ARTICLE pubs.acs.org/IECR
Effect of Domain Perturbations on the Critical Condition for Steady-State Thermal Explosions W. R. Batson, L. E. Johns, and R. Narayanan* Department of Chemical Engineering, University of Florida, Gainesville, Florida 32611, United States ABSTRACT: A region heated by a temperature-dependent source is said to be stable if the heat conducted to the surroundings can balance the heat generated in the region. The critical value of the Frank-Kamenetskii number gives the limit of stability, the point at which a steady temperature field is no longer possible. Assuming that the critical Frank-Kamenetskii number can be obtained for a symmetric region where the problem is one-dimensional, we present a method for determining how it changes as the region is displaced into a less-symmetric region. We find that the displacement can be either stabilizing or destabilizing. The case of exponential heat generation is fairly well-predicted by the case of linear heat generation.
1. INTRODUCTION All of the known solutions to thermal runaway problems are one-dimensional, having been derived for regions of high symmetry (viz., regions bounded by parallel plane walls, concentric circular cylinders, concentric spheres, etc.), and they all correspond to infinite activation energy (see Frank-Kamenetskii1). Yet, most cases of interest are not one-dimensional, and twodimensional problems can lead to unexpected results. We can anticipate some of these two-dimensional results by determining the effect on the critical condition as the boundary of a symmetric region, e.g., a circle, is given a small displacement, whereupon a one-dimensional problem becomes a two-dimensional problem. The available solutions for exponential heating (i.e., Arrhenius heating at infinite activation energy) all appear as indicated in Figure (1), where the greatest temperature in the region is plotted against the Frank-Kamenetskii number (λ). There are two solution branches, upper and lower, where the lower branch ending in a turning point is called the nose of the curve, as λ increases to its critical value, denoted λcrit. A review of what is known about one-dimensional problems can be found in Gray and Scott.2 Some results on displacing symmetric domains are given by Adler3 and by van de Velde and Ward.4 Like van de Velde and Ward, our aim is to determine what happens to the critical value of λ as the boundary of a simple twodimensional domain is given a small perturbation. Our first job is to establish a criterion by which the critical value of λ can be identified, no matter the domain of interest. We then deal with a domain where the displacement may be either stabilizing or destabilizing, and we show that linear heating is a useful guide. 2. IDENTIFYING THE CRITICAL VALUE OF λ ON AN ARBITRARY DOMAIN Our plan, to identify the turning point depicted in Figure 1, is to advance λ along the lower branch and at each value of λ where there is a solution, and there is a solution at λ = 0, ask: What prevents us from advancing u as we advance λ? The answer is that, at some point, du/dλ fails to remain finite and, at that point, we have λ = μ1(λ), where μ1 is the fundamental eigenvalue of the homogeneous part of the du/dλ problem. r 2011 American Chemical Society
The scaled temperature u g 0 on a domain D is assumed to satisfy ∇2 u þ λFðuÞ ¼ 0
on D
ð1Þ
where, on a part, S 1 , of the boundary of D u¼0
ðisothermalÞ
whereas, on the remainder, S 2 , if any n 3 ∇u ¼ 0 B
ðinsulatedÞ
and where the value of the function F is monotonically increasing as u increases, and the scaling is such that Fð0Þ ¼ 1 ¼ F 0 ð0Þ The part S 2 of the boundary cannot be the entire boundary, because then there would be no steady-state solution for any positive value of λ. For any domain D , given a solution u*, λ* on the lower branch in Figure 1, we wish to determine whether or not λ* is the turning point, i.e., upon advancing λ by a small amount beyond λ*, we wish to know whether or not a solution u exists near u*. To this end, we introduce u_ , which is defined as u_ t du/ dλ, whereupon if u_ exists at λ*, the solution branch extends beyond λ*, whereas if u_ is not finite at λ*, λ* must be the critical value of λ. Differentiating eq 1 with respect to λ, we find that u_ satisfies the relation 0 ∇2 u_ þ λ F ðuÞ_u ¼ FðuÞ
u_ ¼ 0 on S 1 nB 3 ∇_u ¼ 0 on S 2
on D ð2Þ
Special Issue: Ananth Issue Received: March 8, 2011 Accepted: July 13, 2011 Revised: July 11, 2011 Published: July 13, 2011 13244
dx.doi.org/10.1021/ie200458r | Ind. Eng. Chem. Res. 2011, 50, 13244–13249
Industrial & Engineering Chemistry Research
ARTICLE
ϕ1(λ), ϕ2(λ), ... for any value of λ short of λcrit, where λcrit = μ1(λcrit), we find that u_ (λ) becomes infinite as λ approaches λcrit and, hence, the nose is indeed a turning point. At λ = 0, μ1(0), the lowest heat conduction eigenvalue, is positive and hence μ1(λ) is greater than λ at λ = 0. In addition, upon advancing λ, μ1(λ) does not increase as λ increases and, hence, at some point (i.e., the critical point), we have μ1(λ) = λ. To see that μ·1(λ), for any λ on the lower branch, is not positive, we differentiate eq 3 with respect to λ to obtain 0 0 00 ∇2 ϕ_ 1 þ μ1 F ðuÞϕ_ 1 ¼ μ_ 1 F ðuÞϕ1 μ1 F ðuÞ_uϕ1 ϕ_ 1 ¼ 0 on S 1 n 3 ∇ϕ_ 1 ¼ 0 on S 2 B
Figure 1. The base solution for exponential heating, F(u) = eu, calculated for a circular cross section. Also shown is the curve F(u) = 1 + u.
This is a linear problem in u_ , and, to solve it, we introduce the eigenvalue problem ∇2 ϕ þ μF 0 ðuÞϕ ¼ 0
on D
ϕ ¼ 0 on S 1 nB 3 ∇ϕ ¼ 0 on S 2
ð3Þ
and denote its solutions ϕ1 , ϕ2 , ::: corresponding to the eigenvalues 0 < μ1 e μ2 ::: where the ϕ's and μ's depend on λ* and where ϕ1 is singly signed (see Courant and Hilbert5). If λ* = 0, we have u* = 0 and F0 (u*) = 1; hence, the μ's, at λ* = 0, are the pure heat conduction eigenvalues on the domain D , corresponding to the assigned isothermal and adiabatic conditions along its boundary. They are positive and, with the idea of advancing λ* along the lower branch, we start with λ* < μ1(λ*) at λ* = 0. Then, as long as λ* remains less than μ1(λ*), the homogeneous part of eq 2, viz., 0 ∇2 u_ þ λ F ðuÞ_u ¼ 0
on D
u_ ¼ 0 on S 1 nB 3 ∇_u ¼ 0 on S 2
ð4Þ
has only the solution zero, whence eq 2 can be solved with the result that the solution to eq 1 can be advanced along the lower branch at λ*. However, if λ* = μ1(λ*), then eq 4 has the solution ϕ1(λ*) and eq 2 cannot be solved for u_ at λ*, unless a solvability condition is satisfied. However, this condition, viz., Z Z ϕ1 ðλÞFðuÞ dx dy ¼ 0 D
cannot be satisfied, because ϕ1 and F(u*) are singly signed. Hence, u_ does not exist at λ*, the solution to eq 1 cannot be advanced through λ*, and λ* must be the nose of the curve (viz., λcrit). In addition, if u_ (λ) is expanded in the eigenfunctions
ð5Þ
and we observe that, for 0 e λ e λcrit, eq 5, with its right-hand side set to zero, has the solution ϕ1, which is not zero. Then, because eq 5 has a solution, viz., the derivative of ϕ1, the solvability condition must be satisfied and this determines μ·1 as Z Z F 00 ðuÞ_uϕ21 dx dy μ1 μ_ 1 ¼ Z Z F 0 ðuÞϕ21 dx dy whence we have μ·1 e 0, because F0 , F00 , and u_ are all non-negative on the lower branch. In the case of linear heating, viz., FðuÞ ¼ 1 þ u we have μ·1 = 0; hence, μ1(λ) = μ1(0) and, therefore, μ1(0), which is the lowest heat conduction eigenvalue on the domain, is the critical value of λ. Thus, for any region, the critical value of λ for linear heating is the lowest heat conduction eigenvalue and this exceeds the critical value of λ for any F where F00 > 0 (e.g., for exponential heating; see Figure 1). Due to this connection to the linear heating problem, and because they are readily available for regions of many shapes, the heat conduction eigenvalues will be of interest to us. The linear heating problem differs from the nonlinear problem in many important ways. It does not have a turning point; where it has finite solutions, they are unique. However, for λ approaching the eigenvalues μ1(0), μ2(0), ..., the solution becomes unbounded. This is illustrated in Figure 1 as λ approaches μ1(0). Evidently, with a little encouragement, it would turn back.
3. THE EFFECT OF DOMAIN PERTURBATIONS ON THE CRITICAL FRANK-KAMENETSKII NUMBER If we could solve eq 1, we could find the critical value of λ by inspection, and if we then solved eq 3, we would find that λcrit = μ1(λcrit). Our aim is to assume that we have a solution to eq 1 on a simple domain and then to follow the critical value of λ as the domain is perturbed. In doing this, we write λ in place of μ1, viz, λ = μ1(λ) in eq 3 and denote ϕ1 by ψ. To illustrate our method and to get results of some physical interest, we denote by D 0 , a two-dimensional domain where eq 1 has a one-dimensional solution, denoted, at critical conditions, by u0 and λ0. Specifically, D 0 is the region bounded by two concentric circles of radii R0 and kR0, where k < 1, and our aim is to 13245
dx.doi.org/10.1021/ie200458r |Ind. Eng. Chem. Res. 2011, 50, 13244–13249
Industrial & Engineering Chemistry Research
ARTICLE
determine how the critical value of the Frank-Kamenetskii number changes as the shape of the domain is given a small displacement. Hence, at the critical condition, u0 and λ0 satisfy ∇2 u0 þ λ0 Fðu0 Þ ¼ 0 u0 ¼ 0 at r ¼ R0 u0 ¼ 0 or B n 3 ∇u0 ¼ 0
at r ¼ kR0
ð6Þ
and eq 3 on D 0 becomes
n 3 ∇u2 ¼ 0 u2 ¼ 0 or B
ψ0 ¼ 0
and where we will see that ψ2 is not needed. Because ψ0, not zero, satisfies the homogeneous part of eq 8, in order for u1 to exist we must have Z Z ψ0 Fðu0 Þ dx dy ¼ 0 λ1 whence λ1 = 0, because ψ0 and F(u0) are singly signed. Hence, we can write, for u1,
at r ¼ R0
ψ0 ¼ 0 or B n 3 ∇ψ0 ¼ 0
at r ¼ kR0 ð7Þ
We hold the inner circle fixed at r = kR0 and give the outer circle a small displacement, viz., r ¼ RðθÞ ¼ R0 þ εR1 ðθÞ þ
1 2 ε R2 ðθÞ þ ::: 2
Thus, expanding λ, u, and ψ, the solutions to eqs 1 and 3 at the critical condition, via λ ¼ λ0 þ ελ1 þ
1 2 ε λ2 þ ::: 2
u ¼ u0 þ εu1 þ
1 2 ε u2 þ ::: 2
and 1 2 ε ψ2 þ ::: 2
where u0 and ψ0 depend only on r, we have, at first order in ε, ∇2 u1 þ λ0 F 0 ðu0 Þu1 ¼ λ1 Fðu0 Þ
ð8Þ
and ∇2 ψ1 þ λ0 F 0 ðu0 Þψ1 ¼ λ1 F0 ðu0 Þψ0 λ0 F00 ðu0 Þu1 ψ0
u1 ¼ Cψ0 þ
D0
where F00 (u0) and ψ0 depend only on r. Substituting for u1, and integrating over θ, this solvability condition reduces to Z F 00 ðu0 ÞCψ0 3 r dr ¼ 0 Hence, C must be zero, because F00 (u0) and ψ0 are singly signed. Because C is zero, we can find λ2, knowing only u0, ψ0, and u1. Thus, we need ψ0, because it appears in all solvability conditions, and we need the ψ1 equation, because it determines the constant C in u1. However, after C is found to be zero, we do not need to solve for ψ1, nor do we need the ψ2 equation. To obtain a formula for λ2, which is the first nonzero correction to λ0 as the domain is displaced, we multiply eq 10 by ψ0, eq 7 by u2, and subtract and integrate over the domain, whereupon we have Z
ð9Þ
Z 0
at r ¼ R0
n 3 ∇u1 ¼ 0 B n 3 ∇ψ1 ¼ 0 B
0 2π
2π Z R0 kR0
F 00 ðu0 Þu21 ψ0 r dr dθ
u2 B n 3 ∇ψ0 R0 dθ
ð11Þ
where the second integral on the right-hand side is to be carried out at r = R0, whence it is # Z 2π " 2 dψ0 ∂u1 du0 2 d u0 ðR0 Þ ðR0 Þ R1 ðR0 Þ dθ R0 2R1 ðR0 Þ R2 dr ∂r dr2 dr 0
at r ¼ kR0
At second order in ε, we have ∇2 u2 þ λ0 F0 ðu0 Þu2 ¼ 2λ1 F 0 ðu0 Þu1 λ0 F 00 ðu0 Þu21 λ2 Fðu0 Þ
ð10Þ on D 0 , where u2 ¼ 2R1
kR0
Z
Fðu0 Þψ0 r dr dθ ¼ λ0 þ
and u1 ¼ 0 or ψ1 ¼ 0 or
2π Z R0
λ2
on D 0 , where du0 ) dr dψ0 ψ 1 ¼ R1 dr
∞
∑ Am ðrÞ cos mθ m¼1
where the Am's are not zero if and only if the expansion of R1 includes a term in cos mθ. Then, with λ1 = 0, to determine ψ1, we must satisfy the solvability condition Z Z F 00 ðu0 Þu1 ψ0 2 r dr dθ ¼ 0
0
u1 ¼ R1
at r ¼ kR0
D0
∇2 ψ0 þ λ0 F 0 ðu0 Þψ0 ¼ 0
ψ ¼ ψ0 þ εψ1 þ
and
∂u1 d2 u0 du0 R12 2 R2 ∂r dr dr
at r ¼ R0
Two points can be made. First, in Appendix B, Adler’s problem is solved using the method outlined above and Adler’s result is obtained. Second, the method presented is not limited to the displacement of plane annular regions. Given any region D 0 on which eq 1 has a one-dimensional solution, the critical value of λ can be obtained as this region is displaced to a nearby region, where the solution would be two-dimensional if it could be obtained. Equations 6 and 7 must be solved on D 0 , both one-dimensional problems, satisfying whatever boundary conditions are specified at the surface of D 0 . Most likely, this must be done in some approximation. An arbitrary first-order displacement 13246
dx.doi.org/10.1021/ie200458r |Ind. Eng. Chem. Res. 2011, 50, 13244–13249
Industrial & Engineering Chemistry Research
ARTICLE
would then specify R1, for example a function of θ say, in a series of surface harmonics, and set R2 to zero. The problem for u1 and ψ1 would be one-dimensional were it not for the appearance of R1 in the boundary conditions at the surface of D 0 . Again, the solvability conditions for u1 and ψ1 will require λ1 and C to be zero, whence u1 can be expanded in a series of surface harmonics whose coefficients must be determined, a series of one-dimensional problems where solutions must be obtained in some approximation. Then, λ2 is obtained via Z Z Z
λ2
Z Z Z
D0
Fðu0 Þψ0 dV ¼ λ0 Z Z þ
∂D 0
D0
F 00 ðu0 Þu21 ψ0 dV
dA½u2 B n 3 ∇ψ0 ψ0
n 3 ∇u2 B
where the boundary conditions satisfied by u2 and ψ0 reduce the second integrand on the right-hand side to terms that are dependent on u0, u1, and R1, e.g., for Dirichlet conditions, the integrand would be ! 2 ∂u1 d u dψ0 0 R1 2 2 2R1 ∂r dr dr The Case of Linear Heating. In the case of linear heating, the critical value of λ is the fundamental heat conduction eigenvalue. Hence, on any domain D , we must solve
∇2 ψ þ λψ ¼ 0
on D
2π Z R0
1 2 ε λ2 2
0
kR0
ψ0 2 r dr dθ ¼ R0
and R2 ¼
3 1 ðcos 4θÞ cos 2θ 2 2
R1 ¼ cos θ
and find that λ1 is zero and that λ2 is given by Z
R1 ¼ cos 2θ
on S 2
Solutions to this problem are known for domains of many shapes. However, our interest is simply in how the fundamental eigenvalue, λ, on a base domain D 0 changes as the boundary of D 0 is perturbed. For a base domain perturbed by displacing the outer circle, we write
λ2
In the first case, with the major axis of the ellipse taken to be R0(1 + ε), the minor axis is then R0(1 ε + ε2 + 3 3 3 ), and we have
In the second case, we have
ψ ¼ 0 on S 1 ψ ¼ 0 or B n 3 ∇ψ ¼ 0
λ ¼ λ0 þ ελ1 þ
Figure 2. Correction to the critical value of λ for linear heating.
dψ0 ðR0 Þ dr
R1 2
Z
0
2π
∂ψ1 ðR0 Þ 2R1 dr
d ψ0 dψ0 ðR0 Þ dθ ðR0 Þ R2 dr 2 dr 2
ð12Þ where ψ0 is the fundamental eigenfunction on the base domain and ψ1 and ψ2 are its corrections upon displacement. This equation can be obtained by setting F0 = 1, F00 = 0 in eq 11 or by perturbing eq 7 with F0 = 1.
4. LINEAR HEATING IN AN ANNULAR DOMAIN The base domain lies between two circles: one at r = R0, and the other at r = kR0 (where k < 1). The outer circle is isothermal at zero temperature, and the inner circle is either isothermal at zero temperature or adiabatic. The outer circle is displaced in two ways, both of which leave the area of the domain unchanged. First, we turn the outer circle into an ellipse, then we move the outer circle off-center.
and R2 ¼
1 1 ðcos 2θÞ 2 2
In Figure 2, we present the linear heating results for four cases: two displacements of the outer circle and both Neumann and Dirichlet conditions at the inner circle. In the case where the outer circle is displaced into an ellipse and the inner circle is adiabatic, we see that displacement stabilizes small k annuli and destabilizes large k annuli. The dividing value of k is ∼45/ 100. We will see that this is also true for exponential heating, i.e., sometimes a displacement makes a region less dangerous, and sometimes it makes it more dangerous. We can advance an explanation for this. The displacement of the outer circle, which is the heat sink, into an ellipse of the same area affects the heat conduction eigenvalues in two ways. First, there is an increase in the length of the curve along which the domain contacts the sink and, second, there is a redistribution of the area of the domain, some of it now lying further from the sink, some closer to the sink (cf. Figure 3). The first of these effects strengthens the heat loss and is independent of k; the second effect weakens the heat loss and is strengthened by increasing k until, as k approaches 1, it dominates the conduction of heat out of the domain. The remaining curves in Figure 2 correspond, first, to replacing the adiabatic inner circle with an isothermal sink. This strengthens the effect of the area redistribution moving the dividing value of k to the left. Two curves then are shown which correspond to displacing the outer circle off-center. Only area redistribution plays a role in this case; hence, such a 13247
dx.doi.org/10.1021/ie200458r |Ind. Eng. Chem. Res. 2011, 50, 13244–13249
Industrial & Engineering Chemistry Research
ARTICLE
Figure 3. Sketch showing the displacement creating regions closer to and further from the sink. Figure 4. Correction to the critical value of λ for exponential heating.
displacement is always dangerous, the more so the larger the value of k. Exponential Heating in an Annular Domain. In the case of exponential heating, we have
where ^u1 ¼
at r ¼ R0
and
FðuÞ ¼ eu
^u1 ¼ 0 or
and to determine the effect on the critical Frank-Kamenetskii number of displacing the outer circle, we must solve d2 u0 1 du0 þ λ0 eu0 ¼ 0 þ 2 r dr dr d2 ψ0 1 dψ0 þ λ0 eu0 ψ0 ¼ 0 þ 2 r dr dr and ∂2 u1 1 ∂u1 1 ∂2 u1 þ þ þ λ0 eu0 u1 ¼ 0 r ∂r r 2 ∂θ2 ∂r 2 on the domain kR0 e r e R0, where u0 ¼ 0 ¼ ψ0 du0 u1 ¼ R1 dr
du0 dr
) at r ¼ R0
and where either u0, ψ0, and u1 are all zero at r = kR0 or du0/dr, dψ0/dr, and ∂u1/∂r all vanish at r = kR0. We can obtain u0 and λ0 by using formulas given by Kearsley,6 and then we can obtain approximations to ψ0 using the method of weighted residuals. To determine u1, we write either u1 ¼ ^u1 ðrÞ cos 2θ or u1 ¼ ^u1 ðrÞ cos θ depending on whether R1 is cos 2θ or cos θ, and then solve d2 ^u1 1 d^u1 4ðor 1Þ þ ^u1 þ λ0 eu0 ^u1 ¼ 0 r dr r2 dr 2
d^u1 ¼0 dr
at r ¼ kR0
and where approximations to ^u1 can be obtained using the method of weighted residuals. Equation 11 can then be used to determine λ2. Results are presented in Figure 4. In the case where the outer circle is displaced into an ellipse, we have two possibilities, depending on the value of k. At low values of k, the base domain is more dangerous than the displaced domain (i.e., λ2 is positive) and λ shifts to the right upon displacement (i.e., displacement is stabilizing). At high values of k, displacement is destabilizing. As previously discussed, the competition is between the perimeter increase and the area redistribution, with the area redistribution being strengthened by increasing k and by replacing Neumann conditions by Dirichlet conditions. In the case where the outer circle is moved off-center, the displacement is always dangerous. We see that the effect of a displacement on the lowest heat conduction eigenvalue gives a good qualitative account of the effect on the critical Frank-Kamenetskii number. This is also true in the case of Arrhenius heating, which is presented in Appendix A.
5. CONCLUSION Solutions are known to one-dimensional thermal explosion problems in domains of high symmetry. There are no known solutions to two-dimensional problems in regions of less symmetry. Our aim is to present some information about two-dimensional regions by determining the effect of a small domain perturbation on the critical value of the FrankKamenetskii number. We find, first, that perturbations make some domains more dangerous and other domains less dangerous, and second, the lowest heat conduction eigenvalue, which is a solution to the Helmholtz equation on the domain, gives a reliable guide to the direction in which the 13248
dx.doi.org/10.1021/ie200458r |Ind. Eng. Chem. Res. 2011, 50, 13244–13249
Industrial & Engineering Chemistry Research
ARTICLE
critical Frank-Kamenetskii number moves upon perturbation of the domain.
’ APPENDIX A: THE CASE OF ARRHENIUS HEATING In the case of Arrhenius heating, we have FðuÞ ¼ eβu=ðβ þ uÞ where β is the scaled activation energy and where F(u) can be approximated by eu if β is much greater than the largest value of u on the domain. In this case, F(u) and F0 (u) are always positive, whereas F00 (u) is positive for small u, negative for large u, being zero at u = (1/2)β(β 2). Then, we have Fð0Þ ¼ 1 ¼ F 0 ð0Þ and
Figure A1. Correction to the critical value of λ for Arrhenius heating.
2 F ð0Þ ¼ þ 1 β 00
and it remains to obtain ψ0, where
For small values of β, F00 (u) is negative for almost all values of u (i.e., almost all u's are large), and there is no critical value of λ. For higher values of β, F00 (u) is positive for almost all u and, thus, a low critical value of λ is found (ignition point) but then, at higher values of u, F00 (u) becomes negative and the base curve turns around for the same reason why the low β curves never turn back. Hence, we have a second critical value of λ (extinction point). Both points can be found by setting μ1(λ) = λ. For the outer circle displaced into an ellipse, and the inner circle adiabatic, we find two critical points for all values of k for β = 5 and 7. The results, plotted as λ2/λ0 vs k are given in Figure A1. The ignition curves in this figure do not differ by much from the corresponding curve in Figure 4, i.e., when λ is scaled by λ0 to see the effect of the displacement, β = 5 and 7 do not differ by much from β = ∞.
’ APPENDIX B: ADLER’S RESULT We obtain Adler’s problem in our notation by setting k = 0, R0 = 1, R1 = cos θ, and R2 = 0.3 Then, with λ1 = 0, λ2 is given by Z 1 Z 1 u0 e ψ0 r dr ¼ λ0 eu0 ^u1 2 ψ0 r dr 2λ2 0 0 ( ) dψ0 d^u1 d2 u0 þ ð1Þ 2 ð1Þ 2 ð1Þ dr dr dr We obtain u0 and λ0, at critical, from Kearsley’s work, as 6
λ0 ¼ 2 u0 ¼ 2 ln
2 1 þ r2
Whence, we have eu0 ¼
4 ð1 þ r 2 Þ2
and ^u1 ¼
du0 4r ¼ 1 þ r2 dr
d 2 ψ0 1 dψ0 4 þ λ0 þ ψ0 ¼ 0, r dr dr 2 ð1 þ r 2 Þ2
ψ0 ¼ 0 at r ¼ 1
The fundamental solution to this eigenvalue problem is λ0 ¼ 2 and ψ0 ¼
1 r2 1 þ r2
whereupon we find λ2 ¼ 2 and, hence, we have λ ¼ λ0 þ
1 2 ε λ2 ¼ 2 ε 2 2
which is Adler’s result.3 The perturbation is destabilizing due to an area increase with no increase in perimeter.
’ AUTHOR INFORMATION Corresponding Author
*E-mail: ranga@ufl.edu.
’ REFERENCES (1) Frank-Kamenestkii, D. Diffusion and Heat Transfer in Chemical Kinetics; Plenum Press: New York, 1969. (2) Gray, P. M.; Scott, S. K. Chemical Oscillations and Instabilities: NonLinear Chemical Kinetics; Oxford University Press: Oxford, U.K., 1994. (3) Adler, J. Criticality in a Nearly Circular Cylinder. Proc. R. Soc. London A 1987, 411, 413. (4) van de Velde, E. F.; Ward, M. J. Criticality in Reactors Under Domain or External Temperature Perturbations. Proc. R. Soc. London A 1991, 434, 341. (5) Courant, R.; Hilbert, D. Methods of Mathematical Physics, Vol. 1; WileyInterscience: New York, 1953. n (6) Kearsley, E. A. Solution of the Equation ψxx + (1/x)ψx + Kx eψ = 0. J. Res. Natl. Bur. Stand., Part B 1963, 67B (4), 245. 13249
dx.doi.org/10.1021/ie200458r |Ind. Eng. Chem. Res. 2011, 50, 13244–13249
