Graphics Reference
In-Depth Information
This is called the
continuity equation
. For an incompressible fluid the
material derivative of density
Dρ/Dt
is zero, i.e.,
∂ρ
∂t
+
u
·∇
ρ
=0
.
Subtracting this from the continuity equation gives
ρ
∇·
u
=0,ormore
simply
∇·
u
=0
,
which is termed the
incompressibility condition
. Note that this is indepen-
dent of density, even for problems where fluids of different densities mix
together.
We can apply the same process to the rate of change of momentum:
=
Ω
∂ P
∂t
∂
(
ρu
)
∂t
.
Momentum can change in two ways: the transport of fluid across the bound-
ary and a net force
F
applied to region. The transport of momentum with
the fluid is the boundary integral of momentum
ρu
times the speed of the
fluid through the boundary:
−
(
ρu
)
u
·
n.
∂
Ω
(The negative sign comes from the fact that the normal is outward-
pointing.) There are two forces in play for an inviscid fluid: pressure
p
on the boundary and gravity
ρg
throughout the region:
pn
+
Ω
F
=
−
ρg.
∂
Ω
(Again, we get a negative sign in front of the pressure integral since the
normal is outward-pointing.) Combining all these terms we get
pn
+
Ω
∂
(
ρu
)
∂t
=
−
(
ρu
)
u
·
n
−
ρg.
Ω
∂
Ω
∂
Ω
Transforming the boundary integrals into volume integrals with the Fun-
damental Theorem of Calculus and rearranging gives
+
u
)+
p
=
Ω
∂
(
ρu
)
∂t
Ω
∇·
(
ρu
⊗
Ω
∇
ρg.
Ω
