unit 1.pdf

DRAFT and INCOMPLETE

Table of Contents

from

A. P. Sakis Meliopoulos

Power System Modeling, Analysis and Control

Chapter 5 _____________________________________________________________ 2

Modeling - Synchronous Machines ________________________________________ 2 5.1 Introduction____________________________________________________________ 2 5.2 The Construction of a Synchronous Machine ________________________________ 3 5.3 The Dynamical Equations of a Machine ____________________________________ 10 5.4 Park's Transformation __________________________________________________ 19 5.5 The Per Unit System ____________________________________________________ 26

5.5.1 The Per Unit System for a Synchronous Machine __________________________________ 29 5.5.2 Selection of Base Quantities ___________________________________________________ 29

5.6 Equivalent Circuits of a Synchronous Machine______________________________ 31 5.7 Synchronous Machine Torque Equation ___________________________________ 38 5.8 State Space Model of a Synchronous Machine_______________________________ 41 5.9 Steady State Analysis ___________________________________________________ 43 5.10 Synchronous Machine Performance Under Faults __________________________ 51

5.10.1 Subtransient and Transient Inductances_________________________________________ 52 5.10.2 Time Constants of a Synchronous Machine _____________________________________ 56 5.10.3 Summary of Synchronous Machine Parameters __________________________________ 58

5.11 Synchronous Machine Simplified Models__________________________________ 60 5.11.1 The Steady State Model_____________________________________________________ 60 5.11.2 Constant Main Field Winding Flux Model ______________________________________ 63 5.11.3 Constant Main Field and Damper Windings Flux Model ___________________________ 66 5.11.4 Summary of Simplified Models_______________________________________________ 70 5.11.5 One Axis Synchronous Machine Model ________________________________________ 71

5.12 Exciter Model and Voltage Control_______________________________________ 74 5.13 Synchronous Generator Capability Curves ________________________________ 74 5.14 Summary and Discussion _______________________________________________ 74

Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos

5.14 Problems ____________________________________________________________ 75

Chapter 5

Modeling - Synchronous Machines

5.1 Introduction The synchronous machine is almost exclusively used for bulk generation of electric power. The synchronous machine is one of the most complex components of an electric power system as well as an important component of an electric power system. In this chapter we examine the synchronous machine and develop appropriate models for various applications. Synchronous machines are manufactured from small sizes to very large sizes (up to 1,250 MVA) in order to capitalize on economies of scale inherent in large power generation plants. The growth in size brought about a growth in design sophistication. In addition, the concern about the synchronous generator performance during steady state operation and transients provided the impetus for application of sophisticated control schemes for the synchronous machine. For example, during steady state operation, a control loop is active to maintain terminal voltage at specified values, while during transients another control loop (power system stabilizer) is activated to stabilize the machine. The study of these phenomena and the control requirements dictate the need for appropriate mathematical models of the synchronous machine. The objective of this chapter is to fill in this need. A brief outline of the chapter is as follows: First, the construction of the synchronous machine is examined. The mechanism of energy conversion is explained in terms of interacting magnetic fields produced by coils. This analysis leads to the general dynamical equations. Then it is shown that an appropriate selection of a per unit system will lead to equivalent circuit representation of a synchronous machine. The equivalent circuits together with the torque equation lead to a nonlinear state space model of a synchronous machine. The particular state space model employed here selects as states the electric current flowing in the machine. This model is then utilized to study a) the steady state performance of the machine, b) the performance under faults, and c) the performance during transients. In the final section of the chapter, we investigate the relationship between the state space model of the synchronous machine and various simplified models which may be used for specific studies, i.e. the classical model, etc. The simplified models are presented in terms of the assumptions which must be applied

Page 2 Copyright © A. P. Sakis Meliopoulos – 1990-2006


to the general model for their derivation. This analysis provide insight into the applicability of the simplified models.

Figure 5.1 Design Details of a Salient Pole Synchronous Machine

5.2 The Construction of a Synchronous Machine There are two basic designs of a synchronous machine: (a) the salient pole machine and b) the cylindrical rotor machine. These two designs are illustrated in Figures 5.1 and 5.2 respectively. In general salient pole machines are characterized with a large number of

poles. The speed of the machine is dependent upon the number of poles, em Pωω 2

= and

therefore the larger the number of poles the lower the speed will be. These units are suitable with slow speed prime movers, for example, a hydroturbine. Cylindrical rotor machines, see Figure 5.2, are characterized with a small number of poles (2 or 4) and are used with high speed prime movers, for example, steam turbines. The rotor is a cylinder on which slots have been engraved to accept the sides of the field coil. For the illustrated two pole machine two sets of slots are required. In general for a P-pole machine P-sets of slots will be required. The field coil is distributed in the rotor slots in such a way that a dc current through the coil will produce a magnetic flux density in the air gap which has approximately sinusoidal distribution. The maximum of the magnetic flux occurs along the d-axis. In normal operating conditions the rotor, the field

Copyright © A. P. Sakis Meliopoulos – 1990-2006 Page 3


coil, and the d-axis are rotating with the synchronous mechanical speed. The rotor and the field winding are schematically represented with the inductance Lf.

Figure 5.2 Design Details of Three-Phase Synchronous Turbogenerator In large synchronous machines the rotor bears another set of windings, the damper windings. These are short circuited windings which are placed in slots cut on the surface of the rotor. During normal operating conditions these windings are inactive because, as it will be shown, they link a constant magnetic flux and thus no voltage will be induced in these coils. However during transient conditions, electric voltage and current is induced in these windings. A torque is generated in a direction such as to bring the system back to synchronous speed. Therefore they introduce damping to the system, which is extremely beneficial for the performance of the machine during transients. Slots similar to those of the rotor are engraved in the stator as illustrated in Figure 5.2. Three sets of windings are placed in these slots: phase A, phase B and phase C windings. The figure illustrates the sets of slots which carry the phase A windings. Note that each set occupies 600 of the stator. In reality, the design of the stator is more complicated than the one illustrated in Figure 5.2. For example the set of slots of phase A overlaps with the sets of slots of phases B and C. This is done in such a way as to achieve sinusoidal distribution of the magnetic flux inside the air gap due to the current flowing in any one of the three phases. For simplicity, however, this aspect of the design is not shown.



(stationary reference)

rotor magnetic axis( or d-axis)

Figure 5.3 Schematic Representation of the Main Windings of A Synchronous

Machine Figure 5.3 summarizes the design of a synchronous machine in a schematic fashion. All three phase windings are illustrated. Also the field winding is illustrated. Note also the magnetic axes of the magnetic field generated by the various windings. It is expedient to examine the operation of the machine during steady state conditions. For this purpose the position of the rotor will be measured from the magnetic axis of phase A which is stationary. Specifically, θ m will define the angle between the phase A magnetic axis and the rotor magnetic axis or d-axis. Since the rotor rotates, the angle θm is a function of time. We postulate that the rotor position is given with:

P

ttt mmsmπδωθ ++= )()( (5.1)

Where msω is the synchronous angular speed (mechanical) of the generation rotor and P is the number of magnetic poles. Note that at steady state the rotor rotates with constant speed ωms . In this case the angle is a constant and independent of time, i.e. mm t δδ =)( . In this case, the variable mδ is the angle between phase A magnetic axis and rotor quadrature axis at time zero, since

( )PP mmmπθδπδθ −=→+= 0)0( .



Under steady state conditions, the following relationships hold

esms Pωω 2

= , fes πω 2=

where esω is the synchronous electrical angular speed, f is the frequency of the generated voltage, and P is the number of magnetic poles of the synchronous machine. Upon multiplication of equation (5.1) by (P/2), equation (5.2) is obtained:

2

)()( πδωθ ++= ttt eese (5.2)

where )(2

)( tPt me δδ =

)(2

)( tPt me θθ =

In subsequent discussions, we use above transformed equation. For simplicity, we drop the subscript e. In summary, in Figure 5.3 two references have been defined. The spatial reference is defined to be the stationary magnetic axis of phase A. Spatial angles are measured from this reference in the direction of rotation. Also a time reference has been selected with equation (5.1). Figure 5.4 illustrates the magnetic flux distribution in the air gap of a synchronous machine due to phase A electric current (part a), phase B electric current (part b), phase C electric current (part c), resultant magnetic flux due to phases A, B, and C currents (part d), and main field winding current (part e). In the Figure the rim of the stator has been sketched in a straight line. Thus point A is identical to point B. Consider Figure 5.4a. When electric current ia(t) flows through phase A, a magnetic flux is established in the air gap which has the indicated spatial distribution. Analytically, the magnetic flux density is expressed with αα cos)(),( tkitB aas = (5.3)

( ) ⎟⎠⎞

⎜⎝⎛ −=

32cos)(, παα tkitB bbs

( ) ⎟⎠⎞

⎜⎝⎛ −=

34cos)(, παα tkitB ccs

where



),( tBas α is the magnetic flux density in position α of the air gap and time t, due to electric current, , at time t in phase A )(tia

),( tBbs α is the magnetic flux density in position α of the air gap and time t, due to electric current, , at time t in phase B )(tib

),( tBcs α is the magnetic flux density in position α of the air gap and time t, due to electric current, , at time t in phase C )(tic

k is a proportionality constant depending on the design of the synchronous machine and especially the length of the air gap.

Note that α, the spatial angle is measured from the phase A magnetic axis. If the currents

, , and are sinusoidal currents, the magnetic flux density )(tia )(tib )(tic ),( tBas α , ),( tBbs α , and ),( tBcs α will be also a function of time. Under sinusoidal steady state

and balanced conditions, the electric currents will be: ( )Ima tIti ϕθ += )(cos)(

⎟⎠⎞

⎜⎝⎛ −+=

32)(cos)( πϕθ Imb tIti (5.4)

⎟⎠⎞

⎜⎝⎛ −+=

34)(cos)( πϕθ Imc tIti

where Iϕ is the phase of the phase A electric current phasor. Now consider the magnetic flux due to current, , of phase A. From equation 5.3 it is

obvious that the points of zero magnetic flux are stationary and occur at

)(tia

2πα ±= . Thus

the magnetic flux density ),( tBas α is a pulsating wave. The figure illustrates a snapshot taken at time when the electric current is positive maximum, i.e. θ(t1t )( 1tia 1) + ϕI = 0. Similarly the pulsating magnetic flux density waveforms of phases B and C are derived and shown in Figures 5.4b and 5.4c.



( a )

( b )

( c )

( d )

( e )

Figure 5.4 Illustration of Synchronous Machine Magnetic Fluxes It should be mentioned that what is shown in Figure 5.4 is the fundamental component of the magnetic flux density. In reality the waveform does have harmonics because it is a practical impossibility to design a winding which will produce a perfectly sinusoidal waveform. The resultant magnetic flux density in the gap is computed by superposition of the individual contributions: ( ) ( ) ( )tBtBtBtB csbsass ,,,),( αααα ++= (5.5) Direct substitution and after some trigonometric manipulations:



( παϕθα −−+= Is ttB )(cos23),( ) (5.6)

The last equation represents a sinusoidal waveform which rotates with speed

sdttd

dtd ωθα

==)( (5.7)

In conclusion, the three pulsating magnetic flux waveforms generate a rotating sinusoidal magnetic flux waveform. The speed of rotation is the synchronous speed sω . Now consider the field winding. A dc electric current flows through it under steady state conditions. This current generates a magnetic flux density waveform of sinusoidal shape. The analytic expression for this flux is ( ) ( )( )παθα −−= tRtBr cos, . This waveform is stationary and constant with respect to the rotor. But since the rotor rotates with synchronous speed, so does the field magnetic flux waveform.

armature windingproduced magneticfield ( all phases)

rotor winding producedmagnetic field

air gap

A

α

Figure 5.5 Illustration of the Stator and Rotor Produced Magnetic Fluxes

Above results are summarized in Figure 5.5. In the figure, α measures the angle of a point on the stator from the spatial reference and )(tθ measures the position of the d-axis from the reference. Two magnetic flux waveforms are generated, one due to stator currents and the other due to the field winding current. Both waveforms rotate with synchronous speed. These waveforms try to align with each other (think of them as two electromagnets). This action generates the electromagnetic torque. In the next section will analyze the same phenomena in a quantitative way leading to a set of dynamical equations describing the machine.



5.3 The Dynamical Equations of a Machine Qualitative analysis of a synchronous machine, similar to the one in the previous section, provides insight in the operation of such machines but does not enable engineering analysis and design. For this purpose, analytical models of synchronous machines are needed for precise quantitative analysis of the peormeance these machines. For model development, two approaches can be taken: a) Field Approach: Given the design of the synchronous machine, the magnetic

flux densities can be computed as well as their interaction leading to the computation of the electromagnetic torque. Also an appropriate mathematical model can be developed for the electrical system of the machine.

b) Circuit Approach: The synchronous machine can be viewed as a set of mutually

coupled inductors, which interact among themselves to generate the electromagnetic torque. Straightforward circuit analysis leads to the derivation of an appropriate mathematical model.

Both approaches are equivalent, leading to equivalent mathematical descriptions of a synchronous machine. The circuit approach however is simpler and will be followed here. Figure 5.6 illustrates the windings of a synchronous machine: three phase windings, A, B, and C, a field winding, f, and two damper windings D, Q acting along the d- and q- axes respectively. The winding of phase A is schematically represented with the inductance La. Similarly phases B and C are schematically represented with the inductors Lb and Lc. In general the three phase winding may be delta or wye connected. Figure 5.6 illustrates a wye connection. For generality it will be assumed that the neutral of the synchronous machine is grounded through an impedance consisting of inductance Ln and resistance rn. Note that with the exception of the inductor Ln, all other inductors are mounted on the same magnetic circuit and thus they are all magnetically coupled.



va(t)

vb(t)

vc(t)ic(t)

ib(t)

in(t)

ia(t)

phase amagnetic axis reference

θ

d-axis

q-axisra

rbrc

iD(t)

iQ(t)

ω

if(t)

vf(t)+_

vn(t)

Figure 5.6 Representation of a Synchronous Machine as a Set of Mutually

Coupled Windings Application of Kirchoff's voltage law to the circuit of Figure 5.6 yields

dt

tdtirtv aaaa

)()()( λ−−=

dt

tdtirtv b

bbb)(

)()(λ

−−=

dt

tdtirtv c

ccc)(

)()(λ

−−=

dt

tdtirtv f

fff

)()()(

λ−−=−

dt

tdtir DDD

)()(0 λ−−=



dt

tdtir Q

QQ

)()(0

λ−−=

In above equations λa(t) is the magnetic flux linkage of phase A λb(t) is the magnetic flux linkage of phase B λc(t) is the magnetic flux linkage of phase C λf(t) is the magnetic flux linkage of the field winding, f λD(t) is the magnetic flux linkage of the D-damper winding λQ(t) is the magnetic flux linkage of the Q-damper winding. All other variables are defined in Figure 5.6. It is expedient to write above equations in compact matrix notation. To this purpose define:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

)()()(

)(tvtvtv

tv

c

b

a

abc

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

00

)()(

tvtv

f

fDQ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

c

b

a

abc

rr

rR

000000

Rabc = rI, if ra = rb = rc = r, here I is the 3x3 identity matrix.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

Q

D

f

fDQ

rr

rR

000000

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

111

13



⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

)()()(

)(tititi

ti

c

b

a

abc

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

)()()(

)(tititi

ti

Q

D

f

fDQ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

)()()(

)(ttt

t

c

b

a

abc

λλλ

λ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

)()()(

)(ttt

t

Q

D

f

fDQ

λλλ

λ

The voltage equations now can be written in the following compact form:

)()()()( 3 tdtdtiRtvtv abcabcabcnabc λ−−=− 1

)()()( tdtdtiRtv fDQfDQfDQfDQ λ−−=

In above equations the magnetic flux linkages are complex functions of the rotor position and electric currents flowing in the various winding of the machine. The general expressions are: Stator Windings )()()()()()()( tiLtiLtiLtiLtiLtiLt QaQDaDfafcacbabaaaa +++++=λ )()()()()()()( tiLtiLtiLtiLtiLtiLt QbQDbDfbfcbcbbbabab +++++=λ )()()()()()()( tiLtiLtiLtiLtiLtiLt QcQDcDfcfcccbcbacac +++++=λ Rotor Windings )()()()()()()( tiLtiLtiLtiLtiLtiLt QfQDfDfffcfcbfbafaf +++++=λ



)()()()()()()( tiLtiLtiLtiLtiLtiLt QDQDDDfDfcDcbDbaDaD +++++=λ )()()()()()()( tiLtiLtiLtiLtiLtiLt QQQDQDfQfcQcbQbaQaQ +++++=λ The notation in above equations is obvious. Lii is the self inductance of winding i, while Lij is the mutual inductance between windings i and j. Many of the inductances in above equations are dependent on the position of the rotor which is time varying. Thus these inductances are time dependent. In matrix notation above equations read

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡)()(

)()()()(

)()(

titi

tLtLtLtL

tt

fDQ

abc

rrrs

srss

fDQ

abc

λλ

where

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

ccbcac

bcbbab

acabaa

ss

LLLLLLLLL

tL )(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

cQcDcf

bQbDbf

aQaDaf

sr

LLLLLLLLL

tL )(

)()( tLtL T

srrs =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

QQDQfQ

DQDDfD

fQfDff

rr

LLLLLLLLL

tL )(

The dependence of the inductances on rotor position is explained next. Stator Self Inductances Laa, Lbb, and Lcc in general depend on rotor position. An approximate expression of this dependence is:

( ))(2cos tLLL msaa θ+=

⎟⎠⎞

⎜⎝⎛ −+=

32)(2cos πθ tLLL msbb (5.8)

⎟⎠⎞

⎜⎝⎛ −+=

34)(2cos πθ tLLL mscc



where )(tθ is the direct axis position relative to the phase A magnetic axis. Note that as

the rotor rotates, its position )(tθ is described with equation 2

)()( πδωθ ++= ttt s . A

typical variation of Lii is shown in Figure 5.7

0

L

LsLm

π θ

ii

Figure 5.7 Typical Variation of Phase A Self Inductance as a Function of θ Rotor Self Inductances Lff, LDD, and LQQ are approximately constants and can be symbolized with: QQQDDDfff LLandLLLL === ,, (5.9) Stator Mutual Inductances Lab, Lac, and Lbc are negative. They are functions of the rotor position . Approximate expressions for these functions are: θ( )t

⎟⎠⎞

⎜⎝⎛ +−−==

6)(2cos πθ tLMLL msbaab

⎟⎠⎞

⎜⎝⎛ −+−−==

32

6)(2cos ππθ tLMLL mscbbc (5.10)

⎟⎠⎞

⎜⎝⎛ −+−−==

34

6)(2cos ππθ tLMLL msacca

A typical variation of Lij is shown in Figure 5.8



0

-MsLm

π θ

Lba

Figure 5.8 Typical Variation of the Mutual Inductance Between Two Phase Windings

Rotor Mutual Inductances LfD, LfQ, and LDQ are constants, independent of , because the rotor windings are stationary with each other.

θ( )t

RDffD MLL ==

(5.11) 0== QffQ LL 0== QDDQ LL Stator to Rotor Mutual Inductances Laf, Lbf, and Lcf are dependent upon the rotor position θ(t) as follows.

( ))(cos tMLL Ffaaf θ==

⎟⎠⎞

⎜⎝⎛ −==

32)(cos πθ tMLL Ffbbf (5.12)

⎟⎠⎞

⎜⎝⎛ −==

34)(cos πθ tMLL Ffccf

Similarly

( ))(cos tMLL DDaaD θ==

⎟⎠⎞

⎜⎝⎛ −==

32)(cos πθ tMLL DDbbD (5.13)

⎟⎠⎞

⎜⎝⎛ −==

34)(cos πθ tMLL DDccD



The damper winding Q is orthogonal to the D winding. Therefore

⎟⎠⎞

⎜⎝⎛ −==

2)(cos πθ tMLL QQaaQ

⎟⎠⎞

⎜⎝⎛ −−==

32

2)(cos ππθ tMLL QQbbQ (5.14)

⎟⎠⎞

⎜⎝⎛ −−==

34

2)(cos ππθ tMLL QQccQ

Actually the inductances are perturbed from sinusoidal variation with harmonics. Generally speaking these harmonics are kept low with the use of distributed coils, double layers and fractional pitch. In this textbook, these harmonics will be neglected.

Since 2

)()( πδωθ ++= ttt s , it is obvious that the inductance matrix is time dependent

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡)()(

)()()()(

titi

LtLtLtLt

fDQ

abc

rrrs

srss

fDQ

abc

λλ

or )()()( titLt =λ In summary the voltage equations read:

dt

tdtiRtvtv abcabcabcnabc

)()(1)()( 3λ

−−=−

dt

tdtiRtv fDQ

fDQfDQfDQ

)()()(

λ−−=

In above equations, the matrices Lss(t), Lsr(t), Lrs(t) and Lrr(t) should be substituted with the expressions (5.8) through (5.10). The motion of the generator rotor is determined by the motion equation which in the general form is

emm TT

dttd

J −=2

2 )(θ

where J is the rotor moment of inertia,



Tm is the mechanical torque applied to the rotor, Te is the electromagnetic torque developed by the generator, and

θm(t) is the rotor position.

position can be substituted with the electrical rotor position (see equation 5.2) ielding

The rotory

eme TT

dttd

PJ

−=2

2 )(2 θ

The mechanical torque Tm is determined by the prime mover. The electrical torque is determined by the amount of power converted from mechanical into electrical. It can be omputed by using the principle of virtual work displacementc :

m

fielde

WT

θ∂∂

=

where Wfield is the total energy stored in the magnetic field of the synchronous machine, nd θm is the position of the rotor.

he total magnetic energy stored in the windings of the synchronous machine is given by

a T

[ ] [ ] ⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

)()(

)()()(

)()(21

)()(

)()(21

titi

LtLtLtL

titititi

ttWfDQ

abc

rrrs

srssTfDQ

Tabc

fDQ

abcTfDQ

Tabcfield λλ

An alternative way is to compute the total power converted from mechanical into electrical and to divide this power by the speed of the rotor. Both approaches give the same answer. We use here the latter. The total power converted from mechanical into electrical, , is: )(tPem

dttd

tW

dttd

tTtP m

m

fldmeem

)()(

)()()(

θθ

θ∂

∂==

omputed with:

c

)())(

()())(()( tidt

tdti

dttdtP fDQ

TfDQabc

Tabcem

λλ+= , or

nd

)()()()()( titetitetP fDQTfDQabc

Tabcem +=

a

)()(

2)()(

)(ttPP

ttP

tT em

m

eme ωω

==



Obviously, the equations become quite complex. In the next section will introduce a transformation which simplifies above equations. Later, the model will be further simplified with the application of the perunit system leading to equivalent circuits.

5.4 Park's Transformation Park’s idea is based on the observation that the synchronous machine circuit equations are tremendously simplified if a transformation T is applied to the model that will make the matrix L(t) a constant matrix independent of time. Let this transformation be T, defined with: )('=(t)or )()(' -1 tTtTt λλλλ = )('=(t)or )()(' -1 tiTitTiti = Upon substitution of and i(t) in the generator equations: λ( )t (t)i' )()(' 1-1 −= TtLtT λ Premultiplication of the equation with T yields (t)i' )()(' 1−= TtTLtλ or (t)i' ')(' Lt =λ where: 1' )( −= TtTLL Park’s transformation T is selected in such a way as to make the matrix 'L a constant matrix. This transformation is defined below

⎥⎦

⎤⎢⎣

⎡=

IP

T0

0

where



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

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−=

αθαθθαθαθθ

2)(sin)(sin)(sin2)(cos)(cos)(cos

21

21

21

32

ttttttP

3

2πα =

I the 3x3 identity matrix 0 a 3x3 zero matrix

2

)()( πδωθ ++= ttt s (position of d-axis)

The above selected matrix P has the following property (which can be easily checked): TPP =−1

The proof that the matrix L’ is constant is obtained by direct substitution and subsequent manipulations. First observe that the inverse of Park’s transformation is

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

−−

IP

IP

TT

00

001

1

Recall that the matrix L(t) is:

⎥⎦

⎤⎢⎣

⎡=

rrTsr

srss

LLLL

tL )(

Substitution into the expression for the transformed inductance matrix L yields

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

rrT

rs

srT

ssT

rrrs

srss

LPLPLPPL

IP

LLLL

IP

L0

00

0'

By direct substitution, the following is obtained

invariant) ( tan00

0000

timetconsL

LL

PPL

q

d

oT

ss =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=



invariant) ( tan00

0000

timetconskM

kMkMPL

Q

DFsr =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

invariant) ( tan00

0000

timetconskM

kMkM

PL

Q

D

FT

rs =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

invariant) ( tan00

00

timetconsL

LMML

L

Q

DR

RF

rr =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

where: ss MLL 20 −=

mssd LMLL ⎟⎠⎞

⎜⎝⎛++=

23

mssq LMLL ⎟⎠⎞

⎜⎝⎛−+=

23

23

=k

To complete the model, consider the generator voltage equations:

dt

tdtRitvtv n)()()(

01

)( 3 λ−−=⎥

⎦

⎤⎢⎣

⎡−

Upon substitution of the currents and flux linkage with )(')( 1 tiTti −= )(')( 1 tTt λλ −= and premultiplication of the resulting equation with T, the following is obtained

dt

tdTTtiTRTtvtvT n)()(')(

01

)('1

13 λ−− −−=⎥

⎦

⎤⎢⎣

⎡−

Now define:



)()()(

)()()(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

tvtvtv

Ptvtvtv

c

b

a

q

d

o

⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡−

)()(

00

)()(v)()(v

=)(01

)(=(t)v' q

0

3

tvtv

tvttvt

tvtvTfDQ

odq

f

d

n

The voltage equations of the generator can be expressed in terms of the electric currents, yielding the so called current model or they can be expressed in terms of the magnetic flux linkages, yielding the so called flux model. Both model are given below. Electric Current Model

))(''()(')(' 11 tiLTdtdTtiTRTtv −− −−=

)('')(' tiLt =λ Magnetic Flux Model

))('()('')(' 111 tTdtdTtLTRTt λλ −−− −−=v

)('')(' 1 tLti λ−= Both models can be further simplified by the following results which are obtained by straightforward manipulations:

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=−

Q

D

f

rr

rr

rr

TRT

000000000000000000000000000000

1

dt

tdtdt

dTTdt

tdTTtdt

dTTtTdtdT )(')(')(')('))('(

11

11 λλλλλ +=+=

−−

−−

By direct computation



⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−=

−

0000000000000000000000)(0000)(00000000

1 tt

dtdTT

ωω

, dt

tdt )()( θω =

Thus

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−=−

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

000

)()()()(

0

))('( 1

tttttt

dtdtt

tt

tTdtdT

Q

D

f

q

d

o

d

q

λλλλλλ

λωλω

λ

Upon direct substitution of above results and rearranging the equations the current model is:

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

=−

Q

q

D

f

d

o

Q

DFd

D

f

Qq

Q

q

D

f

d

o

iiiiii

rrkMkML

rr

kMLrr

vv

vv

vv

00000000000000000

00000000

0

0ωωω

ωω

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

Q

q

D

f

d

o

QQ

Qq

DRD

Rff

Dfd

iiiiii

dtd

LkMkML

LMkMMLkMkMkML

L

00000000

000000000000000

where k = 32

and the flux model is:



⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

Q

q

D

f

d

Q

q

D

f

d

o

Q

DFd

D

f

Qq

Q

q

D

f

d

o

dtd

rrkMkML

rr

kMLrr

vvvv

vv

λλλλλλ

λλλλλλ

ωωω

ωω0

00000000000000000

00000000

Another useful form of the generation model equations is in the state space form dxdt

f x u= ( , ) which is obtained as follows.

Electric Current Model. Define the following [ ]v v v v vG

To d f q= − 0 0

[ ]i i i i i i iG

To d f D q Q=

[ ]λ λ λ λ λ λ λG

To d f D q Q=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

Q

D

f

rr

rr

rr

R

000000000000000000000000000000

1

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

=

000000000000000000000

0000000000

2

DFd

Qq

kMkML

kML

R



⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

QQ

Qq

DRD

Rff

Dfd

o

eq

LkMkML

LMkMMLkMkMkML

L

L

00000000

00000000000000

Now the current model is written as follows:

dt

tdiLtiRRtv GeqGG

)()()()( 21 −+−= ω

)()( tiLt GeqG =λ or

)()()()( 121

1 tvLtiRRLdt

tdiGeqGeq

G −− −+−= ω

)()( tiLt GeqG =λ The flux model is obtained by eliminating the electric current from above model, i.e.

)()()()( 121 tvtLRR

dttd

GGeqG −+−= − λω

λ

)()( 1 tLti GeqG λ−=

Above models of a synchronous machine are illustrated in Figure 5.9.



-

+Vf

rf

if

Lf

rD

iD

LD

rQ

iQ

LQ

+

-

oV

io

- +

+

-ed

r

dV

id

r Lo

-+

+

-eq

r

qV

iq

Ld

kM f

kM D

kM Q

M R

Lq

RotorStator

Figure 5.9. Equivalent Synchronous Machine Circuits QQqqd ikMiLe ωω +=

DDffddq ikMikMiLe ωωω ++= The equivalent synchronous machine circuits of Figure 5.9 can be further simplified with the introduction of the per unit system. This is discussed next.

5.5 The Per Unit System Many times it is expedient to work with normalized (per unitized) quantities instead of physical quantities. For this purpose, an appropriate system of units is introduced and all quantities are expressed as multiples of the corresponding unit. Judicious selection f the units can result in: (a) the equations expressing physical laws, such as ohm’s law, Kirchoff’s laws, etc. remain of the same form in the per unit system, (b) some models are



simplified, and (c) the resulting numbers are of the same order of magnitude and therefore easier to compute. In computer based analysis procedures, numerical round off errors are much lower.

Table 5.1 Physical Quantities, Symbolism and Units.

Symbol Units Voltage v V Current i A Power p, q or S W, VAr or VA

Flux Linkage λ Wb Resistance r Ω Inductance L H

Time t sec Angular Velocity ω (sec)-1

Angle δ rad The normalization procedure is quite simple and mimics the basic procedure by which the various unit systems have been developed for example the metric system. Consider for example the physical quantities of Table 5.1. Base quantities can be arbitrarily selected for each one of the listed physical quantities. For example 10 Volts for the voltage base quantity, 5 Amperes for the current base quantity, etc. Then the per unitized value of a quantity will equal the true value of this quantity divided by the base value. To continue with the example a 5 volt voltage will equal 0.5 (= 5/10) in per unit etc. Thus the per unitization procedure is very simple. However if all base quantities are arbitrarily selected, equations describing physical laws will have to be modified in the new base system. In general equations describing physical laws in an arbitrarily selected per unit system are complex. To avoid this complication it is necessary to select the base quantities in such a way that physical laws are expressed with the same equations in terms of per unitized quantities and a physical system of units such as the metric (MKSA) system for example. In order to clarify the point assume that we selected from the quantities of Table 5.1 the base for power, voltage and speed arbitrarily as SB, VB, and

while the remaining are selected with: ω B

B

Bt ω1

= (seconds)

B

BB V

SI = (Amperes)

BBB tV=λ (Webers)

B

BB I

L λ= (Henrys)



B

BB I

VR = (Ohms)

Then a good number of physical laws will be expressed with the same equations in both the perunit system and metric system of units. The following example illustrates this simple concept

Example E5.1: Consider the equation dtdiLriv −−= which is expressed in the metric

system. That is v is measured in volts, r in ohms, I in Amperes, L in henries, t in seconds. Develop the per unitized version of this equation. Solution: Divide each term of the equation by VB or any of its equivalent expressions:

B

BB

B

BBBB t

ILt

IRV ===λ

The result is:

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−=

B

B

BBBB

ttd

Iid

LL

Ii

Rr

Vv

or

u

uuuuu dt

diLirv −−=

where the subscript u denotes per unitized quantities. Obviously the per unitized equation is of identical form as the original equation. In multicircuit networks which are not interconnected, but magnetically coupled, there is a certain degree of freedom in selecting the base quantities for the per unit system. One should exercise this freedom towards simplification of the equations and possible physical interpretation of mathematical models. This stated objective can be defined as follows: a) The form of the system voltage equations should be exactly the same whether the

equations are expressed in pu or metric system units. b) The form of the system power equations must be invariant (same in pu or actual

metric units). c) The per unit system should be selected in such a way that mutual inductances can be

represented with T-equivalent circuits, after per unitization.



In chapter four the per unit model of a power transformer was developed. Here we apply the same procedure to develop the per unit model of a generator. 5.5.1 The Per Unit System for a Synchronous Machine The concepts developed in Chapter 4 are readily applicable to the normalization of the synchronous machine circuits. There are four magnetically coupled circuits. Normalization of the synchronous machine equations requires the following steps: Step 1: Select a base for the armature windings SB usually the one phase rated power of the machine VB usually the phase to neutral rated voltage of the machine usually the synchronous angular frequency ω B

Step 2: Select the base for the remaining windings (main field, D-damper, Q-damper) as follows:

same SB and ω B

VB on the assumption of equal mutual flux Above procedure guarantees that the form of the equations describing the synchronous machine remains the same in the per unit system. This leads to the simple procedure of simply replacing the actual quantities with their per unit quantities. Subsequently the above steps are described in more detail. Eventually the results of steps 1 and 2 will be summarized in Table 5.3. The procedure leads to the generator equivalent circuit of Figure 5.12. 5.5.2 Selection of Base Quantities Assume that base quantities have been selected for the stator winding as described in step 1. Then consider the magnetic flux linkage equations: )()( tiLt GeqG =λ In general dmdd LL l+= qmqq LL l+= L L f mf= + l f

D

Q

L L D mD= + l L L Q mQ= + l Notice that always



kM L Lf md= mf

kM L LD md m= D

kM L LQ mq m= Q

M L LR mf m= D The selection of the base quantities is done on the basis of equal mutual flux linkages as follows: Define:

kM

kMLkM

kMLf

R

D

mf

f

f

md( )= = =

kMkM

LkM

kMLD

R

f

mD

D

D

md( )= = =

kLkM

kMLQ

mQ

Q

Q

mq= =

Then R k RfB f B= 2

L k LfB f B= 2

R k RDB D B= 2

L k LDB D B= 2

R k RQB Q B= 2

L k LQB Q B= 2

etc. Above equations complete the selection of the base quantities. The results are summarized in Table 5.2.

Table 5.2. Summary of a Synchronous Machine Per unit System

1. Select Stator Circuit Base Quantities - per phase power base (rated power/3) BS Bω - angular speed base (rated, electrical) - phase to line voltage base (rated L-L voltage/sBV 3 ) Then the derived based quantities are: BBt ω/1=



sBBsB VSI /= BsBsB tV=λ sBsBsB IL /λ= sBsBsB IVR /= 2. Define the following constants:

md

f

f

mf

D

Rf L

kMkML

kMM

k ===

md

D

D

mD

f

RD L

kMkML

kMMk ===

mq

Q

Q

mQQ L

kMkML

k ==

Above results are summarized in Table 5.2.

Table 5.2. Summary of a Synchronous Machine Per unit System

Armature Field Circuit D-Axis Damper Circuit

Q-Axis Damper Circuit

sBBB VS ,,ω sBfBB VkS ,,ω sBDBB VkS ,,ω sBQBB VkS ,,ω

BBt ω/1=

sBBsB VSI /=

BsBsB tV=λ

sBsBsB IL /λ=

sBsBsB IVR /=

BBt ω/1=

fsBfB kII /=

sBffB k λλ =

sBffB LkL 2=

sBffB RkR 2=

BBt ω/1=

DsBDB kII /=

sBDDB k λλ =

sBDDB LkL 2=

sBDDB RkR 2=

BBt ω/1=

QsBQB kII /=

sBQQB k λλ =

sBQQB LkL 2=

sBQQB RkR 2=

3. Mutual Inductance Base sBfsBfBfsB LkLLM ==

sBDsBDBDsB LkLLM ==

sBQsBQBQsB LkLLM ==

fBDBDfB LLM =

5.6 Equivalent Circuits of a Synchronous Machine



With the defined per unit system, equivalent circuits for the representation of a synchronous machine can be developed. The procedure will be demonstrated for the development of the q-axis equivalent circuit which has been selected for its simplicity. The results will be extended for the development of the d-axis equivalent circuit. The 0-axis equivalent circuit is quite simple. q-Axis Equivalent Circuit: Consider the voltage equations for the q-axis voltage and currents:

dtdi

kMdtdi

LriikMikMiLv QQ

qqqDDffddq −−−++= ωωω

dtdi

Ldtdi

kMir QQ

qQQQ −−−=0

Divide the first equation by , remembering that sBV BQBQsBBsBsBsBsBDBDsBBfBfsBBsBsBBsB tIMtILIRIMIMILV // ====== ωωω The result will be

)(

)(

)(

)(

B

QB

Q

QsB

Q

B

sB

q

sB

q

sBsB

q

DBDsBB

dD

fBfsBB

ff

sBsBB

dd

sB

q

ttd

Ii

d

MkM

ttd

Ii

d

LL

IRri

IMikM

IMikM

ILiL

Vv

−−−++=ωω

ωω

ωω

or

u

QuuQ

u

ququmququuququ dt

dikM

dtdi

Lirev )()( −+−−= l

Similarly divide the second equation by , remembering that sBV BQBQBBsBQsBQBQBsB tILtIMIRV // === The result will be

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−=

B

QB

Q

QB

Q

B

sB

q

QsB

Q

QBQB

QQ

ttd

Ii

d

LL

ttd

Ii

d

MkM

IRir

0

or



u

QuQumQu

u

quQuQuQu dt

diL

dtdi

kMir )(0 l+−−−=

A summary of above results is given below.

u

QuuQ

u

ququmququuququ dt

dikM

dtdi

LireV )()( −+−−= l

u

QuQumQu

u

quQuQuQu dt

diL

dtdi

kMir )(0 l+−−−=

DuDuufufuududuuqu ikMikMiLe ωωω ++= Now it is obvious that the circuit of Figure 5.10 corresponds exactly to above two

-+

+

-equ

LAQu

r lqu

rQu

lQuquV

iquu

Figure 5.10 q-Axis equivalent Circuit in p.u. Now consider the d-axis equations: A similar procedure on these equations yields the perunitized equations:

u

dudnbufudu

umduduududu dt

diiii

dtdLirev l−++−−−= )(

u

dufuDufudu

umdufufufu dt

diiii

dtdLirv l−++−−=− )(

u

duDuDufudu

umduDuDu dt

diiiidtdLir l−++−−= )(0

Where: QuQuuduquuqu ikMiLe ωω += Above equations represent the equivalent circuit of Figure 5.11



-

-

+

+Vfu

+

-e

r r

r

L

l

ll

fu

fuDu

Du

du

du

mdu

V

idu

du

u

Figure 5.11 d-Axis Equivalent Circuit in p.u. The results of the q-axis and d-axis equivalent circuit are summarized in Figure 5.12 which also includes the 0-axis equivalent circuit.



-

-

+

+Vf

-+

r L0io

Vo

+

+

+

-

-

-

e

r r

r

L

l

ll

f

fD

D

d

d

d

AD

V

id

eq

LAQ

r lq

rQ

lQ

qV

iq

(a)

(b)

(c)

Figure. 5.12 Equivalent Circuits of Synchronous Machine with Two damper

Windings in p.u. (a) 0-axis Equivalent Circuit (b) d-axis Equivalent Circuit (c) q-axis Equivalent Circuit

The per unitized model of a synchronous machine will be illustrated with an example. Example E5.2: A 1300MVA, 60 Hz, 4 pole, 25 kV generator has the following parameters:



H = 2.8 sec r = 0.002 ohms rD = 0.00623 ohms rQ = 0.00737 ohms rf = 0.0058 ohms Ld = 2.7035 mH Lq = 2.474 mH l ld q= = 0 32. mH Lf = 35.5056 mH l f = 4 1029. mH MR = 7.8729 mH kMQ = 1.4863 mH Lo = 0.723 mH lD = 0 2. mH l mH Q = 013. Use the following base system for the stator circuit

MVASB 31300

= 1sec 602 −= πωB VoltsVB 325000

=

and compute the per unit values of the following parameters: r, rD, rQ, rf, LAD, LAQ, , , , , ,Lld lq l f lD lQ o

Solution: The derived bases are computed using the equations in Table 5.2. First we compute some of the parameters. mHLL ffmf 4027.31=−= l

mHLL ddmd 3835.2=−= l

mHLLkM mfmdf 65149.8==

62975.3==md

ff L

kMk

mHkM

kMf

RD 16899.2==

mHLL qqmq 154.2=−= l

Now 91.0==md

DD L

kMk ⇒ ,9737.1)( mHkMkL DDmD ==

mHLL mDDD 196.0=−=l

And 69.0==mq

QQ L

kMk ⇒ ,0254.1)( mHkMkL QQmQ ==

mHLL mQQQ 1359.0=−=l

From above quantities, the base quantities for all circuits are computed as follows: A. Stator:



SB = 433 33. MVA 377 s

SB = 0.0026525

SB = 0.001275s

. Main Field

kf = 3.62975 MVA

fB = 52395258

rs

fB =0.0168019 s

. D - Circuit

kD = 0.91 33 MVA

DB = 1313

ers

DB =0.0010560s

. Q - Circuit

kQ = 0.69 33 MVA

QB = 9959

ers

QB =0.000607s

−1ec ωB = V Volts B = 14433 76. t 8 seconds ISB = 30022.18 amperes λSB = 38 2867. Wb L 28 H RSB = 0.4807698 ohm B SfB = 433.33 ω fB = 377 sec-1

V 0.94 V tfB = tSB = 0.0026 sec IfB = 8271.1426 ampe λ fB = 138 971. Wb L H RfB = 6.33418 ohm C SDB = 433. ωDB = 377 sec-1

V 4.72 V tDB = 0.00265258 sec IDB = 32991.406 amp λDB = 34 84089. Wb L 59 H RDB = 0.398125 ohm D SQB = 433. ωQB = 377 sec-1

V .294 V tQB = 0.00265258 sec IQB = 43510.4058 amp λQB = 26 4178. Wb L 16 H RQB = 0.2288945 ohm



E. Mutual Inductance

HLLM SBfBfSB 00462894.0==

HLLM SBDBDSB 0011605.0==

HLLM SBQBQSB 00087994.0==

HLLM fBDBDfB 004211234.0== F. Torque Base

mNS

TSB

SBB .446.416,149,1==

ω

hen the per unit values of the machine parameters are:

r = 0.00416 rD = 0.01565

9

Lo

.7 Synchronous Machine Torque Equation

torque and power one has to analyze the electromagnetic field and thus

he output electrical power is:

p(t) = va(t) ia(t) + vb(t) ib(t) + vc(t) ic(t) = vo(t) io(t) + vd(t) id(t) + vq(t) iq(t)

d q o o

T rQ = 0.0322 rf = 0.0009156 LAD = 1.869 LAQ = 1.689 ld = lq = 0.250 l f = 0.2442 l = 0.1856 = 0.2238 D lQ

= 0.5669

5 To compute thecompute the energy transferred through the air gap. If this energy is known one can compute the torque. The same result can be obtained as follows: T

p(t)=-r(i 2+i 2)-ri 2-L iodidt

o +Ldiddidt

d +kMfiddidt

f +kMDiddidt

D +Lqiqdidt

q +kMQiqdidt

Q ]

d d +kMf f MD D q +ω (L i i +k i )i - ω (Lq q QiQ)id

r

p(t) = Pohmic + Pst. magn. + Ptrnf.

here

i +kM o w



Pohmic : ohmic losses on resistors

gnetic field energy

sing the principle of virtual work displacement

Pst. Magn. : rate of change of stator ma Ptrnf. : power transferred across air gap U we obtain

TW P

emfld fld= =

∂∂θ

∂∂ω

⇒

em = (Ldid +kMfif +kMDiD)iq - (Lqiq +kMQiQ)id

bove equation yields the total electromagnetic field torque. In vector notation

he power transferred through the air gap is the one provided with the voltage source e

T A

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−=

Q

q

D

f

d

o

dQdqqDqfqdem

iiiiii

ikMiLikMikMiLT 0

T d, and eq. Pem = -edid + eqiq

nd the electromechanical torque is A

)(2)()()(

tieieP

ttPtT qqdd

m

emem ωω

+−==

he generator rotor motion equation is

T

emmm TT

dttdJ −=2

2 )(θ

or

emm TTdtdJ

P−=

ω2

ote that above equation is the so-called swing equation. This equation can be written in Np.u. For this purpose let SB the per phase base power. The 3-Phase power will be 3SB. And the 3-phase base torque



B

BB

SPTω3

23 =

pon division of the swing equation with the above base torque, one obtains:

U

23

2

PJ

Sddt

T TB

Bm e

⎛⎝⎜

⎞⎠⎟

= −ω ω m

here

Tm is normalized mechanical torque rque

ote that

w Tem is normalized electromagnetic to N

23

12

2 23

22 2

PJ

S td d

dt tJ

P Sddt

Hddt

B

B

B

B

B

B

B B

B

u

uB

u

u

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

=ω ω ω ω ω ω ω

ωω( / )

( / )( )

et . Thus the normalized swing equation becomes L τ ωj BH= 2

)()()( tTtTdt

tdemumj −=

ωτ

Now consider the equation 2

)()( πδωθ ++= ttt s . The derivative of this equation is

dt

tdtdt

tds

)()()( δωωθ+==

ω B (note ωω

s

B= 1) Normalize equation by dividing by

uBu dt

tdtd

tdt )(1)(

)(1)( δωδω +=+= , or

1)()(−= t

dttd ωδ

n summary, the state space swing equation in p.u. for a generator is given by

I



[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−+=

Q

q

D

f

d

o

dQdqqDqfqdjj

m

iiiiii

ikMiLikMikMiLtT

dttd 01)()(

ττω

1)()(−= t

dttd ωδ

Above two equations are known as the swing equation of a synchronous generator.

5.8 State Space Model of a Synchronous Machine In this section we summarize the model of a synchronous machine. Recall that two models were developed: (a) the current model and (b) the flux model. A summary of the six voltage-current equations and the swing equations is given below (current model). Synchronous Machine Electric Current Model (Summary)

( ) ( ) ( ) tvLtiRRL

dttdi

GeqGeqG 1

211 −− −+−= ω ( )

( ) ( ) ( ) ( )( )mGTT

Gj

TtDtiRtidt

td+−= ω

τω

21

( ) ( ) 1−= tdt

td ωδ

where

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

Q

q

D

f

d

o

G

iiiiii

i , ,

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−=

0

0

q

f

d

o

G

v

vvv

v

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

QQ

Qq

DRD

Rff

Dfd

o

eq

LkMkML

LMkMMLkMkMkML

L

L

00000000

00000000000000

[ ]dQdqqDqfqdT ikMiLikMikMiLC −−−= 0



⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

Q

D

f

rr

rr

rr

R

000000000000000000000000000000

1

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

=

000000000000000000000

0000000000

2

Dfd

Qq

kMkML

kML

R

where Bj Hωτ 2= The actual voltage or current phase quantities may be obtained from

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

)()()(

P=)()()(

1-

tvtvtv

tvtvtv

q

d

o

c

b

a

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

)()()(

P=)()()(

1-

tititi

tititi

q

d

o

c

b

a



5.9 Steady State Analysis In this section we examine the operation of a synchronous machine in steady state. This analysis provides insight into the operation of a synchronous machine and the quiescent operation point of the machine. At steady state conditions, the current and voltage will be (assuming balanced operation conditions)

)2

cos(2)( πϕδω +++= Isa tIti

)3

22

cos(2)( ππϕδω −+++= Isb tIti

)3

42

cos(2)( ππϕδω −+++= Isc tIti

)2

cos(2)( πϕδω +++= Vsa tVtv

)3

22

cos(2)( ππϕδω −+++= Vsb tVtv

)3

42

cos(2)( ππϕδω −+++= Vsc tVtv

if(t) = If (dc) vf(t) = Vf (dc) vn = 0

2

)( πδωθ ++= tt s

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

)cos(3)sin(3

0

)()()(

)()()(

V

V

c

b

a

q

d

o

VV

tvtvtv

Ptvtvtv

ϕϕ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

)cos(3)sin(3

0

)()()(

)()()(

I

I

c

b

a

q

d

o

II

tititi

Ptititi

ϕϕ

Observe that the o-d-q axes currents and voltages are constant at steady state, balanced conditions

Thus 0)(=

dttdiodq



0)(=

dttdvodq

If above relationships are substituted into the synchronous machine model, we obtain: vo = 0 vd = -rid -ω Lqiq - kMω QiQ vq = -riq + Lω did + kMω fif +ω kMDiD -vf = -rfif 0 = -rDiD 0 = -rQiQ From the last two equations, it is apparent that: iD = 0 iQ = 0 Thus vo = 0 vd = -rid -ω Lqiq vq = -riq + Lω did - kMω fif -vf = -rfif Note that: vd, vq, vf, id, iq, if represent dc-quantities. From those if, vf are actually dc-quantities while the others are projected (transformed) quantities. To obtain actual phase quantities Park’s transformation is applied. Specifically, the phase A voltage and current is obtained in terms of the o-d-q quantities.

))(sin)(cos(32)( tvtvtv qda θθ +=

))(sin)(cos(32)( tititi qda θθ +=

2

)( πδωθ ++= tt s

All other phases are displaced by 1200. A geometric interpretation can be given to the previous equations. Consider the fact that at steady state the rotor rotates with synchronous speed and it’s position is defined with

( )2

)( πδωθ ++= ttt s . The quantities id, vd represent current and voltage on the d-axis of

the rotor, and the quantities iq, vq represent current and voltage at the q-axis of the rotor. The angle (t) indicates the position of the q-axis. This is shown in Figure 5.13. θ



d-axis q-axis

referenceq

di

i

θ ω δπ

= + +st 2

ω δst +

( Phase A Magnetic axis )

Figure 5.13. Schematic Representation of the d-q Axes of a Synchronous Machine Rotor and Electric Current

It is customary to draw the diagram of Figure 5.13 at t = 0 and to consider the two dimensional diagram as the complex plane where the reference is the real axis. Then currents and voltages becomes complex quantities which are expressed as :

)

2( δπ+→

=j

dd eii

)(δjqq eii =

→

)

2( δπ+→

=j

dd evv

)(δjqq evv =

→

Above conclusions can be justified analytically by using the usual definition of phasors. Now, consider the phase A voltage:

)sincos(32)( θθ qda vvtv +=

Upon substitution of the voltages and converting the sines into cosines, we have:

)]cos()()2

cos()[(32)( δωωωδπωω +++−+++−−= tikMiLritiLritv ffddqqqda

Note that the rms phasor representation of va(t) is:



δδπ

ωωω jffddq

j

qqda eikMiLrieiLriV )(3

1)(3

1~ )2

(++−++−=

+

This equation can be written in the form EIjxIjxIIrV qqddqda

~~~)~~(~ +−−+−= where xd = ω Ld d-axis synchronous reactance xq = ω Lq q-axis synchronous reactance

EM if f=

ω2

generated voltage

)

2(

31~ δ

π+

=j

dd eiI

δjqq eiI

31~ =

Same analysis can be repeated for phases B and C. The final results are as follows: EIjxIjxIIrV qqddqda

~~~)~~(~ +−−+−=

32

32

32

32 ~~~)~~(~ ππππ jj

qq

j

dd

j

qdb eEeIjxeIjxeIIrV−−−−

+−−+−=

34

34

34

34 ~~~)~~(~ ππππ jj

qq

j

dd

j

qdc eEeIjxeIjxeIIrV−−−−

+−−+−= The previous analysis for the voltage Va can be repeated for the current Ia. Specifically, the time function of the current ia(t) is:

)]cos()2

cos([32)sincos(

32)( δωδπωθθ ++++=+= titiiiti qdqda

The rms phasor representation of the current ia(t) is :

qdj

q

j

da IIeieiI ~~3

13

1~ )2

(+=+=

+ δδπ

The derived equations are summarized in Figure 5.14.



d-axis q-axis

reference

E

V

II

II

IV

V jx

jx

a

a

q

q q q

d

d

d d

δ

r( Id + Iq )

δ'

Figure 5.14. Phasor Diagram of Phase A of a Synchronous Machine at Steady State

Using the derived steady state model of a synchronous machine we address the following fundamental problem. Given the generator terminal quantities Va, Ia, P, etc. determine (compute) the transformed quantities id, iq and the position of the axes (δ ). From the given terminal conditions it is always possible to compute the phasors ~Va and ~Ia . Then, the phasors can be obtained from the solution of the following equations.

~Id and ~Iq

EIjxIjxIrV qqddaa

~~~~~ +−−−= (5.15)

qda III ~~~ += An alternative way is to solve for through a graphical solution. The graphical solution is based on rewriting the equation (5.15) in the form

~ ,~I Id q

EIxxjIjxIrV dqdaqaa

~~)(~~~ +−−=++ Observe that the phasor is along the q-axis and equals the phasor: EIxxj dqd

~~)( +−− aqaa IjxIrVA ~~~~ ++= which is computable. Thus, one can first compute the phasor and define the location of the q-axis from the phase of the phasor

~A~A



The graphical solution is shown in Figure 5.15. The steps of the graphical construction are as follows. Step 1: From the tip of the phasor ~Va draw the phasor r Ia

~ . Step 2: Draw the phasor jx Iq a

~ . The line OA defines the q-axis, i.e. the angle δ . Step 3:

Compute ~Id , as projections of on the d and q-axes and complete the diagram. The procedure will be illustrated with an example.

~Iq~Ia

d-axis q-axis

reference

E

II

II

I

Vjx

jx

a

a

q

q q

d

d d

Ijxq a

A

r Ia

δ

O

Figure 5.15 Graphical Solution for Determining the Position of the q-Axis Example E5.3: A 150 MVA generator, 15 kV, wye-connected delivers 6158.4 A at 0.85 power factor lagging and rated voltage. Determine the steady state operating conditions. The unit is connected with a transmission line of 0.015 +j 0.14 ohms to an infinite bus. The following data are given: if = 926 A Ld= 6.341x10-3 H Lq= 6.118x10-3 H r(1250 C) = 1.542x10-3 ohms When the machine operates unloaded with rated voltage at the terminals, the field current is if = 365 A (open circuit test). Solution: Compute the following xd = ω Ld = 2 39. Ω xq = ω Lq = 2 30. Ω



co s . .φ φ= ⇒ = −085 31780

rIa = 9.496 V xqIa = 14164.32 V From above quantities, a plot is drawn.

d-axis q-axis

reference

E

II

I

I

I

V jx

jx

a

a

q

q q

d

d d

Ijxq a

31.78

36.75o

o

From the plot we measure: δ = 36.750

Vd = -5182 vd = -8975 Vq = 6939 vq = 12019 Also Iq = Ia cos(31.78 + 36.75) = 2254 A Id = -Iasin(31.78 + 36.75) = -5731 A id = -9926.4 A iq = 3904 A And jxdId = j13,697.1 V jxqIq = j5,184.2 V By completing the graph and measuring E we obtain E = 20,700 V The problem can be also solved analytically. Specifically, first we compute the vector A:



voltsejIjxIrVA jaqaa

073.3602.124,2076.035,1214.128,16~~~~ =+=++= Above computation provides the angle of the q-axis. Then the d- and q-axes currents are computed as projections of the phase current on the d- and q-axes. Knowing these ciurrents, the generated voltage E is computed as the vector sum of the components indicated in the Figure. The result is: VeE j 073.36700,20~ = Example E5.4: The generator of example E5.3 is connected to an infinite bus through a series compensated transmission line. The parameters of the line are: r = 0.02 ohms, L = 0.001 henries, and C = 650 µF If the generator outputs 100 MVA at rated terminal voltage and 0.9 power factor lagging, compute the steady state voltages vd, vq, the current id, iq and the generated voltage E. Also compute the voltages ed, and eq at the infinite bus. Solution: Compute vector (phasor). ~A ~ ~ ~ ~A V r I jx I Aea a q a

j= + + = δ

where ~ . .V Vea

j j= =1443376 100 0e ~ . .. .I Ae ea

j j= =− −2309398 0 769230 451 0 451 xq = ω =1.9399 Lq

~ . .A e j= 2 12918 0 6817

Iq = 0.76923 cos(0.451 + 0.6817) = 0.32632 Id = -0.76923 sin(0.451 + 0.6817) = -0.69658 iq = 3 Iq = 0.5652 id = 3 Id = -1.2065 vq = 3 (1.0)cos(0.6817) = 1.3433 vd = - 3 (1.0)sin(0.6817) = -1.0901 E = 2.12918 + (xd -xq) |Id|= 2.2546 The transmission line steady state equations read (transformed using Park’s transformation): vd - vdL = 0.02id +ω (0.001)iq vq - vqL = 0.02iq -ω (0.001)id



or after normalization vd - vdL = 0.04159id +0.78414ω iq vq - vqL = 0.04195iq - 0.78414ω id ⇒ vdL = -1.4831 vqL = 0.37373 The series capacitor steady state equations, after normalization, are: Zc = 2.7186 pu id = 2.7186ω vqL - 2.7186ω eq iq = -2.7186ω vdL + 2.7186ω ed ed = -1.2752 eq = 0.81751 E ∞ = 087454.

5.10 Synchronous Machine Performance Under Faults A typical synchronous generator, when faulted, exhibits three distinct phenomena. Immediately upon the application of the fault, the fault currents are relatively high and decay fast towards a steady state value. One can identify at least two distinct rates of decay. This is shown in Figure 5.16 for a three phase fault and the phase A rms value of current is shown in Figure 5.17. It is apparent from this Figure that the synchronous machine impedance changes with time. The three distinct time periods are named (a) subtransient (b) transient and (c) synchronous as it illustrated in Figure 5.17. In this section we examine the parameter and time constants for these three time regions.



Figure 5.16 Synchronous Machine Short Circuit Currents Following a Three Phase Fault

1 2Time(seconds)

Is

I'o

I"o

I'

Transient Synchronous

Subtransient

Figure 5.17 Illustration of the Decaying rms Value of the Phase A Current

5.10.1 Subtransient and Transient Inductances



The synchronous machine comprises a number of coupled magnetic circuits. The magnetic flux in these circuits has a certain inertia. When sudden changes occur in the machine, the magnetic fluxes start changing. Typical construction characteristics of synchronous machines are such that the time constants associated with the magnetic flux changes due to a sudden change are on the order of few cycles for the damper winding magnetic flux and on the order of seconds for the main field winding flux. Thus, in determining the initial dehaviour of a synchronous machine, we can make the following assumptions: (a) initially, all magnetic fluxes remain constant to a sudden change of the external conditions. This assumption provides the subtransient parameters of the synchronous machine. (b) Few cycles after a disturbance, the damper winding magnetic flux assumes steady state values, while the main field winding flux remains constant. This assumption provides the transient parameters of the synchronous machine. We will use this approach to evaluate the subtransient and transient parameters of a synchronous machine. For simplicity, we neglect the resistances. d-axis subtransient & transient inductances. These inductances will be derived in this paragraph. The result is given by:

)(

2" 2

22

fDADDf

fDADd

ADDf

ADfDADdd L

LLLLL

LLLLLL

llll

ll

++

+−=

−

−+−=

f

ADdd L

LLL

2' −=

Where: is the d-axis subtransient inductance, and dL" is the d-axis transient inductance. '

dL The d-axis subtransient and transient inductances are defined under the following conditions: (1) Assume machine rotates with synchronous speed, (2) Assume all field circuits are shorted, vf = 0, and (3) Assume voltage in suddenly applied to the terminals and such that vo = 0 vd = 3 Vu( )t vq = 0 where u(t) is the step function. In this case the machine will behave initially as an inductance, of value L”d - subtransient inductance (d-axis), and after few cycle will behave as an inductance of different value, L’d - transient inductance (d-axis). Specifically, assuming that the main field winding flux and the damper winding flux will remain constant, will yiled the subtransient d-axis inductance. Ignoring the damper winding flux and assuming that the main field winding flux remains constant will yield the transient inductance. The procedure for obtaining the subtransient and transient d-axis inductances is given below.



The derivation of above equations is as follows. Since the generator field winding is shorted, then at t = 0+, the magnetic flux in the main field winding and the d-axis dampoer winding will be: λ f f d f fkM i L i M i= = + +0 R D

R f λD D d D DkM i L i M i= = + +0 Solution of above equations for the currents if, iD in terms of id yields:

ikM L kM M

L L Mif

f D D R

f D Rd= −

−− 2

ikM L kM M

L L MiD

D f f R

f D Rd= −

−− 2

The magnetic flux in the d-axis, λ d , is computed from ffddDDd ikMiLikM ++=λ Substituting the filed current and the d-axis damper winding current, yields:

λ d d d dD f f D f D R

f D RdL i L

kM L kM L kM kM ML L M

i= = −+ −

−" (

( ) ( ))

2 2

22

Recall that when the synchronuous machine parameters are expressed in pu, then: RfDAD MkMkML === Thus:

22 2

"ADDf

ADfDADdd LLL

LLLLLL

−

−+−=

If the damper winding D is neglected (MR = 0, kMD = 0)

λ d d d df

fdL i L

kML

i= = −' (( )

)2

Thus:

f

ADdd L

LLL

2' −=



q-axis subtransient and transient inductances. These inductances are computed in similar way as the d-axis inductances. The final result for the q-axis subtransient and transient inductances is:

Q

AQqq L

LLL

2

" −=

qq LL ='

Where: is the q-axis subtransient inductance, and qL"

is the q-axis transient inductance. 'qL

The q-axis subtransient and transient inductances are defined under the following conditions: (1) Assume machine rotates with synchronous speed, (2) Assume all field circuits are shorted, vf = 0, and (3) Assume voltage in suddenly applied to the terminals and such that vo = 0 vd = 0 vq = 3 Vu( )t where u(t) is the step function. In this case the machine will behave as an inductance of value L”q - at fast varying currents subtransient inductance (q-axis) and as an inductance of different value L’q - at moderately varying currents transient inductance (q-axis). The derivation of the above equations is as follows. Since the generator field winding is shorted, then at t = 0+

QQqQQ iLikM +== 0λ Solution of above equations for the currents iQ1, iQ2 yields:

qQ

QQ i

LkM

i −=

The q-axis magnetic flux for this condition will be:

qQ

Qqqqq i

LkM

LiL ))(

(2

"" −==λ

Thus



Q

AQq

Q

Qqq L

LL

LkM

LL22

"" )(−=−=

The of q-axis transient reactance is derived by neglecting the q-axis damper winding. Then: qq LL ='

5.10.2 Time Constants of a Synchronous Machine The time constants are computed by considering the conditions (subtransient or transient) and determining the time constants involved. As an example consider the q-axis equivalent circuit of a generator. If the q-axis circuit is open, then the only loop in the circuit is the one consisting of rQ, and LlQ AQ. The time constant for this circuit is

τqoAQ Q

Q

Lr

" =+ l

which is the open circuit q-axis subtransient time constant A more systematic approach to determine the time constants of a generator is to compute the Laplace transform of the equivalent circuit. The roots of the Laplace function provide the time constants. This procedure is tedious and it is omitted. Instead, approximate value of the various time constants have been computed and tabulated in Table 5.3. An example provides typical value of time constants.

Table 5.3 Approximate Time Constant of a Synchronous Generator d-axis

τ subtransient time constant (open circuit) doD R f

D

L M Lr

" /=

− 2

τ τd dod

d

LL

" ""

'= subtransient time constant (short circuit)

τdof

f

Lr

' ≅ transient time constant (open circuit)

τ τd dod

d

LL

' ''

= transient time constant (short circuit)

q-axis



τqoQ

Q

Lr

" = subtransient time constant (open circuit)

τ τq qoq

q

L

L" "

"

'= subtransient time constant (short circuit)

τ τq qoq

q

L

L' '

'

=

Armature Time Constant: It describes the decay characteristics of the 3-

phase short circuit rms currents (DC of stator or rms of field current)

τaLr

= 2 L2 = L Ld q+

2

Example E5.5: The following parameters have been computed for the generator of Example E5.3. r = 0.00416 LAQ = 1.689 rD = 0.01565 l ld q= = 0 2509. rQ = 0.0322 l f = 0 2442. rf = 0.0009165 lD = 01865. LAD = 1.869 lQ = 0 2238. Lo = 0.5669 Compute the per unit values of the d- and q- axes subtransient and transient reactances L”d, L’d, L”q, L’q Also compute the following time constants τ τ τ τ τ τ τ τ τdo d do d qo q qo q

" " ' ' " " ' ', , , , , , , and a

Solution:

L L LL L L

L L LL L

Ld d ADD f AD

f D ADd AD

D f

f D AD D f

"( )

.= −+ −

−= −

+

+ +=2

222

0 3507l l

l l l l

L’d = Ld - LL

AD

f

2

=0.4669

L LL

Lq qAQ

Q AQ

" .= −+

=2

10 4485

l



L’q = Lq = 1.9399

τdoD RL M L

r" /

=− f

D.=

2

25 66 or 0.0686 sec

τ τd dod

d

LL

" ""

' .= = 19 27

or 0.05112 sec

τdof

f

Lr

' ≅ = 2307.99 or 6.122 sec

τ τd dod .508 32 or 1.3483 secd

LL

' ''

= =

τqoQ

Q

Lr

" .= = 59 403 or 0.15757 sec

τ τq qoq

q

L

L" "

"

' .= = 13734 or 0.0364 sec

or this m

τ τq qo' ', meaningless f odel.

τ aLr

= =2 289 27. or 0.7673 sec

of Synchronous Machine Parameters

.4. T short able f m the

ecently, techniques have been developed by which these parameters can

Manufacturer Regulator

5.10.3 Summary A comprehensive list of synchronous machine parameters is given in Table 5 hecircuit ratio (SCR) is defined in Figure 5.21. All these parameters are avail romanufacturer. Rbe estimated with real time monitoring systems.

Table 5.4 List of Generator Parameters

Short Circuit Ratio(SCR)

Unit Rated MVA Unit Rated KV Unit Rated PF x” , x” , r , xd q a 2 x’ , x’ , x , xd q l o



xd , xq , r2

τ τ τdo d" " , ,

τ τd qo' " , , τ

τ τ τ

do

q

qo q

'

"

' '

, , a

rf

ad

H

Vf full lo

Field Current

Vol

tage

o a b g o'

deh

scc

occfair-gap

linec

SCR ObOg

= air gap length

Figure 5.18 Definition of the SCR for a Synchronous Machine



5.11 Synchronous Machine Simplified Models Many times, a simplified model will suffice for stability studies under certain conditions. For example if the first swing of the system is to be studied or the performance of synchronous machine stabilizer. Traditionally, simplified models have been employed for these studies. This section presents such models. In addition it tries to add perspective to these models towards increasing the intuition of the reader. For example the usability and limitations of these models are explicitly explained. The approach taken here is that of determining the electric power output/input of the machine versus slow varying variables such as generated voltage, infinite bus voltage, etc. Thus a simplified expression for the electric power of the swing equation results. It is however important to realize the limitations of these models. The simplified models are introduced on the basis of the main assumptions leading to the simplified model. As such, three simplified models will be introduced. a) Steady State Model. b) Constant Main Field Winding Flux Model. c) Constant main Field and Damper Windings Flux Model. The leading assumptions themselves can provide information regarding the applicability of the resulting simplified models. For example the steady state model can be employed whenever the transitions from one operating point to another are smooth. Similarly the constant main field winding flux model should be employed to study phenomena which lasts less than the time constant of the main field winding, etc. Elaboration on the usability and applicability of the models is further provided in the individual section describing the model. 5.11.1 The Steady State Model The derivation of this model is based upon the assumption of steady state conditions for the synchronous machine. Thus the applicability of this model extends to cases in which the synchronous machine operates in steady state or slowly changes its operating point. One such case is a generator placed under governor control with a slow ramp. For this model it is further assumed the prevailing conditions are balanced. The equations describing this mode are: vd = -rid - ω Lqiq (5.8) vq = -riq +ω Ldid +ω kMfif (5.9) vf = -rfif (5.10) Pe = vdid + vqiq (5.11)



dqqde ivivQ −= In above equations the o -axis equation and the damper winding equations are neglected for obvious reasons. Figure 5.19 illustrates the associated phasor diagram of the synchronous machine. In this diagram the generated voltage phasor has magnitude ~E

E = ωM if f

2 (5.12)

d-axis q-axis

reference

E

V

II

II

IV

V

jx

jx

a

a

q

q q q

d

d

d d

δ

Figure 5.19 Phasor Diagram of a Synchronous Machine Steady State Model Note that saturation can be accounted for in the usual way by observing that the inductances Lq and Ld and kMf are functions of the saturation level, that is Lq = Lq(iq) (5.13) Ld = Ld(id + if) (5.14) kMf = kMf(id + if) (5.15) From the equations describing this model the electric power at the terminals of the machine can be expressed in terms of the terminal voltage Va and the generated voltage E. For this purpose recall that v Vd a= − 3 sinδ (5.16) v Vq a= 3 cosδ (5.17)



Then solve equations (5.8) and (5.9) for id and iq to obtain

i rr x x

vx

r x xv

x

r x xEd

d qd

q

d qq

q

d q

= −+

++

−+2 2 2 3 (5.18)

ix

r x xv r

r x xv r

r x xEq

d

d qd

d qq

d q

= −+

−+

++2 2 2 3 (5.19)

Upon substitution into equation (5.11)

δδδ sin3

cos32sin233

222

22

2 EVxxr

xEV

xxrrV

xxrxx

Vxxr

rP aqd

qa

qda

qd

qda

qde +

++

++

−+

+−= (5.20)

Usually the resistance of a synchronous machine is very small as compared to xd or xq. Thus it can be neglected. In this case equation (5.19) is simplified to :

δδ 2sin)11(2

3sin3 2

dq

a

d

ae xx

Vx

EVP −+= (5.21)

Also the reactive power is computed to be Qe = vdiq - vqid (5.22) Upon substitution we obtain:

)cossin(3)cossin(32

222

2

δδδδ qqd

aqd

qd

ae xr

xxrEVxx

xxrVQ +−

+++

+−= (5.23)

If the resistance is neglected

QV

x xx x

V Exe

a

d qd q

a

d= − + +

3 22 2( sin cos ) cosδ δ

3δ (5.24)

Example E5.6: A 625 MVA, 60 Hz, 18 kV synchronous generator has the following parameters:

puVpuEpuxpuxpur qd 105.1,03387.2,42231.1,5645.1,001066.0 ===== Compute the simplified steady state model of this unit under the assumption of constant generated voltage. Solution: By direct computation a) If the armature resistance is neglected



Pe = 4.3095 sinδ +0.117 sin 2δ Qe = -2.57544 sin2 δ -2.3413 cos 2δ + 4.3095 cosδ b) If the armature resistance is not neglected. Pe = -0.00175 + 0.11703sin 2δ + 0.2154 cosδ + 4.3095sin δ = -0.00175 + 0.11703sin 2δ + 4.3148sin (δ +0.04994) Qe = -2.57544sin2 δ - 2.34137cos 2δ + 4.30954cosδ - 0.00323sin δ 5.11.2 Constant Main Field Winding Flux Model The leading assumption in the deviation of this model is that the magnetic flux linkage of the main field winding remains constant, i.e. tconsf tan=λ (5.25) Since the time constant of the main field winding is in the order of seconds, it is obvious that this model is appropriate whenever phenomena which last less than a second are to be studied. Such an example is the classical transient stability problem. It was postulated earlier that the stability of the power system is determined from the first swing of the system following the disturbance. Since the first swing lasts less than a second, the classical model utilizes an assumption which is equivalent to constant main field winding flux. As we shall see this assumption is equivalent to the assumption of the classical model “constant voltage behind transient reactance”. For this reason this model will be also referred to as “the classical transient model”. An additional assumption for this model is that the damper winding are neglected. The equations describing this model are derived from the general equation by utilizing the stated assumptions:

vd = -rid - ω Lqiq - Ldidt

kMdidtd

df

f− (5.26)

vq = -riq +ω Ldid +ω kMfif - Ldidtq

q (5.27)

=Lλ f fif +kMfid (5.28) Pe = vdid + vqiq (5.29) Qe = vdiq - vqid (5.30) Equation (5.28) yields



df

f

f

ff i

LkM

Li −=

λ (5.31)

and since λ is assumed constant: f

dtdi

LkM

dtdi d

f

ff −= (5.32)

Upon substitution of if, didt

f in equations (5.26) and (5.27)

vd = -rid - ω Lqiq -dtdi

LkM

L d

f

fd )

)((

2

− (5.33)

vq = -riq +ω ( LkM

Ldf

f−

( )2

)id+(ω kMfλ f)/Lf - Ldidtq

q (5.34)

In above equations the term LkM

Ldf

f−

( )2

can be recognized as the d-axis transient

inductance, L’d:

f

fdd L

kMLL

2' )(

−= (5.35)

The transient reactance will be x’d = (5.36) ωLd

'

A further approximation can be introduced into the model by observing that for the range of applicability of this model the d-axis and q-axis current id, and iq vary slowly. In this

case the terms Ldidtd

d' and Ldidtq

q are very small compared to other terms of the

equation. Thus, they may be omitted. Finally, since λ f is assumed to be constant the term ( kMω fλ f)/Lf is also constant. Define

f

ff

L

ME

2'

λω= (5.37)

which shall be called the generated voltage behind transient reactance.



d-axis q-axis

reference

E'

I

I

I

V

jx'

jx'

a

a

q q

d d

Figure 5.20 Phasor Diagram of the Transient Model Now equations (5.33) and (5.34) become vd = -rid - xqiq (5.38) vq = -riq +x’did + 3E' (5.39) In phasor representation above equations read '~~~~~ ' EIjxIjxIrV qqddaa +−−−= (5.40) The phasor diagram is illustrated in Figure 5.23. The power equations are derived in a straightforward manner as in the previous model. The result is

δδδ sin'

'3cos

''3

2sin'

'2

3'

3222

2

2

2

qd

aq

qd

a

qd

qda

qd

ae xxr

EVxxxr

ErVxxr

xxVxxr

rVP

++

++

+

−+

+−= (5.41)

)cossin('

'3)cossin'(

'3

222

2

2

δδδδ qqd

aqd

qd

ae xr

xxrEV

xxxxr

VQ +−

+++

+−=

(5.42) Again if the resistance is neglected the power equations are simplified to:

δδ 2sin)'11(

23

sin'

'3 2

dq

a

d

ae xx

Vx

EVP −+= (5.43)



δδδ cos'

'3)cossin'(

'3 22

2

d

aqd

qd

ae x

EVxx

xxV

Q ++−= (5.44)

Example E5.7: Consider a 625 MVA, 60 Hz, 18 kV synchronous machine with the following parameters

puVpuEpuxpuxpur qd 105.1,03387.2,42231.1,5645.1,001066.0 =====

And puEpuxd 072.1,2776.0 '' ==Compute the simplified model under the assumption of constant field winding flux. Solution: By direct substitution: V = 1.105 pu λ f f f f dL i kM i= + = 870 63. E’q = 9283 V or E’q = 1.072 pu x’d = 0.3904 ohms or x’d = 0.2776 pu xq = 1.4223 pu Pe = 12 8 531 2. sin . sinδ δ− or Pe = − + + −0 00989 12 8 0 00959 5494 2. . sin . cos . sinδ δ δ 5.11.3 Constant Main Field and Damper Windings Flux Model The leading assumption in the derivation of this model is that the magnetic flux linkage of the main field winding and the damper windings remain constant, i.e. (5.45) λ f cons t= tan (5.46) λD cons t= tan (5.47) λQ cons t= tan From the assumption, the range of application of this model is obvious: phenomena which last less than the smallest time constant of the three windings: main field, D and Q damper windings. This time constant is in the order of few 60 Hz cycles. Thus this model is suitable for the determination of the electric currents immediately following a fault. This is important in the application of modern switchgear which operate in 2 to 3 cycles. The equations describing this model are derived from the general equations:

vd = -rid - ω Lqiq - kMω QiQ - Ldidt

kMdidt

kMdidtd

df

fD

D− − (5.48)



vq = -riq + Lω did + kMω fif +ω kMDiD - Ldidt

kMdidtq

qQ

Q− (5.49)

=Lλ D DiD +kMDid + MRif (5.50) =LλQ QiQ +kMQiq (5.51) =Lλ f fif +kMfid + MRiD (5.52) Pe = vdid + vqiq (5.53) Qe = vdiq - vqid (5.54) Equations (5.50), (5.51) and (5.52), solved for the electric currents if, iD, iQ yield the following

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

q

d

Q

D

f

Q

DR

Rf

Q

D

f

Q

DR

Rf

Q

D

f

ii

kMkMkM

LLMML

LLMML

iii

000

0000

0000

11

λλλ

(5.55)

and upon differentiation assuming constant λ λ λf D, and Q :

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

q

d

Q

D

f

Q

DR

Rf

Q

D

f

ii

dtd

kMkMkM

LLMML

iii

dtd

000

0000

1

(5.56)

Substituting for the currents if, iD and iQ in equations (5.48) and (5.49) one obtains:

vd = -rid - ω ( LkM

LqQ

Q−

( )2

)iq -ω

λkML

Q

QQ

- (( ) ( )

)LL kM L kM M kM kM

Adidtd

D f f D R f D−+ −2 2 2 d (5.57)

vq = -riq +ω(( ) ( )

)LL kM L kM M kM kM

Aid

D f f D R f Dd−

+ −2 2 2

- (( )

)LkM

Ldidtq

Q

Q

q−2

+ω λL kM M kM

AD f R D

f− +

ωL kM M kM

Af D R f

D−

λ (5.58)

where A = LfLD - M2

R



In above equations the subtransient inductances and reactances can be identified:

L”d = LL kM L kM M kM kM

AdD f f D R f−

+ −( ) ( )2 2 2 D (5.59)

L”q = LkM

LqQ

Q−

( )2

(5.60)

x”d = ω L”d (5.61) x”q = ω L”q (5.62) Since, by assumption λ λ λf D, and Q remain constant, equation (5.57) and (5.58) contain

three constants. The constant ω

λkML

Q

QQ represents a voltage generated along the d-axis,

while the constants ω λL kM M kM

AD f R D

f− and ω λ

L kM M kMA

f D R fD

− represent

voltages generated along the q-axis. Define

E”1 = E”d = ω

λM

LQ

QQ2

(5.63)

E”2 = ωL M M M

AD f R D

f−2

λ (5.64)

E”3 = ωL M M M

Af D R f

D−2

λ (5.65)

E”q = E”2 + E”3 (5.66) Substituting above definitions into equations (5.57) and (5.58) to obtain

vd = -rid - ω L iq" q - 3 1E" - Ldidtd

d" (5.67)

vq = -riq +ω - L id d" Ldidtq

q" + 3 E”2 + 3 E”3 (5.68)

One final approximation is introduced by observing that the terms Ldidtd

d" and Ldidtq

q"

are small compared to other terms, yielding vd = -rid - ω L iq" q - 3E d" (5.69)

vq = -riq +ω +L id d" 3 E”q (5.70) The phasor representation of above equations is:



(5.71) ~ ~ ~ ~ ~ ~" " "V r I jx I jx I E Ea a d d q q q= − − − + + "

d

The phasor diagram for this model is illustrated in Figure 5.21.

d-axis q-axis

reference

E"

I

I

I

jx"

jx"

a

q q

d dIar

E"q

E"d

Figure 5.21 Phasor Diagram of the Subtransient Model The power equations are derived in a straightforward manner as for the previous model. Because of the complexity of the results only the case of negligible armature resistance is presented.

PV Ex"

Vx" x"

V Ex"e

a q

d

a

q d

a d

q= + − −

3 32

1 1 232"

sin ( ) sin"

cosδ δ δ (5.72)

QV

x" x"x" x" V

Ex"

Ex"e

a

d qd q a

q

d

d

q= − + + +

33

22 2( sin cos ) (

"cos

"sin )δ δ δ δ (5.73)

Example E5.8: A synchronous generator delivers rated power, at rated terminal voltage and power factor equal 1.0. Suddenly a three phase fault is placed at the terminals of the generator. Compute the rms value of the fault current immediately after the fault. Neglect the dc component. The parameters of the generator are: Ld = 1.71 r = 0 Lq = 1.57 LAD = 1.49 LD = 1.79 LAQ = 1.35 LQ = 1.8 Lf = 1.77



Solution: The subtransient model will be employed. The subtransient inductance are computed to be: L”d = 0.35199 L”q = 0.5575 Then the initial conditions are computed using the steady state model. ~ ~ ~ ~ . .A V r I jx I ea a q a

j= + + = 18614 1 003655

~ ~ ( )~ . .E A j x x I ed q dj= + + = 1 97948 1 003655

δ = 1 003655. rad The generated voltage behind the subtransient reactance is: ~" ~ ~ ~ ~ ." " .E V rI jx I jx I ea a d d q q

j= + + + = 0 995789 0 42578 Immediately after fault initiation, E” remain constant, and Va = 0 Thus ~" ~ ~" "E r I jx I jx Iaf d df q qf= + +

~

Let ~I A eqf

j= 1δ

~ ( )I A edf

j=

+

22

δπ

Substitute into equation to obtain 0 = -(0.35199) A e + j(0.5575) 995789 0 42578. .e j j

2δ A ej

1δ

Note that immediately after fault initiation, δ = 1003655. . Upon solution of above equation for A1 and A2: A1 = -0.975679 A2 = -2.369668 ⇒ i Adf = = −3 4104382 . pu i Aqf = = −3 168991 . pu Iaf = 2.5626 pu 5.11.4 Summary of Simplified Models



The three simplified models presented in the previous paragraphs are summarized with the following equations qqddaa IjxIjxIrVE ~~~~~ +++= steady state model

qqddaa IjxIjxIrVE ~~~~'~ '' +++= transient model

qqddaa IjxIjxIrVE ~~~~"~ "" +++= subtransient model

δδ 2sin)11(2

3sin

3 2

dq

a

d

ae xx

Vx

EVP −+=

δδ 2sin)'11(

23

sin'

'3 2

dq

a

d

ae xx

Vx

EVP −+=

δδδ cos"

"32sin)

"1

"1(

23

sin"

"3 2

q

da

dq

a

d

qae x

EVxx

Vx

EVP −−+=

where in deriving the Pe equations, the resistance r has been neglected. The three models are known as: a) Steady state model b) Transient model and c) Subtransient model. Figure 5.22 summarizes the phasor diagrams for all models.

d-axis

q-axis

reference

E

VI

I

II

IV

Vjx

jx'

a

a

q

q q q

dd

d d

E'

I

I

jx'

jxq q

d d

E"

I

I

jx"

jx"q q

d d

E"q

E"d

(t = 0)θ ω δ π= + +s t 2

Figure 5.22 Summary of Simplified Models 5.11.5 One Axis Synchronous Machine Model



A further simplification of the presented models is the so called one axis models. The leading assumption for these models is that the reactances of a synchronous machine are equal along the d and q-axes. That is Steady state model xd = xq = x Transient model x’d = x’q = x’ Subtransient model x”d = x”q = x” Utilizing these assumptions, the model equations for each one of the three models are: Steady state model ~ ~ ~ ~V rI jxIa a a= − − + E

PV E

xea= sinδ

Transient model ~ ~ ~ ~''V rI jx Ia a a= − − + E

PV E

xea=

''

sinδ

Subtransient model ~ ~ ~ ~""V rI jx I Ea a a= − − +

PV E

x"ea=

"sinδ

From above equations it is obvious that an equivalent circuit can be derived for each model. These circuits are illustrated in Figure 5.23. It should be mentioned that many times we refer to the one axis transient model as the classical model.



r x

E

r x'

E'

r x"

E"

(a)

(b)

(c)

Figure 5.23 One Axis Synchronous Machine Simplified Models (a) Steady State Model

(b) Transient Model (c) Subtransient Model

The assumptions of the one axis models provide insight into the degree of approximation performed. For example since the d- and q- axes synchronous reactances of a cylindrical rotor machine have numerical values approximately equal (within 10% for example), it is concluded that the one axis steady state model of a cylindrical rotor machine is a reasonable approximation. This can not be claimed for a salient pole synchronous machine. Regarding the one axis subtransient model one can observe that most synchronous machines have d- and q- axes subtransient reactances approximately equal. Thus the one



axis subtransient model can be considered to be a reasonable approximation. This claim, however, can not be extended to the one axis transient model. In general the d axis transient reactance is much smaller than the q-axis transient reactance.

5.12 Exciter Model and Voltage Control (to be added)

5.13 Synchronous Generator Capability Curves (to be added)

5.14 Summary and Discussion This chapter presented the general synchronous machine model. The physical construction of the synchronous generator was discussed and the model describing the internal construction of the machine was introduced. The model was manipulated to yield several model under specific simplified assumptions. These model are valid for studying the performance of the machine for certain phenomena. The derivation also provides insight into the overall behavior of a synchronous generator.



5.14 Problems Problem 5.1 A 3-phase, 60 Hz, 500 MVA, 15 kV (Line to Line) generator has the following parameters:

ohmsxd 75.0= ohmsxq 70.0= HenrieskM f 023.0= ohmsr 0005.0= HenrieskMQ 00065.0= HenrieskM D 001.0=

Henriesd 0001976.0=l Henriesq 00021.0=l

Express in per unit the following equation of the mathematical model of above generator

dtdi

kMdtdi

LriikMikMiLV QQ

qqqDDffddq −−−++= ωωω

Problem 5.2 One of the assumptions in the classical model of synchronous generators for transient analysis (constant voltage source behind transient reactance) is the following: “The mechanical angle of the synchronous machine rotor (q-axis) coincides with the electrical phase angle of the voltage behind transient reactance.” Show that this assumption is equivalent to the following condition: “The q-axis transient reactance is equal to the q-axis synchronous reactance.” Problem 5.3 A three phase, 60Hz, 900MVA, 15kV (line to line, rms) synchronous generator has the following parameters (in a per unit system with SB = 300 MVA, ω B = 2π 60sec-1 , and VB = 15000/ 3 Volts for the stator)

0,77.1,8.1,79.1,35.1,57.1,49.1,71.1

===

=====

randpuLpuL

puLpuLpuLpuLpuL

fQ

DAQqADd

(a) Compute the following: Transient reactance, d-axis Subtransient reactance, d-axis Transient reactance, q-axis Subtransient reactance, q-axis (b) If the generator delivers rated power, at rated terminal voltage and power factor equal to 1.0, compute:

The generated voltage E The voltage behind transient reactance, E’ The voltage behind subtransient reactance, E’’

Assume that a three phase fault is suddenly applied to the terminals of the generator. Prefault conditions are same as stated above. Compute the rms value of the fault current immediately following the fault. Neglect the so-called dc-component.



Problem 5.4 The equivalent o-d-q axis network representation of a synchronous 60 Hz, 15kV (line to line, rms), 810 MVA generator is given in Figure P5.4. All quantities are expressed in a per unit system with the following base quantities for the stator: (a) SB

=270 MVA (per phase), (b) =ω 2π 60 sec-1 , VB = 15,000/ 3 Volts. (a) Compute the per unit values of the following quantities: Synchronous inductance, d-axis Synchronous inductance, q-axis Transient inductance, d-axis Transient inductance, q-axis Mutual Inductances MR, kMf, kMD, kMQ Main Field Winding self inductance, Lf D-winding self inductance, LD Q-Winding self inductance, LQ

(b) Define the minimum amount of information required to compute above quantities in actual MKSA units (metric system).

0.001 j0.21

0.007

j0.35

0.02

j0.28 j1.48

-

-

+

+Vf

ωλqd-axis circuit

0.001 j0.21

0.04

j0.45j1.37

-+ωλq-axis circuit d

0.001 j1.1

o-axis

Figure P5.4



Problem 5.5 A 3-phase, 160 MVA, 60 Hz, synchronous generator delivers 150 MW of real power to an infinite bus through a 3-phase transformer and a transmission line, as in Figure P5.5. The power factor at the terminals of the generator is 0.95, current lagging. The infinite bus is maintained to a voltage of 225 kV line to line. Determine the steady state conditions of the system.

Infinite Bus3-Phase Line

G

Figure P5.5

Additional Information

(a) The positive sequence series impedance of the transmission line is: . ohmsjz 0.500.5 +=

(b) The transformer is a 160 MVA, 15kV:220 kV, 60 Hz, 3-phase with the following impedance: on the transformer ratings. pujz 078.0004.0 +=

(c) The parameters of the synchronous generator are: HkMpuxpuxpur fqd 080.0,0.2,2.2,0015.0 ====

Problem 5.6 Compute the per-unitized voltage equations of the generator of problem 5.5. The generator is a 160 MVA, 15 kV (line to line) unit. Additional information is as follows:


Henriesd 0005022.0=l Henriesq 0005022.0=l HenriesM R 090.0=

ohmsCrf 37.0)125( 0 = ohmsrD 0175.0= ohmsrQ 0195.0=

HenriesL f 0.2= HenriesLD 0055.0= HenriesLD 00145.0=

HenriesL 001856.00 = Problem 5.7 The equivalent circuit representation of a synchronous machine is given in Figure P5.7 (in per unit). Note that the resistances are neglected. The machine is connected to an infinite bus through a transmission line which has j0.35 pu reactance and negligible resistance. It delivers rated power at rated terminal voltage and power factor 1.0. 1. Compute the steady state operating conditions. Specifically compute the quantities

(rotor angle, generated voltage, voltage behind transient reactance and voltage at the infinite bus). δ α, , ' E E and E∞e j

2. Compute an expression of the transient power versus angle δ under the assumption of constant voltage E’.



j 0 . 2 5

j 0 . 2 4j 0 . 1 9 j 1 . 8 7

- +

V f

q

j 0 . 2 5

j 0 . 2 3j 1 . 6 9

-+

de

e

d - a x i s

q - a x i s

j 0 . 5 6 6 9

o - a x i s

Figure P5.7 Problem 5.8 A three phase, 60 Hz, 500 MVA, 15 kV (line to line), two pole generator has the following parameters:


Henriesd 0002.0=l Henriesq 0002.0=l

Consider the q-axis voltage equation

v L i kM i kM i ri Ldidt

kMdidtq d d f f D D q q

qQ

Q= + + − − −ω ω ω

Write the per unitized version of above equation (all parameters should be substituted with their numerical values).



Problem 5.9 Consider the electric power system of Figure P5.9a. The steady state operating conditions of the system are defined with the power flow solution of Table P5.9. To be determined is the steady state conditions of generator G2. The d- and q-axes equivalent circuits of this 100 MVA, 60 Hz, 2 poles generator are given in Figure P5.9b. a) Compute the transient and subtransient inductances of the generator G2. b) Compute the steady state conditions of generator G2, i.e., id, iq, if, E, E’, E”, and δ .

G1 G2

x=6% x=6%

V=15kV V=15kV

100MW + j30 MVAR

PG2

= 40MW

12 3

4

5

Figure P5.9a Electric Power System

Table P5.9 Power flow Solution

~V1 = ~V2 = ~V3 = ~V4 = ~V5 =



0 . 0 0 1 j 0 . 2

0 . 0 0 2

j 0 . 3 2

0 . 0 3

j 0 . 2 5 j 1 . 5

-

-

+

+V f

ω λ q

d - a x i s c i r c u i t

0 . 0 0 1 j 0 . 2

0 . 0 4

j 0 . 3 5j 1 . 4

- +ω λ

q - a x i s c i r c u i t

d

Figure P5.9b d- and q-axes Circuits of Generator G2 Problem 5.10 The per unit parameters of a 450 MVA, 60 Hz, 4-pole synchronous generator are given in Table P5.10a. It is desirable to compute the ohmic losses of the synchronous machine for a variety of operating conditions. For this purpose complete the Table P5.10b. For each entry of Table P5.10b, compute the steady state operating condition of the synchronous machine and the ohmic losses in the stator and rotor. Examine your solution and state your observations. Table P5.10a. Synchronous Machine Parameters in pu

pur 00416.0= purD 01565.0= purQ 0322.0=

pud 2509.0=l puq 2509.0=l purf 0009156.0= puf 2442.0=l puD 1856.0=l puQ 2238.0=l

puL 5669.00 = puLAD 869.1= puLAQ 689.1=



Table P5.10b

Operating Condition

Rotor Position Angle δ

Generated Voltage

(pu)

Stator Losses

(W)

Rotor Losses

(W) S=450 MVA Rated Voltage PowerFactor=1.0

S=225 MVA Rated Voltage PowerFactor=1.0

S=450 MVA Rated Voltage PowerFactor=0.9 (lagging)

S=450 MVA Rated Voltage PowerFactor=0.9 (leading)


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