419900

EG/94/710

FACULTY OF ELECTRICAL ENGINEERING

Group Electrical Energy Systems

RESTRICTED EARTH FAULT

DIFFERENTIAL PROTECTION

Master Thesis of

Pierre Raphael Bermejo

EG/94/710.A.

The Faculty of Electrical Engineering of theEindhoven University of Technology does notaccept any responsibility for the contents oftraining or terminal reports.

M. Sc. graduation report coached by:Prof.dr.-ing. H. Rijanto (EUT)Ir. P. Bertrand (Group SCHNEIDER)Eindhoven, April 1994.

EINDHOVEN UNIVERSITY OF TECHNOLOGY

SUMMARY

The Restricted Earth Fault (REF) differential protection is a zone protection that has to beable to detect currents to ground (zero sequence current). The principal of the RestrictedEarth Fault is based upon the distinction between a fault current inside the protected zoneand a fault current outside the protected zone. The REF is applied to protect e.g. powertransformer.

For faults inside the protected zone the protection has to react (switch-off) and for faultsoutside the protected zone we do not want a reaction.

Further on the REF has to be able to recognise effects like saturation of a currenttransformer -in consequence of a large short-circuit current or in consequence of aninrush current from a power transformer- to avoid undesirable swith-off command of theprotection. Current transformers are used like a measuring instrument to reduce current toan acceptable level for the hard-and software inside the computer.

The quantities diffusion current (id) and through current (it) are defined to distinguish afault current inside the protected zone from a fault current outside the protected zone.

As current transformers are not perfect (the can get saturated) the proportional current(ip) is introduced.

To discover an inrush current in relation to an undesirable switch-off command of theprotection the detection of second harmoniC has been applied, however, without succes.

From measurements we learn that in case there is no inrush-effect there is a possibility todetect second harmonics only for faults inside the protected zone and not for faultsoutside the protected zone.

Another possibility to perceive the inrush-effect is by detecting fifth harmonics becausethey are typically descending from power transformers (as we learn from literature). Thislast possibility has not been examined because the research had to be stopped. Neverthe-less an algorithme can be designed with the help from the flowchart diagram on page 37.

Conclusion:

The realised protection algorithme (see appendex page 92-94) functions if the inrush-effect is left out of consideration.

Further the Restricted Earth Fault protection algorithme can not detect a Three-phase toground fault because the zero sequence current 10 is equal to zero.

CONTENTS

page

INTRODUCTION

I SIMULATION OF THE S.E.P.A.M.2000

1.1 What is SEPAM 20001

1.2 Treatment of the input signal

1.2.1 The Rogowski coil

1.2.1.1 Model of the Rogowski coil

1.3 Treatments of the signal with a digital filter

1.3.1 The FIR filter

II SIMULATION OF AN ELECTRICAL NETWORKS

2.1 What is EMTP ?

2.2 Model of an electrical network

1

2

2

2

3

3

6

6

8

8

9

III CALCULATION OF A FAULT CURRENT IN STEADY STATE CONDITION 10

3.1 Symmetrical components method 10

3.1.1 First case: (Phase-to ground fault) 10

3.1.2 Second case: (Twophase-to ground fault) 13

3.2 Method with the value of the reactance 15

3.2.1 Third case:(Threephase-to ground fault) 15

IV THE RESTRICTED EARTH FAULT DIFFERENTIAL PROTECTION 16

4.1 The differential earth fault system or restricted earth fault protection 16

4.1.1 Principal of the REF 16

Pierre BERMEJO

restricted earth fault protectionUniversity of Technology in Eindhoven (The Netherlands)

15/04/94

1

4.1.2 The differential current

4.1.3 The through current

17

17

4.2 Differential current and through current in-and outside the protected zone 18

v THE SATURATION EFFECT 22

5.1 Output signals produced by EMTP for a phase-to ground fault inside theprotected zone 22

5.2 Output signals produced by EMTP for a phase-to ground fault outside theprotected zone 24

5.3.1 Saturation of the CT in consequence of very large fault current 26

5.3.2 The proportional current 28

5.3.3 Saturation in consequence of inrush current 30

5.3.3.1 Inrush phenomena 30

5.4 Choice of the harmonic 31

5.4.1 The second harmonic 31

5.4.2 The third harmonic 31

5.4.3 Higher harmonics 31

5.5 Detection of the second harmonic 31

5.5.1 Second harmonic from CT saturation in consequenceof large fault current 32

5.5.1.1 Analyse of the second harmonic when a fault appear insidethe protected zone in case of a phase-to earth faults. 32

5.5.1.2 Analyse of the second harmonic when a fault appear outsidethe protected zone in case of a phase-to earth faults. 33

5.5.2 Second harmonic from inrush current 34

5.5.2.1 Analyse of the second harmonic when a transformer is energized 34

Pierre BERMEJO


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5.5.2.1 Analyse of the second harmonic when a transformer is energized 34

VI RESULT OF THE PROTECTION ALGORITHME 35

6.1 Result of the protection algorithme for a fault inside the protected zone 35

6.2 Result of the protection algorithme for a fault outside the protected zone 35

VII CONCLUSION 37

VIII RECOMMENDATION FOR FURTHER WORKS 39

BIBLIOGRAPHY 40

Pierre BERMEJO


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INTRODUCTION

Nowadays there is an increasing tendancy in protecting electrical networks, by usingcomputers. Microprocessor technology is ideal for implementing and integrating protecti-ve functions to provide a lower cost per function. Such implementations improve theprecision and quality of classical protective functions and at the same time provideadvanced features including self diagnostics, metering and event recording at no additio-nal cost. Another important gain of the application of computers is reliability. The accentlies on the safety-aspect for people and apparatus.

For this reason the division "Protection Control and Command (PCC) ", of Merlin Gerin(France) wants to develop in the near future a protection algorithme for transformerscalled "Restricted earth fault (REF)differential protection", to satisfy the customers need.

The REF is a zone protection that has to be able to continuously detect current to ground.For faults outside its zone it is necessary that the REF undertakes no action, for faultsinside its zone action has to be took upon.

Purpose of the experiment:Acquise insight in the application of the REF (advantages and disadvantages of theseprotection) .

Pierre-Raphael BERMEJORestricted Earth Fault differential protection

University of Technology in Eindhoven (The Netherlands)

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I SIMULATION OF THE S.E.P.A.M.2000

1.1 What is SEPAM 20oo?

SEPAM is a digital protection unit. It is used for protecting electrical installations.Mastering electrical power calls for the use of units with the capacity to continuouslymonitor electrical networks and equipment, and to trigger the appropriate action forprotecting and controlling them.

It ensures all the protections, measurements and automation functions required for themost diversified applications. It is enhanced by a serial communication interface options.It is especially well-adapted for centralized control of electrical networks.

SEPAM 2000

Figl: Treatment of a input signal (lin) inside the SEPAM 2000

1.2 Treatments of the input signal

The treatments of the input signal, see Fig1, is as following. A current transformermeasures a signal. After this operation the signal must be clear from any noise and gothrough two Low Pass Filters. After filtering the analog signal is converted into a digitalform with the use of an analog - into digital convertor. After convolution of the signalwith a Finit Impuls Response (FIR) filter and the technics of the Discret Fourier Trans-form (DFT) the signal is ready to be analysed by the Central Processing Unit (CPU).



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Merlin Gerin used two different method to measure the input signal (lin)' The first one iswith the uses of a current transformer the other one is with the uses of a Rogowski coil.

Merlin Gerin used the Rogowski coil for measuring current waveforms containing fasttransients. For the simulation of this component we need a model.

1.2.1 The Rogowski coil

1.2.1.1 Model of the Rogowski coil

A Rogowski coil is an air cored solenoidal winding of small cross section wich can bereadily looped around a conductor[1]. If formed into a closed loop then the voltage E(t)induce in the coil is directly proportional to the rate of change of current i(t) passingthrough the loop according to the equation. It is relatively inexpensive to make, providesisolated measurement and being air cored it does not suffer from saturation.

There are obvious advantages in a measuring coil wich does not have a ferromagneticcore. The core may also be made flexible so that it can be strapped around a conductorwithout having to disconnect the primary circuit.

The principal of the Rogowski coil is the application of the induction law of Faraday[2].

with

~ = L. i

we get

E(t) = -L:ti(t)

further we know that

For the Rogowski coilN NffBndA: =E BiA = AE B i = A (Bl +B2 + ..... +BN- l +BN )~=l ~=l

(1)

(2)

(3)

(4)

(5)

where B, = B2 = = BN B, is the magnetic field from a winding of the Rogowskicoil. For N windings



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(6)

(7)

(8)

with

B1 (t)

and

(9)

(area of a winding) (10)

(11)

In this case the frequency and the capacitors of the Rogowski coil are very low f < 1 kHzthus Do is constant thus d/dt(constant) =0the magnetic flux Ht(t)is equal to:

2TtR

fill'idl = H( R) f dl = H(R) 2nR = Il~H(R) =1=0

(12)

(13)

E( t)

where

= _ N r 2 d I (t)2R Jlo dt 1 (14)

N r 2= --Jlo2 R

(15)

andJ.to=4?r.1O-7 [Him]



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Further the Rogowski coil has a proper Rogowski coil resistance from the windings R1and finaly a capacitance C2 of the windings_ The Rogowski coil is connected to theSEPAM 2000 We have to take into consideration the presence of the input resistance (R2)of the SEPAM-The result scheme is as follows:

E{t) u{t)

Fig2: Model of the Rogowski coilWith the help of fig2 we can develop the following equations

1 ) E ( t) = i 1 ( t) R1 + L 1 ddt i 1 ( t) + u ( t) (16)

d= C2 dt U (t)

= u (t)R2

(17)

(18)

The differential equation of this scheme is:

E ( t) =[i R (t) + i c (t) ] R1 + L 1 dd [i R (t) + i c (t)] + u ( t)2 2 t 2 2



(19)

(20)

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E( t) ::: (R +L~)[U(t) +C2ddt U(t)] +U(t)1 1 dt R2 (21)

d 2 eRR +L d R +R 1--u(t)+( 212 l)_U(t)+( 1 2 )u(t)=--E(t)d t 2 1R2 C2 d t 1R2 C2 1 C2

(22)

To simulate this second order differential equation with a computer we have to transformthis equation into an equation of difference .This equation of difference enables us to simulate the conduct of the Rogowski coilduring short-current situation in the electrical network.

u(nTs ) -2u(nTs -Ts ) +u(nTs -2Ts ) (u(nTs)) -u(nTs-Ts)) )--......;;;...---......;;;...----:;'-----....:;;....-~ +A +Bu (nTs

=CeT/ Ts

(23)The software implementation of the Rogowski coil and the filters inside the SEPAM 2000is shown on the annexe on pages 83-86.

The protection algorithme has to take into account the phenomenons of saturation of theCT so in the first instance the simulation of the Rogowski coil won't be used.

1.3 Treatments of the signal with a digital filter

After treatment of the signal with a finit impuls response (FIR) filter and the technics ofthe Discret Fourier Transform (DFT), we are capable to detect multiple of the groundfrequency of 50 Hz component from the saturation effect[3].

1.3.1 The F.I.R filter

The SEPAM 2000 uses two digital filters DFT(n.50Hz) for the treatment of the signalsfrom the analog to digital convertor. Because the signal f(nTs) is respectively convolutewith a cosinus function (Fi_c) and a sinus function (Fi_s) of (n.50 Hz) frequency. Withthese filters we are capable to detect multiples of the 50Hz ground harmonic from thefault current. We need this information eventually for preventing false trips duringenergization of a power transformer[4]. These filters are known as FIR (Finit ImpulsResponse) filters.



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FIR

f(nTS)~C(nTS)FI_cf(t) ADC

f(nTS~Wr~1 Js(nTs)Fig3: Digltal treatment of the slgnal f(nTs) wlth Fl_C and Fl_S fllters

f(t) is the input signal sampled at 12 points per sequence.fc(nTs), fs(nTs) are the ouputs of the filters(Fi_c and Fi_s).

FILTER Fi sm

fsm(nTs) = L an'f-nn=O

FILTER Fi cm

fcm(nTs) = L an'f-nn=O

(23)

where Ts = lIfs (fs sample frequency) and where hk is the digital impuls response of thefilters



(25)

(26)

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II SIMULATION OF AN ELECTRICAL NETWORKS

For the testing of the algorithme we need to modelise an electrical power network. This ispossible by using EMTP.

2.1 What is EMTP?

With EMTP it is possible to modelise electrical power networks as functions of time,typically following some disturbance such as the switching of a circuit breaker, or a fault.It also is used by those who specialize in power electronics.

For testing the protection algorithme we need the simulation of an electrical network.Inside this network it is possible to create situations where earthfaults occur, for example:

-Phase-to ground fault-Twophase-to ground fault-Threephase-to ground fault

In faults without ground contact we are not interested because the algorithme of therestricted earth fault works with the current wich flows through the ground.The network to be simulated with EMTP is shown in Fig4 here under.



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2.2 Model of an electrical network

Z

YyO; 62,5/21 kV36MVA; 16,45%

SW2

Zn2

L. ~ REF1 ~---------~=j~~~):=:::L _J -

Dy1 ;62,5/36,08 kV500MVA;1,64%

SW1G "v+----------i

Fig4: Model of an electrical network

In this figure above the Dy! power transformer is dimentioned in a way such that he cannot get satureted because we have not used the saturable element 96 from EMTP.The other power transformer used this element because we want to know wat can happenin situations of energizing it, to study the protection algorithme.

The Dyl power transformer has on the secondary side an impedance (Znl) between thestar point of the transformer and the earth. For the other power transformer the primaryside have a impedance to ground (zn2). With Zn2 we can simulate different forms ofgrounding.

The electrical network is dimensionned in a way such that if a earth fault occurs, half thefault current will flow respectively through Znl and Zn2. With the use of the twomeasuring switches (SWI and SW2) we can distinguish a fault inside and a fault outsidethe protected zone. About this subject we will talk later.



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With the method of the symmetrical components, it is possible to verify the calcules ofEMTP.

III CALCULATION OF FAULT CURRENT IN STEADY STATE CONDITION

3.1 Symmetrical components method

With the methode of the symmetrical components it is possible to calculate the currentsand voltages on the faultplace.

Further we are interested in the current which flows through the ground because we needit for the working of the restricted earth fault differential protection.

3.1.1 First case: (Phase-to ground fault)

From the symmetrical components[5] we know that for a phase-to ground faultcounts:

10=11 =12

In addition

10 the zero sequence overcurrent11 the direct sequence overcurrent12 the inverse sequence overcurrent

further

UO+U1+U2=O

In addition

UO the zero sequence overvoltageU1 the direct sequence overvoltageU2 the inverse sequence overvoltage

For each component of the electrical network we give the complex value. For somecomponents these value are only reactif. There is no contribution of the resistif partbecause this one can be neglected. Its value is too low in comparaison to the reactif part.



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further we know that for the voltage of each phase equal to:

UA = a2U 1 + aU2 + UoUB = aU 1 + a2U2 + UoUe = Ui + U2 + Uo

further we know that for the current of each phase equal to:

IA = a2II + aI2 + 10IB = all + a2I 2 + 10Ie = Ii + 12 + 10remarque: a2 = -112 -j (1I2).v3; a = -112 + j (1I2).V3

The result scheme is as follows:

Xs Xt1 XLI F XLr Xt2

IU1 ZLt 12Xs Xt1 XLI F XLr Xt2

U2 ZL 11-12-10

Xt10 XUO lOt XLrO Xt20

X01 t UO X02 X03 OL~-

-

Fig5: Phase-to ground fault

The value of each component is as following:

S = 36 kVXs = j6,5 ohm (reactance of the source s)Xtl = < < j I ohmXLI = j4 ohm (reactance of the area line at the leftside of the fault)



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XLr = j4 ohm (reactance of the area line at the rightside of the fault)Xt2 = j17,85 ohmZL = infiniteX01 = 3*j40 = j 120 ohmXtlO = < < j 1 ohmXLlO = j4 ohmXLrO = j4 ohmXt20 = j125,8 ohmX02 = 1 E-4 ohmX03 = infiniteZOL = infinite

After any calculation the value of successively

Zd = j 10,5 ohm (total value of the direct network)Zi = j 10,5 ohm (total value of the inverse network)Zh = j63,61 omh (total value of the homopolare network)

With the help of the values we found it is possible to give a simplification of fig 5.Using this new scheme makes it easy to calculate the values of currents and voltages.The definitive scheme is like:

zd L-sr- i1 t IU1zi 12 tl-[ - t U2 11-12-10zh loll _-I~ - t UOI

-

-

Fig6: Phase-to ground fault



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further

UO = -Zh.1O = -j63,63.-j426,67 103 = -27,14kVU2 = -Zi.I2 = -Zi.IO = -jlO,5.-j426,67 103 = -4,48kVUl = - U2 - VO = 31,62kV

thus

UA = -40,7 1

The equivalent scheme is like:

i1-

zd-

I

s~ I U1 i2zi

-

-I

- I U2 iOZh --

I- I UO

Fig7: Twophase-to ground fault

v, = S.Zp/(Zp+Zd) with Zp=(ZLZh)/(Zi+Zh)= j9 ohm= 36103 * 0,46 = 16,65kV

VA = a2V, + aV, + V, = (a2 + a + 1 )V, = 0VB = aV, + a2V, + V, = (a + a2 + 1 )V, = 0Vc = V, + V, + V, = 3VI

thus

Vc = 3*16,65kv = 50kV ;

I, = (S-V,)/Zd = -j 1,85kA12 = -V/Zi = -V/Zi = jl,58kA10 = -Vo/Zh = -V,/Zh = j 261,75A

IA = a2I1 + al2 + 10 = -2,97kA + j 396,75AIB = all + a2I2 + 10 = 2,97kA + j 396,75AIe = I, + 12 + 10 = 0 A

EMTP (file P3B. dat) shows these results en thus are in accordance with the calculationof the method of the symmetrical components.



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3.2 Method with the value of the reactance

3.2.1 Third case:(Threephase-to ground fault)

For this case we do not need the technique of the symmetrical components methodebecause we have a symmetrical short circuit and it is not necessary to find the equivalentscheme of each phase. Only one phase is enough because the current through the differentphases is the same. Of course the network is symmetrical for all phases.

The equivalent scheme is shown below.

Dy1 ;62,5/36,08 kVSOOMVA;1,64%

Xt XI

Xg

Fig8: Threephase- to ground fault

~ S 36,08kV~



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= (38,08)2. 65 =2,41062,5 '

(38,08)2Xtrafo = 1,64% 500 = 0,0470

x = (36,08)2'4=1,330L 62,S

The total value of the reactance is:

Xtotal = j2,41 + jO.047 + jl.33 = j3,780I = S

F 13 XTotal= 36,08 = 5, 5kA/phase13 3,78

EMTP (file P3C. dat) shows these results en thus are in accordance with the calculationwith the method of the reactance value.

IV THE RESTRICTED EARTH FAULT DIFFERENTIAL PROTECTION

4.1 The differential earth fault system or Restricted Earth Fault protection

4.1.1 Principal of the REF

The principal of the REF is based on the detection of zero-sequence current. Thisdetection is only possible in case of fault(s) to ground. The system is operative for faultswithin the region between current transformers[6]. The system will remain stable for allfaults outside this zone.



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CTZn2~---------~-r-IIIPT

CT

iaibic

iO\--.....----+ REF

Fig9: Detection of the zero sequence current 10

For the algorithme we need in the first instance two values called:

-the differential current(id)-the through current(it)

4.1.2 The differential current:

The differential current (id) is defined as the difference between the zerosequence current(ia + ib + ic) and the ground-current (in).

id = (ia + ib + ic) -in (27)

4.1.3 The through current:

The through current (it) is defined as the addition of the zerosequence current and theground-current.

it =(ia + ib + ic) + in

2(28)



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where

i.(t) = i.1i cos(wt + ~.) = 2rcf; f = 50Hz

.In

Fig 10: The differential current (id) and the through current (it)

For the correct working of the restricted earth fault we must distinguish a fault inside theprotected zone from a fault outside the zone.

4.2 Differential current and throu~h current in-and outside the protected zone

The principal of the Restricted Earth Fault is based upon the distinction between a faultcurrent inside the protected zone and a fault current outside the protected zone. This thereason that the three line currents inside the protected zone are called: ial, ib 1, icl andthe three line currents outside the protected zone are called: ia2, ib2, ic2.

For a phase-to ground fault we give this following scheme



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in tZn2

ib2

ia2

SW2

-

-ic2

-

ia1-

SW1

ib1-

-ic1

Zn1TRANSF01

Fig!!: Current in-and outside the protected zone

where at t=t1 the amplitude of the current inside respectively outside the protected zoneis equal to:

ia1=1,5kAib1 =292Aic1=292A

ia2=292Aib2=292Aic2=876A

For a fault inside the protected zone it is necessary that id 0 and it=O

After the moment of appearance of a fault inside the protected zone to ground thecurrents (id) and (it) can take the following values at any time.



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idl = [l500cos(wt + cpj + 292cos(wt + /Ph) + 292cos(wt + CPJ] - 876cos(wt) =1,79kA ;cO

itl = [l500cos(wt + cpj + 292cos(wt + /Ph) + 292cos(wt + CPJ] + 876cos(wt) =0

After the moment of appearance of a fault outside the protected zone to ground thecurrents (id) and (it) can take the following values at any time.

For a fault outside the protected zone it is necessary that id=O and it;cO

id2=[292cos(wt + CPJ + 292cos(wt + 'Ph) + 292cos(wt + CPJ] - 876cos(wt) =0

it2=[292cos(wt + CPJ + 292cos(wt + /Ph)+ 292cos(wt + CPJ] + 876cos(wt) = -1,75kA;cO

These results are in accordance with the plot of idl and id2 from file p3A.dat. see Figl2

calcul P3A2000,------,-------.---.:...:.----r------........------.,

oI---...L'---I-lr---'--f--\---f-+--H--\--+--+--+--+-'-+_t_-/--\---H-----___;

-2000 \-----\--~__t+_-+--_t_-I--t-+-'---+-f---'r+---I'r-t---_'d_--4d-;..-----___;

0.250.20.150.10.05-4000 L..- ---I.... .....I.- -JI....- -.l... ---J

o

idlx10--41,------,---,~-_rr_-"1Tr-__m_-----.T""'""""r--~-_mr-___,_----____,

0.250.20.150.10.05-1 '-----_---'----"~ -...._ __WL_ __U...IL.-"------'Lll..__....Lll.._--lL- ----l

o

-0.5

-< 0\---__

id2,

Figl2: Differential current (id), inside respectively outside the protected zone



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conclusion:

We can already conclude by now that concerning the algoritme we only have to look atthe value of (id) and (it) to make a distinction between a fault inside te protected zone anda fault outside the protected zone.

Sometimes- under extreme conditions- the current-transformers can get satisfied inconsequence of an inrush-current or very large fault current. We will return to thissubject later on.

Let us suppose that the primary current is measured bij a Rogowski-coil. Because itsphysical property (the coil doesn't contain iron) it is not possible to get satisfied. So it islinear.

With this simple criteria we can imagine a flowchart diagram for the protection algo-rithme like:

yTRIPPING)

Fig13: flowchart diagram of the protection algoritme for a case without saturation

Of course we have not finished the determination of the protection algorithme because weneed more knowledge about the comportments of the signals (id), (it) in case of saturationand also we need to know how we can recognise a signal inside the protected zone withsaturation from a signal outside the protected zone with saturation?

If the primary current is measured by a current transformer the possibility to saturation isrealistic, for example when a fault occurs where the line current is very large or in caseof energizing of a power transformer where magnetizing inrush current appears[7].



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V THE SATURATION EFFECT

5.1 Output si~nals produced by EMTP for a phase-to ~round fault inside the protectedzone.

Between time t1 and t2 a fault to ground has been simulated. At the point of time t1, asinglephase-to ground sets in inside the protected zone (see fig14).

The output signals of the three line currents and the ground current produced by EMTPand measured by the CT are shown under. see Fig14.

2000

a

=D1) Ii::l

I"0

0.5 f-aE0 1

0te en [sec]

courant point neutre, fondamental

j!I

0.05 0.1 0.15 0.2

III

0.25te en [sec]

13

courant H2 dans Ies phases et Ie neutre2 j~ 1.5~c:: i1) !1) 1 i-"3"0

0.50S

00 0.05 0.1 0.15 0.2

JIII

I0.25

te en [sec]courant point neutre, fondamental

/~,"~//~'-/_-~\

\///_--\,-~--.Jr------J'------./3

r~/I

~ 2 I!r::11) ,

c:: 1 !c:: Ii

A' -la 0.05 0.1 0.15

II

I-l

te en [sec]

courant H2 dans Ies phases et Ie neutre defaut externe1.5,---------,---------,----------,--------,-------,

I

10.250.20.15

,,,,,

"

1 " '., ,,

, ,

, ', ,

0.5 ),(\\1/ \1I I \\ _

al n __ ~l~\

courant H2 dans Ies phases et Ie neutre defaut interne, p3A.dat

----- ---------~- -., ::::.:;.;..;::;.:.:.;.:.;::::c==>=c::::::=:c==+

2,

~ 1.5 ~c I:lJ~ If-='

""::l ic 0.5 f-a0'

0 0.05 0.1 "' 0.15 0.2

--J

IJiI

--I

I0.25

te en [sec]courant point neutre, fondamental

3 1I ~"~"i~ I J2l-c i !,:lJ Ic 1I- Ic i

i III /

0 1 .-J I0 0.05 0.1

te en [sec]0.15 0.2

15

1J

0.25

condition de declencheIl1ent P3A,defaut interneBr----------r-------,------,..----------,-------,

inn7 + . ........................................ .............~................- .

6 , _ ~ .

1 .

,-----------~---------------------~---------------------~---------------------~r! lid5rj..r1

- --.-1

4 .....................................................:.... ..,r.. . , r- -- : .f' ..: ...................................................i _ 1 ~

3 --r + ..f f!P 1........................ . .

2 . . , ..t.. .._T.... ; _ .!"'._._._._._._._._.~.-._._._._._._._._._._._._._._.-!--.- ._._._._._._,_._._._._._._._.'f._._._._.-._._._._._._.-._._._._.,

1Iij

.., ..;i

...! .. .. ~ ~ . ..........................~g .

0.250.20.150.10.050'---------1. ........... .L.- ----1 ---'o

16

condition de declenchement P3A,defaut externeB.---------.--------,-------...,.-------..----------,

7 ~ ~.P.,J! 1... r........................ ..................................................................................I

6 f- . ..............................................................................~ i .

,---------------t---------------------15 ~ ; i...~ .L. 1 .1 .1. .i ~! ! j

_____________________~--------------------- .1 i !4 .. . ~ c .

I3----------!'----------~p+-F~~~~~:::t=~===:::~:=:::l-------------..............................................................'!.................~..........................,............:

! j2 ,........................................ h ,.~, !' h + .1.. ...... : !

i :...._._._._._._._._._._._.1._._._._._._._._._._._._._._. _._.~Ii !1 , ,,, , : t'! ..ig.. H ;! -

; i._._._._._ ....._._._._._._._....._.-\_._._....._._._._._._._._._._._._..;......_._.; ,

0.250.20.150.10.05O'----------'---------'-------...L-------'----------'o

17

199419 janvier

Cx

6.56.56.5

R

COPT EPSILN TOLHAT TSTART

- NEUTRE j40 SUR TFO AMONT- NEUTRE DIRECT Rn=1.E-4 ohm SUR TRANSFO AVAL- DEFAUT BIPHASE/TERRE FRANC

P. BermejoGROUPE SCHNEIDERP.C.C.

+----------------------------------------------------------------------+

BEGIN NEV DATA CASEC +----------------------------------------------------------------------+C P3B.DATC ----------------------------------C ETUDE D'UNE PROTECTION DE TERRE RESTREINTEC ----------------------------------CCCCCCCCCCCCCCC DECLARATION DU NOM DU FICHIER DE SORTIE; ICI : P3B.PL4CSOPEN, UNIT=4 FILE=P3B.PL4 FORM=FORMATTEDCC pas de calcul : 0.0222 ms ; pas de sortie: 0.111 ms soit 9.009 kHzC CE SONT LES MISC. DATA CARDS II-B-1 ,II-B-2C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-POYER FREQUENCY, 50.C DELTAT !MAX XOPT

.222E-4 2.OE-1 50.0ClOUT IPLOT IDOUBL KSSOUT HAXOUT IPUN MEMSAV ICAT NERERG IPRSUP

10000 5 0 1 1 0 0 2 0 0C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-C ***********************C MODELISATION DE LA PCCC ***********************C BRANCH CARDS TYPE 0 IV.A.2.1C NODE NAMES NODE NAMESC BUS1 BUS2 BUS3 BUS4o GENA PRIA .65o GENB PRIB .65o GENC PRIC .65

C *********************************C MODELISATION DU DEPART EN DEFAUTC *******************************C cable 1C ----------C BRANCH CARDS TYPE 0 IV.A.2.1C --BUS1--BUS2--BUS3--BUS4-----R-----XC 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-

SVA1CABLA1 .20 4.0SVB1CABLB1 .20 4.0SVC1CABLC1 .20 4.0

CABLA1 SVA2 .20 4.0CABLB1 SVB2 .20 4.0CABLC1 SVC2 .20 4.0

1.E-31.E-31.E-3

C *************************C MODELISATION DE LA CHARGEC *************************C SEA2C SEB2C SEC2C *******************C IMPEDANCE DU DEFAUT

18

C *******************CABLA1 DEFA1CABLB1 DEFB1

0.10.1

.2884 11940.1781221430.0 6892.6134700553

.02361 3978.96790399350.0 -5957.9167178570.0 -3439.376150684

.2884 11940.1781221430.0 -3439.3761506840.0 -1985.4830642670.0 6892.6134700553

.02361 3978.96790399350.0 -5957.9167178570.0 -3439.3761506840.0 -5957.9167178570.0 -3439.376150684

.2884 11940.1781221430.0 -3439.3761506840.0 -1985.4830642670.0 -3439.3761506840.0 -1985.4830642670.0 6892.6134700553

.02361 3978.9679039935

221421. 353382000.0221421.353382000.00.0221421. 35338200

SECC SECN3

6 SECC SECN

4 SECB SECN

5 PRIC PRIB

3 PRIB PRIA

USE RL$UNITS, 0.50E+02 , O.

1 PRIA PRIC2 SECA SECN

C *******************************C HODELISATION DU TRANSFORHATEUR1C *******************************C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-$VINTAGE, 1,

1 SECA SECN2 SECB SECN

19.2884 59581.657584189

0.0 20008.958586089.02361 6721.4129175

0.0 -29727.20628132

40.0

16692.100203564-8290.06214476716692.100203564-8290.062144767-8290.06214476716692.100203564

SEC2 SECN23

3 PRB2 PRN

USE RL$UNITS, 0.50E+02 , o.

1 PRA2 PRN2 SEA2 SECN2

$VINTAGE, 0,$UNITS, -1. ,-1.

USE RLC *************************************************C HODELISATION DE L'IHPEDANCE DE NEUTRE DU TRANSF01C *************************************************C BRANCH CARDS TYPE 0 IV.A.2.1C NODE NAMES NODE NAMES-----R-----X-----CC 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-C BUS1 BUS2 BUS3 BUS4

SECNC *******************************C HODELISATION DU TRANSFORHATEUR2C *******************************$VINTAGE, 1,

1 SEA2 SECN22 SEB2 SECN2

4 SEB2 SECN2

5 PRC2 PRN

6 SEC2 SECN2

0.0 -9985.69862399.2884 59581.657584189

0.0 -9985.698623990.0 -3354.397653070.0 20008.958586089

.02361 6721.41291750.0 -29727.206281320.0 -9985.698623990.0 -29727.206281320.0 -9985.69862399

.2884 59581.6575841890.0 -9985.698623990.0 -3354.397653070.0 -9985.698623990.0 -3354.397653070.0 20008.958586089

.02361 6721.4129175$VINTAGE, 0,$UNITS, -1.,-1.

USE RLC 3456789-123456789-123456789-123456789-123456789-123456789-96 SEA2 SECN2 8888.

-2.3400E+00 -6.2431E+01-8.7949E-01 -6.0787E+01-5.8633E-01 -6.0419E+01-2.6385E-01 -5.9139E+01-1.1727E-01 -5.7856E+01-4.3975E-02 -5.6759E+01

1.4658E-02 -5.4562E+015. 1304E-02 -5. 1999E+018.5017E-02 -4.7604E+011.0261E-01 -4.0281E+011.1727E-01 -2.9295E+011.4658E-01 1.9591E+011.6124E-01 2.7097E+012.0522E-01 3.6619E+012.6385E-01 4.3941E+013.1954E-01 4.7604E+014. 1776E-01 5. 1267E+015.7168E-01 5.4562E+017.8422E-01 5. 7124E+011.0261E+00 5.8957E+011.4658E+00 6.0787E+012.3453E+00 6.2252E+013.5181E+01 6.2617E+01

9999.96 SEB2 SECN2 SEA2 SECN2 8888.96 SEC2 SECN2 SEA2 SECN2 8888.C *************************************************C MODELISATION DE L'IMPEDANCE DE NEUTRE DU TRANSF02C *************************************************C BRANCH CARDS TYPE 0 IV.A.2.1C NODE NAMES NODE NAMES R X CC 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-C BUS1 BUS2 BUS3 BUS4

SVN 1.E-4SECN2 1.E+6

BLANK CARD TERMINATING BRANCHESC ***********************C HODELISATION DU DEFAUTC ***********************C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456~-o DEFA1 20.0E-3 1.0E1o DEFB1 20.0E-3 1.0E1

C *******************C SVITCHES DE ME SUREC *******************C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-

SECA SVA1 MEASURING 1SECB SVB1 MEASURING 1SECC SVC1 MEASURING 1SVA2 PRA2 MEASURING 1SVB2 PRB2 MEASURING 1SVC2 PRC2 MEASURING 1

PRN SVN MEASURING 1BLANK CARD TERMINATING SVITCHESC **************************C MODELISATION DE LA SOURCEC **************************C STATIC ELECTRIC NETVORK SOURCES VII.C.4 TYPE 14C 3456789-123456789-123456789-123456789-123456789-123456789-123456789-123456789-C NAMEST AMPLITUDE FREQUENCY PHASE Al TSTART TSTOP14 GENA 51.0E3 50.0 090.0 -1.0014 GENB 51.0E3 50.0 -030.0 -1.0014 GENC 51.0E3 50.0 +210.0 -1.00BLANK CARD TERMINATING SOURCESC NAM1 NAM2 NAM3 NAM4 NAM5 NAM6 NAM7 NAM8 NAM9C GENA GENB GENC PRIA PRIB PRIC SECA SECB SECCC PRA2 PRB2 PRC2 SEA2 SEB2 SEC2CABLA1CABLB1CABLC1C PRN SECN2BLANK CARD TERMINATING OUTPUTBLANK CARDBEGIN NEV DATA CASE

21

calcul P3B5000 r------,.---------r-------,--------.------------,

0.250.20.150.10.05

01----.

-5000

-10000 l-- ----l. ---'- ----'- --'-- ---'o

ial

10000 r------,.---------r------,-------.------------,

0.250.20.150.1

................j. ..1.. .. ..............r .... ....! i

0.05

01--------'

-5000 L-- ------' ----1. ----L.. ----'-- ---'o

ibl

22

400 calcul P3B

200

200 calcul P3B

100

-< 0 ..................-..............

-100

-2000 0.05 0.1 0.15 0.2 0.25

ia2

200

100

-< 0

-100

-200 --0 0.05 0.1 0.15 0.2 0.25

ib2

24

400 calcul P3B

200

-< 0

-200

-4000 0.05 0.1 0.15 0.2 0.25

ic2

1000

500

-< 0 ......................... _..,.......

-500

-10000 0.05 0.1 0.15 0.2 0.25

in

25

courant differentiel et traversant sature2,--------,--------r---------,----------,,--------,

1.5 ,--....-...- ....-._.....-.-.. ""-'- '-"'-- .-....-----, '-_...- r ..---~-----:-:a_

1 1-....... ,............. . -. ,........... -. . - -.-. .~ _..~

0.,.-.....-f--4:---3~f-4_+_+-44~f-_f__++-4-4~t--_t- ........--..- ............................-'-H'-~-'+"-"- _. --' - ~.- ,-- "-~~--$ _ -.. _ _..;._ _ - .._ _ - -

o.5 ..- - _ ~- -.. : - - - ...---.; '- :..__.._. - - ..- ---- .t-. .- -. ..-

- 1 1- _.......... . - .._ ~~---'3~-::.._- ._._._; _ __._ _ __._-

........-. ..- :-.~ _._-_.-.._._ _ _......_ .

.;.-_...

+

~o...

~C;l....

::l~C;ltiltilQ)Z. -0.5as~til...

"t:ltil...

- 1.5 ~ -... .. -;11l''';-'''''-''-'iI! ..~_.........- ..._ _-., - .._ _ _ - - - ..!

0.250.20.150.10.05-2 '-------.........-------'----------'-------'----------'

o

t en [sec]

26

courant differentiel et traversant sature5,.........---------,.--------.-------,....---------,--------,

'---.;:;;;0.250.20.150.10.05

- a.5 1\-1VJ..--i - ~ ! ; - .

-1 L..- --'-- ---'-- ---' -'---- ---'a

ic2

0.250.20.150.10.05

\.... + _ , + +- .

0.1

0.05

< a

-0.05

-0.1a

in

43

!l

0.250.20.it50.10.05

X10-5 courant differentiel et traversant non sature:defaut interne:p3c8 '-1-----------,-------------,----------,,.------------.------------,

I61 ++-/y\++ \+ \4 ,\

[ + J+/ +\++\ --, ~--2 r+++ * ++ \ / ~, -~-------------------------- .

1++++ + / +++Y-d-+~~IIIIIIIIII'IIIIIIIIII,IIIIIIIIIIIIIIII'IIIIIIIIIIIIII Lf0' -if(1o

~i

I0.250.20.150.10.05

te en (sec]courant proportionnel non sature:defaut interne:p3c

3001II

........,

200 ~*..........

C In.l 100~0-

I

oL0

te en (sec]

44

courant differentiel et traversant sature:defaut interne:p3c

0.05 0.1 0.15 it 0.2 0.25

,,

~ 2001~ I.& 100 t-

ii

te en [sec]courant proportionnel sature:defaut interne:p3c

300.-,----------------.,------------------.,--------,

J ~o,-I__.1-1 ---'-- -------'- --'----- ------'--,--------'1o 0.05 0.1 0.15 0.2 0.25

te en [sec]

45

0.250.20.150.10.05a

xl0-4 courant differentiel et traversant non sature:defaut externe:p3c1 'I-------,-------------,-----------r-------~------l

I + I

L -F+ +~~ :

0.5 1 +\ id 1I + + ++I ++ ++~+. ~~-++t~ 1111111111111111 \11111111\1111111111\ 11\11\ 111111111++++++l--q.+ +*+-tt-'

ai+ + + it

r::Q)

---J

JI

i0.250.2

=~0.150.10.05

te en [sec]xlO-3 courant proportionnel non sature:defaut externe:p3c41,---------r--------,-------~----------,-------_

:~1 ~o1\ /~"-a

c:Q)

0..

te en [sec]

46

xl0-4 courant differentiel et traversant sature:defaut exteme:p3c1 '--1--+-,-----------,-------------,.---------,-----------,1

1

-P-+ +++ -l+

it+

a 0.05 0.1 0.15 0.2 0.25

0.05 0.1te en [sec]

0.15 0.2

47

courant H2 dans Ies phases et Ie neutre defaut interne:p3c5,-------.----------,----------,-------------,-------,

0.25

i

~0.25

1

0.21

0.15

0.5

!r\"" ..(\ :-" (,~, ,-r~~ / \.' I J \ ... X-/ \-\Q.) I' 1\ \ \

I ,. I \1 II II i! \ \./,,\\~: ::' '-~

L__L.:)~__~l ----' .J....'_--~_-=:-==:==':=:::-~:2._ ~o r, , ! =o 0.05 0.1 0.15 0.2

te en [sec]xlO-4 courant point neutre, fondamental

1 ~,--------,----------,-------,---------,---------,

j!

te en [sec]

48

xl0-4 courant H2 dans Ies phases et Ie neutre defaut externe:p3c8

1

I I I I

6rJ\. !ML , '\;: ! \\41~{! C'\' f \\2 / ./ \--..

, ' \I ' \I I,J \a~~_/----'-,'/_--~

a 0.05 0.1 0.15 0.2te en [sec]

1

i--j

~0.25

0.20.150.1

X 10-4 courant point neutre, fondamental1~--------------------'---------'---------'i

IJ

0.25te en [sec]

49

courant differentiel et traversant sature15 r--------r------......,....--------,-------,....-----------,

10 1- -- _.- - ..-.-+ id------.~ ----.-.. ~-------~--~_------------4--------.--- -~ ---

5 1--,,--,,-,--- ... __....J _+_ _ ,_,_,_ . ._ _..___._ _._..___ _.

o.,..-_~+~:......~... '-- .f....:t_ ..

- 5 r--....---.---. -_. #-----111'-

- 1 0 --..--- --- - -.- i--'''-'''''-- .. - - -. .._.__~---.-.------ ..-~.- -- -.-.---..----l._-.---.._-..--.._.-.---

- 15 _._..- .._- _ _ -.-- ..- ,.-_._.._ _-- - -.- , - -._......--..--..-f..-...._.. ..- ~ - -- -

0.250.20.150.10.05-20 '-- '-- '-- .1....- .1....- ----'

o

t en [sec]

50

0.250.20.150.1

id

courant differentiel et traversant sature

0.05

-_~____--------t-___--_~__-------__-_---+._ _---- -.- -..----.

1'''T'i' ---'r-----------!---------------;. _ _ - _-.

I-:--::~rlr-ii--~-------------_-_----..j.------_-------,--.---.---- ..---.-..- .....

x10-42

1.5,.......,

(ones(inn)*O.05/sqrt(2));% sur le courant diffc2=id> (ones(id)*O.05/sqrt(2));% sur le courant prop.c3=ip> (ones(ip)*5);% globalc4 = cl.*c2.*c3;

%traceclgplot(te,cl+6.5,te,c2+4.5,te,c3+2.5,te,c4+0.5);gridtitle('condition de declenchement P3D,defaut interne ');gridgtext ( , inn' )gtext ( , id ' )gtext ( , i P , )gtext ( , i g , )print

94

VoorbladSummaryContentsIntroduction1. Simulation of the S.E.P.A.M. 2000.2. Simulation of an electrical networks.3. Calculation of fault current in steady state condition.4. The restricted earth fault differential protection.5. The saturation effect.6. Result of the protection algorithme.7. Recommendation for further works.BibliographyAppendix

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