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 (a) 26.7 (b) 42.5 (c) 80.7 (d) 142.9

Answer 1

26.667

Step-by-step explanation:

Given Data

For Tool A

Life exponent
`{\ n_1}`=0.45

Constant
`{C_1}`=90

For tool B

Life exponent
`{n_2}`=0.3

Constant
`{C_2}`=60

and tool life equation is

`VT^(n)=c`

`VT_(A)^(0.45)=90`

`T_(A)^(0.45)=(90)/(V)`

`T_(A)=(90)/(V)^{(1)/(0.45)}`

`For Tool B`

`VT_(A)^(0.3)=60`

`T_(B)^(0.3)=(60)/(V)`

`T_(B)=(60)/(V)^{(1)/(0.3)}`

`T_(A)>T_(B)`

`(90)/(V)^{(1)/(0.45)}>(60)/(V)^{(1)/(0.3)}`

`V>26.667`