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Assymmetry of temperature and concentration profiles in porous catalyst particles
218
RAY
Y. K. YANG,
L. I’ADMANAHHAN
LEON
LAI’IDUS
RELATION BETWEEN UNIQUENESS AND SYMMETRY
The Asymmetry of Temperature and Concentration Profiles in Porous Catalyst Particles The character of asymmetric solutions in porous catalysts involving internal and external transport limitations is analysed. Whenever uniqueness of the solution holds, symmetry exists. When solutions have multiplicity, a sujicient condition for symmetry is that the Sherwood and Nusselt numbers must be equal. This includes the simple case of uniformly constant surface concentrations and temperature. However, in the more generaland realistic case of unequal Sherwood and Nusselt numbers, it is suggested that some of the solutions may be highly cisymmetrical. It is proposed that the d@erence between these numbers may be used as a measure of asymmetry.
AND
Using a one-dimensional slab geometry we consider a porous catalyst of unit width with faces exposed to constant bulk concentration C, of the reactant and temperature TB. The steady-state equations for mass and energy transport within the particle are
d2y
d2z dX2 == -/IR(y,
= R(y, z) dx2
In the study of transport phenomena coupled with chemical reaction in porous catalyst particles it has always been a priori taken for granted that the temperature and concentration profiles are symmetric within the particle. This approach has recently been questioned by Horn et al.,’ who observe that the assumption of symmetry is justifiable only when uniqueness of the solution is known, and that when multiple solutions can exist, asymmetry is possible. In fact, these authors exhibited an asymmetric solution for a deliberately chosen set of physical parameters. Although they gave no systematic procedure for detecting such asymmetry in a given nevertheless their demonstration is of system, sufficient impact to urge caution while generalising the available results on uniqueness, multiplicity and stability of solutions for the catalyst problem. If asymmetric solutions can exist even under uniform boundary conditions (by this we mean constant ambient conditions), then the logical and natural question is whether one can infer their existence, or what is no less useful, their absence. In this short communication, we attempt to answer this two-fold question, concentrating mainly on the second feature. Specifically, we show that in certain special cases (e.g. when Sherwood and Nusselt numbers are equal) every solution is symmetric. The importance of this demonstration is transparent when we realise that most of the theoretical studies on the catalyst particle problem pertain to those special cases and thus the results therein obtained under the assumption of symmetry can be used with confidence.
(1)
where y and z are the dimensionless concentration and temperature variables normalised with respect to the bulk quantities C, and TB. R(y, z) is the chemical reaction rate in a dimensionless form and has the property: R(0, z) = 0 for irreversible
INTRODUCTION
z)
reaction,
R(y_,, z,J
= 0
at equilibrium of a reversible reaction, otherwise R(y, z) > 0. /I is the dimensionless heat effect parameter and signifies the temperature rise accompanying the reaction. The intra-particle phenomena depicted above are coupled to the inter-particle processes via the boundary conditions which describe the mass and energy transfer across the surface film: ~YYO)
=
-(I - y(0));
& z’(0) &Y'(l)
= 1 - y(l);
j&z'(l)
Here Sh and Nu are Sherwood respectively.
= -(I
- z(0))
= 1 - z(l)
(2)
and Nusselt numbers
Now we merely comment that uniqueness of the solution of eqns. (1) and (2) implies symmetry. The proof can be ascertained from the argument sketched by Horn et al.’
SOME FEATURES OF ASYMMETRIC SOLUTIONS At this juncture let us note that by(x) + z(x) is at most linear in x. This follows from integrating twice the equation by” + z” = 0, which is obtainable from eqn. (1). Thus we can write: By(x) + Z(X) = I + px
(3)
TEMPERATURE
AND CONCENTRATION
PROFILES
where I and p are arbitrary constants. We shall now evaluate this from the boundary conditions.
Applying
/?Y(O) + z(0) = 1
(4)
PY(l) + z(l) = 1 + /r
(5)
fi Sh (I - y(l)) Elimination yields :
+ Nu (I - z(l)) = /I
of y(O) and y(l)
= (6)
/1 Sh + Nu - A Sh + z(l) x (Sh - Nu) = /‘(I + Sh)
(9)
which leads to:
(4 1 - ztO))(Sh - W Sh + 2
(10)
On the other hand, eqns. (4) and (5) give: /I = B(Y(l) - Y(0)) + z(l) - z(0)
(11)
eqns. (10) and (11):
~( =- KY(l)
- y(O))(Nu Nu + 2
- Sh)
(12)
which is similar to eqn. (10). These different modes of expressing p will prove to be useful in the sequel. The following relation between the boundary values can be easily obtained from eqns. (I I) and (12):
4) - 40) = -KY(l)
+
= , +
(S)
- Y(O))
z(o)
pJ,to)
(13)
which shows that if y(l) > Y(O), then z(l) < z(0) for the asymmetric state. Next we shall express z(l) and z(0) in terms of y(l), y(0) and the physical parameters. Adding eqns. (6) and (7) we find:
Ptv(I)
+
- Y(l) - Y(0))
(14)
-
ytQ)(Nu Nu + 2
-
Sh)X
NY(~)- Y@)WU- Wx
+
(2 - y(0) - y(l)) + (S)
(Y(l) - Y(O))]
(17)
where, in the last step, we have used eqn. (16). From this we conclude that a solution is asymmetric, if 11 # 0, i.e., the coefficient of x in eqn. (17) is non-zero. Since p = 0 whenever z(l) = z(0) as seen from eqn. (IO), in order to detect asymmetry experimentally, one needs only to measure the surface temperatures. If the surface temperature is uniform, one can confidently assert that only symmetric profiles can exist. In this context, we mention that Maymo and Smith” did observe local variations in surface temperature of the particle immersed in a uniform environment.
BOUNDS
ON THE SURFACE
TEMPERATURE
We first note that Y, the reactant concentration has the following bounds 0 I Y(x) _< Y, I I, where Y, = Y(0) or Y(l), the surface concentration. For z, we have I 5 z, < z(x) if the reaction is exothermic and 0 _< z(x) 2 z, < I, if it is endothermic. Here z, is, of course, the surface temperature. Both of these can be combined in:
oakB
z(l) + z(0) = 2 + E(2
219
PARTICLES
Nu + 2
from eqns. (4) to (7)
(8)
If we combine
I?r,(o)
(7)
p Sh + Nu - II Sh + z(O)(Sh - Nu) = -p
~ _
CATALYST
We can now write PJJ, + z in terms of the position coordinate and the surface concentrations. Thus, from eqns. (4) and (12) we obtain:
eqn. (2) to 0~ + z’ = p, we obtain:
p Sh (I - y(0)) + Nu (1 - z(0)) = -p
IN POROUS
1) <- 44 - 1 P
We shall now obtain sharper Equation (15) can be rewritten as:
bounds
on
z,.
Then eqns. (13) and (14) yield: o
z(l) = 1 + a ?!I. (2 - y(l) 2 ] Nu -
& -
(&$)
- y(0))
(Y(l) - Y(O))]
(15)
d
_..
2tzu~ - 1) = B
(
$$
1
g
(2 - Y(l) - y(0))
(Y(l) - Y(O))
(2 - y(0) - y(1))
($$)
(Y(O) - Y(l))]
(16)
+ Y(O) [(s)
- g]
(18)
KAY
220
Y. K. YANG,
L. I’ADMANAHtlAN
Noting that [(Sh + 2)/(Nu + 2)) - Sh/Nu ; 0 according as Sh ; Nu, and using the bounds on J’,, we arrive at : Sh 0 5 0 < G +
Sh -
max
Any solution 11 = 0. Then By+z=-1
of eqns. (1) and (2) is symmetric
+fl;+fly(o)(l
-E)
+P&+/?y(O)(l
(20)
Then
-2)
Conversely, let fly + z be constant, Then from eqn. (12), -
Nu+2
Sh)
Thuseithery(l) = y(O), or (Nu - Sh)/(Nu + 2) = 0, the latter being equivalent to Nu = Sh. Here we have two possibilities: first the trivial case of Sh = Nu = 00, which implies y(0) = y(l) from the boundary condition (eqn. (2)): the second possibility is that Sh = Nu < 03. We shall now show that y(l) = y(O) in this case as well. From eqn. (17), we have by + z = 1 + j?. Thus z = i + /i’ - By. Inserting this in eqn. (I), we obtain: By!
Suppose now, y(I) > _I(()). Thus the LHS of eqn. (23) is positive. On the other hand, since P(1) 5Y(O)
Al)
I
Y(O) -
R(y) dy > 0
and 2 > y(1) + y(O), the RHS is negative, which is a contradiction. Thus y(1) > y(O). In a similar manner, y(O) + ~(1). Thus y(1) = y(0). Thus: 1)
== y(o)
--+ y’(l) = -y’(O)
(24) (25)
using eqn. (2). Now the function g(x) 2 y(I - X) has the initial value y(l) and initial slope -y’(I) and satisfies the differential equation (eqn. (21)). Since R(y) E C’(y), it is Lipshitzian and hence the initial value problem j” = i?(j), g(O) = y(l), p’(O) = --y’(I) has a unique solution. On the other hand, y(x) also satisfies the same equation and by eqns. (24) and (25) assumes the same initial value and initial slope as P(X). Thus, y(x) is also a solution. By virtue of uniqueness, y(x) = c(x) = ~(1 - x), or, y(x) is symmetric. As fly(x) + z(x) is constant when p = 0, z(x) is also symmetric.
y’(O)’ Sh*(2 - y(l)
- Y(O))(Y(OJ - v( 1))
Many of the theoretical studies in the past on the catalyst particle have been confined to the case Sh = Nu, for which we have shown that symmetry holds. In fact, it is very frequently assumed that y(O) = y(I) = I = z(0) = z( 1) which corresponds to Sh = Nu = 03. Exceptions are’s3e4 where, however, the authors have assumed the symmetry of the solution. Their conclusions are to be re-examined. We showed that if Sh=Nu, By+z=fi+ 1. The converse also holds. To see this, we set p = 0 and ), = I + /?, which corresponds to fly + z = /I + I, in eqns. (8) and (9) to obtain: (Sh -
(21)
Multiplying by 2y’, integrating from x = 0 to x = 1 and using boundary condition (eqn. (2)), we arrive at: 2
(23)
Remarks o
1 + P -
- J’(0))
i.e., 11 --= 0.
=
& R(Y,
R(J~dy
Y(O) - Si12(? - y(I)
_I,(1) - _tlO)
if
and :
R(Y)
)‘( 0
2
11 = 0 --+ J’(
Let (y(x), z(x)) be a symmetric solution. y( 1) = y(O). From eqn. (17) we conclude that:
Y” =
Thus
ON SYMMETRY
(Y(I) - Y@NW
I Al’l1)lJS
(19)
Nu’ Nu $- 2
Proof
j?y+z=l
l.I!ON
Sh t- 2 ~
where .zs = 1 + (/?@/2). This provides bounds for the surface temperature in both the cases of exothermic and endothermic reactions. To obtain estimates of the maxima of (z(x) - 1)/p one needs to use the specific reaction rate data. We will not pursue this here.
A THEOREM
ANI)
(22)
Nu)(z(O) -
(Sh - Nu)(z(l)
-
1) = 0 1) == 0
(26)
Similarly: (Sh - Nu)(l
- y(O)) = 0
(Sh - Nu)(l
- y(l)) = 0
(27)
Therefore it follows that either Sh = Nu < 00, or y(0) = y(l) = z(0) = z(1). The latter is equivalent to Sh =: Nu = 00.
TEMPERATURE
A MEASURE
AND
CONCENTRATION
PROFILES
the larger p is, the more asymmetric the solution will be. To obtain a ‘feel’ for the conditions under which asymmetry is enhanced, let us fix bounds on p. From eqn. (12), noting that 0 I J(O), y(l) < 1 when Sh < 03, we obtain:
On the other obtain : ,p, <
hand,
INu - Sh’ & c> Nu + 2 from
eqns.
(10) and
(19) we
lb-1INu - W 2(Sh + 2) x {$
Combining
+ max (g,
s)}
2. C,
CATALYST
221
PARTICLES
ACKNOWLEDGEMENT
OF ASYMMETRY
We have shown that if 11 = 0, the solution is symmetric. This suggests that perhaps 11 may be a measure of asymmetry. Indeed, since
lpl < IpI
IN POROUS
(29)
The authors wish to acknowledge the support of this work from National Science Foundation Grant NSF-GP-2858.
NOMENCLATURE
CB bulk concentration
c>
bound
NZ
Nusselt
R
reaction
Sh
Sherwood
on ~1 number rate number
TB
bulk temperature
X
distance
Y z
dimensionless
concentration
dimensionless
temperature
coordinate
the two, we obtain Greek
IPI <
c,, according
as Sh : Nu
In any case,*we see that the bounds are smaller as Sh approaches Nu. Thus, we conclude the less Sh differs from Nu, the less asymmetric the solutions will be. It is of interest to note that Carberry’ has estimated that the magnitude of Sh is much larger than that of Nu for gas-solid systems.
symbols
B
heat effect parameter
Y
activation energy parameter arbitrary constant arbitrary
Thiele modulus temperature
CONCLUSION In this paper, we have attempted to analyse the character of asymmetric solutions. A result of some significance is that asymmetry cannot exist when the Sherwood and Nusselt numbers are equal; this is the most widely studied case in the literature. In the more important case of Sh # Nu, there is enough reason to believe that asymmetry can arise; in that case, asymmetric solutions always occur in pairs, i.e., if y(x) is an asymmetric solution, ~(1 - x) is another. Which one prevails in a given physical system perhaps depends on the history of the system reaching that steady state. Sufficient conditions for asymmetry, stability of asymmetric states and uniqueness criteria in the general case of Sh # Nu, are all inviting subjects for further research.
constant
differential
REFERENCES I. BESKOV, V. S., AND MALINOVSKAYA,0. A., Fourfh hirernational Congress on Catalysis, Moscow: 1968. 2. CARBERRY,J. J., Ind. Eng. Chem., 1966 58
p. 10. p. 267.
3. CRESSWELL,D. L., Chem. Eng. Sci., 1970 25
4. HATFIELD, B., AND ARK, R., Chem. Eng. Sci., 1969 24 p. 1213. 5. HORN, F. J. M., JACKSON,R., MARTEL, E., AND PATEL, C.. Chem. Eng. J., 1970 1 p. 79. 6. MAYMO, J. A., AND SMITH, J. M., A.1.Ch.E.J.. p. 845.
RAY Y. K. YANG, L. PADMANABHANAND LEON LAPIDUS Department Princeton Princeton,
of Chemical
Engineering,
University, New Jersey
08540 (USA)
1966 12
RAY
Y. K. YANG,
L. I’ADMANAHHAN
LEON
LAI’IDUS
RELATION BETWEEN UNIQUENESS AND SYMMETRY
The Asymmetry of Temperature and Concentration Profiles in Porous Catalyst Particles The character of asymmetric solutions in porous catalysts involving internal and external transport limitations is analysed. Whenever uniqueness of the solution holds, symmetry exists. When solutions have multiplicity, a sujicient condition for symmetry is that the Sherwood and Nusselt numbers must be equal. This includes the simple case of uniformly constant surface concentrations and temperature. However, in the more generaland realistic case of unequal Sherwood and Nusselt numbers, it is suggested that some of the solutions may be highly cisymmetrical. It is proposed that the d@erence between these numbers may be used as a measure of asymmetry.
AND
Using a one-dimensional slab geometry we consider a porous catalyst of unit width with faces exposed to constant bulk concentration C, of the reactant and temperature TB. The steady-state equations for mass and energy transport within the particle are
d2y
d2z dX2 == -/IR(y,
= R(y, z) dx2
In the study of transport phenomena coupled with chemical reaction in porous catalyst particles it has always been a priori taken for granted that the temperature and concentration profiles are symmetric within the particle. This approach has recently been questioned by Horn et al.,’ who observe that the assumption of symmetry is justifiable only when uniqueness of the solution is known, and that when multiple solutions can exist, asymmetry is possible. In fact, these authors exhibited an asymmetric solution for a deliberately chosen set of physical parameters. Although they gave no systematic procedure for detecting such asymmetry in a given nevertheless their demonstration is of system, sufficient impact to urge caution while generalising the available results on uniqueness, multiplicity and stability of solutions for the catalyst problem. If asymmetric solutions can exist even under uniform boundary conditions (by this we mean constant ambient conditions), then the logical and natural question is whether one can infer their existence, or what is no less useful, their absence. In this short communication, we attempt to answer this two-fold question, concentrating mainly on the second feature. Specifically, we show that in certain special cases (e.g. when Sherwood and Nusselt numbers are equal) every solution is symmetric. The importance of this demonstration is transparent when we realise that most of the theoretical studies on the catalyst particle problem pertain to those special cases and thus the results therein obtained under the assumption of symmetry can be used with confidence.
(1)
where y and z are the dimensionless concentration and temperature variables normalised with respect to the bulk quantities C, and TB. R(y, z) is the chemical reaction rate in a dimensionless form and has the property: R(0, z) = 0 for irreversible
INTRODUCTION
z)
reaction,
R(y_,, z,J
= 0
at equilibrium of a reversible reaction, otherwise R(y, z) > 0. /I is the dimensionless heat effect parameter and signifies the temperature rise accompanying the reaction. The intra-particle phenomena depicted above are coupled to the inter-particle processes via the boundary conditions which describe the mass and energy transfer across the surface film: ~YYO)
=
-(I - y(0));
& z’(0) &Y'(l)
= 1 - y(l);
j&z'(l)
Here Sh and Nu are Sherwood respectively.
= -(I
- z(0))
= 1 - z(l)
(2)
and Nusselt numbers
Now we merely comment that uniqueness of the solution of eqns. (1) and (2) implies symmetry. The proof can be ascertained from the argument sketched by Horn et al.’
SOME FEATURES OF ASYMMETRIC SOLUTIONS At this juncture let us note that by(x) + z(x) is at most linear in x. This follows from integrating twice the equation by” + z” = 0, which is obtainable from eqn. (1). Thus we can write: By(x) + Z(X) = I + px
(3)
TEMPERATURE
AND CONCENTRATION
PROFILES
where I and p are arbitrary constants. We shall now evaluate this from the boundary conditions.
Applying
/?Y(O) + z(0) = 1
(4)
PY(l) + z(l) = 1 + /r
(5)
fi Sh (I - y(l)) Elimination yields :
+ Nu (I - z(l)) = /I
of y(O) and y(l)
= (6)
/1 Sh + Nu - A Sh + z(l) x (Sh - Nu) = /‘(I + Sh)
(9)
which leads to:
(4 1 - ztO))(Sh - W Sh + 2
(10)
On the other hand, eqns. (4) and (5) give: /I = B(Y(l) - Y(0)) + z(l) - z(0)
(11)
eqns. (10) and (11):
~( =- KY(l)
- y(O))(Nu Nu + 2
- Sh)
(12)
which is similar to eqn. (10). These different modes of expressing p will prove to be useful in the sequel. The following relation between the boundary values can be easily obtained from eqns. (I I) and (12):
4) - 40) = -KY(l)
+
= , +
(S)
- Y(O))
z(o)
pJ,to)
(13)
which shows that if y(l) > Y(O), then z(l) < z(0) for the asymmetric state. Next we shall express z(l) and z(0) in terms of y(l), y(0) and the physical parameters. Adding eqns. (6) and (7) we find:
Ptv(I)
+
- Y(l) - Y(0))
(14)
-
ytQ)(Nu Nu + 2
-
Sh)X
NY(~)- Y@)WU- Wx
+
(2 - y(0) - y(l)) + (S)
(Y(l) - Y(O))]
(17)
where, in the last step, we have used eqn. (16). From this we conclude that a solution is asymmetric, if 11 # 0, i.e., the coefficient of x in eqn. (17) is non-zero. Since p = 0 whenever z(l) = z(0) as seen from eqn. (IO), in order to detect asymmetry experimentally, one needs only to measure the surface temperatures. If the surface temperature is uniform, one can confidently assert that only symmetric profiles can exist. In this context, we mention that Maymo and Smith” did observe local variations in surface temperature of the particle immersed in a uniform environment.
BOUNDS
ON THE SURFACE
TEMPERATURE
We first note that Y, the reactant concentration has the following bounds 0 I Y(x) _< Y, I I, where Y, = Y(0) or Y(l), the surface concentration. For z, we have I 5 z, < z(x) if the reaction is exothermic and 0 _< z(x) 2 z, < I, if it is endothermic. Here z, is, of course, the surface temperature. Both of these can be combined in:
oakB
z(l) + z(0) = 2 + E(2
219
PARTICLES
Nu + 2
from eqns. (4) to (7)
(8)
If we combine
I?r,(o)
(7)
p Sh + Nu - II Sh + z(O)(Sh - Nu) = -p
~ _
CATALYST
We can now write PJJ, + z in terms of the position coordinate and the surface concentrations. Thus, from eqns. (4) and (12) we obtain:
eqn. (2) to 0~ + z’ = p, we obtain:
p Sh (I - y(0)) + Nu (1 - z(0)) = -p
IN POROUS
1) <- 44 - 1 P
We shall now obtain sharper Equation (15) can be rewritten as:
bounds
on
z,.
Then eqns. (13) and (14) yield: o
z(l) = 1 + a ?!I. (2 - y(l) 2 ] Nu -
& -
(&$)
- y(0))
(Y(l) - Y(O))]
(15)
d
_..
2tzu~ - 1) = B
(
$$
1
g
(2 - Y(l) - y(0))
(Y(l) - Y(O))
(2 - y(0) - y(1))
($$)
(Y(O) - Y(l))]
(16)
+ Y(O) [(s)
- g]
(18)
KAY
220
Y. K. YANG,
L. I’ADMANAHtlAN
Noting that [(Sh + 2)/(Nu + 2)) - Sh/Nu ; 0 according as Sh ; Nu, and using the bounds on J’,, we arrive at : Sh 0 5 0 < G +
Sh -
max
Any solution 11 = 0. Then By+z=-1
of eqns. (1) and (2) is symmetric
+fl;+fly(o)(l
-E)
+P&+/?y(O)(l
(20)
Then
-2)
Conversely, let fly + z be constant, Then from eqn. (12), -
Nu+2
Sh)
Thuseithery(l) = y(O), or (Nu - Sh)/(Nu + 2) = 0, the latter being equivalent to Nu = Sh. Here we have two possibilities: first the trivial case of Sh = Nu = 00, which implies y(0) = y(l) from the boundary condition (eqn. (2)): the second possibility is that Sh = Nu < 03. We shall now show that y(l) = y(O) in this case as well. From eqn. (17), we have by + z = 1 + j?. Thus z = i + /i’ - By. Inserting this in eqn. (I), we obtain: By!
Suppose now, y(I) > _I(()). Thus the LHS of eqn. (23) is positive. On the other hand, since P(1) 5Y(O)
Al)
I
Y(O) -
R(y) dy > 0
and 2 > y(1) + y(O), the RHS is negative, which is a contradiction. Thus y(1) > y(O). In a similar manner, y(O) + ~(1). Thus y(1) = y(0). Thus: 1)
== y(o)
--+ y’(l) = -y’(O)
(24) (25)
using eqn. (2). Now the function g(x) 2 y(I - X) has the initial value y(l) and initial slope -y’(I) and satisfies the differential equation (eqn. (21)). Since R(y) E C’(y), it is Lipshitzian and hence the initial value problem j” = i?(j), g(O) = y(l), p’(O) = --y’(I) has a unique solution. On the other hand, y(x) also satisfies the same equation and by eqns. (24) and (25) assumes the same initial value and initial slope as P(X). Thus, y(x) is also a solution. By virtue of uniqueness, y(x) = c(x) = ~(1 - x), or, y(x) is symmetric. As fly(x) + z(x) is constant when p = 0, z(x) is also symmetric.
y’(O)’ Sh*(2 - y(l)
- Y(O))(Y(OJ - v( 1))
Many of the theoretical studies in the past on the catalyst particle have been confined to the case Sh = Nu, for which we have shown that symmetry holds. In fact, it is very frequently assumed that y(O) = y(I) = I = z(0) = z( 1) which corresponds to Sh = Nu = 03. Exceptions are’s3e4 where, however, the authors have assumed the symmetry of the solution. Their conclusions are to be re-examined. We showed that if Sh=Nu, By+z=fi+ 1. The converse also holds. To see this, we set p = 0 and ), = I + /?, which corresponds to fly + z = /I + I, in eqns. (8) and (9) to obtain: (Sh -
(21)
Multiplying by 2y’, integrating from x = 0 to x = 1 and using boundary condition (eqn. (2)), we arrive at: 2
(23)
Remarks o
1 + P -
- J’(0))
i.e., 11 --= 0.
=
& R(Y,
R(J~dy
Y(O) - Si12(? - y(I)
_I,(1) - _tlO)
if
and :
R(Y)
)‘( 0
2
11 = 0 --+ J’(
Let (y(x), z(x)) be a symmetric solution. y( 1) = y(O). From eqn. (17) we conclude that:
Y” =
Thus
ON SYMMETRY
(Y(I) - Y@NW
I Al’l1)lJS
(19)
Nu’ Nu $- 2
Proof
j?y+z=l
l.I!ON
Sh t- 2 ~
where .zs = 1 + (/?@/2). This provides bounds for the surface temperature in both the cases of exothermic and endothermic reactions. To obtain estimates of the maxima of (z(x) - 1)/p one needs to use the specific reaction rate data. We will not pursue this here.
A THEOREM
ANI)
(22)
Nu)(z(O) -
(Sh - Nu)(z(l)
-
1) = 0 1) == 0
(26)
Similarly: (Sh - Nu)(l
- y(O)) = 0
(Sh - Nu)(l
- y(l)) = 0
(27)
Therefore it follows that either Sh = Nu < 00, or y(0) = y(l) = z(0) = z(1). The latter is equivalent to Sh =: Nu = 00.
TEMPERATURE
A MEASURE
AND
CONCENTRATION
PROFILES
the larger p is, the more asymmetric the solution will be. To obtain a ‘feel’ for the conditions under which asymmetry is enhanced, let us fix bounds on p. From eqn. (12), noting that 0 I J(O), y(l) < 1 when Sh < 03, we obtain:
On the other obtain : ,p, <
hand,
INu - Sh’ & c> Nu + 2 from
eqns.
(10) and
(19) we
lb-1INu - W 2(Sh + 2) x {$
Combining
+ max (g,
s)}
2. C,
CATALYST
221
PARTICLES
ACKNOWLEDGEMENT
OF ASYMMETRY
We have shown that if 11 = 0, the solution is symmetric. This suggests that perhaps 11 may be a measure of asymmetry. Indeed, since
lpl < IpI
IN POROUS
(29)
The authors wish to acknowledge the support of this work from National Science Foundation Grant NSF-GP-2858.
NOMENCLATURE
CB bulk concentration
c>
bound
NZ
Nusselt
R
reaction
Sh
Sherwood
on ~1 number rate number
TB
bulk temperature
X
distance
Y z
dimensionless
concentration
dimensionless
temperature
coordinate
the two, we obtain Greek
IPI <
c,, according
as Sh : Nu
In any case,*we see that the bounds are smaller as Sh approaches Nu. Thus, we conclude the less Sh differs from Nu, the less asymmetric the solutions will be. It is of interest to note that Carberry’ has estimated that the magnitude of Sh is much larger than that of Nu for gas-solid systems.
symbols
B
heat effect parameter
Y
activation energy parameter arbitrary constant arbitrary
Thiele modulus temperature
CONCLUSION In this paper, we have attempted to analyse the character of asymmetric solutions. A result of some significance is that asymmetry cannot exist when the Sherwood and Nusselt numbers are equal; this is the most widely studied case in the literature. In the more important case of Sh # Nu, there is enough reason to believe that asymmetry can arise; in that case, asymmetric solutions always occur in pairs, i.e., if y(x) is an asymmetric solution, ~(1 - x) is another. Which one prevails in a given physical system perhaps depends on the history of the system reaching that steady state. Sufficient conditions for asymmetry, stability of asymmetric states and uniqueness criteria in the general case of Sh # Nu, are all inviting subjects for further research.
constant
differential
REFERENCES I. BESKOV, V. S., AND MALINOVSKAYA,0. A., Fourfh hirernational Congress on Catalysis, Moscow: 1968. 2. CARBERRY,J. J., Ind. Eng. Chem., 1966 58
p. 10. p. 267.
3. CRESSWELL,D. L., Chem. Eng. Sci., 1970 25
4. HATFIELD, B., AND ARK, R., Chem. Eng. Sci., 1969 24 p. 1213. 5. HORN, F. J. M., JACKSON,R., MARTEL, E., AND PATEL, C.. Chem. Eng. J., 1970 1 p. 79. 6. MAYMO, J. A., AND SMITH, J. M., A.1.Ch.E.J.. p. 845.
RAY Y. K. YANG, L. PADMANABHANAND LEON LAPIDUS Department Princeton Princeton,
of Chemical
Engineering,
University, New Jersey
08540 (USA)
1966 12