ACTIVE AND REACTIVE POWER DELIVERED FROM A GENERATOR
3.5 ACTIVE AND REACTIVE POWER DELIVERED FROM A GENERATOR
3.5.1 A General Case
If the steady state, transient and sub-transient phasors in Figure 3.1 are considered separately, then there is seen to be a similar structure. The terminal voltage V is resolved into its two-axis components
V d and V q . The emfs E, E ′ and E ′′ can also be resolved into their components; E d ,E q ,E ′ d ,E ′ q ,
E ′′ d and E ′′ q . In practical machines E d does not exist (except for an interesting prototype built for the CEGB in approximately 1970, called the Divided Winding Rotor generator, see References 12 and 13). E d would require a second exciter to produce it.
The variables can be regarded as ‘sending-end’ and ‘receiving-end’ variables. The sending-end variables are the emfs E, E d and E q , whilst the receiving-end ones are V , V d and V q . The current
I , resolved into I d and I q , is common to both ends. The emfs, voltages and volt drops along each axis can be equated as,
For the d-axis
( 3.9) For the q-axis
E q = V q + I q R q + I d X d ( 3.10) Where R d and R q are the resistances present in their respective axis, usually both are equal
to R a the armature resistance.
SYNCHRONOUS GENERATORS AND MOTORS
To distinguish between the sending-end and the receiving-end the subscripts ‘s’ and ‘r’ are introduced for the δ angles between E and E q , and V and V q respectively. Hence their compo- nents are:-
I d = −I sin(Ø + δ r )
I q = I cos(Ø + δ r ) Equations (3.9) and (3.10) can be transposed to find I d and I q , (E q −V q )X q + (E d −V d )R q
Active and reactive power leaving the terminals of the ‘receiving-end’ and received by the load are,
Also the active and reactive power leaving the shaft and the exciter are,
P s = Real part of (EI ∗ ) =I q cos(δ r + Ø) + E d sin(δ r + Ø)
Where I ∗ denotes the conjugate of the phasor I.
70 HANDBOOK OF ELECTRICAL ENGINEERING
Q s = Imaginary part of (EI ∗ ) =I q sin(δ r + Ø) − E d cos(δ r + Ø)
The active and reactive power losses are,
P loss =I d 2 R d +I q 2 R q Q loss =I d 2 X d +I q 2 X q
From which the summations of powers are,
P S =P r +P loss Q S =Q r +Q loss
The equations above are shown for the steady state. However they apply equally well for the transient and sub-transient states provided the substitutions for E ′ d ,E ′ q ,E ′′ d ,E ′′ q ,X ′ d ,X ′ q ,X ′′ d and
X ′′ q are made systematically. Such substitutions are necessary in the digital computation of transient disturbances in power systems, those that are often called ‘transient stability studies’.
3.5.2 The Particular Case of a Salient Pole Generator
The first simplification is to assume R d = R q = R a which is very practical. In addition the steady state variables E d and δ r can be assumed to be zero. Hence the equations in sub-section 3.5.1 become.
V d = V sin δ
V q = V cos δ
I d =− I sin(φ + δ)
I q = I cos(φ + δ) (E q − V q )X q − V d R a
X d X q + R a 2 (E q − V q )R a − V d X d
P r 1 + P r 2 P r = DEN
Q r 1 + Q r 2 Q r = DEN
SYNCHRONOUS GENERATORS AND MOTORS
Where,
P r 1 = V sin δ(E q X q )+
sin 2δ(X d −X q )
2 P r 2 = V cos δ(E q R a )−V 2 R a
Q r 1 = V cos δ(E q X q )
Q r 2 = V sin δ(−E q R a )−V 2 (X d sin 2 δ+X q cos 2 δ)
DEN = X d X q +R a 2
The sending-end variables become,
P S =IE q cos(δ + φ) Q S =IE q sin(δ + φ)
3.5.3 A Simpler Case of a Salient Pole Generator
Most practical generators have an armature resistance R a that is much less in value than the syn- chronous reactances X d and X q . Consequently the equations in sub-section 3.5.2 can be further simplified without incurring a noticeable error. They become,
I d = −I sin(φ + δ)
I q = I cos(φ + δ)
E q −V q
X q P r 1 +P r 2
P r = DEN
Q r 1 +Q r 2 Q r = DEN
Where,
1 = V sin δ(E q X q )+
sin δ(X d −X q )
2 P r 2 =0
Q r 1 = V cos δ(E q X q )
72 HANDBOOK OF ELECTRICAL ENGINEERING
Q r 2 = −V 2 (X d sin 2 δ+X q cos 2 δ) DEN = X d X q
The sending-end variables remain the same. These equations are of the same form as those found in most textbooks that cover this subject.