Electrical Power Systems

164 Electrical Power Systems
|Y 12| = 22.36, q 12 = 116.56°
|Y 13| = 31.62, q 13 = 108.4°
\ P 1 = (1.05) 2 × 53.85 × cos (–68.2°) + 1.05 × 0.98265 × 22.36 cos (116.56° – 0 – 3.048° )
+ 1.05 × 1.00099 × 31.62 cos (108.4° – 0 – 2.68°)
\ P 1 = 22.048 – 9.2038 – 9.004
\ P 1 = 3.84 pu MW. = 3.84 × 100 = 384 MW
rom eqn. (7.42)
3
å
k = 1
Q1 = – | V|| V || Y |sin( q - d + d )
1 k 1k 1k 1 k
2
1 11 11 1 2 12 12 1 2
\ Q1 = –| V| | Y |sin q -| V|| V || Y |sin ( q - d + d )
-| V|| V || Y |sin ( q - d + d )
1 3 13 13 1 3
\ Q 1 = –|1.05| 2 × 53.85 × sin (–68.2°) – 1.05 × 0.98265 × 22.36 sin (116.56° – 3.048° )
– 1.05 × 1.00099 × 31.62 sin (108.4° – 2.68°)
\ Q 1 = 55.1238 – 21.1552 – 31.99
\ Q 1 = 1.9786 pu MW. = 197.86 MW
Step-5: Calculation of line flows and line losses.
rom eqn. (7.34)
P ik = –|V i | 2 |Y ik | cos q ik +|V i ||V k ||Y ik | cos (q ik – d i + d k )
\ P 12 = –|V 1| 2 |Y 12| cos q 12 +|V 1||V 2||Y 12| cos (q 12 – d 1 + d 2)
\ P 12 = –(1.05) 2 × 22.36 cos (116.56°) + 1.05 × 0.98265 × 22.36 cos (116.56° – 0 – 3.048° )
\ P 12 = +11.0227 – 9.2038 = 1.8189 pu MW
\ P 13 = –|V 1 | 2 |Y 13 | cos (q 13 ) +|V 1 ||V 3 ||Y 13 | cos (q 13 – d 1 + d 3 )
\ P 13 = –(1.05) 2 × 31.62 cos (108.4°) + 1.05 × 1.00099 × 31.62 cos (108.4° – 0 – 2.68° )
\ P 13 = +11.0038 – 9.0042 = 2.0 pu MW
\ P 23 = –|V 2| 2 |Y 23| cos q 23 +|V 2||V 3||Y 23| cos (q 23 – d 2 + d 3 )
\ P 23 = –(0.98265) 2 × 35.77 × cos (116.6°) + 0.98265 × 1.00099
× 35.77 cos (116.6° + 3.048° – 2.68°)
\ P 23 = 15.4654 – 15.9557
\ P 23 = –0.4903 pu MW