A 500 kV, 60 Hz uncompensated three-phase transmission line is 500 km long. The

Question

A 500 kV, 60 Hz uncompensated three-phase transmission line is 500 km long. The line has three ACSR 1113-kcmil (Finch) conductors per phase with parameters: A = |A| alpha = 0.8137 1.089 degree B = |B| beta = 164.6 85.42 degree Ohm V_R = |V_R| 0^degree = 500 degree kV V_S= |V_S| delta = 500 delta kV Current carrying capability of single ACSR 1113-kcmil (Finch) conductor is 1,110 A. Calculate the theoretical maximum (steady state stability limit) real power that this line can deliver and compare with the thermal limit of the line. Assume V_s = V_R = 1.0 per unit and unity power factor at the receiving end.

Explanation / Answer

DATA OF FINCH CONDUCTORS : CURRENT CARRYING CAPACITY=1110 AMPERES

REAL POWER = (Vs * Vr/ X ) Sin delta

Inductive reactance per mile = 0.380 , For 500 km we have 310.686 miles .

Total reactance = 0.380*310.686 = 118 OHMS

FORMAXIMUM STEADY STATE LIMIT DELTA=90 DEGREE

REAL POWER TRANSFERRED WILL BE = 500*500* SIN90/ 118 = 2118MW

NOW CURRENT at upf WILL BE = 2118 MW / (1.7325*500)= 2564 A

THE CURRENT EXCEEDS THE THERMAL RATING , VOLTAGE LEVEL MUST BE INCREASED TO AVOID THERMAL BREAKDOWN.

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