-
123
6Fault Current Analysis
6.1 Introduction
Transformers should be designed to withstand various possible
faults, such as a short to ground of one or more phases. The high
currents accompany-ing these faults, approximately 1030 times
normal, produce high forces and stresses in the windings and
support structure. Also, depending on the fault duration,
significant amounts of heat may be generated inside the unit. The
design must accommodate the worst-case fault, which can occur from
either a mechanical or thermal standpoint.
The first step when designing transformers to withstand faults
is to deter-mine the fault currents in all the windings. Since this
is an electrical prob-lem, a circuit model that includes leakage
impedances of the transformer and relevant system impedances is
necessary. The systems are typically rep-resented by a voltage
source in series with an impedance or by just an imped-ance, since
we are not interested in detailed fault currents within the system,
external to the transformer. The transformer circuit model
considered in this chapter is that of a two- or three-terminal
per-phase unit with all pairs of ter-minal leakage reactances given
from either calculations or measurement. We ignore the core
excitation because its effects on the fault currents of modern
power transformers are negligible.
In this chapter we consider three-phase units and the following
fault types: three-phase line to ground, single-phase line to
ground, line-to-line, and double line to ground. These are the
standard fault types and are important because they are the most
likely to occur on actual systems. The transformer, or rather each
coil, must be designed to withstand the worst (i.e., the high-est
current) fault it can experience. Each fault type refers to a fault
on any of the single-phase terminals. For example, a three-phase
fault can occur on all the high-voltage (HV) terminals (H1, H2, and
H3), all the low-voltage (LV) terminals (X1, X2, and X3), or all
the tertiary-voltage terminals (Y1, Y2, and Y3), and so forth for
the other fault types. Faults on a single-phase system can be
considered three-phase faults on a three-phase system so that these
are auto-matically included in the analysis of faults on
three-phase systems.
2010 by Taylor and Francis Group, LLC
-
124 Transformer Design Principles
Since the fault types considered can produce unbalanced
conditions in a three-phase system, they can be treated most
efficiently by using symmetri-cal components. In this method, an
unbalanced set of voltages or currents can be represented
mathematically by sets of balanced voltages or currents, called
sequence voltages or currents. These can then be analyzed using
sequence circuit models. The final results are obtained by
transforming the voltages and currents from the sequence analysis
into the voltages and cur-rents of the real system. This method was
introduced in Chapter 5, and we will use those results here. For
easy reference, we will repeat the transforma-tion equations
between the normal phase abc system and the symmetrical component
012 system:
V
V
V
V
Va
b
c
a
a
=
1 1 111
2
2
0
11
2Va
(6.1)
V
V
V
a
a
a
0
1
2
2
2
13
1 1 111
=
VV
V
V
a
b
c
(6.2)
The circuit model calculations to be discussed are for
steady-state condi-tions, whereas actual faults would have a
transient phase in which the cur-rents can exceed their
steady-state values for short periods of time. These enhancement
effects are included using an asymmetry factor, which takes into
account the resistance and reactance present at the faulted
terminal and is considered to be conservative from a design point
of view. This enhance-ment factor was calculated in Chapter 2. The
following references are used in this chapter: [Ste62a], [Lyo37],
[Blu51].
6.2 Fault Current Analysis on Three-Phase Systems
We assume that the system is balanced before the fault occurs,
that is, any prefault (pf) voltages and currents are
positive-sequence sets. Here, we con-sider a general electrical
system as shown on the left side of Figure 6.1. The fault occurs at
some location on the system where fault phase currents Ia, Ib, and
Ic flow, which are shown in Figure 6.1 as leaving the system. The
volt-ages to ground at the fault point are labeled Va, Vb, and Vc.
The system, as viewed from the fault point or terminal, is modeled
using Thevenins theorem. First, however, we resolve the voltages
and currents into symmetrical compo-nents, so that we need only
analyze one phase of the positive-, negative-, and
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 125
zero-sequence sets. This is indicated on the right side of
Figure 6.1, where the a-phase sequence set has been singled
out.
In Thevenins theorem, each of the sequence systems can be
modeled as a voltage source in series with an impedance, where the
voltage source is the open-circuit voltage at the fault point and
the impedance is found by shorting all voltage sources and
measuring or calculating the impedance to ground at the fault
terminal. When applying Thevenins theorem, note that the fault
point together with the ground point is the terminal that is
associ-ated with the open-circuit voltage before the fault. When
the fault occurs, the load is typically set to zero but may include
a fault impedance. The resulting model is shown in Figure 6.2. No
voltage source is included in the negative- and zero-sequence
circuits because the standard voltage sources in power systems are
positive-sequence sources.
System a
Va
Ia
Phase a
System b
Vb
Ib
Phase b
System c
Vc
Ic
Phase c
System 1
Va1
Ia1
Positivesequence
System 2
Va2
Ia2
Negativesequence
System 0
Va0
Ia0
Zerosequence
Figure 6.1Fault at a point on a general electrical system.
(a)
Va2 Va0
Ia2
(b)
Ia0
(c)
Z1 Ia1
Va1E1
Z2 Z0
Figure 6.2Thevenin equivalent sequence circuit models: (a)
positive sequence, (b) negative sequence, and (c) zero
sequence.
2010 by Taylor and Francis Group, LLC
-
126 Transformer Design Principles
The circuit equations for Figure 6.2 are as follows:
V E I V I V Ia a a a a a1 1 1 1 2 2 2 0 0 0= = = Z Z Z, ,
(6.3)
Since E1 is the open-circuit voltage at the fault terminal, it
is the voltage at the fault point before the fault occurs and is
labeled Vpf. We omit the label 1 because it is understood to be a
positive-sequence voltage. We will also con-sider this a reference
phasor and use ordinary type for it. Thus
V I V I V Ia pf a a a a a1 1 1 2 2 2 0 0 0= = = V Z Z Z, ,
(6.4)
If there is some impedance in the fault, it can be included in
the circuit model. However, because we are interested in the
worst-case faults (highest fault currents), we assume that the
fault resistance is zero. If there is an imped-ance to ground at a
neutral point in the transformer, such as, for example, at the
junction of a Y-connected set of windings, then 3 its value should
be included in the single-phase zero-sequence network at that point
because all three zero-sequence currents flow into the neutral
resistor but only one of them is represented in the zero-sequence
circuit.
The types of faults considered and their voltage and current
constraints are shown in Figure 6.3.
System a
Va = 0
Ia
System b
Vb = 0
Ib
System c
Vc = 0
Ic
System a
Va = 0
Ia
System b
Vb
Ib = 0
System c
Vc
Ic = 0
System a
VaIa = 0
System b
VbIb
System c
VcIc = Ib
Vb = Vc Vb = Vc = 0
I + Ib c
(a) (b) (c) (d)
System a
VaIa = 0
System b
VbIb
System c
VcIc
Figure 6.3Standard fault types on three-phase systems: (a)
three-phase line-to-ground fault, (b) single-phase line-to-ground
fault, (c) line-to-line fault, and (d) double line-to-ground
fault.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 127
6.2.1 Three-Phase line-to-ground Fault
Three-phase faults to ground are characterized by
V V Va b c= = = 0 (6.5)
as shown in Figure 6.3a. From Equations 6.5 and 6.2, we find
V V Va a a0 1 2 0= = = (6.6)
Therefore, from Equation 6.4, we get
I I Iapf
a a11
2 0 0= = =V
Z, (6.7)
Using Equations 6.7 and 6.1 applied to currents, we find
I I I I I Ia a b a c c= = =1 2 1 1, , (6.8)
Thus, as expected, the fault currents form a balanced
positive-sequence set of magnitude Vpf/Z1. This example can be
carried out without the use of sym-metrical components because the
fault does not unbalance the system.
We can add the pf-balanced currents to the fault currents in the
abc system. Although this can be done generally for all the faults,
we will not explicitly do this for simplicity.
6.2.2 Single-Phase line-to-ground Fault
For a single-phase line-to-ground fault, we assume, without loss
of general-ity, that the a-phase is faulted. Thus
V I Ia b c= = =0 0, (6.9)
as indicated in Figure 6.3b. From Equation 6.2 applied to
currents, we get
I I II
a a aa
0 1 2 3= = = (6.10)
From Equations 6.9, 6.10, and 6.4, we find
V V V V Ia a a a pf a= + + = + +( ) =0 1 2 1 0 1 2 0V Z Z
Zor
I I Iapf
a a10 1 2
2 0= + +( ) = =V
Z Z Z (6.11)
2010 by Taylor and Francis Group, LLC
-
128 Transformer Design Principles
Using Equations 6.9 and 6.10, we get
I I Iapf
b c= + +( ) = =3
00 1 2
V
Z Z Z, (6.12)
We will keep both Z1 and Z2 in our formulas when convenient even
though for transformers Z1 = Z2.
Using Equations 6.1, 6.4, and 6.9 through 6.11, we can also find
the short-circuited phase voltages:
V Va bpf= =
+ +( ) +
+0 32
32
30 1 2
0 2, V
Z Z Zj Z j Z
=+ +( )
Vcpf
V
Z Z Zj Z
0 1 20
32
32
jj Z3 2
(6.13)
We can see that if all the sequence impedances were the same
then, in mag-nitude, Vb = Vc = Vpf. However, their phases would
differ.
6.2.3 line-to-line Fault
A line-to-line fault can be assumed, without loss of generality,
to occur between lines b and c, as shown in Figure 6.3c. The fault
equations are
V V I I Ib c a c b= = =, , 0 (6.14)
From Equation 6.2 applied to voltages and currents, we get
V V I I Ia a a a a1 2 0 2 10= = = , , (6.15)
Using Equations 6.4 and 6.15, we find
V V V V I
I
a a a pf a
apf
or0 1 2 1 1 2
11
0 0= = = = +( )
=+
, Z Z
V
Z Z222 0 0( ) = =I Ia a,
or
V V V V I
I
a a a pf a
apf
or0 1 2 1 1 2
11
0 0= = = = +( )
=+
, Z Z
V
Z Z222 0 0( ) = =I Ia a, (6.16)
Using Equation 6.1 applied to currents, Equations 6.15 and 6.16,
we get
I I Ia c pf b= = +( ) = 03
1 2
, j
Z ZV (6.17)
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 129
We can also determine the short-circuited phase voltages by the
aforemen-tioned methods.
V V VV
a pf b ca=
+
= = V ZZ Z
22
2
1 2
, (6.18)
The fault analysis does not involve the zero-sequence circuit,
that is, there are no zero-sequence currents. The fault currents
flow between the b and c phases. Also for transformers Z1 = Z2, so
that Va = Vpf.
6.2.4 Double line-to-ground Fault
The double line-to-ground fault, as shown in Figure 6.3d, can be
regarded as involving lines b and c without loss of generality. The
fault equations are
V V Ib c a= = =0 0, (6.19)
From Equations 6.19 and 6.2, we find
V V VV
I I Ia a aa
a a a0 1 2 0 1 230= = = + + =, (6.20)
Using Equations 6.4 and 6.20, we get
I I I V
V
a a apf
a
or
0 1 21
10 1 2
01 1 1+ + = = + +
V
Z Z Z Z
aa pf10 2
0 1 0 2 1 2
=+ +
VZ Z
Z Z Z Z Z Z
(6.21)
or
I I I V
V
a a apf
a
or
0 1 21
10 1 2
01 1 1+ + = = + +
V
Z Z Z Z
aa pf10 2
0 1 0 2 1 2
=+ +
VZ Z
Z Z Z Z Z Z
so that, from Equation 6.4:
I
I
a pf
a pf
10 2
0 1 0 2 1 2
20
= ++ +
=
VZ Z
Z Z Z Z Z Z
VZ
Z00 1 0 2 1 2
02
0 1 0 2 1
Z Z Z Z Z
VZ
Z Z Z Z Z Z
+ +
= + +
Ia pf22
(6.22)
Substituting Equation 6.22 into Equation 6.1 applied to
currents, we get
2010 by Taylor and Francis Group, LLC
-
130 Transformer Design Principles
I Ia b pf
3= =
+ +
+ +0
32
320 2
0 1 0 2 1
, Vj Z j Z
Z Z Z Z Z ZZ
V
j Z j Z
Z
2
0 232
32
=
Ic pf
3
00 1 0 2 1 2Z Z Z Z Z+ +
(6.23)
Using Equations 6.1, 6.4, and 6.22, the fault-phase voltages are
given by
V V Va pf b c= + +
= =V Z ZZ Z Z Z Z Z
300 2
0 1 0 2 1 2
, (6.24)
We see that if all the sequence impedances are equal, then Va =
Vpf.
6.3 Fault Currents for Transformers with Two Terminals per
Phase
A two-terminal transformer can be modeled by a single leakage
reactance zHX, where H and X indicate HV and LV terminals. All
electrical quantities from this point on will be taken to mean
per-unit quantities and will be writ-ten with small letters. This
will enable us to better describe transformers, using a single
circuit without the ideal transformer for each sequence.
The HV and LV systems external to the transformer are described
by system impedances zSH and zSX and voltage sources eSH and eSX.
In per-unit terms, the two voltage sources are the same. The
resulting sequence circuit models are shown in Figure 6.4.
Subscripts 0, 1, and 2 are used to label the sequence circuit
parameters, since they can differ, although the positive and
negative circuit parameters are equal for transformers but not
necessarily for the systems. A fault on the X terminal is shown in
Figure 6.4. The H ter-minal faults can be obtained by interchanging
subscripts. In addition to the voltage source on the H system, we
also see a voltage source attached to the X system. This can be a
reasonable simulation of the actual system, or it can simply be
regarded as a device for keeping the voltage va1 at the fault point
at the rated per-unit value. This also allows us to consider cases
in which zSX is small or zero in a limiting sense.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 131
In order to use the previously developed general results, we
need to compute the Thevenin impedances and pf voltage at the fault
point. From Figure 6.4, we see that
z
z z zz z z
zz z
11 1 1
1 1 12
2=+( )
+ +=SX HX SH
HX SH SX
SX H, XX SHHX SH SX
SX HX SH
2 2
2 2 2
00 0 0
+( )+ +
=+( )
zz z z
zz z zzHHX SH SX0 0 0+ +z z
(6.25)
and, since we are ignoring pf currents,
vpf SH= =e 1 1 (6.26)
where all the pf quantities are positive sequence. The pf
voltage at the fault point will be taken as the rated voltage of
the transformer and, in per-unit terms, is equal to one. Figure 6.4
and Equation 6.25 assume that both terminals are connected to the
HV and LV systems. The system
ia1
(a)
(b)
(c)
va1
eSX1eSH1
zSH1 zSX1zHX1
iHX1 iSX1
ia2
va2zSH2 zSX2zHX2
iHX2 iSX2
ia0
va0zSH0 zSX0zHX0
iHX0 iSX0
Figure 6.4Sequence circuits for a fault on the low-voltage X
terminal of a two-terminal per phase trans-former, using per-unit
quantities: (a) positive sequence, (b) negative sequence, and (c)
zero sequence.
2010 by Taylor and Francis Group, LLC
-
132 Transformer Design Principles
beyond the fault point is not to be considered attached, then
the Thevenin impedances become
z z z z z z z z z1 1 1 2 2 2 0 0 0= + = + = +HX SH HX SH HX SH,
, (6.27)
Using the computer, this can be accomplished by setting zSX1,
zSX2, and zSX0 to large (i.e., open circuited) values in Equation
6.25. In this case, all the fault cur-rent flows through the
transformer. Thus, considering the system not attached beyond the
fault point amounts to assuming that the faulted terminal is open
circuited before the fault occurs. In fact, the terminals
associated with all three phases of the faulted terminal or
terminals are open before the fault occurs. After the fault, one or
more terminals are grounded or connected depending on the fault
type. The other terminals, corresponding to the other phases,
remain open when the system is not connected beyond the fault
point.
The system impedances are sometimes set as zero. This increases
the severity of the fault and is sometimes required for design
purposes. It is not a problem mathematically for the system
impedances on the nonfaulted termi-nal. However, with zero system
impedances on the faulted terminal, Equation 6.25 shows that all
the Thevenin impedances will equal zero, unless we con-sider the
system not attached beyond the fault point and use Equation 6.27.
Therefore, in this case, if we want to consider the system attached
beyond the fault point, we must use very small system impedance
values. Continuity is thereby assured when transitioning from rated
system impedances to very small system impedances. Otherwise, a
discontinuity may occur in the fault current calculation if one
abruptly changes the circuit model from one with a system connected
to one not connected beyond the fault point just because the system
impedances become small.
Much simplification is achieved when the positive and negative
system impedances are equal, if they are zero, or if they are
small, since then z1 = z2 because zHX1 = zHX2 for transformers. We
will now calculate the sequence cur-rents for a general case in
which all the sequence impedances are different. We can then apply
Equation 6.1, expressed in terms of currents, to determine the
phase currents for the general case. However, in this last step, we
assume that the positive and negative impedances are the same. This
will simplify the formulas for the phase currents and is closer to
the real situation for transformers.
To obtain the currents in the transformer during the fault,
according to Figure 6.4 we need to find iHX1, iHX2, and iHX0 for
the standard faults. Since ia1, ia2, and ia0 have already been
obtained for the standard faults, we must find the transformer
currents in terms of these known fault currents. From Figure 6.4,
using Equation 6.26, we see that
v v i z z v i z za pf HX HX SH a HX HX2 SH21 1 1 1 2 2= +( ) =
+( , ))= +( )v i z za HX0 HX SH0 0 0
(6.28)
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 133
Substituting the per-unit version of Equations 6.4 into 6.28, we
get
i i
zz z
i iz
z zHX a HX SHHX a
HX S1 1
1
1 12 2
2
2
=+
=+
,HH
HX aHX SH
2
0 00
0 0
=+
i iz
z z
(6.29)
6.3.1 Three-Phase line-to-ground Fault
For this fault case, we substitute the per-unit version of
Equations 6.7 into 6.29 to get
iv
z zi iHX
pf
HX SHHX HX1
1 12 0 0= +( ) = =, (6.30)
Then, using Equation 6.1 applied to currents, we find
i i i i i iHXa HX HXb HX HXc HX= = =1 2 1 1, , (6.31)
that is, the fault currents in the transformer form a
positive-sequence set as expected.
6.3.2 Single-Phase line-to-ground Fault
For this type of fault, we substitute the per-unit versions of
Equation 6.11 into 6.29 to get
iv
z z zz
z z
iv
HXpf
HX SH
HX
10 1 2
1
1 1
2
=+ +( ) +( )
= ppfHX SH
HXpf
z z zz
z z
iv
z z
0 1 2
2
2 2
00 1
+ +( ) +
=+ ++( ) +
z
zz z2
0
0 0HX SH
(6.32)
If the system beyond the fault point is neglected, using
Equation 6.27, Equation 6.32 can be written as
iv
z z zi iHX
pfHX HX1
0 1 22 0= + +( ) = = (6.33)
2010 by Taylor and Francis Group, LLC
-
134 Transformer Design Principles
Assuming the positive and negative system impedances are equal,
as is the case for transformers or if they are small or zero, and
substituting Equation 6.32 into 6.1 applied to currents, we obtain
the phase currents
iv
z zz
z zz
z zHXapf
HX SH HX SH
2=+( ) +( ) + +(0 1
0
0 0
1
1 12 ))
=+( ) +( ) iv
z zz
z zz
zHXbpf
HX SH HX0 1
0
0 0
1
2 11 1
0 1
0
0 02
+( )
=+( ) +(
z
iv
z zz
z z
SH
HXcpf
HX SH ))
+( )
zz z
1
1 1HX SH
(6.34)
If we ignore the system beyond the fault point and substitute
Equation 6.27 into 6.34, then
iv
z zi iHXa
pfHXb HXc= +( ) = =
3
20
0 1
, (6.35)
This is also seen more directly using Equations 6.33 and 6.1.
This makes sense because, according to Equation 6.12, all of the
fault current flows through the transformer and none is shared with
the system side of the fault point.
If the system impedance beyond the fault point is not ignored,
fault current occurs in phases b and c inside the transformer even
though the fault is on phase a. These b and c fault currents are of
a lower magnitude than the phase a fault current.
6.3.3 line-to-line Fault
For this type of fault, we substitute the per-unit versions of
Equation 6.16 into 6.29 to get
iv
z zz
z z
iv
z
HXpf
HX SH
HXpf
H
11 1
1
1 2
2
=+( ) +
= XX SH
HX
2 2
2
1 2
0 0
+( ) +
=z
zz z
i
(6.36)
If we ignore the system beyond the fault point, this becomes
iv
z zi iHX
pfHX HX1
1 22 0 0= +( ) = = , (6.37)
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 135
For the phase currents, using Equation 6.1 applied to currents
and Equation 6.36 and assuming the positive and negative system
impedances are equal or zero, we get
i
i jv
z z
i jv
z
HXa
HXbpf
HX SH
HXcpf
HX
=
= +( )
=
0
32
32
1 1
11 1+( )zSH
(6.38)
If we ignore the system beyond the fault point, Equation 6.38
becomes
i i jv
ziHXa HXc
pfHXb= = = 0 3 2 1
, (6.39)
In this case, with the fault between phases b and c, phase a is
unaffected. Also, as we saw from Equations 6.36 and 6.37, no
zero-sequence currents are involved in this type of fault.
6.3.4 Double line-to-ground Fault
For this fault, we substitute Equation 6.22 expressed in
per-unit terms into Equation 6.29:
iv
z zz z z z
z z z z z zHXpf
HX SH1
1 1
0 1 1 2
0 1 0 2 1 2
=+( )
++ +
= +( ) + +i
v
z zz z
z z z z z zHXpf
HX SH2
2 2
0 2
0 1 0 2 1 22
00 0
0 2
0 1 0 2
=+( ) + +i
v
z zz z
z z z z zHXpf
HX SH
11 2z
(6.40)
If we ignore the system beyond the fault point, Equation 6.40
becomes
i vz z
z z z z z z
i vz
HX pf
HX pf
10 2
0 1 0 2 1 2
2
= ++ +
= 000 1 0 2 1 2
02
0 1 0 2
z z z z z z
i vz
z z z z
+ +
= + +HX pf zz z1 2
(6.41)
2010 by Taylor and Francis Group, LLC
-
136 Transformer Design Principles
Using Equation 6.1 applied to currents, and Equation 6.40 and
assuming equal positive and negative system impedances or zero
system impedances, we get for the phase currents
iv
z zz
z zz
z zHXapf
HX SH HX SH
=+( ) +( ) +( )2 0 1
1
1 1
0
0 0
=+( )
+( )i
v
z zz j z
z zHXbpf
HX SH23
0 1
21 0
1 1
+( )
=+( )
+
zz z
iv
z zz j
0
0 0
0 1
1
23
HX SH
HXcpf zz
z zz
z z0
1 1
0
0 0HX SH HX SH+( )
+( )
(6.42)
If we ignore the system beyond the fault point, using Equations
6.27 and 6.41, Equation 6.42 becomes
i
iv
z zj j
zz
HXa
HXbpf
=
=+( ) +
+
0
232
32
30 1
0
1
=+( ) +
+i
v
z zj j
zzHXc
pf
232
32
30 1
011
(6.43)
6.3.5 Zero-Sequence Circuits
Zero-sequence circuits require special consideration because
certain trans-former three-phase connections, such as the delta
connection, block the flow of zero-sequence currents at the
terminals and hence provide an essentially infinite impedance to
their passage; this is also true of the ungrounded Y connection.
This occurs because the zero-sequence currents, which are all in
phase, require a return path in order to flow. The delta connection
pro-vides an internal path for the flow of these currents,
circulating around the delta, but blocks their flow through the
external lines. These considerations do not apply to positive- or
negative-sequence currents, which sum to zero vectorially and so
require no return path.
For transformers, since the amp-turns must be balanced for each
sequence, in order for zero-sequence currents to be present they
must flow in both windings. For example in a grounded Y-delta unit,
zero-sequence currents can flow within the transformer but cannot
flow in the external circuit con-nected to the delta side.
Similarly, zero-sequence currents cannot flow in either winding if
one of them is an ungrounded Y.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 137
Figure 6.5 shows some examples of zero-sequence circuit diagrams
for different transformer connections. These can be compared with
Figure 6.4c, which applies when both windings have grounded Y
connections. Where a break in a line occurs in Figure 6.5, imagine
that an infinite impedance is inserted. Mathematically, we should
let the impedance approach infinity or a very large value as a
limiting process in the formulas.
No zero-sequence current can flow into the terminals for the
connection in Figure 6.5a because the fault is on the delta side
terminals. The Thevenin impedance, looking in from the fault point,
is z0 = zSX0. In this case, no zero-sequence current can flow in
the transformer and only flows in the external circuit on the LV
side.
We find for the connection in Figure 6.5b that z0 =
zHX0zSX0/(zHX0 + zSX0), that is, the parallel combination of zHX0
and zSX0. In this case, zero-sequence cur-rent flows in the
transformer, but no zero-sequence current flows into the system
impedance on the HV side.
In Figure 6.5c, because of the ungrounded Y connection, no
zero-sequence current flows in the transformer. This is true
regardless of which side of the transformer has an ungrounded Y
connection. In Figure 6.5d, no zero-sequence current can flow into
the transformer from the external circuit, and thus, no
zero-sequence current flows in the transformer.
(a)
(b)
(c)
(d)
zSH0 zHX0 zSX0
zSH0 zHX0 zSX0
zSH0 zHX0 zSX0
zSH0 zHX0 zSL0
Figure 6.5Some examples of zero-sequence impedance diagrams for
two-terminal transformers: (a) Yg/delta, (b) delta/Yg, (c) Yg/Yu,
and (d) delta/delta. The arrow indicates the fault point. Yg =
grounded Y, Yu = ungrounded Y.
2010 by Taylor and Francis Group, LLC
-
138 Transformer Design Principles
Another issue is the value of the zero-sequence impedances
themselves when they are fully in the circuit. These values tend to
differ from the positive-sequence impedances in transformers
because the magnetic flux patterns associated with them can be
quite different from the positive-sequence flux distribution. This
difference is taken into account by multi-plying factors that
multiply the positive-sequence impedances to produce the
zero-sequence values. For three-phase core form transformers, these
multiplying factors tend to be 0.85; however, they can differ for
dif-ferent three-phase connections and are usually found by
experimental measurements.
If there is an impedance between the neutral and ground, for
example at a Y-connected neutral, then three times this impedance
value should be added to the zero-sequence transformer impedance.
This is because identical zero-sequence current flows in the
neutral from all three phases, so in order to account for the
single-phase voltage drop across the neutral impedance, its value
must be increased by a factor of three in the single-phase circuit
diagram.
6.3.6 Numerical example for a Single line-to-ground Fault
As a numerical example, consider a single line-to-ground fault
on the X terminal of a 24-MVA three-phase transformer. Assume that
the H and X terminals have line-to-line voltages of 112 and 20 kV,
respectively, and that they are connected to delta and grounded-Y
windings. Let zHX1 = 10%, zHX0 = 8%, zSH1 = zSX1 = 0.01%, zSH0 =
0%, and zSX0 = 0.02%. The system impedance on the HV side is set to
zero because it is a delta winding. It is perhaps better to work in
per-unit values by dividing these imped-ances by 100, however if we
work with per-unit impedances in percent-ages, we must also set vpf
= 100%, assuming that it has its rated value. This will give us
fault currents per-unit (not percentages). Then, assum-ing the
system is connected beyond the fault point, we get from Equation
6.25, z1 = 9.99 103% and z2 = 1.995 102%. Solving for the currents
from Equation 6.34, we get iHXa = 11.24, iHXb = 3.746, and iHXc =
3.746. These cur-rents are per-unit values and are not percentages.
For a 45-MVA three-phase transformer with a HV delta-connected
winding with a line-to-line voltage of 112 kV, the base MVA per
phase is 15 and the base winding volt-age is 112 kV. Thus, the base
current is (15/112) 103 = 133.9 A. The fault currents on the HV
side of the transformer are IHXa = 1505 A, IHXb = 501.7 A, and IHXc
= 501.7 A.
If we assume that the system is not connected beyond the fault
point, then we get z1 = 10.01% and z0 = 8% from Equation 6.27, so
that from Equation 6.35, we get iHXa = 10.71 and IHXa = 1434 A. The
other phase currents are zero. This a-phase current is slightly
lower than the a-phase current, assuming the sys-tem is connected
beyond the fault point.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 139
6.4 Fault Currents for Transformers with Three Terminals per
Phase
A three-terminal transformer can be represented in terms of
three single-winding impedances. Figure 6.6 shows the sequence
circuits for such a trans-former, where H, X, and Y label the
transformer impedances and SH, SX, and SY label the associated
system impedances. Per-unit quantities are shown in Figure 6.6. The
systems are represented by impedances in series with voltage
sources. The positive sense of the currents is into the transformer
terminals. Although the fault is shown on the X-terminal, by
interchanging subscripts, the formulas that follow can apply to
faults on any terminal. We again label the sequence impedances with
subscripts 0, 1, and 2, even though the positive- and
negative-sequence impedances are equal for transformers. However,
they are not necessarily equal for the systems. The
zero-sequence
zSH1 zH1
zX1 zSX1
zY1 zSY1
iH1
iX1 iSX1Va1
ia1iY1
(a)
(b)
(c)
eSYeSH1
eSX
zSH2 zH2
zX2 zSX2
zY2 zSY2
iH2
iX2 iSX2Va2
ia2iY2
zSH0 zH0
zX0 zSX0
zY0 zSY0
iH0
iX0 iSX0Va0
ia0iY0
Figure 6.6Sequence circuits for a fault on the X terminal of a
three-terminal per phase transformer, using per-unit quantities:
(a) positive-sequence circuit, (b) negative-sequence circuit, and
(c) zero-sequence circuit.
2010 by Taylor and Francis Group, LLC
-
140 Transformer Design Principles
impedances usually differ from their positive-sequence
counterparts for transformers as well as for the systems.
We will calculate the sequence currents assuming possible
unequal sequence impedances; however, we will calculate the phase
currents assum-ing equal positive and negative system impedances.
This is true for trans-formers and transmission lines. This is also
true for cases in which small or zero system impedances are
assumed, a situation that is often a requirement in fault analysis.
Equation 6.1 can be applied to the sequence currents to get the
phase currents.
From Figure 6.6, looking into the circuits from the fault point,
the Thevenin impedances are
zz z wz z w
wz z
11 1 1
1 1 11
1 1=+( )
+ +=
+(SX XX SX
H SHwhere)) +( )
+ + +
=+(
z zz z z z
zz z w
Y SY
H Y SH SY
SX X
1 1
1 1 1 1
22 2 2 ))+ +
=+( ) +( )
z z ww
z z z zzX SX
H SH Y SY
H
where2 2 2
22 2 2 2
22 2 2 2
00 0 0
0 0 0
+ + +
=+( )
+ +
z z z
zz z wz z w
Y SH SY
SX X
X SX
whhere H SH Y SYH Y SH S
wz z z zz z z z0
0 0 0 0
0 0 0
=+( ) +( )+ + + YY0
(6.44)
If the system beyond the fault point is ignored, the Thevenin
impedances become
z z w z z w z z w1 1 1 2 2 2 0 0 0= + = + = +X X X, , (6.45)
where the ws remain the same. This amounts mathematically to set
zSX1, zSX2, and zSX0 equal to large values. When the system
impedances must be set to zero, then the system impedances must be
set to small values if we wish to consider the system connected
beyond the fault point, as was the case for two-terminal
transformers.
At this point, we will consider some common transformer
connections and conditions that require special consideration,
including changes in some of the above impedances. For example, if
one of the terminals is not loaded, this amounts to setting the
system impedance for that terminal to a large value. The large
value should be infinity, but since these calculations are usually
performed on a computer, we just need to make it large when
compared with the other impedances. If the unloaded terminal is the
faulted terminal, this results in changing z1 and z2 to the values
in Equation 6.45. In other words, it is equivalent to ignoring the
system beyond the fault point. If the unloaded terminal is an
unfaulted terminal, then w1 and w2 will be changed when zSH1, zSH2
or zSY1, zSY2 are set to large values. This amounts to having an
open circuit replace that part of the positive or negative system
circuit associated with the open terminal. If the open terminal
connection is a Y connection, then the
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 141
zero-sequence system impedance should also be set to a large
value and the zero-sequence circuit associated with the unloaded
terminal becomes an open circuit and w0 will change. However, if
the open circuit is delta connected, as for a buried delta
connection, then the zero-sequence system impedance should be set
to zero. This will also change w0 but only slightly, because
zero-sequence currents can circulate in delta-connected windings
even in an open terminal situation, whereas zero-sequence currents
cannot flow in Y-connected windings if the terminals are
unloaded.
Working in the per-unit system, the pf voltages are given by
v e e epf SH SX SY= = = =1 1 1 1 (6.46)
We ignore pf currents, which can always be added later. We
assume all the pf voltages are equal to their rated values or to
one in per-unit terms. We also have
i i iH X Y+ + = 0 (6.47)
Equation 6.47 applies to all the sequence currents. From Figure
6.6,
v e i z z i z e i z za SH H H SH X X SY Y Y1 1 1 1 1 1 1 1 1 1=
+( ) + = + SSY X Xa H H SH X X Y
1 1 1
2 2 2 2 2 2 2
( ) += +( ) + =
i z
v i z z i z i zz z i z
v i z z i zY SY X X
a H H SH X X
2 2 2 2
0 0 0 0 0 0
+( ) += +( ) + == +( ) +i z z i zY Y SY X X0 0 0 0 0
(6.48)
Solving Equation 6.48, together with Equations 6.44, 6.46, 6.47,
and 6.4 expressed in per-unit terms, we obtain the winding sequence
currents in terms of the fault sequence currents:
i iz
z ww
z zi iH a
X H SHH1 1
1
1 1
1
1 12= +
+
=, aaX H SH
H a
22
2 2
2
2 2
0 00
zz w
wz z
i iz
z
+
+
=XX H SH
X aX
0 0
0
0 0
1 11
1
+
+
= +
ww
z z
i iz
z w112 2
2
2 2
0 00
= +
=
, i iz
z w
i iz
z
X aX
X aX00 0
1 11
1 1
1
1 1
+
=+
+
w
i iz
z ww
z zY a X Y SY
=+
+
, i iz
z ww
z zY a X Y SY2 2
2
2 2
2
2 2
=+
+
i iz
z ww
z zY a X Y SY0 0
0
0 0
0
0 0
(6.49)
2010 by Taylor and Francis Group, LLC
-
142 Transformer Design Principles
For Equation 6.49, if we ignore the system beyond the fault
point and use Equation 6.45, we get
i iw
z zi i
wz zH a H SH1
H aH SH
1 11
12 2
2
2 2
=+
=+
,
=+
= =
,
,
i iw
z z
i i i
H aH SH
X a X
0 00
0 0
1 1 2 ii i i
i iw
z zi i
a2 X a
Y aY SY
Y a
,
,
0 0
1 11
1 12
=
=+
= 22 22 2
0 00
0 0
wz z
i iw
z zY SYY a
Y SY+
=+
,
(6.50)
We will use these equations, together with the fault current
equations, to obtain the currents in the transformer for the
various types of fault. Equation 6.1 applied to currents may be
used to obtain the phase currents in terms of the sequence
currents.
6.4.1 Three-Phase line-to-ground Fault
For this type of fault, we substitute the per-unit version of
Equations 6.7 into 6.49 to get
iv
z ww
z zi i
i
Hpf
X H SHH H
X
11 1
1
1 12 0 0= +( ) +
= =,
111 1
2 0
11 1
1
0=+( ) = =
=+( )
v
z wi i
iv
z ww
z
pf
XX X
Ypf
X
,
YY SYY Y
1 12 0 0+
= =z
i i,
(6.51)
If we ignore the system beyond the fault point, we get the same
sequence equations as Equation 6.51. Thus, the system beyond the
fault has no influ-ence on three-phase fault currents. The phase
currents corresponding to Equation 6.51 are
iv
z ww
z zi i iHa
pf
X H SHHb Ha= +( ) +
=1 1
1
1 1
2, , HHc Ha
Xapf
XXb Xa Xc Xa
=
= +( ) = =
i
iv
z wi i i i
1 1
2, , == 0
1 1
1
1 1
2iv
z ww
z zi iYa
pf
X Y SYYb Y= +( ) +
=, aa Yc Ya, i i= = 0
(6.52)
These form a positive-sequence set as expected for a three-phase
fault.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 143
6.4.2 Single-Phase line-to-ground Fault
For this fault, we substitute the per-unit version of Equation
6.11 into Equation 6.49 to get
iv
z z zz
z ww
z zHpf
X H SH1
0 1 2
1
1 1
1
1 1
=+ +( ) +
+
=+ +( ) +
+
iv
z z zz
z ww
z zHpf
X H2
0 1 2
2
2 2
2
2 SSH
Hpf
X
2
00 1 2
0
0 0
0
=+ +( ) +
iv
z z zz
z ww
zz z
iv
z z zz
z w
H SH
Xpf
X
0 0
10 1 2
1
1 1
+
= + +( ) +
= + +( ) +
=
iv
z z zz
z w
iv
Xpf
X
X
20 1 2
2
2 2
0ppf
X
Ypf
z z zz
z w
iv
z z z
0 1 2
0
0 0
10 1 2
+ +( ) +
=+ +( )) +
+
=+
zz w
wz z
iv
z z
1
1 1
1
1 1
20
X Y SY
Ypf
11 2
2
2 2
2
2 2
0
+( ) +
+
=
zz
z ww
z z
iv
X Y SY
Ypff
X Y SYz z zz
z ww
z z0 1 20
0 0
0
0 0+ +( ) +
+
(6.53)
We can simplify these equations by omitting the second term on
the right side of Equation 6.53, if the system beyond the fault
point is ignored:
iv
z z zw
z z
iv
z
Hpf
H SH
Hpf
10 1 2
1
1 1
20
=+ +( ) +
=+ zz z
wz z
iv
z z zw
1 2
2
2 2
00 1 2
0
+( ) +
=+ +( )
H SH
Hpf
zz z
iv
z z zi i
i
H SH
Xpf
X X
Y
0 0
10 1 2
2 0
+
= + +( ) = =
110 1 2
1
1 1
20
=+ +( ) +
=+
v
z z zw
z z
iv
z z
pf
Y SY
Ypf
11 2
2
2 2
00 1 2
0
+( ) +
=+ +( )
zw
z z
iv
z z zw
z
Y SY
Ypf
YY SY0 0+
z
2010 by Taylor and Francis Group, LLC
-
144 Transformer Design Principles
iv
z z zw
z z
iv
z
Hpf
H SH
Hpf
10 1 2
1
1 1
20
=+ +( ) +
=+ zz z
wz z
iv
z z zw
1 2
2
2 2
00 1 2
0
+( ) +
=+ +( )
H SH
Hpf
zz z
iv
z z zi i
i
H SH
Xpf
X X
Y
0 0
10 1 2
2 0
+
= + +( ) = =
110 1 2
1
1 1
20
=+ +( ) +
=+
v
z z zw
z z
iv
z z
pf
Y SY
Ypf
11 2
2
2 2
00 1 2
0
+( ) +
=+ +( )
zw
z z
iv
z z zw
z
Y SY
Ypf
YY SY0 0+
z
(6.54)
Using the per-unit version of Equation 6.1 and assuming equal
positive and negative system impedances, we can obtain the phase
currents from Equation 6.53:
iv
z zz
z ww
z zHapf
X H SH
=+( ) +
+0 1
1
1 1
1
1 122
+
+
+
zz w
wz z
i
0
0 0
0
0 0X H SH
Hbbpf
X H SH
=+( ) +
+
vz z
zz w
wz z0 1
1
1 1
1
1 12 +
+
+
zz w
wz z
i
0
0 0
0
0 0X H SH
Hc == +( ) +
+
v
z zz
z ww
z zpf
X H SH0 1
1
1 1
1
1 12 +
+
+
=
zz w
wz z
i
0
0 0
0
0 0X H SH
Xa +( ) +
+
+
v
z zz
z wz
z wpf
X X0 1
1
1 1
0
0 022
= +( ) +
+i
v
z zz
z wz
zXbpf
X X0 1
1
1 1
0
2 00 0
0 1
1
1 12
+
= +( ) +
w
iv
z zz
z wXcpf
X+
+
=+( )
zz w
iv
z z
0
0 0
0 122
X
Yapf zz
z ww
z zz
z w1
1 1
1
1 1
0
0 0X Y SY X+
+
+
+
+
=+( )
wz z
iv
z zz
0
0 0
0 12
Y SY
Ybpf 11
1 1
1
1 1
0
0 0z ww
z zz
z wX Y SY X+
+
+
+
+
=+( )
wz z
iv
z zz
0
0 0
0 1
1
2
Y SY
Ycpf
zz ww
z zz
z wX Y SY X1 11
1 1
0
0 0+
+
+
+
+
wz z
0
0 0Y SY
(6.55)
We have used the fact that 2 + = 1, along with w1 = w2 and z1 =
z2 to get Equation 6.55.
For cases in which the system beyond the fault point is ignored,
again, restricting ourselves to the case where the positive- and
negative-sequence impedances are equal, using Equation 6.54, the
phase currents are given by
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 145
iv
z zw
z zw
z zHapf
H SH H SH
=+( ) + + +
0 1
1
1 1
0
0 022
iiv
z zw
z zw
z zHbpf
H SH H SH
=+( ) + + +
0 1
1
1 1
0
0 02
=+( ) + + +
i
v
z zw
z zw
z zHcpf
H SH H SH0 1
1
1 1
0
0 02
= +( ) = =
=+
iv
z zi i
iv
z
Xapf
Xb Xc
Yapf
3
20 0
2
0 1
0
, ,
zzw
z zw
z z
iv
z
1
1
1 1
0
0 0
0
2( ) + + +
=+
Y SY Y SY
Ybpf
22 11
1 1
0
0 0
0
zw
z zw
z z
iv
z
( ) + + +
=
Y SY Y SY
Ycpf
++( ) + + +
2 1
1
1 1
0
0 0zw
z zw
z zY SY Y SY
(6.56)
6.4.3 line-to-line Fault
For this fault condition, we substitute the per-unit version of
Equation 6.16 into 6.49 to get
iv
z zz
z ww
z zHpf
X H SH1
1 2
1
1 1
1
1 1
=+( ) +
+
= +( ) +
+
iv
z zz
z ww
z zHpf
X H2 SH2
1 2
2
2 2
2
2
=
= +( ) +
=
i
iv
z zz
z w
i
H
Xpf
X
X
0
11 2
1
1 1
2
0
vv
z zz
z w
i
iv
z z
pf
X
X
Ypf
1 2
2
2 2
0
11 2
0
+( ) +
=
=+( )) +
+
= +
zz w
wz z
iv
z
1
1 1
1
1 1
21
X Y SY
Ypf
zzz
z ww
z z
i2
2
2 2
2
2 2
0 0( ) +
+
=X Y SY
Y
(6.57)
2010 by Taylor and Francis Group, LLC
-
146 Transformer Design Principles
If we ignore the system beyond the fault point, Equation 6.57
becomes
iv
z zw
z z
iv
z z
Hpf
H SH
Hpf
11 2
1
1 1
21 2
=+( ) +
= +(( ) +
=
= +( )
wz z
i
iv
z zi
2
2 2
0
11 2
2
0H SH
H
Xpf
X, == +( ) =
=+( ) +
v
z zi
iv
z zw
z z
pfX
Ypf
Y SY
1 20
11 2
1
1 1
0,
= +( ) +
=
iv
z zw
z z
i
Ypf
Y SY
Y
21 2
2
2 2
0 0
(6.58)
Assuming equal positive and negative system impedances and using
the per-unit version of Equation 6.1, we can obtain the phase
currents from Equation 6.57:
i i jv
z ww
z zHa Hbpf
X H SH
= = +( ) +
= 0 3
2 1 11
1 1
, ii
i i jv
z wi
i i
Hc
Xa Xbpf
XXc
Ya Yb
= =+( ) =
= =
03
2
0
1 1
,
, +( ) +
= j
v
z ww
z zi
32 1 1
1
1 1
pf
X Y SYYc
(6.59)
For cases in which we ignore the system beyond the fault point,
we can obtain the phase currents from Equation 6.58:
i i jv
zw
z zi
i
Ha Hbpf
H SHHc
Xa
= = +
= 0 3
2 11
1 1
,
== = =
= =
03
2
03
2
1
1
1
,
,
i jv
zi
i i jv
zw
z
Xbpf
Xc
Ya Ybpf
YY SYYc
1 1+
=
zi
(6.60)
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 147
6.4.4 Double line-to-ground Fault
For this fault, we substitute the per-unit version of Equation
6.22 into 6.49, to get
i vz z z
z ww
z zH pf X H SH1
0 2 1
1 1
1
1 1
= + +
+
= +
i vz z
z ww
zH pf X H2
0 2
2 2
2
2 ++
= +
z
i vz z
z w
SH
H pfX
2
02 0
0 0ww
z z
i vz z z
z
0
0 0
10 2 1
1
H SH
X pfX
+
= + + ww
i vz z
z w
i
1
20 2
2 2
0
= +
=
X pfX
X
vvz z
z w
i vz z
pfX
Y pf
2 0
0 0
10 2
+
= + +
+
=
zz w
wz z
i vz
1
1 1
1
1 1
2
X Y SY
Y pf 00 22 2
2
2 2
0
+
+
=
zz w
wz z
i
X Y SY
Y
+
+
vz z
z ww
z zpf X Y SY2 0
0 0
0
0 0
(6.61)
where = z z z z z z0 01 2 1 2+ + .For the system not attached
beyond the fault point, by using Equation 6.45,
Equation 6.61 becomes
i vz z w
z z
i v
H pfH SH
H pf
10 2 1
1 1
2
= + +
=
zz wz z
i vz
0 2
2 2
02
+
=
H SH
H pfww
z z
i vz z
i v
0
0 0
10 2
2
H SH
X pf X
+
= + =
, ppf X pf
Y pf
zi v
z
i vz z
00
2
10 2
=
= +
,
+
=
wz z
i vz w
z
1
1 1
20 2
Y SY
Y pfY 22 2
02 0
0 0
+
= +
z
i vz w
z z
SY
Y pfY SY
2010 by Taylor and Francis Group, LLC
-
148 Transformer Design Principles
i vz z w
z z
i v
H pfH SH
H pf
10 2 1
1 1
2
= + +
=
zz wz z
i vz
0 2
2 2
02
+
=
H SH
H pfww
z z
i vz z
i v
0
0 0
10 2
2
H SH
X pf X
+
= + =
, ppf X pf
Y pf
zi v
z
i vz z
00
2
10 2
=
= +
,
+
=
wz z
i vz w
z
1
1 1
20 2
Y SY
Y pfY 22 2
02 0
0 0
+
= +
z
i vz w
z z
SY
Y pfY SY
(6.62)
For cases in which the system is connected beyond the fault
point and we assume equal positive- and negative-sequence
impedances, we obtain the phase currents from Equations 6.61 and
6.1 applied to currents:
iv
z zz
z ww
z zHapf
X H SH
=+ +
+
2 0 1
1
1 1
1
1 1
+
+
=
zz w
wz z
iv
0
0 0
0
0 0X H SH
Hbppf
X H23
0 1
2 0
1
1
1 1
1
1z zj
zz
zz w
wz+
+
++
+
+z
zz w
wz zSH X H SH1
0
0 0
0
0 0
=+
+
+
iv
z zj
zz
zz wHc
pf
X23
0 1
0
1
1
1 1
+
+
wz z
zz w
wz
1
1 1
0
0 0
0
H SH X HH SH
Xapf
X
0 0
0 1
1
1 12
+
= + +
z
iv
z zz
z w
+
= +
zz w
iv
z zj
0
0 0
0 1
2
23
X
Xbpf zz
zz
z wz
z w0
1
1
1 1
0
0 0
+
+
X X
= +
+
+
iv
z zj
zz
zz wXc
pf
X23
0 1
0
1
1
1 1
+
=+
zz w
iv
z zz
z
0
0 0
0 1
1
12
X
Yapf
X ++
+
+
w
wz z
zz w
w
1
1
1 1
0
0 0
0
Y SY X zz z
iv
z zj
zz
Y SY
Ybpf
0 0
0 1
2 0
123
+
=+
+
+
zz w
wz z
z11 1
1
1 1
0
X Y SY zz ww
z z
iv
z
X Y SY
Ycpf
0 0
0
0 0
02
+
+
=+ zz
jzz
zz w
wz z1
0
1
1
1 1
1
1 1
3 +
+
+X Y SY
+
+
zz w
wz z
0
0 0
0
0 0X Y SY
(6.63)
For cases in which the system is not connected beyond the fault
point, Equation 6.63 becomes, using Equation 6.45,
i
v
z zw
z zw
z zHapf
H SH H SH
=+ +
+2 0 1
1
1 1
0
0 0
=+
iv
z zj
zz
wzHb
pf
H23
0 1
2 0
1
1
1
++
+
=
zw
z z
iv
SH H SH
Hcpf
1
0
0 0
2zz zj
zz
wz z
wz0 1
0
1
1
1 1
03+
+
+
H SH HH SH
Xa
Xbpf
0 0
0 1
2 0
0
23
+
=
= +
z
i
iv
z zj
zzz
iv
z zj
zz
1
0 1
0
1
1
23
= +
+
Xc
pf
=+ +
1
2 0 11
1 1
0iv
z zw
z zw
Yapf
Y SY zz z
iv
z zj
zz
Y SY
Ybpf
0 0
0 1
2 0
123
+
=+
+
+
wz z
wz z
1
1 1
0
0 0Y SY Y SY
=+
+
+
iv
z zj
zz
wz zYc
pf
Y SY23
0 1
0
1
1
1 1
+
wz z
0
0 0Y SY
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 149
iv
z zw
z zw
z zHapf
H SH H SH
=+ +
+2 0 1
1
1 1
0
0 0
=+
iv
z zj
zz
wzHb
pf
H23
0 1
2 0
1
1
1
++
+
=
zw
z z
iv
SH H SH
Hcpf
1
0
0 0
2zz zj
zz
wz z
wz0 1
0
1
1
1 1
03+
+
+
H SH HH SH
Xa
Xbpf
0 0
0 1
2 0
0
23
+
=
= +
z
i
iv
z zj
zzz
iv
z zj
zz
1
0 1
0
1
1
23
= +
+
Xc
pf
=+ +
1
2 0 11
1 1
0iv
z zw
z zw
Yapf
Y SY zz z
iv
z zj
zz
Y SY
Ybpf
0 0
0 1
2 0
123
+
=+
+
+
wz z
wz z
1
1 1
0
0 0Y SY Y SY
=+
+
+
iv
z zj
zz
wz zYc
pf
Y SY23
0 1
0
1
1
1 1
+
wz z
0
0 0Y SY
(6.64)
6.4.5 Zero-Sequence Circuits
Figure 6.7 lists some examples of zero-sequence circuits for
three-terminal transformers. An infinite impedance is represented
by a break in the cir-cuit. When substituting into the preceding
formulas, a large-enough value must be used. Although there are
many more possibilities than are shown in Figure 6.7, those shown
can serve to illustrate the method of accounting for the different
three-phase connections. The zero-sequence circuit in Figure 6.6c
represents a transformer with all grounded Y terminal
connections.
In Figure 6.7a, we have zX0 , so that no zero-sequence current
flows into the transformer. This can also be seen in Equation 6.49.
In Figure 6.7b, we see that zH0 since no zero-sequence current
flows into an ungrounded Y winding. This implies that w0 = zY0 +
zSY0, that is, the parallel combination of the H and Y impedances
is replaced by the Y impedances. In Figure 6.7c, no zero-sequence
current flows into or out of the terminals of a delta winding, so
the delta terminal system impedance must be set to zero. This
allows the current to circulate within the delta. In Figure 6.7d,
we see that zero-sequence current flows in all the windings but not
out of the HV delta winding termi-nals. We just need to set zSH0 =
0 in all the formulas, and this only affects w0.
6.4.6 Numerical examples
At this point, we will calculate the fault currents for a single
line-to-ground fault on the X terminal. This is a common type of
fault. Consider a grounded-Y (H), grounded-Y (X), and buried-delta
(Y) transformer, which has a three-phase MVA = 45 (15 MVA/phase).
The line-to-line terminal voltages are
2010 by Taylor and Francis Group, LLC
-
150 Transformer Design Principles
H: 125 kV, X: 20 kV, Y: 10 kV. The base winding currents are
therefore: IbH = 15/72.17 103 = 207.8 A, IbX = 15/11.55 103 =1299
A, and IbY = 15/10 103 = 1500 A. The winding-to-winding leakage
reactances are given in Table 6.1.
From these, we calculate the single-winding leakage reactances
as
z z z z
z z z z
H HX HY XY
X HX XY HY
= + ( ) =
= + ( ) =
12
11
12
1
%
%
zz z z z
z z z z
Y HY XY HX
H HX HY XY
= + ( ) =
= +
12
14
120 0 0 0
%
(( ) =
= + ( ) =
=
9 35
12
0 85
12
0 0 0 0
0
. %
. %z z z z
z
X HX XY HY
Y zz z zHY XY HX0 0 0 16 4+ ( ) = . %
(b)
(a)
(c)
(d)
zSH0
zSH0
zH0
zSH0
zSH0
zH0
zH0
zY0 zSY0
zSX0
zY0
zY0
zY0
zX0
zX0
zX0
zSY0
zSY0
zSY0
zSX0
zSX0
zSX0
Figure 6.7Some examples of zero-sequence circuit diagrams for
three-terminal transformers. The arrow indicates the fault point.
(a) Yg/Yu/Yg, (b) Yu/Yg/Yg, (c) Yg/Yg/delta, and (d) delta/Yg/Yg Yg
= grounded Y, Yu = ungrounded Y.
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 151
z z z z
z z z z
H HX HY XY
X HX XY HY
= + ( ) =
= + ( ) =
12
11
12
1
%
%
zz z z z
z z z z
Y HY XY HX
H HX HY XY
= + ( ) =
= +
12
14
120 0 0 0
%
(( ) =
= + ( ) =
=
9 35
12
0 85
12
0 0 0 0
0
. %
. %z z z z
z
X HX XY HY
Y zz z zHY XY HX0 0 0 16 4+ ( ) = . %
We drop the subscript 1 in these calculations for the
positive/negative reac-tances. We assume that the zero-sequence
reactances are 0.85 times the positive/negative reactances. This
factor can vary with the winding connec-tion and other factors and
is usually determined by experience. The zero-sequence reactances
can also be measured.
Because the Y winding is a buried delta, no positive- or
negative-sequence currents can flow into the terminals of this
winding. Therefore, it is neces-sary to set zSY to a large value in
the formulas. Zero-sequence current can circulate around the delta
so zSY0 is set to zero.
We do not want to consider system impedances in our
calculations. However, if the system is connected beyond the fault
point, we need to set the system impedances to very small values.
Considering a system not attached beyond the fault point amounts to
having the faulted terminal unloaded before the fault occurs. We
will perform the calculation both ways.
We will assume system impedances of
z z z
z zSH SX SY
SH SX
= = == =
0 001 1 000 0000 000 0
. %, , , %. 22 00%, %zSY =
Again, the large value for zSY and zero value for zSY0 are
necessary because it is a buried delta. We assume here that the
system zero-sequence reactances are a factor of two times the
system positive-or negative-sequence reactances. This is a
reasonable assumption for transmission lines because these
consti-tute a major part of the system [Ste62a]. Using these
values, from Equation 6.44, we get
Table 6.1
Leakage Reactances for a Yg, Yg, Delta Transformer
Winding 1 Winding 2Positive/Negative Leakage
Reactance (%)Zero-Sequence Leakage
Reactance (%)
H X 12 zHX 10.2 zHX0H Y 25 zHY 21.25 zHY0X Y 15 zXY 12.75
zXY0
2010 by Taylor and Francis Group, LLC
-
152 Transformer Design Principles
w w1 011 00088 5 23663= =. %, . %
For the system connected beyond the fault point, from Equation
6.44, we get
z z1 00 001 0 002= =. %, . %
and for the system not connected beyond the fault point, from
Equation 6.45, we get
z z1 012 00088 6 08663= =. %, . %
Using these values, along with the impedances in Table 6.1, we
can calculate the per-unit phase and phase currents in the
transformer from Equations 6.53 through 6.56 for the two cases.
Since we are working with percentages, we must set vpf = 100%.
These currents are given in Table 6.2 for the two cases.
The per-unit values in the Table 6.2 are not percentages. They
should sum to zero for each phase. These values would differ
somewhat if different mul-tiplying factors were used to get the
zero-sequence winding impedances
Table 6.2
Per-Unit and Phase Currents Compared for System Connected or Not
Connected Beyond the Fault Point and Small System Impedances
Phase A Phase B Phase C
System connected beyond fault
Per-unit currentsH 8.766 2.516 2.516X 12.380 6.130 6.130Y 3.614
3.614 3.614
Phase currentsH 1822 523 523X 16082 7963 7963Y 5422 5422
5422
System not connected beyond fault
Per-unit currentsH 8.508 1.463 1.463X 9.971 0 0Y 1.463 1.463
1.463
Phase currentsH 1768 304 304X 12952 0 0Y 2194 2194 2194
2010 by Taylor and Francis Group, LLC
-
Fault Current Analysis 153
The currents are somewhat lower for a system not connected
beyond the fault point compared with the currents for a system
connected beyond the fault point. In fact, the delta winding
current is significantly lower. These results do not change much as
the system impedances increase toward their rated values in the 1%
range. Thus, the discontinuity in the currents between the two
cases remains as the system sequence impedances increase. If the
system is considered connected while its impedance has its rated
value, it would be awkward to consider the system disconnected when
the system impedances approach zero.
These currents do not include the asymmetry factor, which
accounts for initial transient effects when the fault occurs.
6.5 Asymmetry Factor
A factor multiplying the currents calculated above is necessary
to account for a transient overshoot when the fault occurs. This
factor, called the asym-metry factor, was discussed in Chapter 2,
and is given by
Krx
= +
2 12
exp sin + (6.65)
where x is the reactance looking into the terminal, r is the
resistance, and = ( )tan 1 x r in radians. The system impedances
are usually ignored when calculating these quantities, so that for
a two-terminal unit
xr
zz
= ( )( )ImRe
HX
HX
(6.66)
and for a three-terminal unit with a fault on the X terminal
x zz z
z zr z= ( ) + ( ) ( )( ) + ( ) = ( )Im
Im ImIm Im
, ReXH Y
H YX ++
( ) ( )( ) + ( )
Re ReRe Re
z zz z
H Y
H Y
(6.67)
with corresponding expressions for the other terminals should
the fault be on them. When K in Equation 6.65 multiplies the rms
short-circuit current, it yields the maximum peak short-circuit
current.
2010 by Taylor and Francis Group, LLC
Chapter 6 Fault Current Analysis6.1 Introduction6.2 Fault
Current Analysis on Three- Phase Systems6.2.1 Three- Phase Line-
to- Ground Fault6.2.2 Single- Phase Line- to- Ground Fault6.2.3 L
ine- to- Line Fault6.2.4 Double Line- to- Ground Fault
6.3 Fault Currents for Transformers with Two Terminals per
Phase6.3.1 Three- Phase Line- to- Ground Fault6.3.2 Single- Phase
Line- to- Ground Fault6.3.3 L ine- to- Line Fault6.3.4 Double Line-
to- Ground Fault6.3.5 Zero- Sequence Circuits6.3.6 Numerical
Example for a Single Line- to- Ground Fault
6.4 Fault Currents for Transformers with Three Terminals per
Phase6.4.1 Three- Phase Line- to- Ground Fault6.4.2 Single- Phase
Line- to- Ground Fault6.4.3 Line- to- Line Fault6.4.4 Double Line-
to- Ground Fault6.4.5 Zero- Sequence Circuits6.4.6 Numerical
Examples
6.5 Asymmetry Factor