Question

In a production turning operation, an engineer wants a single pass must be completed on the cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement? Also calculate the material removal rate for this operation in cubic mm/seconds.

Answer 1

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
`\mathbf{R_(MR) = 2513.2 \ mm^3/sec}`

Step-by-step 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 =
`3 * 10^(-4 )` 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:

`T_m = (\pi D_o L)/(vF)`

where;

`T_m` = time

`D_o` Diameter

L = length

v = cutting speed

F = Feed value

Making v the subject; we have:

`v = (\pi D_o L)/(T_mF)`

`v = (\pi * 0.15 * 0.4)/(5 * 3.0 * 10^(-4))`

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
`R_(MR)` for this operation is :

`R_(MR) = vFd`

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

`R_(MR) = 2094.33 * 0.3 * 4`

`\mathbf{R_(MR) = 2513.2 \ mm^3/sec}`