A 1 mm thick cylindrical tube, 100 mm in diameter, is orthogonally turned such that the entire wall thickness of the tube is cut in a single pass. The axial feed of the tool is 1 m/minute and the specific cutting energy (u) of the tube material is 6 J/mm3 . Neglect contribution of feed force towards power. The power required to carry out this operation is _________ kW (round off to one decimal place).

The correct answer is 31.4 (or 31.416).

To understand why this is the correct answer, let us break down the physical process and perform the calculation step-by-step.

1. Understanding the Given Parameters:
We are given the following values for the orthogonal turning of the cylindrical tube:
- Tube diameter (_d) = 100 mm
- Tube wall thickness (_t) = 1 mm
- Axial feed rate (_v) = 1 m/minute
- Specific cutting energy (_u) = 6 J/mm³

2. Conversion of Feed Rate to Standard Units:
To keep all units consistent, we convert the axial feed rate from meters per minute to millimeters per second:
$$v=1\text{ m/min}=\frac{1000\text{ mm}}{60\text{ s}}=\frac{1000}{60}\text{ mm/s}$$

3. Calculating the Cross-Sectional Area of the Cut:
Since the entire wall thickness of the cylindrical tube is cut in a single pass, the cross-sectional area of the tube being cut (_A) is given by:
$$A=\pi \cdot d\cdot t$$
Substituting the given values:
$$A\approx 3.14159\cdot 100\text{ mm}\cdot 1\text{ mm}\approx 314.16{\text{ mm}}^2$$

4. Calculating the Material Removal Rate (MRR):
The Material Removal Rate is the volume of material removed per unit time, which is the product of the cross-sectional area of the cut and the feed rate:
$$\text{MRR}=A\cdot v$$
Substituting the values:
$$\text{MRR}=314.16{\text{ mm}}^2\cdot \frac{1000}{60}\text{ mm/s}\approx 5235.99{\text{ mm}}^3\text{/s}$$

5. Calculating the Cutting Power:
The power (_P) required for machining is the product of the specific cutting energy (_u) and the Material Removal Rate (MRR):
$$P=u\cdot \text{MRR}$$
Substituting the values:
$$P=6{\text{ J/mm}}^3\cdot 5235.99{\text{ mm}}^3\text{/s}\approx 31415.9\text{ J/s (or Watts)}$$
To convert the power to kilowatts (kW):
$$P=\frac{31415.9}{1000}\approx 31.416\text{ kW}$$
Rounding to one decimal place gives 31.4 kW.