614 ROTATING MACHINES(a) The field current is doubled, with the ar-
mature terminal voltage and the load torque
remaining the same.
(b) The armature terminal voltage is halved,
with the field current and load torque remain-
ing the same.
(c) The field current and the armature terminal
voltage are halved, with the horsepower out-
put remaining the same.
(d) The armature terminal voltage is halved,
with the field current and horsepower output
remaining the same.
(e) The armature terminal voltage is halved and
the load torque varies as the square of the
speed, with the field current remaining the
same.
13.2.1A balanced three-phase, 60-Hz voltage is applied
to a three-phase, two-pole induction motor. Cor-
responding to a per-unit slip of 0.05, determine
the following:
(a) The speed of the rotating-stator magnetic
field relative to the stator winding.
(b) The speed of the rotor field relative to the
rotor winding.
(c) The speed of the rotor field relative to the
stator winding.
(d) The speed of the rotor field relative to the
stator field.
(e) The frequency of the rotor currents.
(f) Neglecting stator resistance, leakage reac-
tance, and all losses, if the stator-to-rotor
turns ratio is 2:1 and the applied voltage is
100 V, find the rotor-induced emf at standstill
and at 0.05 slip.
13.2.2No-load and blocked-rotor tests are conducted
on a three-phase, wye-connected induction motor
with the following results. The line-to-line volt-
age, line current, and total input power for the
no-load test are 220 V, 20 A, and 1000 W; and
for the blocked-rotor test they are 30 V, 50 A, and
1500 W. The stator resistance, as measured on a
dc test, is 0.1per phase.
(a) Determine the parameters of the equivalent
circuit shown in Figure 13.2.5(c).
(b) Compute the no-load rotational losses.*13.2.3A three-phase, 5-hp, 220-V, six-pole, 60-Hz in-
duction motor runs at a slip of 0.025 at full load.
Rotational and stray-load losses at full load are
5% of the output power. Calculate the power
transferred across the air gap, the rotor copper
loss at full load, and the electromagnetic torque
at full load in newton-meters.
13.2.4The power transferred across the air gap of a two-
pole induction motor is 24 kW. If the electromag-
netic power developed is 22 kW, find the slip.
Calculate the output torque if the rotational loss
at this slip is 400 W.
13.2.5The stator and rotor of a three-phase, 440-V,
15-hp, 60-Hz, eight-pole, wound-rotor induction
motor are both connected in wye and have the
following parameters per phase:R 1 = 0. 5 ,
R 2 = 0. 1 ,Xl 1 = 1. 25 , andXl 2 = 0. 2 .
The magnetizing impedance is 40and the core-
loss impedance is 360, both referred to the sta-
tor. The ratio of effective stator turns to effective
rotor turns is 2.5. The friction and windage losses
total 200 W, and the stray-load loss is estimated
as 100 W. Using the equivalent circuit of Figure
13.2.5(a), calculate the following values for a
slip of 0.05 when the motor is operated at rated
voltage and frequency applied to the stator, with
the rotor slip rings short-circuited: stator input
current, power factor at the stator terminals, cur-
rent in the rotor winding, output power, output
torque, and efficiency.
13.2.6Considering only the rotor equivalent circuit
shown in Figure 13.2.2 or 13.2.3, find:
(a) TheR 2 for which the developed torque would
be a maximum.
(b) The slip corresponding to the maximum
torque.
(c) The maximum torque.
(d)R 2 for the maximum starting torque.
*13.2.7A three-phase induction motor, operating at its
rated voltage and frequency, develops a starting
torque of 1.6 times the full-load torque and a
maximum torque of 2 times the full-load torque.
Neglecting stator resistance and rotational losses,
and assuming constant rotor resistance, deter-
mine the slip at maximum torque and the slip
at full load.
13.2.8A three-phase, wye-connected, 400-V, four-pole,
60-Hz induction motor has primary leakage
impedance of 1+j 2 and secondary leakage
impedance referred to the primary at standstill of
1 +j 2 . The magnetizing impedance isj 40