Page 110 - Electric Drives and Electromechanical Systems
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Chapter 3   Power transmission and sizing  103


                     The table is driven through a 20:1 gear box, which has a moment of inertia of 3 10  4
                         2
                     kg m referred to its input shaft. The gearbox friction is considered to be negligible.
                     The table is required to index 120.0 degrees in 0.75 s, and then dwell for a further 2.25 s.
                     A polynomial speed profile is required.
                     The selection process starts with the determination of the peak load speed and accelera-
                   tion, using equation Eq. (2.27), the maximum speed occurs at t ¼ 0.375 s and maximum
                   acceleration occurs at t ¼ 0s.

                                           _       6qt  6qt 2         1
                                          qð0:375Þ¼        ¼ 4:19 rad s
                                                   t m 2  t m
                                                         3
                                               € qð0Þ¼  6q  ¼ 22:34 rad s  2
                                                    t 2
                                                     m
                   Referring all variables to the motor side of the table, the required motor torque can be
                   estimated by using the peak acceleration, and the load and friction torques, giving

                                                     1:5
                                                  4
                                    T peak ¼  3   10  þ    22:34   20  ¼ 1:81 Nm
                                                     20 2
                                                         1
                   The peak motor speed required is 800 rev min . As a first approximation we require a motor
                   that is capable of providing a peak torque in excess of 1.8 Nm, a possible motor and drive have
                   the following specifications:
                     Peak speed: 5000 rev min  1   10% at a terminal voltage of 95 V
                     Continuous stall torque: 0.5 Nm. Peak torque: 2.0 Nm
                     Voltage constant: 0.18 V s rad  1  Current constant: 0.18 Nm A  1
                     Moment of inertia: 1.7   10  4  kg m 2
                     Drive continuous current: 5 A. Drive peak torque: 10 A
                     Drive voltage: 95 V

                     If the motor’s moment of inertia is now included in the calculation, the peak torque
                   requirement is 1.87 Nm which is well within the capabilities of the motor and drive. The
                   required peak current is 10.4 A. The r.m.s. torque can now be calculated with reference to Fig
                   2.11:
                                                    s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                                                      t movement
                                           T rms ¼ T peak    ¼ 0:54 Nm
                                                      3   t cycle
                   While the T rms is acceptable, the peak current is just over the drive’s maximum, thus the
                   system’s specification cannot be achieved. Thus, a case can be made to change the motor,
                   drive or gearbox, this decision would be made after careful consideration of the application.
                     If the gear ratio was changed to 40:1, the motor’s peak speed requirement increases to 1600
                   rev min  1  and the r.m.s. torque drops to 0.36 Nm, and a peak torque of 1.26 Nm. The peak
                   torque requires the drive to provide 7A, which is well within its specification. These figures are
                   well within the specification of motor and drive.
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