The pressure indicates the normal force per unit area at a given point acting on a given plane. Since there is no shearing stresses present in a fluid at rest - the pressure in a fluid is independent of direction.
For fluids - liquids or gases - at rest the pressure gradient in the vertical direction depends only on the specific weight of the fluid.
How pressure changes with elevation can be expressed as
dp = - γ dz (1)
where
dp = change in pressure
dz = change in height
γ = specific weight
The pressure gradient in vertical direction is negative - the pressure decrease upwards.
[h=Specific Weight]3[/h] Specific Weight can be expressed as:
γ = ρ g (2)
where
γ = specific weight
g = acceleration of gravity
In general the specific weight - γ - is constant for fluids. For gases the specific weight - γ - varies with the elevation.
[h=Static Pressure in a Fluid]3[/h] For a incompressible fluid - as a liquid - the pressure difference between two elevations can be expressed as:
p2 - p1 = - γ (z2 - z1) (3)
where
p2 = pressure at level 2
p1 = pressure at level 1
z2 = level 2
z1 = level 1
(3) can be transformed to:
p1 - p2 = γ (z2 - z1) (4)
or
p1 - p2 = γ h (5)
where
h = z2 - z1 difference in elevation - the dept down from location z2.
or
p1 = γ h + p2 (6)
[h=Example - Pressure in a Fluid]4[/h] The absoute pressure at water depth of 10 m can be calulated as:
p1 = γ h + p2
= (1000 kg/m3) (9.81 m/s2) (10 m) + (101.3 kPa)
= (98100 kg/ms2 or Pa) + (101300 Pa)
= 199.4 kPa

where
ρ = 1000 kg/m3
g = 9.81 m/s2
p2 = pressure at surface level = atmospheric pressure = 101.3 kPa
The gauge pressure can be calulated setting p2 = 0
p1 = γ h + p2
= (1000 kg/m3) (9.81 m/s2) (10 m)

= 98.1 kPa
[h=The Pressure Head]3[/h] (6) can be transformed to:
h = (p2 - p1) / γ (6)
h express the pressure head - the height of a column of fluid of specific weight - γ - required to give a pressure difference of (p2 - p1).
[h=Example - Pressure Head]4[/h] A pressure difference of 5 psi (lbf/in2) is equivalent to
(5 lbf/in2) (12 in/ft) (12 in/ft) / (62.4 lb/ft3)
= 11.6 ft of water
(5 lbf/in2) (12 in/ft) (12 in/ft) / (847 lb/ft3)
= 0.85 ft of mercury
when specific weight of water is 62.4 (lb/ft3) and specific weight of mercury is 847 (lb/ft3).
Heads at different velocities are indicated in the table below:
Velocity
(ft/sec)		 Head Water
(ft)
0.5 0.004
1.0 0.016
1.5 0035
2.0 0.062
2.5 0.097
3.0 0.140
3.5 0.190
4.0 0.248
4.5 0.314
5.0 0.389
5.5 0.470
6.0 0.560
6.5 0.657
7.0 0.762
7.5 0.875
8.0 0.995
8.5 1.123
9.0 1.259
9.5 1.403
10.0 1.555
11.0 1.881
12.0 2.239
13.0 2.627
14.0 3.047
15.0 3.498
16.0 3.980
17.0 4.493
18.0 5.037
19.0 5.613
20.0 6.219
21.0 6.856
22.0 7.525


  • 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd