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Adding in some soot of density ρ soot and treating the concentration s
as a volume fraction, we can extend this to
ρ = ρ air 1+ s ρ soot
ρ air
ρ air
(5.3)
= ρ air (1 + αs ) ,
where we treat the relative density difference α =( ρ soot
ρ air ) air as a
constant—i.e., the soot has the same thermal expansion as the air. This
is of course false, but we consider this error negligible relative to other
modeling errors.
At this stage, the buoyancy model from the previous section can be
recovered by linearizing Equation (5.3) around standard conditions
ρ 0 1+ αs
T amb )
1
T amb ( T
ρ
= ρ 0 [1 + αs
β ( T
T amb )] ,
where ρ 0 is the smoke-free air density at ambient temperature and β =
1 /T amb . Plugging it into the momentum equation (where we have multi-
plied both sides by ρ )gives
β Δ T ) Du
Dt
ρ 0 (1 + αs
+
p = ρ 0 (1 + αs
β Δ T ) g.
The hydrostatic pressure for constantdensityfluidatrestis p = ρ 0 g
x ;
write the actual pressure as the sum of this hydrostatic pressure plus a
pressure variation p so that
·
p . This simplifies the momentum
p = ρ 0 g +
equation to
β Δ T ) Du
Dt
p = ρ 0 ( αs
ρ 0 (1 + αs
β Δ T ) g.
+
We now make the Boussinesq approximation that, assuming
|
αs
β Δ T
|
1, we can drop the density variation in the first term, leading to
ρ 0 Du
Dt
p = ρ 0 ( αs
+
β Δ T ) g.
Dividing through by ρ 0 gives the buoyancy form of the momentum equation
from the previous section. It then becomes clear what α and β “should”
be chosen as, though of course these can be left as tunable parameters. (It
also makes it clear that the pressure we solve for in the buoyancy model is
not the full pressure but actually just the variation above the hydrostatic
pressure.)
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