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A SIMULATION MODEL FOR TRANSFORMER INTERNAL FAULTS, BASE FOR THE
STUDY OF PROTECTION ANDMONITORINGSYSTEMS
P. Bertrand *, A. Devalland **, P. Bastard **** Merlin-Gerin, **
France-Transfo, *** Ecole Sue rie ure dElectrici t6.
FranceINTRODUCTIONThe s tudy of in temal faul t s in transformers,
i .e. thecalculation of the fault current and i ts external
detection(phase currents) as a function of the location and
theamplitude of the fault, is of great interest from severalpoints
of view :- evaluation of protection devices efficiency in order
to
get improvement.diagnosis of a true fault using cu rrent
recordings.preventive action when manufacturing transformers,and at
f i rs t heavy duty ones (e .g . arc furnacetransformers), consist
ing of reinforced insulation inmeas where faults may be difficult
to detect.
--
Our main purpose was to develop a transformer
differentialprotection, with improved internal fault detection. It
is thereason why we have tried to get a deep insight into
thephenomenon.Although several measuring results appear in
publications((11, [2]), the studies are recent and the papers
remainincomplete [31.This situation has led us to manufacture a
special, multipleoutput transformer. This device, designed on the
basis of a100 kVA M V LV transformer, has enabled us to cany
outnumerous turn-to-turn and turn-to-earth short-circuits.In
association with these tests we have developed a digitalmodel of
the t ransformer , val idated by the actualmeasurements.This model,
when applied to a HV network transformer, hasenabled the efficiency
of the usual protection devices to beexamined, in particular that
of the transformer differentialprotection.1.CARRIEDOUT M E A S U R
E EDescriotion of the test transformerThis transformer was
manufactured by France-Transfo onthe basis of a distribution
transformer for which thespecification is as follows :
rated power : 100 kVAtransformation ratio : 5500V / 410Vcoupling
: y nshort-circuit voltage : 3.96%LV winding : 67 turns wound in 2
layers
-The external coil is fitted with taps as shown in figure 1.
HV winding : 1556 turns wound in 8 layers
layer : 1 2 3 4 5 6 7f igure : detai l s of a n H V wi n d i n
gRecordind methodThe short circuits were simulated by closing a
contactor.The tested transformer was supplied at reduced voltage by
anautot ransformer wi red in ser ies wi th an i so lat
ingtransformer.Eight measurement channels connected to a PC
enabledrecording of the supply voltages, the phase currents and
thefault current, triggered by the contactor closing.Turn-to-tum
faults
f i g u re 2 : c u r r e n t d i s t r ib u t i o n i n t h e w
i n d i n g sThe recorded currents are reduced to the rated
voltage, thenexpressed in per unit values, taking the primary
ratedcurrent as reference.At first, the origin of the fault is
fixed as being on a phaseterminal of the transformer and the number
of short-circuited turns varies. The current in one phase and
thecurrent in the fault are shown in figure 3.When a largenumber of
turns are short-circuited, the current is mostlyl imi ted by the
short c i rcui t power supplying thetransformer; the high values of
recorded current show thatthis short-circuit power is high.
. L I .
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x In I6
4
200
20 4 6 80 100span of the fault
figure 3 : turn-to-turn fault : span influenceThen, the span of
the fau lt remains fixed to half a layer (thatis 6 25 of the
winding), and its origin moves. It can benoticed that the fault
position in the winding has very littleeffect, at least for this
type of winding technology.
xInso
4
30
2010
2 4 6 8 100location of the fault
figure 4 : turn-to-turn fault : location influence
20 40 60 80 100location of the faultf i g ure : t urn- t o - ea
r t h f a u l t : l o c a t i o ni n f l u e n c eConclus ienThe
results presented here give a synthetic view of themeasurements
carried out. Nevertheless, it is difficult toreach any practical
conclusion because the results depend alot both o n the testing
conditions (short circuit power ofthe test circuit) and on the
transformer (low power, thushigh winding resistance). It is
essential to generalize theresults by calculation.2. MODEL OF I
NTERN LFAULTPurpose of the modelThis digital model enables the
results obtained with the 100kVA test transformer to be extended to
HV networktransformers.It has been validated by the measurements.
It thus fits wellto tranformers wound in l o n g layers. It could
also beextended to other types of windings, through an
additionalexperimental validation operation.
Tum-to-earth faults Two requirements have steered us in the
production of themodel :- ease of use, which implies first the use
of available data
o n l y ,compatibility with a standard transient program.
Consequently, the model produced is an extension of
thetransformer model of EMTP (Electromagnetic Transient
2 E Program).Calculation nnnc&EMTP gives the opportunity of
modeling the transformerby coupled circuits. An auxil iary routine,
BCTRAN,calculates the [R.L] matrix from the results of the
no-loadand short circuit tests of the transformer [4]
. . .
ET$E
figure 5 : current distribution in the windingsThe position of
the fault along the winding moves. Thefault current and the current
in one phase are shown infigure 6.
The principle used to calculate a turn-to-turn or
turn-to-earthfault is to split the faulty winding. Thus the initial
matrixrepresenting a three phase transformer (6 windings) is a
1 . 2 1 . 2
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6x6 matrix;it becom es a 7x7 matrix to study a
turn-to-earthfault and an 8x8 matrix for a turn-to-turn fault.
Oncecalculated, these matrices are directly used by E M F , as
anyother element.
of a windine.
In order to explain the calculationcarried out, we chose as an
example tosplit winding nO1.The initial values calculated from the
no-load and shortcircuit tests of the transformer are the self
inductances ofthe w indings, L 1 et L2. a s well as the mutual
inductanceM12. Splitting winding 1 involves calculating the new
selfinductances, L1' et L1 and mutual inductances, M1'1 ,M i 7 3 M
i 2 .These inductances can be calculated from two leakagefactors;
these factors are of great importance ; the precisionof the model
is directly linked to the precision of theirevaluation.- 01'1 is
the leaka ge factor between windings 1' and 1 .- 01'2 is the
leakage factor between the largest of the twowindings 1' and 1 and
the secondary winding.Using these factors, calculations can be
carried on, basedon two simple principles :- self-consistency : the
connection in series of windings
1' and 1 must enable the initial results to be met.-
proportionality : the transformation ratio between
windings 1' and 1 is equal to the turn ratio (goodapproximation
in equation 5 below).
M;ls, dispersions1 q , l , ,= l - -L1, .LIS.
Table o f init ia l equat ions
1 0 M y , = M l z - M r ~Table of f inal equat ions
The leakage inductance between two coils can be calculatedfrom
the electromagnetic energy stored in the coils. This issimplified
by the following hypotheses :
- no saturation phenomena occurs.
current density is constant in the windings,field H is parallel
to the axis of the core,field H is symmetric in relation to the
core axis,
This calculation has been extensively detailed. We givebelow a
reminder of the way to get the leakage inductancebetween the
primary and the secondary winding of atransformer:
first of all, the shape of field is plotted (figure 8).- the
stored energy in the windings can then be calculated
depending on their dimensions :W = h j j j H 2 d v + ~ ~ j j H Z
d v + h j j j H 2 d v ,where va,vb and vc are the respective
volumes of theinternal winding, the inter-winding space and of
theexternal winding.- f inally the total leakage inductance reduced
towinding 1, LCC, is calculated using the equationW =-Lcc .i?,
where il is the current in winding 1 usedto plot the shape of field
H .
a v b C
12
figure 8 : f ie ld shape created by two concentricc o i l
sUsually, the result of this calculation, LC C, s corrected by
acoefficient. Here, we can determine exactly this coefficient,since
the short-circuit test has enabled the actual value ofLcc to be
calculated.k1 = Lcct,,/Lccis thus a first corrective factor.When
the windings are not of the same height, a secondcorrective factor
is applied [ 6 ] )
a b c
1 . 2 1 . 3
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Using the tools described above, it becomes possible toevaluate
the leakage factors that we need. As an example,consider the
leakage factor between the two parts 1' and 1of the external
winding 1 created by a turn-to-earth fault.
0,020 -
I
0,015
figure 9 : f ie ld shape created during a turn-to-earth
fault
L n /c
The plot of the field shape (figure 9) enables the storedenergy
W in the faulty winding to be calculated and thus thetotal leakage
inductance Lccl' reduced to winding 1'. Thisinductance is corrected
by the factor k l previouslycalculated and by the factor k2 if one
of the coils 1' and 1spans over less than a layer.The leakage
factor between the two coils 1 ' et 1 iscalculated by (J,,,~~2 he
self inductance of portion1' being evaluated from the self
inductance of the wholewinding : L . =Ll
he result of this calculation applied to the test transformeris
shown in figure 10. The leakage factor rapidly increaseswhen the
fault occurs on an outer layer. Moreover, there is aslight decrease
in leakage when the 2 winding parts areoverlapped.
LCCL .
[nr :n l , , r
0 00 0.20 0.40 0,60 0.80 1.00location of the fault
figure 1 : calculated leakage factorComoarison between
calculation and measurementBy reprodu cing 16 turn-to-earth fau lts
and several tens ofturn-to-turn faults, we have noticed a
difference between thecalculated currents and the recorded currents
which neverexceeds 10 in modulus and 10 in phase. The
correlationbetween tests and simulation is even better when the
fault
only involves windings spanning at least once the fullwinding
height.3. APPLICATION : S T v R K U U U R Y FAULTS IN A
.escriDtioIL;rated power : 15 MVAshort-circuit voltage :
9.8%
- no-load current : 0,46%- no-load losses : 11.3 kW-
short-circuit losses : 98.5 kW- rated voltage : 33000V & 12%
with 17 taps
8 layers of 8 0 turns + 1 layer of 80 turns (tap changer)-
internal diameter : 590 mm, radial thickness : 77 mm,height : 780
mm- rated voltage : 16100V- 5 layers of 36 turns- internal diameter
: 436 mm, radial thickness : 64 mm,height : 780 mm
HV winding : (delta)
LV winding : (star)
Turn-to-earth faultsThe differential current in a phase and the
residual current(which is also the fault current) are shown in
figures 11 and12 for a fault with 1 Q resistance and for two
differentearthing systems :- for a directly earthed system, the
zero-sequence current
depends only few on the fault position and isessentially limited
by the zero-sequence short circuitpower (150 MVA for this
calculation).for a neutral earthed through an impedance of
highenough value (fault current limited to 500 A or less),
thezero-sequence current varies between 0.5 and 1 time thelimited
current ; t is well known that a zero-sequenceovercurrent
protection is the best way of detecting thisfault.
-
l8
0 20 40 60 80 100location of the faultfigure 11 : turn-to-earth
fault ; direct ly eartheds y s t e m
1 . 2 1 . 4
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.location of the faultf igure 12 : turn-to-earth fault : neutral
currentl imited to 500ATun-to-turn faultsThe turn-to- tum fault
current has been calculated by varyingthe fault span, the short
circuit power of the system and thefault resistance.As soon as the
number of short-circuited turns is large, thevalue of differential
current depends basically on thesystem's short-circuit power. When
the fault involves lessthan 10 of the winding, this current is very
sensitive tothe fault resistance , the value of w hich is not
controlled.A turn-to-turn fault is usually detected by a
differentialprotection relay. Indeed, this is the main reason for
usingthis protection. Its sensitivity, of around 30 ,
enablesdetection, in the simulated case, of faults involving at
least13 turns.This example shows how useful the protection is : the
faultbetween two layers is easily detected. It also shows
itsperformance limit : detection of a fault between two turnsseems
impossible. It would, however, be interesting toimprove the
sensitivity of such a relay. Indeed, the shortcircuiting of one
adjusting step, i.e. 1.5 of the winding,creates a differential
current of 20 to 50 of rated current,depending on the fault
resistance ; n order to ensure correctdetection, a sensitivity of
10% would be required.
- curve 1 :R fault = O.OOlQ infinite SC pow ercurve 2 : R fault
= 0,OOlQ. SC power = 15OMVA
~ cuwe 3 : R fault = 0,lQ ; infinite SC powercurve 4 : R fault =
0,lQ SC power = 15OMVA
-2 -./ 4. /. /.7
r ///.
1 2 3 4I
span of the faultfigure 14 : detail for faults of small
spanCONCLUSIONThe calculation of currents during an internal fault
in atransformer requires a model validated by measurements. Inorder
to be of general use , his model must be compatiblewith a standard
transient software and only require datawhich are available. The
model presented meets all theserequirements. It would be worth
improving it, in order totake in to account d i f ferent t
ransformer windingtechnologies.In the third part of this report, we
have shown with anexample the use that could be made of such a
model : itenables quant i t a t ive evaluat ion of the qual i ty of
atransformer protection. Other uses are planned such as theevaluat
ion by the t ransformer manufacturer of theconsequences of possible
faults, and the diagnosis of actualfaults by the
user.BIBLIOGRAPHY
[I ] GEC ALSTHOM MEASUREMENTS Protective RelayApplication Guide.
1987[2] S AUSTEN STIGANT A.C. FRANKL IN The J&P
Transformer Book. Newnes-Butterworths[3] J .L. BINARD J .C .
MAUN Power Transformer
Simulation including Inrush C urrents and Internal Faults3 me
confkrence internationale IMACS-TCI P O Nancy,France 1990
[4] V. BRADWADJN H.W. DOMM EL 1.1 DOMM ELMatr ix Representat ion
of Three-phase N-WindingTransformers. IEEE Transactions, vol
PAS-I01 n06 ,1 9 8 2
[ ] M. DENIS-PAPIN La pratique Indus tr iel l e
desTransformateurs. Editions Alibin Michel. 1951
figure 13 : differential current in a turn-to-turnfault of
variable span
1 . 2 1 . 5
