Turning Moment Diagram and Flywheel

A turning moment diagram (crank-effort diagram) is a plot of the turning moment (torque) on the crankshaft against the crank angle.

Four-stroke engine diagram

A cycle takes two revolutions ($720^\circ$): suction, compression, expansion (power) and exhaust.

 T
 ^        /\
 |       /  \          Expansion
 |  ____/    \___
 +-/---------------\-----------------> theta
 |  Suction   Compr.   \_/ Exhaust
 |  (-ve small) (-ve large)
 0     180      360     540      720

• Expansion stroke: large positive torque (gas pressure drives the piston).
• Suction and exhaust: small negative torque (the crank must push the piston).
• Compression: large negative torque (work is absorbed).
• The mean torque line is where the area above equals the area below; its height is $T_{mean} = \dfrac{\text{work per cycle}}{4\pi}$.

Torque above the mean line accelerates the shaft; below it, the shaft slows down. The flywheel absorbs and returns this energy.

Definitions

• Fluctuation of energy ($\Delta E$): the difference between the maximum and minimum kinetic energy of the flywheel in one cycle, i.e. the largest swing of the cumulative area between the torque curve and the mean torque line.
• Coefficient of fluctuation of energy ($C_E$):

$$C_E = \frac{\Delta E}{\text{work done per cycle}}$$

• Coefficient of fluctuation of speed ($C_s$):

$$C_s = \frac{\omega_{max}-\omega_{min}}{\omega_{mean}} = \frac{N_{max}-N_{min}}{N_{mean}}, \qquad \omega_{mean} = \frac{\omega_{max}+\omega_{min}}{2}$$

Its reciprocal is the coefficient of steadiness. Typical $C_s$: 0.01 to 0.02 for alternators, about 0.03 for machine tools, 0.2 for crushers.

Data: mean speed $N = 240$ rpm, $C_s = 0.03$, rim mean radius $R = 0.9$ m.

Energy at each point

Take the energy at A as zero and add the areas in order (positive area = gain of kinetic energy, negative = loss).

Maximum energy = 4.8 kJ, minimum energy = -1.1 kJ.

$$\Delta E = E_{max} - E_{min} = 4.8 - (-1.1) = 5.9\ \text{kJ} = 5900\ \text{J}$$

Moment of inertia

$$\omega = \frac{2\pi \times 240}{60} = 25.13\ \text{rad/s}$$ $$\Delta E = I\omega^2 C_s \Rightarrow I = \frac{\Delta E}{\omega^2 C_s} = \frac{5900}{(25.13)^2 \times 0.03} = 311.4\ \text{kg·m}^2$$

Mass of the rim

Treating the rim as a thin ring, $I = mR^2$:

$$m = \frac{I}{R^2} = \frac{311.4}{0.9^2} = 384\ \text{kg}$$

Answer: $\Delta E = 5.9$ kJ; mass of rim $\approx 384$ kg.

Data: $d = 0.03$ m, $t = 0.01$ m, $\tau_u = 350$ MPa, $N = 300$ rpm, $R = 0.6$ m, $C_s = 0.03$ (since $\pm1.5\%$ means a total variation of $3\%$).

Energy for one punching

Maximum punching force:

$$F = \pi d t\,\tau_u = \pi \times 0.03 \times 0.01 \times 350\times10^{6} = 329867\ \text{N}$$

The force falls from $F$ to zero over the stroke, so the average force is $F/2$:

$$E = \frac{F}{2}\times t = \frac{329867}{2}\times 0.01 = 1649\ \text{J}$$

Motor power

The motor supplies this energy uniformly over every revolution (one revolution takes $60/300 = 0.2$ s):

$$P = E \times \frac{N}{60} = 1649 \times 5 = 8247\ \text{W} = 8.25\ \text{kW}$$

Fluctuation of energy

During the punching (20% of a revolution) the motor supplies only $0.2E$:

$$E_{motor} = 0.2 \times 1649 = 330\ \text{J}$$

The flywheel must supply the rest:

$$\Delta E = E - E_{motor} = 1649 - 330 = 1319\ \text{J}$$

Flywheel

$$\omega = \frac{2\pi \times 300}{60} = 31.42\ \text{rad/s}$$ $$I = \frac{\Delta E}{\omega^2 C_s} = \frac{1319}{(31.42)^2 \times 0.03} = 44.56\ \text{kg·m}^2$$ $$m = \frac{I}{R^2} = \frac{44.56}{0.6^2} = 123.8\ \text{kg}$$

Answer: Motor power $\approx 8.25$ kW; flywheel rim mass $\approx 123.8$ kg.

Derivation

Let the flywheel have moment of inertia $I$, maximum speed $\omega_1$, minimum speed $\omega_2$ and mean speed $\omega = \dfrac{\omega_1+\omega_2}{2}$.

Maximum kinetic energy: $E_{1} = \tfrac12 I\omega_1^2$. Minimum kinetic energy: $E_{2} = \tfrac12 I\omega_2^2$.

$$\begin{aligned} \Delta E &= E_1 - E_2 = \tfrac12 I(\omega_1^2-\omega_2^2) \\ &= \tfrac12 I(\omega_1-\omega_2)(\omega_1+\omega_2) \\ &= I\,\omega\,(\omega_1-\omega_2) \end{aligned}$$

since $\omega_1+\omega_2 = 2\omega$. Using $C_s = \dfrac{\omega_1-\omega_2}{\omega}$, we get $\omega_1-\omega_2 = C_s\omega$, so

$$\Delta E = I\omega^2 C_s = 2E\,C_s$$

where $E = \tfrac12 I\omega^2$ is the mean kinetic energy.

Why the mass is in the rim

For a required $\Delta E$, speed and $C_s$, the value of $I$ is fixed. Since $I = mk^2$, a large radius of gyration $k$ gives the same $I$ with less mass. So the material is placed at the rim, which keeps the weight and cost low. The limit is the rim stress $\sigma = \rho v^2$, which restricts the rim speed.

Rimmed ring vs solid disc

For the same mass and outer radius, the rim stores about twice the energy of a solid disc.