A 200 mm diameter work piece is plain turned over a length of 600 mm in one cut with a tool having Taylor's tool life equation $V T^{0.35} = 75$ . The tool changing time is 15 min and feed used is 0.1 mm/rev.

The optimum cutting velocity required for machining the given component using maximum production rate criteria machining is

84.6 m/min

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68.3 m/min

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47.5 m/min

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23.4 m/min

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Solution

The correct option is D 23.4 m/min
$V T^{0.35} = 75$

$n = 0.35, c = 75$

$T_{c} = 15 min, f = 0.1 mm/rev$

$V_{o p t} = C {(\frac{n}{1 - n} \frac{1}{T_{c}})}^{n}$

$= 75 {(\frac{0.35}{1 - 0.35} \times \frac{1}{15})}^{0.35}$

$V_{o p t} = 23.4 m/min$

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The correct option is D 23.4 m/minVT0.35=75 n=0.35, c=75 Tc=15 min,f=0.1 mm/rev Vopt=C(n1−n1Tc)n =75(0.351−0.35×115)0.35 Vopt=23.4 m/min

The correct option is D 23.4 m/min
$V T^{0.35} = 75$

$n = 0.35, c = 75$

$T_{c} = 15 min, f = 0.1 mm/rev$

$V_{o p t} = C {(\frac{n}{1 - n} \frac{1}{T_{c}})}^{n}$

$= 75 {(\frac{0.35}{1 - 0.35} \times \frac{1}{15})}^{0.35}$

$V_{o p t} = 23.4 m/min$