The cutting speed is reduced by 75%. Using the Taylor equation $(V T^{n} = C)$ for tool life, the percentage increase in tool life is (Take n = 0.5 and C = 400)

400%

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800%

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1500%

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1600%

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Solution

The correct option is C 1500%
$V_{2} = V_{1} - 0.75 V_{1} = 0.25 V_{1}$

From Taylor equation,

$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$

$\Rightarrow {(\frac{T_{2}}{T_{1}})}^{n} = \frac{V_{1}}{0.25 V_{1}} = 4$

$\Rightarrow \frac{T_{2}}{T_{1}} = {(4)}^{2} = 16$

$∴$ % increase in tool life $= (\frac{T_{2}}{T_{1}} - 1) \times 100 = 15 \times 100 = 1500$ %

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V T^{n} = C

Using Taylor equation V T^{n} = C, what is teh percentage increase in tool life when the cutting speed is reduced by 75%. Tool life exponent is \frac{2}{3} .

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