Pressure Head (h = P/ρg)

h=Pρgh = \frac{P}{\rho g}

Worked example: 98.0665 kPa of water → h = 10 m — press Try an example to run it live, then adjust anything.

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Pressure Head (h = P/ρg) explained

Pρh

A column of fluid of height hh presses on its base with ρgh\rho g h. Turn that around and any pressure can be quoted as the height of the column that would produce it: h=P/ρgh = P/\rho g. Hydraulic engineers prefer to work this way because head translates straight into the geometry of a building — a number in metres can be compared against the distance from the mechanical room to the top floor, which a number in pascals cannot.

One atmosphere is 101 325/(1000×9.80665)=10.33101\,325/(1000\times9.80665) = 10.33 m of water. That single figure explains why no suction pump anywhere on Earth can draw water up more than about 10 m: atmospheric pressure is doing the pushing, and it has only 10.33 m of head to spend. A practical case from a mechanical room: filling a hydronic system whose highest point is 30 m above the fill valve requires P=1000×9.80665×30=294P = 1000\times9.80665\times30 = 294 kPa (42.7 psi) merely to reach the top, and standard practice adds another 35 kPa or so to keep the high point positively pressurised — so the fill regulator gets set near 330 kPa, about 48 psi. Get that number wrong and the top-floor coils fill with air rather than water.

The same arithmetic with mercury gives the barometer: 101 325/(13 595×9.80665)=0.760101\,325/(13\,595\times9.80665) = 0.760 m, which is where 760 mmHg comes from. It is also why water towers exist. Elevation is stored pressure, and every 10 m of tank height banks another atmosphere available to the distribution network with no pump running at all.

Head is only a pressure once you say what fluid it is in. Ten metres of water is 98 kPa; ten metres of 40% propylene glycol, at about 1045 kg/m³, is 102 kPa; ten metres of mercury is 1330 kPa. This has a consequence that catches people constantly: a centrifugal pump produces head, not pressure. Put a denser fluid through the same pump at the same speed and it delivers the same metres of head, but a higher pressure and a proportionally higher motor load. A pump curve is drawn in metres for exactly this reason, and reading it as though the vertical axis were kPa is a fair way to undersize a motor.

Three further cautions. Head computed from a gauge reading is head above atmospheric, so a calculation that mixes it with an absolute pressure is out by 10.33 m of water. Density moves with temperature — water at 80 °C is about 972 kg/m³, so the same pressure buys 6% more head than it does at 20 °C, which is small but not nothing when the margins are thin. And do not confuse this static head with what a pump must actually produce. In an open system the pump lifts the fluid and the elevation is a real cost, but in a closed loop the down leg returns everything the up leg spent, so the static height cancels completely and the pump is sized for friction alone. The building's height sets the fill pressure, not the pump head — two different questions that use the same equation and are routinely answered with each other's number.

Pressure Head (h = P/ρg) formula

h=Pρgh = \frac{P}{\rho g}
Where
  • hh= Pressure head (m)
  • PP= Pressure (kPa)
  • ρ\rho= Fluid density (kg/m³)