Electrical Machines LSEGG216A 9080V. Synchronous Motors Week 14.
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Transcript of Electrical Machines LSEGG216A 9080V. Synchronous Motors Week 14.
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Electrical Machines
LSEGG216A9080V
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Synchronous Motors
Week 14Week 14
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IntroductionIntroduction• State the principles of operation of a synchronous motor.• Identify the main parts of a synchronous
alternator/motor.• List the methods used to provide the excitation of a
synchronous alternator/motor.• List the starting methods of synchronous
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Motor TypesMotor Types3 Phase 1 Phase
• Reluctance
• Hysteresis
• Permanent Magnet
• Inductor
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CharacteristicsCharacteristics
• High operating efficiency• Smooth constant starting & accelerating torque• Versatile power factor control• Constant speed• Considerably more expensive than induction motors• Zero starting torque
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StatorStatorSame as an induction motor’s stator
Some books may call this the Armature
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RotorRotor• Wound simular to a wound rotor induction motor
• When operating DC voltage is placed across this coil to produce an electromagnet
Some books may call this the Field Windings
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Motor SpeedMotor Speed• Magnetic force is obtained from an external source
(In an induction machine rotor’s magnetism is induced from the stator)
• Rotor poles lock onto the RMF• Rotor operates at synchronous speed
P
f120N sync = Nrotor
Called “Excitation”
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Torque AngleTorque Angle
No Load
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Torque AngleTorque Angle
Full Load
Torque AngleDependant on:• Load torque• Excitation
• The magnetic link between the stator & the rotor can be thought of as a connecting spring.
• The excitation can be used to strengthen the spring
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If the load Torque fluctuatesOr
Changes rapidly• The magnetic “Spring” will bounce
• Causing large fluctuations in supply current
• Amortisseur windings are added to the rotor
• Also known as hunting
Torque AngleTorque Angle
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Amortisseur Amortisseur WindingsWindings
• When relative movement between the stator and the rotor poles occurs
• Voltage is induced into these windings.
• Subsequent induced magnetic field tends to slow movement and act like a “shock absorber”
• Can be used to aid starting in a simular way to that of the squirrel cage conductors
Similar to the squirrel cage found in induction machines
Also Known as “Damper “ windings
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StartingStartingZero starting Torque
Number of methods:
• Pony Motor
• Low Frequency
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StartingStartingPony Motor
An auxiliary smaller motor is used to spin the main motor up to or near Synchronous speed
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StartingStartingLow Frequency
• The rotor’s excitation windings are short circuited
• Act like a wound rotor induction motor
• Supply is applied at reduced voltage & frequency
• Rotor builds up speed
• Excitation is then applied to windings and rotor locks in
Tumit 3 and the Shoalhaven hydro schemes use this system
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Motor UsesMotor Uses•Can be used as a standard motor similar
to an induction motor
•Main use as a power factor correction device
As an induction machine is cheaper it is seldom used just as a motor
Sometimes called a rotary capacitor
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Power Factor Power Factor CorrectionCorrection
As we increase the excitationThe
sta
tor
curr
ent
will
dro
p
And the Power Factor Improves
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Power Factor Power Factor CorrectionCorrection
If we continue to increase the excitation
The
sta
tor
curr
ent
will
incr
ease
And the Power Factor detieriates
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Power Factor Power Factor CorrectionCorrection
These curves are known as “Vee Curves”
Curve A = Stator Current
Curve B = Power Factor
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Power Factor Power Factor CorrectionCorrection
These curves are only applicable for a set load torque
A different load will produce a different set of curves
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Power Factor Power Factor CorrectionCorrection
Lagging
Unity
Leading
Under Excited Over Excited
If the bride is over excited she will lead you to the marriage bed
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Power Factor Power Factor CorrectionCorrection
A motor has full load of 100A and an excitation current of 8A what will be:
• The stator current?• PF of the motor?
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Power Factor Power Factor CorrectionCorrection
38%
PF =
X 100A = 38A
Stator Current
0.9 Lagging
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ExampleExample• A load of 450 kVA operates at a power factor of 0.65
lagging. • An additional synchronous motor is added having an
input power of 90 kW and a maximum power factor of 0.85 leading.
• Determine reactive power and the overall power factor
450kVA
49.5
342kVar
450 x sin 49.5=
450 x cos 49.5=
292.5kW
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ExampleExample• A load of 450 kVA operates at a power factor of 0.65
lagging. • An additional synchronous motor is added having an
input power of 90 kW and a maximum power factor of 0.85 leading.
• Determine reactive power and the overall power factor
450kVA
49.5
342kVar
292.5kW
90kW
31.8
90 x tan 31.8 =
55.8kVar342 – 55.8 = 286kVar
292.5 + 90 = 382.5kWtan-1 x 286/382.5 = 36.8
PF = 0.8 Lag