Turning moment diagrams and flywheel design

Turning moment diagrams, mean torque and power, maximum fluctuation of energy from loop areas, coefficient of fluctuation of speed and flywheel sizing including rim stress and punching presses.

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Why it matters

An engine's torque swings from strongly positive on the power stroke to negative on the compression stroke, while the load on the crankshaft is nearly steady. Without a flywheel the crankshaft speed would rise and fall within every cycle, giving rough idle, gear rattle and possible stalling. The turning moment diagram tells you how much energy must be stored and released each cycle, and from that you size the flywheel.

Key ideas

Turning moment (crank effort) diagram. A plot of the torque on the crankshaft against crank angle over one working cycle (2π rad for a two-stroke engine, 4π rad for a four-stroke). It is built from the slider-crank force analysis, T = F_T·r, at many crank angles. The area under the curve is work (N·m × rad = J).

Mean torque. If the load torque is constant, it equals the mean engine torque at steady speed: T_mean = (area under the T–θ curve over a cycle) / (cycle angle). Power = T_mean·ω.

Excess energy and the flywheel. Where engine torque is above the mean, the excess work speeds the flywheel up; where it is below, the flywheel slows down and supplies the deficit. Over a full cycle the net excess is zero. The flywheel limits the change of speed within a cycle; it does not control the mean speed when load changes (that is the governor's job).

Maximum fluctuation of energy ΔE. Mark the intercepted areas above (+) and below (−) the mean-torque line, in order. Starting from an arbitrary energy E at the beginning, add the areas one by one: E + a₁, E + a₁ − a₂ and so on. The largest value minus the smallest value is ΔE. It is not simply the largest single area, and not (T_max − T_min) × 2π.

Coefficient of fluctuation of speed C_s. C_s = (N_max − N_min) / N_mean. "Speed within ±1% of the mean" means C_s = 0.02. Typical values: 0.002 to 0.005 for engines driving generators, 0.01 to 0.03 for vehicle engines, up to 0.2 for punching presses and crushers.

Coefficient of fluctuation of energy C_E. ΔE divided by the work done per cycle. It describes the engine (fewer cylinders, larger C_E); a single-cylinder four-stroke engine has a much larger C_E than a six-cylinder one, so it needs a much larger flywheel.

Flywheel size. With N_mean ≈ (N_max + N_min)/2, the change of kinetic energy between maximum and minimum speed is ½I(ω_max² − ω_min²) = I·ω²·C_s. Setting this equal to ΔE gives the moment of inertia needed.

Rim-type flywheel. Most of the inertia is in the rim, so I ≈ m·R², where R is the mean rim radius. The rim radius is limited by the hoop (centrifugal) stress σ = ρ·v², where v = ωR is the rim speed; for cast iron v is limited to about 30 to 35 m/s, for steel considerably more. For a given ΔE a larger radius gives a lighter flywheel, until stress limits it.

Punching presses. The motor runs continuously while the punch takes energy only during a short part of the cycle; the flywheel supplies almost all the punching energy and slows down. ΔE ≈ energy for punching × (1 − θ_punch / 2π), where θ_punch is the crank angle during which punching happens.

Formulas

T_mean = W_cycle / θ_cycle

  • T_mean: mean torque (N·m); W_cycle: work done per cycle (J); θ_cycle: cycle angle (rad): 2π for a two-stroke, 4π for a four-stroke engine.

P = T_mean·ω = 2π·N·T_mean / 60

  • P: power (W); ω: mean angular speed (rad/s); N: mean speed (rpm).

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

  • C_s: coefficient of fluctuation of speed (dimensionless).

ΔE = I·ω²·C_s = 2·E·C_s

  • ΔE: maximum fluctuation of energy (J); I: flywheel moment of inertia (kg·m²); ω: mean speed (rad/s); E = ½·I·ω²: mean kinetic energy of the flywheel (J).

I = m·k² (general) and I ≈ m·R² (rim type)

  • m: mass (kg); k: radius of gyration (m); R: mean radius of the rim (m). A solid disc has k² = R²/2.

σ = ρ·v²

  • σ: hoop stress in a thin rotating rim (Pa); ρ: density (kg/m³); v: rim peripheral speed (m/s).

ΔE = E_punch·(1 − θ_p / 2π)

  • E_punch: energy needed per punching operation (J); θ_p: crank angle during which punching takes place (rad). Assumes a motor supplying energy uniformly.

Worked examples

Example 1 (standard: torque given as a function). The turning moment of an engine is T = 1000 + 300·sin 2θ N·m, θ in radians, and the load torque is constant. The mean speed is 600 rpm and the speed must stay within ±1% of the mean. Find the power and the flywheel moment of inertia.

  1. Mean torque: the sin 2θ term averages to zero over a cycle, so T_mean = 1000 N·m.
  2. ω = 2π × 600 / 60 = 62.83 rad/s. Power = 1000 × 62.83 = 62.8 kW.
  3. Excess torque = 300 sin 2θ, positive from θ = 0 to π/2.
  4. ΔE = ∫₀^(π/2) 300 sin 2θ dθ = 300 × [−cos 2θ / 2]₀^(π/2) = 300 × (½ + ½) = 300 J.
  5. C_s = 0.02 (±1%).
  6. I = ΔE / (ω²·C_s) = 300 / (3947.8 × 0.02) = 3.80 kg·m².

Example 2 (GATE level: areas and a rim flywheel). The intercepted areas between a turning moment diagram and the mean-torque line, taken in order over one cycle, are +52, −124, +92, −140, +85, −68 and +103 mm². The scales are 1 mm = 600 N·m and 1 mm = 3°. The mean speed is 300 rpm with ±1.5% fluctuation. Find the flywheel moment of inertia, the rim mass for a mean radius of 1 m, and the hoop stress for a cast-iron rim (ρ = 7200 kg/m³).

  1. Check: positive areas 52 + 92 + 85 + 103 = 332; negative 124 + 140 + 68 = 332. The net is zero, as it must be.
  2. Energy levels from E: E, E + 52, E − 72, E + 20, E − 120, E − 35, E − 103, E.
  3. Maximum E + 52; minimum E − 120; ΔE = 172 mm².
  4. Scale: 1 mm² = 600 × 3 × π / 180 = 31.42 J, so ΔE = 172 × 31.42 = 5404 J.
  5. ω = 2π × 300 / 60 = 31.42 rad/s; C_s = 0.03.
  6. I = 5404 / (987.0 × 0.03) = 182.5 kg·m².
  7. Rim mass: m = I / R² = 182.5 kg (taking all the inertia in the rim).
  8. Rim speed v = 31.42 × 1 = 31.4 m/s; σ = 7200 × 31.42² = 7.11 × 10⁶ Pa = 7.1 MPa, acceptable for cast iron.

Common mistakes

  • Taking ΔE as the largest single loop area instead of the difference between the highest and lowest cumulative energy levels.
  • Reading ±1% as C_s = 0.01. It is 0.02.
  • Using N in rpm in ΔE = I·ω²·C_s.
  • Using 2π as the cycle angle for a four-stroke engine; it is 4π.
  • Forgetting the area scale: mm² × (N·m per mm) × (rad per mm).
  • Believing a flywheel keeps the mean speed constant when the load changes. That needs a governor.

For GATE ME

Typical questions give T(θ) as a simple function or as a set of loop areas and ask for ΔE, the moment of inertia for a given C_s, the speed fluctuation for a given flywheel, the mean power, or the rim mass. Punching-press flywheels are another common type. Practise the cumulative-energy table for area problems and integration of sin 2θ type torque functions.

Quick check

  1. Speed limits 1010 and 990 rpm: C_s?
  2. ΔE = 2 kJ, ω = 50 rad/s, C_s = 0.02. Required I?
  3. Why does a single-cylinder engine need a larger flywheel than a four-cylinder engine of the same power?
  4. What is the cycle angle of a four-stroke engine's turning moment diagram?
  5. Hoop stress in a steel rim (ρ = 7800 kg/m³) at 40 m/s?

Answers: 1. 20/1000 = 0.02. 2. 2000 / (2500 × 0.02) = 40 kg·m². 3. Its torque fluctuates far more (larger C_E), so ΔE is larger. 4. 4π rad (720°). 5. 7800 × 1600 = 12.5 MPa.

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 design of machines?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 analyzing the fluctuations in torque, which can lead to vibrations and affect the smooth operation of the machine. By understanding these fluctuations, engineers can design flywheels to minimize the effects and ensure efficient operation.

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

    A flywheel is a heavy rotor on the crankshaft that stores kinetic energy. During the power stroke the engine torque exceeds the load torque and the excess energy speeds the flywheel up slightly; during the other strokes it gives the energy back and slows down. This limits the cyclic speed fluctuation within each cycle to a chosen coefficient C_s, using ΔE = I·ω²·C_s, and carries the crank through the dead centres at idle. It does not hold the mean speed when the load changes; that is the governor's or throttle's job.

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

    What matters is the moment of inertia I = m·k², so mass placed at a large radius (in the rim) is most effective; for a given energy fluctuation ΔE the speed fluctuation is C_s = ΔE/(I·ω²). A heavier flywheel gives smoother running and easier starting from rest under load, but it slows the engine's response to the throttle, adds vehicle mass and loads the crankshaft bearings. That is why petrol car engines use relatively light flywheels and large diesels and single-cylinder engines use heavy ones.

  4. 4.Why is it necessary to have a turning moment diagram for multi-cylinder engines?Application

    In a multi-cylinder engine the torque curves of the cylinders are phase-shifted by the firing interval, and the turning moment diagram of the engine is their sum. Adding them fills in the gaps between power strokes, so the fluctuation about the mean torque and the maximum fluctuation of energy are far smaller than for one cylinder. Drawing the combined diagram gives ΔE, and therefore the flywheel size needed; a six-cylinder engine needs a much smaller flywheel than a single-cylinder engine of the same power.

  5. 5.What happens if a flywheel is not used in an engine?Application

    Without a flywheel, the crankshaft speed would rise sharply on each power stroke and fall on the compression stroke, so the speed fluctuation within a cycle would be large. A single-cylinder engine might not even get through compression and the dead centres at low speed and would stall at idle. The torque pulsations would also hammer the gears and drivetrain and make the engine rough; multi-cylinder engines suffer less but still need some inertia.

  6. 6.Calculate the energy stored in a flywheel with a moment of inertia of 5 kg·m² rotating at 3000 RPM.Numerical

    To calculate the energy stored in the flywheel, we use the formula: E = 0.5 * I * ω², where I is the moment of inertia and ω is the angular velocity in rad/s. First, convert RPM to rad/s: ω = 3000 * (2π/60) = 314.16 rad/s. Then, E = 0.5 * 5 * (314.16)² = 246,740.16 J (joules).

  7. 7.What is the effect of increasing the diameter of a flywheel on its moment of inertia?Application

    For a fixed mass, I = m·k², and k is proportional to the radius (k² = R²/2 for a solid disc, about R² for a rim), so doubling the diameter quadruples I. If the thickness and material are kept the same the mass also grows with R², so I rises with R⁴. The limit is the rim's hoop stress σ = ρ·v², which rises with the square of rim speed, so the diameter is chosen as large as the stress and space allow and the mass is then reduced.

  8. 8.A flywheel modelled as a solid disc of mass 10 kg and radius 0.5 m rotates at 2000 rpm. Calculate its kinetic energy.Numerical

    For a solid disc I = ½·m·r² = 0.5 × 10 × 0.5² = 1.25 kg·m². ω = 2π × 2000 / 60 = 209.44 rad/s. KE = ½·I·ω² = 0.5 × 1.25 × 209.44² = 0.625 × 43 865 = 27 416 J, about 27.4 kJ.

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