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:
Show Hint
When solving for the maximum depth of cut, ensure that all units are consistent and use the correct formula for metal removal rate.
Step 1: Formula for metal removal rate.
The metal removal rate \( MRR \) is given by:
\[
MRR = \text{Cutting Speed} \times \text{Feed Rate} \times \text{Depth of Cut}
\]
Where:
- Cutting speed \( V_c = 38 \, \text{m/min} \),
- Feed rate \( f = 0.32 \, \text{mm/rev} \).
The power required for turning is given by:
\[
P = \frac{\text{Power per unit volume} \times MRR}{1000}
\]
Step 2: Relate power and MRR.
Given that the power per unit volume is 0.1 kW/cm³/min, and the maximum power available at the machine spindle is 4 kW, we can calculate the maximum \( MRR \) and then solve for the depth of cut.
\[
4 = \frac{0.1 \times MRR}{1000}
\]
\[
MRR = 48 \, \text{cm}^3/\text{min}
\]
Step 3: Calculate the depth of cut.
Now, substitute the values for \( MRR \), cutting speed, and feed rate into the metal removal rate formula:
\[
48 = 38 \times 0.32 \times \text{Depth of Cut}
\]
\[
\text{Depth of Cut} = 3.29 \, \text{mm}
\]
Final Answer:
\[
\boxed{48 \, \text{cm}^3/\text{min} \, \text{and} \, 3.29 \, \text{mm}}
\]
