A single-phase load is supplied through a 35 kV feeder whose impedance is 115 + j380 Ohms and a 35 kV, 2400 V transformer whose equivalent impedance is 0.26 + j1.21 Ohms referred to its low voltage side. The load is 180 kW at 0.87 leading power factor and 2320 V.
(a) Compute the voltage at high-voltage terminals of the transformer.
(b) Compute the voltage at the sending end of the feeder.
(c) Compute the power and reactive power input at the sending end of the feeder.© BrainMass Inc. brainmass.com October 9, 2019, 8:02 pm ad1c9bdddf
Pl = Vl * Il * (p.f.)
180,000 = 2320 * Il * 0.87
Hence, Load Current = Il = 89.2 Amps @ 30 deg. (Lead)
Load Impedance = Zl = Vl/Il = 26 @ -30 deg. (Capacitive Load since Leading p.f.)
Turns ratio = a = 35,000/2400 = 14.583
a^2 = 212.7
Impedance = Zt = 0.261+j1.21 Ohms (referred to low voltage, secondary side)
Hence, impedance referred to high-volgae, primary side = ...
This solution computes the voltage at two difference points and the power and reactive power input.