Turning moment diagrams and flywheel design

Turning moment diagrams for engines and presses, mean torque and maximum fluctuation of energy, coefficient of fluctuation of speed, and sizing a rim or disc flywheel, with worked numericals.

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Why it matters

A reciprocating engine, compressor or punching press does not deliver or demand torque evenly: the torque swings widely within each cycle while the load (or the motor) wants it steady. A flywheel absorbs the surplus energy and returns it during the deficit, holding the speed within a set band. Sizing it from the turning moment diagram is a standard design calculation and a regular exam problem.

Key ideas

Turning moment (crank effort) diagram. A plot of crankshaft torque T against crank angle θ over one cycle, built from the force analysis at many angles (previous topic).

  • Single-cylinder four-stroke engine: one cycle covers 720°. The power stroke gives a large positive loop; suction, compression and exhaust are mostly negative (compression strongly so). This gives very large fluctuations and needs a heavy flywheel.
  • Single-cylinder two-stroke: one cycle per 360°; smaller fluctuation than a four-stroke of the same power.
  • Multi-cylinder engines: the loops of the cylinders overlap, so the resultant diagram is much flatter and needs a smaller flywheel.
  • Mean torque: T_mean = work per cycle / angle of the cycle. Areas of the diagram are energies: 1 N·m × 1 rad = 1 J.

Fluctuation of energy. With a steady load equal to T_mean, the flywheel speeds up while T > T_mean and slows while T < T_mean. Mark the areas between the T curve and the T_mean line, alternately positive (excess) and negative (deficit). Starting from an arbitrary energy E at the beginning of the cycle, add the areas cumulatively; the highest total gives the maximum speed and the lowest gives the minimum speed.

  • Maximum fluctuation of energy ΔE = E_max − E_min.
  • Coefficient of fluctuation of energy C_E = ΔE / (work done per cycle).

Coefficient of fluctuation of speed. C_s = (ω_max − ω_min) / ω_mean, the total (not ±) band. Typical allowances (take from your data book): around 0.1–0.2 for punching presses and crushers, about 0.03 for general machinery, and 0.01 or less for electric generators and spinning machines.

Flywheel energy relation. Since E = ½·I·ω², ΔE = ½·I·(ω_max² − ω_min²) = I·ω_mean·(ω_max − ω_min) = I·ω_mean²·C_s, using ω_mean = (ω_max + ω_min)/2.

Rim-type flywheel. Most of the inertia comes from the rim: I ≈ m·R² (arms and hub add about 5–10%, often ignored or allowed for by a factor). The rim's hoop stress from rotation is σ = ρ·v² with v the rim speed, which limits rim speed (about 25–30 m/s for cast iron). Disc flywheels use I = ½·m·R².

Punching presses and similar. The motor runs continuously at the average power, while the operation takes a large energy in a short part of the cycle. The flywheel supplies the difference between the energy needed during the operation and the energy the motor delivers in that time.

Flywheel versus governor. A flywheel controls the speed variation within one cycle; it cannot control the mean speed. A governor adjusts the fuel supply to keep the mean speed constant when the load changes over many cycles.

Formulas

T_mean = W_cycle / θ_cycle (N·m; θ in rad: 2π for a two-stroke or a 360° cycle, 4π for a four-stroke)

ΔE = E_max − E_min (J; from cumulative areas, with the scale: 1 mm² = (torque per mm) × (rad per mm) J)

ω_mean = (ω_max + ω_min)/2 ; C_s = (ω_max − ω_min) / ω_mean = (N_max − N_min) / N_mean

ΔE = ½·I·(ω_max² − ω_min²) = I·ω_mean²·C_s = 2·E_mean·C_s

  • I = flywheel moment of inertia (kg·m²), ω in rad/s, E_mean = ½·I·ω_mean².

I = m·k² ; rim: I ≈ m·R² ; solid disc: I = ½·m·R² (m in kg, R and k in m)

ΔE = m·v²·C_s (rim, v = ω_mean·R in m/s)

σ_hoop = ρ·v² (Pa; ρ in kg/m³)

Worked examples

Example 1 (standard): engine flywheel from a turning moment diagram. Given: the areas between the T curve and the mean-torque line over one cycle are, in order, +60, −40, +80, −70, +40, −70 mm². Scales: 1 mm = 500 N·m and 1 mm = 3° of crank angle. Mean speed 300 rpm; total speed fluctuation 2% (C_s = 0.02). Rim mean radius 0.8 m.

  1. Energy per mm²: 500 × (3 × π/180) = 26.18 J.
  2. Cumulative energies from E: E, E + 60, E + 20, E + 100, E + 30, E + 70, E (closes, as it must).
  3. Max = E + 100 and min = E, so ΔE = 100 mm² × 26.18 = 2618 J.
  4. ω_mean = 2π × 300 / 60 = 31.42 rad/s
  5. I = ΔE / (ω_mean²·C_s) = 2618 / (987.0 × 0.02) = 132.6 kg·m²
  6. Rim mass m = I / R² = 132.6 / 0.64 = 207 kg; rim speed v = 31.42 × 0.8 = 25.1 m/s; for cast iron (ρ ≈ 7200 kg/m³) σ = 7200 × 25.1² = 4.55 MPa, which is acceptable. Answer: ΔE ≈ 2.62 kJ, I ≈ 133 kg·m², rim mass ≈ 207 kg.

Example 2 (GATE level): punching press. Given: each punch needs 6 kJ and occupies one quarter of a revolution of the flywheel shaft; the shaft makes one revolution per punch. The motor supplies energy uniformly. The flywheel speed must stay between 240 and 216 rpm.

  1. Motor energy per revolution = energy per punch = 6000 J, supplied uniformly; during the punch (¼ rev) it supplies 6000 / 4 = 1500 J.
  2. The flywheel supplies the rest: ΔE = 6000 − 1500 = 4500 J.
  3. ω_max = 240 × π / 30 = 25.13 rad/s, ω_min = 216 × π / 30 = 22.62 rad/s; ω_max² − ω_min² = 631.7 − 511.6 = 120.0 rad²/s².
  4. I = 2·ΔE / (ω_max² − ω_min²) = 9000 / 120.0 = 75.0 kg·m²
  5. Motor power at the mean speed of 228 rpm (228 punches/min): P = 6000 × 228 / 60 = 22.8 kW (before losses). Without the flywheel the motor would need to supply 6000 J in a quarter revolution, four times this power. Answer: I ≈ 75 kg·m²; motor ≈ 22.8 kW.

Common mistakes

  • Taking ΔE as the largest single area instead of E_max − E_min from the cumulative sums.
  • Using ±C_s (half band) where the formula needs the total band, or vice versa; read the question.
  • Using rpm instead of rad/s in ΔE = I·ω²·C_s.
  • Writing ΔE = I·Δω or I = ΔE/Δω²; these are dimensionally wrong.
  • Using 2π for the cycle of a four-stroke engine; it is 4π.
  • Thinking a flywheel holds the mean speed constant under changing load; that is the governor's job.

For GATE ME

Common question types: flywheel moment of inertia from given ΔE and speed limits; ΔE from a turning moment diagram given as areas or as a simple torque function (for example T = T₀ + a·sin 2θ, where ΔE is the integral of T − T_mean between its zero crossings); punching-press flywheels; rim mass and stress. Practise cumulative-area tables and integrating simple torque functions.

Quick check

  1. Mean speed 600 rpm, max 612, min 588. What is C_s?
  2. I = 20 kg·m², ω_mean = 50 rad/s, C_s = 0.02. Find ΔE.
  3. Over what crank angle does one cycle of a four-stroke engine repeat?
  4. Does a flywheel reduce the mean speed change when load increases permanently?
  5. A torque of 400 N·m acts over π/2 rad. How much work is that?

Answers: 1. 24/600 = 0.04. 2. 20 × 2500 × 0.02 = 1000 J. 3. 720° (4π rad). 4. No; a governor does that. 5. 400 × 1.571 = 628 J.

Interview questions

All Theory of Machines and Vibrations 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 the analysis of engines?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 because it helps in understanding the variations in torque during the cycle, which is crucial for designing components like flywheels to ensure smooth engine operation.

  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, it helps to smooth out the power delivery by storing energy during periods of excess torque and releasing it during periods of low torque. This reduces fluctuations in engine speed and ensures a more consistent output.

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

    The mass of a flywheel affects its moment of inertia, which is a measure of its resistance to changes in rotational speed. A heavier flywheel has a higher moment of inertia, which means it can store more energy and smooth out larger fluctuations in engine speed. However, it also makes the engine less responsive to changes in speed.

  4. 4.Why is it necessary to have a turning moment diagram for designing a flywheel?Application

    A turning moment diagram is necessary for designing a flywheel because it provides detailed information about the variations in torque during an engine cycle. This information is crucial for determining the amount of energy the flywheel needs to store and release to maintain a consistent engine speed, which directly influences the flywheel's size and mass.

  5. 5.What would happen if a flywheel is 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 could lead to inefficient operation, increased wear and tear on engine components, and potentially cause the engine to stall during low torque phases.

  6. 6.Describe how the shape of a turning moment diagram can influence the design of a flywheel.Application

    The shape of a turning moment diagram indicates the magnitude and duration of torque fluctuations during an engine cycle. A diagram with large peaks and troughs suggests significant torque variations, requiring a flywheel with a higher moment of inertia to smooth out these fluctuations. Conversely, a more uniform diagram would require a less massive flywheel.

  7. 7.What is the significance of the area under the turning moment diagram?Concept

    Because torque times angle in radians is energy, the area under the turning moment diagram over one cycle is the work done per cycle, and dividing it by the cycle angle gives the mean torque. For flywheel design the important areas are those between the torque curve and the mean-torque line: each is an energy surplus or deficit. Adding them cumulatively gives the energy levels through the cycle, and the difference between the highest and lowest level is the maximum fluctuation of energy the flywheel must absorb.

  8. 8.A flywheel must absorb a maximum fluctuation of energy of 500 J while its speed varies between 21 and 19 rad/s. What moment of inertia is needed?Numerical

    Use ΔE = ½·I·(ω_max² − ω_min²) = I·ω_mean·(ω_max − ω_min). Here ω_mean = 20 rad/s and ω_max − ω_min = 2 rad/s, so I = 500 / (20 × 2) = 12.5 kg·m². Writing I = ΔE/Δω² is dimensionally wrong; the mean speed must appear, which is also why a flywheel on a faster shaft can be much lighter.

  9. 9.If an engine has a turning moment diagram with a maximum torque of 200 Nm and a minimum torque of 50 Nm, what is the torque fluctuation?Numerical

    The torque fluctuation is the difference between the maximum and minimum torque values. Therefore, the torque fluctuation is 200 Nm - 50 Nm = 150 Nm.

  10. 10.Does a flywheel improve an engine's efficiency or power?Application

    No. A flywheel stores and returns kinetic energy within each cycle, so it reduces cyclic speed fluctuation, smooths the torque delivered to the load and lets the engine pass through compression strokes and idle without stalling, but it cannot add energy. An over-sized flywheel slows the engine's response to throttle and adds mass, so it is sized only for the allowed coefficient of fluctuation of speed. Control of mean speed under changing load is done by a governor, not by the flywheel.

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