A batch of $500$ jobs of diameter $50 m m$ and length $100 m m$ is to be turned at $200 r e v / m i n$ and feed $0.2 m m / r e v$

The number of tool changes required to machine the whole batch is

138

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270

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62

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61

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Solution

The correct option is B $270$
Given data:
$No.of jobs = 500, D = 50 and L = 100 m m$
$N = 200 r p m, f = 0.2 m m / r e v$

Cutting velocity:
$V = π D N / 1000 = 31.4 m / m i n$

Taylor's tool life equation:
$V T^{0.25} = 160$
$T = {(\frac{160}{31.4})}^{1 / 0.25} = 674 m i n$

$Time/part = \frac{L}{f \times N} = \frac{100}{0.2 \times 200} = 2.5 m i n$

No. of tool charges $= \frac{674}{2.5} = 269.6$

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A batch of 500 jobs of diameter 50mm and length 100mm is to be turned at 200rev/min and feed 0.2mm/rev The number of tool changes required to machine the whole batch is

The correct option is B 270Given data: No.of jobs=500,D=50andL=100mm N=200rpm,f=0.2mm/rev Cutting velocity: V=πDN/1000=31.4m/min Taylor's tool life equation: VT0.25=160 T=(16031.4)1/0.25=674min Time/part =Lf×N=1000.2×200=2.5min ​​​​​​​ No. of tool charges =6742.5=269.6

A batch of $500$ jobs of diameter $50 m m$ and length $100 m m$ is to be turned at $200 r e v / m i n$ and feed $0.2 m m / r e v$

The number of tool changes required to machine the whole batch is