Flywheel for a punching machine

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.