Reduction and analysis of the solutions of the problem of the crystallization of filtered solutions

U.S.S.R. Comput.Maths.Math.Phys.,Vol.27,No.4,pp.71-77,1987

0041-5553/87 $10.00+0.OO 0 1 9 8 8 Pergamon Press plc

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REDUCTION AND ANALYSIS OF THE SOLUTIONS OF THE PROBLEM OF THE CRYSTALLIZATION OF FILTERED SOLUTIONS" G.G. ELENIN, V.V. KRYLOV and N.V. SOSNIN

A mathematical model of crystallization from a supersaturated solution which is filtered in a semi-infinite dense layer of particles of a solid phase is investigated. A reduction of the initial problem is carried out. Certain properties of the solutions are investigated and a.priori estimates are obtained. i. F o r m u l a t i o n of the problem. In this paper a mathematical model of the process of crystallization from a supersaturated solution is investigated. Let c(z~ t) be the concentration of the solution at the instant of time t at a point z, u(z, t) the velocity of motion of the crystallized substance due to the growth of crystals and ](z, r, t) the size distribution function of the crystals such that ~(z, r, t)dr is the number of crystals with sizes in the range (r, r-~dr) per unit volume. Let us make the following physical assumptions : i) the diffusion of the solution and fluctuations in the growth of the crystals can be neglected. 2) no new particles are formed in the filtration process. 3) the change in the volume fraction of the solid phase is negligibly small (a conclusion drawn from experiments on crystallization in a dense layer). The differential equations for the material balance and the kinetics of crystallization then have the following form: 0×

0~

.-~+ U,~z+EMz×=O ,

O:O,

3I a , i Ol O_T+~z (u/)±--~-a×-~r = 0 ,

O
.(1.1)

t>O, O
M3=I--e.

(t.21 (t.3)

Here x(z, t)=c(z, t)--c, is the excess concentration over the concentration of the saturated solution c,; us(t) is the specified velocity of motion of the solution, JW.(z, t) are the moments of the distribution function:

M. (z, t) = 1"r~/(z, r, t) dr;

'(1.4)

o

and ~ , E and c~ are specified positive parameters. Moreover, all quantities are in dimension. less units and referred to scales in which c,=I and E=8-'. Eqs. (i.1)-(1.3) are supplemented by initial and boundary conditions. The distribution function and the excess concentration are specified at t=0:

l(z, r, O)=lo(z,

r),

(1.5)

x (z, O) = r e ( z ) . When

r=0

and

r-~oo , the distribution function satisfies the conditions

i(z, O, t)=O, On the boundary

,(1.6)

z=0,

l=O(e-~'),

r-,.oo, '~>0.

(1.7)

one specifies the excess concentration

~(0, t) =xr(t),

(1.8)

the velocity of motion of the crystals

u(O, t)=ur(t) and the distribution function, which is defined, when

1(0, r, t) =/r (r, t)

(1.9) Ur(t)>0, by the condition (l.10)

and, when Ur(t)~<0, by Eq.(1.2). Eqs. (1.1)-(1.3), together with the initial conditions (1.5) and (1.6) and the boundary conditions (1.7)-(1.10) define the initial boundary value problem. The following system of equations for the moments follows from Eq. (1.2) after multiplication by r" and integration by parts using condition (1.7):

*Zh.vychisl.Mat.mat.Fiz.,27,7,1068-1077,1987

71

72

c~Mo + a _ (uM+)=O,

(HI)

Ot Oz c)M~ + O_ 1 Ot Oz (uM.)=~-nu×M.-,. In particular,

when

n=3,

it

follows

from

Eqs.(1.12)

On --

Oz

=

(1.12) and

(1.3) that

1 e' = - - .

ae'×M2,

(1.t3)

t-e

The initial formulation of the problem is now split up into the following problems.

Problem A. It is required to find ×, u, M0, Mt and Mz from Eqs. (i.i), (i.ii), (1.12) and, when n=1, 2, (1.13) with specified boundary conditions (1.8)-(1.10) and the initial condition (1.6) and specified initial values of M0(z, 0), M,(z, 0) and Mz(z, 0) which are determined from (1.5). Problem B. It is required to determine f from Eq. (1.2) with initial condition boundary conditions (1.7) and (i.iO) when the functions × and u are known.

(1.5) and

2. L a g r a n g i a n variables (coordinates of the number of p a r t i c l e s ) In the subsequent analysis it is convenient to consider an analogue of Lagrangian mass coordinates by taking Eq. (1.11) as a basis. Let us define the transformation (z, t)-+(%',T) by the relationships z

v (z, t) =Vo (t) + ~ Mo (z', t) dz',

(2.t)

o

where t

(2.2)

vo(t) = - ~ Mo(O, t')ur(t')dt'. o

For the derivatives, by virtue of

(1.11),

(2.1) and

O/at=a/O~-uMocg/cgv, from which, in particular,

(2.2), we have the relationships

cg/az=Mocg/av,

the equations for z:

Oz/O~=u,

Oz/Ov=V,

(2.3)

follow, where

V=Mo -~

(2.4)

is a quantity which is proportional to the average volume occupied by a single particle. Eq. (i.ii) in (v, r) coordinates is written in the form

~)u/c~v=0 V/Or which necessarily follows from Eq. (2.3). The region (0~0) is transformed into a region

(2.5) v0(T)~0.

where

No = ] Mo(O,z')dz' o

and

v0(x) is defined,

in correspondence with

v , , ( O = - ~M,, r (T') ur (T')d~',

(2.2), by the equation

Mo,r('c)=V-'(vo(x),x)>O,

(2.6)

o

where the function ,%/0.r(T) occurring under the integral sign is defined by the boundary condition when Ur>0 and by the solution of the problem when Ur<0 (see (2.4), (1.4) and (i.iO)). When Up=0 , we have v0(T)=0 by virtue of (2.6). We note that the condition that the boundary velocity Ur is constant implies the monotonicity of the function v0(T) and the definition of the function T0(v) which is inverse to it:

v=-

] Mo,r(x')Ur(r')dx'

(2.7)

o

(see Fig.l, where the domains of the solution in the variables (v,T) are presented: corresponds to graph 1 and Ur<0 corresponds to graph 2). Eqs. (2.3) define the transformation which is the inverse of (2.1) :

ur>O

73

•r

/

Z/ ] II

z(v,x)= i V(v','r)dv',

(2.8)

t (~,, ~) =*.

~o(x)

I

// '

/

v > O , one may also write

When

i

/

Ii

/ / Zo(~)

z(v,x)=zo(v)+

iu(v,x')dx',

(2.9)

O
where v

Fig. 1

(2.10)

ZO(V) = f V (%", O) dv'o

Returning to Eq. (1.2), let us introduce the function

(2.1~)

¢=/v. Then,

(1.2) in the variables

(% T)

is transformed to the form

(see (2.4) and

(2.5))

0q~+ I 8q~ - - ~ x × - - = 0.

Ox

Unlike

3

(2.12)

Or

(1.2), this is an equation with derivatives of just two variables.

3. T h e s y s t e m o f m o m e n t s . Let us define the moments m~ of a function

~:

m. = ir"~dr=VM..

(3.1)

o

By analogy with

(1.11) and

(1.12), we have from (2.12) the equations

Omo/O'~=O,

Om,/Ox='/so~×nm,_,,

n=l,

2. . . . .

(3.2)

The first of these equations is obvious since, by virtue of (3.1) and Let us now define a function R(% ~,T) by the equation

(2.4), we have

ORtOx='l~a×. Integration of system (3.2) then yields with respect to n, we obtain

m0=1.

(3.3)

m0=i, m,=a,+R, m,=a~+2a,R + R ~ etc. and, by induction n

m, (v, ~) = ~.~aC2a,_~(v)R ~(v, r)------(~+R)"

(3.4)

h=¢

(the symbolic binomial notation (a)t------a, is used). Here a~(v) are the integrals of (3.2) for which the relationships a,~=ra0,a~=m,--R, a 2 = m z - 2 m t R + R 2 , etc. follow from (3.4). Induction with respect to l then yields t

a, (v) = Z

( - 1 ) he?m,- ~( v, T)Rk( "¢, x) -~ (m-R)'

(3.5)

(the same symbolic notation (m)"--m. is used). In the case of the integral al there is a certain arbitrariness associated with the definition of R but, by subsequently putting a,(v)= 0, we have m,=R by virtue of (3.4). We note a useful inequality for the moments m. which follows from the Cauchy-HOlder inequality:

m.,+.,>m.,m,=.

For instance, when

ni=l

and

n2=2

and account is taken of

(3.1) and

(1.3), we have

M,M2~ (l-e)Mo. Returning to the moments

R

and, conversely,

from

M.

M,

Mo'

as

(3.4) and

(3.6)

of the function f, we have from

=M,_(M,)' Mo XM~l

,

Ms

as=---3

M,,

MzMt I M , \ ' / MJ + 2 / - -XMo+

(3.1), we have

Mo=V -I,

MI=RV -~,

(3.5)

M==(a2+Rz) V -',

.

~(3.7)

74 where V=e'(as+3a~R+R +) is a relationship which follows from (3.4) when n=3, from (2.8) and from (1.3). As a result a one-to-one correspondence is established between the sets of functions (JI0~M ~ Mi) and (H, ai, aa) which enables one to formulate the initial problem in an equivalent manner in terms of (R, ai, as). We emphasize that the quantities a2 and a3 in Lagrangian coordinates represent the integrals of motion of the initial p r o b l e m (i.i)-(i.iO) and the function R satisfies Eq. (3.3). 4. S o l u t i o n of the problem for the distribution function. Let the solution of Problem A be known. Then, by virtue of (3.3), the solution of Eq. (2.12) is written in the form

~ = ¢ (v, p), where q) and Re a r e a r b i t r a r y

functions.

p=r--t=l(v,~)+R~(~), For f , in accordance w i t h (2.14), we have

/(v, ~)=V-l(v, T)O(v, r-R(v, ~)+R°(v)). Let us define the functions •

and

Rc

(4.1)

in the following manner:

, / O,(v,p), O(v'P/=/ 0,

p>0, p<0,

/ Ro(Zo(V)),

v~>0,

R+(v) = ' [ Hr (.to(V)),

(4.2a)

vo(~) ~
(4.2b)

where

(¢o(Zo(V),p),

v~>O,

O, = t Cr (To(V), p),

vo(~) ~ v < O .

(4.3)

Here, T0(v) and z0(v) are functions which are determined from (2.7) and (2.10). The domains of specification of the initial and boundary conditions, which determine the structure of (4.3), correspond to curve 1 in Fig.l. The functions R0, Rr, ~0, O r are defined by the initial condition (1.5) and the b o u n d a r y condition (i.iO) :

Ro(z)=M~(z, O)/Mo(z, 0),

Rr(t)=Mt(O, t)/Mo(O, t),

• o=Vo-'(zo(v))]o(zo(v), p), Cr=Vr-'(To(v))fr(To(V), p),

(4.4a)

~0,

(4.4b)

V<0,

(4.4C)

where Vo(z)=Mo-1(z,O), Vr(t)=Mo-i(O,t). NOW, both the initial condition as well as the boundary conditions (1.5), (1.7) and (i.iO) are satisfied in the case of the function f. Hence, a solution of Problem B for the d i s t r i b u t i o n function has been obtained. This solution is defined by formulae (4.1)-(4.4) and is written above in Lagrangian coordinates. In order to make the tansition to the coordinates (z, t), it suffices to make use of the transformation (2.8) (see, also, (2.9), (2.10) and (2.1), (2.2)). 5. C e r t a i n properties of the solutions. Assuming that the functions us(t), Xr(t), Ur(t), ×0(z) are continuous and bounded, we start out from the existence of a continuous solution of Problem A, which is defined by the functions ×(z. t), u(z, t), M0(z, t), M,(z, t)~ M~(z, t) (or R(~, 7), ai(~), a3(v), (see paragraph 3) in a domain G: ( 0 < z < ~ , t>0~ with boundary aG: (z=0, t=0). The two families of characteristic curves p l a y a n important role in the investigation of the properties of the solutions. The following lemma holds.

Lemma i. Through each points which is defined by the equation

(z0, to) of the domain G, there passes a unique curve S

dz/dt=a,(t).

(5.1)

This curve intersects the boundary O G at a certain point (z~, t,) (zs=0 or over, there exists a point of intersection with t,
u+(t)>O

t+=0)

and, more-

Vt>O,

(5.2)

is satisfied, then t,
dz/dt=a(z, t).

(5.3)

This curve intersects the boundary OG at a certain point (re, to) (zc~0 or over, there exists a point of i n t e r s e c t i o n with t~
u(z, t)>O

V(Z, t)

from

G,

to=0) and, more-

(5.4)

is satisfied, then t¢
75 satisfied (and correspondingly, z~
Lemma 2. Let Q be an arbitrary non-empty subset of the domain G. Then, there exists an s-characteristic, the connected part of which intersects the boundary 0G of the domain G and belongs to the exterior of Q with the exception of a unique point q0 (boundary for Q). Let q*=(¢', t') be an arbitrary point of the set Q. By virtue of Lemma i, an s-characteristic t

z(t)=z~ q- u,(t') dt' , ts

passes through this point, intersecting the boundary 0G of the domain G at a certain point (z~, t~) and, moreover, t~
to~inf T.

Let

zo=Z(to)

such that q0=(z0, to)~S According to the defintion of the point to, in any neighbourhood of this point there are points which belong, as well as points which do not belong, to T. Correspondingly, in any neighbourhood of the point q0, there are points of the set Q as well as of its complement in G. Hence q0 is a boundary point for Q, a limit point or an isolated point. Let us consider a connected set So (connected by virtue of the continuity of z(t))

g~So-*~-Cg=(z(t), t), t~[t,, to]), i.e. So is a segment of a characteristic S specified parametrically on the segment [t,,to]. It follows immediately from the definition of the point to that, when t~t
Theorem i.

Let

xo>O,

xr>O.

(5.5)

Then, everywhere in the domain G

×>0.

(5.6)

In order to prove this, let us consider Eq. (l.1) in the domain G with the additional conditions (1.6) and (1.8) on the boundary aG:

-~-~x(z, tl+u,(t) ~-zX(Z,t)+EM~(z,t)×(z,t)=O, ×(z, O)=xo(Z),

×(0, t)=x~(t).

Let us assume that the oppostite of (5.6) applies. Then, in G, there exists a non-empty set Q at each point of which x~<0. By virtue of Lemma 2, there exists an interval of an s-characteristic S~ which joins a point g,=(z~,t~) of the boundary aG to a certain boundary point q0=(z0, to) of the set Q and, moreover, t~<~to. By virtue of the continuity of x , we have

×(Zo, to)~0

(5.7)

at the point q~. Further, the equality t~=to is excluded since, then, z(zs, t~)=×(z0, t0)=0, which contradicts (5.5). Now, let S'=So\{qo}. Since q0~&~, the set S' is non-empty and, by virtue of Lenrna 2, belongs to the exterior of Q. Let us now encompass each point of the set S' by a neighbourhood belonging to G but not containing any points of Q and let G' be the combination of these neighbourhoods. Then S'cG' and the inequality ~ > 0 is satisfied everywhere in G'. By taking the time t'=t as the parameter on the characteristic and assuming that the quantity t

~ = z - j" u, (t') dr"

(5.8)

o is a parameter of the characteristic such that the domain G' in the form

0

Ot'

×(~,t')

I n - x(Lt,(g))

O/Ot=@/Ot'--u~O/@~, a/az=0/@~, we write Eq. (i.i) from

EM2(~,t').

(5.9)

76 This representation is possible, in fact, by virtue of the condition that (z, t)~G" Now, by integrating (5.9) along the characteristic ~=COllSt, we

x(z, t)>0 obtain

×(zo-O,to-O)=x(z.t,)exp[-E ~ M2(~,t')dt']>O

for

(5.io)

ts

and consequently,

by virtue of the continuity of x

and the boundedness of

M2,

to

×(zo,to)=X(z.,t,)exp[--E ~M2(~,t')dt'] >0,

(5.i~)

ta

which contradicts (5.7). Hence, inequality (5.6) is satisfied everywhere in the domain G and the theorem is proved. In the subsequent discussion it is assumed that conditions (5.5) are satisfied such that x(z,t)>0 everywhere in the domain of definition.

Corollary i. (The maximum principle). For each value of x(z0,t0) within the domain G, there exists a value ×(z~,L) on the boundary OG which exceeds it. In particular, the maximum value of the function x is attained on the boundary. This follows from Lemma 1 and relationship (5.11) in which E>0, M 2 > 0 and t~
A priori

Theorem 2.

estimates. Let the estimate

×(z,t)~×(z.,t.)exp[

E(i--e) (t--t.)],

(6.1)

/~mln

be satisfied in the case of the function x, where (z~,t,),t~
(6.2) ts

Let us estimate the integral under the exponent sign. The estimate M=~<(I-e)/R holds in the case of the function M..(z,t) by virtue of (3.6) and (3.7). Together with the s-characteristic, we draw the c-characteristic through the point (z, t) up to its intersection with the point (z¢, to) (zc=0 and t~=0), in which in every case t=
(i-,) l(t.t)=E ~M2(~,t')dt"< EB--~'~. (t-t,) ts

(E>O.M~>O,t>t,),

we obtain

Corollary 3.

(6.1) from

Let the condition

(6.2).

0
be satisfied.

Then, when

t>Z/am,., the

estimate

x(z,t)~Cr(t,)exp( is satisfied, where

E(I--8) z .), ttmin R~n

t, is defined by the relationship

i a.(t')dt'=z.

ts

(6.3)

77 u,(t)>0

In fact, according to Lemma I, we have z,z/um,,. Since the inequality

Z--Z,>~Um,,(t--t,), is also valid,

(6.3) follows from

Theorem 3.

(6.1) and

together with

t,
(6.4)

(6.4).

Let

xr~O,

~¢,>~0.

(6.5)

Then, everywhere in domain G

×I>0.

(6.6)

Following the proof of Theorem l, let us consider a set Q in the domain G, at each point of which x~0 and, by repeating the proof of Theorem i, we arrive at a contradiction. Consequently, the equality to--t, is satisfied from which it follows that z0=z, and ~(z,, t,)=~(zo, to)=0. Let us now consider Eq. (i.I) on the characteristic ~=const (see (5.8)). In the case of the function of one variable ×,(t')--x(~,t') (~=c0nst), we have

d×,/dt'=-EM~×,,

(6.7)

and ×,(t,) =0. Let us show that

×.(t')=O

Vt'>t,.

(6.8)

Let us assume that ×,(t')~0 for a certain t'>t, and let To be the set of those t' for which t,~t'
×,(t,) =0.

(6.9)

At the same time, by virtue of the condition t'ET, when z,(t')~0, we have the ti
d,

Ix.(t')l

--in ]-dt' ×.(t')

=-EM~,

tl
and integrating from t" to t,+0, we obtain, by virtue of the boundedness of M2 and the continuity of x,, z,(tl)~0 which contradicts (6.9). Hence, (6.8) is satisfied. Let us now suppose that the function M is negative at a certain point (z°,t') of domain G and draw the s-characteristic through this point. We arrive at a contradiction to (6.8): ~(z*, t')=0. Hence, (6.6) is satisfied at each point of the domain G and the theorem is proved. The following corollary is established in the process of proving Theorem 3.

Corollary 4. If the function x is equal to zero at a certain point, it is equal to zero on the whole of the s-characteristic passing through this point. The authors express their gratitude to A.A. Samarskii for his interest and also to I.V. Melikhov and B.M. Dolgonosov for formulating the physical problem.

Translated by E.L.S.