EFFECT OF THE MAGNETIC FIELD AND HEAT GENERATION ON THE FREE CONVECTION FLOW IN A TALL CAVITY FILLED WITH A POROUS MEDIUM ‡

. A analytical study of the steady magnetohydrodynamics (MFD) free convection in an tall cavity ﬁlled with a ﬂuid-saturated porous medium and with internal heat generation has been performed. It is considered that the Darcy law model is used. It is assume that a uniform magnetic ﬁeld normal to the walls of the cavity is externally imposed. The values of the governing parameters are as follows: Hartmann number Ha = 0 , 10 and 50, Rayleigh number Ra = 10 3 , and the aspect ratio a = 0 . 01. The velocity and temperature proﬁle are determined. These proﬁles are presented graphically at the center line of the cavity. It is found that the analytical solution is in very good agreement with the numerical solution which is obtained by solving partial diﬀerential equations using a ﬁnite-diﬀerence method.


INTRODUCTION
In many engineering applications, enclosures with fluid-saturated porous media are encountered in grain storage, high-performance insulation of buildings, geophysical problems, biological tissues, in the thermal insulation of buildings, in geothermal energy convection, in petroleum reservoirs, etc.These applications can be found in the recent books and review papers, such as, Nield and Bejan [1], Ingham and Pop [2], Vafai [3], Bejan et al. [4], etc. Buoyancy induced flow and heat transfer in enclosures of different geometries and filled with a fluid-saturated porous medium has been widely studied in literature using different numerical technique (Manole and Lage, [5]; Goyeau et al., [6]; Saeid and Pop, [7]; Varol et al., [8]).Recent demands in heat transfer engineering have requested researchers to develop various new types of heat transfer tgrosan@math.ubbcluj.ro,pop.ioan@yahoo.co.uk.
‡ The work of the first author has been supported by grant 2-CEx 06-11-96 and the work of the second author has been supported by grant Ceex 90/2006.equipments with superior performance, especially compact and light-weight ones.Increasing the need for small-size units, focuses have been cast on the effects of the interaction between the fluid flow in a cavity and a magnetic field which usually affects the flow development.This type of treatment is completely justified in many practical problems of the magnetohydrodynamics (MHD) generators.There have recently been published a number of research papers which investigate MHD flows of conducting fluids in cavities.
This paper presents an analytical solution of the effect of a magnetic field on the steady free convection in a tall cavity filled with a porous medium saturated with an electrically conducting fluid using a semi-analytical method proposed by Joshi et al., [9].This analytical solution is compared with the numerical solution obtained in a previous paper by Groşan et al., [10].This type of problem arises in geophysics when a fluid saturates the earth's mantle in the presence of a geomagnetic field.

BASIC EQUATION
The conservation equations for mass, momentum under the Darcy approximation, energy and electric transfer are give by (1) ∇.V = 0, where V is the fluid velocity vector, T is the fluid temperature, p is the pressure, B is the external magnetic field, I is the electric current, φ is the electric potential, g is the gravitational acceleration vector, K is the permeability of the porous medium, α m is the effective thermal diffusivity, ρ is the density, µ is the dynamic viscosity, β is the coefficient of thermal expansion, c p is the specific heat at constant pressure, σ is the electrical conductivity, ρ 0 is the reference density and −∇φ is the associated electric field.As discussed by Garandet et al. [11], Eqs. ( 4) and ( 5) reduce to ∇ 2 φ = 0.The unique solution is ∇φ = 0 since there is always an electrically insulating boundary around the enclosure.Thus, it follows that the electric field vanishes everywhere (see, Alchaar et al., [12]).
Eliminating the pressure term in Eq. ( 2) in the usual way, the governing equations ( 1) to (3) can be written as (7) ∂u ∂x + ∂v ∂y = 0, 8) and ( 9) can be written as ( 12) We consider a long cavity with h l, with heat generation.In this case, it can be assumed that in the central region, the flow is only in vertical direction, that is, u = 0, υ = υ(x) and ψ = ψ(x).Under this assumption, Eqs. ( 12) and (13) reduce to ( 14) The left hand side of Eq. ( 14) is function of x only.Therefore, the right hand side term Ra ∂θ/∂x must be a functional x only.Thus, ∂θ ∂x = h(x) and ( 16) Substituting ( 16) into ( 14) and (15), we get (17) − ag (y) dψ dx = f (x) + a 2 g (y) + 1. Differentiating these equations with respect to x, we obtain Since in this problem v = − dψ dx is not constant on the vertical walls of the cavity it implies that d 2 ψ dx 2 = 0 for all values of x.Hence, from Eq. (20), one can write The left hand side of this equation is a functional of y only, while the right hand side is a function of x only.Therefore, we must have where γ is a constant.It can be conducted from Eqs. ( 16) and ( 21) that the temperature gradient in the vertical direction is constant.Using ( 21) and (18), Eqs.(19) reduced to It should be noted that in the present problem v 2 (0) = 0 and v 1 (1) = 0 so that we have to use other conditions to determine the constants c 1 and c 2 .A condition is the conservation of mass Am .Using the symmetry of the velocity profile about the centerline, some of the constants, can be delimited, we have Solving the system of Eqs. ( 28) and ( 29) we obtain the constant c 1 and c 1 : . In order to determine θ we use Eqs.( 16), ( 17), ( 21) and ( 23).Thus, from these equations, we have (30) where δ 1 is a constant.Then, from (21), we get where δ 2 is a constant.Therefore, θ given by ( 16), can be written as θ(x, y) = −Av 1 + Am 2 aRa y + δ, where δ is a constant.Substitute v 1 from (25), we get (31) θ(x, y) = −A(c 1 e mx + c 1 e −mx ) + Am 2 aRa y + 1 m 2 + δ.Since ψ = ψ(x) is not valid near the boundaries of the cavity, the expression (33) for θ cannot satisfy all the boundary conditions.The value of δ can be calculated in terms of m by substituting θ(x, y) = 0 at the bottom corner of the cavity x = 0 and y = 0 in Eq. (33).This gives In this analysis, the value of m, which depend on the vertical temperature gradient γ, remains undetermined.It can be, however, determined from the fact that under natural convection, the maximum temperature attained must be lower than maximum temperature in conduction region, that is (32) ∂y 2 + 1 = 0, in the central of the cavity the boundary conditions is (33) θ = 0 at x = 0 and x = 1.
Solution at this problem is given by We notice that the temperature is maximum at x = 1/2 and the maximum value Similarly, the temperature cannot be less than the wall temperature, i.e. θ min = 0. Therefore, substituting θ max = 1/8 in (33), we get . This expression gives the value of m as a function of a 2 Ra and A.

RESULTS AND DISCUSSIONS
The obtained analytical solutions in the centerline on the cavity is shown in Figs. 2 to 5 for the temperature and velocity profiles.The values of the parameters considered are the Rayleigh number Ra = 10 3 , Hartmann number Ha = 0, 10, 50 and aspect ration a = 0.01.These solution is shown by circle.In additions the numerical results are included in these figures, which are shown by full line.It is seen that there is a very good agreement between the analytical and numerical solutions obtained.Therefore, we are confident that the present analytical solution is accurate.We notice that the effect of the Hartmann number is not significant for the temperature profile, while it's effect is more significative for the velocity profiles.This is because the magnetic term thus not directly enter in the energy equation.Figs. 3 to 5 shows that the vertical component of the velocity is positive in the center of the cavity and negative close to the vertical walls.This is because of the conservation of mass, Eq. (26).

CONCLUSION
In this paper analytical expression are obtained to describe the free convection in a tall cavity filled with the porous medium and with an applied magnetic field and volumetric heat generation.Because of the volumetric generation, the central region is of high temperature, and therefore the flow is driven upwards due to buoyancy, in the central part of the cavity and driven downwards in the relatively colder boundary regions.

A d 2 v 1 dx 2 −
aγ 1 v 1 − 1 = 0, the general solution of this equation is of the from (25) v 1 (x) = c 1 e mx + c 2 e −mx − 1 Am 2 , where c 1 and c 2 are constants of integration and m is given by is used and the fact that ψ = ψ(x) is not valid near the boundary of the cavity.Using (27) it gives (28) c 1 (e m − 1) − c 2 e −m − 1 = 1 e −m ) −2e −m/2 + e −m + Am 2 a 2 Ra