A cylinder of 25 mm diameter and 100 mm length is turned with a tool, for which the relation $V T^{0.25}$ = 55 is applicable. The cutting velocity is 22 m/min. For a tool feed of 0.046 mm/rev, the number of tool regrinds required to produce 425 cylinders is

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Solution

The correct option is D 85
Given data:
D = 25, L = 100 mm, $V T^{0.25}$ = 25
$V_{C}$ = 22 m/min and f = 0.046 mm/rev.

$N = \frac{1000 V}{π D} = \frac{1000 \times 22}{π \times 25}$
$= 280 r p m$

$Time/piece = \frac{L}{t \times N} = \frac{100}{0.046 \times 280}$

$= 7.76 min$

$Total time = 425 \times 7.76 = 3299.7 min$

$T = {(\frac{50}{V})}^{V θ .25} = {(\frac{55}{22})}^{4} = 39$

$No. of tool regrinds = \frac{3299.7}{39} = 85$

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A cylinder of 25 mm diameter and 100 mm length is turned with a tool, for which the relation VT0.25 = 55 is applicable. The cutting velocity is 22 m/min. For a tool feed of 0.046 mm/rev, the number of tool regrinds required to produce 425 cylinders is

The correct option is D 85Given data: D = 25, L = 100 mm, VT0.25 = 25 VC = 22 m/min and f = 0.046 mm/rev. N=1000VπD=1000×22π×25 =280rpm Time/piece=Lt×N=1000.046×280 =7.76min Total time =425×7.76=3299.7min T=(50V)Vθ.25=(5522)4=39 No. of tool regrinds=3299.739=85

A cylinder of 25 mm diameter and 100 mm length is turned with a tool, for which the relation $V T^{0.25}$ = 55 is applicable. The cutting velocity is 22 m/min. For a tool feed of 0.046 mm/rev, the number of tool regrinds required to produce 425 cylinders is