Numerical calculations in a problem with heat conduction andheat productionCitation for published version (APA):Geldrop - van Eijk, van, H. P. J., van Ginneken, C. J. J. M., & Gelder, van, D. W. (1973). Numerical calculationsin a problem with heat conduction and heat production. (EUT report. WSK, Dept. of Mathematics and ComputingScience; Vol. 73-WSK-05). Technische Hogeschool Eindhoven.

Document status and date:Published: 01/01/1973

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Abstract

. The temperature u of an exothermic reacting chemical in a vessel, governed

by the equation

uu = /iu + Ae ,t

is studied. With the initial condition u(~,O) = uo and the boundary condi­

tion u(~,t) = 0 at the wall of the vessel, a critical initial value u~ is

calculated such that if Uo < u~, the temperature remains bounded and other­

wise, if Uo > u~, an explosion occurs. The cases that the vessel is a slab,

an infinite cylinder, or a sphere are considered. The possible steady

states, together with the questions of their stability are studied; this

study should be considered as a personal presentation of some results, which,

in principle, can be extracted from the literature.

The problem is solved numerically, using standard techniques for solving

partial differential equations (Crank Nicolson).

A number of numerical results are presented in order to justify the supposi­

tions that are fundamental for the algorithm.

AMS Subject Classification (1970): 6SPOS

- I -

O. Statement of the problem

A vessel is filled with a chemical having the homogeneous temperature uO'

An exothermic reaction takes place. The wall of the vessel is kept at con­

stant temperature. Therefore the temperature of the reacting mass changes

by heat conduction to the wall and by heat production due to the reaction.

The problem is to determine a critical value u~ such that, if uo < u~' the

temperature of the chemical remains bounded and that, if Uo > uO' the tem­

perature tends to infinity (explosion).

The cases that the vessel is a slab, an infinite cylinder, or a sphere are

considered.

I. Mathematical model

The temperature of the reacting mass is governed by the equation ([1, ch. 2J)

ut = div(k grad u) + W(u)/(pCv) ,

where

t time

k coefficient of thermal conductivity

p density

Cv specific heat at constant volume

W heat production per unit time and per unit volume.

The heat production term W has the following form (Arrhenius' law)

where

( I • I)

(I.2)

E internal energy

R gas constant

AO constant depending on the reacting mass.

The linearization of (1.2) around uo' supposing that k is a constant and

choosing suitable dimensionless quantities, transforms (1.1) into

ut = t.u + Ae u • ( 1• 3)

- 2 -

Hence, the problem can be formulated as follows. Let V be the reactant and

let S denote the wall. Let u be the solution of

u> a

)ut = l\u + Ae , x EO V, t-u(=.,O) = uo , X EO V ( I. /f)-u(=., t) = a , x EO S, t > a •-

. * . * . . .Determ1ne Uo such that, 1f Uo < uo' u(x,t) rema1ns bounded; otherw1se, 1f

uo > u;, u(x,t) tends to infinity.

If V is a slab, an infinite cylinder, or a sphere, (1.4) transforms (after

appropriate scaling and because of symmetry) into

ut = u + ~ u + Aeu a < x < I ,t > a )xx x x

u(x,O) = Uo 0 < x < 1

u(I,t) = 0, u (O,t) =0, t > a ,x

(I.5)

where n = a if v is a slab, n = 1 if V is a cylinder, n = 2 if V 1S a sphere.

2. Some properties of the model

In this section some - for our purpose essential - properties of the model

will be discussed.

In principle, one can extract these properties from the li terature([ I,

ch. 3 + references], [2J).

The following should be considered as a result of the study of the litera­

ture, with a personal presentation.

2. I. First, we study the steady state solutions of equation (1.5).

These satisfy

d2v n dv v--+--+ Ae = 0,dx2 x dx

dVI = adx x=O

v(l) = a .

a < x < 1)

(2. 1)

- 'J -

We write the solutions of (2.1). if any, as

vex) = v(O) + fey) •

where

y = lAev(O) /2 x •

It follows that f satisfies

d2

f + ~ .!!i + e f = 0dy2 y dy

dfl = 0dy y=O

f(O) = 0 .

(2.2)

(2.3)

The relation between v(O) and A fol10ws from the boundary condition for

x =

(2.4)

Lemma. (2.4). considered as equation in A with v(O) fixed, has exactly one

solution.

Proof. Multiply (2.3) by yn to find

or

n d2

f n-] df + n fy dy2 + ny dy y e = o ,

d (n df) =dy Y dy

n f-y e

Since :;!y=O = O. we obtain

dfdy = n

y

dfWe conclude that dy < 0, so f is decreasing and, clearly. lim fey) = -~.

y-+co

Thus. with v(O) fixed, there is exactly one vaLue of A satisfying (2.4).0

- 4 -

Conversely. considering (2.4) as equation 1n v(O) with A fixed. it turns

out that (2.4) has one or more solutions if A ~ A and no solution ifcrA > A • We now develop this assertion for n = O. 1 or 2.cr

(i) n = o.The solution f of (2.3) is given by

f(y) = -2 log(cosh(y/I:2» •

Hence. (2.4) reduces to

v'1' cosh (S) = S ,

(2.5)

"' fA v(O) /2where S =~- e •2 . 2

Consequently. A = 2/ (sinh (S ». where S satisfies S tanh(S) = 1.cr cr crNumerical values:

A ~ 0.88. v(O) ~ 1.19 •cr cr

The dependence of A on v(O) is illustrated in the following diagram.

A

0.5

0=-----.--'-4.---2....---------r

3---.....4---....S------r

6--- v(0)

V(O)cr: 1.19

(ii) n = J.

The solution f of (2.3) is now given by

2f(y) = -2 log(l + y /8) •

Using (2.6), (2.4) can be written as

8S = A(1 + S) 2 ,

where S - A v(O)- 8' e •

(2.6)

- 5 -

Consequently, A = 2, v(O) = log 4 ~ 1.39 •ct' cr

The dependence of A on v(O) is illustrated in the following diagram.

A

1.5

o,s

'-----r---+----r---~---....,.....---...,......-'----........,-_ V (0)o 1 2

V(O)cr: 1.393 4 5 6

iii) n = 2.

In this case the solution of (2.3) is not analytically known. From

numerical calculations it follows that A ~ 3.32, v(O) ~ 1.61 and, cr crthe following diagram.

A

2

o 2 4v (Q)cr: 1.61

6 8 10 12 14

- 6 -

2.2. We are now going to investigate how a small perturbation of a steady state

solution varies with time.

Let

u(x,t) = vex) + ~(x,t)

be a solution of

u = u + ~ u + Aeu

}t xx x x

u (O,t) = 0, u(I,t) = ° ,x

(2. 7)

(2.8)

where vex) is a steady state solution and ~ a small perturbation of it.

Substituting (2.7) in (2.8) and linearizing we obtain

CPt = CPxx + i ~x + Aev(x)~ = OJ

cP (O,t) = 0, cp(I,t) = 0 .x

(2.9)

We look for non-trivial solutions of (2.9) by applying the technique of

.separation of variab les.

Supposing that

cp(x,t) = F(x).G(t)

we find

dG -Atdt = -AG, so G(t) = e

2~ + ~ ~ + (Aev(x) + A)F = 0dx2 x dx

(2.10)

(2. I I)

dFI - 0dx x=O - , F(l) = 0 ,

where A is a constant such that (2. II) has solutions that are not identical­

ly zero. A is called an eigenvalue and F(x) the eigenfunction belonging to

it.

From the theory of Sturm-Liouville we have

a) The eigenvalues are real and can be numbered such that

with A -+ 00.n

- 7 -

b) The eigenfunction "'i (x) belonging to Ai has exactly i zeros in (0,1).

c) The [unctions 10' (x) form a complete set, so rp(x,t) can be written asn

00

<p(x,t) = Ln=O

-A tna F (x)e

n n

Consequently, if 1..0

> 0, the perturbation cp decreases (v(x) is stable);

otherwise, if 1..0 < 0, cp increases (v(x) is unstable).

Therefore, to investigate whether vex) is stable or unstable we try to

determine the sign of 1..0

,

For this purpose we look at the following problem that arises if we drop

the condition F(I) = 0 in (2.11) and put A = O. Hence,

d2F n dF ( )__ + __ + Aev x F = 0di x dx

(2. 12)

*Ix=o = 0 •

The solutions of (2.12) can be written as F(x) = CH(x), with H(O) = I and C

a constant.

Let E: be the first zero of H(x). Then it follows from the properties a) and

b) that zero is the smallest eigenvalue of

(Aev(x) + A)F = 0

:~Ix=o = 0, F(E:) = 0 •

Therefore,

if ~ > I, then(2.11)has only positive eigenvalues;

- if ~ < I, then(2.II)has at least one negative eigenvalue.

This follows from the fact that, according to the theory of Sturm-Liouville,

the smallest eigenvalue of

d2F n dF v( )- + - - + (Ae x + A)F = 0dx2 x dx

dFI = 0 F(a) = 0 ,dx x=O '

15 a decreasing function of a.

- H .-

o Summarizing, the problem of investigating whether v(x) is stable or un­

stable reduces to the location of the first zero ~ from the solutions of

*Ix=o =0 ,

if t; > 1 , then v(x) is stable,

if t; < I , then v(x) is unstable.

The solutions of (2.13) are

F(x) = C(x :: + 2) ,

where C is a constant.

From (2.2) it follows that

dfF(x) = C(y dy + 2) ,

where y = ~ev(0)/2x.

For the first zero ~ of F(x) we have

~ = lAev (0) /2 '

where YO is the smallest root of

df + 2 0Y dy =.

(2. 13)

o

(2.14)

(2.15)

(2.16)

The function f is decreasing so it follows from (2.4) h IiJ v (0)/2 is ant at e

increasing and t; in (2.15) is a decreasing function of v(O).* of v(O) = 1, then it follows thatLet v (0) be the value such that t;

if v(O) < v*(O) then v(x) 1.5 stable,

-if v(O) > v* (0) then v(x) is unstable.

* is determined fromv (0)

- 9 -

or

*

·1v (0) = -f(yO)

and (2.17)** 2/ v (0)A YO e

Finally we show that A, considered as a function of v(O) given by (2.4),

*has its first extreme value in v (0).

From (2.4) it follows that the extreme values of A considered as a function

of v(O) satisfy

+ 2 = 0 •lAev(0)/2

f(lAev (0)/2) = -v{O)

lAev(0)/2 ~Idy .

y

(2.4)

(2. 18)

So we find lAev (0)/2 =tain

y., where y.~ ~

is a root of (2.16). Using (2.4) we ob-

v(O) = -f(y.) •~

The smallest value, v(O) ,introduced ~n section 2.1, where A is extreme,crequals

v(O) = -f(yO)cr*v (0) •

2.3. We conclude by making some suppositions that are useful for the algorithm

which we shall describe in the next section.

Only those cases are examined where an unstable steady state solution exists.

Let u(x,t,uO) denote the solution of (I.S). We suppose that, if at some time

to u(x,to'uO

) >vt(x) (0 < x < I), then U o > u~; otherwise if at some time

to u(x,to'uO) <vt (x) (0 < x < I), then Uo < u~.

Indeed, in all cases that were examined there was a time to where either

u(x,to'uO) >vt(x) or u(x,to'uO) <vt(x).

With this supposition u; can be determined by successive halving the inter­

val (O,vt(O».

- 10 -

3. Numerical method

To solve the problem numerically, we start by replacing (1.5) by a finite

difference approximation (Crank Nicolson).

We introduce a rectilinear grid with sides parallel to the x- and t-axes,

hand k being the grid spacings in the

x- and t-directions respectively.

The grid points are:

t

1 lk

_1---+_-1-_1x·

1.1. * h, 1. O. I J •• • ,M, Mh =

t. =j *k, J =0,1 •••••J _h-l -x·

We write down the differential equation of (1.5) for the points

x. = i *h. i = O,I •••••M-l1.

tj+~ = (j + Dk, j = 0,1, •••

We have, denoting f (x. ,t.) by f. .,1. J 1.,]

u .. I - u. . 2( ) 1. ,] + 1. , J + 0 (k ) ,u t . . I = k1.,J+2

k t 0 (3. I)

u. . - 2u. . + u. . u. 1 . I - 2u. .+ 1= ~[ 1.+1.J 1.,J 1.-I,J + 1.+ ,J+ 1.,J

h2

h2

+ u. .1.-1 ,]+1] +

~f i :f 0

(3.2)

u. . - u. l' u. 1 . 1 - u. 1 '+1n I[ 1.+1,J 1.- .J + 1+ .J+ 1.- ,J ] += Ih 2 2h 2h

(3.3)

If i 0, then we have, because u (O.t) = 0,x

(3.4)

- II -

Furthermore,

It follows from the boundary conditions that

= 0

(u) . = 0 = u I , j - u-I , j + 0 (h2) h '" 0 • Jx O,J 2h '

(3.5)

(3.6)

If the order terms are neglected the finite difference approximation of

(1.5) arises by substituting (3. I) to (3.5) incl. in the differential

equation of (1.5) for the points

U I . and uM . are eliminated by means of (3.6) •. - ,J ,]We formulate the above with matrices as follows.

We introduce the M-dimensional vectors

j = 0,1, ...

f (U.)- -J

kA

the tridiagonal M x M-matrices Band C

- 12 -

B :=

where

Uo . kaO := 1 + A(n + I) - ~Ake ,], A -

- h Z

U.•a. := I + A - ~Ake ~,J, i = I, ••• ,M-I~

nSo := -A(n + I), Si = -~A(I + Zi)' i = 1•••••M-2

Yi = -P.. (I - ~i)' i = 1, ••• ,M-1 •

The matrix C is found from B by replacing h by -h.

As an approximation of u.. in (1.5) we take U.. which satisfies~,J ~,J

BU. 1 = CD. + f (U . ), j = 0,1, •••-J+ -J - -J

(3.7)

UoGiven U. we can calculate U. 1 from (3.7) solving the tridiagonal-system.-J -J+If we look at the approximation in a fixed point (x, t), we expect, because

of (3.1) - (3.7), that

Z ZU(x,t) = u(x,t) + CI (x,t)h + CZ(x,t)k +

Z Z+ o(h ) + o(k ), h ~ 0, k ~ 0 • (3.8)

We make the following suppositions (cf. section 2.3), which in all cases

examined appeared to be satisfied.

- 13 -

(i) The difference equation ~n (3.7) has (with appropriate A and h) at

least two steady state solutions, a stable one US and an unstable oneR.

U •

(ii) Successive solving of (3.7) with uO' h, k and A given, we find a jo

h . < UR. R.suc that elther U. or U. > U •-J O -J O

*Let UO(h,k,A) be defined so that, if Uo(3.7) successively, a U. <_uR., and if

-Jtively. 0

uri (h,k,A) is taken as an approximation of u~.

find when solving

U.. > UR. respec­-J O

*then it follows that either Uo > UO(h,k,A) or

*Now UO(h,k,A) ~an be calculated by successive halving of the intervalR.

I := (O,(U )0)'

*Given h, k and A, we use the following algorithm to calculate UO(h,k,A).

I) Calculate UR..

2) Determine uo by halving I.

3) Solve (3.7) successively;

*Uo < UO(h,k,A).

4) Adapt I and go back to 2.

Sub 1. To calculate Ui we proceed as follows:

A steady state solution of (3.7) satisfies

(B - C)E. = ! (.£) •

To solve this non-linear system we use Newton's method

a,s.(E - - )Un+ 1

aUu=un -

where

a~ .n £. n= ~(U ) - (-- )U , n =£. - au n -u=u.... -

(3.9)

E = t (B - C) ,

- 14 -

If QO is appropriate, we have lim ~n = ~£. As QO

we take the unstablen-HlO

steady state solution of the differential equation in (1.5). If n = 0

or n = I, then this solution is analytically known. If n = 2, we deter­

mine it numerically.

We expect that, analogously to (3.8),

* * 2 2 2 2UO(h,k,A) = uo + C1h + C2k + o(h ) + o(k ), h '" 0, k '" O. (3.10)

Numerical experimentation has supported this supposition and thus we can

apply extrapolation, if desired so.

4. Numerical results

As an illustration of the numerical method we present some intermediate re­

sults for the case n = 0, A = 0.5.

Steady state solutions of the difference equation (3.7).

h = 1/8.

Stable solution US Unstable l· £x. so ut10n U1

0 0.329 301 216 8 2.891 333 774

o. 125 0.323 871 543 1 2.820 953 600

0.25 0.307 641 324 7 2.619 379 233

0.375 0.280 784 442 2.310 561 119

0.5 0.243 582 495 9 1.922 992 386

0.625 o. 196 413 271 8 1.481 975 217

0.75 0.139 736 002 8 1.006 570 302

0.875 0.074 074 591 7 0.509 788 821

0 0

- 15 -

I ISolving (3.7) successively, h = 8 • k = 64 •

We find with Uo = 2.272 672 213 04

2.889 856 0

2.819 574 4

2.618 258 3

2.309 770 7

!!30 = I. 922 516 7

I. 481 740 8

1.006 483 6

0.509 768 6

0

and with U o = 2.273 705 115 69 we find

!!21 =

2.892 422 8

2.822 482

2.622 006

2.314 421 5

1.927.668 0

1.486 669 I

1.010 365 8

0.5119070

*APproximations of uo•

k = h2

*h Uo(h ,k .A)

1/4 2.28612

1/6 2.27592

1/8 2.27283

1/ 10 2.27154

1/12 2.27089

1/16 2.27029

Extrapolated values

2.26776

2.26886

2.26925

2.26941

2.26952

- 16 -

k = h

*h Uo (h,k ,A)

1/6 2.32287

1/8 2.30475

1/12 2.28844

1/16 2.28145

Extrapolated values

2.28145

2.27539

2.27246

To choose a value of k, when h is given, we have to take into account a

(3. 10).

hand, it appears from the table that a

of uO• This means that Ic21 » Ic] I in

IC1 IAn optimum value of k is about rc;T h.

I c2 1In this case we have~ - 10.

number of contradictory demands. On the one hand we want a large k to reach

quickly the time to' whereupon either U. > Ui or u. < Ui. On the other

-J O -J Olarge k leads to a bad approximation

* 1As an illustration we tabulate the approximations uo for k = 3 h.

*h Uo(h,k ,A)

1/4 2.29147

1/6 2.28146

1/8 2.27702

1/12 2.27331

1/16 2.27184

Extrapolated values

2.27345

2.27131

2.27034

2.26995

For practical reasons we have dropped the idea of trying to determine an op­2timum value of k in all cases. We always took k = h •

*Finally, we present a table and a plot of uO• The tabulated values were ob-

tained by extrapolation and have a relative error of at most approx. 0.001.

Our algorithm was not applicable to small values of A for n = 2. These cases

were not further examined because there was no need for them.

- 17 -

*Table of uo·

* *n = 0 A Uo n = 1 A UoO. 1 4.30 0.3 3.726

0.2 3.48 0.5 3.143

0.3 2.976 0.7 2.742

0.4 2.591 0.9 2.429

0.5 2.270 1• 1 2. 162

0.6 1.979 1.3 1.923

0.7 1.700 1.5 1. 697

0.8 1. 395 1.7 1.467

0.85 1. 197 I .9 1. 194

n = 2 *A Uo1.8 2.039

1.9 1. 970

2.0 1.900

2.2 1.771

2.4 1.645

2.5 1.583

2.6 1.520

2.8 1. 392

3.0 1.254

3.2 1.082

3.25 1.025

- 18 -

*U

f4,8

4,0

3.2

2.4

1.6

.8n: 0

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

-A

*U

f4.0

3.2

2.4

1.6

.8

n : 1

0 .2 .4 .6 .8 to 1.2 1.4 1.6 1.8 2.0

-A

*Uo

t 2.0

1.8

1.6

1.4

1.2

1,0

1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2

----- A

- 19 -

References

[IJ Tipper, C.F.H.

Oxidation and Combustion Reviews. Vol. 2.

Elseviers Publ. Comp., Amsterdam, 1967.

[2J Istratov, A.G. and Librovich, V.B.

On the stability of the solutions in the steady theory of a

thermal explosion.

Prikl. Mat. Meh. 27 (1963), 343-347 (Russian).

Translated as J. Appl. Math. Mech. ~ (1963), 504-512.