0195136047.pdf

13.2 INDUCTION MACHINES 575

For squirrel-cage-rotor machines, the problem is to keep down the starting current while
maintaining adequate starting torque. The input current, for example, can be no more than 6 times
the full-load current, while the starting torque may be about 1.5 times the full-load torque. Depend-
ing on the capacity of the available supply system,direct-on-line startingmay be suitable only
for relatively small machines, up to 10-hp rating. Other starting methods includereduced-voltage
startingby means ofwye–delta starting, autotransformer starting,orstator-impedance starting.
For employing the wye–delta starting method, a machine designed for delta operation is
connected in wye during the starting period. Because the impedance between line terminals for
wye connection is three times that for delta connection for the same line voltage, the line current
at standstill for wye connection is reduced to one-third of the value for delta connection. Since the
phase voltage is reduced by a factor of

√
3 during starting, it follows that the starting torque will
be one-third of normal. For autotransformer starting, the setting of the autotransformer can be
predetermined to limit the starting current to any desired value. An autotransformer, which reduces
the voltage applied to the motor toxtimes the normal voltage, will reduce the starting current in
the supply system as well as the starting torque of the motor tox^2 times the normal values.
Stator-impedance starting may be employed if the starting-torque requirement is not severe.
Series resistances (or impedances) are inserted in the lines to limit the starting current. These
resistances are short-circuited out when the motor gains speed. This method has the obvious
disadvantage of inefficiency caused by the extra losses in the external resistances.

EXAMPLE 13.2.3

An induction motor has a starting current that is 6 times the full-load current and a per-unit full-
load slip of 0.04. The machine is to be provided with an autotransformer starter. If the minimum
starting torque must be 0.3 times the full-load torque, determine the required tapping on the
transformer and the per-unit supply-source line current at starting.

Solution

From Equation (13.2.5), with slip at starting equal to 1, we have

Ts

Tfl

=

(

Is

Ifl

) 2

Sfl

where subscriptssandflcorrespond to starting and full-load conditions.
Substituting values, 0. 3 =(Is/Ifl)^2 × 0 .04, from which
Is/Ifl=

√

0. 03 / 0. 04 = 2. 74

orIs= 2 .74 per unit, withIfltaken as 1 per unit.
The applied voltage to the motor must reduce the current from 6 per unit to 2.74 per unit,
i.e., the tapping must be sufficient to reduce the voltage to 2.74/6=0.456 of the full value.
The supply-source line current will then be 2. 74 × 0. 456 = 1 .25 per unit, where 0.456 is the
secondary-to-primary turns ratio.

Single-Phase Induction Motors

For reasons of simplicity and cost, a single-phase power supply is universally preferred for
fractional-horsepower motors. It is also used widely for motors up to about 5 hp. Single-phase