The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is 200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min) is

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Solution

$t = \frac{L}{f N} a n d V = π D N$

$∴ t = \frac{L π D}{f V} = \frac{900 \times π \times 200}{0.25 \times 300 \times 1000}$

$= \frac{9 π \times 2}{30 \times 0.25} min = \frac{6 π}{2.5} min$

$= 7.54 min$

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The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is 200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min) is

t=LfNandV=πDN ∴t=LπDfV=900×π×2000.25×300×1000 =9π×230×0.25min=6π2.5min =7.54min

$t = \frac{L}{f N} a n d V = π D N$

$∴ t = \frac{L π D}{f V} = \frac{900 \times π \times 200}{0.25 \times 300 \times 1000}$

$= \frac{9 π \times 2}{30 \times 0.25} min = \frac{6 π}{2.5} min$

$= 7.54 min$