Flywheel Run-Down and Rim Stress
A workshop plans to use a steel flywheel as a short-term energy store for a small elevator. The flywheel is a solid disc of radius 0.35 m and mass 42 kg, mounted on a horizontal shaft and connected to a generator through a gearbox. In a test the flywheel is spun up to 3200 rev min−1, then disconnected from the drive and allowed to run down through the generator and bearings; its speed falls to 2800 rev min−1 in 95 s. The steel has density 7.85×10^3 kg m−3, tensile yield strength 250 MPa and Poisson ratio 0.30. The manufacturer warns that the disc should not exceed 80% of yield stress at any point, and that repeated cycling may initiate cracks at the keyway. The gearbox is rated for a continuous input torque of 2.0 N m. The workshop asks whether the observed run-down is consistent with bearing friction alone and whether the flywheel is safe at 3200 rev min−1.
1. Calculate the moment of inertia, the rotational kinetic energy lost, and the average torque on the flywheel during the 95 s run-down. State any assumption you make.
Read the worked answer
For a solid disc, I = ½ m r² = ½ × 42 × 0.35² = 2.57 kg m².
Convert speeds: ω₁ = 3200 × 2π/60 = 335 rad s−1; ω₂ = 2800 × 2π/60 = 293 rad s−1.
E₁ = ½Iω₁² = ½ × 2.57 × 335² ≈ 1.44×10⁵ J; E₂ = ½ × 2.57 × 293² ≈ 1.11×10⁵ J. Energy lost ≈ 3.4×10⁴ J.
Average torque from angular momentum change: τ = I(ω₁−ω₂)/t = 2.57 × 41.9 / 95 ≈ 1.13 N m. This assumes constant angular deceleration and that all retarding torque is lumped together. The torque is below the gearbox continuous rating of 2.0 N m, but the gearbox sees the generator load, not just bearing friction.
2. Estimate the maximum tensile stress in the spinning disc at 3200 rev min−1 using σmax = (3+ν)ρω²R²/8. Compare with the 80% yield criterion and comment on the keyway.
Read the worked answer
Using ω = 335 rad s−1, R = 0.35 m, ρ = 7.85×10³ kg m−3:
σmax = (3.30/8) × 7.85×10³ × 335² × 0.35² ≈ 4.45×10⁷ Pa = 44.5 MPa.
The 80% limit is 0.80 × 250 = 200 MPa, so the smooth disc has a safety factor of about 4.5. The keyway is a stress raiser: a factor of 2–3 would raise the local stress to roughly 90–135 MPa, still below 200 MPa at this speed but much closer. Repeated cycling also raises fatigue concerns, so the keyway should be radiused and inspected. A 20% overspeed would increase stress by 1.2² = 1.44, giving about 64 MPa nominal and potentially near the limit locally.
3. An engineer claims bearing friction alone explains the run-down. Describe a simple test to separate bearing friction from generator load, and explain how the torque–speed graph would distinguish viscous friction from air drag.
Read the worked answer
Run the flywheel down with the generator mechanically or electrically disconnected, under the same bearing conditions. Repeat with the generator connected. The difference between the two decelerations gives the generator torque; the remaining torque is bearing plus air drag.
A useful diagnostic is to measure ω(t) and compute τ = −I dω/dt at several speeds. If τ is roughly constant, dry bearing friction or a constant brake load is indicated. If τ ∝ ω, viscous bearing lubrication is likely. If τ ∝ ω², air drag or a fan-type load is likely. The observed 1.13 N m average is therefore not enough by itself: the speed dependence, not just the average, identifies the mechanism. A run-down in a partial vacuum would reduce air drag and test that contribution.