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Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC component 48582 - P OWER S YSTEM ANALYSIS AND DESIGN L ECTURE 1-S YNCHRONOUS MACHINE MODEL DR.GERMANE XATHANASIUS School of Electrical, Mechanical and Mechatronic Systems UNIVERSITY OF TECHNOLOGY SYDNEY

PSAD Slides Lec1 Synchronous Machine Model

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Page 1: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

48582 - POWER SYSTEM ANALYSIS ANDDESIGN

LECTURE 1 - SYNCHRONOUS MACHINEMODEL

DR. GERMANE X ATHANASIUS

School of Electrical, Mechanical and Mechatronic Systems

UNIVERSITY OF TECHNOLOGY SYDNEY

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Page 2: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Lecture Outline

1 Synchronous machine modelPer phase equivalent circuit

2 Two axis modelPark’s TransformationTransient modelBalanced three phase fault

3 Simplified representation for transient analysis

4 Short circuit current

5 DC components of stator currents

6 Fault on a loaded generator

Page 3: PSAD Slides Lec1 Synchronous Machine Model

Synchronous Generators

Page 4: PSAD Slides Lec1 Synchronous Machine Model

Synchronous GeneratorsSynchronous Generators

• Synchronous generators or alternators are used to convert mechanical power derived from steam, gas, or hydraulic-turbine to ac electric powerto ac electric power

• Synchronous generators are the primary source of electrical energy we consume todayenergy we consume today

• Large ac power networks rely almost exclusively on synchronous generatorsgenerators

Page 5: PSAD Slides Lec1 Synchronous Machine Model

Construction

Basic parts of a synchronous generator:

• Rotor dc excited winding• Rotor - dc excited winding • Stator - 3-phase winding in which the ac emf is generated

The manner in which the active parts of a synchronous machine are cooled determines its overall physical size and structure

Page 6: PSAD Slides Lec1 Synchronous Machine Model

Various Types

Salient-pole synchronous machine

Cylindrical or round-rotor synchronous machine

Non-uniform air-gap

Uniform air-gapUniform air gap

Page 7: PSAD Slides Lec1 Synchronous Machine Model

Salient-Pole Synchronous Generator

StatorStator

Page 8: PSAD Slides Lec1 Synchronous Machine Model

Cylindrical-Rotor Synchronous Generator

Stator

Cylindrical rotor

Page 9: PSAD Slides Lec1 Synchronous Machine Model

Operation PrincipleOperation Principle

The rotor of the generator is driven by a prime-mover

A dc current is flowing in the rotor winding which produces a rotating magnetic field within the machine

The rotating magnetic field induces a three-phase e otat g ag et c e d duces a t ee p asevoltage in the stator winding of the generator

Page 10: PSAD Slides Lec1 Synchronous Machine Model

Electrical Frequency

Electrical frequency produced is locked or synchronized toElectrical frequency produced is locked or synchronized to the mechanical speed of rotation of a synchronous generator:

120m

enP

f =

where f = electrical frequency in Hz

120

where fe = electrical frequency in HzP = number of polesn = mechanical speed of the rotor in r/minnm= mechanical speed of the rotor, in r/min

Page 11: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Schematic diagram

Figure: Schematic diagram of a synchronous generator

Page 12: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Flux linkages

The speed of the machine (N) in rpm is given by

N =120f

P(1)

Stator coils self inductance, Ls = Laa′ = Lbb′ = Lcc′

Mutual inductance −Ms = Lab = Lbc = LcaThe mutual inductance between field coil and stator coil, if themaximum value of mutual inductance is Msf , then mutualinductance for an angle of θd is given by,

Lfa = Msf cos θd

Lfb = Msf cos(θd − 1200

)Lfc = Msf cos

(θd − 2400

)(2)

Self inductance of the field winding Lff ′ is constant

Page 13: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Flux linkages

The flux linkage of stator and rotor coil is given by,λaλbλcλf

=

Laa′ Lab Lac LafLba Lbb′ Lbc LbfLca Lcb Lcc′ LcfLfa Lfb Lfc Lff ′

iaibicif

(3)

For a balanced three phase system, ia + ib + ic = 0 andia = −(ib + ic), ib = −(ia + ic) and ic = −(ia + ib). λa

λbλc

=

Ls + MsLs + MsLs + Ms

iaibic

+

LafLbfLcf

if (4)

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(Ls + Ms)
Page 14: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Flux linkages

If the field current has a constant magnitude If and the angularvelocity of the rotor is ω then we have

dθd

dt= ω and θd = ωt + θd0 (5)

Substituting in (4) and combining (2) and (4) we get, λaλbλc

=

Ls + MsLs + MsLs + Ms

iaibic

+

Msf cos (ωt + θd0)Msf cos

(ωt + θd0 − 1200)

Msf cos(ωt + θd0 − 2400)

If (6)

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(Ls + Ms)
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Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Induced emf

If the coil a has a resistance R, then the emf across the coil isgiven by,

ea = −Ria −dλa

dt

= −Ria − (Ls + Ms)diadt

+ ωMsf If sin (ωt + θd0) (7)

The last term in (7) represents the internal emf induced in coila by the field current and is given by,

e′a =√

2|Ea| sin (ωt + θd0) (8)

where |Ea| is the rms magnitude and is given by

|Ea| =ωMsf If√

2(9)

e′a will be the emf across the coil a when ia = 0. This voltageis known as open circuit / noload / synchronous internal /generated voltage.

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Page 16: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Induced emf

θd0 is the angle with reference to d − axis, if δ is the angle withreference to the q − axis, then δ =

(θd0 − 900),

∴ θd = ωt + θd0 = ωt + δ + 900 (10)

Now equation (8) becomes,

e′a =√

2|Ea| sin(ωt + δ + 900

)=

√2|Ea| cos (ωt + δ) (11)

substituting in (7)

ea = −Ria − (Ls + Ms)diadt

+√

2|Ea| cos (ωt + δ) (12)

Page 17: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Induced emf

Equation (12) can be diagrammatically represented as inFigure 2.

Figure: Equivalent circuit of phase a

Similarly we can find λb, λc , e′b and e′c for other two stator coilsalso.

Page 18: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

If ia lags e′a by θl , we can write,

ia =√

2|Is| cos (ωt + δ − θl) (13)

Similarly for other phases we can write as,

ib =√

2|Is| cos(ωt + δ − θl − 1200

)ic =

√2|Is| cos

(ωt + δ − θl − 2400

)(14)

Now considering the flux linkage with the field circuit andsubstituting for Laf , Lbf and Lcf from (2) into (3) we get,

λf = Lff ′ If + Msf

[ia cos θd + ib cos

(θd − 1200

)+ ic cos

(θd − 2400

)](15)

Page 19: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

Now substituting for ia and θd using (10) and (13) we get,

ia cos θd =√

2|Is| cos (ωt + δ − θl) cos(ωt + δ + 900

)(16)

Using the trigonometric identity2 cos A cos B = cos(A + B) + cos(A− B) we can write,

ia cos θd =|Is|√

2{− sin θl − sin [(2ωt + δ)− θl ]} (17)

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Page 20: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

Similarly we can write,

ib cos(θd − 1200

)=

|Is|√2

{− sin θl − sin

[(2ωt + δ)− θl − 1200

]}ic cos

(θd − 2400

)=

|Is|√2

{− sin θl − sin

[(2ωt + δ)− θl − 2400

]}(18)

In the above equations terms with 2ωt represent a balancedthree phase second harmonic components whose three phasesum will be zero. Now adding (17) and (18) we get,

ia cos θd + ib cos(θd − 1200

)+ ic cos

(θd − 2400

)=−3√

2Is sin θl (19)

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Page 21: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

Substituting (19) in (15)

λf = Lff ′ If +

√32

Msf id (20)

where

id =

√23

[ia cos θd + ib cos

(θd − 1200

)+ ic cos

(θd − 2400

)]=

√3Is sin θl (21)

From the above discussion we conclude the field flux linkagedue to time varying currents ia, ib and ic produce a constantflux linkages and does not vary with time. We can representthis flux linkage as one coming from a fictitious coil with asteady DC current id and have an axis coinciding with daxis andthe two coils synchronously rotate and have a mutualinductance of Msf as represented in Figure 3.

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sqrt(3/2)*Msf
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Page 22: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

f

f

d

d

sf

Figure: Stator equivalent d − axis coil.

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Page 23: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Field flux linkages

Now the field voltage,

Vff ′ = If Rf +dλ

dt

= If Rf sincedλ

dtdoes not vary

The current id depends on Is and θl . For lagging power factorsid will be negative causing demagnetising effect. So If has tobe increased to counteract. For leading power factors id will bepositive causing magnetising effect and If is decreased. Thiseffect is called armature reaction and this principle is used inexcitation system control.

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Page 24: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Per phase equivalent circuit

Per phase equivalent circuit

Figure: Per phase equivalent circuit.

Figure: Phasor diagram.

Page 25: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

Need for the model:1 The synchronous machine model we developed so far is

based on round rotor theory. This model will be sufficientto analyse the machine under steady state conditions.

2 For transient studies we need to use two axis model.3 In salient pole machines the air gap is small above pole

faces and large in the interpolar regions. These aspectsneed to be considered.

4 Also the rotor has damper windings. These windings arerepresented as Daxis and Qaxis windings which are shortcircuited coils.

Page 26: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

b c

a

a

cb

d

f

f

Figure: Synchronous machine schematic diagram with damperwindings.

Page 27: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

We can rewrite equation (3) including the damper windings as,λaλbλcλfλDλQ

=

Laa′ Lab Lac Laf LaD LaQLba Lbb′ Lbc Lbf LbD LbQLca Lcb Lcc′ Lcf LcD LcQLfa Lab Lac Lff ′ LfD LfQLDa LDb LDc LDf LDD′ LDQLQa LQb LQc LQf LQD LQQ′

iaibicifiDiQ

(22)

The elements of the above matrix can be defined as follows:stator self inductances:

Laa′ = Ls + Lm cos 2θd

Lbb′ = Ls + Lm cos(

2θd −2π

3

)Lcc′ = Ls + Lm cos

(2θd +

3

)(23)

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Page 28: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

stator mutual inductances:

Lab = Lba = −Ms − Lm cos(

2θd +π

6

)Lbc = Lcb = −Ms − Lm cos

(2θd −

π

2

)Lca = Lac = −Ms − Lm cos

(2θd +

6

)(24)

rotor mutual inductances:

LfD = LDf = Mr

LQf = LfQ = 0LDQ = LQD = 0 (25)

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Page 29: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

stator-rotor mutual inductances:

Laf = Lfa = Msf cos θd

Lbf = Lfb = Msf cos(

θd −2π

3

)Lcf = Lfc = Msf cos

(θd −

3

)(26)

stator-rotor daxis damper mutual inductances:

LaD = LDa = MsD cos θd

LbD = LDb = MsD cos(

θd −2π

3

)LcD = LDc = MsD cos

(θd −

3

)(27)

Page 30: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Two axis model

stator-rotor qaxis damper mutual inductances:

LaQ = LQa = MsQ cos θd

LbQ = LQb = MsQ cos(

θd −2π

3

)LcQ = LQc = MsQ cos

(θd −

3

)(28)

Page 31: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Park’s Transformation

Features of Park’s Transformation

1 The inductance elements given by (23) to (28) aredependant on the angular position of the rotor which will bevarying continuously so these inductance will be timevarying. Analysing the model with time varying inductancewill be difficult.

2 Using Park’s transformation these inductances can bemade time invariant for the purpose of analysis.

3 The stator a, b, c variables are transformed to direct axis(d), quadrature axis (q) and zero sequence (0) quantitiesusing the transformation matrix P.

4 The matrix P has the orthogonality property ie., P−1 = PT .Any stator variable can be transformed to dq0 axis bymultiplying with P.

Page 32: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Park’s Transformation

Park’s Transformation

For example stator currents are transformed as, idiqi0

= P

iaibic

(29)

where

P =

√23

cos θd cos(θd − 1200) cos

(θd − 2400)

sin θd sin(θd − 1200) sin

(θd − 2400)

1√2

1√2

1√2

(30)

Page 33: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Equation (22) can be represented in short form as,[λabcλfDQ

]=

[LSS LSRLRS LRR

] [iabcifDQ

](31)

Using Park’s transformation we can transform the stator timevarying inductance to rotor reference frame without modifyingthe rotor quantities. Now the transformation to do that is,[

λdq0λfDQ

]=

[P 00 I

] [λabcλfDQ

](32)

where I is 3 × 3 identity matrix.

Page 34: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

From (32) we can write,[λabcλfDQ

]=

[P−1 0

0 I

] [λdq0λfDQ

][

P−1 00 I

] [λdq0λfDQ

]=

[LSS LSRLRS LRR

] [iabcifDQ

]=

[LSS LSRLRS LRR

] [P−1 0

0 I

] [idq0ifDQ

][

λdq0λfDQ

]=

[P 00 I

] [LSS LSRLRS LRR

][

P−1 00 I

] [idq0ifDQ

](33)

Page 35: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Now substituting for P and P−1 in (33) we get,λdλqλ0λfλDλQ

=

Ld 0 0 kMsf kMsD 00 Lq 0 0 0 kMsQ0 0 L0 0 0 0

kMsf 0 0 Lff ′ Mr 0kMsD 0 0 Mr LDD′ 0

0 kMsQ 0 0 0 LQQ′

idiqi0ifiDiQ

(34)

where

k =

√32

Ld = Ls + Ms +32

Lm

Lq = Ls + Ms −32

Lm

L0 = Ls − 2Ms (35)

Page 36: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Similarly using (32) we can find the transformations for currentsand voltages without modifying the rotor quantities as,[

idq0ifDQ

]=

[P 00 I

] [iabcifDQ

](36)[

vdq0vfDQ

]=

[P 00 I

] [vabcvfDQ

](37)

Page 37: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

We can write the voltage equation using (7) as,

vavbvcvfvDvQ

= −

Ra 0 0 0 0 00 Ra 0 0 0 00 0 Ra 0 0 00 0 0 Rf 0 00 0 0 0 RD 00 0 0 0 0 RQ

iaibicifiDiQ

− ddt

λaλbλcλfλDλQ

(38)

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Page 38: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Using simplified notations and transformations we get,[P−1 0

0 I

] [vabcvfDQ

]=

−[

Rabc 00 RfDQ

] [P−1 0

0 I

] [idq0ifDQ

]−d

dt

[P−1 0

0 I

] [λdq0λfDQ

][

vabcvfDQ

]= −

[P 00 I

] [Rabc 0

0 RfDQ

][

P−1 00 I

] [idq0ifDQ

]−[

P 00 I

]ddt

[P−1 0

0 I

] [λdq0λfDQ

]

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Page 39: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

On simplification we get,[vabcvfDQ

]= −

[Rabc 0

0 RfDQ

] [idq0ifDQ

]−[

PdP−1

dt 00 I

] [λdq0λfDQ

](39)

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Vdq0
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l
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Page 40: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Now using (39),

PdP−1

dt= P

dtdP−1

dθ= ωP

dP−1

=23ω

cos θd cos(θd − 1200) cos

(θd − 2400)

sin θd sin(θd − 1200) sin

(θd − 2400)

1√2

1√2

1√2

×

cos θd sin θd1√2

cos(θd − 1200) sin

(θd − 1200) 1√

2cos

(θd − 2400) sin

(θd − 2400) 1√

2

= ω

0 1 0−1 0 00 0 0

(40)

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Page 41: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Transient model

Transient model

Combining (34), (39) and (40) we get,

2666664vdvqv0−vf

00

3777775 =

2666664Ra ωLq 0 0 0 ωkMsQωLd Ra 0 ωkMsf ωkMsD 0

0 0 Ra 0 0 00 0 0 Rf 0 00 0 0 0 RD 00 0 0 0 0 RQ

3777775

2666664idiqi0ifiDiQ

3777775

2666664Ld 0 0 kMsf kMsD 00 Lq 0 0 0 kMsQ0 0 L0 0 0 0

kMsf 0 0 Lff ′ Mr 0kMsD 0 0 Mr LDD′ 0

0 kMsQ 0 0 0 LQQ′

3777775ddt

2666664idiqi0ifiDiQ

3777775(41)

The zero sequence voltage v0 is not coupled with other equations so it can be treated

separately. In (41) all the matrix coefficients are constants if ω is assumed to be

constant.

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Page 42: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Balanced three phase fault

Balanced three phase fault

Consider a three phase fault on the terminals of a three phasegenerator, which is operating with constant speed andexcitation. Before fault we assume the phase currents ia, ib, icare zero, this will make the currents id , iq, i0 are also zero.Initial field current will be given by,

if =Vf

Rf(42)

During fault the terminal voltages will be zero, ie.,va, vb, vc = 0 so vd , vq, v0 = 0.

Page 43: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Balanced three phase fault

Balanced three phase fault

We shall use (41) to solve for the fault currents. Since i0 = 0, we caneliminate it from (41) and write as,

vdvq−vf00

=

Ra ωLq 0 0 ωkMsQωLd Ra ωkMsf ωkMsD 0

0 0 Rf 0 00 0 0 RD 00 0 0 0 RQ

idiqifiDiQ

Ld 0 kMsf KsD 00 Lq 0 0 kMsQ

kMsf 0 Lff ′ Mr 0KsD 0 Mr LDD′ 00 kMsQ 0 0 LQQ′

ddt

idiqifiDiQ

(43)

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Page 44: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Balanced three phase fault

Balanced three phase fault

Equation (43) can be represented in short form as,

v = −Ri− Ld idt

d idt

= −L−1Ri− L−1v (44)

Equation (43) is a set of linear first order differential equationwhich can be solved for analytical solutions or by usingcomputer integration methods.

Page 45: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Simplified representation for transient analysis

During fault conditions, there will be a sudden rise in currentwhich will not be backed up by instantaneous changes in fluxlinkages in the rotor circuits and armature reaction effects. Itwill take a few cycles for the flux linkages to settle for steadystate values. These initial period following the fault is termed assub-transient and transient periods.Using equation (34) we get,

∆λd = Ld∆id + kMsf ∆if + kMsD∆iD (45)∆λf = kMsf ∆id + Lff ′∆if + Mr∆iD (46)∆λD = kMsD∆iD + Mr∆if + LDD′∆id (47)

The ∆ quantities indicate incremental changes.

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Page 46: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Simplified representation for transient analysis

The field and damper fluxes cannot change instantaneously so∆λf = 0 and ∆λD = 0. Substituting in (46) and (47) we get,

∆if = −(

kMsf LDD′ − kMsDMr

Lff ′LDD′ −M2r

)∆id (48)

∆iD = −(

kMsDLff ′ − kMsf Mr

Lff ′LDD′ −M2r

)∆id (49)

Substituting (48) and (49) in (45) we get,

∆λd

∆id= L′′d = Ld − k2

(M2

sf LDD′ + M2sDLff ′ − 2Msf MsDMr

Lff ′LDD′ −M2r

)(50)

∆λd∆id

is the flux linkage change for unit current whichsub-transient d-axis inductance L′′d and sub-transient d-axisreactance X ′′

d which is less than steady state reactance Xd .

Page 47: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Simplified representation for transient analysis

After the sub-transient period the effect of damper windings canbe neglected, so the flux linkage equations become,

∆λd = Ld∆id + kMsf ∆if (51)∆λf = kMsf ∆id + Lff ′∆if (52)

Equating ∆λf to zero gives,

∆if = −(

kMsf

Lff ′

)∆id (53)

substituting in (51),

∆λd

∆id= L′d = Ld −

(kMsf )2

Lff ′(54)

where L′d is known as transient d-axis inductance and transientd-axis reactance X ′

d = ωL′d which is less than steady statereactance Xd . The reactance will values will beX ′′

d < X ′d < Xd .

Page 48: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Simplified representation for transient analysis

1 From this we can conclude that immediately after the faultthe reactance X ′′

d and the current decays with the timeconstant τ ′′d and has a typical value of around 0.03 s.During this period the reactance is X ′′

d .2 Once the effect of the damper becomes negligible, the

machine reactance raises to X ′d and the fault current

decays with the time constant τ ′d during this period the τ ′ddepends inversely on the field resistance Rf .

3 In steady state conditions Xd = ωLd and Xq = ωLq forsalient pole machines and Xd = ωLd for round rotormachines.

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Page 49: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Short circuit current

1 The behavior of the short circuit current is similar to R − Lcircuit behavior when voltage is applied suddenly but in amore complex manner.

2 The short circuit currents contain dc components whichmakes the three phase currents asymmetrical.

3 If we remove the dc component of the current, we canrepresent the ac component of the fault current as,

Iac = |Ea|1

Xd+ |Ea|

(1

X ′d− 1

X ′d

)e− t

τ ′d + |Ea|(

1X ′′

d− 1

X ′d

)e− t

τ ′′d (55)

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sin(wt+\delta)
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Page 50: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Short circuit current

From (55) it can be seen that the fault current Iac has onesteady state component and two decaying components withtime constants τ ′d and τ ′′d representing transient andsub-transient periods.

Figure: Fault current ac component

Page 51: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Short circuit current

The rms value of the sub-transient current is given by,

|I′′| = oz√2

=|Ea|X ′′

d(56)

The rms value of the transient current is given by,

|I′| = oy√2

=|Ea|X ′

d(57)

The rms value of the steady state fault current is given by,

|I| = ox√2

=|Ea|Xd

(58)

Page 52: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

DC components of stator currents

Similar to RL circuit switching transients, during fault on thegenerator a dc component will be superimposed on the acwave. The dc component value depends on the instantaneousstator voltage Ea and the rotor angle δ at the time of fault. Thetime constant associated with dc component decay is given by

τdc =X

′′

d + X′′q

2Ra(τdc is usually around 0.05 - 0.175 s)(59)

and most of the dc decay occurs during sub-transient period.The magnitude of the dc component will be different fordifferent phases since it depends on the instantaneous voltage.The dc component for the phase a is given by,

Idc−a =√

2|Ea|X ′′

dsin δe

tτdc (60)

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Page 53: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

DC components of stator currents

The net asymmetrical stator current is obtained bysuperimposing dc over ac wave. This asymmetrical current forphase a is given by,

iasym−a =√

2|Ea| ×[1

Xd+

(1

X ′d− 1

X ′d

)e− t

τ ′d +

(1

X ′′d− 1

X ′d

)e− t

τ ′′d +1

X ′′d

sin δet

τdc

](61)

Worst possible transient condition will occur if δ = 900. Themaximum dc component of fault current is given by,

Idc−max =√

2|Ea|X ′′

d(62)

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Page 54: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

DC components of stator currents

The maximum rms value of the asymmetrical fault current isgiven by,

Iasym−max =√

I ′′2 + I2dc

=

√√√√(Ea

X ′′d

)2

+

(√2Ea

X ′′d

)2

=√

3Ea

X ′′d

=√

3 I′′ (63)

Iasym−max is used to determine the max asymmetrical breakingcapacity of the circuit breaker.

Page 55: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Fault on a loaded generator

If the generator is loaded at the time fault and delivering a loadcurrent of Il , we have three internal voltages associated withsub-transient, transient and steady state periods viz., E ′′

a , E ′a

and Ea respectively. These voltages are given by,

E ′′a = Va + X ′′

d Il , E ′a = Va + X ′

d Il and Ea = Va + Xd Il(64)

Page 56: PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model Two axis model Simplified representation for transient analysis Short circuit current DC components of stator currents Fault on a loaded generator

Fault on a loaded generator

Figure: Fault on a loaded generator