Determine the total active power:
Circuit is introduced in the Face 1. It is about a single-wires face for a three-phase system. The voltage of the AC source is of 69kV. The source is a balanced source of 60 Hz.The voltage phasor of phase ahasa zero angle:VPhase a = (69/√3) ∠ 0o.
For a given load RL1, we obtain the following three currents for the AC source.The currents are measured so that the direction of the currents is from the bar SRC towards the ground (neutral) but not towards the bar BUS1. The board below indicates values crests
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Module (A)
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Phase (degre)
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SRCa
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1.844847 x 103
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1.5325 x 102
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SRCb
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1.844847 x 103
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3.325 x 101
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SRCc
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1.844847 x 103
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- 8.675 x 101
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1-Is the circuit seen by the source balanced? Explain why.
2-Calculate the total apparent power S (expressed in phasor) produced (from the source to the circuit) by the three-phase source. Feel free to use the shortest way of calculation.
3-What is the total active power (P) produced by the three-phase source?
4-What is the total reactive power (Q) produced by the three-phase source?
5-What is the power factor of the circuit seen by the source?
6-Calculate the voltages (phasors, line-neutral) of phases a, b and c in the bar BUS1.
7. Calculate the voltage phasor of direct sequence (line-neutral) in the bar BUS1.
8-Calculate the voltage phasorof homopolarsequence in the barBUS1
9-Calculate the impedance of direct sequence at 60 Hz seen by the source.