Aps_chap6 (Syn Generator)

7/30/2019 Aps_chap6 (Syn Generator)

http://slidepdf.com/reader/full/apschap6-syn-generator 1/67

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



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



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



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



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



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



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



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



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



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

+=

+=



EQUIVALENT qd0 CIRCUIT IN ROTOR REF. FRAME



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



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



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



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



SIMULATION OF A SYNCHRONOUS MACHINE

Overall Diagram




abc to rotor qd0 block from oscillator block




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 −==



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



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



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




qd0 stator and rotor block




q-axis stator block



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



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=



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

++−=

+−−=

+=



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




Rotor block

Tem ωr



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 ===== −−




qd0 rotor frame to abc block




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

+=

+=



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



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



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



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.



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.



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



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



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



SYNCHRONOUS MACHINE TRANSIENT ANALYSIS

Generator transient behavior sub-transient,

transient,

steady state



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



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

’



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



SUB-TRANSIENT INDUCTANCES

Equivalent circuit: Lls in series with (Lmd // Lff ’// Llkd’)

LlsLmd Llf ’ Llkd’Ld

”

TRANSIENT TIME CONSTANT



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



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



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

’



C l l ti M hi P t



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



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


C l l ti M hi P t



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


C l l ti M hi P t



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


C l l ti M hi P t




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




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

−−−=





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





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



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





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




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





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





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



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



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



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




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




Electric circuit on synchronous machine

machine equations

Project 6-3: Simulation of Permanent MagnetS h M hi



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

Aps_chap6 (Syn Generator)

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Transcript of Aps_chap6 (Syn Generator)