The motion of a binary mixture with a cylindrical free boundary at small Marangoni numbers Текст научной статьи по специальности «Математика»

УДК 532.516

Mixture with a Cylindrical Marangoni Numbers

Victor K. Andreev*

Institute of Computational Modelling RAS SB Akademgorodok, 50/44, Krasnoyarsk, 660036 Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Natalya L. Sobachkina^

Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 19.08.2017, received in revised form 28.08.2017, accepted 06.12.2017 We studied the problem of axisymmetric motion of a binary mixture with a cylindrical free boundary at small Marangoni numbers. Using Laplace transformation properties the exact analytical solution is obtained. It is shown that a stationary solution is the limiting one with the growth of time if satisfy certain conditions imposed on the external temperature. Some examples of numerical reconstruction of the velocity, temperature and concentration fields are considered, which correspond well with the theoretical results.

Keywords: binary mixture, free boundary, stationary solution, Laplace transformation, the Marangoni number.

DOI: 10.17516/1997-1397-2018-11-2-194-205.

Introduction

It is well known that at low Reynolds numbers the momentum equation can be simplified by discarding the convective acceleration. Such movement is called creeping. The same simplification can be obtained for equations of energy transfer and concentration. One of such problems considered in the work [1,2], associated with the unsteady motion of a drop (bubble). As the Reynolds number was taken as the number Marangoni, which is small both due to the radius of the droplet, and due to the physical parameters. In another task (the motion of fluid in a cylindrical tube) the small parameter was the product of the number Grashof on the Prandtl number [3].

In the present work we considered a similar problem. It is associated with unsteady motion of a binary mixture with a cylindrical free boundary. Here the analogue of the Reynolds number acts the Marangoni number. This problem is reduced to inverse initial-boundary value problem for parabolic equations. It is solved in quadratures, and it is proved that at certain conditions, its solution tends to a stationary regime with a growth time. Numerical treatment of the Laplace transform obtained quantitative results for a model compound, which confirm the theoretical conclusions.

* andr@icm.krasn.ru tsobachkinanat@mail.ru © Siberian Federal University. All rights reserved

The Motion of a Binary Free Boundary at Small

1. Statement of the problem

Let us turn to the problem of motion of a binary mixture. Axisymmetric motion of an incompressible viscous heat-conducting mixture is considered. Denote by u(x,t), p(x,t), respectively, the velocity vector and the pressure, e(x,t) and c(x,t), respectively, deviations from the average values of temperature and concentration. Then the system of equations of thermodiffusion motion in the absence of external forces (g = 0) has the form:

du 1

--\— Vp = vAu, div u = 0,

dt p

dd dc ,

— = \Ad, — = dAc + adAe, dt dt

where p is the average density, v is the kinematic viscosity, x is the thermal diffusivity, d is the diffusion coefficient, a is the thermodiffusion coefficient (the coefficient Soret); d/dt = d/dt+uV.

After the transition to dimensionless variables (as the scale of length, time, velocity, pressure, temperature and concentration are taken of values h0, h0/v, œ1Ah0/^, œ1A, Ah0, Ah0p1/ftz) will receive task:

du/dt + Mau •Vu + Vp = Au, (2)

div u = 0, (3)

Pr (de/dt + Mau •Ve)=Ae, (4)

Sc (dc/dt + Mau • Vc) = Ac - eAc. (5)

Here introduced dimensionless parameters: Ma = i£1Ahz/^v is the thermal Marangoni number, Pr = v/x is Prandtl number, Sc = v/d is Schmidt number, e = -apz/p1 is thermodiffusion parameter; A is temperature gradient, œ1 is some constant. We assume that the Marangoni number is many times less than one Ma ^ 1 (the creeping motion), then in the momentum equations, heat transfer and concentration convective terms can be discarded. With the set physical parameters of a liquid, this condition is carried out if sufficiently small value Ah2. Formally expanding functions u, p, e, c in series of Ma, to get a first approximation task (2)-(5) with Ma = 0.

Let u(r,z,t), w(r,z,t) be the projections of the velocity vector on the axis of a cylindrical coordinate system r and z. Suppose that the free boundary in this system is described by the equation r = h(z,t) (see Fig. 1). Then the conditions will take [3]:

(1 - hZ)(uz + wr) + 2hz(ur - wz) = — z pv

ht + whz — u = 0, (6)

(7)

du da '

(hzer + ez) de + (hzcr + cz) dc

pgas - p + 2pvL [ur - hz (v,z + wr ) + hzwz] = 2aH, (8)

kL-1 (er - hz ez ) + 7 (e - egas) = Q, (9)

cr - hzcz + a(er - hzez) = 0, (10)

where L = (1 + h?z)1^z; a(e, c) is the coefficient of surface tension of the mixture and for most real liquids, it is well approximated by a linear dependence

a(e, c)= a0 - œ1(e - e0) - œz(c - c0), (11)

where œ1 > 0 is the temperature coefficient, œz is the concentration coefficient (typically, œz < 0, since the surface tension increases with increasing concentration), a0,e0,c0 are some constants

average values; pgas and 9gas are the pressure and the temperature of the ambient gas, which is considered passive; y is the coefficient of interphase heat transfer, k is the thermal conductivity coefficient.

In equation (8) H is the average curvature of the free boundary

H _ hhzz - Kj - 1

H _ 2h(1 + h2)3/2 ■ (12)

r-h(z.t) gas

Fig. 1. Current Flow Diagram

Remark 1. In the recording of the Marangoni number is present the parameter included in the dependence of surface tension on temperature and concentration (11) (Fig. 1).

Consider the subgroup generated by four operators dz,tdz + dw, do,dc,dp. It is easy to check that it is allowed by a system of equations of thermodiffusion. Its invariants t, r, u, then partially invariant solutions with respect to this subgroup are of the form [5]:

u = u(r,t), w = w(r,z,t), p = p(r,z,t), 6 = 6(r,z,t), c = c(r,z,t). (13)

In this case, from the equation of conservation of mass it follows that w is a linear function of z. Let

w = zv(r,t). (14)

General view of the invariant manifolds with respect to the considered subgroup {r, z, t} is r = h(t) with an arbitrary function h(t). Let the dependence a(6,c) has the form (11), then from boundary conditions (7) we get that s1(6 — 60) — (c — c°) is a quadratic function of z. It is therefore natural to assume that

6(r,z,t) = a(r,t)z2 + b(r,t), c(r,z,t)= l(r,t)z2 + m(r,t), (15)

Interpretation of the solution of (13)-(15) is as follows. Let an axisymmetric heated long enough cylinder of a binary mixture of external temperature on the boundary has a maximum of (a < 0) or minimum (a > 0) at point z = 0. Then in a neighbourhood of this point the external temperature can be approximated by a parabolic law.

The substitution of the solution (13)-(15) to the system of equations of thermodiffusion and separation of a variable z leads to the next task in the region t > 0, 0 < r < h(t)

1 u

ut + Pr = urr + - ur--, (16)

rr2

zvt + Pz = z(vrr + , (17)

ur + - + V = 0, (18) r

at = Pi (arr + lar) , (19) - 196 -

bt = Pr (brr + rbr +2ay (20)

lt = SC (lrr + 1 lr) - S^c (^rr + 1 ar) , (21)

1 ( 1 „A e A 1

mt = — (mrr +— mr + 2l^\--(brr +— br + 2^ . (22)

Sc V r J Sc V r J

The boundary conditions at r = h(t) as follows

ht = u, (23)

vr = -2a - 2Ml, (24)

pgas - p + 2v = We - (a - 9°) - M(l - c0), (25)

a.r + Bi(a - agaS)=0, (26)

br + Bi(b - bgas)=0, (27)

lr - ear = 0, (28)

mr - ebr = 0. (29)

Here introduced dimensionless parameters: Bi = 7h°/k is the number of Bio, We = Ah°

is the Weber number, M = Ma = where Mc = x2Ah°^1/^v^2 is the concentration

Marangoni number;

In addition, it is necessary to require the boundedness of functions on the cylinder axis when r=0

|u| < x, |v| < x, \p\ < x, |a| < x, \b\ < x, |l| < x, \m\ < x, (30)

and specify initial conditions at t = 0

u = u°(r), v = v°(r), a = a°(r), b = b°(r), l = l°(r), m = m°(r), h(0) = h° = const > 0. (31)

Note that the functions u° and v° are related by equation (18), v°, a°, l° by condition (25), l°, a° by condition (29), a m°,b° by condition (30).

Suppose that at the initial moment of time the free surface satisfies the initial condition: r = h(0) = h° = const > 0, and decompose the function h(t) in a series of Marangoni. Then the variable r will vary in the range 0 < r < h° + Mah(t). We introduce dimensionless variables: r' = r/h°, z' = z/h°, t' = vt/ho2. The kinematic condition (6) in dimensionless variables has the form (touches for convenience omitted):

Maht = Mau(1 + Mah(t),t). (32)

Assuming that Ma ^ 1, from the last equation in Ma ^ 0 we find that the kinematic condition at r = 1 gets converted to

ht = u(1,t). (33)

The function u is expressed from the continuity equation (18) in the following way

1 rr

u = —

rvdr. (34)

From equations (16), (17) we express the pressure gradient (pr,pz)

1u

pr = urr +— ur--2 - ut, (35)

r

0

Pz = z(vrr + 1 Vr - v^j . (36)

1

r

Differentiating equation (35) for the variable z, and the equation (36) for the variable r, and given that a function u is not depends on z, we get the equation of compatibility: prz = pzr = 0. Whence it follows that the function v(r,t) is determined from the equation

and the pressure is given by

Vt = Vrr + 1 Vr + q(t), (37)

r

z2

p = — — q(t) + s(r,t), (38)

where q(t) is arbitrary function, and the derivative in the variable r of a function s(r, t) is precisely the right side of equation (35).

Suppose that in the equation (34) when r = 1 running

I rvdr = 0. (39)

J 0

It follows that u(1,t) = 0. Then, returning to the kinematic condition (33), we have: h(t) = 1. So a free boundary does not change with time and remains constant value. Converse is also true: if h(t) = 1, then condition (39).

Remark 2. In fact, certain functions q(t), i.e. the "gradient pressure" along the axis z, can be found as follows. Multiply equation (38) on r and integrating on r from zero to 1, reducing it thereby to a "loaded" equation

d f1 f1 1

— J rvdr = J (rVr)r dr + 2q(t). (40)

Integral, the left-hand side of the equation equal to zero according to the condition (39). It follows that

q(t) = -2vr (1,t), (41)

where the right part is set as a boundary condition (24).

Therefore, first solve the equation (19) on the function a(r,t) with boundary conditions (26), (30) and the initial condition (31); next we define the function l(r,t) from the equation (21) with conditions (28), (30), (31), while function b(r,t), m(r,t) are calculated similarly. Thereby recovering the temperature and concentration mixture according to formulas (15). Then solve the problem on determination of the axial components of the velocity vector v(r,t), satisfy the equation (38) and the boundary conditions (24), (30), (31). As the result is to find the radial component of u(r, t) of equation (34) with boundary conditions (30), (31), the function q(t) with (41) and the pressure p from equation (38) and the boundary conditions (8).

2. Stationary solution

For this decision all the unknown functions do not depend on time; let them through u0(r), v0(r), p0(r), a0(r), b0(r), l0(r), m0(r). In addition, q(t) = q0 = const. Write out the corresponding boundary value problem for 0 < r < 1:

1 n u

0

u°rr. + -u0 — = P0, (42)

0 i 1 0 pz 0 i-AO\

Vrr + - V0 = — = —q , (43)

0

r

rz

u° + 1 u° + v° = 0, (44)

rr

P (a°r + 1 a°°) = 0, (45)

b°r + 1b° + 2a0) =0, (46)

^ (C + 11°) = 0, (47)

(^l<°r + r

(m°r + 1 m° + 2l°^j

Sc(m°r + 1 mr + 2l0 ) =0, (48)

General solution of system (42)-(48) when Pr = 0, Sc = 0 are easily found (taken into account the conditions of constraints of (30)):

0 2 0 2 a0 r2 ea0 r2

0 o 1.0 gas .7,0 . 0 /n c i o* — 1 \ ;0 0 0 gas „ , .„n a = agas, b =--+ bgas + agas(0.5 + Bi ^ l = eagas, m =--^- + C (49)

where a0gas, b^gas, C = COnst.

Remark 3. The constant C remains arbitrary, it is not surprising, as the task for the concentration contains only derivatives. On the other hand, this constant can be determine if ask the

i ea°as average concentration in the cross section of z = 0, i.e. f cr0(r)dr = 0. From this C = ——.

0 4

Thus, we have the following representation for m0

22

The other functions are as follows

m0 = ea2as(2 - r2) . (50)

u0 = ^r(r2 - 1), v0 = - r2) , p0 = q2 (z2 - £), q0 = 4a°Bas(1 + Me). (51)

The obtained solutions satisfy the boundary and initial conditions (23)-(31).

3. Determination of the temperature field

For solution of nonstationary problem is used Laplace transform. Believe (assuming the existence of a, ar, arr, agas)

a(r,p)= a(r,t)e-pt dt (52)

0

then the problem for a(r, t) is reduced to boundary problem for ordinary differential equations

arr +1 ar - Pr pa = - Pr a0 (r), 0 < r < 1, (53)

ar + Bi(a - agas)=0, r = 1, (54)

\a(0,p)\ < x. (55)

The General solution of the equation (52) it is easy to write (taken into account the conditions of constraints (55))

C1I0 ( VPp r) +

+ J Pryao(y) /0^ ^Pr P V^jKo^ v^r P ^ - /o ^ VPp r^Ko^ v^r P yj

dy, (56)

with a constant Ci, determined from the boundary conditions:

1

Ci =

agas +

ypTP /^ ^/Pp) + Bi/o^ yPT^^ |Bii

+ yPTP J Pryao(y) /^yPTPy^Ki^v^) + /^yPT^^K^yPTPy)

- Bi J Pryao(y) /^yprp^K^^PT^ - /o^yPT^^K^yprp^ dyj. (57)

dy-dy

The task is to determine the image of b(r,p) exactly coincides with the task (53)-(55) with the replacement right parts: —Prao(r) for —Pr bo(r) — 2a. Thus, this function is given by

b = C/

( ,_ x Ci r/A VPrp r)

r)--TP^+

/r

Pry(bo(y) + 2ao(y)) [/o^VPP y)Ko(y/Prp r) - /o(VPP ^^(^/Pp y)]dy (58)

with a constant C2, determined from the boundary conditions:

- i

C2 +

dy-

V/Prp+ Bi/o(v/Pip) ] {Bi bgas + Cir/o (v/Prpr) + /■ i

V/PTp yo Pry(bo(y) + 2oo(y)) [/^^v/Prpy^^v'Pp) + /^^v/Prp)K^^v/Prpy) /■ i

- Biyo Pry(bo(y) + 2ao(y)) [^(v^y)Ko(v/Pp) - /o (v^)Ko (v^y)] dy}, (59)

where /j, Kj are Bessel functions of the first and the third kind of imaginary argument. It is possible to show, using the explicit formula (56), (57) that

lim a(r,t) = lim pa(r,p) = ag„s, t^^ ' p^o K ' gas

where aoas is stationary solution for the function a(r,t). In the derivation we must assume the existence of the limit lim agas(t) = limpaqas(p) = a°as. Note that when t ^ 0 [6]:

/o(t) - 1 + 4 , Ko(t)--/o(t)lnt,

t t3

1t

h(t)---1--, Ki(t)---+ - Int --.

2 16' t 2 4

From (56), (57) after long calculations we obtain that pa(r,p) ^ aPqas when p ^ 0.

2

Similarly, it is shown that

Km b(r, t) = Jim pb(r,p) = -0.5a°gasr2 + b0gas + a0gas (0.5 + Bi-1) ,

where the right side of the equality is a stationary solution for the function b(r,t), assuming the

existence of a limit lim baas(t) = lim pbaas(p) = bg„s.

y aas

Lemma 1. Problem solving for the functions a(r,t), b(r,t) are defined by the inverse Laplace transformation according to the formulas (56), (58), and with the growth of time, they reach a stationary regime, if agas(t) ^ a°aas, bgas(t) ^ b0gas when t ^ <x.

4. Determination of the concentration of the mixture

Applying to the initial-boundary problem for the concentration the mixture of Laplace transform, obtain for the image I(r,p) task

lrr + -lr — Scpl = —Sclo(r) — ePrao(r) + ePrpa, 0 <r < 1, r

lr — ear = 0, r = 1,

(60) (61)

\l(0,p)\ < (62) The General solution of the equation (60) with Pr = Sc is represented as follows

1 = C3Ig (VScpr) +

pr

+ y( Sc lg(y)+ £ Pr ag (y)) I^v/Scp y^Kg(y/Scp r) - Ig (V/Sc"p r^ Kg (y/Scp y) dy+

+

ePr Pr - Sc

C\lJ V/Prp r

+

/r

Pryao(y) Io (VPrpy) Ko (VPrpr) — Io (yPpr) Ko (\fPrpyj with a constant C3, determined from the boundary conditions (61):

C3 = [vscpI^^x/Scp^1 { scr—^PrC^v/PrpIi(v/Pr"p) +

_ 1 _ _ _ _

v/ScP y(Sc lo(y) + e Pr ao(y)) [^(v^^K^^/Sc^) + Ii (y^) Ko (y^y)

dy (63)

+

dy—

Prp f Pryao(y) [lo (VPp^Ko (TPTp) — Io (VPrp) Ko (^PTpy)]dyj. (64)

£Sc Sc - Pr

The task is to determine the image of m(r,p) exactly coincides with the task (60)-(62) with the replacement right parts: —Sc lg(r) — £Pr ag(r) + £Prp a for —Sc mg(r) — £Pr bg(r) + £Prp b. It is possible to show, using explicit formulas (63), (64), that

lim l(r,t) = lim pi(r,p) = £a°aas,

where £aPgas is stationary solution for the function l(r,t). If the conclusion is again necessary to

assume the existence of the limit lim aaas(t) = lim panas(p) = aPnn,.

p^g y nas

Similarly, it is shown that lim m(r, t) = lim pm(r,p) = m0(r), where m0(r) is stationary solu-

t^^ p^0

tion for the function m(r, t), assuming the existence of a limit lim baas(t) = lim pbaas(p) = b°„„.

p^0 y aas

Thus, the fair

Lemma 2. Problem solving for the functions l(r,t), m(r,t) are determined by the inverse Laplace transformation according to the formulas (63), (64), and with the growth of time, they reach a stationary regime, if aaas(t) ^ agas, baas(t) ^ b<0as when t ^ <x.

5. Determination of velocity field

The application of the Laplace transform to the problem for velocity reduces it to a boundary problem for ordinary differential equations

vrr +— vr — Prpv = —v0(r) — q(p), 0 < r < 1, (65)

r

rv dr = 0, r = 1, (66)

0

vr = —2a — 2M I, r = 1, (67)

\V(0,p)\ < x>. (68)

The General solution of the equation (65) is written as follows

v = C5I0 (VPTpr) + S&l + f yv0(y) [I0 (Upy) K0 (Upr) — I0 (Upr) K0 (Upy)] dy (69)

p0

with a constant C5, determined from the boundary conditions (67):

C5 = [Upii (Upr1 { —2a — 2mI+

+Up 0 yv0(y) [I0 (Upy) Ki (Up) + ii (Up) K0 (Upy)] dy^, (70)

where functions a(r,p), I(r,p) are defined by formula (55), (63) when r = 1.

In addition, from the boundary conditions (66), (67) defines a function q(p), namely

q(p) = — 2vr (1,p).

Working similarly to the proof of the convergence of the obtained solutions to stationary solutions of temperature and concentration of the expressions (69), (70) we can deduce the equality

lim pv(r,p) = v0(r), (71)

p^0

where v0(r) is stationary velocity distribution of (51). In the derivation of (71) we must assume

the existence of the limit lim aaas(t) = ag^*.

aas

Thus, the fair

Lemma 3. The solution of the problem for the function v(r,t) is determined by the inverse Laplace transformation formula (69), (70), and with increasing time it comes on stationary regime, if aaas(t) ^ a^ when t ^ <x.

6. Numerical results

The formulas (56), (63), (69) in the image Laplacian were used to numerical finding of the fields of velocity, temperature and concentration mixture under certain conditions imposed on the temperature outside agas(t). This was done by the method of numerical inverse the Laplace transform [7], which was obtained quantitative results for the model system with the following parameter values: o°gas = 0.2, e = 0.1, M = 1000, Bi = 2, Pr = 0.2, Sc = 0.1, A = 10, w = 10.

In Fig. 2-4 shows the evolution of the dimensionless profiles temperature, concentration and velocity of the mixture. In Fig. 2a presented the profile of the dimensionless 'temperature" a(r,t), provided that agas(t) = aggas + exp(—At) sin(wt). The function agas(t) has a finite limit when t ^ to, agas. In this case, there is a convergence to stationary temperature distribution. In Fig. 2b presented the dependence a(r,t), when agas(t) = sin(wt), i.e. agas(t) has no limit when t ^ to. As can be seen from graphics, solution with increase in time does not converge to a stationary regime. The situation is similar with other functions. The results of the numerical inverse Laplace transformation confirmed the conclusion that the stationary solution is the limit when large values of time. Thus, by changing the temperature environment, more precisely the function agas(t), can be set various modes of movement of the mixture in the cylinder with a free boundary.

Fig. 2. Temperature profile at different points in time: 1 — t = 0.21, 2 — t = 0.32, 3 — t = 0.68,

4 — t = oo

Conclusion

Studied partially-invariant solution of the axisymmetric motion of a binary mixture with a cylindrical free the boundary at small Marangoni numbers. The problem is reduced to initial-boundary value problem for parabolic equations. The images by Laplace obtained the exact analytical solution. Found the stationary solution of the problem and proved that it is the limit when t ^ to, if you satisfy certain conditions imposed by on the temperature outside. The examples of numerical reconstruction of fields the velocity, temperature and concentration, confirming the solution in stationary regime.

This work was supported by a grant from the Russian Foundation for basic research no. 1401-00067.

Fig. 3. Concentration profile at different points in time: 1 — t = 0.21, 2 — t = 0.32, 3 — t = 0.68,

4 — t = oo

Fig. 4. Velocity profile at different points in time: 1 — t = 0.21, 2 — t = 0.32, 3 — t = 0.68,

4 — t = oo

References

[1] L.K.Antonovsky, B.K.Kopbosynov, Unsteady thermocapillary drift of a viscous liquid drop, Zhurnal Prikladnoi Mekhaniki i Tekhnicheskoi Fiziki, (1986), no. 2, 59-64 (in Russian).

[2] O.V.Admaev, V.K.Andreev, Unsteady motion of the bubble under the action of thermocap-illary forces, Mathematicheskoe modelirovanie v mehanike, Krasnoyarsk, CVAC SB RAS, 1997, Dep. VINITI, 4-9 (in Russian).

[3] V.K.Andreev, About single invariant solution of the equations of diffusion, Materials of the 63rd scientific conference, "Gertsenovskie chteniya", Spb. RGPU im. A.I.Herzena, 2010, 8-17 (in Russian).

[4] V.K.Andreev, V.E.Zakataev, E.A.Rybicki, Thermocapillary instability, Novosibirsk, Nauka, 2000 (in Russian).

[5] L.V.Ovsyannikov, Group analysis of differential equations, Moscow, Nauka, 1978 (in Russian).

[6] V.K.Andreev, N.L.Sobachkina, Properties of solutions of initial-boundary value tasks arising from the movement of a binary mixture in cylindrical the pipe, Preprint no. 1. Krasnoyarsk, ICM SB RAS, 2009 (in Russian).

[7] M.Abramovic, I.Stigan, Handbook of special functions, Moscow, Nauka, 1979 (in Russian).

[8] V.I.Krylov, N.S.Skoblya, Methods of approximate Fourier transformation and conversion of the Laplace transform, Moscow, Nauka, 1974 (in Russian).

Движение бинарной смеси с цилиндрической свободной границей при малых числах Марангони

Виктор К. Андреев

Институт вычислительного моделирования СО РАН Академгородок, 50/44, Красноярск, 660036 Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041,

Россия

Наталья Л. Собачкина

Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041

Россия

Исследуется задача об осесимметрическом движении бинарной смеси с цилиндрической свободной границей при малых числах Марангони. В изображениях по Лапласу получено точное аналитическое решение. Найдено стационарное решение задачи и доказано, что оно является предельным с ростом времени, если выполнены определенные условия, налагаемые на внешнюю температуру. Даны примеры численного восстановления полей скорости, температуры и концентрации, которые хорошо соотносятся с теоретическими результатами.

Ключевые слова: бинарная смесь, свободная граница, стационарное решение, преобразование Лапласа, число Марангони.