chapter 16 transformers and equipment · 2021. 1. 6. · commercially available ceati transizetm...

EPRI Underground Distribution Systems Reference Book

CHAPTER 16 Transformers and EquipmentAuthors: Stephen L. Cress, Reviewers: Gordon Hayslip, Snohomish PUD

Kinectrics. Inc. John Igielski, Northeast UtilitiesAli Naderian, Kinectrics, Inc. Ken Ochs, We Energies

Joseph Somma, Consolidated Edison

Abstract:This chapter reviews several aspects related to transformers, including transformerlosses, loading characteristics, selection criteria for pad-mounted transformers,transformer cooling, interpretation of tests on transformers and oil, and capacitors.

Stephen L. Cress graduated in 1976 from The University of Torontowith a Bachelor of Applied Science in Electrical Engineering. He is cur-rently Department Manager – Distribution Asset Management at Kine-ctrics Inc. Stephen has over 33 years’ experience in specialized technicalinvestigations, research, testing, and applications in the power distribu-tion field based on his work at Federal Pioneer, Ontario HydroResearch Division, and Kinectrics Inc. He has conducted major proj-ects for North American distribution utilities dealing with: transformer

loading and sizing, transformer losses and efficiency, asset management, asset conditionassessment, life extension, distribution protection, equipment failure analysis (transform-ers, switchgear, fuses, capacitors), standard testing, distribution modeling, and develop-ment of utility-oriented engineering software. Stephen is the holder of a U.S. patent onhigh-voltage current limiting fuses. He is a co-author of the CEATI reference booksApplication Guide for Distribution Fuses and Engineering Guide for Distribution Overcur-rent Protection. Stephen’s work in the development of probabilistic methods for calculat-ing transformer loss evaluation, loss-of-life, and loading have been incorporated in thecommercially available CEATI TRANSIZETM computer program. He has publishedpapers with international organizations such as IEEE, CIRED, and INTER-RAM, andarticles in power industry magazines. He is the Chair of the harmonized CSA andCNC\IEC TC32 Committees dealing with High Voltage Fuses, and a Professional Engineerin the Province of Ontario.

Ali Naderian received his B.Sc. and M.Sc. degrees from Sharif Univer-sity of Technology in 1998 and the University of Tehran in 2000, respec-tively. During his studies, his part-time employment experience includedISC (1997-1999) for testing of switchgear and circuit breakers, and ITS(1999-2000) for designing and manufacturing of HV power transform-ers. He was co-designer of a 3*300-kV cascade HV testing transformer.He compared commercially available RTV coatings for outdoor insula-tors in his PhD thesis during his research at the University of Waterloo,

Ontario (2003-2006). He has been a project manager of high-voltage testing at Kinectrics,Inc. (formerly Ontario Hydro Research) since 2007, working on diagnostics of powertransformers, high-voltage cables, and outdoor insulators. He performs on-line and off-line PD measurements for HV apparatus. His research interests include high-voltage testtechniques, dielectric frequency response, and partial discharge. He has published severalpapers, is actively involved in IEEE transformer working groups, and is a registered engi-neer in the Province of Ontario.

16-1

Chapter 16: Transformers and Equipment EPRI Underground Distribution Systems Reference Book

16.1 INTRODUCTION

This edition of the Bronze Book covers only a subset ofwhat will be a comprehensive look at underground dis-tribution transformers. Included here are sections ontransformer losses, loading characteristics, pad-mounted transformer selection criteria, interpretationof tests on transformers and oil, as well as a discussionon capacitors.

The next edition will also include an overview of trans-former types by application, unit components and coreconstruction, installation options, and insulation types.Additional topics will be transformer cooling, testingand monitoring, and typical examples of failure rootcauses.

The reader is encouraged to refer to other sources morebroadly covering this topic, including the Electric PowerDistribution Handbook by Tom Short (2004), PowerTransformers, Principles and Applications by John Wind-ers (2002), and the ABB Distribution Transformer Guide(2002).

16.2 TRANSFORMER LOSSES

Losses in distribution transformers are categorized asload and no-load losses. Load losses vary with thesquare of the load on the transformer, whereas no-loadlosses are continuous and constant regardless of load.

16.2.1 No-Load Loss

No-load losses (or excitation losses, iron losses, or corelosses) are inherent to the excitation of the transformer.No-load losses are associated with the core design. Theyinclude core loss, dielectric loss, and the loss in the wind-ings due to exciting current. For distribution transform-ers at 27.6 kV and below, the dielectric loss is negligible.

The no-load loss in the transformer core is a function ofthe magnitude, frequency, and waveform of theimpressed voltage. No-load losses are affected by volt-age fluctuations. When an AC voltage is applied to theterminals of the transformer, magnetizing current flowsthrough the winding, and a magnetic flux appears in thecore. The predominant component is core loss, which iscomposed of hysteresis and eddy current losses.

The hysteresis loss is proportional to the frequency anddependent on the area of the hysteresis loop in the B-Hdiagram, and, therefore, characteristic of the materialand a function of the peak flux density.

The variable magnetic flux induces current running inpaths perpendicular to the direction of the flux. The

induced current, called eddy current, produces losses inthe core plates. The eddy current loss can be calculatedby Equation 16.2-1

16.2-1Where: σ is the core conductivity. ƒ is frequency. d is the core thickness.B is the peak value of the flux density.V is the core volume.

As per Equation 16.2-1, eddy current is controlled byusing laminated core to cut large current loops at thecross section of the core. The no-load loss is the sum ofhysteresis and eddy current losses, as shown in Equation16.2-2.

16.2-2

16.2.2 Load Losses

Load losses (or copper losses or resistive losses) are pri-marily a function of the winding design of the trans-former. They result from the load current flowing in theprimary and secondary windings.

Components of load loss are I2R and stray losses. For adistribution transformer, I2R is in the range of 92-99%of the load loss. The proportion is lower for larger kVAsizes. Load loss is affected by:

• number of turns of winding

• mean length of the primary and secondary turns

• conductor cross section

• material of the conductor—i.e., copper or aluminum

Stray losses vary inversely as the temperature, therebymaking necessary the calculation of load loss at a spe-cific temperature such as 85°C. Stray losses have threecomponents: conductor eddy currents, conductor circu-lating currents, and stray currents in the core wall andcore clamps.

The current, which is applied to the windings, createslosses due to the winding resistance. The losses of atransformer are losses incident to a specified load car-ried by the transformer. Load losses in distribution-classtransformers mainly include I2R loss in the windingsdue to load current.

Load loss follows Ohm’s law and can be decreased byreducing the number of winding turns, by increasing thecross-sectional area of the turn conductor, or by a com-

22 2 2

6eddyP f d B Vπ σ=

0 eddy hP P P= +

16-2

EPRI Underground Distribution Systems Reference Book Chapter 16: Transformers and Equipment

bination of both. However, reducing the number ofturns requires an increase of the flux—i.e., an increasein the core cross-section, which increases the iron weightand iron loss. Therefore, a tradeoff has to be madebetween the load loss and the no-load loss.

16.2.3 Total Loss

The following summarizing relationships are usefulwhen considering the losses in distribution transformers:

No Load Loss α Flux Density α 1/# Turns α 1/ CoreCross Section

Load Loss α # Turns

Impedance α Reactance α (# Turns)2

Initial Cost α Core Material α Winding Material

Table 16.2-1 summarizes the components of load andno-load losses.

Figure 16.2-1 and Table 16.2-2 provide some typicalload and no-load loss values for distribution transform-ers. Figure 16.2-1 illustrates how load losses vary withload on the transformer.

16.2.4 Transformer Efficiency

Transformer efficiency is related to the amount of wattslosses that occur when the transformer is in operation.

Transformer efficiency (η) is the ratio of a transformer’suseful power output to its total power input as indicatedin Equation 16.2-3 (IEEE 2006).

16.2-3

Where: η is the efficiency. Pin is the input power. Pout is the output power. Ploss is the total power loss of the transformer to

be introduced.

In many jurisdictions, government energy agencies havemandated minimum efficiency levels for liquid-filled anddry-type distribution transformers (DOE 2007; NEMA2002). Table 16.2-3 provides an example of the acceptedefficiency levels for liquid-immersed distribution trans-

Table 16.2-1 Components of Transformer Load and No-Load Loss

Type No-Load Lossa

a. Where I represents current, and I2R is the current squared times the conductor resistance.

Load Lossa

Electric I2R from No-load II2R from load II2R from I supply-

ing Losses

Magnetic

Core Hysteresis LossCore Eddy Current LossStray Eddy Current Loss in

Internal components

Conductor Eddy Current from Leakage fields

Dielectric Dielectric Loss

Table 16.2-2 Typical Losses for Power Distribution Transformers

RatingKVA Load Loss Watts

No-Load LossWatts

Efficiencya

at 50% load

a. Calculated using Equation 16.2-4.

250 3800 880 0.9925

400 5500 1200 0.9932

667 7900 1700 0.9941

1000 11000 2300 0.9945

1500 15000 3000 0.9950

2500 23000 5000 0.9954

Figure 16.2-1 Typical load and no-load losses of distribution transformers.

out in loss

in in

P P P

P Pη

−= =

16-3


formers. An example of the minimum efficiency for dry-type distribution transformers is shown in Table 16.2-4(DOE 2007). These efficiency values are computed at50% of nameplate-rated load.

Efficiency can be expressed directly as a function of theload and no-load losses as in Equation 16.2-4 (NEMA2002). The efficiency values computed using this for-mula are provided alongside the load and no-load lossesin the examples in Table 16.2-2.

16.2-4

Where: KVA is the transformer rated power. PL is the load loss. P0 is the no-load loss. Lp.u. is the per-unit load (the ratio of actual load

to the rated full load).

Present distribution transformers are, for the most part,between 98% and 99.5% efficient. For the new trans-formers, the guideline from (DOE 2007), presented inTables 16.2-3 and 16.2-4, should be followed. Becausevirtually all-electric energy passes through distributiontransformers, losses in these devices, though small, areestimated to constitute as much as 2 to 3% of all energygenerated.

Generally transformers are at maximum efficiency whenthey are 50% loaded. When transformers are lightlyloaded, the no-load losses form a large percentage of thepower utilized, and therefore, the efficiency is low. Asthe transformer is loaded to higher levels, the load lossesdominate the efficiency. The maximum efficiency pointis the optimal point of lowest load and no-load losses. Itis determined by the design of the transformer and theo-retically could be designed to occur at any load percent-age. It typically is designed to occur at 50%, because theaverage load tends to be about 50% of the peak load.However, transformers with high no-load losses are

Table 16.2-3 Standard Levels of Efficiency for Liquid-immersed Distribution Transformers (DOE 2007)

Table 16.2-4 Standard Levels of Efficiency for Dry-type Distribution Transformers (DOE 2007)

. .

2. . 0 . .

p u

p u p u L

KVA L

KVA L P L Pη

×=

× + + ×

16-4


most efficient at 60%-80% load, and transformers withlow no-load losses are most efficient at about 40% load.(See Figure 16.2-2.)

16.2.5 Reduction of Transformer Losses

Reduction of transformer losses and improvement inefficiency can be achieved by reduction of either load orno-load losses. For any given set of core and windingmaterials, reduction of load losses often leads to anincrease in no-load losses and vice versa.

Many factors of core design affect no-load losses andcan be altered to reduce these losses. Higher magneticflux density leads to higher losses. Larger gaps in cutcores lead to higher losses. These gaps can be reducedby manufacturing techniques. The thickness of theenamel insulation on the winding conductors affects thesize of the core. High-quality enamel can be used in verythin layers to reduce core size and no-load losses.Mechanical arrangement of the windings and taps alsoaffects the efficient use of space and the size of the core.

Traditionally cores have been made from grain-orientedsilicon steel formed into thin sheets and wrapped into arectangular shape. The loss decreases as the thickness ofthe sheets decrease. Standard grades are M-2 at 0.18mm, M-3 at 0.23 mm, M-4 at 0.27 mm, and M-6 at 0.35mm. Losses also depend on the permeability of the steelalloy. Higher permeability leads to lower losses. Thepermeability depends upon the alloy and the orienta-tions of the grains.

A large advance in technology occurred in the 1980swith the development of amorphous steel cores. Thesecores are also built up by wrapping thin sheets or rib-bons, but the steel itself (such as Co-Fe-Si-B alloy) isquenched during manufacture to ensure that no grainsare formed in the steel. This process increases the effec-

tive permeability of the steel, thus reducing the losses,but it also decreases the saturation magnetic flux den-sity, which increases the amount of material required inthe core. Together, these effects reduce the no-load lossof the core, but the amorphous steel cores are larger,heavier, and more costly to produce. (Permeabilityincreases by a factor of 4, but saturation flux densitydecreases by a factor of 0.75, requiring 1.3 times asmuch material in the core, so overall loss is lower by afactor of 3.) On average, amorphous core loss values areabout 30% of that for high-efficiency silicon steel, andonly 15% of that for older, less efficient steels.

Numerous questions have arisen regarding the mechani-cal robustness and long-term mechanical performanceof amorphous metals. Short-term testing programs havenot substantiated these beliefs, but the concern persists.

More recently nano-crystalline steel has become avail-able for use in transformer cores.

The best are based on an Fe-Zr-B alloy that is formed inan amorphous state and then annealed to produce verysmall grain sizes. This approach makes the material lessbrittle and thereby decreases production costs. This steelhas even higher permeability and also higher saturationinduction than the amorphous materials, but it is notyet available in manufactured transformer cores. Thenew steel has 17 times the permeability of steel and 0.89of the saturation flux density; so losses should bereduced by a factor of 15.

Load losses are caused primarily by the heating of thewindings by the passage of current (I2R losses). The cur-rent is determined by the impedance of the load on thetransformer and the voltage levels and so is not underthe control of the transformer designer. The resistancedepends on the material used in the winding, the cross-sectional area of the wires, and the number of turns.

Transformer windings are made of either copper or alu-minum in round wires, square wires, or flat sheets. Theresistivity of aluminum is about 1.6 times larger thanthat of copper, but aluminum has a lower cost. Manydifferent alloys of aluminum and copper are available.In general, the lower resistance alloys are more expen-sive and harder to work with in the manufacturing pro-cesses, leading to higher initial costs.

In addition to choice of material, load losses areaffected by the cross-sectional area of the wire used.Larger wires produce lower load losses, but then thewindings are larger, and this requires a larger core,which increases the no-load losses.Figure 16.2-2 Transformer efficiency as a function of load.

16-5


Some load loss is caused by induced currents from adja-cent windings. These currents can be reduced by usingcontinuously transposed conductor in the winding andthus reducing load losses. This approach also leads tohigher initial costs.

16.2.6 Transformer Short-circuit Impedance

When provided a customer’s cost of no-load and loadlosses, transformer manufacturers will use software thatperforms hundreds of iterations, varying core, winding,and tank options, to arrive at a transformer with anoptimal balance of losses and initial cost.

The short-circuit impedance of a transformer is used tocalculate the maximum short-circuit current and isneeded for sizing circuit breakers, fuses, cables, andother equipment connected to the secondary of thetransformer.

Transformer impedance (or short-circuit impedance orimpedance voltage) is the percent of per unit voltagethat must be applied to the primary side of a trans-former, so that the rated current flows when the second-ary terminals are short-circuited. This impedance isformulated as Equation 16.2-5.

16.2-5

As the no-load test result is available, the ohmic part ofthe impedance can be calculated using Equation 16.2-6,and therefore, the inductive part of the impedance canbe derived by Equation 16.2-7.

16.2-6

16.2-7

In a transformer having a tapped winding, the short-cir-cuit impedance is referred to a particular tap. Unlessotherwise specified, the nominal tap applies and is theimpedance (Z%) that is marked on the nameplate. Theimpedance voltage of distribution transformers withrated power below 630 kVA is usually 4% or less, andthis value is usually around 6% for 630 kVA up to2.5 MVA distribution transformers.

For parallel operation of two or more transformers,short-circuit impedance is critical. If paralleled trans-formers do not have the same short-circuit impedance,the load will be shared in an unbalanced way such thatone transformer can be overloaded and the transformercan be underloaded.

16.2.7 Cost-of-Losses Formula

The lifetime cost of a transformer depends on the capi-tal cost of the transformer and the cost of the load andno-load losses during its lifetime. The present valuemethod is often employed to express the lifetime cost interms of a dollar value in the present year. Losses fromdistribution transformers are a significant contributionto distribution system losses, and their reduction repre-sents an opportunity for improving energy efficiency

A cost-of-losses formula for purchasing purposes isoften employed to determine the lifetime costs for vari-ous transformer options available to utilities. Compari-sons can then be made between more capital intensivelow-loss transformers and less expensive higher-losstransformers.

The following paragraphs describe the general formula-tion of a cost-of-losses formula. Table 16.2-5 defines thequantities used in these equations.

% 100Z

P

UZ

Z= ×

3

63

% 100.10

load lossPR

MVAϕ

ϕ

− −= ×

2 2% % %X Z R= −

Table 16.2-5 Definition of Symbols for Cost of Losses Formula

CAP Capital cost ($)

CLL Present value of cost of load losses ($/W)

CLL(m) Cost of load losses for month “m” ($/kW)

CLY(y) Cost of load losses for year “y” ($)

CNLL Present value of cost of no-load losses ($/W)

CNLL(m) Cost of no-load losses for month “m” ($/kW)

D Demand charge, monthly ($/kW)

D(m) Demand charge for month “m” ($/kW)

E Energy charge, monthly (¢/kWh)

EOP(m) Energy charge off-peak for month “m” (¢/kWh)

EP(m) Energy charge on-peak for month “m” (¢/kWh)

FYG(y) Factor for yearly load growth accumulated to year “y”

g(y) Growth of load for year “y” (%/100)

HOP(m) Hours off-peak for month “m” (h)

HP(m) Hours on-peak for month “m” (h)

i(y) Interest rate for year “y” (%/100)

j(y) Inflation rate for year “y” (%/100)

PVLC Present value of lifetime cost ($)

LL Load losses (W)

LSF Loss factor (average loss/peak loss)

NLL No-load losses (W)

NY Number of years in economic study period

p(y) Growth of power costs for year “y” (%/100)

PVF Present value factor for a period of years

PVF(y) Present value factor for year “y”

RATL Rated load for transformer (kVA)

RFResponsibility factor (load at system peak/peak load)2

UF Utilization factor (peak load/rated load)

16-6


The basic form of the cost of losses formula, providingthe present value of the lifetime cost (PVLC) of a trans-formers, is as expressed in Equation 16.2-8.

16.2-8

Where:CAP is the capital cost or initial purchase price

of the transformer. NLL is the no-load losses that occur continu-

ously when the transformer is energized, regardless of the loading.

CNLL is the cost of no-load losses and is inde-pendent of the loading and dependent on the demand and energy charges. Time-of-use energy charges can be considered by using on-peak and off-peak energy charges, and considering the hours that the transformer is on-peak or off-peak.

LL is the load loss at rated load. The value of load loss at rated load is a measured parameter, and load losses at other load-ings are derived from this value.

CLL is the cost of the load losses, and depends on the demand and energy charge rates as well as on the loading of the transformer throughout its life.

Common cost-of-losses equations use flat-rate demandand energy charges and fixed annual economic factors,such as interest rate, to evaluate the lifetime costs ofload losses (CLL) and cost of no-load losses (CNLL).The concepts of load factor, loss-factor, utilization fac-tor, and responsibility factor are used to describe theloads on the transformer. A load growth factor can beused to include the influence of rising loads on thetransformer losses over the transformer’s lifetime. Notethat the load growth factor is 1 in the first year, and thenchanges to a fixed factor at the start of the second year.The present value factor includes the influence of eco-nomic factors such as inflation of the cost of power andinterest rates. The growth in power costs factor is 1 inthe first year and then changes to a fixed factor in thesecond year. The rate of interest starts in the first year.Note that there are 8760 hours in a year.

16.2-9

16.2-10

16.2-11

16.2-12

16.2-13

Where:UF, the utilization factor, is defined as the ratio of the peak load to the transformer rated load. It repre-sents the portion of the transformer rated load that is utilized when the transformer is at its highest loading.

16.2-14

RF, the peak responsibility factor, is used to adjust the load toreflect the proportion of the asset load that actually contrib-utes to the peak load of the utility as a whole. That is, it indi-cates how much the load loss of the particulartransformer contributes to the total demand. Theresponsibility factor is the ratio of the transformer loadat system peak to the peak load, all squared.

16.2-15

LSF, the loss factor, is the ratio of the average loss to thepeak loss. The loss factor can be derived from the loadfactor. The load factor is a single value that character-izes the load profile. The load factor is the ratio of theaverage load to the peak load.

Load and loss factors are dependent on the shape of theload profile. Loading profiles are different for indus-trial/commercial, urban residential, and rural residentialtransformers. Industrial/commercial loads are steadierboth over the day and over the week. A typical load fac-tor is 0.85. Residential loads are more variable, with typ-ical load factors of 0.4 for a single transformer. Urbanresidential transformers tend to be more heavily loadedthan rural transformers.

Theoretically the loss factor may have a value betweenthe value of the load factor and the load factor squared,depending on the load profile shape. A common for-

PVLC CAP NLL CNLL LL CLL= + ∗ + ∗

112 8760

1000 100E

CNLL D PVF⎛ ⎞= + ∗⎜ ⎟⎝ ⎠

[ ]{ }2

1

112 8760

1000 100

( ) ( )NY

y

ECLL D RF LSF

UF FYG y PVF y=

⎛ ⎞= ∗ + ∗⎜ ⎟⎝ ⎠

∗ ∗ ∗∑1(1 )

( )(1 )

y

y

pPVF y

i

−+=

+

1

( )NY

y

PVF PVF y=

= ∑1( ) (1 )yFYG y g −= +

peak load rated load

UF =

2load at system peak

peak loadRF

⎛ ⎞= ⎜ ⎟

⎝ ⎠

16-7


mula that has been used to calculate loss factor fromload factor is as shown in Equation 16.2-16.

16.2-16

WhereLDF is the load factor of the daily load profile.

PVF, the present value factor, accounts for the changingvalue of money and expresses the present worth of dol-lars spent in the future.

Economic factors, of course, are generally not fixed overlong time periods of time. Further, there is considerableutility interest in applying variable or time-of-use rates.With addition of several parameters, the cost of lossesformula can be modified to consider these variable eco-nomic inputs.

Energy and demand charges can be expressed as beingdependent on the time of use, either on-peak or off-peak. Economic factors and the load growth can beallowed in the equation to vary from year to year. Notethat either p(y) or j(y) must be set to zero for all yearsy. To use the initial cost of power in the first year, setp(1) to zero [or j(1) to zero, if you are not using p(y)].

Therefore the components of the cost of losses formulacan be further expressed as:

16.2-17

16.2-18

16.2-19

16.2-20

16.2-21

16.2-22

16.2-23

16.2-24

16.2-25

Where the variables are as defined in Table 16.2-5.

With computer assistance the cost of losses formula canbe further expanded to replace the use of the load factorconcept and determine loads directly from daily andmonthly profiles.

Figure 16.2-3 shows a general graph of costs versustransformer mass for a typical distribution transformer.There is an optimum value for total cost. If the loss eval-uation figures are submitted to the transformer manu-facturers in the request for quotation, they can design atransformer with an optimal cost from the end userpoint of view. The result of this process should be thecheapest transformer in the useful life period—i.e., withthe lowest total owning cost, optimized for a givenapplication.

20.85 * 0.15 *LSF LDF LDF= +

[ ]{ }

12

1

2

1

1( )

1000

( ) ( )

m

NY

y

CLL CLL m

UF FYG y PVF y

=

=

⎡ ⎤= ∗⎢ ⎥⎣ ⎦

∗ ∗ ∗

∑

∑( ) ( )

( ) ( )( ) ( )

100 100

CLL m D m RF

EP m EOP mHP m HOP m

LSF

= ∗

⎡ ⎤+ ∗ + ∗⎢ ⎥⎣ ⎦∗

12

1

1( )

1000 m

CNLL CNLL m PVF=

⎡ ⎤= ∗ ∗⎢ ⎥⎣ ⎦

∑

( ) ( ) ( )

( )( )

100( )

100

CNLL m D m HP m

EP mHOP m

EOP m

= +

∗ +

∗

[1 ( )] [1 ( )]( ) ( 1)

[1 ( )p y j y

PVF y PVF yi y

+ ∗ += ∗ −

+

[ ] [ ][ ]

1 (1) 1 (1)(1)

1 (1)

p jPVF

i

+ ∗ +=

+

1

( )NY

y

PVF PVF y=

= ∑

( ) [1 ( )] ( 1)FYG y g y FYG y= + ∗ −

(1) 1 (1)FYG g= +

Figure 16.2-3 Transformer mass vs. transformer lifetime cost.

16-8


16.3 LOAD CHARACTERISTICS FOR TRANSFORMERS

One of the main considerations for selecting the appro-priate transformer is the characteristic of the load. Notonly the number and type of loads, but the load patternneeds to be considered.

Because load is a function of human behavior and life-style variables, as well as the type and size of electricequipment and weather changes, load forecasting hassome level of uncertainty.

16.3.1 Load Types

Several types of loads occur on a distribution systems:

• Domestic (residential): Mainly lights, fans, heaters,refrigerators, air conditioners, ovens, small pumps,and other household appliances.

• Commercial: Lighting of shops, air-conditioning,heating, and shop appliances.

• Industrial: Medium and large motors.

• Municipal (Public): Street lights, and traffic signals.

• Agricultural: Motors and pumps.

Commercial loads typically have a dedicated trans-former; however, multiple residences are usually servedby a single or three-phase transformer. Public loads usu-ally need their own dedicated transformer due to the loadsize. The daily load profiles of these three load categoriesare not usually matched. Commercial and industrialloads may at times e served on a spot network of multipletransformers in parallel. Some service areas, mainly inmetropolitan areas of loads including residential andcommercial loads are serviced from distributed grids ofmany transformers in parallel via network protectors.

Distribution transformers serving primarily residentialloads regularly carry average loads that are only 15 to25% of the transformer's rated capacity but also must bedesigned to support peak morning and evening loads.Because of the wide gap between peak and non-peakloads, and the relatively limited amount of time that thetransformer is peak-loaded, average transformer load-ing tends to be fairly low.

16.3.2 Load Profiles

Transformer loads generally follow cycles that repeatdaily, and may have seasonal variation during the yearand yearly growth. The daily load variation for manyutilities repeats every 24 hours and has two commonforms: a single hump shape (as shown in Figure 16.3-1)or a double hump shape. A multistep load cycle calcula-tion can be used to describe the load (IEEE 1995b). The

24-hour load profile is modeled by a series of constantloads of a short duration, usually 1 hour. The equivalentload during the short time steps is determined by usingthe maximum peak load during the short-time periodunder consideration. An equivalent two-step overloadcycle can be used for determining emergency overloadcapability, as shown in Figure 16.3-1. The equivalenttwo-step load cycle consists of a prior load and a peakload. A constant load that generates total losses thesame as a fluctuating load is assumed to be an equiva-lent load from a temperature standpoint. Equivalentload for a specific part of daily load is expressed byEquation 16.3-1.

16.3-1

Where: Li is various load steps in% or per unit. N is the total number of load steps. ti is the duration of each load step.

16.3.3 Peak Load

Equivalent peak load is the rms load obtained by Equa-tion 16.3-1 for the limited period over which the majorpart of the actual peak exists. If the peak load durationis over-estimated, the rms peak value may be consider-ably below the maximum peak demand. To protectagainst overheating due to high, brief overloads duringthe peak overload, the rms value for the peak loadperiod should not be less than 90% of the integrated ½hour maximum demand.

Besides daily peak load, seasonal peak load needs to betaken into account. Depending on the geographic loca-tion, and due to weather conditions, a winter peak orsummer peak can be expected.

An example of a daily load profile with two peak loadsis given in Figure 16.3-2.

Figure 16.3-1 Example of actual load cycle and equivalent load cycle of IEEE C57.91.

2

1

1

N

i ii

eq N

ii

L tL

t

=

=

=∑

∑

16-9


16.3.4 Average Load

According to IEEE C57.91, the average continuous loadis the rms load obtained by Equation 16.3-1 over a cho-sen period of the day. A period of 12 hours precedingand following the peak is suggested to be considered forthe time interval of average load calculation. Time inter-vals (t) of 1 hour are suggested as a further simplifica-tion of the equation, which for a 12-hour periodbecomes Equation 16.3-2. The dashed line in Figure16.3-2 shows the average load cycle constructed fromthe actual load cycle.

16.3-2

In fact, the average load determines the kWh billing rev-enue that will be obtained from serving the load,whereas peak load determines how much system capac-ity is required to serve that particular load group.

16.3.5 Load Factor

The ratio of the average demand over a time interval tothe maximum demand over the same time interval is theload factor.

16.3-3

Load factor can be calculated daily, monthly, and annu-ally based on the load profile.

16.3.6 Load Diversity, Diversity Factor, and Demand Factor

Load diversity is the difference between the sum of theindividual maximum demands of loads on a system and

the actual maximum demand on the system as describedin Equation 16.3-4.

16.3-4

Diversity factor in a distribution system is the ratio ofthe sum of the individual maximum demands of the var-ious subdivisions of a system to the maximum demandof the whole system under consideration (see Equation16.3-5). Loads do not normally all peak at the sametime. Therefore, the sum of the individual peak loads isgreater than the peak load of the composite system.Therefore, diversity factor is usually more than one.

16.3-5

Demand factor is the ratio of the maximum demand ofa system, or part of a system, to the total connectedload on the system. Demand factor is always less thanone. “Demand factor” is a percentage by which the totalconnected load on a service or feeder is multiplied todetermine the greatest probable load that the feeder willbe called upon to carry. For example, in hospitals,hotels, apartment complexes, and dwelling units, it isnot likely that all of the loads are connected to everybranch-circuit served by a service or feeder would be“on” at the same time. Therefore, instead of sizing thefeeder to carry the entire load on all of the branches, apercentage can be applied to this total load, and thecomponents sized accordingly. Equation 16.3-6 formu-lates the size of a distribution transformer consideringthe incorporated factors:

16.3-6

Where:S is the rated power of transformer. N is the number of loads (appliances) of the

same type. kW is the rated power of each load. DF is demand factor. LF is the load factor. PF is the power factor of each load. M is the number of different type of loads. DivF is the diversity factor.

Table 16.3-1 suggests typical values for load factor,diversity factor, and demand factor of loads (Pabla2004).

Figure 16.3-2 Morning and evening peak loads (from Pabla 2004).

122

1

(12 ) 0.29average ii

L h L=

= ∑

( )% 100

( )Average Demand Power kW

LDFPeak Load kW

= ×

Load Diversity Individual Maximum Demands

System Maximum Demand

=

−∑

F

Individual Maximum DemandsD

System Maximum Demand= ∑

1( )

Mi i i i

i i

N kW DF LFPF

S kVADivF

=

× × ×

=∑

16-10


16.3.7 Load Growth

Estimating load growth includes an element of specula-tion. Load growth for each year into the future may beestimated from known factors such as planned installa-tion and geographically related load patterns.

If the annual rate of load growth is available, the loadgrowth can be calculated for the transformer useful life-time interval. The modified transformer rating is asshown in Equation 16.3-7.

16.3-7

Where: S is the calculated power from Equation

16.3-6. i is the annual growth rate. n is the typical expected transformer life.

16.3.8 Load Diversity Charts

Load diversity considerations account for the fact thatnot all loads connected to the distribution transformerwill be drawing power at the same time. Many individualloads are thermostatically controlled or cycling andtherefore are not likely to be turned on at the sametime—that is, not coincident. Transformer loadingneeds to accommodate the diversified or coincident loadas opposed to the total connected load.

For the purpose of characterizing loads on a distribu-tion transformer, it is useful to determine the maximumpeak load that is likely to occur when a group of similarload types are connected to the transformer. Forinstance, in practice, it is useful to know the ultimatepeak load that will result from connecting a number ofsimilar electrically heated residences to a distributiontransformer. The total diversified or coincident load onthe transformer will be less than the sum of the maxi-mum peak demand of all the residences.

In general, the method to determine the maximumdiversified load of a number of houses consists of thefollowing steps:

• Define the type of houses based on major electricalusage, such as space heating, water heating, and airconditioning.

• Identify all loads in the type of home being consid-ered.

• Determine the value of all Connected Loads (Lk) andthe Maximum Non-Coincident Demand (MNCD).

• Determine the maximum peak load for each housetype.

• Use demand factors to determine the MaximumCoincident Demand (MCD) for groups of similartypes of houses.

• Develop charts of number of kW per home vs. num-ber of homes, and total kW vs. number of homes.

Residential loads can be analyzed to determine the typeof electrical equipment and its electrical load that wouldbe connected in typical homes. Electrical equipmentused in residential homes may be general (e.g., clotheswasher, microwave oven, stereo, hair dryer, etc.), high-energy consuming (e.g., electric clothes dryer), or ther-mostatically controlled (e.g., refrigerator, air-condition-ing, heating). For an average home, major appliancesconsume the most electrical energy (10.3-kWh/day).Lighting would consume an average of 4.1-kWh/day.Homes with air-conditioning units would utilize7.3-kWh for cooling and motor blower. Houses withelectric heating use, would utilize on average about120 kWh/day, and average houses with electric waterheating consume 14.7-kWh/day.

Based on the electrical energy equipment and load,houses can be classified into different major categoriessuch as:

• Natural gas heated with no air conditioning

• Natural gas heated with air conditioning

• Natural gas heated with air conditioning and electricwater heating

• Natural gas heating and cooking with air condition-ing

• Central electrical heating, electrical water heatingwith no air conditioning

The definitions and relations for maximum coincidentload, maximum noncoincident load, connected loads,

Table 16.3-1 Typical values for Demand Factor, Diversity Factor, and Load Factor

Demand Factor%

Diversity Factor Load Factor%

Domestic 70-100 1.2-1.3 10-15

Commercial 90-100 1.1-1.2 25-30

Industrial (less than 500 kW)

70-80 - 60-65

Industrial (Above 500 kW)

85-90 - 70-80

Municipal 100 1 25-30

Agricultural 15-20 1-1.5 90-100

(1 )nTS S i= +

16-11


diversity factor, and demand factor are expressed inEquations 16.3-8 and 16.3-9.

16.3-8

16.3-9

Where:DF is the demand factor.DF1 is the demand factor for one house (ratio of

Maximum Demand to Total Connected Load for one house).

Lk is the total connected load. N is the number of houses.P is the probability that one house has the

same Coincident Loads as other houses within the same time period.

With these relations, demand factors for different condi-tions can be established.

A second method for developing diversity charts isusing the “diversity factor” and the relation as shown inEquation 16.3-10.

16.3-10

Where:Σ kWn is the sum of the maximum non-diversified

load.

Table 16.3-2 is an example of a diversity chart for 1 to 20houses for different scenarios including air conditioned,electric heating, natural gas appliances, etc. The refer-ence size of the house is a range of 1250 to 1750 squarefeet. Larger or smaller homes or with a mix of loadswould require appropriate adjustments to these loadfactors. Utilities should develop their own diversitycharts based on their regional loading data.

The demand factor approach was used in Table 16.3-2,where:

Demand factor for N = 1 is 0.64.

Probability factor is 0.7.

( ) . kMaximum Coincident Demand MCD DF L=

( )( )

[ ]

N1

N

DF . 1 PDF

N 1 P

⎡ ⎤−⎣ ⎦=−

( ) ( )n

Maximum Diversified kW

Coincident Demand Div Fact

Σ=

n

Table 16.3-2 Diversity Chart for 1 to 20 Detached Houses

Transformer Peak Load (kW) for Detached Houses (1250 to 1750 ft2)

PeakSeason Number of Houses

1 2 3 4 5 6 7 8 9 10

Demand Factor 0.64 0.55 0.47 0.41 0.36 0.31 0.28 0.25 0.23 0.21

Natural Gas Heated – No A/C Summer 8.6 14.9 19.4 22.7 25.2 27.1 28.6 29.9 30.9 31.8

Natural Gas Heated – Central A/C

Summer 10.2 17.9 23.9 28.9 32.9 36.4 39.4 42.1 44.8 47.2

Natural Gas Heated – Natu-ral Gas Stove - Central A/C

Summer 8.8 15.5 20.9 25.2 28.9 32.2 35.0 37.7 40.2 42.5

Natural Gas Heated – electric Water Heater - Central A/C

Summer 12.8 22.9 31.2 38.1 44.1 49.5 54.4 59.0 63.4 67.6

Electric Space and Water Heat Winter 14.8 27.7 39.1 49.6 59.4 68.8 77.8 68.6 95.2 103.7

16-12


16.4 PAD-MOUNT TRANSFORMER SELECTION

16.4.1 Loading Criteria and Transformer Rating

The rated kVA of a transformer is the output that canbe delivered for the time specified at rated secondaryvoltage and rated frequency without exceeding the spec-ified temperature-rise limitations and within the limitsestablished in the design spec.

Selection of a transformer with an appropriate rating toserve to load should be done by considering several fac-tors, including:

• Transformer internal temperatures, such as hottestspot in the winding, top oil temperature, and averagewinding temperatures,

• Transformer loss of life, and

• Total lifetime cost of the transformer

Hottest-spot, Top Oil Temperatures, and Average Winding TemperatureTransformer loading causes heat to be generated due tothe winding and core losses, which results in a tempera-ture rise of the oil and solid insulation. In addition, ele-vated loading increases the presence of oxygen, moisture,and their byproducts, and will accelerate the process ofinsulation aging. It is, therefore, important to ensure thatthe temperature rise is kept within the design limits. It ispossible to relate normal and abnormal loading to thetransformer hottest-spot temperature in order to under-stand how loading affects the life of the insulation.

The hot-spot winding temperature is the principal factorin determining the degradation of the transformer dueto loading and hence has major bearing on the trans-former life. The hottest-spot temperature can be consid-ered as the sum of the temperature of the coolingmedium, the average temperature rise of the copper, andthe hot-spot allowance. It is given by Equation 16.4-1

16.4-1

Where: θΑ is the average ambient temperature. ΔθΤ is the top-oil rise over ambient temperature.ΔθΗ is the winding hottest-spot rise over top-oil

temperature.

It is not possible to measure the hottest-spot tempera-ture directly in a traditional transformer because of thehazards in placing a temperature detector at the properlocation. Standard allowances for hottest-spot rise overtop-oil temperature have been obtained from laboratorytests. A hottest-spot allowance at rated load of 15°C has

been used for transformers with 65°C average windingtemperature rise.

Top oil temperature alone should not be used as a guidein loading transformers, because the difference betweentop oil and hot-spot copper temperatures varies withdifferent designs and with load. Transformers may beoperated above average continuous hottest-spot temper-atures (95°C for 55°C rated transformers and 110°C for65°C rated transformers) for short times, provided theyare operated over much longer periods at temperaturesbelow 95°C and 110°C, respectively. According to Equa-tion 16.4-1, 110°C is the sum of the following: averagewinding rise (65°C), ambient (30°C), and hot spot rise(15°C).

Two characteristic modes of operation can be identifiedwith respect to the aging of insulation:

• Normal operation—corresponds to the normal lifeexpectancy where the deterioration under varyingconditions of load and ambient temperature is nor-mal.

• Overload operation—which is permitted when neces-sary without risking the reliability of the transformer.

Loading of transformers above nameplate is a contro-versial subject. Transformers, at some time, may have tobe overloaded during power system emergencies, inorder to preserve system reliability. The maximum con-tinuous load-carrying capacity of the transformerdepends on its rating, on the temperature of the coolingmedium, ambient temperature, and the level of acceptedinsulation aging governed by the effect of temperatureand time.

Overload capacity of a transformer is the maximumload for which the transformer can be subjected for aparticular duration and considering a particular ambi-ent temperature.

The overload capacity depends on the average windingtemperature rise that has been used to design the trans-former. This temperature can be 55°C or 65°C, depend-ing on the standard or request of end user at purchasetime.

When transformer purchase specifications include over-loadability requirements for specific load profiles, induration, frequency, and magnitude of overload, themanufacturer will adjust the design accordingly to guar-antee such overload operation as normal, and can alsodo so with no loss of life as specified. This design adjust-ment usually results in a more substantial design and/orlower loss unit.

H A T H

T T A

θ θ θ θ

θ θ θ

= + Δ + Δ

Δ = −

16-13


According to IEEE C57.91, normal life expectancy willresult from operating continuously with hottest-spotconductor temperature of 110°C or with an equivalentdaily transient cycle. Distribution transformer tests indi-cate that the normal life expectancy at a continuous hot-test-spot temperature of 110°C is 20 years.

Long-term and Short-time Emergency OverloadsThe permissible loading of transformers for normal lifeexpectancy depends on the design of the particulartransformer, its temperature rise at rated load, tempera-ture of the cooling medium, duration of the overloads,the load factor, and the altitude above sea level if air isused as the cooling medium. ANSI-IEEE C57.92(ANSI/IEEE 1981) has developed several permissibleoverload graphs for different types of transformers withrespect to a number of factors. Figure 16.4-1 shows atypical overload capability curve for oil-immersed trans-formers from ANSI C57.92 for ambient temperature of30oC and oil temperature rise of 65oC. For example, aliquid-filled transformer with a 50% continuous equiva-lent base load at 30°C ambient temperature could beloaded to 120% of full load nameplate rating for fivehours without excessive loss of insulation life.

Overloading of transformers should not be practicedwithout investigation of the various limitationsinvolved, other than winding and oil temperature. Oilexpansion; pressure in sealed- type units; heating of

bushings, leads, soldered connections, and tap changers;and heating of associated equipment such as cables, cir-cuit breakers, fuses, disconnecting switches, and currenttransformers are examples of associated equipment.Any one of these may constitute the practical limit inload-carrying ability.

If the loading strategy is based on the average windingtemperature, as a typical value, for each degree Celsiusin excess of 5°C that the average winding test tempera-ture rise is below 65 °C, the transformer load may beincreased above rated kVA by 1.0%. The 5°C margin istaken to provide a tolerance in the measurement of tem-perature rise. The load value thus obtained is the kVAload, which the transformer can carry at 65°C rise.

For a very short-time loading that is less than ½ hour, itis possible to load transformers up to 300%, with themaximum hottest spot of 200oC and top-oil temperatureof 120oC. If the high loading factor continues more than½ hour, the insulation aging takes place. It should beclearly understood that, while the insulation aging rateinformation is considered to be conservative and helpfulin estimating the relative loss of life due to loads abovenameplate rating under various conditions, this infor-mation is not intended to furnish the sole basis for cal-culating the normal life expectancy of transformerinsulation. The uncertainty of service conditions andthe wide range in ratings covered should be consideredin determining a loading schedule. As a guide, utilitiesconsider an average loss of life of 4% per day in any oneemergency operation to be reasonable.

Percent Loss-of-Life due to LoadingAging or deterioration of insulation is a function oftime and temperature. When cellulose ages, the cellulosechains are cut in a process called chain scission, reduc-ing the average length of the cellulose chains and result-ing in shorter fibers. This can be measured by Degree ofPolymerization, or so-called DP. The rate of degrada-tion is very slow at room temperature. At elevated tem-peratures, however, the rate of degradation increasesexponentially, effectively doubling for approximatelyevery 8°C increase in temperature. Because the tempera-ture distribution in most apparatus is not uniform, thepart that is operating at the highest temperature willordinarily undergo the greatest deterioration. Therefore,it is usual to consider the effects produced by the highesttemperature, or the hottest spot.

Traditionally NEMA developed graphs of % of loss oflife of transformers versus the hottest spot temperature,as shown in Figure 16.4-2. The basis of the aging factormodeled by IEEE is the exponential curve of aging ver-sus temperature.

Figure 16.4-1 Permissible overload for varying periods of time for oil-filled transformers with 65oC rise based on the initial load, normal life expectancy, ambient = 30oC (ANSI C57.92).

16-14


IEEE C57.91-1995 (IEEE 1995) has a well-definedmodel for transformer aging and life of insulation. Itincludes a per unit life model to calculate the aging oftransformers, as shown in Equation 16.4-2.

16.4-2

where θΗ is the winding hottest spot in °C, A = 2 × 10−18

and B is a constant equal to 15,000 for most insulation

types. This standard defines “insulation aging rate”,FAA, as shown in Equation 16.4-3.

16.4-3

Where FAA is the insulation aging rate, θHS,R is the refer-ence hot spot temperature for the insulation, and θHS isthe hot spot temperature at which aging is evaluated.

A curve of FAA versus hottest-spot temperature for a65°C rise insulation system is shown in Figure 16.4-3.FAA has a value greater than 1 for winding hottest-spottemperatures greater than the reference temperature110°C and less than 1 for temperatures below 110°C.

Reduced Life Expectancy with Heavy LoadingIEEE C57.91 has defined a method to calculate thereduced life expectancy based on “aging accelerated fac-tor”, FAA as shown in Figure 16.4-3. The reduced lifeexpectancy, RLF , is calculated from Equation 16.4-4.

16.4-4

16.4-5

Where:Feq is equivalent aging factor for the total time

period.N is the total number of intervals. FAAn is aging acceleration factor for the tempera-

ture that exists during the time interval Δtn.

t is the time period in hours.

Figure 16.4-2 Loss of life versus temperature for different time periods, 65oC rise time (NEMA TR-98-1964).

273H

B

PerUnit Life Aeθ +=

, 273 273HS R HS

A A

AAF e θ θ⎛ ⎞⎜ − ⎟⎜ ⎟+ +⎝ ⎠=

Figure 16.4-3 Insulation’s aging acceleration factor (IEEE C57.91-1995).

% 100eqLF

F tR

Normal Life

×= ×

1

1

( )n

N

AA nn

eq N

nn

F tF

t

=

=

Δ=

Δ

∑

∑

16-15


Normal Life is defined by manufacturer. As a bench-mark for a distribution transformer, normal life is 20years for a well-dried, oxygen-free 65oC average windingtemperature rise insulation at a reference temperature of110oC.

Unusual Service Condition A number of factors related to transformer loading areconsidered unusual service conditions such as:

• Increase of ambient temperature

• Installation in a height more than 1000 m (3300 ft)

The design of distribution transformers usually consid-ers ambient temperature of 30oC. If the average of ambi-ent temperature increases, the loading should belowered to keep the normal life expectancy. A guidelineprovided by IEEE C57.91 suggests a load de-rating of1.5% for each oC up to 50oC. The load is allowed toincrease by 1% for each oC lower than 30oC. Averageambient temperatures can be considered to cover24-hour periods. The maximum ambient temperature in24 hours should not be more than 10°C above the aver-age temperature.

The effect of the decreased air density due to high alti-tude is to increase the temperature rise of transformers,because they are dependent upon air for the dissipationof heat losses. If the transformer is installed at a heightof 1000 m (3300 ft) above sea level, a de-rating factorneeds to be considered as shown in Figure 16.4-4.

Note that if enough information has been delivered tothe transformer designer, the effect of de-rating due tohigh ambient temperature or high altitude level is usu-ally considered by the designer. Therefore, the name-plate ratings do not need to be de-rated.

For transformers installed in subsurface manholes andvaults of minimum size with natural ventilation through

roof gratings, a higher ambient temperature than theoutdoor air is expected. The amount of increasedepends on the design of the manholes and vaults, netopening area of the roof gratings, and the adjacent sub-surface structures. Therefore, the increase in effectiveambient temperature for expected transformer lossesmust be determined before loading limitations can beestimated.

Total Lifetime Cost As discussed in Section 16.2.7, “Cost of Loss Formula,”the transformer cost has three components: capitalinvestment, no-load loss, and load loss. If the end-userprovides the energy price with the purchase request, thedesigner can develop a transformer design that will min-imize the total lifetime cost including the cost of losses.The result of this process is the cheapest transformer inthe useful life period—i.e., with the lowest total owningcost—optimized for a given application.

The following considers the total cost of losses for trans-formers loaded at different fractions of their rating.Typically a transformer is designed to have a minimumloss when operated at about 50% of rating. However, alarger transformer operated at a lower fraction of rat-ing, may have a smaller cost of losses than a smaller unitoperated at 50% of rating. This circumstance will beparticularly true in situations with significant annualload growth.

The present value of the total cost of losses can be cal-culated by calculating the loss in each of the next 40years and then applying a discount factor to account forinflation, and the cost of capital or the expected rate ofreturn on capital investment. The losses in any one yearare calculated as the sum of load and no-load losses.The no-load power loss is simply the no-load lossexpressed as a percent of rating times the rating of thetransformer. Because the no-load loss is constant, thispower loss is simply multiplied by the hours in a year toobtain the energy loss.

The load losses in a transformer vary with the load. Themanufacturer usually states load losses at rated load asa percentage of transformer rating. The value of loss atother loads can be estimated by multiplying by the ratioof the loads squared, because the loss increases with thesquare of the current. This procedure ignores thedecrease in loss at lower temperatures caused by thedecrease in resistance as the temperature decreases(approximately 25% from 90ºC to 20ºC), because thisdecrease is small compared to the quadratic decrease.The peak power loss is calculated at the peak load, andthe energy loss is calculated at the peak loss (at peakload) multiplied by the loss factor to give the energy

Figure 16.4-4 Permissible KVA loading and ambient temperature for altitude above 1000 m (ANSI C.57.12.00).

16-16


loss. The loss factor can be an input parameter, or it canbe calculated from the load factor using an assumedload profile by the empirical equation LF = 0.85(LD2) +0.15 LD, where LF is the loss factor and LD is the loadfactor. If the exact load profile of a transformer isknown, such as hourly load for a year, then the loss fac-tor can be calculated from the load data, and the losscalculation will be exact.

The input parameters to the calculation procedure are:

• Load loss for each transformer rating

• Noload loss for each transformer rating

• Cost of losses (kW and kWh)

• Real discount rate

• Annual load growth rate

• Load factor

• Loss factor

Figure 16.4-5 shows the present value of the cost oflosses over a 40-year life versus peak load in the firstyear. Single–phase, 4-kV polemount transformer dataare used in this example for sizes ranging between 10and 100 kVA to provide the widest data coverage. Thelowest losses are often for a transformer that is severelyundersized. To make a reasonable limit on the loading,the “thermal limits” are shown as vertical dashed linesbased on IEEE C57 – Distribution, Power and Regulat-ing Transformers (Table 7, 2-hour peak load duration at10°C and 30°C [winter and summer operation] 65°Crise.). For winter and summer operation, the peak limitwas set at 1.87 and 1.57, respectively. This is not a firmlimit, because the loss of life of a transformer depends

on the time for which the peak load occurs, the previousload condition, and the thermal time constant of thetransformer. Short time peaks of up to 200% of ratingcan be justifiable.

Optimal transformer sizing can be determined using theDiversity Chart and Figure 16.4-5. Using the peak loadcalculation from Table 7 of IEEE C57, the first verticalintercept with a transformer plot determines the mostoptimum size in terms of lifetime ownership cost.

Transformer size selection, at any specific load level, iscontrolled by the thermal load limit, not by the cost oflosses. This conclusion depends on the ratio of no-loadloss to load loss for the particular set of transformers. Itwill be true as long as the difference in no-load loss fromone transformer size to the next is larger than the loadloss of the smaller size transformer when loaded near itsrating.

The overall conclusion is that a utility cannot reducetransformer losses by going to a larger size transformerthat will have lower load losses. The minimum loss costsare achieved if the smallest possible transformer isselected based on thermal loading limits.

16.4.2 Other Parameters for Transformer Selection

Selection of the appropriate transformer should alsoinclude consideration of:

• Preferred power ratings

• Short-circuit capacity

• Noise level

Figure 16.4-5 Cost of ownership vs. initial load – 4 kV pole transformers – single phase.

16-17


Preferred Power RatingsDespite the selection of an exact power rating that maybe optimal for an application, distribution transformersare generally produced in a number of preferred ratings.

Preferred continuous kVA ratings of single-phase andthree-phase distribution and power transformers basedon an average winding rise by resistance of 65oC aredefined as following:

Single-Phase (kVA): 5, 10, 15, 25, 37.5, 50, 75, 100, 167,250, 333, 500, 800,1250, 1600, 2500, 3300

Three-Phase (kVA): 15, 30, 45, 75, 112.5, 150, 225, 300,500, 750, 1000, 1500, 2000, 2500, 3750, 5000

To reduce inventory, some utilities seek to further limitthe ratings of the transformers that they purchase.

Short-circuit CapacityAnother of the important factors for selecting a trans-former is the short-circuit capacity. Transformersshould be designed and constructed to withstand themechanical and thermal stresses produced by externalshort circuits. The external short circuits shall includethree-phase, single line-to-ground, double line-to-ground, and line-to-line faults on any one set of termi-nals at a time.

IEEE Std C57.12.00 limits determine the short-circuitcurrent duration of distribution transformers as shownin Equation 16.4-6.

16.4-6

Accordingly, the above standard has determined theshort-circuit withstand capability of distribution trans-formers based on the symmetrical short-circuit currentshown in Table 16.4-1.

The symmetrical short-circuit current can be calculatedas follows:

16.4-7

Where: ISC is the symmetrical short-circuit current. IR is the rated current. ZS is the system impedance connected to the

transformer. ZT is the transformer short-circuit impedance.

Based on IEEE Std C57.12.00, multi-winding trans-formers shall be considered to have system fault powersupplied at no more than two sets of un-faulted termi-nals rated greater than 35% of the terminal kVA of thehighest capacity winding.

Noise LevelTransformers in service cause sound, which may causediscomfort to people in the environment in the longterm. This is mainly the problem of power transformers.However, it can be an issue for large distribution trans-formers too. Sound can be defined as the pressure varia-tion in air that the human ear can detect. The normalrange of hearing of a healthy young person is fromapproximately 20 Hz to 20 kHz. The weakest sound thatan ear can detect is dependent on the frequency.

Sound pressure level, LP, expressed in dB, is defined inEquation 16.4-8

16.4-8

Where: po is the reference level equal to 20μPa. p is the sound pressure measured by a micro-

phone.

2

1250500

2 500

s

s

t if S kVAI

t if S kVA

= ≤

= >

Table 16.4-1 Short-circuit Withstand Capability (ANSI/IEEE C57.12.00)

Single Phase (kVA) Three Phase (kVA)

Withstand Capability per Unit of Base Current(Symmetrical)

5-25 15-75 40

37.5-110 112.5-300 35

167-500 500 25

Above 500 kVA Should be calculated using trans-former impedance only.

3( )

% % %

SC

S T

R R

S T T

UI

Z Z

I I

Z Z Z

=+

= ≅+

2

20

10 logP

pL

p=

16-18


To provide a feeling, a quiet living area has a soundpressure level of about 45 dB, and a city street withheavy traffic can have 95 dB sound pressure.

The dominant generating source of transformer soundis core magnetization. When the magnetic flux changes,the magnetic domains change their directions. There-fore, when excited by a sinusoidal flux, the core sounds.In three-phase cores, the changes of magnetic domainfor each core limb do not occur simultaneously, whichmeans that the whole core is subjected to pulsating dis-tortions. Comprehensive investigations are made to cor-relate human perception of loudness at variousfrequencies and sound pressure. To imitate the responsecurves of the human ear, three different filters areinserted in the measuring equipment, named A-weighted, B-weighted, and C-weighted filters. They imi-tate the curves going through 40, 70, and 100 dB,respectively. For transformers, the frequency spectra ofthe audible sound consists primarily of the even har-monics of the power frequency; thus, for a 60-Hz powersystem, the audible sound spectra consists of tones at120 Hz, 240 Hz, 360 Hz, 480 Hz, etc. A transformer“hum” is usually in the range of 100 Hz to 300 Hz.Depending on other nearby ambient noise, the trans-former sounds might not be noticeable.

The noise of a transformer is defined as the A-weightedsound pressure level measured in dB at a specified mea-suring surface with a sound level meter, and then con-verted to a sound power, LW, with the formula shown inEquation 16.4-9.

16.4-9

Where: LS is the measuring surface level in dB.

Table 16.4-2 can be used as a guideline for the noiselevel of distribution transformers up to 5 MVA. As an

option, it is suggested to order transformers designed at3 dB below NEMA standard sound levels.

Methods are available to the transformer designer tocontrol the transformer noise:

• Reducing the core flux density from 1.5 T - 1.6 T to arange of 1.2 T-1.3 T. This can be done either byincreasing the core cross section, or by increasing thenumber of turns in the winding.

• Making a heavier framework for the core

• Inserting pad of damping material between core lay-ers, or between active part and tank

Dimensions and Relation between KVA and SizeThere is a certain fundamental relationship between theKVA rating of transformers and their physical size. Arather obvious relationship is the fact that large trans-formers of the same voltage have lower loss than smallerunits.

As a typical scaling rule, the length, width, and height

are scaled as . Where D represents all

directions of the dimension.

To overcome the limitation of the transformer size,manufacturers have several options, some of whichresult in a tradeoff in transformer performance:

• Reducing the size of core by using Hi-B material orchanging the flux density design value, which resultshigher core loss and noise

• Reducing the space between windings

• Reducing the oil volume by using thermally upgradedinsulation

Each of the above solutions may affect other designparameters, which need to be fully evaluated beforemanufacturing.

W P SL L L= +

Table 16.4-2 Average Sound Power Level for Distribution Transformers (NEMA TR-1, 1993)

Power(kVA)/Sound Power (dB) 0-50 51-100 101-300 301-500 700-1000 1600 2000 2500 3000 4000 5000

Oil-Type 48 51 55 56 57 60 61 62 63 64 65

Dry-TypeSelf-cooled (open) 50 55 58 60 64 66 66 68 68 70 71

Dry-typeSelf-cooled (sealed) 50 55 57 59 63 65 65 66 66 68 69

16-19


16.5 TRANSFORMER COOLING

16.5.1 Mineral Oils and Alternative Ester Oils

Traditionally, transformer dielectric insulating fluid hasbeen a refined naphthenic mineral oil that is stable athigh temperatures and has excellent electrical insulatingproperties. Transformers for indoor use either have beena dry type, or have used a less-flammable liquid.

Up to the 1970s, polychlorinated biphenyls (PCBs) wereused as a dielectric fluid, because they are not flamma-ble. PCBs are toxic, and under incomplete combustion,can form highly toxic products such as furans. Startingin the early 1970s, concerns about the toxicity of PCBsled to their being banned in many countries. Recentlynontoxic, stable silicon-based or fluorinated hydrocar-bons have been used, where the added expense of a fire-resistant liquid offset the additional building cost for atransformer vault.

In the early 20th century, there was interest in seed-oil-based coolants, but compared to mineral oils, these had ahigher pour point and inferior resistance to oxidation.Synthetic esters found specialty applications where highflash point and lower pour point were desired. However,the high cost of synthetic esters limited widespread use.In the early 1990s, natural esters were revisited due toenvironmental regulations. The natural ester productsdeveloped, shared many of the desirable products of thesynthetic esters, and were more economical. Combus-tion-resistant vegetable oil-based dielectric coolants andsynthetic pentaerythritol tetra fatty acid esters are becom-ing increasingly common as alternatives to mineral oil.

Transformer insulating fluids can be compared based onfeatures such as: availability, their effect on losses, heat

transfer properties, flash and fire points, dielectricbreakdown, oxidative stability, decomposition, watersolubility, long-term aging, sludging, climatic effects,economics, and maintenance relative to standardapproved mineral oils. Table 16.5-1 provides a compari-son of many of these parameters for mineral oils andnatural ester oils.

As is evident from Table 16.5-1, natural ester dielectricoils offer several advantages over mineral oils. Theseadvantages include their availability from renewabledomestic sources, their nontoxicity, their being readilybiodegradable, and their being non-carcinogenic. Natu-ral esters have a higher flash point (i.e., lower volatility),superior thermal conductivity, and no sulphur content,and offer a significant reduction in damage to celluloseinsulation.

Any two adjacent conductors form a capacitor. In anideal capacitor, the phase difference between an appliedAC voltage and the current is 90°, and the power dissi-pated is zero. If the dielectric between the conductors isless than ideal, the phase difference will be less than 90°,and some power dissipation will occur. To keep this losslow, it is desirable to have the dielectric as near to idealas is practical.

For insulating oils, the value for this characteristic iscalled the power factor or loss tangent (dissipation fac-tor) and is expressed as a percentage at a specified tem-perature. These values are determined experimentallyand represent trigonometric functions of the angle ofphase difference. With the particular functions used, avalue of zero would represent a 90° phase difference andthe ideal condition; therefore, low values are desirable.In Table 16.5-1, it can be seen that the natural esters

Table 16.5-1 Transformer Oil Comparison

Typ

e

Po

wer

Fac

tor

Dis

sip

atio

n F

acto

r A

ST

M D

924

% @

25o

C/1

00oC

Sp

ecifi

c G

ravi

ty

Sp

ecifi

c H

eat

Cal

/g/o

C @

100o

C

Th

erm

al C

on

du

ctiv

ity

cal/(

cm.s

ec.

oC

Fla

sh P

oin

t A

ST

M D

93(c

lose

d c

up

) oC

Fire

Po

int

AS

TM

D92

(op

en c

up

)oC

Die

lect

ric

Bre

akd

own

AS

TM

D18

16M

inim

um

/gap

/ im

pu

lse

kV

Slu

dg

ing

(O

xid

atio

n S

tab

ility

)A

ST

M D

2440

72 h

/164

h%

Neu

tral

izat

ion

Nu

mb

erA

ST

M D

974

mg

. KO

H/g

Aq

uat

icB

iod

egra

dat

ion

Wat

er C

on

ten

tp

pm

@ 1

5oC

Lo

ng

Ter

m A

gin

g,

Pro

ject

ed L

ife

Year

s

Clim

atic

Eff

ects

, Po

ur

Po

int

AS

TM

D97

oC

Natural Ester Oil 0.15/3.0 0.92 0.6 4.0 x 10-4 330 ≥350 ≥30/≥20/NA ≤0.2/≤0.2 0.04 120%±33%

after 28 days 75 20+ -18

Mineral Oil 0.003/0.06 0.885

0.45@ 20oC 3.0 x10-4 145 160 >35/>28/>180 0.01/0.01 ≤0.03 28% to 49%

after 28 days 45 20+ -47

16-20


have higher power dissipation factor values than themineral oils.

Other disadvantages of the natural esters are higher oxi-dation, pour points, and water retention. Oxidation andsludging are the weakest points of ester oils.

Exposure to atmospheric oxygen can lead to sludging,acid by-products, and finally polymerization of the oils.Natural esters are often supplemented with anti-oxi-dants to address this limitation. In North America,transformers are normally sealed, which limits exposureto oxygen. Once manufactured, the oils are shipped withnitrogen blanketing in the container void to prevent oxi-dation during transport and storage. Oxidation andcontamination of oil can cause the power factor of anoil to rise, so determination of this property may pro-vide useful information about used electrical insulatingoil. Because these values vary with temperatures, com-parisons must always be made at the same temperature.When oils are applied properly, oxidation is a low con-cern. When specifying ester oils, one should confirmthat the transformer is sealed.

The higher temperature pour point is not deemed aproblem for small, sealed transformers because thedielectric properties are maintained. As the oil warmsup after the transformer is energized, its fluid propertiesare restored. For outdoor transformers, use in trans-formers with mechanical oil circulation or internalswitches may be an issue in very cold climates. Indoortransformers with controlled ambient above the pourpoint do not have these restrictions.

Water content is used to monitor a dielectric fluid’squality. It is an indicator of possible oil deterioration,which could, for instance, lead to dielectric breakdown.The values used are based on the relative saturation ofwater in the dielectric fluid. The relative saturation isbased on the amount of water dissolved in the oildivided by the total amount of water that the oil couldhold at that temperature. The dielectric strength of oilstarts to fall when saturation reaches about 50%. Forpetroleum-based oils, 50% at room temperature is 30 to35 mg/kg. Esters hold 500-600 mg/kg water at roomtemperature. In a closed system, the affinity of the esteroils for water has been observed to be a desirable trait.Mineral oil lacks this property, leaving water to migrateto the kraft insulating papers. Moisture in the papercauses it to age. Residual acid in the paper catalyzeshydrolysis and degradation of the cellulose results.

The natural ester oils may not meet some criteria ofstandards such as pour point, water content, and sludg-ing. A separate set of acceptance criteria may be neededfor these oils or limits of application (e.g., outdoortransformers not employing external cooling radiators,circulating pumps, etc.).

Compared to standard mineral transformer oil, esteroils are more costly. The capital cost of a new trans-former filled with the new oils is estimated at 1.25 to1.30 times the same transformer containing mineral oil.For this price differential, a number of advantages arecited, usually the higher flash point and lower life cycleenvironmental cost (i.e., spills and end-of-life disposal).

As with any transformer asset, periodic sampling andanalysis of the oil are recommended as a preventativemeasure and would be part of the life-cycle cost. Eventhough toxicity of ester oils is low, the rules for cleanupof spills are the same as any other substance. The onlydifference is the cleanup cost should be lower becausespecial precautions are not needed compared to hazard-ous substances. In terms of medical issues, the MSDSsheets call only for standard precautions when workingwith the eyes —to avoid getting oil in the eyes, inhalingthe mists, or handling oil if hot.

Information about long-term aging of the ester oils isnot well known, because the products have not been onthe market long. The longest time in service is about 10years. However, some aging tests have been performed,and field-sampling tests have been conducted by theU.S. EPA. Since 1996, more than 17,000 transformershave been built with natural ester fluid, primarily distri-bution low-power, pad-mounted, and pole-mountedtypes, ranging from 10 kVA up to 10 MVA. In 2001, thefirst medium-power transformer (50 MVA) was retro-filled with natural ester oil. Accelerated aging tests perIEEE C57.100™ (IEEE 1999) show that the paper-aging range is significantly slower when aged in naturalesters vs. mineral oil. Full-scale tests per C57.100resulted in units lasting between three and four timesthe required standard average life. Based on theseresults, it has been calculated that the natural estertested has a 21oC higher thermal index than mineral oil.The improved thermal index means longer life at a giventemperature or the ability to operate at higher tempera-tures for a given life. An ASTM standard and an IEEEmaintenance guide have been developed for ester oils.

16-21


16.6 INTERPRETATION OF TEST RESULTS

A number of measurements and tests can be performedon distribution transformers to assess the condition ofthe oil, the solid insulation, the windings, and the trans-former internal construction. Though many of the testsare relatively simple to perform, interpretation of themeaning of the test results requires some expertise.

16.6.1 Oil Tests Interpretation

Standard methods are available to assess the quality ofoil; however, these oil tests are not commonly used onsmall rating distribution transformers. IEEE, ASTM,and other standards do not specify interpretation of theoil test results specifically for distribution transformers,such as pad-mounted or network transformers. The sug-gested numbers in Table 16.6-1 provide a guideline forinterpreting oil test results.

Table 16.6-1 summarizes the oil quality test standardsand recommended limits according to the standards for“service-aged insulating oil.”

Dissolved Gas AnalysisA simple interpretation method for dissolved gas analy-sis (DGA) results is the “Key Gas Method,” as shown inTable 16.6-2. It should be noted that small amounts ofH2, CH4, CO, and CO2 are generated by normal aging.

IEEE Std. C57.104-1991, “IEEE Guide for the Interpre-tation of Gases Generated in Oil-Immersed Transform-ers” (IEEE 1991) introduces a four-condition DGAguide to classify risks to transformers with no previousproblems. This guide uses combinations of individualgases and total dissolved combustible gas concentration(TDCG). Table 16.6-3 summarizes the DGA key gaslimits suggested by IEEE. However, these numbers havebeen generated based on power transformer units, andno data is available for distribution transformers in thisor other standards. Table 16.6-3 assumes that no previ-ous tests on the transformer for dissolved gas analysishave been made or that no recent history exists. If a pre-vious analysis exists, it should be reviewed to determineif the situation is stable or unstable.

Table 16.6-1 Recommended Oil Quality Tests for Service-aged Insulating Oil (IEEE Std C57.106-2006)

Test Dielectric Strength Dissipation FactorInterfacial ten-

sion (IFT)Neutralization

number (acidity) PCB

Standard

ASTM D1816 -97 (1 mm gap)

ASTM 877[min]

ASTM D924-99 @ 25 0C

[max]

ASTM D-971-91[min]

ASTM D974-92[max]

ASTM 4059-91[max]

Limit (new oil) 1816:23kV877: 26 kV 0.1% 35 mN/m 0.03 mg KOH/g 2

Limit (service aged oil) 1mm gap:23kV 0.5% 24 mN/m 0.2 mg KOH/g 50

Table 16.6-2 Key Gas Interpretation Method

Key Gas Secondary Gas Fault Pattern Possible Root Cause

H2 CH4 and minor C2H6 and C2H4Low-energy partial

discharge

Aging of insulation, possible carbon particles in oil, poor grounding of metal objects, loosed lead, float-

ing metal, or contamination

C2H4 CH4 and minor H2 and C2H6 Oil overheating Paper insulation destroyed, metal discoloration, oil heavily carbonized.

C2H2 H2 and minor CH4 and C2H4 High energy Arcing Poor contacts in leads, weakened insulation from aging, carbonized oil.

CO CO2

If the fault involves and oil-impreg-nated structure CH4 and C2H4

Conductor overheating Overloading or cooling problem, bad connection in leads, stray magnetic flux, discoloration of paper.

Table 16.6-3 IEEE Dissolved Key Gas Concentration Limits (in ppm)

Status H2 CH4 C2H2 C2H4 C2H6 CO CO2 TDCG

Condition 1 100 120 35 50 65 350 2500 720

Condition 2 101-700 121-400 36-50 51-100 66-100 351-570 2500-4000 721-1920

Condition 3 701-1800 401-1000 51-80 101-200 101-150 571-1400 4001-10000 1921-4630

Condition 4 >1800 >1000 >80 >200 >150 >1400 >10000 >4630

16-22


Condition 1: TDCG below 720; satisfactory operation;any individual combustible gas exceeding specified levelsshould prompt additional investigation.

Condition 2: Action should be taken to establish a trendquarterly.

Condition 3: High level of decomposition; immediateaction should be taken to establish a trend monthly.

Condition 4: Continued operation could result in failureof the transformer. Immediate action required toremove the transformer from service.

In interpreting DGA, relative gas concentrations arefound to be more useful than actual concentrations. If apossible fault is suspected, a scheme developed by Rog-ers (IEEE 1991) and later simplified by the IEEE, can beused to define transformer condition. The three-ratioversion of the Rogers Ratio Method uses the followingratios: R1 = C2H2/ C2H4, R2 = CH4/H2, R3 = C2H4/C2H6.

Figure 16.6-1 is the flowchart recommended by IEEE tointerpret the Rogers Ratio Method. It is important tomention that the gas ratio method is for determining thepossible fault type, not for detecting the presence of afault. The validity of this method is based on correlation

of the results of a number of failure investigations withthe gas analysis for each case. Another ratio method isthe “Doernenburg method,” which is very similar to theRogers method with 5 ratios. Another DGA interpreta-tion technique proposed by IEC 60599 is based on theDuval triangle. This method provides a coded list offaults detectable by DGA of a faulty transformer.

CIGRE (International Council on Large Electric Sys-tems), one of the leading worldwide organizations onelectrical power systems, has reported phenomenacalled “stray gassing,” which occurs when some types ofinsulating oils are heated at relatively low temperatures(100 to 120°C), producing hydrogen or hydrocarbons.This gas formation seems to reach a plateau after sometime and then stops. Under certain conditions, straygassing may interfere with DGA evaluation. CIGREhas found that at 120°C, the main gas produced, in gen-eral, is hydrogen, followed by methane. The productionof hydrogen is temperature dependent.

Development of saturated hydrocarbons without fault isa common issue that can easily be misinterpreted usingthe Rogers or Duval methods. Typical for these cases isthe production of ethane, ethylene, and methane in highamounts. The ratio of ethane to ethylene, and especiallyethylene to propylene, may be higher than 10. Ethane,ethylene, and methane increase steadily in the first years

Figure 16.6-1 IEEE recommendation for Rogers Ratio Method.R1= C2H2/ C2H4 , R2= CH4/H2 , R3= C2H4/ C2H6

16-23


after commissioning, while the amounts of hydrogenand ethylene stay constant and low. Such behavior hasbeen observed in new transformers as well as in oldones. The interpretation of DGA usually indicates a hotspot below 150°C; however, the transformers are failure-free (Duval 2004).

Dielectric TestThe dielectric test measures the voltage at which oilbreaks down electrically. This test can give a good indi-cation of the amount of contaminants such as dirt,water, and oxidation particles. The IEEE guide for insu-lating oil equipment prefers the ASTM D-1816 (ASTM2004) dielectric test method rather than the ASTM D-877 (ASTM 2002), because the electrodes are closer tothose in real application, and the test is more sensitive tomoisture than the ASTM D-877. If ASTM D 877 isused instead of ASTM D1816 for dielectric strength, thelimit is 26 kV rather than 23 kV. If a 2 mm gap is usedfor ASTM D1816, 40 kV is recommended.

It should be noted, however, that high dielectric strengthis no guarantee that the oil is not contaminated. Testson oil from a failed transformer are not indicative of theoil quality just before failure, because carbon and debrisfrom the failure will be suspended in the oil.

Although rarely performed, carbon and other particu-late matter can be removed by filtration methods priorto dielectric testing.

Power Dissipation FactorThe dissipation factor is a measure of the power lostwhen an electrical insulating liquid is subjected to an acfield. The power is dissipated as heat within the fluid. Alow-value dissipation factor means that the fluid willcause little of the applied power to be lost. The test isused as a check on the deterioration and contaminationof insulating oil because of its sensitivity to ionic con-taminants. ASTM D924 (ASTM 2008a) is a referencefor this test. This test may be satisfactorily performed inthe field, as well as in a laboratory environment. Avisual check should be performed to ensure that thesample does not contain air bubbles due to agitationduring transport.

The maximum recommended levels of percent powerfactor for different categories of new and service agedoils are shown in Table 16.6-4, according to IEEE Std62-1995 (IEEE 1995a).

High levels of power factor (>0.5% @ 25 °C) in oil areof concern, because contaminants can collect in areas ofhigh electrical stress in the winding. Very high power

factor (>1.0% at 25 °C) in oil can be caused by the pres-ence of free water, which could be hazardous to theoperation of a transformer. Oxidation, free water, wetparticles, contamination, and material incompatibilityare all possible sources of high power factor in oil.

Polychlorinated Biphenyl (PCB)According to IEEE Std 62-1995, low polychlorinatedbiphenyl (PCB) concentration (<50 ppm) generally indi-cates an extremely low risk (according to the U.S. EPA),and the oil is classified as noncontaminated. A moder-ate PCB concentration (50 ppm to 500 ppm) causes theoil to be classified as contaminated. Any concentrationabove 500 ppm is considered as if it were pure PCB.Local governmental regulations and environmental leg-islations may require specific values of even lower than50 ppm. Some regulations do not allow moderate con-centration (50 ppm to 500 ppm) near sensitive areassuch as a hospital, food or feed processing plant, seniorcare facility, pre-school/daycare, or a school. The term“Non-PCB” means PCB free from origin of manufac-ture and tested out at less than 1 ppm PCB.

If a high level of PCBs was detected, the oil needs to beretrofilled. To reduce the PCB concentration in the coreand coil of a PCB-contaminated transformer, the con-taminated oil is drained out, and new replacement oil isput in its place—a process called “retrofilling.” The onlytime that it would be logical to retrofill a transformer toreclassify it to non-PCB status is if the transformer has areasonable life expectancy. As a routine, all transformersthat come out of service should be sampled and analyzedfor PCBs before they are repaired, disposed, or recycled.Retrofills cannot reach a level of 1 ppm; more likely, lessthan 50 ppm PCB is more reasonable due to leach-backfrom 10% typical retained oil volume in saturated insula-tion. Less than 50 ppm PCB is not the same as non-PCB,and in some ways is handled differently.

Table 16.6-4 Maximum Suggested Dissipation Power Factors for Different Categories of New and Service Aged Oils (IEEE Std 62)

Type of OilPower Factor

@25 oCPower Factor

@100 oC

New oil as received 0.05 0.3

New oil in new trans-former 0.15 1.5

New oil after filling the transformer, prior to ener-gizing

0.1 -

Service-aged oil 0.5 -

16-24


Acid Number and Inter Facial Tension (IFT)ASTM D974 (ASTM 2008b) is the traditional color-change indicator method of titrating the acids with amild (0.1 N) KOH solution. On some service-aged liq-uids, the color may be so dark as to impair the ability ofthe technician to determine the indicator color changein ASTM D974, so ASTM D664 (ASTM 2009) is usedinstead. IEEE maximum acceptance value for acid num-ber is 0.2 mg KOH/g.

Acceptable limits for IFT vary with operating voltage.For a service-aged oil, the minimum acceptance value is24 mN/m. For oils in service, a decreasing value indi-cates the accumulation of contaminants, oxidationproducts, or both.

16.6.2 Transformer Tests Interpretation

Insulation Resistance and Polarization IndexThe purpose of the transformer insulation resistance testis to measure the condition of a “major” insulation sys-tem—i.e., the insulation between a winding and ground(core) or between two windings. IEEE C57.125-1991 rec-ommends 500 V, 1000 V, or 2500 V DC to be applied tothe transformer winding. The resistance of each mea-

surement should not be smaller than . R is

in MΩ measured at 20 0C, and UW is the winding voltagein kV. If the winding is Y-connected, then UW is thephase-to-ground voltage. If it is Delta-connected, thenUW is equal to phase-to-phase voltage. KVA is the ratedpower of the winding under the test. Megaohm metertest results below this minimum value would indicateprobable insulation breakdown.

If a transformer passes the insulation resistance test,before applying any overvoltage test, it is recommendedto do a Polarization Index (PI) test. The polarizationindex is a ratio of the Megohm resistance at the end of a10-minute test, to that at the end of a 1-minute test at aconstant voltage. Another common way for PI calcula-tion is the ratio of resistance readings that are taken 15and 60 seconds after connecting the voltage. Table16.6-5 is a guide to interpretation of the PI test results.

Power FactorIn general, power factor measurement equipment comeswith three basic modes of operation: grounded speci-men test, grounded specimen with guard, andungrounded. The three measurement modes allow mea-surement of the current leaking back to the test set oneach lead, individually and together. In general, a powerfactor of less than 1% is considered good; 1-2% is ques-tionable; and if it exceeds 2%, action should be taken.Practically, the evaluation is not only based on a singlepower factor data point but is also based on the historyof the change in power factor. Values obtained at thetime of the original tests are used as benchmarks todetermine the amount of insulation deterioration onsubsequent tests.

The power factor of an insulation system should notincrease with an increase in applied ac voltage. If it doesincrease as the ac voltage is increased, there is a problemin the insulation system. Another value of the power fac-tor measurement is that it will detect voids in the insula-tion system that may be causing high partial discharges.Table 16.6-6 is a guideline to interpret the insulationpower factor test. The tests can be done, respectively, onhigh-voltage winding to ground, high- to low-voltagewinding, and low-voltage winding to ground.

Turns RatioThe purpose of a turn-ratio test basically is to diagnosea problem in the winding turn-to-turn or shorted multi-turn insulation system in a transformer. This test detectsprimarily inner winding short circuits. A very low volt-age ac source is used to determine the turn ratio. Twowindings on one phase of a transformer are connectedto the instrument, and the internal bridge elements arevaried to produce a null indication on the detector, withexciting current also being measured in most cases.Measured ratios should compare with ratios calculatedfrom nameplate voltage to within 0.5%, but shouldcompare even closer to actual benchmark values. Out-of-tolerance readings should be compared with priortests. The turn-ratio test may also detect high-resistanceconnections in the lead circuitry or high contact resis-tance in tap changers by higher excitation current and adifficulty in balancing the bridge.

Table 16.6-5 Test Interpretation

Polarization Index Insulation Condition

Less than 1 Dangerous

1.0 - 1.1 Poor

1.1 - 1.25 Questionable

1.25 - 2.0 Fair

Above 2.0 Good

1.5 WUR

KVA=

Table 16.6-6 Power Factor Test Interpretation

Power Factor Insulation Condition

Above 2.0% Dangerous wet transformer

1.0 – 2.0 Investigate

0.5 – 1.0 Deteriorated

Less than 0.5 Good

16-25


Winding ResistanceWinding resistance is used to indicate the winding con-ductor and tap changer contact condition. The testrequires an ohmmeter capable of accurately measuringresistance in the range of 20Ω down to fractions of anohm. Resistance measurements can be used to check forproper connections and to determine if an open-circuitcondition or a high-resistance connection exists in par-allel conductor windings. On three-phase transformers,measurements are made on the individual windingsfrom phase to neutral, when possible. On delta connec-tions, there will always be two windings in series, whichare in parallel with the winding under test. Therefore, ona delta winding, three measurements must be made tobe able to calculate each individual winding resistance.Winding resistance varies with oil temperature. Becausethe resistance of copper varies with temperature, all testreadings must be converted to a common temperatureto give meaningful results. Most factory test data is con-verted to 85oC. This has become the most commonlyused temperature. Variations of more than 5% may indi-cate a damaged conductor in a winding.

Partial DischargeFor large power transformers, the partial discharge (PD)test is performed in the laboratory as a routine test,although a PD test is not required for quality control ofdistribution transformers. However, the PD test is wellknown as a diagnostics tool and can be employed todetect minor and progressing problems leading to a cat-astrophic fault inside a transformer. The two commonlyused PD detection methods are: detection of the acous-tic signals, and measurement of the electrical signalsproduced by the PD. The acceptable PD limits for newtransformers are dependent on the voltage and size ofthe transformers and range from 100 to 500 pC. PDpulses generate mechanical stress waves that propagatethrough the surrounding oil. To detect these waves,acoustic emission sensors are mounted on the trans-

former tank wall. If multiple sensors are used, the PDcan be located based on the arrival time of the pulses atthe sensors.

In the field, the test can be done on-line or off-line. Forthe off-line test, a three-phase source is required toapply the voltage. On-line PD measurement can beemployed using acoustic sensors, via busing tap, orthrough high-frequency current transformers (HFCT)located either on bushing tap or in the neutral of trans-former. Figure 16.6-2 shows a PD resolved pattern onthe left, recorded using an HFCT sensor via neutralcable. A classification technique is employed to separatethe contributions of PD from those generated by distur-bances. Each PD pulse waveform is acquired, and theso-called equivalent time-length and bandwidth areevaluated and plotted on the TF map, as shown in Fig-ure 16.6-2 (b).

16.7 CAPACITORS

16.7.1 Purpose of Capacitors

Loads on electric utility systems include two compo-nents: active power (measured in kilowatts) and reactivepower (measured in kilovars). Active power is gener-ated, whereas reactive power can be provided by eithergeneration or capacitance. Distribution systems haveVAR requirements, because distribution power lines andloads are primarily inductive. Uncompensated distribu-tion systems operate at lagging power factor, drawingreactive power from generation.

Fixed and switched capacitors are inexpensive means ofproviding VAR compensation for distribution systemsand thus correcting power factor and reducing systemlosses. Shunt capacitors supply reactive current tooppose the out-of-phase component of the currentrequired by an inductive load. A shunt capacitor draws

Figure 16.6-2 PD measurement using HFCT.

16-26


leading current, which counteracts the lagging compo-nent of the current at the point of its installation.

When shunt capacitors are applied, the magnitude ofthe source current can be reduced, the power factor canbe improved, and the voltage drop can also reduced.Capacitors can provide effective cost-reduction bydeferring or eliminating investment in new plant.Capacitors aid in minimizing operating expenses andallow utilities to serve new loads and customers with aminimum system investment Advantages of installingshunt capacitors in distribution systems are as follows:

• Released system capacity. The installation of shuntcapacitors decreases the reactive power demand fromthe generation. Thus, generation, transmission, anddistribution substation capacities are released.

• Reduction in losses. The reactive components of linecurrents are reduced from the point of the capacitorinstallation back to the generator. Dollar savings arerealized from peak power and energy loss reductions.

• Improvement in voltage regulation. The demandcapacity of distribution feeders is often limited by thevoltage drop along the line rather than by the thermalampacity of the conductor. The installation of shuntcapacitors will improve the voltage profile of thefeeder. An additional benefit from improving thevoltage profile is the ability to practice conservativevoltage reduction (CVR) or peak shaving from whichfurther demand savings can be achieved.

Depending on the uncorrected power factor of the sys-tem, the installation of capacitors can increase substa-tion capability for additional load by as much as 30%,and can increase individual circuit capability, from thevoltage regulation point of view, approximately 30 to100%. Furthermore, the current reduction for trans-formers, distribution lines, and equipment can reducethe load on these kilovoltampere-limited apparatus andconsequently delay new facility installations.

The preceding benefits can be achieved by both fixedand switched capacitors. With a variable capacitor, thebenefits can be further enhanced by closely matchingthe VAR requirements of the load. If control of a vari-able capacitor can be achieved quickly, transient-freeswitching and voltage flicker reduction are additionalbenefits.

16.7.2 Description of Capacitors

Distribution capacitors are typically housed in rectan-gular, sealed, metal cans, which can be made of stainlesssteel. The cans contain rolled packs of aluminum foil,with layers of insulating paper, and/or plastic film,

between the foil. Recently manufactured capacitors haveall-film-insulating layers. The rolls are packed tightly inthe can, and the can is filled with a dielectric fluid. Thepacks are connected in series and parallel using tabsconnected to the foil to obtain the desired capacitance.Connection to capacitor elements is generally by meansof mechanical crimps or ultrasonic welds.

Terminal leads are connected to the tabs and exitthrough the bushings to form the exterior connections.Capacitor bushings are generally processed porcelainand are welded to the top of the case and the hermeti-cally sealed system.

Capacitors nameplates generally include the followinginformation:

• Name of manufacturer

• Unique serial number

• Catalog number

• Year of manufacture

• Rated capacitance

• Rated rms voltage

• Number of poles

• Rated frequency Rated BIL

• Amount of fluid, indicate flammable or not flamma-ble

Capacitors are rated for line-to-line voltage in the eventthat they are applied on ungrounded or poorlygrounded systems.

Capacitor units are capable of continuous operationover an ambient range of -50°C to +55°C, provided thatthe following limitations are not exceeded:

• 135% of nameplate KVAR

• 110% of rated voltage rms, including harmonics

• 180% of rated current rms, including fundamentaland harmonic currents

16.7.3 Application of Capacitors at Stations and on Feeders

Capacitors are used in distribution stations or on distri-bution feeders. Station capacitors are rack mounted inlarge banks. Capacitors installed on feeders are usuallyin pole-top banks with necessary group fusing. Themaximum bank sizes are about 1800 KVAR at 15 kVand 3600 KVAR at higher voltage levels. Usually, utili-ties do not install more than four capacitor units oneach feeder. Approximately 60% of capacitors are

16-27


applied to feeders, 30% to the substation bus, and 10%on the transmission systems.

Capacitors can be applied as fixed or switched units.Switched units have capacitor bank controllers thatswitch several capacitor banks. Such controllers canswitch capacitor banks at the point of installation orbased on a user-specified time schedule. There are alsocontrollers that switch capacitors on the zero crossing ofvoltage in order to reduce transients. The componentsrequired for a switched capacitor installation operating onVAR conditions are as follows:

• Capacitors

• Oil switch

• Surge arrester

• Current transformer

• Potential transformer

• Transducers to convert CT and PT values into suit-able signals for capacitor controller

• Capacitor bank status

• Local/remote switch status

• Local/remote relay control

Capacitors can be applied at almost any voltage level.Individual capacitor units can be added in parallel toachieve the desired kilovar capacity and can be added inseries to achieve the required kilovolt voltage.

A three-phase capacitor bank on a distribution feedercan be co nnec ted in d e l ta , g rou nded -wye, orungrounded-wye. The type of connection used dependsupon:

• System type—i.e., whether it is a grounded or anungrounded system

• Fusing requirements

• Capacitor-bank location

• Telephone interference considerations

A resonance condit ion may occur in delta andungrounded-wye banks when there is one- or two-lineopen-type fault that occurs on the source side of thecapacitor bank. The resonance occurs due to the main-tained voltage on the open delta, which backfeeds anytransformers located on the load side of the open condi-tion through the series capacitor. As a result of this con-dition, the single-phase of distribution transformers onfour-wire systems may be damaged. Therefore,ungrounded-wye capacitor banks are not recommendedunder the following conditions:

• On feeders with light load, where the minimum loadper phase beyond the capacitor bank does not exceed150% of the per-phase rating of the capacitor bank

• On feeders with single-phase breaker operation at thesending end

• On fixed-capacitor banks

• On feeder sections beyond a sectionalizing-fuse orsingle-phase recloser

• On feeders with emergency load transfers.

Usually, grounded-wye capacitor banks are employedonly on four-wire, three-phase primary systems. Other-wise, if a grounded-wye capacitor bank is used on athree-phase, three-wire ungrounded-wye or delta sys-tem, it furnishes a ground current source that may dis-turb ground relays.

The optimum amount of capacitor kilovars to employ isgenerally the amount at which the economic benefitsobtained from the addition of the last kilovar equals theinstalled cost of the kilovars of capacitors. The methodsused by the utilities to determine the economic benefitsderived from the installation of capacitors vary fromcompany to company, but usually they all determine thetotal installed cost of a kilovar of capacitance.

The economic benefits that can be derived from capaci-tor installation can be itemized as:

• Released generation capacity.

• Released transmission capacity

• Released distribution substation capacity

• Reduced energy (copper) losses

• Reduced voltage drop and consequently improvedvoltage regulation

• Released capacity of feeder and associated apparatus

• Postponement or elimination of capital expendituredue to system improvements and/or expansions

• Revenue increase due to voltage improvements

The total yearly benefit due to the installation of capaci-tor banks can be summarized as

16.7-1Where:Δ$G = annual benefit from generation capacity

released above capacity at original pf, ($/yr).

Δ$T = annual benefit from transmission capacity released above capacity at original pf, ($/yr).

$ $ $ $ $ $ $G T DS DF LR ECΔ = Δ + Δ + Δ + Δ + Δ + Δ∑

16-28


Δ$DS = annual benefit from distribution station capacity released above that at original pf, ($/yr).

Δ$DF = annual benefit from distribution feeder capacity released above that at original pf, ($/yr).

Δ$LR = annual benefit from reduction in energy losses, ($/yr).

Δ$CE = annual additional revenue from increased consumption by voltage improvement, ($/yr).

The total benefits obtained should be compared againstthe annual equivalent of the total cost of the installedcapacitor banks. The total cost of the installed capacitorbanks can be found from

16.7-2

Where: $TIC = annual equivalent of total cost of installed

capacitor banks, $/yr.ΔQc = required amount of added capacitance,

KVAR.$IC = cost of installed capacitor banks, $/KVAR.ic = annual fixed charge rate applicable to

capacitors.

If only fixed-type capacitors are installed, the utility willexperience an excessive leading power factor and volt-age rise at low-load conditions. Therefore, some of thecapacitors should be installed as switched-capacitorbanks, so they can switched off during light-load condi-tions. Thus, the fixed capacitors are sized for light loadand connected permanently. Switched capacitors can beswitched as a block or in several consecutive steps as thereactive load becomes greater from light-load level topeak load, and sized accordingly.

A system analysis is required in choosing the type ofcapacitor installation. As a result of load flow programruns on feeders or distribution substations, the system’slagging reactive loads (i.e., power demands) can bedetermined, and the reactive power in KVARs can beplotted against time of day. This plot is called the reac-tive load duration curve, and is the cumulative sum of thereactive loads (e.g., fluorescent lights, household appli-ances, and motors) of consumers and the reactive powerrequirements of the system (e.g., transformers and regu-lators). Once the daily reactive load duration curve isobtained, then the size of the fixed capacitors can bedetermined to meet the minimum constant reactive loadrequirements. The remaining kilovar demands of theloads are met by the generator or preferably by theswitched capacitors. Switched capacitor sizes can be

selected to match the remaining load characteristicsfrom hour to hour.

A rule of thumb is often used to determine the size ofthe switched capacitors. Switched capacitors are addeduntil:

16.7-3

The kilovars needed to raise the voltage at the end of thefeeder to the maximum allowable voltage level at mini-mum load is the size of the fixed capacitors that shouldbe used. On the other hand, if more than one capacitorbank is installed, the size of each capacitor bank at eachlocation should have the same proportion, that is:

16.7-4

The resultant voltage rise must not exceed the light-loadvoltage drop. The approximate value of the percent volt-age rise is:

16.7-5

Where:% VR = percent voltage rise.Qc⋅3φ = three-phase reactive power due to fixed

capacitors applied, KVAR.X = line reactance, Ω.l = length of feeder from sending end to fixed-

capacitance location, mile.VL-L = line-to-line voltage, kV.

If the fixed capacitors are applied to the end of thefeeder, and if the percent voltage rise is already deter-mined, the maximum value of the fixed capacitors inKVARs can be determined from:

16.7-6

The %voltage rise equation above can also be used tocalculate the percent voltage rise due to the switchedcapacitors. Therefore, once the percent voltage rises dueto both fixed and switched capacitors are found, thetotal percent voltage rise can be calculated as:

16.7-7

Where: Σ % VR = total percent voltage rise.% VRF = percent voltage rise due to fixed capac-

itors.% VRsw = percent voltage rise due to switched

capacitors.

$ $TIC c IC cQ i= Δ ⋅ ⋅

var0.70

vark from switched fixed capacitors

k of peak reactive feeder load+

≥

varvar

k of load center kVA of load centerk of total feeder kVA of total feeder

=

3

2%

10c

L L

Q XlVR

Vϕ⋅

−

=⋅

2

3

10(% ) L Lc

VR VMax Q

Xlϕ−

⋅ =

F % VR = % VR + %VRSW∑

16-29


Another rule of thumb sometimes used is that: The totalamount of fixed and switched capacitors for a feeder isthe amount necessary to raise the receiving-end feedervoltage to maximum at 50% of peak feeder load.

16.7.4 Capacitor Location

Once the kilovars of capacitors necessary for the systemare determined, the location of the capacitors needs tobe determined. The rule of thumb for locating the fixedcapacitors on feeders with uniformly distributed loads isto locate them approximately at two-thirds of the dis-tance from the substation to the end of the feeder. Forthe uniformly decreasing loads, fixed capacitors arelocated approximately halfway out on the feeder. Thelocation of switched capacitors is often determined byvoltage regulation requirements, and they are usuallylocated on the last one-third of the feeder away from thesource.

The best location for capacitors can be found by opti-mizing power loss and voltage regulation. A feeder volt-age profile study is required to determine the mosteffective location for capacitors and a voltage that iswithin recommended limits. Usually, a 2-V rise on cir-cuits used in urban areas and a 3-V rise on circuits usedin rural areas are the maximum voltage changes that areallowed when a switched-capacitor bank is placed intooperation. A general iteration process is summarized asfollows:

1. Obtain circuit and load information:

• kilovoltamperes, kilovars, kilowatts, and loadpower factor for each load

• desired corrected power of circuit

• feeder circuit voltage

• a feeder circuit map that shows locations of loadsand presently existing capacitor banks

2. Determine the kilowatt load of the feeder and thepower factor.

3. Determine the kilovars per kilowatts of load neces-sary to correct the feeder-circuit power factor fromthe original to the desired power factor.

4. Determine the individual kilovoltamperes and powerfactor for each load or group of loads.

5. Determine the kilovars on the line.

6. Determine the line loss in watts per thousand feet dueto the inductive loads determined in steps 4 and 5above. Multiply these line losses by their respectiveline lengths in thousands of feet. Repeat this processfor all loads and line sections, and add them to findthe total inductive line loss.

7. If there are capacitors presently on the feeder, per-form the calculation of step 6, but subtract the capac-

itive line loss from the total inductive line loss. Usethe capacitor kilovars determined in steps 3 and 6,and find the line loss in each line section due tocapacitors.

8. To find the distance to capacitor location, divide totalinductive line loss by capacitive line loss per thou-sand feet. If this quotient is greater than the line sec-tion length:

• Divide the remaining inductive line loss by capaci-tive line loss in the next line section to find the loca-tion.

• If this quotient is still greater than the line sectionlength, repeat step 8a.

9. Construct a voltage profile for the feeder. If the pro-file shows that the voltages are inside the recom-mended limits, then the capacitors are installed at thelocation of minimum loss. If not, then use engineer-ing judgment to locate them for the most effectivevoltage control applications.

Some summary rules that can be used in the applicationof capacitor banks include the following:

1. The location of fixed shunt capacitors should bebased on the average reactive load.

2. There is only one location for each size of capacitorbank that produces maximum loss reduction.

3. One large capacitor bank can provide almost as muchsavings as two or more capacitor banks of equal size.

4. When multiple locations are used for fixed-shunt-capacitor banks, the banks should have the same rat-ing to be economical.

5. For a feeder with a uniformly distributed load, afixed-capacitor bank rated at two-thirds of the totalreactive load and located at two-thirds of the distanceout on the feeder from the source gives an 89% lossreduction.

6. The result of the two-thirds rule is particularly usefulwhen the reactive load factor is high. It can beapplied only when fixed shunt capacitors are used.

7. In general, particularly at low reactive load factors,some combination of fixed and switched capacitorsgives the greatest energy loss reduction.

8. In actual situations, it may be difficult, if not physi-cally impossible, to locate a capacitor bank at theoptimum location; in such cases, the permanentlocation of the capacitor bank ends up being sub-optimum.

16.7.5 Capacitor Protection Considerations

The main function of capacitor protection is to electri-cally remove failed capacitors from the distribution sys-tem.

16-30


The protection must isolate a faulted bank or individualshunt capacitors without interrupting service on theremainder of the circuit. When the capacitor does fail,the protection should rapidly remove it from the systemto avoid case rupture. If the protection has beenproperly coordinated, it should also operate before anyother upstream protective devices. While fulfilling thisfault-clearing role, the capacitor protection must alsoremain immune to a number of “normal” transientconditions such as energizing inrush, discharging/out-rush, parallel switching outrush, and lightning surges.

To ensure that capacitor protection will fulfill thesefunctions, a number of issues must be considered as out-lined in the following sections.

Location ConstraintsWithin substations, capacitors are usually individuallyfused. Capacitor fuses will typically be installed on out-door steel structures, which permits the use of any out-door protection option. However, it is also possible topurchase capacitor banks with under-oil fuses installedinside each capacitor unit, and fuses available in encap-sulated designs may be specified for this application.

For feeder installations, capacitors are most oftenlocated on overhead systems, due to the inherent capaci-tance of underground cable systems, so the protectionequipment can be located at a pole-top location. Out-door protection options can, therefore, be specified,ranging from distribution cutouts through solid mate-rial power fuses to current-limiting fuses.

Bank Configuration Although capacitor units can be connected in severaldifferent configurations, the majority of power capaci-tor equipment installed on primary distribution feedersis connected three-phase, either grounded-wye orungrounded-wye (delta and single-phase connectionsare usually made only on low-voltage circuits). A num-ber of advantages can be derived from the grounded-wye type of connection. With the grounded-wye con-nection, tanks and frames are at ground potential,which provides additional personnel safety. Grounded-wye connections facilitate faster operation of the seriesfuse in the event of a capacitor failure. Grounded capac-itors can divert some line surges to ground and, there-fore, exhibit a certain degree of self-protection fromtransient voltages and lightning surges. The grounded-wye connection also provides a low-impedance path forharmonics.

In general, phase-neutral rated capacitors should beused on grounded capacitor banks, and phase-phaserated banks should be used on ungrounded-wye or deltasystems.

The number of capacitor groups in series is an impor-tant factor in determining the appropriate type of fuse.The impedance of the series groups limits the currentinto a faulted unit and thus determines the magnitudeof the available fault current into a single shorted can.As a general rule, the fault current through the fuse,when the unit that it is protecting becomes shorted,should not be less than 10 times the rated capacitor cur-rent. This available fault current is also affected bywhether or not the neutral is grounded.

The number of series units in a capacitor installationalso affects the overvoltage that healthy units areexposed to after the short-circuit failure of one seriesunit. This factor is discussed later in the section entitled“Overvoltage Protection.”

The number of cans in a parallel group is also an impor-tant consideration in choosing appropriate protection.Energy stored in the capacitors in parallel with a faultedwill be discharged into the faulted unit. This dischargemust be withstood by the fuses on the good cans. Whena large bank is desired, it may be better to use a double-Y construction so as to retain the use of expulsion fuses.There is also a minimum number of capacitors that canbe connected safely in parallel in a group. Below thiscritical number, individual capacitor fuses must be ratedat such a large percentage of the total phase current that,in the event of failure of a unit, the magnitude of thefault current is insufficient to produce rapid fuse clear-ing. Figure 16.7-1 illustrates the effect of the number ofseries sections and the number of parallel units in a sec-tion on the available current through a shorted unit.

Figure 16.7-1 Current through a shorted unit versus the number of units per section for a grounded-wye bank.

16-31


Individual versus Group Protection Capacitor protection practices at distribution voltagescan be divided into two basic protection techniques:individual protection and group protection.

Individual protection is commonly used for large capac-itor banks, which are normally located at the distribu-tion substation. In these installations, each capacitorunit is protected by its own individual fuse; backup pro-tection in the form of a circuit breaker or higher-ratedfuse is normally provided to protect capacitors againstbus faults ahead of the individual fuses. Fuses in thiscase are of the bus-mounted type.

Group protection is commonly used to protect pole-mounted capacitor banks, which are normally locatedon the primary feeders. In this case, only one fuse perphase is used, and each fuse protects all capacitor unitsthat are located in that phase.

Continuous CurrentAlthough capacitors are considered constant currentdevices, in actual operation they are subject to overcur-rents. These are caused by overcapacitance, operation athigher than rated voltage, and system harmonics.ANSI/IEEE Standard C37.99 (ANSI/IEEE 1990)allows a manufacturing tolerance of +15% on the ratedreactive power of capacitors at rated voltage and fre-quency at 25oC. Also, capacitor banks can be operatedat up to 10% overvoltage (though typical system volt-ages do not exceed 6% of nominal voltage). These twofactors may combine to increase continuous current byup to 25%. Harmonic currents depend on system condi-tions and are difficult to predict; however, practice dic-tates that an allowance of 5 to 10% of rated currentshould be used. Ungrounded-wye or delta connectedcapacitors need less margin for harmonic currents,because there is no path for third or multiples of thirdharmonic current.

In capacitor application, it is common to consider thatthe continuous current may be equivalent to 135% ofthe capacitor rated current for grounded-wye connectedbanks, and 125% for ungrounded-wye banks. Thisaccounts for the effect of overvoltage conditions, capac-itance variations, and harmonic currents.

The continuous current for an individual grounded-wyeconnected capacitor can be calculated as follows:

16.7-8

For a three-phase grounded-wye application, the con-tinuous current will be:

16.7-9

Whenever possible, the lowest rated fuse that can con-tinuously carry this current should be selected, becausethis provides maximum sensitivity for high impedancefaults and the greatest protection against tank rupture.However, a fuse selected in this way will be more vulner-able to transient surges. Note that the continuous cur-rent-carrying capability is not necessarily the same asthe rated current of the fuse. Some fuses will continu-ously carry currents above their rating. K and T links,for instance, will carry 150% of their rated current.

Transient CurrentsA capacitor fuse must withstand, without damage, thetransient currents and voltages due to lightning surges,as well as transient currents during energization and de-energization of the capacitor bank. In addition, it mustwithstand discharge currents and parallel switchingtransients.

Figure 16.7-2 illustrates the various types of transientcurrents and provides a reference for the symbols usedin the equations that follow. Note that the capacitors onone phase are shown with individual protection.

CLN

1 kVAR = 1.35 I

kV

ϕ

C

LL

3 kVAR = 1.35 I

3 kV

ϕ

Figure 16.7-2 Illustration of parameters used in capacitor transient current equations.

16-32


In individually fused substation capacitor banks, tran-sients due to lightning surges will typically be of littleconcern, because of the reliable substation shielding andbecause the large number of capacitors in parallel willeffectively share the transient current. Currents generatedby switching transients are also controlled in substationapplications through the use of current-limiting reactorsand switch-closing resistors. Individually fused units aregenerally exposed to only one significant form of tran-sient current—that is, discharge or outrush currents.

For feeder capacitors, where group capacitor protectionis typically used, a number of transient considerationsare of concern, including energizing transients or inrushcurrents, parallel switching (outrush) transients, de-energizing transients, and transients due to lightningsurges.

The various forms of transient currents experienced bycapacitors in normal operation are described in thefollowing paragraphs.

Discharge or Outrush Into Faulted CapacitorDischarge currents in capacitor banks occur when oneparallel unit fails and the remaining good capacitorsdischarge into the faulted unit. To prevent spurious fuseblowing and the disruptive failure of the capacitor case,the fuses on the healthy units must be capable ofwithstanding these outrush currents. A typical dischargetransient waveform is shown in Figure 16.7-3.

The approximate I2t for the outrush current (Io in Fig-ure 16.7-3) from a capacitor to a failed unit can be esti-mated by the following:

16.7-10

Where:R1 = resistance for an individually fused capaci-

tor unit, (ohms).C = capacitance of a single unit, (F).V = voltage, (V).

Table 16.7-1 provides some typical I2t values for singlecapacitors discharging from full voltage. When capaci-tors are connected in parallel, the actual discharge I2tfrom healthy units into a failed unit is typically 66% ofthe tabulated values.

To prevent excessive outrush into faulted capacitors, thetotal parallel-stored energy should not exceed theenergy capability or joule rating of either the capacitorunit or the fuse. According to ANSI C37.99 and tomanufacturing recommendations, the calculated energyof the bank must not exceed 15,000 Joules (4650 KVARin parallel) for all-film capacitors or 10,000 Joules (3100KVAR) for paper-film capacitors. In cases when thecalculated value of the parallel-stored energy surpassesthe limitation capability of expulsion fuses, two possiblesolutions are suggested: reconfiguration of the capacitorbank, or the use of current-limiting fuses (CLFs).

Inrush CurrentWhen a capacitor is energized, there is an initial inrushcurrent (Iin in Figure 16.7-3). This is a short-duration,high-frequency damped sinusoidal current whose char-acteristics depend on the capacitor size, the point on thevoltage wave at which energization occurs, and theimpedance of the supply circuit.

For adequate protection, the melting I2t of the fuse mustbe higher than that of the capacitor inrush current.

Figure 16.7-3 Typical discharge transient waveform.

22

1

1 CVt = . I2 2R

Table 16.7-1 I2t for Capacitor Discharge

I2t (times 1000 A2s)

UnitVolts

100KVAR

150KVAR

200KVAR

300KVAR

2400 10.4 18.6 25.0 -

4160 8.9 15.9 21.0 -

4800 8.5 15.0 20.2 -

7200 6.9 12.3 16.6 25.4

8320 6.3 11.1 15.4 23.2

12470 4.5 8.1 11.5 17.8

13280 4.2 7.5 10.9 17.2

13800 4.0 7.3 10.6 16.8

14400 3.9 6.9 10.3 16.2

16000 3.5 6.1 9.2 15.0

19920 2.5 4.5 7.2 12.5

16-33


With acceptable accuracy, the I2t of the inrush currentcan be calculated using the following relationship:

16.7-11

Where:Isc = fault current at capacitor bank location

(kA).IL = capacitor bank line current (A).K = X/R at the bank location.

As can be seen from this equation, the inrush I2t is afunction of capacitor phase current, available short-cir-cuit current at the point of application, and the X/Rratio of the source impedance at that point.

If N parallel capacitors on a phase are individuallyfused, then the inrush current through each fuse wouldbe the value from the inrush I2t equation divided by N.

Figure 16.7-4 shows graphically the inrush I2t as a func-tion of capacitor phase current for a number of systemshort circuit currents and X/R ratios.

Parallel Switching TransientsParallel switching transients, which are also commonlyreferred to as back-to-back switching transients, occurwhen de-energized capacitor banks are switched intoservice in the vicinity of a previously energized capacitorbank. The energized capacitors discharge high-magni-tude, high-frequency currents (Ip) into the unit beingswitched on, over a period lasting several millisecondsafter the parallel is established. These discharge currentsare only significant when individual capacitor units areinstalled in close proximity on the same distributionfeeder. The high-frequency transient outrush currentfrom the already energized capacitor bank is solely

dependent on the surge impedance of the dischargepath, which is a function of the equivalent capacitanceof the two banks, the total inductance of the dischargepath (the inductance of the conductors between the twobanks and the inductance of the capacitor banks them-selves), and the magnitude of the voltage at the instantthe second bank is energized.

The minimum equivalent circuit inductance (inductanceof the discharge path) L required between the twocapacitor banks to prevent spurious fuse operation canbe calculated using the expression shown in Equation16.7-12.

16.7-12

Where:K = constant equal to 3.7, which represents a

typical inrush damping factor of 0.81.V = peak value of the line-to-ground voltage

when the capacitor bank is energized, (V).Ce = equivalent capacitance of the discharge

path, (Farads).

In Equation 16.7-12, the equivalent capacitance of thedischarge path Ce may be derived from Equation 16.7-13.

16.7-13

Where: KVAR= single-phase KVAR.Vlg = line-to-ground voltage, (V).

I2t= high-frequency surge withstand capability of thecapacitor bank, defined in Equation 16.7-14.

16.7-14Where:FPLD = preload adjustment factor. FHFSW= high-frequency surge-withstand I2t factor.

For a specific conductor size and configuration, with aknown inductance per unit length, the correspondingminimum line length between the two capacitor bankscan be calculated as shown in Equation 16.7-15.

16.7-15

Where:Lb = self inductance of the two banks in paral-

lel supplied by the capacitor manufac-turer, (H).

Lc = conductor inductance per foot (H/ft).

2 2 2L sct = 2.65 1+ sI I I K A

Figure 16.7-4 Energizing inrush I2t for grounded-wye banks.

( )2 4 3

e22

K V CL = Henrys

I t

e 2lg

C 1 2.65 kVAR = = C

2 2 V•

2 2PLD HFSW60Hzt = ( t )I I F F• •

b

c

L - LDistance = L

16-34


When capacitor banks are separated by several pole-spans, nuisance operation of capacitor fuses due toparallel switching is not a major concern, and the calcu-lation shown in Equation 16.7-15 can be used to con-firm this.

Figure 16.7-5 provides an example of the capacitor dis-charge I2t for various KVAR units that are spaced avarying number of 150-foot spans on a line with 795ACSR conductor.

De-Energizing TransientsDe-energizing transient currents (Id) can occur whenopening a capacitor switch. When the capacitor switch isbeing opened, the capacitor tries to maintain the poten-tial that it had before the contacts where opened. If theswitch restrikes, the oscillatory current discharge has ahigh peak value. De-energizing transients are morelikely to occur in circuits with voltages above 25 kV. Inthese cases, restrike-free switches must be installed.

De-energizing transients can be estimated with Equa-tion 16.7-16.

16.7-16

Where:Isc = fault current at capacitor bank location

(kA).IL = capacitor bank line current (A).K = X/R at the bank location.

High-Frequency TransientsCapacitor fuses are commonly exposed to high-fre-quency transients due to lightning surges. These surgesare more likely to damage low-current rated links. Whengroup protection is employed, spurious fuse blowing canbe reduced by utilizing a slow-clearing T tin link of up

to 25 A. In addition, the location of the arrester betweenthe fuse cutout and the capacitor must be avoided.

Available Fault Current

The system fault current available at the capacitor loca-tion, the type of connection (such as delta or wye, neu-tral grounded, or ungrounded), the number of seriesgroups, kVA rating of the bank, and the number ofcapacitors in parallel are all factors that should be takeninto consideration by the protection engineer whendetermining the proper protection for the capacitorbank.

When capacitors are connected grounded-wye or delta,any capacitor failure will cause the system fault currentto flow through the faulted capacitor. The capacitormust withstand the short-circuit current flow until thecircuit is interrupted by the fuse. When multiple-seriesgroups of capacitors are used, as a general rule systemfault current will not flow through a faulted capacitor,and expulsion fuses can be employed.

With the wye configuration, the neutral can be eithergrounded or floating. When grounded, the fault currentthrough a failed capacitor is the available system line-to-ground fault current. For the delta connection, line-to-line system fault current will flow through the failedcapacitor.

In an ungrounded-wye capacitor bank, the fault currentis limited to three times normal line current. Availablefault current to the failed unit and interrupting duty onthe fuse are, therefore, reduced. The fuse, however, mustbe small enough to detect this low-level fault current.Furthermore, while the faulted capacitor is in the cir-cuit, the neutral shift causes the voltage across thecapacitor in the unfaulted phases to increase to 1.73times the rated voltage. Operation under these condi-tions will result in failure of the healthy capacitors in ashort time. The fuse must operate as quickly as possibleto remove this overvoltage.

It must be ensured that the available fault current doesnot exceed the interrupting rating of the selected fuse.The available fault current, along with the following con-siderations of capacitor rupture hazard, are used todetermine whether CLFs are required for an application.

Capacitor Case Rupture HazardCapacitor case rupture will occur if the total energyapplied to the capacitor under short-circuit conditionsis greater than the ability of the capacitor case to with-stand such energy.

Figure 16.7-5 I2t from parallel switching of capacitors on a distribution feeder.

2 2 2L sct = 10.6 1+ sI I I K A

16-35


A capacitor unit internally consists of a number of seriesgroups of parallel-connected packs. Capacitor failureusually starts with the breakdown of one pack, whichthen shorts out the group. The capacitor currentincreases, as does the voltage in the remaining seriesgroups. This increased voltage will eventually lead to thedielectric failure of another pack, causing anotherincrease in current and voltage across the remaininggood groups. This process will continue until all thegroups have failed, and the capacitor acts as a boltedfault. The process may take hours or longer, duringwhich time current escalates in discrete steps. It is desir-able that the capacitor fuse operate before all the seriesgroups have failed, because the then remaining goodgroups will limit the fault current and the possibility ofcase rupture will be minimized.

The cause of capacitor case failure is attributable to thedevelopment of excessive internal pressure sufficient tostress the capacitor case beyond its mechanical limits.When a capacitor dielectric fails, the resulting arc cre-ates internal pressure from heat and a gas generated inthe liquid dielectric of the unit. The pressure variesdepending on the magnitude of the fault current to thefailed unit and the time that it is allowed to flow. Thisforce can swell the sides of the capacitor case or rupturethe case—that is, anything from opening a seam orbushing seal, to violent bursting, endangering adjacentequipment.

Case-rupture curves are essential to the correct selectionof fuse links for overcurrent protection of any capacitorinstallation. These curves, which are available fromcapacitor manufacturers and standards, illustrate theprobability of case rupture for various time and currentrelationships.

Capacitor case rupture for newer all-film capacitordesigns is generally defined by a single-case rupturecurve (see Figure 16.7-6). This is possible, because all-film units fail to short circuit in a more predictable man-ner, and thus have a more well-defined rupture thresh-old than older paper-film capacitors.

Capacitor fuses must have a time-current clearing char-acteristic that will ensure rapid isolation of a faultedcapacitor without case rupture. For adequate protec-tion, the fuse total-clearing time current curve (TCC)must lie to the left of the single case-rupture curve of thecapacitor. For high fault currents, case-rupture curvesmust be compensated for asymmetry.

Capacitor case rupture must be considered, not only byusing the TCCs, but by ensuring that the maximum let-

through I2t of the fuse must always be less than the min-imum rupture I2t of the capacitor. I2t coordination ofthe capacitor minimum rupture I2t curve and the fusetotal clearing I2t curve will determine whether expulsionfuses are suitable to protect against case rupture at highfault levels, or whether CLFs are necessary.

Capacitor manufacturers supply the minimum ruptureI2t information for their units. Some typical values areprovided in Table 16.7-2. Comparison of the tabulatedvalues with I2t curves for expulsion fuses, as illustratedin Figure 16.7-7, can be used to determine the fault cur-rent limit for expulsion fuse protection. If a capacitorminimum rupturing I2t is about 1,000,000 A2s, as illus-trated in Figure 16.7-7, then expulsion fuses will provideprotection with fault currents up to 8000 A. For acapacitor with a lower minimum rupturing I2t of100,000 A2s, expulsion fuses would only provide protec-

Figure 16.7-6 Case-rupture curves for shunt capacitors (150, 200, and 300 KVAR all-film capacitors).

Table 16.7-2 Typical All-film Capacitor Minimum Rupture I2t

Capacitor Unit

Voltage

All-film Units100 KVAR

All-film Units150, 200, 300 KVAR

Above 9000 V 112,500 A2s 450,000 A2s

Below 9000 V 250,000 A2s 1,000,000 A2s

16-36


tion to about 3000 A. Note from Figure 16.7-7 that aCLF can be used for protection at higher fault levels.

Overvoltage ProtectionThe fuse link must protect healthy capacitors againstthe possibility of being operated at excessive overvoltageduring failure of an adjacent unit. Capacitors aredesigned to operate normally at specific 60-Hz nominalvoltage, which is listed on the unit nameplate. However,a 10% overvoltage is allowed without causing any dam-age to the capacitor.

Table 16.7-3 illustrates the overvoltage that is experi-enced by good units during the failure of a capacitor onanother phase. The table illustrates that the overvoltageis also a function of the number of units in series on thefaulted phase. In ungrounded-wye installations consist-

ing of a single capacitor per phase, a failure on onephase increases the voltage on the other phases to 1.73times the rated voltage. If the faulted capacitor is notremoved promptly, this overvoltage can cause a secondcapacitor failure.

IEEE Standards 18 and 1036 provide recommendedlimits for the duration of power frequency overvoltageson capacitors in service. These limits range from a dura-tion of six cycles for an overvoltage 2.2 times rated volt-age, to a duration of 30 minutes for an overvoltage of1.25 times rated voltage. The fuse on a faulted unit mustoperate fast enough to limit the duration of an overvolt-age on the healthy units. One general rule for selectingfuses is to require the fuse to operate within 5 minutes at95% of the fault current. K-type links operate morequickly at high currents than equally rated T-links andthus offer reduced overvoltage durations.

Fuse manufacturers generally publish selection tablesthat reflect consideration of all the factors mentionedabove, and permit the direct selection of the capacitorbank fuse, thereby eliminating the need to perform com-plex calculations or graphical studies. However, for thepurpose of explanation, a step-by-step procedure toselect fuses for capacitor protection is provided in thefollowing section.

16.7.6 Application of Capacitor Fuses

In summary of the preceding sections, a capacitor fusemust be selected to:

• carry continuous capacitor current,

• isolate a faulted capacitor,

• withstand transient currents,

• have sufficient interrupting capacity to interrupt themaximum fault current at the point of application,

• limit the energy let-through to a faulted unit to mini-mize the possibility of capacitor case rupture,

• protect healthy units against prolonged overvoltages.

Fuse TypeThe required interrupting duty of the protection devicecan be established through an assessment of the avail-able system fault current and the capacitor bankconfiguration. Reviewing the available fault current andthe appropriate case-rupture curves for the capacitorunits will dictate whether 1/2-cycle fault clearing is ade-quate or if fractional-cycle clearing is necessary.

The fuse-interrupting duty is established based on thephase configuration (grounded wye versus ungroundedwye versus delta), the capacitor unit arrangement

Figure 16.7-7 Limit for capacitor protection by expulsion fuse.

Table 16.7-3 Overvoltage on Healthy Capacitor Units During Short-Circuit Failure of a Series Unit

Number ofSeriesGroups

Voltage on Each Phase During a Short-Circuit Failure of One Series Unit on Phase "a" (per-unit

nominal phase voltage)

Grounded Wye Ungrounded Wye

Va Vb Vc Va Vb Vc

12345

-2.001.501.331.25

1.001.001.001.001.00

1.001.001.001.001.00

-1.501.291.201.15

1.731.151.081.051.04

1.731.151.081.051.04

16-37


(series-parallel combinations), and the protectionarrangement (group versus individual).

The phase configuration determines the maximum levelof fault current that will flow for the condition where allseries branches of a bank phase are shorted.

The capacitor unit arrangement is used to determine theproportion of this fault current that will flow for thefailure of a single capacitor unit (impedance dividerprinciple), along with the total I2t discharged into a sin-gle unit from the unfaulted parallel units. If the total I2tavailable from the system and from parallel units overthe first half-cycle is found to exceed the case-ruptureI2t, then current-limiting fuses will be required. Thisstep will, therefore, determine whether a distributioncutout, a solid-material power fuse or a current-limitingfuse is to be used.

Fuse Current Rating and Speed

The idealized goal for capacitor fuse application is theselection of the fastest fuse that will avoid damage fromthe range of transient conditions. It is evident that theseare conflicting requirements that must be reconciledwhen choosing the fuse rating and speed ratio. With-standing continuous current and transients would sug-gest use of a slow speed fuse, whereas reducing caserupture and overvoltage require a fast fuse. Some typi-cal practices are provided as examples in the followingparagraphs.

In group protection applications requiring high continu-ous current ratings (above 25 A), K-type links provideadequate withstand to transient currents, while keepingthe melting time as short as practical and providingmaximum protection against case rupture. Good coor-dination is obtained using a K fuse link having con-tinuous current capability of at least 165% of thecapacitor current rating for grounded-wye banks and150% capacitor current rating for ungrounded-wyebanks. In group protection applications where a fusewith a low current rating is required (below 25 A),slower T-links may be preferred. The small T-links havehigher immunity to transients and lightning surges andmay be particularly advantageous in areas with expo-sure to lightning activity (rural areas with little tree shel-ter and high iso-keraunic levels). If power fuse orcurrent-limiting protection is required, the slowest avail-able E-rated or C-rated fuses should be used.

For individual capacitor protection, where low-current-rated links are generally required, T-links have theadvantage of withstanding greater outrush current.Where higher current ratings are required, K-links offerfaster clearing for improved case-rupture protection.

In parallel with selecting the fuse speed is selection ofthe fuse rating. This step is completed by comparing theTCCs of short-listed fuses with the fuse withstand char-acteristics for outrush into faulted units, parallel switch-ing outrush, and energizing inrush as appropriate, andwith the case-rupture curves for the specified capacitorunits. Where current-limiting fuses are specified, theminimum-melting I2t of the various fuse ratings can becompared directly with the relevant withstand values,and the total let-through I2t of the fuses can be com-pared directly with the case-rupture I2t. The fuse-melt-ing TCC must remain above and to the right of therelevant withstand curve throughout the current range,and should be selected to stay as close as possible to thiscurve. By virtue of this selection preference, the designercan ensure that the probability of case rupture is as lowas possible.

Example Capacitor Protection

A three-phase, 600-KVAR, grounded-wye connectedcapacitor bank is installed in a pole-top configuration.The capacitor bank is configured with two single-phase,all-film construction, 100-KVAR capacitor units in par-allel connected in each phase. The capacitor units havevoltage ratings of 7.2 kV. Assume the maximum line-to-ground fault current is 800 A rms symmetrical. The per-phase load current of the capacitor bank is 27.8 A.Select a primary protective device for this application.

According to the capacitor manufacturer, the capacitorbank can be adequately protected with expulsion fuses ifthe maximum available fault current does not exceed 3.1kA rms symmetrical, which is the case.

To accommodate the highest anticipated capacitor bankcurrent, the fuse continuous current is selected based onthe following bank tolerances: 10% overvoltage, 15% incapacitance, and 10% in harmonics. Consequently, theminimum continuous current that the fuse must carry isdetermined as shown in Equation 16.7-17.

16.7-17

Because K and T links are 150% rated, the continuouscurrent value must be divided by 1.5. As a first approxi-mation, a 25T fuse link mounted on a 200 A distribu-tion cutout is selected to protect each group ofcapacitors. The cutout has a rated voltage of 7.2 kV, andan interrupting capability of 10 kA rms symmetrical(based on X/R ratio of 4), which exceeds the maximumavailable short-circuit current at the capacitor banklocation.

The ability of the fuse to withstand the energizinginrush currents associated with the capacitor bank is

cI = 1.35 x 27.8 = 37.53 A

16-38


determined by comparing the unloaded high-frequencysurge-withstand I2t capability of the fuse with the I2t ofthe transient inrush current. The I2t of the inrush cur-rent when the capacitor bank is energized is calculatedas shown in Equation 16.7-18.

16.7-18

Data available from the fuse manufacturer indicates thatthe unloaded high-frequency surge-withstand I2t for sil-ver-copper eutectic element links is approximately 45%of the 60-Hz minimum melting I2t value. Using the min-imum-melting TCC of the selected fuse (25T), the cur-rent at 1 second is 200 A, which gives a minimum-melting I2t of approximately 40,000 A2s. The fuse high-frequency, surge-withstand I2t is calculated as shown inEquation 16.7-19.

16.7-19

This means that the transient inrush current associatedwith energizing an isolated capacitor bank will notcause nuisance blowing of the expulsion link selected.

The next step is to verify that the selected fuse can effec-tively protect the individual capacitor units against caserupture. This step is accomplished by comparing thetotal-clearing TCC of the fuse with the case-rupturecurve of the capacitor unit. Figure 16.7-8 shows a plotof the case-rupture curve of a 100-KVAR all-film capac-itor unit along with the fuse total-clearing TCC.

The fuse total-clearing TCC lies below and to the left ofthe capacitor case-rupture curve for all current valuesup to approximately 3900 A. This crossover point indi-cates the maximum short-circuit current for capacitorbank protection. Because the maximum available faultcurrent at capacitor location (800 A) is lower than thefault-current value at the point of intersection of thetwo curves, the fuse will always clear the circuit prior tocase rupture. The selected fuse will clear the fault cur-rent in approximately 0.07 seconds.

If the maximum available fault current was greater thanthe maximum fault-current for capacitor protection, theselected expulsion fuse would not provide adequate pro-tection to the capacitor bank, and one of the followingalternatives would be considered:

• Move the capacitor bank to a location where theavailable fault current is lower.

• Use larger capacitor units.

• Individually fuse the capacitor units.

• Use a partial-range CLF in series with the expulsionfuse, or use a full-range CLF.

Table 16.7-4 provides some examples of fuse link ratingsselected for individual protection of all-film capacitorswith expulsion fuses. Table 16.7-5 provides some exam-ples of CLFs selected to protect individual all-film2 2 2t = 2.65 27.8 0.793 = 241 sI 1+ 4 A• • •

2 2hfI t = 0.45 x 40,000 = 18,000 A s

Figure 16.7-8 Example of capacitor protection.

Table 16.7-4 Typical Individual Expulsion Fuse Ratings for All-Film Capacitors

CapacitorUnit

VoltageRating

Fuse VoltageRating

(kV) Capacitor Unit Rating (KVAR)

50 100 150 200 300 400

Fuse Link Rating

2400 8.7 20T 40K 65K 80K - -

4800 8.7 12T 20T 30T 40T - -

7200 8.7 10T 15T 20T 25T 40T 50T

7960 8.7 10T 15T 20T 25T 40T 50T

8320 8.7 10T 15T 20T 25T 40T 50T

14400 15.0 a

a. For high-voltage 50-KVAR units, fuses with appro-priate current ratings will not withstand the out-rush I2t.

10T 15T 20T 25T 30T

16-39


capacitors. Table 16.7-6 provides examples of expulsionfuses for group protection of all-film capacitors.

16.8 HIGHLIGHTS

Efficiency and Components of Transformer Loss

• Loss in transformers is due to two causes: load lossand no-load loss. Physically, two main components oftransformer loss are: electric (I2 R) and magnetic(core hysteresis and core eddy current loss). Trans-former efficiency is related to the amount of wattslosses that occur when the transformer is in opera-tion. The percentage of power that is available on thesecondary side of the transformer, as a percentage ofthe power input on the primary, is termed the effi-ciency.

Table 16.7-5 Typical Individual CLF Ratings for All-Film Capacitors

CapacitorUnit

VoltageRating

Fuse Voltage Rating Capacitor Unit Rating (KVAR)

50 100 150 200 300 400

CLF Current Rating

2400 8.3 30 65 90a

a. Indicates parallel fuses.

- - -

4800 8.3 18 30 45 65 - -

7200 8.3 18 25 30 40 65 80a

7960 8.3 18 18 30 40 65 80a

8320 15.5 10 18 25 35 50a 80a

14400 15.5 - 10 18 25 30 50a

Table 16.7-6 Typical Group Protection for All-Film Capacitors

SystemLine-to-Line

Voltage

CapacitorLine

Voltage

Three-Phase-Bank

KVAR

Rated LineCurrent inAmperes Typical Link Size

Grounded Wye Ungrounded Wye

4160 2400

150300450600

20.841.662.583.3

20 T40 K65 K80 K

20 T40 K65 K80 K

8320 4800

15030045060090012001350

10.420.831.241.762.583.393.8

10 T20 T30 K40 K65 K80 K80 K

10 T20 T30 K40 K65 K80 K80 K

12480 7200

1503004506009001200135018002400

6.913.920.827.841.752.559.078.7105

8 T15 T20 T25 T40 K50 K65 K80 K100 K

6 T12 T 20 T25 T40 K50 K65 K80 K100 K

13800 7960

1503004506009001200135018002400

6.312.618.825.137.750.256.575.4100.5

6 T12 T20 K25 K40 K50 K50 K80 K100 K

6 T12 T20 K25 K40 K50 K50 K65 K100 K

24900 14400

300450600900120013501800240027003600

6.910.413.920.827.831.241.755.662.583.3

8 T10 T15 T20 T25 K30 K40 K50 K65 K80 K

6 T10 T12 T20 T25 K30 K40 K50 K65 K80 K

16-40


• Generally transformers are at maximum efficiencywhen they are 50% loaded. When transformers arelightly loaded, the no-load losses form a large per-centage of the power utilized, and, therefore, the effi-ciency is low. As the transformer is loaded to higherlevels, the load losses dominate the efficiency. Themaximum efficiency point is the optimal point oflowest load and no-load losses. It is determined bythe design of the transformer and, theoretically,could be designed to occur at any load percentage. Ittypically is designed to occur at 50%, because theaverage load tends to be about 50% of the peak load.

• Regulations by Energy Departments often mandateminimum efficiency levels for liquid-filled and dry-type distribution transformers.

Reduction of Transformer Losses

• Reduction of transformer losses and improvement inefficiency can be achieved by reduction of either loador no-load losses. For any given set of core and wind-ing materials, reduction of load losses often leads toan increase in no-load losses and vice versa.

• More recently, nano-crystalline steel has becomeavailable for use in transformer cores. The best ofthese steels are based on an Fe-Zr-B alloy that isformed in an amorphous state and then annealed toproduce very small grain sizes. This process makesthe alloy less brittle and, thereby, decreases produc-tion costs. The alloy has even higher permeability andalso higher saturation induction than the amorphousmaterials, but it is not yet available in manufacturedtransformer cores.

• Transformer windings are made of either copper oraluminum in round wires, square wires, or flat sheets.The resistivity of aluminum is about 1.6 times largerthan that of copper, but aluminum has a lower cost.Many different alloys of aluminum and copper areavailable. In general, the lower-resistance alloys aremore expensive and harder to work with in the manu-facturing processes, leading to higher initial costs.

• In addition to choice of material, load losses areaffected by the cross-sectional area of the wire used.Larger wires produce lower load losses, but then thewindings are larger, and this requires a larger core,which increases the no-load losses.

Long-term and Short-time Emergency Overloads

• The permissible loading of transformers for normallife expectancy depends on the design of the particu-lar transformer, its temperature rise at rated load,temperature of the cooling medium, duration of theoverloads, the load factor, and the altitude above sealevel if air is used as the cooling medium. ANSI-

IEEE C57.92 has developed several permissible over-load graphs for different types of transformers withrespect to a number of factors. For example, a liquid-filled transformer with a 50% continuous equivalentbase load at 30°C ambient temperature could beloaded to 120% of full load nameplate rating for fivehours without excessive loss of insulation life.

Total Lifetime Cost

• The transformer cost has three components: capitalinvestment, no-load loss, and load loss. If the end-user provides the energy price with the purchaserequest, the designer can develop a transformerdesign that will minimize the total lifetime cost,including the cost of losses. The result of this processis the cheapest transformer in the useful life period—i.e., with the lowest total owning cost, optimized for agiven application.

• Typically a transformer is designed to have a mini-mum loss when operated at about 50% of rating.However, a larger transformer operated at a lowerfraction of rating may have a smaller cost of lossesthan a smaller unit operated at 50% of rating. Thislatter case will be particularly true in situations withsignificant annual load growth.

• Transformer size selection, at any specific load level,is controlled by the thermal load limit, not by thecost of losses. This conclusion depends on the ratio ofno-load loss to load loss for the particular set oftransformers. It will be true as long as the differencein no-load loss from one transformer size to the nextis larger than the load loss of the smaller size trans-former when loaded near its rating.

• The overall conclusion is that a utility cannot reducetransformer losses by going to a larger size trans-former that will have lower load losses. The minimumloss costs are achieved if the smallest possible trans-former is selected based on thermal loading limits.

Polarization Index Test

• If a transformer passes the insulation resistance test,before applying any overvoltage test, it is recom-mended to do a Polarization Index (PI) test. Thepolarization index is a ratio of the Megohm resis-tance at the end of a 10-minute test, to that at the endof a 1-minute test at a constant voltage. Anothercommon way for PI calculation is the ratio of resis-tance readings that are taken 15 and 60 seconds afterconnecting the voltage. The following table is a guideto interpreting the PI test results.

16-41


Power Factor Test

• In general, power factor measurement equipmentcomes with three basic modes of operation: groundedspecimen test, grounded specimen with guard, andungrounded. The three measurement modes allowmeasurement of the current leaking back to the testset on each lead, individually and together. In gen-eral, a power factor of less than 1% is consideredgood; 1-2% is questionable; and if a power factorexceeds 2%, action should be taken. Practically, theevaluation is not only based on a single power factordata point, but is also based on the history of thechange in power factor. Values obtained at the timeof the original tests are used as benchmarks to deter-mine the amount of insulation deterioration on sub-sequent tests.

Purpose of Capacitors

• Fixed and switched capacitors are inexpensive meansof providing VAR compensation for distribution sys-tems and thus correcting power factor and reducingsystem losses.

• Advantages of installing shunt capacitors in distribu-tion systems are as follows:

—Released system capacity

—Reduction in losses

—Improvement in voltage regulation

• Depending on the uncorrected power factor of thesystem, the installation of capacitors can increase thesubstation capability for additional load by as muchas 30%, and can increase individual circuit capability,from the voltage regulation point of view, approxi-mately 30 to 100%.

Description of Capacitors

• Distribution capacitors are typically housed in rect-angular, sealed metal cans, which can be made ofstainless steel. The cans contain rolled packs of alu-minum foil with layers of insulating paper, and/orplastic film, between the foil. Recently manufacturedcapacitors have all-film-insulating layers.

• Capacitors are rated for line-to-line voltage in theevent that they are applied on ungrounded or poorlygrounded systems.

Application of Capacitors at Stations and on Feeders

• Capacitors are used in distribution stations or on dis-tribution feeders. Station capacitors are rackmounted in large banks. Capacitors installed on feed-ers are usually in pole-top banks with necessarygroup fusing. Capacitors can be applied as fixed orswitched units. Switched units have capacitor bankcontrollers that switch several capacitor banks.

• A three-phase capacitor bank on a distributionfeeder can be connected in delta, grounded-wye, orungrounded-wye. The type of connection useddepends upon:

— System type—i.e., whether it is a grounded or anungrounded system

— Fusing requirements

— Capacitor-bank location

— Telephone interference considerations

• Ungrounded-wye capacitor banks are not recom-mended under the following conditions:

— On feeders with light load, where the minimumload per phase beyond the capacitor bank doesnot exceed 150% of the per-phase rating of thecapacitor bank

— On feeders with single-phase breaker operation atthe sending end

— On fixed-capacitor banks

— On feeder sections beyond a sectionalizing-fuse orsingle-phase recloser

— On feeders with emergency load transfers

• Usually, grounded-wye capacitor banks areemployed only on four-wire, three-phase primary sys-tems. Otherwise, if a grounded-wye capacitor bank isused on a three-phase, three-wire ungrounded-wye ordelta system, it furnishes a ground current sourcethat may disturb ground relays.

• The optimum amount of capacitor kilovars toemploy is generally the amount at which the eco-nomic benefits obtained from the addition of the lastkilovar equals the installed cost of the kilovars ofcapacitors.

• The economic benefits that can be derived fromcapacitor installation can be itemized as:

— Released generation capacity

— Released transmission capacity

— Released distribution substation capacity

— Reduced energy (copper) losses

— Reduced voltage drop, and consequently, improved voltage regulation

Polarization Index Insulation Condition

Less than 1 Dangerous

1.0 - 1.1 Poor

1.1 - 1.25 Questionable

1.25 - 2.0 Fair

Above 2.0 Good

16-42


— Released capacity of feeder and associated appa-ratus

— Postponement or elimination of capital expendi-ture due to system improvements and/or expan-sions

— Revenue increase due to voltage improvements

• If only fixed-type capacitors are installed, the utilitywill experience an excessive leading power factor andvoltage rise at low-load conditions. Therefore, someof the capacitors should be installed as switched-capacitor banks so they can switched off during light-load conditions.

• A rule of thumb is often used to determine the size ofthe switched capacitors. Switched capacitors areadded until:

• The kilovars needed to raise the voltage at the end ofthe feeder to the maximum allowable voltage level atminimum load is the size of the fixed capacitors thatshould be used. On the other hand, if more than onecapacitor bank is installed, the size of each capacitorbank at each location should have the same propor-tion—that is:

• The resultant voltage rise must not exceed the light-load voltage drop. The approximate value of the per-cent voltage rise is:

Where:% VR = percent voltage rise.Qc⋅3φ = three-phase reactive power due to fixed

capacitors applied, KVAR.X = line reactance, Ω.l = length of feeder from sending end to fixed-

capacitance location, mile.VL-L = line-to-line voltage, kV.

• Another rule of thumb sometimes used is that: Thetotal amount of fixed and switched capacitors for afeeder is the amount necessary to raise the receiving-end feeder voltage to maximum at 50% of peak feederload.

Capacitor Location

• The rule of thumb for locating the fixed capacitors onfeeders with uniformly distributed loads is to locate

them approximately at two-thirds of the distancefrom the substation to the end of the feeder. For theuniformly decreasing loads, fixed capacitors arelocated approximately halfway out on the feeder. Thelocation of switched capacitors is often determinedby voltage regulation requirements, and they are usu-ally located on the last one-third of the feeder awayfrom the source.

• The best location for capacitors can be found by opti-mizing power loss and voltage regulation. A feedervoltage profile study is required to determine themost effective location for capacitors and a voltagethat is within recommended limits. Usually, a 2-V riseon circuits used in urban areas and a 3-V rise on cir-cuits used in rural areas are the maximum voltagechanges that are allowed when a switched-capacitorbank is placed into operation.

• Some summary rules that can be used in the applica-tion of capacitor banks include the following:

— The location of fixed shunt capacitors should bebased on the average reactive load.

— There is only one location for each size of capaci-tor bank that produces maximum loss reduction.

— One large capacitor bank can provide almost asmuch savings as two or more capacitor banks ofequal size.

— When multiple locations are used for fixed-shunt-capacitor banks, the banks should have the samerating to be economical.

— For a feeder with a uniformly distributed load, afixed-capacitor bank rated at two-thirds of thetotal reactive load and located at two-thirds of thedistance out on the feeder from the source givesan 89% loss reduction.

— The result of the two-thirds rule is particularlyuseful when the reactive load factor is high. It canbe applied only when fixed shunt capacitors areused.

— In general, particularly at low reactive load fac-tors, some combination of fixed and switchedcapacitors gives the greatest energy loss reduc-tion.

— In actual situations, it may be difficult, if notphysically impossible, to locate a capacitor bankat the optimum location; in such cases, the per-manent location of the capacitor bank ends upbeing sub-optimum.

Capacitor Protection Considerations

• The main function of capacitor protection is to elec-trically remove failed capacitors from the distributionsystem.

var0.70

vark from switched fixed capacitors

k of peak reactive feeder load+

≥

varvar

k of load center kVA of load centerk of total feeder kVA of total feeder

=

3

2%

10c

L L

Q XlVR

Vϕ⋅

−

=⋅

16-43


• The protection must isolate a faulted bank orindividual shunt capacitors without interrupting ser-vice on the remainder of the circuit. When the capac-itor does fail, the protection should rapidly remove itfrom the system to avoid case rupture. If the protec-tion has been properly coordinated, it should alsooperate before any other upstream protective devices.While fulfilling this fault-clearing role, the capacitorprotection must also remain immune to a number of“normal” transient conditions such as energizinginrush, discharging/outrush, parallel switching out-rush, and lightning surges.

• A capacitor fuse must be selected to:

— carry continuous capacitor current,

— isolate a faulted capacitor,

— withstand transient currents,

— have sufficient interrupting capacity to interruptthe maximum fault current at the point of appli-cation,

— limit the energy let-through to a faulted unit tominimize the possibility of capacitor case rupture,

— protect healthy units against prolonged overvolt-ages.

• To ensure that capacitor protection will fulfill thesefunctions, of the following issues must be considered:

— Location constraints

— Bank configuration

— Individual versus group protection

— Continuous current

— Transient currents, including discharge or outrushinto faulted capacitor, inrush current, parallelswitching transients, de-energizing transients, andhigh-frequency transients

— Available fault current

— Capacitor case-rupture hazard

— Overvoltage protection

Application of Capacitor FusesSelection of the appropriate fuse involves considerationof the following:

• Fuse Type. Through an assessment of the availablesystem fault current and the capacitor bankconfiguration, the required interrupting duty of theprotection device can be established. Reviewing theavailable fault current and the appropriate case-rup-ture curves for the capacitor units will dictatewhether 1/2-cycle fault clearing is adequate or if frac-tional-cycle clearing is necessary. This review will alsodetermine whether an expulsion of a current-limitingfuse is required.

• Fuse Current Rating and Speed. The idealized goal forcapacitor fuse application is the selection of the fast-est fuse that will avoid damage from the range oftransient conditions. These conflicting requirementsmust be reconciled when choosing the fuse rating andspeed ratio. Withstanding continuous current andtransients would suggest use of a slow speed fuse,whereas reducing case rupture and overvoltagerequire a fast fuse.

16-44


REFERENCES

ABB. 2002. ABB Distribution Transformer Guide. ABB Distribution Transformer Division.

ABB. 2007. ABB Transformer Handbook. ABB Man-agement Services Ltd.

ANSI/IEEE 1981. ANSI/IEEE Std C57.92-1981. “IEEE Guide for Loading Mineral-Oil-Immersed Power Transformers Up to and Including 100 MVA with 55 Degree C and 65 Degree C Average Winding Rise.”

ANSI/IEEE. 1990. ANSI/IEEE C37.99. “Guide for Protection of Shunt Capacitor Banks.”

ANSI/IEEE. 2000. ANSI/IEEE C57.12.00-2000. “Gen-eral Requirement for Liquid-Immersed Distribution, Power, and Regulating Transformers.”

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