Fluids: Liquids and Gases - Computer Animation: Algorithms and Techniques - page 274

Using the divergence operator in Equation 8.33 produces Equation 8.35 .
@r

@t ¼rðrvÞ

(8.35)

If the fluid is incompressible, then the density does not change. This means that @r
@t

is zero and the

constant density can be divided out, resulting in Equation 8.36 .
0 ¼r v (8.36)

Conservation of momentum
The momentum of an object is its mass times its velocity. The momentum of a fluid is its density times
the volume of the fluid times the average velocity of the fluid.
A change in momentum dðmvÞ
dt

f¼ma ¼ m dv
dt

). The force in the
element of a flow field is either a global force, such as viscosity and gravity, or a change in pressure
across the element, dp
dx

must be induced by a force (i.e.,

. As noted previously, effects of viscosity are commonly ignored in order to sim-
plify the equations; we will also ignore gravity for now.
For an element, the change in momentum is the change of momentum inside the element plus the
difference between the momentum entering the cell and the momentum leaving the element. The mass
inside the cell is rV and, thus, the momentum inside the element is rVv . The rate of change of momen-
tum inside the element is @ðpVvÞ
@t

.
Given a three-dimensional element, consider flow only in the x -direction. The mass flowing out one
surface is density

area of the surface times velocity in the x -direction, v x . The momentum out one
surface is the mass flowing out the surface times its velocity, r Av x v . The force on the element in
the x -direction is the pressure difference across the element from x to xþdx . The force is equal to
the negative of the change in momentum Equation 8.37 .

¼ @ðrVvÞ

A
xþdx pj

A
x

pj

@t þðrv x AÞvj xþdx ðrv x AÞvj x

(8.37)

Following the steps above in developing the conservation of mass equation by replacing
the area, A , with its definition in terms of element dimensions, dydz , replacing the volume, V ,
with dxdydz , dividing through by dxdydz , and putting everything in differential form gives

Equation 8.38 .

@p

@x ¼ @ðrv x v

Þ

þ @ðr

v Þ

(8.38)

@x @t

Now, considering flow in all three directions, and separating out momentum only in the x direction,
produces Equation 8.39 .

2
x

@t ¼ @ rv

þ @ rv x v y

@p þ @ rv x v z

ð Þ

þ @ rvðÞ
@t

(8.39)

@x @y @z

If desired, viscosity and other forces can be added to the pressure differential force on the left-hand
side of the equation. Similar equations can be derived for the y and z directions.