Question

A constant torque motor of 0.25 kW drives a riveting machine. The mass of the moving parts including the flywheel is 125 kg at 700 radius of gyration. One riveting operation absorbs 1 kJ of energy and takes one second. Speed of the flywheel is 240 rpm before riveting. The number of rivets closed per hour, and reduction in speed after the riveting operation are

• A. 900 holes, 4.7 rpm
• B. 600 holes, 2.5 rpm
• C. 750 holes, 4.7 rpm
• D. 900 holes, 2.7 rpm

Answer 1

The correct option is A 900 holes, 4.7 rpm


Area, A + B + C

Given Energy supplied by motor in 1 sec = 250 J

Energy required for 1 rivet operation in 1sec = 1000 J

Excess energy required = 100-250 = 750 J

Total time required for motor to supply 1000 J

$=\frac{1000}{250}=4sec$

So non- machining time = 4 - 1 = 3 sec.

Energy supplied by motor in 3sec is stored in flywheen andthis energy is released during riveting operation which takes place in 1 sec.

No. of rivet closed per hour

$=\frac{\text{time of execution}}{\text{total time required per hole}}$

$=\frac{1\times 60\times 60}{4}=900holes$

$I=m{k}^2=125\times {0.7}^2=61.25kg{m}^2$

Now, $\Delta E=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)=750$

$=\frac{1}{2}\times 61.25\times \left(\frac{2\pi }{60}\right)({240}^2-N_2^2)=750$

$N_2=235.3rpm$

Reduction in speed = 240 - 235.3 = 4.7 rpm