Page 169 - Introduction to Computational Fluid Dynamics

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2D CONVECTION – CARTESIAN GRIDS
the fluid inside the channel is stagnant. However, in the present case, because of the
density gradient caused by the vapour-pressure difference, a mass transfer buoyancy
force will induce fluid motion. The objective, therefore, is to examine the range of
mass transfer Grashof numbers Gr m for which the stagnant flow assumption may
be reasonably justified. Such an inquiry has been undertaken by McBain et al. [47]
in which the inner channel is a circular tube placed inside a cubical enclosure. We
have modified this 3D configuration to accommodate a 2D analysis in Cartesian
coordinates.
∗
∗
∗
We define L = L/l, H = H/l, h = h/l, and t = t/l. In this case equa-
∗

tions for = u 1 , u 2 ,ω, and p must be solved. Invoking the Boussinesq approx-
imation, except for the gravity-affected source term in the u 2 -momentum equa-
tion, we assume the density will be constant. Also viscosity and mass diffusivity
are assumed constant. Thus, the governing equations can be nondimensionalised
2
∗
using u = u i /(ν/l), p = (p + ρ gx 2 )/ρ (ν/l) , ω = (ω − ω 0 )/(ω 1 − ω 0 ), and
∗
∗
i
∗
x = x i /l. The relevant source terms are
i
∂p ∗ ∂p ∗
∗
S u =− , S u =− + Gr m ω , S ω = 0, (5.125)
∗
∗
∗
1 ∂x ∗ 2 ∂x ∗
1 2
2
3
∗
where Gr m = g β m (ω 1 − ω 0 )l /ν and β m = ρ −1 ∂ρ/∂ω .
The boundary conditions are
∂ω ∗
∗
u = 0, = 0 on all walls, (5.126)
i
∂n ∗
∗
where n is normal to the walls. The x = 0 line is the symmetry boundary and
1
computations are performed over the domain to the right of the symmetry line. The
mass transfer boundary conditions on the floor (x = 0) are
∗
2
∗
u = 0,
1

∗
∗ −1
∂ω
∗
u = Sc −1 (ω − ω ) , ω = ω (water),
∗
∗
∗
2 1 T 1
∂x ∗
∗
2 x =0
2

∂ω ∗
∗ −1
∗
∗
∗
∗
u = Sc −1 (ω − ω ) : , ω = ω (brine), (5.127)
2 0 T 0
∂x ∗
∗
2 x =0
2
∗
∗
where ω = 1 and ω = 0.
1 0
These specifications indicate that in the present mass transfer problem, the
momentum equations are coupled with the mass transfer equation in two ways,
∗
firstly, through the source term Gr m ω and, secondly, through the floor boundary
condition. The dimensionless total evaporation flux is, therefore, given by
1/2 ∗
∗
∗
F conv = 2 Sc −1 (1 − ω ) ∂ω dx . (5.128)
T 1
∗
0 ∂x 2 x =0
∗
2