Electrical Power Systems

182 Electrical Power Systems
( 2)
d = ( 1)
( 1)
3 d3 3
+Dd = –3.2° – 0.15° = –3.35°
|V 2 | (2) = |V 2 | (1) + D|V 2 | (1) = 1.0384 + 0.008 = 1.0464
7.15 AST DECOUPLED LOAD LOW
The diagonal elements of J 1 described by eqn. (7.57) may be written as:
d
P i
i
n
å
k = 1
Using eqns. (7.65) and (7.51), we get
\
d
P i
d
i
P i
i
= | V|| V || Y | sin(qik – di + dk) – |Vi| 2 |Yii| sin qii i k ik
= –Q i – |V i| 2 |Y ii| sin q ii
= –Q i – |V i| 2 B ii
...(7.65)
...(7.66)
where B ii = |Y ii | sin q ii is the imaginary part of the diagonal elements of the bus admittance
matrix. In a practical power system, B ii >> Q i and hence we may neglect Q i . urther simplification
is obtained by assuming |V i | 2 » |V i |, which gives,
d
P i
i
= –|V i|B ii
...(7.67)
Under normal operating conditions, d k – d i is quite small. Therefore, q ik – d i + d k » q ik and
eqn. (7.58) reduces to
d
P i
Assuming |V k | » 1.0
d
k
P i
k
= –|V i||V k|B ik
= –|V i|B ik
Similarly, the diagonal elements of J 4 as given by eqn. (7.59) may be written as:
| i|
= –|V n
i||Yii|sin qii – å| Vi|| Vk|| Yik|
k = 1
Qi
V
Using eqns. (7.69) and (7.51), we get,
\
Qi
| V|
= –|Vi||Yii|sin qii + Qi i
Qi
| V|
= –|Vi|Bii + Qi Again B ii >> Q i, Q i may be neglected.
\
i
Qi
| V|
= –|Vi|Bii i
...(7.68)
sin(q ik – d i + d k) ...(7.69)
...(7.70)
...(7.71)