102
CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS
4.1.1
Equations of fluid statics
The equations of fluid statics are usually derived by a simple force balance under the assumption
of mechanical equilibrium (pressure independent of time in a motionless—
i.e.
, static—fluid) for
the fluid system. Here, however, we obtain them as a special case of the N.–S. equations. This is
actually much easier; and now that we have these equations available, such an approach, among
other things, emphasizes the unity of fluid mechanics in general as embodied in the N.–S. equations.
Thus, we begin by writing the 3-D incompressible N.–S. equations in a Cartesian coordinate
system with the positive
z
direction opposite that of gravitational acceleration. This gives
u
x
+
v
y
+
w
z
= 0
,
(4.1a)
u
t
+
uu
x
+
vu
y
+
wu
z
=
−
1
ρ
p
x
+
ν
∆
u ,
(4.1b)
v
t
+
uv
x
+
vv
y
+
wv
z
=
−
1
ρ
p
y
+
ν
∆
v ,
(4.1c)
w
t
+
uw
x
+
vw
y
+
ww
z
=
−
1
ρ
p
z
+
ν
∆
w
−
g .
(4.1d)
Now because the fluid is assumed to be motionless, we have
u
=
v
=
w
≡
0, implying that all
derivatives of velocity components also are zero. In particular, there can be no viscous forces in
a static fluid. Clearly, the continuity equation is satisfied trivially, and the momentum equations
collapse to
p
x
= 0
,
(4.2a)
p
y
= 0
,
(4.2b)
p
z
=
−
ρg
=
−
γ ,
(4.2c)
where
γ
is the specific weight defined in Chap. 2.
The first two of these equations imply that in a static fluid the pressure is constant throughout
planes aligned perpendicular to the gravitational acceleration vector;
i.e.
,
p
x
= 0 implies
p
is not a
function of
x
, and similarly for
y
. We now integrate Eq. (4.2c) to formally obtain
p
(
x, y, z
) =
−
γz
+
C
(
x, y
)
,
where
C
(
x, y
) is an integration function
.
(When integrating a partial differential equation we
usually obtain integration functions instead of integration constants.) But we have already noted
that Eqs. (4.2a) and (4.2b) imply
p
cannot be a function of either
x
or
y
; so in this special case
C
is a constant, and we then have
p
(
z
) =
−
γz
+
C .
(4.3)
We remark here that the first liquid considered in this context was water, and as a result the
contribution
γz
=
ρgz
to the pressure in a static fluid is often termed the
hydrostatic pressure
.
The value of
C
must now be determined by assigning a value to
p
, say
p
0
, at some reference
height
z
=
h
0
. This leads to
p
0
=
p
(
h
0
) =
−
γh
0
+
C
⇒
C
=
p
0
+
γh
0
,
and from this it follows that
p
(
z
) =
p
0
+
γ
(
h
0
−
z
)
.
(4.4)
This actually represents the solution to a simple
boundary-value problem
consisting of the first-
order differential equation (4.2c) and the boundary condition
p
(
h
0
) =
p
0
, and we will emphasize
this viewpoint in the sequel as we introduce several examples to demonstrate use of the equations
of fluid statics.