A cast iron block of 200 mm length is being shaped in a shaping machine with a depth of cut of 4 mm, feed of 0.25 mm/stroke and the tool principal cutting edge angle of 30°. Number of cutting strokes per minute is 60. Using specific energy for cutting as $1.49 J / m m^{3},$ the average power consumption (in watt) is
298

Open in App

Solution

The correct option is A 298
The volume rate of material removal is:

Q = fdLN
where f(feed) = 0.25 mm/stroke
d (depth of cut) = 4 mm
L (length of block) = 200 mm

N(number of cutting stroke per minute) = 60

$Q = \frac{0.25 \times 4 \times 200 \times 60}{60} (\frac{m m^{3}}{s})$
$= 200 m m^{3} / s$
$U_{c} = 1.49 J / m m^{3}$

$P_{a v e r a g e} = U_{c} \times Q = 1.49 \times 200 J / s = 298 W$

Suggest Corrections

Similar questions

Q. In a shaping process, the number of double strokes per minute is

30

and the quick return ratio is

0.6

. If the length of the stroke is

250 m m

the average cutting velocity in

m m / m i n

Q. A

60 d e g

symmetrical

V

tool is used in shaping at workpiece with a depth of cut of

0.1 m m

and feed of

0.1 m m / s t r o k e .

The theoretical peak to valley height, in

m m

, of the surface produced is

Q. Turning is done using a tool of 5-5-8-8-8-25-0.1 mm, geometry, with a feed of 0.1 mm/stroke and depth of cut of 2 mm.

If shear angle is

30^{\circ}

, the plan area of cut in

m m^{2}

Q. Turning is done using a tool of 5 - 5 - 8 - 8 - 8 - 25 - 0.1 mm, geometry, with a feed of 0.1 mm/stroke and depth of cut of 2 mm.

If shear angle is

30^{\circ}

, the plan area of cut in

m m^{2}

Common Data:
A disc of $200 m m$ outer and $80 m m$ inner diameter is faced at a feed of $0.1 m m / r e v$ with a depth of cut of $1 m m$ . The facing operation is undertaken at a constant cutting speed of $90 m / m i n$ in a CNC lathe The main (tangential) cutting force is $200 N$

Neglecting the contribution of the feed force towards cutting power, the specific cutting energy in $J / m m^{3}$ is

Join BYJU'S Learning Program

The correct option is A 298The volume rate of material removal is: Q = fdLN where f(feed) = 0.25 mm/stroke d (depth of cut) = 4 mm L (length of block) = 200 mm N(number of cutting stroke per minute) = 60 Q=0.25×4×200×6060(mm3s) =200 mm3/s Uc=1.49 J/mm3 Paverage=Uc×Q=1.49×200 J/s=298 W