genopz synchronous machine model - genopz sychronous m… · of machine modeling presently in use...

genopz

Synchronous Machine Model

John UndrillApril 23, 2020

Model development starts with dynamics and statics of fluxes - this is the general form

Assume Laf = Lak = Lfk etc Leads to gensal and genrou

Assume Lak/Lkk = Laf/Lkf Leads to gentpf and gentpj

genrou / gensal

Stator current affects field current instantaneously

Field flux affects stator flux instantaneously

Assume Laf = Lak = Lfk etc Leads to genrou and related models

gentpf / gentpj

Stator current affects field current AFTER subtransient time lag

Field flux affects stator flux AFTER subtransient time lag

Assume Lak/Lkk = Laf/Lkf Leads to gentpf and gentpj

-0.05 0 0.05 0.1 0.15 0.2

-5

0

5C

urre

nt, p

er u

nit

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-5

0

5

Cur

rent

, per

uni

t

-0.05 0 0.05 0.1 0.15 0.2Time, sec

0

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2

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rent

, per

uni

t

Sudden short circuit test reveals the d-axis characteristic

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rent

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rent

, per

uni

t

Ld

L’d

L’’d

T’’do

T’do

Efd(s) = 0

∆id(s) =∆vq(s)

Ld(s)

Ld(s) = (Ld − Ll)

⎛

⎝

(

1 +(L′

d−Ll)

(Ld−Ll)T ′

dos

)(

1 + (L”d−Ll)(L′

d−Ll)

T”dos)

(1 + T ′

dos)(1 + T”dos)

⎞

⎠+ Ll

G(s) =1 + L”d−Ll)

(L′

d−Ll)T”dos

(1 + sT ′

do)(1 + sT”do)

Current interruption test reveals generator d-axis characteristics

Current interruption test when done at 0pf this test reveals the d-axis characteristic

∆vq(s) = Ld(s)∆id(s)

Thevenin generator model stated in Laplace operator terms

In general v(s) = G(s)Efd − L(s) i(s)

For d-axis vq(s) = G(s)Efd − Ld(s) id(s)

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rent

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rent

, per

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132 CHAPTER 7. SYNCHRONOUS MACHINES

To develop the two transfer functions we must enter the ’culture’ of synchronous machinetheory.

7.4.4 Sudden short circuit

The unidirectional and alternating currents produced by a synchronous machine when thestator winding is short circuited suddenly from an initial condition of rated voltage and opencircuit take the form illustrated by figures 7.4 and 7.5.

The features of the alternating component of short circuit current behavior seen in figures 7.4and 7.5 can be described by the transfer function model given in the following section. The timeconstants in the numerators of the transfer functions are those of the decay of the waveformenvelopes shown in the bottom of figure 7.4 and in figure 7.5. The paramters Ld, L0

d, L”d are thereciprocals of waveform amplitudes as identified in figure 7.5.

7.4.5 Operational Impedances

When a machine is producing reactive power, but no real power, at its terminals its behavior,as can be observed in various tests including a sudden short circuit test, can be describedcompletely emperically by the transfer function relationship

vq(s) = Yd(s) = G(s)Ef d(s)� Ld(s)Id(s) (7.4)

in which

vq is the amplitude of the positive sequence AC voltage at the terminals the machineYd is the amplitude of the flux wave linking the stator windingEf d is the voltage applied to the field windingId is the amplitude of the positive sequence current in the stator winding

The transfer functions have the form

Ld(s) = (Ld � Ll)

0

@1 + (L0

d�Ll)(Ld�Ll)

T0dos

1 + T0dos

1

A

0

@1 + (L”d�Ll)

(L0d�Ll)

T”dos

1 + T”dos

1

A+ Ll (7.5)

G(s) =✓

11 + sT0

do

◆0

@1 + (L”d�Ll)

(L0d�Ll) T”dos

1 + sT”do

1

A (7.6)

Figure 7.6 shows the form of the transfer function, Ld(jw).

The transfer function expression Ld(s) can be written by direct reference to figures 7.4 and 7.5.This transfer function can also be derived from the basic electromagnetic equations describing


To develop the two transfer functions we must enter the ’culture’ of synchronous machinetheory.

7.4.4 Sudden short circuit

The unidirectional and alternating currents produced by a synchronous machine when thestator winding is short circuited suddenly from an initial condition of rated voltage and opencircuit take the form illustrated by figures 7.4 and 7.5.

The features of the alternating component of short circuit current behavior seen in figures 7.4and 7.5 can be described by the transfer function model given in the following section. The timeconstants in the numerators of the transfer functions are those of the decay of the waveformenvelopes shown in the bottom of figure 7.4 and in figure 7.5. The paramters Ld, L0

d, L”d are thereciprocals of waveform amplitudes as identified in figure 7.5.

7.4.5 Operational Impedances

When a machine is producing reactive power, but no real power, at its terminals its behavior,as can be observed in various tests including a sudden short circuit test, can be describedcompletely emperically by the transfer function relationship

vq(s) = Yd(s) = G(s)Ef d(s)� Ld(s)Id(s) (7.4)

in which

vq is the amplitude of the positive sequence AC voltage at the terminals the machineYd is the amplitude of the flux wave linking the stator windingEf d is the voltage applied to the field windingId is the amplitude of the positive sequence current in the stator winding

The transfer functions have the form

Ld(s) = (Ld � Ll)

0

@1 + (L0

d�Ll)(Ld�Ll)

T0dos

1 + T0dos

1

A

0

@1 + (L”d�Ll)

(L0d�Ll)

T”dos

1 + T”dos

1

A+ Ll (7.5)

G(s) =✓

11 + sT0

do

◆0

@1 + (L”d�Ll)

(L0d�Ll) T”dos

1 + sT”do

1

A (7.6)

Figure 7.6 shows the form of the transfer function, Ld(jw).

The transfer function expression Ld(s) can be written by direct reference to figures 7.4 and 7.5.This transfer function can also be derived from the basic electromagnetic equations describing


7.14 Generator dynamic models for simulations

7.14.1 Initial deveopment of transfer function model

While the synchronous modeling described in section 7.5 is straightforward and can accommodatecomprehensive treatment of magnetic saturation it is not used directly in the large scale dynamicsimulation programs presently in widespread use. Among the reasons for this:

The early development of electric machine modeling preceded the development of digitalcomputers and calculations that are routine today were impractical with the tools thenavailable; approximations and simplifications were essential. Many of the implementationsof machine modeling presently in use owe their form to such simplifications.

The inductance coefficients appearing in equations (7.7)-(7.17) are not readily availableand do not directly describe characteristics of the machine that can be measured in tests.

The long-established tests of synchronous machines yield values of the parameters appearingin the operational impedance, equation (7.5). The modeling used in production simulationassumes that the following parameters are available for the direct and quadrature axes:

L synchronous reactanceL0 transient reactanceL” subtransient reactanceT0 transient open circuit time constantT” subtransient open circuit time constant

Construction of a transfer function starts with breaking the operational impedance (7.5) into itspartial fractions. We use a and b to represent the ratios as follows:

a = (L0 � Ll)/(L � Ll)b = (L” � Ll)/(L0 � Ll)

Then (7.5) can be written as

L(s) =(L � Ll)(1 + saTa)(1 + sbTb)

(1 + sTa)(1 + sTb)+ Ll (7.80)

and the partial fraction expansion proceeds to yield

L(s) = (L � Ll)

✓a +

1 � a1 + sTa

◆✓b +

1 � b1 + sTb

◆+ Ll (7.81)

L(s) = (L � Ll)

✓ab +

(1 � a)b1 + sTa

+a(1 � b)1 + sTb

+(1 � a)(1 � b)

(1 + sTa)(1 + sTb)

◆+ Ll (7.82)










a = (L0 � Ll)/(L � Ll)b = (L” � Ll)/(L0 � Ll)


L(s) =(L � Ll)(1 + saTa)(1 + sbTb)

(1 + sTa)(1 + sTb)+ Ll (7.80)


L(s) = (L � Ll)

✓a +

1 � a1 + sTa

◆✓b +

1 � b1 + sTb

◆+ Ll (7.81)

L(s) = (L � Ll)

✓ab +

(1 � a)b1 + sTa

+a(1 � b)1 + sTb

+(1 � a)(1 � b)

(1 + sTa)(1 + sTb)

◆+ Ll (7.82)

Ld(s) = (Ld − Ll)

⎛

⎝

(

1 +(L′

d−Ll)

(Ld−Ll)T ′

dos

)(

1 + (L”d−Ll)(L′

d−Ll)

T”dos)

(1 + T ′

dos)(1 + T”dos)

⎞

⎠+ Ll

G(s) =1 + L”d−Ll)

(L′

d−Ll)T”dos

(1 + sT ′

do)(1 + sT”do)

Current interruption test reveals generator d-axis characteristics

7.13. SUDDEN SHORT CIRCUIT TEST 167

We now have the generator equations, for the special case of a simple impedance load, in thestandard form

dyrdt

= Ayr (7.79)

The dynamic behavior of the generator in a short circuit is defined by the eigenvalues of thematrix A, with the note that analysis by the direct use of A will be an approximation to theextent that it ignores the variation of generator inductances due to magnetic saturation.

(a)Initial

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.40

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time, sec

(b)Full

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time, sec

Figure 7.33: Simulation of three phase sudden short circuit - direct axis variablesRed - stator current

Blue - field winding currentMagenta - Iron circuit current

Black - terminal voltage

Figure 7.33 shows the result of using the differential equation (7.77) to simulate the behaviorof a generator when subjected to a three phase short circuit, from an initially open circuittest condition. The stator current (red) has the expected form of an initial rapid decay, amuch slower decay, and a final steady value. The rotor currents (blue, magenta) are seento be associated with the variation of the stator current. At the moment the fault is appliedthe currents in the field winding (blue) and iron circuit (magenta) jump to positive values toprovide the magnetomotive force needed to maintain the flux linkages in the rotor circuits.Then the current in the iron circuit, whose resistance is relatively high, decays rapidly. As themmf provided by the ’iron’ current disappears the current in the field winding increases tomake up for the quick loss of mmf. Note that current, small but not zero, continues to flowin the iron circuit as long as stator and field winding currents are changing. At the end of thetransient the field current has returned to its original value and the current in the iron circuithas returned to zero.

Sudden short circuit according to operational impedance, Ld(s)

7.14. GENERATOR DYNAMIC MODELS FOR SIMULATIONS 169

Similar partial fraction expansion can be used for the transfer function, G(s), in equation (7.6)to yield.

G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)

Then a transfer function diagram implementing equations (7.4), (7.5), and (7.6) with the partialfraction relationships can be drawn ’by inspection’. One possible form is shown in figure 7.34.

Figure 7.34: Transfer function block diagram corresponding to partial-fraction expression (7.83).

In the absence of saturation the abbreviations, a and b in the figure represent constants; recognitionof saturation makes them variables that must be reevaluated at each step of a numerical integrationprocess.

Expanding the expressions involving a and b in figure 7.34 results in figure 7.35.

Figure 7.35: Transfer function block diagram 7.34 with expressions in a, b expanded.

It should be noted that the development of figures 7.34 and 7.35 is completely emperical;it requires no knowledge of internal details of the machine, only values describing featuresobserved in test results. This point is important in commerce because specifications of machineryshould, wherever possible, be stated in terms of characteristics that can be measured directlyin normal operation or in practical tests.



G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)
















a = (L0 � Ll)/(L � Ll)b = (L” � Ll)/(L0 � Ll)


L(s) =(L � Ll)(1 + saTa)(1 + sbTb)

(1 + sTa)(1 + sTb)+ Ll (7.80)


L(s) = (L � Ll)

✓a +

1 � a1 + sTa

◆✓b +

1 � b1 + sTb

◆+ Ll (7.81)

L(s) = (L � Ll)

✓ab +

(1 � a)b1 + sTa

+a(1 � b)1 + sTb

+(1 � a)(1 � b)

(1 + sTa)(1 + sTb)

◆+ Ll (7.82)

L(s) = (L− ll)

!

(1 + asTa)

(1 + sTa)

(1 + bsTb)

(1 + sTb)

"

+ Ll



G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)









G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)









G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)
















a = (L0 � Ll)/(L � Ll)b = (L” � Ll)/(L0 � Ll)


L(s) =(L � Ll)(1 + saTa)(1 + sbTb)

(1 + sTa)(1 + sTb)+ Ll (7.80)


L(s) = (L � Ll)

✓a +

1 � a1 + sTa

◆✓b +

1 � b1 + sTb

◆+ Ll (7.81)

L(s) = (L � Ll)

✓ab +

(1 � a)b1 + sTa

+a(1 � b)1 + sTb

+(1 � a)(1 � b)

(1 + sTa)(1 + sTb)

◆+ Ll (7.82)



G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)









G(s) =✓

b1 + sTa

+(1 � b)

(1 + sTa)(1 + sTb)

◆(7.83)







Ld =Ldu − Ll

(1 + S(ψ))+ Ll

L′

d =L′

du− Ll

(1 + S(ψ))+ Ll

L”d =L”du − Ll

(1 + S(ψ))+ Ll

a =

L′

d− Ll

L′

d− Ll

b =L′′

d− Ll

L′

d− Ll

Magnetic saturation saturated inductance parameters are continually varying functions of flux

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

0 200 400 600 800 1000 1200 1400 1600

Field Amps

Sta

tor

Vo

ltag

e, p

u

VsTest

AirGapLine

Ifd-Parabolic

Ifd-Exponential

Figure 5.1.1

Gas Turbine Magnetization Curve

Se = S(ψd) = S(Ifd)

Se = S(ψag) = S(|Vs + jXlIs|)

Se = S(ψ′′) = S(|Vs + jX ′′Is|)

Se = S(ψ) = S(ψwhere?)

ψ teeth = ψag +KisIs

How to characterize saturation ???

Se = S(ψ) Theoretical

Devil is in the details

Open circuit operation

Flux behind subtransient reactance genrou and family

Flux in air gap gentpf

Stator current affects saturation

Se = S(ψ teeth) Flux in stator teeth gentpj and genopz

0 1 2 3 4 5 6 7 8 9 100.5

0.6

0.7

0.8

0.9

1

1.1

Volta

ge, p

u

VtopzVttpjVtrou

0 1 2 3 4 5 6 7 8 9 10Time, sec

40

60

80

100

120

Pow

er, M

W

PopzPtpjProu

Red genopzGreen genrouBlue gentpj

Transmission fault

Constant Efd

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

Vt

0 5 10 15 20 25 30Time, sec

0.2

0.4

0.6

0.8

1

Ifdz

Ifdj I

fdr

CurrentInterruption

Constant Efd


0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.40.85

0.9

0.95

1

Vt

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4Time, sec

0.2

0.4

0.6

0.8

1

Ifdz

Ifdj I

fdr

CurrentInterruption

Constant Efd


10-4 10-3 10-2 10-1 100 101 1020

0.5

1

1.5

2Am

plitu

de

10-4 10-3 10-2 10-1 100 101 102

Frequency, Hz

-60

-50

-40

-30

-20

-10

0

Phas

e, d

eg


∆vq(jω)

∆id(jω)= Ld(jω)

-0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8Reactive Power, per unit

0

0.5

1

1.5

2

2.5

Fiel

d C

urre

nt, p

er u

nit

V curve McNary 4

Red - Ifd test Black - Ifd model

Blue - voltage Magenta - power

Mbase = 86.00 MVA

Ld = 0.620 Lq = 0.400

s1 = 0.120 S12= 0.380

Ll = 0.175 Kis= 0.000

Ra = 0.0000 Afag= 620.0

If0 = 0.000

-0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8Reactive Power, per unit

0

0.5

1

1.5

2

2.5

Fiel

d C

urre

nt, p

er u

nit

V curve McNary 4

Red - Ifd test Black - Ifd model

Blue - voltage Magenta - power

Mbase = 86.00 MVA

Ld = 0.620 Lq = 0.400

s1 = 0.120 S12= 0.380

Ll = 0.175 Kis= 0.075

Ra = 0.0000 Afag= 620.0

If0 = 0.000

-0.2 -0.1 0 0.1 0.2 0.3 0.4Reactive Power, per unit

0

0.5

1

1.5

2

2.5

3

3.5

Fiel

d C

urre

nt, p

er u

nit

Magenta - Ifd test R,G,B - Ifd model / Kis = 0.0, 0.05, 0.1Cyan - voltage Black - powerMbase = 1559.00 MVALd = 1.990 Lq = 1.860s1 = 0.142 S12= 0.738Ll = 0.220 Kis= 0.100Ra = 0.0000 Afag=1800.0

ci-auto.pdf

Red genrouGreen gentpjBlue genopz

CurrentInterruption

AVR Auto

10 seconds

ci-auto-s.pdf


CurrentInterruption

AVR Auto

1 second

avrstep-offline.pdf


AVR referencestep

Off line

avrstep-online.pdf


AVR referencestep

On line

Ld(s) = (Ld − Ll)

⎛

⎝

1 +(L′

d−Ll)

(Ld−Ll)T ′

dos

1 + T ′

dos

⎞

⎠

⎛

⎝

1 + (L”d−Ll)(L′

d−Ll)

T”dos

1 + T”dos

⎞

⎠+ Ll

L(s) = (L− Ll)

(

ab+(1− a)b

1 + sTa

+a(1− b)

1 + sTb

+(1− a)(1− b)

(1 + sTa)(1 + sTb)

)

+ Ll

gopz dynamic model

Same saturation treatment as gentpj

Saturation is a function of BOTH air gap flux and stator current

Does not depend on assumed relationships among inductance coefficients

Is readily related to basic observable performance characteristics

Is simple to explain

Notes on generator models

ALL of our generator models are approximationswith regard to

magnetic saturationtorque developed during transmission faults. . . . . .

Which one of our models is best suited for any particular application is dependent on

the electrical configuration in questionthe disturbance under considerationthe time frame of the post disturbance event

Thank you

genopz synchronous machine model - genopz sychronous m… · of machine modeling presently in use...

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