Question

In a machine operation with a turning tool, the tool life (T) is related to cutting speed v (m/s), feed f (mm), and depth of cut d (mm) as T = C v^-2.5 f^-0.9d^-0.15

Where C is a constant. The suggested values for the cutting parameters are: v = 1.5 m/s, f = 0.25 mm and d = 3 mm for normal rough turning. If the operation is performed at twice the cutting speed and the other parameters remain unchanged, the corresponding percentage change (magnitude) in tool life is _______.

Answer 1

Concept:

T = C v-2.5 f-0.9d-0.15

Where C = constant, v = velocity (m/s), f = feed (mm), d = depth of cut (mm), T = tool life 

Calculation:

Given:

v = 1.5 m/s, f = 0.25 mm, d = 3 mm

∴ T₁ = C v-2.5 f-0.9d-0.15

T₁ = C × (1.5)^-2.5  × (0.25)^-0.9 × (3)^-0.15

T₁ = 1.07166 C

Now according to the question the cutting speed has becomes twice and other parameters remain unchanged.

v₂ = 3 m/s

∴ T2 = C v-2.5 f-0.9d-0.15

T2 = C × (3)-2.5  × (0.25)-0.9 × (3)-0.15

T₂ = 0.18944 C

\(Percentage \ change \ in\ the\ tool \ life =\frac{T_1-T_2}{T_1}\times 100\)

\(Percentage \ change \ in\ the\ tool \ life =\frac{1.07166 \ C\ -\ 0.18944\ C}{1.07166\ C}\times 100\)

⇒ 82.32 %