Transcript Induction-Motor Starting

Induction-Motor Starting
• Connect directly across full-voltage using
a contactor for each phase.
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Full-Voltage Starting
• Major disadvantage is the high in-rush current.
Large voltage drops occur, dimming lights and
forcing computers off-line.
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Reduced-Voltage Starting
• Autotransformer starting
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“S”, or “start” contactor – close to start the motor at a
reduced voltage – the autotransformers are set to 50, 60,
or 80 percent of full-voltage.
At rated speed, open the S contactors and close the “R”,
or run contactors – apply full voltage.
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After coming up to speed, “Running”
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Example 5.18
• A three-phase, 15-hp, 460-V, 150-A, 60Hz, six-pole, 1141 r/min, design B motor
with a code letter H is to be started at
reduced voltage using an autotransformer
with a 65-% tap. Determine
• a) locked-rotor torque and expected
average in-rush current to the stator if the
motor is started at rated voltage
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Example 5.18 continued
• b) repeat part (a) assuming the motor is
started at reduced voltage using an
autotransformer with a 65% tap
• c) the in-rush current when starting at
reduced voltage.
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Example 5.18 Solution
Tn
T (1141)
P
 125 
5252
5252
Trated  575.37lb  ft
Check Table 5.1 for the minimum lockedrotor torque for a 125-hp, design B, sixpole motor
Tlr ,460  125%(Trated )
Tlr ,460  1.25(575.37)  719.2lb  ft
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Example 5.18 Solution continued
• From Table 5.9, the average locked-rotor
kVA/hp for a code H motor is
6.3  7.1
 6.70kVA / hp
2
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Example 5.18 Solution continued
• The expected average in-rush current to
the stator is
kVA / hp 1000  hp  3Vline I line
6.7 1000 125  3  460  I line
I line,460  1051A
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Example 5.18 Solution continued
• With the 65% tap, the voltage across the
stator at locked-rotor is
V2  0.65  460  299V
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Example 5.18 Solution continued
The average in-rush current to the stator will
be proportional to the voltage applied to
the stator since the locked-rotor input
impedance is constant.
I  1051 0.65  683A
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Example 5.18 Solution continued
• The locked-rotor torque is proportional to
the square of the voltage.
2
 V2 
 0.65  460 
T2  T1    719.2  
  303.9lb  ft
 460 
 V1 
2
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Current Curves
Transition from low to rated
voltage is at slip = 0.125.
shaft speed = ns(1-s)
shaft speed = 1200(1-0.125)
shaft speed = 1050 r/min
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Locked-Rotor Torque Curves
Transition from low to rated
voltage is at slip = 0.125.
shaft speed = ns(1-s)
shaft speed = 1200(1-0.125)
shaft speed = 1050 r/min
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The in-rush line current
a
VHS ,line
I HS
I LS

 68.3  0.65  444 A
a
VLS ,line

VHS ,line
0.65 VHS ,line
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
0.65
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