dyn_of_sync_mac

Dynamics of the synchronous machine

Dynamics of the synchronous machine

ELEC0047

October 2012

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Dynamics of the synchronous machine Modelling the synchronous machine

Modelling the synchronous machine

Stator and rotor windings

f : field windingd1, q1: damper (or amortisseur) windingsq2: equivalent circuit to account for eddy currents in round-rotor machineno q2 winding in the model of a salient-pole machine

2 / 75


Basic stator equations

va(t) = −Raia(t)− dψa

dt

vb(t) = −Raib(t)− dψb

dt

vc (t) = −Raic (t)− dψc

dt

In matrix form: vT = −RT iT −d

dtψT with RT = diag(Ra Ra Ra)

3 / 75


Basic rotor equations

vf (t) = Rf if (t) +dψf

dt

0 = Rd1id1(t) +dψd1

dt

0 = Rq1iq1(t) +dψq1

dt

0 = Rq2iq2(t) +dψq2

dt

In matrix form: vr = Rr ir +d

dtψr with Rr = diag(Rf Rd1 Rq1 Rq2)

4 / 75


Inductances

Assumption: non saturated material

[ψT

ψr

]=

[LTT (θ) LTr (θ)LT

Tr (θ) Lrr

] [iTir

]

LTT and LTr vary with the rotor position θ

but not Lrr

the elements of LTT and LTr are obviously periodic functions of θ

magnetic field created by any coil assumed to be radial in the air gap and itsmagnitude to vary sinusoidally with θ → sinusoidal machine

orthogonal coils not coupled magnetically.

5 / 75


LTT (θ) = L0 + L1 cos 2θ −Lm − L1 cos 2(θ + π6 ) −Lm − L1 cos 2(θ − π

6 )−Lm − L1 cos 2(θ + π

6 ) L0 + L1 cos 2(θ − 2π3 ) −Lm − L1 cos 2(θ + π

2 )−Lm − L1 cos 2(θ − π

6 ) −Lm − L1 cos 2(θ + π2 ) L0 + L1 cos 2(θ + 2π

3 )

Lo , L1, Lm > 0 depend on the geometry and permittivity of material

6 / 75


LTr (θ) = Laf cos θ Lad1 cos θ Laq1 sin θ Laq2 sin θLaf cos(θ − 2π

3 ) Lad1 cos(θ − 2π3 ) Laq1 sin(θ − 2π

3 ) Laq2 sin(θ − 2π3 )

Laf cos(θ + 2π3 ) Lad1 cos(θ + 2π

3 ) Laq1 sin(θ + 2π3 ) Laq2 sin(θ + 2π

3 )

Laf , Lad1, Laq1, Laq2 > 0 depend on geometry of machine and permittivity ofmaterial

Lrr =

Lff Lfd1 0 0Lfd1 Ld1d1 0 0

0 0 Lq1q1 Lq1q2

0 0 Lq1q2 Lq2q2

7 / 75

Dynamics of the synchronous machine Park transformation and equations

Park transformation and equations

Park transformation

Three-phase (a, b, c) stator variables transformed into Park (d , q, o) variables:

vP = P vT

ψP = P ψT

iP = P iT

where P =

√2

3

cos θ cos(θ − 2π3 ) cos(θ + 2π

3 )sin θ sin(θ − 2π

3 ) sin(θ + 2π3 )

1√2

1√2

1√2

vP =[

vd vq vo

]TψP =

[ψd ψq ψo

]TiP =

[id iq io

]TIt is easily shown that P PT = I ⇔ P−1 = PT

8 / 75


Interpretation

The d and q variables relate to fictitious windings:

attached to the rotor and thus rotating with the latter

d in the direct axis, q in the quadrature axis

producing a magnetic field proportional to the one produced by the statorwindings a, b, c .

The o fictitious winding:

is magnetically not coupled with the d and q windings

plays a role in unbalanced operating conditions only

In steady-state balanced operating conditions:

currents id and iq are constant

current io is zero.

9 / 75


Park equations of the synchronous machine

vT = −RT iT −d

dtψT

P−1 vP = −RaI P−1 iP −d

dt(P−1 ψP )

vP = −RaPP−1iP − P (d

dtP−1)ψP − PP−1

d

dtψP

= −RP iP − θPψP −d

dtψP

with: RP = RT P =

0 1 0−1 0 00 0 0

Decomposing: vd = −Raid − θψq −

dψd

dt

vq = −Raiq + θψd −dψq

dt

vo = −Raio −dψo

dt

θψ : speed voltages (or emf) dψ/dt : transformer voltages (or emf)10 / 75


Park inductance matrix[ψT

ψr

]=

[LTT LTr

LTTr Lrr

] [iTir

][P−1ψP

ψr

]=

[LTT LTr

LTTr Lrr

] [P−1iP

ir

][ψP

ψr

]=

[PLTTP−1 PLTr

LTTrP−1 Lrr

] [iPir

]=

[LPP LPr

LrP Lrr

] [iPir

]

[LPP LPr

LrP Lrr

]=

Ldd Ldf Ldd1

Lqq Lqq1 Lqq2

Loo

Ldf Lff Lfd1

Ldd1 Lfd1 Ld1d1

Lqq1 Lq1q1 Lq1q2

Lqq2 Lq1q2 Lq2q2

Ldd = L0 + Lm +

3

2L1 Lqq = L0 + Lm −

3

2L1 Loo = L0 − 2Lm

Ldf =

√3

2Laf Ldd1 =

√3

2Lad1 Lqq1 =

√3

2Laq1 Lqq2 =

√3

2Laq2

11 / 75


Park equations a bit more detailed (o winding left aside)

For the (d , f , d1) subset we have: vd

−vf

0

= −

Ra

Rf

Rd1

idifid1

− θψq

00

− d

dt

ψd

ψf

ψd1

and for the (q, q1, q2) subset: vq

00

= −

Ra

Rq1

Rq2

iqiq1iq2

+

θψd

00

− d

dt

ψq

ψq1

ψq2

with the following relationships between magnetic fluxes and currents: ψd

ψf

ψd1

=

Ldd Ldf Ldd1

Ldf Lff Lfd1

Ldd1 Lfd1 Ld1d1

idifid1

ψq

ψq1

ψq2

=

Lqq Lqq1 Lqq2

Lqq1 Lq1q1 Lq1q2

Lqq2 Lq1q2 Lq2q2

iqiq1iq2

12 / 75

Dynamics of the synchronous machine Energy, power and torque

Energy, power and torque

Power balance of stator:

pT + pJs +dWms

dt= pr→s

pT instantaneous power leaving the statorpJs Joule losses in statorWms magnetic energy accumulated in stator windingspr→s power transferred from rotor to stator (mechanical / electrical ?)

Instantaneous power leaving the stator:

pT (t) = vaia + vb ib + vc ic = vTT iT = vT

PPPT iP = vTP iP = vd id + vq iq + vo io

= − (Rai2d + Rai2q + Rai2o )︸︷︷︸pJs

− (iddψd

dt+ iq

dψq

dt+ io

dψo

dt)︸︷︷︸

dWms/dt

+θ(ψd iq − ψq id )

⇒ pr→s = θ(ψd iq − ψq id )

13 / 75


Power balance of rotor:

Pm + pf = pJr +dWmr

dt+ pr→s +

dWc

dt

Pm mechanical power provided by the turbinepf instantaneous power entering the field windingpJr Joules losses in rotorWmr magnetic energy accumulated in rotor windingsWc kinetic energy of rotating masses.

Instantaneous power entering the field winding:

pf = vf if = vf if + vd1id1 + vq1iq1 + vq2iq2

= (Rf i2f + Rd1i2d1 + Rq1i2q1 + Rq2i2q2)︸︷︷︸pJr

+ ifdψf

dt+ id1

dψd1

dt+ iq1

dψq1

dt+ iq2

dψq2

dt︸︷︷︸dWmr/dt

Pm −dWc

dt= θ(ψd iq − ψq id ) = power transmitted through torque

= θTe ⇒ Te = ψd iq − ψq id14 / 75


Torque components

Te = Ldd id iq + Ldf if iq + Ldd1id1iq − Lqq iq id − Lqq1iq1id − Lqq2iq2id

(Ldd − Lqq) id iq : reluctant synchronous torque

exists only in salient-pole machines

even without excitation (if = 0), the rotor tends to align its direct axis withthe rotating field created by stator currents (minimum reluctance)

' 10 - 20 % of total torque.

Ldd1 id1iq − Lqq1iq1id − Lqq2iq2id : damping torque

vanishes in steady state

Ldf if iq :

largest part of total torque

in steady state, if constant → synchronous torque due to excitation

during transients contributes to damping torque.

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Dynamics of the synchronous machine The machine in steady-state operating conditions

The machine in steady-state operating conditions

three-phase balanced currents in stator windings

three-phase balanced voltages at the stator terminals

all with angular frequency ωN

constant (DC) voltage Vf applied to field winding

current in field winding:

if =Vf

Rf

rotor rotates at synchronous angular speed:

θ = θo + ωN t

no current in other windings:

id1 = iq1 = iq2 = 0

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Open-circuit operation

ia = ib = ic = 0 ⇒ id = iq = io = 0

Magnetic fluxes in d and q windings: ψd = Ldf if ψq = 0

Park equations:

vd = 0

vq = ωNψd = ωN Ldf if

Going back to stator (phase a for instance):

va(t) =

√2

3ωN Ldf if sin(θo + ωN t) =

√2Eq sin(θo + ωN t)

Eq =ωN Ldf if√

3= emf proportional to field current = open-circuit voltage.

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Under-load operation

va(t) =√

2V cos(ωN t + φ) ia(t) =√

2I cos(ωN t + ψ)

vb(t) =√

2V cos(ωN t + φ− 2π

3) ib(t) =

√2I cos(ωN t + ψ − 2π

3)

vc (t) =√

2V cos(ωN t + φ+2π

3) ic (t) =

√2I cos(ωN t + ψ +

2π

3)

Currents and voltages in d and q windings:

id =√

3I cos(θo − ψ) iq =√

3I sin(θo − ψ) io = 0

vd =√

3V cos(θo − φ) vq =√

3V sin(θo − φ) vo = 0

Magnetic fluxes in d and q windings: ψd = Ldd id + Ldf if ψq = Lqq iq

Park equations:

vd = −Raid − ωN Lqq iq = −Raid − Xq iq

vq = −Raiq + ωN Ldd id + ωN Ldf if = −Raiq + Xd id +√

3Eq

Xd and Xq : direct- and quadrature-axis synchronous reactances, respectively18 / 75


Phasor diagram

19 / 75


Relationship between voltage and current phasors

V cos(θo − φ) = −RaI cos(θo − ψ)− XqI sin(θo − ψ)

V sin(θo − φ) = −RaI sin(θo − ψ) + Xd I cos(θo − ψ) + Eq

are the projections on axes d et q of the equation:

Eq = Va + Ra Ia + jXd Id + jXq Iq

Particular case: round-rotor machine Xd = Xq = X

Eq = Va + Ra Ia + jX (Id + Iq) = Va + Ra Ia + jX Ia

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Complex, active and reactive powers

It is easily shown that:

S = 3Va I ?a = (vd − j vq)(id + j iq)

and hence:P = vd id + vq iq Q = vd iq − vq id

Neglecting Ra and expressing P and Q as a function of V , Eq and the internalangle ϕ:

P = 3EqV

Xdsinϕ+

3V 2

2(

1

Xq− 1

Xd) sin 2ϕ Q = 3

EqV

Xdcosϕ−3V 2(

sin2 ϕ

Xq+

cos2 ϕ

Xd)

In the case of a round-rotor machine:

P = 3EqV

Xsinϕ Q = 3

EqV

Xcosϕ− 3

V 2

X

21 / 75

Dynamics of the synchronous machine Inductances and time constants of the synchronous machine

Inductances and time constants of the synch. machine

Over ' 0.1 s after a disturbance:

rotor speed cannot change significantly due to inertia

transients essentially of electromagnetic nature (changing magnetic fluxes inwindings)

Assumptions:

θ constant

non saturated material

Objectives

define accurately the time constants and inductances involved in theshort-circuit current of the synchronous machine (see course ELEC0029)

use these expressions to obtain inductances and resistances of Park theoryfrom measurements

introduce the ”quasi-sinusoidal” approximation through an example.

22 / 75


Laplace transform of Park equations

Vd (s) + θψq(s)−Vf (s)

0

= −

Ra + sLdd sLdf sLdd1

sLdf Rf + sLff sLfd1

sLdd1 sLfd1 Rd1 + sLd1d1

︸︷︷︸

Rd + sLd

Id (s)If (s)Id1(s)

+Ld

id (0)if (0)id1(0)

Vq(s)− θψd (s)

00

= −

Ra + sLqq sLqq1 sLqq2

sLqq1 Rq1 + sLq1q1 sLq1q2

sLqq2 sLq1q2 Rq2 + sLq2q2

︸︷︷︸

Rq + sLq

Iq(s)Iq1(s)Iq2(s)

+Lq

iq(0)iq1(0)iq2(0)

23 / 75


Time constants and inductances

Eliminating If , Id1 , Iq1 and Iq2 yields:

Vd (s) + θψq(s) = −Zd (s)Id (s) + sG (s)Vf (s)

Vq(s)− θψd (s) = −Zq(s)Iq(s)

with:

Zd (s) = Ra + sLdd −[

sLdf sLdd1

] [ Rf + sLff sLfd1

sLfd1 Rd1 + sLd1d1

]−1 [sLdf

sLdd1

]= Ra + s`d (s) `d (s) : d-axis operational inductance

Zq(s) = Ra + sLqq −[

sLqq1 sLqq2

] [ Rq1 + sLq1q1 sLq1q2

sLq1q2 Rq2 + sLq2q2

]−1 [sLqq1

sLqq2

]= Ra + s`q(s) `q(s) : q-axis operational inductance

24 / 75


Knowing the properties of RL circuits, `d (s) and `q(s) can be factorized into:

`d (s) = Ldd(1 + sT

′

d )(1 + sT′′

d )

(1 + sT′d0)(1 + sT

′′d0)

with 0 < T′′

d < T′′

d0 < T′

d < T′

d0

`q(s) = Lqq

(1 + sT′

q)(1 + sT′′

q )

(1 + sT′q0)(1 + sT

′′q0)

with 0 < T′′

q < T′′

q0 < T′

q < T′

q0

Limit values:

lims→0

`d (s) = Ldd d-axis synchronous inductance

lims→∞

`d (s) = L′′

d = LddT

′

d T′′

d

T′d0 T

′′d0

d-axis subtransient inductance

lims→0

`q(s) = Lqq q-axis synchronous inductance

lims→∞

`q(s) = L′′

q = Lqq

T′

q T′′

q

T′q0 T

′′q0

q-axis subtransient inductance

25 / 75


Direct derivation of L′′

d :

elimin. of f and d1

Rd + sLd −→ Ra + s`d (s)

s →∞ ↓ ↓ s →∞

sLd −→ sL′′

d

elimin. of f and d1

L′′

d = Ldd −[

Ldf Ldd1

] [ Lff Lfd1

Lfd1 Ld1d1

]−1 [Ldf

Ldd1

]= Ldd −

L2df Ld1d1 + Lff L2

dd1− 2Ldf Lfd1Ldd1

Lff Ld1d1 − L2fd1

and similarly for the q axis.

26 / 75


Transient inductances

If damper winding effects are neglected, the operational inductances become:

`d (s) = Ldd1 + sT

′

d

1 + sT′d0

`q(s) = Lqq

1 + sT′

q

1 + sT′q0

with the limit values:

lims→∞

`d (s) = L′

d = LddT

′

d

T′d0

d-axis transient inductance

lims→∞

`q(s) = L′

q = Lqq

T′

q

T′q0

q-axis transient inductance

Using the same derivation as for L′′

d , one easily gets:

L′

d = Ldd −L2

df

LffL

′

q = Lqq −L2

qq1

Lq1q1

27 / 75


Typical values

machine with machine withround rotor salient poles round rotor salient poles

(pu) (pu) (s) (s)

Ld 1.5-2.5 0.9-1.5 T′

d0 8.0-12.0 3.0-8.0

Lq 1.5-2.5 0.5-1.1 T′

d 0.95-1.30 1.0-2.5

L′

d 0.2-0.4 0.3-0.5 T′′

d0 0.025-0.065 0.025-0.065

L′

q 0.2-0.4 T′′

d 0.02-0.05 0.02-0.05

L′′

d 0.15-0.30 0.25-0.35 T′

q0 2.0

L′′

q 0.15-0.30 0.25-0.35 T′

q 0.8

T′′

q0 0.20-0.50 0.04-0.15

T′′

q 0.02-0.05 0.02-0.05Tα 0.02-0.60 0.02-0.20

inductances in per unit on the machine nominal voltage and apparent power

28 / 75


Comments

in the direct axis: pronounced “time decoupling”:

T′

d0 T′′

d0 T′

d T′′

d

subtransient time constants T′′d and T

′′d0: short, originate from damper winding

transient time constants T′d and T

′d0: long, originate from field winding

in the quadrature axis: less pronounced time decoupling (windings of quitedifferent nature)

salient-pole machines: single winding in q axis ⇒ no value for L′

q,T′

q and T′

q0.

29 / 75

Dynamics of the synchronous machine An example of electromagnetic transients: short-circuit of a machine. . .

An example of electromagnetic transients:short-circuit of a machine operating at no load

Assumptions:

for t < 0 : stator windings opened. Voltage magnitude E oq =

ωN Ldf iof√

3for t ≥ 0: stator windings short-circuited: va = vb = vc = 0

rotor speed constant and equal to synchronous speed : θ = ωN

(not valid for long durations !)

constant field voltage vf = v of = Rf io

f

damper windings ignored (for simplicity of derivation)

30 / 75


Deriving the expression of short-circuit currents

Field winding: Vf (s) =Rf i0f

s

Park equiv. windings: Vd (s) = Vq(s) = Vo(s) = 0

The o winding can be discarded. Indeed:

Raio + Loodiodt

= 0 and io(0) =

√2

3

1√2

(ia(0) + ib(0) + ic (0)) = 0

Laplace transform of Park eqs. with windings d1, q1 and q2 omitted:

(Ra + sLdd )Id (s) + sLdf If (s) + ωN LqqIq(s)− Lfd i0f = 0 (1)

sLdf Id (s) + (Rf + sLff )If (s)− Lff i0f =Rf i0f

s(2)

(Ra + sLqq)Iq(s)− ωN (Ldd Id (s) + Lfd If (s)) = 0 (3)

31 / 75


From (2):

If (s) = − sLfd

Rf + sLffId (s) +

i0fs

Substituting If (s) in (1):

[Ra + s`d (s)] Id (s) + ωN LqqIq(s) = 0 (4)

with `d (s) = Ldd1 + sT

′

d

1 + sT′do

and T′

do =Lff

Rf

Substituting If (s) in (3):

− ωN`d (s)Id (s) + [Ra + sLqq] Iq(s) =

√3E 0

q

s(5)

Let’s solve (4) and (5) to obtain Id (s) and Iq(s) (2 eqs. with 2 unknowns).

32 / 75


Determinant:

∆ =

[R2

a

`d (s)Lqq+ Ra

(1

Lqq+

1

`d (s)

)s + s2 + ω2

N

]`d (s)Lqq

Simplifications accounting for the small value of Ra:

assume Ra = 0: Too simplified (some components of currents would notvanish as observed)

assume R2a = 0 only: still heavy and unnecessarily detailed

Belgian compromise: R2a ' 0 and `d (s) ' lim

s→∞`d (s) = L

′

d .

∆ '[

s2 + Ra

(1

Lqq+

1

L′d

)s + ω2

N

]`d (s)Lqq =

[s2 +

2

Tαs + ω2

N

]`d (s)Lqq

with Tα =2

Ra

11

L′d

+ 1Lqq

(6)

33 / 75


Solving with respect to Id (s) and Iq(s):

Id (s) = −ωN

√3E 0

q

s`d (s)(

s2 + 2Tα

s + ω2N

) = −ωN

√3E 0

q (1 + sT′

do)

Ldd s(1 + sT′d )(

s2 + 2Tα

s + ω2N

)Iq(s) =

(Ra + s`d (s))√

3E 0q

s`d (s)Lqq

(s2 + 2

Tαs + ω2

N

) ' √3E 0

q

Lqq

(s2 + 2

Tαs + ω2

N

)

s2 +2

Tαs + ω2

N :

discriminant:4

T 2α

− 4ω2N ' −4ω2

N

factorized into:

(s +

2

Tα+ jωN

)(s +

2

Tα− jωN

)

34 / 75


Using inverse Laplace transform, neglecting small terms compared to ωN andtaking (6) into account:

id (t) = −√

3E 0q

Xd+√

3E 0q

(1

Xd− 1

X′d

)e−t/T

′d +

√3E 0

q

X′d

e−t/Tα cosωN t

iq(t) =

√3E 0

q

Xqe−t/Tα sinωN t

Using inverse Park transform iT = P−1iP = PT iP :

ia(t) = −√

2E 0

q

Xdcos(ωN t + θo) +

√2E 0

q

(1

Xd− 1

X′d

)e−t/T

′d cos(ωN t + θo)

+√

2E 0

q

X′d

e−t/Tα cos(ωN t + θo) cosωN t

+√

2E 0

q

Xqe−t/Tα sin(ωN t + θo) sinωN t

35 / 75


Using some trigonometric identities:

ia(t) = −√

2E 0q

[1

Xd+

(1

X′d

− 1

Xd

)e−t/T

′d

]cos(ωN t + θo)

+√

2E 0q

1

2

(1

X′d

− 1

Xq

)e−t/Tα cos(2ωN t + θo)

+√

2E 0q

1

2

(1

X′d

+1

Xq

)e−t/Tα cos θo

For the field current:

If (s) =

√3ωN Lfd E 0

q

Rf Ldd (1 + sT′d )(

s2 + 2Tα

s + ω2N

) +i0fs

if (t) = i0f +Xd − X

′

d

X′d

i0f e−t/T′d − Xd − X

′

d

X′d

i0f e−t/Tα cosωN t

Refer to Chapter 12 of ELEC0029 for the interpretation of the various terms interms of magnetic fields.

36 / 75


More accurate expression taking into account the effect of the damper windings:

ia(t) = −√

2E oq

[1

Xd+

(1

X′d

− 1

Xd

)e−t/T

′d +

(1

X′′d

− 1

X′d

)e−t/T

′′d

]cos(ωN t + θo)

+√

2E oq

1

2

(1

X′′d

− 1

X ′′q


+√

2E oq

1

2

(1

X′′d

+1

X ′′q

)e−t/Tα cos θo

Tα =X

′′

d

Ra

37 / 75


Numerical example :

E oq = 1, θo = 0

Xd = Xq = 2 X′

d = 0.3 X′′

d = X′′

q = 0.2 Ra = 0.005 pu

T′

do = 9 T′′

d = 0.0333 s

from which one obtains:

T′

d = T′

do

X′

d

Xd= 1.35 s Tα =

X′′

d

Ra= 40 pu =

40

2π50= 0.127 s

38 / 75


39 / 75

Dynamics of the synchronous machine Rotor motion

Rotor motion

θm angular position of rotor, i.e. angle between one axis attached to the rotorand one attached to the stator. Linked to “electrical” angle θ through:

θ = p θm p number of pairs of poles

ωm mechanical angular speed: ωm =d

dtθm

ω electrical angular speed: ω =d

dtθ = pωm

Basic equation of rotating masses (friction torque neglected):

Id

dtωm = Tm − Te

I moment of inertia of all rotating masses

Tm mechanical torque provided by prime mover (turbine, diesel motor, etc.)

Te electromagnetic torque developed by synchronous machine40 / 75


Motion equation expressed in terms of ω:

I

p

d

dtω = Tm − Te

Dividing by the base torque TB = SB/ωmB :

IωmB

pSB

d

dtω = Tmpu − Tepu

Defining the speed in per unit:

ωpu =ω

ωN=

1

ωN

d

dtθ

the motion eq. becomes:

Iω2mB

SB

d

dtωpu = Tmpu − Tepu

Defining the inertia constant:

H =12 Iω2

mB

SB

the motion eq. becomes:

2Hd

dtωpu = Tmpu − Tepu

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Inertia constant H

called specific energy

ratio kinetic energy of rotating masses at nominal speedapparent nominal power of machine

has dimension of a time

with values in rather narrow interval, whatever the machine power.

Hthermal plant hydro plant

p = 1 : 2 − 4 s 1.5 − 3 sp = 2 : 3 − 7 s

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Relationship between H and launching time tl

tl : time to reach the nominal angular speed ωmB when applying to the rotor,initially at rest, the nominal mechanical torque:

TN =PN

ωmB=

SB cosφN

ωmB

PN : turbine nominal power (in MW) cosφN : nominal power factor

Nominal mechanical torque in per unit: TNpu =TN

TB= cosφN

Uniformly accelerated motion: ωmpu = ωmpu(0) +cosφN

2Ht =

cosφN

2Ht

At t = tl , ωmpu = 1 ⇒ tl =2H

cosφN

Remark. Some define tl with reference to TB , not TN . In this case, tl = 2H.

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Compensated motion equation

In some simplified models, the damper windings are neglected. To compensate forthe neglected damping torque, a correction term can be added to the motion eq. :

2Hd

dtωpu + D(ωpu − ωsys) = Tmpu − Tepu D ≥ 0

where ωsys is the system angular frequency (defined more precisely in Chapter 3).

Electromagnetic torque expression

Te = p(ψd iq − ψq id )

In per unit:

Te

TB= p

ωB

pSB(ψd iq − ψq id ) ⇒ Tepu = ψdpu iqpu − ψqpu idpu

In matrix form:

Te = ψTP P iP where P =

0 1 0−1 0 00 0 0

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Dynamics of the synchronous machine The Quasi-sinusoidal (or phasor) approximation

The Quasi-sinusoidal (or phasor) approximation

Standard approximation in the analysis of short- and long-term dynamicsmain difference with respect to electromagnetic transient studiesconsists in neglecting transformer voltages dψ/dt in stator equations.

Revisiting the example of a short-circuit of a synchronous machineoperating at no load

[Ra + s`d (s)] Id (s) + ωN LqqIq(s) = 0

−ωN`d (s)Id (s) + [Ra + sLqq] Iq(s) =

√3E 0

q

s

Id (s) = −ωN

√3E 0

q (1 + sT′

do)

Ldd s(1 + sT′d )(

s2 + 2Tα

s + ω2N

)Iq(s) =

(Ra + s`d (s))√

3E 0q

s`d (s)Lqq

(s2 + 2

Tαs + ω2

N

) ' 0

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ia(t) = −√

2E 0q

[1

Xd+

(1

X′d

− 1

Xd

)e−t/T

′d

]cos(ωN t + θo)

+√

2E 0q

1

2

(1

X′d

− 1

Xq


+√

2E 0q

1

2

(1

X′d

+1

Xq

)e−t/Tα cos θo

system dynamics approximated by a succession of sinusoidal regimeswith amplitude varying in time

In this simple example, the rotor speed was assumed constant.In reality the various machine speeds vary.

system dynamics approximated by a succession of sinusoidal regimes, theamplitude and phase angle of the alternating quantities varying with time

v(t) =√

2 V (t) cos (ωN t + φ(t))

i(t) =√

2 I (t) cos (ωN t + ψ(t))

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limit of validity:

V (t), φ(t), I (t), ψ(t), etc. . . must vary “smoothly” with respect to the periodTN = 2π/ωN

do not include in the model phenomena whose time constant is smallcompared to TN ! Illusory gain in accuracy !

notions and techniques relative to sinusoidal regime can be extended to thequasi-sinusoidal regime, for instance:

rotating vectors and phasors:

v(t) =√

2 V (t) cos (ωNt + φ(t))

V (t) = V (t) e jωN t+jφ(t) −→ v(t) =√

2 re(V (t)

)active, reactive and complex powers, which vary with time.

“phasor” approximation

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Extensions of results relative to steady state

Assuming balanced three-phase voltages of the form:

va =√

2V (t) cos(ωN t + φ(t))

vb =√

2V (t) cos(ωN t + φ(t)− 2π

3)

vc =√

2V (t) cos(ωN t + φ(t) +2π

3)

let’s show that vd is still the projection on the d axis of the rotating vectorrelative to voltage va.

vd =

√2

3

[cos θ va + cos(θ − 2π

3) vb + cos(θ +

2π

3) vc

]= . . .

=√

3V (t) cos(θ − ωN t − φ(t))

and in per unit: vd = V (t) cos(θ − ωN t − φ(t) QED.

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Constant rotor speed assumption θ ' ωN

Examples:

1 oscillation of θ with a magnitude of 90o and period of 1 second superposedto the uniform motion at synchronous speed:

θ = θo + 2πfN t +π

2sin 2πt ⇒ θ = 100π+π2 cos 2πt ' 314 + 10 cos 2πt

at its maximum, it deviates from nominal by 10/314 = 3 % only.

2 in a large interconnected system, after primary frequency control, frequencysettles at f 6= fN . |f − fN | = 0.1 Hz is already a large deviation. In this case,machine speeds deviate from nominal by 0.1/50 = 0.2 % only.

3 a small isolated system may experience larger frequency deviations. But evenfor |f − fN | = 1 Hz, the machine speeds deviate from nominal by 1/50 = 2 %only.

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Model with transformer voltages neglected and speed set to nominal value

vP = −RP iP − ωNPψP (7)

ψP = LPP iP + LPr ir (8)

ψr = LTPr iP + Lrr ir (9)

vr = Rr ir +d

dtψr (10)

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Dynamics of the synchronous machine Machine behaviour just after a voltage change

Machine behaviour just after a voltage change (inparticular a short-circuit)

fluxes in rotor windings do not changerotor currents vary significantly (to keep fluxes constant!)remark: the stator fluxes do not change either but with the transformervoltages neglected, the model allows them to change instantaneously.

Obtaining ir from (9):ir = L−1rr ψr − L−1rr LT

Pr iP

and substituting into (8):

ψP = LPP iP + LPr (L−1rr ψr − L−1rr LTPr iP )

= (LPP − LPr L−1rr LTPr )iP + LPr L−1rr ψr (11)

It is easily shown that:

LPP − LPr L−1rr LTPr = L

′′

P =

Loo 0 0

0 L′′

d 0

0 0 L′′

q

(12)

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Introducing (11) and (12) into (7):

vP = −(RP + ωNPL′′

P )iP − ωNPLPr L−1rr ψr

Decomposing into d and q components (o left aside):

vd = −Raid − X′′

q iq + e′′

d vq = −Raiq + X′′

d id + e′′

q

with

0

e′′

d

e′′

q

= −ωNPLPr L−1rr ψr

e′′

d , e′′

q are the e.m.f. behind subtransient reactances

e′′

d , e′′

q being proportional to magnetic fluxes:

are continuous (rotor fluxes evolving according to diff. eqs.) ⇒ can becomputed from the pre-disturbance situationdo not vary much after a disturbance ⇒ can be considered constant over' 3T

′′do

the above equations are similar to the Park eqs. in steady state, except forthe presence of an emf in the d axis.

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In practice X′′

d ' X′′

q and the eqs. can be combined into:

E′′

= Va + Ra Ia + jX′′

Ia

which leads to the following equivalent circuit:

used in short-circuit calculations.

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Dynamics of the synchronous machine Case study

Case study

generator at bus 2:

synchronous machine modelled with 4 rotor windings

constant excitation and constant mechanical torque

fault on line 1-3:

takes place very near bus 3 ⇒ applied to bus 3

applied at t = 1 s, cleared at t = 1.1 s by opening one circuit of transm. line

load at bus 5 : exponential model (α = 1.5, β = 2.5)54 / 75


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Dynamics of the synchronous machine “Classical” model of the synchronous machine

“Classical” model of the synchronous machine

Very simplified model used:

in some analytical developmentsin qualitative reasoningfor fast assessment of transient (angle) stability.

Only dynamics = rotor motion. Electrical part simplified as follows:1 damper windings d1 et q2 ignored2 magnetic flux in f and q1 windings assumed constant⇒ model valid over no more than ' 1 second

3 same transient reactances in both axes: X ′d = X ′q = X ′

4 stator resistance neglected.

vd = −X ′iq + e′dvq = X ′id + e′q

which can be combined into:

V + jX ′ I = E ′ = E ′∠δ57 / 75

Dynamics of the synchronous machine “Classical” model of the synchronous machine

Modified rotor motion equation

e′d and e′q constant

⇒ E ′ fixed with respect to d and q axes⇒ δ differs from θ by a constant

1

ωN

d

dtθ = ωpu ⇔ 1

ωN

d

dtδ = ωpu

Rotor motion equation involving powers instead of torques:

2H ωpud

dtωpu = ωpu Tmpu − ωpu Tepu

ωpuTmpu = mechanical power Pm produced by prime mover

ωpuTepu = power transmitted through torque = pr→s

= active power Ppu produced by the machine (quasi-sinusoidal approx.,balanced operating conditions, Ra neglected).

Assuming ωpu ' 1:

2Hd

dtωpu = Pmpu − Ppu

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Dynamics of the synchronous machine Per unit system for the synchronous machine model

Per unit system for the synchronous machine model

Recall on per unit systems

Consider two magnetically coupled coils with:

ψ1 = L11i1 + L12i2 ψ2 = L21i1 + L22i2

For simplicity, we take the same time base in both circuits: t1B = t2B

In per unit: ψ1pu =ψ1

ψ1B=

L11

L1B

i1I1B

+L12

L1B

i2I1B

= L11pu i1pu +L12I2B

L1B I1Bi2pu

ψ2pu =ψ2

ψ2B=

L21I1B

L2B I2Bi1pu + L22pu i2pu

In Henry, one has L12 = L21. We request to have the same in per unit:

L12pu = L21pu ⇔ I2B

L1B I1B=

I1B

L2B I2B⇔ S1B t1B = S2B t2B ⇔ S1B = S2B

A per unit system with t1B = t2B and S1B = S2B is called reciprocal59 / 75


in the single phase in each in each rotorcircuit equivalent to of the d , q winding,

stator windings windings for instance f

time tB =1

ωN=

1

2πfN

power SB = nominal apparent 3-phase

voltage VB : nominal (rms)√

3VB VfB : to be chosenphase-neutral

current IB =SB

3VB

√3IB

SB

VfB

impedance ZB =3V 2

B

SB

3V 2B

SB

V 2fB

SB

flux VB tB

√3VB tB VfB tB

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The equal-mutual-flux-linkage per unit system

For two magnetically coupled coils, it is shown that (see theory of transformer):

L11 − L`1 =n21

RL12 =

n1n2

R

L22 − L`2 =n22

R

L11 self-inductance of coil 1

L22 self-inductance of coil 1

L`1 leakage inductance of coil 1

L`2 leakage inductance of coil 2

n1 number of turns of coil 1

n2 number of turns of coil 2

R reluctance of the magnetic circuit followed by the magnetic field lines whichcross both windings; the field is created by i1 and i2.

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Assume we choose V1B and V2B such that:

V1B

V2B=

n1

n2

In order to have the same base power in both circuits:

V1B I1B = V2B I2B ⇒ I1B

I2B=

n2

n1

We have:

(L11 − L`1)I1B =n21

RI1B =

n21

Rn2

n1I2B =

n1n2

RI2B = L12I2B (13)

The flux created by I2B in coil 1 is equal to the flux created by I1B in the part ofcoil 1 crossed by the magnetic field lines common to both coils.

Similarly in coil 2:

(L22 − L`2)I2B =n22

RI2B =

n22

Rn1

n2I1B =

n1n2

RI1B = L12I1B (14)

This per unit system is said to yield equal mutual flux linkages (EMFL)

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Alternative definition of base currents

From respectively (13) and (14) :

I1B

I2B=

L12

L11 − L`1

I1B

I2B=

L22 − L`2L12

A property of this pu system

L12pu =L12I2B

L1B I1B=

(L11 − L`1)

L1B= L11pu − L`1pu

L21pu =L21I1B

L2B I2B=

(L22 − L`2)

L2B= L22pu − L`2pu

In this pu system, self-inductance = mutual inductance + leakage reactance.Does not hold true for inductances in Henry !

The inductance matrix of the two coils takes on the form:

L =

[L11 L12

L12 L22

]=

[L`1 + M M

M L`2 + M

]63 / 75


Application to synchronous machine

we have to choose a base voltage (or current) in each rotor winding.Let’s first consider the field winding f (1 ≡ f , 2 ≡ d)

we would like to use the EMFL per unit system

we do not know the “number of turns” of the equivalent circuits f , d , etc.

instead, we can use one of the alternative definitions of base currents:

IfB√3IB

=Ldd − L`

Ldf⇒ IfB =

√3IB

Ldd − L`Ldf

Ldd , L` can be measuredLdf can be obtained by measuring the no-load voltage Eq produced by aknown field current if :

Eq =ωNLdf√

3if ⇒ Ldf =

√3Eq

ωN if(15)

the base voltage is obtained from VfB =SB

IfB

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What about the other rotor windings ?

one cannot access the d1, q1 and q2 windings to measure Ldd1, Lqq1 et Lqq2

using formulae similar to (15)

one may assume there exist base currents Id1B , Iq1B et Iq2B leading to theEMFL per unit system, but their values are not known

hence, we cannot compute voltages in Volt or currents in Ampere in thosewindings (only in pu)

not a big issue in so far as we do not have to connect anything to thosewindings (unlike the excitation system to the field winding). . .

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Dynamics of the synchronous machine Dynamic equivalent circuits of the synchronous machine

Dynamic equivalent circuits of the synchronous machine

In the EMFL per unit system, the Park inductance matrices take on the simplifiedform:

Ld =

L` + Md Md Md

Md L`f + Md Md

Md Md L`d1 + Md

Lq =

L` + Mq Mq Mq

Mq L`q1 + Mq Mq

Mq Mq L`q2 + Mq

For symmetry reasons, same leakage inductance L` in d and q windings

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Dynamics of the synchronous machine Dynamic equivalent circuits of the synchronous machine

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Dynamics of the synchronous machine Exercises

Exercises

Exercise 1

A machine has the following characteristics:

nominal frequency: 50 Hznominal apparent power: 1330 MVAstator nominal voltage: 24 kVXd = 0.9 Ω (value of per phase equivalent)X` = 0.1083 Ω (value of per phase equivalent)field current giving the nominal stator voltage at no-load: 2954 A

choose the base power, voltage and current in the stator windings

choose the base power, voltage and current in the field winding, using theEMFL per unit system

compute Xd , X` and Ldf in per unit.

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Dynamics of the synchronous machine Exercises

Exercise 2

A 1330 MVA, 50 Hz machine has the following characteristics (values in pu on themachine base):

X` = 0.20 pu Ra = 0.004 puXd = 2.10 pu Xq = 2.10 puX ′d = 0.30 pu X ′q = 0.73 pu

X′′

d = 0.25 pu X′′

q = 0.256 pu

T′′

do = 0.03 s T′′

qo = 0.20 sT ′do = 9.10 s T ′qo = 2.30 s

Determine the inductances and resistances of the Park model, using the EMFL perunit system. Check your answers by computing X

′′

d and X′′

q from the Parkinductance matrices.

Hints:

time constants must be converted in per unit !

identify first the parameters of the equivalent circuits.

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Dynamics of the synchronous machine Modelling of material saturation

Modelling of material saturation

Saturation of magnetic material modifies:

the machine inductances

the initial operating point (in particular the rotor position)

the field current required to obtain a given stator voltage.

Notation

parameters with the upperscript u refer to unsaturated values

parameters without this upperscript refer to saturated values.

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Open-circuit magnetic characteristic

Machine operating at no load, rotating at nominal angular speed ωN .Terminal voltage Eq measured for various values of the field current if .

saturation factor: k =OA

OB=

O ′A′

O ′A< 1

Eq = ωN Md if with Md = k Mud

Example of saturation model: k =1

1 + m(Eq)n

The above characteristic is measured in the d axis (field due to if only).

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Leakage and air gap flux

The flux linkage in the d winding is decomposed into:

ψd = Lìd + ψad

Lìd : leakage flux, not subject to saturation (path mainly in the air)ψad : direct-axis component of the air gap flux, subject to saturation.

Expression of ψad :

ψad = ψd − Lìd = Md (id + if + id1)

Expression of ψaq:

ψaq = ψq − Lìq = Mq(iq + iq1 + iq2)

Considering that the d and q axes are orthogonal, the air gap flux is given by:

ψag =√ψ2

ad + ψ2aq (16)

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Saturation characteristic in loaded conditions

saturation is different in the d and q axes, especially for a salient polemachine (air gap larger in q axis !). Hence, different saturation factors (kd

and kq) should be considered

we assume that only the direct-axis saturation characteristic is available,which is often the case in practice

in this case, the same saturation factor k is assumed in both axes

k is obtained from the open-circuit saturation characteristic as follows.

In no-load conditions:

ψad = Md if and ψaq = 0 ⇒ ψag = Md if

k =1

1 + m(Eq)n=

1

1 + m(ωN Md if )n=

1

1 + m(ωNψag )n(17)

In loaded conditions: (17) is assumed to hold true with ψag given by (16).

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Complete model

ψd = Lìd + ψad

ψf = L`f if + ψad

ψd1 = L`d1id1 + ψad

ψad = Md (id + if + id1)

Md =Mu

d

1 + m(ωN

√ψ2

ad + ψ2aq

)n

vd = −Raid − ωNψq

d

dtψf = vf − Rf if

d

dtψd1 = −Rd1id1

2Hd

dtω = Tm − (ψd iq − ψq id )

ψq = Lìq + ψaq

ψq1 = L`q1if + ψaq

ψq2 = L`q2if + ψaq

ψaq = Mq(iq + iq1 + iq2)

Mq =Mu

q

1 + m(ωN

√ψ2

ad + ψ2aq

)n

vq = −Raiq + ωNψd

d

dtψq1 = −Rq1iq1

d

dtψq2 = −Rq2iq2

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Dynamics of the synchronous machine Exercise

Exercise

Bring the model of the previous slide in the form:

d

dtx = f(x, y,u)

0 = g(x, y,u)

with the vector of differential states:

x = [ψf ψd1 ψq1 ψq2 ω]T ,

the vector of algebraic states:

y = [id iq ψad ψaq]T ,

and the vector of “inputs”:

u = [vd vq vf Tm]T

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