Electrical Power Systems

\
Also
d
P 3
d
2
P 3
3
=– G 32 sind 23 – B 32 cosd 23 = 0.5 sind 23 – 5 cosd 23
=– G 31sind 31 + B 31cosd 31 + G 32sind 23 + B 32cosd 23
P3 d 3 = sind31 + 10 cosd31 – 0.50 sind23 + 5 cosd23
d
P 1
2
=0
P1 d
=– G13 sind31 – B13 cosd31 = sind31 – 10 cosd31 3
or this problem, eqn. (16.76) reduces to the form
L
NM
P2 P
d d
2
3
2
P2 P
d d
3
3
3
O L
QP
P
1 -
P
NM
1
L
g2
O
QP
= –
L
M
NM
0
d
Next calculate d 23 and d 31 corresponding to a particular P g2.
P 1
Let initial value of P g2 = P g2 0 = 1.0 pu
\ P 2 = P g2 0 = 1.0 pu
3
O
P
QP
Optimal System Operation 435
...(xii)
rom eqn. (ix), we have,
0.50 – 0.5 cosd 23 + 5 sind 23 = 1.0
\ 5 sind 23 – 0.50 cosd 23 = 0.50 ...(xiii)
Solving above equation, iteratively, we get,
d 23 = 11.5 0
Now
P 3 = – 2.0 pu
\ rom eqn. (x), we have,
1.50 – cosd 31 + 10 sind 31 – 0.5 cos (11.5 0 ) – 5 sin (11.5 0 ) = –2
\ 10 sind 31 – cosd 31 = – 2.0132
\ d 31 » – 5.85 0
Now P 1 = P g1
rom eqn.(viii), we have,
P g1 = 1.0 – cosd 31 – 10 sind 31
\ P g1 = 1.0 – cos (– 5.85) – 10 sin (– 5.85 0 )
\ P g1 = 1.024 pu