ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
0.1 Variables
S TEADY S TATE A NALYSIS OF S ALIENT-P OLE S YNCHRONOUS
G ENERATORS
This paper is intended to provide a procedure for calculating the internal voltage of a salientpole 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. Assumptions 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., I˜a = Ia e jθV I .
Variables without a tilde over the top are magnitudes. The subscripts on angles specify between what phasors the angles span; e.g., the angle θV I is 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., I˜a = Ia e jθV I .
1.2 Terms
The direct axis is defined as the direction along the rotor that the field winding current causes
magnetic flux to flow. The quadrature axis is defined as the axis located π2 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
ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
S the total three phase complex power
E˜q a phasor representing the “a” phase line-to-neutral stator internal voltage
V˜a a phasor representing the “a” phase line-to-neutral terminal voltage
I˜a a phasor representing the “a” phase line current
I˜q a phasor representing the quadrature axis component of the “a” phase line current
The quadrature component is in phase with the stator internal voltage Ẽq .
I˜d a phasor representing the direct axis component of the “a” phase line current
The direct component leads or lags the quadrature component by π=2 radians.
the angle of V˜a with repect to I˜a
θEI the angle of E˜q with repect to I˜a
δ the angle of the E˜q with repect to V˜a
θV I
2 Assumptions
It is assumed that the voltage due to the stator winding resistance is negligible; i.e., the stator
winding resistance is zero. Magnetic saturation will be ignored. The “a” phase line-to-neutral
voltage, V˜a is the reference for angles; i.e., V˜a = Va .
3 Power Delivered by a Salient-Pole Synchronous Generator
EqVa
V2
P=3
sin(δ) + a
Xd
2
Xd Xq
sin(2δ)
Xd Xq
Remove the 3 in Eq. 1 if calculations are being performed in per unit.
Page 2/7
(1)
ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
4 Procedure for Calculating E˜q Given V˜a and S
is
ax
j(X
is
-ax
q-
)I d
-d X q
Eq
d
jX q
E xq
jX q
Iq
delta
theta VI
Iq
Ia
jX d
Id
Va
Ia
Id
S
I˜a =
3Ṽa
θV I
=
I˜a = Ia e
jθV I
angle(I˜s )
Ẽxq = Ṽa + jXq I˜a
Id =
)
(2)
(3)
)
Ẽxq = Exq e jδ
(4)
δ = angle(Ẽxq )
(5)
Ia sin(θV I + δ)
(6)
I˜d
= Id e
Ẽq = E˜xq + j(Xd
j(δ+ π2 )
)
I˜d = Id e j(δ+ 2 )
(7)
Xq )I˜d
)
Ẽq = Eq e jδ
(8)
π
Remove the 3 in Eq. 2 if calculations are being performed in per unit.
Page 3/7
ECE 523
Law, J.D.
Symmetrical Components
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 different magnetic flux 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: Xd and Xq . The quadrature component is in
phase with the stator internal voltage, Ẽq . The direct component leads or lags the quadrature
component by π=2 radians.
Equations 9 and 10 are used to determine I˜a and Ẽq given I˜q , I˜d , and the terminal voltage, Ṽa .
I˜a = I˜q + I˜d
(9)
Ẽq = Ṽa + jXd I˜d + jXq I˜q
(10)
The relationships between the phasors described by Eq. 9 and 10 are displayed in the phasor
diagram shown below.
is
ax
q-
Eq
xis
d-a
jX q
Iq
delta
theta VI
jX d
Va
Ia
Id
Page 4/7
Id
Iq
ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
Typically I˜q and I˜d are not initially known. Usually the line current, I˜a , and the terminal
voltage, Ṽa , are known. To decompose the stator current, I˜a , in to I˜q and I˜d the phase of the
stator internal voltage, Ẽq , must be know. To solve the problem of needing the phase of Ẽq to
calculate Ẽq , Eq. 10 can be modified as shown in the following three equations. First zero in
the form of jXd I˜d jXd I˜d is added to the right hand side of Eq. 10.
Ẽq = Ṽa + jXd I˜d + jXq I˜q + jXq I˜d
jXq I˜d
(11)
Rearranging terms yields:
Ẽq = Ṽa + jXq (I˜q + I˜d ) + j(Xd
Xq )I˜d
(12)
Recognizing that I˜a = I˜q + I˜d yields:
Ẽq = Ṽa + jXq I˜a + j(Xd
Xq )I˜d
(13)
A new voltage, Ẽxq , is defined as the portion of Eq. 13 that is in terms of only Ṽa and I˜a .
Ẽxq Ṽa + jXq I˜a
)
δ = angle(Ẽxq )
(14)
Equation 13 can be rewritten in terms of the newly defined voltage Ẽxq .
Ẽq = Ẽxq + j(Xd
Xq )I˜d
(15)
Observing that the voltage j(Xd Xq )I˜d is in phase with Ẽq means that the voltage Ẽxq is also
in phase with Ẽq . That is, the voltages j(Xd Xq )I˜d , Ẽxq and Ẽq are all in phase.
The result of the above observation is that Ẽxq can be calculated with Eq. 14 using only Ṽa
and I˜a . By calculating Ẽxq the angle of Ẽq with respect to Ṽa is known. The angle of Ẽq with
respect to Ṽa is defined as δ. With δ known, I˜a can be decomposed in to I˜q and I˜d using Eq. 16
and 17.
I˜q = Ia cos(δ + θV I )e jδ
I˜d =
(16)
π
Ia sin(δ + θV I )e j(δ+ 2 )
(17)
Page 5/7
ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
6 Procedure for Calculating E˜q Given Ṽsys and Ssys
This section presents a procedure for calculating Ẽq given the system voltage, Ṽsys , and the
complex power delivered to the system, S sys , through a reactance Xth .
6.1 Development
Solving Eq. 10 for Ṽa yields:
Ṽa = Ẽq
jXd I˜d
jXq I˜q
(18)
Kirkhoff’s voltage law yields
Ṽa = Ṽsys + jXth I˜a
(19)
Equating Eq. 18 and 19 yields
Ẽq
jXd I˜d
jXq I˜q = Ṽsys + jXth (I˜q + I˜d )
(20)
Rearranging terms yields:
Ẽq = Ṽsys + j(Xq + Xth )I˜q + j(Xd + Xth )I˜d
(21)
Ẽq = Ṽsys + jXqeq I˜q + jXdeq I˜d
(22)
where
Xdeq Xd + Xth
(23)
Xqeq Xq + Xth
(24)
Page 6/7
ECE 523
Law, J.D.
Symmetrical Components
Fall 2007
Salient Pole S/G
S19 Revised
6.2 Procedure
I˜a =
θV I
=
Ssys
)
3Ṽsys
=
(26)
)
E˜xq = Exq e jδsys
(27)
δsys = angle(Ẽxq )
(28)
Ia sin(θV I + δsys )
(29)
π
I˜d = Id e j(δsys + 2 )
E˜q = E˜xq + j(Xdeq
"
(25)
angle(I˜a )
E˜xq = Ṽsys + jXqeq I˜a
Id
jθV I
I˜a = Ia e
Xqeq )I˜d
2
Vsys
EqVsys
Psys = 3
sin(δsys ) +
Xdeq
2
)
I˜d = Id e j(δsys + 2 )
(30)
)
E˜q = Eq e jδsys
(31)
π
#
Xdeq Xqeq
sin(2δsys )
Xdeq Xqeq
Remove the 3 in Eq. 25 and 32 if calculations are being performed in per unit.
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(32)