitted. How- ever, the arguments leading to these conclusions depend on rm being sufficiently large, and p and n not being exactly equal. For smaller r,,, more complicated addition policies for B may be optimal, and when p = n it appears that there may no longer be a unique optimal control policy. The effect of reducing r, is illustrated by Figs.


P’3.0

(ii)

Fig. 7. (i) Optimum temperature policy with all B added at t = 0. (ii) Corresponding adjoint variables: n = 2, a(0) = 0, = 1.0, b(O) = Q = 2.5, k,m= = 1 .O, krm= = 2.0, k,“‘” = km‘”

=0,7=4-o,p=3*0.

8, 9, and 10, which show the optimal r(t) for isothermal operation at the maximum temperature, with successively more severe restrictions on r,. Apart from the values of r,, they correspond to the same problem parameters as the policy shown in Fig. 3. In these cases we are less certain that the policies indicated are optimal than in the case of Fig. 3, since the arguments of the Appendix break down as r, is decreased, but we do know that the Maximum Principle is satisfied in each case, and there is strong independent evidence from hill-climbing computa- tions that they are, indeed, optimal. It is interesting to see that when r,,, is reduced to 2.0, the form of the optimal r(t) changes and a second interval with r = r, appears late in the batch. A further restriction of r,,, to the value 1-O extends this interval through to t = r.

857


Fig. 8. r(t) for n = 2, a, = 1.0, Q = 2.5, k, = k,““” = 1 .O. T = 4.0 and r, = 5.0.

1 0 t/r I 0

Fig. 9. r(f) for n = 2, a, = 1.0, Q = 2.5, k, = k,max = 1.0, r = 4.0 and r,,, = 2.0.

IC

L

I-

.

L

0 t/T

Fig. 10. r(t) for n = 2, a, = 1 .O, Q = 2.5, k, = k,““” = 1.0, r and k, for fixed values

of the state variables ki velocity constant for reaction i; i = I,2

ki, frequency factor for velocity constant k,;i= 1,2

858


k.““” 2

k.min

M;Q) n

h’(Q) P 4

Q

r

R t

T pax

maximum permissible value of ki

order of the reaction rate with respect

minimum permissible value of ki upper bound for 1 dA,/dt 1

Tmin minimum permissible value of the reactor temperature

to distributed reactant B CY upper bound for IdA,/dtl Y ratio of the activation energies, EJE, Aa amount of B added to the mixture from

t=Ototimet MQ) total amount of B added to thL.mixture hb

during the total reaction time instantaneous rate of addition of B to At,@)

the mixture AC ideal gas constant time coordinate I-L uniform temperature in reactor at time t maximum permissible value of the reac- 7

tor temperature

Greek symbols

final value of the adjoint variable p adjoint

km/k%

variable corresponding to physical variable a

upper bound for 1 A,1

adjoint variable corresponding to physical variable b

upper bound for I Abl

adjoint variable corresponding to physical variable c

REFERENCES

adjoint variable corresponding to physical variable q

total length of time for the batch operation

[l] HORN F., Chem. Engng Sci. 1961 14 77. [2] VAN DE VUSSE J. G. and VOETTER H., Gem. Engng Sci. 1961 14 90. [3] JACKSON R. and SENIOR M. G., Chem. Engng Sci. 1968 23 97 I. [4] OBANDO R., M.S. Thesis, Rice University, Houston, Texas 1970.

APPENDIX Mathematical formulation of the problem and necessary

conditions for optimality

The physical variables are continuous functions which satisfy the following differential equations (except at points of discontinuity of the right hand sides) together with the associated boundary conditions.

da z = - k,ab - k,ab” ;a(O) = a, (1)

$=-k,ab-nk,ab’+r ;b(O) = 0 (2)

dc z = k,ab ; c(0) = 0 (3)

dq

z=r : q(O) = 0, q(7) = Q (4)

where q(t) is the total amount of substance B added to the reactor between times zero and t. The velocity constants kI and k2 are Arrhenius functions of the temperature and it is convenient to regard k, itself as a control variable in place of the temperature, writing

k, = ak,’

where (Y = k,,/k,,f’ and p = ET/E,, the ratio of the activation energies.

Then it is required to find the piecewise continuous functions k,(t), r(t) in [0, T], subject to the constraints

0 < k,“‘” s k, =S k,“““, 0 < r L rm (5)

such that Eqs. (l-4) are satisfied and C(T) is maximized. Such a pair of functions k,(t), r(t) will be referred to as an optimal control policy and it is assumed here that such a policy exists. With this assumption, we show that the form of the optimal policy is determined by the necessary condition provided by Pontryagin’s Maximum Principle. Introducing adjoint variables A,, he. A,, p satisfying the following differential equations and boundary conditions

$=k,b(A,+A,-l)+k,b”(A,+nA,):A,(~)=0 (6)

dAb z= k,a(A,,+Ab- l)+nk,ab”-‘(A,+nAJ: Ab(7) = 0 (7)

dhc dr = 0; A,(T) = 1

d$ = 0; CL(T) = y(unspecitied initially)

and constructing the Hamiltonian

H = (Ab+Y)r+kI ab(l -A,-Ah,)-k, abYA,+nA,,).

(8)

(9)

(10)

859

We have the following necessary conditions for optimality of the policy r(t), k,(t).

(a) For each tr[O, r], if H is regarded as a function of r and kI in the rectangle of the (r,k&plane defined by (5). it takes its maximum value at the point (r(t), kr(t)).

(b) The corresponding maximum value of H is independent of time.

We proceed to derive a series of results which enable the form of the optimum policy to be determined without com- putation in certain circumstances, assuming its existence. It is convenient to state these results as a series of lemmas and theorems, as the chain of reasoning is quite long Results which make some statement about the form of the optimum policy will be called theorems, while auxiliary results needed in the derivation of the theorems will be referred to as lemmas.

Lemma I The total length of all intervals in which r = r, may not

exceed Q/r,,,.

Proof This follows immediately from the fact that Q is the total

amount of B added to the system.

Lemma 2 a, b, c, A,,, and Ab are continuous on [O,T] and have piece-

wise continuous first derivatives in the open interval (0,r).

Proof The continuity follows from the definition of these vari-

ables. The tint derivatives are piecewise continuous because of the postulated piecewise continuity of r(t) and k,(t).

Lemma 3 In [O,T] a is bounded above by a,,, b is bounded above by

Q, and bounds A.(Q) and &(Q) for 1 A.1 and 1 &,I can be found which are independent of r, for a given value of Q.

Proof The bounds a < a0 and b s Q follow immediately from

Eqs. (l), (2), and (4), and the fact that a, b, kl, and h are non-negative.

The existence of the bounds A,,(Q) and 4(Q) follows from a general result for homogeneous linear equations with continuous bounded coefficients. For equations and boundary conditions of the form

xJ(tO) = XJO, be [O, 71 J-1

where I&(t)] s A (ah i,j, r), it can be shown that each Ix&)1 is bounded in [O, T] by a bound depending only on A.

Theorem 1 (a) For an optimal policy, k,(r) is continuous at all points

where r < r,,,. (b) When p > 1, kl(t) is continuous at all points of an

optimum policy, provided r,,, is sufficiently large.

Indication of proof For an optimal policy k, is chosen so as to maximize the


Hamiltonian at each time. But the Hamiltonian depends on k, through the terms

fi = k, ab(l - A. - A& - ak/‘aby A,, + nAb)

and the factors a&l -A.- Ab) and abYA,,+ nA,) are continuous functions of time, by Lemma 2. Thus, k,can change discontinuously only if the Hamiltonian has two equal maximum values at different points kll, klP of the interval [kImIn, k,““]. E xamination of all possible forms of the de- pendence of i? on kl reveals that, in all cases where this can occur, Amar d 0 at the point of discontinuity of kl.

Now when r < r,,, at this point, either 0 < r < r,,,, in which case Ab+ y =, 0, or r = 0. In either case (A*+-y)r = 0, and hence H = H when both are evaluated on the optimal trajectory. It therefore follows that H,, c 0 at any point of discontinuity of kl where r < r,,,.

However, at t = 7

H(T) = yr + kla(+o)

so

H,,(T) = Y sgn’Wrm + klmexa(d btd > 0

where

sgn’(y) = 1 for y > 0

=Ofory s 0.

But according to the maximum principle, H, takes the same value for all t on an optimal trajectory, so we have a contradiction and must conclude that there are no points of discontinuity of kl where r < r,,,, which is part (a) of the theorem.

Part (b) of the theorem is proved in a similar way, by estab- lishing that H,,, < I?,,,.= 6 0 at points of discontinuity of k, where r = r,,,,, provided p > 1 and r, is sufficiently large. This again leads to a contradiction when H,,, is compared with Hi.,(r).

A complete exposition of the proof of this theorem is given elsewhere [4].

Lemma 4

For an optimal control policy A, and A, have piecewise continuous derivatives in (0, T) and bounds M(Q) and N(Q) for IdA./dtl and IdA,/dtl can be found which aremdependent of r, for a given value of Q. Step discontinuities of dA,/dt and dA,/dt may occur only at points of discontinuity of k,.

Proof The piecewise continuity of dA,/dr and dA,/dt follows

from Eqs. (6) and (7), Lemma 2 and Theorem 1. Furthermore, using Lemma 3, Eqs. (6) and (7) yield

I&,/d4 s k,“““Q(il, + & + I)

+ ~(ktm9%!Yn, + 6) = M(Q)

IdA,/dtl s klmaxao(&+ Anb+ 1)

+ na(klmax)P~p”-‘(& + n&J = N(Q).

Theorem 2 For an optimal policy k,(r) has a piecewise continuous

860

first derivative. The discontinuities in dk,/dt occur at points of discontinuity of r, at points of discontinuity of k,, and at points where k, = k, ma or kI = k,“l”. The left and right hand limits of dk,/dt exisi at the points of discontinuity.

Proqf dk,/dt does not

we are concerned exist at points of discontinuity of k,. so only with the behavior of dk,/dt at points

of continuity of k,. At interior points of intervals where k, = k,mln or k,max. dk,/dt is certainly continuous, as it vanishes identically. At interior points of intervals where k,“‘” < k, < k,m”x, the Hamiltonian is maximized with respect to k, at a stationary value, for which aH/ak, = 0. Hence.

k,P-1 = I-&---A*

ap bn-,(A, + nA,)’

Differentiating, and noting that the denominator cannot vanish if k, is to lie in (klm’“, klmax), and that dA,/dt and dA,/dt are continuous, since we are concerned with points of continuity of k,. it is seen that dk,/dt is continuous except at discontinuities of dbldt. i.e., at discontinuities of r.

Apart from points of discontinuity of k, and r, therefore, dk,/dr can have discontinuities only where an interval with aH/ak, = 0 meets an interval with k, = k,m’” or k, - k,max.

It is clear that the left and right hand limits of dk,/dt exist at all the discontinuities identified above.

Lemma 5 For an opti&l control policy Ab has a piecewise con-

tinuous second derivative in (0, 7). Discontinuities of d2Ab/dt2 may occur at discontinuities of r and discontinuities of dk,/dt. The left and right hand limits of d*A,/dt* exist at the points of discontinuity.

Proof Direct differentiation of Eq. (7) is possible at all points

of continuity of r and dk,/dt. Thus, d*AJdP exists at all these points. It follows that discontinuities of d*A,/dt* must coincide with discontinuities of r or dk,/dt, and hence that d*A,/dt* is piecewise continuous. Since the left and right hand limits of both r and dk,/dt exist at their points of discontinuity, it follows that the same is true for d*A,/dP.

Lemma 6 For an optimal policy

Y s QN(Q)/r,,,.

Proof We need only consider y > 0 since the result is certainly

true for y s 0. Then Ab+ y > 0 at t = 7 and, since A, + y is continuous and d/dr(Ab+ y) is smaller in magnitude than N(Q), it follows that (Ab+ 7) > 0 throughout some interval lt’.rl. where r-t’ 3 y/N(Q). But when A,+y > 0, H is maximized by r = r,. Thus:; = r,,, throughout ihis interval, and it follows from Lemma I that

y/N(Q) s T-t’ s Q/r,,, or y G QN(Q)/r,,,.

Lemma 7 For an optimal control policy there is no value of &(O,T)

such that k,min < k,(t) < k,““” and dA,/dt = 0, unless n = p.


Proof For an optimal policy k,“‘” < k,(t) < k,“” only if k,

corresponds to a stationary maximum value of H, or

ab(A,, + Ab - 1) + ap k,P-, ab”(A. + nhb) = 0. (11)

If dAJdt = 0, on the other hand, it follows from Eq. (7) that

k,a( A, + A* - 1) + an k,*ab”-,( A, + nh,) = 0. (12)

A single value of k, satisfies both these equations only if

(i) A,+A,-1 =O,or (ii) n = p.

Case (i) gives k, = 0, which does not lie in the specified range, so only case (ii) is relevant, and the result follows.

Lemma 8 For an optimal policy, with p # n, and for sufficiently

large values of r,,,. the function A&) may not have a local maximum value at an interior point of a time interval in which r=r,.

Proof It is first necessary to dispose of the possibility that Ab

may have a local maximum at a point of discontinuity of k,. When p > 1, kl is continuous for sufficiently large r,,, (Theorem 1). so this question does not arise. When p S 1 detailed consideration of the possible forms of the Hamil- tonian shows that the limits of dA,/dt on approaching the point of discontinuity of k, from opposite sides have the same sign, thus precluding the possibility that A, has an extremum there. The reasoning is given in full elsewhere [4].

Any local maximum of Ab at an interior point of a time interval in which r = r,,, must therefore be a stationary point. Then by Lemma 7, k, = k,“‘” or k, = k,““” at this point, provided p # n. But at any such point dk,/dt vanishes, or, if the point coincides with an end point of an interval in which k, = k,“‘” or k, = k,max, the limit of dk,/dt on approaching the point from within this interval vanishes.

Consider the value of d2Ab/dtf at the stationary point, or, when appropriate as above, the limit of d”A,/dt’ on approaching the stationary point from within an adjacent interval of constant k,. This derivative can be found by differentiation of Eq. (7). Using the fact that dk,/dt = 0 at the stationary point, or at the points considered in approaching the stationary point as a limit, we obtain

= a(k, + nk,b”-,) 2+ n(n - l)k,ab”-2(A, + nA& $

after using the condition dA,/dt = 0 at the stationary point. Again, the left hand side is to be interpreted as a limit if necessary. From Eqs. (6) and (7) it also follows that

%=-k%(n-l)byA.+nA,)

where dA,/dt = 0. Thus,

= (n - l)bab”-*(A,+ nA,)(nr, - nk,ab - n*k%ab” I

-k,b*-&b”+,}. (13)

861

For sufficiently large values of r,,,. for example

r,,, > k,n’axaoQ + nk2max aoQn + k,““” Q' -+klmaxQn+l. (14)

n

the term in braces in (13) is positive. Since n > 1 and k, > 0,

it follows that (d2Ab/dtZ), has the sign of (A, + nh,). which we now consider.

Since the stationary point lies within an interval where r = rm and the control policy is optimal, (A, + y) 2 0 at the stationary point. (Otherwise r = r, would not maximize H.) Thus, using Lemma 6

A,?=-y 2 - QN/r,

Now suppose (A, + nh,) c 0. Then A,, + Ab c - (n - I)A, or, using the above inequality

A,+hb G (n-l)QN. rm

Thus.

I-A,-A, * l-(n-~)@f. rm

Recalling that N does not depend on r,,, (Lemma 3). it follows that 1 - A, - Ab > 0, provided

r, > (n-1)QN. (1%

But at a stationary point of A,, Eq. (7) shows that either

I- A, - Ab = an k,p-‘b”-‘( A, + nA,), (k, # 0) (16)

or k, = 0. As k, = 0 does not lie in the permitted interval, we need only consider (16), which shows that (A,+ nh,) has the same sign as (1 -A, - A& and hence that (A, + nh,) > 0, which contradicts our initial hypothesis that (A,, + nh,) =s 0.

We conclude that (A,+nA,) > 0 and hence, from (13), that (d2A,/dt2), > 0. Thus, the stationary value of A,(t) cannot be a local maximum.

Lemma 9 For an optimal policy, with p Z n, the function A&)

may not have a local minimum value at an interior point of a time interval in which r = 0.

Proof According to Theorem I.(a), there are no discontinuities

of k, in an interval where r = 0. Thus, the reasoning leading to Eq. (13) in Lemma 8 is still valid in this case, except that r = r,,, must be replaced by r = 0 in Eq. (13) giving

= -(n- l)k,ab”-‘(A, + nAb){nk,a + n2k,abn-’

+klb+nk,b”}. (17)

The term in braces is clearly positive so, since n > 1 and k, > 0, (d2A,/dtP), has the opposite sign to (A, + nh,).

At t = r the Hamiltonian is given by

H = yr + k,a(T) b(T)


and it is clear that its maximum value, Hmax(r), with respect to r and k,. is positive, irrespective of the sign of y. Further- more, from the maximum principle, H,,, has the same value at all points of an optimum trajectory and, in particular, equating its value at a point where r = 0 to its final value

Hmax(~) = k,ab( 1 - A, - AJ - k2 a&‘( A0 + nh,).

At a stationary point of Ab, using Eq. (7), this reduces to

Hmax(7) = (n - l)k,ab”(A, + nh,) > 0

whence it follows that (A,+nA,) > 0. Thus, (d*A,/dr*), < 0 and the stationary point cannot be a minimum.

Theorem 3 Whenp # n: (i) For an optimal control policy there exists no interval

(tr. t2), 0 < t,, t, < 7, such that r = 0 for all te(l,, t,) and A&) + y = A&J + y = 0.

(ii) For an optimal control policy, provided r, is sufficiently large, there exists no interval (tr, r,), 0 < ti, r, < 7. such that r = r,,, for all t&i, tt) and A&) + y = A&) + y = 0.

Proof (i) (A,+ y) is continuous in [tl, f2], so it attains its mini-

mum value in this interval. Furthermore, this minimum value is negative, since

(a) A,+y is nowhere positive in (ti, t2), otherwise r would take the value r, in some neighborhood of such a point

(b) Ab +y does not vanish identically in [t,, rZ], otherwise r would take values appropriate to a singular segment, as given in Theorem 6 below, rather than the specified value zero.

Since (A* + y) = 0 at t = tl, fZ, the minimum occurs at an interior point. But this contradicts Lemma 9. and we conclude that no interval of the type described may exist.

(ii) The proof of the second part follows exactly similar lines, within “minimum” replaced by “maximum”, the sign of (A,+ y) reversed, and r = 0 replaced by r = r,,,. Thus the existence of an interval as described is shown to imply the existence of a local maximum of Ab at an interior point, and this contradicts the result of Lemma 8 for sufficiently large r,.

Theorem 4 For an optimal policy, provided r, is sufficiently large.

there exists no interval (r1,7]. 1, > 0, such that r = r, for all te(t,, 71.

Proof If there exists such an interval at all. there exists such an

interval with Ab(tl)+ y = 0, since the interval with r = r, cannot have length 7, for sufficiently large rm, and (Ab + y) must vanish at its left terminal point.

(A,+y) is continuous in [rl, r], so it attains its maximum value and, as in Theorem 3, the fact that r = r, throughout (1.71 implies that this maximum value is positive. It cannot therefore occur at fi. Furthermore, d/dt(A,+y) = dA,/dr = - k,a < 0 at r = 7, so the maximum cannot occur at r = 7. It must, therefore, occur at an interior point, which again contradicts Lemma 8.

862

Corollary For sufficiently large r,,,, y 5 0 for an optimal policy.

Proof From Theorem 4 there exists a closed interval with r

as its right hand end point on which A,+ y < 0. Since h*(t) = 0 and IdA,/dt( is bounded by N, with a value independent of r,,,. it follows that-y s 0.

Lemma 10 For given values of Q. r,, k,“‘” and k,“““. the maximum

value of c(T) is a monotone increasing function of 7.

Proof

interval:-as seen from Eq. (3), so c(rJ > c(r), while Q and

Suppose r is chosen optimally in [0, T], with _fX r dt = Q.

r, are unchanged. Thus, there exists a policy in [0, T,] such that c(rJ > max(c(r)) when r1 > r, and it follows a fortiori

In (T, r,], r, > r, set r = 0. dc/dt > 0 everywhere in this

that max (~(7~)) > max(c(r)).

Theorem 5 For an optimal policy there exists no interval (0. t,),

0 < t, < r. such that r = Ofor all tc(0, tr).

Proof In such an interval b - 0. so a&) = a, and b(t,) = 0.

Thus, the initial conditions are simply transferred to I = t,. and the control policy, which is constrained by the same values of Q, r,, k,“‘“, and kimax, must now operate during the time interval r-t, < r. It follows from Lemma IO that the final value of c is less than max (c(r)), with the policy freely available over the whole interval [O. T]. Thus, a policy with an initial interval r = &I cannot be an optimal policy.

Theorem 6 If an optimal policy includes an interval of non zero length

in which 0 < r < r, everywhere, then

(a) Al=-Y,A,= k, + y(k, + tik,b”-‘)

k, + nk#-’

(b) (n- l)M ab”[l -r(n- I)1 = k, + r&&n-i

kima”a(r)b(r) + r,Y sgn’ (Y)

where . . . .

and

sgn’(y) = 1 f0r.y > 0 =0 forys0

(c) r = (k, + nk,b”-‘)&a + b/n).

(The part of the optimum trajectory corresponding to such an interval will be called a singular segment.)

Proof hb = -y implies that d&,/dt - 0, and substituting these

into Eq. (7) and solving for A, establishes result (a) above. Using these values of A. and A, in Eq. (IO) an expression

for the Hamiltonian can be found in terms of the physical variables only. Equating the value of this on the trajectory to the maximum value of the Hamiltonian at t = 7, as required by the maximum principle, result (b) follows.


To prove part (c) we need to refer to the proofs of Theorems 9 and IO below, where it is shown that k, takes the constant value k,min or k,ma” on a singular segment, depending on the sign of (n-p), and this is shown without reference to the result we are at present attempting to prove.

We may thus differentiate with respect to time the expression given for A, in result (a), treating k, and k, as constants. Substituting for dA,/dt in the result from Eq. (6), and for db/dr from Eq. (2) and using the values of A, and Ab already found. the desired expression for r is obtained.

Theorem 7

(ii) The final point t = r may not belong to a singular seg-

(i) The initial point t = 0 may not belong to a singular segment.

ment.

Proof (i) At the initial point a = a0 and b = 0. But examination

of Theorem 6(b) shows that the curve representing the singular segment parametrically in the (a,b)-plane does not pass through (ao, 0).

(ii) dA,/dt = - k,a(T) at t = r. Thus, dA,/dr < 0 at t = r and. since dA,/dt is piecewise continuous, there exists a non-zero interval adjacent to t = T in which dA,/dr < 0. But it is necessary that dA,/dt = 0 on a singular segment, so this interval may not be part of such a segment.

Theorem 8 For an optimal policy, provided r, is sufficiently large,

the function r(t) must take one of the two forms indicated in Fig. A. I.

Proof By Theorems 5 and 7(i), an optimal policy must start

with an interval in which r = r,,,. This must terminate at some t1 < r provided rmr > Q. Then A&) + y = 0 if the maximum principle is to be satisfied. Adjacent to the above interval in t > t, must be either an interval with r = 0, or an interval occupied by a singular segment. In the former case, the interval with r = 0 must extend to t = r, for if it terminated at some f, < 7, A&,)+ y = 0, and Theorem 3(i) would be violated. Thus, we have the control policy shown in Fig. (A.l,a).

In the latter case, the singular segment must terminate at some t2 < T, or Theorem 7(ii) would be violated. It cannot be followed by an interval in which r = r,, for if this interval terminated at some t, < 7, Theorem 3(ii) would be violated, while if it continued to t = 7, Theorem 4 would be violated. Thus, the singular segment must be followed by an interval with r = 0 and, if Theorem 3(i) is not to be violated, this must extend to I = 7, giving the policy shown in Fig. (A. I, b).

Theorem 9 For an optimal policy, when r,,, is sufficiently large and

p < n, k, = k,““” for all r > t,, where tr is the terminus of the initial interval with r = r,. (See Fig. A. I.)

Proof In the general expression for JH/Jk, obtained by differen-

tiating Eq. (IO), substitute the values of A, and Ab on the sin-

863


P

0

r

(a)

T t

(b)

J *

Fig. A. 1. Optimum r-policy.

gular segment, as given by Theorem 6(a). Then

aH krab”[l -r(n- l)](n-p) -= akr (k, + r&b”-‘) ’

But II > 1, and for sufficiently large r, we know that y s 0, so aH/ak, > 0 when p < n. It follows that the singular segment can form part of an optima1 policy only if k, = klm” everywhere on it.

Consider now the final interval with r = 0, i.e., the interval (t,, T] in Fig. (A.l, a) or the interval (f2, T] in Fig. (A.l, b). Denote the initial point of this interval by t’, where t’ = t, or t, in the cases represented by Figs. (A.l, a) and (A.1, b), respectively. We know that A&‘) + y = 0; A*(r) + y d 0, &It’, ~1; A&) = 0 and that y 4 0 for sufficiently large r,,,. From Theorem l(a), k, is continuous in (t,,~), so Ar is con- tinuously differentiable. It foilows that dA*/dr s 0 at t = t’, and t’ may not be the left hand end point of an interoal in which dA,/dt = 0, or this interval would correspond to a singular segment, contrary to the hypothesis that r=O in (t’,~]. Thus, recalling that dA&t is continuous, there exists some interval (t’, t’ + c) in which dA,/dt c 0. Then, if dA*/dt 2 0 at any point f” of (t’,T), there exists a value t”’ of t, t”’ e(t’, t”], such that A&(t) has a stationary minimum value at t’“. But this violates Lemma 9, so we conclude that dAddr < 0 for all h(t’,T). This implies that 0 C Ab < -y for all te(t’, 7).

Since the interval (t’, T] is part of an optimal policy, H,=(t)

= H,,,-(T) > 0 for all t&‘, T] and, since r = 0, this gives

k&A.+Ab-l)+kpbYA,+nAl) < Oforallt&‘,T].

Hence, from Eq. (6), dA./dt < 0 for ah t&‘.T), so A,(t)

> AJT) = 0 for all t&‘, T). From this inequality, and the corresponding inequality

previously found for A*(t), it is seen that

(A,+nAi,) > 0 fOrL3lltt$t’,T)

and, incidentally, A,+nA* decreases monotonically to zero in this interval.

Since dA,/dt < 0 in (t’, 7). Eq. (7) shows that

a&l - A, - A,) > F abYA, + nh,,) for all t&‘, 7). (18) 1

FromEq. (10)

g= ob(l-A._-A,)-p$ab~A,,+nA,) 1 1

and, using (18) above

aH (n-p% - > - ab”(A. + nA& ah k,

for all cc@‘, 7).

But we have shown that (A,+nA& > 0, so dH/akl > 0 when p < n, for ah te(t’, 7). Thus, this interval can form part of an optimal policy only if k, = klm” everywhere on it. This completes the proof.

Theorem 10 When p > n, for sufficiently large r, and sufficiently small

k,*, the optimal policy contains no singular segment.

Proof

On a singular segment it has already been shown that

aH Lab”[l-y(n-l)](n-p) -= ak, (kl + r&ab”-‘) .

For sufficiently large r,,,, y s 0 (Corollary to Theorem 4), so for p > n > 1, aH/akl < 0 on the singular segment, which can therefore form part of an optimal policy only if k, = k,* everywhere on it.

The equation of the singular segment in the (a, b)-plane can then be written (Theorem 6(b))

(n- I)ab”[l -r(n- l)] = kl*+n~*b”-l

klmaxd+o) k,tin tin ki

1 + na(klwn)P-lbn-L = a(kl*)P ’

(19)

We proceed to establish a contradiction by showing that the left hand side is bounded above by a bound independent of k,*, while the right hand side increases without bound as kl* + 0.

It is easy to show that b”-‘, b/b(T) and a/U(T) are bounded

864 i


above by bounds independent of k,m’“. Now consider the factor 1 --&I- 1). Since y < 0 for sufficiently large r,, this is equal to 1+ Iy((n-- l), so we need a bound for 1~1. But we are concerned with points on a singular segment, where ~,+y = 0, so Jy( = Jhb(. In Lemma 3 it was shown that IhbJ is bounded above by a bound independent of r, for a given value of Q. and the same argument also establishes the existence of a bound independent of /rImin. This is also a bound for IyI, so the left hand side of Eq. (19) is indeed bounded above by a bound whose value is independent of /c,~‘“. For sufficiently small values of k,“‘“, therefore, Eq. (19) cannot be satisfied and we conclude that a singular segment may not form part of an optimal policy.

For sufficiently large values of r,,,, in particular, in the limit as r,,, + m, Theorems 9 and 10 effectively determine the nature of the control policy. When p < n, Theorem 9 shows that it is optimal to operate isothermally at the maximum possible temperature and control through except pos-

in an initial whose length shrinks zero as + m.

p > Theorem 10 that it is optimal to add all the reactant B as quickly as possible at the beginning of the batch (i.e., dump in all the B instantaneously at t = 0, in the limit as rm -+ m), then exercise control through k,(r) or, equivalently, through the temperature.

R&m& On considkre le contr6le de rCactions chimiques concurrentes qui sont d’un ordre difft%ent par rapport B l’un des rCactifs. A la fois la temptrature et le taux d’apport des r&ctifs sont consid&% comme des variables de contr6le et il est demontrt que la nature de la politique optimale de contrBle dCpend de la relation entre le rapport des Cnergies d’activation des r&actions concurrentes et le rapport de leurs ordres en r&action avec le rhctif distribut. II y a un changement discontinu dans la nature de la politique optimale quand ces deux quantites sont 6gales.

Zusammenfassung- Die Steuerung von im Wettstreit befindlichen chemischen Reaktionen, die in Bezug auf einen der Reaktionsteilnehmer verschiedener Ordnung sind, wird behandelt. Es wird sowohl die Temperatur als such die Geschwindigkeit der Zugabe der Reaktionsteilnehmer als Steuer- variable in Erwiigung gezogen, und es wird gezeigt, dass das Wesen des optimalen Steuervorganges von der Beziehung zwischen dem VerhLltnis der Aktivierungsenergien der konkurrierenden Reaktionen und dem Verhlltnis ihrer Ordnungen in Bezug auf den verteilten Reaktionsteilnehmer abhlngt. Wenn diese beiden Griissen gleich sind erfolgt eine diskontinuierliche Anderung im Wesen dieses Optimalvorganges.

865

CES Vol. 26 No. 6-H

on the control of competing chemical reactions

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