Pressure in fluids

Pressure tells us how strongly a force is concentrated over an area. In a fluid at rest, pressure acts in every direction and increases with depth because deeper layers support more fluid above them.

Pressure is force per area

p = F_⊥/A and 1 Pa = 1 N m⁻²

Only the force component perpendicular to the surface contributes directly to pressure. The same force on a smaller area creates a larger pressure—that is why a sharp needle penetrates more easily than a blunt object.

In a fluid at rest, pressure at a point has no preferred direction: a tiny surface placed there experiences a normal pressure force regardless of its orientation.

This is different from a single force vector. Pressure is a scalar field; the direction of the force on a surface comes from the surface normal. Curved surfaces therefore receive pressure forces in many directions, which is important in buoyancy and pressure-vessel design.

Pressure increases with depth

In a stationary liquid of uniform density, pressure rises linearly with depth. Points at the same depth in the same connected fluid have the same pressure.

For a liquid of approximately constant density ρ, the pressure difference between the surface and a point at depth h is:

Δp = ρgh so p = p_surface + ρgh

The result depends on depth, density and gravity—not on the shape of the container.

Pressure with depth

Calculate gauge pressure Δp = ρgh for a stationary liquid.

Density ρ (kg/m³) Depth h (m)

Gauge pressure and absolute pressure

Absolute pressure is measured relative to a vacuum. Gauge pressure is measured relative to local atmospheric pressure. A tire gauge reading of 220 kPa means the pressure inside is about 220 kPa above atmospheric pressure, not 220 kPa absolute.

Pascal's principle and hydraulics

Hydraulics trade distance for force. The pressure is transmitted through the fluid, so a larger piston can provide a larger force while moving a shorter distance.

A pressure change applied to an enclosed, nearly incompressible fluid is transmitted throughout the fluid. For pistons at the same height, an ideal hydraulic system gives F₁/A₁ = F₂/A₂.

The large force does not create free energy: the small piston moves farther. Ideally, input work F₁d₁ equals output work F₂d₂.

What the simple depth formula assumes

The relation p = p₀ + ρgh assumes a fluid at rest and nearly constant density. It works very well for modest depths in liquids. For gases over large height ranges, density changes substantially and a more complete model is needed.

Moving fluids add another effect. Along a streamline for steady, incompressible, negligible-viscosity flow, Bernoulli's equation links pressure, speed and height: p + ½ρv² + ρgy = constant. This does not mean “faster always means lower pressure” in every flow; the assumptions and the points being compared matter.

Worked examples

1. Force on a small area

A 600 N force is spread uniformly over 0.020 m². What pressure is produced?

Solution

p = F/A = 600/0.020 = 3.0 × 10⁴ Pa = 30 kPa.

2. Water pressure at depth

Find the gauge pressure 12 m below the surface of freshwater, taking ρ = 1000 kg m⁻³.

Solution

Δp = ρgh = 1000 × 9.81 × 12 = 1.18 × 10⁵ Pa, about 118 kPa.

3. Hydraulic lift

A small piston has area 4.0 cm² and a large piston 200 cm². If 80 N is applied to the small piston, what ideal output force is available?

Solution

F₁/A₁ = F₂/A₂, so F₂ = 80 × (200/4.0) = 4000 N. Real systems lose some energy to friction and fluid effects.