Balancing: two planes for a rotor, a compromise for a piston
Introduzido| English | Significado |
|---|---|
| static balance | Balance in which the centre of mass lies on the rotation axis, so there is no net rotating force. |
| dynamic balance | Balance in which both the net rotating force and the net rotating couple are zero. |
| two planes | The two correction planes in which masses are added to balance a rotor dynamically. |
| reciprocating mass | The mass that moves back and forth in a straight line, such as a piston and part of its connecting rod. |
| primary force | The unbalanced inertia force of a reciprocating engine at crank frequency. |
| secondary force | The unbalanced inertia force of a reciprocating engine at twice crank frequency, caused by the connecting rod's angle. |
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.
Static, dynamic or reciprocating?
Classify each balancing problem.
A long rotor balanced statically only still has:
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
The moments of the masses along the shaft remain.
The pump shaft
- 0.5 kg at 60 mm, 0°, 100 mm along; 0.3 kg at 80 mm, 90°, 250 mm along. Correction planes at 0 and 400 mm, radius 50 mm.
- Plane B: $\mathbf{0.335}$ kg at $\mathbf{243°}$. Plane A: $\mathbf{0.485}$ kg at $\mathbf{202°}$.
- Before balancing, $38.4\ \mathrm{g\cdot m} \times 150.8^2 = \mathbf{873}$ N shook the bearings.
Rotating force of 38.4 g m at 150.8 rad/s (N, whole number)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
m r omega squared.
Plane-B correction mass at 50 mm for 16.8 g m (three decimals, kg)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
16.8 over 50 mm.
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.
Primary force amplitude for 0.8 kg at 40 mm and 300 rpm (one decimal, N)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
m r omega squared.
Secondary amplitude with n = 4 (one decimal, N)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
Primary over n.
A counterweight can cancel a reciprocating force completely.
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
It trades a force along the stroke for one across it.
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.
G 6.3 at 150.8 rad/s on a 30 kg rotor: permitted residual unbalance (two decimals, g·m)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
$e = 6.3/150.8 = 0.0418$ mm; $30 \times 0.0418 = 1.25$ kg·mm $= 1.25$ g·m.
A well-balanced rotor has zero residual unbalance.
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
Every rotor keeps some; the balance grade sets how much is acceptable for its speed.
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.
Original 100 µm at 30°; with 10 g trial at 0°, 70 µm at 80°. Correction mass (one decimal, g)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
$\mathbf E = 70\angle80° - 100\angle30° = 76.8\angle165.7°$; $|\mathbf O/\mathbf E| = 1.302$; $1.302 \times 10 = 13.0$ g.
Three balancing slips
- Static balance called done. The couple remains.
- Adding $mr$ as numbers. They are vectors.
- A counterweight expected to cancel a piston. It only trades direction for size.
Recap
- Two planes: 0.485 kg at 202° and 0.335 kg at 243° on the pump shaft.
- The unbalance was 873 N at 1440 rpm.
- Primary 31.6 N, secondary 7.9 N for the dosing piston.