How this calculation works
Use vertical depth in a stationary constant-density fluid. Keep the rise relative to the fluid surface distinct from absolute pressure, especially when the surface is not at atmospheric pressure.
Pressure rise = density × gravity × vertical depth; absolute pressure = surface absolute pressure + riseA worked example
For 1,000 kg/m³, 10 m depth and g = 9.80665 m/s², the pressure rise is 98.0665 kPa. With an entered 101.325 kPa surface absolute pressure, depth pressure is 199.3915 kPa absolute.
Good to know
A single constant-density fluid at rest with constant gravity. No layered/compressible fluids, flow pressure, vessel strength, live atmosphere or diving/medical guidance.
Reference: Reference: model, definitions and assumptions (opens in a new tab) ↗
Pressure rise depends on vertical depth
The default 1,000 kg/m³ fluid at 10 m vertical depth gives 1,000 × 9.80665 × 10 = 98,066.5 Pa, or 98.0665 kPa, of pressure rise. A sloping pipe length is not substituted for that vertical depth.
With constant density and gravity, 5 m gives half the rise and 20 m gives twice the rise. The comparison table holds the supplied surface pressure fixed. A narrow tube and a broad tank can therefore have the same static pressure at the same depth under these assumptions; their total fluid quantities can be very different.
The pressure calculator divides a supplied uniform force by area. Hydrostatic pressure here varies with vertical depth, so an extended surface may require an integrated force calculation.
A pressure reference should travel with the number
At an entered surface absolute pressure of 101.325 kPa, adding the default rise gives 199.3915 kPa absolute at depth. The 98.0665 kPa rise is relative to the supplied fluid surface. It is only also a gauge pressure relative to local air when that surface reference actually matches the relevant local air pressure.
For a sealed vessel with an entered 200 kPa absolute surface pressure, the same fluid and depth give 298.0665 kPa absolute. The fluid-column contribution has not changed. Labeling the rise as absolute pressure would omit the supplied surface contribution.
Stationary-fluid pressure and flowing-fluid power are separate questions
The hydraulic pump power calculator needs flow and an appropriate total pressure rise across a stated boundary. A stationary-fluid column does not by itself determine pumping losses or pump input power. This page does not include changing velocity, pipe friction or layers with different densities.
The pressure calculator uses force divided by area for a uniform-pressure relationship. Applying one pressure to a large vertical surface whose depth varies would require accounting for that variation. Hydrostatic pressure here is the value at the entered depth, not an integrated force on a wall or a vessel-strength assessment.
A few questions, answered
Is 101.325 kPa the current pressure at my location?
No. It is an editable example surface reference. No atmospheric data is fetched.
Does container shape change the calculated pressure at one depth?
Not in the constant-density stationary-fluid model with the same surface pressure and gravity.
Can I use the length of a sloping pipe as depth?
Use vertical depth below the fluid surface. Sloping path length is a different quantity.
Does this include flow losses or layered fluids?
No. It uses one fluid density at rest and constant gravity.