s 19 calculating ef

8/11/2019 s 19 Calculating Ef

1/7

ECE 523

Law, J.D.

Symmetrical Components

Fall 2007

Salient Pole S/G

S19 Revised

0.1 Variables

STEADYSTATEA NALYSIS O FS ALIENT-POLESYNCHRONOUS

GENERATORS

This paper is intended to provide a procedure for calculating the internal voltage of a salient-

pole synchronous generator given the terminal voltage and the complex power delivered.

First the notation, terms, parameters, and variables used in the procedure will be defined. As-

sumptions will be listed. An equations for the power delivered by a salient-pole synchronous

generator is given. The actual procedure will be presented followed by an development of the

procedure. A modified procedure is given for a generator connected through a reactance to

an infinite bus.

1 Definitions

1.1 Notation

Phasors will be designated by putting a tilde over the top of a variable; e.g., Ia Iae

jV I.

Variables without a tilde over the top are magnitudes. The subscripts on angles specify be-

tween what phasors the angles span; e.g., the angle V Iis the angle of the voltage with respect

to the current. A superscript * denotes the complex conjugate of a phasor or complex num-

ber; e.g., Ia

IaejV I

.

1.2 Terms

Thedirect axisis defined as the direction along the rotor that the field winding current causes

magnetic flux to flow. The quadrature axisis defined as the axis located 2

electrical radians

behind the direct axis of the rotor. The stator internal voltage is that portion of the terminal

voltage due to the flux caused exclusively by the field current.

1.3 Parameters and Variables

Xd direct axis reactance

Xq quadrature axis reactance

Page 1/7


2/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

S

the total three phase complex power

Eq a phasor representing the a phase line-to-neutral stator internal voltage

Va a phasor representing the a phase line-to-neutral terminal voltage

Ia a phasor representing the a phase line current

Iq a phasor representing the quadrature axis component of the a phase line current

The quadrature component is in phase with the stator internal voltage Eq.

Id a phasor representing the direct axis component of the a phase line current

The direct component leads or lags the quadrature component by

2 radians.

V I the angle ofVawith repect to Ia

EI the angle of Eq with repect to Ia

the angle of the Eq with repect to Va

2 Assumptions

It is assumed that the voltage due to the stator winding resistance is negligible; i.e., the statorwinding resistance is zero. Magnetic saturation will be ignored. The a phase line-to-neutral

voltage, Va is the reference for angles; i.e., Va Va.

3 Power Delivered by a Salient-Pole Synchronous Genera-

tor

P

3

EqVa

Xd sin

V2a2

Xd Xq

XdXq

sin

2

(1)

Remove the 3 in Eq. 1 if calculations are being performed in per unit.

Page 2/7


3/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

4 Procedure for Calculating EqGiven VaandS

thetaVI

delta

q-axis

Va

E q

Ia

I d

I q

d-ax

is

jX qI q

jX dI d

Exq

j(X d-X q

)I d

jX qI a

Ia S

3Va

Ia Iae

jV I (2)

V I angle Is (3)

Exq Va jXq Ia Exq Exqej (4)

angle Exq (5)

Id Iasin V I (6)

Id

Idej

2

Id

Idej

2

(7)

Eq Exq j Xd Xq Id Eq Eqej (8)

Remove the 3 in Eq. 2 if calculations are being performed in per unit.

Page 3/7


4/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

5 Development of the Procedure

The components of the stator current acting along the direct and quadrature axes result in dif-

ferent magneticflux per ampere due to the non-uniform airgap of a salient-pole synchronous

generator. The salient-pole steady state model accounts for this effect by decomposing the

stator current in to a direct, Id, and quadrature, Iq, component. The direct and quadrature

components act through two different reactances: XdandXq. The quadrature component is in

phase with the stator internal voltage, Eq. The direct component leads or lags the quadrature

component by

2 radians.

Equations 9 and 10 are used to determine Iaand Eqgiven Iq, Id, and the terminal voltage, Va.

Ia Iq Id (9)

Eq Va jXdId jXq Iq (10)

The relationships between the phasors described by Eq. 9 and 10 are displayed in the phasor

diagram shown below.

thetaVI

delta

q-axis

Va

E q

Ia

I d

I q

d-ax

is

jX qI q

jX dI d

Page 4/7


5/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

Typically Iq and Idare not initially known. Usually the line current, Ia, and the terminal

voltage, Va, are known. To decompose the stator current, Ia, in to Iq and Idthe phase of the

stator internal voltage, Eq, must be know. To solve the problem of needing the phase of Eq to

calculate Eq, Eq. 10 can be modified as shown in the following three equations. First zero in

the form of jXdId jXdIdis added to the right hand side of Eq. 10.

Eq Va jXdId jXq Iq

jXq Id jXq Id

(11)

Rearranging terms yields:

Eq Va jXq Iq Id j Xd Xq Id (12)

Recognizing that Ia Iq Idyields:

Eq Va jXq Ia j Xd Xq Id (13)

A new voltage, Exq, is defined as the portion of Eq. 13 that is in terms of only Va and Ia.

Exq Va jXq Ia angle Exq (14)

Equation 13 can be rewritten in terms of the newly defined voltage Exq.

Eq Exq j Xd Xq Id (15)

Observing that the voltage j Xd Xq Idis in phase with Eqmeans that the voltage Exq is also

in phase with Eq. That is, the voltages j Xd Xq Id, Exq and Eq are all in phase.

The result of the above observation is that Exq can be calculated with Eq. 14 using only Vaand Ia. By calculating Exq the angle of Eq with respect to Va is known. The angle of Eq with

respect to Vais defined as. Withknown, Iacan be decomposed in to Iqand Idusing Eq. 16and 17.

Iq Iacos V I ej (16)

Id Iasin V I ej

2 (17)Page 5/7


6/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

6 Procedure for Calculating EqGiven VsysandSsys

This section presents a procedure for calculating Eq given the system voltage,

Vsys, and thecomplex power delivered to the system,Ssys, through a reactanceXth .

6.1 Development

Solving Eq. 10 for Vayields:

Va Eq jXdId jXq Iq (18)

Kirkhoffs voltage law yields

Va Vsys jXth Ia (19)

Equating Eq. 18 and 19 yields

Eq jXdId jXq Iq Vsys jXth Iq Id (20)

Rearranging terms yields:

Eq Vsys j Xq Xth Iq j Xd Xth Id (21)

Eq Vsys jXqeq Iq jXdeq Id (22)

where

Xdeq Xd Xth (23)

Xqeq Xq Xth (24)

Page 6/7


7/7

ECE 523

Law, J.D.


Fall 2007

Salient Pole S/G

S19 Revised

6.2 Procedure

Ia Ssys

3Vsys

Ia Iae

jV I (25)

V I angle Ia (26)

Exq Vsys jXqeqIa Exq Exqejsys (27)

sys angle Exq (28)

Id Iasin V I sys (29)

Id Idej

sys 2

Id Idej

sys 2 (30)

Eq Exq j Xdeq Xqeq Id Eq Eqejsys (31)

Psys 3

EqVsys

Xdeqsin sys

V2sys

2

Xdeq Xqeq

XdeqXqeq

sin 2sys

(32)

Remove the 3 in Eq. 25 and 32 if calculations are being performed in per unit.

Page 7/7

s 19 calculating ef

Documents