Diploma report Implementation and verification of a simple ... - LPAS

have
∫
d z2
∫ x2
ρ(x, z, t)dxdz = F 1 (z, t) − F 2 (z, t) + G 1 (x, t) − G 2 (x, t), (6)
dt z 1 x 1
where F i (z, t) are the flux functions at the lateral boundaries x i , i = 1, 2, and G i (x, t) the
fluxes at the top and bottom boundaries z i , i = 1, 2. Note that F 1 (z, t) corresponds to the
flux to the right and −F 2 (z, t) represents the flux to the left.
Equation (6) represents the integral from of a conservation law for the mass and the
flux functions are to be related to ρ(x, z, t) in order to obtain an equation that might be
solvable for ρ. The mass flux through boundary x i is associated with the mass transported
with velocity u through x i so that F is given by ρu. Similarly, G is given by ρw. The integral
form of the conservation law becomes then
d
dt
∫ z2
∫ x2
z 1
x 1
ρ(x, z, t)dxdz = −ρu∣ x 2
− ρw∣ z 2
. (7)
x 1 z 1
Assuming that ρ, u and w are smooth functions, the integral form of the conservation law can
be transformed into a partial differential equation (see appendix A). The differential form of
the conservation law for ρ is then
ρ t + (ρu) x + (ρw) z = 0. (8)
This equation called the continuity equation models the conservation of mass.
In addition to ρ(x, z, t), the velocity components u(x, z, t) and w(x, z, t) are also to be determined
and two equations are needed for this. The velocities themselves are not conserved
quantities, but the x-momentum ρu and the z-momentum ρw are. Each of these momenta
are transported with the flow and, by analogy with the density flux in (8), this will give
contributions of ((ρu)u) x + ((ρu)w) z and ((ρw)u) x + ((ρw)w) z to the x- and z-momentum
respectively. Beside this, there are additional microscopic terms for the momentum fluxes
that are due to the pressure in the fluid. These microscopic fluxes are caused by the fact that
all nearby molecules are not moving at the same speed in a gas. Pressure variation in the
x-direction, measured by p x appears in the equation for ρu, while p z appears in the equation
for ρw.
In the vertical direction, there are external gravity forces acting on the gas that accelerate
downward individual molecules. These forces must be integrated over the domain
and this contribution added to the integral conservation law for z-momentum (if z is aligned
with the gravitational acceleration). Thus the differential form of the z-momentum equation
will contain a source term of s = −ρ∇Φ, where Φ = gz is the gravitational potential.
The differential forms of the conservation laws for momentum are then
(ρu) t + (ρu 2 + p) x + (ρuw) z = 0, (9)
(ρw) t + (ρwu) x + (ρw 2 + p) z = −ρ∇Φ. (10)
6