In a static liquid (a body of liquid at rest), the pressure difference between any two points is in direct proportion to the vertical distance between two points. This pressure difference is due to the weight of the liquid and can be calculated by multiplying the vertical distance by the fluid density (or vertical distance x density of water x specific gravity of the fluid). In commonly used units:
| P (static) = | Z x Density H2O x S.G.
A |
| Unit of Measure | Metric | | English |
| P | kg/cm2 | | psi |
| Z | meters | | feet |
| density of H20 | 1000 kg/m3 | | 62.4 lb/ft |
| A | 10,000 cm2/m2 | | 144 in2/ft2 |
Example: Calculate pressure difference between two points. Vertical distance = 5.49 m (18 ft),
specific gravity = 1.23. | Metric | | English |
| P = Z x | 1000
10,000 | x S.G | | P = Z x | 62.4
144 | x S.G |
| P = 5.49 x 0.1 x 1.23 | | P = 18 x .433 x 1.23 |
| P = 0.6753 kg/cm2 | | P = 9.59 psi |
To obtain pressure in elevation units, the equation is rearranged:
| Z = | P static
Density H20 x S.G. | x A |
Example:
A pressure gauge reads 5.97 kg/cm2 (85 psi). The fluid has specific gravity of 0.95. What is the height of the equivalent column of fluid that would produce that same pressure?
| Metric | | English |
| Z = | 5.97 x 10,000
1000 x .95 | = 62.9 m | | Z = | 85 x 144
62.4 x .95 | = 206.5 ft |
Static headThe hydraulic pressure when the liquid is at rest.
Friction headThe pressure loss due to frictional losses in flow.
Velocity headThe energy in a fluid due to its velocity (e.g. head unit).
Pressure headA pressure measured in equivalent head units.
Discharge headThe outlet pressure of a pump in operation.
Total headThe total pressure difference between the inlet and outlet of a pump in operation.
Inlet headThe positive pressure on a pump inlet.
Suction liftThe negative pressure on a pump inlet.
