B. A punching press is required to punch 30 mm diameter holes in a plate of 20 mm thickness at the rate of 20 holes/mm. It requires 6Nm of energy / mm2 of sheared area. If the punching takes place in 1/10 of sec and rpm of the flywheel varies from 160 to 140, determine the weight of the flywheel having radius of gyration is 1m.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: May 2016

Give data:-

d=30 mm.

t=20mm

rate of holes= 20 holes/min

Energy Required= 6N-m

Radius of gyration k=1m

T=0.15

$n_1 =160 rpm \\ n_2 = 140 rpm$

To find weight,

Calculating the max fluctuation of energy ΔEmax

Energy req. Per hole is given by

$E= E_1$ *sheared area per hole

E$=6*Π*dt \\ =6*Π*30*20$

E=11309.73 J

Energy required for punching operation per sec

=E* no. of holes/sec

=11309.73*20/60

=3769.91 J/s

Energy supplied by the motor in 1/10 of sec

$E_2=3769.91/10=376.991 J/s$

ΔE $=E_1 – E_2 \\ =11309.73-376.991$

ΔE=10932.73 J

Mean speed $=( N_1+N_2)/2 \\ ={160+140}/2 \\ =150 rpm$

ii. Calculation of coefficient of fluctuation of speed

$w= (2ΠN)/60 =2*Π*150/60=15.7079 rad/sec \\ w_1 =2*Π*N1/60=2*Π*160/60=16.7551 rad/sec \\ w_2=2*Π*N2/60=2*Π*140/60=14.6607 rad/sec \\ Cs = (w_1-w_2)/w=(16.7551-14.6607)/15.7079=0.1333$

$ΔE_{\max}=Iw_2Cs \\ 10932.73=I*(15.7079)^2*0.1333 \\ I=10932.73/(15.7079)^2 *0.1333 \\ I=332.400kgm^2 \\ I=mk^2 \\ 332.400=m*12 \\ m=332.40 Kg \\ w=m*g \\ =332.40*9.81 \\ w=3260.85≈3261N$