Recall from chapter 1 that the foundation of
modern hydraulics was established when Pascal
discovered that pressure in a fluid acts equally in
all directions. This pressure acts at right angles
to the containing surfaces. If some type of
pressure gauge, with an exposed face, is placed
beneath the surface of a liquid (fig. 2-6) at a
specific depth and pointed in different directions,
the pressure will read the same. Thus, we can say
that pressure in a liquid is independent of
Pressure due to the weight of a liquid, at any
level, depends on the depth of the fluid from the
surface. If the exposed face of the pressure gauges,
figure 2-6, are moved closer to the surface of the
liquid, the indicated pressure will be less. When
the depth is doubled, the indicated pressure is
doubled. Thus the pressure in a liquid is directly
proportional to the depth.
Consider a container with vertical sides
(fig. 2-7) that is 1 foot long and 1 foot wide. Let
it be filled with water 1 foot deep, providing 1
cubic foot of water. We learned earlier in this
chapter that 1 cubic foot of water weighs 62.4
pounds. Using this information and equation 2-2,
P = F/A, we can calculate the pressure on the
bottom of the container.
Since there are 144 square inches in 1 square foot,
This can be stated as follows: the weight of a
column of water 1 foot high, having a cross-
sectional area of 1 square inch, is 0.433 pound.
If the depth of the column is tripled, the
weight of the column will be 3 x 0.433, or 1.299
pounds, and the pressure at the bottom will be
1.299 lb/in2 (psi), since pressure equals the force
divided by the area. Thus, the pressure at any
depth in a liquid is equal to the weight of the
column of liquid at that depth divided by the
Figure 2-6.Pressure of a liquid is independent of direction.
cross-sectional area of the column at that depth.
The volume of a liquid that produces the pressure
is referred to as the fluid head of the liquid. The
pressure of a liquid due to its fluid head is also
dependent on the density of the liquid.
If we let A equal any cross-sectional area of
a liquid column and h equal the depth of the
column, the volume becomes Ah. Using equation
2-4, D = W/V, the weight of the liquid above area
A is equal to AhD.
Figure 2-7.Water pressure in a 1-cubic-foot container.