Flywheels, Balancing and Rotors
Energy storage and speed fluctuation, flywheel sizing and stress, balance grades and balancing of rotating masses.
What a flywheel does
A flywheel is a rotating mass that stores kinetic energy, ½ I ω². In a machine where the load torque varies over each cycle (a press, a shear, a piston engine or compressor) it supplies the peak energy and takes back the excess, so the shaft speed stays nearly constant and the motor can be sized for the average power, not the peak. The same idea powers kinetic energy storage systems, and (with the motor’s help) lets the hydraulic accumulator of Circuits and Accumulators do the same job in a different medium.
Sizing a flywheel
ΔE = ½ I (ωmax² − ωmin²) = Cs I ω² I = ΔE ÷ (Cs ω²) Cs = (ωmax − ωmin) ÷ ωmean
| Machine | Coefficient of fluctuation Cs |
|---|---|
| Punch and shear presses | 0.10 to 0.20 |
| Crushers, pumps, reciprocating compressors | 0.05 to 0.15 |
| Machine tools | 0.03 to 0.05 |
| Textile machinery, AC generators | 0.01 to 0.02 |
| Engines (automotive) | 0.01 to 0.05 |
| Direct-drive turbines, electric generators | 0.002 to 0.005 |
Worked example: the punch press
Worked example: a 3 kJ stroke every second at 1 500 rpm
- A punch press takes ΔE = 3 000 J during a 0.12 s stroke once per second. The average power is 3 000 W but the peak during the stroke is 25 kW: 8 times the average. Without a flywheel the motor (and the supply) would have to deliver the peak: here the 11 kW motor of Induction Motors and Drives is more than enough on the average.
- Allow a fluctuation of Cs = 0.10 at ω = 157.1 rad/s: I = 3 000 ÷ (0.10 × 24 674) = 1.22 kg·m². The speed swings between 1 575 and 1 425 rpm (the figure shows the simulated speed against the motor’s slip curve).
- The physical size: a solid steel disc has I = ½ m r², so for r = 250 mm: m = 2 I ÷ r² = 38.9 kg and a thickness t = m ÷ (π r² ρ) = 25.2 mm. Stored energy at full speed ½ I ω² = 15 kJ: the 3 kJ stroke takes 20 percent of it.
- Halving the fluctuation (Cs = 0.05) doubles the inertia to 2.43 kg·m² (the energy ΔE is fixed): a tighter speed costs more mass and more starting energy.
- The flywheel is on the shaft of highest speed: the energy stored is proportional to ω², so a flywheel at the motor speed (1 500 rpm) is 16 times smaller than the same flywheel at one quarter of that speed behind a reduction gear.
Rim stress and speed limits
A rotating ring or disc is stressed by its own centrifugal force. For a thin rim of mean radius r the hoop stress is σ = ρ v², with v = ω r the rim speed: for this disc v = 157.1 × 0.25 = 39.3 m/s, so σ = 7 850 × 1 542 = 12.1 MPa (5.0 MPa at the center of a solid disc: (3 + ν) ÷ 8 × ρ ω² r²). The stress is independent of the size of the wheel and depends only on the rim speed squared: a steel wheel at 40 m/s is under 13 MPa, at 100 m/s about 79 MPa, and cast iron bursts at about 120 to 150 m/s rim speed (limit it to 30 to 40 m/s in service). Hold the rim speed to a safe fraction of the burst speed, with a safety factor of 3 on speed (9 on stress) for a cast iron flywheel, and fit a guard; a flywheel that bursts releases its energy as shrapnel. Composite and steel rotors for energy storage run at 300 to 1 000 m/s in vacuum housings.
Unbalance and balance grades
A rotor is balanced when its mass is distributed symmetrically about its axis of rotation. An unbalance U = m e (kg·mm or g·mm) is the product of the mass and its distance from the axis (or the eccentricity of the center of mass). The centrifugal force it makes is F = U ω², which grows with the square of the speed. ISO 21940-11 sets balance quality grades G (in mm/s), the product of the allowed eccentricity and the angular speed: eper = G ÷ ω.
| Grade | Rotor type | Permitted eccentricity at 1 500 rpm |
|---|---|---|
| G 4 000 | rigidly mounted slow marine diesels | 25.46 µm |
| G 630 | large four-stroke engines on rigid mounts | 4 010.70 µm |
| G 100 | complete engines for cars, trucks | 636.62 µm |
| G 16 | drive shafts, crushers, agricultural machinery | 101.86 µm |
| G 6.3 | pumps, fans, electric motor rotors, machine parts | 40.11 µm |
| G 2.5 | turbines, compressors, machine-tool drives | 15.92 µm |
| G 1 | grinding machine drives, gyroscopes | 6.37 µm |
| G 0.4 | precision grinder spindles | 2.55 µm |
Balancing a rotor
Worked example: balancing a 20 kg rotor to G6.3
- Allowed eccentricity at 1 500 rpm: e = G ÷ ω = 6.3 ÷ 157.1 = 40.1 µm; allowed unbalance U = m e = 20 kg × 0.0401 mm = 802.14 kg·mm (802 141 g·mm).
- A casting error leaves a 100 g excess at a radius of 120 mm: U = 0.10 × 120 = 12.0 kg·mm, 0 times the allowance. The force at speed is F = U ω² = 296 N: more than the weight of the rotor (196 N).
- Correction: remove or add mass opposite: mc = U ÷ rc = 12.0 ÷ 150 = 80 g at a radius of 150 mm, 180° from the heavy spot (drill the heavy side or add a weight on the light side).
- For G2.5 the allowance falls to 318.31 kg·mm (e = 15.9 µm): 2.5 times tighter: ground, precisely made rotors only.
- Static unbalance (a heavy spot in one plane, the center of mass off the axis) is fixed in one plane; couple (dynamic) unbalance (two opposite heavy spots in different planes) needs correction in two planes on a balancing machine. Long rotors (length more than about 5 times the diameter) and anything over about 1 000 rpm should be balanced in two planes; disc-shaped rotors with a width under about a fifth of the diameter need one.
Balancing reciprocating machines
The reciprocating mass of a piston and its rod end creates a primary shaking force at crank speed, m r ω² cos θ, and a secondary force at twice the speed, m r ω² (r ÷ l) cos 2θ (see Linkages and Mechanisms). A 0.5 kg reciprocating mass on a 50 mm crank at 1 500 rpm gives 617 N of primary force. A rotating counterweight can balance the primary force in one direction but adds an equal force at right angles unless it balances only about half of it (a common compromise). Multi-cylinder arrangements cancel the forces by symmetry (an inline four cancels the primary force and moment, but not the secondary force; a flat-four or a boxer cancels almost everything). Balance shafts (two shafts rotating at twice engine speed) cancel the secondary force of an inline four.
Related
- Previous: Clutches, Brakes and Couplings. Next: Machine Design Case Study: the 80 kN Press Frame.
- Sizing a Motion Axis for inertia matching.
- Machine Design: the whole guide.
FAQ
Why put the flywheel on the fast shaft?
Because the energy ½ I ω² grows with the square of the speed: for the same energy a faster flywheel needs a much smaller inertia.
What is the coefficient of speed fluctuation?
The ratio of the speed swing to the mean speed. 0.1 to 0.2 for presses, 0.002 to 0.005 for generators.
Can I balance a rotor by hand?
To a modest grade (G16, G40) with static methods (a knife edge) yes. For G6.3 and finer, a balancing machine measures the unbalance and the angle, and is needed above a few hundred rpm.
Why does the vibration go away at one speed and return at another?
The balance is correct at one speed, but the shaft is flexible: near a critical speed the unbalance grows a great deal. Balance at the service speed on a flexible rotor, and avoid operating near the critical.
References
- Shigley, J. E., Mischke, C. R., Budynas, R. G., Mechanical Engineering Design, McGraw-Hill (flywheels); Norton, R. L., Machine Design and Design of Machinery, McGraw-Hill (balancing).
- ISO 21940-11 (rotors in a constant (rigid) state: balance tolerances), ISO 21940-12 (flexible rotors), ISO 1940-1 (previous edition).
- Rao, S. S., Mechanical Vibrations, Pearson; Foster, K., Parker, G. A., Mechanics of Machines; Genta, G., Kinetic Energy Storage.
Last reviewed: October 2026. Rim speeds and coefficients are typical; check the burst margin for any real flywheel.