analyzing of relevant fault type for calculation of

1IntroductionUvod

In many text books and papers such as [1]-[5], it ispossible to find descriptions about shunt fault calculationsand about the influence of the three-phase transformerconnection Dyn on the fault currents distribution throughdistribution transformers. In [6] is described a method fordigitally simulating a three-phase power system which issubjected to a fault at one location with blown fuse or openconductors at another location. The method is applicablewhen the simultaneous unbalances are on opposite sides of adelta-grounded star transformer. From these books andpaper it can be concluded, that in the common case of a deltaprimary-grounded star secondary connection (Dyn), thecurrents on the source side of the transformer differ from thecorresponding currents on the fault side for the unbalancedsecondary shunt fault and for the simultaneous fault.

This paper analyzes influence of three-phasetransformer connections (Dyn5 and Yzn5) and several typeshort circuits and simultaneous faults in determining ofrelevant fault type for calculation of minimum fault currentmagnitude on the distribution transformer primary side,which will be relevant for the selection of fuse-linksminimum breaking current of high-voltage fuses fortransformer circuit applications. In the carried out analysis itis assumed, that a secondary short circuit (shunt) faultoccurs between the transformer terminals and the low-voltage side protective device and that the fault must becleared by the high-voltage side fuses, which are used astransformer primary-side protective devices (incomingprotection). In this paper it is considered that a simultaneousfault represents a multiple fault which involves at the same

3

V Boras, T Bari , A Muharemovi. . ć . ć

ISSN 1330-3651

UDC/UDK 621.314 : 621.316.933

ANALYZING OF RELEVANT FAULT TYPE FOR CALCULATION OF MINIMUM FAULTCURRENT MAGNITUDE ON DISTRIBUTION TRANSFORMER PRIMARY SIDE

Vedran Boras, Tomislav Bari , Alija Muharemović ć

The paper analyses the influence of three-phase transformer connections and several types of bolted shunt faults (short circuits), which occur on a distributiontransformer secondary side as well as simultaneous faults in determining of relevant fault type for calculation of minimum fault current magnitude on thetransformer primary side, which will be relevant for the selection of fuse-links minimum breaking current of high-voltage fuses for transformer circuitapplications. The two transformer connections: delta primary-grounded star secondary (denoted as Dyn) and star primary-grounded interconnected starsecondary, also known as zig-zag (denoted as Yzn) are enclosed in the carried out analysis. In this paper a simultaneous fault involves the shunt fault on thetransformer secondary side with blown fuse in any phase on the transformer primary side. Calculation of these faults is carried out by means of the symmetricalcomponents method in per-unit.

Keywords: fuse-links minimum breaking current, high-voltage fuses, shunt fault, simultaneous fault, three-phase transformer connection

Original scientific paper

U ovom radu je analiziran utjecaj spoja namota trofaznog transformatora i više tipova izravnih kratkih spojeva, koji se pojavljuju na sekundarnoj strani nekogdistribucijskog minimalnog iznosa struje kvara na primarnoj stranitransformatora, koji a

na primaru - uzemljena zvijezda na sekundaru ( u oznaci Dyn) i zvijezda na primaru –uzemljena izlomljena zvijezda na sekundaru poznata i pod nazivom kao cik-cak (u oznaci Yzn). U ovom radu simultani kvar obuhvata kratki spoj nasekundarnoj strani transformatora uz istodobno pregorjeli osig

transformatora kao i simultanih kvarova u određivanju mjerodavnog tipa kvara za proračunće biti mjerodav n u odabiru minimalne prekidne struje topljivih umetaka visokonaponskih osigurača za zaštitu transformatora. Ovom

analizom obuhvaćene su dvije grupe spoja namota transformatora: trokut

urač u jednoj od faza na primarnoj strani transformatora. Proračun ovih kvarova izveden jeprimjenom metode simetričnih komponenti u sustavu jediničnih vrijednosti.

Ključne riječi: minimalna prekidna struja topljivog umetka, , spoj namota trofaznog transformatora,poprečni kvar, simultani kvar visokonaponski osigurači

Izvorni znanstveni članak

Analiziranje mjerodavnog tipa kvara za proračun iznosa minimalne struje kvarana primarnoj strani distribucijskog transformatora

Analiziranje mjerodavnog tipa kvara za proračun iznosa minimalne struje kvara na primarnoj strani distribucijskog transformatora

Technical Gazette 17, (2010), 31 -15

time a secondary shunt fault and any blown fuse on thetransformer primary side. In [7] is noted that thesimultaneous fault can lead to a situation where electricalequipment may be overstressed due to the short-circuitduration and thus analyzing of such fault type can be veryuseful.

According to the IEC Standards [8]-[9], one of ratingswhich should be specified relating to the protection of aHV/LV transformer circuit is minimum breaking current.This current is defined as minimum value of prospectivecurrent that a fuse-link is capable of breaking at a statedvoltage under prescribed conditions of use and behaviour.According to [8], fuse-links should be selected so that thevalue of minimum breaking current is appropriate to theparticular application concerned. It should be stressed thatthe use of a fuse-link having too high a value of minimumbreaking current could, under certain circumstances, resultin disruptive failure of the fuse-link and consequentdamage.

There are two possible cases:a) For applications where low fault levels are unlikely tooccur, suitable Back-Up fuses may be used. In this case, it isnecessary to ascertain that the rated minimum breakingcurrent of the fuse-link is less than the smallest short-circuitcurrent likely to appear upstream of the low-voltageprotecting device (referred to high-voltage side).b) For applications where very low values of fault currentoccur on the distribution system, the fuse should haveminimum breaking current, which is as low as possible.Thus, general purpose fuses should be used for suchapplications.

Only for the sake of determining the relevant type offault current, in this paper are made some assumptions,which considerably simplify the longhand calculations.

When the relevant type of fault current is selected, the realminimum fault current magnitude should be determinedwithout these assumptions, taking into account all relevantand required parameters (for example equivalentdistribution system impedance etc ). Like this fault currentmagnitude can be used by protection engineers to assess thetime-current characteristics of high-voltage fuses asincoming protection on the high-voltage side of adistribution transformer.

This paper focuses on three-phase small liquid-immersed distribution transformers with three-legged core-form construction.

The method presented in this paper employssymmetrical component techniques to represent the shuntfaults and the faults with blown fuse conditions. A numberof methods, which use the method of symmetricalcomponents for computing simultaneous faults involvingtwo faults, have been studied in [1]-[4]. In this paper, eachsequence network is described mathematically inappendices in accordance with Kirch off's current andvoltage law, and the interconnections between the sequencenetworks at the unbalance locations are expressed in theform of constraint equations. For the calculation of theunknown sequence current and voltages duringsimultaneous faults, linear matrix equations are presented inappendices.

This paragraph analyses bolted three-phase fault,bolted phase-to-phase fault and bolted phase-to neutral faultat the fault point F on the secondary side of a Dyn5 or of anYzn5 transformer in the vicinity of the transformerterminals, as shown in Fig. 1.

, .

h

2Shunt faults (short circuits)P ratki spojevi)oprečni kvarovi (k

4

Analyzing of relevant fault type for calculation of minimum fault current magnitude on distribution transformer primary side V Boras, T Bari , A Muharemovi. . ć . ć

Tehni ki vjesnikč , 317, 1(2010) -15

positione) In the carried out per-unit analysis, the transformervoltage ratio is assumed 1:1, i.e. it will be for by aDyn5 transformer and for by an Yzn5 transformer.f) Conversion of the transformer impedance magnitude toper-unit value, i.e. will be made on a base power

. Phase angle can be easy tocalculate from data designated by the transformermanufacturer and it lies in range from 47° to 78°. Infollowing examples will be used =60° and correspondingratioh) The neutral conductor impedance from transformerterminal to fault location is neglected.I) Zero-sequence reactances of three-phase smalldistribution Dyn5 and Yzn5 transformers with three-leggedcore-form construction are lower than their correspondingpositive-sequence reactances and usually are equal:- for Dy transformers and

forYz transformers.j) Zero-sequence resistances of three-phase smalldistribution Dyn5 and Yzn5 transformers with three-leggedcore-form construction are:

for Dy transformers andforYz transformers.

k) Zero-sequence current is equal to zero, because thedelta connection on the high-voltage side of the Dyn5transformer or the ungrounded star connection on the high-voltage side of theYzn5 transformer does not allow the flowof zero-sequence current.

In many text books it is possible to find descriptionabout shunt fault calculations by means of the method ofsymmetrical components. Thus, in this paragraph we willuse only final equations of analyzed shunt faults.

Let us consider a bolted three-phase fault at the faultpoint F on the secondary side of a Dyn5 or of an Yzn5transformer, as shown in Fig. 1. The method of symmetricalcomponents will allow us to express the bolted three-phasefault current as follows:

.

-

or .g) In carried fault current calculations it is necessary totake into account both the resistance and reactance values ofsmall distribution transformer impedance. Carried outanalysis will be simplified by using the polar representationof the impedance

.

-

--

φ

φ

2.1Bolted three-phase faultIzravni tropolni kratki spoj

Figure 1Slika 1.

Transformer secondary-side shunt faults (short circuits)Poprečni kvarovi (kratki spojevi) na sekundaru transformatora

The basic assumptions, which considerably simplifythe longhand calculations and comparison, are made:a) The fault currents calculation will be made using per-unit method under no-load conditions neglecting all shuntreactances (loads, charging and magnetizing reactances)and neglecting all mutual reactancesb) Only for the sake of practical examples in this paper, theequivalent distribution system impedance up to transformermedium voltage bus will be neglected. Thus, the shortcircuit currents will be calculated on the transformer low-voltage side in the vicinity of the transformer terminalsassuming that primary winding is connected to an infinitivebus (a system with zero impedance,c) The source voltage will be assumed to be 1,0 for faultcalculations using per-unit method.d) The transformers will be analyzed on nominal tap

.

).pupuZ S 0j01 ��

3�n23�n

puZ T 11 �

kTb uSS � knSCC uII �_3

�� TT ZZ

577,0�TT XR

1T0T XX 0,95�1T0T XX 0,1�

1T0T RR �1T0T R4,0R �

0AI

0.;0; 0211

1_3 ��

�� aaTS

aSCC IIZZ

VII (1)

From (1) it can be concluded that only the positive-sequence network is involved in the three-phase fault.

Substituting above mentioned values band in (1) we get the

following:

y ,pujV 1�puZT �� 6011 pujZS 01 �

pu.II _SCCa ��

�� 301601

90131 (2)

Substituting the values of and from (1) and(2) in (A6) we obtain secondary line (winding) currents.

21, aa II a0I

The sequence currents flowing in the leads to the deltawinding of the Dyn5 transformer are given in relation to thepositive- and negative-sequence secondary line currents(i.e. star winding currents) by means of matrix equation(A9) and (A10). Substituting these in (A7) we obtain Dyn5transformer primary line currents. Similarly, the sequencecurrents flowing in the leads to the star winding of the Yzn5transformer are given in relation to the sequence secondaryline currents (i.e. interconnected star winding currents) bymeans of matrix equation (A17) and (A18). Substitutingthese in (A7) we obtain Yzn5 transformer primary linecurrents. The fault analysis results, for the bolted three-phase fault on the secondary side of a Dyn5 or an Yzn5transformer, are shown in Table 1.

5Technical Gazette , 317, 1(2010) -15

Thus, we can conclude that the bolted phase-to-phasefault is 0,866 of the bolted three-phase fault.

In the same way as in the case of the bolted three-phasefault, we can determine primary line currents of a Dyn5 or ofanYzn5 transformer for this fault type.

The fault analysis results, for the bolted phase to -phase fault on the secondary side of a Dyn5 or of an Yzn5transformer, are shown in Table 2.

b -c

V Boras, T Bari , A Muharemovi. . ć . ć Analiziranje mjerodavnog tipa kvara za proračun iznosa minimalne struje kvara na primarnoj strani distribucijskog transformatora

In the event o having and ,and substituting this and the basic assumptionsin (4) we get the following:

f

Table 1Tablica 1

Three-phase fault currents analysisAnaliza struja trofaznog kratkog spoja.

Transformer

connections

Primary line

currents (pu)

Secondary line

currents (pu)

�� 1800,1AI �� 300,1aI

�� 600,1BI �� 2700,1bIDyn5

Yzn5�� 3000,1CI �� 1500,1cI

Table 1 shows, that for a three-phase secondary fault,the per-unit currents on the primary equal those on thesecondary (with the actual currents related by the turns ratioof the transformer). The currents on the primary lead to thecorresponding currents on the secondary by 150°.

Let us consider a bolted phase-to-phase fault, involvingphases and , at a fault point F on the secondary side of aDyn5 or of anYzn5 transformer, as shown in Fig. 1.

The method of symmetrical components will allow usto express the bolted phase -to-phase current as follows:

2.2Bolted phase-to-phase faultIzravni dvopolni kratki spoj

b c

b c

� � 11

22

21

112

212

21a

3

3

0I

aaaac

aaaab

aa

IjIaaIaIaI

IjIaaIaIaI

II

��

��

��

(3)

where

0.22

011

21 ��

�� a

TS

aa I;ZZ

VII (4)

pujV 1� puZT �� 6011

pujZS 01 �

��

�� 305,0602

90121 aa II (5)

Substituting (5) in (3), we will get the following:

pu.I

puI

puI

c

b

a

1202

3

602

3

0

��

��

�

(6)

Table 2Tablica 2.

The phase b-to-phase c fault current analysisAnaliza dvopolnog kratkog spoja (faze b-c)

Transformer

connections

Primary line

currents (pu)

Secondary line

currents (pu)

�� 1205,0AI 0,0�aI

�� 1205,0BI �� 60867,0bIDyn5

Yzn5�� 1200,1CI �� 120867,0cI

It can be seen from Table 2 that bolted phase – to-phase secondary fault produce one large primary linecurrent and two smaller primary line currents by bothtransformer connections. These reduced primary linecurrents can be a problem in protective relaying. The largerprimary line current is approximately 16 % higher than theper-unit secondary line current. If fuse blows in the phasewith larger primary line current, then fault will be cleared.The two smaller primary line fault currents are less than thesecondary line current. Analyzing these results it can beconcluded, that secondary of an Yzn5 transformer, whenseen from transformer primary side, behaves as deltaconnection.

Let us consider a bolted single phase to-neutral fault atthe fault point F on the secondary side of a Dyn5 or of anYzn5 transformer, as shown in Fig. 1. In this faultcalculation, the faulted phase is assumed to be phase ,because of the comparison with results of followingsimultaneous fault analysis. In the event of havingand the technique of symmetrical componentswill allow us to express the fault currents as follows:

bc

-

b

2.3Bolted phase-to-neutral faultIzravni jednopolni kratki spoj

,21 SS ZZ �

21 TT ZZ �

011

2

12

0212

22

33

0,0

TTSab

aaa

ca

ZZZ

VaIaI

IIaIa

II

��

��

��

(7)

If we substitute the basic assumption in (7), wecan derive from this expression the following:

pu01 �SZ

1

01

2

2

3

T

TTb

Z

ZZ

VaI

��

(8)

,

.

.

.

6

This expression shows that the bolted single phase to-neutral fault current on line is the function of: the three-phase bolted fault current and the ratio of the zero-sequence impedance with the positive-sequenceimpedance.

Using assumptions under bullets i) and j) it is easy toobtain the following expression for a Dyn5 transformer:

-b

The fault analysis results, for the bolted phase to -neutral fault on the secondary side of a Dyn5 or of an Yzn5transformer, are shown in Table 3.

Therefore, it can be concluded that the bolted singlephase–to-neutral fault currents on the secondary side of aDyn5 or of an Yzn5 transformer can reach approximately1,013 or 1,377 times the bolted three-phase fault currentvalue, respectively.

Table 3 shows clearly that the bolted single phase-to-neutral fault on secondary side of a Dyn5 or of an Yzn5transformer produces the per unit primary-side line currentsless than the per-unit secondary line currents.

Thus, it can be concluded that the transformer will notbe well protected in this case; because primary-side fuses inphases and take longer to clear the bolted phase-to-neutral fault on secondary.

In this paragraph have been analysed simultaneousfaults which involve single blown fuse on the primary sideof a Dyn5 or of an Yzn5 transformer with one of thefollowing shunt faults on the secondary:- bolted three-phase fault,- any bolted double phase-to-neutral and- bolted phase -to-neutral.

In [7], it is noted that simultaneous faults which involveany phase-to-phase faults without neutral connection on thetransformer secondary side with blown fuse in phase onthe transformer primary side cause merely low currents andthus it will not be taken into consideration in this paper.

In this paragraph are used the same basic assumptionsas in paragraph .

Let us consider a simultaneous fault, which involves abolted three-phase fault at the fault point F located on thesecondary side of a Dyn5 or of an Yzn5 transformer, asshown in Fig. 2 and Fig. 3, with blown fuse in phase on thetransformer primary-side.

When medium-voltage phase is the reference phase,solving the linear matrix equation (B4) by using the basicassumptions stated above we get for this fault type thefollowing positive- and negative-sequence currents:

b -

B C

b

A

A

A

3Simultaneous faults

3.1Bolted three-phase fault on the transformersecondary-side with blown fuse in phase

Simultani kvarovi

Izravni tropolni kratki spoj na sekundarnoj stranitransformatora

2

A

s pregorjelim osiguračem u fazi A

Analyzing of relevant fault type for calculation of minimum fault current magnitude on distribution transformer primary side

Tehni ki vjesnikč , 317, 1(2010) -15

V Boras, T Bari , A Muharemovi. . ć . ć

1TZV

��

��

��

��

��

��

1

12

1

11

0 1

1

05,01

T

T

T

TT

T

X

Rj

X

RZ

Z

(9)

Analogously, the following expression can be obtainedfor anYzn5 transformer:

��

��

��

��

��

��

1

12

1

11

0 1

1

3,04,0

T

T

T

TT

T

X

Rj

X

RZ

Z

(10)

,,

Substituting (9) and (10) in (8) can be obtained valuesof the bolted single phase-to-neutral fault current on line ,for a Dyn5 and anYzn5 transformer respectively.

In the event of havingphase angle =60°, i.e. , and

substituting these values in (9) and afterwards in (8) we willget the fault current for phase of a Dyn5 transformer, asfollows:

b

b

φpuZpujV T �� 601,1 1

pujZ S 01 � 577,011 �TT XR

23 1 301,0126 270

2,9625 0,0216b

aI pu.

j

� � ��

�(1 )1

,, ,

Analogously, substituting the valuesphase angle =60°, i.e.

in (10) and afterwards in (8), we will getthe fault current for phase of an Yzn5 transformer, asfollows:

φ

b

pujV 1�puZ T �� 6011 pujZ S 01 �577,011 �TT XR

pu.j

aIb 273377,1

13,0175,2

3013 2

��

�� (11a)

Thus, it can be concluded that the bolted phase-to-neutral fault current is higher than the bolted three-phasefault current.

In the same way as in the case of the bolted three-phasefault, we can determine primary line currents of a Dyn5 or ofanYzn5 transformer for this fault type.

Table 3Tablica 3.

The phase b-to-neutral fault current analysisAnaliza jednopolnog kratkog spoja (faza b-nulti vodič)

Transformer

connections

Primary line

currents (pu)

Secondary line

currents (pu)

�� 00,0AI �� 00,0aI

�� 4,90585,0BI �� 270013,1bIDyn5

�� 4,270585,0CI �� 00,0cI

�� 00,0AI �� 00,0aI

�� 6,91795,0BI �� 273377,1bIYzn5

�� 6,271795,0BI �� 00,0cI

� 1 2

1 1

1 1

2 2

0 0

1 900,5 30

2 1 602

150 0,5 120

150 0,5 180 0,5

0,0

A A

S T

a A

a A

A a

VI I pu

Z Z

I I pu

I I pu

I I pu.

� ��

� � ��

� � � � � �

� � � � � � �

� �

(12)

, and in,

The primary line currents , and can be obtainedsubstituting equation (A7), and thesecondary line currents and can be obtained

AI BI CI

1AI 2AI 0AI

aI bI cI

.

.

�

7

substituting , and in equation (A6).Fig. 2 shows results of the simultaneous fault currents

on both sides of a Dyn5 transformer.

Using the basic assumptions stated above for an Yzn5transformer, the following positive-, negative-, and zero-sequence per-unit currents can be obtained from (13) and(14):

Technical Gazette , 317, 1(2010) -15

Analiziranje mjerodavnog tipa kvara za proračun iznosa minimalne struje kvara na primarnoj strani distribucijskog transformatoraV Boras, T Bari , A Muharemovi. . ć . ć

Fig. 2 and Fig. 3 show that primary-side fuses see lowercurrents during this simultaneous fault type than thosewhich occur during the three-phase fault only, and thus theytake longer to clear this simultaneous fault. However, partialshort-circuit currents at the medium-voltage side may not betoo small to operate other two medium-voltage fuses.

a) First, let us consider bolted double phase-to-neutralfault, involving phases and , at the fault point F located onthe secondary side of a Dyn5 or of an Yzn5 transformer, asshown in Fig. 4 and Fig. 5, with blown fuse in phase on thetransformer primary-side.

When medium-voltage phase is the reference phaseand solving the linear matrix equation (C6), the followingpositive-, negative-, and zero-sequence currents can beobtained:

3.2Double phase-to-neutral fault on the secondary sidewith blown fuse in phase A

Dvopolni kratki spoj s istodobnim spojem s neutralnimvodičem na sekundarnoj strani i s pregorjelim osiguračemu fazi A

a c

A

A

1aI 2aI 0aI

Figure 2

Slika 2.

Per-unit fault currents on both sides of a Dyn5 transformerfor three-phase fault on transformer secondary-side with blown fuse

in phase A on transformer primary-sideJedinični iznosi struja kvara na obje strane nekog Dyn5

transformatora kod tropolnog kratkog spoja na sekundaru transformatoras pregorjelim osiguračem u fazi A na primaru

The fault analysis results for the bolted secondarythree-phase fault of an Yzn5 transformer with blown fuse inphase on primary-side, are shown in Fig. 3.A

Figure 3

Slika 3.

Per-unit fault currents on both sides of an Yzn5 transformerfor three-phase fault on transformer secondary-side with blown fuse

in phase A on transformer primary-sideJedinični iznosi struja kvara na obje strane nekog Yzn5

transformatora kod tropolnog kratkog spoja na sekundarutransformatora s pregorjelim osiguračem u fazi A na primaru

� 1 2

1 1 0

1 1

2 2 1

20 1 2

1 1 0

2 4

150

150 330

90

2

A A

S T T

a A

a A A

a a a

S T T

VI I

Z Z Z

I I

I I I

VI a I aI

Z Z Z

� ��

� � �

� � � � � �

� ��

� �

(13)

Substituting the basic assumption in (13) andrearranging, the next expression can be derived:

pu0,01 �SZ

��

��

��

�

��

��

��

��

1

0

3

1

01

212

122

12T

T

SCC

T

TT

AA

Z

Z

I

Z

ZZ

VII (14)

where are defined by (9) and (10).Using the basic assumptions stated above for a Dyn5

transformer, the following positive-, negative-, and zero-sequence per-unit currents can be obtained from (13) and(14):

10 TT ZZ

��

��

��

15,593418,00,0

85,0171,085,210171,0

15,119171,085,30171,0

00

22

11

aA

aA

aA

II

II

II

(15)

The primary and secondary line currents can beobtained substituting corresponding positive-, negative,and zero-sequence currents from (15) in (A7) and (A6),respectively. Fig. 4 shows results of these simultaneousfault currents on both sides of a Dyn5 transformer. In thiscase, partial short-circuit currents on the medium-voltageside will be too small to operate other two medium-voltagefuses and it can lead to overstressing of electricalequipment.

Figure 4

Slika 4.

Per-unit fault currents on both sides of a Dyn5 transformerfor the simultaneous fault, which involves phase a-phase c-to-neutral

fault on the star side and blown fuse in phase A on the delta side

a cJedinični iznosi struja kvara na obje strane nekog Dyn5

transformatora kod simultanog kvara koji obuhvata spoj faza i sneutralnim vodičem na strani zvijezda namota i pregorjeli osigurač

u fazi A na strani namota u trokutu

��

��

��

9,587275,00,0

1,1364,01,211364,0

9,118364,01,31364,0

00

22

11

aA

aA

aA

II

II

II

(16)

,

.

.

.

The primary and secondary line currents can beobtained substituting corresponding positive-, negative,and zero-sequence currents from (16) in (A7) and (A6),respectively. Fig. 5 shows results of these simultaneousfault currents on both sides of anYzn5 transformer.


b) Let us consider bolted double phase-to-neutralfault, involving phases and , at the fault point F located onthe secondary-side of a Dyn5 or of an Yzn5 transformer, asshown in Fig. 6 and Fig. 7, with blown fuse in phase on thetransformer primary-side.

When medium-voltage phase is the reference phaseand solving the linear matrix equation (C6), the followingpositive-, negative-, and zero-sequence currents can beobtained:

a b

A

A

Substituting the basic assumption in (17)and rearranging, the next expression can be derived:

pu

8 Tehni ki vjesnikč , 317, 1(2010) -15

c) Let us consider bolted double phase-to-neutralfault, involving phases and , at a point F located on thesecondary side of a Dyn5 or of an Yzn5 transformer, withblown fuse in phase on the transformer primary-side.

When medium-voltage phase is the reference phase,by solving the linear matrix equation (C6) can be obtainedthe positive-, negative-, and zero-sequence per-unitcurrents, which are equal to those in previous case, i.e. theyare defined by means of equation (17), too. Thus, allprimary and secondary currents are equal to those shown inFig. 6 and Fig. 7.

b c

AA

Figure 5

Slika 5.

Per-unit fault currents on both sides of an Yzn5 transformerfor the simultaneous fault, which involves phase a-phase c-to-neutral

fault on the secondary-side and blown fuse in phase A on the transformerprimary-side

g kvara koji obuhvata spoj faza a i cJedinični iznosi struja kvara na obje strane nekog Yzn5

transformatora kod simultano sneutralnim vodičem na sekundarnoj strani i pregorjeli osigurač u fazi A

na primarnoj strani transformatora

�

� 0112

210

122

11

01121

2

90

330150

150

2

TTSaaa

AAa

Aa

TTSAA

ZZZ

VIaIaI

III

II

ZZZ

VII

��

��

��

��

��

(17)

0,01 �SZ

1

0

3

1

01

21

22T

T

SCC

T

TT

AA

Z

Z

I

Z

ZZ

VII

��

��

��

��

�� (18)

Using the basic assumptions stated above for a Dyn5transformer, the following positive-, negative-, and zero-sequence per-unit currents can be obtained from (17) and(18):

��

��

��

4,1203375,00,0

4,03375,04,2103375,0

6,1193375,04,303375,0

00

22

11

aA

aA

aA

II

II

II

(19)

The primary and secondary line currents can beobtained substituting corresponding positive-, negative,and zero-sequence currents from (19) in (A7) and (A6),respectively. Fig. 6 shows results of these simultaneousfault currents on both sides of a Dyn5 transformer.

Figure 6

Slika 6.

Per-unit fault currents on both sides of a Dyn5 transformerfor the simultaneous fault, which involves phase a-phase b-to-neutral


ltanog kvara koji obuhvata spoj faza a i bJedinični iznosi struja kvara na obje strane nekog Dyn5

transformatora kod simu sneutralnim vodičem na strani zvijezda namota i pregorjeli osigurač

u fazi A na strani namota u trokutu

Using the basic assumptions stated above for an Yzn5transformer, the following positive- , negative-, and zero-sequence per-unit currents can be obtained from (17) and(18):

��

��

��

4,123459,00,0

4,3459,04,213459,0

6,116459,04,33459,0

00

22

11

aA

aA

aA

II

II

II

(20)

The primary and secondary line currents can beobtained substituting corresponding positive-, negative,and zero-sequence currents from (20) in (A7) and (A6),respectively. Fig. 7 shows results of these simultaneousfault currents on both sides of anYzn5 transformer.

Figure 7

Slika 7.

Per-unit fault currents on both sides of an Yzn5 transformerfor the simultaneous fault, which involves phase a-phase b-to-neutral

fault on the secondary-side and blown fuse in phase Aon the transformer primary-side

a i bJedinični iznosi struja kvara na obje strane nekog Yzn5

transformatora kod simultanog kvara koji obuhvata spoj faza sneutralnim vodičem na sekundarnoj strani i pregorjeli osigurač u fazi A

na primarnoj strani transformatora

.

.

.

.

,

Fig. 8 and Fig. 9 show results of the simultaneous faultcurrents on both sides of a Dyn5 and an Yzn5 transformer,respectively.

From previous analysis we can conclude that the faultcurrent on each side of a Dyn5 and an Yzn5 three-phasetransformer can differ in magnitude and phasing, dependingon the type of fault and the fault location. Excepting three-phase secondary fault and blown fuse in phase , theanalyzed simultaneous fault currents can be equal to orlower than the minimum shunt fault current and thus theycan directly affect coordination of transformer protectivedevices.

For a delta-grounded star (Dyn5) transformer,substantially the lowest fault current on the transformerprimary-side occurs for double phase-to-neutral fault on thesecondary-side, involving phases and , with blown fusein phase . This current magnitude equals 0,296 .

For a star-grounded zig-zag (Yzn5) transformer,substantially the lowest fault current on the transformerprimary-side occurs for double phase-to-ground fault on thesecondary-side, involving phases and , with blown fusein phase . This fault current magnitude equals 0,63 .

Finally, we can conclude that:- when faults occur on the secondary-side of a delta-grounded star or a star-grounded zig-zag transformer, theper-unit fault current magnitude in each phase depends onthe type of faults,- Dyn5 and Yzn5 distribution transformers could not bewell protected, if we do not take into consideration theseminimum simultaneous currents,- determining minimum simultaneous fault currents couldbe necessary for selection of fuse-links minimum breakingcurrent of high-voltage fuses for transformer circuitapplications- taking the minimum simultaneous fault currents intoaccount we can be sure that the high-voltage fuses canadequately protect the transformer for the various faulttypes and ensure proper coordination.

Final expression for the simultaneous fault currents onboth sides of a Dyn5 or anYzn5 transformer, obtained in thispaper, confirms the generalized equation given in [7]. Thesefinal expressions for the simultaneous fault currents havebeen checked and are confirmed by means of the faultsimulation on the three-phase transformer, model 61-107manufactured from Feedback, too.

The authors suggest that the operating current of thehigh-voltage fuse in transformer applications should bedetermined so that it is less than the minimum (shunt orsimultaneous) fault current of the transformer as limited bythe combination of distribution system impedance andtransformer impedance. But, this operating current shouldnot be so low to cause circuit interruption due to the inrushexcitation current of the transformer or normal currenttransients in the secondary circuits.

According to the analysis carried out in this paper, it isshown that protection schemes of utility substations couldbe designed by taking into account minimum fault currentsduring simultaneous faults, too.

4ConclusionZaključak

A

a cA pu

a cA pu



3.3Phase -to-neutral fault on the secondary side withblown fuse in phase

b

A

Kratki spoj faze s neutralnim vodb ičem na sekundarnojstrani i s pregorjelim osiguračem u fazi A

In [1] it is noted, that in case of single phase- -to-neutral fault and blown fuse in phase or in the case ofsingle phase- -to-neutral fault and blown fuse in phase ,the resulting low currents are stipulated by the transformeropen-circuit impedances and they may be neglected. Thus,we will consider bolted phase- -to-neutral fault at the faultpoint F located on the secondary-side of a Dyn5 or an Yzn5transformer, as shown in Fig. 8 and Fig. 9, with blown fusein phase on the transformer primary-side.

When medium-voltage phase is the reference phaseand solving the linear matrix equation (D6), the followingpositive-, negative- and zero-sequence currents can beobtained:

aA

c A

b

AA

�

� 0111

20

122

11

01121

2

90

330150

150

2

TTSaa

AAa

Aa

TTSAA

ZZZ

VIaI

III

II

ZZZ

VII

��

��

��

��

��

(21)

Once when the sequence currents are determined bymeans of equation (D6), the primary and secondary linecurrents can be determined from (A7) and (A6),respectively.

Figure 8

Slika 8.

Per-unit fault currents on both sides of a Dyn5 transformerfor the simultaneous fault, which involves phase b-to-neutral


bJedinični iznosi struja kvara na obje strane nekog Dyn5

transformatora kod simultanog kvara koji obuhvata spoj faze sneutralnim vodičem na strani zvijezda namota i pregorjeli

osigurač u fazi A na strani namota u trokutu

Figure 9

Slika 9.

Per-unit fault currents on both sides of an Yzn5 transformerfor the simultaneous fault, which involves phase b-to-neutral fault

on the secondary-side and blown fuse in phase A on the primary-side

bJedinični iznosi struja kvara na obje strane nekog Yzn5

transformatora kod simultanog kvara koji obuhvata spoj faze sneutralnim vodičem na sekundarnoj strani i pregorjeli osigurač

u fazi A na primarnoj strani transformatora

.



Appendix ADodatak A

a) Sequence phase shift through three-phase two windingtransformer

a) Fazni pomaci sustava kroz trofaznidvonamotni transformator Dyn5

simetričnih komponenti

According to [2] and [3], two basic rules of transformerpolarity are used in this paper for making and analyzing theproper three-phase transformer connections.

Dyn5

Let us consider a Dyn5 three-phase two windingtransformer as shown in Fig.A1.

, and, and

According to the first rule of transformer polarity, as itis noted in [2] and [3], and in accordance with Fig. 1, it canbe seen that delta currents are in phasewith corresponding winding currents onthe transformer secondary side. Thus, the positive- andnegative-sequence primary line currents can be given inrelation to the positive- and negative-sequence secondaryline currents by the next matrix equations:

Fig A1Sl A1

ureika .

Transformer phase relationships- Dyn5 connectionUzajamni odnos faza transformatora-Dyn5 spoj

Throughout this analysis, three distinct sets ofsymmetrical components are defined: positive-, negative-,and zero-sequence for both currents and voltages. Positive-sequence sets consist of the balanced (symmetrical) three-phase currents or three-phase line-to-neutral voltages,which can be designated like these:- for medium-voltage currents or voltages as

1111

12

112

1

11

ACAC

ABAB

AA

VaVIaI

VaVIaI

VI

��

�� (A1)

and for low-voltage currents or voltages as

1111

12

112

1

11

acac

abab

aa

VaVIaI

VaVIaI

VI

��

�� (A2)

where a is unit phasor, so that:

0.1

2

35,02401

2

35,01201

2

2

��

��

��

aa

ja

ja

(A3)

Negative-sequence sets are also balanced (symmetrical) with three-phase currents or three-phase line-to-neutralvoltages, but with the phase rotation or sequence reversed inrelation to positive-sequence sets. Negative-sequence setscan be designated like these:

-

It is well known, that any unbalanced currents orvoltages can be determined from sequence componentsfrom the following basic equations:

- for medium-voltage currents or voltages as

22

222

2

2222

22

ACAC

ABAB

AA

VaVIaI

VaVIaI

VI

��

�� (A4)

and for low-voltage currents or voltages as

22

222

2

2222

22

acac

abab

aa

VaVIaI

VaVIaI

VI

��

�� (A5)

022

1022

1

0212

0212

021021

aaacaaac

aaabaaab

aaaaaaaa

VVaVaVIIaIaI

VVaVaVIIaIaI

VVVVIIII

��

��

��

(A6)

022

1022

1

0212

0212

021021

AAACAAAC

AAABAAAB

AAAAAAAA

VVaVaVIIaIaI

VVaVaVIIaIaI

VVVVIIII

��

��

��

(A7)

From the connection of Fig. A1, it can be written that:

Dy

ABaBAACA

n

VVVIII

�� (A8)

BAI CBI

aIACI

bI cI

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

1

1

1

270

30

150

1

1

1

2

2

1

12

1

1

1

1

31

11

110

011

1011

a

a

a

j

j

j

Dya

a

a

Dy

a

a

a

DyC

B

A

I

I

I

e

e

e

nI

I

I

aa

a

a

n

Ia

Ia

I

nI

I

I

(A9)

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

2

2

2

90

30

150

2

2

2

2

2

22

2

2

2

2

2

31

11

110

011

1011

a

a

a

j

j

j

Dya

a

a

Dy

a

a

a

DyC

B

A

I

I

I

e

e

e

nI

I

I

aa

a

a

n

Ia

Ia

I

nI

I

I

(A10)

,

.

.

.

.

.

.

�

�

lag with respect to the corresponding negative-sequencecurrents and voltages on low-voltage side (LV) by 150°.

In the same way, we can find the phase shift betweenpositive-sequence primary and positive-sequencesecondary phase-to-neutral voltages and the phase shiftbetween negative-sequence primary and negative-sequencesecondary phase-to-neutral voltages of a Dyn5 or an Yzn5transformer.

Let us consider a simultaneous fault, which involves abolted three-phase fault at the fault point F located on thesecondary side of a Dyn5 or an Yzn5 transformer withblown fuse in any phase on the primary.

The calculation of this simultaneous fault is carried outby means of per-unit method selecting the phase with blownfuse as the reference.

Appendix BDodatak B

Three-phase short circuit on transformer secondary withany blown HV fuseTropolni kratki spoj na sekundaru transformatora i pregorjeli bilokoji VN osigurač



The delta connection on the medium-voltage side of theDyn5 transformer and the ungrounded star connection onthe medium-voltage side of the Yzn5 transformer do notallow the flow of zero-sequence current, i.e. it will be equalto zero ( ).

According to Fig.A2 we get the following:

b) Sequence phase shift through three-phase two windingtransformer

b) Fazni pomaci kroz trofaznidvonamotni transformator Yzn5

sustava simetričnih komponentiYzn5

Let us consider an Yzn5 three-phase transformer asshown in Fig.A2.

According to (A1)-(A5) and (A14)-(A16), the positive-and negative-sequence primary line currents can be given inrelation to the positive- and negative-sequence secondaryline currents by the next matrix equation, respectively:

puIII CBA 0,0000 ��

Yz

ABa

Yz

aaA

n

VVV

n

III

�

�� (A11)

Yz

BCb

Yz

bbB

n

VVV

n

III

�

�� (A12)

Yz

CAc

Yz

ccC

n

VVV

n

III

�

�� (A13)

From the connection of Fig. A2, it can be written that:

caaa IIII �� , (A14)

abbb IIII �� , (A15)

bccc IIII �� , (A16)

Transformer phase relationships- Yzn5 connectionFig A2ureika .Sl A2 Uzajamni odnos faza transformatora-Yzn5 spoj

��

�

�

��

�

�

��

�

�

��

�

�

��

��

�

�

��

�

�

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

1

1

1

270

30

150

1

2

1

1

1

2

2

1

12

1

1

1

1

2

31

11

110

011

1011

a

a

a

j

j

j

a

a

a

Yz

a

a

a

YzC

B

A

I

I

I

e

e

e

N

N

I

I

I

aa

a

a

n

Ia

Ia

I

nI

I

I

(A17)

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

2

2

2

90

30

150

2

2

2

2

2

22

2

2

2

2

2

31

11

110

011

1011

a

a

a

j

j

j

Yza

a

a

Yz

a

a

a

YzC

B

A

I

I

I

e

e

e

nI

I

I

aa

a

a

n

Ia

Ia

I

nI

I

I

(A18)

Equations (A9) and (A17) show that for the Dyn5 andthe Yzn5 transformer connections, the positive-sequencecurrents on medium-voltage side (MV) lead tocorresponding positive-sequence currents on low-voltageside (LV) by 150°. Equations (A10) and (A18) show that forthe Dyn5 and the Yzn5 transformer connections, thenegative-sequence currents on medium-voltage side (MV)

Fig. B1

Sl. B1

Sequence networks interconnection for bolted three-phasefault on the transformer low-voltage side with any blown fuse

on the transformer primary-sideInterkonekcija sustava simetričnih komponenti za izravni tropolni

kratki spoj na niskonaponskoj strani transformatora i pregorjelimosiguračem u bilo kojoj fazi na primarnoj strani transformatora

.

.

.

�

.

�



This shunt fault location and the fuse location are onopposite sides of the Dyn5 or the Yzn5 transformer and thusthe calculation of this simultaneous unbalance shouldinvolve the phase shift caused by these transformers asexplained in Appendix A. Also, for this phase shift, ideal(perfect) transformers T , and T are required in the

sequence connection diagram, as shown in Fig. B1.Since the sequence networks are interconnected at

more than one location to represent such simultaneousunbalances, the connections must satisfy the sequenceconstraint equations at both unbalance points.

To draw the sequence network for any blown fusesituation (open phase), two constraints have to embed intothe network. The first of these is a voltage constraint and thesecond is a current constraint.

The sequence voltages between points x and y are givenin matrix form by:

1 2

Note that the ideal transformers T and T in Fig. B1 are

not necessary, but they are shown because they are requiredin general for the other fault types. According to [6], Fig. C1shows the interconnection of the resulting equivalentsequence networks for this simultaneous fault type. Fromsequence networks interconnection as is shown in Fig. B1, itcan be concluded that current can flow through thetransformer without connected load.

If the three sequence networks of Fig. B1 are describedmathematically in accordance with Kirch off's current andvoltage law, we can resolve the interconnected sequencenetworks for the unknown sequence currents and voltagesby solving the next linear matrix equation :

6 7

h

(B4)

If we assume that blown fuse is in phase (i.e. -phaseis open), then values for and are equal to zero, and

substituting these values in (B1) we obtain the sequencevoltages and in relation to Similar

consideration can be carried out for the other two cases.According to the condition that the current in open

phase must be equal to zero, the current constraint can be

A AV V

V , V , V V .

Bxy Cxy

1xy 2xy 0xy Axy

Once when the sequence currents are determined fromequation (B4), the primary and secondary line currents canbe determined from (A7) and (A6), respectively.

The calculation of this simultaneous fault is carried outin this paper by means of per-unit method selecting thephase with blown fuse as the reference.

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

�

�

Cxy

Bxy

Axy

xy

xy

xy

V

V

V

aa

aa

V

V

V

111

1

1

3

1 2

2

0

2

1

(B1)

0.02211 �� AAA IICIC (B2)

The value of the ideal transformer constants, andwill depend upon the phase in which the fuse is blown.These values are given in the table in Fig. B1. Fig. B1 showsthat the two ideal transformers T and T are connected

between points x and y in the positive- and negative-sequence network, because they implement the two voltageand current constraints, which are given by means of (B1)and (D1), respectively. When medium-voltage phase isthe reference phase, the sequence constraint equations forthree-phase short circuit at a fault point F on low-voltageside are:

4 5

A

1C 2C

pu.,VVV aaa 00021 �� (B3)

expressed by means of (B2):

blownisCphaseinfusewhen

blownisBphaseinfusewhen

blownisAphaseinfusewhen

0

0

0

0

0

0

0

0

010000000

001501000000

000150100000

000010000

000001015010

000000101501

0000000

30100000

30010000

*

*

*

*

0

2

1

0

2

1

2

1

21

111

211

Cxyxy

Bxyxy

Axyxy

xy

a

A

A

a

a

a

A

A

TS

TS

VV

VV

VV

V

V

V

V

V

I

I

I

I

I

CC

CZZ

CZZ

�

�

�

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

��

��

��

�

�

(B4)

Appendix CDodatak C

Double phase-to-neutral fault on transformer secondarywith any blown HV fuseDvopolni kratki spoj s istodobnim spojem s neutralnim vodičemna sekundaru transformatora i pregorjeli bilo koji osiguračVN

Let us consider any bolted double phase-to-neutral faultat the fault point F located on the secondary side of a Dyn5or of an Yzn5 transformer with blown fuse in any phase onthe primary.

.

.

.

h

(C6)

Fig. C1 shows values of the andWhen making the connections to the positive-,

negative-, and zero-sequence networks, ideal transformersT , T , and T with the input-output relationships must be

used to implement voltage and current constraints given in(C2)-(C4). The value of the ideal transformer constants,and will depend upon phases, which are faulted and theirvalues are given in Fig. C1.

If the three sequence networks of Fig. C1 are describedmathematically in accordance with Kirch off's current andvoltage law, we can resolve the interconnected sequencenetworks for the unknown sequence currents and voltagesby solving the next linear matrix equation :

6 7 8



This shunt fault location and the fuse location are onopposite sides of the Dyn5 or the Yzn5 transformer and thusthe calculation of this simultaneous unbalance shouldinvolve the phase shift caused by these transformers asexplained in Appendix A. Also, for this phase shift, ideal(perfect) transformers T , and T are required in the

sequence connection diagram, as shown in Fig. C1.According to [6], Fig. C1 shows the interconnection of

the resulting equivalent sequence networks for thissimultaneous fault type. This sequence networksinterconnection shows that current can flow through thetransformer without load and that the zero-sequencenetwork is connected in parallel with positive- andnegative-sequence network at the fault point F.

Since the sequence networks are interconnected atmore than one location to represent such simultaneousunbalances, the connections must satisfy the sequenceconstraint equations at both unbalance points.

To draw the sequence network for any blown fusesituation (open phase), two constraints into the networkhave to be embed. The first of these is a voltage constraintand the second is a current constraint.

Equation (B1) gives the sequence voltages betweenpoints x and y, which fulfil the voltage constraint.

According to the condition that the current in openphase must be equal to zero, the current constraint can beexpressed by means of (C1):

1 2

1 1H 2 2H 0HC I C I I 0,� � � (C1)

where values of the and are given in Fig. C1.The two ideal transformers T and T , which are

connected between points x and y in the positive- andnegative- sequence network as shown in Fig. C1, are used toimplement the two voltage and current constraints.

On the other side, the sequence constraint equations,which must be satisfied at the shunt fault point for a doublephase-to-neutral on any two low-voltage phases, are:

4 5

1C 2C

Fig C1

Sl C1

ure

ika .

Sequence networks interconnection for any double phase-to-neutral fault on the transformer low-voltage side with any blown fuse

n the transformer primary-sideInterkonekcija sustava simetričnih komponenti za dvopolni

kratki spoj s neutralnim vodičem na niskonaponskoj stranitransformatora i pregorjelim osiguračem u bilo kojoj fazi

na primarnoj strani

1 1L 2 2L 0LK I K I I 0� � � (C2)

1 1L 2 2LK V K V� (C3)

1 1L 0LK V V� (C4)

0L 0L 0TV I Z� (C5)

1K 2K

1K

2K




0

0

0

0

0

0

0

0

01000000

01015000000

0015015000000

0000100150150

000001015010

000000101501

0000000

30100000

30010000

*

*

*

*

0

2

1

0

2

1

2

1

0

1

21

21

21

111

211

Cxyxy

Bxyxy

Axyxy

xy

a

A

A

a

a

a

A

A

T

TS

TS

VV

VV

VV

V

V

V

V

V

I

I

I

I

I

Z

K

KK

KK

CC

CZZ

CZZ

�

�

�

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

��

��

��

��

�

�

(C6)

.

.



Once when the sequence currents are determined fromequation (C6), the primary and secondary line currents canbe determined from (A7) and (A6), respectively.

The calculation of this simultaneous fault is carried outin this paper by means of per-unit method selecting thephase with blown fuse as the reference.

Let us consider any bolted single phase- to-neutral faultat a Fault F on the secondary of a Dyn5 or of an Yzn5distribution transformer with blown fuse in any phase on thetransformer primary-side. This shunt fault location and thefuse location are on opposite sides of the Dyn5 or the Yzn5transformer and thus the calculation of this simultaneousunbalance should involve the phase shift caused by thesetransformers as explained in Appendix A. Also, for thisphase shift, ideal (perfect) transformers T and T are

required in the sequence connection diagram, as shown inFig. D1. It should be pointed out, that ideal transformersmust be used with positive- and negative-sequencenetworks at both the shunt fault and fuse locations.

Since the sequence networks are interconnected atmore than one location to represent such simultaneousunbalances, the connections must satisfy the sequenceconstraint equations at both unbalance points.

To draw the sequence network for any blown fusesituation (open phase), two constraints into the networkhave to be embed. The first of these is a voltage constraintand the second is a current constraint. Equation (B1) givesthe sequence voltages between points x and y, which fulfilthe voltage constraint. According to the condition that thecurrent in open phase must be equal to zero, the currentconstraint can be expressed by means of (D1):

Appendix DDodatak D

Single phase-to-neutral fault on transformer secondarywith any blown HV fuseKratki spoj jedne faze s neutralnim vodičem na sekundarutransformatora i pregorjeli bilo koji VN osigurač

1 2

1 1H 2 2H 0HC I C I I 0.� � � (D1)

Values of the and in equation (D1) are given inFig. D1. Fig. D1 shows that two ideal transformers T and T

are connected between points x and y in the positive- andnegative- sequence network, because they implement thevoltage and the current constraints, which are given bymeans of (B1) and (D1), respectively.

On the other side, the sequence constraint equations,which must be satisfied at the shunt fault point for a single

4 5

1C 2C

Fig. D1

Sl. D1

Sequence networks interconnection for a single phase-to-neutral fault on the transformer low-voltage side with any blown fuse

on the transformer primary-side

eu bilo kojoj fazi na primarnoj strani

Interkonekcija sustava simetričnih komponenti za kratki spojjedn faze s neutralnim vodičem na niskonaponskoj strani i pregorjelim

osiguračem

0

0

0

0

0

0

0

0

01000000

00000010150

0000000150150

0115015000000

000001015010

000000101501

0000000

30100000

30010000

*

0

2

1

0

2

1

2

1

0

1

21

21

21

111

211

xy

a

A

A

a

a

a

A

A

T

TS

TS V

V

V

V

V

I

I

I

I

I

Z

K

KK

KK

CC

CZZ

CZZ

��

�

�

��

�

�

�

��

�

�

��

�

�

��

�

�

��

�

�

��

��

��

��

��

�

�

(D6)

1 1L 2 2L 0LK V K V V 0� � � (D2)

1 1L 2 2LK I K I� (D3)

1 1L 0 LK I I� (D4)

0L 0L 0TV I Z�� (D5)

h

(D6)

When making the connections to the positive-,negative-, and zero-sequence networks, ideal transformersT , T , and T with the input-output relationships must be

used to implement voltage and current constraints given in(D2)-(D4). The value of the ideal transformer constants,and will depend upon phase which is faulted and theirvalues are given in Fig. D1.

If the three sequence networks of Fig. D1 are describedmathematically in accordance to Kirch off's current andvoltage law, we can resolve the interconnected sequencenetworks for the unknown sequence currents and voltagesby solving the next linear matrix equation :

6 7 8

phase-to-neutral on any low-voltage phase, are :(D2)

1K

2K

.

.



,

,

,

,

-

Once when the sequence currents are determined fromequation (C6), the primary and secondary line currents canbe determined from (A7) and (A6), respectively.

The calculation of this simultaneous fault type iscarried out by means of per-unit method selecting the phasewith blown fuse as the reference.

- Per unit line-to neutral driving voltage- General unbalanced three-phase currents on the

transformer secondary side- General unbalanced three-phase currents

- Three-phase short circuit current- Transformer three-phase full-load current on the

transformer primary-side- General unbalanced three-phase low-voltage

phase-to-neutral voltages- General unbalanced three-phase high-voltage

phase-to-neutral voltages- Positive-sequence equivalent network per unit

impedance- Negative-sequence equivalent network per unit

impedance- Positive-sequence transformer per unit impedance- Negative-sequence transformer per unit impedance- Zero-sequence transformer per unit impedance- Per unit fault impedance- Base power in kVA- Rated apparent power of distribution transformer in

kVA- Short-circuit voltage at rated current in per unit

- Number of turns in primary winding set

- Number of turns in secondary winding set

2 - Number of turns in each of six secondary winding in

zig zag winding set- Actual winding turns ratio by a Dyn5

transformer

-Actual winding turns ratio by anYzn5transformer

- Transformer ratio ( and – phase-to-neutral

voltages), where

List of SymbolsPopis oznaka

S

u

N

N

N /

n =N /N

N=V /V V V

T

k

A a A a

1

2

2

Dy 1 2




*

*

*

Cxyxy

Bxyxy

Axyxy

VV

VV

VV

�

�

�

V

a bI I, cI

A BI I, CI

SCC3I

nI

a bV V, cV

A BV V, CV

S1Z

S2Z

T1Z

T2Z

T0Z

FZ

bS

22

1

N

NnYz �

1 1 1, ,A B CI I I

2 2 2, ,A B CI I I

1 1 1a b cI I I, ,

2 2 2, ,a b cI I I

000 ,, cba III

111 ,, CBA VVV

222 ,, CBA VVV

111 ,, cba VVV

222 ,, cba VVV

ReferencesLiteratura

[1] Das, J. C. Power SystemAnalysis, Marcel Dekker, Inc., 2002.[2] Blackburn, J. Lewis. Symmetrical Components for Power

Systems Engineering, Marcel Dekker, Inc., 1993.[3] Blackburn, J. Lewis. Protective Relaying-Principles and

Applications, Second Edition, Marcel Dekker Inc., 1998.[4] Anderson, P. Analysis of Faulted Power Systems, Iowa State

University Press,Ames, IA, 1973.[5] Short, T. A. Electric Power Distribution Handbook, CRC

Press LLC, 2004.[6] Smith, D. R. Digital Simulation of Simultaneous Unbalances

Involving Open and Faulted Conductors. // IEEETransactions on Power Apparatus and Systems, Vol. PAS-89,No. 8, November/December 1970, pp 1826-1835.

[7] IEC Std. 60909 (1988): Short-circuit current calculation inthree-phaseA.C. systems.

[8] IEC STANDARD Publication 787, 1983: Application Guidefor the selection of fuse-links of high-voltage fuses fortransformer circuit applications.

[9] IEC Std. 282-1, 1984: High-voltage fuses-Part 1: Current-limiting fuses.

[10] IEEE Std. 242-2001: IEEE Recommended Practice forProtection and Coordination Industrial and CommercialPower Systems.

[11] ANSI/IEEE Std. C37.91-1985: Guide for protective relayApplications to power transformers.

- Positive-sequence high-voltage currents

- Negative-sequence high-voltage currents

- Positive-sequence low-voltage currents

tive-sequence low-voltage currents

- Zero-sequence low-voltage currents

- Negative-sequence low-voltages

- Nega

- Positive-sequence high-voltages

- Negative-sequence high-voltages

- Positive-sequence low-voltages

Authors' addressesAdrese au orat

Vedran Boras, D.Sc. Associate Professor

, D.Sc. Assistant Professor

, Ph.D

University of SplitFaculty of Natural Sciences and mathematicsDepartment for PolytechnicTeslina 1221000 Split, Croatia

University of OsijekFaculty of Electrical engineeringKneza Trpimira 2b31000 Osijek, Croatia

University of SarajevoFaculty of Electrical EngineeringZmaja od Bosne bb71000 Sarajevo, Bosnia and Herzegovina

Tomislav Barić

Alija Muharemović

analyzing of relevant fault type for calculation of

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