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

a. Tm = 0.3192min.

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

Step-by-step 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 = (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= (V)/(\pi *D)`

`N= (25000)/(3.142*12.7)`

`N= (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 = (L)/(Fr)`

`Tm = (60)/(187.97)`

Tm = 0.3192min.

Therefore, the cutting time is 0.3192 minutes.

For the material removal rate (MRR);

`MRR = (\pi *D^(2)Fr)/(4)`

`MRR = (3.142*12.7^(2)*187.97)/(4)`

`MRR = (3.142*161.29*187.97)/(4)`

`MRR = (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.