Example - Fault Analysis

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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:
 = 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
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