Answer:
So % increment in tool life is equal to 4640 %.
Explanation:
Initially n=0.12 ,V=130 m/min
Finally C increased by 10% , V=90 m/min
Let's take the tool life initial condition is [tex]T_1[/tex] and when C is increased it become [tex]T_2[/tex].
As we know that tool life equation for tool
[tex]VT^n=C[/tex]
At initial condition [tex]130\times (T_1)^{0.12}=C[/tex]------(1)
At final condition [tex]90\times (T_2)^{0.12}=1.1C[/tex]-----(2)
From above equation
[tex]\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}[/tex]
[tex]T_2=47.4T_1[/tex]
So increment in tool life =[tex]\dfrac{T_2-T_1}{T_1}[/tex]
=[tex]\dfrac{47.4T_1-T_1}{T_1}[/tex]
So % increment in tool life is equal to 4640 %.
