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Дата индексирования: Sat Dec 22 18:23:30 2007
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(), .. , .. ,
, , . . , . , - . . , ( ), . : , , , , , . The process of dissolution of a stagnant drop in a capillary in the presence of a micellar solution is modeled and the effects of the main physico-chemical properties of the fluid system on the dissolution rate are studied. Using the method of integral moments an approximate one-dimensional analytical solution is derived. The dissolution rate, the positions of the drop/solvent interface and the concentration front as functions of time, as well as the time for complete dissolution of the drop are calculated. It has been found that the time for complete dissolution of the drop increases as the diffusion coefficient decreases, the drop solubility (equilibrium concentration of the dissolving substance in the solvent) decreases, the constant of solubilization increases or the drop size increases. Key words: diffusion; dissolution; drop; capillary, solubilization rate, micelles.

1. - / , ( ), , , , , , , . . [1-2]. , , , . , / . [3], - . 14

()- « », - , [4-6]. , () , . () « », - [2, 7-10]. () , , . , () , ( ) . () « », , , . , / -

. . , 2004, 3 (23)

. . , . .

, «» [1113]. : (Ostwald ripening) [1113]. . [14, 15], . , . , , . . , [15], , , (), . , . [4], . n, [3]. , [3] , . () [16] . ,

. [9] , (-) . , , . . . / . (Ostwald ripening) [4-5] , n- -- () . [17], [18] . (Ostwald ripening) [5], , , . ( ) . , , . , , 15

. . , 2004, 3 (23)

( , ). 2. 2 0 2 L0 , ( ) Lc >> L0 . 1 (. 1). . 2 , 2 , [19-22]. , L(t) (. 1). , - 2 . 1. , . - . , 2 , a 2 . a , . ,

c ( L1 ( t ) , t ) = ceq . -

c a 2c + =D 2 , t t x

L1(t)
(1)

[4]. c 2 , ceq ( ) 2 L1(t) -, .. t (. 1). , [3] ,

D 2 , x ( ). c(x, t) a(x, t) c(L(t),t) = 0; ( c / x )x= L (t) = 0; c(L1(t),t) = ceq a(L(t),t)=0; a(L1(t),t) = 0; (2)

c - x

= (c eq - c( L1 (t ), t )) , - x = L1 ( t )

-. [3, 4], , () : Mn +O M n+1, n = 0,1,2,...,N. O Mn , , , n , N , .

c( L0 ,0) = 0; a( L0 ,0) = 0
, a(L1(t),t) = 0, , (). , ( ) , , , 2 -

16

. . , 2004, 3 (23)

. . , . .

( 2 ). , [3, 4],

x c( x, t ) = A(t ) 1 - L

2

(6)

a = ka c t
k a = k + c m , k
+

c(x,t), (2)

(3)
m

, c

-

1 - c( x, t ) = c eq (1 -

z l y

)

2

2

(7)

. , (1)-(3) . L1(t) ( ) . , ( , ) , L1/(t) c = 0 [23]. x x= L1 ( t ) (1)-(3), [19-22], c(x,t)

: l = L / L0 , l1 = L1 / L0 , z = x / L0 , y = l1 / l . 2 (1)-(3)

0 l1 +

l

l
1

(a + c)dz =

0

(8)

(8) , a(x,t), (2) (1)

k

dl

1

d

=-

dv d

- ka R v

(9)

c( x, t ) A(t ) + B(t ) x + E (t ) x

2

(4)

k = 3 0 /ceq = l - l1 . = t /
2

R

A(t), B(t), E(t) , , , (1) , . (2),

, R = L0 / D . l l1 [19-22] , (1) w( z ) = 1

A(t ) = - B (t ) L + E (t ) L2 , E (t ) = - B(t ) / 2 L

l (t

)

l1 ( t

)

. , -

(5)

(1)

, (4)
. . , 2004, 3 (23)

l1

l

(

c + a ) dz =

l

1

l

2 2

z

c ( z , ) dz

(10) 17

(10) (3) (7),

dl1 6 - d (k - 3)

1 1- e
- 2

(16)

3

dl

1

d

=-

dv d

- ka R v +

6 v

(11)

(9) (11) , , l1 (0) = 1 .

, ,

dl d

= ( -

6e

2

(k - 3)

)

e

-2 -2

1- e

(17)

l - l1 = u

-2 ka t

(12)

l1 ( c ) = l1 (t c ) = 0 . , (13) , tc, , ..

u = 1 - e

= 1- e

-2

, = ka R ,

= 6k /(k - 3) .
, (11) (12)

tc = ln(cosh ) / k
= ( k - 3

a

(18)

)

/6.

l1 = 1 -

3 (k - 3)

ln

1+ u 1- u

(13)

(12) (13) l1 ( 0 ),
l1 1 - 12 k (k - 3)

>> 1 (, 2 ) tc

tc =

R (k - 3)k
6k
a

>> 1

(19)

l l1 +

12k (k - 3)

0

(14)

3. . 2 -

,

t .

t >> 1 /(2k a ) (13)
l1 1 - 6ln 2 (k - 3) -

t >> 1/(2k ) a (k - 3)

6

(15)

(15) . (13) 18

. 2. ka (D = 10-6 2/, L0 = 10-3 , k = 3 0 / ceq = 100).

. . , 2004, 3 (23)

. . , . .

, . ka. ceq (, , k = 3 0 / ceq ) (. 3). ka ceq . 4 5. , tc ka, / ceq. , .., ka=0, tc,0, (18)

. 3. k (D = 10-6 2/, L0 = 10-3 , ka = 0,1 -1).
2500
k a= 0 .1 s
-1
-1
-1

2000 1500
t,s
c

k a= 0 .5 s
k a= 1 .0 s

t

c ,0

= k0 2 R /12
,0

(20) ceq
,0

k0 = 3 0 / ceq

1000 500 0

0 .0

0 .5

1 .0

R

1 .5
=L
2 0

2 .0
/D , s

2 .5

3 .0

. 4. R = L0 / D ka (D = 10-6 2/, L0 = 10-3 , k = 3 0 / ceq = 1000).
2

tc -

. [12, 23]. tc tc,0, , , , , . r , .. r = ceq ,0 / ceq < 1 . ,

12. 0 10. 0 8. 0
t x 10 s
-5

k a = 0 .1 s

-1

k a = 0 .5 s

-1

k0 = 3 0 / ceq ,0 = 3 0 / ( r ceq ) = k / r t0 = tc / t

c ,0

k a = 1 .0 s

-1

-

6. 0 4. 0 2. 0 0. 0

t0 =
0. 0 2 .0 4 .0 6 .0
-5

tc 12 ln ( cosh ) 2 = r, tc ,0 ka k 2 R

(21)

c

8. 0

10. 0

k x 10

. 5.
tc k = 3 0 / ceq ka (D = 10-6 2/, L0 = 10-3 ).

. t t . -

- . . 6 t0 (21). , (, ka) , ka. , (, ceq). 19

. . , 2004, 3 (23)

0 .0 1 0 0 .0 0 8 0 .0 0 6
t
0

k= 50 k=100 k=200

0 .0 0 4 0 .0 0 2 0 .0 0 0

ka / . ( INCOCOPERNICUS, PL979098) ( 01-01-00276).
1 .0

0 .0

0 .2

0 .4
k a, s
-1

0 .6

0 .8

1.

. 6. ka k (D = 10-6 2/, L0 = 10-3 , r = 0,1).

2.

3.

4. , . , , ( , , , ..) . , , , . : ·
4. 5. 6.

7.

8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

t t . · ka ( ka) / ceq. , ka . , / .

·

21. 22. 23.

·

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