In a laboratory test on the shaper, feed, depth of cut, and length of stroke are 2 mm/stroke, 4 mm, and 300 mm respectively. If the specific power consumption is 0.05 kW/cm³ per minute and the number of working strokes per minute is 20, the Power consumption is

Concept:

Velocity is given by:

(V) = f × N

where, f = feed (mm/stroke), N = number of cutting strokes per minute

Area of the workpiece (A) = d × L

where, d = depth of cut, L = length of the workpiece

Material removal rate (MRR) is given by:

\(Q = \frac{{f \times N \times d \times L}}{{60}}\)

where Q is in \(\frac{{{\rm{m}}{{\rm{m}}^3}}}{{\rm{s}}}\)

Average power consumption is given by:

Pavg = Q × U

where U is specific cutting energy.

Calculation:

Given:

f = 2 mm/stroke, N = 20, d = 4 mm, L = 300 mm and U = \(0.05\frac{KW}{{c{m^3}}} per \min=3\times10^{-3}\frac{KJ}{{m{m^3}}}\) 

Material removal rate (Q)

\(Q = \frac{{f \times N \times d \times L}}{{60}} = \frac{{2 \times 20 \times 4 \times 300}}{{60}} = 800\frac{{m{m^3}}}{s}\)

Average power (Pavg)

Pavg = Q × U =\(800\times 3\times10^{-3}\)  = 2.4 KW