Question

For tool A, Taylor’s tool life exponent (n) is 0.45 and constant (K) is 90. Similarly, for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is

1. 26.7
2. 42.5
3. 80.7
4. 142.9

Answer 1

Correct Answer - Option 1 : 26.7

Concept:

The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is found by equating the tool life of both the tools.

V_AT_A^nA = K_A

\(T_A=\left(\frac{K_A}{V_A}\right)^{\frac{1}{nA}}\)

V_BT_B^nB = K_B

\(T_B=\left(\frac{K_B}{V_B}\right)^{\frac{1}{nB}}\)

Calculation:

Given:

n_A = 0.45, K_A = 90

nB = 0.30, KB = 60

\({V_A}T_A^{0.45} = 90 \Rightarrow {T_A} = {\left( {\frac{{90}}{{{V_A}}}} \right)^{\frac{1}{{0.45}}}}\)

\({V_B}T_B^{0.3} = 60 \Rightarrow {T_B} = {\left( {\frac{{60}}{{{V_B}}}} \right)^{\frac{1}{{0.3}}}}\)

At limiting conditions i.e.

T_A = T_B, put V_A = V_B = V

\({\left( {\frac{{90}}{V}} \right)^{\frac{1}{{0.45}}}} = {\left( {\frac{{60}}{V}} \right)^{\frac{1}{{0.3}}}} \Rightarrow V = 26.67\;m/s\)