Electrical Power Systems

434 Electrical Power Systems
Assuming d i – d K = d iK, we have
P i =
n
å Vi VK
[GiKcosdiK + BiKsindiK] ...(i)
K=1
where Y iK cosq iK = G iK, Y iK sinq iK = B iK
and Y iK = G iK + jB iK
\ or the sample power system shown in ig. 16.9,
V 1 = V 2 = V 3 = 1.0 pu and n = 3.
3
\ Pi = å GiK cos diK + BiK
sin diK
K=1
\ P 1 = G 11 + G 12 cosd 12 + B 12sind 12 + G 13cosd 13 + B 13 sind 13 ...(ii)
P 2 = G 21 cosd 21 + B 21 sind 21 + G 22 + G 23 cosd 23 + B 23 sind 23
P 3 = G 31 cosd 31 + B 31 sind 31 + G 32 cosd 32 + B 32 sind 32 + G 33
But d 12 = – d 21 = 0.0, d 32 = – d 23 , d 13 = – d 31
...(iii)
...(iv)
\ P 1 = G 11 + G 13 cosd 31 – B 13 sind 31 ...(v)
P 2 = G 22 + G 23 cosd 23 + B 23 sind 23
P3 = G33 + G31 cosd31 + B31 sind31 + G32 cosd23 – B32 sind23 Now from the YBUS matrix, we have
G11 = 1.0, G22 = 0.50, G33 = 1.50,
G 13 = G 31 = – 1.0, G 23 = G 32 = – 0.50,
B 13 = B 31 = 10.0, B 23 = B 32 = 5.0,
...(vi)
...(vii)
\ P 1 = 1.0 – cosd 31 – 10 sind 31 ...(viii)
P 2 = 0.50 – 0.50 cosd 23 + 5 sind 23
P3 = 1.50 – cosd31 + 10 sind31 – 0.50 cosd23 – 5 sind23 Power loss expression can be obtained as:
P L =
3
å Pi = P1 + P2 + P3 i=1
\ P L = 1.0 – cosd 31 – 10 sind 31 + 0.50 – 0.50 cosd 23 + 5 sind 23 + 1.50 – cosd 31 +
10 sind 31 – 0.50 cosd 23 – 5 sind 23
...(ix)
\ P L = 3.0 – 2 cosd 31 – cosd 23 ...(xi)
Now
P2 d 2 =– G23sind23 + B23cosd23 = 0.5 sind23 + 5 cosd23 P2 d3
= G23sind23 – B23cosd23 = – 0.5 sind23 – 5 cosd23 ...(x)