Respuesta :
Answer:
Gross power perform this operation is 9.07 kW
Explanation:
Given :
Cutting speed [tex]v = 200 \times 10^{3}[/tex] [tex]\frac{mm}{min}[/tex]
Feed [tex]f = 0.25[/tex] mm
Depth of cut [tex]d = 3.5[/tex] mm
Mechanical efficiency [tex]\eta = 0.90[/tex]
First find the material removal rate,
[tex]R = v \times f \times d[/tex]
[tex]R = 200 \times 10^{3} \times 0.25 \times 3.5[/tex]
[tex]R = 175000[/tex] [tex]\frac{mm^{2} }{min}[/tex]
[tex]R = \frac{175000}{60} = 2916.67[/tex] [tex]\frac{mm^{2} }{sec}[/tex]
From the standard table,
Specific cutting energy [tex]U = 2.8 \frac{J}{mm^{2} }[/tex]
Cutting power,
[tex]P = 2916.67 \times 2.8[/tex]
[tex]P = 8166.67[/tex] W
We know that gross power is given by,
[tex]P_{gross} = \frac{P}{\eta}[/tex]
[tex]P_{gross} = \frac{8166.67}{0.90}[/tex]
[tex]P_{gross} = 9074.07[/tex] W
[tex]P_{gross} = 9.07[/tex] kW
Therefore, gross power perform this operation is 9.07 kW
The amount of power that the lathe will draw in performing the given operation at the given mechanical efficiency is; 9.074 kW
We are given;
- Cutting speed; v = 200 m/min = 200000 mm/min
- Feed; f = 0.25 mm/rev
- Depth of cut; d = 3.5 mm
- Mechanical efficiency; η = 90% = 0.9
- Formula for material removal rate is;
R = vfd
Thus;
R = 200000 × 0.25 × 3.5
R = 175,000 mm³/min
Converting to mm³/s gives;
R = 2916.67 mm³/s
- From the table attached and by interpolation, the specific cutting energy is; U = 2.8 J/mm³
Thus;
Gross power = RU/η
Gross power = 2916.67 × 2.8/0.9
Gross power = 9074.08 W
Gross power = 9.074kW
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