Aps_chap6 (Syn Generator)
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Transcript of Aps_chap6 (Syn Generator)
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Basics of Electric Machines (Rotating field)
Rotating field fundamental space mmf established by a single phase
winding with a current, i=Iacosωt, is Fa1=Fm1cosωt cosθa,where Fm1 is the peak value of fundamental mmf Fm1=(4/π)kw(Nph/P)Ia
Fa1 is the combination of two counter-revolving mmf waves Fa1=Fm10.5cos(θa-ωt) +F m10.5cos(θa +ωt)
FbFf
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Rotating field In a three phase machine, the axes of the three phase
windings are 120o
apart, assume the phase currentia=imcosωtib=imcos(ωt-120o)ic=imcos(ωt+120o)
fundamental air gap mmf of three phases
Fa1=(1/2)Fm1{cos(θa-ωt) +cos(θa +ωt)}Fb1=(1/2)Fm1{cos(θa-ωt) +cos(θa +ωt-240o)}Fc1=(1/2)Fm1{cos(θa-ωt) +cos(θa +ωt+240o)}
Fa1 +Fb1 +Fc1 =(3/2)Fm1cos(θa-ωt)the resultant airgap mmf is a constant amplitude,
sinusoidal shape, revolving wave with a speed of ω in adirection corresponding to the sequence of the phasecurrent
it is related to the synchronous machine voltage andmechanical speed:ωe=(P/2)ωm
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Developed torque of a uniformairgap machine
The torque and coenergy relation
T=∂W’fld /∂θm where W
’fld =λi- Wfld
λi=LSSiS2+Lrrir
2+2Lsrircosδsr
Wfld= (1/2)LSSiS2+(1/2)Lrrir
2+Lsrisircosδsr
W’fld= λi- Wfld =(1/2)LSSiS2+(1/2)Lrrir
2+Lsrircosδsr J
Developedtorque:
the negative sign indicates the developed torque is in the direction oppositeto that of increasing δsr
m N sin2
2 ;
''
⋅−=
=∂
∂
∂
∂=
∂
∂=
sr r ssr
srmsr
srm
sr
sr
fld
srm
fld
ii LP
T
PW W T
δ
δ δ δ
δ
δ δ
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FLUX LINKAGE RELATIONS
mmfs of rotor windings are along d-q axis. Axis of north
pole is in d-axis, stator internal voltage is in q-axis
Field winding is mainly in d-axis, damper winding are ind,q axis
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FLUX LINKAGE RELATIONS
For a phase current flowing into the stator, the a
phase mmf Fa produces flux components ψd, ψq
along the d- and q- axes
The inductance representation has θr variable
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MATH MODEL OF SYNCHRONOUS MACHINE (abc frame)
Voltage equation of the stator and rotor windings
Flux linkage equation of the stator and rotor windings
Lss, Lsr, Lrr are shown in Eq.7.7 to Eq.7.9 (pp.264) Computational difficulty in abc frame
The inductances in Lss and Lsr are time dependent variablespresent computational difficulty when voltage equations are beingsolved
To obtain the phase current quantities, the inverse of the time-varying inductance matrix will have to be computed at each timestep, this would be time consuming and could produce numerical
instability
ΛΛ+
=
r
s
r
s
r
s
r
s
dt
d
i
i
r
r
v
v
0
0
[ ] r rr s
t
sr r
r sr ssss
i Li L
i Li L
+=Λ
+=Λ
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TRANSFORMATION TO ROTOR REFERENCE FRAME
Transform the stator quantities to rotor qd0 reference
frame, then the voltage equation has time-invariantcoefficient
Rotor voltage equation already attached to q- and d- axes,qd0 transformation only need be applied to the statorwinding
For transformation, we denote
where Tqd0(θr) is as Eq. (7.11) Apply transformation Tqd0(θr) into stator voltage equation,
the stator voltage equation becomes
)(00
)(00
)(00
)(
)(
)(
abcsr qd qd
abcsr qd qd
abcsr qd qd
T
iT i
vT v
Λ=Λ
=
=
θ
θ
θ
( ) ( ) ( )01
000
1
00)(0)(00 qd qd qd qd qd sqd abcsqd abcssqd qd T
dt
d T iT r T
dt
d T ir T v Λ+=
Λ+= −−
[ ]
+
−
+
−
=
2
1
2
1
2
1
3
2in
3
2in sin
3
2cos
3
2cos cos
3
2)(0
π θ
π θ θ
π θ
π θ θ
θ r r r
r r r
r qd ssT
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TRANSFORMATION TO ROTOR REFERENCE FRAME
The first term of the transforming equation on
flux linkage (pp.266)
The second term of the transforming equation onflux linkage (pp.266)
The stator voltage equation of qd0 rotor
reference frame
00
1
00
0 0 0
0 0 1
0 1 0
qd r qd qd qd
T dt
d T Λ
−=Λ
− ω
00
1
00 qd qd qd qd dt
d
dt
d T T Λ=Λ−
0000
0 0 0
0 0 1-
0 1 0
qd qd r qd sqd dt
d ir v Λ+Λ
+= ω
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FLUX LINKAGE IN TERMS OF WINDING CURRENTS
Flux linkage Λqd0 and qd0 current can be obtained from
stator quantities
The stator qd0 flux linkage equations
The rotor qd0 flux linkage equations
( ) r qd sr qd sqd qd ssqd qd s i LT iT LT 000
1
000, +=Λ −
00
0
0
)(2
3
)(2
3
i L
i Li Li L L L
i Li Li L L L
lss
skd skd sfd sfd d mslsds
skqskqsfqsfqqmslsqs
=
++
−+=
++
−+=
λ
λ
λ
kqkqkq fq fqkqqskqkq
kq fqkq fq fqfqqsfq fq
kd kdkd f fkd d skd kd
kd fkd f ff d sfd fd
i Li Li L
i Li Li L
i Li Li L
i Li Li L
++=
++=
++=
++=
2
3
2
3
2
32
3
λ
λ
λ
λ
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VOLTAGE EQUATIONS IN ROTOR REFERENCE FRAME
Voltage equations of
synchronous machine
dt
d
ir v
dt
d ir v
dt d ir v
dt
d ir v
dt
d ir v
dt
d
dt
d ir v
dt
d
dt
d ir v
kq
kqkqkq
fq
fq fq fq
kd kd kd kd
fd
fd fd fd
s
r q
d d sd
r d
q
qsq
'
'''
'
'''
'
'''
'
'''
000
λ
λ
λ
λ
λ
θ λ
λ
θ λ
λ
+=
+=
+=
+=
+=
−+=
++=
Flux linkage equations
in terms of L and i
where
''''
''''
''''
''''
00
''
''
kqkqkq fqmqqmqkq
kqmq fq fqfqqmq fq
kd kdkd fd md d md kd
fd fdfd kd md d md fd
ls
kd md fd md d d d
kqmq fqmqqqq
i Li Li L
i Li Li L
i Li Li L
i Li Li L
i L
i Li Li L
i Li Li L
++=
++=
++=
++=
=
++=
++=
λ
λ
λ
λ
λ
λ
λ
md lsd
mqlsq
L L L
L L L
+=
+=
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EQUIVALENT qd0 CIRCUIT IN ROTOR REF. FRAME
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CURRENT EQUATIONS IN ROTOR REFERENCE FRAME
Current equations in
terms of λs, q-axis(from Fig.7.4)
mqlkqlfqls MQ
kq
lkq
MQ
fq
lfq
MQ
q
ls
MQ
mq
kq fqqmqmq
mqkq
lkq
kq
mq fq
lfq
fq
mqq
ls
q
L L L L L
L
L
L
L
L
L
iii Lwhere
Li
Li
Li
11111
)(
)(1
)(1
)(1
''
'
'
'
'
''
'
'
'
''
'
+++=
++=
++=
−=
−=
−=
λ λ λ λ
λ
λ λ
λ λ
λ λ
Current equations in
terms of λs, d-axis(from Fig.7.4)
md lkd lfd ls MD
kd
lkd
MD f
lfd
MDd
ls
MDmd
kd fd d md md
md kd
lkd
kd
md fd
lfd
fd
md d
ls
d
L L L L L
L
L
L
L
L
L
iii Lwhere
Li
Li
Li
11111
)(
)(1
)(1
)(1
''
'
'
'
'
''
'
'
'
''
'
+++=
++=
++=
−=
−=
−=
λ λ λ λ
λ
λ λ
λ λ
λ λ
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CURRENT EQUATIONS IN ROTOR REFERENCE FRAME
Matrix current equations in terms of flux linkage
the above matrix can be substituted into voltage and torqueequation to obtain the math model of synchronous machine
the synchronous machine model derived from above use fluxlinkages of rotor windings as state variables
−
−−
−
=
'
'
'''''
'''''
''
'
'
1)1( - -
- 1
)1(
- - 1
)1(
kd
fd
d
lfd lfd
MD
lkd lfd
MD
lkd ls
MD
lkd lfd
MD
lfd lfd
MD
lfd ls
MD
lkd ls
MD
lfd ls
MD
lsls
MD
kd
fd
d
L L
L
L L
L
L L
L
L L
L
L L
L
L L
L
L L
L
L L
L
L L
L
i
i
i
λ
λ
λ
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ELECTROMAGNETIC TORQUE
Electromagnetic torque is obtained from the component
of input power that transferred across airgap Stating from the input power into the machine:
Pin=vaia+ vbib +vcic+ vfdifd+ vfqifq
Stator quantities transformed into qd0 reference frame
Pin=(3/2)(vqiq+ vdid)+3v0i0+ vfdifd+ vfqifq
Eliminating ohmic losses and rate of change of magneticenergy (Eq.7.33 in pp.271) Pem=(3/2)ωr(λdiq - λqid) W
For a P-pole machine, ωr=(P/2)ωrm
Pem=(3/2)(P/2)ωrm(λdiq - λqid) W
Dividing the Pem by ωrm, the torque developed by P-polemachine
Tem=(3/2)(P/2)(λdiq - λqid) N.m
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SIMULATIONS OF 3 PHASE SYNCHRONOUS MACHINE
The winding equation of Syn. mach. model could useterminal voltage as input and currents as output
Main inputs: stator abc phase voltage, excitation voltage tofield winding, applied mechanical torque (load)
Main outputs: stator abc phase current
Modeling procedure Stator winding voltage must be transformed to qd0 reference
frame attached to rotor
)(3
1
)70.7( )(3
13
1
3
1
3
2
:
0 cba
bc
s
d
cba
s
q
vvvv
vvv
vvvv
StepFirst
++=
−=
−−=
(7.72) rad.elect. )0()()(
)(cos)(sin
(7.71) )(sin)(cos
0∫ +=
+=
−=
t
r r r
r
s
d r
s
a
r
d
r
s
d r
s
q
r
q
dt t t
where
t vt vv
t vt vv
StepSecond
θ ω θ
θ θ
θ θ
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SIMULATION OF A SYNCHRONOUS MACHINE
Overall Diagram
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SIMULATION OF A SYNCHRONOUS MACHINE
abc to rotor qd0 block from oscillator block
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SIMULATION OF A SYNCHRONOUS MACHINE
Variable-frequency oscillator the equation of the oscillator
convert 2nd-order equation to 2 1st-orderequation:
rewrite the above equations into integral forms:
assume y1 is sinωt, y2 is cosωt
1
2
2
1
2
ydt
yd ω −=
∫∫ =−= dt y ydt y y 2112 , ω ω
121
2 ,1
ydt
dy
dt
dy y ω
ω −==
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MODEL OF VARIABLE FREQUENCY OSCILLATOR
The cosωt, sinωt signals could be
obtained by integrating each other Variable y1 , y2 could be obtained
by variable ωt The variable oscillation block is
shown as below
( )br r ω ω θ /cos ×br ω ω /
( )br r ω ω θ /sin ×
r r ω θ ×− sinr θ cos
r θ sin
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SIMULINK RESULT OF OSCILLATOR
plot result: y1 and y2 (from oscillator block), ω isconstant
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-5
0
5
y 1
a n d
y 2
y2
y1
starting reference
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TERMINAL VOLTAGE AND FLUX LINKAGE EQUATIONS
Modeling procedure (relate input variables to state variables) Stator: relate terminal voltages (input variables) with flux linkage of
stator and rotor windings (state variables) (Eq.7.74) Rotor: In this case, only flux linkages of field winding in d-axis and a
pair of damper windings in d- and q- axes are related together(Eq.7.74)
)74.7(
)(
)(
000 dt x
r v
dt x
r v
dt x
r v
ls
sb
d md
ls
sq
b
r d bd
qmq
ls
sd
b
r qbq
∫
∫
∫
Ψ−=Ψ
Ψ−Ψ+Ψ−=Ψ
Ψ−Ψ+Ψ−=Ψ
ω
ω
ω
ω
ω
ω
ω
(7.74)
)}({
)(
)(
'
'
'
'
'
'
'
'
'
'
''
dt x
x E
x
r
dt x
r
dt x
r
fd md
lf
md
f
md
kqb
fd
kd md
lkd
kd b
kd
kqmq
lkq
kqb
kq
∫
∫
∫
Ψ−Ψ+=Ψ
Ψ−Ψ=Ψ
Ψ−Ψ=Ψ
ω
ω
ω
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SIMULATION OF A SYNCHRONOUS MACHINE
qd0 stator and rotor block
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SIMULATION OF A SYNCHRONOUS MACHINE
q-axis stator block
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STATOR AND ROTOR FLUX LINKAGE
whererotor flux linkage stator flux linkage
mqkqlkqkq
md kd lkd kd
md fd lf fd
f
f
md f
fd kqd md
kqqmq
i x
i x
i x
r
v x E
iii x
ii x
Ψ+=Ψ
Ψ+=ΨΨ+=Ψ
=
++=Ψ
+=Ψ
'''
'''
'''
'
'
''md
'mq
)(
)(
00 i x
i x
i x
ls
md d lsd
mqqlsq
=ΨΨ+=Ψ
Ψ+=Ψ
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WINDING CURRENTS
Modeling procedure (relate state variables to outputvariables)
relate flux linkages (state variables) with windingcurrents (output variables)
Stator winding current Rotor winding current
ls
md d d
ls
mqq
q
xi
xi
Ψ−Ψ=
Ψ−Ψ=
'
'
'
'
'
'
'
'
'
lfd
md fd
fd
lkq
mqkq
kq
lkd
md kd kd
x i
x i
x i
Ψ−Ψ=
Ψ−Ψ
=
Ψ−Ψ=
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WINDING CURRENTS
Modeling procedure (dq0 to abc transformation) transform qd0 rotor frame currents to stationery frame
transform stationery qd0 currents to abc currents
)(cos)(sin
)(sin)(cos
t it ii
t it ii
r
r
d r
r
q
s
d
r
r
d r
r
q
s
q
θ θ
θ θ
+−=
+=
0
0
0
3
1
2
1
31
21
iiii
iiii
iii
s
d
s
qc
sd
sqb
s
qa
++−=
+−−=
+=
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TORQUE EXPRESSION
Electromechanical torque developed with P-polesin motoring or generating
The net accelerating torque:
generatingfor (-)motoring,for )(:Therew
m N )(22
3
m N )(22
3
em +
⋅−=
⋅−==
d qqd
b
d qqd
rm
emem
iiP
iiPP
T
ψ ψ ω
λ λ ω
( )
( )∫ −+=−
−===−+
t
dampmechemer
er r rmdampmechem
dt T T T J
Pwt
dt
wt d
P
J
dt
t d
P
J
dt
t d J T T T
02)(
)(2)(2)(
ω
ω ω ω
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SIMULATION OF A SYNCHRONOUS MACHINE
Rotor block
Tem ωr
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PER-UNIT EXPRESSION FOR TORQUE AND SPEED
Per-unit variables:
( ){ }bbm
ber
r
b
pudamp pumech puem
S J H dt
wt d H
dt t d
P J
T T T T
2
)()()(
2
1 where pu,
/)(2
pu )(21
ω ω ω
ω
=−
=
=−+
P
S T
I
V Z
V
S I V V bbm
bm
bb
b
bb
b
bblinetolineb
2 , , ,
3
2 ,
3
2ω ω
ω ===== −−
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SIMULATION OF A SYNCHRONOUS MACHINE
qd0 rotor frame to abc block
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SIMULATION OF A SYNCHRONOUS MACHINE
VIPQ block
( )( ){ }
( )( ){ }qd d qd qd q
d d qqd qd q
iviv jii jvvQ
iviv jii jvvP
−=−−=
+=−−=
*
*
Im
Re
22
22
dsqst
dsqst
I I I
V V V
+=
+=
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MINIMIZE STARTUP TRANSIENTS
Startup transient occurs whenever the state variables of
simulation (such as ψ’s) are not properly initialized To wait for the startup transient is rather time consuming
especially when the time constants of field and machineryare high
To minimize startup transients, we need to initialize the
corresponding states with good estimates The initial values of integrators in the model should be set
close to the desired operating condition to minimize theinitial simulation transients
The desired operating condition for the variables
initialization could be obtained from steady-state analysis Startup transients can also be minimized by setting high
damping coefficient Dω, until steady-state is established,then Dω is reset to the previous value
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Project. 6-1 Project homework
(Homework): With the same machine parameters givenin Table 7.2 (pp.317), you are asked to build up asynchronous generator model has the following source:v1=1√2sin(120πt+0) puv2=1√2sin(120πt-2π /3) pu
v3=1√2sin(120πt+2π /3) pu1) With source voltage connected in 1pu, excitation voltage Ef =
1pu, change the mechanical torque from 1pu to 0 pu at t=10 sec,and then change the mechanical torque again from 0pu to -1puat t=15 sec. Observe and plot the following responses from 0~20
sec1) the Pgen, Tem, ω, δ, in one figure2) Qgen, If ’, id, iq in one figure3) ia, ib, ic in one figure4) discuss what you see on the plots (ex. observe transient in
field current, 3 phase current, Q, ω, etc)
h k
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Project. 6-1 Project homework
2. With Excitation voltage = 1pu, Tmech = 0.5 pu (mechanicaltorque), short the 3-phase source terminal voltage toground from 1pu to 0 pu at t=10 sec, fault clear att=10.25 sec, observe and plot
1) vq, vd, iq, id, in one figure
2) ia, ib, ic in one figure3) Pgen, Tem, δ, ω in one figure4) Qgen, If ’, in one figure5) discuss what you see on the plots (ex.observe transient in field
current and qd, abc current)6) calculate the critical clearing time that machine could run out of
synchronism, observe your result
Suggestion:In the case 2, the figure time scale can be shown starting from t=9 secthrough the time when system becomes stable after the fault cleared
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MACHINE PARAMETER
Machine data from manufacturers are in form of reactance time constants resistances they are derived from measurement of stator windings
The parameters of the rotor winding from statormeasurements are taken from a method of shortcircuited oscillogram Short circuited oscillogram: stator is initially open-
circuited, keep field excitation constant, then applythree phase short-circuit ground on stator and observe3 phase stator current decay period
The rate of stator current decay could tell usinformation of machine’s time constant, reactance, etc.
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SYNCHRONOUS INDUCTANCES
Inductance is defined as L=λ /I Peak of rotating mmf aligned with d-axis, then
Ld= λd /I, same manner as Lq= λq /I, Ld and Lqcould be obtained in the synchronous operation.
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SYNCHRONOUS INDUCTANCE AND TRANSIENTINDUCTANCE
Balanced steady state operation rotor and stator mmf are in synchronous speed, relative
speed of mmf=0, flux linkage does not vary with time,no voltage is induced in rotor circuits
generator is represented with a constant emf behindsynchronous reactance Xs=Xd, as figure below
Transient operation short circuit at the generator terminal will result in
varying flux linkages with rotor, induce rotor transientcurrent and in turn reacts on stator armature transient analysis must use transient machine
parameters transient parameters must be obtained by test of
oscillogram
Efd
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MACHINE PARAMETER
Short circuit oscillogram the stator’s short-circuited transient current decay
exhibits two distinctly different decay period
sub-transient: the first few cycles of short circuit whencurrent decay is fast, mainly attribute from the changes
of currents in damper winding transient: rate of current decay is slower and attribute
to the current changes in field winding
Oscillogram cause
sub-transient: change of current in the outer damperwinding limit the stator-induced flux from penetratingthe rotor
transient: change of current in the field winding reactthe same manner
SYNCHRONOUS MACHINE TRANSIENT
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SYNCHRONOUS MACHINE TRANSIENT ANALYSIS
Generator transient behavior sub-transient analysis: lasting for only a few cycles right after the
disturbance
transient analysis: lasting longer time than sub-transient period
steady state analysis: transient phenomena die out
SYNCHRONOUS MACHINE TRANSIENT
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SYNCHRONOUS MACHINE TRANSIENT ANALYSIS
Generator transient behavior sub-transient,
transient,
steady state
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TRANSIENT INDUCTANCES
Induced current in the damper decay is more rapid the damper winding resistance >> field winding resistance
In the transient period, we assume damper current decay is over induced field current still changes to opposite the change in flux
linkage by stator current
Procedure for transient inductance estimation stator initially opened, then short circuit is applied keep rotor field excitation constant
Change of flux linkage in d-axis in Stator winding Δλd=Ld Δid+Lmd Δ if ’ (7.96)
No change of flux linkage in rotor winding Δλfd= Lmd Δid+Lff
’ Δif ’ = 0 (7.97) where Lff
’=Lmd+L’f ’, Lkd’=Lmd+Lkd’,
Substitute Δ if ‘ in (7.96) into (7.97), we have Δλd= (Ld -Lmd
2 /Lff ) Δid
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TRANSIENT INDUCTANCES
d-axis and q-axis transient inductance Ld
’= Δλd/ Δid= (Ld -Lmd2 /Lff ’) (7.99)
Lq’= Δλq/ Δiq= (Lq -Lmq
2 /Lgg’) (7.100)
d,q axis transient inductance could beconsidered as Lls in series with (Lmd//Lff ’) orLls in series with (Lmq//Lgg’)
LlsLmd Llf ’Ld
’
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SUB-TRANSIENT INDUCTANCES
Procedure for sub-transient inductance estimation stator initially opened, then short circuit is applied keep rotor field excitation constant
For the sub-transient period, rotor flux linkagekeeps initially constant, thus Δλf ’ =0, Δλkd’ = 0 Δλf ’ = Lmd Δid+Lff ’ Δif ’+Lmd Δikd’ = 0 Δλkd’ = Lmd Δid+Lmd Δif ’+Lkdkd’ Δikd’ = 0 (7.104)
Corresponding change in d-axis stator flux linkage
Δλd = Ld Δid+Lmd Δif ’+Lmd’ Δikd’ (7.106) Rearrange Δif ’ = Δikd’ in (7.104) and substitute into
(7.106), we obtain (7.107) Ld’’ = Δλd / Δid
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SUB-TRANSIENT INDUCTANCES
Equivalent circuit: Lls in series with (Lmd // Lff ’// Llkd’)
LlsLmd Llf ’ Llkd’Ld
”
TRANSIENT TIME CONSTANT
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TRANSIENT TIME CONSTANT
Procedure for open circuit transient time constant
estimation stator initially open circuited, then apply field excitation
voltage change
Transient time constant Tdo’
larger value observe field winding current decay
Procedure for open circuit sub-transient timeconstant estimation stator initially open circuited, then short field winding
Sub-transient time constant Tdo’’ smaller value observe damper winding current decay
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Open/Short Circuit Transient (Field) Time Constant
Open circuit transient (field) time constant Tdo’ observe change in field currents in response to a change in
excitation voltage when stator is open circuited
Tdo’ = Lff ’ / rf ’
Lff ’ equivalent circuit: Llf ’ in series with Lmd
Tqo’ = Lgg’ / rg’
Typically, Tdo’ is of order of 2 to 11 secs.
Short circuit transient (field) time constant Td’ observe change in field currents in response to a change in
excitation voltage when stator is short circuited, Td’
ratio of Td’ and Tdo’ equals to ratio of Ld’ and Ld, that is Td’ / Tdo’ = Ld’ / Ld
The transient time constant can also be expressed as
(7.113)
−
−=
'
2
'
' )(1
d d
lsd
f
do L L
L L
r T
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Open circuit Sub-transient Time Constants
Open circuit sub-transient time constant Tdo’’
time constant of kd damper winding current when stator windings are open circuited field windings are then shorted
also defined as time in second required for d-axiscomponent of stator voltage decrease to (1/e) value
after a short circuited on the stator winding is suddenlyremoved in rated speed
During the initial decay of the sudden open circuitstator voltage, the effective inductance of kdwinding Lkdo’’=Llkd’+(Lmd Llf ’)/(Lmd + Llf ’) Lkdo
’’ equivalent circuit: Llkd’ in series with Lmd // Llf
’
Therefore Tdo’’ = Lkdo’’ / rkd
’ (7.118)
By symmetry Tqo’’ => Tqo’’ = Lkqo’’ / rkq
’
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C l l ti M hi P t
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Calculating Machine Parameters
Procedure to calculate the parameters for the
developed model Normally, the parameters given by the
manufacturers are the stator parameters stator leakage reactance, xls
sometimes x0
= xls
is given instead of xls
d,q axis reactance, xd, xq
Rotor parameters needs to be determined Procedure 1: Calculate xmd, xmq
xmq
= xq
– xls
xmd = xd – xls
Procedure 2: Obtain transient and sub-transientparameters
C l l ti M hi P t
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Determine transient reactance Xd’: obtain rms steady short circuit current after
approximately the 10th cycle
use the equation: ( )( ) t mct I I ior
e I I i
d d d
t
d d d
''/ln'ln
'
''
/''
−=−−=∆
−=∆ −
τ
τ
Calculating Machine Parameters
C l l ti M hi P t
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Determine transient parameters
rms transient current
transient reactance
transient time constant
Determine sub-transient parameters
obtain rms short circuit current of the first 2 cycle
use the following equation
d
c
d I e I += ''
'
0'
d d I
E
X =
'
1'
md =τ
( )
( )t mct I I ior
e I I i
d d d
t
d d d
''''/ln''ln
''
'''''
/'''''
−=−−=∆
−=∆ −
τ
τ
Calculating Machine Parameters
C l l ti M hi P t
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Determine sub-transient parameters
rms sub-transient current
sub-transient reactance
sub-transient time constant
'''''
d
c
d I e I +=
''
0''
d
d I
E X =
''
1''
md =τ
Calculating Machine Parameters
C l l ti M hi P t
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Calculating Machine Parameters
Procedure 3: Calculate xlf ’
when xd’ is obtained from Procedure 2
from (7.99), calculate xlf ’
)()( '
'
'
lsd md
lsd md lf
x x x x x x x−− −=
Calc lating Machine Pa amete s
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Calculating Machine Parameters
Procedure 4: calculate xlkd’
when xd’ is obtained from Procedure 2, from Eq. (7.107) calculate
Procedure 5: calculate xlkq’
Use Lq’’ given in Eq. (7.108) by setting Llg->∞, no fieldwinding in q axis
if xmq >> xlkq’
xlkq’≈ xq’’-xls
))((
)(''''
'''
'
lf md lsd md lf
lf md lsd
lkd x x x x x x
x x x x x
+−−
−=
'
lf md '''
'''
' xxif
)(
)(>>
−−
−=
lsd lf
lf lsd
lkd
x x x
x x x x
)(
)(''
'''
lsqmq
mqlsq
lkq x x x
x x x x
−−−=
Calculating Machine Parameters
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Calculating Machine Parameters
Procedure 6: calculate rf ’
Use Td0’ and Eq. (7.138)
Procedure 7: calculate rkd’
Use Td0’’ and Eq. (7.141)
Procedure 8: calculate rkq
’
use Tq0’’ and Eq. (7.143)
)(1 ''
''
'
lsd lkd
dob
kd x x xT
r −+=ω
)(1 '
'
'
md lf
dob
f x xT
r +=ω
)(1 '
''
'
mqlkq
qob
kq x xT
r +=ω
Calculating Machine Parameters
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Calculating Machine Parameters
Or using short-circuit time constants to calculate rf ’, rkd
’,
rkq’
)(1
''
'
'
''
'
lf lslf md lsmd
lf lsmd
lkd
d b
kd x x x x x x
x x x xr
+++=
τ ω
)(1 '
'
'
lsmd
lsmd lf
d b
f x x
x x xr
++=
τ ω
)(1 '
''
'
lsmq
lsmq
lkq
qb
kq x x
x x xr
++=
τ ω
Project 6 2: Synchronous Machine Parameter Test
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Project 6-2: Synchronous Machine Parameter Test
(Homework): You need to build up your
synchronous machine model according toparameters given from table 7.2 and then applythe short circuit test or open circuit test to obtain:
1) xd’, xd’’, xq’, xq’’, Td’, Td’’, rf ’, rkd’, rkq’ by short
circuit test assume only rs, xd, xq, xls areavailable from Table 7.2
2) assume Tdo’, Tdo’’ are available from Table 7.2by open circuit test, and obtain rf ’, rkd’, rkq’,
3) verify your results with parameters in Table7.2
4) you may need to calculate rotor parametersthat is similar to 7.7.1 to build up your model
Project 6-2: Synchronous Machine Parameter Test
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Project 6-2: Synchronous Machine Parameter Test
Table 7.2 % parameters of three-phase synchronous machine Set 1 Perunit = 1 % parameters given in per unit of machine base Frated = 60; Poles = 4; Pfrated= 0.9; Vrated =18e3; Prated=828315e3; rs = 0.0048;
xd = 1.790; xq = 1.660; xls = 0.215; x’d = 0.355; x’q = 0.570; x”d = 0.275; x”q = 0.275; T’do = 7.9; T’qo = 0.410; T”do = 0.032; T”qo = 0.055; H = 3.77; Domega = 0; % mechanical damping coeff
Permanent Magnet Synchronous Motors
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Permanent Magnet Synchronous Motors
The dc excitation field winding can be replaced with
permanent magnet Difference of replacing the dc excitation with permanent
magnet is the elimination of copper losses
Magnet arrangement:
Surface: high magnetic material, leakage of magnet is small, noteasy to demagnetize, smaller volume of permanent magnetinterior: need a longer length of magnet, flux focusing is needed
Permanent Magnet Synchronous Motors
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Permanent Magnet Synchronous Motors
PMSM advantage: simple construction,lower weight and size for sameperformance
PMSM disadvantage: high price, magnetchange with time
Motor start:
line-start: have a rotor cage to help start motoron a fixed frequency supply
inverter-fed: variable frequency supply to startmotor to synchronous speed
Permanent Magnet Synchronous Motors
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Permanent Magnet Synchronous Motors
The circuit model used is similar to the
synchronous model we had before, the differenceis the rotor field winding Rotor field winding is replaced by a permanent
magnet inductance Lrc and equivalentmagnetizing current im’ in d-axis only, no q-axis
field winding Some difference of dq0 equations between
Synchronous and PM Synchronous motors flux linkage:
λd = Ld id +Lmdikd’+Lmdim’
λkd’ = Lmd id +Lkdkdikd’+Lmdim
’
λmd = Lmd (id +ikd’+im
’) = LMD (λd /Lls +λkd’ /Lkd’+ im’)
Tem = (3/2)(P/2) (λd iq - λq id)
Basics of Magnetics
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Basics of Magnetics
Recoil line H (magnetic field) vs. B (flux density)
F (mmf) vs. φ (flux)
i (current) vs. λ (flux linkage)
φ, F relationship of the Recoil line
B, φ, λ
H, F, i(-F
o, 0)
(0, φr1)
rc
o
orcr
rco
PF F
F P
PF F
φ
φ
φ
−=
=
−=
1
)(
DC
Prc
FFo
φ
+
-
Basics of Magnetics
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Basics of Magnetics
Equivalent magnetic circuit thevenin’s equivalent
norton’s equivalent
F P
PF F
F F P
rcr
rc
o
r oorcr
−=
−=
=
1
11
getwe,in
withreplaceto
φ φ
φ
φ φ
DC
Prc
FFo
φ
+
-
rc
o
orcr
rco
PF F
F P
PF F
φ
φ
φ
−=
=
−=
1
)(
FPrc
+
-
φr1
φ
Basics of Magnetics
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Basics of Magnetics
Math dual between magnetic and electric circuit norton equivalent
thevenin equivalent
DC
Prc
FFo
φ
+
-
VLrc
+
-
io
i
FPrc
+
-
φr1
φ
DC
Lrc
VVr
i
+
-
P t M t S h M t
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Permanent Magnet Synchronous Motors
Electric circuit on synchronous machine the permanent magnetic part will be represented by a
fixed current source im’ in parallel with inductor Lrc
DC
im’
Lrc
Lmd
equivalent electriccircuit for PM part
in rotor
stator part
P t M t S h M t
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Permanent Magnet Synchronous Motors
Electric circuit on synchronous machine
machine equations
Project 6-3: Simulation of Permanent MagnetS h M hi
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Synchronous Machine
(Homework): You need to build up the permanent
magnet synchronous generator (PMSG) model accordingto parameters given in Table 7.6 (im
’ =1.6203pu) . ThePMSG is connected to a Y balanced resistive load(Ω=0.1pu). When a torque is applied at the generator,obtain the following simulation result for the simulation
run time of 10 sec.1) A constant torque 0.5pu is applied to the generator,
obtain d-q terminal voltages, stator currents,electromagnetic torque, rotor speed, input power,resistive loss, and real power output.
2) A series torque is applied as the following:[0 0 1 1 0.5 0.5 1 1] for the corresponding time [5 5.45.4 5.8 5.8 6.2 6.2 6.5], obtain d-q terminal voltages,stator currents, electromagnetic torque, rotor speed,input power resistive loss and real power output