cap_3

Core und coil assetrlblies of rr rhree-plruse 20.3 k VA/345

k V Y step-up ~ransfortrrer. This oil-inl~tnersed

tra~ls/ormer is rated 325 A1 VA sey-cooled ( OA)/542 MVA forced oil, forced air-

cooled (FOA)/607M VA fbrced oil, forced air-cooled

(FOA) (Courtesy of Generol Electric)

POWER TRANSFORMERS

T h e power transformer is a major power system component that pesn~its economical power transmission with high efficiency and low series-voltage drops. Since electric power is proportional to the product of voltage and current, low current levels (and therefore low I'R losses and low IZ voltage drops) can be maintained for given power lcvels via high voltages. Power transformers transform ac voltage and current to optimum levels for genera- tion, transn~ission, distribution, and utilization of electric power.

The development in 1885 by William Stanley of a comn~ercially pn~cti- cal transformer was what made ac power systems more attractive than dc power systems. Thc ac system with a transformer overcame voltage problems encountered in dc spstcms as Icxid lcvcls anc! transmission distances incrciiscd. Today's modern power transformers have nearly 100%) efficiency, with ratings up to and beyond 1300 MVA.

CHAPTER 3 POWER TRANSFORMERS

In this chapter, we review basic transformer theory and develop equivalent circuits for practical transformers operating under sinusoidal-steady- state conditions. We look at models of single-pliase two-winding, three-phase two-winding, and three-phase three-winding transformers, as well as auto- transformers and regulating transformers. Also, the per-unit system, which simplifies power system analysis by eliminating the ideal transfornler winding in transformer equivalent circuits, is introduced in this chapter and used throughout the remainder of the text.

C AS E S T U D Y The following article illustrates what electric utility companies look for in power and distribution transformers. The author discusses transformer price, losses, noise, reliability. and manufacturer's support He also describes transformer evaluation programs of four utilities [8].

How Electric Utilities Buy Quality When They Buy Transformers JOHN R M O N

Because transformers are passive devices with few moving parts, it is difficult to evaluate the quality of one over another. But today, when the lifetime cost of transformer losses far exceeds the initial transformer purchase price and a significant percentage of transformer purchases is to replace units that have failed in service, utilities need a mechanism to weigh one manufacturer's offering against another's-often well before the transformer is actually built

Power and distribution transformers present entirely different problems to the purchasing engineers charged with evaluating quality. Power transformers are generally custom-built and today they are often very different from any transformer the utility has bought before. Power transformers should be evaluated according to a wide range of quality factors, each of which has a different importance or weight, depending on the purchasing utility.

In contrast, diswibution transformers are purchased in bulk and, provided detailed failure records are kept, the quality can be rather easily determined from computerized statistical programs.

Reprinted wid, permission from Electrical World.

LOW LOSSES MEAN HIGH QUALITY

One factor in the engineer" hvor is that high- quality transformers are also low-loss transformers. In a sense, the cost of high quality is automatically paid for in the first few years of transformer life by reduced losses. To this benefit is added the fact that the lifetime of a transformer built today will actually be significantly longer than that of a transformer built only a few years ago.

Losses are divided into load and no-load losses and various formulas andor computer programs are available to evaluate their lifetime impact. When individual utilities plug their cost factors into the formulas, the lifetime impacts they calculate vary widely. For example, the ratio of estimated costs of no-load to load losses can vary by a factor as much as 10 to one. The relative cost of load and no-load losses can also vary from year to year as regulatory pressures push utility management to emphasize different needs.

Noise is becoming an increasingly important factor in transformer selection. Again, this factor varies widely from utility to utility. The greatest need for a

low-noise transformer is felt by utilities in highly de- veloped areas where substations must be located close to residential neighborhoods.

Transformer noise is generated from three sources: (I) the magnetosvictive deformation of the core, (2) aerodynamic noise produced by cooling fans, and (3) the mechanical and flow noise from the oil-circulating pumps. The radiated core noise, con- sisting of a 120-Hz tone, is the most difficult to reduce and is also the noise that generates the most complaints from residents living near the transformer.

Fortunately, improved core-construction techniques and lower-loss core steel both tend to reduce transformer core noise. If further reduction in core noise is needed, it can only be achieved by increasing the cross-sectional area of the core to reduce the flux density. This design change increases the construction cost of the transformer and decreases the core losses. However, a point of diminishing returns is reached at which the cost of increasing core size outweighs the savings in reduced losses (Fig. I).

the trend is for the manufacturer to assemble and fill the transformer on site, rather than leave it to the utility. This provides assurance that the transformer is correctly installed and minimizes the cost of lost parts, misunderstanding, etc.

Manufacturing facilities provide a key indica- tion of the product quality (Fig. 2). Most utilities use plant visits as the f i rs t step in their evaluation process. Facility review should include the manufacturer's quality-assurance program, in-service and test reliability records, contract administration and order support, and technical strength.

Coating systems, especially for pad-mount transformers, are becoming increasingly important since the life of the transformer tank may be the

! ' tially disassembled and without oil in the tank. Today, i

CASE STUDY

I

I Installation costs are significant because a I power transformer must generally be delivered par- bQ - A

total cost to purchase sum of price of transformer plus iron loss cost

!! m - -

0 n

cost of iron loss

Iron loss and/or noise

1 Figure 1 " Noise reduction requirements may require that iron

losses are reduced below the optimum economic level. i

Figure 2 Manufacturing facilities are a good indicator of producr qualityi Here, electrical connections are made on core form power transformer.

74 CHAPTER 3 POWER TRANSFORMERS

limiting factor in transformer life. The problem of evaluating and comparing coating systems on pad- mount transformers from different manufacturers was eliminated with the introduction of ANSI Stan- dard C57.12.28- 1988. This is a functional standard that does not dictate to manufacturers how they should coat transformers, but prescribes a series of tests that the coating must withstand to meet the standard. A companion standard, C57.12.3 1 for poletop transformers, is now under development.

Tests prescribed by the standards include: Scratching to bare metal and exposing to salt spray for 1500 hours; cross-hatch scratching to check for adhesion, humidity exposure at 1 13 "C, impact of 160 in.-lb with no paint chipping, oil and ultraviolet resistance, and 3000 cycles of abrasion resistance.

In response to this standard, most manufacturers have revamped or rebuilt their painting processes- from surface preparation (Fig. 3) through application of primers, to finished coating systems. The most advanced painting processes now use electrodeposition methods-either as a dip process or with paint applied as a dry powder. These processes not only ensure a uniform coating system to every part of the transformer tank but also, because they eliminate traditional solvent-based paints, more easily meet the Clean Air Act Amendments of 1 990.

Figure 3 Coating systems are an important part of transformer quality. Here, tank is chemically cleaned prior to electrodeposition of coating.

Hard evaluation factors are set down in the purchaser's technical specifications, which form the primary document to ensure that all suppliers' products meet a minimum standard. Technical specifications generally include an evaluation formula for no-load and load losses, price, noise level, and delivery date. Technical assistance during installation, warranty assistance, and the extent of warranty are additional hard evaluation factors.

Soft factors do not have a precise monetary value, but also may be important in comparing suppliers' bids. The [following] l is t suggests soft factors for buyers to include in a transformer-purchase de- cision. While they do not have a direct dollar value, it is valuable to assign a fixed dollar value or a percentage of bid value to these factors so that they can be used in comparing suppliers' bids. A well-written specification places all potential suppliers on an equal footing.

SOFT FACTORS THAT SHOULD INFLUENCE CHOICE OF SUPPLIER

Wide choice of designs

Computer-aided design procedures

R&D directed at product improvement

Participation in long-term R&D projects through industry groups Clean-room assembly facilities

Availability of spare and replacement parts

Wide range of field services

Application assistance/coordination

Ongoing communication with users

Tony Hartfield, ABB Power T&D Co., Power Transformer Div., St. Louis, Missouri, says it is important to review technical specifications in detail with prospective suppliers before a request for bids is issued. "We attempt to resolve ambiguous terms such as 'substantial,' 'long-lasting.' or 'equal-to,' and replace them with functional requirements that clearly define what must be supplied.

"Many times, items are added to a specification

CASE STUDY 75

to prevent recurrence of past problems. These can be counterproductive, particularly if the technology has advanced to a point where the source of the problem has been eliminated."

GOOD IN-SERVICE RECORDS VITAL Distribution transformers are purchased in large quantities under very competitive conditions where a unit-price change of a few cents can affect the choice of supplier. As a result, the most sophisti- cated programs used to guide purchasing policy are based on statistical records of units in service.

One example of a systematic failure-analysis program is that conducted by Wisconsin Public Service Corp. (Electn'col World, September 199 1, p 73). All transformers purchased by the utility since the mid

f 1980s and all transformer failures are entered into a computerized record-keeping system. Failure rates

1 and equivalent costs are calculated for each manufacturer on a 4-year rolling window. According to Senior Standards Engineer Michael Radke, the system has substantially reduced failure rates, improved communications with transformer vendors, reduced costs, and reduced outages. The system has even helped some manufacturers to reduce failure rates.

Georgia Power Co.5 vendor evaluation program has been in place for about 5 years. This program looks at supplier and product separately, judg- ing each according to pre-established criteria. The scores for each criterion are weighted and the over- all score used to calculate a numerical multiplier, which is applied to initial bid price. David McClure, research manager, quality and support, explains that the program involves four departments: engineering, materials, quality assurance, and procurement Each department is responsible for a portion of the evaluation and the results from each are entered into a computer program.

The evaluation involves objective and subjective factors. Compliance, for example, can be measured objectively, but customer service must be evaluated

t: subjectively. Even so, reviewers follow a well-

i defined procedure to determine scores for each factor. This approach ensures that ratings are applied consistently to each vendor.

Public Service Co. of Colorado (PSC) uses a numerical multiplier that is applied to the bid price. The multiplier incorporates several factors-including historical failure rate, delivery, and quality. O f these factors, historical failure rate is by far the most important, accounting for more than half of the multiplier penalty. For example, the average multiplier for . . , - , pole-mounted transformers adds 6.3%, of which failure rate accounts for 4.9%; the average multiplier F + * I

- ' I for single-phase pad-mount transformers adds 5.3%, ',::; , ;' of which failure rate accounts for 3.6%. C J ,dk .c ;

Failure rate is calculated using a computer pro- J"ei-*; gram supplied by General Electric Co., Transformer pi Business Dept, Hickory, North Carolina. It is based ,@ ., on failures of transformers purchased in the last 10 years. The cost of failure includes the cost of a replacement unit and the costs of changeout and downtime.

A delivery penalty is calculated by PSC, based on the difference in weeks between promised and actual delivery dates. Significantly, this penalty is calculated equally for early as for late delivery. Early delivery is considered disruptive. John Ainscough, senior engineer, automation analysis and research, reports that his department is planning to modify this factor to encourage both short lead-times and on- time delivery. Currently, the delivery factor does not incorporate the supplier's manufacturing cycle time.

PSC's quality factor is based on the percentage . A . of an order that must be repaired or returned to the manufacturer; the accuracy with which prod- 3' " ucts conform to the original specifications, including losses and impedance; and the number of days re-


former is a Lotus-compatible worksheet for evaluating distribution transformers offered by ABB Power T&D Co. The worksheet adjusts first-cost figures by a value factor that comprises criteria for reliability, quality, delivery/availability, and support. The lower the value factor, the lower is the effective f i rs t cost of the transformer. To the adjusted f i rs t cost is added the cost of losses, yielding a life-cycle cost for the transformer.

Suggested weightings, based on surveys of utilities, are provided for each criterion, but users can easily modify these criteria in light of their own ex- perience and needs. According to ABB's Dorman Whitley, this ensures that the worksheet does not favor any one manufacturer. Users can also incorporate soft criteria (such as supplier's long-term commitment t o the industry, o r level of investment in R&D).

LOSSES INFLUENCE RELIABILITY As in the case of power transformers, the higher the quality of distribution transformers, the lower the losses. But the relationship between losses and reliability is even more pronounced: Low-loss transformers, because of the reduced internal heating, tend to be more reliable.

Utah Power & Light Co. (UP&L) began monitor- ing transformer failures in 1973 and in 1985 established a rigorous auditing procedure to keep track

of the failure rates of different manufacturers' products. Unsatisfactorily high failure rates were uncovered-including a rate of 33% for one manufacturer's 3-phase, pad-mounted unit. In this case, the high failure rate was eliminated by switching to amorphous-core, low-loss transformers.

Distribution Engineering Manager Dennis Hor- man reports that the highest failure rates occurred in seasonally loaded transformers serving irrigation pumps. These were shut off during the winter, when they were not in use, to conserve no-load losses. However, the action of shutting off the transformers apparently caused condensation which resulted in failures when the units were put back in service.

When these transformers were replaced with amorphous-core units, the utility found that they had such low losses that it was economical to keep them energized year-round. This avoided the problem of failures caused by condensation and eliminated the manpower cost of energizing and de-energizing them. Horman says that visits to three manufacturers of amorphous-core transformers and tear- downs of delivered transformers convinced engineers at UP&L that all three were producing high-quality transformers. The utility now routinely purchases amorphous-core transformers from the three manufacturers and now accepts bids from other manufacturers without going through test procedures or plant visits.

T H E IDEAL TRANSFORMER

'Figure 3.1 shows a basic single-phase two-winding transformer, where the two windings are wrapped around a magnetic core [ I , 2. 3). It is assumed here that the transformer is operating under sinusoidal-steady-state excita- tion. Shown in the figure are the phasor voltages El and E2 across the windings, and the phasor currents II entering winding 1 , which has N I turns, and I2 leaving winding 2, which has N2 turns. A phasor flux a,. set up in the core and a magnetic field intensity phasor H, are also shown. The core has a cross-sectional area denoted A,, a mean length of the magnetic circuit I,., and a magnetic permeability p,., assumed constant.

SECTION 3.1 THE IDEAL TRANSFORMER

FIGURE 3.1 Basic single-phase two-

winding transformer

Core permeabrlrty p,

,--Core cross-sect~onal area, A, /

of the magnetic

For an ideal transformer, the following are assumed: 1. The windings have zero resistance; thcscfore, the I'R losses in the

windings are zero. 2. The core permeabilily p, is infinite, which corresponds to zcro core

reluctance. 3. There is no leakage Ilux; that is, the cntire flux 0,. is confined to the

core and links both windings. 4. There are no core losses.

A schematic reprcsentation of a two-winding transfomn~er is shoi+.n in Figure 3.2. Ampere's and Fasnday's laws can be used along with the preced- ing assumptions to derive the idcal transformer relationships. Ampcre's law states that the tangential component of the magnetic field intensity vector in- tegrated along a closed path ccluals the net current enclosed by that path; that is,

I Htan = [enclosed If the core center line shown in Figure 3.1 is selected as the closed path,

and if H, is constant along the path as well as tangent to the path, then (3.1.1) becomes

FIGURE 3.2 Sl- -- s2 Schenlatic representation

of a single-phase two- winding transformer

'I; - 211~ ; 'l N1 4


Note that the current II is enclosed N1 times and Ir is enclosed Nz times, one time for each turn of the coils. Also, using the right-hand rule*, current II contributes to clockwise Rux but current I2 contributes to counter- clockwise flux. Thus, in (3.1.2) the net current enclosed is NI II - N212. For constant core permeability p,, the magnetic flux density B, within the core, also constant. is

and the core flux a,, is

Using (3.1.3) and (3.1.4) in (3.1.2) yields

We define core reluctance R, as

Then (3.1.5) becomes

Equation (3.1.7) can be called "Ohm's law" for the magnetic circuit, wherein the net magnetomotive force rnrnf = NI I[ - N212 equals the product of the core reluctance R, and the core flux a,. Reluctance R,, which impedes the establishment of flux in a magnetic circuit, is analogous to resistance in an electric circuit. For an ideal transformer, p, is assumed infinite, which, from (3.1.6), means that R, is zero, and (3.1.7) becomes

In practice, power transformer windings and cores are contained within enclosures, and the winding directions are not visible. One way of conveying winding information is to place a dot at one end of each winding such that when current enters a winding at the dot, it produces an rnmf acting in the sanie direction. This dot convention is shown in the schematic of Figure 3.2. The dots are conventionally called polarity rztarks.

Equation (3.1.8) is written for current I, entering its dotted terminal and current I2 leaving its dotted terminal. As such, I, and I2 are in phase, since II = (N2/Nl)12. If the direction chosen for I2 were reversed, such that both currents entered their dotted terminals, then II would be 180" our of pitase with 12.

Faraday's law states that the voltage e( t ) induced across an N-turn winding by a time-varying flux d( t ) linking the winding is

*The right-hand rule for a coil is as follows: Wrap the fingers of your right hand around the coil in the direction of the current. Your right thumb then points in the direction of the Bux.

SECTION 3.1 THE IDEAL TRANSFORMER 79

Assuming a sinusoidal-steady-state flux with constant frequency w, and representing e( t ) and +([) by their phasors E and @, (3.1.9) becomes

For an ideal tsansfor~ner, the entire flux is assumed to be confincd to the core, linking both windings. From Faraday's law, the induced voltages across the windings of Figure 3.1 are

Dividing (3.1.1 1) by (3.1.12) yields

The dots shown in Figure 3.2 indicate that the voltages El and E,, both of which have their + polarities at the dotted terminals, are in phase. I f the polarity chosen for one of the voltages in Figure 3.1 were reversed, then El would be 180" out of phase with E2. ~ k g ~ ~ a t i o ~ ~ ~ i S 4 ~ d 1 1 ~ f c i a s follows:

Using u, in (3.1.8) and (3.1.14), the basic relations for an ideal single-phase two-winding transformer are

Two additional relations concerning colnplex power and impedance can be derived from (3.1.16) and (3.1.17) as follows. The complex power entering winding 1 in Figure 3.2 is

SI = El I; Using (3.1.16) and (3.1.17),


As shown by (3.1.19), the complex power Sj entering winding 1 equals the complex power S2 leaving winding 2 . That is, an ideal transformer has no real or reactive power loss.

If an impedance Z2 is connected across winding 2 of the ideal transformer in Figure 3.2, then

This impedance, when measured from winding 1, is

Thus, the impedance Z2 connected to winding 2 is referred to winding 1 by multiplying Z2 by a;, the square of the turns ratio.

EXAMPLE 3.1 Ideal, single-phase two-winding transformer

A single-phase two-winding transformer is rated 20 kVA, 4801120 V, 60 Hz. A source connected to the 480-V winding supplies an impedance load connected to the 120-V winding. The load absorbs 15 kVA at 0.8 p.f. lagging when the load voltage is 118 V. Assume that the transformer is ideal and calculate the following:

a. The voltage across the 480-V winding

b. The load impedance

c. The load impedance referred to the 480-V winding d. The real and reactive power supplied to the 480-V winding

SOLUTION

a. The circuit is shown in Figure 3.3, where winding 1 denotes the 480-V winding and winding 2 denotes the 120-V winding. Selecting the load voltage E2 as the reference,

E 2 = 118/00 V

FIGURE 3.3 Circuit for Example 3.1

SECTION 3.1 THE IDEAL TRANSFORMER 8 1

The turns ratio is, from (3.1.13),

n - NI Elrated -480 - - - 4 1-K=E2riltcd 120

and the voltage across winding 1 is

b. The complex power S2 absorbed by the load is SZ = Ell; = 1181,' = 1 5 , 0 0 0 / ~ 0 ~ ~ ' ( 0 . 8 ) = 15,000/36.87" VA

Solving, the load current 11 is I? = 127.12/-36.87O A

The load impedance Z2 is

c. From (3.1.2 I), the load impcdatlce referred to the 480-V winding is Z; = afz2 = (4)2(0.9283/36.870) = 14.85/36.87" R

d. From (3.1.19) SI = S2 = 15,000/36.87 = 12,000 + j9000

Thus, the real and reactive powers suppliecl to the 480-V winding arc PI = Re SI = 12,000 W = 12 kW Q , = Im SI = 9000 var = 9 kvar

Figure 3.4 shows a schematic of a col~ceptual single-phase, phase-shifting transformer. This transformer is not an idealization of an actual transSol.mer since it is physically impossible to obtain a complex turns ratio. I t will be used

FIGURE 3.4 St - -- s2

Schematic representation of a conceptual single- 'l"l1C - c "

phase, phase-shifting transformer el* 1


later in this chapter as a mathematical model for representing phase shift of three-phase transformers. As shown in Figure 3.4, the complex turns ratio N, is defined for the phasc-shifting transformer as

wherc 4 is the phasc-shift anglc. Tlic tsansfonncr relations are thcn

Note that the phase angle of El leads the phase angle of E2 by 4. Similarly, II leads I2 by the angle 4. However, the magnitudes are unchanged; that is, [Ell = IE21 and 1111 = [I1).

From thcse two relations, the following two additional relations are derived:

Thus, impedance is unchanged when it is referred from one side of an ideal phase-shi fting transformer to the other. Also, the ideal phase-shifting transformer has no real or reactive power losses since S1 = S2.

Note that (3.1.23) and (3.1.24) for the phase-shifting transformer are the same as (3.1.16) and (3.1.17) for the ideal physical transformer except for the complex conjugate (*) in (3.1.24). The cornplex conjugate for the phase- shifting transformer is required to make SI = S2 (complex power into winding I equals complex power out of winding 2), as shown in (3.1.25).

EQUIVALENT CIRCUITS FOR PRACTICAL TRANSFORMERS

Figure 3.5 shows an equivalent circuit for a practical single-phase two-winding transformer, which differs from the ideal transformer as follows:

1. The windings have resistance. 2. The core permeability jr,. is finite. 3. The magnetic flux is not entirely confined to the core. 4. There are real and reactive power losses in the core.

SECTION 3.2 EQUIVALENT CIRCUITS FOR PRACTICAL TRANSFORMERS

FIGURE 3.5 Eq~~ivalent circuit of a practical single-phase

two-winding transformer

The resistance R I is included in series with winding 1 of the figure to account for I'R losses in this winding. A reactance X I , called the leahage reactance of winding 1, is also included in series with winding 1 to account for the leakage flux of winding 1. This leakage flux is the component of the flux that links winding 1 but docs not link winding 2; it causes a voltage drop I l ( j X I ) , which is proportional to I, and leads II by 90". There is also a reactive power loss l f x l associated with this leakage reactance. Similarly, there is a resistance Rz and a leakage reactance X2 in series with winding 2.

Equation (3.1.7) s h o ~ ~ s that for finitc core pern~eability k i , , tlic total mmf is not zero. Dividing (3.1.7) by NI and using (3.1.1 I), we get

Defining the term on the sight-hand side of (3.2.1) to be I,,, called ~~~t ryne t i z - ing current, it is evident that I,,, lags El by 90, and can be represented by a

shunt inductor with susceptance B,,, = (3) mhos.* However, in rcality there is an additional shunt branch, represented by a resistor with conduc- tance G,. mhos, which carries a current I,., called the core 1o.u cumcnl. 1, is in phase with E l . When the core loss current I, is included, (3.2.1) becomes

The equivalent circuit of Figure 3.5, which includes the shunt branch with admittance (G, - jB,,,) mhos, satisfies the KCL equation (3.2.2). Note that when winding 2 is open (I? = 0) and when a sinusoidal voltage 1'1 is applied to winding 1, then (3.2.2) indicates that the current II will have two components: the core loss current I,. and the magnetizing current I,,,. .4sso- ciated with I,. is a real power loss I'/G, = E ~ G , W. This real power loss accounts for both hysteresib and eddy current losscs within the core. Hysteresis loss occurs because a cyclic variation of flux within the core requires energy dissipated as heat. As such. hysteresis loss can bc reduced by the usc of'spe- cia1 high grades of alloy stccl a h core mi~torial. Eddy current losb occurs

*The units of :~dn~ittancc. cl)nduc.t,~~~cc. ant1 SLISCL'I)[;IIICC. W I I ~ C I I in ~ h c SI system arc sicnlsns (with synlbol S), arc i ~ l x ~ callctl ~nllob (ui th synlbol U) (11 . ulllns ' (will1 symbol l2 ' ) .

CHAFTER 3 POWER TRANSFORMERS

FIGURE 3.6 Equivalent circuits for a

practical single-phase two-winding transformer

because induced currents called eddy currents flow within the magnetic core perpendicular to the flux. As such, eddy current loss can be reduced by con- structing the core with laminated sheets of alloy steel. Associated with I,,, is a reactive power loss I;,/B,,, = E:B,, var. This reactive power is required to magnetize the core. The phasor sum (I, + I,,,) is called the exciting current I,.

Figure 3.6 shows three alternative equivalent circuits for a practical single-phase two-winding transformer. In Figure 3.6(a), the resistance Rz and leakage reactance X2 of winding 2 are referred to winding 1 via (3.1.21). In Figure 3.6(b), the shunt branch is omitted, which corresponds to neglecting the exciting current. Since the exciting current is usually less than 5% of rated current, neglecting it in power system studies is often valid unless transformer efficiency or exciting current phenomena are of particular concern. For large power transformers rated more than 500 kVA, the winding resistances, which are small compared to the leakage reactances, can often be neglected, as shown in Figure 3.6(c).

Thus, a practical transformer operating in sinusoidal steady state is equivalent to an ideal transformer with external impedance and admittance branches, as shown in Figure 3.6. The external branches can be evaluated from short-circuit and open-circuit tests, as illustrated by the following example.

Nl N2

(a) R2 and X2 are referred to winding 1

(b) Neglecting exciting current

(c) Neglecting exciting current and I2R winding loss


EXAMPLE 3.2 Transformer short-circuit and open-circuit tests

A single-phase two-winding transformer is rated 20 kVA, 4801120 volts, 60 Hz. During a short-circuit test, where rated current a t rated frequency is applied to the 480-volt winding (denoted winding I), with the 120-volt winding (winding 2) shorted, the following readings are obtained: V1 = 35 volts, PI = 300 W. During an open-circuit test, where rated voltage is applied to winding 2, with winding 1 open, the following readings are obtained: I? = 12 A, P2 = 200 W.

a. From the short-circuit test, determine the equivalent series impedance Zeq1 = RqI + jXeqI referred to winding 1. Neglect the shunt admittance.

b. From the open-circuit test, determine the shunt admittance Y,,, = G, - jB,,, referred to winding 1. Neglect the series impedance.

SOLUTION

a. The equivalent circuit for the short-circuit test is shown in Figure 3.7(a), where the shunt admittance branch is neglected. Rated current for winding 1 is

ReqI, ZeqI, and XeqI are then determined as follows:

FIGURE 3.7 / I = llrared Circuits for Example 3.2

Vl q---j - 11 lL1 480 : 120

(a) Short-c~rcu~t test (neglecting shunt adm~ttance)

CHAPTER 3 P O W E R TRANSFORMERS

b. The equivalent circuit for the open-circuit test is shown in Figure 3.7(b), where the series impedance is neglected. From (3.1.16),

G,, Y,, and B,,, are then determined as follows:

B,, = 4- = \(0.00625)~ - (0.000868)2 = 0.006 19 S Y,,,=G,- jB,,=0.000868- j0.00619=0.00625/-82.02 S Note that the equivalent series impedance is usually evaluated at

rated current from a short-circuit test, and the shunt admittance is evaluated at rated voltage from an open-circuit test. For small variations in transformer operation near rated conditions, the impedance and admittance values are often assumed constant.

The following are not represented by the equivalent circuit of Figure 3.5:

1. Saturation 2. Inrush current 3. Nonsinusoidal exciting current 4. Surge phenomena

They are briefly discussed in the following sections.

SATURATION

In deriving the equivalent circuit of the ideal and practical transformers, we have assumed constant core permeability p, and the linear relationship B, = ,LL~H,. of (3.1.3). However, the relationship between B and H for ferro- magnetic materials used for transformer cores is nonlinear and multivalued. Figure 3.8 shows a set of B-H curves for a grain-oriented electrical steel typically used in transformers. As shown, each curve is multivalued, which is

B I l curvcs for M-5 grain-oriented electrical

steel 0.012 in. thick (Armco Inc.)


FIGURE 3.8

caused by hysteresis. For many engineering applications, the B-H curves can be adequately described by the dashed line drawn through the curves in Figure 3.8. Note that as H increases, the core becomes saturated; that is, the curves flatten out as B increases abovc 1 wb/m2. If the magnitude of the voltage applied to a transformer is too large, the core will saturate and a high magnetizing current will flow. In a well-designed transformer, the applied peak voltage causes the peak flux density in steady state to occur at the knee of the B-H curve, with a corresponding low value of magnetizing current.

INRUSH CURRENT

When a transformer is first energized, a transient current much larger than rated transformer current can flow for several cycles. This current, called ill- rush currenf, is nonsinusoidal and has a large dc component. To understand the cause of inrush, assume that before energization, the transformer core is magnetized with a rcsidual flux density B(0) = 1.5 wb/m2 (near the knee of the dotted curve in Figure 3.8). If the transfornler is then energized when the source voltage is positive and increasing, Faraday's law, (3.1.9), will cause the flux density B(t) to increase further, since

As B(1) moves into the saturiltion region of the B--H curve, large values of H(t) will occur, kinti, from Amprc 's law, (3.1. I ) , corresponding large values of current ;(I) will flow for sc\cr;~l cyclcs until i l hils dissipated. Sincc no1.1na1 inrush currents can be as Iargc as abnormal short-circuit currents in trans-


formers, transformer protection schemes must be able to distinguish between these two types of currents.

NONSINUSOIDAL EXCITING CURRENT

When a sinusoidal voltage is applied to one winding of a transformer with the other winding open, the flux d ( t ) and flux density B(t) will, from Faraday's law, (3.1.9), be very nearly sinusoidal in steady state. However, the magnetic field intensity H(t) and the resulting exciting current will not be sinusoidal in steady state, due to the nonlinear B-H curve. If the exciting current is measured and analyzed by Fourier analysis techniques, one finds that it has a fun- damental component and a set of odd harmonics. The principal harmonic is the third, whose rms value is typically about 40% of the total rms exciting current. However, the nonsinusoidal nature of exciting current is usually neglected unless harmonic effects are of direct concern, because the exciting current itself is usually less than 5% of rated current for power transformers.

SURGE PHENOMENA

When power transformers are subjected to transient overvoltages caused by lightning or switching surges, the capacitances of the transformer windings have important effects on transient response. Transformer winding capacitances and response to surges are discussed in Chapter 12.

T H E PER-UNIT SYSTEM

Power-system quantities such as voltage, current, power, and impedance are often expressed in per-unit or percent of specified base values. For example, if a base voltage of 20 kV is specified, then the voltage 18 kV is (18120) = 0.9 per unit or 90%. Calculations can then be made with per-unit quantities rather than with the actual quantities.

One advantage of the per-unit system is that by properly specifying base quantities, the transformer equivalent circuit can be simplified. The ideal transformer winding can be eliminated, such that voltages, currents, and external impedances and admittances expressed in per-unit do not change when they are referred from one side of a transformer to the other. This can be a significant advantage even in a power system of moderate size, where hun- dreds of transformers may be encountered. The per-unit system allows us to avoid the possibility of making serious calculation errors when referring quantities from one side of a transformer to the other. Another advantage of the per-unit system is that the per-unit impedances of electrical equipment of

SECTION 3.3 THE PER-UNIT SYSTEM 89

similar type usually lie within a narrow nun~erical range when the equipment ratings are used as base values. Because of this, per-unit impedance data can be checked rapidly for gross errors by someone Pdmiliar with per-unit quantities. In addition, manufacturers usually specify the impedances of machines and transformers in per-unit or percent of nalneplate rating.

Per-unit quantities arc calculated as follows: actual quantity per-unit quantity = basc value of quantity

where L I C ~ L ~ C I ~ q 1 1 ~ 1 1 1 t i l j ~ is the value of the q ~ i ~ l ~ t i t ~ in the actual units. The base value has the same units as the actual quantity, thus making the per-unit quantity dimensionless. Also, thc basc value is always a real number. There- fore, the angle of the per-unit quantity is the same as the angle of the actual quantity.

Two independent base values can be arbitrarily selected at one point in a power system. Usually the base voltage VbitbcLN and base complex power Sbase14 are selected for either a single-phase circilit or for one phase of a three- phase circuit. Then, in order for elcctrical laws to be valid in the per-unit system, the following relations IIILIS~ be used for other base values:

1 Ybilst: = Gbase = Bbase = - (3.3.5) Z b.1~

In (3.3.2)-(3.3.5) the subscripts LN and 14 denote "line-to-neutral" and "per-phase," respectively, for three-phase circuits. These equations are also valid for single-phase circuits, where subscripts can be omitted.

By convention, we adopt the following two rules for base quantities:

1. The value of Sbusc14 is the same for thc entire power system of concern.

2. The ratio of the voltage bases on either side of a transformer is selected to be the same as the ratio of the transformer voltage ratings.

With these two rules, per-unit inlpcdi~ncc remains unchanged when referred from one side of a transformer to the other.

EXAMPLE 3.3 Per-unit impedance: single-phase transformer

A single-phase two-winding tri~nsl'ormcr is ri~ted 20 kVA, 480/120 volts, 60 Hz. The equivalent Icakagc iml>cdancc of the transformer referred to the 120-volt winding, clenotccl ciincling 3, is Z,,,? - 0.0525/78.13 R. Using the


transformer ratings as base values, determine the per-unit leakage impedance referred to winding 2 and referred to winding 1.

SOLUTION The values of Sbase, Vbawl, and Vbase2 are, from the transformer ratings,

SbaE = 20 kVA, Vbasel = 480 volts, Vbase2 = 120 volts Using (3.3.4), the base impedance on the 120-volt side of the transformer is

Then, using (3.3.1), the per-unit leakage impedance referred to winding 2 is 2-72 Zeq2p.u. = - -

Zbase2 0'0525a = 0.0729/78.13O per unit 0.72

If Zeq2 is referred to winding 1,

= 0.84/78.13O i2

The base impedance on the 480-volt side of the transformer is

(480)~ 'basel =

Zbasel = - = 11.52 !2 S base 20,000 and the per-unit leakage reactance referred to winding 1 is

Z,, 0.84/78.13O - Z e q ~ p . u . = - - = 0.0729/78.13O per unit = Zeq2p.u.

Zbase 1 1 1.52 Thus, the per-unit leakage impedance remains unchanged when referred from winding 2 to winding 1. This has been achieved by specifying

Figure 3.9 shows three per-unit circuits of a single-phase two-winding transformer. The ideal transformer, shown in Figure 3.9(a), satisfies the per-unit relations = EZp.u., and Ilp.u. = 12p.u., which can be derived as follows. First divide (3.1.16) by Vbasel :

Then, using V ~ S ~ I /Vbase2 = Vratedl /Vrated2 = NI /N2,

SECTION 3.3 THE PER-UNIT SYSTEM 9 I

FIGURE 3.9 Per-unit equivalent

circuits of a single-phase two-winding transformer

(a) Ideal transformer

(b) Neglect~ng exc~t~ng current

(c) Complete representation

Similarly, divide (3.1.17) by Ibaxl :

Thus, the ideal transformer winding in Figure 3.2 is eliminated from the per-unit circuit in Figure 3.9(a). The per-unit leakage impedance is included in Figure 3.9(b), and the per-unit shunt admittance branch is added in Figure 3.9(c) to obtain the complete representation.

When only one component, such as a transformer, is considered, the nameplate ratings of that component are usually selected as base values. When several components are involved, however, the system base values may be different from the nameplate ratings of any particular device. It is then necessary to convert the per-unit impedance of a device from its nameplate ratings to the system base values. To convert a per-unit impedance from "old" to "new" base values, use


or, from (3.3.4),

EXAMPLE 3.4 Per-unit circuit: three-zone single-phase network

FIGURE 3.10 Circuits for Example 3.4

Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones are connected by transformers TI and T2, whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected.

SOLUTION First the base values in each zone are determined. Sbase = 30 kVA is the same for the entire network. Also, Vbasel = 240 volts, as specified for zone 1. When moving across a transformer, the voltage base

Zone 1 I

I v, = 220/0" volts I

Zone 2 Zone 3

30 kVA 2401480 volts

X, = 0.10 p.u.

20 kVA 46011 15 volts

X, = 0.10 p.u.

(a) Single-phase circuit

+ : - L - 1 j0.10 P.U. 10.2604 p.u. 1j0.1378

vmu. = I I I

I P.u. 1 t0adp.u =

0.9167m p.u. I I I I 1.875 + 10.4167 p.u I I I I I I

-

I I Zone 1 I Zone 2 I Zone 3 I

I I Vbss., = 240 volts V D ~ S ~ P = 480 volts I V,,,, = 120 volts

I 1

(b) Per-unit circuit

SECTION 3.3 THE PER-UNIT SYSTEM

is changed in proportion to the transformer voltage ratings. Thus,

and

Vbawl = (g) (480) = 120 volts The base impedances in zones 2 and 3 are

and

and the base current in zone 3 is

Next, the per-unit circuit impedances are calculated using the systenl base values. Since Sbilse = 30 kVA is the same as the kVA rating of transformer T I , and Vbasel = 240 volts is the same as the voltage rating of the zone 1 side of transformer T I , the per-unit leakage reactance of T i is the same as its nameplate value, XTlpu. = 0.1 per unit. However, the per-unit leakage reactance of transformer Tz must be converted from its nameplate rating to the system base. Using (3.3.1 1 ) and VbaseZ = 480 volts,

460 30,000 XT2p.u. = (0.10) (=) (rn) = 0.1378 per unit

Alternatively, using Vbase3 = 120 volts,

1 15 ' 30,000 XTB.u. = (0.10) (rn) (rn) = 0.1378 per unit

which gives the same result. The line, which is located in zone 2, has a per- unit reactance

Xlinc - 7 - - Xlinep.~. = - - = 0.2604 per unit Zbase2 7.68

and the load, which is located in zone 3, has a per-unit impedance Zl,X,'l - 0.9 + .lo.?

Zloadp.u. = - - r= 1.875 + j0.4167 per unit Zbase3 0.48

The per-unit circuit is shown in Figure 3.10(b), where the base values for each zone, per-unit impedances, and the per-unit source voltage are


shown. The per-unit load current is then easily calculated from Figure 3. I O(b) as follows:

= 0.43951-26.01' per unit

The actual load current is

Note that the per-unit equivalent circuit of Figure 3.10(b) is relatively easy to analyze, since ideal transformer windings have been eliminated by proper selection of base values. 1

Balanced three-phase circuits can be solved in per-unit on a per-phase basis after converting A-load impedances to equivalent Y impedances. Base values can be selected either on a per-phase basis or on a three-phase basis. Equations (3.3.1)-(3.3.5) remain valid for three-phase circuits on a per-phase basis. Usually Sbase3$ and VbaseLL are selected, where the subscripts 34 and LL denote "three-phase" and "line-to-line," respectively. Then the following relations must be used for other base values:

EXAMPLE 3.5 Per-unit and actual currents in balanced three-phase networks

As in Example 2.5, a balanced-Y-connected voltage source with Eat, = 480@ volts is applied to a balanced-A load with ZA = 3 0 m a. The line impedance

SECTION 3.3 THE PER-UNIT SYSTEM

FIGURE 3.1 1 Circuit for Example 3.5

between the source and load is ZL = 1& I2 for each phase. Calculate the per-unit and actual c~lrrellt in phasc ( I of the line using Sbase3# = 10 kVA and VbaseLL = 480 volts.

SOLUTION First, convert ZA to an equivalent Z y ; the equivalent line- to-neutral diagram is shown in Figurc 2.17. Thc base impedance is, from (3.3.16),

The per-unit line and load impedances are ZL -

2 L p . u . = - - - = 0.04340/85" per unit Zbase 23.04 and

Also,

V ~ ; , ~ L L - - 480 Vbax~N = - -- - 277 volts

J3 JS and

E,,,, - 277/ -30" Elolp.u. = - = 1 .O[-3W per unit

V h s e ~ N 277 The per-unit equivalent circuit is shown in Figure 3.1 1. The per-unit line current in phase N is then

= 2.1471-73.78" per unit


The base current is

and the actual phase a line current is I, = (2.1471 -73.7g0)(12.03) = 25.83 f -73.78" A

-- - - -

THREE-PHASE TRANSFORMER CONNECTIONS A N D PHASE SHIFT

Three identical single-phase two-winding transformers may be connected to form a three-phase bank. Four ways to connect the windings are Y-Y, Y-A, A-Y, and A-A. For example, Figure 3.12 shows a three-phase Y-Y bank. Figure 3.12(a) shows the core and coil arrangements. The American standard for marking three-phase transformers substitutes HI , H2, and H3 on the high-voltage terminals and XI, X2, and X3 on the low-voltage terminals in place of the polarity dots. Also, in this text, we will use uppercase letters ABC to identify phases on the high-voltage side of the transformer and lowercase letters abc to identify phases on the low-voltage side of the transformer. In Figure 3.12(a) the transformer high-voltage terminals HI, H2, and H3 are connected to phases A , B, and C, and the low-voltage terminals XI, X2, and X3 are connected to phases a , h, and c, respectively.

Figure 3.12(b) shows a schematic representation of the three-phase Y- Y transformer. Windings on the same core are drawn in parallel, and the phasor relationship for balanced positive-sequence operation is shown. For example, high-voltage winding HI-N is on the same magnetic core as low- voltage winding XI-n in Figure 3.12(b). Also, V A N is in phase with V,,. Fig- ure 3.12(c) shows a single-line diagram of a Y-Y transformer. A single-line diagram shows one phase of a three-phase network with the neutral wire omitted and with components represented by symbols rather than equivalent circuits.

The phases of a Y-Y or a A-A transformer can be labeled so there is .no phase shift between corresponding quantities on the low- and high-voltage windings. However, for Y-A and A-Y transformers, there is always a phase shift. Figure 3.13 shows a Y-A transformer. The labeling of the windings and the schematic representation are in accordance with the American standard, which is as follows:

In either a Y-A or A-Y transformer, positive-sequence quantities on the high-voltage side shall lead their corresponding quantities on the low-voltage side by 30".

As shown in Figure 3.13(b), V A N leads V, by 30".

SECTION 3.4 THREE-PHASE TRANSFORMER CONNECTIONS AND PHASE SHIFT

FIGURE 3.12 Three-phase two-

winding Y-Y transformer bank

(a) Core and cod arrangements (c) S~ngle-l~ne d~agrarn

(b) Schemat~c representation showlng phasor relationsh~p for posltlve sequence operation

The positive-sequence phasor diagram shown in Figure 3.13(b) can be constructed via the following five steps, which are also indicated in Figure 3.13:

STEP I Assume that balanced positive-sequence voltages are applied to the Y winding. Draw the positive-sequence phasor diagram for these vo1t:iges.

STEP 2 Move phasor A - N next to terminals A-N in Figure 3.13(a). Identify the ends of this line in the same manner as in the phasor diagram. Sinlilarly, move phasors B-N and C-N next to terminals B N and C'-N in Figure 3.13(a).

STEP 3 For each singlc-phase transformer, the voltage across the low- voltage winding nus st be in phase with the voltage across the


FIGURE 3.13 Three-phase two-

winding Y-A transformer bank

a (Step 3Ia N b-

(a) Core and coil arrangement

(Step 5)

(b) Positive-sequence phasor diagram

high-voltage winding, assuming an ideal transformer. There- fore, draw a line next to each low-voltage winding parallel to the corresponding line already drawn next to the high-voltage winding.

STEP 4 Label the ends of the lines drawn in Step 3 by inspecting the polarity marks. For example, phase A is connected to dotted terminal HI, and A appears on the right side of line A-N. Therefore, phase a, which is connected to dotted terminal XI, must be on the ricght side, and b on the left side of line a-h. Similarly, phase B is connected to dotted terminal H2, and B is down on line B-N. Therefore, phase b, connected to dotted

SECTION 3.4 THREE-PHASE TRANSFORMER CONNECTIONS AND PHASE SHIFT

terminal X2, must be ~ l o ~ ~ . n on line b-c. Similarly, c is up on line c.-rt.

STEP 5 Bring the three lines labeled in Step 4 together to complcte the phasor diagram for the low-voltage A winding. Note that liN leads Kt;,, by 30" in accordance with the American standard.

EXAMPLE 3.6 Phase shift in A-Y transformers

Assume that bala~lced negative-sequence voltitges are applied to the high- voltage windings of the Y-A tncnsformer shown in Figure 3.13. Determine the negative-sequence phase shift of this transfonner.

SOLUTION The negative-sequence diagram, shown in Figure 3.14, is constructed from the following fivc steps, as outlined above:

FIGURE 3.14 Example 3.6-

Construction of nega tive-sequence

phasor diagram for Y-A transformer bank

L

(Step 5)


STEP I Draw the phasor diagram of balanced negative-sequence voltages, which are applied to the Y winding.

STEP 2 Move the phasors A-N, B-N, and C-N next to the high- voltage Y windings.

STEP 3 For each single-phase transformer, draw a line next to the low-voltage winding that is parallel to the line drawn in Step 2 next to the high-voltage winding.

STEP 4 Label the lines drawn in Step 3. For example, phase B, which is connected to dotted terminal H2, is shown irp on line B-N; therefore phase b, which is connected to dotted terminal X2, must be zip on line b-c.

STEP 5 Bring the lines drawn in Step 4 together to form the negative- sequence phasor diagram for the low-voltage A winding.

As shown in Figure 3.14, the high-voltage phasors lay the low-voltage phasors by 30". Thus the negative-sequence phase shift is the reverse of the positive- sequence phase shift.

The A-Y transformer is commonly used as a generator step-up transformer, where the A winding is connected to the generator terminals and the Y winding is connected to a transmission line. One advantage of a high-voltage Y winding is that a neutral point N is provided for grounding on the high- voltage side. With a permanently grounded neutral, the insulation requirements for the high-voltage transformer windings are reduced. The high-voltage insulation can be graded or tapered from maximum insulation at terminals ABC to minimum insulation at grounded terminal N. One advantage of the A winding is that the undesirable third harmonic magnetizing current, caused by the nonlinear core B-H characteristic, remains trapped inside the A winding. Third harmonic currents are (triple-frequency) zero-sequence currents, which cannot enter or leave a A connection, but can flow within the A. The Y-Y transformer is seldom used because of difficulties with third harmonic exciting current.

The A-A transformer has the advantage that one phase can be removed for repair or maintenance while the remaining phases continue to operate as a three-phase bank. This open-A connection permits balanced three-phase operation with the kVA rating reduced to 58% of the original bank (see Problem 3.24).

Instead of a bank of three single-phase transformers, all six windings may be placed on a common three-phase core to form a three-phase transformer, as shown in Figure 3.15. The three-phase core contains less iron than the three single-phase units; therefore it costs less, weighs less, requires less floor space, and has a slightly higher efficiency. However, a winding failure would require replacement of an entire three-phase transformer, compared to replacement of only one phase of a three-phase bank.

SECTION 3.5 PER-UNIT EQUIVALENT CIRCUITS OF THREE-PHASE TRANSFORMERS

FIGURE 3.15 Transformer core

coufigurations

coils

laminated sheet core

(a) Single-phase core type


(bl Single-phase shell type


(c) Three-phase, three-legged core type

(d) Three-phase shell type

PER-UNIT EQUIVALENT CIRCUITS OF BALANCED THREE-PHASE T W O - W I N D I N G TRANSFORMERS

Figure 3.16(a) is a schenlatic representation of an ideal Y-Y transforn-rer grounded through neiltral impedances Z,\. and Z,,. Figure 3.16(b) shows the per-unit equivalent circuit of this idcul transformer for balanced three-phase operation. Throughout the relnuinder of this text, per-unit quantities will be used unless otherwise indicated. Also, the subscript "p .~ . , " used to indicate a per-unit quantity, will be omittcd in most cases.

By convention, we adopt thc following two rules for selecting base quantities:

I . A common Sbilg is sclcctcd For both tlic 1-1 and X terminals.

2. The ratio of thc voltage bnscs VbLISCII/Vt,.,bCM is selected to be equal to the ratio or thc rated li~ic-to-line voltages Vl,llcdl.iLL/VrdtedXLL.

When balanced threc-phi~sc ci~rsc~ltb arc ;~pplieiI to the transformer, the I ~ ~ L I - tral currents are zero and thcrc ill-c 110 ~o1 ta .g~ dsops across the neutral impcd- ances. Therefore, the per-unit ccllri\alcnt circuit of the ideal Y-Y transformcr, Figure 3.16(b), is thc si~inc 'I> the pi-uilit singlc-ph;~se ideal transformer, I ig- use 3.9(a).

The per-unit ccl~~iv:~lcnt cil.ciii[ 01' ;I pr;~ctic;~I Y Y tri~llsfc~r~ncr is shou 11


FIGURE 3.16 Ideal Y-Y transformer

(a) Schematic representation

-

(b) Per-unit equivalent circuit for balanced three-phase operation

in Figure 3.17(a). This network is obtained by adding external impedances to the equivalent circuit of the ideal transformer, as in Figure 3.9(c).

The per-unit equivalent circuit of the Y-A transformer, shown in Fig- ure 3.17(b), includes a phase shift. For the American standard, the positive- sequence voltages and currents on the high-voltage side of the Y-A transformer lead the corresponding quantities on the low-voltage side by 30". The

FIGURE 3.17 Per-unit equivalent circuits of practical Y-Y, Y-A, and A-A transformers for balanced three-phase operation

Single-line diagram

Per-unit equivalent

circuit

+F

P P VH - -

-+E-

P A - -

elm : 1

-3t a A

!T - -

SECTION 3.5 PER-UNIT EQUIVALENT CIRCUITS OF THREE-PHASE TRANSFORMERS

phase shift in the equivalent circuit of Figure 3.17(b) is represented by thc phase-shifting transformer of Figure 3.4.

The per-unit equivalent circuit of the A-A transformer, shown in Figure 3.17(c), is the same as that of the Y-Y transformer. It is assumed that the windings are labeled so there is no phase shift. Also, the per-unit impedances do not depend on the winding connections, but the base voltages do.

EXAMPLE 3.7 Vol ta~e calculations: balanced Y-Y and A-Y transformers

FIGURE 3.18 Per-unit network for

Example 3.7

Three single-phase two-winding transformers, each rated 400 MVA, 13.81 199.2 kV, with leakage reactance Xeq = 0.10 per unit, are connected to form a three-phase bank. Winding resistances and exciting current are neglected. The high-voltage windings are connected in Y. A three-phase load operating under balanced positive-sequence conditions on the high-voltage side absorbs 1000 MVA at 0.90 p.f. lagging, with = 199.2/0" kV. Determine the voltage V,,, at the low-voltage bus if the low-voltage windings are connected (kt) in Y, (b) in A.

SOLUTION The per-unit network is shown in Figure 3.18. Using the transformer bank ratings as base quantities, Sbasr3& = 1200 MVA, VbasefILL = 345 kV, and IbaseH = 1200/(345&) = 2.008 kA. The per-unit load voltage and load current are then

V A N = 1.0/00 per unit

I,., = 1000/(345fi)/-~os -'O.!I = O.8333/-25.840 per unit 2.008

-I - (a) Y-connected low-voltage wrndings

I, = 0 8 3 3 3 /-2584" m a -

van v = j [-'-o+vAN = 1 .o /g (b) A-connected low-voltaye w~ndrngs


a. For the Y-Y transformer, Figure 3.18(a),

I, = IA = 0.8333/-25.84" per unit

Kt , , = %N + ( j & q ) I ~ = 1 . O w + (jO.lO)(O.8333/-25.84") = 1 .O + 0.08333/64.16" = 1.0363 + j0.0750 = 1.039/4.139" = 1.039/4.139" per unit

Further, since VbaseXLN = 13.8 kV for the low-voltage Y windings, V,, = 1.039(13.8) = 14.34 kV, and

b. For the A-Y transformer, Figure 3.18(b),

Em =e-i300&N = 1.01-30' per unit

I, = e-j3001A = 0.8333/-25.84" - 30" = 0.8333/-55.84" per unit G,, = EAN + ( jXeq)IA

= 1 .O/-30" + (jO.10)(0.8333/-55.84") V, = 1.039/-25.861" per unit

Further, since VbnseXLN = 1 3 . 8 / 6 = 7.967 kV for the low-voltage A windings, Van = (1.039)(7.967) = 8.278 kV, and

EXAMPLE 3.8 Per-unit voltage drop and per-unit fault currene: balanced three-phase transformer

A 200-MVA, 345-kVAl34.5-kV Y substation transformer has an 8% leakage reactance. The transformer acts as a connecting link between 345-kV trans-

'

mission and 34.5-kV distribution. Transformer winding resistances and exciting current are neglected. The high-voltage bus connected to the transformer is assumed to be an ideal 345-kV positive-sequence source with negligible source impedance. Using the transformer ratings as base values, determine:

a. The per-unit magnitudes of transformer voltage drop and voltage at the low-voltage terminals when rated transformer current at 0.8 p.f. lagging enters the high-voltage terminals

b. The per-unit magnitude of the fault current when a three-phase-to- ground bolted short circuit occurs at the low-voltage terminals

SECTION 3.5 PER-UNIT EQUIVALENT CIRCUITS OF THREE-PHASE TRANSFORMERS 105

SOLUTION In both parts (a) and (b), only balanced positive-sequence current will flow, since there are nu imbalances. Also, because we are interebted only in voltage and current magnitudes, the A-Y transformer phase shift can be omitted. a. As shown in Figure 3.19(a),

vdrop = ImtedXcq = (1.0)(0.08) = 0.08 per unit and

V,, = VAN - ( jXeq) lratsd = I . O N - (jO.O8)(1 .O/-36.87") = 1.0 - (j0.08)(0.8 - j0.6) = 0.952 - j0.064 = 0.9541-3.85O per unit

b. As shown in Figure 3.19(b). VAN 1.0 I~,=---- - - 12.5 per unit X,, 0.08

Under rated current conditions (part (a)], the 0.08 per-unit voltage drop across the transformer leakage reactance causes the voltage at the low-voltage ternli- nals to be 0.954 per unit. Also, under three-phase short-circuit conditions [part (b)], the fault current is 12.5 times the rated transformer current. This example illustrates a compromise in the design or specification of transformer leakage reactance. A low value is desired to minimize voltage drops, but a high value is desired to limit fault currents. Typical transformer leakage reactances are given in Table A.2 in the Appendix.

FIGURE 3.19 Circuits for Example 3.8 +

Van

(a) Rated current

(b) Short-c~rcu~t current

1 06 CHAPTER 3 POWER TRANSFORMERS

THREE-WINDING TRANSFORMERS

Figure 3.20(a) shows a basic single-phase three-winding transformer. Thc ideal transformer relations for a two-winding transformer, (3.1.8) and (3.1.14), can easily be extended to obtain corresponding relations for an ideal three- winding transformer. In actual units, these relations are

where I I enters the dotted terminal, Iz and I3 leave dotted terminals, and E l , El, and E3 have their + polarities at dotted terminals. In per-unit, (3.6.1) and (3.6.2) are

where a common Sbase is selected for all three windings, and voltage bases are selected in proportion to the rated voltages of the windings. These two per- unit relations are satisfied by the per-unit equivalent circuit shown in Figure 3.20(b). Also, external series impedance and shunt admittance branches are included in the practical three-winding transformer circuit shown in Figure

(a) Basic core and coil configuration (b) Per-unit equivalent circuit-ideal transformer

(c) Per-unit equivalent circuit-practical transformer FIGURE 3.20 Single-phase three-winding transformer


Because of the varying voltage drops caused by changing loads, LTCs are often operated to automatically regulate a bus voltage. This is particularly truc whcn thcy arc used as stcp-down transformers. To place the example transformer on automatic control, click on the "Manual" field. This toggles the transformer control mode to automatic. Now the transformer manually changes its tap ratio to maintain the load voltage within a specified voltage range, between 0.995 and 1.005 per unit in this case. To see the LTC in automatic operation use the load arrows to vary the load, noting that the LTC changes to keep the load's voltage within the specified deadband.

The three-phase regulating transformers shown in Figures 3.26 and 3.27 can be modeled as transformers with off-nominal turns ratios. For the voltage- magnitude-regulating transformer shown in Figure 3.26, adjustable voltages AV,, A VbN, and A V,, which have equal magnitudes AV and which are in phase with the phase voltages Kt!,, VI,,,, and V,, are placed in the series link between buses a-a', b-b', and c-c'. Modeled as a transformer with an off- nominal turns ratio (see Figure 3.25), c = (1 + AV) for a voltage-magnitude increase toward bus abc, or c = (1 + AV)-' for an increase toward bus cr'b'c'.

For the phase-angle-regulating transformer in Figure 3.27, the series voltages AV,,, AVb,, and AV, are +90 out of phase with the phase voltages V,, Vbn, and V,. The phasor diagram in Figure 3.27 indicates that each of the bus voltages &I,,, Vht,, and &I,, has a phase shift that is approximately proportional to the magnitude of the added series voltage. Modeled as a transformer with an off-nominal turns ratio (see Figure 3.25), c z ll/a for a phase increase toward bus ahc or c x l k for a phase increase toward bus a'b'c'.

FIGURE 3.26 An example of a

voltage-magnitude- regulating transformer

SECTION 3.8 TRANSFORMERS WITH OFF-NOMINAL TURNS RATIOS

FIGURE 3.27 An example of a phase-angle-reguliitit~g rransfor~ncr. Windings drawn in parallel arc on the same core

EXAMPLE 3.13 Voltage-regulating and phase-shifting three-phase transformers

Two buses abc and ~~'h 'c . ' are connected by two parallel lines L1 and L2 with positive-sequence series reactances XL1 = 0.25 and XL2 = 0.20 per unit. A regulating transformer is placed in series with line L1 at bus a'b'c'. Deter- mine the 2 x 2 bus admittance matrix when the regulating transformer (a) provides a 0.05 per-unit increase in voltage magnitude toward bus a'b'c' and (b) advances the phase 3" toward bus cr'b'c'. Assume that the regulating transformer is ideal. Also, the series resistance and shunt admittance of the lines are neglected.

SOLUTION The circuit is shown in Figure 3.38. I a. For the voltage-magnitudc-regulating transformer, c = (1 + AV)- =

(1 .05)~ ' = 0.9524 per unit. From (3.7.5)-(3.7.8), the admittance parameters of the regulating transfornler in series with line L1 are

For line L2 alone,


FIGURE 3.28 10.20 Positive-seauence circuit r""----1 1.

for ~ x a r n ~ l e 3.13

Combining the above admittances in parallel,

Y12 = YZ1 = YIZLl + YIZL2 = j3.810 + j5.0 = j8.810 per unit b. For the phase-angle-regulating transformer, c = I& = 1/-3". Then, for

this regulating transformer in series with line L1,

The admittance parameters for line L2 alone are given in part (a) above. Combining the admittances in parallel,

YZI = -0.2093 + j3.9945 + j5.0 = -0.2093 + j8.9945 per unit An animated view of this example is provided in PowerWorld Simulator case Example 3-13. In this case, the transformer and parallel transmission are assumed to be supplying power from a 345-kV generator to a 345-kV load, with an initial phase angle of 3 degrees. Click on the arrows next to the phase angle to change the angle in one-degree steps, or by the tap field to change the LTC tap in 0.625% steps. Notice that changing the phase angle primarily changes the real power flow, whereas changing the LTC tap changes the reactive power flow. In this example, the line flow fields are showing the absolute value of the real or reactive power flow; the direction of the flow is indicated with arrows. Traditional power flow programs usually indicate power flow direction using a convention that flow into the line or other device is assumed to be positive. Using this convention with the figure, the flows on the left side of the line would be positive and those on the right would be negative. You can display results in PowerWorld Simu-

PROBLEMS

m ramld- cs%mrrwb LPm 5 -1EWj 3E-i,Ed*!Joda Bun Hale 1 Log , S t d r SoCnr -- - -- --A - - --

m

1 Phase Sh i f t ing Transf o-r

I Screen for Example 3.13 lator using this convention by unchecking the "Use Absolute Values Sor MW/Mvar Line Flows" field on the Display Options page of the Oneline Display Options dialog. H

Note that a voltage-magnitude-regulating transformer controls the re- acrioe power flow in the series link in which i t is installed, whereas a phase- angle-regulating transformer controls the rctrl powcr flow (see Problem 3.44).

SECTION 3.1 3.1 Consider an ideal transformer with ,Vl - 3000 and ,V2 - 500 turns. Let winding I be

connected to a source whose voltagc is c1(1) - 100(1 111) volts for - 1 I I 5 1 and el(l) = 0 for 111 > I second. A 3-hrud capacitor is conncctccl across winding 2. Sketch el ( I ) , ( , , ( I ) , il ( t ) , and i2(1) versus time I.


3.2 A single-phase 100-kVA, 24001240-volt, 60-Hz distribution transformer is used as a stepdown transformer. The load, which is connected to the 240-volt secondary winding, absorbs 80 kVA at 0.8 power factor lagging and is at 230 volts. Assuming an ideal transformer, calculate the following: (a) primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.

3.3 Rework Problem 3.2 if the load connected to the 240-V secondary winding absorbs 110 kVA under short-term overload conditions at 0.85 power factor leading and at 230 volts.

3.4 For a conceptual single-phase, phase-shifting transformer, the primary voltage leads the secondary voltage by 30". A load connected to the secondary winding absorbs 50 kVA at 0.9 power factor leading and at a voltage E2 = 277/00 volts. Determine (a) the primary voltage, (b) primary and secondary currents, (c) load impedance referred to the primary winding, and (d) complex power supplied to the primary winding.

3.5 Consider a source of voltage v ( t ) = 10fi sin(2t)V, with an internal resistance of 1800 R. A transformer that can be considered as ideal is used to couple a 50-Q resistive load to the source. (a) Determine the transformer primary-to-secondary turns ratio required to ensure maximum power transfer by matching the load and source resistances. (b) Find the average power delivered to the load, assuming maximum power transfer.

3.6 For the circuit shown in Figure 3.29, determine v,,,(t).

ldeal transformer

ldeal transformer

FIGURE 3.29 Problem 3.6

SECTION 3.2 3.7 The following data are obtained when open-circuit and short-circuit tests are per-

formed on a single-phase, 50-kVA, 24001240-volt, 60-Hz distribution transformer.

VOLTAGE CURRENT POWER (volts) (amperes) (watts)

Measurements on low-voltage side with high-voltage winding open 240 5.97 213

Measurements on high-voltage side with low-voltage winding shorted 60.0 20.8 750

(a) Neglecting the series impedance, determine the exciting admittance referred to the high-voltage side. (b) Neglecting the exciting admittance, determine the equivalent

PROBLEMS

series impedance referred to the high-voltage side. (c) Assuming equal series impedances for the primary and referred secondary, obtain an equivalent T-circuit referred to the high-voltage side.

3.8 A single-phase 100-kVA, 2400/240-colt, 60-Hz distribution transformer has a 2-ohm equivalent leakage reactance and a 6000-ohm magnetizing reactance referred to the high-voltage side. If rated voltage is applied to the high-voltage winding, calculate the opencircuit secondary voltage. Neglcct I ~ R and GIV losses. Assume equal series leakage reactances for the primary and rcfcrrcd seco~~dary.

3.9 A single-phase 50-kVA, 2400/240-bolt, 60-I-IL distribution transfonllcr is used as a step-down transformer at the load erid ul' a 2400-volt feeder hhose series impedance i h ( 1.0 + j2.0) ohms. The equi\alcn~ series ~mpcclancc of ~ h c transfornwr is ( I .O + j2.5 J ohms referred to the high-voltage (primary) side. The tralzsfornler is delivering rated load at 0.8 power factor lagging ;rnd at rated secondary voltage. Neglecting the transformer exciting current, dctzrniiuc (a) the voltage at the transformer primarb terminals, (b) the voltage at the sending end of the Sccder, and (c) the real and rcacticc power delivered to the scnding end 01' [he fceder.

3.10 Rework Problem 3.9 if the transformer is delivering rated load at rated secondarl voltage and at (a) unity powcr b~ctor, (b) 0.8 power factor leading. Compare the re- s ~ ~ l t s with those of Problem 3.9.

3.1 1 A single-phase, 50-kVA, 2300/240-V. 00-1 IL distribution transformer has the following parameters:

Resistance of the 2400-V winding: R = 0.75 i2

Resistance of the 940-V winding: R1 = 0.0075 Q

Leakage reactancc of the 9400-V winding: X 1 = 1.0 R

Leakage reactance of the 140-V winding: X2 = 0.01 R

Exciting adn~ittiincc on the 230-V sitlc ; 0.003 -- j0.02 S (a) Draw the equivalent circuit refcrrcd to the high-voltage side of the transformer. (b) Draw the equivalent circuit refcrrcd to the low-voltage side of the transformer. Show the numerical values of impddances on the equivalent circuits.

3.12 The transformer of Problem 3.1 1 is supplying a rated load of 50 kVA at a rated secondary voltage of 240 V and at 0.8 powcr factor lagging. Neglecting the transformer exciting current, (a) determine the input terminal voltage of the transformer on thc high-voltage side. (b) Sketch the corresponding phasor diagram. (c) If the transfornler is used as a step-down transformer at the load end of a feeder whose impedance is 0.5 + j3.0 R, find the voltage Vs and the powcr factor at the sending end of the feeder. SECTION 3.3

3 . Using the transformer ratings as basc qua~ntitics, work Problem 3.8 in per-unit. 3.14 Using the transformer fillings as baht cluanlities, work Problem 3.9 in per-unit. 3.15 Using base values of 90 kVA and 1 I5 ~ o l t s 111 ronc 3, rcwork Example 3.4. 3.16 Rework Example 3.5, using Sa,,3+ - I00 kVA and Vb,l,cLL -- 600 volts. 3.17 A bal:lnccd Y-conncctcil boltage hoi~rcc uitll E,,,, - 277/0" volt5 is applicd to ;I

biiliinced-Y loud in parallel with ii h;~laocccl-A lo~rd, ~vhcrc Zy = 30 + j10 ;rnd ZA --


45 - j25 ohms. The Y load is solidly grounded. Using base values of SbaScl4 = 5 kVA and VbarLN = 277 volts, calculate the source current I, in per-unit and in amperes.

3.18 Figure 3.30 shows the one-line diagram of a three-phase power system. By selecting a common basc of 100 MVA and 33 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given as follows:

G: $0 MVA 22 kV x = 0.18 per unit TI : 50 MVA 221220 kV x = 0.10 per unit T2: 40 MVA 22011 1 kV x = 0.06 per unit T3: 40 MVA 2211 10 kV x = 0.064 per unit T4: 40MVA IlO/llkV x=O.O8perunit M: 66.5 MVA 10.45 kV x = 0.185 per unit

Lines 1 and 2 have series reactances of 48.4 and 65.43 0, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.

FIGURE 3.30 Problem 3.18 Line 1 I

3.1 9 For Problem 3.18, the motor operates at full load, at 0.8 power factor leading, and at a terminal voltage of 10.45 kV. Determine (a) the voltage at bus 1, the generator bus, and (b) the generator and motor internal EMFs. SECTION 3.4

3.20 Determine the positive- and negative-sequence phase shifts for the three-phase transformers shown in Figure 3.3 1.

3.21 Consider the three single-phase two-winding transformers shown in Figure 3.32. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in A, place the polarity marks, and label the terminals n, 6, and c in accordance with the American standard. (b) Relabel the terminals n', b', and c' such that GiAN is 90' out of phase with GI,, for positive sequence.

3.22 Three single-phase, two-winding transformers, each rated 450 MVA, 20 kV/288.7 kV, with leakage reactance & = 0.10 per unit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in A with American standard phase shift, (b) in Y with an opep neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current arc neglected.

3.23 Consider a bank of three single-phase two-winding transformers whose high-voltage terminals are connected to a three-phase, 13.8-kV feeder. The low-voltage terminals

PROBLEMS

N turns

(a) Y-A-1 transformer

FIGURE 3.3 1 Problems

FIGURE 3.32 Problem 3.2 1

N turns lJ-3 N turns

N turns

2N turns

N turns

(b) Y-1-z~g-zag transformer (c) Extended A autotransformer

3.20 and 3.37 (Coils drawn on the same vertical line are on the same core)

are connected to a three-phase subst i~~io~l load rated 3.4 M V A and 1.3 kV. Detcmminc the required voltage, current, ancl M V A ratings of both windings of each transfor~l.ler, when the high-volt:tgc/low-oltiigl: windings arc connected (a) Y A, (b) A Y. (c) Y--Y, and (d) A-A.


3.24 Three single-phase two-winding transformers, each rated 25 MVA, 34.5/13.8 kV. arc connecled to form a threc-phase A-A bank. Balanced positive-sequence voltages are applied to thc high-voltagc tcnninals, anti a bali~nccd, rcsistivc Y load conncctcd to the low-voltage terminals absorbs 75 MW at 13.8 kV. If one of the single-pllase transformers is removed (resulting in an open-A connection) and the balanced load is simultaneously reduced to 43.3 MW (57.7%) of the original value), determine (a) thc load voltngcs K,,,, I f / , , , , and I;,,: (b) lonrl currents I,,. I,,. and I,.; and (c) the MVA si~p- plied by each of the remaining two transformers. Arc balanced voltagcs still applicd to the load? Is the opend transformer overloaded?

3.25 Three single-phase two-winding transformers, each rated 25 MVA, 38.113.81 kV. are connected to form a three-phase Y-A bank with a balanced Y-connected resistive load of 0.6 Q per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 66 kV (line-to-line) for the high voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance RL in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base.

3.26 Consider a three-phase gcncrator rated 300 MVA, 23 kV, supplying a system load of 240 MVA and 0.9 power factor lagging at 230 kV through a 330 MVA, 23 A/ 230 Y-kV step-up transformer with a leakage reactance of 0.1 1 per unit. (a) Neglect- ing the exciting current and choosing base values at the load of 100 MVA and 230 kV, find the phasor currents I*, I B . and Ic supplied to the load in per unit. (b) By choosing the load terminal voltage V A as reference, specify the proper base for the generator circuit and determine the generator voltage V as well as the phasor currents I,, It,, and I,, from the generator. (Notc~: Take into account the phase shift of the transformer.) (c) Find the generator terminal voltage in kV and the real power supplied by the generator in MW. (d) By omitting the transformer phase shift altogether, check to see whether you get the same magnitude of generator ternlinal voltage and real power delivered by the generator.

SECTION 3.5 3.27 The leakage reactance of a three-phase, 500-MVA, 345 Y/23 A-kV transformer is 0.09

per unit based on its own ratings. The Y winding has a solidly grounded neutral. Draw the per-unit equivalent circuit. Neglect the exciting admittance and assume American standard phase shift.

3.28 Choosing system bases to be 360124 kV and 100 MVA, redraw the per-unit equivalent circuit for Problem 3.27.

3.29 Consider the single-line diagram of the power systcm shown in Figure 3.33. Equip- ment ratings are:

generator 1: generator 2: synchronous motor 3:

750 MVA, 18 kV, X" = 0.2 per unit 750 MVA, I8 kV, X" = 0.2 1500 MVA, 20 kV, X" = 0.2

3-phase A-Y transformers 750 MVA, 500 kV Y/20 kV A, X = 0.1 T I , T2, T3, T4:

3-phase Y-Y transformer Ts: 1500 MVA, 500 kV Y/20 kV Y, X = 0.1

PROBLEMS 125

FIGURE 3.33

Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of 100 MVA and 500 kV for the 40-ohm line. Determine the per-unit reactances.

Problems 3.29 and 3.30

3.30 For the power system in Problem 3.29, the synchronous motor absorbs 1200 MW at 0.8 power factor leading with the bus 3 voltage at 18 kV. Determine the bus 1 and bus 2 voltages in kV. Assume that generators I and 2 deliver equal real powers and equal reactive powers. Also assume a balanced three-phase system with positive-sequence sources.

3.31 Three single-phase transformers, each rated 10 MVA, 66.4112.5 kV, 60 Hz, with an equivalent series reactance of 0.12 per unit divided equally between primary and secondary, are connected in a three-phase bank. The high-voltage windings are Y connected and their terminals are directly connected to a 1 15-kV three-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y connected, (b) A connected.

3.32 A 100-MVA, 13.2-kV three-phase generator, which has a positive-sequence reactance of 1.2 per unit on the generator base, is connected to a 110-MVA, 13.2 A/115 Y-kV step-up transformer with a series impedance of (0.005 + jO.l) per unit on its own base. (a) Calculate the per-unit generator reactance on the transformer base. (b) The load at the transformer terminals is 80 MW at unity power factor and at 115 kV. Choosing the transformer high-side voltage as the reference phasor, draw a phasor diagram for this condition. (c) For the condition of part (b), find the transformer low-side voltage and the generator internal voltage behind its reactance. Also compute the generator output power and power factor.

3.33 Figure 3.34 shows a one-line diagram of a system in which the three-phase generator is rated 300 MVA, 20 kV with a subtransient reactance of 0.2 per unit and with its neutral grounded through a 0.4-0 reactor. The transmission line is 64 km long with a series reactance of 0.5 Q/km. The three-phase transformer TI is rated 350 MVA, 3301 20 kV with a leakage reactance of 0.1 per unit. Transformer Tr is composed of three single-phase transformers, each rated 100 MVA, 137/13.2 kV with a leakage reactance of 0.1 per unit. Two 13.2-kV inotors A!, and M I will1 ;I bi~btr:~~lsic[it rca~lancc of 0.2 per unit for each motor reprcselit the load. Adl has il rated input of 200 MVA with its neutral grounded through n 0.4-Q current-limiting reactor. M2 has a rated input of

I26 CHAPTER 3 POWER TRANSFORMERS

FIGURE 3.34 Problems 3.33 and 3.34

100 MVA with its neutral not connected to ground. Neglect phase shifts associated with the transformers. Choose the generator rating as base in the generator circuit and draw the positive-sequence reactance diagram showing all reactances in per unit.

3.34 The motors M I and M2 of Problem 3.33 have inputs of 120 and 60 MW, respectively, at 13.2 kV, and both operate at unity power factor. Determine the generator terminal voltage and voltage regulation of the line. Neglect transformer phase shifts.

SECTION 3.6 3.35 A single-phase three-winding transformer has the following parameters: Z1 = Zz =

Z3 = 0 + j0.06, G, = 0, and B,, = 0.2 per unit. Three identical transformers, as described, are connected with their primaries in Y (solidly grounded neutral) and with their secondaries and tertiaries in A. Draw the per-unit sequence networks of this transformer bank.

3.36 The ratings of a three-phase three-winding transformer are: Primary (I): Y connected, 66 kV, 20 MVA Secondary (2): Y connected, 13.2 kV, 15 MVA Tertiary (3): A connected, 2.3 kV, 5 MVA

Neglecting winding resistances and exciting current, the per-unit leakage reactances are:

XI, = 0.08 on a 20-MVA, 66-kV base XI3 = 0.10 on a 20-MVA, 66-kV base

X23 = 0.09 on a 15-MVA, 13.2-kV base

(a) Determine the per-unit reactances XI , X2, Xj of the equivalent circuit on a 20-MVA, 66-kV base at the primary terminals. (b) Purely resistive loads of 12 MW at 13.2 kV and 5 MW at 2.3 kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per-unit impedance diagram, showing the per-unit impedances on a 20-MVA, 66-kV base at the primary terminals.

3.37 Draw the per-unit equivalent circuit for the transformers shown in Figure 3.30. In- clude ideal phase-shifting transformers sliowing phase shifts determined in Problem 3.20. Assume that all windings have the same kVA rating and that the equivalent leakage reactance of any two windings with the third winding open is 0.10 per unit. Neglect the exciting admittance.

3.38 The ratings of a three-phase, three-winding transformer are: Primary: Y connected, 66 kV, 15 MVA Secondary: Y connected, 13.2 kV, 10 MVA Tertiary: A connected, 2.3 kV, 5 MVA

PROBLEMS

Neglecting resistances and exciting current, the leakage reactances are:

XPS = 0.07 per unit on a 15-MVA. 66-kV base XpT = 0.09 per unit on a 15-MVA, 66-kV base Xsr = 0.08 per unit on a 10-MVA, 13.3-kV base

Determine the per-unit reactances of rhc pcr-phase equivalent circuit using a base of 15 MVA and 66 kV for the primary.

3.39 An infinite bus, which is a constant voltage source, is connected to the primary of thc three-winding transformer of Problem 3.38. A 7.5-MVA, 13.2-kV synchronous motor with a subtransient reactance of 0.3 per unit is connected to the transformer secondary. A 5-MW, 2.3-kV three-phase resistive load is connected to the tertiary. Choosing a base of 66 kV and 15 MVA in thc primary, draw the impedance diagram of the system showing per-unit impedances. Neglect transformer exciting current, phase shifts. and all resistances except the resistive load.

SECTION 3.7

3.40 A single-phase 15-kVA, 1400/240-volt, 60-HL two-winding distribution transformer is connected as an autotransfornler to step up the voltage from 2400 to 2640 volts. (a) Draw a schematic diagram of this arrangement, showing all voltages and currcntb when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransfornler if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How ~tluch of this kVA rating is transformed by magnetic induction? (c) The following data arc obtained from tests carried out on the transformer when it is connected as a two-winding transfornler:

Open-circuit test with the low-voltage terminals excited: Applied voltage = 240 V, Input current = 0.68 A, Input power = 105 W. Short-circuit test with the high-voltage terminals excited: Applied voltage = 120 V, Input current = 6.35 A, Input power = 330 W

Based on the data, compute the efficiency of the autotransformer corresponding to fill1 load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransfornler than as a two-winding transformer.

3.41 Three single-phase two-winding transfor~ners, each rated 3 LVA, 22011 10 volts, 60 Hz, with a 0.10 per-unit leakage reacvance, are connected as a three-phase extended A autotransformer bank, as shown in Figure 3.3 1 (c). The low-voltage A winding has a 1 10 volt rating. (a) Draw the positive-sequence phasor diagram and show that the high- voltage winding has a 479.5 volt rating. (b) A thrce-phase load connected to the low- voltage terminals absorbs 6 kW at 110 volts and at 0.8 power factor lagging. Draw 1l1c per-unit impedance diagram and calculate the volti~gc and current at the high-voltagc terminals. Assume positive-scquencc operation.

3.42 A two-winding single-phase transl'ormcr rated 60 kVA, 140/1200 V, 60 Hz, has an efficiency of 0.96 when operated at rated load, 0.8 power Factor lagging. T h ~ s transfornler is to be utilircd us ;I l440/1100-V stcp-down a~itotransfor~ner in a power distribution system. (21) Fintl the pcrnlissible LVA rating of the autotri1115- fo

cap_3

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