Machining Time of Facing Operation in a CNC Turning Centre

Estimation of Machining Time in Facing in a CNC Turning Centre using constant spindle speed and constant cutting velocity

Estimation of Machining Time in Facing in a CNC Turning Centre using constant spindle speed and constant cutting velocity

Facing is the machining process of removing metal from the end of a workpiece to produce a flat surface. It is done to create a smooth, flat face perpendicular to the axis of a cylindrical part.

Machining time is the time needed to process or machine a material under some operation(s). The machining time needs to be determined or assessed from time to time for several purposes, such as, assessment of productivity, evaluation of machining cost per piece, determination of man-machine hour cost per piece, assessment of relative performance or capability of any machining methods, machine tool, cutting tool, or any specific techniques in terms of saving in machining time.

The machining time required for a particular operation can be determined either by approximately calculation (estimation) or by actual measurement. Measurement generally provides accurate and precise information but it is tedious and expensive. On the other hand, though estimation by simple calculation may not be accurate, is much more simple, quick and inexpensive. Therefore, machining time is regularly evaluated for different purposes by simple calculation.

Facing of a disc in a CNC turning centre can be done in two cases. One with constant spindle speed and another with constant cutting velocity.

Consider a cylindrical shell with outer diameter d₂ and inner diameter d₁ and other parameters as shown in figure.

(a) Facing with constant Spindle Speed (n):

If the spindle speed (n) is constant, then the machining time (t_m) can be estimated neglecting the approach and over-travel as:

 Where d₂ (mm) is outer diameter of the cylinder workpiece, d₁ (mm) is the inner diameter of the workpiece, s is feed in (mm/rev) and n is spindle speed in (rpm).

(b) Facing with constant Cutting Velocity:

If the spindle speed is automatically compensated by the CNC system and the cutting velocity is kept constant, then the machining time (t’_m) can be estimated as:

Also, we can estimate this machining time using another method as shown below:

Let for a infinitesimally small change of time, dt, the change in length be –dR =  –dD/2

Here, t_m is machining time in minutes, d₂ (mm) is outer diameter of the cylinder workpiece, d₁ (mm) is the inner diameter of the workpiece, s is feed in (mm/rev) and v_c is cutting velocity in mm/min.

Flywheel Tutorial Problem & Solution (Dynamics of Machines, ME22004)

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The resisting torque on the crank of a riveting machine is shown below. The duration of cycle is 2 sec. The motor driving the machine has a speed of 1500 rpm and it delivers constant torque. The crank-shaft of the machine is geared to the motor shaft. Neglecting frictional losses, determine: 

a) the power of the motor 

b) the moment of inertia of the flywheel mounted on the motor shaft to keep the speed fluctuation within 2% of the average speed. 

c) the maximum angular acceleration of the flywheel Solution: In this problem, torque supplied is constant and demand torque is fluctuating. 

The demand torq  ue is shown in figure. a) The energy required per cycle 3 1 200 1600 (1600 200) 200 2 4 2 4 2 4 2904.5 Nm The duration of cycle is 2 seconds 2904.5 Energy required per second 1452.25 2 Power of motor required, p 1.452 kW 1452.25 W ) 2904.5 The average torque 462.5 Nm 2 2 462.5 Nm The shaded portion is fluctuation of energy. Therefore, 3 (1600 462.5) (1600 462.5) (1600 462.5) 4 4 2 2 (1600 200) 1618.45 Nm 2 2 1500 157.08 rad/s 60 60 2 ( 2) Coefficient of speed fluctuation, 0.04 100 But we have, 1256.33 0.04 (157.08) 1.64 kgm Moment of inertia of the flywheel is 1.273 kgm c) Maximum s s N k E Ik I I max acceleration shall occur where torque is maximum and minimum acceleration will occur where torque is minimum Maximum excess torque ( ) 1600 462.5 1137.5 Nm Let angular acceleration be ' ' excess 2 max 2 max 2 1137.5 893.6 rad/s 1.273 893.6 rad/s Maximum angular acceleration of the flywheel is 893.6 rad/s excess excess 

The resisting torque on the crank of a riveting machine is 200 Nm for first 90o, from 90o to 135o is 1600 Nm then it drops linearly to 200 Nm upto 180o and remains the same upto 360o. The duration of cycle is 2 sec. The motor driving the machine, however, has a speed of 1450 rpm and it delivers constant torque. The crank shaft of the machine is geared to the motor shaft. The speed fluctuation is limited to  2% of mean speed. Determine : 133 (a) power of the motor, and Flywheel (b) moment of inertia of the flywheel mounted on the motor shaft. Solution In this problem, torque supplied is constant and demand torque is fluctuating. The demand torque is shown in Figure 4.14. (a) The energy required per cycle 3 1 200 1600 (1600 200)  3 200 1600 1400 200 2 4 8 4 700 300 400 50 (750 175) 925 4= 2904.5 Nm The duration of the cycle is 2 seconds  Energy required per second 2904.5 1452.25 2  Power of motor required, p = 1.452 kW. Figure 4.14 : Figure for Example 4.5 (b) The average torque 2904.5 2 2 av E M or, Mav = 462.5 Nm The shaded portion is fluctuation of energy. Therefore, (1600 462.5) 3 (1600 462.5) 4 (1600 462.5) 4 2 2 (1600 200) E 2 1137.5 1137.5 4 1400 4  (1137.5 924.22) 1618.45 Nm 4 1600 462.5 200 Mav M 900 1350 134 Theory of Machines  2 2 1450 151.76 r/s 60 60  Coefficient of speed fluctuation 2 2 0.04 100  2 1618.45 151.76) or, 1618.45 2 1.76 kg m 0.04 23032.1

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