Question

A drilling operation is to be performed with a 12.7 mm diameter drill on cast iron. The hole depth is 60 mm and the drill point angle is 118∘. The cutting speed is 25 m/min and the feed is 0.30 mm/rev. Calculate:___________.
a) The cutting time (min) to complete the drilling operation
b) Material removal rate (mm3/min) during the operation, after the drill bit reaches full diameter.

Answer 1

Answer:

a. Tm = 0.3192min.

b. MRR = 396.91mm^{3}/s.

Explanation:

Given the following data;

Drill diameter, D = 12.7mm

Depth, L = 60mm

Cutting speed, V = 25m/min = 25,000m

Feed, F = 0.30mm/rev

To find the cutting time;

Cutting time, Tm =?

$Tm = \frac{L}{Fr}$ .......eqn 1

We would first solve for the feed rate (F);

$Fr = NF$ .......eqn 2

But we need to find the rotational speed (N);

$N= \frac{V}{\pi *D}$

$N= \frac{25000}{3.142*12.7}$

$N= \frac{25000}{39.90}$

N = 626.57rev/min.

Substiting N into eqn 2;

$Fr = NF$

Fr = 626.57 * 0.30

Fr = 187.97mm/min.

Substiting F into eqn 1;

$Tm = \frac{L}{Fr}$

$Tm = \frac{60}{187.97}$

Tm = 0.3192min.

Therefore, the cutting time is 0.3192 minutes.

For the material removal rate (MRR);

$MRR = \frac{\pi *D^{2}Fr}{4}$

$MRR = \frac{3.142*12.7^{2}*187.97}{4}$

$MRR = \frac{3.142*161.29*187.97}{4}$

$MRR = \frac{95258.16}{4}$

$MRR = 23814.54mm^{3}/min$

Time in seconds, we divide by 60;

MRR = 23814.54/60 =396.91mm^{3}/s.

Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.