12.1 Hydraulic Fundamentals: Pascal’s Law, Pressure, Flow & Power

Major Work Activity E — Services fluid power systems carries 21 of the 135 questions on the Interprovincial exam, and Task E-21 (Services hydraulic systems) alone carries 12 questions — the single largest task on the entire exam. Nearly every one of those questions rests on the four relationships covered here. Millwrights install, diagnose, maintain and repair hydraulic power units on presses, injection moulders, mobile equipment, mill gates, log haul-ups, and clamping systems, so the fundamentals must be automatic.

Pascal’s Law and Confined Fluid

Pascal’s law states that pressure applied to a confined fluid at rest is transmitted undiminished in all directions and acts with equal force on equal areas, at right angles to the containing surfaces. Three consequences drive the whole trade:

1. A small input force acting on a small area can generate a very large output force on a large area — hydraulic force multiplication.
2. Pressure is uniform throughout a static connected circuit, so a gauge anywhere downstream of the pump (with no restriction between) reads the same value.
3. Liquids are effectively incompressible (mineral oil compresses roughly 0.5% per 1,000 psi), so motion transfers almost instantly — unlike air, which stores energy by compressing.

Pressure Definition and Units

Pressure is force per unit area. Canadian shops work in both systems, and the exam uses both:

Gauge pressure (psig) reads zero at atmosphere; absolute pressure (psia) adds the 14.7 psi atmospheric column. Pump inlet (suction) calculations must use absolute pressure, because a pump inlet can only ever pull down toward a vacuum — atmospheric pressure is what actually pushes fluid into the pump.

Pumps Make Flow — Resistance Makes Pressure

This is the single most-tested concept in hydraulics, and the most common apprentice misconception.

A hydraulic pump is a flow generator. It moves a fixed volume of oil per revolution from inlet to outlet. If that outlet were piped straight back to an open reservoir, the gauge would read only the few psi needed to push oil through the return line. Pressure is generated by resistance to flow — a load on a cylinder, an orifice, a closed valve, or a relief valve setting.

        PUMP OUTPUT = FLOW (GPM / L/min)          RESISTANCE = PRESSURE (psi / kPa)

        +--------+     Q      +--------------+       Load 20,000 lb
        |  PUMP  |----------->|   CYLINDER   |=========[]============
        +--------+            +--------------+
             |                        ^
             |  Relief valve          |  Pressure rises only until the
             +--> set 3,000 psi       |  load moves, or the relief opens

Practical consequences a millwright is expected to state:

• A machine that will not build pressure is usually leaking flow internally (worn pump, bypassing cylinder seal, relief valve stuck open), not "short of pressure."
• A machine that builds full pressure but moves slowly has lost flow (worn pump, restricted filter, partly closed flow control), not force.
• Loss of force points to pressure; loss of speed points to flow. Memorize that pairing — it decides many exam troubleshooting questions.

Force, Area and Mechanical Advantage

$F = P \times A$

For a round cylinder bore, area is:

$A = \frac{\pi D^2}{4} = 0.7854 D^2$

Worked example — extension force. A cylinder with a 3.0 in bore is fed 2,000 psi.

$A = 0.7854 \times 3.0^2 = 7.069\text{ in}^2$ $F = 2{,}000 \times 7.069 = 14{,}137\text{ lbf}$

Worked example — retraction force. The same cylinder has a 1.5 in rod. On retraction, oil acts on the annulus (bore area minus rod area):

$A_{\text{ann}} = 0.7854 \times (3.0^2 - 1.5^2) = 0.7854 \times 6.75 = 5.301\text{ in}^2$ $F = 2{,}000 \times 5.301 = 10{,}603\text{ lbf}$

A standard differential cylinder therefore always pushes harder than it pulls, and always retracts faster than it extends at the same flow rate. Presses, balers and clamps are oriented so the working stroke is the extend stroke.

Flow, Velocity and Actuator Speed

Flow rate (Q) is volume per unit time — GPM (US gallons per minute) or L/min. One US gallon equals 231 cubic inches, a conversion the exam expects you to use unprompted.

Cylinder speed:

$v\ (\text{in/min}) = \frac{231 \times \text{GPM}}{A\ (\text{in}^2)}$

Worked example. Feeding the 3.0 in bore cylinder above with 8 GPM:

$v = \frac{231 \times 8}{7.069} = \frac{1{,}848}{7.069} = 261.4\text{ in/min} \approx 21.8\text{ ft/min}$

Retracting the same cylinder with the same 8 GPM on the 5.301 in² annulus gives 348.6 in/min — about 1.33 times faster, exactly the inverse of the force ratio. Energy is conserved: what you gain in speed you lose in force.

Fluid Velocity in Conductors

Line sizing controls turbulence, heat and noise. Accepted industrial design velocities:

Undersized suction line is the classic root cause of pump cavitation and its distinctive high-pitched rattling noise.

Hydraulic Power

$\text{Hydraulic HP} = \frac{P\ (\text{psi}) \times Q\ (\text{GPM})}{1714}$

$\text{Hydraulic kW} = \frac{P\ (\text{bar}) \times Q\ (\text{L/min})}{600}$

Worked example — motor sizing. A power unit runs 10 GPM at 3,000 psi.

$\text{HHP} = \frac{3{,}000 \times 10}{1714} = 17.5\text{ hp}$

At a combined pump and drive efficiency of about 85%, the electric motor must deliver roughly 20.6 hp, so a 25 hp motor is selected. Every hydraulic horsepower that does not do useful work becomes heat: 1 hp = 2,545 BTU/hr. A relief valve dumping 10 GPM at 3,000 psi continuously adds about 44,500 BTU/hr to the reservoir — which is why a machine left deadheaded on relief overheats within minutes.

Exam Traps

• Pressure gauge reads zero, motor runs. The pump may be fine — look for an unloaded circuit, an open-centre valve, or a relief valve stuck open.
• Hydraulic oil is not a coolant. Any pressure drop across a valve, orifice or relief converts directly to heat in the tank.
• Do not confuse head with pressure: 1 psi lifts water 2.31 ft, but oil at 0.87 specific gravity gives 2.65 ft per psi.