Liquid Pressure and Hydraulics: Worked Examples

Liquid pressure and hydraulics: choose the depth and area

A pressure calculation begins by identifying what produces the pressure and where it is measured. A liquid column requires vertical depth and density. A piston requires force and contact area. Combining these ideas is useful, but switching formulas without describing the physical situation often produces an answer with the correct unit and the wrong meaning.

These calculations of liquid pressure and hydraulic force use stationary, uniform-density liquids and ideal pistons unless otherwise stated. Take g = 10 N kg⁻¹. Ignore piston weight, friction and fluid compressibility in the ideal hydraulic examples. The numbers are supplied classroom data, not construction or operating instructions for pressure equipment.

Three relationships, with their limits

Quantities needed before choosing a pressure equation
Relationship	Meaning	Required condition
p = F/A	Normal force per unit area	For a uniform pressure; otherwise it gives an average
p = p_surface + ρgh	Pressure at vertical depth h	Liquid at rest, constant density ρ
F₁/A₁ = F₂/A₂	Equal transmitted gauge pressure at the pistons	Ideal pistons at equal elevation with the same external pressure

Pressure is measured in pascals, Pa, where 1 Pa = 1 N m⁻². Density is in kg m⁻³, depth in metres and area in m². Pressure is not force: the same pressure acting over different areas produces different forces.

Absolute pressure is measured relative to a vacuum. Gauge pressure is measured relative to the surrounding reference pressure, normally atmospheric pressure. For an open tank, p_absolute = p_atmospheric + ρgh, while p_gauge = ρgh. An absolute pressure cannot be treated as gauge pressure when calculating a net force against the surrounding atmosphere.

Worked example 1: an open water tank

Setup A, text diagram: An open water surface is the depth origin, h = 0. Point P lies vertically 1.60 m below that surface. Point Q lies at the same horizontal level as P, near a sloping tank wall. Water density is 1,000 kg m⁻³; atmospheric pressure is 100,000 Pa.

At P, gauge pressure = ρgh = 1,000 × 10 × 1.60 = 16,000 Pa = 16.0 kPa. Absolute pressure = 100,000 + 16,000 = 116,000 Pa = 116 kPa. Giving 16 kPa as the absolute pressure would leave out the atmospheric contribution.

Q has the same pressure as P because the points are at equal depth in the same connected stationary liquid. Distance along the sloping wall is irrelevant. If that sloping distance were 2.00 m, substituting it as h would overestimate the liquid pressure. Vertical depth means the vertical distance below the free surface, not the length of a pipe or container wall.

The equal-depth statement needs its conditions. Separate vessels can have different pressures at equally deep points if different pressures act on their surfaces. Different liquids can also produce different pressures at equal depths. “Equal depth means equal pressure” is incomplete unless the liquid and surface-pressure conditions are specified.

Worked example 2: force on a small horizontal cover

A small horizontal cover at P has area 25.0 cm². Water acts on its inside; air at the stated atmospheric pressure acts on its outside. Find the net force due to the pressure difference, ignoring the cover’s weight.

First convert area: 1 cm² = (0.01 m)² = 0.0001 m². Therefore 25.0 cm² = 0.00250 m². The net force is F = ΔpA = 16,000 × 0.00250 = 40.0 N, directed from the water side towards the air side.

The tempting conversion 25 cm² = 0.25 m² is wrong because an area conversion squares the length conversion. It would give 4,000 N, one hundred times too large. Using absolute pressure would instead calculate 290 N on the water side alone; the outside atmosphere provides an opposing 250 N. Subtracting those gives the same 40 N net force.

A vertical cover extending over a substantial depth would not experience uniform liquid pressure. This horizontal-cover example deliberately avoids that complication. State whether your result is a one-sided force or a resultant force before applying pA.

Worked example 3: an ideal hydraulic press

Setup B, text diagram: Two horizontal pistons enclose a connected liquid. Their liquid-contact faces are at the same elevation. The small piston has area 4.0 cm² and receives a downward applied force of 60 N. The large piston has area 100 cm². Both outer faces are exposed to the same atmosphere.

Convert the small area: 4.0 cm² = 0.00040 m². The applied gauge pressure is 60/0.00040 = 150,000 Pa. The large area is 0.0100 m², so its ideal supported load is 150,000 × 0.0100 = 1,500 N.

An equivalent method uses the area ratio: F₂ = F₁(A₂/A₁) = 60 × (100/4) = 1,500 N. Areas may remain in cm² in this ratio because the identical units cancel. They must be in m² when calculating pressure in pascals.

The pressure is the same at the two faces; the forces are not. If diameters rather than areas are given, square the diameter ratio. Doubling a circular piston’s diameter multiplies its area by four, not two. Pascal’s principle concerns a transmitted pressure change, and does not remove the ordinary pressure variation with height.

Worked example 4: displacement and work check

In Setup B, the small piston moves down 20 cm. Assume the face-height difference during motion is negligible for this ideal calculation. How far does the large piston rise, and has the machine created energy?

The displaced liquid volumes are equal: A₁d₁ = A₂d₂. Therefore d₂ = (4/100) × 20 = 0.80 cm = 0.0080 m. Input work = 60 × 0.20 = 12 J. Output work = 1,500 × 0.0080 = 12 J. The larger force moves through a proportionally smaller distance.

Setup C, unequal-height boundary case: If the output face is 0.50 m higher than the input face in liquid of density 800 kg m⁻³, the static pressure at the higher face is 800 × 10 × 0.50 = 4,000 Pa lower. At that position, 150,000 Pa input gauge pressure corresponds to 146,000 Pa output gauge pressure, supporting 1,460 N on 0.0100 m². Simply equating the two pressures would overlook the height difference.

Three solved practice tasks

Practice 1: identify the reference pressure

Oil of density 850 kg m⁻³ is in an open tank. A point is 0.80 m below its surface. Atmospheric pressure is 101 kPa. Find gauge and absolute pressures.

Solution: Gauge pressure = 850 × 10 × 0.80 = 6,800 Pa = 6.8 kPa. Absolute pressure = 101 + 6.8 = 107.8 kPa. Add quantities in matching units; adding 101 directly to 6,800 would mix kilopascals and pascals.

Practice 2: use diameters correctly

An ideal equal-level hydraulic system has circular piston diameters 2 cm and 8 cm. An 80 N input acts on the smaller piston. Find the output force and output movement for a 12 cm input movement.

Solution: Area ratio = (8/2)² = 16. Output force = 80 × 16 = 1,280 N. Output movement = 12/16 = 0.75 cm. Input work is 80 × 0.12 = 9.6 J; output work is 1,280 × 0.0075 = 9.6 J.

Practice 3: compare tank shapes

Two open tanks contain the same liquid. Their depths are both 0.60 m, but one tank widens towards the top. Must their base pressures and total base forces be equal?

Solution: Their gauge base pressures are equal because ρ, g and h are equal. Their total base forces are equal only if their base areas are equal. Shape alone does not supply either missing area or an additional pressure term.

Extend the pressure comparison to a solid and a gas

The liquid and hydraulic cases above remain unchanged. These two supplied-data cases distinguish the average pressure at a solid’s contact face from pressure produced by gas-particle collisions. They are paper exercises, with no pressurised-device construction, heating or operating instructions.

Worked solid-contact case

Text diagram: A rigid rectangular block rests on a horizontal surface. Its weight is 120 N, directed downwards. The surface provides an equal upward normal force. Orientation A has a 20 cm × 10 cm contact rectangle; orientation B has a 10 cm × 5 cm contact rectangle. No extra vertical load, acceleration, partial contact or deformation is included.

For A, area = 200 cm² = 0.0200 m². Average contact pressure = F/A = 120/0.0200 = 6000 Pa. For B, area = 50 cm² = 0.00500 m², giving 120/0.00500 = 24,000 Pa. The force is unchanged and the contact area is one quarter as large, so the average pressure is four times as large. These are average normal contact stresses due to the block, not an absolute gas pressure. Local stress may vary over a real contact face.

Worked gas-pressure case with fixed temperature

State map: A sealed, fixed amount of ideal gas has equilibrium state 1: volume 300 cm³, absolute pressure 100 kPa, temperature 300 K. Equilibrium state 2 has volume 200 cm³ and the same temperature 300 K. No gas leaks, reacts or condenses. The surrounding reference pressure is 100 kPa in both states.

With fixed amount and temperature, pV = constant, so p₂ = p₁V₁/V₂ = 100 × 300/200 = 150 kPa absolute. Matching volume units cancel in the ratio. State 2 gauge pressure is 150 − 100 = 50 kPa. State 1 gauge pressure is zero, but its absolute pressure is not zero.

At fixed temperature, the model’s mean translational kinetic energy stays unchanged. The same particles in a smaller volume strike the walls more frequently per unit area, increasing pressure. Do not claim that they necessarily move faster: that would imply a changed temperature. The calculation compares settled states at the same temperature; a rapid compression that warms a real gas does not satisfy this assumption. Use absolute pressure in Boyle’s law. Multiplying the initial gauge value of zero by 300/200 would incorrectly predict zero throughout.

Independent checks

A 90 N block has complete contact over 150 cm². Find average pressure. What happens if the area alone doubles?
A fixed ideal-gas sample changes from 250 cm³ at 120 kPa absolute to 400 cm³ at the same temperature. Atmospheric pressure stays 100 kPa. Find the final absolute and gauge pressures, and explain the negative gauge value.

Checked answers: (1) 150 cm² = 0.0150 m²; pressure = 90/0.0150 = 6000 Pa. Doubling area at fixed normal force halves it to 3000 Pa. (2) p₂ = 120 × 250/400 = 75 kPa absolute; gauge pressure = 75 − 100 = −25 kPa. The gas is 25 kPa below the surrounding reference, not below a vacuum. The stated final volume must be supported by an unspecified boundary arrangement; do not assume a freely moving piston would remain in that state against atmosphere alone.

Principle check: OpenStax, the ideal gas law. The fixed-amount, fixed-temperature restriction gives Boyle’s law from pV = nRT; no new hydraulic model is required.

Frequently asked questions

Is liquid pressure always gauge pressure?

No. The expression ρgh gives the increase below the reference surface. Include the surface pressure when absolute pressure is required.

Why can atmospheric pressure cancel?

It cancels in a pressure difference when the same atmosphere acts on both relevant external faces. It does not disappear from absolute pressure.

Does a narrow container give higher pressure?

At the same depth in the same stationary liquid with equal surface pressure, no. Container width does not occur in ρgh.

Does a hydraulic press multiply pressure?

An ideal system transmits the applied pressure change. The larger piston multiplies force through its larger area.

Why is the real output smaller?

Friction and other losses can reduce useful output. Piston weight and height differences also need accounting when the ideal assumptions do not apply.

What is the fastest numerical check?

Check area conversion, pressure reference and force ratio. If force increases, an ideal equal-level output movement should decrease by the same factor.

Further revision

Use the existing Physics topical revision collection, which requires login for its contents. The spelling of its address is retained exactly. Background principles are explained in OpenStax’s pressure-depth section and Pascal’s principle section.