Question

During the turning operation, the cutting speed = 2.95 m/s, the feed f = 0.40 mm/rev, the depth of cut, d = 1.83 mm, the rake angle α = 5° and the deformed chip thickness= 0.58 mm. The workpiece material has volumetric specific heat = 2.96 x10-3 J/mm3- °C, thermal diffusivity = 0.783 cm2/s and specific energy = 1.036 J/mm3, obtained at feed equal to 0.25 mm/rev. The correction factor for the specific energy is CF = 0.75 to-0.21, when (to) is in mm. Calculate the cutting temperature in °C (ambient temperature = 25°C).

Answer 1

0-333 T = DAU [vto Jo Where, T= cutting temperature, oc U= specific energy in operation, N-m/mm3 V = cutting speed, mis to a chip thickness before the cut, (m) Pc = volumetric specific heat of work Material, J/mmc K = Thermal diffusivity of The Work Material , m7/A. given v= 2.95m la .. to = 0.21 mm = 2.1x10 m K = 0.783 (mil. = 0.783 X10 m/ Pc = 2096 X10 J/mm? oc. U = 1.036 J/mm3 Te 0-4X 1-036 x 52.95 * 2-1820 0.4x 1.036 T= | 2.958 2.187670.333 2.95 x 2.1846 2.968103 0.783 X10 7 = 278.6°C] = 27