In a turning operation the feed rate t is .13 mm (.005 in.) and the depth of cut

Question

In a turning operation the feed rate t is .13 mm (.005 in.) and the depth of cut normal to the plane of the paper is 2.5mm (.100 in). The cutting speed is 245 m/min (800 fpm). The cutting force F_c is found to be 1780N (400lb) and the normal force F_t is 890 N(200lb). The rake angle of the tool is +8 degrees.

find:

a) the power required for the cut in KW (hp)

b) the rate of metal removal in cm^3/min (in^3/min)

c) the unit power in kW/cm^3/min (hp/in^3/min)

answers are: a) 7.22 kW, b) 78.65 cm^3/min, c) .0917

**Please answer all questions with correct answers for five stars. Please show the process for each step.**

Explanation / Answer

a)Power required for the cut can be given by the formula = F_c * v

where v is in m/sec 245 m/min = 245/60 m/sec

Power = 1780* 245/60 =7268 W = 7.2 KW

b) Rate of metal removal = D* f * v = D ( in cm ) * 0.13 * 10^ -1 * 245 * 10 ^ 3 = 78.65

c) Power required/ rate of removal = 7.2/ 78.65 = 0.91 kw-min/ cm^ 3

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