SYNCHRONOUS MACHINE THEORY AND MODELING

Ref. P. Kundur, “ Power System Stability and Control”, McGraw-Hill, Inc., 1994

Synchronous generators form the principal source of electric energy in power systems.

Many large loads are driven by synchronous motors.

Synchronous condensers are sometimes used as a means of providing reactive power compensation and controlling voltage.

These devices operate on the same principle.

In this chapter, we will develop in detail the mathematical model of a synchronous machine and briefly review its steady-state and transient performance characteristics.

Fig. 1 Schematic diagram of a three-phase synchronous machine

1. PHYSICAL DESCRIPTION

Fig. 1 shows the schematic of the cross section of a three-phase synchronous machine with one pair of field poles. The machine consists of two essential elements: the field and the armature. The field winding carries direct current and produces a magnetic field which induces alternating voltages in the armature windings.

1.1 Armature and field structure

The armature windings usually operate at a voltage that is considerably higher than that of the field and require more space for insulation. They are also subject to high transient current and must have adequate mechanical strength. Therefore, normal practice is to have the armature on the stator.

The three-phase windings of the armature are distributed 1200 apart in space so that, with uniform rotation of the magnetic field, voltages displaced by 1200 in time phase, will be produced in the windings. Because the armature is subjected to a varying magnetic flux, the stator core is built up of thin laminations to reduce eddy current losses.

When carrying balanced three-phase currents, the armature will produce a magnetic field in the air-gap, rotating at synchronous speed. The magnetic field produced by the direct current in the rotor winding, revolves with the rotor. For production of steady torque, the magnetic fields of stator and rotor must rotate at the same speed. Therefore, the rotor must be run precisely at the synchronous speed. The synchronous speed is given by

n = fp

f120 (1)

where n is the speed in rpm, f is the frequency in Hz and pf is the number of field poles.

Fig. 2 Salient pole rotor construction

There are two basic rotor structures namely, Salient pole rotor and cylindrical rotor. When turbine speed is low, salient pole rotors are used. Large number of poles are required to produce the rated frequency. Salient pole rotors often have damper windings or amortisseurs at the poles face. They are intended to damp out speed oscillations. Fig. 2 shows the salient pole rotor construction,

Fig. 3 Solid round rotor construction

Steam or gas turbines, on the other hand, operate at high speeds. Their generators have cylindrical rotors made up of solid steel forgings. They have two or four field poles, formed by distributed windings placed in rotor slots. The solid steel rotor offers path for eddy currents which have the effect equivalent to amortisseurs currents. Fig. 3 shows the solid round rotor construction details.

Fig. 4 Synchronous machine with two pairs of rotor poles rotor construction

1.2 Machines with multiple pole pairs

Machines with more than one pair of field poles will have stator windings made up of multiple set of coils. Schematic diagram of synchronous machine with two pairs of rotor poles is shown in Fig. 4. Its armature has two sets of coils.

For purpose of analysis, it is convenient to consider only a single pair of poles and recognize that conditions associated with other pole pairs are identical to those for the pair under consideration. Angles are normally measured in electrical radians or degrees. The angle covered by one pole pair is 2π electrical radians or 360 electrical degrees. The angle covered by one revolution is 2π mechanical radians or 360 mechanical degrees. The relationship between angle θ in electrical unit and the corresponding angle θm in mechanical unit is

θ = 2p f θm (2)

Fig. 5 MMF waveform due to a single coil

1.3 MMF Waveforms

In practice, the armature windings and field windings of round rotor machine are distributed in many slots so that the resulting mmf waveforms have nearly sinusoidal space distribution. In case of salient pole machines, which have field windings concentrated at the poles, pole faces are shaped suitably to minimize the harmonics in the flux produced.

Let us first consider the mmf waveform due to the armature winding only. Th e mmf produced by the current flowing in only one coil in phase a, is illustrated in Fig. 5, in which the cross section of the stator is cut open and rolled out in order to develop a view of mmf wave.

Fig. 6 MMF waveform due to number of coils

By adding more coils, mmf wave distribution shown in Fig. 6 may be obtained. We see that, as more coils are added, mmf waveform is progressing from a square wave towards a sine wave. It is reasonable to assume that each phase winding produces a sinusoidally distributed mmf wave.

Fig. 7 Spatial mmf wave of phase a

Rotating magnetic field

Let us determine the net mmf wave due to the three-phase windings in the stator. Fig. 7 shows the mmf wave form of phase a.

With γ representing the angle along the periphery of the stator with respect to the centre of phase a, the mmf wave due to the three phases may be described as follows:

MMFa = K ia cos γ; MMFb = K ib cos (γ - 3π2 ); MMFc = K ic cos (γ +

3π2 )

MMFa = K ia cos γ; MMFb = K ib cos (γ - 3π2 ); MMFc = K ic cos (γ +

3π2 )

where ia, ib and ic are the instantaneous values of the phase currents and K is a constant. The three mmf waves due to three phases are displaced 120 electrical degrees apart in space.

Each phase winding produces a stationary mmf wave whose magnitude changes as the instantaneous value of the current through the winding changes. With balanced phase currents, and time origin arbitrarily chosen as the instant when ia is maximum, we have

ia = Im cos (ωs t); ib = Im cos (ωs t - 3π2 ); ic = Im cos (ωs t +

3π2 ) (3)

where ωs is the angular frequency of stator currents in electrical red. / sec.

The total mmf due to the three phases is given by

MMFtotal = MMFa + MMFb + MMFc

= K Im (cos (ωs t) cos γ + cos (ωs t - 3π2 ) cos (γ -

3π2 ) +

cos (ωs t + 3π2 ) cos (γ +

3π2 )

Simplifying using trigonometric relations

MMFtotal = 23 K Im cos (γ - ωs t) (4)

This is the equation of a travelling wave. At any instant in time, the total mmf has a sinusoidal spacial distribution. It has a constant amplitude and a space-phase angle ωs t, which is a function of time.

Thus, the entire mmf wave moves at the constant angular velocity of ωs electrical rad. / sec. Denoting ωsm as the mechanical angular velocity, corresponding speed of rotation of the stator field is:

ns = 2πω60 sm rpm =

2π60

fp2 ωs rpm (5a)

Thus ns = 2π60

fp2 2 π f =

fpf120 rpm (5b)

This is the same as the synchronous speed of the rotor given by equation 1. Therefore, for balanced operation, the mmf wave due to stator currents is stationary with respect to the rotor.

Fig. 8 Stator and rotor mmf wave shapes phase a

The stator and rotor mmf waves are shown in Fig. 8 relative to the rotor structure, again with both stator and rotor cross sections rolled out.

The magnitude of the stator mmf wave and its relative angular position with respect to the rotor mmf wave depend on the synchronous machine load (output). The electromagnetic torque on the rotor acts in a direction so as to bring both the magnetic fields into alignment. If the rotor field leads the armature field, the machine functions as a generator. On the other hand if the rotor field lags the armature field, the machine functions as a motor.

Fig. 1 Schematic diagram of a three-phase synchronous machine

1.4 Direct and quadrature axes

The magnetic circuits and all rotor windings are symmetrical with respect to both polar axis and the inter-polar axis. Therefore, for the purpose of identifying synchronous machine characteristics, two axes are defined as shown in Fig. 1:

Fig. 1 Schematic diagram of a three-phase synchronous machine

§ The direct (d) axis, centered magnetically in the centre of the north pole;

§ The quadrature (q) axis, 90 electrical degrees ahead of the d-axis.

The position of the rotor relative to the stator is measured by the angle θ between the d-axis and the magnetic axis of phase a winding.

Fig. 9 Stator and rotor circuits of synchronous machine

Rotation

ib

ia

ic

eb

ea

ec

ψb ψa

ψc

Stator

e fd i fd

i kq

i kd

d-axis q-axis

Axis of phase a

θ

Rotor

ωr elec. rad / sec

2. MATHEMATICAL DESCRIPTION OF A SYNCHRONOUS MACHINE

a, b,c: Stator phase windings fd: Field winding

kd: d-axis amortisseurs circuit kq: q-axis amortisseurs circuit

θ: Angle by which d-axis leads the magnetic axis of phase a winding, electrical rad.

ωr: Rotor angular velocity, electrical rad. / sec.

ib

ia

ic

eb

ea

ec

ψb ψa

ψc

Stator

e fd i fd

i kq

i kd

d-axis q-axis

Axis of phase a

θ

Rotor

ωr elec. rad / sec

Fig. 9 shows the circuits involved in the analysis of a synchronous machine. The stator circuit consists of three-phase armature windings carrying alternating currents. The rotor circuits comprise field and amortisseurs windings. The field winding is connected to a source of direct current. The currents in amortisseurs windings may be assumed to flow in closed circuits. θ is the angle by which the d-axis leads the axis of phase a winding. Since the rotor is rotating with respect to stator, angle θ is continuously increasing and is related to the rotor angular velocity ωr an time t as θ = ωr t. The electric performance equations of a synchronous machine can be developed by writing equations of the coupled circuit identified in Fig. 9.

Fig. 10 Variation of permeance for the stator flux with respect to rotor position

2.1 Basic Equations of a Synchronous Machine

While deriving equations of synchronous machine, we use generator convention for polarities so that the positive direction of stator winding current is assumed to be out of the machine. The positive direction of field and amortisseurs currents is assumed to be into the machine.

The flux produced by a stator winding follows a path through the stator iron, across the air-gap, through the rotor iron, and back across the air-gap. Variations in permeance of this flux path as a function of the rotor position can be approximated as shown in Fig. 10.

Fig. 10 Variation of permeance for the stator flux with respect to rotor position

P = P0 + P2 cos 2α (6)

Here, α is the angular distance from the d-axis along the periphery. A double frequency variation is produced, since the permeance of the north and south poles are equal and they occur in 180 electrical degrees apart. Higher order harmonics of permeance exist; but are small enough to be neglected.

The following notations are used in writing equations for the stator and rotor circuits:

ea, eb, ec = instantaneous stator phase to neutral voltages

ia, ib, ic = instantaneous stator currents in phases a, b,c

e fd = field voltage

i fd, i kd, i kq = field and amortisseurs circuit currents

R fd, R kd, R kq = rotor circuit resistances

ℓ aa, ℓ bb, ℓ cc = self-inductances of stator windings

ℓ ab, ℓ bc, ℓ ca = mutual inductances between stator windings

ℓ ffd, ℓ kkd, ℓ kkq = self-inductances of rotor circuits

ℓ afd, ℓ akd, ℓ akq = mutual inductances between stator and rotor windings

Ra = armature resistance per phase

p = differential operator

Stator circuit equations

The voltage equations of the three phases are

ea = dt

dψa - Ra ia = p ψa - Ra ia (7)

eb = p ψb – Ra ib (8)

ec = p ψc – Ra ic (9)

The flux linkage in the phase a winding at any instant is given by

ψa = - ℓ aa ia - ℓ ab ib - ℓ ac ic + ℓ afd i fd + ℓ akd i kd + ℓ akq ikq (10)

Similar equations apply to the flux linkages of phase b and phase c windings. They are collectively represented as

c

b

a

ψψψ

=

ckqckdcfdcccbca

bkqbkdbfdbcbbba

akqakdafdacabaa

kq

kd

fd

c

b

a

iiiiii

(10 a)

The negative sign associated with the stator winding currents is due to their assumed direction. All the inductances in the above matrix equation are functions of rotor position and hence time-dependence.

Fig. 11 Phase a mmf wave and its components

Stator self-inductances

The self-inductance ℓ aa is equal to the ratio of flux linking phase a winding to the current ia, with all other circuit currents equal to zero. The inductance is directly proportional to the permeance, which as indicated earlier has a second order harmonic variation. Thus the inductance ℓ aa will be a maximum for θ = 00, a minimum for θ = 900, a maximum again for θ = 1800 and so on.

The mmf of phase a has a sinusoidal distribution in space with its peak centered on phase a axis. The peak amplitude of the mmf wave is equal to Na ia, where Na is the effective turns per phase. As shown in Fig. 11, this can be resolved into two other sinusoidally distributed mmf’s, one centered on the d-axis and the other on the q-axis.

The peak values of the two components waves centered on d- and q- axis are

peak MMFad = Na ia cos θ (11)

peak MMFaq = Na ia cos (θ + 900) = - Na ia sin θ (12)

The reason for resolving the mmf into the d- and q-axis components is that each acts on specific air-gap geometry of defined configuration. Air-gap fluxes per pole along the two axes are

Φ gad = (Na ia cos θ) Pd (13)

Φ gaq = ( - Na ia sin θ) Pq (14)

In the above, Pd and Pq are the permeance coefficients of the d- and q-axis, respectively. In addition to the actual permeance, they include factors required to relate the flux per pole with the peak value of mmf wave.

Total air-gap flux linking phase a is

Φ gaa = Φ gad cos θ + Φ gaq cos (θ+90) i.e.

Φ gaa = Φ gad cos θ - Φ gaq sin θ (15)

= Na ia (Pd cos2θ + Pq sin2θ) = Na ia [ 2Pd (1 + cos 2θ) +

2Pq (1 - cos 2θ)]

= Na ia [ 2PP qd

+2

PP qd cos 2θ] (16)

The self-inductance ℓ gaa of phase a due to air-gap flux is

ℓ gaa = a

gaaa

iφN

= Na2 [

2PP qd

+2

PP qd cos 2θ]

Separating the fundamental and the second harmonic terms

ℓ gaa = Lg0 + Laa2 cos 2θ (17)

The total self inductance ℓ aa is obtained by adding to the above, the leakage inductance Lal, which represents the leakage flux nor crossing the air-gap. Thus

ℓ aa = Lal + Lg0 + Laa2 cos 2θ i.e.

ℓ aa = Laa0 + Laa2 cos 2θ (18)

where

Laa0 = Na2

2PP qd

+ Lal (19)

Laa2 = Na2

2PP qd

(20)

Fig. 12 Variation of self-inductance of phase a with θ

Since the windings of phases b and c are identical to that of phase a and are displaced from it by 1200 and 2400 respectively, we have

ℓ bb = Laa0 + Laa2 cos 2 (θ - 3π2 ) (21)

ℓ cc = Laa0 + Laa2 cos 2 (θ + 3π2 ) (22)

The variation of ℓ aa with θ is shown in Fig. 12.

Stator mutual inductances

It is to be noted that the mutual inductance ℓ ab is equal to the mutual inductance ℓba. The mutual inductance ℓ ab can be found either by evaluating the air-gap flux Φ gab linking phase a, when only phase b is excited or by evaluating the air-gap flux Φgba linking phase b, when only phase a is excited. We prefer the latter, since we can easily find Φgba from the already known Φgaa. As we wish to find the flux

linking phase b due to mmf of phase a, θ is replaced by θ - 3π2 in equation (15).

Φ gaa = Φ gad cos θ - Φ gaq sin θ (15)

Thus Φgba = Φ gad cos (θ - 3π2 ) - Φ gaq sin (θ -

3π2 )

Substituting for Φ gad and Φ gaq from equations (13) and (14)

Φ gad = (Na ia cos θ) Pd (13) Φ gaq = ( - Na ia sin θ) Pq (14)

Φgba = Na ia [ Pd cos θ cos (θ - 3π2 ) + Pq sin θ sin (θ -

3π2 ) ]

Knowing cos θ cos (θ - 3π2 ) = -

41 +

21 cos ( 2θ -

3π2 ) and

sin θ sin (θ - 3π2 ) = -

41 -

21 cos ( 2θ -

3π2 ), above equation becomes

Φgba = Na ia [ - 4PP qd

+ 2

PP qd cos (2θ -

3π2 ) ] (23)

The mutual inductance between phases a and b due to the air-gap flux is

ℓ gba = a

gbaa

iφN

= Na2 [ -

4PP qd

+ 2

PP qd cos (2θ -

3π2 ) ]

= - 21 Lg0 + Lab2 cos (2θ -

3π2 )

The total mutual inductance ℓ ba is obtained by adding to the above, a very small leakage inductance which represents the leakage flux nor crossing the air-gap. Thus the mutual inductance between phases a and b can be written as

ℓ ab = ℓ ba = - Lab0 + Lab2 cos (2θ - 3π2 )

Knowing that cos δ = - cos (δ + π)

ℓ ab = ℓ ba = - Lab0 - Lab2 cos (2θ + 3π ) (24)

In the above Lab0 is nearly equal to Laa0 / 2 and Lab2 = Laa2

Fig. 13 Variation of mutual inductance ℓ ab with θ

Now, other mutual inductances ℓ bc and ℓ ca can be obtained by replacing θ by

θ - 3π2 and θ +

3π2 .

[ 2(θ - 3π2 ) +

3π = 2θ – π and 2(θ +

3π2 ) +

3π = 2θ +

3π5 = 2θ -

3π ]

Therefore

ℓ ab = ℓ ba = - Lab0 - Lab2 cos (2θ + 3π ) (24)

ℓ bc = ℓ cb = - Lab0 - Lab2 cos (2θ - π) (25)

ℓ ca = ℓ ac = - Lab0 - Lab2 cos (2θ - 3π ) (26)

The variation of mutual inductance between phases a and b as a function of angle θ is illustrated in Fig. 13.

Mutual inductance between stator and rotor windings

With the variation in air-gap due to stator slots neglected, the rotor circuits see a constant permeance. Therefore the situation in this case is not one of variation of permeance; instead, the variation in the mutual inductance is due to the relative motion between the stator and rotor windings.

When the rotor winding is lined up with a stator winding, the flux linking the two windings is maximum and the mutual inductance is maximum. When the two windings are displaced by 900, no flux links the two circuits and the mutual inductance is zero. (Refer Fig. 9) Therefore

ℓ afd = ℓ fda = L afd cos θ (27)

ℓ akd = ℓ kda = L akd cos θ (28)

ℓ akq = ℓ kqa = L akq cos (θ + 2π ) = - L akq sin θ (29)

Corresponding to phase b and phase c windings

ℓ bfd = ℓ fdb = L afd cos (θ - 3π2 ) (30)

ℓ bkd = ℓ kdb = L akd cos (θ - 3π2 ) (31)

ℓ bkq = ℓ kqb = - L akq sin (θ - 3π2 ) (32)

ℓ cfd = ℓ fdc = L afd cos (θ + 3π2 ) (33)

ℓ ckd = ℓ kdc = L akd cos (θ + 3π2 ) (34)

ℓ ckq = ℓ kqc = - L akq sin (θ + 3π2 ) (35)

We now have the expressions for all the inductances that appear in the equation for flux linkage in phase a stator winding. On substituting the expressions for these inductances into equation (10), we obtain

ψa = - ℓ aa ia - ℓ ab ib - ℓ ac ic + ℓ afd i fd + ℓ akd i kd + ℓ akq ikq

= - ia [ Laa0 + Laa2 cos 2θ ] + ib [ Lab0 + Lab2 cos (2θ + 3π ) ]

+ ic [ Lab0 + Lab2 cos (2θ - 3π ) ] + i fd L afd cos θ + i kd L akd cos θ - ikq L akq sin θ

(36)

Similarly for phase b and c we have

ψb = - ℓ ba ia - ℓ bb ib - ℓ bc ic + ℓ bfd i fd + ℓ bkd i kd + ℓ bkq ikq

= ia [Lab0 + Lab2 cos (2θ + 3π ) ] - ib [ Laa0 + Laa2 cos 2 (θ -

3π2 ) ]

+ ic [ Lab0 + Lab2 cos (2θ - π) ] + i fd L afd cos (θ - 3π2 ) + i kd L akd cos (θ -

3π2 )

- ikq L akq sin (θ - 3π2 ) (37)

ψc = - ℓ ca ia - ℓ cb ib - ℓ cc ic + ℓ cfd i fd + ℓ ckd i kd + ℓ ckq ikq

= ia [ Lab0 + Lab2 cos (2θ - 3π ) + ib [Lab0 + Lab2 cos (2θ - π) ]

- ic [ Laa0 + Laa2 cos 2 (θ + 3π2 ) ] + i fd L afd cos (θ +

3π2 )

+ i kd L akd cos (θ + 3π2 ) - ikq L akq sin (θ +

3π2 ) (38)

Rotor circuit equations

The rotor circuit voltage equations are:

e fd = pψ fd + R fd i fd (39)

0 = pψ kd + R kd i kd (40)

0 = pψ kq + R kq i kq (41)

kq

kd

fd

ψψψ

=

kkqkdckqbkqa

kkdkdfkdckdbkda

fkdffdfdcfdbfda

0000

kq

kd

fd

c

b

a

iiiiii

(41 a)

The rotor circuit see constant permeance because of the cylindrical structure of the stator. Therefore, the self-inductances L ffd, L kkd and L kkq and the mutual inductance L fkd between each other do not vary with rotor position. Only the rotor to stator mutual inductances vary periodically with θ as given by equations (27) to (36). Thus the rotor circuit flux linkages can be expressed as follows:

ψ fd = - ℓ afd ia - ℓ bfd ib - ℓ cfd ic + ℓ ffd ifd + ℓ fkd ikd

ψ fd = - ℓ afd ia - ℓ bfd ib - ℓ cfd ic + ℓ ffd ifd + ℓ fkd ikd

Substituting for the various inductances

ψ fd = –(L afd cos θ) ia – [L afd cos (θ-3π2 )] ib - [L afd cos (θ+

3π2 )] ic + L ffd ifd + L fkd ikd

= – L afd [ ia cos θ + ib cos (θ- 3π2 ) + ic cos (θ+

3π2 ) ] + L ffd ifd + L fkd ikd (42)

Similarly

ψ kd = – L akd [ ia cos θ + ib cos (θ- 3π2 ) + ic cos (θ+

3π2 ) ] + L fkd ifd + L kkd ikd (43)

ψ kq = L akq [ ia sin θ + ib sin (θ- 3π2 ) + ic sin (θ+

3π2 ) ] + L kkq ikq (44)

3. THE dq0 TRANSFORMATION

Equations (7) to (9) (for ea, eb, ec) and equations (36) to (38) (for ψa, ψb, ψc) associated with the stator circuits, together with equations (39) to (44) (for efd, ekd, ekq, ψfd, ψkd, ψkq) associated with rotor circuits, completely describe the electrical performance of synchronous machine. However, these equations contain inductance terms which vary with angle θ which in turn varies with time. This introduces considerable complexity in solving machine and power system problems. A much simpler form, leading to a clearer physical picture is obtained by appropriate transformation of stator variables.

We see from equations (42) to (44) that the stator currents combine into convenient form in each axis. This suggests the transformation of the stator phase currents into new variables as follows:

id = kd [ ia cos θ + ib cos (θ- 3π2 ) + ic cos (θ+

3π2 ) ] (45)

iq = - kq [ ia sin θ + ib sin (θ- 3π2 ) + ic sin (θ+

3π2 ) ] (46)

The constants kd and kq are arbitrary and their values may be chosen to simply numerical coefficients in the performance equations. We shall take values of kd and kq as 2 / 3.

With kd = kq = 2 / 3, for balanced sinusoidal stator currents, the peak values of id and iq are equal to the peak value of the stator current as shown below.

For the balanced condition,

ia = Im sin ωst ; ib = Im sin (ωst - 3π2 ); ic = Im sin (ωst +

3π2 )

Substitution of the above in equation (45) yields

id = kd Im [ sin ωst cos θ + sin (ωst - 3π2 ) cos (θ-

3π2 )

+ sin (ωst + 3π2 ) cos (θ+

3π2 ) ]

= kd Im 23 sin ( ωst - θ )

With kd = 32 , the peak value of id is equal to Im, the peak value of stator current.

Similarly, with balanced sinusoidal stator currents

iq = - kd Im [ sin ωst sin θ + sin (ωst - 3π2 ) sin (θ-

3π2 )

+ sin (ωst + 3π2 ) sin (θ+

3π2 ) ]

iq = - kd Im 23 cos ( ωst - θ )

Again kq = 32 , the peak value of iq is equal to Im, the peak value of stator current.

To give a complete degree of freedom, a third current component must be defined so that the three-phase currents are transformed into three variables. Since, as seen from equations (42) to (44), the two current components id and iq together produce a field identical to that produced by the original set of phase currents, the third component must produce no space field in the air-gap. Therefore, a convenient third variable is the zero sequence current i0, associated with the symmetrical components.

Thus i0 = 31 ( ia + ib + ic ) (47)

Under balanced conditions, ia + ib + ic = 0 and, therefore i0 = 0.

The transformation from the abc phase variables to the dq0 variables can be written in the following matrix form:

0

q

d

iii

= 32

21

21

21

)3π2θ(sin)

3π2(θsinθsin

)3π2(θcos)

3π2(θcosθcos

c

b

a

iii

(48)

The inverse transformation is given by

c

b

a

iii

=

1)3π2θ(sin)

3π2θ(cos

1)3π2(θsin)

3π2-θ(cos

1θsin -θcos

0

q

d

iii

(49)

The above transformations also apply to stator flux linkages and voltages.

Stator flux linkages in dq0 components

Using the expressions for ψa, ψb and ψc given by equations (36), (37) and (38), transforming the flux linkages and currents into dq0 components and with suitable reduction of terms involving trigonometric terms, we obtain the following expressions:

ψd = - ( L aa0 + L ab0 + 23 L aa2 ) id + L afd ifd + L akd ikd (50)

ψq = - ( L aa0 + L ab0 - 23 L aa2 ) iq + L akq ikq (51)

ψ0 = - ( L aa0 - 2 L ab0 ) i0 (52)

Defining the following new inductances

Ld = L aa0 + L ab0 + 23 L aa2 (53)

Lq = L aa0 + L ab0 - 23 L aa2 (54)

L0 = L aa0 - 2 L ab0 (55)

the flux linkages equations become

ψd = - Ld id + L afd ifd + L akd ikd (56)

ψq = - Lq iq + L akq ikq (57)

ψ0 = - L0 i0 (58)

The above three equations in matrix form is:

0

q

d

ψψψ

=

000L00L000L0

0LL00L

0

akqq

akdafdd

kq

kd

fd

0

q

d

iiiiii

(59)

It is seen that the dq0 components of stator flux linkages are related to the components stator and rotor currents through constant inductances.

Rotor flux linkages in dq0 components

Substitution of expressions for id and iq in equations (42) to (44)

ψ fd = – L afd [ ia cos θ + ib cos (θ- 3π2 ) + ic cos (θ+

3π2 )] + L ffd ifd + L fkd ikd (42)

ψ kd = – L akd [ ia cos θ + ib cos (θ- 3π2 ) + ic cos (θ+

3π2 )] + L fkd ifd + L kkd ikd (43)

ψ kq = L akq [ ia sin θ + ib sin (θ- 3π2 ) + ic sin (θ+

3π2 )]+ L kkq ikq (44)

straight away gives

ψ fd = L ffd ifd + L fkd ikd – 23 L afd id (60)

ψ kd = L fkd ifd + L kkd ikd – 23 L akd id (61)

ψ kq = L kkq ikq - 23 L akq iq (62)

ψ fd = L ffd ifd + L fkd ikd – 23 L afd id (60)

ψ kd = L fkd ifd + L kkd ikd – 23 L akd id (61)

ψ kq = L kkq ikq - 23 L akq iq (62)

Writing in matrix form, the above three equations yield

kq

kd

fd

ψψψ

=

kkqakg

kkdfkdakd

fkdffdafd

L00L230

0LL0L23

0LL0L23

kq

kd

fd

q

d

iiiii

(63)

Again it is seen that inductances are constants and independent of rotor position.

It is interesting to note that the current i0 does not appear in the rotor flux linkage equations. This is because zero sequence component of armature current does not produce net mmf across the air-gap.

While the dq0 transformation has resulted in constant inductances, the mutual inductances between stator and rotor quantities are NOT RECIPROCAL. For example, mutual inductance between the flux linking the field winding and the

current id is - 23 L afd, whereas the mutual inductance associated with the flux

linking the d-axis due to field current ifd is L afd. This problem will be overcome by appropriate choice of per unit system for the ROTOR QUANTITIES.

Stator voltage equations in dq0 components

Equations ea = p ψa - Ra ia (7) eb = p ψb – Ra ib (8) ec = p ψc – Ra ic (9)

are the basic equations for phase voltages in terms of phase linkages and currents. Using dq0 transformation, we can get ed, eq and e0 in terms of ψd, ψq, ψ0, id, iq and i0.

We know ψa = ψd cos θ - ψq sin θ + ψ0 [Refer eq.(49)]

p ψa = - ψd pθ sin θ + pψd cos θ - ψq pθ cos θ – pψq sin θ + p ψ0 (64)

Knowing that

23 ed = cos θ ea + cos (θ-

3π2 ) eb + cos (θ+

3π2 ) ec

= cos θ [ p ψa – Ra ( id cos θ – iq sin θ + i0 ) ] + other cyclic terms

= - ψd pθ sin θ cos θ+ pψd cos2 θ - ψq pθ cos2 θ – pψq sin θ cos θ

+ p ψ0 cos θ – Ra id cos2 θ + Ra iq sin θ cos θ - Ra i0 cos θ + other cyclic terms

= 0 + 23 pψd -

23 ψq pθ + 0 + 0 -

23 Ra id + 0 + 0

Thus ed = pψd - ψq pθ - Ra id (65)

Similarly

23 eq = - sin θ ea - sin (θ-

3π2 ) eb - sin (θ+

3π2 ) ec

= - sin θ [ p ψa – Ra ( id cos θ – iq sin θ + i0 ) ] + other cyclic terms

= ψd pθ sin2 θ - pψd sin θ cos θ + ψq pθ sin θ cos θ + pψq sin2 θ - p ψ0 sin θ

+ Ra id sin θ cos θ - Ra iq sin2 θ + Ra i0 sin θ + other cyclic terms

= 23 ψd pθ + 0 + 0 +

23 pψq + 0 + 0 -

23 Ra iq

Thus eq = pψq + ψd pθ - Ra iq (66)

and 3e0 = ea + eb + ec

= p ψa – Ra ( id cos θ – iq sin θ + i0 ) + other cyclic terms

= - ψd pθ sin θ + pψd cos θ - ψq pθ cos θ – pψq sin θ + p ψ0

– Ra id cos θ + Ra iq sin θ – Ra i0 + other cyclic terms

= 0 + 0 + 0 + 0 + 3 p ψ0 – 3 Ra i0

Thus e0 = pψ0 - Ra i0 (67)

Collectively

ed = pψd - ψq pθ - Ra id (65)

eq = pψq + ψd pθ - Ra iq (66)

e0 = pψ0 - Ra iq (67)

ed = pψd - ψq pθ - Ra id (65)

eq = pψq + ψd pθ - Ra iq (66)

e0 = pψ0 - Ra iq (67)

The term pθ in the above equations represents the angular velocity ω r of the rotor. For a 50 Hz system, under steady state condition pθ = ω r = ωs = 2πx50 = 314 electrical rad. / sec.

The above equations have a form similar to those of static coil, except for the ψqpθ and ψd pθ terms. They result from stationary to rotating reference frame and represent the fact that a flux wave rotating in synchronism with the rotor will create voltages in the stationary armature coils. The terms ψq pθ and ψd pθ are referred as speed voltages (due to flux change in space) and the terms pψd and pψq as transformer voltages (due to flux change in time).

Electrical power and torque

The instantaneous three-phase output of the stator is

Pt = ea ia + eb ib + ec ic (68)

Using the dq0 transformation

Pt = [ed cos θ – eq sin θ + e0] [id cos θ – iq sin θ + i0] + [ed cos (θ -3π2 )

– eq sin (θ- 3π2 ) + e0] [id cos (θ -

3π2 ) – iq sin (θ-

3π2 ) + i0] +

[ed cos (θ +3π2 ) – eq sin (θ +

3π2 ) + e0] [id cos (θ +

3π2 ) – iq sin (θ+

3π2 ) + i0]

= 23 ed id +

23 eq iq + 3 e0 i0 Thus

Pt = 23 ( ed id + eq iq + 2 e0 i0 ) (69)

Under balanced operation, e0 = i0 = 0 and the expression for power is given by

Pt = 23 ( ed id + eq iq )

Substituting equations (65) to (67) in eq. (69)

Pt = 23 [ id (pψd - ψq pθ - Ra id) + iq (pψq + ψd pθ - Ra iq) + 2 i0 (pψ0 - Ra iq) ]

= 23 [ (id pψd + iq pψq + 2 i0 pψ0) + ωr (ψd iq – ψq id) – Ra (i2d + i2q + 2 i20) ] (70)

= Rate of change of armature magnetic energy

+ power transferred across the air-gap – armature resistance loss

The air-gap torque Te is obtained by dividing the power transferred across the air-gap (i.e., power corresponding to the speed voltage) by the rotor speed in mechanical radians per second.

Te = 23 (ψd iq – ψq id)

mech

r

ωω =

23 (ψd iq – ψq id)

2Pf (71)

The flux-linkage equations

ψd = - Ld id + L afd ifd + L akd ikd (56) ψ fd = L ffd ifd + L fkd ikd – 23 L afd id (60)

ψq = - Lq iq + L akq ikq (57) ψ kd = L fkd ifd + L kkd ikd – 23 L akd id (61)

ψ0 = - L0 i0 (58) ψ kq = L kkq ikq - 23 L akq iq (62)

associated with the stator and rotor circuits, together with the stator and voltage equations

ed = pψd - ψq pθ - Ra id (65) e fd = pψ fd + R fd i fd (39)

eq = pψq + ψd pθ - Ra iq (66) 0 = pψ kd + R kd i kd (40)

e0 = pψ0 - Ra iq (67) 0 = pψ kq + R kq i kq (41)

and the torque equation

Te = 23 (ψd iq – ψq id)

mech

r

ωω =

23 (ψd iq – ψq id)

2Pf (71)

describe the electrical performance of the machines in terms of the dq0 components. These equations are known as Park’s equations and the dq0 transformation is known as Park transformation.

Physical interpretation of dq0 transformation

In Section 1.3, we saw that the combined mmf wave due to the currents in three armature phases, travel along the periphery of the stator at a velocity of ωs rad. / sec. The angular velocity of the rotor also equals to ωs rad. / sec. Therefore, for balanced synchronous operation, the armature mmf wave appears stationary with respect to the rotor and has a sinusoidal space distribution. Since a sine function can be expressed as a sum of two sine functions, the mmf due to stator windings, can be resolved into two sinusoidally distributed mmf waves stationary with respect to the rotor, so that one has its peak over the d-axis and the other has its peak over the q-axis.

Therefore, id may be interpreted as the instantaneous current in a fictitious armature winding which rotates at the same speed as the rotor, and remains in such a position that its axis always coincides with the d-axis. The value of the current in this winding is such that it results in the same mmf on the d-axis as do actual phase currents flowing in the armature windings. A similar interpretation applies to iq, except that it acts on the q-axis instead of d-axis.

The mmf’s due to id and iq are stationary with respect to the rotor and act on path of constant permeance. Therefore, the corresponding inductances Ld and Lq are constant.

For balanced steady-state conditions, the phasor currents may be written as follows:

ia = Im sin (ωst + Φ) (72)

ib = Im sin (ωst + Φ - 3π2 ) (73)

ic = Im sin (ωst + Φ + 3π2 ) (74)

Using the dq0 transformation, id = 32 [ia cos θ + ib cos (θ -

3π2 ) + ic cos (θ +

3π2 )]

id = 32 Im [ sin (ωst + Φ) cos θ + sin (ωst + Φ -

3π2 ) cos (θ -

3π2 )

+ sin (ωst + Φ + 3π2 ) cos (θ +

3π2 ) ] =

32 Im

23 sin (ωst + Φ – θ) i.e.

id = Im sin (ωst + Φ – θ) (75)

id = Im sin (ωst + Φ – θ) (75)

Similarly

iq = - Im cos (ωst + Φ – θ) (76)

i0 = 0 (77)

For synchronous operation, the rotor speed ωr, is equal to the angular frequency ωs of the stator currents. Hence θ = ωr t = ωs t. Therefore,

id = Im sin Φ = constant; iq = - Im cos Φ = constant

Thus, for balanced steady-state operation, id and iq are constant. In other words, alternating phase currents in the abc reference frame appear as direct currents in dq0 reference frame. The dq0 transformation may be viewed as a means of referring the stator quantities to the rotor side, with a major advantage of constant inductances in the dynamic performance equations.

4. PER UNIT REPRESENTATION

Per unit representation is commonly used in power system problems. In the case of synchronous machine, the per unit system is used to remove arbitrary constants and simply mathematical equations so that they may be expressed in terms of equivalent circuits. The basis for selection of per unit system for the stator is straightforward, whereas it requires careful consideration for the rotor.

4.1 Per unit System for the Stator Quantities

Let us choose the following base quantities for stator (denoted by subscript s):

es base = peak value rated line-to-neutral voltage, V

is base = peak value of rated line current, A

fbase = rated frequency, Hz

The base values of the remaining quantities are automatically set and depend on the above as follows:

ωbase = 2π fbase, electrical radians / sec.

ωm base = ωbase (fp

2 ) mechanical radians / sec.

Zs base = es base / is base, ohms

Ls base = Zs base / ωbase, henrys

ψs base = Ls base is base = (Zs base / ωbase) is base = es base / ωbase, weber-turns

3-phase VAbase = 3 ERMS base IRMS base = 3 (es base / 2 ) (is base / 2 )

= 23 es base is base, volt-amperes

Torquebase = base m

base

ω VAphase -3 =

23 ψs base ωbase is base

base mω1

= 23

2p f ψs base is base, newton-meters

(78)

4.2 Per Unit Stator Voltage Equations

From equation (65)

ed = pψd - ψq ωr - Ra id. Dividing throughout by es base and noting that

es base = Zs base is base = ψs base ωbase

Expressed in per unit notation

ed pu = baseω1 p ψd pu - ψq pu ωr pu - Ra pu id pu (79)

The unit of time in the above equation is seconds. Time can also be expressed in per unit, taking the base time as the time required for the rotor to move through one electrical radian at synchronous speed.

base sbase s

da

basebase s

rq

basebase s

d

bases

d

iZiR

ωψωψ

)ωψ

ψ(pe

e

Knowing ωbase is the electrical radian per second

tbase = baseω1 ; Then ppu =

putdd =

tdd tbase =

baseω1 p

The above equation becomes

ed pu = ppu ψd pu - ψq pu ωr pu - Ra pu id pu (80)

Comparing eq.(65) and eq.(80), we see that the form of the original equation is unchanged, when all quantities involved are expressed in per unit. Similarly the per unit form of equations (66) and (67) are

eq pu = ppu ψq pu + ψd pu ωr pu - Ra pu iq pu (81)

e0 pu = ppu ψ0 pu - Ra pu iq pu (82)

4.3 Per Unit Rotor Voltage Equations

From eq.(39) e fd = pψ fd + R fd i fd, dividing throughout by

efd base = Zfd base ifd base = ψfd base ωbase, the per unit field voltage equation may be written as

e fd pu = ppu ψ fd pu + R fd pu i fd pu (83)

Similarly, the per unit forms of equations (40) and (41) are

0 = ppu ψ kd pu + R kd pu i kd pu (84)

0 = ppu ψ kq pu + R kq pu i kq pu (85)

The above equations show the form of rotor circuit voltage equations. However, we have not yet developed a basis for the choice of the rotor base quantities.

4.4 Per Unit Stator Flux Linkage Equations

Stator flux linkage equations are:

ψd = - Ld id + L afd ifd + L akd ikd (56) ψq = - Lq iq + L akq ikq (57) ψ0 = - L0 i0 (58)

Dividing by ψs base and noting that ψs base = Ls base is base the per unit form of equations (56), (57) and (58) may be written as

ψd pu = - Ld pu id pu + L afd pu ifd pu + L akd pu ikd pu (86)

ψq pu = - Lq pu iq pu + L akq pu ikq pu (87)

ψ0 pu = - L0 pu i0 pu (88)

where by definition

L afd pu = basesbase s

basefdafd

iLiL

(89)

L akd pu = basesbase s

basekdakd

iLiL

(90)

L akq pu = basesbase s

basekqakq

iLiL

(91)

4.5 Per Unit Rotor Flux Linkage Equations

Rotor flux linkage equations are:

ψ fd = L ffd ifd + L fkd ikd –23 L afd id (60)

ψ kd = L fkd ifd + L kkd ikd –23 L akd id (61)

ψ kq = L kkq ikq - 23 L akq iq (62)

Dividing eq. (60) by ψfd base and noting that ψfd base = Lfd base ifd base, dividing eq. (61) by ψkd base and noting that ψkd base = Lkd base ikd base and dividing eq. (62) by ψkq base and noting that ψkq base = Lkq base ikq base the per unit form of equations (60), (61) and (62) may be written as

ψ fd pu = L ffd pu ifd pu + L fkd pu ikd pu – L fda pu id pu (92)

ψ kd pu = L kdf pu ifd pu + L kkd pu ikd pu –L kda pu id pu (93)

ψ kq pu = L kkq pu ikq pu - L kqa pu iq pu (94)

where by definition

L fkd pu = basefdbase fd

basekdfkd

iLiL

(rotor to rotor) (95)

L fda pu = 23

basefdbase fd

base safd

iLiL

(rotor to stator) (96)

L kdf pu = basekdbase kd

basefdfkd

iLiL

(rotor to rotor) (97)

L kda pu = 23

basekdbase kd

basesakd

iLiL

(rotor to stator) (98)

L kqa pu = 23

basekqbase kq

basesakq

iLiL

(rotor to stator) (99)

By appropriate choice of per unit system, we have eliminated the factor 3/2 in the rotor flux linkage equations. However, we have not yet tied down the values of the rotor base voltages and currents, which we will proceed to do next.

4.6 Per Unit System for the Rotor

The rotor circuit base quantities will be chosen so as to make the flux linkage equations simple by satisfying the following:

(a) The per unit mutual inductances between different windings (stator, fd, kd and kq) are to be reciprocal i.e.

L afd pu = L fda pu; L akd pu = L kda pu; L akq pu = L kqa pu; L fkd pu = L kdf pu

This will allow the synchronous machine model to be represented by equivalent circuits.

(b) All per unit mutual inductances between stator and rotor circuits in each axis are to be equal; i.e. Lad pu = Lafd pu = Lakd pu and Laq pu = Lakq pu

To satisfy L afd pu = L fda pu, referring to equations (89) and (96)

basesbase s

basefdafd

iLiL

= 23

basefdbase fd

base safd

iLiL

i.e.

Lfd base (ifd base)2 = 23 Ls base (is base)2

Multiplying by ωbase and noting that ωLi = e

efd base ifd base = 23 es base is base = 3-phase VA base for stator (100)

Similarly, to make L akd pu = L kda pu and L akq pu = L kqa pu, required conditions are

ekd base ikd base = 23 es base is base (101)

ekq base ikq base = 23 es base is base (102)

Finally, to make L fkd pu = L kdf pu, referring to equations (95) and (97)

basefdbase fd

basekdfkd

iLiL

= basekdbase kd

basefdfkd

iLiL

i.e.

Lkd base (ikd base)2 = Lfd base (ifd base)2

Multiplying by ωbase and noting that ωLi = e

ekd base ikd base = efd base ifd base (103)

Thus to make the mutual impedances to be reciprocal, it must be ensured that the Volt Ampere base for all rotor circuits must be same and equal to the stator three-phase VA base.

So far, we have specified only the product of base voltage and base current for the rotor circuits. The next step is to specify either the base voltage or the base current for those circuits.

The stator self inductances Ld pu and Lq pu are associated with the total flux linkages due to currents id and iq respectively. They can be split into two parts: the leakage inductance due to flux that does not link any rotor circuit and the mutual inductance due to flux that links the rotor circuits. Thus

Ld pu = Lℓ pu + Lad pu and (104)

Lq pu = Lℓ pu + Laq pu (105)

In order make all the per unit mutual inductances between the stator and circuits in the d-axis equal, i.e. to make Lad pu = Lafd pu = Lakd pu , it follows that

Lad pu = base s

ad

LL = Lafd pu =

basesbase s

basefdafd

iLiL

= Lakd pu = basesbase s

basekdakd

iLiL

. Therefore

ifd base = afd

ad

LL is base (106)

ikd base = akd

ad

LL is base (107)

Similarly, to make the per unit mutual inductances between the stator and circuit in q-axis equal i.e to make Laq pu = Lakq pu , it follows that

Laq pu = base s

aq

LL

= Lakq pu = basesbase s

basekqakq

iLiL

Thus

ikq base = akq

aq

LL

is base (108)

This completes the choice of rotor base quantities.

4.7 Per Unit Power and Torque

From equation (69), the instantaneous power at the machine terminal is

Pt = 23 ( ed id + eq iq + 2 e0 i0 ) (69)

Dividing by the three-phase VA = (3/2) es base is base, the expression for the per unit power may be written as

Pt pu = ed pu id pu + eq pu iq pu + 2 e0 pu i0 pu (109)

Similarly, with the base torque = 23

2p f ψs base is base, the per unit form of equation

(71), given by, the air-gap torque Te = 23 (ψd iq – ψq id)

2Pf , is

Te pu = ψd pu iq pu – ψq pu id pu (110)

4.8 Summary of Per Unit Equations

Stator Base quantities

3-phase VA base = volt-ampere rating of machine, VA

es base = peak line-to-neutral rated voltage, V

is base = peak rated line current = bases

base

e(3/2) VAphase-3

A

Zs base = es base / is base, ohms

fbase = rated frequency, Hz

ωbase = 2π fbase, electrical radians / sec.

ωm base = ωbase (fp

2 ) mechanical radians / sec.

Ls base = Zs base / ωbase, henrys

ψs base = Ls base is base = es base / ωbase, weber-turns

Rotor Base quantities

ifd base = afd

ad

LL is base A; ikd base =

akd

ad

LL is base A; ikq base =

akq

aq

LL

is base A

efd base = basefd

base

i VAphase-3

V

Zfd base = basefd

basefd

ie

= 2basefd

base

)i( VAphase-3

ohms

Zkd base = 2basekd

base

)i( VAphase-3

ohms

Zkq base = 2basekq

base

)i( VAphase-3

ohms

Lfd base = base

basefd

ωZ

H; Lkd base = base

basekd

ωZ

H; Lkq base = base

basekq

ωZ

H

tbase = baseω1 sec.

Tbase = base m

base

ω VAphase -3 N-m

Complete set performance equations in per unit

In view of the per unit system chosen, in per unit

Lafd = Lfda = Lakd = Lkda = Lad

Lakq = Lkqa = Laq

Lfkd = Lkdf

This means that we will have only Lad, Laq and Lfkd as mutual inductances.

In the following equations, two q-axis amortisseur circuits are considered, and the subscripts 1q and 2q are used (in place of kq) to identify them. Only one d-axis amortisseur circuits is considered and it is identified by the subscripts 1d. Since all quantities are in per unit, we drop the subscript pu.

Per unit stator and rotor voltage equations

ed = p ψd - ψq ωr - Ra id (111) e fd = p ψ fd + R fd i fd (114)

eq = p ψq + ψd ωr - Ra iq (112) 0 = p ψ 1d + R 1d i 1d (115)

e0 = p ψ0 - Ra i0 (113) 0 = p ψ 1q + R 1q i 1q (116)

0 = p ψ 2q + R 2q i 2q (117)

Per unit stator and rotor flux linkage equations

ψd = - (Lad + Lℓ) id + Lad ifd + L ad i1d (118) ψ fd = L ffd ifd + L f1d i1d – L ad id (121)

ψq = - (Laq + Lℓ) iq + Laq i1q + L aq i2q (119) ψ 1d = L f1d ifd + L 11d i1d – L ad id (122)

ψ0 = - L0 i0 (120) ψ 1q = L 11q i1q + Laq i2q - L aq iq (123)

ψ 2q = L aq i1q + L22q i2q - L aq iq (124)

The above equations in matrix form are

0

q

d

ψψψ

=

0000L00LL000)L(L000LL00)L(L

0

aqaqaq

adadad

2q

1q

1d

fd

0

q

d

iiiiiii

(124 a)

2q

1q

1d

fd

ψψψψ

=

22qaqaq

aq11qaq

11df1dad

f1dffdad

LL00L0LL00L000LL0L00LL0L

2q

1q

1d

fd

q

d

iiiiii

(124 b)

Per unit air-gap torque equation

Te = ψd iq – ψq id (125)

In writing equations (123) and (124), we have assumed that the per unit mutual inductance L12q is equal to Laq. This implies that the stator and rotor circuits in the q-axis all link a single mutual flux represented by Laq. This is acceptable because the rotor circuits represent the overall rotor body effects, and actual windings with physically measurable voltages and currents do not exist.

Per unit reactances

If the frequency of the stator quantities is equal to the base frequency, the per unit reactance of a winding is numerically equal to the per unit inductance. For example, Xd = 2 π f Ld ohms. Dividing both sides by Zs base = 2 π fbase Ls base,

bases

d

ZX =

basefπ2fπ2

bases

d

LL

When f = fbase, bases

d

ZX =

bases

d

LL i.e. Xd pu = Ld pu

5. EQUVALENT CIRCUITS FOR DIRECT AND QUADRATURE AXES

While equations (111) to (124) can be used directly to determine synchronous machine performance, it is a common practice to use equivalent circuits to provide a visual description of the machine model.

Before we develop an equivalent circuit to represent complete electrical characteristics of the machine, first let us consider only d-axis flux linkage. d-axis stator and rotor flux linkage equations are:

ψd = - (Lad + Lℓ) id + Lad ifd + L ad i1d (118)

ψ fd = L ffd ifd + L f1d i1d – L ad id (121)

ψ 1d = L f1d ifd + L 11d i1d – L ad id (122)

Lad

id ifd i1d

Lffd – Lf1d

Lℓ Lf1d - Lad

L11d – Lf1d

ψd

ψ1d ψfd

Fig. 14 The d-axis equivalent circuit illustrating ψ - i relationship

The matrix form of above three equations is:

1d

fd

d

ψψψ

=

11df1dad

f1dffdad

adadad

LLLLLLLL)LL(

1d

fd

d

iii

(126)

Figure 14 shows the equivalent circuit that represent the above three equations. The currents id, ifd and i1d appear as loop currents

A similar equivalent circuit can be developed for q-axis flux linkage and current relationship. At this point, it is helpful to introduce the following rotor circuit per unit inductances.

Lfd = Lffd - Lf1d (127)

L1d = L11d - Lf1d (128)

L1q = L11q - Laq (129)

L2q = L22q - Laq (130)

Equivalent circuits representing the complete characteristics, including the voltage equations, are shown in Fig. 15. In these equivalent circuits, voltages as well as flux linkages appear. Therefore, flux linkages are shown in terms of their time derivatives.

In the d-axis equivalent circuit, the series inductance Lf1d – Lad represents the flux linking both the field winding and the amortisseurs, but not the armature. It is a common practice to neglect this series inductance on the grounds that the flux linking the damper circuit is very nearly equal to that linking the armature, because the damper windings are near the air-gap. In such case

ωr ψq

ed pψd

pψ1d pψfd

Lf1d - Lad

Lℓ

Ra

+ +

+

+ +

+

-

- - - - -

id ifd i1d

(a) d-axis equivalent circuit

ωr ψd

eq pψq

pψ1q pψ2q

Lℓ Ra

+

+

+

+ +

-

- - - -

iq i2q i1q

Laq

Lad

L1d Lfd

Rfd R1d

efd

R1q

L1q

R2q

L2q

(b) q-axis equivalent circuit

Fig. 15 Complete d-and q-axis equivalent circuits

Lℓ

Lf1d = Lad (130 a)

Then from eq. (127) Lffd = Lfd + Lad (130 b)

Example 1

A 555 MVA, 24 kV, 0.9 p.f., 60 Hz, 3-phase, 2 pole synchronous generator has the following inductances and resistances associated with the stator and field windings:

ℓ aa = 3.2758 + 0.0458 cos (2θ) mH

ℓ ab = - 1.6379 – 0.0458 cos (2θ + π/3) mH

ℓ afd = 40 cos θ mH

Lffd = 576.92 mH

Ra = 0.0031 Ω

Rfd = 0.0715 Ω

a. Determine Ld and Lq in henrys.

b. If the stator leakage Lℓ is 0.4129 mH, determine Lad and Laq in henrys.

c. Using the machine rated values as the base values for the stator quantities,

determine the per unit values of the followings:

Lℓ, Lad, Laq, Ld, Lq, Lafd, Lffd, Lfd, Ra and Rfd.

Solution

a.

Ld = Laa0 + Lab0 + 23 Laa2 = 3.2758 + 1.6379 +

23 x 0.0458 = 4.9825 mH

Lq = Laa0 + Lab0 - 23 Laa2 = 3.2758 + 1.6379 -

23 x 0.0458 = 4.8451 mH

b.

Lad = Ld - Lℓ = 4.9825 – 0.4129 = 4.5696 mH

Laq = Lq - Lℓ = 4.8451 – 0.4129 = 4.4322 mH

c.

3-phase VA base = 555 x 106 VA

ebase = 3

10x2x24 3

= 19.596 x 103 V;

ibase = bases

base

e(3/2) VAphase-3

= 3

6

10x19.596X1.510x555 = 18881.5 A

Zbase = 18881.4

10x19.596 3

= 1.0378 Ω

ωbase = 2 π x 60 = 377 elec, rad. / sec.

Ls base = 377

10x1.0378 3

= 2.753 mH

ifd base = afd

ad

LL is base =

404.5696 x 18881.5 = 2158 A

efd base = 2158

10x555 6

= 257.183 x 103 V

Zfd base = 2158

10x257.183 3

= 119.18 Ω

Lfd base = 377

119.18 x 103 = 316.12 mH

The per unit values are:

Lℓ pu = 2.7530.4129 = 0.15; Lad pu =

2.7534.5696 = 1.66; Laq pu =

2.7534.4322 = 1.61

Ld pu = Lad pu + Lℓ pu = 1.66 + 0.15 = 1.81

Lq pu = Laq pu + Lℓ pu = 1.61 + 0.15 = 1.76

L afd pu = basesbase s

basefdafd

iLiL

= 18881.5 x 2.7532158 x 40 = 1.66

Lffd pu = base fd

ffd

LL =

316.12576.92 = 1.825

Lfd pu = Lffd – Lad Refer eq. (130 b)

= 1.825 – 1.66 = 0.165

Ra pu = 1.037840.0031 = 0.003; Rfd pu =

119.180.0715 = 0.0006

6.1 Voltage, Current and Flux Linkage Relationships

As has been shown in Section 3, the dq0 transformation applied to balanced steady-state armature phase currents results in steady direct currents. This is also true for stator voltages and flux linkages. Since rotor quantities are also constant under steady state, all time derivatives terms drop out of machine equations. In addition, zero-sequence components are absent and ωr = ωs = 1 pu.

With pψ terms set to zero in equations (115), (116) and (117),

Rid i1d = Riq i1q = R2q i2q = 0

Therefore, all amortisseurs currents are zero. The per unit machine equations (111) to (124), under balanced steady-state conditions, become

ed = - ψq ωr - Ra id (131)

eq = ψd ωr - Ra iq (132)

e fd = R fd i fd (133)

ψd = - Ld id + Lad ifd (134)

ψq = - Lq iq (135)

ψ fd = L ffd ifd – L ad id (136)

ψ 1d = L f1d ifd – L ad id (137)

ψ 1q = ψ 1q = - L aq iq (138)

Field current

From equation (134),

ifd = ad

ddd

LiLψ ; But from eqn.(132), ψd =

r

qaq

ωiRe

. Therefore,

ifd = ad

ddr

qaq

L

iLω

iRe

= adr

ddrqaq

LωiLωiRe

Replacing the product of synchronous speed and inductance L by the corresponding reactance X,

ifd = ad

ddqaq

XiXiRe

(139)

The above equation is useful in computing the steady-state value of the field current for any specified operating condition. The inductances/reactances appearing in equations (131) to (139) are saturated values.

6.2 Phasor Representation

For balanced steady-state operation, the stator phase voltages may be written as

ea = Em cos(ωst + α) (140)

eb = Em cos(ωst - 3π2 + α) (141)

ec = Em cos(ωst + 3π2 + α) (142)

where ωs is the angular frequency and α is the phase angle of ea with respect to time origin.

Applying dq transformation gives

ed = Em cos(ωst + α- θ) (143)

eq = Em sin(ωst + α- θ) (144)

The angle θ by which d-axis leads the axis of phase a is given by

θ = ωrt + θ0 (145)

where θ0 is the value of θ at t = 0.

With ωr equals to ωs at synchronous speed. substitution for θ in eqns. (143) and (144) yields

ed = Em cos(α- θ0) (146)

eq = Em sin(α- θ0) (147)

In the above two equations, Em is the peak value of phase voltage. In steady-state analysis, we are more interested in rms values and phase displacements. Using Et to denote per unit rms value of armature voltage and noting that in per unit, rms and peak values are equal,

ed = Et cos(α- θ0) (148)

eq = Et sin(α- θ0) (149)

Fig. 16 Representation of dq components of armature voltage and current as phasors

(a) Voltage components (b) Current components

d-axis

q-axis

ed

eq Et

δi

α-θ d-axis

q-axis Et

Φ

iq

id

It

It is seen that dq components of armature voltage are scalar quantities. However, in view of the trigonometric relationship between them, they can be expressed as phasors in complex plane having d- and q-axes as coordinates. This is illustrated in Figure 16. Thus the armature terminal voltage may be expressed in complex form as

Et = ed + j eq (150)

By denoting δi as the angle by which the q-axis leads the phasor Et , equations (148) and (149) become

ed = Et sin δi (151)

eq = Et cos δi (152)

Similarly, the dq components of the armature terminal current It can be expressed as phasors. If Φ is the power factor angle, we can write

id = It sin(δi + Φ) (153)

iq = It cos(δi + Φ) (154)

and

It = id + j iq (155)

From the above analysis, it is clear that in phasor form with dq axes as reference, the rms armature phase current and voltage can be treated the same way as is done with phasor representation of alternating voltages and currents. This provides the link between the steady-state values of dq components of armature quantities and the phasor representation used in conventional ac circuit analysis.

The relationships between dq components of armature terminal voltage and current are defined by equations (131), (132), (134) and (135).

ed = - ψq ωr - Ra id (131) ψd = - Ld id + Lad ifd (134)

eq = ψd ωr - Ra iq (132) ψq = - Lq iq (135)

Thus ed = - ωr ψq – Ra id = ωr Lq iq – Ra id = Xq iq – Ra id (156)

eq = ωr ψd – Ra iq = - ωr Ld id + ωr Lad ifd – Ra iq = - Xd id + Xad ifd – Ra iq (157)

The reactances Xd and Xq are called the direct- and quadrature-axis synchronous reactances, respectively.

We have not yet developed a means of identifying the d- and q-axis positions relative to Et. In order to assist us in this regard, let us define a voltage Eq as

Eq = Et + (Ra + jXq) It = (ed + j eq) + (Ra + jXq) (id + j iq) (158)

Using eqs. (156) and (157), the above can be simplified as

Eq = Xq iq – Ra id + j(- Xd id + Xad ifd – Ra iq) + Ra id – Xq iq +j(Xq id + Ra iq)

= j[ Xad ifd – ( Xd – Xq) id ] (159)

d-axis

q-axis

Et

It

Eq

δi

Ra It

Xq It

Fig. 17 Phasor Eq in dq complex plane

This means that Eq is in line with q-axis. The corresponding phasor diagram is shown in Fig. 17.

As seen by eq. (158) Eq = Et + (Ra + jXq) It the position of q-axis with respect to Et can be identified by computing Eq , the voltage behind Ra + jXq.

6.3 Rotor Angle

Under no load or open-circuit condition, id = iq = 0. Substituting in equations

ψd = - Ld id + Lad ifd (134)

ψq = - Lq iq (135)

ψd = Lad ifd and ψq = 0 and hence from equations

ed = - ψq ωr - Ra id (131)

eq = ψd ωr - Ra iq (132),

ed = 0 and eq = Lad ifd ωr = Xad ifd.

Therefore

Et = ed + j eq

= j Xad ifd (160)

Thus under no-load conditions, Et has only q-axis component and hence as seen from Fig. 17, δi = 0. As the machine is loaded, δi increases. Therefore, the angle δi is referred to as the internal rotor angle or load angle.

The angle δi represents the angle by which the q-axis leads the stator terminal voltage phasor Et , and is given by (Fig. 16 a)

= 900 - (α – θ0) (161)

where α is the phase angle of ea and θ0 is the value of θ with respect to the time origin. Therefore, δi depends on the angle between the stator and rotor magnetic fields. For any given machine power output, either α or θ0 may be arbitrarily chosen, but not both.

0t 0E Ra Xs

iq δE

Fig. 18 Steady-state equivalent circuit with saliency neglected

6.4 Steady-State Equivalent Circuit

If saliency is neglected, Xd = Xq = Xs where Xs is the synchronous reactance. Therefore,

Eq = Et + (Ra + jXs) It (162)

With Xd = Xq , from equation Eq = = j[ Xad ifd – ( Xd – Xq) id ] (159)

the magnitude of Eq is given by

qE = Xad ifd (163)

The corresponding equivalent circuit is shown in Fig. 18. The resistance Ra is usually very small and may be neglected.

The voltage qE may be considered as effective internal voltage. Since it is equal

in magnitude to Xad ifd it represents the excitation voltage. The synchronous reactance Xs accounts for the flux produced by the stator circuit, i.e., the armature reaction. For the round rotor machine, Xd is nearly equal to Xq and therefore the above equivalent circuit provides satisfactory representation.

For salient pole machine, Xd is not equal to Xq. The effect of saliency is, however, not very significant so far as the relationships between terminal voltage, armature current, power and excitation over the normal operating range are concerned. The approximate equivalent circuit often provides sufficient insight into steady-state characteristics. The effect of saliency become significant, only at small excitation.

Active and reactive power

Complex power S = Et It* = (ed + jeq) ( id – jiq). Thus

Active power Pt = ed id + eq iq (164)

Reactive power Qt = eq id – ed iq (165)

Steady-state torque is given by

Te = ψd iq – ψq id

= (ed id + eq iq) + Ra (id2 + iq2) = Pt + Ra It2 (166)

6.5 Procedure for Computing Steady-State Values

The following steps summarize the procedure for computing the initial steady-state values of machine variables as a function of specified terminal quantities. It is assumed that all quantities are expressed in per unit.

(a) Normally, terminal active power Pt , reactive power Qt , and magnitude of Et

are specified. The corresponding terminal current tI and power factor angle Φ

are computed as follows:

tI = t

2t

2t

EQP

Φ = cos-1 ( tt

t

IEP )

Xq It

Ra It It

δi

Φ

Φ

Eq q-axis

d-axis Fig. 19 Steady-state phasor diagram

Et

(b) The next step is to compute the internal rotor angle δi. Since Eq lies along the q-axis, as illustrated in Fig. 19, the internal angle is given by

δi = tan-1 (φsinIXφcosIRE

φsinIRφcosIX

tqtat

tatq

)

(c) Knowing δi , the dq components of stator voltage and currents are given by

ed = tE sin δi eq = tE cos δi

id = tI sin (δi + Φ) iq = tI cos (δi + Φ)

(d) The remaining machine quantities, in per unit, are computed as follows:

ψd = eq + Ra iq

ψq = - ed - Ra id

ifd = ad

ddqaq

XiXiRe

efd = Rfd ifd

ψfd = (Lad + Lfd) ifd – Lad id

ψ1d = Lad ( ifd – id )

ψ 1q = ψ 2q = - L aq iq

Te = Pt + Ra 2

tI

Example 2

The following are the parameters in per unit on machine rating of a 555 MVA, 24 kV, 0.9 p.f., 60 Hz, 3600 rpm turbine generator:

Lad = 1.66 Laq = 1.61 Lℓ = 0.15 Ra = 0.003

Lfd = 0.165 Rfd = 0.0006 L1d = 0.1713 R1d = 0.0284

L1q = 0.7252 R1q = 0.00619 L2q = 0.125 R2q = 0.02368

Lfkd is assumed to be equal to Lad.

(a) When the generator is delivering rated MVA at 0.9 p.f. lagging and rated terminal voltage, compute the following:

(i) Internal angle δi in electrical degrees

(ii) Per unit values of ed, eq, id, iq, i1d, i1q, i2q, ifd, efd, ψfd, ψ1d, ψ1q, ψ2q.

(iii) Air-gap Te in per unit and newton-meters.

Assume that the effect of magnetic saturation at the given operating condition is to reduce Lad and Laq to 83.5% of the values given above.

(b) Compute the internal angle δi and the field current ifd for the above operating condition, using the approximate equivalent circuit. Neglect Ra.

Solution

(a) With the given operating condition, the per unit values of terminal quantities are

Pt = 0.9, Qt = 0.436, tE = 1.0, tI = 1.0, Φ = 25.840

The saturated values of the inductances are

Lad = 0.835 x 1.66 = 1.386; Laq = 0.835 x1.61 = 1.344

Ld = Lad + Lℓ = 1.386 + 0.15 = 1.536; Lq = Laq + Lℓ = 1.344 + 0.15 = 1.494

(i) δi = tan-1 (φsinIXφcosIRE

φsinIRφcosIX

tqtat

tatq

)

= tan-1 (0.436 x 1.0 x 1.4940.9 x 1.0 x 0.0031.0

0.436 x 1.0 x 0.0030.9 x 1.0 x 1.494

) = tan-1 (0.812)

= 39.1 electrical degrees

ii) ed pu = tE sin δi = 1.0 x sin 39.10 = 0.631

eq pu = tE cos δi = 1.0 x cos 39.10 = 0.776

id pu = tI sin (δi + Φ) = 1.0 x sin (39.1 + 25.84) = 0.906

iq pu = tI cos (δi + Φ) = 1.0 x cos (39.1 + 25.84) = 0.423

ifd pu = ad

ddqaq

XiXiRe

= 1.386

0.906 x 1.5360.423 x 0.0030.776 = 1.565

efd pu = Rfd ifd = 0.0006 x 1.565 = 0.000939

ψfd pu = (Lad + Lfd) ifd – Lad id = (1.386 + 0.165) x 1.565 – 1.386 x 0.906 = 1.17

ψ1d pu = Lad ( ifd – id ) = 1.386 ( 1.565 – 0.906) = 0.913

ψ 1q pu = ψ 2q pu = - L aq iq = - 1.344 x 0.423 = - 0.569

Under steady state, i1d = i1q = i2q = 0

(iii) Air-gap torque Te pu = Pt + Ra 2

tI = 0.9 + 0.003 x 1.02 = 0.903

Tbase = base m

base

ω VAphase -3 N-m =

60 x 210 x 555 6

= 1.472 x 106 N-m

Te = 0.903 x 1.472 x 106 = 1.329 x 106 N-m

(b) Using the saturated value of Xad

qE = Xad x ifd = 1.386 x ifd

and Xs = Xad + Xℓ = 1.386 + 0.15 = 1.536

With Et as reference

Eq = Et + j Xs It = 1.0 + j 1.536 (0.9 – j 0.436) = 1.67 + j 1.382 = 2.17 039.6

δi = 39.6 electrical degrees

Therefore ifd pu = 1.3862.17 = 1.566

The values of δi and ifd pu computed using the approximate representation are seen to be in good agreement with the accurate calculation. This is to be expected, since Xq is nearly equal to Xd and we are considering rated operating condition.

Trigonometric identities

sin (A+B) = sin A cos B + cos A sin B

sin (A-B) = sin A cos B - cos A sin B

cos (A+B) = cos A cos B – sin A sin B

cos (A-B) = cos A cos B + sin A sin B

2 sin A cos B = sin (A+B) + sin (A-B)

2 cos A sin B = sin (A+B) - sin (A-B)

2 cos A cos B = cos (A+B) + cos (A-B)

2 sin A sin B = cos (A-B) - cos (A+B)

sin 2A = 2 sin A cos A

cos 2A = cos2 A – sin2 A

2 cos2 A = 1 + cos 2A

2 sin2 A = 1 - cos 2A

sin A + sin (A - 3π2 ) + sin (A +

3π2 ) = 0

cos A + cos (A - 3π2 ) + cos (A +

3π2 ) = 0

sin A cos A + sin (A - 3π2 ) cos (A -

3π2 ) + sin (A +

3π2 ) cos (A +

3π2 ) = 0

sin2 A + sin2 (A - 3π2 ) + sin2 (A +

3π2 ) =

23

cos2 A + cos2 (A - 3π2 ) + cos2 (A +

3π2 ) =

23

sin A cos B + sin (A - 3π2 ) cos (B -

3π2 ) + sin (A +

3π2 ) cos (B +

3π2 )

= 23 sin (A – B)

sin A sin B + sin (A - 3π2 ) sin (B -

3π2 ) + sin (A +

3π2 ) sin (B +

3π2 )

= 23 cos (A – B)

cos A cos B + cos (A - 3π2 ) cos (B -

3π2 ) + cos (A +

3π2 ) cos (B +

3π2 )

= 23 cos (A – B)