Fault Current Analysis

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


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.



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.



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)



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)



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



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.



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.



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)



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)


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.


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)



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.


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)



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.


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)


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)



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.



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.



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.



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.



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



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)



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.


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.




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



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



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)


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)




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)




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



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



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



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.



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



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



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.


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

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Fault Current Analysis