Question

Q3. (30) Consider the energy distribution network of fig.2.a where E1 and E2 generatorsaiven as 1.0+j0.0 pu and 0.90+j0.0 pu, respectively, on a 500 kV base. After the source transformation, the equivatdiagram is redrawn as in fig.2.b. a) (03) On the diagram of fig.2.b, write down the values of the bus currents l1, 12 and admittances yij. b) (07) Write down the bus admittance matrix Ybus c) (10) Find the bus voltage vector Vbus-[Vi Vsing Ibus-YbusVbus d) (10) Eind the line power flow S12 flowing from bus1 to bus2. 1 212 0.01+j0.01 2 Z10 j0.01 220 j0.02 E1 E2 fig.2.a. 12 1 12 y10 y20 fig.2.b.

Answer 1

Soluion 2 1 2 172 12 20 J 02 i jo 001-Jod ะ 50-Jr。

2 2 0 l 2 yes 2 i 一50 tjto

(tto-Jiss) ( 5.jn') _ (-sotin)(-5.jnj 111.8じ62 . y 3 70.7に心 )4893 L13-77