Respuesta :
Answer:
The value of the cutting speed = 84.85 m/min
a.)The value of the spindle speed = 270.08 rpm
b.)The value of the Power for the unit force = 1.414 W
c.)[tex]u_{max}[/tex] = 0.35, [tex]u_{min}[/tex] = 0.25
d.)The rate of maximum force to minimum force = 1.4
Explanation:
As given, T₁ = 0.8 min
T₂ = 0.2 min
V₁ = 60 m/min
V₂ = 120 m/min
Now, as we know that Taylor's tool life equation -
VTⁿ = k
where V = speed in m/min
T = time in min
n = Index
k = constant
Now,
V₁T₁ⁿ = V₂T₂ⁿ
⇒60 (0.8)ⁿ = 120(0.2)ⁿ
⇒ (0.8)ⁿ = 2(0.2)ⁿ
⇒ ([tex]\frac{0.8}{0.2}[/tex] )ⁿ = 2
⇒4ⁿ = 2
Taking log both side , we get
⇒n log(4) = log(2)
⇒n = [tex]\frac{log(2)}{log(4)}[/tex] = 0.5
Now,
As given , T = 0.4 min
We know,
VTⁿ = V₁T₁ⁿ
⇒V (0.4)ⁿ = 60(0.8)ⁿ
⇒V = 60([tex]\frac{0.8}{0.4}[/tex] )ⁿ
⇒V = 60(2)ⁿ
⇒V = 60([tex]2^{0.5}[/tex])
⇒V = 84.85 m/min
∴ The value of the cutting speed = 84.85 m/min
a.)
As we know ,
V = π×d×N
Let d = 100 mm = 0.1 m
⇒ 84.85 = π×0.1×N
⇒ N = [tex]\frac{84.85}{0.1\pi } = 270.08[/tex] rpm
∴ The value of the spindle speed = 270.08 rpm
b.)
As we know,
P = F×V
⇒P = 1×[tex]\frac{84.85}{60} = 1.414[/tex] W per unit force
∴ The value of the Power for the unit force = 1.414 W
c.)
As given - The friction between tool and workpiece varies 0.25-0.35 during the process.
⇒[tex]u_{max}[/tex] = 0.35
[tex]u_{min}[/tex] = 0.25
d.)
Rate of maximum force , [tex]F_{max}[/tex] = [tex]u_{max}[/tex] × Normal force
Rate of minimum force , [tex]F_{min}[/tex] = [tex]u_{min}[/tex] × Normal force
∴ we get
[tex]\frac{F_{max} }{F_{min} } = \frac{u_{max} }{u_{min} } = \frac{0.35}{0.25}[/tex] = 1.4
∴ we get
The rate of maximum force to minimum force = 1.4
