Page 680 - Mechatronics with Experiments

666 MECHATRONICS
It should be quickly pointed out that these performance characteristics are only for the
motor when controlled from a line voltage supply directly without any active commutation
by a drive. The current-torque characteristics of an AC motor can be made to behave like
a DC motor using the so called “field oriented vector control” algorithm which is typically
implemented in the drive. The performance of a motor should always be evaluated together
with the drive it is used in. Depending on the drive type used, the performance characteristics
of the motor-drive combination can be quite different.

Example Consider an AC induction motor driven by a line supply frequency of w =
e
60 Hz. Assume that it is a two-pole motor, P = 2. The motor design is such that at the
maximum load, the slip is 20% of the synchronous speed, s = 20%. Determine the speed
of the motor for the following conditions: (a) no-load speed, (b) speed at maximum load,
(c) speed at a load that is 50% of the maximum load. Determine the speed variation
(sensitivity) due to the variation of load as percentage of its maximum value. Let maximum
load torque be 100 lb ⋅ in.
Referring to Figure 8.41, let us assume that the curve connecting the no-load speed
and maximum-load speed points is a linear line. In steady-state, the actual motor speed is
determined by the intersection of the torque-speed curve with the load torque. As long as
the load torque is less than the maximum load torque, the motor speed will be somewhere
between the no-load speed and the speed at maximum load. As the load varies up to the
maximum load torque, the steady-state speed of the motor variation follows the linear
torque-speed line.
The no-load speed of the motor is
w e 60 Hz
w rm = = = 60 rev∕s = 3600 rev∕min (8.250)
P∕2 2∕2
At the maximum load, the motor specifications indicate that it has 20% slip,
w syn − w rm
s = = 0.2 (8.251)
w
syn
w = w − 0.2 ⋅ w = 0.8 ⋅ w = 2880 rev∕min (8.252)
rm syn syn syn
When the load is 50% of maximum load, the slip will be 50% of the maximum slip.
Therefore, the steady-state rotor speed is
w syn − w rm
s = = 0.1 (8.253)
w syn
w rm = w syn − 0.1 ⋅ w syn = 0.9 ⋅ w syn = 3240 rev∕min (8.254)

The speed varies from synchronous speed at no-load to 20% slip at maximum load, hence
w − ((1 − s) ⋅ w )
ΔV syn syn
= (8.255)
ΔT 100
l
s ⋅ w syn
= (8.256)
100
= 7.2rpm∕(lb ⋅ in) (8.257)

8.5.2 Drives for AC Induction Motors
The drive controls the electrical variables, which are the voltage and current, in the stator
windings of an AC induction motor in order to obtain the desired behavior in mechanical
variables, which are torque, speed, and position. In particular, frequency and magnitude
of the voltage control are of interest. Major drive types are discussed below (Figure 8.43)