Question

A steel workpiece is to be milled. Metal removal rate is 30 cm³/min. Depth of cut is 5 mm and width of cut is 100 mm. The rate of feed in mm/min is}

• A. 30 mm/min
• B. 60 mm/min
• C. 75 mm/min
• D. 150 mm/min

Hint:

In milling, the metal removal rate is \( \text{MRR} = \text{depth} \times \text{width} \times \text{feed rate} \). Convert units to mm for consistency when calculating.

Answer 1

The Correct Option is B

Step 1: Understand the milling operation and metal removal rate.
In milling, the metal removal rate (MRR) is the volume of material removed per unit time. It is given by: \[ \text{MRR} = \text{depth of cut} \times \text{width of cut} \times \text{feed rate}, \] where:
Depth of cut (\( d \)): The depth of material removed in one pass,
Width of cut (\( w \)): The width of the material being removed,
Feed rate (\( f \)): The speed at which the workpiece moves relative to the cutter (in mm/min).
Step 2: Identify the given values.
Metal removal rate: \( \text{MRR} = 30 \, \text{cm}^3/\text{min} = 30 \times 10^3 \, \text{mm}^3/\text{min} \) (since \( 1 \, \text{cm}^3 = 1000 \, \text{mm}^3 \)),
Depth of cut: \( d = 5 \, \text{mm} \),
Width of cut: \( w = 100 \, \text{mm} \).
We need to find the feed rate \( f \) in mm/min. Step 3: Set up the equation for MRR.
\[ \text{MRR} = d \times w \times f, \] \[ 30 \times 10^3 = 5 \times 100 \times f, \] \[ 30,000 = 500 \times f. \] Step 4: Solve for the feed rate.
\[ f = \frac{30,000}{500}, \] \[ f = 60 \, \text{mm/min}. \] Step 5: Evaluate the options.
(1) 30 mm/min: Incorrect, as the calculated feed rate is 60 mm/min. Incorrect.
(2) 60 mm/min: Matches the calculated feed rate. Correct.
(3) 75 mm/min: Incorrect, as the calculated value is lower. Incorrect.
(4) 150 mm/min: Incorrect, as the calculated value is much lower. Incorrect.
Step 6: Select the correct answer.
The rate of feed is 60 mm/min, matching option (2).