Balancing: two planes for a rotor, a compromise for a piston

Remove the cause, not the effect

• An unbalanced shaft at 1440 rpm shakes the frame, loads the bearings and fatigues the pipework. Balancing removes the cause.
• A mass $m$ at radius $r$ turning at $\omega$ pulls outwards with $mr\omega^2$. Static balance makes $\sum m_ir_i = 0$; dynamic balance also makes $\sum m_ir_iz_i = 0$.

Why two planes are enough

• Any set of rotating unbalances is one force plus one couple. Moments about plane A give the correction in plane B; the force balance then gives plane A.
• Two unknown vectors, two vector equations: two planes always suffice for a rotor.

The pump shaft

The reciprocating mass

• A reciprocating mass gives a primary force $mr\omega^2\cos\theta$ and a secondary force $mr\omega^2\cos2\theta/n$ at twice the speed. For 0.8 kg at 40 mm and 300 rpm: $\mathbf{31.6}$ N and $\mathbf{7.9}$ N.
• A counterweight for half the mass leaves 15.8 N along the stroke but adds 15.8 N across it: a smaller rotating force instead of a large one in one direction.

How good is good enough?

• Balance quality is graded by $G = e\omega$ in mm/s (ISO 21940). Pumps are usually G 6.3, so at 150.8 rad/s the permitted mass-centre offset is $6.3/150.8 = \mathbf{41.8}$ µm.
• For a 30 kg rotor that allows $30 \times 0.0418 = \mathbf{1.25}$ g·m of residual unbalance: about the 1.2 g·m the mounts lesson assumes.

Worked example: balancing in the field with a trial mass

• Running as found, the bearing reads 100 µm at 30°. A 10 g trial mass at 0° changes it to 70 µm at 80°, so the trial's own effect is the vector difference, 76.8 µm at 165.7°.
• The correction must cancel the original: $-\mathbf O/\mathbf E = 1.30$ at $44.3°$, so fit $\mathbf{13.0}$ g at $44°$ instead of the trial. For a narrow rotor, one run and one trial find the correction even when the unbalance itself cannot be measured; a long rotor needs a trial in each plane.

Three balancing slips

Recap