Question

The power required for turning a mild steel rod is found to be 0.1 kW/cm³/min. The maximum power available at the machine spindle is 4 kW. Assuming a cutting speed of 38 m/min and feed rate of 0.32 mm/rev, the maximum metal removal rate and depth of cut are respectively:

• A. 48 cm³/min and 3.29 mm
• B. 40 cm³/min and 3.29 mm
• C. 40 cm³/min and 4.29 mm
• D. 48 cm³/min and 4.29 mm

Hint:

When solving for the maximum depth of cut, ensure that all units are consistent and use the correct formula for metal removal rate.

Answer 1

The Correct Option is A

Step 1: Metal Removal Rate Formula.
The metal removal rate \( MRR \) is calculated as follows: \[ MRR = \text{Cutting Speed} \times \text{Feed Rate} \times \text{Depth of Cut} \] Where: - Cutting speed \( V_c = 38 \, \text{m/min} \), - Feed rate \( f = 0.32 \, \text{mm/rev} \). The power \( P \) needed for turning is: \[ P = \frac{\text{Power per unit volume} \times MRR}{1000} \] Step 2: Power and MRR Relationship. Given a power per unit volume of 0.1 kW/cm³/min, and a maximum machine spindle power of 4 kW, we determine the maximum \( MRR \) to find the depth of cut. \[ 4 = \frac{0.1 \times MRR}{1000} \] \[ MRR = 48 \, \text{cm}^3/\text{min} \] Step 3: Depth of Cut Calculation. Substitute the \( MRR \), cutting speed, and feed rate values into the \( MRR \) formula: \[ 48 = 38 \times 0.32 \times \text{Depth of Cut} \] \[ \text{Depth of Cut} = 3.29 \, \text{mm} \] Final Answer: \[ \boxed{48 \, \text{cm}^3/\text{min} \, \text{and} \, 3.29 \, \text{mm}} \]