Synchronous generator dynamics
Source: BPA
Source: O. Elgerd
Olof Samuelsson
Outline
– Synchronous generator at steady state
– Synchronization
– Swing equation
– Transient angle stability
– The Equal Area Criterion
– Small-signal stability
– Frequency control
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Synchronous machine
– Rotor
• One field winding
g fed with DC current
– Stator
• Three windings 120°
120 apart in space
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Loaded synchronous generator
• “Armature
Armature reaction”
reaction flux from load current in stator
• Adds to field flux to form air gap flux
St t flux
fl also
l includes
i l d lleakage
k
flflux
• Stator
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Equivalent circuit 1
Indices:
a
armature
t
d
d-axis
f
field
g
l
q
r
generator
lleakage
k
q-axis
resulting
g
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Equivalent circuit 2
Eq(If)
I
V
internal voltage (also Ef)
stator current
terminal voltage
Xd=Xa+Xl≈Xa
Eq=V+(Ra+jXd)I
Indices:
a
armature
d
d-axis
f
field
l
leakage
q
q-axis
I lags V by angle (g)
Eq leads V byy angle
g (g)
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Load angle 
 is a spatial angle between field and air gap flux
and
a phase angle between Eq and V+(Ra+jXl)I
Note
  is given relative to rotating reference (rotor)
  is a spatial coordinate for a mass – the rotor
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Steady state operation
Pe
Ra neglected
Eq

jXdI


Eq V
Xd
V
I
Scale by V/Xd

VI

V2
Xd
Qe
Components of VI:
Vertical
EqV/Xdsin=VIcos=Pe
Horizontal
EqV/Xdcos-V2/Xd=VIsin=Qe
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Two control inputs
Pe2
Pe1
Eq


V
Qe2
Turbine P  Pe
Eq constant
Qe1
P affects Q
Pe
Eq2
Eqq1

V

Qe2
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Field current If  Eq
P constant
Q >0 or <0
Qe1
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Synchronous generator rotor types
– Round rotor = ”Turbo”
Turbo rotor
• Two poles
q-axis
• High speed - 3000 rpm @ 50Hz
d-axis
• Used with steam turbines (e.g. nuclear)
– Salient pole rotor (Swe: utpräglade poler)
p
• Manyy poles
q-axis
• Lower speed - e.g. 150 rpm @ 50 Hz
d-axis
• Used with hydro turbines
• Gear ratio with more poles: mechanical=electrical·(2/p)
• IEA lab
l b generators
t
h
have ffour-pole
l salient
li t pole
l rotors
t
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Salient pole rotor
q-axis
q-axis
d-axis
Iq
Vq
d-axis
Eq
q-axis

jXq Iq

Vd
d- and q-axis different
•Geometry
Geometry
•Flux
•Inductance
•Currents and voltages
Eq=V+jXdId +jXqIq
V
jXd Id
Id
d-axis
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P and Q for salient pole rotor
Pe
Pe+j Qe=(Vd+jVq)(Id+jIq)*
Vd+jV
jVq=V(sin+jcos)
V(sin jcos)
Id=(Eq-Vq)/Xd
Iq=V
Vd/Xq

EqV
V 2  1
1 


 sin 2  Pfield  Preluctance

Pe 
sin  
 

Xd
2 Xq Xd 


EqV
2 sin  cos  
Qe 
cos  V 

Xd 
Xd
 Xq

2
2
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Try Xd=X
Xq!
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Synchronization
– Connect to an energized network (think Thévenin equiv.)
1. Control prime mover to reach correct speed  right el
2. Magnetize field (and armature) winding
3. Make V close to Vsystem (magnitude and angle!)
4 Connect!
4.
– Aim
d t t no-load
l d situation
it ti
• St
Steady-state
– Careless synchronization
• High currents and high mechanical stress
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Synchronization conditions
– V close to Vsystem if the voltages have
• Same phase order (= wiring correct)
• Same frequency (= speed correct)
• Same magnitude (= right magnetization)
• Same phase (= right timing of connection)
– Think generator Eq and network V as rotating
three-phase phasors:


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The swing equation
d m
J
 Tm  Te
dt
Torque balance for rotor
p magnetic rotor poles
2
 m mech. rad/s   e elec. rad/s
p
Multiply
p y torque
q balance by
y m
Use e as state and e≈s,e
Divide by Sbase to get p
u
p.u.
2H d e
 Pm  p.u.  Pe  p.u.
 s,e dt
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The inertia constant H
1 2
J m
2
Sbase
Kinetic energy of rotating masses
H=
g
Generator MVA rating
Unit: Ws/VA=s
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H on different MVA bases
– Machine base
• Steam turbines
4-9 s
• Gas turbines
3-4 s
• Hydro turbines
2-4
2
4s
• Synchronous compensator
1-1.5 s
– Common
C
b
base
Narrow
range!
• H ~ generator size (kW-GW)
• Infinite bus has infinite H
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Single Machine Infinite Bus
E’q
E
V0
X
Pm
X
H, X’d
”Classical
Classical model”:
model :
”Infinite
Infinite bus”
bus generator:
•Fixed E’q behind X’d
•Infinite H
•Constant Pm
•Zero impedance
•No damping
•Fixed voltage V0
•No saliency
Xeq=Xline+X’d
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”Classical” dynamic generator model
Synchronous generator connected to infinite bus:
2H d e
 Pm  Pe  


dt
 s,e
d

 e   s,e



dt
Pe
Pm
omega
wnom/2/H
1
1
s
s
Pmax
sin
•  in rad,, e in rad/s,, s,e
yp
y 100 rad/s
s etypically
•E’q and X’d in Pe () for slow transients
•Second
Second order ssystem
stem with
ith poor damping
•Electro-mechanical or “swing” dynamics
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delta
Two equilibrium points
E 'q V
Pm  Pe   
sin 
X eq
Two solutions for :

 Pm X eq 

 0  arcsin

 
E
'
V
q



180   0

Pe
Pm

• Synchronizing torque dPe/d
•dPe/d>0 for <90° - stable equilibrium
q
•dPe/d<0 for >90° - unstable equilibrium
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Dynamic response
*
Temporary short-circuit
near generator, Pe zero
during fault
Response?
Pe
Pm
1. Second order system
2 No damping
2.

3. Oscillator!  and  oscillate
(roughly sinusoidally)
4.
(t) will lag (t)
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Demo
sm.mdl
tcl=0 05
tcl=0.05
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Second order response
Pe zero at short-circuit near gen
Pe
Step in Pm-Pe
Mechanical states slow
Start at 0 and Pe((0)
Pm
Acceleration during fault
F lt removed
Fault
d att =
 1

Overshoot to 2 and Pe(2)
Oscillate around equilibrium 0 so Pe(0)=Pm
0
2
1
Si l ti tcl=0.05,
Simulation
t l 0 05 0.1
01
PW Example 11.5
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Angle stability
0 must be less than steady state limit 90º
2 also has limit – transient angle stability limit
Questions:
How large can 2 be?
What happens when it becomes too large?
What is the largest disturbance that is OK?
Simulation
tcl=0.15,
0 1505 0
151
0.1505,
0.151
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The Equal Area Criterion
Pe
DA
Short-circuit: Pe=zero
Mark areas between Pe() and
Pm in interval 0 to 2
AA
Pm
Accelerating Area: Below Pm
Decelerating Area : Above Pm

For stable system AA=DA
0 2
1
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Transient stability limit
Pe
DA
More severe disturbance:
AA
AA larger
Pm
g
Greater 2 makes DA larger
Maximum DA at 2=180º-0

0 1
For larger 2 only AA grows
grows…
2
Simulation
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Beyond stability limit
•
d/dt never becomes zero
•
Rotor accelerates even more
•
Machine transiently unstable = loses synchronism
•
Must disconnect and resynchronise
Demo
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Equal Area Criterion
•
Stability check for known disturbance
Use EAC for 2 and check 2<UEP
•
Max disturbance from stability limit
Determine disturbance for 2=UEP
•
Typical disturbances
Loss of line
line, generator or load
Short-circuit
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Stability analysis tools
Analytical – the Equal Area Criterion
•
Simple, can be done by hand, but approximate
•
Formulated before 1930 by Ivar Herlitz, KTH (First
Swedish PhD in engineering)
Time simulation
•
Computer application since the beginning
•
Voltages and currents as phasors or waveforms
•
Multi-machine model with Differential Algebraic Equations
•
Set of Differential equations for each generator
•
Power flow for Algebraic network equations
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Small-signal angle stability
•
Linearize at steady state (0, 0, Pm0)
•
State space: dx/dt=Ax+Bu
•
Compute eigenvalues i of A
•
Compute right eigenvectors i of A
•
Applies also to multi-machine
multi machine models
•
Popular application of control theory
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Eigenvalues and eigenvectors
•
Eigenvalue i:
Im(i)=resonance oscillation frequency (e.g. 0.35 Hz)
Re(i)=resonance oscillation damping
0 for all i system is small-signal stable
>0
0 ffor any i system
t
is
i small-signal
ll i
l unstable
t bl
•
Right eigenvector i:
Which generators participate in mode (resonance) i
those in NO and DK
E.g.
g Generators in FI against
g
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Small-signal damping
•
Low >0 for
f uncontrolled system
•
Negative damping from controllers
•
•
Automatic Voltage Regulators
•
HVDC controllers
Damping added by dedicated controls
•
P
Power
System
S t
Stabilizers
St bili
(PSS) on generator
t
•
Power Oscillation Damper (POD) on HVDC or FACTS
FACTS=MW size p
power electronic devices
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System frequency
One eigenvector shows all generator speeds vary together
The rigid body mode – the dynamics of system frequency
All generators synchronize to same , but which one?
 system
H 

H
i
i
i
i
i
Large generators dominate
Infinite bus is extreme case
H= so that system=infbus
frequency like center or gravity!
Also center of inertia frequency,
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System frequency dynamics
All generators modeled as one with:
=system
H   Hi
i
Pm
Pe,tot
+
–
nom
2H
system

1
2
fsystem
D
This is single machine, but no infinite bus (to relate  to)
Electrical load is frequency dependent
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Frequency event without control
G
Generator
is suddenly disconnected...
•
•
Step reduction of Pm
system
Pm
Unbalance: Pm<Pe
+
–
Pe,tot
nom
•
decreases
•
Decrease stops when Pe is reduced to Pm
•
Error in 
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
2H
1
2
D
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fsystem
Turbine governor
Proportional frequency
f
control law:
Pm= Pref+f/R
f=fnom–fsystem
R is speed droop, Hz/MW or p.u./p.u.
fnom
fsystem
+
–
f
+
+

K
1/R
R
P
–
+
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Pm
Pref
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R on machine base
All generators usually have the same R given in p.u. on machine
base
A disturbance
di t b
gives
i
same f everywhere
h
All generators do same p.u. contribution
Typical R value is 5%
f=0.05 p
p.u. g
gives Pm=1 p
p.u.
PW Example 12.4
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R on common base
R for entire system on common base:
1
Rtotal

i
1
Ri
More generators give greater 1/Rtotal
In Nordel 1/Rtotal≈6000 MW/Hz
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Frequency error tolerance
Instantaneous value of f:
± 0.01
0 01 Hz in US
± 0.1 Hz in Nordel
± 0.2 Hz in Ireland
Time integral of :
Time error on clocks <10s in Nordel
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600 MW step excites f and angle dynamics
Deviation from
50 Hz at LTH
Phase angle
differences
(degrees)
LTU=Luleå
13:39 13/11: 600 MW generator in Denmark disconnected
F
Frequency
dip
di and
dN
North-South
th S th angle
l oscillations
ill ti
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Conclusions
• S
Steady state
– P and Q for round and salient pole rotor
• Transient
T
i t angle
l stability
t bilit
– Equal Area Criterion and simulations
• Small-signal
Small signal stability
– Eigenvalues and eigenvectors
• Frequency dynamics
– All generators like one
– Fair sharing: All generators respond equally in p.u. on
machine base if same p.u. R
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