-
Appendix C
Transformers and Reactors
A power transformer is an important component of the power
system. Thetransformation of voltages is carried out from
generating voltage level to trans-mission, subtransmission,
distribution, and consumer level. The installed capacityof the
transformers in a power system may be seven or eight times the
generatingcapacity. The special classes of transformers include
furnace, converter, regulat-ing, rectier, phase shifting, traction,
welding, and instrument (current andvoltage) transformers. Large
converter transformers are installed for HVDCtransmission.
The transformer models and their characteristics are described
in the relevantsections of the book in various chapters. This
appendix provides basic concepts, anddiscusses autotransformers,
step-voltage regulators, and transformer models notcovered
elsewhere in the book.
C.1 MODEL OF A TWO-WINDING TRANSFORMER
We represented a transformer model by its series impedance in
the load ow andshort-circuit studies. We also developed models for
tap changing, phase shifting, andreactive power ow control
transformers. Concepts of leakage ux, total ux, andmutual and
self-reactances in a circuit of two magnetically coupled coils
aredescribed in Chap. 6 and Eq. (6.32) of a unit transformer are
derived. These canbe extended and a matrix model can be written
as
12n
r11 r12 r1nr21 r22 r2n rn1 rn2 rnn
i1i2in
L11 L12 L1nL21 L22 L2n Ln1 Ln2 Lnn
d
dt
i1i2in
C:1
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
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A two-winding transformer model can be derived from the circuit
diagramshown in Fig. C.1(a) and the corresponding phasor diagram
(vector diagram)shown in Fig. C-2. The transformer supplies a load
current I2 at a terminal voltageV2 and lagging power factor angle
2. Exciting the primary winding with voltage V1produces changing ux
linkages. Though the coils in a transformer are tightly coupled
Transformers and Reactors 757
Figure C-1 (a) Equivalent circuit of a two-winding transformer;
(b), (c), (d) simplicationsto the equivalent circuit.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
by interleaving the windings and are wound on a magnetic
material of high perme-ability, all the ux produced by primary
windings does not link the secondary. Thewinding leakage ux gives
rise to leakage reactances. In Fig. C-2, m is the main ormutual ux,
assumed to be constant. The emfs induced in the primary windings is
E1which lags m by 908. In the secondary winding, the ideal
transformer produces anemf E2 due to mutual ux linkages. There has
to be a primary magnetizing currenteven at no load, in a time phase
with its associated ux, to excite the core. Thepulsation of ux in
the core produces losses. Considering that the no-load currentis
sinusoidal (which is not true under magnetic saturation, see Chap.
17), it must havea core loss component due to hysteresis and eddy
currents:
I0 I2m I2e
qC:2
where Im is the magnetizing current, Ie is the core loss
component of the current, andI0 is the no-load current; Im and Ie
are in phase quadrature. The generated emfbecause of ux m is given
by
E2 4:44fn2m C:3where E2 is in volts when m is in Wb/m
2, n2 is the number of secondary turns, and fis the frequency.
As primary ampe`re turns must be equal to the secondary
ampe`returns, i.e., E1I1 E2I2, we can write:
E1=E2 n1=n2 n andI1=I2 n2=n1 1=n
C:4
758 Appendix C
Figure C-2 Vector diagram of a two-winding transformer on
load.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
The current relation holds because the no-load current is small.
The terminalrelations can now be derived. On the primary side, the
current is compoundedto consider the no-load component of the
current, and the primary voltage is equalto E1 (to neutralize the
emf of induction) and I1r1 and I; x1 drop in the primarywindings.
On the secondary side the terminal voltage is given by the induced
emfE2 less I2r2 and I2x2 drops in the secondary windings. The
equivalent circuit istherefore as shown in Fig. C-1(a). The
transformer is an ideal lossless transformerof turns ratio n.
In Fig. C-1(a) we can refer the secondary resistance and
reactance to theprimary side or vice-versa. The secondary windings
of n turns can be replacedwith an equivalent winding referred to
the primary, where the copper loss in thewindings and the voltage
drop in reactance is the same as in the actual winding.We can
denote the resistance and reactance of the equivalent windings as r
02and x 02:
I21 r02 I22 r2 r 02 r2
I22I21
! r2
n1n2
n2r2
x 02 x2I2E1I2E2
x2
n1n2
n2x2
C:5
The transformer is an ideal transformer with no losses and
having a turns ratioof unity and no secondary resistance or
reactance. By also transferring the loadimpedance to the primary
side, the unity ratio ideal transformer can be eliminatedand the
magnetizing circuit is pulled out to the primary terminals without
appreci-able error, Fig. C-1(b) and (c). In Fig. C-1(d) the
magnetizing and core loss circuit isaltogether omitted. The
equivalent resistance and reactances are
R1 r1 n2r2X1 x1 n2x2
C:6
Thus, on a simplied basis the transformer positive or negative
sequence model isgiven by its percentage reactance specied by the
manufacturer, on the transformernatural cooled MVA rating base.
This reactance remains fairly constant and isobtained by a
short-circuit test on the transformer. The magnetizing circuit
compo-nents are obtained by an open circuit test.
The expression for hysteresis loss is
Ph KhfBsm C:7where Kh is a constant and s is the Steinmetz
exponent, which varies from 1.5 to 2.5,depending on the core
material; generally, it is =1.6.
The eddy current loss is
Pe Kef 2B2m C:8where Ke is a constant. Eddy current loss occurs
in core laminations, conductors,tanks, and clamping plates. The
core loss is the sum of the eddy current and hyster-esis loss. In
Fig. C-2, the primary power factor angle 1 is > 2.
Transformers and Reactors 759
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
C.1.1 Open Circuit Test
Figure C-3 shows no-load curves when an open circuit test at
rated frequency andvarying voltage is made on the transformer. The
test is conducted with the secondarywinding open circuited and
rated voltage applied to the primary winding. For high-voltage
transformers, the secondary winding may be excited and the primary
wind-ing opened. At constant frequency Bm is directly proportional
to applied voltage andthe core loss is approximately proportional
to B2m.The magnetizing current risessteeply at low ux densities,
then more slowly as iron reaches its maximum perme-ability, and
thereafter again steeply, as saturation sets in.
From Fig. C-1 the open circuit admittance is
YOC gm jbm C:9This neglects the small voltage drop across r1 and
x1. Then:
gm P0V21
C:10
where P0 is the measured power and V1 is the applied voltage.
Also,
bm Q0V21
S20 P20
qV21
C:11
where P0,Q0, and S0 are measured active power, reactive power,
and voltampe`res onopen circuit. Note that the exciting voltage E1
is not equal to V1, due to the drop thatno-load current produces
through r1 and x1. Corrections can be made for this drop.
C.1.2 Short-Circuit Test
The short-circuit test is conducted at the rated current of the
winding, which isshorted and a reduced voltage is applied to the
other winding to circulate a full-rated current:
Psc I2scR1 I2scr1 n2r2 C:12
760 Appendix C
Figure C-3 No-load test on a transformer.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
where Psc is the measured active power on short-circuit and Isc
is the short-circuitcurrent.
Qsc I2scX1 I2scx1 n2x2 C:13
Example C.1
The copper loss of a 2500 kA, 13.8-4.16 kV deltawye connected
three-phase trans-former is 18 kW on the delta side and 14 kW on
the wye side. Find R1, r1, r2, and r
02
for phase values throughout. If the total reactance is 5.5%, nd
X1, x1 x2, and x02,
assuming that the reactance is divided in the same proportion as
resistance.The copper loss per phase on the 13.8 kV side = 18/3 = 6
kW and the current
per phase = 60.4 A. Therefore, r1 = 1.645 ohms. Similarly for
the 4.16 kV side, thecopper loss = 14/3 = 4.67 kW, the current =
346.97 A, and r2 = 0.0388 ohms; r2referred to the primary side = r
02 = (0.0388)(13.8
p3/4.16)2= 1.281 ohms, and
R1 = 2.926 ohms. A 5.5% reactance on a transformer MVA base of
2.5 = 12.54ohms on the 13.8 kV side, then x1 = (12.54)(1.645)/2.926
= 7.05 ohms andx 02 = 5.49 ohms. Referred to the 4.16 kV side x2 =
0.166 ohms. The transformerX=R ratio = 4.28, which is rather
low.
Example C.2
The transformer of Example C.1 gave the following results on
open circuit test: opencircuit on the 4.16 kV side, rated primary
voltage and frequency, input = 10 kWand no-load current = 2.5 A.
Find the magnetizing circuit parameters.
The active component of the current Ie = 3.33/13.8 = 0.241 A per
phase.Therefore,
gm 10 103
3 13:8 1032 0:017 103mhos
The magnetizing current is
Im I20 I2e
q
1:442 0:2412
p 1:42A
The power factor angle of the no-load current is 9.638, and bm
from Eq. (C.11) is0:103 103 mhos per phase.
C.2 TRANSFORMER POLARITY AND TERMINAL CONNECTIONS
C.2.1 Additive and Subtractive Polarity
The relative direction of induced voltages, as appearing on the
terminals of thewindings is dependent on the order in which these
terminals are taken out of thetransformer tank. As the primary and
secondary voltages are produced by the samemutual ux, these must be
in the same direction in each turn. The load current in
thesecondary ows in a direction so as to neutralize the mmf of the
primary. How theinduced voltages will appear as viewed from the
terminals depends on the relativedirection of the windings. The
polarity refers to the denite order in which theterminals are taken
out of the tank. Polarity may be dened as the voltage vector
Transformers and Reactors 761
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
relations of transformer leads as brought out of the tank.
Referring to Fig. C-4(a) thepolarity is the relative direction of
the induced voltage from H1 to H2 as comparedwith that from X1 to
X2, both being in the same order. The order is important in
thedenition of polarity.
When the induced voltages are in the opposite direction, as in
Fig. C-4(b), thepolarity is said to be additive, and when in the
same direction, as in Fig. C-4(a), it issaid to be subtractive.
According to the ANSI standard all liquid immersed powerand
distribution transformers have subtractive polarity. Dry-type
transformers alsohave subtractive polarity.
762 Appendix C
Figure C-4 (a) Polarity and polarity markings, subtractive
polarity; (b) additive polarity.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
When the terminals of any winding are brought outside the tank
and markedso that H1 and X1 are adjacent, the polarity is
subtractive. The polarity is additivewhen H1 is diagonally located
with respect to X1, Fig. C-4(b). The lead H1 isbrought out as the
right-hand terminal of the high-voltage group as seen when
facingthe highest voltage side of the case. The polarity is often
marked by dots on thewindings. If H1 is dotted, then X1 is dotted
for subtractive polarity. The currents arein phase. Angular
displacement and terminal markings for three-phase transformersand
autotransformers are discussed in Ref. [1].
C.3 PARALLEL OPERATION OF TRANSFORMERS
Transformers may be operated in parallel to supply increased
loads, and for relia-bility, redundancy, and continuity of the
secondary loads. Ideally, the followingconditions must be
satised:
. The phase sequence must be the same.
. The polarity must be the same.
. Voltage ratios must be the same.
. The vector group, i.e., the angle of phase displacement
between primary andsecondary voltage vectors, should be the
same.
. Impedance voltage drops at full load should be the same, i.e.,
the percentageimpedances based on the rated MVA rating must be the
same.
It is further desirable that the two transformers have the same
ratio of percen-tage resistance to reactance voltage drops, i.e.,
the same X=R ratios.
With the above conditions met, the load sharing will be
proportional to thetransformer MVA ratings. It is basically a
parallel circuit with two transformerimpedances in parallel and a
common terminal voltage:
I1 IZ2
Z1 Z2I2
IZ1Z1 Z2
C:14
where I1 and I2 are the current loadings of each transformer and
I is the total current.In terms of the total MVA load, S, the
equations are
S1 SZ2
Z1 Z2S2
SZ1Z1 Z2
C:15
While the polarity and vector group are essential conditions,
two transformers maybe paralleled when they have:
. Unequal ratios and equal percentage impedances
. Equal ratios and unequal percentage impedances
. Unequal ratios and unequal percentage impedances
It is not a good practice to operate transformers in parallel
when:
. Either of the two parallel transformers is overloaded by a
signicant amountabove its rating.
Transformers and Reactors 763
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
. When the no-load circulating current exceeds 10% of the
full-rated load.
. When the arithmetical sum of the circulating current and load
current is>110%.
The circulating current means the current circulating in the
high- and low-voltagewindings, excluding the exciting current.
Example C.3
A 10-MVA, 13.84.16 kV transformer has a per unit resistance and
reactance of0.005 and 0.05, respectively. This is paralleled with a
5-MVA transformer of thesame voltage ratio, and having per unit
resistance and reactance of 0.006 and 0.04,respectively. Calculate
how these will share a load of 15 MVA at 0.8 power
factorlagging.
Convert Z1 and Z2 on any common MVA base and apply Eqs. (C.14)
and(C.15). The results are:
10 MVA transformer S1 = 9.255
-
This can be written as
V1
Z1 1Z2
1ZL
E1
Z1 E2Z2
C:18
For a given load, the caculation is iterative in nature, as
shown in Example C.4.
Example C.4
In Example C.3, the transformers have the same percentage
impedances and thesame X=R ratios; the secondary voltage of the
10-MVA transformer is 4 kV andthat of the 5-MVA transformer is 4.16
kV. Calculate the circulating current at noload.
We will work on per phase basis. The 10-MVA transformer
impedance referredto 4 kV secondary is 0:008 j0:08 ohms, and the
5-MVA transformer impedance at4.16 kV secondary is 0:0208 j0:1384
ohms; Z1 Z2 0:0288 j0:2184. Assumethat the load voltage is 4 kV;
then, on a per phase basis, the load is 5 MVA at 0.8power factor
and the load impedance is 0:853 j0:64 ohms.
E1Z1
E2Z2
40003
p 0:008 j0:08 4160
3p 0:0208 j0:1384 5409 j45:550 kA
Also,
1
Z1 1Z2
1ZL
3:048 j20:003
From Eq. (C.18), the load voltage is 2260 j74:84 volts
phase-to-neutral.From Eq. (C.16), the 10-MVA transformer load
current is 980:04 j445:347 andthat of the 5-MVA transformer is
672:17 j874:42. The total load current is1652:2 j1319:75 and the
single-phase load MVA is 3.645 MW and 3.112Mvar. This is much
different from the desired loading of 4 MW and 3 Mvar.This is due
to assumption of the load voltage. The calculation can be
repeatedwith a lower estimate of load voltage and recalculation of
load impedance.
C.4 AUTOTRANSFORMERS
The circuit of an autotransformer is shown in Fig. C-6(a). It
has windings commonto primary and secondary, i.e., the input and
output circuits are electrically con-nected. The primary voltage
and currents are V1 and I1 and the secondary voltageand current are
V2 and I2. If the number of turns are n1 and n2, as shown,
thenneglecting losses:
V1V2
I2I1 n1 n2
n2 n C:19
Transformers and Reactors 765
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
766 Appendix C
Figure C-6 (a) Circuit of an autotransformer, step-down
conguration shown; (b) vectordiagram of an autotransformer on load;
(c) equivalent circuit of an autotransformer.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
The ampe`re turns I1n1 oppose ampe`re turns I2n2 and the common
part of the wind-ing carries a current of I2 I1. Consequently, a
smaller cross-section of the con-ductor is required. Conductor
material in the autotransformer as a percentage ofconductor
material in a two-winding transformer for the same kVA output
is
MautoMtwowinding
I1n1 I2 I1n2I1n1 n2 I2n2
1 2n I2=I1 1 V2
V1
C:20
The savings in material cost are most effective for
transformation voltages close toeach other. For a voltage ratio of
2, approximately 50% savings in copper could bemade.
The vector diagram is shown in Fig. C-6(b) and the equivalent
circuit in Fig. C-6(c). Neglecting the magnetizing current:
V1 E1 I1r1 cos x1 sin I2 I1r2 cos x2 sin C:21where is the load
power factor. Note that the impedance drop in the commonwinding is
added, because the net current is opposed to the direction of I1.
Theequation for the secondary voltage is
V2 E2 I2 I1r2 cos x2 sin C:22Combining these two equations, we
write:
V1 nV2 I1r1 n 12r2 cos x1 n 12x2 sin C:23This means that the
equivalent resistance and reactance corresponding to a two-winding
transformer are:
R1 r1 1 n2r2X1 x1 1 n2x2
C:24
An autotransformer can be tested for impedances exactly as a
two-winding trans-former. The resistance and reactance referred to
the secondary side is:
R2 R1=n2 r2 r1=n2X2 X1=n2 x2 x1=n2
C:25
The kVA rating of the circuit with respect to the kVA rating of
the windings is n1=n A 1-MVA, 3322 kV autotransformer has an
equivalent two-winding kVA of1:5 1=1:5 0:333 1000 333 kVA. The
series impedance is less than that of atwo-winding transformer.
This is benecial from the load-ow point of view, as thelosses and
voltage drop will be reduced, however, a larger contribution to
short-circuit current results.
A three-phase autotransformer connection is shown in Fig.
C-7(a). Such banksare usually Y-connected with a grounded neutral,
and a tertiary winding is added forthird-harmonic circulation and
neutral stabilization (see Fig. 17-6). This circuit isakin to that
of a three-winding transformer, and the positive and zero
sequence
Transformers and Reactors 767
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
circuits are as shown. The T-circuit positive sequence
parameters are calculated byshorting one set of terminals and
applying positive sequence voltage to the otherterminals and
keeping the third set of terminals open circuited.
ZHZXZY
1=2
1 1 11 1 11 1 1
ZHXZHYZXY
pu C:26
This is identical to Eq. (1.54).
768 Appendix C
Figure C-7 (a) Circuit of a three-phase autotransformer with
tertiary delta; (b), (c) positiveand zero sequence circuits.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
The zero sequence impedance is given by
ZX0ZH0Zn0
1
2
1 1 1 n 1=n1 1 1 n 1=n21 1 1 1=n
ZHXZHYZXY6Zn
C:27
where n is dened in Eq. (C.19). If the neutral of the
autotransformer is ungrounded,the current in the secondary winding
is balanced by circulating currents in thetertiary and no current
ows in the primary winding.
C.4.1 Scott Connection
Two autotransformers with suitable taps can be used in a Scott
connection, forthree-phase to two-phase conversion and ow of power
in either direction. Thearrangement is shown in Fig. C-8. The line
voltage V appears between terminalsC and B and also between
terminals A and B and A and C. The voltage between Aand S is V
3
p =2; the second autotransformer, called the teaser transformer,
has 3p =2 turns. The two secondaries having equal turns produce
voltages equal inmagnitude and phase quadrature. The neutral of the
three-phase system can belocated on the second or teaser
transformer. The neutral must have a voltage ofV= 3p
to terminal A, i.e., the neutral point can be trapped at V 3p =2
1= 3p 0:288n1 turns from S. It can be shown that the three-phase
side is balanced for a two-
Transformers and Reactors 769
Figure C-8 (a) Circuit of a Scott-connected transformer; (b)
three-phase primary and two-phase secondary voltages.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
phase balanced load, i.e., if the load is balanced on one side
it will be balanced on theother.
C.5 STEP-VOLTAGE REGULATORS
Step-voltage regulators [2] are essentially autotransformers.
The most commonvoltage regulators manufactured today are of
single-phase type with reactiveswitching, resulting in 10% voltage
regulation in 32 steps, 16 boosting and16 bucking. The rated
voltages are generally up to 19920 kV (line to neutral),150 kV BIL
(basic insulation level), and the current rating ranges from 5
to2000 A (not at all voltage levels). The general application is in
distribution sys-tems and three single-phase voltage regulators can
be applied in wye or deltaconnection to a three-phase three-wire or
three-phase four-wire system. The wind-ing common to the primary
and secondary is designated as a shunt winding, andthe winding not
common to the primary and secondary is designated as a
serieswinding. The series winding voltage is 10% of the regulator
applied voltage. Thepolarity of this winding is changed with a
reversing switch to accomplish buck orboost of the voltage. When
the voltage regulation is provided on the load side itis called a
type-A connection, Fig. C-9(a). The core excitation varies as the
shuntwinding is connected across the source voltage. In a type-B
connection, Fig. C-9(b), the regulation is provided on the load
side and the source voltage is appliedby way of series taps. Figure
C-9(d) shows the schematic of a tap-changing circuitwith
current-limiting reactors and equalizer windings.
C.5.1 Line Drop Compensator
The step regulators are controlled through a line drop
compensator; its schematiccircuit is shown in Fig. C-9( c). The
voltage drop in the line from the regulator to theload is simulated
in a R 0X 0 network in the compensator. The settings on
theseelements are decided on the basis of load ow prior to
insertion of the regulator,i.e., the voltage and current at the
point of application give the system impedance tobe simulated by R
0 and X 0 in the line drop compensator.
C.6 EXTENDED MODELS OF TRANSFORMERS
A transient transformer model should address saturation,
hysteresis, eddy current,and stray losses. Saturation plays an
important role in determining the transientbehavior of the
transformer. Extended transformer models can be very involvedand
these are not required in every type of study. At the same time, a
simplemodel may be prone to errors. As an example, in distribution
system load ow,representing a transformer by series impedance alone
and neglecting the shunt ele-ments altogether may not be proper, as
losses in the transformers may be consider-able. For studies on
switching transients, it is necessary to include capacitance of
thetransformers as high-frequency surges will be transferred more
through electrostaticcouplings rather than through electromagnetic
couplings. For short-circuit calcula-tions, capacitance and core
loss effects can be neglected. Thus, the type of selectedmodel
depends on the nature of the study. There are many approaches to
the models,some of which are briey discussed.
770 Appendix C
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
Transformers and Reactors 771
Figure C-9 (a), (b) Circuits of type-A and type-B connection
step-voltage regulators; (c)schematic of line-drop compensator; (d)
reactance-type tap changer with reversing switch.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
The equivalent circuit of the shunt branch of a transformer for
nonlinearitycan be drawn as shown in Fig. C-10. The excitation
current has half-wave sym-metry and contains only odd harmonics
(see Appendix E). We may consider theexcitation current as composed
of two components, a fundamental frequencycomponent and a
distortion component. The fundamental frequency componentis broken
into two components, ie and im as discussed before, which give rise
toshunt components gm and bm, Fig. C-1(a). The distortion component
may beconsidered as a number of equivalent harmonic current sources
in parallel withthe fundamental frequency components, each of which
can be represented in thephasor form, Iei < i. To consider the
effect of variation in the supply systemvoltages, the model
parameters at three voltage levels of maximum, minimum,and rated
voltage can be approximated by quadratic functions of the
supplysystem voltage:
W a bV cV2W Rm;Xm; Ie3; Ie5; . . . ; 3; 5; . . .
C:28
where Ie3, Ie5, . . ., 3, 5, . . . are the harmonic currents and
their angles. The coef-cients a, b, and c can be found from
abc
1 Vmin V2min
1 Vrated V2rated
1 Vmax V2max
W0W1W2
C:29
whereW0,W1,W2 are measured values ofW for Vmin, Vrated, and
Vmax, respectively.
C.6.1 Modeling the Hysteresis Loop
A model of the hysteresis loop can be constructed, based on
measurements. Thelocus of the midpoints of the loop is obtained by
measurements at four points and itsdisplacement by a consuming
function, whose maximum value is ob, and ef changesperiodically by
half-wave symmetry (Fig. C-11). The consuming function can
772 Appendix C
Figure C-10 Transformer shunt branch model considering
nonlinearity.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
be written as f x ob sin!t. The periphery can be then
represented by 16 linesegments [3]:
i ik mkk mk ob sin!tk1 < jj4kk 1; 2; . . . 16
C:30
C.6.2 EMTP Models
Figure C-12(a) shows a single-phase model; R remains constant,
and it is calculatedfrom excitation losses. The nonlinear inductor
is modeled from transformer excita-tion data and from its nonlinear
VI characteristics. In modern transformers thecores saturate
sharply and there is a well-dened knee. Often a two-slope
piecewiselinear inductor is adequate to model such curves. The
saturation curve is not sup-plied as a uxcurrent curve, but as a
rms voltagerms current curve. The saturaroutine in EMTP [4]
converts voltagecurrent input into uxcurrent data. FiguresC-12(b)
and (c) show an example of this conversion for a 750-MVA, 42027 kV
ve-leg, core type, wyedelta connected transformer. The nonlinear
inductance should beconnected between the windings closest to the
iron core. The input data are pre-sented in per unit values with
regard to the winding connections and the base currentand voltage.
S base = 250MVA, V base = 27kV and I base = 9259A
Imag I2ex Pex=Vex21=2 C:31There is linear interpolation between
the assumed values and nite-differenceapproximation to sinusoidal
excitation. The hysteresis is ignored.
The EMTP model hysdat represents a hysteresis loop in 45 points
to 2025points for a specic core material. The positive and negative
saturation points, as
Transformers and Reactors 773
Figure C-11 Piece wise hysteresis loop curve tting.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
774 Appendix C
Figure C-12 (a) EMTP model satura; (b), (c) conversion of VI
characteristics into Icharacteristics.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
dened in Fig. C-11, need only to be specied. Figure C-13 shows
EMTP simulationof the excitation current in a single-phase, 50-kVA
transformer.
C.6.3 Nonlinearity in Core Losses
Figure C-14 shows a frequency domain approach and considers that
winding resis-tance and leakage reactance remain constant and the
nonlinearity is conned to thecore characteristics [5]. The core
loss is modeled as a superimposition of lossesoccurring in ctitious
harmonic and eddy current resistors. The magnetizing
char-acteristics of the transformer is dened by a polynomial
expressing the magnetizingcurrent in terms of ux linkages:
iM A0 A1 A22 A33 C:32Only a specic order of harmonic currents ow
to appropriate Gh resistors in Fig.C-14. From Eqs. (C.7) and (C.8)
the core loss equation is
Pfe Ph Pe KhBsf KeB2f 2 C:33
Transformers and Reactors 775
Figure C-13 Simulation of inrush current of a 50-kVA
transformer.
Figure C-14 Nonlinear shunt model with superimposition of
harmonic currents in resistors.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
For a sinusoidal voltage, this can be written as:
Pfe kh f 1sEs keE2 kh 6 Kh and ke 6 Ke C:34This denes
two-conductance Gh for hysteresis loss and Ge for eddy current
loss,given by
Gh kh f 1sEs2;Ge ke C:35
C.7 HIGH-FREQUENCY MODELS
For the response of transformers to transients a very detailed
model may includeeach winding turn and turn-to-turn inductances and
capacitances [6]. Consider adisk-layer winding or pancake sections
as shown in Fig. C-15(a). Each numberedrectangular block represents
the cross-section of a turn. The winding line terminal isat A and
winding continues beyond E. Each section can be represented by a
series ofinductance elements with series and shunt capacitances as
shown in Fig. C-15(b).Though the model looks complex, the mutual
inductances are not shown, resistancesare not represented, and no
interturn capacitances are shown. This circuit will beformidable in
terms of implementation. For most applications, representation
ofeach turn is not justied and by successive lumping a much simpler
model isobtained, Fig. C-15(c). A further simplied model is shown
in Fig. 19-4.
Consider the circuit in Fig. C-16. A 7.5-Mvar capacitor bank is
switched at the13.8-kV bus and the resulting switching overvoltages
on the secondary of a 2.5-MVA, 13.80.48 kV transformer connected
through 400 ft 500-KCMIL, 15-kVcable are simulated using EMTP. The
transformer model shown in Fig. 19-4 isused. The results are shown
in Fig. C-17. This gure shows high-frequency compo-nents, due to
multiple reections in the connecting cable, and the peak
secondary.Voltage is 3000 volts. This is very high for a 480-volt
system. Surge arresters andcapacitors applied at transformer
terminals will appreciably reduce this voltage.
The switching of transformers can give rise to voltage
escalation inside thetransformer windings. The windings have
internal ringing frequencies and certainswitching operations can
excite these frequencies creating excessive intra-windingstresses.
A snubber circuit (usually a capacitor in series with a resistor)
connectedphase to ground can limit these voltage peaks. See also
Refs. [7,8].
C.8 DUALITY MODELS
Duality-based models can be used to represent transformers.
These models are basedon core topology and utilize the
correspondence between electric and magneticcircuits, as expressed
by the principle of duality. Voltage, current, and inductancein
electrical circuits correspond to ux, mmf, and reluctance,
respectively:
I VR
MMFl=0a
MMFS
C:36
where l is the length of the magnetic path, a is the
cross-sectional area, and S is thereluctance, which is analogous to
resistance in an electrical circuit and determines the
776 Appendix C
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
magnetomotive force necessary to produce a given magnetic ux.
Permeance is thereverse of reluctance.
Figure C-18(a) shows electrical equivalent circuit of a
three-winding core-typetransformer portraying magnetic coupling in
three- and ve-limbed transformers [9].Non-linear inductances
correspond to iron ux paths in the magnetic circuit, permit-ting
each core limb to be modeled separately. Each Lk represents top and
loweryokes and each Lb represents a wound limb; L0 represents the
ux path through
Transformers and Reactors 777
Figure C-15 (a) Winding turns in a pancake coil; (b) circuit of
winding inductance andcapacitances; (c) simplied circuit model.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
778
Figure C-16 Circuit for simulation of capacitor switching
transient.
Figure C-17 EMTP simulation of transient overvoltage on 480-V
secondary of 2.5-MVAtransformer in Fig. C-16.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
the air, outside the core and around the windings. Finally, the
ladder networkbetween linear inductances L0 and Lb represents
winding leakages through air.Inductances Lh, Ly represent unequal
ux linkages between turns due to nite wind-ing radial build and
these are small compared to L0 and Lb. This model is simpliedas
shown in Fig. C-18(b). The various inductances are calculated from
short-circuittests.
Duality models can be used for low-frequency transient studies,
such as short-circuits, inrush currents, ferroresonance, and
harmonics [10].
C.9 GIC MODELS
Geomagnetically induced currents (GICs) ow through the earths
surface due tosolar magnetic disturbances and these are typically
0.001 to 0.1 Hz and can reach
Transformers and Reactors 779
Figure C-18 (a) Duality based-circuit model of a core-type,
three-winding transformer;(b) simplied circuit derived from
(a).
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
peak values of 200 A. These can enter transformer windings
through groundedneutrals, Fig. C-19(a), and bias the transformer
core to half-cycle saturation. As aresult the transformer
magnetizing current is greatly increased. Harmonics increaseand
these could cause reactive power consumption, capacitor overload,
false opera-tion of protective relays, etc. [11].
A model for GIC is shown in Fig. C-19(b). Four major ux paths
are included.All R elements represent reluctances in different
branches. Subscripts c, a, and tstand for core, air, and tank,
respectively, and 1, 2, 3, and 4 represent major branchesof ux
paths. Branch 1 represents the sum of core and air uxes within the
excitationwindings, branch 2 represents the ux path in yoke, and
branch 3 represents the sumof uxes entering the side leg, part of
which leaves the side leg and enters the tank.Branch 4 represents
ux leaving the tank from the center leg. An iterative program
isused to solve the circuit of Fig. C-19 so that nonlinearity is
considered.
C.10 REACTORS
We have discussed the following applications of reactors in
various chapters of thisbook:
. Current-limiting reactors, mainly from the standpoint of
limiting the short-circuit currents in a power system. These can be
applied in a feeder, in a tie-
780 Appendix C
Figure C-19 (a) GIC entering the grounded neutrals of the
wye-connected transformers;(b) a transformer model for GIC
simulation.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
line, in synchronizing bus arrangements (Fig. 7-16 ), and as
generator reac-tors (Fig. 13-20).
. Shunt reactors for reactive power compensation.
. Reactors used in static var controllers, i.e., SVCs, TCRs,
TSCs, and dis-charge reactors in a series capacitor (Chap. 13).
. Harmonic lter reactors and inrush current-limiting
reactors
. Line reactors to limit notching effects and dc reactors for
ripple currentlimitation in drive systems.
Further applications are:
. Smoothing reactors are used in series with an HVDC line or
inserted into adc circuit to reduce harmonics on the dc side.
Filter reactors are installed onthe ac side of converters. Radio
interference and power-line carrier lterreactors are used to reduce
high-frequency noise propagation.
. Reactors are installed in series in a medium-voltage feeder
(high-voltage sideof the furnace transformer) to improve efciency,
reduce electrode con-sumption, and limit short-circuit
currents.
. An arc suppression reactor, called a Peterson coil, is a
single-phase variablereactor that is connected between the neutral
of a transformer and groundfor the purpose of achieving a resonant
grounding system, though suchgrounding systems are not in common
use in the USA, but prevalent inEurope. The inductance is varied to
cancel the capacitance current of thesystem for a single
line-to-ground fault.
. Reactors are used in reduced-voltage motor starters to limit
the startinginrush currents.
. Series reactors may be used in transmission systems to modify
the powerow by changing the transfer impedance.
We will discuss a duplex reactor, which can sometimes be
usefully applied tolimit short-circuit currents and at the same
time improve steady-state performanceas compared to a conventional
reactor [12]. It consists of two magnetically coupledcoils per
phase. The magnetic coupling, which is dependent on the
geometricproximity of the coils, is responsible for desirable
properties of a duplex reactorunder short-circuit and load-ow
conditions. The equivalent circuit is shown inFigs C-20 (a) and (b)
and its application in Fig. C-20( c). The mutual couplingbetween
coils is
Lm L11 L1L22 L2
p k
L11L22
p kL C:37
where k is the coefcient of coupling. Note that the inductance
of sections 1 to 4 inthe T equivalent circuit, for direction of
current ow from source to loads,becomes negative kL. Between
terminals 2 and 4, and 3 and 4 it is L. Theterminal 4 is ctitious
and the source terminal is 1, while the load terminals are 2and 3.
Thus, the effective inductance between the source to a load
terminal is1 kL, where L is the inductance of each winding and k is
the coefcient ofcoupling. Effective reactance to load ow is reduced
by a factor of k and voltagedrops will also be reduced by the same
factor. For a short-circuit on any of theload buses, the currents
in one of the windings reverses and the effective induc-tance is
2L1 k. The short-circuit currents will be more effectively limited.
The
Transformers and Reactors 781
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
782 Appendix C
Figure C-20 (a), (b) Equivalent circuits of a duplex reactor;
(c) a circuit showing applicationof a duplex reactor.
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
-
limitations are that k is dependent on the geometry of the coils
and is almostindependent of the current loading of the coils.
Cancellation of the magneticeld under load ow occurs only when the
second winding is loaded. The advan-tages of a duplex reactor are
well exploited if the loads are split equally on twobuses,
Fig-C.20(c).
REFERENCES
1. ANSI. Terminal Markings and Connections for Distribution and
Power Transformers,1978 (Revised 1992). Standard C57.12.70.
2. IEEE. Standard Requirements, Terminology and Test Code for
Step-VoltageRegulators, 1999. Standard C57.15.
3. CE Lin, JB Wei, CL Huang, CJ Huang. A NewMethod for
Representation of Hysteresis
Loops. IEEE Trans Power Deliv 4: 413-419, 1989.4.
Canadian/American EMTP User Group. ATP RuleBook. Portland: Oregon,
1987-1992.5. JD Green, CA Gross. Non-Linear Modeling of
Transformers. IEEE Trans Ind Appl 24:
434-438, 1988.
6. WJ McNutt, TJ Blalock, RA Hinton. Response of Transformer
Windings to SystemTransient Voltages. IEEE Trans PAS 9:457-467,
1974.
7. PTM Vaessen. Transformer Model for High Frequencies. IEEE
Trans Power Deliv 3:
1761-1768, 1988.8. T Adielson, et al. Resonant Overvoltages in
EHV Transformers-Modeling and
Application. IEEE Trans PAS 100: 3563-3572, 1981.
9. X Chen, SS Venkta. A Three-Phase Three-Winding Core-Type
Transformer Model forLow-Frequency Transient Studies. IEEE Trans PD
12: 775-782, 1997.
10. A Narang, RH Brierley. Topology Based Magnetic Model for
Steady State andTransient Studies for Three-Phase Core Type
Transformers. IEEE Trans PS 9: 1337-
1349, 1994.11. S Lu, Y Liu, JDR Ree. Harmonics Generated from a
DC Biased Transformer. IEEE
Trans PD 8: 725-731, 1993.
12. JC Das, WF Robertson, J Twiss. Duplex Reactor for Large
Cogeneration DistributionSystem-An Old Concept Reinvestigated.
TAPPI, Engineering Conference, Nashville,637-648, 1991.
Transformers and Reactors 783
Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.
POWER SYSTEM ANALYSISCONTENTSAPPENDIX C: TRANSFORMERS AND
REACTORSC.1 MODEL OF A TWO-WINDING TRANSFORMERC.1.1 OPEN CIRCUIT
TESTC.1.2 SHORT-CIRCUIT TEST
C.2 TRANSFORMER POLARITY AND TERMINAL CONNECTIONSC.2.1 ADDITIVE
AND SUBTRACTIVE POLARITY
C.3 PARALLEL OPERATION OF TRANSFORMERSC.4 AUTOTRANSFORMERSC.4.1
SCOTT CONNECTION
C.5 STEP-VOLTAGE REGULATORSC.5.1 LINE DROP COMPENSATOR
C.6 EXTENDED MODELS OF TRANSFORMERSC.6.1 MODELING THE HYSTERESIS
LOOPC.6.2 EMTP MODELSC.6.3 NONLINEARITY IN CORE LOSSES
C.7 HIGH-FREQUENCY MODELSC.8 DUALITY MODELSC.9 GIC MODELSC.10
REACTORSREFERENCES