Static Pressure vs. Head
Static pressure vs. pressure head in fluids.
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 in a fluid can be expressed as
Δp =  γ Δh (1)
where
Δ p = change in pressure (Pa, psi)
Δ h = change in height (m, in)
γ = specific weight of fluid (N/m^{3}, lb/ft^{3} )
The pressure gradient in vertical direction is negative  the pressure decrease upwards.
Specific Weight
Specific Weight of a fluid can be expressed as:
γ = ρ g (2)
where
ρ = density of fluid (kg/m^{3}, slugs /ft^{3} )
g = acceleration of gravity (9.81 m/s^{2}, 32.174 ft/s^{2})
In general the specific weight  γ  is constant for fluids. For gases the specific weight  γ  varies with elevation (and compression).
The pressure exerted by a static fluid depends only upon
 the depth of the fluid
 the density of the fluid
 the acceleration of gravity
Static Pressure in a Fluid
For a incompressible fluid  as a liquid  the pressure difference between two elevations can be expressed as:
Δ p = p_{2} p_{1}
=  γ (h_{2} h_{1} ) (3)
where
p_{2}= pressure at level 2 (Pa, psi)
p_{1} = pressure at level 1 (Pa, psi)
h_{2}= level 2 (m, ft)
h_{1} = level 1 (m, ft)
(3) can be transformed to:
Δ p = p_{1}  p _{ 2 }
= γ (h_{2} h_{1} ) (4)
or
p_{1}  p_{2}= γ Δ h (5)
where
Δ h = h_{2} h_{1} = difference in elevation  the dept down from location h_{2}to h_{1} (m, ft)
or
p_{1} = γ Δ h + p_{2}(6)
Example  Pressure in a Fluid
The absolute pressure at water depth of 10 m can be calculated as:
p_{1} = γ Δ h + p_{2}
= (1000 kg/m^{3} ) (9.81 m/s^{2}) (10 m) + (101.3 kPa)
= (98100 kg/ms^{2}or Pa) + (101300 Pa)
= 199400 Pa
= 199.4 kPa
where
ρ = 1000 kg/m^{3}
g = 9.81 m/s^{2}
p_{2}= pressure at surface level = atmospheric pressure = 101.3 kPa
The gauge pressure can be calculated by setting p_{2}= 0
p_{1} = γ Δ h + p_{2}
= (1000 kg/m^{3} ) (9.81 m/s^{2}) (10 m)
= 98100 Pa
= 98.1 kPa
Pressure vs. Head
(6) can be transformed to:
Δ h = (p_{2} p_{1} ) / γ (7)
Δ h express head  the height difference of a column of fluid of specific weight  γ  required to give a pressure difference Δp = p_{2} p_{1} .
Example  Pressure vs. Head
A pressure difference of 5 psi (lb_{f} /in^{2}) is equivalent to head in water
(5 lb_{f} /in^{2}) (12 in/ft) (12 in/ft) / (62.4 lb/ft^{3} )
= 11.6 ft of water
or head in Mercury
(5 lb_{f} /in^{2}) (12 in/ft) (12 in/ft) / (847 lb/ft^{3} )
= 0.85 ft of mercury
Specific weight of water is 62.4 (lb/ft^{3} ) and specific weight of mercury is 847 (lb/ft^{3} ) .
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