Question

Taylor's tool life equation is used to estimate the life of a batch of identical HSS twist drills by drilling through holes at constant feed in 20 mm thick mild steel plates. In test 1, a drill lasted 300 holes at 150 rpm while in test 2, another drill lasted 200 holes at 300 rpm. The maximum number of holes that can be made by another drill from the above batch at 200 rpm is (correct to two decimal places.)

Answer 1

Time required for drilling a hole
$t=\frac{L}{fN}$

Here L and f are cosntant and N is variable.

$V_1=\pi D\times 150m/min$

$T_1=300\times \frac{L}{f\times 150}min$

$V_2=\pi D\times 300min$

$T_2=200\times \frac{L}{f\times 300}min$

$V_3=\pi D\times 200min$

$T_3=x\times \frac{L}{f\times 200}min$

[No. of hole is 'x']

$V_1{T}_1^n=V_2{T}_2^n$

$\pi D\times 150\times {(300\times \frac{L}{f\times 150})}^n$

$=\pi D\times 300\times {(200\times \frac{L}{f\times 300})}^n$

$or,2^n=2{\left(\frac{2}{3}\right)}^n$

$or,3^n=2$

$n=0.63093$

$V_1{T}_1^n=V_3{T}_3^n$

$\pi D\times 150\times {(300\times \frac{L}{f\times 150})}^{0.63093}$

$=\pi D\times 200\times {(x\times \frac{L}{f\times 200})}^{0.63093}$

$3\times 2^{0.63093}=4\times {\left(\frac{x}{200}\right)}^{0.63093}$

$\Rightarrow x=253.53$