Question

During turning a metallic rod at a given condition, the tool life was found to increase from 25 min to 50 min, when cutting velocity V_c alone was reduced from 141.4 m/min to 100 m/min. How much will be the life of that tool if machined at 353.5 m/min?

1. 20 min
2. 15 min
3. 10 min
4. 4 min

Answer 1

Correct Answer - Option 4 : 4 min

Concept:

Tool life for a tool is given by,

VTⁿ = C

V₁T₁ⁿ = V2T2n

where, V = Cutting velocity of tool, m/min

T = Tool life of a tool, min

n = Taylor’s tool life exponent

C = constant

 Values of both ‘n’ and ‘C’ depend mainly upon the tool-work materials and the cutting environment (cutting fluid application).

Calculation:

Given:

V1 = 141.4 m/min, V₂ = 100 m/min, V₃ = 353.5 m/min

T₁ =  25 min, T₂ = 50 min, T₃ = ?

V1T1n = V2T2n

141.4 × 25ⁿ = 100 × 50n

1.414 = 2ⁿ

ln 1.414 = n × ln 2

n = 0.5

V2T2n = V3T3n

100 × 50^0.5 = 353.5 × T₃0.5

\(\frac{100}{353.5} = \left (\frac{T_{3}}{50} \right )^{0.5}\)

\(\left (\frac{100}{353.5} \right )^{2} = \frac{T_{3}}{50} \)

T₃ = 4 min