advertisement

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: = 1 p. u. • The equivalent impedance is the sum of the transformer and line impedances: = = = 0.15 + 0.05 = 0.2 p. u. In the original phase domain the circuit is described by the following equations: = 2 − 0 0 Roberto Ferrero – ELEC 301/401 – 2018/2019 0 0 0 0 3 Step 2: Definition of fault Mathematical constraints representing the fault Definition of fault: • VY = VB = 0 • IR = 0 0 1 = 1 1 1 2 1 2 0 1 2 0 + 1 + 2 = 0 = 2 − = 2 − 2 1 1 + 2 2 + 0 0 = 0 = − = − 1 1 + 2 2 2 + 0 0 = 0 − 1 2 2 + 1 − 0 0 = 0 Roberto Ferrero – ELEC 301/401 – 2018/2019 2 2 = 0 0 4 Step 2: Definition of fault Impedance matrix in the sequence domain = 0 0 0 0 0.2 0 0 0 0.2 0 p. u. 0 = 0 0 0 0.2 • The impedance contributions arise from the transformer and line only: 1 = 2 = = = = 0.2 p. u. • The transformer is grounded on both sides and the current path through the ground has negligible resistance: 0 = 1 = 2 = 0.2 p. u. 0 0 0 1 = − 0 2 0 0 Roberto Ferrero – ELEC 301/401 – 2018/2019 0 1 0 0 0 2 0 1 2 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: = 1 p. u. 0 = 1 = 2 = 0.2 p. u. So the circuit can be solved: 1 = = −3.33 p. u. 1 + 2 0 / 2 +0 … and the fault current in the original domain can be calculated: 2 = + = 20 − 1 − 2 = 30 = −31 = 5 p. u. 2 +0 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: = 5 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: = 1 p. u. 1 = 2 = 0.2 p. u. So the circuit can be solved: 1 = = −2.5 p. u. 1 + 2 … and the fault current in the original domain can be calculated: = = 2 − 1 = −4.33 p. u. Roberto Ferrero – ELEC 301/401 – 2018/2019 9