12/1/2012
Overview
•
•
•
•
•
35-Synchronous Motors
Text: 8.1 to 8.7
Principle of Operation
Equivalent Circuit-Round Rotor
Equivalent Circuit-Salient Pole
Power Expressions
Excitation Effects
ECEGR 450
Electromechanical Energy Conversion
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Dr. Louie
Introduction
Introduction
Important characteristics of synchronous motors:
• rotate at synchronous speed
• not self-starting
• can be operated at leading, unity or lagging PF
• physically identical to synchronous generator
• Synchronous motors rotate at synchronous speed
 Constant speed-torque characteristic
• Nm = Ns
 Synchronous speed dependent on number of poles
and frequency of applied voltage:
Ns
Nm
120f
P
• Applied load torque must be within the capability
of the motor, otherwise the motor stops
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Principle of Operation
Principle of Operation (Starting)
• Rotor is similar to synchronous generator
 contains field winding or PM
 portions of the rotor act like north and south poles
• Applied AC source acts like rotating magnet that “pulls” the
rotor at synchronous speed
•
•
•
•
Synchronous motors do not self-start
Synchronous speed of rotating field is usually very fast (e.g. 3600
rpm)
Rotor has a large inertia J
Before rotor rotates significantly, the rotating field has switched
polarity
A.
B.
A.
m
stator
stator
p
Rotor flux attracted to (tries to align with)
stator flux. Torque on rotor will be in
direction that aligns rotor flux to stator flux
with shortest amount of rotation.
stator
t = 0s
torque on rotor CCW
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p
p
Stator flux rotates CCW
Torque on rotor CCW
•
Stator flux advances ~180o in
~1/120 s
Torque on rotor is now CW
B. Rotor has started rotating
CCW
•
t = 0.0083s
torque on rotor CW
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2-pole synchronous motor
Rotor is at standstill
•
•
Rotor cannot rotate far enough
before torque switches direction.
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Principle of Operation (Starting)
Principle of Operation
• Add induction (armortisseur) winding to rotor
• Induced current will cause rotor to revolve with
stator flux
 Exactly like induction motor
• Synchronous motors are started as induction
motors
• Near synchronous speed the field winding “locks”
the rotor in at synchronous speed
dc source
 Current no longer induced in induction winding
induction
winding
field
winding
• Field winding should not be disconnected during
start-up
 Very large voltage can be induced
• Under load, induction winding prevents “hunting”
 Hunting: small variations in motor speed due to
sudden changes in load
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Equivalent Circuit (Round Rotor)
Equivalent Circuit (Round Rotor)
• Per-phase equivalent circuit is similar to the
synchronous generator
• Synchronous Motors (all PFs):
 Ea lags Va
 Power angle
 Phase current in opposite direction
is negative
• Circuit equations
Va Ea IaR a
Ia
jIaXs
Unity PF
Va Ea
R a jXs
Ra
Ea
jXs
IaRa
+
Ia V
a
-
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Ia
Ea
jIaXs
Va Ea IaRa
jIaXs
Synchronous Motor
Ia defined as leaving Ea
Ia defined as entering Ea
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Td
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Pd
s
• Power out is:
Po
vfif
• Copper losses
Pd Pr Psl
• Output power is mechanical, so it is common to
express in horsepower
3 | Ia |2 R a v fif
• Power developed:
Pd Pin Pcu 3 | Va || Ia | cos
Ea
• Torque developed by the motor is:
• Input power:
Pcu
jIaXs
Va
Va IaRa
Va IaRa
Equivalent Circuit (Round-Rotor)
 stator and rotor are connected to an external
circuit
PF
jIaXs
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• Like synchronous generators, synchronous
motors are doubly fed
3 | Va || Ia | cos
Ia
Synchronous Generator
Equivalent Circuit (Round-Rotor)
Pin
Unity PF
Ea
 1hp = 745.7 W
3 | Ia |2 R a
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Power Expressions (Round Rotor)
Power Expressions (Round Rotor)
• First consider the round (cylindrical) rotor
• Circuit equations
Va Ea IaR a
Ia
• From the equivalent circuit:
Ia
jIaXs
E V | E |2
Pd 3 Re{EaIa} 3 Re{ a a * a }
Zs
E V*Z
3 Re{ a a 2 s
| Zs |
jXs
Ia
Ea
Va Ea
Zs
• Power developed:*
Va Ea
R a jXs
Ra
Va Ea
R a jXs
+
Va
-
| Ea |2 R a
| Zs |2
Note that:
Pd
| E |2 X
j a 2 s}
| Zs |
3 Re{EaIa} 3 | Ea || Ia | cos(
PF
)
 using:
Z s | Zs |
z
Z*s | Zs |
1
Z*s
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Recall that dividing by a phasor
means dividing by the magnitude
and subtracting the angle
z
1 Z
Z*s Z
*
s
*
s
| Zs |
| Zs |2
| Zs |
| Zs |2
z
2
z
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Zs
| Zs |2
z
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Power Expressions (Round Rotor)
Power Expressions (Round Rotor)
• Let the applied voltage be the reference:
Power can be written as:
Va | Va | 0
E V*Z
Pd 3 Re{ a a 2 s
| Zs |
• Then:
Ea | Ea |
| Ea |2 R a
| Zs |2
j
| Ea |2 Xs
}
| Zs |2
E V*Z
| Ea |2 R a
3 Re{ a a 2 s }
| Zs |
| Zs |2
| Ea | cos( ) j | Ea | sin( )
E V* (R jX ) | Ea |2 R a
3 Re{ a a a 2 s }
| Zs |
| Zs |2
|E| (cos
3 Re{ a
jsin )|V|*a (R a
| Zs |2
3 | Ea || Va |
(R a cos
| Zs |2
Xs sin )
jXs )
}
| Ea |2 R a
| Zs |2
3 | Ea |2 R a
| Zs |2
Important result
How does this compare to the power output of a synchronous generator?
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Dr. Louie
Power Expressions (Round Rotor)
Power Expressions (Round Rotor)
• Ignoring armature resistance:
Pd
3 | Ea || Va | sin
Xs
Similar expression as synchronous generator
power output. Negative sign is a consequence of
assumed current direction.
• Torque ignoring armature resistance:
Td
• Ignoring armature resistance:
Pd
3 | Ea || Va | sin
Xs
• And the torque:
3 | Ea || Va | sin
Xs s
• What happens with
16
Td
= 0 degrees?
3 | Ea || Va | sin
Xs s
• What happens with
= 0 degrees?
 Torque equals 0
 Motor stalls
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Power Relationship Summary
Pin
3 | Va || Ia | cos
Power Relationship Example
Pin
v fif (input electrical power)
PF
vfif (field winding loss)
Pcu 3 | I2a | R a (copper loss in armature)
3 | Ea || Va |
3 | Ea |2 R a
(R a cos
Xs sin )
| Zs |2
| Zs |2
Pd
*
a a
3Re{E I }
3 | Ea || Ia | cos(
PF
)
developed electrical power
3 | Va || Ia | cos
Let:
vf=14V
if = 5A
Pr + Psl= 230W
Zs = 0.1 + j5
= 30o
Va = 130V
|Ea| = 230V
PF
v fif
8928.53W
Vfif = 70W (field winding loss)
Pcu
Pd
3 | I2a | R a
261.05W
3 | Ea || Va |
(R a cos
| Zs |2
3 | Ea |2 R a
| Zs |2
Xs sin )
8642.47W
Pr + Psl = 230W
Pr + Psl (rotational and stray load loss)
Po= 8412.47W
Po=Ts
s
(output mechanical power)
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Equivalent Circuit (Salient Pole)
Equivalent Circuit (Salient Pole)
• Similar equivalent circuit as salient pole
generators
• Continuing:
 Phase current is in opposite direction
Ea
Va IaR a
jIdXd
jIqXq
Ea
Va IaR a
jIaXq
jId (Xd
jIdXd jIqXq
+ +
+
Ea
Ea
Va IaR a
jIaXq
Ea'
Va IaR a
jIaXq
Ea Ea'
Xq )
+
Va
-
 For lagging:
| Ea' | | Ea | | Id | (Xd Xq )
 For leading/unity: | Ea' | | Ea | | Id | (Xd Xq )
Ra
-IaRa
Iq
Ea
-IqXq
E’a
E’a
• The salient pole power expressions are derived
from the round rotor by inspection
jXq
Ia
+
Va
-
Pd
| Va | cos( ) | Ea |
Xd
3 | Ea || Va | sin
Xs
Pd
Ea
Ia
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3 | Va || Ea' | sin
Xq
jXq
Va
E’a
round rotor
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Va
22
jXs
Note: | Id |
-
Salient Pole Power Expressions
• Example phasor diagram of lagging power factor
Ia
+
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Equivalent Circuit (Salient Pole)
Id
jXq
Ia
E’a
jId (Xd Xq )
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Va
Ra
jId (Xd Xq )
• E’a is the effective excitation voltage
• E’a, Ea have the same phase angle ( )
• Note that E’a = Ea when Xd = Xq (round rotor)
Ra
Ia
-
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Ia
Va
salient pole
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Power Expressions (Salient Pole)
• Recall that: | E
'
a
Power Expressions (Salient Pole)
• Power developed by salient pole motor:
| | Ea | | Id | (Xd Xq )
 + for lagging PF
 - for leading or unity PF
Pd
• Through substitution:
'
Pd
Pd
3 | Va || Ea | sin
Xq
3 | Va || Id | sin [
3 | Va || Ea | sin
Xd
3 | Va | sin [
Xd Xq
Xq
Xd Xq
XdXq
Td
3 | Va |2 sin(2 )[
Pd
m
](| Ea | | Id | Xd )
Xd Xq
2XdXq
3 | Va || Ea | sin
Xd m
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3 | Va |2 sin(2 )[
• Note that if Ea = 0 and
Pd
3 | Va || Ea | sin
Xd
]
3 | Va |2 sin(2 )[
3 | Va |2 sin(2 )[
Xd Xq
2XdXq
]
m
How do these expressions compare to power
and torque expressions for synchronous generators?
Xd Xq
2XdXq
Effect of Excitation
• Power and torque developed by the motor
depends on |Ea|
]
0
Xd Xq
2XdXq
Pd
3 | Ea || Va | sin
Xs
• |Ea| is a function of the excitation (field) current
• We can control the motor through excitation
] 0
• Salient pole motors will rotate without a field
winding connection
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Power Expressions (Salient Pole)
• From Pd
2XdXq
] Important result
Dr. Louie
3 | Va || Ea | sin
Xd
Xd Xq
]
Va cos( )
3 | Va || Ea | sin
Xd
3 | Va |2 sin(2 )[
• Torque developed:
3 | Va || Ea | sin
Xq
using | Ea | | Id | Xd
3 | Va || Ea | sin
Xd
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Effect of Excitation
Effect of Excitation
• We next examine the effect of excitation on
synchronous motor operation
• Assumptions:
• From the equivalent circuit if an ideal motor has
no load, then it draws no current
 Ea = Va
 = 0o
 No armature resistance
 No rotational losses
 Connected to infinite bus
Ra
Ea
Ea
Va
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Dr. Louie
jXs
+
Ia V
a
-
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Effect of Excitation
Over-Excited Operation
• Normal excitation: When the field current is
adjusted so that |Ea| = |Va|
• Over-excitation: field current is greater than what
is needed for normal excitation |Ea| > |Va|
• Under-excitation: field current is less than what is
needed for normal excitation |Ea| < |Va|
• When over-excited and under no load:
 |Ea| > |Va|
 = 0o
Ia
Va Ea
| Ia | 90 A
jXs
Po
Re{VaI*a}
Ia
0VAR (motor supplies reactive power)
0
jXs
Ea
0W
Im{VaI*a}
Qo
+
Va
-
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jXs
Va - Ea
Ia
Qo
Ea
Ea
Va
Ia leads Va
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Over-Excited Operation
Under-Excited Operation
• When a motor operates in this mode, it is also
known as a synchronous condenser
• Improves the power factor of a facility
• When under-excited and under no load:
 |Ea| < |Va|
 = 0o
Ia
Va Ea
| Ia |
jXs
Po
*
a a
Qo
Re{V I }
90 A
0W
Im{VaI*a}
0
0VAR (motor consumes reactive power)
jXs
Va - Ea
Qo
Ia
Ea
Va
Ea
Ia lags Va
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• Armature current also changes due to:
Ia
Xs
cannot be zero
• Assume that Va and Pd are constant
• If Ea changes due to a change in field current,
then |Ea|sin must be constant to equal Pd
Va Ea
R a jXs
• The power into the motor should not change
Pin 3 | Va || Ia | cos
PF
• Therefore, |Ia|cos
must decrease as |Ea| increases
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Effect of Excitation
• Now consider a motor under a load
3 | Ea || Va | sin
• From:Pd

+
Ia V
a
-
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Effect of Excitation

+
Ia V
a
-
PF
should be constant
 As |Ia| changes in response to Ea changing, the
power factor changes
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Phasor Diagrams
V Curves
Each diagram has same real power developed, differing excitation
|Ia|cos
PF
Ia
PF
|Ia|cos
PF
Ia
Va
|Ea|sin
jIaXs
Ea
Va
jIaXs
Ia
Ea
Leading PF
(over-excited)
P
PF
Va
PF
jIaXs
Ea
Unity PF
Leading PF
(under-excited)
P
P, Q
Increasing power
developed
armature current (A)
|Ia|cos
Lagging PF
(over-excited)
unity PF
Leading PF
(under-excited)
field current (A)
Q
Increasing |Ea|
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Excitation (infinite bus)
Power Effects
• For synchronous motors connected infinite bus,
adjusting excitation:
• How does varying the power output effect , Ia,
p for a constant excitation |Ea| and terminal
voltage Va?
• To increase power:
3 | Va || Ea | sin
Po
 |Ea|sin increases
Xs
 |Ia|cos PF increases
Po 3 | Va || Ia | cos PF
 Does not affect real power consumption if power
angle is zero, put reactive power will be affected
 Increases real power consumption if power angle is
held constant (non-zero)
 May not affect real power consumption if power
angle is adjusted such that |Ea|sin is constant, but
reactive power will be affected
Dr. Louie
• Increasing Po also increases power angle
• Real part of Ia also increases
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Phasor Diagrams
Each diagram has same excitation, differing power
• How do we change the real power developed?
• Power described by:
Pin
3 | Va || Ia | cos(
|Ia|cos
PF
PF
)
(Pd = Pin with Ra ignored)
• If excitation is held constant, increasing the
power developed by the motor results in:
Va
|Ia|cos
PF
PF
Va
Ia
Va
PF
jIaXs
Ea
Ia
jIaXs
Ea
 Increase in power angle magnitude (more
negative)
 Increase in |Ia|cos PF (real part of armature
current)
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|Ia|cos
PF
Ia
|Ea|sin
3 | Va || Ea | sin
Xs
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Mechanical Power
Pd
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jIaXs
Ea
Increasing Power
Leading PF
(over-excited)
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Unity PF
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Lagging PF
(under-excited)
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Synchronous Machines
Generator Operation
positive
P, Q
Summary
Motor Operation
negative
• Synchronous motors are physically similar to
synchronous generators
• Synchronous motors do not self-start without
induction winding
• Armature current is minimum when it is in-phase
with the terminal voltage
P
over-excited
Q
P
P
P
P, Q
 Unity power factor
• When under-excited (|Ea| < |Va|) the motor behaves
as an inductor (lagging power factor)
• When over-excited (|Ea| > |Va|) the motor behaves
as a capacitor (leading power factor)
under-excited
Q
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