A hole of 30 mm diameter and 45 mm depth is to be drilled in mild steel component- Lip angle of twist drill is 118- - The cutting speed can be taken of 55 m-min and feed rate as 0-25 mm-rev- The machining time will be s-

22.21 Drilling time =L/fN L=L_1+ Break through distance Break through distance, A=D/2 tan ( α/2 )=30/2tan59^∘=9.012 mm L = 45 + 9.012 = 54.012 mm Spindle s ...

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Standard VII

Physics

Bernoulli's Principle

A hole of 30 ...

Question

A hole of 30 mm diameter and 45 mm depth is to be drilled in mild steel component. Lip angle of twist drill is $118^{\circ} .$ The cutting speed can be taken of 55 m/min and feed rate as 0.25 mm/rev. The machining time will be _______s.

22.21

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Solution

The correct option is A 22.21
22.21

Drilling time $= \frac{L}{f N}$

$L = L_{1} +$ Break through distance

Break through distance, $A = \frac{D}{2 t a n (\frac{α}{2})} = \frac{30}{2 t a n 59^{\circ}} = 9.012 m m$

L = 45 + 9.012 = 54.012 mm

Spindle speed, $N = \frac{1000 \times 55}{π \times 30} = 583.568 r e v / m i n$

Drilling time $= \frac{54.012}{0.25 \times 583.568} = 0.3702 m i n$

= 22.21 sec

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A hole of 30 mm diameter and 45 mm depth is to be drilled in mild steel component. Lip angle of twist drill is 118∘. The cutting speed can be taken of 55 m/min and feed rate as 0.25 mm/rev. The machining time will be _______s. 22.21

The correct option is A 22.2122.21 Drilling time =LfN L=L1+ Break through distance Break through distance, A=D2tan(α2)=302tan59∘=9.012mm L = 45 + 9.012 = 54.012 mm Spindle speed, N=1000×55π×30=583.568rev/min Drilling time =54.0120.25×583.568=0.3702min = 22.21 sec

A hole of 30 mm diameter and 45 mm depth is to be drilled in mild steel component. Lip angle of twist drill is $118^{\circ} .$ The cutting speed can be taken of 55 m/min and feed rate as 0.25 mm/rev. The machining time will be _______s.

22.21