Physics · Fluids

Hydrostatic pressure in fluids at rest

In a static liquid of constant density, local pressure rises linearly with vertical depth; add pressure at the reference surface.

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Formula

Calculate

p = p_{0} + ρ g h

p and p₀ in Pa; ρ in kg/m³; g in m/s²; h in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Absolute pressure at depth	Pa
$p₀$	Pressure at the surface	Pa
$ρ$	Fluid density	kg/m³
$g$	Gravitational acceleration	m/s²
$h$	Vertical depth below the surface	m

LaTeX

p = p_0 + \rho g h

QUANTA STUDYquanta-study.de

Liquid column with a free surface and depth marker h; sideways pressure arrows grow longer with depth. — The hydrostatic pressure difference grows with depth h: Δp = ρgh; p₀ is absolute pressure at the free surface. The pressure arrows show the depth-proportional increment Δp.

Hydrostatic pressure calculator

Find the pressure increase, absolute pressure, depth or density in a liquid at rest with constant density, including below two liquid layers.

Calculation model

Find

Density ρ

Density of the liquid, e.g. 1000 kg/m³ for water, about 1025 kg/m³ for seawater.

Gravitational acceleration g

Gravitational acceleration, 9.81 m/s² on Earth.

m/s²

Depth h

Vertical depth below the surface, e.g. 10 m when diving.

Filled with example values. Type your own and the result updates straight away.

Result: Pressure increase Δp

Δp = 98.1 kPa

With ρ = 1000 kg/m³, g = 9.81 m/s² and h = 10 m, Δp = 98.1 kPa.

Working · rearrange for Δp, substitute, evaluate

1. Rearrange for Δp
$\Delta p=\rho gh$
2. Substitute values (in SI units)
$1000\cdot 9.81\cdot 10=98100$
3. Result
Δp = 98.1 kPa

Quick tips

Rule of thumb for water: every 10 m of depth adds about 1 bar (exactly 0.981 bar). At 10 m the total pressure is roughly twice atmospheric.
Only the vertical depth counts, not the shape or width of the container: 1 m below the surface of a thin water pipe the pressure equals that 1 m deep in a lake.

Assumptions and limits

Static fluid of constant positive density and constant g>0; h≥0 is vertical depth. Δp is the pressure increase from the same reference surface, not automatically atmospheric gauge pressure.

Derivation & background

An imaginary static column obeys pA − p₀A = ρAhg. Cancelling area gives p − p₀ = ρgh. p and p₀ need the same reference. Equal elevation gives equal pressure within the same connected static liquid. Vessel shape and total weight do not determine this local increment; different surface pressures or layers need explicit treatment.

Exam blueprint

Validity range

Static fluid, constant density and g; h is vertical depth. Specify surface pressure separately. Equal elevation gives equal pressure in the same connected static liquid, not automatically in separate vessels.

Derivation steps

An imaginary static liquid column balances pressure forces with its weight; area cancels.

1Specify vertical depth h below a reference surface and pressure p₀ there.
2Consider an imaginary column of cross-sectional area A: V = Ah.
3At constant density, column mass is m = ρAh.
4Bottom pressure gives upward pA; top pressure p₀A and weight ρAhg act downward.
5Write equilibrium: pA − p₀A − ρAhg = 0.
6Divide by A: p = p₀ + ρgh; p and p₀ use the same reference.
7Varying density or g requires integration; equal depth means equal pressure only under matching connected conditions.

Rearrangements

Depth from pressure difference

h=\frac{p-p_0}{\rho g}

Use pressure difference p−p₀ with the same reference. It is atmospheric gauge pressure only when p₀ is ambient pressure; do not mix absolute and relative values.

Task variant

What is the gauge pressure 3.0 m below water? Use ρ=1000 kg/m³ and g=9.81 m/s².

$\Delta p=\rho gh=1000\cdot9.81\cdot3.0\,\mathrm{Pa}=29430\,\mathrm{Pa}\approx29.4\,\mathrm{kPa}$ .

Common mistakes

Including vessel shape or measuring h along a sloped wall.

Only vertical depth matters. Equal pressure at equal elevation applies within the same connected stationary fluid under matching boundary conditions; separate surface pressures may differ.

Exam context

Pressure on a diver, fluid columns, manometers and comparisons between depths.

Formula cluster

Fluids at rest and in motion

Connects pressure, density, buoyancy and flow.

p = F/A

Pressure

ρ = m/V

Density

F_A = ρ·V·g

Buoyant Force (Archimedes)

p + ½ρv² + ρgh = const.

Bernoulli Equation

Worked example

At 5.0 m depth in water, Δp=1000·9.81·5.0=49,050 Pa≈49.1 kPa. With p₀=101.3 kPa, absolute pressure is about 150.4 kPa. Dimensional check: ρgh has units kg/(m·s²)=Pa.

Applications

Manometers, diving, liquid tanks and pressure distribution in stationary fluids.

How do you calculate with Hydrostatic pressure?

Here is how to work through a typical Hydrostatic pressure (p = p₀ + ρgh) task step by step:

1
Task
What is the gauge pressure 3.0 m below water? Use ρ=1000 kg/m³ and g=9.81 m/s².
Solution path
$\Delta p=\rho gh=1000\cdot9.81\cdot3.0\,\mathrm{Pa}=29430\,\mathrm{Pa}\approx29.4\,\mathrm{kPa}$ .

Understand and apply the equation

Why does pressure increase as you dive deeper? In a stationary liquid, the pressure gradient must balance the fluid’s weight. Derive the depth relation and distinguish pressure increase from total pressure, including cases with multiple liquid layers.

How do you calculate pressure increase and absolute pressure at depth?

Imagine a small column of fluid within water. It remains at rest despite its downward weight. Pressure force at its lower face must therefore exceed pressure force at its upper face by exactly that weight. This balance explains the increase with depth. It does not require a physical column above every measurement point: vessel shape may differ while the fluid remains stationary in the same gravitational field.[1]

$p=p_0+\rho gh$ gives local pressure in a static liquid of constant density with constant g. h is positive vertical depth below a chosen reference surface. p₀ is pressure there, not automatically standard atmospheric pressure. An open container uses actual pressure above its liquid; a closed one may have another value.[1]

\begin{aligned}&p(h)=p_0+\rho gh\\&\Delta p=p(h)-p_0=\rho gh\end{aligned}

p and p₀ must use the same pressure reference; their difference is reference-independent.

To obtain absolute p, insert absolute p₀. The term ρgh is a pressure difference. If p₀ is ambient pressure, it is gauge pressure relative to that environment. h is neither pipe length nor an inclined distance.

Balancing a static liquid column

Consider an imaginary vertical column of area A and height h. Bottom pressure produces upward force pA; top pressure contributes downward p₀A. Column mass is ρAh and its weight is ρAhg. Side-pressure forces on this column have no vertical component. Static equilibrium requires zero net force.[1]

\begin{aligned}pA-p_0A-\rho Ahg&=0\\p-p_0&=\rho gh\end{aligned}

A cancels; pressure increase is independent of the imaginary column’s area.

\begin{aligned}&\frac{\mathrm dp}{\mathrm dz}=-\rho g\\&\frac{\mathrm dp}{\mathrm dh}=\rho g\end{aligned}

z points upward and h downward. Positive density makes pressure increase downward.

The imaginary column is a local balance device. A real straight liquid column need not exist above every wall point. In connected static regions, the pressure field also accommodates inclined walls and irregular vessels.

Rearrange using pressure differences

\begin{aligned}&h=\frac{p-p_0}{\rho g}\\&\rho=\frac{p-p_0}{gh}\\&[\rho gh]=\mathrm{kg\,m^{-1}\,s^{-2}}=\mathrm{Pa}\end{aligned}

Rearrangements require positive ρ,g or h respectively; use matching pressure units.

For water modelled at $\rho=1000\,\mathrm{kg\,m^{-3}}$ with $g=9.81\,\mathrm{m\,s^{-2}}$ , each metre adds $9810\,\mathrm{Pa}=9.81\,\mathrm{kPa}$ . Ten metres adds 98.1 kPa, approximately but not exactly one bar. Absolute pressure without known p₀ does not uniquely determine depth.

Examples: pressure in seawater and beneath two liquid layers

Pressure at eight metres depth

Given: Static water with assumed density 1025 kg/m³; h = 8 m; g = 9.81 m/s²; surface pressure p₀ = 101.3 kPa absolute.
Find: Find added pressure and absolute pressure.

Model: Water is stationary and density and g are constant. Calculate the pressure increase first, then add absolute surface pressure.
$\begin{aligned}\Delta p&=\rho gh\\p&=p_0+\Delta p\end{aligned}$
Approach: At constant density use $\Delta p=\rho gh$ , then $p=p_0+\Delta p$ . Calculate the hydrostatic difference.
$\Delta p=1025\cdot9.81\cdot8=80442\,\mathrm{Pa}$
Match units before addition.
$\Delta p=80.442\,\mathrm{kPa}$
Add surface pressure.
$p=101.3+80.442=181.742\,\mathrm{kPa}$

Result: Added pressure is 80.442 kPa; absolute pressure is 181.742 kPa in this model.

Check: $h=\frac{p-p_0}{\rho g}=\frac{181742-101300}{1025\cdot9.81}\,\mathrm m=8\,\mathrm m$ . Density is a problem assumption, not a universal seawater value.

Oil above water in an open vessel

Given: Two static immiscible layers: 0.30 m oil at 800 kg/m³ above 0.70 m water at 1000 kg/m³. p₀ = 101.3 kPa absolute; g = 9.81 m/s². Neglect capillary pressure.
Find: Find pressure at the interface and bottom.

Model: Each stationary layer has its own constant density. With capillary pressure neglected, pressure is continuous at the interface.
$\begin{aligned}\Delta p_i&=\rho_i gh_i\\p_{\mathrm{interface}}&=p_0+\Delta p_{\mathrm{oil}}\\p_{\mathrm{bottom}}&=p_0+\Delta p_{\mathrm{oil}}+\Delta p_{\mathrm{water}}\end{aligned}$
Approach: Each layer adds $\Delta p_i=\rho_i gh_i$ ; bottom pressure is $p=p_0+\sum_i\Delta p_i$ . First traverse the upper layer.
$\Delta p_{\mathrm{oil}}=800\cdot9.81\cdot0.30=2354.4\,\mathrm{Pa}$
Use interface pressure as the next reference.
$p_{\mathrm{interface}}=103654.4\,\mathrm{Pa}$
Add the water-layer contribution.
$p_{\mathrm{bottom}}=103654.4+1000\cdot9.81\cdot0.70=110521.4\,\mathrm{Pa}$

Result: Interface pressure is 103.6544 kPa; bottom pressure is 110.5214 kPa absolute.

Check: Total added pressure 9221.4 Pa is below 9810 Pa for one metre of pure water. A lighter upper layer reduces the increment at equal total depth.

View diagram: pressure across two fluid layers

Pressure increase in oil and water

QUANTA STUDYquanta-study.de

Oil extends from the surface to 0.30 m depth, with water below it to 1 m. The curve is continuous at the interface: pressure increase is 2.3544 kPa at 0.30 m and 9.2214 kPa at the base. Pressure increases more per metre of depth in water.[1]

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Equal depth, vessel shape and connected regions

Points at equal elevation in the same connected static liquid of uniform density have equal pressure; otherwise a horizontal pressure difference would accelerate the fluid. Separate vessels with different surface pressures can have different pressure at equal depth. Depth alone is insufficient without boundary conditions. Buoyancy arises from the sum of pressure forces on an immersed body; in the appropriate model its magnitude equals the weight of displaced fluid.[4]

Pressure force on a horizontal flat bottom is pA for uniform p. It need not equal total liquid weight. Inclined walls in widening or narrowing vessels carry vertical pressure-force components, and external pressure also acts. This resolves the hydrostatic paradox.

On a vertical wall, pressure increases with depth; bottom pressure times the entire wall area is incorrect. Integrate over the surface. For a vertical rectangular wall in uniform liquid, the mean hydrostatic pressure difference equals its value at mean depth $\frac{h_{\mathrm{top}}+h_{\mathrm{bottom}}}{2}$ . If the top edge lies at the reference level, this is half the bottom-edge depth.

Manometers and added surface pressure

A U-tube manometer connects pressures through static liquid columns. Trace the tube: add ρgΔh downward and subtract it upward. A single uniform manometer liquid gives pressure difference ρgΔh for level difference Δh. Include gas density and additional liquid layers when their contributions matter. Pressure must always be stated with its measurement reference.[2]

Increasing pressure over an enclosed liquid by 20 kPa raises pressure at every unchanged depth by 20 kPa in the ideal static model. The depth increment ρgh remains. Separating uniform pressure change from height-dependent baseline pressure is Pascal’s principle.[3]

Variable density, acceleration and flow

p(h)-p_0=\int_0^h\rho(\xi)g(\xi)\,\mathrm d\xi

Integrate varying quantities rather than multiplying by an arbitrary single value.

Gas density may vary strongly with pressure and height. Stratified liquids also may require more than one density. Accelerating or rotating vessels change the equilibrium field; their pressure distributions cannot simply use unchanged vertical ρgh.

The derivation assumes rest. In flowing pipes, speed changes, friction, pumps and valves also affect pressure. A static pipe pressure inferred from elevation therefore does not automatically give pressure with an open tap.

Five problems with solutions

Calculate · Foundation

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m³ and g = 9.81 m/s².

Show hint

Calculate only the increment ρgh.

Learning goal: Calculate hydrostatic pressure increase from water depth.

Show solution and calculation

Approach: Only the pressure increase relative to the surface is requested.
For stationary water of constant density, use the hydrostatic increment without adding surface pressure.
Result: 24.525 kPa.

Check: The increase is 9.81 kPa per metre; multiplying by 2.5 gives the same value. Absolute pressure remains unknown without surface pressure.

\begin{aligned}\Delta p&=\rho gh\\&=1000\cdot9.81\cdot2.5\,\mathrm{Pa}\\&=24525\,\mathrm{Pa}=24.525\,\mathrm{kPa}\end{aligned}

[1]

Rearrange · Foundation

For a liquid at ρ = 800 kg/m³, p − p₀ = 15.696 kPa. Find depth with g = 9.81 m/s².

Show hint

Use the pressure difference in pascals.

Learning goal: Find depth from the known pressure increase relative to surface pressure and fluid density.

Show solution and calculation

Approach: Solve the pressure difference for depth.
Convert 15.696 kPa to 15 696 Pa and divide this difference by density times g.
Result: 2 m below the reference surface.

Check: This liquid adds 7.848 kPa per metre, giving 15.696 kPa for two metres. Vessel shape is not needed.

\begin{aligned}h&=\frac{\Delta p}{\rho g}\\&=\frac{15696}{800\cdot9.81}\,\mathrm m=2\,\mathrm m\end{aligned}

[1]

Calculate · Application

A water vessel has p₀ = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m³ and g = 9.81 m/s².

Show hint

Use the actual surface pressure.

Learning goal: Correctly combine hydrostatic increase with absolute surface pressure.

Show solution and calculation

Approach: Absolute pressure equals absolute surface pressure plus the hydrostatic increase.
Result: 149.43 kPa absolute.

Check: Subtracting surface pressure leaves 29.43 kPa for three metres of water. Use the stated vessel pressure of 120 kPa; replacing it with a standard atmospheric value would describe a different system.

\begin{aligned}p&=p_0+\rho gh\\\rho gh&=1000\cdot9.81\cdot3\,\mathrm{Pa}=29.43\,\mathrm{kPa}\\p&=(120+29.43)\,\mathrm{kPa}=149.43\,\mathrm{kPa}\end{aligned}

[1]

Interpret · Application

Two open vessels contain stationary water of equal density 1000 kg/m³ to 1.20 m above a horizontal base. Base areas are 20 cm² and 50 cm²; g=10 m/s² and ambient pressure is identical. Compare gauge pressure and the additional base force due to the water column. Does equal gauge pressure imply equal forces?

Show hint

Gauge pressure depends on depth. Multiplication by loaded area is needed to obtain force.

Learning goal: Distinguish equal pressure from different resultant bottom forces.

Show solution and calculation

Approach: calculate the common gauge pressure first, then convert each area to square metres and use it separately.
Both bases have 12 kPa gauge pressure, but additional forces are 24 N and 60 N.

Check: force increases in the area ratio. Only the additional pressure loading is requested; atmospheric contributions and the complete vessel force balance must be considered separately.

\begin{aligned}\Delta p&=\rho gh=12\,000\,\mathrm{Pa}\\F_1&=\Delta pA_1=(12\,000)(0.0020)\,\mathrm N=24\,\mathrm N\\F_2&=\Delta pA_2=(12\,000)(0.0050)\,\mathrm N=60\,\mathrm N\end{aligned}

[1][2]

Error analysis · Application

A stationary water layer 0.80 m deep has 0.30 m of oil above it. Densities: water 1000 kg/m³, oil 800 kg/m³; g=10 m/s², top surface open. A calculation uses water density over the entire 1.10 m depth and obtains 11 kPa gauge pressure at the base. Correct the value and quantify the absolute error.

Show hint

Add each layer’s pressure increase using its own density; pressure is continuous at the interface.

Learning goal: Balance layered fluids with piecewise constant density.

Show solution and calculation

Approach: write gauge pressure as the sum of the two hydrostatic contributions.
Oil adds 2.4 kPa and water 8.0 kPa, giving 10.4 kPa.
The incorrect calculation is 0.6 kPa too high.

Check: lighter oil must yield less pressure than an equally deep all-water column. Absolute pressure additionally requires adding ambient pressure.

\begin{aligned}\Delta p&=g(\rho_{\mathrm{oil}}h_{\mathrm{oil}}+\rho_{\mathrm{water}}h_{\mathrm{water}})\\&=10(800\cdot0.30+1000\cdot0.80)\,\mathrm{Pa}=10\,400\,\mathrm{Pa}\\p_{\mathrm{wrong}}-p_{\mathrm{correct}}&=11\,000-10\,400=600\,\mathrm{Pa}\end{aligned}

[1]

Key points

In a fluid at rest, pressure rises with depth. At constant density and g, the added pressure is $\rho gh$ ; include surface pressure to obtain absolute pressure. For layered liquids, add the contribution of each layer.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Distinguished pressure increase from absolute pressure in the heading; added paragraph connections explaining measurement reference and buoyancy, with a primary source.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

Cite this reference page

Quanta. Hydrostatic pressure. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/hydrostatischer-druck

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Frequently asked questions about Hydrostatic pressure

What does hydrostatic pressure describe physically?+

In a static fluid, the pressure gradient balances weight. Constant density and g therefore make pressure rise with vertical depth: p = p₀ + ρgh. p₀ is pressure at the chosen reference surface. For water at 1000 kg/m³ with g = 9.81 m/s², each metre adds 9.81 kPa. This increment is a pressure difference, not complete absolute pressure. An open vessel uses actual ambient pressure for p₀; a closed system may have another surface pressure. Vessel shape does not appear in this local relation, but fluid density and matching boundary conditions do matter.

How can I check a calculation involving hydrostatic pressure?+

First check that h is vertical depth below the same reference surface where p₀ is specified. Calculate Δp = ρgh, then add p₀ after matching units. For 8 m, density 1025 kg/m³ and g = 9.81 m/s², Δp = 80442 Pa. With p₀ = 101.3 kPa absolute, p = 181.742 kPa absolute. Back-substitution in h = (p − p₀)/(ρg) must recover 8 m. At h = 0, pressure must equal p₀. Doubling depth doubles only the added pressure, not automatically total absolute pressure. Record assumed density separately from measured pressure.

What assumptions and limits apply to hydrostatic pressure?+

The simple linear form assumes static fluid, constant density and approximately constant gravity. h is vertical depth; p and p₀ need the same pressure reference. Add separate contributions for multiple layers and integrate ρg over depth for continuously varying density. Constant density is often unsuitable for gases over large elevation changes. Accelerating or rotating vessels also change the equilibrium field. Flowing pipes may have additional pressure changes from speed, friction, valves or pumps. Elevation difference alone therefore does not generally determine pressure at an open tap or inside a moving vessel. State which of these effects the chosen model neglects.

How do I check units and scale for hydrostatic pressure?+

Use density in kg/m³, g in m/s² and depth in metres. Then ρgh has units kg/(m·s²), or pascals. One kPa is 1000 Pa and one bar is 100000 Pa. Ten metres of water modelled at 1000 kg/m³ adds 98100 Pa, or 98.1 kPa, for g = 9.81 m/s². “One bar per ten metres” is thus an approximation, not an exact conversion. Compare pressures with the same reference. Absolute pressure includes surface pressure; pressure increment relative to the surface does not. Convert centimetre depths to metres before substituting.

When should I combine hydrostatic pressure with another relation?+

For pressure force on a planar area, combine the pressure field with F⊥ = ∫p dA. Pressure is constant over a horizontal bottom but increases with depth on a vertical wall. Buoyancy adds pressure forces on all relevant body surfaces, yielding displaced fluid weight in the appropriate model. A manometer connects measurement points through static columns; add ρgΔh with the correct signs. Bottom force need not equal total liquid weight because inclined walls can carry vertical pressure-force components. Flow may require an extended energy or momentum balance rather than ρgh alone. Choose the additional relation to match the actual system boundary.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §14.1: Fluids, Density, and Pressure

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F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

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E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Hydrostatic pressure in fluids at rest

In a static liquid of constant density, local pressure rises linearly with vertical depth; add pressure at the reference surface.

BasicExam-relevantLower secondaryUpper secondaryUniversity/college

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

p

Absolute pressure at depth

p₀

Pressure at the surface

ρ

Fluid density

kg/m³

g

Gravitational acceleration

m/s²

h

Vertical depth below the surface

Derivation & background

Exam blueprint

Validity range

Derivation steps

An imaginary static liquid column balances pressure forces with its weight; area cancels.

1Specify vertical depth h below a reference surface and pressure p₀ there.
2Consider an imaginary column of cross-sectional area A: V = Ah.
3At constant density, column mass is m = ρAh.
4Bottom pressure gives upward pA; top pressure p₀A and weight ρAhg act downward.
5Write equilibrium: pA − p₀A − ρAhg = 0.
6Divide by A: p = p₀ + ρgh; p and p₀ use the same reference.
7Varying density or g requires integration; equal depth means equal pressure only under matching connected conditions.

How do you calculate with Hydrostatic pressure?

Here is how to work through a typical Hydrostatic pressure (p = p₀ + ρgh) task step by step:

Task

What is the gauge pressure 3.0 m below water? Use ρ=1000 kg/m³ and g=9.81 m/s².

Solution path

$\Delta p=\rho gh=1000\cdot9.81\cdot3.0\,\mathrm{Pa}=29430\,\mathrm{Pa}\approx29.4\,\mathrm{kPa}$ .

How do you calculate pressure increase and absolute pressure at depth?

\begin{aligned}&p(h)=p_0+\rho gh\\&\Delta p=p(h)-p_0=\rho gh\end{aligned}

p and p₀ must use the same pressure reference; their difference is reference-independent.

Balancing a static liquid column

\begin{aligned}pA-p_0A-\rho Ahg&=0\\p-p_0&=\rho gh\end{aligned}

A cancels; pressure increase is independent of the imaginary column’s area.

\begin{aligned}&\frac{\mathrm dp}{\mathrm dz}=-\rho g\\&\frac{\mathrm dp}{\mathrm dh}=\rho g\end{aligned}

z points upward and h downward. Positive density makes pressure increase downward.

Rearrange using pressure differences

\begin{aligned}&h=\frac{p-p_0}{\rho g}\\&\rho=\frac{p-p_0}{gh}\\&[\rho gh]=\mathrm{kg\,m^{-1}\,s^{-2}}=\mathrm{Pa}\end{aligned}

Rearrangements require positive ρ,g or h respectively; use matching pressure units.

Examples: pressure in seawater and beneath two liquid layers

Pressure at eight metres depth

Given: Static water with assumed density 1025 kg/m³; h = 8 m; g = 9.81 m/s²; surface pressure p₀ = 101.3 kPa absolute.
Find: Find added pressure and absolute pressure.

Model: Water is stationary and density and g are constant. Calculate the pressure increase first, then add absolute surface pressure.
$\begin{aligned}\Delta p&=\rho gh\\p&=p_0+\Delta p\end{aligned}$
Approach: At constant density use $\Delta p=\rho gh$ , then $p=p_0+\Delta p$ . Calculate the hydrostatic difference.
$\Delta p=1025\cdot9.81\cdot8=80442\,\mathrm{Pa}$
Match units before addition.
$\Delta p=80.442\,\mathrm{kPa}$
Add surface pressure.
$p=101.3+80.442=181.742\,\mathrm{kPa}$

Result: Added pressure is 80.442 kPa; absolute pressure is 181.742 kPa in this model.

Check: $h=\frac{p-p_0}{\rho g}=\frac{181742-101300}{1025\cdot9.81}\,\mathrm m=8\,\mathrm m$ . Density is a problem assumption, not a universal seawater value.

Oil above water in an open vessel

Given: Two static immiscible layers: 0.30 m oil at 800 kg/m³ above 0.70 m water at 1000 kg/m³. p₀ = 101.3 kPa absolute; g = 9.81 m/s². Neglect capillary pressure.
Find: Find pressure at the interface and bottom.

Model: Each stationary layer has its own constant density. With capillary pressure neglected, pressure is continuous at the interface.
$\begin{aligned}\Delta p_i&=\rho_i gh_i\\p_{\mathrm{interface}}&=p_0+\Delta p_{\mathrm{oil}}\\p_{\mathrm{bottom}}&=p_0+\Delta p_{\mathrm{oil}}+\Delta p_{\mathrm{water}}\end{aligned}$
Approach: Each layer adds $\Delta p_i=\rho_i gh_i$ ; bottom pressure is $p=p_0+\sum_i\Delta p_i$ . First traverse the upper layer.
$\Delta p_{\mathrm{oil}}=800\cdot9.81\cdot0.30=2354.4\,\mathrm{Pa}$
Use interface pressure as the next reference.
$p_{\mathrm{interface}}=103654.4\,\mathrm{Pa}$
Add the water-layer contribution.
$p_{\mathrm{bottom}}=103654.4+1000\cdot9.81\cdot0.70=110521.4\,\mathrm{Pa}$

Result: Interface pressure is 103.6544 kPa; bottom pressure is 110.5214 kPa absolute.

Check: Total added pressure 9221.4 Pa is below 9810 Pa for one metre of pure water. A lighter upper layer reduces the increment at equal total depth.

View diagram: pressure across two fluid layers

Pressure increase in oil and water

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Equal depth, vessel shape and connected regions

Manometers and added surface pressure

Variable density, acceleration and flow

p(h)-p_0=\int_0^h\rho(\xi)g(\xi)\,\mathrm d\xi

Integrate varying quantities rather than multiplying by an arbitrary single value.

Five problems with solutions

Calculate · Foundation

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m³ and g = 9.81 m/s².

Show hint

Calculate only the increment ρgh.

Learning goal: Calculate hydrostatic pressure increase from water depth.

Show solution and calculation

Approach: Only the pressure increase relative to the surface is requested.
For stationary water of constant density, use the hydrostatic increment without adding surface pressure.
Result: 24.525 kPa.

Check: The increase is 9.81 kPa per metre; multiplying by 2.5 gives the same value. Absolute pressure remains unknown without surface pressure.

\begin{aligned}\Delta p&=\rho gh\\&=1000\cdot9.81\cdot2.5\,\mathrm{Pa}\\&=24525\,\mathrm{Pa}=24.525\,\mathrm{kPa}\end{aligned}

[1]

Rearrange · Foundation

For a liquid at ρ = 800 kg/m³, p − p₀ = 15.696 kPa. Find depth with g = 9.81 m/s².

Show hint

Use the pressure difference in pascals.

Learning goal: Find depth from the known pressure increase relative to surface pressure and fluid density.

Show solution and calculation

Approach: Solve the pressure difference for depth.
Convert 15.696 kPa to 15 696 Pa and divide this difference by density times g.
Result: 2 m below the reference surface.

Check: This liquid adds 7.848 kPa per metre, giving 15.696 kPa for two metres. Vessel shape is not needed.

\begin{aligned}h&=\frac{\Delta p}{\rho g}\\&=\frac{15696}{800\cdot9.81}\,\mathrm m=2\,\mathrm m\end{aligned}

[1]

Calculate · Application

A water vessel has p₀ = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m³ and g = 9.81 m/s².

Show hint

Use the actual surface pressure.

Learning goal: Correctly combine hydrostatic increase with absolute surface pressure.

Show solution and calculation

Approach: Absolute pressure equals absolute surface pressure plus the hydrostatic increase.
Result: 149.43 kPa absolute.

\begin{aligned}p&=p_0+\rho gh\\\rho gh&=1000\cdot9.81\cdot3\,\mathrm{Pa}=29.43\,\mathrm{kPa}\\p&=(120+29.43)\,\mathrm{kPa}=149.43\,\mathrm{kPa}\end{aligned}

[1]

Interpret · Application

Two open vessels contain stationary water of equal density 1000 kg/m³ to 1.20 m above a horizontal base. Base areas are 20 cm² and 50 cm²; g=10 m/s² and ambient pressure is identical. Compare gauge pressure and the additional base force due to the water column. Does equal gauge pressure imply equal forces?

Show hint

Gauge pressure depends on depth. Multiplication by loaded area is needed to obtain force.

Learning goal: Distinguish equal pressure from different resultant bottom forces.

Show solution and calculation

Approach: calculate the common gauge pressure first, then convert each area to square metres and use it separately.
Both bases have 12 kPa gauge pressure, but additional forces are 24 N and 60 N.

Check: force increases in the area ratio. Only the additional pressure loading is requested; atmospheric contributions and the complete vessel force balance must be considered separately.

\begin{aligned}\Delta p&=\rho gh=12\,000\,\mathrm{Pa}\\F_1&=\Delta pA_1=(12\,000)(0.0020)\,\mathrm N=24\,\mathrm N\\F_2&=\Delta pA_2=(12\,000)(0.0050)\,\mathrm N=60\,\mathrm N\end{aligned}

[1][2]

Error analysis · Application

A stationary water layer 0.80 m deep has 0.30 m of oil above it. Densities: water 1000 kg/m³, oil 800 kg/m³; g=10 m/s², top surface open. A calculation uses water density over the entire 1.10 m depth and obtains 11 kPa gauge pressure at the base. Correct the value and quantify the absolute error.

Show hint

Add each layer’s pressure increase using its own density; pressure is continuous at the interface.

Learning goal: Balance layered fluids with piecewise constant density.

Show solution and calculation

Approach: write gauge pressure as the sum of the two hydrostatic contributions.
Oil adds 2.4 kPa and water 8.0 kPa, giving 10.4 kPa.
The incorrect calculation is 0.6 kPa too high.

Check: lighter oil must yield less pressure than an equally deep all-water column. Absolute pressure additionally requires adding ambient pressure.

\begin{aligned}\Delta p&=g(\rho_{\mathrm{oil}}h_{\mathrm{oil}}+\rho_{\mathrm{water}}h_{\mathrm{water}})\\&=10(800\cdot0.30+1000\cdot0.80)\,\mathrm{Pa}=10\,400\,\mathrm{Pa}\\p_{\mathrm{wrong}}-p_{\mathrm{correct}}&=11\,000-10\,400=600\,\mathrm{Pa}\end{aligned}

[1]

Key points

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Distinguished pressure increase from absolute pressure in the heading; added paragraph connections explaining measurement reference and buoyancy, with a primary source.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

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Quanta. Hydrostatic pressure. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/hydrostatischer-druck

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Frequently asked questions about Hydrostatic pressure

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Hydrostatic pressure in fluids at rest

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Fluids at rest and in motion

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Applications

How do you calculate with Hydrostatic pressure?

How do you calculate pressure increase and absolute pressure at depth?

Balancing a static liquid column

Rearrange using pressure differences

Examples: pressure in seawater and beneath two liquid layers

Pressure at eight metres depth

Oil above water in an open vessel

Pressure increase in oil and water

Equal depth, vessel shape and connected regions

Manometers and added surface pressure

Variable density, acceleration and flow

Five problems with solutions

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m3 and g = 9.81 m/s2.

For a liquid at ρ = 800 kg/m3, p − p0 = 15.696 kPa. Find depth with g = 9.81 m/s2.

A water vessel has p0 = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m3 and g = 9.81 m/s2.

A stationary water layer 0.80 m deep has 0.30 m of oil above it. Densities: water 1000 kg/m3, oil 800 kg/m3; g=10 m/s2, top surface open. A calculation uses water density over the entire 1.10 m depth and obtains 11 kPa gauge pressure at the base. Correct the value and quantify the absolute error.

Key points

Sources for this explanation

Frequently asked questions about Hydrostatic pressure

Subject references and further reading

More formulas: Mechanics

Retain Hydrostatic pressure for exams

Hydrostatic pressure in fluids at rest

Variables & units

Hydrostatic pressure calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Fluids at rest and in motion

Worked example

Applications

How do you calculate with Hydrostatic pressure?

How do you calculate pressure increase and absolute pressure at depth?

Balancing a static liquid column

Rearrange using pressure differences

Examples: pressure in seawater and beneath two liquid layers

Pressure at eight metres depth

Oil above water in an open vessel

Pressure increase in oil and water

Equal depth, vessel shape and connected regions

Manometers and added surface pressure

Variable density, acceleration and flow

Five problems with solutions

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m3 and g = 9.81 m/s2.

For a liquid at ρ = 800 kg/m3, p − p0 = 15.696 kPa. Find depth with g = 9.81 m/s2.

A water vessel has p0 = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m3 and g = 9.81 m/s2.

A stationary water layer 0.80 m deep has 0.30 m of oil above it. Densities: water 1000 kg/m3, oil 800 kg/m3; g=10 m/s2, top surface open. A calculation uses water density over the entire 1.10 m depth and obtains 11 kPa gauge pressure at the base. Correct the value and quantify the absolute error.

Key points

Sources for this explanation

Frequently asked questions about Hydrostatic pressure

Subject references and further reading

More formulas: Mechanics

Retain Hydrostatic pressure for exams

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m³ and g = 9.81 m/s².

For a liquid at ρ = 800 kg/m³, p − p₀ = 15.696 kPa. Find depth with g = 9.81 m/s².

A water vessel has p₀ = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m³ and g = 9.81 m/s².

Find the pressure increase at 2.5 m water depth with ρ = 1000 kg/m³ and g = 9.81 m/s².

For a liquid at ρ = 800 kg/m³, p − p₀ = 15.696 kPa. Find depth with g = 9.81 m/s².

A water vessel has p₀ = 120 kPa absolute. Find p at h = 3 m with ρ = 1000 kg/m³ and g = 9.81 m/s².