Question

Problem 2:Three phase droult (42 pts) A three-phase wye connected generator ﹃plies two paral el loads through a feeder. tad is matalaeeday connected load and load 2 is a balanced delta connected loed. System data: Generator: Rating S-5oo MVA; Line Voltage: Veu-25KV; Feeder impedance: 7-(0.08 jo 62) O/mile, length-8 miles Generator reactance Xe- 1108: P.-75MW pf c-o.78 (all the power factors are lagging) (a) Draw the one-line diagram and a three-phase equivalent circuit (8pts) (b) Calculate the generator and feeder line impedances (8pts) (c) Calculate the load i line currents (zpes) (d) Calculate the load 2 line currents (not the current in the delta connected load phases) (zpts) (e) Calculate each of the generator line currents. (8pts) (1) Caleulate one of the actual generator line to neutral voltages needed under these load conditions 4)

Answer 1

Given that, generator S_g = 600mva xgll = 25kv xg = 118% feeder: z = (0.08 + j0.62)ohm Vertical-bar mile length = 8miles load: vll = 24kv Pa = 75mw Pb = 75mw Pc = 75mw Pc-a = 0.8 Pf-b = 0.84 p\Pf-c = 0.78 load 2: vll = 24kv P total = m225mw Pf = 0.8 one line diagram is SPa generator impledance: - given Xg = 118% Xg = j%Xg + (kv)_2/mva = J118/100 times (25)^2/600 Generator impedance = j1.229 ohm Transmission line impedance: - ZL = 0.08 Vertical-bar n + j0.6^2 = 0.08 times 8 + j0.62 ZL = 0.64 + j0.62 ohm cor loop 1 wye is connected so line current will be equal to phase current but given it is an unbalanced load so line current will be different for each V_phase = 24 times 10^3/squareroot 3 = 13856.406V Ia1 = 75/10^6/13856.406 times 0.8 (cos^-1(0.8)) = 6765.82angle-36.86 degree A Ib1 = 75 time 10^6/13856.406 times 0.84 (-cos_-1 (0.84)) = 6443.64 angle -32.86 degree A I_C1 = 75 times 10_6/13856.406 times 0.78_(-cos^-1(0.78)) = 6939.3 angle-38.73 degree A