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ELEC 301/401 – Power Generation, Transmission and Distribution Example – Fault Analysis Dr Roberto Ferrero Email: [email protected] Telephone: 0151 7946613 Office: Room 506, EEE A block Roberto Ferrero – ELEC 301/401 – 2018/2019 1 Unbalanced fault analysis Problem definition A three-phase line is connected to an infinite busbar through a star-starconnected transformer with grounded neutrals on both sides (primary side: busbar – secondary side: line). The equivalent transformer and line impedances for each phase are j0.15 p.u. and j0.05 p.u., respectively, and any mutual inductive coupling between phases is neglected. Also, the zero-sequence impedances of both transformer and line can be assumed to be equal to the corresponding positive- and negativesequence impedances. • Calculate the fault current produced by a double-line-to-earth short circuit, assuming that no load is connected to the line. • How would the fault current change if the transformer primary circuit were delta-connected? And if, on the contrary, the secondary circuit were deltaconnected? Roberto Ferrero – ELEC 301/401 – 2018/2019 2 Step 1: Equivalent circuit Thevenin equivalent circuit in the point of fault • The network is operated at rated voltage: 𝐄 = 1 p. u. • The equivalent impedance is the sum of the transformer and line impedances: 𝐙𝑅 = 𝐙𝑌 = 𝐙𝐵 = 𝑗0.15 + 𝑗0.05 = 𝑗0.2 p. u. In the original phase domain the circuit is described by the following equations: 𝐕𝑅 𝐙𝑅 𝐄 𝐕𝑌 = 𝐚2 𝐄 − 0 𝐕𝐵 0 𝐚𝐄 Roberto Ferrero – ELEC 301/401 – 2018/2019 0 𝐙𝑌 0 0 0 𝐙𝐵 𝐈𝑅 𝐈𝑌 𝐈𝐵 3 Step 2: Definition of fault Mathematical constraints representing the fault Definition of fault: • VY = VB = 0 • IR = 0 0 1 𝐈𝑌 = 1 𝐈𝐵 1 1 𝐚2 𝐚 1 𝐚 𝐚2 𝐈0 𝐈1 𝐈2 𝐈0 + 𝐈1 + 𝐈2 = 0 𝐕𝑌 = 𝐚2 𝐄 − 𝐙𝑌 𝐈𝑌 = 𝐚2 𝐄 − 𝐚2 𝐙1 𝐈1 + 𝐚𝐙2 𝐈2 + 𝐙0 𝐈0 = 0 𝐕𝐵 = 𝐚𝐄 − 𝐙𝐵 𝐈𝐵 = 𝐚𝐄 − 𝐚𝐙1 𝐈1 + 𝐚2 𝐙2 𝐈2 + 𝐙0 𝐈0 = 0 𝐚 − 1 𝐙2 𝐈2 + 1 − 𝐚 𝐙0 𝐈0 = 0 Roberto Ferrero – ELEC 301/401 – 2018/2019 𝐙2 𝐈2 = 𝐙0 𝐈0 4 Step 2: Definition of fault Impedance matrix in the sequence domain 𝐙𝑅𝑌𝐵 𝐙𝑅 = 0 0 0 𝐙𝑌 0 𝑗0.2 0 0 0 𝑗0.2 0 p. u. 0 = 0 𝐙𝐵 0 0 𝑗0.2 • The impedance contributions arise from the transformer and line only: 𝐙1 = 𝐙2 = 𝐙𝑅 = 𝐙𝑌 = 𝐙𝐵 = 𝑗0.2 p. u. • The transformer is grounded on both sides and the current path through the ground has negligible resistance: 𝐙0 = 𝐙1 = 𝐙2 = 𝑗0.2 p. u. 𝐙0 𝐕0 0 𝐕1 = 𝐄 − 0 𝐕2 0 0 Roberto Ferrero – ELEC 301/401 – 2018/2019 0 𝐙1 0 0 0 𝐙2 𝐈0 𝐈1 𝐈2 5 Step 3: Fault current calculation Solution in the sequence domain and transformation in the original domain The voltage source and the impedances in the equivalent circuit are known: 𝐄 = 1 p. u. 𝐙0 = 𝐙1 = 𝐙2 = 𝑗0.2 p. u. So the circuit can be solved: 𝐈1 = 𝐄 = −𝑗3.33 p. u. 𝐙1 + 𝐙2 𝐙0 / 𝐙2 +𝐙0 … and the fault current in the original domain can be calculated: 𝐙2 𝐈𝑓 = 𝐈𝑌 + 𝐈𝐵 = 2𝐈0 − 𝐈1 − 𝐈2 = 3𝐈0 = −3𝐈1 = 𝑗5 p. u. 𝐙2 +𝐙0 Roberto Ferrero – ELEC 301/401 – 2018/2019 6 Effects of the transformer connection Delta-star connection A zero-sequence current can flow in the secondary circuit of the transformer because of its connection to earth and because a corresponding zerosequence current can flow within the delta in the primary circuit. • The equivalent circuit representing the fault in the sequence domain is the same as in the previous case: The fault current is unchanged: 𝐈𝑓 = 𝑗5 p. u. Roberto Ferrero – ELEC 301/401 – 2018/2019 7 Effects of the transformer connection Star-delta connection (1) In this case, zero-sequence currents can flow in the primary circuit and within the delta of the secondary circuit, but not outside it, so not in the line. • The zero-sequence circuit is open: The fault is now a line-to-line short circuit, without any current flowing through the ground. Roberto Ferrero – ELEC 301/401 – 2018/2019 8 Effects of the transformer connection Star-delta connection (2) The circuit parameters are the same as in the previous case, except for Z0: 𝐄 = 1 p. u. 𝐙1 = 𝐙2 = 𝑗0.2 p. u. So the circuit can be solved: 𝐈1 = 𝐄 = −𝑗2.5 p. u. 𝐙1 + 𝐙2 … and the fault current in the original domain can be calculated: 𝐈𝑓 = 𝐈𝑌 = 𝐚2 − 𝐚 𝐈1 = −4.33 p. u. Roberto Ferrero – ELEC 301/401 – 2018/2019 9