Hydraulic Jack

What is modelled

Pressure in a closed liquid is the same everywhere (Pascal), so a small piston pushing with a modest force makes a large ram push with a large force, by the ratio of the two areas. The price is distance: the ram moves by the same ratio less per stroke, so lifting takes many strokes. Seal friction and slip past the pump valve waste a part of the work, and the oil itself is slightly compressible, so the first strokes only raise the pressure.

Equations

• Pressure p = F1 / A1 = F2 / A2, with the handle giving F1 = F Lr.
• Pressure needed p = M g / (A2 (1 − cf)); pressure available F Lr (1 − cf) / A1.
• Lift per stroke h = A1 s (1 − ℓ) / A2, with the slip ℓ growing with the pressure.
• Oil compressibility ΔV = V0 (1 − e−p/K), with bulk modulus K = 1.5 GPa.

Checks run in the page

• The same pressure on both pistons and the work in against the work out, for an ideal jack.
• One stroke of the stepping model against the volume balance.
• The compression volume found by small pressure steps against the closed form.
• The number of strokes to lift 10 cm from the stepping model against the formula.

Honest simplifications

Seal friction is a fixed fraction of the force on each piston, the slip is proportional to the pressure, the check valves are ideal, the hoses do not swell beyond the single trapped-oil volume, the ram has a fixed travel of 35 cm and heating of the oil is ignored. Use it for the trends, not for rating a jack.

• Press Car jack and watch the pressure build, then the ram move a fraction of a millimetre per stroke.
• Press Too heavy for the hand and see the handle stall; raise the hand force or the handle ratio.
• Press Worn seals and soft hoses to see the efficiency fall and extra strokes spent compressing oil.