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7. Solution of the Navier-Stokes Equations
would like the discretization to retain this property. Let us see how this
might happen.
If Gip represents the numerical approximation to the ith component of
the pressure gradient, then, when the discretized ui-momentum equation is multiplied by ui, the pressure gradient term gives a contribution
C ui Gip A n . Energy conservation requires that this contribution be equal
to (cf. Eq. (7.9)):
where the subscript N indicates that the sum is over all CVs (grid nodes),
Sb is the boundary of the solution domain, vn is the velocity component
normal to the boundary and Diui is the discretized velocity divergence
used in the continuity equation. If this is so, Diui = 0 at each node, so
the second term in the above equation is zero. The equality of the left and
right hand sides can then be ensured only if Gi and Di are compatible in
the following sense:
N
C ( U ~
Gip + p Diui) Afl = surface terms .
(7.11)
i=l
This states that the approximation of the pressure gradient and the divergence of the velocity must be compatible if kinetic energy conservation is
to hold. Once either approximation is chosen, the freedom to choose the
other is lost.
To make this more concrete, assume that the pressure gradient is approximated with backward differences and the divergence operator with forward
differences (the usual choice on a staggered grid). The one-dimensional version of Eq. (7.11) on a uniform grid then reads:
The only two terms that remain when the sum is taken are the "surface
terms" on the right hand side. The two operators are therefore compatible
in the above sense. Conversely, if forward differences were used for the
pressure gradient, the continuity equation would need to use backward
differences. If central differences are used for one, they are required for the
other.
The requirement that only boundary terms remain when the sum over all
CVs (grid nodes) is taken applies to the other two conservative terms, the
convective and viscous stress terms. Satisfaction of this requirement is not
easy in any case and is especially difficult for arbitrary and unstructured
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