The Chemical Engineering Journal, Elsevier Sequoia S.A., Lausanne. Printed in The Netherlands

Necessary Conditions for the Existence of Asymmetric Composition and Temperature Profiles in Catalyst Pellets

R. JACKSON

Rice University, Houston, Texas (USA)

and

F. J. M. HORN

University of Rochester, Rochester, New York (USA)

(Received 11 December, 1970; in final form 28 December, 1970)

Abstract

For a single irreversible chemical reaction in an infinite catalyst slab with a uniform environment, it is shown that temperature and composition profiles unsym- metric about the center plane of the slab may occur only when there are concentrated heat and mass trans- fer coefficients, the effective thermal conductivity and diffusion coefficient in the slab, and the heat of reac- tion.

In particular, for a conventional irreversible reaction, it is found that asymmetric profiles may exist only if the reaction is exothermic and Sh’ > Nu: where Sh’ and Nu’ are the modified Sherwood and Nusselt numbers.

INTRODUCTION

Recently, Horn et al. 1 pointed out that solutions of the equations of diffusion and chemical reaction within a catalyst pellet need not necessarily have the symmetry of the pellet, even when the environment is uniform. In the particular case of an infinite catalyst slab, with Newtonian resistances to heat and mass transfer at the boundaries, they demonstrated the existence of solutions unsymmetric with respect to reflection across the center plane and solutions not invariant against translations parallel to the faces. In view of this it is interesting to identify conditions under which asymmetric solutions may occur or, alternatively, conditions sufficient to ensure

Chem. Eng. J., 3 (1972)

symmetry. In the case of reflection symmetry across the center plane of a slab, Yang et al.2 have recently shown that a sufficient condition for symmetry is Sh’ = Nu’, where Sh’ and Nu’ are the modified Sher- wood and Nusselt numbers defined by Sh’ = la/D and Nu’ = ma/K. Here 1 and m are the mass and heat transfer coefficients at the slab faces, D and K are the effective diffusion coefficient and thermal con- ductivity in the slab, and a is the half thickness.

In this note we derive some necessary conditions for the existence of asymmetric solutions, which contain the results reported by Yang et al. as a

particular case. For example, for the case of a con- ventional irreversible reaction, it is shown that the reaction must be exothermic and that Sh’ must be greater than Nu’ if asymmetric solutions are to exist.

MATHEMATICAL DESCRIPTION OF THE PROBLEM

We consider a single reaction A = B in an infinite slab of catalyst, with internal diffusion and heat con- duction and Newtonian resistances to heat and mass transfer at the faces. Outside the slab the temperature and concentration of reactant A have the uniform values Tb and cb. Then, considering only solutions which are invariant against translations parallel to the slab faces, the concentration of A and the temperature satisfy :

(1)

ASYMMETRIC COMPOSITION AND TEMPERATURE PROFILES 83

K $= -Qr(c,T) (2)

in an interval (x0,x r ), where x is measured per- pendicular to the slab faces, and xo and xr are the coordinates of the faces (xe < xr). Y(C, r) denotes the rate of reaction, D and K are the effective diffusion coefficient and thermal conductivity, respectively, and Q is the heat of reaction, counted positive if the reaction is exothermic. The following boundary conditions at the faces must also be satisfied:

= @(x0) - Cb)

and

= - m(T(xl) - Tb)

where I and m are the mass and heat transfer coefficients, respectively, at the boundaries. Hence- forth it will frequently prove convenient to use primes to denote differentiation with respect to x, and the contracted notation CO, TO, cl, T1 for c(xe), 7(x0), c(xr) and T(a).

It will be assumed throughout that r(c,T) is positive for all non-zero values of its arguments.

SOME SIMPLE PROPERTIES OF THE

SOLUTION

Sincer>O,eqn.(l)showsthatc”>Oin(xo,xr), so c’(x r ) > c’(xo). Using boundary conditions (3) and (4) it follows that cr + co < 2 cb. Now suppose c’(xe) > 0. Then from eqn. (3) CO > cb. Also c’(x) > 0 throughout (xo,xr), since c” > 0, so cr > co > cb. Consequently, CO t cr > 2 cb, and we have a contradiction. We conclude that C’(XO) < 0, and a similar argument shows that c’(xr ) > 0. From eqns. (3) and (4) it then follows that ce < cb and cr < cb and, since c” > 0 in (x0,x1), c(x) is unimodal in this interval, with a stationary minimum at some interior point xm . &o c < cb throughout [x0,x1].

Chem. Eng. J., 3 (1972)

Similar arguments based on eqn. (2) and boundary conditions (5) and (6) show that T(x) is also unimodal in [x0,x1] with a single stationary minimum, and that T < Tb everywhere in this interval when the reaction is endothermic (Q < 0). For an exothermic reaction, on the other hand, T(x) is found to have a single stationary maximum at an interior point of [xe,xr] , and T > Tb everywhere in this interval.

From eqns. (1) and (2):

QDc” t KT” = 0

and integrating twice gives:

QDc’ t KT’ = a.2

and

(7)

QDctKT=cul ta2x (8)

where ol and o2 are constants. Substituting for c’ and T’ from eqns. (3) to (6) into eqn. (7):

QZ(co - CL,) + m(To - Tb) = CYZ

and:

Ql(cl - CL,) + m(T1 - Tb) = - CY~

Taking the difference of these:

QZ(cl - CO) + m(T1 - TO) = - 2~12

But, from eqn. (8) applied at x = xo and xr

(9)

QD(c1 - co)+K(Tl- To)=c&-xo) 00)

From eqns. (9) and (10) by elimination of T1 - TO and cr - CO, respectively, we obtain the following expressions for ff2 :

(y2 = Q(mD - W (cl- CO) 2K + m(xl -x0) (11)

and

o12 = (ZK - mD) (Tl - To)

w t Z(x1 -x0) (12)

84 R. JACKSON AND F. J. M. HORN

NECESSARY CONDITIONS FOR THE

EXISTENCE OF AN ASYMMETRIC SOLUTION

We will derive conditons which are necessary for the existence of solutions c(x) and T(x) which are not symmetric about the center plane of the slab. The converse of these conditions, of course, provides sufficient conditions for symmetry of the solutions.

First note that T can be eliminated from eqn. (1) using eqn. (8), giving:

(13)

Any solution of the problem defined by eqns (1) to (6) satisfies eqn. (13) with suitable values of (~1 and (~2. Indeed, a complete solution of the problem defined by eqns. (1) to (6) can be obtained by solving eqn. (13) subject to the boundary conditions (3) and (4) generating T(x) from the resulting c(x) by use of eqn. (8) and adjusting the parameters 011 and a2

until the boundary conditions (5) and (6) are also satisfied.

We can now prove:

Theorem 1 If c(x) and T(x) satisfy eqns. (1) to (6) and c(x) satisfies eqn. (13) with (~2 = 0, then both c(x) and T(x) are symmetric about the center plane of the slab.

Proof: Since o2 = 0, the right-hand side of eqn. (13) is a function of c only, so c(x) is symmetric about x = x,,, where c takes its minimum value. It has already been shown that c has a unique minimum in (x0,x1), and it remains only to show that this is located at the center plane of the slab.

Replacing xr by 2xm - xi in boundary condition (4) and using the symmetry of c(x) about xm , we find:

Dc’Pm - xl) = I [c(zxm - xl> - cb]

so condition (3) is satisfied at x = 2x, - x r . Further- more, c’(x) decreases monotonically from zero as x decreases from xm, while 1 [c(x) - cb] increases mono- tonically and without bound from a negative value l [c(x,) - cb] , so it follows that the condition DC’ = I(c - cz,) is satisfied at one and only one point in x < x,,, . This point is the slab face x = xo, so 2x, -xi=xe,orxm = $(x0 + xi), showing that xm corresponds to the center plane.

Chem. Eng. .I., 3 (1972)

Having established that c(x) is symmetric about this plane, it follows immediately from eqn. (8) that T(x) is also symmetric when (~2 = 0, thus completing the proof.

This theorem was also established by Yang et al. 2 using a somewhat different argument and, as shown by these writers, it follows almost immediately that all solutions are symmetric when ZK = mD. The solutions are also necessarily symmetric if ce = ci or ifTe=Ti.

Theorem 2 If T(C, T) is a monotone increasing function of T, necessary conditions for the existence of an asymmetric solution of the problem defined by eqns. (1) to (6) are:

(a) 1K > mD when the reaction is exothermic (Q > 0)

(b) IK < mD when the reaction is endothermic

<Q < 0)

PLoof: From Theorem 1, the existence of an asymmetric solution of the problem defined by eqns. (1) to (6) implies the existence of an asymmetric c(x) satisfying eqn. (13) and the boundary conditions (3) and (4), with o2 # 0. In particular, suppose a2 > 0. It is known that c(x) has a unique stationary minimum at some xm E (x0,x1), so c satisfies eqn. (13) and the conditions:

c(xm>=cm, c’(x,) = 0 (14)

We introduce a comparison function y(x) satisfying

Dy” = r Y, 011 +~~xrn - Q0-u

K 1 and the conditions

(15)

rem) = cm > Y'(Xm ) = 0 (16)

Since r > 0 and the right-hand side of eqn. (15) is a function of y only, y(x) is symmetric about a minimum at x = xm . Again, since r > 0, eqn. (13) shows that c is a monotone increasing function of x in x > Xm , SO we

can write:

X=f(C)inx>Xm

expressing x as a single valued function of c.

(17)

ASYMMETRIC COMPOSITION AND TEMPERATURE PROFILES 85

Multiplying eqn. (13) by 2c’, integrating, and using eqn. (17) we obtain:

II = 2 i

=m r (u, ” ’ ol2F’ - @‘) du

(inx>xm) (18)

while similarly

D(r’)2 = 2 _f r (u, a’ ’ a2F - Qfi) du

%?I

(in x > xm)

Setting y = c and subtracting (19) from (18):

D t(c’)* - tY’>21 y=c

=2 (~1 + cyzf(u) - QLh

K

(19)

a1 +oL2xm - QDu K II du (20)

But ~12 > 0, f(u) > xm in the interval of integration, and r(c, T) is a monotone increasing function of T, so the integrand on the right-hand side of eqn. (20) is positive everywhere in (cm,c)., Since c > cm, it follows that the integral is positive, so:

[(C’)2 - (r’)2] y== > 0 (in X > Xm)

or, since both c’ and y’ are positive in x > xm,

c’ > y’ for all pairs of points with c = 7 (21)

An exactly similar argument shows that the inequality (21) is also satisfied for all pairs of points in x < xm with c = y, SO the forms of C(X) and y(x)

are as indicated by the continuous and broken curves, respectively, in Fig. 1.

Point A in Fig. 1 represents the right-hand face of the slab atx=xr, where:

Dc’(xr) = I [Cb - C(Xl)]

But from (21) y (x”r) < c’(xr), where XI such that rc?r) = c(xr). Thus,

Dy’(?r) < 2 [Q, - y(&)] (Point B)

Chem. Eng. J., 3 (1972)

(4)

(>Xm) iS

Fig. 1. Comparison of c(x) and r(x) used in proof of Theorem 2.

and hence, since y’ increases monotonically and cb - y decreases monotonically and without bound as x increases, there exists a unique Xl > 21, such that:

as indicated by Point C in Fig. 1. Since y(x) is symmetric about x = xm, it then follows that:

&‘(km - XI) = 1 [‘y(am - ?I) - Cb] (Point D)

But from (21) c’(x*) > r’(2xm - Xl), where X*

(< xm) is such that c(x*) = y(;?xm - ?I). Thus:

&‘(X*) > l [C(X*) - cb] (Point E) (22)

However, at the left-hand face of the slab:

Dc’(xo) = I [c(xo) - Cb] (3)

and, since c’ decreases monotonically and c increases monotonically and without bound as x decreases from xm, comparison of (3) and (22) shows that xe <x*, as indicated by Point F in Fig. 1. Then:

C(X0) > C(X*) = *2Xm - Xl) = y(Zl) > y(x"l)=C(Xl)

86 R. JACKSON AND F. J. M. HORN

or cr - co < 0. Referring to eqn. (11) and recalling that (1~2 > 0, by hypothesis, it is seen that:

lK>mD ifQ>O (23)

and

lK<mD ifQ<O (24)

which is the desired result. Starting from the hypothesis (~2 < 0, rather than

(~2 > 0, we would have deduced, in the same way, that cr - co > 0, leading, once again, to conditions (23) and (24).

Theorem 3 If r(c,T) is a monotone increasing function of c, the condition IK > mD is necessary for the existence of an asymmetric solution of the problem defined by eqns. (1) to (6) for both exothermic and endothermic reactions.

Indication of proofi This result is established by arguments closely analogous to those used in deriving Theorem 2, so it will suffice to indicate the line of reasoning.

The proof of Theorem 2 was based on eqn. (13) obtained by eliminating T from the original differ- ential equations (1) and (2). Alternatively, eqn. (8) may be used to eliminate c from these equations, giving:

KT” = - Qr (~1 + (1~2~ - KT

QD ‘T (25)

subject to the boundary conditions (5) and (6). T(x) has a unique maximum or unique minimum at some xm E (x0,x1), for Q > 0 and Q < 0, respectively, so:

T(xm) = Tm; T’(xm) = 0 for some Xm E (x0,x1)

(26) which may be compared with conditions (14) satisfied by c(x).

We may now introduce a comparison function r(x), satisfying:

(27)

and the conditions

T(xm)=Tm, 7’(xm) = 0 (28)

and proceed to compare the functions T(x) and r(x), just as c(x) and y(x) were compared in proving Theorem 2. It is necessary to reason separately for the cases Q > 0 and Q < 0 because of the way Q enters eqns. (25) and (27) but in each case the reasoning follows the same lines as the proof of Theorem 2, except that it uses the monotonicity of r with respect to c, rather than T. In both cases an initial hypothesis that (~2 > 0 leads to the conclusion that TI > To, and it then follows from eqn. (12) that IK > mD. Taking (~2 < 0, one finds TI < TO, and, hence, the same result.

Theorems 2 and 3 are the main results of this work, but by introducing further assumptions about the form of r(c,T), corresponding to commonly-used kinetic expressions, stronger statements can be made. For example, a conventional kinetic expression for an irreversible reaction A + B is of the form:

r(c,T) = k(T) h(c)

where k(T) is a monotone increasing function of T and h(c) is a monotone increasing function of c. The fact that r is monotone in both c and Tin such a case leads immediately to a further result.

Theorem 4 If r(c,T) is a monotone increasing function of both its arguments, necessary conditions for the existence of an asymmetric solution of the problem defined by eqns. (1) to (6) are:

(a) Q > 0 (i.e. the reaction is exothermic)

and (b) s> 1

Proof: When Q < 0, the necessary conditions of Theorems 2 and 3 are mutually exclusive, so an asymmetric solution is impossible.

When Q > 0, both Theorem 2 and Theorem 3 yield the further necessary condition IKfmD > 1.

In terms of the modified Sherwood and Nusselt numbers, we have shown that asymmetric solutions cannot occur unless the reaction is exothermic and Sh’ > Nu’. In practice, of course, 1KfmD is usuall) larger than one because of the enhancement of effective thermal conductivity by the solid material in the slab, and indeed, in such a case Horn et al. 1 have demonstrated the existence of an asymmetric solution.

Note that it cannot be concluded that the conditions of Theorem 4 apply to all irreversible

Chem. Eng. J., 3 (1972)

ASYMMETRIC COMPOSITION AND TEMPERATURE PROFILES

reactions; for example, with kinetics of the Langmuir- Hinshelwood form, r may not be monotone in both c and T. It is for this reason that we referred to “conventional” irreversible reactions above and earlier in this paper.

It is tempting to speculate that the results may also be applicable in the case of reversible reactions, when I is not positive for all values of its arguments, but before any such conclusion can be firmly established, it must be proved that r retains a positive value throughout the catalyst.

ACKNOWLEDGMENT

The authors would like to thank Professor L. Lapidus for an opportunity to see reference 2 before pub- lication, and for permission to quote the results of this work.

The present work was supported, in part, by the National Science Foundation under grant No. GK 12522.

NOMENCLATURE

a half thickness of catalyst slab c concentration of reactant co, Cl values of c(xo), c(xi) cb concentration of reactant outside catalyst

slab Cm minimum value of c in slab D effective diffusion coefficient of reactant

in slab

K

1 m Nu’ Q r Sh’ T To, TI Tb T??l

X

X0, Xl

X??I

modified Sherwood number, la/D absolute temperature value of T(xo), T(xI)

temperature outside catalyst slab minimum/maximum value of Tin slab for endothermic/exothermic reaction coordinate measured perpendicular to slab faces coordinates of slab faces coordinate of stationary value of c(x) or T(x)

&X1,X* points in (xo, x 1) introduced in proof of Theorem 2

87

effective thermal conductivity in catalyst slab mass transfer coefficient at face of catalyst heat transfer coefficient at face of catalyst modified Nusselt number, ma/K heat of reaction, positive when exothermic reaction rate

Greek symbols Ql, cuz integration constants introduced in eqns.

(7) and (8) Y function of x defined by eqns. (15) and

(16) 7 function of x defined by eqns. (27) and

(28)

REFERENCES

1 F. J. M. Horn, R. Jackson, E. Mattel and C. Patel, The Chem. Eng. J., I (1970) 79.

2 Y. K. Yang, L. Padmanabhan and L. Lapidus, The Chem. Eng. J., 2 (1972) 222.

Chem. Erzg. J., 3 (1972)