Answer:
the cutting speed that must be used to meet this machining time requirement is 125.66 m/min
the material removal rate for this operation is [tex]\mathbf{R_{MR} = 2513.2 \ mm^3/sec}[/tex]
Explanation:
Given that:
Time = 5mins
Length of the cylindrical workpiece = 400 mm = 0.4 m
Diameter of the cylindrical workpiece = 150 mm = 0.15 m
Feed value = 0.30 mm/rev = [tex]3 \times 10^{-4 }[/tex] m/rev
Depth of cut = 4.0
what cutting speed must be used to meet this machining time requirement?
The cutting speed can be estimated by using the formula:
[tex]T_m = \dfrac{\pi D_o L}{vF}[/tex]
where;
[tex]T_m[/tex] = time
[tex]D_o[/tex] Diameter
L = length
v = cutting speed
F = Feed value
Making v the subject; we have:
[tex]v = \dfrac{\pi D_o L}{T_mF}[/tex]
[tex]v = \dfrac{\pi \times 0.15 \times 0.4}{5 \times 3.0 \times 10^{-4}}[/tex]
v = 125.66 m/min
Hence; the cutting speed that must be used to meet this machining time requirement is 125.66 m/min
Also calculate the material removal rate for this operation in cubic mm/seconds.
The material removal rate [tex]R_{MR}[/tex] for this operation is :
[tex]R_{MR} = vFd[/tex]
where ;
v = 125.66 m/min
to mm/sec : we have.
= ((125.66 × 1000 )/60 ) mm/sec
= (125660/60) mm/sec
= 2094.33 mm/sec
F = 0.30 mm/rev
d = depth of the cut = 4.0 mm
[tex]R_{MR} = 2094.33 \times 0.3 \times 4[/tex]
[tex]\mathbf{R_{MR} = 2513.2 \ mm^3/sec}[/tex]
