Pressure & Head
Converting between pressure and the height of a liquid column - pumps, tanks and gauges.
Why this matters
Where it occurs
Pump curves are written in metres of head while gauges read in kPa or psi. Tradespeople convert between them constantly.
Why calculate it
Pressure at the bottom of a liquid column depends on density, gravity and height - the same physics that feeds a fire hose or stresses a tank wall.
What decision it supports
Is the pump producing the head its curve promises? Will the fixture on the third floor get the pressure it needs?
What happens if it is wrong
Confusing gauge pressure with absolute, or metres with kilopascals, leads to wrong pump selection and undersized systems.
The concept
The pressure produced by a column of liquid is P = ρ x g x h. For water, every 10 m of height adds roughly 98 kPa.
Rearranged, pressure converts to head: h = P / (ρ x g). A gauge reading of 350 kPa on water is about 35.7 m of head.
Gauge pressure reads above atmospheric; absolute pressure adds atmospheric on top. For pump and pipe work, gauge values are what you work with.
Formula
Hydrostatic Pressure
P = rho x g x h
- P = Pressure (pascals (Pa))
- ρ = Density (kg/m³ (water ≈ 1000))
- g = Gravity (9.81 m/s²)
- h = Height of liquid (metres (m))
Worked example
Problem: A tank of water is 20 m tall. What pressure exists at the bottom?
- Use P = ρ x g x h.
- Substitute: P = 1000 x 9.81 x 20.
- Calculate: P = 196,200 Pa.
- Convert: 196.2 kPa.
196.2 kPa - about 98 kPa for every 10 m of water column.
Quick solve: practice it
A gauge reads 350 kPa on a water line. What head does it represent?
Quick-solve habit: calculate, then ask yourself - does this answer make sense in real units?
Games that use this mathematics
Apply the calculation inside a workplace simulation.