Power, MRR, cost in a turning operation. Refer to Fig. P8.ll. Mean diameter d_m

Question

Power, MRR, cost in a turning operation. Refer to Fig. P8.ll. Mean diameter d_m = 75 mm, length L = 200 mm, depth of cut a = 5 mm, feed per revolution f_r = 0.25 mm. depth of cut a = 5 mm, feed per revolution f_r = 0.25 mm. Specific force K_3 = 2000 N/mm62, cutting speed v = 150 m/min. tool life equation v^3 f_v^2 T = 8 times 10^+ [v (m/min), f_r (mm), T(mim)][. C_tool edge = $4.0, tool change = 8 min, machine rate r = $0.4/mm. Determine the power consumed (kW), machining time t_m, cost per part C/p.

Explanation / Answer

Given: Mean dia = 75 mm , depth of cut , d= 5mm . Hence Dia of the rod , D= 75 +5=80 mm

           L=200 mm , feed f= 0.25mm/rev , cutting speed ,v= 150m/min, specific force,Ks= 2000N/mm^2

Power consumed

Ks= Fc ( cutting force)/ fd

   Therefore, Fc= Ks X fd = 2000 x 5X .25 = 2500 N

     Power consumed during cutting = Fc X V (cutting speed m/s)

                                                        = 2500 X 150/60 N-m/s = 6250 J/s= 6.25 KW

Machining time

Machining time= T= Length of cut / (feed x rpm) = L / (f X N)   min ; where f= feed in mm/rev   ................1

Cutting speed = V= (pi) DN/ 1000 m/min     ;      here D= starting dia of work or can be taken as mean dia in mm......2

From 1 &2,

T= (pi)DL/ 1000 fV min

= (3.14 * 75 *200) / (1000 *0.25*150) min = 1.256 min (Ans.)

         

         

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