154 6 Hydrostatics and Aerostatics
The general equations of momentum can be expressed as
ρ
∂U
j
∂t
+ U
i
∂U
j
∂x
i
= −
∂P
∂x
j
−
∂τ
ij
∂x
i
+ ρg
j
(6.5)
and its special form is deduced for a fluid at rest (U
j
=0andthe
molecular-dependent momentum transport τ
ij
) as the following system of
partial differential equations, which represents the set of basic equations of
hydrostatics and aerostatics:
∂P
∂x
j
= ρg
j
(j =1, 2, 3) (6.6)
or, written out for all three directions j = 1, 2, 3,
∂P
∂x
1
= ρg
1
,
∂P
∂x
2
= ρg
2
,
∂P
∂x
3
= ρg
3
(6.7)
In this section, the pressure distribution in a fluid, mainly defined by the field
of gravity, will be considered more closely. Restrictions are made concerning
the possible fluid properties; the fluid is assumed to be incompressible for
hydrostatics, i.e. ρ = constant. This condition is in general fairly well fulfilled
by liquids, so that the following derivations can be considered as valid for
liquids, see also refs. [6.1] to [6.8].
For the derivation of the pressure distribution in a liquid at rest, a rect-
angular Cartesian coordinate system is introduced, whose position is chosen
such that the mass acceleration {g
i
} given by the field of gravity shows only
one component in the negative x
2
direction, i.e. the following vector holds:
{g
i
} =
⎧
⎨
⎩
0
−g
0
⎫
⎬
⎭
. (6.8)
Then the differential equations (6.7), given above generally for the pressure,
can be written as follows:
∂P
∂x
1
=0,
∂P
∂x
2
= −ρg,
∂P
∂x
3
=0 (6.9)
From ∂P/∂x
1
= 0, it follows that P = f(x
2
,x
3
)andfrom∂P/∂x
3
=0
it follows that P = f(x
1
,x
2
). Thus a comparison yields P = f(x
2
)andthis
shows that the pressure of a fluid within a plane is constant, P (x
2
)=const,
when it is perpendicular to the direction of the field of gravity. The free
surface of a fluid stored in a container is a plane of constant pressure and
all planes parallel to it are also planes of constant pressure. The pressure
increases in the direction that was defined by g
j
, i.e. in the direction of the
gravitational acceleration.