Pascal's Principle & Hydraulics

In a fluid at rest the pressure difference between two points depends only on their height difference: P₁ − P₂ = ρg(y₂ − y₁). Nothing in that relation knows about a piston. So if you push on one boundary and raise the pressure there by ΔP, every other point must rise by the same ΔP, or the height relation would no longer hold. Pascal's principle is therefore a corollary of hydrostatics plus incompressibility, not a separate law. Notice exactly what is transmitted: the increment. The bottom of a hydraulic reservoir still sits at higher absolute pressure than the top, by ρgh, before the pump runs and after it.

Pressure acts normally on every surface it touches, so a piston of area A held against gauge pressure P feels F = PA. Two pistons in the same enclosed oil at the same height share one pressure, and their forces differ purely by area: F₂/F₁ = A₂/A₁. A garage lift with a 25 cm output piston has A₂ = π(0.125 m)² = 4.91 × 10⁻² m². Holding a 1400 kg car, weight 1.37 × 10⁴ N with g = 9.81 m s⁻², needs a gauge pressure of 13 734/0.0491 = 2.80 × 10⁵ Pa, about 2.8 bar. Feed that same pressure to a 2.5 cm input piston of area 4.91 × 10⁻⁴ m² and only 137 N is needed. A diameter ratio of 10 gives an area ratio of 100, and so a force ratio of 100.

Oil leaving the small cylinder has to appear in the large one, so A₁d₁ = A₂d₂. The same factor of 100 that multiplied the force divides the displacement. Raising that car by 1.8 m sweeps 0.0491 × 1.8 = 0.0884 m³, about 88 litres, which the small piston must supply over a stroke of 180 m — some 600 pump strokes of 0.30 m each. Multiply out both sides: 137.3 N × 180 m = 24.7 kJ in, and 13 734 N × 1.8 m = 24.7 kJ out. A hydraulic system is a force amplifier and an energy conduit. Any claim of amplified work is an arithmetic error.

Divide the work equality by time and it becomes a power equality: F₁v₁ = F₂v₂. Since v = Q/A at each piston, both sides equal ΔP × Q. The area ratio fixes the force; the pump's volume flow fixes the speed. A machine at 20 MPa fed by a 12 L min⁻¹ pump — 2.0 × 10⁻⁴ m³ s⁻¹ — delivers 2.0 × 10⁻⁴ × 20 × 10⁶ = 4.0 kW to the load, whatever the cylinder areas are. That is why a hand jack replaces one impossible 180 m stroke with hundreds of short ones, check valves holding the gain between strokes, and why an industrial press is specified by a pressure and a flow rate together.

Four assumptions carry everything above: incompressible fluid, rigid walls, no leaks, pistons at equal height. Hydraulic oil has a bulk modulus near 1.5 GPa, so at 20 MPa a fraction ΔV/V = ΔP/B ≈ 1.3% of the fluid volume disappears into compression. Trapped air is far worse: a 1.0 cm³ bubble at 100 kPa, squeezed slowly to 5 MPa, shrinks to about 0.02 cm³ and swallows 0.98 cm³ of piston sweep — through a 25 mm master cylinder of area 4.91 cm² that is 2.0 mm of dead piston travel, and several times that at the pedal pad once the lever ratio is counted: the spongy feel bleeding cures. Hoses swell, seals rub, and a height offset adds ρgΔh: 1.2 m of oil at 870 kg m⁻³ is 10.2 kPa, negligible beside 20 MPa but 3.7% of the 280 kPa lift above.