Question

Tool life of a $10$ hours is obtained when cutting with a single point tool at $63m/min$. If Taylor's constant $C=257.35,$ tool life on doubling the velocity will be

• A. $5$ hours
• B. $25.7$ min
• C. $38.3$ min
• D. unchanged

Answer 1

The correct option is B $25.7$ min

Given data:
$V_1=63m/min,T_1=10hrs$
$C=257.35$

Taylor's tool life equation:
$V_1{T}_1^n=C$

$T_1^n=\frac{C}{V_1}$
$n=\frac{ln\left(\frac{C}{V_1}\right)}{ln{T}_1}=\frac{ln\frac{257.35}{63}}{ln600}=0.22$

If cutting velocity is doubled
$V_2=2\times V_1=126m/min$
Tool life for the corresponding double cutting velocity :
$T_2={\left(\frac{257.35}{126}\right)}^{1/0.22}$

$={\left(\frac{257.35}{126}\right)}^{4.54}=25.6min$