Numerical Simulation of Three-Dimensional Viscous Flows with ...

1334
ISMAIL-ZADE et al.
E 0 = 2 × 10 4 J mol –1 , V 0 = 4 × 10 –6 m mol –1 , and R = 8.3 J mol –1 K –1 . As a result, we obtain Fr = 0.95 × 10 –24 ,
La = 2.52 × 10 3 , Ra = 16.96, Di = 0.24 × 10 –4 , and He = 0.27 × 10 8 Q 0 . The actual (effective) values of these
parameters can substantially differ from those listed here, because the problem is solved for the density ρ
and viscosity µ calculated by using formulas (4) and (5), or their dimensionless counterparts. The resulting
effective value of the Rayleigh number is sufficiently high to be indicative of an unstable flow. Thus, since
the real processes dealt with in geodynamics problems are slow, and the inertia-to-gravity force ratio is a
small quantity, one can set Fr = 0 and drop the terms on the left-hand side of the Navier–Stokes equations.
Note also that, since the parameters c, k, α, and κ characterize fluid properties, they are carried by fluid
particles in the same manner as the thermally unperturbed density and viscosity are. Therefore, they must
satisfy transport equations of the form
∂ [] ·
-------- + 〈 ∇ []u · , 〉 = 0.
∂t
These parameters may be complicated functions of temperature, pressure, and other flow variables. To simplify
analysis and avoid unnecessary solution of additional equations, we assumed that c, k, α, and κ are
constant parameters. Accordingly, transport equations for these parameters were not included in the set of
relations that determine the flow dynamics. Assuming also that the gravitational acceleration, activation
energy, and activation volume are constant parameters, we set c = 1, k = 1, α = 1, κ = 1, g = 1, E = 1, and
V = 1 in the governing equations. The heat equation was considered in the Boussinesq approximation.
The simplifications and assumptions introduced above lead to the following simplified system of equations
for the desired dimensionless flow variables:
La∇p
=
div ( µe ij )–
Laρe 3 ,
(7)
divu = 0,
∂
( ρ* T) + 〈 u,
∇ρ (
∂t
*
T)
〉 = ∆T + DiµΦ + HeρQ,
ρ( t,
x) = ρ *
( t,
x) { 1–
α 0 T 0 [ Tt ( , x) – 1]
},
E
µ ( t,
x) µ *
( t,
x)
0 + ρ ρ * 0 g 0 x 3 l 0 V 0 E
--------------------------------------------- 0 + ρ 0 g 0 l 0 V = exp – -------------------------------- 0
⎝
⎛ RTT 0
RT 0
⎠
⎞ ,
(8)
(9)
(10)
(11)
∂ρ ∂µ
--------- * + 〈 ∇ρ , u〉
= 0, --------- * + 〈 ∇µ (12)
∂t * ∂t *
, u〉
= 0.
To eliminate the incompressibility condition divu = 0 and pressure p, we define the vector potential
y = (ψ 1 , ψ 2 , ψ 3 ) by the relation u = curly and apply the curl operator to Eq. (7). Using the identities
curl(∇p) = 0 and div(curly) = 0, we derive the following equations from (7) and (8):
3
∂ 2 ( µe i3 ) ∂ 2 ( µe
-------------------
i2 )
∑ – ------------------- [ La( 1 + α
∂x 2 ∂x i ∂x 3 ∂x 0 T 0 )–
RaT] ∂ρ ∂T
=
--------- * – Raρ
i
∂x 2 *
-------,
∂x 2
i = 1
3
∂ 2 ( µe i3 ) ∂ 2 ( µe
-------------------
i1 )
∑ – ------------------- [ La( 1 + α
∂x 1 ∂x i ∂x 3 ∂x 0 T 0 )–
RaT] ∂ρ ∂T
=
--------- * – Raρ
i
∂x 1 *
-------,
∂x 1
i = 1
(13)
(14)
3
∂ 2 ( µe i2 ) ∂ 2 ( µe
-------------------
i1 )
∑ – ------------------- = 0.
∂x 1 ∂x i ∂x 2 ∂x i
i = 1
The pressure p can be determined from (7) up to a constant. The components of u = (u 1 , u 2 , u 3 ) can be
calculated by using the equation u = curly as
(15)
∂ψ
u 3 ∂ψ
1 -------- 2 ∂ψ
– -------- , u 1 ∂ψ
=
∂x 2 ∂x 2 = -------- – -------- 3
, u
3 ∂x 3 ∂x 3 =
1
∂ψ
-------- 2
∂x 1
∂ψ
– -------- 1
.
∂x 2
(16)
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS Vol. 41 No. 9 2001