Page 89 - Handbook of Electrical Engineering

SYNCHRONOUS GENERATORS AND MOTORS 69

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:-

V d = V sin δ r
V q = V cos δ r
E d = E sin δ s
E q = E cos δ s
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
I d =
X d X q + R d R q
And
(E q − V q )R d − (E d − V d )X d
I q =
X d X q + R d R q
Active and reactive power leaving the terminals of the ‘receiving-end’ and received by the
load are,

P r1 + P r2
P r =
DEN
Q r1 + Q r2
Q r =
DEN
Where,

V 2
P r1 = V sin δ r (E q X q + E d R q ) + sin 2δ r (X d − X q )
2
2 2
P r2 = V cos δ r (E q R d − E d X d ) − V (R q sin δ r + R d cos δ r )
V 2
Q r1 = V cos δ r (E q X q + E d R q ) + sin 2δ r (R d − R q )
2
2 2 2
Q r2 = V sin δ r (E d X d − E q R d ) − V (X d sin δ r + X q cos δ r )
DEN = X d X q + R d R q

Also the active and reactive power leaving the shaft and the exciter are,
∗
P s = Real part of (EI )

= I E q cos(δ r + Ø) + E d sin(δ r + Ø)
∗
Where I denotes the conjugate of the phasor I.