The tool life equation for HSS tool is $V T^{0.14} f^{0.7} d^{0.4} = Constant .$

The tool life (T) of 30 min is obtained using the following cutting conditions: V = 45 m/min, feed = 0.35mm, d= 2.0 mm. If feed (f) and depth of cut (d) are decreased individually by 25% and speed (V) is increased by 25%, the tool life (in min) will be:

1.06

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58.42

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45.84

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Solution

The correct option is C 58.42
Given data:
$T_{1} = 30 min, f_{1} = 0.35 mm,$

$d_{1} = 2.0 mm, V_{1} = 45$

$V_{2} = 1.25 V_{1}, f_{2} = 0.75 f_{1}$

$d_{2} = 0.75 d_{1}, C = 45.8425$

$V_{1} T_{1}^{0.14} f_{1}^{0.7} d_{1}^{0.14} = C = V_{2} T_{2}^{0.14} f_{2}^{0.7} d_{2}^{0.4}$

$(1.25 \times 30) \times T_{2}^{0.14} (0.75 \times 0.35)^{0.7} \times (0.75 \times 2)^{0.4} = 45.8425$

$\Rightarrow T_{2} = 58.42 min$

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The correct option is C 58.42Given data: T1=30 min, f1=0.35 mm, d1=2.0 mm, V1=45 V2=1.25V1, f2=0.75 f1 d2=0.75d1, C=45.8425 V1T0.141f0.71d0.141=C=V2T0.142f0.72d0.42 (1.25×30)×T0.142(0.75×0.35)0.7×(0.75×2)0.4=45.8425 ⇒T2=58.42 min