A lathe is used to turn a 25mm diameter bar. The value of n = 0.16 and tool life is 45 min at speed of 100 m/min. If a length of 50 mm per component is machined and the feed rate is 0.25 mm/rev. The number of pieces that can be produced between tool changes if spindle should run to obtain a tool of 5 hours is _____.
1410

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Solution

The correct option is A 1410
$V \times (5 \times 60)^{0.16} = 100 \times (45)^{0.16}$
$V = 73.82 m / m i n$

Spindle speed $= \frac{73.82 \times 1000}{π \times 25} = 939.90543 r e v / m i n$

Cutting time per piece $= \frac{50 \times 60}{939.90543 \times 0.25} = 12.7672 s e c$

No. of components per tool change $= \frac{50 \times 60 \times 60}{12.7672} = 1409.858 = 1410$

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The correct option is A 1410V×(5×60)0.16=100×(45)0.16 V=73.82m/min Spindle speed =73.82×1000π×25=939.90543rev/min Cutting time per piece =50×60939.90543×0.25=12.7672sec No. of components per tool change =50×60×6012.7672=1409.858=1410