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Reverse flow in magnetoconvection of two
immiscible fluids in a vertical channel
Alessandra Borrelli
Dipartimento di Matematica e Informatica,
Universita degli Studi di Ferrara,
via Machiavelli 30, 44121 Ferrara, Italy
Email: [email protected]
Giulia Giantesio
Dipartimento di Matematica e Fisica,
Universita Cattolica del Sacro Cuore,
via Musei 41, 25121 Brescia, Italy
Email: [email protected]
Maria Cristina Patria
Dipartimento di Matematica e Informatica,
Universita degli Studi di Ferrara,
via Machiavelli 30, 44121 Ferrara, Italy
Email: [email protected]
ABSTRACT
This paper concerns the study of the influence of an external magnetic field on the reverse
flow occurring in the steady mixed convection of two Newtonian immiscible fluids filling a verti-
cal channel under the Oberbeck-Boussinesq approximation. The two isothermal boundaries are
kept either at different or at equal temperatures. The velocity, the temperature and the induced
magnetic field are obtained analytically. The results are presented graphically and discussed for
various values of the parameters involved in the problem (in particular the Hartmann number
and the buoyancy coefficient) and are compared with those for a single Newtonian fluid. The
occurrence of the reverse flow is explained and carefully studied.
Nomenclature
bi constant fluid thermal diffusivity
C constant such that Pi = −Cx1 + p0i, i = 1, 2
d1 +d2 channel width
d =d2
d1ratio between the channel widths
Ei electric field
g = ge1 gravity acceleration
Hi total magnetic field
hi(y) dimensionless function describing the induced magnetic field defined in (14)
H0e2 external uniform magnetic field (H0 > 0)
H1(x2) induced magnetic field component in the x1−direction
ki constant fluid thermal conductivity
k =k2
k1ratio between the fluid thermal conductivities
M2i Hartmann number defined in (14)
Pi difference between the hydromagnetic pressure and the hydrostatic pressure
p0i arbitrary constant
Ti = Ti(x2) temperature
T0 reference temperature
Tw1,Tw2 uniform temperatures (Tw2 ≥ Tw1)
vi velocity field
ui(y) dimensionless function describing the velocity defined in (14)
V0 characteristic velocity
yi dimensionless transverse coordinate defined in (14)
Greek symbols
αi thermal expansion coefficient
γ constant defined in (15)
ηi magnetic diffusivity
ϑi(yi) dimensionless temperature defined in (14)
λi buoyancy coefficient defined in (14)
µi Newtonian viscosity coefficient (µi > 0)
µ =µ2
µ1ratio between the viscosities
µe magnetic permeability
ρ0i mass density at the temperature T0
σi electrical conductivity, 1σi
is called resistivity
σ =σ2
σ1ratio between the electrical conductivities
1 Introduction
The mixed magnetoconvection of an electrically conducting fluid in a channel under the effect of a
uniform transverse magnetic field is of relevant interest because of its many applications in industrial
and engineering processes such as geothermal and petroleum reservoirs, nuclear reactors, thermal insu-
lation, filtration, electronic packages and others. There are many studies of such flows in the literature
concerning a single Newtonian or non-Newtonian fluid as for example [1–3].
However, most of the problems relating to the practical applications involve multi-fluid flow situations
so that in the recent years several papers have been devoted to investigate the two-fluid flow in many
physical conditions ( [4–6]) and in particular in vertical, inclined and horizontal channels ( [7–14]).
Aim of our research is to solve the problem of the mixed convection of two immiscible Boussinesquian
electrically conducting fluids which fill a vertical channel (fully developed flow). A uniform magnetic
field is applied normal to the walls of the channel that are maintained at constant temperatures. As in
fully developed flow of a single fluid there is a constant pressure drop in the direction parallel to the
walls. We take into account the induced magnetic field unlike most of the papers concerning Mag-
netohydrodynamics. Since the dissipation terms are neglected in the energy equation we are able to
determine the analytical solution of the problem. In the statement of our study the interface retains its
plane shape because of the form of the velocities. Even if the geometry and the conditions of the problem
are simple, the study of this case of a more general problem is significant, not only for academic reasons,
but also from a practical point of view because it allows to obtain analytical solution and it can be the
starting point for stability analysis. Actually, the present paper shows the occurrence of the reverse flow
which is studied in details comparing the results with those concerning two fluids in the absence of the
external magnetic field and with those concerning a single fluid. The reverse flow phenomenon that is
relevant in many physical and engineering situations is related to the presence of the buoyancy forces.
It has been firstly discovered for a single Newtonian fluid in [15] and it occurs when the velocity of the
fluid considered changes its direction in the presence of a pressure drop (i.e. C 6= 0) for asymmetric
heating. In our analysis we obtain that the reverse flow is prevented by the external magnetic field. Few
Authors treat this effect ( [1, 9]) and find that the reverse flow increases as the strength of the magnetic
field increases. In [2, 3, 16] the opposite effect is discovered. This discrepancy is due to the fact that the
problem in [2, 3, 16] is solved determining the total magnetic field, unlike to the other papers present in
the literature. Actually, the constant electric field associated to the magnetic field in the Lorenz forces
is not neglected. This constant term has opposite sign of the buoyancy forces and contributes to prevent
the reverse flow near the coldest wall if C > 0 and near the hottest wall if C < 0.
The organization of the paper is the following.
In Section 2 we describe the physical situation and deduce the dimensionless boundary value problem
of ordinary differential equations governing the flow.
Section 3 is devoted to the symmetric heating, i.e. the case where the walls are kept at the same constant
temperature. We find that the temperature is constant (there is not heat transfer) and we determine the
analytical expressions of the velocities and the induced magnetic fields of the fluids.
The solution when the heating is asymmetric in the cases of natural convection and mixed convection is
given in section 4.
In section 5 we study the reverse flow: we begin by studying the motion of a single fluid in order to
understand this particular phenomenon. The buoyancy parameter λ (which has the same sign of C) plays
a fundamental role: if λ > 0, then the reverse flow appears near the coldest wall provided λ is greater
than a critical value λ∗ which is a strictly increasing function in the Hartmann number, while if λ < 0,
then the reverse flow arises near the hottest wall provided λ <−λ∗. These results show that the external
magnetic field tends to prevent the reverse flow for any strength of H0, as usual also in other physical
situations ( [2,3,16–18]). Then, we consider the double channel (d1 = d2) in the absence of the external
magnetic field. We find that for particular values of the material parameters the reverse flow occurs as
for a single fluid, but there is the possibility of a new situation: the reverse flow may occur near the to
the coldest/hottest wall if the buoyancy parameter is negative/positive.
The presence of the external magnetic field is considered in subsection 5.3; we find that in general the
reverse flow is prevented as for a single fluid with the exception of some cases depending on the values
of the material parameters and on the value of the Hartmann number. This new effect is characteristic
of the two-fluid model.
The trend of the solutions is plotted and commented for particular values of the material parameters in
sections 3, 4 and 5. Section 6 summarizes the results.
2 Problem description and governing equations
Let us consider the region S between two infinite rigid, fixed, non-electrically conducting vertical
plates Π1, Π2 separated by a distance d1 +d2 (see Figure 1).
We assume the regions outside the planes to be a vacuum (free space). The coordinate axes are fixed
in order to have
S = {(x1,x2,x3) ∈ R3 : (x1,x3) ∈ R
2, x2 ∈ [−d1,d2]},
Πi = {(x1,x2,x3) ∈ R3 : (x1,x3) ∈ R
2, x2 = (−1)idi}, i = 1,2 (1)
and x1-axis is vertical upward.
Our aim is to study the steady mixed convection in the fully developed flow of two immiscible New-
tonian electrically conducting fluids filling S in the Boussinesq approximation under the action of an
external uniform magnetic field H0e2, (H0 > 0), normal to isothermal planes Π1,2 kept at a temperature
Twi, i = 1,2, Tw1 ≤ Tw2.
Fluid 1 occupies the part S1 of S such that −d1 ≤ x2 ≤ 0, while fluid 2 occupies the part S2 of S
such that 0 ≤ x2 ≤ d2.
The flow in the absence of free electric charges under the Oberbeck-Boussinesq approximation is gov-
erned by
ρ0ivi ·∇vi =−∇Pi +µi△vi +µe(∇×Hi)×Hi −ρ0iαi(Ti −T0)g, (2)
∇ ·vi = 0, (3)
ηi△Hi = ∇× (Hi ×vi), (4)
∇ ·Hi = 0, (5)
∇Ti ·vi = bi△Ti, in Si, i = 1,2. (6)
All the material parameters are positive constants and µe is equal to the magnetic permeability of free
space. As it is usual in the Boussinesq approximation, in equation (6) the dissipative terms have been
neglected.
We search the velocities and the induced magnetic fields parallel to the rigid plates Πi; moreover
vi,Hi,Ti, i = 1,2 are such that:
vi = vi(x2)e1, Hi = H0e2 + Hi(x2)e1, Ti = Ti(x2), i = 1,2, (7)
where (e1,e2,e3) is the canonical base of R3. Under these hypotheses the interface retains its plane
shape.
We choose the reference temperature T0 =Tw1 +Tw2
2. If Tw1 = Tw2, then the heating is called sym-
metric, if Tw2 > Tw1, then the heating is called asymmetric.
By virtue of (7) and (2), we deduce Pi = Pi(x1) =−Cix1 + p0i (Ci, p0i some constants). We assume
that both fluids have a common modified pressure gradient so that C1 = C2 = C. We note that this
assumption is not restrictive because the features of the motion are not modified in a relevant manner.
Then the governing equations in Si become:
µiv′′i +µeH0H ′
i +ρ0iαi(Ti −T0)g = −C, (8)
ηiH′′i +H0v′i = 0, (9)
T ′′i = 0. (10)
To these equations we have to adjoin suitable boundary and interface conditions.
The boundary conditions are
v1(−d1) = v2(d2) = 0, H1(−d1) = H2(d2) = 0, T1(−d1) = Tw1, T2(d2) = Tw2 (11)
and the interface conditions have the form
v1(0) = v2(0), H1(0) = H2(0), T1(0) = T2(0), (12)
µ1v′1(0) = µ2v′2(0), H ′1(0) = H ′
2(0), k1T ′1(0) = k2T ′
2(0). (13)
Conditions (11) are standard; conditions (12) indicate that both fluids have the same velocity, the same
induced magnetic field, the same temperature at the interface x2 = 0 and conditions (13) express the
continuity of shear stress, curl of magnetic field and heat flux across the interface.
In the case of asymmetric heating we put
yi =x2
di
, ui(yi) =vi(diyi)
V0, hi(yi) =
Hi(diyi)
V0√
σ1µ1, ϑi(yi) =
Ti(diyi)−T0
Tw2 −Tw1,
M2i =
σi
µiµ2
eH20 d2
i , λi =ρ0iαig(Tw2 −Tw1)d
2i
µiV0, (14)
where V0 is a suitable reference velocity. If C 6= 0 we choose V0 =Cd2
1
µ1.
It is convenient to introduce the parameters
d =d2
d1, µ =
µ2
µ1, σ =
σ2
σ1, k =
k2
k1, γ =
ρ02α2
ρ01α1, (15)
so that
λ2 =γd2
µλ1, M2
2 =σd2
µM2
1. (16)
Moreover, for the sake of brevity, we omit the index i in y (see (14)); therefore y varies in [−1, 0] in S1
and varies in [0, 1] in S2.
By means of (14), (15) and (16), equations (8), (9), (10) can be written in dimensionless form:
u′′i +(
1√σµ
)i−1
Mih′i +λiϑi +
(d2
µ
)i−1
= 0, (17)
h′′i +(√
σµ)i−1 Miu′i = 0, (18)
ϑ′′i = 0, i = 1, 2. (19)
Of course, (17) holds if C 6= 0, otherwise the nonhomogeneous term disappears.
Boundary and interface conditions in dimensionless form become
u1(−1) = u2(1) = 0, h1(−1) = h2(1) = 0, ϑ1(−1) =−1
2,
ϑ2(1) =1
2, u1(0) = u2(0), u′1(0) =
µ
du′2(0),
h1(0) = h2(0), h′1(0) =1
dh′2(0), ϑ1(0) = ϑ2(0), ϑ′
1(0) =k
dϑ′
2(0). (20)
3 Symmetric heating: analytical solution
In order to solve problem (17), (18), (19), (20), it is convenient to consider some particular cases. In
this section we study the problem assuming that the walls are kept at the same temperature Tw1 = Tw2 =
T0.
The solution of equations (10) together with conditions (11), (12), (13) concerning Ti is
T1 = T2 = T0, (21)
i.e. the temperature of two fluids is not influenced by the motion and by the magnetic field.
By virtue of (21) we have that the buoyancy forces vanish and the motion is studied by means of the
system (17), (18) with λi = 0 together with conditions (20) concerning ui, hi.
Fist of all, we notice that if C = 0 (natural convection) the solution of the problem is
u1 = u2 = 0, h1 = h2 = 0, (22)
i.e. the fluids are at rest and the induced magnetic field is equal to zero.
If there is a pressure drop (C 6= 0), then the solution has the following form
u1(y) =C11
M1+C12 cosh(M1 y)+C13 sinh(M1 y)+
1
M21
,
h1(y) = − y
M1−C12 sinh(M1 y)−C13 cosh(M1 y)+C14, −1 ≤ y ≤ 0,
u2(y) =C21
M2+C22 cosh(M2 y)+C23 sinh(M2 y)+
1
σM21
,
h2(y) = − dy
M1−√
σµ[C22 sinh(M2 y)+C23 cosh(M2 y)−C24], 0 ≤ y ≤ 1, (23)
where the constants are given in Appendix A.
We now turn to analyze the behavior of the solution by using some remarks and some interesting
plots. In the following pictures we show the profiles of the functions in [−1,1], so that u = u1 and h = h1
in [−1,0], while u = u2 and h = h2 in [0,1].
By virtue of the Maxwell equations associated with the simplified Ohm’s law:
Ei =1
σi∇×Hi +µeHi ×vi (24)
and (7) we get that the electric fields associate to the magnetic fields are:
Ei =−µeH0V0
Mi
[(1√σµ
)i−1
h′i +Miui
]e3. (25)
They are orthogonal to the velocities and to the magnetic fields. Moreover from (18) and (23) they are
constant because the terms in square brackets are given by C11 and C21 respectively.
If the two fluids have the same material parameters
µ = σ = 1, M1 = M2 ≡ M
and d = 1, then we find
ui(y) =coshM− cosh(My)
M sinhM,
hi(y) =sinh(My)
M sinhM− y
M,
Ei =µeH0V0
M2
(1− M coshM
sinhM
)e3. (26)
Therefore the solution coincides with the Hartmann flow [19] (see Figures 2(a) and 2(b)), as one might
expect.
Figure 2 shows the influence of H0 and of µ on the velocities ui and on the induced magnetic fields
hi. In Figure 2(a) we see that the velocities decrease as M1 increases as we can expect because of the
Lorentz forces and tend to become constant; the corresponding induced magnetic fields are plotted in
Figure 2(b): the absolute value of h is an increasing function in M1 until a critical value M∗1 , if M1 > M∗
1
then |h| is a decreasing function. This behavior has been already noticed in [2,3]. Figure 2(c) shows that
the velocities decrease as µ increases and the velocity of the more viscous fluid has a grater decrease.
This effect is more pronounced with decreasing of d. In Figure 2(d) we see that the amplitude of the
interval in which h is positive reduces as µ increases. The influence on the motion of the parameter σ is
analogous to that of µ, but is less relevant for the velocities. For the sake of brevity, we omit the pictures
showing the influence of σ and d.
4 Asymmetric heating: analytical solution
In this section we consider the case of asymmetric heating, i.e. Tw2 > Tw1.
4.1 Natural convection
First of all we determine the solution of the problem when C = 0 (natural convection). Therefore we
have to solve system (17), (18), (19) where the nonhomogeneous term vanishes and to add conditions
(20). The solution is the following:
u1(y) = λ1
{C∗
11
M1+C∗
12 cosh(M1 y)+C∗13 sinh(M1 y)+
2ky+ k−d
2M21(k+d)
}, (27)
h1(y) = −λ1
{C∗
12 sinh(M1 y)+C∗13 cosh(M1 y)−C∗
14 +ky2 +(k−d)y
2M1(k+d)
}, (28)
ϑ1(y) =2ky+ k−d
2(k+d), −1 ≤ y ≤ 0, (29)
u2(y) = λ1
{C∗
21
M2+C∗
22 cosh(M2 y)+C∗23 sinh(M2 y)+
γ(2dy+ k−d)
2σM21(k+d)
}, (30)
h2(y) = −λ1√
σµ
{C∗
22 sinh(M2 y)+C∗23 cosh(M2 y)−C∗
24 +dγ[dy2 +(k−d)y]
2√
σµM1(k+d)
}, (31)
ϑ2(y) =2dy+ k−d
2(k+d), 0 ≤ y ≤ 1, (32)
where the constants are given in Appendix B.
The electric fields given by (25) are still constants because the terms in square brackets are now
given by λ1C∗11 and λ1C∗
21 respectively.
If the fluids have the same material parameters and d = 1, then the solution coincides with that given
in [2] for a single fluid in the case of natural convection:
θi(y) =y
2,
ui(y) =−λ[sinh(My)− ysinhM]
2M2 sinhM,
hi(y) =λ[cosh(My)− coshM]
2M2 sinhM+
λ
4M(1− y2),
Ei(y) = 0.
We notice that unlike the case of a single fluid, when the channel is filled by two fluids with different
materials parameters, as we have underlined, the electric fields in general do not vanish because of the
electromagnetic interaction between the fluids.
From Figures 3(a) and 3(b) we see that the velocities point in opposite directions and their absolute
value decreases as M1 increases; the induced magnetic fields are positives and present a maximum,
whose value increases until a critical value of M1 and then decreases ( [2, 3]). Figures 3(c), 3(d) and
3(e), 3(f) show the influence of the parameters k and γ. We notice that the increase of k determines
an increase of the velocities that can change sign; moreover hi have a maximum and minimum. The
influence of γ is analogous. Increases of the parameters d and σ produce an increase of the absolute
value of the velocities and hi are positive and increasing, while an increase of µ produces for the first
fluid a slight increase of the absolute value of the velocity and for the other fluid a significant increase
of the velocity (the pictures are omitted for the sake of brevity).
4.2 Mixed convection
At this stage we can easily determine the solution when C 6= 0 by virtue of linearity of the problem.
Of course the expressions of ϑ1, ϑ2 are given in (29) and (32), while ui,hi are
u1(y) =C11 +λ1C∗
11
M1+(C12 +λ1C∗
12)cosh(M1 y)+(C13 +λ1C∗13)sinh(M1 y)
1
M21
+λ12ky+ k−d
2M21(k+d)
,
h1(y) = −(C12 +λ1C∗12)sinh(M1 y)− (C13 +λ1C∗
13)cosh(M1 y)+(C14 +λ1C∗14)−
y
M1
−λ1ky2 +(k−d)y
2M1(k+d), −1 ≤ y ≤ 0,
u2(y) =C21 +λ1C∗
21
M2+(C22 +λ1C∗
22)cosh(M2 y)+(C23 +λ1C∗23)sinh(M2 y)+
1
σM21
+λ1γ(2dy+ k−d)
2σM21(k+d)
,
h2(y) = −√σµ[(C22 +λ1C∗
22)sinh(M2 y)+(C23 +λ1C∗23)cosh(M2 y)− (C24 +λ1C∗
24)]
− dy
M1−λ1
dγ[dy2 +(k−d)y]
2M1(k+d), 0 ≤ y ≤ 1. (33)
We conclude this section with some remarks.
The electric fields Ei are constant and they are given by summing the fields obtained in section 3
and in subsection 4.1
Outside the planes where there is a vacuum, by virtue of the usual transmission conditions for the
electromagnetic field across Π1,2, we have H = H0e2 and
E =−µeH0V0
M1(C11 +λ1C∗
11)e3 −∞ < x2 <−d1,
E =−µeH0V0
M2(C21 +λ1C∗
21)e3 d2 < x2 <+∞.
If the fluids have the same material parameters and d = 1, then the solution coincides with that given
in [2] for a single fluid. Unlike the case of two fluids, if we have a single fluid then E does not
depend on the buoyancy coefficient λ.
The skin friction τi at both plates are given by
τi = µiV0
diu′i((−1)i)e1.
The expression of τi is related to the occurrence of the reverse flow, as we will see in the next section.
The parameters d, M1 and µ produce the same effects on the flow as in the symmetric heating, so
that we omit the pictures. The parameter σ reduces, but not in a significant way, the velocities (see
Figure 4(a)); from Figure 4(b) we see that h presents maximum and minimum more pronounced when
σ increases. The effect of k on the velocities is the opposite of σ but more relevant. h has maximum
and minimum more pronounced (see Figures 4(c) and 4(d)). The parameter γ has the same effect of
k. Finally, in the Figures 4(e), 4(f) we represent the effect of the positive buoyancy parameter. If λ
increases, then the velocity decreases for the the first fluid and increases for the second one, while hi
increases for both fluids. If λ is negative one has the opposite effects. Figure 4(e) shows that for λ = 10
there is the phenomenon of the reverse flow near the coldest wall. In the next section we will study the
influence of the material parameters on this phenomenon.
We observe that our results are in agreement with those of previous papers and in particular with [7].
The Author of [7] is the only one, to our knowledge, that suppose both fluids electrically conducting,
but he neglects the induced magnetic field.
Finally, we see that if the electrical conductivities σi → +∞ (i.e. the fluid are ideal electrical con-
ductors) then the velocities and the induced magnetic fields tend to zero.
The aim of the previous figures is to show how the functions describing the velocities and the
magnetic fields change when one of the parameters varies. This is very interesting in order to study
the trend of the functions, but in reality when the channel is occupied of two real fluids the val-
ues of ki,αi,µi,µe,ρ01,σi are fixed because they depend on the kind of fluid studied, while we can
change the intensity of the magnetic field and of the temperature of the two walls. The simulation
of the presence of two real fluids is very interesting for real applications, but show similar results to
the one studied in this section and in the following, as can be easily seen in Figure 5. These plots
show the dimensionless profile of the velocities and of the magnetic fields when M1 and λ1 varies
and mercury and seawater occupied Π1 and Π2, respectively. For both of these fluids we have µe =
1.2510−6Hm−1, which is the vacuum permeability, while at T0 = 20◦C we have k1 = 8.34W Km−1, k2 =
0.596W Km−1, α1 = 1.810−4K−1, α2 = 3.210−4K−1
, ρ01 = 13354Kgm−3, ρ02 = 1027Kgm−3, µ1 =
1.5510−3Kgm−1s−1, µ2 = 10−3Kgm−1s−1, σ1 = 106Sm−1, σ2 = 5Sm−1. We notice that since the two
fluids have different parameters Figures 5(c) and 5(d) are not symmetric to 5(e), 5(f).
5 Reverse flow: results and discussion
In this section we examine and discuss the phenomenon of the reverse flow.
The reverse flow occurs when the velocity of the fluid taken into account changes its direction. This
phenomenon has been first discovered for the mixed convection of a single fluid in a vertical channel
with different wall temperatures in [15] and in recent years in [1], [2] when a uniform external magnetic
field is impressed. The reverse flow is a feature of the mixed convection (C 6= 0) in the case of asym-
metric heating .
In order to understand this phenomenon, we begin by studying the simplest case of a single fluid.
As it is well known, in the Poiseuille flow the pressure gradient (∇P =−Ce1) has opposite sign respect
to v, v = ve1. If the walls have different temperatures (Tw1 < Tw2), then the reverse flow can appear
thanks to the presence of the buoyancy forces. The flow is now determined by ∇P+ρ0α(T −T0)g.
Therefore, if C > 0, then near the coldest wall
C+ρ0α(T −T0)g ≡ F
may become negative and v becomes negative so that the reverse flow appears near the coldest wall.
If C < 0 then F may become positive near the hottest wall determining the occurrence of the reverse
flow near the hottest wall.
5.1 Reverse flow for a single fluid
In [2] it has been proved that the velocity of a Newtonian fluid filling a vertical channel in the
physical situation described in section 2, supposing d1 = d2, is given by
u(y) =coshM− cosh(My)
M sinhM− λ
sinh(My)− ysinhM
2M2 sinhM−1 ≤ y ≤ 1. (34)
The occurrence of reverse flow is related to the sign of u(0) and to the sign of u′(−1),u′(1). Precisely,
we have to impose the condition
u′(−1)u′(1)> 0. (35)
In this case we have
u(0) =coshM−1
M sinhM> 0, (36)
u′(−1) = 1−λM coshM− sinhM
2M2 sinhM, u′(1) =−1−λ
M coshM− sinhM
2M2 sinhM. (37)
where λ =ρ0αg(Tw2 −Tw1)
C.
Since u(0)> 0, if u′(−1)< 0 and u′(1)< 0, then the reverse flow can occurs only in (−1, 0), i.e. near
the coldest wall; if u′(−1)> 0 and u′(1)> 0, then the reverse flow can occurs only in (0, 1), i.e. near the
hottest wall. Moreover, the sign of u′(−1),u′(1) depends on the sign of λ so that we have to distinguish
between λ > 0 ⇔C > 0 and λ < 0 ⇔C < 0.
A) λ > 0.
By studying the sign of (37) we obtain that the only possibility of arises of reverse flow when u′(−1)<
0,u′(1)< 0 for all values of λ such that
λ > λ∗ =2M2 sinhM
M coshM− sinhM.
Therefore the reverse flow happens in (−1, 0) (near to the coldest wall).
B) λ < 0.
In this case the reverse flow appears when u′(−1)> 0,u′(1)> 0 for all values of λ such that
λ <−λ∗.
Therefore the reverse flow occurs in (0, 1) (near to the hottest wall).
We notice that λ∗ is a strictly increasing function in M and tends to 6 (the critical value of λ in the
absence of external magnetic field) as M → 0+. Therefore, as well as in [2], we find that the presence of
the external magnetic field tends to prevent the occurrence of the reverse flow. In order to explain this
effect we notice that, to the mechanical force Fe1, it is now necessary to adjoin the component along
Ox1 of the Lorentz forces given by µeσH0(−E0 +µevH0) where
E0 =C
σµeH0
(1−M
coshM
sinhM
).
Taking into account the expression of E0, we deduce that −E0 has the same sign of C. Hence the
presence of this term in the Lorentz forces explains why the reverse flow is prevented. We underline that
this term is present because we have not neglected the induced magnetic field unlike most of the papers
concerning magnetoconvection of a single fluid or two fluids in a vertical channel ( [1, 7–9]). Actually
in the cited papers the constant term −µeσH0E0 does not appear so that the Authors conclude that the
magnetic field favors the reverse flow.
The reverse flow is shown in Figure 6.
5.2 Two fluids in the absence of the external magnetic field
We begin by considering two fluids in the absence the external magnetic field. Since we are inter-
ested to the influence on the reverse flow of the material parameters, we fix the geometry of the problem
by putting d1 = d2.
In this case we have:
u1(y) =1
2(µ+1)[−(µ+1)y2+(1−µ)y+2]+
λ1
12(k+1)(µ+1){(µ+1)[−2ky3+3(1− k)y2]
+ [γ(3k−1)−µ(k−3)]y+ γ(3k−1)+ k−3}
u2(y) =1
2µ(µ+1)[−(µ+1)y2 +(1−µ)y+2µ]+
λ1
12µ(k+1)(µ+1){−γ(µ+1)[2y3 +3(k−1)y2]
+ [γ(3k−1)−µ(k−3)]y+µ[γ(3k−1)+ k−3]}. (38)
and
u′1(−1) =3+µ
2(µ+1)+
λ1
12(k+1)(µ+1)[γ(3k−1)−µ(k+3)−6]
u′2(1) = − 1+3µ
2µ(µ+1)− λ1
12µ(k+1)(µ+1)[γ(6kµ+3k+1)+µ(k−3)]. (39)
Of course (35) is replaced by
u′1(−1)u′2(1)> 0. (40)
Unlike the case of one fluid, the sign of u1(0) = u2(0) in general is not positive, but depends on the
material parameters.
In order to compare the reverse flow of two fluids with that of one only we vary one material parameter
in (38), (39) and fix the others equal to 1.
We proceed analogously to the previous subsection.
Case I: γ = k = 1.
In this case
u1(0) = u2(0) =1
µ+1> 0. (41)
Since u1(0) = u2(0)> 0, we will see that the phenomenon of the reverse flow occurs in the same way
as in subsection 5.1. More precisely:
IA) λ1 > 0.
The study of inequality (40) together with (41) shows that the reverse flow appears in (−1,0) (near the
coldest wall) for all λ1 >3(µ+3)
µ+1=: λ∗
1IA.
It is interesting to observe that if µ = 1 then we find again the result of a single fluid, i.e. there is
the reverse flow for all λ > λ∗ = 6.
IB) λ1 < 0.
In this subcase the reverse flow appears in (0,1) (near the hottest wall) for all λ1 <−3(1+3µ)
µ+1=: λ∗
1IB
and if µ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow for all
λ <−λ∗ =−6.
It is interesting to compare these results with those of a single fluid : if µ2 < µ1 then the reverse flow is
prevented in (−1, 0), while is favored in (0, 1).
Case II: µ = k = 1.
Now
u1(0) = u2(0) =1
24[12+λ1(γ−1)], (42)
so that the sign depends on λ1 and γ.
IIA) λ1 > 0.
The study of inequality (40) together with (42) shows that if u′1(−1) > 0, u′2(1) > 0 then the reverse
flow does not appear. If u′1(−1)< 0, u′2(1)< 0 then we deduce that the reverse flow appears in (−1, 0)
(near the coldest wall, u1(0)> 0) if
γ < 1 and24
5− γ< λ1 <
12
1− γor 1 ≤ γ < 5 and λ1 >
24
5− γ.
If γ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow near the coldest
wall for all λ > 6 and if γ2 > γ1, then the reverse flow is prevented.
Moreover we have a situation which does not appear for a single fluid, i.e. u1(0) < 0 and the reverse
flow occurs in (0, 1) for the second fluid if
γ <1
5and
12
1− γ< λ1 <
24
1−5γor
1
5< γ < 1 and λ1 >
12
1− γ.
IIB) λ1 < 0.
If u′1(−1) < 0, u′2(1) < 0 then the reverse flow does not appear. If u′1(−1) > 0, u′2(1) > 0 the reverse
flow appears in (0, 1) (near the hottest wall, u1(0)> 0) if
γ > 1 and12
1− γ< λ1 <
24
1−5γor
1
5< γ ≤ 1 and λ1 <
24
1−5γ.
If γ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow near the hottest wall
for all λ1 <−6 and if γ2 < γ1 then the reverse flow is prevented.
Moreover also in this subcase we have a situation which does not appear for a single fluid, i.e. u1(0)< 0
and the reverse flow occurs in (−1, 0) for the first fluid if
γ > 5 and24
5− γ< λ1 <
12
1− γor 1 < γ < 5 and λ1 <
12
1− γ.
Finally, me notice that if λ1 =12
1− γthen u1(0) = 0 so that u1 and u2 have opposite sign.
Figures 7 show the reverse flows for different values of γ and λ.
Case III: γ = µ = 1.
Now
u1(0) = u2(0) =1
6(k+1)[3(k+1)+λ1(k−1)]. (43)
IIIA) λ1 > 0.
In this subcase the reverse flow appears in (−1, 0) (near the coldest wall, (u1(0)> 0) if
k < 1 and12(k+1)
5− k< λ1 <
3(k+1)
1− kor 1 ≤ k < 5 and λ1 >
12(k+1)
5− k.
If k = 1 then we find again the result of a single fluid, i.e. there is the reverse flow near the coldest wall
for all λ1 > 6 and if k2 > k1 then the reverse flow is prevented.
Moreover we have again a situation which does not appear for a single fluid, i.e. u1(0) < 0 and the
reverse flow occurs in (0, 1) for the second fluid if
k <1
5and
3(k+1)
1− k< λ1 <
12(k+1)
1−5kor
1
5< k < 1 and λ1 >
3(k+1)
1− k.
IIIB) λ1 < 0.
In this subcase the reverse flow appears in (0, 1) (near the hottest wall, (u1(0)> 0) if
1
5< k ≤ 1 and λ1 <
12(k+1)
1−5kor k > 1 and
3(k+1)
1− k< λ1 <
12(k+1)
1−5k.
If k = 1 then we find again the result of a single fluid, i.e. there is the reverse flow near the hottest wall
for all λ1 <−6 and if k2 < k1 then the reverse flow is prevented.
Further we find u1(0)< 0 and the reverse flow in (−1, 0) for the first fluid if
k > 5 and12(k+1)
5− k< λ1 <
3(k+1)
1− kor 1 < k < 5 and λ1 <
3(k+1)
1− k.
Finally, me notice that if λ1 =3(k+1)
1− kthen u1(0) = 0, so that u1 and u2 have opposite sign.
We omit the figures because the behavior of the reverse flow is analogous to the previous case.
We remark that in Case I, in which only µ varies, the situations are the same as for a single fluid because
the interaction between the two fluid does not modify the sign of u1(0) = u2(0) given by (41). In the
other cases γ and k vary: these parameters are related to the thermal properties of the fluids so that it is
reasonable to expect that they may modify the sign of u1(0) = u2(0) (see (42), (43)).
5.3 Reverse flow for two fluids in the presence of the external magnetic field
Now we take into account the presence of the external magnetic field so that we need
u′1(−1), u′2(1), u1(0):
u′1(−1) = M1(−C12 sinhM1 +C13 coshM1)+λ1
[−M1C∗
12 sinhM1 +M1C∗13 coshM1 +
k
M21(k+d)
],
u′2(1) = M2(C22 sinhM2 +C23 coshM2)+λ1
[M2C∗
22 sinhM2 +M2C∗23 coshM2 +
dγ
M21σ(k+d)
],
u1(0) = u2(0) =C11
M1+C12 +
1
M21
+λ1
[C∗
11
M1+C∗
12 +k−d
2M21(k+d)
], (44)
where the coefficients Ci j,C∗i j, i = 1,2, j = 1,2,3 are given in the Appendices.
We examine the reverse flow in some particular cases.
Case IM: d = k = γ = 1, µ = σ ⇒ M1 = M2 ≡ M.
In this case
u1(0) = u2(0) =2(coshM−1)
(1+µ)M sinhM> 0 (45)
and
u′1(−1) = 1+(µ−1)(1− coshM)coshM
(1+µ)sinh2 M+
λ1
2M2
sinhM−M coshM
sinhM,
u′2(1) =1
µ
[−1+
(µ−1)(1− coshM)coshM
(1+µ)sinh2 M
]+
λ1
2µM2
sinhM−M coshM
sinhM. (46)
IAM) λ1 > 0.
If u′1(−1)> 0, u′2(1)> 0, then the reverse flow does not appear.
If u′1(−1)< 0, u′2(1)< 0, then we deduce that the reverse flow appears in (−1, 0) (near the coldest wall)
for all
λ1 >2M2[2sinh2 M+(µ−1)(coshM−1)]
(1+µ)sinhM(M coshM− sinhM)=: λ∗
1IAM. (47)
If µ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow for all λ > λ∗ =2M2 sinhM
M coshM− sinhM.
IBM) λ1 < 0.
In this subcase the reverse flow appears in (0,1) (near the hottest wall) when u′1(−1)> 0, u′2(1)> 0 for
all
λ1 <2M2[−2µsinh2 M+(µ−1)(coshM−1)]
(1+µ)sinhM(M coshM− sinhM)=: λ∗
1IBM. (48)
and if µ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow for all λ <−λ∗.
In order to study the influence of the external magnetic field on the reverse flow, we compare the critical
value of λ1 given in (47) with that one given in case IA. By studying Λ ≡ λ∗1IAM
−λ∗1IA as function in
M, we see that Λ is positive for all value of M if
µ <27
11= lim
M→0+
2M2(coshM−1)−4M2 sinh2 M+9sinhM(M coshM− sinhM)
2M2(coshM−1)−3sinhM(M coshM− sinhM),
so that the reverse flow is prevented for all value of M.
If µ >27
11, then Λ is positive only if M is sufficiently big i.e. M > M∗, so that H0 prevents the reverse
flow only if its strength is sufficiently big (see Table 1). In the case of reverse flow near the hottest
wall, we have to compare the critical value of λ1 given in (48) with that one given in case IB. From the
comparison between λ∗1IBM
and λ∗1IB, we obtain that if
µ >11
27= lim
M→0+
2M2(coshM−1)−3sinhM(M coshM− sinhM)
2M2(coshM−1)−4M2 sinh2 M+9sinhM(M coshM− sinhM),
then the reverse flow is prevented for all value of M.
If µ <11
27, then H0 prevents the reverse flow only if its strength is sufficiently big, i.e. M > M∗ (see
Table 1).
In Figures 8 some subcases are illustrated.
Case IIM: d = k = µ = σ = 1 ⇒ M1 = M2 ≡ M.
In this case
u1(0) = u2(0) =coshM−1
M sinhM+λ1
γ−1
8M3 sinhM[(M2 +4)(coshM−1)−2M sinhM] (49)
so that the sign depends on the sign of λ1 and (γ−1) because the other terms are positive.
Moreover, from (44) we deduce
u′1(−1) =1+λ1
8M2 sinhM
{γ[(M2 +2)sinhM−2M coshM
]− (M2 −2)sinhM−2M coshM
},
u′2(1) =−1− λ1
8M2 sinhM
{γ[(M2 −2)sinhM+2M coshM
]− (M2 +2)sinhM+2M coshM
}.
(50)
IIAM) λ1 > 0.
We obtain results analogous to those of Case IIA replacing 5 withb
a, 24 with
c
a, 12 with f where
a = (M2 +2)sinhM−2M coshM, b = (M2 −2)sinhM+2M coshM,
c = 8M2 sinhM, f =8M2(coshM−1)
(M2 +4)(coshM−1)−2M sinhM. (51)
We have
0 < a < b < c, 0 < f < 12 ∀M > 0. (52)
If γ = 1 then we find again the result of a single fluid, i.e. there is the reverse flow near the coldest wall
for all λ1 > λ∗ =2M2 sinhM
M coshM− sinhM.
The reverse flow near the coldest wall is prevented by the presence of the external magnetic field be-
causeb
a< 5 and if γ <
b
athen
c
b−aγ>
24
5− γ. Also in the case γ2 > γ1 the reverse flow is prevented
becausec
b−aγ> λ∗.
Since f < 12, we have that if u1(0)< 0 then the presence of the magnetic field tends to favor the reverse
flow for the second fluid.
IIBM) λ1 < 0.
Also in this subcase, with the substitutions mentioned previously, we find again results analogous to
those of subcase IIB.
If γ = 1 then we obtain the result of a single fluid: there is the reverse flow near the hottest wall for all
λ1 <−λ∗.
The reverse flow near the hottest wall is prevented by the presence of the external magnetic field because
a
b>
1
5and if γ >
a
bthen
c
a−bγ<
24
1−5γ. Also in the case γ2 < γ1 the reverse flow is prevented because
c
a−bγ<−λ∗.
Since f < 12, we have that if u1(0)< 0 then the presence of the magnetic field tends to favor the reverse
flow for the first fluid.
Finally, we notice that if λ1 =f
1− γthen u1(0) = 0 so that u1 and u2 have opposite sign.
Figures 9 show some physical situations.
To conclude the section we remark that one can study the case in which k varies and the others parame-
ters are equal to 1, obtaining results analogous to case III. We omit the details for the sake of brevity.
Finally, in addition to the remark on the Lorentz forces at the end of subsection 5.1, we notice that
for two fluids the constant electric fields depend on λ1 (see (25)). This feature, which is typical for
two-fluids flow, explains why for some values of the parameters the external magnetic field favors the
reverse flow, unlike what happens for a single fluid whose electric field is independent of the buoyancy
coefficient.
6 Conclusions
The mixed convection of two immiscible Boussinesquian electrically conducting Newtonian fluids
in a vertical channel is studied. A uniform magnetic field is applied normal to the walls of the channel
that are maintained at constant temperatures. Since we have neglected the dissipation terms in the en-
ergy equation, we have found the analytical solution of the boundary value problem which governs the
motion. The velocities, the temperatures, the induced magnetic and the electric fields for the two fluids
depend on the parameters M1, k, d, σ, µ and γ.
The behavior of the solution is studied and represented graphically for various values of the parameters,
with special attention to the occurrence of the reverse flow. Actually, this phenomenon is studied and
discussed when one of the parameters varies and the other are fixed. The obtained results can be sum-
marized as in Figures 10 and 11, where we have used the following notation:
no r.: no reverse flow
Tw1: reverse flow occurs near the coldest wall
Tw2: reverse flow occurs near the hottest wall
f 1: reverse flow occurs in (−1,0) for the first fluid
f 2: reverse flow occurs in (0,1) for the second fluid
P: the magnetic field tends to prevent the occurrence of the reverse flow
F: the magnetic field tends to favor the occurrence of the reverse flow
Figure 10 refers to the case of a single fluid, two fluids in absence or presence of the external magnetic
field when µ (and M) varies (case I and IM), while Figure 11 summarizes the results of case II and IIM
when γ changes (analogous schematisation can be made when k varies as in case III).
By taking into account the expressions of the buoyancy and Lorentz forces, we have explained why
the reverse flow appears and how the magnetic field influences this phenomenon.
Acknowledgements
This work was supported by National Group of Mathematical Physics (GNFM-INdAM).
The authors are thankful to the compent reviewers for their useful suggestions.
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Appendix A: Constants in Symmetric heating
C11
M1=−σcoshM1C22 +
√σµsinhM1C23 −
1
M21
,
C12 = σC22, C13 =√
σµC23,
C14 =−σsinhM1C22 +√
σµcoshM1C23 −1
M1,
C24 =−√
σ
µsinhM1C22 + coshM1C23 −
1√σµM1
,
C21
M2= (σ−σcoshM1 −1)C22 +
√σµsinhM1C23 −
1
σM21
,
C22 =1+d√σµM1∆
(sinhM2 +√
σµsinhM1),
C23 =1+d√σµM1∆
[1− coshM2 −σ(1− coshM1)],
∆ = 2(1− coshM2)+2σ(1− coshM1)− (1+σ)(1− coshM1)(1− coshM2)−
(1+µ)
√σ
µsinhM1 sinhM2 < 0. (53)
Appendix B: Constants in Asymmetric heating
C∗11
M1=−σcoshM1C∗
22 +√
σµsinhM1C∗23 +
1
2M21
−AcoshM1 +BsinhM1,
C∗12 = σC∗
22 +A, C∗13 =
√σµC∗
23 +B,
C∗14 =−σsinhM1C∗
22 +√
σµcoshM1C∗23 +
d
2M1(k+d)−AsinhM1 +BcoshM1,
C∗24 =−
√σ
µsinhM1C∗
22 + coshM1C∗23 +
d√σµ2M1(k+d)
+−AsinhM1 +B(coshM1 −1)√
σµ,
C∗21
M2= (σ−σcoshM1 −1)C∗
22 +√
σµsinhM1C∗23 +A(1− coshM1)+BsinhM1 +
2kσ− γ(k−d)
2σM21(k+d)
,
C∗22 =− 1
∆{(coshM2 − coshM1)
[A(1− coshM1)+BsinhM1 +
kσ+dγ
M21σ(k+d)
]+
(sinhM2√
σµ+ sinhM1)
[−AsinhM1 +B(coshM1 −1)+
d(1− kγ)
2M1(k+d)
]},
C∗23 =
1
∆{[coshM2 −1+σ(1− coshM1)]
[−AsinhM1 +B(coshM1 −1)+
d(1− kγ)
2M1(k+d)
]1√σµ
+
(sinhM2 +
√σ
µsinhM1)
[A(1− coshM1)+BsinhM1 +
kσ+dγ
M21σ(k+d)
]},
A =(k−d)(γ−1)
2M21(k+d)
, B =µγ−σk
σM31(k+d)
(54)
List of Figures
1 Physical configuration and coordinate system. . . . . . . . . . . . . . . . . . . . . . . 28
2 Symmetric heating (Tw1 = Tw2 = T0): effects of H0 and µ on the motion. . . . . . . . . 29
3 Asymmetric heating - natural convection: influence of M1, k and γ on the velocities and
on the induced magnetic fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4 Asymmetric heating - mixed convection: influence of σ, k and λ on the velocities and
on the induced magnetic fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5 Asymmetric heating - mixed convection: channel occupied by mercury and seawater. . 32
6 Reverse flow for a single fluid. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
7 Reverse flow in case II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8 Reverse flow in case IM when λ1 and M vary. . . . . . . . . . . . . . . . . . . . . . . 34
x1
x2
T w1
T w2
x 2=d 2
x 2=-d 1
S1 S2
H0e2
x3
g
Fig. 1. Physical configuration and coordinate system.
Table 1. Critical value of M starting from which the magnetic field prevents the occurrence of the reverse flow.
λ1 positive λ1 negative
µ M∗ µ M∗
3.00 1.9804 0.05 16.1841
4.00 3.3967 0.10 8.6024
5.00 4.4504 0.20 4.4504
10.00 8.6024 0.30 2.5289
9 Reverse flow in case IIM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
10 Behavior of the reverse flow for a single fluid or when µ varies for two fluids. . . . . . 36
11 Behavior of the reverse flow when γ varies for two fluids. . . . . . . . . . . . . . . . . 36
List of Tables
1 Critical value of M starting from which the magnetic field prevents the occurrence of
the reverse flow. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
y
u
σ =1, µ =1, d =1.
M1 = 1M1 = 2M1 = 4M1 = 10
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.2
−0.15
−0.1
−0.05
0
0.05
0.1
0.15
y
h
σ =1, µ =1, d =1.
M1 = 1M1 = 2M1 = 4M1 = 10
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y
u
M1 =1, σ =1, d =1.
µ = 1µ = 2µ = 3µ = 10
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.06
−0.04
−0.02
0
0.02
0.04
0.06
y
h
M1 =1, σ =1, d =1.
µ = 1µ = 2µ = 3µ = 10
(d)
Fig. 2. Symmetric heating (Tw1 = Tw2 = T0): effects of H0 and µ on the motion.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.03
−0.02
−0.01
0
0.01
0.02
0.03
y
u
σ =1, µ =1, k =1, λ1 =1, d =1, γ =1.
M1 = 1M1 = 2M1 = 4M1 = 10
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10
0.005
0.01
0.015
0.02
0.025
0.03
0.035
y
h
σ =1, µ =1, k =1, λ1 =1, d =1, γ =1.
M1 = 1M1 = 2M1 = 4M1 = 10
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.15
−0.1
−0.05
0
0.05
0.1
y
u
M1 =1, σ =1, µ =1, λ1 =1, d =1, γ =1.
k = 0.1k = 0.5k = 1k = 2
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.015
−0.01
−0.005
0
0.005
0.01
0.015
0.02
0.025
0.03
y
h
M1 =1, σ =1, µ =1, λ1 =1, d =1, γ =1.
k = 0.1k = 0.5k = 1k = 2
(d)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.05
0
0.05
0.1
0.15
0.2
y
u
σ =1, µ =1, k =1, λ1 =1, d =1, M1 =1.
γ = 1γ = 2γ = 3γ = 4
(e)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.01
0
0.01
0.02
0.03
0.04
0.05
y
h
σ =1, µ =1, k =1, λ1 =1, d =1, M1 =1.
γ = 1γ = 2γ = 3γ = 4
(f)
Fig. 3. Asymmetric heating - natural convection: influence of M1, k and γ on the velocities and on the induced magnetic fields.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y
u
M1 =1, µ =1, k =1, λ1 =1, d =1, γ =1.
σ = 0.5σ = 1σ = 2σ = 6
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.4
−0.35
−0.3
−0.25
−0.2
−0.15
−0.1
−0.05
0
0.05
0.1
y
h
M1 =1, µ =1, k =1, λ1 =1, d =1, γ =1.
σ = 0.5σ = 1σ = 2σ = 6
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y
u
M1 =1, σ =1, µ =1, λ1 =1, d =1, γ =1.
k = 0.1k = 0.5k = 1k = 2
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
y
h
M1 =1, σ =1, µ =1, λ1 =1, d =1, γ =1.
k = 0.1k = 0.5k = 1k = 2
(d)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
y
u
σ =1, µ =1, k =1, γ =1, d =1, M1 =1.
λ1 = 1λ1 = 2λ1 = 3λ1 = 10
(e)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
−0.05
0
0.05
0.1
0.15
0.2
0.25
y
h
σ =1, µ =1, k =1, γ =1, d =1, M1 =1.
λ1 = 1λ1 = 2λ1 = 3λ1 = 10
(f)
Fig. 4. Asymmetric heating - mixed convection: influence of σ, k and λ on the velocities and on the induced magnetic fields.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y
u
σ = 5× 10−6, µ =0.645, k =0.0715, λ1 =1, d =1, γ =0.137.
M1 = 1M1 = 2M1 = 5M1 = 10
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
y
h
σ = 5× 10−6, µ =0.645, k =0.0715, λ1 =1, d =1, γ =0.137.
M1 = 1M1 = 2M1 = 5M1 = 10
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1.4
−1.2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
y
u
σ = 5× 10−6, µ =0.645, k =0.0715, γ =0.137, d =1, M1 =1.
λ1 = 1λ1 = 4λ1 = 7λ1 = 12
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.2
−0.15
−0.1
−0.05
0
0.05
0.1
0.15
y
h
σ = 5× 10−6, µ =0.645, k =0.0715, γ =0.137, d =1, M1 =1.
λ1 = 1λ1 = 4λ1 = 7λ1 = 12
(d)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.5
0
0.5
1
1.5
2
2.5
y
u
σ = 5× 10−6, µ =0.645, k =0.0715, γ =0.137, d =1, M1 =1.
λ1 = −1λ1 = −4λ1 = −7λ1 = −12
(e)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
y
h
σ = 5× 10−6, µ =0.645, k =0.0715, γ =0.137, d =1, M1 =1.
λ1 = −1λ1 = −4λ1 = −7λ1 = −12
(f)
Fig. 5. Asymmetric heating - mixed convection: channel occupied by mercury and seawater.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
y
u
M =1.
λ1 = 1λ1 = 6λ1 = 6.4λ1 = 7λ1 = 10
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
y
u
M =1.
λ1 = −1λ1 = −6λ1 = −6.4λ1 = −7λ1 = −10
(b)
Fig. 6. Reverse flow for a single fluid.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−4
−2
0
2
4
6
8
10
12
14
16
y
u
µ =1, k =1, λ1 =50, d =1.
γ = 0.1γ = 0.5γ = 0.76 (u1(0) = 0)γ = 3γ = 6
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−16
−14
−12
−10
−8
−6
−4
−2
0
2
4
y
u
µ =1, k =1, λ1 =-50, d =1.
γ = 0.1γ = 0.5γ = 1.24 (u1(0) = 0)γ = 3γ = 6
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−2.5
−2
−1.5
−1
−0.5
0
0.5
y
u
µ =1, k =1, γ =0.1, d =1.
λ1 = 2λ1 = 7λ1 = 13.3 (u1(0) = 0)λ1 = 20λ1 = 50
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−7
−6
−5
−4
−3
−2
−1
0
1
y
u
µ =1, k =1, γ =5.5, d =1.
λ1 = −1λ1 = −2λ1 = −2.6 (u1(0) = 0)λ1 = −7λ1 = −25
(d)
Fig. 7. Reverse flow in case II.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
y
u
σ =4, µ =4, k =1, γ =1, d =1, M =4.
λ1 = 1λ1 = 4λ1 = 5λ1 = 10
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
y
u
σ =0.2, µ =0.2, k =1, γ =1, d =1, M =4.
λ1 = −1λ1 = −15λ1 = −17λ1 = −20
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.15
−0.1
−0.05
0
0.05
0.1
0.15
0.2
0.25
y
u
σ =4, µ =4, k =1, λ1 =10, d =1, γ =1.
M = 0M = 1M = 4M = 5M = 10
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y
u
σ =1.5, µ =1.5, k =1, λ1 =10, d =1, γ =1.
M = 0M = 1M = 4M = 5M = 10
(d)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
y
u
σ =0.2, µ =0.2, k =1, λ1 =-10, d =1, γ =1.
M = 0M = 1M = 4M = 5M = 10
(e)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
y
u
σ =1, µ =1, k =1, λ1 =-10, d =1, γ =1.
M = 0M = 1M = 4M = 5M = 10
(f)
Fig. 8. Reverse flow in case IM when λ1 and M vary.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
y
u
σ =1, µ =1, k =1, γ =0.1, d =1, M =4.
λ1 = 5λ1 = 10λ1 = 14λ1 = 20
(a)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
y
u
σ =1, µ =1, k =1, γ =3, d =1, M =4.
λ1 = −1λ1 = −5λ1 = −10λ1 = −14
(b)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
y
u
σ =1, µ =1, k =1, λ1 =7, d =1, γ =0.1.
M = 0M = 1M = 4M = 5M = 10
(c)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.1
−0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
y
u
σ =1, µ =1, k =1, λ1 =-3, d =1, γ =3.
M = 0M = 1M = 4M = 5M = 10
(d)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
y
u
σ =1, µ =1, k =1, λ1 =15, d =1, M =4.
γ = 0.01γ = 0.2γ = 0.5γ = 1.5
(e)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
y
u
σ =1, µ =1, k =1, λ1 =-7, d =1, M =4.
γ = 1γ = 2γ = 3γ = 4
(f)
Fig. 9. Reverse flow in case IIM .
Tw1Tw2no r. no r.
0 λ1Aλ1B λ
μ<– : P2711
μ>– and M>M*: P2711
μ>– and M<M*: F2711
μ>– : P27
11
μ<– and M>M*: P
μ<– and M<M*: F27
11
27
11
* *
Fig. 10. Behavior of the reverse flow for a single fluid or when µ varies for two fluids.
Tw1no r. no r.
0 λ
no r.f 2γ<–a
b
b-aγc—–– f—–1-γ
ca-bγ—––
Tw�Tw� no r. no r.
0 �
f 2–<γ<1ab
ca-bγ—––
b-aγc—–– f—–1-γ
Tw�Tw� no r. no r.
0 �
f 11<γ<–
cfa-bγ—––—–1-γ b-aγ
c—––
Tw� no r. no r.
0 �
no r. f 1γ>–
cb-aγ
fa-bγ
c —––—–1-γ—––ab
ab
P F
F
F
F
PP
P P
P
Fig. 11. Behavior of the reverse flow when γ varies for two fluids.