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
5.11.2
5.11.3
5.11.4
5.11.5
The Steady State Model _____________________________________________________ 60
Constant Main Field Winding Flux Model ______________________________________ 63
Constant Main Field and Damper Windings Flux Model ___________________________ 66
Summary of Simplified Models _______________________________________________ 70
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
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
to the general model for their derivation.
applicability of the simplified models.
This analysis provide insight into the
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
2
poles. The speed of the machine is dependent upon the number of poles, ω m = ω e and
P
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
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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.
Page 4
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
rotor magnetic axis
( or d-axis)
(stationary
reference)
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 θ m is a function of
time. We postulate that the rotor position is given with:
θ m (t ) = ω ms t + δ m (t ) +
π
P
(5.1)
Where ω ms 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 ω ms . In this case the angle
is a constant and independent of time, i.e. δ m (t ) = δ m . In this case, the variable δ m is the
angle between phase A magnetic axis and rotor quadrature axis at time zero, since
π
π
θ m ( 0) = δ m +
→ δ m = θ (0) − .
P
P
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 5
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Under steady state conditions, the following relationships hold
ω ms =
2
ω es ,
P
ωes = 2πf
where ω es 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:
θ e (t ) = ω es t + δ e (t ) +
π
2
(5.2)
P
δ m (t )
2
P
θ e (t ) = θ m (t )
2
where δ e (t ) =
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
Bas (α , t ) = kia (t ) cosα
(5.3)
2π ⎞
⎛
Bbs (α , t ) = kib (t ) cos⎜ α −
⎟
3 ⎠
⎝
4π ⎞
⎛
Bcs (α , t ) = ki c (t ) cos⎜ α −
⎟
3 ⎠
⎝
where
Page 6
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Bas (α , t ) is the magnetic flux density in position α of the air gap and time t, due
to electric current, i a (t ) , at time t in phase A
Bbs (α , t ) is the magnetic flux density in position α of the air gap and time t, due
to electric current, ib (t ) , at time t in phase B
Bcs (α , t ) is the magnetic flux density in position α of the air gap and time t, due
to electric current, i c (t ) , at time t in phase C
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
i a (t ) , ib (t ) , and i c (t ) are sinusoidal currents, the magnetic flux density Bas (α , t ) ,
Bbs (α , t ) , and Bcs (α , t ) will be also a function of time. Under sinusoidal steady state
and balanced conditions, the electric currents will be:
ia (t ) = I m cos(θ (t ) + ϕ I )
2π ⎞
⎛
ib (t ) = I m cos⎜ θ (t ) + ϕ I −
⎟
3 ⎠
⎝
4π ⎞
⎛
i c (t ) = I m cos⎜θ (t ) + ϕ I −
⎟
3 ⎠
⎝
(5.4)
where ϕ I is the phase of the phase A electric current phasor.
Now consider the magnetic flux due to current, i a (t ) , of phase A. From equation 5.3 it is
π
obvious that the points of zero magnetic flux are stationary and occur at α = ± . Thus
2
the magnetic flux density Bas (α , t ) is a pulsating wave. The figure illustrates a snapshot
taken at time t1 when the electric current i a (t1 ) is positive maximum, i.e. θ(t1) + 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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 7
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
( 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:
Bs (α , t ) = Bas (α , t ) + Bbs (α , t ) + Bcs (α , t )
(5.5)
Direct substitution and after some trigonometric manipulations:
Page 8
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
B s (α , t ) =
3
cos(θ (t ) + ϕ I − α − π )
2
(5.6)
The last equation represents a sinusoidal waveform which rotates with speed
d α dθ ( t )
=
= ωs
dt
dt
(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 ωs .
A
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 B r (α , t ) = R cos(θ (t ) − α − π ) . 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 winding
produced magnetic
field ( all phases)
α
air gap
rotor winding produced
magnetic field
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 θ (t ) 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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 9
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 qaxes 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.
Page 10
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
phase a
magnetic axis
reference
ia(t)
va(t)
θ
q-axis
ra
d-axis
ω
+
iD(t)
if(t)
vf(t)_
in(t)
vn(t)
iQ(t)
rc
rb
ib(t)
vb(t)
ic(t)
vc(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
v a (t ) = − ra ia (t ) −
dλ a ( t )
dt
v b (t ) = − rb ib (t ) −
dλ b ( t )
dt
v c ( t ) = − rc i c (t ) −
dλ c ( t )
dt
− v f (t ) = − r f i f (t ) −
0 = − rD i D (t ) −
dλ f (t )
dt
dλ D ( t )
dt
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 11
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
0 = − rQ iQ (t ) −
dλQ (t )
dt
In above equations
λa(t)
λb(t)
λc(t)
λf(t)
λD(t)
λQ(t)
is the magnetic flux linkage of phase A
is the magnetic flux linkage of phase B
is the magnetic flux linkage of phase C
is the magnetic flux linkage of the field winding, f
is the magnetic flux linkage of the D-damper winding
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:
⎡v a ( t ) ⎤
v abc (t ) = ⎢v b (t ) ⎥
⎥
⎢
⎢⎣ v c (t ) ⎥⎦
⎡v f (t )⎤
v fDQ (t ) = ⎢ 0 ⎥
⎥
⎢
⎢⎣ 0 ⎥⎦
Rabc
⎡ra
= ⎢0
⎢
⎢⎣ 0
0⎤
0⎥
⎥
rc ⎥⎦
0
rb
0
Rabc = rI, if ra = rb = rc = r, here I is the 3x3 identity matrix.
R fDQ
⎡r f
⎢
= ⎢0
⎢⎣ 0
0
rD
0
0⎤
⎥
0⎥
rQ ⎥⎦
⎡1⎤
13 = ⎢1⎥
⎢⎥
⎢⎣1⎥⎦
Page 12
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
⎡ia (t )⎤
iabc (t ) = ⎢ib (t ) ⎥
⎢
⎥
⎢⎣ic (t ) ⎥⎦
⎡i f ( t ) ⎤
⎢
⎥
i fDQ (t ) = ⎢i D (t )⎥
⎢⎣iQ (t ) ⎥⎦
⎡λ a ( t ) ⎤
λabc (t ) = ⎢λb (t ) ⎥
⎢
⎥
⎢⎣λc (t ) ⎥⎦
⎡λ f (t ) ⎤
⎢
⎥
λ fDQ (t ) = ⎢λ D (t )⎥
⎢⎣λQ (t ) ⎥⎦
The voltage equations now can be written in the following compact form:
v abc (t ) − v n (t )13 = − Rabc iabc (t ) −
v fDQ (t ) = − R fDQ i fDQ (t ) −
d
λabc (t )
dt
d
λ fDQ (t )
dt
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
λa (t ) = Laa ia (t ) + Lab ib (t ) + Lac ic (t ) + Laf i f (t ) + LaD i D (t ) + LaQ iQ (t )
λb (t ) = Lba ia (t ) + Lbb ib (t ) + Lbc ic (t ) + Lbf i f (t ) + LbD i D (t ) + LbQ iQ (t )
λc (t ) = Lca ia (t ) + Lcb ib (t ) + Lcc ic (t ) + Lcf i f (t ) + LcD i D (t ) + LcQ iQ (t )
Rotor Windings
λ f (t ) = L fa ia (t ) + L fb ib (t ) + L fc ic (t ) + L ff i f (t ) + L fD i D (t ) + L fQ iQ (t )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 13
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
λ D (t ) = LDa ia (t ) + LDb ib (t ) + LDc ic (t ) + LDf i f (t ) + LDD i D (t ) + LDQ iQ (t )
λQ (t ) = LQa ia (t ) + LQb ib (t ) + LQc ic (t ) + LQf i f (t ) + LQD i D (t ) + LQQ iQ (t )
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
⎡ λabc (t ) ⎤ ⎡ Lss (t ) Lsr (t )⎤ ⎡ iabc (t ) ⎤
⎢λ (t )⎥ = ⎢ L (t ) L (t )⎥ ⎢i (t )⎥
rr
⎦ ⎣ fDQ ⎦
⎣ fDQ ⎦ ⎣ rs
where
⎡ Laa
Lss (t ) = ⎢ Lab
⎢
⎢⎣ Lac
Lab
Lbb
Lbc
Lac ⎤
Lbc ⎥
⎥
Lcc ⎥⎦
⎡ Laf
⎢
Lsr (t ) = ⎢ Lbf
⎢⎣ Lcf
LaD
LaQ ⎤
⎥
LbQ ⎥
LcQ ⎥⎦
LbD
LcD
Lrs (t ) = LTsr (t )
⎡ L ff
⎢
Lrr (t ) = ⎢ L fD
⎢⎣ L fQ
L fD
LDD
LDQ
L fQ ⎤
⎥
LDQ ⎥
LQQ ⎥⎦
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:
Laa = Ls + Lm cos(2θ (t ) )
2π ⎞
⎛
Lbb = Ls + Lm cos⎜ 2θ (t ) −
⎟
3 ⎠
⎝
4π ⎞
⎛
Lcc = Ls + Lm cos⎜ 2θ (t ) −
⎟
3 ⎠
⎝
Page 14
(5.8)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
where θ (t ) is the direct axis position relative to the phase A magnetic axis. Note that as
π
the rotor rotates, its position θ (t ) is described with equation θ (t ) = ω s t + δ (t ) + . A
2
typical variation of Lii is shown in Figure 5.7
L ii
Ls
Lm
π
0
θ
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:
L ff = L f , LDD = LD , and
LQQ = LQ
(5.9)
Stator Mutual Inductances Lab, Lac, and Lbc are negative. They are functions of the rotor
position θ( t ) . Approximate expressions for these functions are:
π⎞
⎛
Lab = Lba = − M s − Lm cos⎜ 2θ (t ) + ⎟
6⎠
⎝
π 2π ⎞
⎛
Lbc = Lcb = − M s − Lm cos⎜ 2θ (t ) + −
⎟
6
3 ⎠
⎝
π 4π ⎞
⎛
Lca = Lac = − M s − Lm cos⎜ 2θ (t ) + −
⎟
6 3 ⎠
⎝
(5.10)
A typical variation of Lij is shown in Figure 5.8
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 15
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
L ba
π
0
-M s
θ
Lm
Figure 5.8 Typical Variation of the Mutual Inductance Between Two Phase
Windings
Rotor Mutual Inductances LfD, LfQ, and LDQ are constants, independent of θ( t ) , because
the rotor windings are stationary with each other.
L fD = LDf = M R
L fQ = LQf = 0
(5.11)
LDQ = LQD = 0
Stator to Rotor Mutual Inductances Laf, Lbf, and Lcf are dependent upon the rotor position
θ(t) as follows.
Laf = L fa = M F cos(θ (t ) )
2π ⎞
⎛
Lbf = L fb = M F cos⎜ θ (t ) −
⎟
3 ⎠
⎝
4π ⎞
⎛
Lcf = L fc = M F cos⎜θ (t ) −
⎟
3 ⎠
⎝
(5.12)
Similarly
LaD = LDa = M D cos(θ (t ) )
2π ⎞
⎛
LbD = LDb = M D cos⎜θ (t ) −
⎟
3 ⎠
⎝
4π ⎞
⎛
LcD = LDc = M D cos⎜θ (t ) −
⎟
3 ⎠
⎝
Page 16
(5.13)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
The damper winding Q is orthogonal to the D winding. Therefore
π⎞
⎛
LaQ = LQa = M Q cos⎜θ (t ) − ⎟
2⎠
⎝
π 2π ⎞
⎛
LbQ = LQb = M Q cos⎜θ (t ) − −
⎟
2
3 ⎠
⎝
π 4π ⎞
⎛
LcQ = LQc = M Q cos⎜ θ (t ) − −
⎟
2
3 ⎠
⎝
(5.14)
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 θ (t ) = ω s t + δ (t ) +
π
2
, it is obvious that the inductance matrix is time dependent
⎡λ abc (t )⎤ ⎡ Lss (t ) Lsr (t )⎤ ⎡ iabc (t ) ⎤
⎢λ
⎥=⎢
Lrr ⎥⎦ ⎢⎣i fDQ (t )⎥⎦
⎣ fDQ ⎦ ⎣ Lrs (t )
or
λ (t ) = L(t )i (t )
In summary the voltage equations read:
v abc (t ) − v n (t )13 = − Rabc iabc (t ) −
v fDQ (t ) = − R fDQ i fDQ (t ) −
dλ abc (t )
dt
dλ fDQ (t )
dt
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
J
d 2θ m ( t )
= Tm − Te
dt 2
where J is the rotor moment of inertia,
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 17
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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.
The rotor position can be substituted with the electrical rotor position (see equation 5.2)
yielding
2 J d 2θ e ( t )
= Tm − Te
P dt 2
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
computed by using the principle of virtual work displacement:
Te =
∂W field
∂θ m
where Wfield is the total energy stored in the magnetic field of the synchronous machine,
and θm is the position of the rotor.
The total magnetic energy stored in the windings of the synchronous machine is given by
W field =
⎡ iabc (t ) ⎤ 1 T
⎡ L (t ) Lsr (t )⎤ ⎡ iabc (t ) ⎤
1 T
= iabc (t ) i TfDQ (t ) ⎢ ss
λ abc (t ) λTfDQ (t ) ⎢
⎥
Lrr ⎥⎦ ⎢⎣i fDQ (t )⎥⎦
2
⎣ Lrs (t )
⎣i fDQ (t )⎦ 2
[
]
[
]
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, Pem (t ) , is:
dθ (t ) ∂W fld dθ m (t )
Pem (t ) = Te (t ) m
=
dt
∂θ m (t ) dt
computed with:
dλ fDQ (t ) T
dλabc (t ) T
) iabc (t ) + (
) i fDQ (t ) , or
dt
dt
T
Pem (t ) = eabc
(t )iabc (t ) + e TfDQ (t )i fDQ (t )
Pem (t ) = (
and
Te (t ) =
Page 18
Pem (t ) P Pem (t )
=
ωm (t ) 2 ω ( t )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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:
λ ' (t ) = Tλ (t )
or
λ (t) = T -1λ ' (t )
i' (t ) = Ti(t )
or
i(t) = T -1i' (t )
Upon substitution of λ( t ) and i(t) in the generator equations:
T -1λ ' (t ) = L(t )T −1 i' (t)
Premultiplication of the equation with T yields
λ ' (t ) = TL(t )T −1 i' (t)
or
λ ' (t ) = L' i' (t)
where:
L' = TL(t )T −1
Park’s transformation T is selected in such a way as to make the matrix L' a constant
matrix.
This transformation is defined below
⎡ P 0⎤
T =⎢
⎥
⎣0 I⎦
where
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 19
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
1
1
1
⎡
⎤
⎢
⎥
2
2
2
2⎢
cos(θ (t ) ) cos(θ (t ) − α ) cos(θ (t ) − 2α )⎥⎥
P=
⎢
3
⎢ sin(θ (t ) ) sin(θ (t ) − α ) sin(θ (t ) − 2α )⎥
⎢⎣
⎥⎦
α=
2π
3
I
the 3x3 identity matrix
0
a 3x3 zero matrix
θ (t ) = ω s t + δ (t ) +
π
(position of d-axis)
2
The above selected matrix P has the following property (which can be easily checked):
P −1 = P T
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
T
−1
⎡ P −1
=⎢
⎣ 0
0⎤ ⎡ P T
⎥=⎢
I⎦ ⎣ 0
0⎤
⎥
I⎦
Recall that the matrix L(t) is:
⎡L
L(t ) = ⎢ Tss
⎣ Lsr
Lsr ⎤
Lrr ⎥⎦
Substitution into the expression for the transformed inductance matrix L yields
Lsr ⎤ ⎡ P T
Lrr ⎥⎦ ⎢⎣ 0
⎡ P 0⎤ ⎡ Lss
L' = ⎢
⎥⎢
⎣ 0 I ⎦ ⎣ Lrs
0⎤ ⎡ PLss P T
⎥=⎢
I ⎦ ⎣ Lrs P T
PLsr ⎤
⎥
Lrr ⎦
By direct substitution, the following is obtained
⎡ Lo
⎢
PLss P = ⎢ 0
⎢⎣ 0
T
Page 20
0
Ld
0
0⎤
⎥
0 ⎥ = cons tan t (time invariant)
Lq ⎥⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
⎡ 0
⎢
PLsr = ⎢kM F
⎢⎣ 0
kM D
0
⎡0 kM F
⎢
Lrs P = ⎢0 kM D
⎢⎣0
0
0 ⎤
⎥
0 ⎥ = cons tan t (time invariant)
kM Q ⎥⎦
T
⎡ LF
⎢
Lrr = ⎢ M R
⎢⎣ 0
0 ⎤
⎥
0 ⎥ = cons tan t (time invariant)
kM Q ⎥⎦
0
MR
LD
0
0⎤
⎥
0 ⎥ = cons tan t (time invariant)
LQ ⎥⎦
where:
L0 = Ls − 2 M s
⎛ 3⎞
Ld = Ls + M s + ⎜ ⎟ Lm
⎝2⎠
⎛ 3⎞
Lq = Ls + M s − ⎜ ⎟ Lm
⎝2⎠
3
k=
2
To complete the model, consider the generator voltage equations:
⎡1 ⎤
dλ (t )
v (t ) − ⎢ 3 ⎥ v n (t ) = − Ri (t ) −
dt
⎣0⎦
Upon substitution of the currents and flux linkage with
i(t ) = T −1i' (t )
λ (t ) = T −1λ ' (t )
and premultiplication of the resulting equation with T, the following is obtained
⎡1 ⎤
dT −1λ' (t )
T {v (t ) − ⎢ 3 ⎥ v n (t )} = −TRT −1i ' (t ) − T
dt
⎣0⎦
Now define:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 21
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
⎡ v o (t ) ⎤
⎡v a (t )⎤
⎢
⎥
⎢
⎥
⎢v d (t )⎥ = P ⎢ v b (t ) ⎥
⎢⎣ v q (t ) ⎥⎦
⎢⎣ v c (t ) ⎥⎦
⎡ v 0 (t ) ⎤
⎢ v (t ) ⎥
⎥
⎢ d
⎢ v q (t ) ⎥ ⎡ v odq (t ) ⎤
⎡13 ⎤
v' (t) = T {v (t ) − ⎢ ⎥ v n (t )} = ⎢
⎥ = ⎢v ( t ) ⎥
⎣0⎦
⎢ − v f (t )⎥ ⎣ fDQ ⎦
⎢ 0 ⎥
⎥
⎢
⎣ 0 ⎦
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
v ' (t ) = −TRT −1i ' (t ) − T
λ ' (t ) = L' i ' ( t )
d
(T −1 L' i ' (t ))
dt
Magnetic Flux Model
v ' (t ) = −TRT −1 L' −1 λ ' (t ) − T
d
(T −1λ ' (t ))
dt
i' (t ) = L' −1 λ ' (t )
Both models can be further simplified by the following results which are obtained by
straightforward manipulations:
TRT −1
T
⎡r
⎢0
⎢
⎢0
=⎢
⎢0
⎢0
⎢
⎢⎣0
0 0
r
0
0
0
0
0
0 0
r 0
0 rf
0 0
0
0
0
rD
0 0
0
0
0⎤
0⎥
⎥
0⎥
⎥
0⎥
0⎥
⎥
rQ ⎥⎦
d −1
dT −1
dλ ' (t )
dT −1
dλ ' (t )
λ ' (t ) + TT −1
λ ' (t ) +
(T λ ' (t )) = T
=T
dt
dt
dt
dt
dt
By direct computation
Page 22
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
0
0
⎡0
⎢0
ω (t )
0
⎢
0
dT −1 ⎢0 − ω (t )
T
=⎢
0
0
dt
⎢0
⎢0
0
0
⎢
0
0
⎣0
0
0
0
0
0
0
0
0
0
0
0
0
0⎤
0⎥
⎥
0⎥
⎥,
0⎥
0⎥
⎥
0⎦
ω (t ) =
dθ ( t )
dt
Thus
0
⎡ λo ( t ) ⎤
⎡
⎤
⎢ λ (t ) ⎥
⎢ ω ( t )λ ( t ) ⎥
q
⎢ d ⎥
⎢
⎥
⎢ − ω ( t )λ d ( t ) ⎥ d ⎢ λ q ( t ) ⎥
d
T (T −1λ ' (t )) = ⎢
⎥ + ⎢λ (t ) ⎥
0
dt
⎢
⎥ dt ⎢ f ⎥
⎢λ D ( t ) ⎥
⎢
⎥
0
⎥
⎢
⎢
⎥
0
⎢⎣λQ (t ) ⎥⎦
⎣
⎦
Upon direct substitution of above results and rearranging the equations the current model
is:
0
⎡r
⎡ vo ⎤
⎢0
⎢ v ⎥
r
⎢
⎢ d ⎥
0
⎢0
⎢ −vf ⎥
⎥ = −⎢
⎢
0
⎢0
⎢ v D = 0⎥
⎢0 − ωLd
⎢ vq ⎥
⎢
⎥
⎢
0
⎣⎢0
⎣⎢ vQ = 0⎦⎥
⎡ L0
⎢0
⎢
⎢0
−⎢
⎢0
⎢0
⎢
⎢⎣ 0
where k =
0
0
0
0
rf
0
− ωkM F
0
0
rD
− ωkM D
ωLq
0
0
0
0
0
0
0
Ld
kM f
kM D
0
kM f
Lf
MR
0
kM D
MR
LD
0
0
0
0
Lq
0
0
0
kM Q
3
0
0
r
⎤ ⎡ io ⎤
ωkM Q ⎥ ⎢ id ⎥
⎥⎢ ⎥
0 ⎥ ⎢i f ⎥
⎥⎢ ⎥
0 ⎥ ⎢i D ⎥
0 ⎥ ⎢ iq ⎥
⎥⎢ ⎥
rQ ⎦⎥ ⎣⎢iQ ⎦⎥
0
0 ⎤ ⎡ io ⎤
0 ⎥ ⎢ id ⎥
⎥ ⎢ ⎥
0 ⎥ d ⎢i f ⎥
⎥ ⎢ ⎥
0 ⎥ dt ⎢i D ⎥
kM Q ⎥ ⎢ i q ⎥
⎥ ⎢ ⎥
LQ ⎥⎦ ⎢⎣iQ ⎥⎦
2
and the flux model is:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 23
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
0
⎡r
⎡ vo ⎤
⎢0
⎢ v ⎥
r
⎢
⎢ d ⎥
0
⎢0
⎢− v f ⎥
⎥ = −⎢
⎢
0
⎢0
⎢ vD ⎥
⎢0 − ωLd
⎢ vq ⎥
⎢
⎥
⎢
0
⎣⎢0
⎣⎢ vQ ⎦⎥
0
0
0
0
rf
0
− ωkM F
0
0
rD
− ωkM D
ωLq
0
0
0
0
0
r
⎡ λ0 ⎤
⎤ ⎡ λo ⎤
⎢λ ⎥
⎥
⎢
⎥
ωkM Q λ d
⎢ d⎥
⎥⎢ ⎥
0 ⎥ ⎢λ f ⎥
d ⎢λ f ⎥
⎥⎢ ⎥ = − ⎢ ⎥
0 ⎥ ⎢λ D ⎥
dt ⎢λ D ⎥
⎢ λq ⎥
0 ⎥ ⎢ λq ⎥
⎢ ⎥
⎥⎢ ⎥
rQ ⎦⎥ ⎣⎢λQ ⎦⎥
⎣⎢λQ ⎦⎥
0
Another useful form of the generation model equations is in the state space form
dx
= f ( x, u) which is obtained as follows.
dt
Electric Current Model. Define the following
[
v TG = v o
[
i TG = i o
id
[
λTG = λ o
⎡r
⎢0
⎢
⎢0
R1 = ⎢
⎢0
⎢0
⎢
⎢⎣0
if
iD
iq
iQ
λq
λf
λD
0
0
0
r 0
0 rf
0 0
0 0
0
0
rD
0
0
0
0
r
0
0
0
0
0
]
0 vq
λd
0
⎡0
⎢0
0
⎢
0
⎢0
R2 = ⎢
0
⎢0
⎢0 − Ld
⎢
0
⎣0
Page 24
−v f
vd
0
0
0
0
− kM F
0
0
]
λQ
]
0⎤
0⎥
⎥
0⎥
⎥
0⎥
0⎥
⎥
rQ ⎥⎦
0
0
0
0
− kM D
0
0
Lq
0
0
0
0
0 ⎤
kM Q ⎥
⎥
0 ⎥
⎥
0 ⎥
0 ⎥
⎥
0 ⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
⎡ Lo
⎢0
⎢
⎢0
Leq = ⎢
⎢0
⎢0
⎢
⎣⎢ 0
0
0
0
0
Ld
kM f
kM D
0
kM f
Lf
MR
0
kM D
MR
LD
0
0
0
0
Lq
0
0
0
kM Q
0 ⎤
0 ⎥
⎥
0 ⎥
⎥
0 ⎥
kM Q ⎥
⎥
LQ ⎦⎥
Now the current model is written as follows:
v G (t ) = −( R1 + ωR2 )iG (t ) − Leq
λG (t ) = Leq iG (t )
diG (t )
dt
or
diG (t )
−1
−1
= − Leq ( R1 + ωR2 )iG (t ) − Leq v G (t )
dt
λG (t ) = Leq iG (t )
The flux model is obtained by eliminating the electric current from above model, i.e.
dλ G ( t )
= −( R1 + ωR2 ) L−eq1 λG (t ) − v G (t )
dt
iG (t ) = L−eq1 λG (t )
Above models of a synchronous machine are illustrated in Figure 5.9.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 25
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Stator
Rotor
io
Lo
r
+
if
rf
Vf
Vo
-
Lf
+
kM f
id
r
-
+
MR
Ld
iD
Vd
rD
LD
kM D
-
-
+
iq
r
iQ
rQ
ed
+
LQ
kM Q
Lq
Vq
+
eq
-
-
Figure 5.9. Equivalent Synchronous Machine Circuits
ed = ωLq iq + ωkM Q iQ
eq = ωLd id + ωkM f i f + ωkM D i D
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
Page 26
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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.
Voltage
Current
Power
Flux Linkage
Resistance
Inductance
Time
Angular Velocity
Angle
Symbol
v
i
p, q or S
λ
r
L
t
ω
δ
Units
V
A
W, VAr or VA
Wb
Ω
H
sec
(sec)-1
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
ω B while the remaining are selected with:
tB =
1
ωB
IB =
SB
VB
λ B = VB t B
LB =
λB
IB
(seconds)
(Amperes)
(Webers)
(Henrys)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 27
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
RB =
VB
IB
(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
di
Example E5.1: Consider the equation v = − ri − L
which is expressed in the metric
dt
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:
V B = RB I B =
λB
tB
=
LB I B
tB
The result is:
⎛ i ⎞
d ⎜⎜ ⎟⎟
⎛ r ⎞⎛ i ⎞ ⎛ L ⎞ ⎝ I B ⎠
v
⎟⎟⎜⎜ ⎟⎟ − ⎜⎜
⎟⎟
= −⎜⎜
VB
⎝ R B ⎠⎝ I B ⎠ ⎝ L B ⎠ ⎛⎜ t ⎞⎟
d⎜ ⎟
⎝ tB ⎠
or
di u
dt u
where the subscript u denotes per unitized quantities.
v u = − ru i u − Lu
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.
Page 28
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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
ω B usually the synchronous angular frequency
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:
λG (t ) = Leq iG (t )
In general
Ld = Lmd + l d
Lq = Lmq + l q
L f = L mf + l f
L D = L mD + l D
L Q = L mQ + l Q
Notice that always
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 29
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
kM f = L md L mf
kM D = L md L mD
kM Q = L mq L mQ
M R = L mf L mD
The selection of the base quantities is done on the basis of equal mutual flux linkages as
follows:
Define:
MR
L
kM f
) = mf =
kM D
kM f
L md
MR
L
kM D
k D (=
) = mD =
kM f
kM D
L md
L mQ
kM Q
kQ =
=
kM Q
L mq
k f (=
Then
R fB = k 2f R B
L fB = k 2f L B
R DB = k 2D R B
L DB = k 2D L B
R QB = k Q2 R B
L QB = k Q2 L B
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
SB
ωB
V sB
- per phase power base (rated power/3)
- angular speed base (rated, electrical)
- phase to line voltage base (rated L-L voltage/ 3 )
Then the derived based quantities are:
tB = 1/ ωB
Page 30
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
I sB = S B / VsB
λ sB = V sB t B
LsB = λ sB / I sB
RsB = V sB / I sB
2. Define the following constants:
kf =
Lmf
kM f
MR
=
=
kM D kM f
Lmd
kD =
L
MR
kM D
= mD =
kM f
kM D
Lmd
kQ =
LmQ
kM Q
=
kM Q
Lmq
Above results are summarized in Table 5.2.
Table 5.2. Summary of a Synchronous Machine Per unit System
Armature
Field Circuit
S B , ω B , k f V sB
D-Axis Damper
Circuit
S B , ω B , k DVsB
Q-Axis Damper
Circuit
S B , ω B , k QV sB
S B , ω B , VsB
tB = 1/ ωB
I sB = S B / VsB
tB = 1/ ωB
I fB = I sB / k f
tB = 1/ ωB
I DB = I sB / k D
tB = 1/ ωB
I QB = I sB / k Q
λ sB = V sB t B
LsB = λ sB / I sB
λ fB = k f λ sB
λ DB = k D λ sB
λQB = k Q λ sB
LDB = k D2 LsB
RsB = V sB / I sB
L fB = k 2f LsB
LQB = k Q2 LsB
R fB = k 2f R sB
R DB = k D2 R sB
RQB = k Q2 R sB
3. Mutual Inductance Base
M fsB = L fB L sB = k f LsB
M DsB = L DB LsB = k D LsB
M QsB = LQB LsB = k Q LsB
M DfB = LDB L fB
5.6 Equivalent Circuits of a Synchronous Machine
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 31
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 0axis equivalent circuit is quite simple.
q-Axis Equivalent Circuit: Consider the voltage equations for the q-axis voltage and
currents:
diq
diQ
v q = ωLd id + ωkM f i f + ωkM D i D − riq − Lq
− kM Q
dt
dt
diq
diQ
0 = − rQ iQ − kM Q
− LQ
dt
dt
Divide the first equation by V sB , remembering that
V sB = ω B LsB I sB = ω B M fsB I fB = ω B M DsB I DB = R sB I sB = LsB I sB / t B = M QsB I QB / t B
The result will be
vq
V sB
ωkM f i f
ri q
Lq
ωLd i d
ωkM D i d
=
+
+
−
−
ω B LsB I sB ω B M fsB I fB ω B M DsB I DB R sB I sB LsB
iq
)
d(
iQ
)
I QB
kM Q
I sB
−
t
t
M QsB
d( )
d( )
tB
tB
d(
or
v qu = e qu − ru i qu − ( Lmqu + l qu )
di qu
dt u
− ( kM Q ) u
diQu
dt u
Similarly divide the second equation by V sB , remembering that
V sB = RQB I QB = M QsB I sB / t B = LQB I QB / t B
The result will be
0=−
rQ iQ
RQB I QB
−
kM Q
M QsB
⎛ iQ ⎞
⎛ iq ⎞
⎟
d⎜
⎟⎟
d ⎜⎜
⎜I ⎟
L
I
QB
Q
sB
⎠
⎝
⎝
⎠−
L
⎛ t ⎞
⎛ t ⎞
QB
d ⎜⎜ ⎟⎟
d ⎜⎜ ⎟⎟
⎝ tB ⎠
⎝ tB ⎠
or
Page 32
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
0 = − rQu iQu − kM Qu
diqu
− ( LmQu + l Qu )
diQu
dt u
dt u
A summary of above results is given below.
diqu
diQu
Vqu = equ − ru i qu − ( Lmqu + l qu )
− ( kM Q ) u
dt u
dt u
diQu
diqu
− ( LmQu + l Qu )
0 = − rQu iQu − kM Qu
dt u
dt u
equ = ωu Ldu idu + ωu kM fu i fu + ωu kM Du i Du
Now it is obvious that the circuit of Figure 5.10 corresponds exactly to above two
lqu
ru
iqu
+
rQu
lQu
LAQu
Vqu
+
-
-
equ
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:
di
d
(i du + i fu + i bu ) − l dn du
dt u
dt u
di
d
− v fu = − r fu i fu − Lmdu
(i du + i fu + i Du ) − l fu du
dt u
dt u
di
d
0 = − rDu i Du − Lmdu
(idu + i fu + i Du ) − l Du du
dt u
dt u
v du = −e du − ru i du − Lmdu
Where: equ = ωu Lqu idu + ωu kM Qu iQu
Above equations represent the equivalent circuit of Figure 5.11
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 33
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
ldu
ru
rfu
lfu
idu
+
r
Du
lDu
Vdu
Lmdu
Vfu
+
-
-
+
edu
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.
Page 34
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
L0
r
io
+
Vo
-
(a)
ld
r
rf
l
f
id
+
r
D
l
D
Vd
LAD
Vf
+
(b)
-
+
ed
lq
r
iq
+
rQ
l
Q
Vq
LAQ
+
-
-
eq
(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:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 35
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
H = 2.8
rD = 0.00623
rf = 0.0058
Lq = 2.474
sec
ohms
ohms
mH
r = 0.002
ohms
rQ = 0.00737 ohms
Ld = 2.7035 mH
l d = l q = 0.32 mH
Lf = 35.5056
MR = 7.8729
Lo = 0.723
l Q = 013
.
mH
mH
mH
mH
l f = 4.1029 mH
kMQ = 1.4863 mH
mH
l D = 0.2
Use the following base system for the stator circuit
25000
1300
ω B = 2π 60 sec −1
VB =
Volts
SB =
MVA
3
3
and compute the per unit values of the following parameters:
r, rD, rQ, rf, LAD, LAQ, l d , l q , l f , l D , l Q ,Lo
Solution: The derived bases are computed using the equations in Table 5.2. First we
compute some of the parameters.
Lmf = L f − l f = 31.4027mH
Lmd = Ld − l d = 2.3835mH
kM f = Lmd Lmf = 8.65149mH
kf =
kM D
kM f
= 3.62975
Lmd
M
= R = 2.16899mH
kf
Lmq = Lq − l q = 2.154mH
Now
And
kM D
= 0.91
⇒
Lmd
l D = LD − LmD = 0.196mH
kD =
kQ =
kM Q
Lmq
= 0.69
⇒
LmD = k D (kM D ) = 1.9737mH ,
LmQ = k Q (kM Q ) = 1.0254mH ,
l Q = LQ − LmQ = 0.1359mH
From above quantities, the base quantities for all circuits are computed as follows:
A. Stator:
Page 36
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
S B = 433.33 MVA
ω B = 377 sec −1
VB = 14433.76
tSB = 0.00265258
ISB = 30022.18
λ SB = 38.2867
LSB = 0.00127528
RSB = 0.4807698
Volts
seconds
amperes
Wb
H
ohms
B. Main Field
kf = 3.62975
SfB = 433.33
MVA
sec-1
ω fB = 377
VfB = 52390.94
V
tfB = tSB = 0.00265258 sec
IfB = 8271.1426
ampers
Wb
λ fB = 138.971
LfB =0.0168019
H
RfB = 6.33418
ohms
C. D - Circuit
kD = 0.91
SDB = 433.33
ω DB = 377
VDB = 13134.72
tDB = 0.00265258
IDB = 32991.406
λ DB = 34.84089
LDB =0.001056059
RDB = 0.398125
MVA
sec-1
V
sec
ampers
Wb
H
ohms
kQ = 0.69
SQB = 433.33
ω QB = 377
MVA
sec-1
VQB = 9959.294
tQB = 0.00265258
IQB = 43510.4058
λ QB = 26.4178
V
sec
ampers
Wb
LQB =0.00060716
RQB = 0.2288945
H
ohms
D. Q - Circuit
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 37
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
E. Mutual Inductance
M fSB = L fB LSB = 0.00462894 H
M DSB = LDB LSB = 0.0011605H
M QSB = LQB LSB = 0.00087994 H
M DfB = LDB L fB = 0.004211234 H
F. Torque Base
TB =
S SB
ω SB
= 1,149,416.446 N .m
Then the per unit values of the machine parameters are:
r = 0.00416
rQ = 0.0322
LAD = 1.869
l d = l q = 0.2509
rD = 0.01565
rf = 0.0009156
LAQ = 1.689
l f = 0.2442
l D = 0.1856
l Q = 0.2238
Lo = 0.5669
5.7 Synchronous Machine Torque Equation
To compute the torque and power one has to analyze the electromagnetic field and thus
compute 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:
The 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)
di q
di Q
di o
di
di
di
+Ldid d +kMfid f +kMDid D +Lqiq
+kMQiq
]
dt
dt
dt
dt
dt
dt
+ ω (Ldid +kMfif +kMDiD)iq - ω (Lqiq +kMQiQ)id
p(t)=-r(id2+iq2)-rio2-Loio
or
p(t) = Pohmic + Pst. magn. + Ptrnf.
where
Page 38
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Pohmic : ohmic losses on resistors
Pst. Magn. : rate of change of stator magnetic field energy
Ptrnf.
: power transferred across air gap
Using the principle of virtual work displacement we obtain
∂Wfld ∂Pfld
=
⇒
∂θ
∂ω
Tem = (Ldid +kMfif +kMDiD)iq - (Lqiq +kMQiQ)id
Tem =
Above equation yields the total electromagnetic field torque. In vector notation
Tem = [0 Ld iq
kM f iq
kM D iq
− Lq id
⎡ io ⎤
⎢i ⎥
⎢ d⎥
⎢i f ⎥
− kM Q id ]⎢ ⎥
⎢i D ⎥
⎢ iq ⎥
⎢ ⎥
⎣⎢iQ ⎦⎥
The power transferred through the air gap is the one provided with the voltage source ed,
and eq.
Pem = -edid + eqiq
And the electromechanical torque is
Tem (t ) =
Pem (t ) P − ed id + eq i q
=
ω m (t ) 2
ω (t )
The generator rotor motion equation is
J
or
d 2θ m (t )
= Tm − Tem
dt 2
2 J dω
= Tm − Tem
P dt
Note that above equation is the so-called swing equation. This equation can be written in
p.u. For this purpose let SB the per phase base power. The 3-Phase power will be 3SB.
And the 3-phase base torque
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 39
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
3TB =
P 3S B
2 ωB
Upon division of the swing equation with the above base torque, one obtains:
⎛ 2 ⎞ J ω B dω
= Tm − Tem
⎜ ⎟
⎝ P ⎠ 3S B dt
2
where
Tm is normalized mechanical torque
Tem is normalized electromagnetic torque
Note that
2
dω u
⎛ 2 ⎞ J ω B ω B ( d ω / d ω B ) 1 ⎛ 2 ω B ⎞ 2 ω B dω u
= J⎜
)
= 2 Hω B
⎜ ⎟
⎟ (
⎝ P ⎠ 3S B t B ( dt / t B )
2 ⎝ P ⎠ 3S B dt u
dt u
2
Let τ j = 2 Hω B . Thus the normalized swing equation becomes
τj
dω (t )
= Tm (t ) − Temu (t )
dt
Now consider the equation θ (t ) = ω s t + δ (t ) +
π
2
. The derivative of this equation is
dθ (t )
dδ (t )
= ω (t ) = ω s +
dt
dt
Normalize equation by dividing by ω B (note
ω u (t ) = 1 +
ωs
= 1)
ωB
dδ (t )
dδ (t )
=1+
, or
d (tω B )
dt u
dδ (t )
= ω (t ) − 1
dt
In summary, the state space swing equation in p.u. for a generator is given by
Page 40
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
d ω ( t ) Tm ( t ) 1
=
+ [0 − Ld i q
dt
τj
τj
− kM f i q
− kM D i q
Lq i d
dδ ( t )
= ω (t ) − 1
dt
⎡ io ⎤
⎢i ⎥
⎢d⎥
⎢i f ⎥
kM Q i d ]⎢ ⎥
⎢i D ⎥
⎢ iq ⎥
⎢ ⎥
⎣⎢iQ ⎦⎥
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)
di G (t )
= − L−eq1 (R1 + ωR 2 )i G (t ) − L−eq1 v G (t )
dt
dω (t ) 1 T
iG (t )R 2T iG (t ) − Dω (t ) + Tm
=
τj
dt
(
)
dδ (t )
= ω (t ) − 1
dt
where
⎡ io ⎤
⎢i ⎥
⎢ d⎥
⎢i f ⎥
iG = ⎢ ⎥ ,
⎢i D ⎥
⎢ iq ⎥
⎢ ⎥
⎢⎣iQ ⎥⎦
⎡ Lo
⎡ vo ⎤
⎢0
⎢ v ⎥
⎢
⎢ d ⎥
⎢0
⎢− v f ⎥
vG = ⎢
⎥ , Leq = ⎢ 0
⎢
⎢ 0 ⎥
⎢0
⎢ vq ⎥
⎢
⎢
⎥
⎢⎣ 0
⎣ 0 ⎦
C T = [0 − Ld i q
− kM f i q
Copyright © A. P. Sakis Meliopoulos – 1990-2006
− kM D i q
0
0
0
0
Ld
kM f
kM D
0
kM f
Lf
MR
0
kM D
MR
LD
0
0
0
0
Lq
0
0
0
kM Q
Lq id
0 ⎤
0 ⎥
⎥
0 ⎥
⎥
0 ⎥
kM Q ⎥
⎥
LQ ⎥⎦
kM Q id ]
Page 41
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
⎡r
⎢0
⎢
⎢0
R1 = ⎢
⎢0
⎢0
⎢
⎣⎢0
where
0
0
0
0
r 0
0 rf
0 0
0 0
0
0
rD
0
0
0
0
r
0
0
0
0
0⎤
0⎥
⎥
0⎥
⎥
0⎥
0⎥
⎥
rQ ⎦⎥
0
⎡0
⎢0
0
⎢
0
⎢0
R2 = ⎢
0
⎢0
⎢ 0 − Ld
⎢
0
⎣0
0
0
0
0
− kM f
0
0
0
0
0
− kM D
0
0
Lq
0
0
0
0
0 ⎤
kM Q ⎥
⎥
0 ⎥
⎥
0 ⎥
0 ⎥
⎥
0 ⎦
τ j = 2 Hω B
The actual voltage or current phase quantities may be obtained from
⎡ v o (t ) ⎤
⎡v a (t )⎤
⎢ v (t ) ⎥ = P -1 ⎢v (t )⎥
⎢ d ⎥
⎢ b ⎥
⎢⎣ v q (t ) ⎥⎦
⎢⎣ v c (t ) ⎥⎦
⎡ io ( t ) ⎤
⎡ia (t )⎤
⎢i (t ) ⎥ = P -1 ⎢i (t )⎥
⎢d ⎥
⎢b ⎥
⎢⎣i q (t ) ⎥⎦
⎢⎣ic (t ) ⎥⎦
Page 42
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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)
i a (t ) = 2 I cos(ω s t + δ + ϕ I +
π
2
)
2π
)
2
3
π 4π
i c (t ) = 2 I cos(ω s t + δ + ϕ I + −
)
2
3
ib (t ) = 2 I cos(ω s t + δ + ϕ I +
π
v a (t ) = 2V cos(ω s t + δ + ϕ V +
−
π
2
2π
)
2
3
π 4π
v c (t ) = 2V cos(ω s t + δ + ϕ V + −
)
2
3
v b (t ) = 2V cos(ω s t + δ + ϕ V +
if(t) = If
vf(t) = Vf
vn = 0
π
)
−
(dc)
(dc)
θ (t ) = ω s t + δ +
π
2
⎡ v o (t ) ⎤
0
⎡v a (t )⎤ ⎡
⎤
⎥
⎢
⎢
⎥
⎢
⎥
⎢v d (t )⎥ = P ⎢ v b (t ) ⎥ = ⎢ 3V sin(ϕ V ) ⎥
⎢⎣ v q (t ) ⎥⎦
⎢⎣ v c (t ) ⎥⎦ ⎢⎣ 3V cos(ϕ V )⎥⎦
⎡ io (t ) ⎤
⎥
⎢
⎢i d (t )⎥ =
⎢⎣i q (t ) ⎥⎦
⎡i a (t )⎤
P ⎢i b ( t ) ⎥ =
⎢
⎥
⎢⎣i c (t ) ⎥⎦
0
⎡
⎤
⎢ 3I sin(ϕ ) ⎥
I ⎥
⎢
⎢⎣ 3I cos(ϕ I )⎥⎦
Observe that the o-d-q axes currents and voltages are constant at steady state, balanced
conditions
Thus
diodq (t )
dt
=0
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 43
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
dvodq (t )
dt
=0
If above relationships are substituted into the synchronous machine model, we obtain:
vo = 0
vd = -rid - ω Lqiq - ω kMQiQ
vq = -riq + ω Ldid + ω kMfif + 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 + ω Ldid - ω kMfif
-vf = -rfif
Note that: vd, vq, vf, id, iq, if represent dc-quantities. From those if, vf are actually dcquantities 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.
v a (t ) =
2
( v d cos θ (t ) + v q sin θ (t ))
3
2
(i d cos θ (t ) + i q sin θ (t ))
3
π
θ (t ) = ω s t + δ +
2
i a (t ) =
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
π
θ (t ) = ω s t + δ (t ) + . The quantities id, vd represent current and voltage on the d-axis of
2
the rotor, and the quantities iq, vq represent current and voltage at the q-axis of the rotor.
The angle θ (t) indicates the position of the q-axis. This is shown in Figure 5.13.
Page 44
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
d-axis
θ = ωst + δ +
iq
π
2
q-axis
ωst + δ
reference
( Phase A
Magnetic axis )
id
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 :
→
id = id e
π
j ( +δ )
2
→
iq = iq e j (δ )
→
vd = vd e
π
j ( +δ )
2
→
v q = v q e j (δ )
Above conclusions can be justified analytically by using the usual definition of phasors.
Now, consider the phase A voltage:
2
v a (t ) =
( v d cos θ + v q sin θ )
3
Upon substitution of the voltages and converting the sines into cosines, we have:
π
2
v a (t ) =
[( − rid − ωLq i q ) cos(ωt + + δ ) + ( − ri q + ωLd i d + ωkM f i f ) cos(ωt + δ )]
3
2
Note that the rms phasor representation of va(t) is:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 45
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
π
j ( +δ )
1
1
~
Va = −
( rid + ωLq i q )e 2 +
( − ri q + ωLd i d + ωkM f i f )e jδ
3
3
This equation can be written in the form
~
~ ~
~
~ ~
Va = − r ( I d + I q ) − jx d I d − jx q I q + E
where
xd = ω Ld
xq = ω Lq
ωM f i f
E=
2
d-axis synchronous reactance
q-axis synchronous reactance
generated voltage
π
j ( +δ )
1
~
Id =
id e 2
3
1
~
Iq =
i q e jδ
3
Same analysis can be repeated for phases B and C. The final results are as follows:
~
~ ~
~
~ ~
Va = − r ( I d + I q ) − jx d I d − jx q I q + E
~
~ ~ −j
Vb = − r ( I d + I q )e
~
~ ~ −j
Vc = − r ( I d + I q )e
2π
3
4π
3
~ −j
− jx d I d e
~ −j
− jx d I d e
2π
3
4π
3
~ −j
− jx q I q e
~ −j
− jx q I q e
2π
3
4π
3
~ −j
+ Ee
~ −j
+ Ee
2π
3
4π
3
The previous analysis for the voltage Va can be repeated for the current Ia. Specifically,
the time function of the current ia(t) is:
2
π
2
(i d cos θ + i q sin θ ) =
i a (t ) =
[i d cos(ωt + + δ ) + i q cos(ωt + δ )]
3
3
2
The rms phasor representation of the current ia(t) is :
π
j ( +δ )
1
1
~
~ ~
Ia =
id e 2 +
i q e jδ = I d + I q
3
3
The derived equations are summarized in Figure 5.14.
Page 46
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
δ'
d-axis
q-axis
E
Vq
δ
jxq Iq
Iq
jx Id
d
Va
Vd
Ia
reference
r( Id + I q )
Id
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 Id and Iq can be obtained from the solution of the following
equations.
~
~
~
~ ~
Va = − rI a − jx d I d − jx q I q + E
~ ~ ~
Ia = Id + Iq
(5.15)
~ ~
An alternative way is to solve for Id , Iq through a graphical solution. The graphical
solution is based on rewriting the equation (5.15) in the form
~
~
~
~
~
Va + rI a + jx q I a = − j ( x d − x q ) I d + E
~
~
Observe that the phasor − j ( x d − x q ) I d + E is along the q-axis and equals the phasor:
~ ~
~
~
A = Va + rI a + jx q I a
~
which is computable. Thus, one can first compute the phasor A and define the location
~
of the q-axis from the phase of the phasor A
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 47
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 q Ia . The line OA defines the q-axis, i.e. the angle δ . Step 3:
~
~ ~
Compute Id , Iq as projections of Ia on the d and q-axes and complete the diagram. The
procedure will be illustrated with an example.
d-axis
q-axis
E
A
jxq Ia
Iq
Va
O
r Ia
Ia
Id
jxq Iq
δ
jxdId
reference
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 = ωL d = 2.39Ω
xq = ωL q = 2.30Ω
Page 48
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
cos φ = 0.85 ⇒ φ = −3178
. 0
rIa = 9.496 V
xqIa = 14164.32 V
From above quantities, a plot is drawn.
d-axis
q-axis
E
jxq Ia
jxq Iq
Iq
jx Id
d
o
36.75 Va
o
31.78
reference
Ia
Id
From the plot we measure:
δ = 36.750
Vd = -5182
Vq = 6939
vd = -8975
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:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 49
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
0
~ ~
~
~
A = Va + rI a + jxq I a = 16,128.14 + j12,035.76 = 20,124.02e j 36.73 volts
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:
0
~
E = 20,700e j 36.73 V
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 A (phasor).
~ ~
~
~
A = Va + r Ia + jx q Ia = Ae jδ
where
~
Va = 14433.76Ve j0 = 10
. e j0
~
Ia = 23093.98Ae − j0.451 = 0.76923e − j0.451
xq = ωL q =1.9399
~
A = 2.12918e j0.6817
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
Page 50
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
or after normalization
vd - vdL = 0.04159id +0.78414 ω iq
vq - vqL = 0.04195iq - 0.78414 ω id
⇒
vdL = -1.4831
vqL = 0.37373
The series capacitor steady state equations, after normalization, are:
Zc = 2.7186 pu
id = 2.7186 ω vqL - 2.7186 ω eq
iq = -2.7186 ω vdL + 2.7186 ω ed
ed = -1.2752
eq = 0.81751
E ∞ = 0.87454
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 51
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Figure 5.16 Synchronous Machine Short Circuit Currents Following a Three Phase
Fault
Subtransient
Transient
Synchronous
I"o
I'o
I'
Is
1
Time(seconds)
2
Figure 5.17 Illustration of the Decaying rms Value of the Phase A Current
5.10.1 Subtransient and Transient Inductances
Page 52
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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:
L"d = Ld − L2AD
L'd = Ld −
LD + L f − 2 L AD
L f LD − L2AD
= Ld − L2AD
lD +l f
l f l D + L AD (l D + l f )
L2AD
Lf
Where:
L"d is the d-axis subtransient inductance, and
L'd is the d-axis transient inductance.
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 53
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 = 0 = kM f i d + L f i f + M R i D
λ D = 0 = kM D i d + L D i D + M R i f
Solution of above equations for the currents if, iD in terms of id yields:
if = −
iD = −
kM f L D − kM D M R
L f L D − M 2R
kM D L f − kM f M R
L f L D − M 2R
id
id
The magnetic flux in the d-axis, λ d , is computed from
λ d = kM D i D + Ld i d + kM f i f
Substituting the filed current and the d-axis damper winding current, yields:
λd =
L"d i d
= (Ld −
( kM D ) 2 L f + ( kM f ) 2 L D − 2 kM f kM D M R
)i d
L f L D − M 2R
Recall that when the synchronuous machine parameters are expressed in pu, then:
L AD = kM D = kM f = M R
Thus:
L"d = Ld − L2AD
LD + L f − 2 L AD
L f LD − L2AD
If the damper winding D is neglected (MR = 0, kMD = 0)
λ d = L'd i d = ( L d −
( kM f ) 2
)i d
Lf
Thus:
L'd = Ld −
Page 54
L2AD
Lf
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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:
L"q = Lq −
L2AQ
LQ
L'q = Lq
Where:
L"q is the q-axis subtransient inductance, and
L'q is the q-axis transient inductance.
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+
λQ = 0 = kM Qiq + LQiQ
Solution of above equations for the currents iQ1, iQ2 yields:
iQ = −
kM Q
LQ
iq
The q-axis magnetic flux for this condition will be:
( kM Q ) 2
""
λq = Lq iq = ( Lq −
)iq
LQ
Thus
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 55
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
L = Lq −
""
q
( kM Q ) 2
LQ
= Lq −
L2AQ
LQ
The of q-axis transient reactance is derived by neglecting the q-axis damper winding.
Then:
L'q = Lq
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, l Q and LAQ. The time constant for this circuit is
τ "qo =
L AQ + l Q
rQ
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 "do
τ "d
L D − M 2R / L f
subtransient time constant (open circuit)
=
rD
=
τ "do
τ 'do ≅
L"d
L'd
Lf
rf
τ 'd = τ 'do
subtransient time constant (short circuit)
transient time constant (open circuit)
L'd
Ld
transient time constant (short circuit)
q-axis
Page 56
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
τ "qo =
τ "q
τ 'q
LQ
subtransient time constant (open circuit)
rQ
=
τ "qo
=
τ 'qo
L"q
subtransient time constant (short circuit)
L'q
L'q
Lq
Armature Time Constant:
τa =
L2
r
It describes the decay characteristics of the 3phase short circuit rms currents (DC of stator or
rms of field current)
L2 =
Ld + Lq
2
Example E5.5: The following parameters have been computed for the generator of
Example E5.3.
r = 0.00416
rD = 0.01565
LAQ = 1.689
l d = l q = 0.2509
rQ = 0.0322
rf = 0.0009165
LAD = 1.869
l f = 0.2442
l D = 01865
.
l Q = 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 "do , τ "d , τ 'do , τ 'd , τ "qo , τ q " , τ 'qo , τ 'q and τ a
Solution:
L" d = L d − L2AD
L’d = Ld L"q = L q −
L D + L f − 2 L AD
Lf LD −
L2AD
= L d − L2AD
lD + lf
l f l D + L AD ( l D + l f )
= 0.3507
L2AD
=0.4669
Lf
L2AQ
l Q1 + L AQ
Copyright © A. P. Sakis Meliopoulos – 1990-2006
= 0.4485
Page 57
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
L’q = Lq = 1.9399
τ "do =
τ "d
=
L D − M 2R / L f
= 25.66
rD
or
0.0686 sec
= 19.27
or
0.05112 sec
Lf
= 2307.99
rf
or
6.122 sec
or
1.3483 sec
or
0.15757 sec
or
0.0364 sec
or
0.7673 sec
τ "do
τ 'do ≅
τ 'd = τ 'do
τ "qo =
τ "q
=
LQ
rQ
τ "qo
L"d
L'd
L'd
Ld
= 508.32
= 59.403
L"q
L'q
= 13.734
τ 'q , τ 'qo meaningless for this model.
τa =
L2
= 289.27
r
5.10.3 Summary of Synchronous Machine Parameters
A comprehensive list of synchronous machine parameters is given in Table 5.4. The short
circuit ratio (SCR) is defined in Figure 5.21. All these parameters are available from the
manufacturer. Recently, techniques have been developed by which these parameters can
be estimated with real time monitoring systems.
Table 5.4 List of Generator Parameters
Manufacturer
Short Circuit Ratio(SCR)
Regulator
Unit Rated MVA
Unit Rated KV
Unit Rated PF
x”d , x”q , ra , x2
x’d , x’q , x l , xo
Page 58
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
xd , xq , r2
τ "do , τ "d , τ 'do
τ 'd , τ "qo , τ "q
τ 'qo , τ 'q , τ a
H
rf
Vf full load
f
air-gap
line
occ
Voltage
c
scc
o
a
b
g
h
e
d
o'
Field Current
SCR =
Ob
Og
air gap length
Figure 5.18 Definition of the SCR for a Synchronous Machine
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 59
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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)
b)
c)
Steady State Model.
Constant Main Field Winding Flux Model.
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.
equations describing this mode are:
vd = -rid - ω Lqiq
vq = -riq + ω Ldid + ω kMfif
vf = -rfif
Pe = vdid + vqiq
Page 60
The
(5.8)
(5.9)
(5.10)
(5.11)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Qe = vd iq − vqid
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 E has magnitude
E=
ωM f i f
(5.12)
2
d-axis
q-axis
E
Vq
δ
jxq Iq
Iq
jx Id
d
reference
Va
Vd
Id
Ia
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)
Ld = Ld(id + if)
kMf = kMf(id + if)
(5.13)
(5.14)
(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 d = − 3Va sin δ
(5.16)
v q = 3Va cos δ
(5.17)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 61
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Then solve equations (5.8) and (5.9) for id and iq to obtain
xq
xq
r
vd + 2
vq − 2
r + xd xq
r + xd xq
r + xd xq
x
r
r
iq = − 2 d
vd − 2
vq + 2
r + xd xq
r + xd xq
r + xd xq
id = −
2
3E
(5.18)
3E
(5.19)
Upon substitution into equation (5.11)
Pe = −
3 xq
3r
3 xd − xq 2
3r
Va2 +
Va sin 2δ + 2
Va E cos δ + 2
Va E sin δ
2
r + xd xq
r + xd xq
2 r + xd xq
r + xd xq
2
(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 :
Pe =
3Va E
3V 2 1
1
sin δ + a ( − ) sin 2δ
xd
2 x q xd
(5.21)
Also the reactive power is computed to be
Qe = vdiq - vqid
(5.22)
Upon substitution we obtain:
3Va2
3V E
Qe = − 2
( xd sin 2 δ + xq cos 2 δ ) + 2 a
(− r sin δ + xq cos δ )
r + xd xq
r + xd xq
(5.23)
If the resistance is neglected
3Va2
3V E
Qe = −
( x d sin 2 δ + x q cos 2 δ ) + a cos δ
xd
xd xq
(5.24)
Example E5.6: A 625 MVA, 60 Hz, 18 kV synchronous generator has the following
parameters:
r = 0.001066 pu, x d = 1.5645 pu, x q = 1.42231 pu, E = 2.03387 pu, V = 1.105 pu
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
Page 62
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
Pe = 4.3095 sin δ +0.117 sin 2δ
Qe = -2.57544 sin2 δ -2.3413 cos 2 δ + 4.3095 cos δ
b) If the armature resistance is not neglected.
Pe = -0.00175 + 0.11703sin 2δ + 0.2154 cos δ + 4.3095sin δ
= -0.00175 + 0.11703sin 2δ + 4.3148sin ( δ +0.04994)
Qe = -2.57544sin2 δ - 2.34137cos 2 δ + 4.30954cos δ - 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.
λ f = cons tan t
(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:
di d
di
− kM f f
dt
dt
di q
vq = -riq + ω Ldid + ω kMfif - L q
dt
λ f =Lfif +kMfid
vd = -rid - ω Lqiq - L d
(5.26)
(5.27)
(5.28)
Pe = vdid + vqiq
(5.29)
Qe = vdiq - vqid
(5.30)
Equation (5.28) yields
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 63
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
if =
λf
−
Lf
kM f
Lf
(5.31)
id
and since λ f is assumed constant:
di f
dt
=−
Upon substitution of if,
kM f did
L f dt
(5.32)
di f
in equations (5.26) and (5.27)
dt
( kM f ) 2 di d
vd = -rid - ω Lqiq - ( Ld −
)
Lf
dt
vq = -riq + ω ( L d −
(5.33)
di q
( kM f ) 2
)id+( ω kMf λ f)/Lf - L q
Lf
dt
In above equations the term L d −
(5.34)
( kM f ) 2
can be recognized as the d-axis transient
Lf
inductance, L’d:
L'd = Ld −
( kM f ) 2
(5.35)
Lf
The transient reactance will be
x’d = ωL'd
(5.36)
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
di q
di
case the terms L'd d and L q
are very small compared to other terms of the
dt
dt
equation. Thus, they may be omitted. Finally, since λ f is assumed to be constant the
term ( ω kMf λ f)/Lf is also constant. Define
E' =
ωM f λ f
(5.37)
2L f
which shall be called the generated voltage behind transient reactance.
Page 64
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
d-axis
q-axis
E'
jx'q Iq
jx'dId
reference
Va
Ia
Figure 5.20 Phasor Diagram of the Transient Model
Now equations (5.33) and (5.34) become
vd = -rid - xqiq
vq = -riq +x’did + 3E'
(5.38)
(5.39)
In phasor representation above equations read
~
~
~
~ ~
Va = − rI a − jx d' I d − jx q I q + E '
(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
Pe = −
3x qVa E '
3rVa2
3Va2 x ' d − x q
3rV E '
sin 2δ + 2 a
cos δ + 2
sin δ (5.41)
+
2
2
2 r + x' d x q
r + x' d x q
r + x' d x q
r + x' d x q
Qe = −
3V a2
3V E '
( x ' d sin 2 δ + x q cos 2 δ ) + 2 a
( − r sin δ + x q cos δ )
2
r + x' d x q
r + x' d x q
(5.42)
Again if the resistance is neglected the power equations are simplified to:
3Va E '
3Va2 1
1
sin δ +
( −
Pe =
) sin 2δ
2 x q x' d
x' d
Copyright © A. P. Sakis Meliopoulos – 1990-2006
(5.43)
Page 65
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
3V a2
3V E '
( x ' d sin 2 δ + x q cos 2 δ ) + a cos δ
Qe = −
x' d x q
x' d
(5.44)
Example E5.7: Consider a 625 MVA, 60 Hz, 18 kV synchronous machine with the
following parameters
r = 0.001066 pu, x d = 1.5645 pu, x q = 1.42231 pu, E = 2.03387 pu, V = 1.105 pu
And x d' = 0.2776 pu, E ' = 1.072 pu
Compute the simplified model under the assumption of constant field winding flux.
Solution: By direct substitution:
V = 1.105 pu
λ f = L f i f + kM f i d = 870.63
E’q = 9283 V
or
E’q = 1.072 pu
or
x’d = 0.2776 pu
x’d = 0.3904 ohms
xq = 1.4223 pu
Pe = 12.8 sin δ − 5.31sin 2δ
or
Pe = −0.00989 + 12.8 sin δ + 0.00959 cos δ − 5.494 sin 2δ
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.
λ f = cons tan t
λ D = cons tan t
λ Q = cons tan t
(5.45)
(5.46)
(5.47)
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 - ω kMQiQ - L d
Page 66
di d
di
di
− kM f f − kM D D
dt
dt
dt
(5.48)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
vq = -riq + ω Ldid + ω kMfif + ω kMDiD - L q
di q
dt
− kM Q
di Q
dt
(5.49)
λ D =LDiD +kMDid + MRif
λ Q =LQiQ +kMQiq
(5.50)
(5.51)
λ f =Lfif +kMfid + MRiD
(5.52)
Pe = vdid + vqiq
Qe = vdiq - vqid
(5.53)
(5.54)
Equations (5.50), (5.51) and (5.52), solved for the electric currents if, iD, iQ yield the
following
⎡i f ⎤ ⎡ L f
⎢ ⎥ ⎢
⎢i D ⎥ = ⎢ M R
⎢i Q ⎥ ⎢ 0
⎣ ⎦ ⎣
MR
LD
0
0⎤
⎥
0⎥
LQ ⎥⎦
−1
⎡λ f ⎤ ⎡ L f
⎢ ⎥ ⎢
⎢λ D ⎥ − ⎢ M R
⎢λ Q ⎥ ⎢ 0
⎣ ⎦ ⎣
0⎤
⎥
0⎥
LQ ⎥⎦
MR
LD
0
−1
⎡ kM f
⎢
⎢kM D
⎢ 0
⎣
0 ⎤
⎥ ⎡i d ⎤
0 ⎥⎢ ⎥
iq
kM Q ⎥⎦ ⎣ ⎦
(5.55)
and upon differentiation assuming constant λ f , λ D and λ Q :
⎡ Lf
⎡i f ⎤
d ⎢ ⎥
⎢
iD ⎥ = − ⎢ M R
⎢
dt
⎢ 0
⎢iQ ⎥
⎣
⎣ ⎦
MR
LD
0
0⎤
⎥
0⎥
LQ ⎥⎦
−1
⎡ kM f
⎢
⎢kM D
⎢ 0
⎣
0 ⎤
⎥ d
0 ⎥
dt
kM Q ⎥⎦
⎡id ⎤
⎢i ⎥
⎣ q⎦
(5.56)
Substituting for the currents if, iD and iQ in equations (5.48) and (5.49) one obtains:
vd = -rid - ω ( L q −
( kM Q ) 2
LQ
)iq -
ωkM Q
LQ
λQ
L D ( kM f ) 2 + L f ( kM D ) 2 − 2 M R kM f kM D di d
)
- (Ld −
A
dt
L D ( kM f ) 2 + L f ( kM D ) 2 − 2 M R kM f kM D
)i d
vq = -riq + ω ( L d −
A
( kM Q ) 2 di q
L kM f − M R kM D
+ω D
- (Lq −
λf +
)
A
LQ
dt
L kM D − M R kM f
ω f
λD
A
where
(5.57)
(5.58)
A = LfLD - M2R
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 67
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
In above equations the subtransient inductances and reactances can be identified:
L”d = L d −
L”q = L q −
L D ( kM f ) 2 + L f ( kM D ) 2 − 2 M R kM f kM D
A
2
( kM Q )
LQ
x”d = ω L”d
x”q = ω L”q
(5.59)
(5.60)
(5.61)
(5.62)
Since, by assumption λ f , λ D and λ Q remain constant, equation (5.57) and (5.58) contain
ωkM Q
λ Q represents a voltage generated along the d-axis,
LQ
L kM f − M R kM D
L kM D − M R kM f
while the constants ω D
λ f and ω f
λ D represent
A
A
voltages generated along the q-axis.
three constants. The constant
Define
E”2 = ω
E”3 = ω
ωM Q
λQ
2LQ
LDM f − M R M D
E”1 = E”d =
2A
Lf M D − M R M f
2A
(5.63)
λf
(5.64)
λD
(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" q iq - 3E"1 - L" d
vq = -riq + ωL" d i d - L" q
di q
dt
di d
dt
(5.67)
+ 3 E”2 + 3 E”3
One final approximation is introduced by observing that the terms L" d
(5.68)
di q
di d
and L" q
dt
dt
are small compared to other terms, yielding
vd = -rid - ω L" q iq - 3E" d
(5.69)
vq = -riq + ωL" d i d + 3 E”q
(5.70)
The phasor representation of above equations is:
Page 68
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
~
~
~
~ ~
~
Va = − r Ia − jx "d Id − jx "q Iq + E "q + E "d
(5.71)
The phasor diagram for this model is illustrated in Figure 5.21.
d-axis
q-axis
E"q
jx"
q Iq
E"
E"
d
r Ia
jx" I
reference
d d
Ia
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.
Pe =
3Va E" q
Qe = −
x" d
sin δ +
3Va2 1
3V E"
1
−
(
) sin 2δ − a d cos δ
x" q
2 x" q x" d
E" q
3Va2
E"
( x" d sin 2 δ + x" q cos 2 δ ) + 3Va (
cos δ + d sin δ )
x" d x" q
x" d
x" q
(5.72)
(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
Lq = 1.57
LD = 1.79
LQ = 1.8
Copyright © A. P. Sakis Meliopoulos – 1990-2006
r=0
LAD = 1.49
LAQ = 1.35
Lf = 1.77
Page 69
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 = Va + r Ia + jx q Ia = 18614
.
e j1.003655
~ ~
~
E = A + j( x d + x q ) Id = 1.97948e j1.003655
δ = 1.003655 rad
The generated voltage behind the subtransient reactance is:
~ ~
~
~
~
E" = Va + r Ia + jx "d Id + jx "q Iq = 0.995789e j0.42578
Immediately after fault initiation, E” remain constant, and Va = 0
Thus
Let
~
~
~
~
E" = r Iaf + jx "d Idf + jx "q Iqf
~
Iqf = A 1e jδ
π
j( δ + )
~
Idf = A 2 e 2
Substitute into equation to obtain
0.995789e j0. 42578 = -(0.35199) A 2 e jδ + j(0.5575) A1e jδ
.
. Upon solution of above
Note that immediately after fault initiation, δ = 1003655
equation for A1 and A2:
⇒
A1 = -0.975679
A2 = -2.369668
i df = 3A 2 = −410438
.
pu
i qf = 3A 1 = −16899
.
pu
Iaf = 2.5626
pu
5.11.4 Summary of Simplified Models
Page 70
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
The three simplified models presented in the previous paragraphs are summarized with
the following equations
~ ~
~
~
~
E = Va + rI a + jx d I d + jx q I q
~ ~
~
~
~
E ' = Va + rI a + jx d' I d + jx q' I q
~ ~
~
~
~
E " = Va + rI a + jx d" I d + jx q" I q
steady state model
transient model
subtransient model
Pe =
3Va E
3V 2 1
1
sin δ + a ( − ) sin 2δ
2 xq xd
xd
Pe =
3Va E '
3V 2 1
1
sin δ + a ( −
) sin 2δ
2 x q x' d
x' d
Pe =
3Va E " q
x" d
sin δ +
3Va2 1
3V E "
1
(
) sin 2δ − a d cos δ
−
2 x" q x" d
x" q
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
π
θ = ωst + δ +
2
(t = 0)
q-axis
E
E"q
E'
jx'q Iq jxq Iq
Vq
Iq
E" jx"q Iq
Va
jx" I
E"d
jx'dId
jxdId
reference
d d
Ia
Id
Vd
Figure 5.22 Summary of Simplified Models
5.11.5 One Axis Synchronous Machine Model
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 71
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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
Transient model
Subtransient model
xd = xq = x
x’d = x’q = x’
x”d = x”q = x”
Utilizing these assumptions, the model equations for each one of the three models are:
Steady state model
~
~
~ ~
Va = − rIa − jx Ia + E
VE
Pe = a sin δ
x
Transient model
~
~
~ ~
Va = − rIa − jx' Ia + E'
V E'
Pe = a sin δ
x'
Subtransient model
~
~
~ ~
Va = − rIa − jx " Ia + E"
V E"
Pe = a sin δ
x"
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.
Page 72
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
x
r
E
(a)
x'
r
E'
(b)
x"
r
E"
(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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 73
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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.
Page 74
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
5.14 Problems
Problem 5.1 A 3-phase, 60 Hz, 500 MVA, 15 kV (Line to Line) generator has the
following parameters:
x d = 0.75 ohms
r = 0.0005 ohms
x q = 0.70 ohms
kM f = 0.023 Henries
kM Q = 0.00065 Henries
kM D = 0.001 Henries
l d = 0.0001976 Henries
l q = 0.00021 Henries
Express in per unit the following equation of the mathematical model of above generator
Vq = ωLd id + ωkM f i f + ωkM D i D − riq − Lq
diq
dt
− kM Q
diQ
dt
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 =
2 π 60sec-1 , and VB = 15000/ 3 Volts for the stator)
Ld = 1.71 pu, L AD = 1.49 pu, Lq = 1.57 pu, L AQ = 1.35 pu, LD = 1.79 pu,
LQ = 1.8 pu, L f = 1.77 pu, and r = 0
(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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 75
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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
j1.1
o-axis
j0.21
0.001
0.007
0.02
j0.35
j0.28
j1.48
Vf
+
d-axis circuit
+
ωλq
j0.21
0.001
0.04
j0.45
j1.37
+
q-axis circuit
-
ωλ d
Figure P5.4
Page 76
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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.
3-Phase Line
G
Infinite Bus
Figure P5.5
Additional Information
(a) The positive sequence series impedance of the transmission line is:
z = 5.0 + j50.0 ohms .
(b) The transformer is a 160 MVA, 15kV:220 kV, 60 Hz, 3-phase with the following
impedance: z = 0.004 + j 0.078 pu on the transformer ratings.
(c) The parameters of the synchronous generator are:
r = 0.0015 pu, x d = 2.2 pu, x q = 2.0 pu, kM f = 0.080 H
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:
x d = 2.2 ohms
r = 0.0015 ohms
x q = 2.0 ohms
kM f = 0.080 Henries
kM Q = 0.0021 Henries
kM D = 0.004 Henries
l d = 0.0005022 Henries
l q = 0.0005022 Henries
M R = 0.090 Henries
r f (125 0 C ) = 0.37 ohms
rD = 0.0175 ohms
rQ = 0.0195 ohms
L f = 2.0 Henries
LD = 0.0055 Henries
L D = 0.00145 Henries
L0 = 0.001856 Henries
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
δ, E, E' and E ∞ e jα (rotor angle, generated voltage, voltage behind transient
reactance and voltage at the infinite bus).
2. Compute an expression of the transient power versus angle δ under the assumption
of constant voltage E’.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 77
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
j0 .5 6 6 9
o -a x is
j0 .2 5
j0 .2 4
j1 .8 7
j0 .1 9
Vf
ed
d -a x is
+
j0 .2 5
j0 .2 3
j1 .6 9
+
q -a x is
e
q
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:
x d = 0.75 ohms
r = 0.0005 ohms
x q = 0.70 ohms
kM f = 0.023 Henries
kM Q = 0.00065 Henries
kM D = 0.001 Henries
l d = 0.0002 Henries
l q = 0.0002 Henries
Consider the q-axis voltage equation
v q = ωL d i d + ωkM f i f + ωkM D i D − ri q − L q
di q
dt
− kM Q
di Q
dt
Write the per unitized version of above equation (all parameters should be substituted
with their numerical values).
Page 78
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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 .
V=15kV
V=15kV
2
3
1
4
G1
G2
x=6%
x=6%
P = 40MW
G2
5
100MW + j30 MVAR
Figure P5.9a Electric Power System
Table P5.9 Power flow Solution
~
V1
~
V2
~
V3
~
V4
~
V5
Copyright © A. P. Sakis Meliopoulos – 1990-2006
=
=
=
=
=
Page 79
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
j0 .2
0 .0 0 1
0 .0 0 2
0 .0 3
j0 .3 2
j0 .2 5
Vf
j1 .5
+
-
+
ω λq
d -a x is c irc u it
j0 .2
0 .0 0 1
0 .0 4
j0 .3 5
j1 .4
-
+
ω λd
q -a x is c irc u it
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
r = 0.00416 pu
rD = 0.01565 pu
rQ = 0.0322 pu
l d = 0.2509 pu
l q = 0.2509 pu
r f = 0.0009156 pu
l f = 0.2442 pu
l D = 0.1856 pu
l Q = 0.2238 pu
L0 = 0.5669 pu
L AD = 1.869 pu
L AQ = 1.689 pu
Page 80
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 5, Meliopoulos
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)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 81