A 17 mm diameter hole is to be punched in a steel plate 19 mm thick at the rate of 20 holes per minute. The actual punching takes place in 1/5th the interval between punches. The driving motor runs at 1200 rpm and is geared to a countershaft, which runs at 160 rpm and upon which the flywheel is mounted. The countershaft in turn is geared to the crankshaft of the press. The resistance to shear for the plate may be taken as 310 MPa. Find (a) the power required for the motor if no flywheel is used, (b) the power required for the motor assuming a flywheel is used, and (c) the mass of the flywheel rim required assuming that 90% of the effective mass at the rim is due to the rim alone. The average speed at the rim diameter is 20 m/s, and the coefficient of speed fluctuation is 0.10.
Question 10.16: A 17 mm diameter hole is to be punched in a steel plate 19 m...
\text { Time taken for punching a hole }=\frac{60}{20}=3 s .
\text { Actual punching time }=\frac{3}{5}=0.6 s .
\text { Energy required for punching }=\pi \times 17 \times 19 \times 310=314567 Nm .
v=20 m / s , C_{s}=0.1, N _{m}=160 rpm .
\text { (a) Motor power without flywheel }=\frac{314567}{3 \times 10^{3}}=104.85 kW .
\text { (b) Energy required during punching operation }=\frac{314567 \times 0.6}{3}=62913.4 N m .
\text { Energy stored by flywheel }=314567-62913.4=251653.6 Nm .
\text { Motor power with flywheel }=\frac{251653.6}{3 \times 10^{3}}=83.88 kW .
(c) \omega=\frac{2 \pi \times 160}{60}=16.75 rad / s .
D=\frac{2 v}{\omega}=\frac{40}{16.75}=3.388 m .
K^{2}=\frac{D^{2}}{8}=0.713 m ^{2} .
\text { Mass of flywheel, } m=\frac{900 E_{f}}{\pi^{2} K^{2} N_{m}^{2} C_{s}}=\frac{900 \times 251653.6}{\pi^{2} \times 0.713 \times 160^{2} \times 0.1}=12572 kg .