Turning moment diagrams and flywheels

Turning-moment diagrams, maximum fluctuation of energy, coefficients of fluctuation of speed and energy, and sizing rim-type flywheels for engines and punching presses.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

An engine produces torque in pulses, and a punching or shearing press demands it in pulses, while motors and loads prefer steady speed. The flywheel bridges the two by storing kinetic energy when supply exceeds demand and giving it back when demand exceeds supply. Sizing it from the turning-moment diagram is a standard design calculation for presses, compressors and engines.

Key ideas

Turning-moment (crank-effort) diagram. A plot of the torque on the crankshaft against crank angle over one working cycle (360° for a two-stroke or a press, 720° for a four-stroke engine). The area under it is the work done per cycle, because W = ∫T dθ.

  • The mean torque T_mean = W/θ_cycle is the steady load torque the engine can drive at constant mean speed.
  • Where the engine torque is above the mean line the flywheel speeds up (absorbs energy); below it the flywheel slows down (gives energy back).
  • A multi-cylinder engine has a much flatter diagram than a single-cylinder one, so it needs a smaller flywheel.

Fluctuation of energy. Starting from any point with energy E, add the areas (above the mean line positive, below negative) one by one. The largest and smallest values of the running total mark the maximum and minimum speeds. Their difference is the maximum fluctuation of energy ΔE. Over a full cycle the areas must sum to zero.

Coefficients.

  • Coefficient of fluctuation of speed: C_s = (ω_max − ω_min)/ω_mean. Typical values are a few per cent for pumps and presses and much smaller for generators; take the value for your application from a design data book.
  • Coefficient of fluctuation of energy: C_e = ΔE/W_cycle.

Flywheel versus governor. A flywheel limits the speed variation within a cycle at a given load; it does nothing to the mean speed. A governor adjusts the fuel or steam supply to keep the mean speed constant when the load changes from cycle to cycle.

Rim-type flywheel. Most of the inertia is in the rim, so I ≈ m·R² (arms and hub neglected or taken as a small fraction). For a given energy, a larger radius gives a lighter wheel, but rim speed is limited by the hoop (centrifugal) stress σ = ρ·v², which does not depend on the rim cross-section.

Punching press. The motor delivers energy steadily over the whole cycle, while the punch uses it in a short fraction of the cycle. The flywheel supplies the shortfall during punching and is recharged by the motor for the rest of the cycle.

Formulas

E = ½·I·ω² — kinetic energy (J); I moment of inertia (kg·m²); ω angular speed (rad/s).

ΔE = ½·I·(ω_max² − ω_min²) = I·ω_mean²·C_s — maximum fluctuation of energy (J), with ω_mean = (ω_max + ω_min)/2.

ΔE = m·R²·ω²·C_s = m·v²·C_s — rim-type flywheel; m rim mass (kg), R mean rim radius (m), v = ω·R mean rim speed (m/s).

C_s = (ω_max − ω_min)/ω_mean = (N_max − N_min)/N_mean — dimensionless.

C_e = ΔE/W_cycle — dimensionless.

T_mean = W_cycle/θ_cycle — N·m, θ in rad (2π for one revolution, 4π for a four-stroke cycle).

σ = ρ·v² — hoop stress in a thin rotating rim (Pa); ρ density (kg/m³).

Worked examples

Example 1 (standard). The areas between an engine's turning-moment curve and the mean torque line over one cycle, in order, represent +1200, −1800, +1500 and −900 J. The engine runs at a mean 300 rev/min and the total speed fluctuation must be 2 % of mean. Find ΔE, the flywheel inertia and the rim mass for a 0.6 m mean rim radius, and check the rim stress (cast iron, ρ = 7200 kg/m³).

  1. Running energy from E: E + 1200, E − 600, E + 900, E. Max = E + 1200, min = E − 600.
  2. ΔE = 1200 − (−600) = 1800 J.
  3. ω = 2π × 300/60 = 31.42 rad/s; C_s = 0.02.
  4. I = ΔE/(ω²·C_s) = 1800/(987.0 × 0.02) = 91.2 kg·m².
  5. Rim mass: m = I/R² = 91.2/0.36 = 253 kg.
  6. Rim speed v = ω·R = 18.85 m/s; σ = ρv² = 7200 × 18.85² = 2.56 × 10⁶ Pa.
  7. ΔE = 1.8 kJ, I ≈ 91.2 kg·m², m ≈ 253 kg, σ ≈ 2.56 MPa (well within cast iron's strength)

Example 2 (GATE level). A press needs 8 kJ per stroke, and punching occupies 1/6 of each cycle. The flywheel speed may fall from 240 to 216 rev/min during punching. The press makes 20 strokes per minute. Find the motor power and the flywheel inertia.

  1. Motor power (energy supplied uniformly): P = 8000 × 20/60 = 2667 W ≈ 2.67 kW.
  2. Motor energy during punching: 8000/6 = 1333 J. The flywheel supplies the rest: ΔE = 8000 − 1333 = 6667 J.
  3. ω_max = 2π × 240/60 = 25.13 rad/s; ω_min = 2π × 216/60 = 22.62 rad/s.
  4. ω_max² − ω_min² = 631.7 − 511.6 = 120.0 rad²/s².
  5. I = 2ΔE/(ω_max² − ω_min²) = 2 × 6667/120.0 = 111.1 kg·m². Check: ω_mean = 23.88 rad/s, C_s = 2.51/23.88 = 0.105, I = 6667/(23.88² × 0.105) = 111.1 kg·m² ✓.
  6. P ≈ 2.67 kW, I ≈ 111 kg·m²

Common mistakes

  • Taking ΔE as the largest single loop area instead of max minus min of the running total.
  • Using rev/min in ½Iω².
  • Using ±1 % as C_s = 0.01 when the question means a total band of 2 % (C_s = 0.02); read the wording.
  • Forgetting that the motor keeps supplying energy during the punching stroke.
  • Saying a flywheel controls mean speed; that is the governor's job.
  • Using the four-stroke cycle angle as 2π instead of 4π when finding mean torque.

For GATE PI

Expect ΔE from given loop areas or a simple torque function, flywheel inertia for a given C_s, mean torque and power from a turning-moment diagram, and punching-press energy and motor-power problems. Practise the running-total table and the identity ΔE = I·ω²·C_s.

Quick check

  1. A flywheel of I = 10 kg·m² runs at 3000 rev/min. How much energy does it store?
  2. If C_s is halved for the same ΔE and speed, what happens to the required inertia?
  3. ΔE = 2500 J and the work per cycle is 10 000 J. What is C_e?
  4. A four-stroke engine does 4π kJ per cycle. What is the mean torque?

Answers: 1. 493 kJ; 2. it doubles; 3. 0.25; 4. 1000 N·m.

Interview questions

All Theory of Machines and Machine Design interview questions

Try answering each one aloud before you open it.

  1. 1.What is a turning moment diagram, and why is it important in machine design?Concept

    A turning moment diagram, also known as a crank effort diagram, is a graphical representation of the turning moment or torque exerted on the crankshaft during one complete cycle of an engine. It is important in machine design because it helps in analyzing the fluctuations in torque, which can lead to vibrations and affect the performance and longevity of the machine. By understanding these fluctuations, engineers can design flywheels to smooth out the torque variations.

  2. 2.Explain the role of a flywheel in an engine.Concept

    A flywheel is a mechanical device specifically designed to efficiently store rotational energy. In an engine, its primary role is to smooth out the fluctuations in the power output by storing excess energy during periods of high torque and releasing it during periods of low torque. This helps in maintaining a consistent engine speed and reduces the stress on the engine components.

  3. 3.How does the mass of a flywheel affect its performance?Application

    A flywheel's effect depends on its moment of inertia I = m·k², so mass matters but radius matters more: doubling the radius of gyration quadruples I. For a required fluctuation of energy ΔE and speed band C_s, I = ΔE/(ω²·C_s), so a larger I gives a smaller speed variation within the cycle. A heavier wheel also slows acceleration of the engine and loads the bearings, which is why designers put the mass in a large-radius rim, limited by the hoop stress σ = ρv².

  4. 4.Why are flywheels typically made with a high moment of inertia?Application

    Flywheels are designed with a high moment of inertia to maximize their ability to store and release energy. A high moment of inertia means the flywheel can absorb more energy during periods of excess torque and release it during periods of low torque, effectively smoothing out the engine's power output and reducing vibrations.

  5. 5.What would happen if a flywheel were not used in an engine?Application

    Without a flywheel, an engine would experience significant fluctuations in speed due to the varying torque produced during different stages of the engine cycle. This would lead to increased vibrations, reduced efficiency, and potentially more wear and tear on engine components, ultimately shortening the engine's lifespan.

  6. 6.Describe how a turning moment diagram can be used to design a flywheel.Application

    Draw the mean torque line, which is the work per cycle divided by the cycle angle. Take the areas between the torque curve and the mean line in order, positive above and negative below, and keep a running total of energy; the difference between the largest and smallest totals is the maximum fluctuation of energy ΔE. The required inertia is I = ΔE/(ω_mean²·C_s), and for a rim-type wheel the rim mass is about I/R².

  7. 7.What factors should be considered when designing a flywheel?Application

    When designing a flywheel, several factors should be considered: the moment of inertia required to smooth out torque fluctuations, the material used (which affects strength and weight), the flywheel's size and shape (which influence its rotational dynamics), and the operating conditions such as temperature and speed. Additionally, safety factors must be considered to prevent mechanical failure.

  8. 8.Calculate the energy stored in a flywheel with a moment of inertia of 10 kg·m² rotating at 3000 RPM.Numerical

    First, convert the rotational speed from RPM to rad/s: 3000 RPM × (2π rad/60 s) = 314.16 rad/s. The energy stored in a flywheel is given by the formula E = 0.5 × I × ω², where I is the moment of inertia and ω is the angular velocity. Substituting the values, E = 0.5 × 10 kg·m² × (314.16 rad/s)² = 493,480.32 J.

  9. 9.If a flywheel is redesigned to reduce the coefficient of fluctuation of speed from 5% to 2%, what does this imply about its design?Application

    For the same fluctuation of energy and mean speed, I = ΔE/(ω²·C_s), so the inertia must increase by 5/2 = 2.5 times. That can come from more rim mass, a larger rim radius (I grows with R²), or a higher running speed if the flywheel can be placed on a faster shaft. The larger radius option is limited by rim hoop stress ρv².

  10. 10.A flywheel with a radius of 0.5 m and mass of 50 kg is rotating at 200 rad/s. Calculate its kinetic energy.Numerical

    The moment of inertia (I) for a solid disk flywheel is given by I = 0.5 × m × r², where m is the mass and r is the radius. Substituting the values, I = 0.5 × 50 kg × (0.5 m)² = 6.25 kg·m². The kinetic energy (E) is given by E = 0.5 × I × ω². Substituting the values, E = 0.5 × 6.25 kg·m² × (200 rad/s)² = 125,000 J.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousGear trains: simple, compound and epicyclicNext Balancing of rotating and reciprocating masses