Question

ystem Design December 17, 2019 2. (20 pts.) Using the "DC" power flow method find the real power flows in MW between the buses and the power generated in MWW by the generator on bus 1. The impedance values shown for the lines in the figure are in per unit with a base of 100 MVA and 230 kV. PG2 = 250 MW P2 = 50 MW 120 1 3 j0.25 j0.2 j0.4 P3 = 300 MW

Answer 1

voltage angle at vouãge angle at bus 2 S, voltage angle at bus 3- E, bus 1, 8,-0° Real Power injected from bus 2 to the nebwrk, { Pa,- PL2) =(250-50) Mw- 2O Mw P (200 2 pu pawer ryolein thouwn in The from the fig. shown in the quetian), We Com write (32-ds) (82-)+ 23 (82-83) 82) +o.2 0.25 4 82 +5(82-83) (982-583)= [: 20 pu] . Real aPower Iujeeted from bun 3 to the network, pjrj = (3A)+ C8-) - C5) + a (-s) = (15 83-sd2) Pyr-(-Ps)-0-300) nW=roo (7.583-508)= - 3.0 X32 %3D X31 0.4 %3D - 30 p.u 300 sohirg egns W 4 (2), we get, Sz = 0 , 82=- * real powenr flow Calaulation bus I to buo 2, Piz =Câ-da ) O[:= d2 = o rad] -0.4 Tad %3D 저2 to bus 3, P,3 = bus I %3D lo0 MW) bus 2 to bus 3, Py = 0-to 9); 20 pu 200 Mw Cg-) =