Mechanical Power & Energy Storage

Flywheel Energy Calculator

Calculate flywheel mass moment of inertia, stored rotational kinetic energy, usable discharge energy, coefficient of speed fluctuation (Cs), and centrifugal rim tensile stresses.

Flywheel Cross-Section & Inertia Model

I = 0.5 · m · Ro² (Uniform cross-section)

R_o = 400 mm300270 RPMb = 80 mmI = 25.25 kg·m²ΔE = 2,368 Jv_tip = 12.6 m/s

Flywheel Sizing & Material

Geometry, dimensions, and operational speeds

mm

Maximum outer radius of the flywheel.

mm

Face width / rim thickness.

Material Selection
kg/m³

Standard steel: 7,850 kg/m³.

kg

Leave blank to auto-calculate from geometry.

Speed & Duty Cycle

RPM

Maximum operational rotational speed.

RPM

Speed after delivering peak discharge energy.

s

Duration to deliver usable energy (for power kW).

Tip Speed Limits

Safe rim tip speeds:
Cast Iron: < 35 m/s
Ductile Iron: < 50 m/s
Forged Steel: < 100–150 m/s

Usable Discharge Energy (ΔE)

2,368Joules

Equivalent Capacity: 0.658 Wh (2.37 kJ)

Inertia (I)

25.25 kg·m²

Rotor Mass

315.7 kg

Radius of Gyration (k)

282.8 mm

Fluctuation (Cs)

0.105

Tip Speed (v_tip)

12.6 m/s

Hoop Stress (σ)

1.24 MPa

Kinematic & Power Delivery Metrics

Max Stored Energy (E1)

12.46 kJ

Min Stored Energy (E2)

10.09 kJ

Power Delivery Rate

0.95 kW

Max Angular Speed (ω1)

31.42 rad/s

Min Angular Speed (ω2)

28.27 rad/s

Mean Speed (N_mean)

285.0 RPM

Governing Dynamic Formulas

ΔE = 0.5 · I · (ω₁² - ω₂²) = I · ω_mean² · Cs | σ_hoop ≈ ρ · v_tip²
ΔEJ / Wh

Usable discharge kinetic energy

Ikg·m²

Mass moment of inertia

Cs

Coefficient of speed fluctuation (ω1-ω2)/ωm

v_tipm/s

Peripheral rim tip speed

σ_hoopMPa

Centrifugal rim hoop tensile stress

PkW

Average discharge power (ΔE / t)

Engineering Code

Export flywheel energy calculations to Python or MATLAB.

Python
# Flywheel Sizing & Usable Energy Calculation
import math

def calculate_flywheel(R_o_mm, thickness_mm, density_kg_m3, n_max_rpm, n_min_rpm, geometry="solid"):
    R_o_m = R_o_mm / 1000.0
    b_m = thickness_mm / 1000.0
    
    # Mass and Inertia (Solid disk example)
    volume = math.pi * (R_o_m**2) * b_m
    mass = density_kg_m3 * volume
    I = 0.5 * mass * (R_o_m**2) # kg.m^2
    
    # Angular Speeds
    omega1 = 2 * math.pi * n_max_rpm / 60.0
    omega2 = 2 * math.pi * n_min_rpm / 60.0
    omega_mean = (omega1 + omega2) / 2.0
    
    # Energy Capacity
    E_max = 0.5 * I * (omega1**2)
    E_min = 0.5 * I * (omega2**2)
    delta_E = E_max - E_min # Joules
    
    Cs = (omega1 - omega2) / omega_mean
    v_tip = omega1 * R_o_m # m/s
    hoop_stress_MPa = (density_kg_m3 * (v_tip**2)) / 1e6
    
    return {
        "mass_kg": mass,
        "inertia_kg_m2": I,
        "usable_energy_J": delta_E,
        "usable_energy_Wh": delta_E / 3600.0,
        "fluctuation_Cs": Cs,
        "tip_velocity_m_s": v_tip,
        "hoop_stress_MPa": hoop_stress_MPa
    }

# Inputs
res = calculate_flywheel(
    R_o_mm=400,
    thickness_mm=80,
    density_kg_m3=7850,
    n_max_rpm=300,
    n_min_rpm=270
)
print(f"Moment of Inertia: {res['inertia_kg_m2']:.2f} kg·m²")
print(f"Usable Energy (ΔE): {res['usable_energy_J']:.1f} J ({res['usable_energy_Wh']:.3f} Wh)")
print(f"Fluctuation (Cs): {res['fluctuation_Cs']:.3f}")
print(f"Tip Speed: {res['tip_velocity_m_s']:.2f} m/s | Hoop Stress: {res['hoop_stress_MPa']:.2f} MPa")
MATLAB
% Flywheel Kinetic Energy Script
Ro = 400 / 1000;
b = 80 / 1000;
rho = 7850;
N1 = 300;
N2 = 270;

mass = rho * pi * Ro^2 * b;
I = 0.5 * mass * Ro^2;

w1 = 2 * pi * N1 / 60;
w2 = 2 * pi * N2 / 60;

delta_E = 0.5 * I * (w1^2 - w2^2);
v_tip = w1 * Ro;

fprintf('Inertia: %.2f kg.m^2\n', I);
fprintf('Usable Energy: %.1f Joules\n', delta_E);
fprintf('Tip Speed: %.2f m/s\n', v_tip);
Excel Formula
=0.5 * I_kgm2 * (((2*PI()*N_max/60)^2) - ((2*PI()*N_min/60)^2))

Example Calculation: Mechanical Punching Press Flywheel

A solid steel disk flywheel (ρ = 7,850 kg/m³, R_o = 400 mm, thickness b = 80 mm) operates between N₁ = 300 RPM and N₂ = 270 RPM during punching:

m = 7,850 × [π × (0.4)² × 0.08] = 315.65 kg | I = 0.5 × 315.65 × (0.4)² = 25.25 kg·m²
ω₁ = 31.42 rad/s | ω₂ = 28.27 rad/s | Tip Velocity v = 12.57 m/s | Fluctuation Cs = 0.105
ΔE (Usable Energy) = 2,367.6 Joules (0.658 Wh) | Hoop Stress σ = 1.24 MPa

Technical Explanation: Flywheel Mechanics and Energy Sizing

A flywheel is an energy reservoir that stores mechanical energy in the form of rotational inertia. It absorbs excess energy when torque supplied by a prime mover exceeds load requirements and delivers energy when peak demand exceeds continuous motor capacity.

Kinetic Energy & Usable Power Delivery

The total kinetic energy stored in a rotor spinning at angular velocity ω (rad/s) is:

E = 0.5 · I · ω²

When supplying a peak transient load (such as a metal punch stroke or rock crusher impact), the rotor decelerates from ω₁ to ω₂. The usable energy released during this slowdown is:

ΔE = 0.5 · I · (ω₁² - ω₂²) = I · ω_mean² · Cs

Coefficient of Speed Fluctuation (Cs)

The allowable cyclic speed drop is governed by the application:

  • Crushers and Forging Presses: Cs = 0.10 – 0.20 (larger speed drop allowed to minimize motor size).
  • Machine Tools and Pumps: Cs = 0.03 – 0.05 (tight speed stability required).
  • AC Generator Synchronous Drives: Cs = 0.005 – 0.02 (strict grid frequency regulation).

Centrifugal Rim Stress Limits

As a flywheel spins, centrifugal forces create circumferential tensile hoop stress:

σ_hoop ≈ ρ · v_tip² = ρ · (ω · R_o)²

Gray cast iron (Class 30) flywheels must strictly remain below tip speeds of 30–40 m/s to prevent catastrophic burst fragmentation, whereas high-grade forged alloy steels can safely handle tip speeds above 100 m/s.

Real-World Engineering Cases

Catastrophic Cast Iron Flywheel Rim Burst in a Sawmill

A steam-driven sawmill suffered a mechanical governor failure, causing a 2.4-meter cast iron flywheel to overspeed from 180 RPM to 390 RPM. Rim tip speed surged beyond 49 m/s, exceeding the material's tensile limit. The flywheel burst into multiple fragments that punched through concrete masonry walls.

Engineering Lesson

Never exceed maximum allowable tip speed ratings. Cast iron possesses low tensile fatigue resistance under hoop stress. Modern high-energy flywheels must utilize forged steel, ductile iron, or carbon fiber composite rims with overspeed trips.

Undersized Flywheel Causing Severe Motor Overheating in a Stamping Press

A metal stamping plant upgraded a 150-ton mechanical press to thicker sheet stock. The existing flywheel had insufficient inertia (ΔE too low), causing the drive motor to stall and draw locked-rotor currents during each punch cycle. The drive motor burned out within two shifts.

Engineering Lesson

The flywheel must supply 80–90% of the instantaneous punch energy. The electric motor only serves to restore the flywheel's kinetic energy during the non-working return stroke.

Frequently Asked Questions

Why are rim-type flywheels more efficient than solid disks?

Mass located further from the axis of rotation contributes exponentially more inertia (I = m · r²). Concentrating 90% of the mass in an outer rim provides nearly double the inertia per kilogram compared to a solid flat disk.

What is the radius of gyration (k)?

The radius of gyration (k = √(I / m)) is the equivalent radial distance from the rotation center at which the entire mass could be concentrated without altering its moment of inertia.

How do I size the electric motor for a flywheel machine?

Motor power equals usable energy divided by total cycle recovery time (P = ΔE / t_recovery / η_drive). The motor does not need to handle the peak punching force, only the average continuous replenishment power.

Engineering calculations provided by this tool are for educational and preliminary design purposes. Always verify rotational dynamics, centrifugal stress concentrations, keyway shear stresses, dynamic balancing, and statutory containment shielding before fabricating high-speed rotating machinery.