motor sikron

8/19/2019 Motor Sikron

1/146

SYNCHRONOUS MACHINES

Two-axis model

du

f u

f i

di

Di

Qi

quqi

γ

d

q

d

q

Q D

f


2/146

Synchronous machines

8 MW diesel-generator 270 MVA turbo-generator


3/146

Rotor windings

Field winding Damping bar Field winding Damping bar


4/146

PM synchronous machine; 1 MW ship propulsion motor


5/146

ψ ωψ

ψ ωψ

ψ

ψ

ψ

⎧= + −⎪

⎪⎪ = + +⎪⎪⎪

= +⎨⎪⎪ = +⎪⎪⎪ = +⎪⎩

d

d

ddd

d

d0d

d0

d

dd s d q

qq s q d

f f f f

DD D

QQ Q

u R it

u R it

u R it

R it

R it

ψ

ψ

ψ

ψ

ψ

⎧ = + +⎪

⎪ = + +⎪⎪

= + +⎨⎪

= +⎪⎪

= +⎪⎩

d d d df f dD D

f df d f f fD D

D dD d fD f D D

q q q qQ Q

Q qQ q Q Q

L i L i L i

L i L i L i

L i L i L i

L i L i

L i L i

Voltage, flux-linkage and motion equations

( ) ω

ψ ψ

γ ω

⎧

− = +⎪⎪⎨⎪ =⎪⎩

3 d

2 d

d

d

q q md d

J p i i T p t

t


6/146

Space-vector model for induction motor

Rotor frame of reference

Voltage and flux linkage

The axes in the rotor frame of reference denoted by d and q

ψ ωψ

ψ

⎧= + +⎪

⎪⎨⎪

= +⎪⎩

d j

dd

d

r sr r r

s s s sr r r r

r r r

u R it

u R it

ψ

ψ

⎧ = +⎪⎨= +⎪⎩

r r r s s m r s

r r r m s r r r

L i L i

L i L i

⎧ = +⎪⎨

= +⎪⎩

r s sd sq

r r rd rq

i i i

i i i

ψ ωψ

ψ ωψ

ψ

ψ

⎧= + −⎪

⎪⎪ = + +⎪⎪⎨⎪ = +⎪⎪⎪ = +⎪⎩

d

dd

dd

dd

d

sdsd s sd sq

sq

sq s sq sd

rdrd r rd

rq

rq r rq

u R it

u R i t

u R it

u R i t

ψ

ψ

ψ ψ

= +⎧⎪ = +⎪⎨

= +⎪⎪ = +⎩

sd s sd m rd

sq s sq m rq

rd m sd r rd

rq m sq r rq

L i L i

L i L i

L i L iL i L i

i s d

i s d

u s d

u s d

i s q

i s q

u s q

u s q q

d

β

α

( )ψ ψ = −

3

2e sd sq sq sdT p i i


7/146

Load angle

γ

ω =

d

dt

δ

ω ω − =

d

ds tis replaced by

where the load angle δ is the angle

between the excitation voltage vector u p (on q-axis) and stator voltage

vector ur s

δ

δ

=⎧⎪⎨ =⎪⎩

ˆ sin

ˆ cos

d s

q s

u u

u u

For synchronous machines, the position angle of the rotor is defined

differently from the angle γ typically used for induction machine analysis

γ

d

q

r su

pu

δ


8/146

Direct-axis equivalent circuit

= = =df dD fD mdL L L L Assumption

where Lmd is the direct-axis magnetising inductance

ψ ωψ

ψ

ψ

⎧= + −⎪

⎪⎪

= +⎨⎪⎪

= +⎪⎩

d

dd

d

d0 d

dd s d q

f f f f

DD D

u R i t

u R it

R i t

( )

( )

( )

σ

σ

σ

ψ

ψ

ψ

⎧ = + + +⎪⎪

= + + +⎨⎪

⎪ = + + +⎩

d d d m d f D

f f f m d f D

D D D m d f D

L i L i i i

L i L i i i

L i L i i i

( )

( )

( )

σ

σ

σ

ωψ ⎧

= − + + + +⎪

⎪⎪

= + + + +⎨⎪⎪

= + + + +⎪⎩

d d

d dd d

d dd d

0

d d

dd s d q d m d f D

f f f f f m d f D

DD D D m d f D

iu R i L L i i i

t tiu R i L L i i i

t ti

R i L L i i i

t t


9/146

Direct-axis equivalent circuit

di

ψ d

dd

td

dmdL

t

σ

d

d

DL

t

σ

d

d

f L

t

DR f R

+ +d D f i i i Di f i

σ dd

dLt

f u

( )

( )

( )

σ

σ

σ

ωψ ⎧ = − + + + +⎪⎪⎪

= + + + +⎨⎪⎪

= + + + +⎪⎩

d dd d

d d

d d

d d0 d d

ds q m Dd d d d f

f m D f f f f d f

DD D D m Dd f

iu R i L L i i it t

iu R i L L i i i

t t

iR i L L i i it t


10/146

More accurate equivalent circuit for direct axis

Assumption

where Lmd is the direct-axis magnetising inductance

di

ψ d

dd

td

dmdL

t

σ dd

DLt

σ dd

f Lt

DR R

+ +d D f i i i Di f i

σ

d

ddL

t f uσ d

dkL

t

= =


11/146

Quadrature-axis equivalent circuit

Notation

where Lmq is the quadrature-axis magnetising inductance

=qQ mqL L

ψ ωψ

ψ

⎧= + +⎪⎪

⎨⎪ = +⎪⎩

d

dd

0

d

qq s q d

QQ Q

u R it

R i

t

( )

( )

σ

σ

ψ

ψ

⎧ = + +⎪⎨⎪ = + +⎩

q q q m q Q

Q Q Q m q Q

L i L i i

L i L i i

( )

( )

σ

σ

ωψ ⎧

= + + + +⎪

⎪⎨⎪ = + + +⎪⎩

d d

d dd d

0d d

qq s q d q m q Q

QQ Q Q m q Q

iu R i L L i i

t ti

R i L L i it t


12/146

Quadrature-axis equivalent circuit

qi

ψ d

d

q

td

dmqL

t

σ dd

QLt

QR

+q Qi i Qi

σ

d

dqL

t

( )

( )

σ

σ

ωψ ⎧

= + + + +⎪⎪⎨

⎪ = + + +⎪⎩

d d

d dd d

0 d d

qq s q d q m q Q

QQ Q Q m q Q

iu R i L L i i

t ti

R i L L i it t


13/146

Notations and abbreviations

Leakage factors and other abbreviations

σ

σ

σ

σ

= −

= −

= −

= −

2

2

2

2

1

1

1

1

df df

d f

dDdD

d D

fD fD

f D

qQqQ

q Q

L

L L

L

L LL

L L

LL L

µ

µ

= −

= −

1

1

df fD f

dD f

dD fDD

Ddf

L L

L L

L L

L L


14/146

Synchronous machine in steady state

Steady state => a) Space vectors are constants in rotor frame

of reference

= + =

= + =

j constant

j constant

r s d q

r s d q

u u u

i i i

b) The time-derivatives of flux linkages vanish

ψ ψ = =

dd0

d d

qd

t t

c) The currents of the damper windings are zero

= = 0D Qi i


15/146

Space-vector diagram for

a synchronous generator

r si

r su

pu

ω df f L i

ω − q qL i

ω j d dL i

ωψ j r s r s sR i

f id

q

ψ r s

ϕ

δ

di

qi

( )ω − -j r q sdL L i

ω -j r q sL i

ωψ

ωψ

⎧ = −⎪

= +⎨

⎪ =⎩

s qd d

q s q d

f f f

u R i

u R i

u R i

ψ

ψ

= +⎧⎪⎨

=⎪⎩

d d d df f

q q q

L i L i

L i

ω ω

ω

= + − +

= =

j

ˆ j j

r r s s s q q pd d

p p df f

u R i L i L i u

u u L i

( )

( )

δ ϕ

δ ϕ

⎧ = +⎪⎨

= +⎪⎩

ˆ sin

ˆ cos

sd

q s

i i

i i


16/146

Electromagnetic torque

( ) ( )ψ ψ ⎡ ⎤= − = − +⎣ ⎦3 3

2 2e q q q q qd d d d df f T p i i p L L i i L i i

Neglecting the stator resistance

δ ω

δ ω

= = −⎧⎪⎨

= = +⎪⎩

sin

ˆcos

s q qd

q s pd d

u u L i

u u L i u

δ

ω

δ

ω

−⎧=⎪

⎪⎨⎪ = −

⎪⎩

ˆcos

sin

s pd

d

sq

q

u ui

L

ui

L

=>

( )( )δ δ δ

ω ω

δ δ

ω

⎡ ⎤− −⎢ ⎥= − −

⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞= − + −⎢ ⎥⎜ ⎟

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2 2

2

ˆ ˆ ˆ ˆ ˆcos sin sin3

2

ˆ ˆ ˆ3 1 1 sin sin 2

2 2

q s p sd p se

q qd

s p s

qd d

L L u u u u uT p

L L L

u u p u

X X X


17/146

Operator inductances of synchronous machine

Laplace transformation of the voltage and flux-linkage equations

ψ ψ ωψ

ψ ψ

ψ

ψ

ψ ψ ωψ

ψ ψ

⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟

⎝ ⎠⎪⎪ ⎛ ⎞

= + −⎪ ⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞

= + −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞

= + − +⎪ ⎜ ⎟⎝ ⎠⎪

⎪⎛ ⎞⎪ = + −⎜ ⎟

⎪ ⎝ ⎠⎩

0

0

0

0

0

0

0

ds qd d d

f f f f f

DD D D

qq s q q d

QQ Q Q

u R i s

s

u R i ss

R i s s

u R i ss

R i ss

ψ

ψ

ψ

ψ

ψ

⎧ = + +⎪⎪ = + +⎪⎪

= + +⎨⎪= +⎪

⎪= +⎪⎩

d d d df f dD D

f df d f f fD D

D dD d fD f D D

q q q qQ Q

Q qQ q Q Q

L i L i L i

L i L i L i

L i L i L i

L i L i

L i L i


18/146

ψ ψ

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

0 0

d dd d

d df dD f f

f df f fD f

DdD fDD DD D

ii

s sL L Li

L L L is s

L L L ii

s s

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

q qq q

q qQ

qQ QQ QQ Q

iiL Ls sL L i

is s

The flux differences expressed using the current differences

Typical initial values: = = =0

0 0 00; f

D Q f

f

ui i i

R


19/146

Quadrature-axis operator inductance

ψ ψ ψ ψ

⎛ ⎞= + − − = −⎜ ⎟

⎝ ⎠

0 00 =>

Q Q Q QQ Q Q Q

R iR i s

s s s

ψ ψ

⎛ ⎞ ⎛ ⎞⎛ ⎞− = − + = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0 00 =>

q qQ qQQ qQ q Q Q Q q

Q Q

i L s iL i L i i i

s s R L s s

( )ψ

ψ ⎛ ⎞⎛ ⎞ ⎛ ⎞

− = − − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠ ⎝ ⎠⎝ ⎠

20 0 0q qQ q q

q q q q q

Q Q

L s i iL i L s i

s R L s s s

Voltage equation for the damper winding

Flux-linkage equation for the damper winding

ψ

ψ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− = − + = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0 0 0 0q q q qQ q

q q q qQ Q q q qQ qQ Q

i i L s i

L i L i L i L is s s R L s s

=>

Flux-linkage equation for the stator winding


20/146

Quadrature-axis subtransient inductance II

( ) σ →∞ →∞

⎛ ⎞= = − = − =⎜ ⎟

⎜ ⎟+

⎝ ⎠

2 2'' lim lim

qQ qQq q q q qQ q

s s

Q Q Q

L s LL L s L L L

R L s L

At the beginning of a transient process

σ σ σ σ

σ

= + = ++ +'' 11 1

Q mqq s sQ mq

Q mq

L LL L LL L

L L

L’’q is called quadrature-axis subtransient inductance

( ) = −+

2qQ

q qQ Q

L sL s L

R L s


21/146

Quadrature-axis operator inductance III

( )

σ

+ −

= − = − =+

+ +

⎛ ⎞⎜ ⎟+ − +⎜ ⎟ +⎝ ⎠= = =

++ +

+

=+

2 2

2

2

''

''0

''''

''0

1 1

1

1 11 1

1

1

qQ qQQ Qq q

qQ Q Q Q Qq q q

Q QQ Q

Q Q

qQ Q Qq q qQ

Q Q qQq q

Q Q q

Q Q

qq

q

L LL Ls L L s s

L s R R L RL s L L

L LR L ss s

R R

L L LL L s s

L R T sRL L

L L T ss sR R

sT

Ls

T

where σ = = = =

'''' '' '' ''0 0 0 0;

qQq Q q qQ Q Q

Q q

LLT T T T T

R L

open-circuit short-circuittime constant time constant


22/146

Quadrature-axis operator inductance IV

( )

+

⎛ ⎞= = + = + −⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

''0

'' ''

'' '' ''

1

1 1 1 1 1 1

1 1 1

q

q q qq q

q q q

s

T A BsL s s sL LL L

s s s sT T T

An equation needed later for the inverse Laplace transformation


23/146

Short circuit at the terminals of a PM synchronous motor

B

H

r B

cH

Magnetic characteristicof the permanent magnets


24/146

-1000

-800

-600

-400

-200

0

200

400600

800

1000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

L i n e v o l t a g e s [ V ]

-15000

-10000

-5000

0

5000

10000

15000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

L i n e c

u r r e n t s [ A ]

Line voltage and current in a 3-phase short circuit


25/146

-350000

-300000-250000

-200000

-150000

-100000

-500000

50000

100000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

T o r q u e [ N m ]

Torque and minimum flux density in permanent magnets

-1.20

-1.00

-0.80

-0.60

-0.40

-0.20

0.00

0.20

0.40

0.60

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

M i n i m u m

f l u x d e n s i t y [ T ]


26/146

-12000

-10000

-8000

-6000

-4000

-2000

0

2000

4000

6000

800010000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

L i n e c

u r r e n t s [ A ]

-1000

-800

-600

-400

-200

0

200

400600

800

1000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

L i n e v o l t a g e s [ V ]

Line voltage and current in a 2-phase short circuit


27/146

-0.80

-0.60

-0.40

-0.20

0.00

0.20

0.40

0.60

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

M i n i m u m

f l u x d e n s i t y [ T ]

-150000

-100000

-50000

0

50000

100000

150000

0 50 100 150 200 250 300 350 400 450 500

Time [ms]

T o r q u e [ N m ]

Torque and minimum flux density in permanent magnets


28/146

Direct-axis operator inductance

Laplace transformed voltage and flux-linkage equations

ψ ψ ωψ

ψ ψ

ψ ψ

ψ ψ ωψ

ψ

ψ

⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟⎝ ⎠⎪

⎪ ⎛ ⎞= + −⎪ ⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= + −⎨ ⎜ ⎟

⎝ ⎠⎪⎪ ⎛ ⎞

= + − +⎪ ⎜ ⎟⎝ ⎠⎪

⎪ ⎛ ⎞⎪ = + −⎜ ⎟⎪ ⎝ ⎠⎩

0

0

0

0

0

0

0

ds qd d d

f f f f f

DD D D

qq s q q d

Q

Q Q Q

u R i ss

u R i ss

R i ss

u R i ss

R i s s

ψ ψ

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

0 0

d dd d

d df dD

f f f df f fD f

DdD fDD DD D

ii

s sL L L

iL L L is s

L L L ii

s s

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

q qq q

q qQ

qQ QQ QQ Q

iiL Ls s

L L iis s

d-axis flux and field winding current will be solved as functions of

the d-axis current and field winding voltage.


29/146

Direct-axis operator inductance II

( ) ( )

( ) ( )

ψ

ψ

⎛ ⎞⎛ ⎞

− = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

00 0

0 00

+

+

f d d

d d d d f

f f d f d d d f

ui

L s i G s us s s

i uii sG s i F s u

s s s

After eliminating the damper-winding current and field-winding flux linkage,

we get

Ld(s) is the direct-axis operator inductance

( ) ( )

⎧ ⎡ ⎤⎛ ⎞⎛ ⎞⎪⎢ ⎥⎜ ⎟= + − + −⎨ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎪ ⎝ ⎠⎣ ⎦⎩

⎫⎛ ⎞⎛ ⎞+ − ⎪⎜ ⎟⎜ ⎟+ − − ⎬⎜ ⎟⎜ ⎟

− ⎪⎝ ⎠⎝ ⎠⎭

22

0 0

2 2 22

00 2

1 1 1

2 1

df dD Dd d d f D d f d

D fD dD f df df fD dDD f d

D f D f fD

LLL s L sL T T D s L L L L

L L L L L L L Ls T T L

L L L L L


30/146

Direct-axis operator inductance and time constants

The denominator D(s) is

Using time constants, the direct-axis operator inductance can be written

( )( )( )

( )

( )( )

µ +=+ +

+=

+ +

00 ' ''

0 0

0 0

' ''0 0

1

1 1

1

1 1

df D Dd f

f d d

f Dd

f d d

L T sG s T L T s T s

T T sF s

L T s T s

( ) ( ) ⎛ ⎞

⎜ ⎟= + + + −⎜ ⎟⎝ ⎠

22

0 00 01 1 fD

D D f f D f

LD s s T T s T T

L L

( ) ( )( )

( )( )

+ +=

+ +

' ''

' ''0 0

1 1

1 1

d dd d

d d

T s T sL s L

T s T s

The other coefficient operators in the ψ d, i f equations are


31/146

Time constants( )

( )

( )

( )( )

( )( )

+ += =

+ +

' ''

' ''

0 0

1 1

1 1

d dd d

d d

T s T sN sL s L

D s T s T s

σ σ σ

− = ≈ + ≈ =

− = ≈ = =+

'00 0 0

1

00''00

2 00

1

1

f Dd f f

f

D f DDd fD fD fD

D D f

LT T T T

s R

T T LT T

s T T RSubtransient open-circuittime constant

Transient open-circuittime constant

Roots of the denominator

( ) ( ) ⎛ ⎞

⎜ ⎟= + + + − =⎜ ⎟⎝ ⎠

2

0 00 02 2

1 11 0

fDD D f f

D f

LD sT T T T

s L Ls s

( )

σ

⎡ ⎤+ ⎢ ⎥− = = ± −

⎢ ⎥+

⎢ ⎥⎣ ⎦

0 00 0

2

00

11 1 4

2

D D f f fD

D f

T T T T T

s

T T


32/146

Time constants

Roots of the numerator

σ σ σ σ

σ σ σ

σ σ σ σ σ

≈ + ≈ =

≈ ≈ ≈ =+

'0 0 0

''

00 '' '' '' ''0'' 0

0 0

f Dd dD df f df f df

f

dD fD f

D fD fDd d d d d Dd

D DdD df f d df d df d df

LT T T T

RL

T T T L L L T L L

T T T L L L R

Subtransientshort-circuittime constant

Transientshort-circuit

time constant

( ) ( )

( )

( )( )

( )( )

+ += =

+ +

' ''

' ''

0 0

1 1

1 1

d dd d

d d

T s T sN sL s L

D s T s T s

( ) ( )σ σ σ = + + + =''

0 00 02 2

1 10dD DdD df f fD f

dd

N s LT T T T

s Ls L s

( )

σ σ σ

σ σ

⎡ ⎤⎢ ⎥

+ ⎢ ⎥− = = ± −⎢ ⎥

+⎢ ⎥⎢ ⎥⎣ ⎦

''

000 0

2

0 0

11 1 4

2

dD fD f

DdD df f d

DdD df f

LT T

T T LT

s

T T


33/146

Direct-axis subtransient and transient inductances

Subtransient inductance is effective at the beginning of a transient process

( )→∞

+ −

= = − =−

2 2 ' ''

'' 2 ' ''0 0

2lim

DdD f df df fD dD d dd d d d

s D f fD d d

L L L L L L L T T L L s L L

L L L T T

σ −

= ≈ =⎛ ⎞

+ − −⎜ ⎟⎝ ⎠

' '' ''

'' '' '' 0

0 0 ''1

d d dd d d df d

ddd d d

d

T T T L L L L

T L

T T T L

Transient inductance


34/146

( )

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= = + +

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

' ''0 0

''

' '' ' ''

' '' '

' ''

1 1

1 1

1 1 1 1

1 1 1 1 1 1 1

1 1

d d

d d

d d d d

d dd d d

d d

s sT T A B C

sL s sLs s s s s

T T T T

sL LL L Ls sT T

An equation (Heaviside expansion) needed later for the

inverse Laplace transformation

Direct-axis operator inductance


35/146

Typical value s

Turbo-generators(with solid

rotor)

Salient-pole generatorswith damper winding

p < 8 p > 8

Salient-pole generatorswithout damper winding

p < 8 p > 8

Salient-polemotors

with damper

winding

xd (p.u.) 1.5–2.5 0.95–1.78 0.83–1.6 0.98–1.7 0.86–1.5 0.85–2.5

x 'd (p.u.) 0.15–0.35 0.15–0.37 0.23–0.34 0.2–0.35 0.25–0.4 0.22–0.56

x"d

(p.u.) 0.1–0.25 0.08–0.24 0.16–0.24 0.2–0.35 0.25–0.4 0.11–0.32

xq (p.u.) 1.2–2.3 0.46–0.91 0.57–0.89 0.52–0.9 0.45–0.8 0.5–1.5

x"q (p.u.) 0.1–0.25 0.08–0.26 0.17–0.25 0.52–0.9 0.45–0.8 0.11–0.32

T 'd0 (s) 5–15 2–10 4.2–10 2–10 1.5–8 1–7

T 'd (s) 0.6–2.0 0.4–2.5 1.0–2.0 0.5–2.5 0.55–2.5 0.2–1.5

T"d (s) 0.02–0.6 0.02–0.08 0.02–0.05 – – 0.004–0.06

Ta (s) 0.1–0.7 0.04–0.25 0.07–0.15 0.09–0.6 0.1–0.6 0.02–0.15

Reactances and time constants


36/146

Sudden three-phase short circuit

Currentmeasurement

Torquemeasurement

Grid

uf

if


37/146

Sudden three-phase short circuit

Calculation of short-circuit currents is needed for

• design and protection of electrical networks

• dimensioning of switch-gear

• design of electrical machines

Calculation of dynamic forces and torques is needed for

• mechanical dimensioning of the machine, its coupling andfoundations

Two-phase short circuits occur more often than three-

phase short circuits.

Three-phase short-circuit tests are used for determining

machine parameters.


38/146

No-load conditions before the short circuit

Three-phase short-circuit at t =0:

Speed of the rotor is assumed to be constant

Basic conditions

ωψ ω

ψ

= = = =

= = == =

0 0 0 0

0 0 0 0

0 0

0

ˆ0; 0

d q D Q

s q d df f

q d

i i i i

u u L iu

= =

= 0

0d q

f f f

u u

u R i

E ti f t i t l i


39/146

Equations for transient analysis

Laplace transformed voltage and flux-linkage equations

ψ ψ ωψ

ψ ψ

ψ ψ

ψ ψ ωψ

ψ ψ

⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟⎝ ⎠

⎪⎪ ⎛ ⎞= + −⎪ ⎜ ⎟

⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= + −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞

= + − +⎪ ⎜ ⎟⎝ ⎠⎪

⎪

⎛ ⎞⎪ = + −⎜ ⎟⎪ ⎝ ⎠⎩

0

0

0

0

0

0

0

ds qd d d

f f f f f

D

D D D

qq s q q d

QQ Q Q

u R i s

s

u R i ss

R i s s

u R i ss

R i ss

ψ ψ

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

0 0

d dd d

d df dD f f

f df f fD f

DdD fDD DD D

ii

s sL L L iL L L i

s sL L L i

i

s s

ψ ψ

ψ ψ

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

0 0

q qq q

q qQ

qQ QQ QQ Q

iiL Ls s

L L ii

s s

Equations II


40/146

( ) ( )

( )

( ) ( )

ψ ψ

ψ ψ

⎛ ⎞⎛ ⎞− = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞− = −⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

00 0

0 0

0 00

+

+

f d dd d d d f

q qq q q

f f d f d d d f

uiL s i G s u

s s s

iL s i

s si ui

i sG s i F s us s s

Equations II

Stator flux-linkages and excitation current as functions of stator currents

and excitation voltage


41/146

Modelling of the short circuit

Laplace transformed voltage and flux-linkage equations for

stator winding after substituting the zero initial values

ψ

ψ ωψ

ψ ωψ

⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎨

⎪ = + +⎩

0

0

0

d

s qd d

s q q d

R i s sR i s

( )( )

ψ

ψ

ψ

⎧ − =⎪⎨⎪ =⎩

0

d

d d d

q q q

L s isL s i

After substituting the fluxes

( )[ ] ( )

( ) ( )

ω

ω

⎧ + − =

⎪⎨ ⎡ ⎤+ + = −⎪ ⎣ ⎦⎩0

0

ˆ

s q qd d

ss q qd d

R sL s i L s i

uL s i R sL s i

s

S l ti f th t t t


42/146

Solution for the stator currents

whereD(s)

is the “characteristic polynomial”

( )

( )

( )

( )

ω ⎧= −⎪

⎪⎨

+⎪

= −⎪⎩

0

0

ˆ

ˆ

qsd

s ds

q

L sui

s D s

R sL su

i s D s

( ) ( )[ ] ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( )

ω

ω

⎡ ⎤= + + +⎣ ⎦

⎧ ⎫⎡ ⎤⎪ ⎪= + + + +⎨ ⎬⎢ ⎥

⎣ ⎦⎪ ⎪⎩ ⎭

2

22 21 1

s s q qd d

sq sd

qd d d

D s R sL s R sL s L s L s

RL s L s s sR

L s L s L s L s

Si lifi ti ( ) ( )'' ''li liL L L L


43/146

Simplifications

where = =⎡ ⎤ ⎡ ⎤+⎣ ⎦+⎢ ⎥

⎢ ⎥⎣ ⎦

'' ''

'' ''

'' ''

22

1 1

qda

s qds

qd

L LT

R L LR

L L

( ) ( ) ( )ω ω ⎧ ⎫⎛ ⎞⎪ ⎪

qd

aa

D s L s L s sT T

( )

ω

ω

= −⎡ ⎤⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

02

2

ˆ1

sd

da

ui

sL s sT

( ) ω

= −⎡ ⎤⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

02

2

ˆ1

sq

qa

ui

L s sT

( ) ( ) ( ) ω ⎧ ⎫

≈ + +⎨ ⎬⎩ ⎭

2 22qd

a

D s L s L s s sT

Equations for the currents

is the armature short-circuit time constant

( ) ( )→∞ →∞

= =lim , limq qd ds s

L L s L L s


44/146

Inverse Laplace transform for the currents

where

Inverse transformation of a product of two functions is obtained

using the convolution integral

( ) ( ){ } ( ) ( )−

⋅ = ∗ 11 2 1 2L f s f s f t f t

( ) ( ) ( ) ( ) ( ) ( )τ τ τ τ ∗ = − = −∫ ∫1 2 1 2 1 20 0d d

t t

f t f t f f t t f t f t

( ){ }( )

ω

ω

− − −

⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ ⎪ ⎪

= − ∗⎨ ⎬ ⎨ ⎬⎡ ⎤⎩ ⎭ ⎛ ⎞⎪ ⎪⎢ ⎥+ +⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

1 1 10 2

2

1 1ˆL L L

1d s

d

a

i s usL s

sT


45/146

Inverse Laplace transform for the currents II

ω

− − −⎡ ⎤⎛ ⎞⎛ ⎞

= − + − + − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

' ''

0 '' ' ''

1 1 1 1 1 1

ˆ e e e cos'ad dt T t T t T

sd d d d d d di u tX X X X X X

ω −= − 0''

ˆe sinat T sq

q

ui t

X

Current in stator phase a

ϑ ϑ ϑ ω ϑ = − = + 0cos sin ;sa r q r r r di i i t

After the convolution integration, the currents on the d- and q-axis


46/146

Current in phase a

Substituting the rotor angle gives

( )ω ϑ

ϑ

− −

− −

⎧⎡ ⎤⎪⎢ ⎥⎛ ⎞⎛ ⎞⎪

= − + − + − +⎢ ⎥⎨ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎪⎢ ⎥⎪⎣ ⎦⎩

⎛ ⎞ ⎛ ⎞− + − −⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

' ''

0 0'' '

III III

0'' '' '' ''

IV

1 1 1 1 1ˆ e e cos

'

1 1 1 1 1 1 e cos e

2 2

d d

a

t T t T sa s r

d d d d d

t T t T r

q qd d

i u tX X X X X

X X X X ( )ω ϑ

⎫⎪⎪

+ ⎬⎪

⎪⎭

0

V

cos 2a r t

ω ϑ

ω ϑ

− − −

−

⎡ ⎤⎛ ⎞⎛ ⎞= − + − + − −⎢ ⎥⎜ ⎟⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

+

' ''

0 '' ' ''

0''

1 1 1 1 1 1ˆ e e e cos cos

'

ˆ

e sin sin

ad d

a

t T t T t T sa s r

d d d d d d

t T sr

q

i u tX X X X X X

u

tX

ϑ ω ϑ = + 0r r t


47/146

Current in phase a

The components of the phase current

I steady-state short-circuit current

II transient component that is damped by the time constant

III subtransient component that is damped by the time constant

IV DC component that is damped by the time constant T a

V double-frequency component that is damped by the time constant T a

'dT

''

dT

Components of a phase current


48/146

0.00 0.10 0.20 0.30 0.40 0.50

Time [s]

C

u r r e n t

p p

I II

IV

VIII

Subtransient period


49/146

p

0.00 0.01 0.01 0.02 0.02

Time [s]

C

u r r e n t

I II

IV

VIII

Three phase short circuit; field current


50/146

Three-phase short circuit; field current

( )− = −0

f

f d d

ii sG s i

s

ω − −⎡ ⎤⎛ ⎞ ⎛ ⎞

= − − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

'/ /0 ' '

1+ 1 e 1 e cosadt T t T d d f f d d

X X i i t

X X

Substituting the d-axis current and doing the inverse transformation,

we obtain

Three-phase short circuit at the terminals


51/146

Before the short circuit, the generator

is operating at its rated operation point.

S = 8.2 MVA

cosfii = 0.8 capacitiveUs = 6300 V

f = 50 Hz

Uf = 150 V

A three-phase short-circuit occurs at

t =20 ms.

=> Us = 0

Uf = 150 V

of a 8.2 MVA synchronous generator

After the short circuit, the shaft torque is kept constant. The speed of

the rotor is defined by the equation of motion.

10000

Three-phase short circuit; 8.2 MVA generator


52/146

-10000

-8000

-6000

-4000

-2000

0

2000

4000

6000

8000

10000

0 20 40 60 80 100 120 140 160

Time [m s]

L i n e v o

l t a g e s

[ V ]

-15000

-10000

-5000

0

5000

10000

15000

20000

0 20 40 60 80 100 120 140 160

Time [ms]

L i n e

c u r r e n

t s [ A ]

Three-phase short circuit; 8.2 MVA generator


53/146

0

200

400

600

800

1000

1200

1400

1600

1800

0 20 40 60 80 100 120 140 160

Time [ms]

F i e l d c u r r e n t

[ A ]

p ; g

600000

Three-phase short circuit; 8.2 MW generator


54/146

-1000000

-800000

-600000

-400000

-200000

0

200000

400000

600000

0 20 40 60 80 100 120 140 160

Time [ms]

T o r q u e [ N m ]

74

75

76

77

78

79

80

81

82

8384

0 20 40 60 80 100 120 140 160

Time [ms]

S p e

e d [ r a d / s ]

Two-phase short circuit at the terminals of


55/146

a 8.2 MVA synchronous generator

A two-phase short circuit is most

severe, if it occurs at a time instant

when the voltage between the

shorted lines is zero. At this instant,

the stator flux associated with the

shorted branch has a maximum

value. In the short circuit, the flux

“freezes” to this maximum value

and the rotor rotates in a static,slowly decaying flux component.

Two-phase short circuit; 8.2 MVA generator

10000


56/146

-10000

-8000

-6000

-4000

-2000

0

2000

40006000

8000

10000

0 20 40 60 80 100 120 140 160

Time [ms]

L

i n e v o

l t a g e s

[ V ]

-20000

-15000

-10000

-5000

0

5000

10000

15000

20000

0 20 40 60 80 100 120 140 160

Time [m s]

L i n e c u r r e n

t s [ A ]

Two phase short circuit; 8 2 MVA generator


57/146


0

200

400

600

800

1000

1200

1400

1600

0 20 40 60 80 100 120 140 160

Time [ms]

F i e l d c u r r e n

t [ A ]

1000000



58/146

-1500000

-1000000

-500000

0

500000

0 20 40 60 80 100 120 140 160

Time [m s]

A i r - g a p t o r q u e [ N m ]

73

74

75

7677

78

79

80

81

82

83

0 20 40 60 80 100 120 140 160

Time [ms]

R o t a t i o n s p e e d [ r a d / s ]

Asynchronous operation of h hi


59/146

synchronous machines

( )ω

ω = − ≠1 ; 0s M S S p

Currentmeasurement

Torquemeasurement

Grid

u Rif f=

if

R

Asynchronous operation of a synchronous machine


60/146

y p y

Asynchronous operation

• After a drop out of synchronism caused, for instance,by an over load or loss of excitation

• A self-starting synchronous motor when starting

Typically, the field winding is either short-circuited

or connected in series with a resistor.

= 0 f u

Stator voltage in rotor frame of reference


61/146

( )ω ϕ += j

ˆ e s uts

s su u

( )γ ω γ = − + 01 sS t

( )ϕ γ ω −= 0 j jˆ e eu sr S ts su u

( )

( )

ω ϕ γ

ω ϕ γ

= + −⎧⎪⎨

= + −⎪⎩

0

0

ˆ cos

ˆ sin

d s s u

q s s u

u u S t

u u S t

( )ϕ γ ω −= 0 j jˆ e eu sS tsu u=⎧

⎨ = −⎩ jd

q

u u

u u

Constant slip

Stator voltage in stator and rotor frames of reference

d- and q-components

Complex notation (phasors)

Laplace transformed voltage equations andthe definition of the operator inductances


62/146

the definition of the operator inductances

ψ ψ ωψ

ψ ψ ωψ

⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨

⎛ ⎞⎪ = + − +⎜ ⎟⎪ ⎝ ⎠⎩

0

0

ds qd d d

q

q s q q d

u R i ss

u R i s s

( ) ( )

( )

ψ ψ

ψ ψ

⎧ ⎛ ⎞⎛ ⎞− = − −⎪ ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎪ ⎝ ⎠⎨⎛ ⎞⎪ − = −⎜ ⎟⎪⎝ ⎠⎩

00 0

0 0

+ f d d

d d d d f

q qq q q

uiL s i G s u

s s si

L s is s

The initial values of the flux linkages are put to zero, and theLaplace variable s is replaced by jSω s. Field winding is assumedto be short-circuited, u f = 0

Solution of stator fluxes (Rs = 0)


63/146

( )

( )

ω ψ ω ψ

ω ψ ω ψ

= = + − −⎧⎪⎨

= − = + + −⎪⎩

j 1

j j 1

d s d s d s q

q s q s q s d

u u R i S S

u u R i S S

≈ 0sR

Voltage equations

Solution of flux linkages after neglecting the stator resistance

( )

( )

ω ψ ω ψ

ω ψ ω ψ

= − −⎧⎪⎨

= − −⎪⎩

j 1

j 1

s s qd

s s qd

u S S

u S S =>

( )

( )

( )

( )

ω ψ

ω ω

ω ψ

ω ω

−⎧= = −⎪ −⎪

⎨ − −⎪ = = −⎪ −⎩

2

2

1 2 j

j 1 2

j 1 2

j 1 2

sd

ss

sq

ss

S uu

S

S uu

S

( )

ψ ω ϕ γ ω

ψ ω ϕ γ ω

π⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + −⎪⎩

0

0

ˆ cos2

ˆsin

s s uds

sq s u

s

u S t

uS t

Instantaneous flux linkages

( )

( )

ω ϕ γ

ω ϕ γ

= + −⎧⎪⎨

= + −⎪⎩

0

0

ˆ cos

ˆ sin

d s s u

q s s u

u u S t

u u S t =>

Solution of stator currentsC t t f th fl li k


64/146

( ) ( )

( ) ( )

ϕ

ϕ

ω ω ω ω

ω ω ω ω

π⎛ ⎞− +⎜ ⎟⎝ ⎠

−

⎧⎪ = − =⎪⎨⎪ = − = −⎪⎩

j2

j

je

j j

e j j

d

q

ds d s s d s

q s q s s q s

u ui

L S L S

u u

i L S L S

( )

( ) ( )

ω ϕ ϕ γ ω ω

ω ϕ ϕ γ

ω ω

π⎧ ⎛ ⎞= + − − −⎜ ⎟⎪ ⎝ ⎠⎪

⎨⎪ = − + − −

⎪⎩

0

0

ˆcos

2 j

ˆcos

j

sd s u d

s d s

sq s u q

s q s

ui S t

L S

ui S t

L S

( )

( )

ψ ω

ψ ω

⎧ =⎪⎨

=⎪⎩

j

j

d d s d

q q s q

L S i

L S i =>

Where are the arguments of the operator inductances.

Instantaneous currents

Current components from the flux linkages

A phase current

γ γ = −cos sinsa d qi i i

ϕ ϕ qd and

Phase current


65/146

( ) ( )

( ) ( ) ( )

ω ϕ ϕ γ ω γ ω ω

ω ϕ ϕ γ ω γ

ω ω

π⎛ ⎞

= + − − − − +⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠

π⎡ ⎤+ + − − − + −⎢ ⎥⎣ ⎦

0 0

0 0

ˆ

cos cos 12 j

ˆ cos cos 1

2 j

ssa s u d s

s d s

ss u q s

s q s

ui S t S tL S

uS t S t

L S

( ) ( )⎡ ⎤= − + +⎣ ⎦1

cos cos cos cos2

x y x y x y

( ) ( )

( )

( )

ω ϕ ϕ ω γ ϕ ϕ ω ω

ω ϕ ϕ ω γ ϕ ϕ

ω ω

π π⎧ ⎫⎛ ⎞ ⎡ ⎤= + − − + − + − + +⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭

π π⎧ ⎫⎛ ⎞ ⎡ ⎤+ + − − + − + − + −⎨ ⎬⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦⎩ ⎭

0

0

ˆcos cos 1 2 2

2 22 j

ˆ cos cos 1 2 2

2 22 j

ssa s u d s u d

s d s

ss u q s u q

s q s

ui t S t

L S

ut S t

L S

Modification using the identity

The phase current

is of the form ( ) ( )ω ϕ ω ϕ = + + − +⎡ ⎤⎣ ⎦1 1 2 2ˆ ˆcos cos 1 2sa sa s sa si i t i S t

A closer study of the phase current components


66/146

( ) ( )

ϕ ω

ω ω ω

π⎛ ⎞−⎜ ⎟⎝ ⎠

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦

j j2

1

ˆ 1 1e e

2 j j

usts

sas s q sd

ui

L S L S

( ) ( )( )

ϕ γ ω

ω ω ω

π⎛ ⎞− −⎜ ⎟ − −⎝ ⎠⎡ ⎤

= −⎢ ⎥⎢ ⎥⎣ ⎦

0 j 2* j 1 22 2

ˆ 1 1e e

2 j j

usS ts

sas s q sd

ui

L S L S

( ) ( ) ( )

( ) ( ) ( )[ ]

ω ϕ ω ω ω

ω ϕ ω ω ω

= + +

+ − − +

1

2

ˆ 1 1cos2 j j

ˆ 1 1 cos 1 2

2 j j

ssa ss s q sd

ss

s s q sd

ui tL S L S

uS t

L S L S

Fundamental frequency component in phasor notation

Complex conjugate of the other current phasor

Phase current

Electromagnetic torque


67/146

( )ψ ψ = −32e q qd dT p i i

Flux linkages

Currents

( )

ψ ω ϕ γ ω

ψ ω ϕ γ ω

π⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + −⎪⎩

0

0

ˆ cos2

ˆsin

ss ud

s

sq s u

s

u S t

uS t

( )

( ) ( )

ω ϕ ϕ γ ω ω

ω ϕ ϕ γ

ω ω

π⎧ ⎛ ⎞= + − − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + − −⎪⎩

0

0

ˆcos

2 j

ˆcos

j

s

d s u ds d s

sq s u q

s q s

ui S t

L S

ui S t

L S

Electromagnetic torque II


68/146

( )

( )

( )

ψ ψ

ϕ ω ϕ γ ϕ

ω ω

ϕ ω ϕ γ ϕ

ω ω

= −

π π⎧ ⎫⎛ ⎞ ⎡ ⎤= + + + − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭

π π⎧ ⎫⎛ ⎞ ⎡ ⎤− − + + − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭

2

02

2

02

3

2

ˆ3 cos cos 2 2 2

2 24 j

ˆ3 cos cos 2 2 2

2 24 j

e q qd d

ss ud d

s sd

sq s u q

s q s

T p i i

puS t

L S

puS t

L S

( ) ( ) ( ) ( )

ϕ ϕ

ω ω ω ω ω ω

π π⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎡ ⎤⎢ ⎥⎝ ⎠ ⎝ ⎠= + = +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥⎣ ⎦

2 2

1 2 2

cos cosˆ ˆ3 3 1 12 2

Im j j j4 4 j

qds s

es q sdsds sq s

pu puT

L S L SL S L S

( )ω ϕ = + +1 2ˆ cos 2e e e s T T T T S t

The torque is of the form

The average torque is

Operator inductances and slip frequency


69/146

( )

⎛ ⎞= + −⎜ ⎟

⎜ ⎟ ⎛ ⎞⎝ ⎠ +⎜ ⎟⎜ ⎟⎝ ⎠

''

''

1 1 1 1 1

1q q qq

q

sL s sL LLs

T

( )

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

' '' '

' ''

1 1 1 1 1 1 1 1

1 1d d dd d d

d d

sL s sL LL L Ls s

T T

( )

ω ω

ω ω ω

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

' '' '

' ''

j j1 1 1 1 1 1

j 1 1 j j

s s

sd d dd d d

s sd d

S S

L S L LL L LS S

T T

( )

ω

ω

ω

⎛ ⎞⎜ ⎟= + −⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎜ ⎟+

⎜ ⎟⎝ ⎠

''

''

j1 1 1 1

j 1 j

s

q s q qqs

q

S

L S L LLS

T

Direct axis

Quadrature axis

=>

=>

⎡

The average electromagnetic torque


70/146

ω ω ω

ω ω

ω ω

⎡

⎢ ⎛ ⎞ ⎛ ⎞= − + −⎢ ⎜ ⎟ ⎜ ⎟⎢ ⎝ ⎠ ⎝ ⎠+ +⎢⎣

⎤

⎥⎛ ⎞ ⎥+ −⎜ ⎟⎜ ⎟ ⎥+⎝ ⎠ ⎥

⎦

2 ' ''1 2 ' '' '' ''

' ''

''

'' ''''

ˆ3 1 2 1 21 11 18

1 2 1

1

s d de

ds d d ds sd ds sd d

q

q q s qs q

pu L LT LL L LS T S T

S T S T

L

L L S T S T

= + +

+ + +

1 I II IIII II III

I II III

2 2 2e p p p

p p p

p p p

T T T T S S SS S S

S S S S S S

ω ω ω

= = =I II III' '' ''1 1 1

; ; ; p p p

s d s d s q

S S S

T T T

consists of three terms similar to the steady-state torque of aninduction machine

where

Average torque in asynchronous operation


71/146

a) Total torque;Stator resistance is

taken into account

b) Total torque;

Stator resistanceis neglected

c) Torque caused by the

damper winding

d) Torque caused by thefield winding

Influence of an external resistance connectedin series with the field winding


72/146

Without external resistance With an external resistance

The pulsating torque


73/146

( ) ( )

ϕ γ ω

ω ω ω

π⎛ ⎞− −⎜ ⎟⎝ ⎠

⎡ ⎤= −⎢ ⎥

⎢ ⎥⎣ ⎦

02 j 2 2

j22 2 2

ˆ3 1 1e e

j j4

usS ts

es q sds

puT

L S L S

ω = = −

2

2 2̂3 1 1ˆ0 =>4

se

qds

puS T L L

The reason for the pulsating torque is the saliency of the rotor.

If

At the synchronous speed, this component gives the reluctance torque.

Its peak value is

( ) ( )ω ω = => =2ˆ j j 0s q s edL S L S T

Amplitude of the pulsating torqueas a function of slip


74/146

Fast and slow starting

''


75/146

a) Starting time

b) Starting time ≈ ≈'100 100 sdT

≈ ≈''

10 0.5 sdT

Starting of a synchronous motor


76/146

Currentmeasurement

Torque and speedmeasurement

u Rif f=

if

R

Grid

2.5 MW synchronous motor


77/146

Rated values:

U s = 5500 VI s = 290 A

P = 2.5 MW f s = 50 Hz

When starting:

T shaft = 0Rext = 9 R f

4000

6000Starting of a 2.5 MW synchronous motor


78/146

-8000

-6000

-4000

-2000

0

2000

0 100 200 300 400 500

Time [ms]

L i n e c u r r e n t s [ A ]

-800

-600

-400

-200

0

200

400

600

0 100 200 300 400 500

Time [ms]

F i e

l d c u r r e n t [ A ]

150000

200000

Starting of a 2.5 MW synchronous motor


79/146

-150000

-100000

-50000

0

50000

100000

0 100 200 300 400 500

Time [ms]

A

i r - g a p t o r q u e [ N m ]

0

10

20

30

40

50

60

70

0 100 200 300 400 500

Time [ms]

R o t a t i o

n s p e e d [ r a d / s ]

Starting of a 2.5 MW synchronous motor Space vector of stator current


80/146

-7500

-5000

-2500

0

2500

5000

7500

-7500 -5000 -2500 0 2500 5000 7500

Re [A]

I m [

A ]

Asynchronous operation of a 8.2 MVAsynchronous generator


81/146

Rated values:

U s = 6300 V f s = 50 Hz

I s = 940 AI f = 215 A

S = 8.2 MVA

Under fault:

U f = 0 V T shaft = T N

83

84

85

d / s ]

Asynchronous operation of a synchronous generator


82/146

75

76

77

78

79

80

81

82

0.0 0.5 1.0 1.5 2.0 2.5

Time [ms]

R o t a t i o n s p e e d [ r a d

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

0.0 0.5 1.0 1.5 2.0 2.5

Time [ms]

R o t o r a n g l e [ e l . r a d ]

Time [s]



83/146

-300000

-250000

-200000

-150000

-100000

-50000

0

50000

0.0 0.5 1.0 1.5 2.0 2.5

Time [s]

A i r - g a p t o r q u e [ N

m ]

3000

4000

5000



84/146

-5000

-4000-3000

-2000

-1000

0

1000

2000

0.0 0.5 1.0 1.5 2.0 2.5

Time [s]

L i n e c u r r e n t [ A ]

-500

-400

-300

-200

-100

0

100

200

300400

500

0.0 0.5 1.0 1.5 2.0 2.5

Time [s]

F i e

l d c u r r e n t [ A ]



85/146

Operating characteristics

Terminal voltage 6300. V

Supply frequency 50.0 Hz Terminal current 1652.1 A Peak current 3731.7 A Power factor -0.4784 Ind.

Slip-ring voltage 0.00 V Slip-ring current 196.3 A

Rotational speed 755.3 1/min Air-gap torque -99890 Nm

Equations of synchronous machine usingload angle


86/146

Load angle δ is the angle between the

excitation voltage vector (on q-axis)

and stator voltage vector.

In a balanced stiff three-phase network,the space vector of stator voltage is

The angle of the rotor presented in stator frame of reference is

The angular speed of the rotor, its derivative and slip are

( )ω ϕ += jˆ e s uts su u

γ ω ϕ δ π

= + + −2

s ut

γ δ ω γ δ ω ω δ ω ω

ω ω

−= = + = = = = −

2 2

2 2

d d d d d 1 d; ;

d d d dd ds

ss s

St t t tt t

su

du j quγ

δ

q

d

β

α

Equations of synchronous machine usingload angle


87/146

δ

δ

=⎧⎪⎨ =⎪⎩

ˆ sin

ˆ cos

d s

q s

u u

u u

The equation of motion becomes

and the voltage equations are

The dq-components of the balanced three-phase voltage are

( ) δ

ψ ψ − = +2

2

3 d

2 dd q q d m

J p i i T

p t

ψ δ ω ψ ψ

ψ δ ω ψ ψ

⎧ = + − −⎪⎪⎨⎪ = + + +⎪⎩

d d

d dd d

d d

dd s d s q q

qq s q s d d

u R i

t t

u R it t

Small-signal analysis for synchronous machines

A l t i l hi i ld i l t d t t Th i l


88/146

An electrical machine is seldom in a real steady state. There is alwayssome variation in the shaft torque and supply voltage that causes

disturbances to the operation of the machine. If the variations are small in

amplitude, equations can be linearised to the steady state operation point.

The variables are presented as a sum of a steady state value and

small variation, for instance

The voltage equations for the steady-state variables are

∆ ω ω ∆ω = + = +0 ;d d d si i i

ω ψ

ω ψ

= −⎧⎪⎨

= +⎪⎩

0 0 0

0 0 0

d s d s q

q s q s d

u R i

u R i

Equations for the voltage variations

F th di t b d i bl th lt ti


89/146

For the disturbed variables, the voltage equations are

Subtracting the two sets of equations and assuming that the products

of two small variations can be neglected, we get the equations for

the variations

( ) ( )( )

( ) ( ) ( )

∆ψ ∆ ∆ ω ∆ω ψ ∆ψ

∆ψ

∆ ∆ ω ∆ω ψ ∆ψ

⎧+ = + + − + +⎪⎪

⎨⎪

+ = + + + + +⎪⎩

0 0 0

0 0 0

d

dd

d

dd d s d d s q q

q

q q s q q s d d

u u R i it

u u R i i t

∆ψ ∆ ∆ ω ∆ψ ψ ∆ω

∆ψ ∆ ∆ ω ∆ψ ψ ∆ω

⎧= + − −⎪

⎪⎨⎪ = + + +⎪⎩

0

0

d

dd

d

dd s d s q q

qq s q s d d

u R i

t

u R it

Laplace transformation of the voltage variations

The equations are linear and easy to Laplace transform


90/146

The equations are linear and easy to Laplace transform

The corresponding flux-linkage equations are

Combining the equations, we get

∆ ∆ ∆ψ ω ∆ψ ψ ∆ω

∆ ∆ ∆ψ ω ∆ψ ψ ∆ω

⎧ = + − −⎪⎨

= + + +⎪⎩

0

0

d s d d s q q

q s q q s d d

u R i s

u R i s

( ) ( )

( )

∆ψ ∆ ∆

∆ψ ∆

⎧ = +⎪⎨

=⎪⎩

d d d f

q q q

L s i G s u

L s i

( ) ( ) ( )

( ) ( ) ( )

∆ ∆ ω ∆ ∆ ψ ∆ω

∆ ∆ ω ∆ ω ∆ ψ ∆ω

⎧ = + − + −⎡ ⎤⎣ ⎦⎪⎨

⎡ ⎤= + + + +⎪ ⎣ ⎦⎩

0

0

d s d d s q q f q

q s q q s d d s f d

u R sL s i L s i sG s u

u R sL s i L s i G s u

Variation of the torque ( p=1)

The equation for the varying torque is


91/146

The equation for the varying torque is

The corresponding steady-state torque equation is

Subtracting the two equations, we get

( )ψ ψ − =0 0 0 0 03

2 d q q d mi i T

∆ω ψ ∆ ∆ψ ψ ∆ ∆ψ ∆⎡ ⎤+ − − − =⎣ ⎦0 0 0 03 d2 d

d q q d q d d q mi i i i T J t

( )( ) ( )( ) ∆ω

ψ ∆ψ ∆ ψ ∆ψ ∆ ∆⎡ ⎤+ + − + + = + +⎣ ⎦0 0 0 0 0

3 d

2 dd d q q q q d d m mi i i i J T T

t

Variation of the torque II

The Laplace transformation of the torque variation is


92/146

Substituting the flux-linkage equations, the torque becomes

ψ ∆ ∆ψ ψ ∆ ∆ψ ∆ ∆ω ⎡ ⎤+ − − − =⎣ ⎦ 0 0 0 0

3

2 d q q d q d d q mi i i i T sJ

( ) ( ) ( )

{ }ψ ∆ ψ ∆ ∆ ∆ ∆ω ⎡ ⎤ ⎡ ⎤− − − + − =

⎣ ⎦ ⎣ ⎦

0 0 0 0 0

3

2 d d q q q q d d q f mi L s i i L s i i G s u T sJ

Variation of the voltage components

If the machine is connected to a balanced stiff line voltage, the variation

of the load angle causes a variation in the voltage components


93/146

of the load angle causes a variation in the voltage components

∆ ∆δ δ ∆δ

∆ ∆δ δ ∆δ

= =⎧⎪⎨

= − = −⎪⎩

0 0

0 0

ˆ cos

ˆ sin

d q s

q d s

u u u

u u u

( )∆ δ ∆δ δ ∆δ δ ∆δ

∆δ ∆δ ∆δ

+ = + = +

= + ≈ +

0 0 0 0

0 0 0 0

ˆ ˆ ˆsin sin cos cos sin

cos sin

d d s s s

d q d q

u u u u u

u u u u

( )∆ δ ∆δ δ ∆δ δ ∆δ

∆δ ∆δ ∆δ

+ = + = −

= − ≈ −

0 0 0 0

0 0 0 0

ˆ ˆ ˆcos cos cos sin sin

cos sin

q q s s s

q d q d

u u u u u

u u u u

Forced variations of a synchronous machine

The variations are often periodic. This is the case, for instance, for the

shaft torque of a diesel driven generator For sinusoidal disturbance


94/146

shaft torque of a diesel driven generator. For sinusoidal disturbance

( ) ( )λ ϕ ∆ ∆ ∆ += + = j0 ˆRe where e idt

d d d d di i i i i

( )∆ λ ϕ = + +0 ˆ cosm m m T T T T t

If the sinusoidal disturbance in the torque is small, the machine behaves

linearly and the other parameters will also vary sinusoidally. We can use

phasor variables, for instance

The Laplace transformed equations derived previously can be changesto phasor equations by replacing the Laplace variable s with thefrequency jλ . The variation of the rotation speed can be written

γ δ δ ω ω ∆ω ω ω ∆ω λ∆δ = = + = − = =

d d d => => j

d d ds s

t t t

Voltages and flux linkages

The voltage equations, assuming that u f

is constant and Rs

=0, are


95/146

f

The steady-state flux linkages (Rs=0) and voltages are related as

( ) ( )

( ) ( )

∆ λ λ ∆ ω λ ∆ ψ λ∆δ

∆ λ λ ∆ ω λ ∆ ψ λ∆δ

⎧ = − −⎪⎨

= + +⎪⎩

0

0

j j j j

j j j j

d d d s q q q

q q q s d d d

u L i L i

u L i L i

ψ ω

ψ ω

⎧ =⎪⎪⎨⎪ = −⎪⎩

00

00

qd

s

dq

s

u

u

∆ ∆δ

∆ ∆δ

=⎧⎪⎨

= −⎪⎩

0

0

d q

q d

u u

u u

On the other hand

Current phasor

Combining the three previous equations

( ) ( )∆δ λ λ ∆ λ ∆ λ∆δ⎧ 0j j j jduL i L i


96/146

( ) ( )

( ) ( )

∆δ λ λ ∆ ω λ ∆ λ∆δ ω

∆δ λ λ ∆ ω λ ∆ λ∆δ ω

⎧ = − +⎪⎪⎨⎪− = + +⎪

⎩

00

00

j j j j

j j j j

dq d d s q q

s

qd q q s d d

s

uu L i L i

uu L i L i

( ) ( )

( ) ( )

λ ∆δ λ λ ∆ ω λ ∆

ω

λ ∆δ ω λ ∆ λ λ ∆

ω

⎧⎛ ⎞− = −⎪⎜ ⎟

⎝ ⎠⎪⎨

⎛ ⎞⎪− + =⎜ ⎟⎪

⎝ ⎠⎩

00

00

j j j j

j j +j j

dq d d s q q

s

qd s d d q qs

uu L i L i

uu L i L i

=>

From which we can solve the currents

( )

( )

∆ ∆δ ω λ

∆ ∆δ ω λ

⎧ = −⎪⎪⎨⎪ = −⎪⎩

0

0

j

j

dd

s d

qq

s q

uiL

ui

L

Equation of motion in phasor form

The equation of motion in phasor form is

3


97/146

After substituting the fluxes and currents

( )

( )

∆ ∆δ ω λ

∆ ∆δ ω λ

⎧ =⎪⎪⎨

⎪ = −⎪⎩

0

0

j

j

dd

s d

qq

s q

ui

L

uiL

( ) ( ){ }ψ λ ∆ ψ λ ∆ ∆ λ ∆δ ⎡ ⎤ ⎡ ⎤− − − − = −⎣ ⎦ ⎣ ⎦2

0 0 0 03

j j2

d d q q q q d d mi L i i L i T J

( ) ( ) ∆δ ∆ λ ∆δ

ω ω λ ω λ

⎡ ⎤⎢ ⎥− + − − − = −⎢ ⎥⎣ ⎦

2 20 20

0 0 0 03

2 j j

q dq d d q m

s s q s d

u uu i u i T J

L L

ψ ω

ψ ω

⎧=⎪

⎪⎨

⎪ = −⎪⎩

00

00

qd

s

dq

s

u

u

we get

Equation of motion in phasor form II

The term ( )−

0 0 0 0

3

2 q d d qu i u i


98/146

can be expressed with the aid of the steady-state reactive power

The torque equation becomes

( )

( ) ( )

( ) ( )

( ) ( )

δ δ

δ δ φ δ δ φ

δ φ δ δ φ δ

− = −

⎡ ⎤= + − +⎣ ⎦

= + − +⎡ ⎤⎣ ⎦

0 0 0 0 0 0 0 0

0 0 0 0

0 0 0 0

3 3ˆ ˆcos sin

2 23 ˆ ˆˆ ˆ cos sin sin cos23 ˆˆ sin cos cos sin

2

q d d q s d s q

s s s s

s s

u i u i u i u i

u i u i

u i

( )δ φ δ φ = + − = =⎡ ⎤⎣ ⎦0 0 03 3ˆ ˆˆ ˆ sin sin2 2

s s s su i u i Q

( ) ( )λ ∆δ ∆

ω ω λ ω λ

⎡ ⎤⎢ ⎥+ − − =⎢ ⎥⎣ ⎦

2 202 0 0

2 2

3 3

2 2 j j

q dm

s s q s d

uQ u J T

L L

To shorten the notation, a ‘complex synchronisation coefficient’

is defined

Complex synchronisation coefficient


99/146

( ) ( )( ) ( )

λ λ λ ω ω λ ω λ

⎡ ⎤+ = − − −⎢ ⎥

⎢ ⎥⎣ ⎦

2 20 0

01 3 3

j2 j 2 j

q ds d

s s q s d

u uK K Q

L L

and the equation of motion becomes

( ) ( )λ λ λ λ ∆δ ∆⎡ ⎤− + = −⎣ ⎦2

js d mK J K T

where K d represents damping and K s∆δ is the synchronising torque.

K d

and K s

are frequency-dependent but their values can be

calculated for a certain excitation frequency.

Complex synchronisation coefficient II

( ) ( )δ δ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= − + +⎨ ⎬ ⎨ ⎬

2 22 200 02 2

ˆ ˆ3 1 3 1cos Re sin Res ss Q u uK


100/146

( ) ( )δ δ

ω λ λ ω ω ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ + +⎨ ⎬ ⎨ ⎬

⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭0 02 2

cos Re sin Re2 j 2 j

ss q ds s

QK L L

( ) ( )δ δ λ λ λ ω

⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭⎣ ⎦

2

2 20 02

ˆ3 1 1cos Im sin Im2 j j

sdq ds

uK L L

In normal operation,

This means that the properties of the damper winding on the

quadrature axis are important from the point of view of damping.

δ δ δ < >>0 2 2

0 0 030 => cos sin

Numerical study of a synchronous generator using small forced vibrations


101/146

Rated values S = 8.2 MVA

f s = 50 HzU

s

= 6300 VI s = 750 A

cosϕ = 0.80I f = 215 A

Ω Ω Ω λ = +0 ˆ cosm m m t

Forced oscil lation of the mechanical rotation

speed

100

Forced oscil lation of speed and the resultingoscillation in air-gap torque (λ = 50 Hz)


102/146

0

20

40

60

80

300 350 400 450 500

Time [ms]

S p

e e d [ r a d / s ]

-140000

-120000-100000

-80000

-60000

-40000

-20000

0

300 350 400 450 500

Time [ms]

T

o r q u e [ N m ]

Power transfer in the forced oscillation

Variation of speed: λ λ Ω Ω λ =ˆ cos t


103/146

Mass of inertia:

Viscous damping:

Spring constant:

Total torque:

Instantaneous power:

λ

ϕ Ω λ λ = = −

2

J 2

d ˆ sind

T J J tt

λ λ λ

Ω Ω λ λ Ω λ λ

λ = − + +tot

ˆˆ ˆsin cos sinT J t d t k t

λ

λ λ λ λ

Ω

Ω Ω λ Ω λ λ Ω λ λ

λ

=

⎛ ⎞= − + +⎜ ⎟

⎝ ⎠

tot

ˆˆ ˆ ˆ cos sin cos sin

P T

t J t d t k t

λ

ϕ Ω λ = =d

d ˆ cosd

T d d tt

λ Ω

ϕ λ λ = =

k

ˆ

sinT k k t

Average power of the forced oscillation

100

Instantaneous harmonic power (λ = 50 Hz)


104/146

-150

-100

-50

0

50

100

300 350 400 450 500Time [ms]

H a r m o n i c p

o w e r [ k W ]

λ λ λ λ

λ

Ω Ω λ Ω λ λ Ω λ λ

λ

Ω

⎛ ⎞= − + +⎜ ⎟

− ⎝ ⎠

=

∫ave

2

ˆ1 ˆ ˆ ˆcos sin cos sin d

1 ˆ 2

b

a

t

ab t

P t J t d t k t t

t t

d

Is the harmonic power always negative,i.e. is the damping always positive?

If th h i i ti 12


105/146

If the harmonic power is negative,

the machine takes mechanical

power from the shaft forcing the

oscillation. There is damping thattries to reduce the oscillation.

If the harmonic power is positive,

the system tends to increase the

oscillation and the system may beunstable.

From the harmonic power, we can

calculate the damping coefficient

for the system, which should bepositive.

-2

0

2

4

6

8

10

12

0 20 40 60 80 100

Excitation frequency [Hz]

D a m p i n

g c o e f f i c i e n t [ k N m s ]

Damping coefficient after reducing theresistance of the stator winding to zero

12


106/146

-2

0

2

4

6

8

10

0 20 40 60 80 100

Excitation frequency [Hz]

D a m p i n g

c o e f f i c i e n t [ k

N m s ]

Mechanical model for torsional vibrationsof a diesel-generator


107/146

Θ Θ Θ Θ Θ Θ

Θ Θ Θ Θ

+ −−

+ − −

− −+ + +

+ − + − =

2 n n 1 n n 1n nn n n n 12

n n n 1 n 1 n n 1 n

( ) ( )

( ) ( ) ( )

d dd d J B C C dt dt dtdt

K K T t

Equation of motion for a mass of inertia

Synchronousgenerator

Coupling Diesel engine

Torsional vibrations

Θ

2d

Explanations for the terms in the torque balance equation

Inertia torque of mass


108/146

Θ 2

d

dn

n J t

Inertia torque of mass n

Mechanical friction torque acting on mass n

Damping torque component in the shaft

between masses n and n+1

Spring constant of the shaft betweenmasses n and n+1

Active torque on a mass of inertia.Electromagnetic torque in the electrical machine.Piston torque from a cylinder of the diesel engine.

Θ Θ +− 1d( )

dn n

nC t

Θ dd

nnB t

( )nT t

Θ Θ +− 1( )n n nK

80

100

Torque produced by one of the 18 cylinders


109/146

Fundamental frequency is 6.25 Hz

18 cylinders => largest shaft-torque harmonic at 112.5 Hz

-60

-40

-20

0

20

40

60

0.00 0.05 0.10 0.15 0.20

Time [s]

T o r q u e [ k

N m ]

Natural vibration frequencies and modes

2d d

Equation of motion in matrix form


110/146

Θ Θ Θ Θ ⎡ ⎤+ + = ⎣ ⎦

2T

1, 2, 3,2

d d ; ...,

dd N

tt

J C K Τ

Natural vibration modes and frequencies; trial function λ e t θ

λ λ λ λ λ λ λ λ + + + + =2

22

d de e e e e e 0

dd

t t t t t t

ttθ θ θ θ θ θ

J C K J C K

Non-zero solutions for θ possible if λ satisfies the characteristic

equation

( )λ λ + + =2det 0 J C K

λ λ ⎡ ⎤+ + =⎣ ⎦2 0 J C K=>

Natural frequencies and modes

The roots of the characteristic equation are typically complex numbersλ α ω= + ji i i


111/146

ω i is a natural frequency of the mechanical system, α i is associated with

the damping of a vibration mode i. All α i should have negative values,otherwise the system is unstable.

Typically, the damping terms are so small that they can be neglected

if one is only interested in the natural frequencies. In such a case,matrix C is left out of the analysis.

λ α ω = + ji i i

Results from the combined electromechanical model


112/146

Rated values

S = 8.2 MVA f s = 50 Hz

U s = 6300 V

I s = 750 Acosϕ = 0.80I f = 215 A

400

450

500

Torque versus torsion angle characteristicfor the non-linear coupling


113/146

Coupling: Rated torque 110 kNm; Stands 330 kNm

The mass of inertia of the synchronous generator is about twice as

large as the mass of inertia of the diesel engine.

0

50

100

150

200

250

300

350

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

Torsion angle [rad]

T o r q u e [ k N m ]

Two-phase short circuit in the line


114/146

Current and voltagemeasurement


if

Grid

uf

Z

Diesel engine

0

25005000

7500

10000

o l t a g e

[ V ]

Voltages and currents at the terminals of the machine


115/146

-10000-7500

-5000

-2500

0

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

T e r m i n a l v

-8000

-6000

-4000

-2000

0

2000

4000

6000

8000

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

T e r

m i n a l c u r r e n t [ A ]

600

700

800

900

1000

c u r r e n t [ A ]

Field-winding current and the torques


116/146

0

100200

300

400

500

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

M a g n e t i s a t i o n

-800

-600

-400

-200

0

200400

600

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]


Air gap Coupling

Synchronisation trial at phase opposition


117/146

Current and voltagemeasurement


if

Grid

uf

Z

Diesel engine

0

25005000

7500

10000

v o l t a g e

[ V ]

Voltages and currents at the terminals of the machine


118/146

-10000-7500

-5000

-2500

0

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

T e r m i n a l v

-20000

-15000

-10000

-5000

0

5000

10000

15000

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

T e r

m i n a l c u r r e n t [ A ]

1000

12001400

16001800

2000

n c u r r e

n t [ A ]

Field-winding current and the torques


119/146

-200

0200

400600

800

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]

M a g

n e t i s a t i o n

-1500

-1000

-500

0

500

1000

0 20 40 60 80 100 120 140 160 180 200 220

Time [ms]


Air gap Coupling

A complete description of the electromagnetic field requires5 field variables ( E, D, H , B, J ).

Magnetic field analysis for electrical machines


120/146

( , , , , J)

Maxwell’s equations Material conditions

For the frequencies considered,

In addition, we need the boundary conditions for the fields.

ρ ∇ ⋅ =

∇ ⋅ =

∂∇ × = −

∂∂

∇ × = +∂

0

t

t

D

B

B E

D H J

ε

σ

µ

=

=

=

D E

J E

B H

∂

≈∂ 0t

D

Vector potential A and scalar potential φ

∇⎧ H J

∇ × = A B φ ∂

= − − ∇

∂t

A EDefinitions:


121/146

=>

The 5 original field variables and their 15 components have

been compressed to 2 potentials with altogether 4 components.

µ

∇ × =⎧⎪

=⎨

⎪ = ∇ ×⎩

H J

B H

B A µ

⎛ ⎞∇ × ∇ × =

⎜ ⎟⎝ ⎠

1 A J

σ

φ

=⎧⎪

∂⎨= − − ∇⎪ ∂⎩ t

J E

A E

=> σ σ φ ∂

= − − ∇

∂t

A J

σ σ φ

µ

⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟

∂⎝ ⎠

10

t

A A=>

Boundary and symmetry conditions

Vector potential on the outer surfaces of a solution region is

typically ∂= =∂

0 or 0 An

A A


122/146

According to the first condition, the flux through the surface is

zero. The second condition states that the flux density isorthogonal to the surface.

= ± 21 A A

∂

A

Pole 2

Pole 1

− A

= A 0

= A 0

In the analysis of electricalmachines, symmetry of the field is

often utilised. The vector potential

is forced to be periodic on the

boundaries of the solution sector.

A

Ferromagnetic non-linearity of iron

Iron is used to increase the magnetic flux of a machine. Its

permeability is a very complicated function of flux density.=> The field equations become non-linear


123/146

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

-1500 -1000 -500 0 500 1000 1500

H [A/m]

B

[ T ]

Iron sample in alternating field Field in the yoke of electrical machine

Modelling motion

Usually, there is a slotting

on both stator and rotor surfaces, and there is no

reference frame in which


124/146

the motion would not affect

the material properties.

The stator and rotor fieldsmust be solved in their own

reference frames and

forced to be continuous

over the air gap.

σ σ φ µ

⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟ ∂⎝ ⎠

10

t

A A

σ σ φ µ

⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟ ∂⎝ ⎠

1 ' ' ' 0' t

A' A'

In air gap: = A A'

Simplification I – Two-dimensional field

The solution of a time- and motion-dependent, non-linear,

three-dimensional magnetic field requires too muchcomputation power and time.


125/146

The field is usually assumed to be two-dimensional,

independent from the coordinate parallel to shaft of themachine (z-direction).

The solution task is reduced significantly as the vector

potential has only one component and the scalar potential is asimple, linear function of the coordinate parallel to the shaft.

( )φ φ

⎧⎪ =

⎪ =⎨⎪⎪ = − +⎩

0

( , , )

( , , )z

z

A x y t

J x y t

u tz

l

A e

J e=>

( )σ σ

µ

∂⎡ ⎤−∇ ⋅ ∇ + =⎢ ⎥ ∂⎣ ⎦

( , , )1( , , )

u t A x y t A x y t

t l

Circuit equations

Electrical machines are typically supplied from voltage

sources. The voltage induced in a winding should beintegrated along the winding


126/146

motor sikron

Documents