appendix chapter for transformer standard

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.

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.


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.


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.



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.


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



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.


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.



. 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



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.


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



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.


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-


Figure C-8 (a) Circuit of a Scott-connected transformer; (b) three-phase primary and two-phase secondary voltages.


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



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.


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.


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


Figure C-11 Piece wise hysteresis loop curve tting.


774 Appendix C

Figure C-12 (a) EMTP model satura; (b), (c) conversion of VI characteristics into Icharacteristics.


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


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.


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


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


Figure C-15 (a) Winding turns in a pancake coil; (b) circuit of winding inductance andcapacitances; (c) simplied circuit model.


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.


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


Figure C-18 (a) Duality based-circuit model of a core-type, three-winding transformer;(b) simplied circuit derived from (a).


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.


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



782 Appendix C

Figure C-20 (a), (b) Equivalent circuits of a duplex reactor; (c) a circuit showing applicationof a duplex reactor.


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.



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

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