1991_Boshoff_Faults in Sugar Mill

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Proceedings of TheSouth African Sugar Technologists' Association - June 1991

FAULTS IN SUGAR MILL POWER SYSTEMSByT.L. BOSHOFF

Tongaat-Hulett Sugar Ltd, Glenashley

The transformer as source offault currentsEvery powertransformer hasitscharacteristic 'impedancevoltage' statedon the nameplate. Thisis definedasthe voltage which has to beappliedto the pnmarywinding to allow

the full-load current to flow in the primary, when the secondary winding is short circuited. This is expressed as apercentage of the rated phase-to-phase voltage (see Figure2).

Introduction tofault currentsWere a shortcircuittooccuracross theheaterbankshownin Figure I, the fault current would be a function. of thevoltage available fromthe transformer, the internal impedanceof the transformer and the impedance of anycables orother conductors in the circuit.If the fault occurred close to the transformer, the faultcurrentwould be determined only by the impedance of thetransformer.

Under normal conditions the switch has to handle the'making'current, i.e.thecurrentwhich appears at the instantof closing the switch, and hasto conductthe normalcurrenton a continuous basiswithout overheating. It must be ableto 'make' and 'break' these initial and continuous currentsas often as required, and remain reliable.The switch also has to be able to close onto, withstandand interrupt the maximumcurrentwhich can occur in thecircuitunderfaultconditi.ons. Thisprospective faultcurrentmaybe a thousandtimesmore than the normalcontinuouscurrent.

Short-circuit

Secondaryrimary

AbstractAn established method of calculating three-phase symmetrical faultcurrents in alternating current power circuitsis presented. General information is given on the methodsused to limit and control fault currents. The practical approach adoptedmakes the method useful to engineers whoare not specialists in electrical technology.

Discussion

IntroductionUnderfaultconditions the magnitudes ofcurrentsflowingin powercircuits increase greatly, and the switches ~ u s t be

capable of safely interrupting thesefault currents. It IS therefore important that the highest possible fault currents be .calculated.In this paper some simplifications have been applied.However the resulting relatively small inaccuracies causethecalcuiated faultlevelstobe conservative, i.e.on the highside.With switchgear the worst condition possible is a symmetrical three-phase fault, and this is the typeof fault thatoccurs most under plant conditions. Whena single twophase fault occurs, it usually rapidly bums through into athree-phase fault. .In this paper, only s y m m e t r i c ~ l t h r e e - p ~ a s e faults h ~ v ebeenconsidered. Remember that in the design of protectiverelaysystems formajorplantitemsSUCh. as large a l t e r n a ~ o r s ,single and two-phase faults shouldbe ngorously taken intoaccount.With overhead distribution lines, single and two-phasefaults are less likely to develop into three-phase faults.Thecalculation methodpresented isonlyoneof a numberofpossible approaches. Theauthorbelieves it bethe mostpractical approach for the plant engineer IS n?t a specialistin electrical technology and who requires quick, conservative answers. For formal, accurate analyses, othermethods may be superior.

Switchgear ratingsFigure I shows a simpleAC circuit comprising a ~ r a n s former feeding a heater bank, and controlled by a SWItch.

FIGURE 2 Transformer 'impedance voltage' test circuit.

FIGURE 1 Basic circuit under short-circuit conditions.Primary

Transformer Switch

Fault current

Secondary

Short-circuit Taking a typical impedance voltage value of 5% meansthat if 5%, or one twentieth, of the normal voltage was appliedto the transformer primarywinding, theratedfull-loadcurrent would flow in the primary winding when the secondarywinding was short-circuited.It canbe assumed that under theseconditions thecurrentin the short-circuited secondary winding would alsobe close

to itsratedfull-load value. Therefore ifthe full ratedprimaryvoltage was applied, the current in the short-circuited secondarywould be twenty times the rated full-load secondarycurrent of the transformer.188


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Proceedings of TheSouthAfrican Sugar Technologists' Association - June 1991The prospective fault current that can be expected froma transformer therefore depends on the transformer full-loadrating (expressed in secondary amps or in kVA), and ontransformer impedance. This assumes that the power supplyto the transformer would be able to maintain full primaryvoltage for the duration of the fault. Transformer fault current (secondary amps) = rated secondary currenttamps) *100/impedance voltage (%).

orFault level (MVA)= rating (MVA) * 100limpedance voltage (%) . , . (1)This is a simplification which gives slightly conservative(high) values and is applicable to faults located close to thetransformer secondary terminals.Table 1 shows typical impedance voltages of transformersmanufactured to SABS 780.

Table 3Typical cable impedance values (ohms/km at SO Hz) from a manufacturer'scatalogueCond Cable type and insulationcross-sect 600/1 000V PVC 600/1 100V Paper 6,6 kV XLPEmm' Cu Al Cu Al Cu AI25 0,807 1,325 0,807 1,326 0,92335 0,578 0,947 0,578 0,947 0,66350 0,410 0,666 0,410 0,666 0,47070 0,297 0,480 0,297 0,481 0,343 0,54895 0,226 0,358 0,225 0,358 0,260 0,409120 0,185 0,291 0,185 0,291 0,213 0,328150 0,154 0,239 0,155 0,239 0,179 0,267185 0,134 0,202 0,134 0,201 0,153 0,223240 0,113 0,153 0,115 0,167 0,129 0,179

Table 1Impedance voltages of SABS 780 transformers

Table 2Sub-transient reactance values for typical four-pole 6,6 kV sugar millalternators

Wheremore than one transformer is connected in parallel,the total prospective fault current is equal to the sum of theprospective fault currents of the individual transformers.The alternator as source offault currentsThe prospective fault currents of alternators can be cal

culated in a similar manner to transformers. The resistanceof an alternator is low in comparison with its reactance andso the resistance is usually ignored. Instead of the impedance, the reactance then becomes the important quantity infault calculations.Instead of the impedance voltage considered in transformers, the relevant characteristic for alternators is the 'subtransient reactance'. Table 2 shows typical sub-transient reactance values for sugar mill alternators.

Relative impedancesThe percentage impedance of each item of equipment isbased on the apparent power rating (MVA)of that item. Thepercentage impedances of items with different MVA ratingscannot be directly combined. For example, the combinedpercentage impedance of a 10 MVA transformer with 5%impedance, in parallel with a 1 MVA transformer with 5%impedance, is not 2,5%,whichwould be the caseif the transformers were identical. If the equipment items all had thesame MVA rating then the percentage impedances could bedirectly combined. In the method of relative impedances,the MVA ratings of the different items of equipment are allreferred to a common base, usually 100MVA, and the calculated relative percentage impedances can then be directlycombined.Consider a 10MVA transformer with 5%impedance. From(1) the fault level is 10 * 100/5 = 200 MVA. Ifa 100 MVAtransformer is substituted, what must its percentage impedance be to give the same fault level?

From (1), 200 MVA fault level = 100 MVA rating * 100/Xtherefore X = 50%, or 10 times the original impedance.The relative percentage impedance when referred to 100MVA base, is equal to the original percentage impedancemultiplied by the ratio of 100 MVA to the original MVArating. . . . (2)

4,0 to 5,04,5 to 5,55,0 to 6,5

Impedance voltage (%)< 500501 to I 250> I 250

Transformer rating (kVA)

FormulaeThe following formulae apply to the manipulation of relative reactances. The common base has been taken as 100MVA.

Mill Alternator Sub-transientrating reactance(MVA) (%)FX 13125 21AK 500O 20DL 813O 25MS 9060 17,5

Effectofcables on prospective fault currentsUnlike transformers and alternators, cables have substantial resistance as well as reactance. In calculating the reduction in prospective fault current caused by a particular cable

run, the impedance per conductor, being the vector sum ofthe resistance and reactance, must be used. Table 3 providesreactance values for cable types frequently found in sugarmills.

Xp = Percentage reactanceat nameplate MVA rating.Xpr = Relative percentagereactance at 100MVA base.Xo = Reactance in ohms.E = System voltage (kV)(phase-to-phase).N = Actual equipment ratingin MVA.FL = Actual fault level in MVA.From (1), FL = N * 100/XpFrom (2), Xpr = Xp * lOO/N

(3)(4)

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Proceedings of The South African Sugar Technologists' Association - June 1991For cables:Apparent fault power per conductor= phase current (I) phase voltage (Ep)but from Ohm's law, I = Ep/Xotherefore apparent fault power for three conductors= FL = 3 Ep- 2/Xo

but Ep = E / ~ 3therefore FL = E2 /Xo (5)from (5) and (3), Xp = 100 N Xo/E2hence, from (4),Xpr for cables = 100 100 Xo/E2 (6)From 3-phase power circuit basics:Fault current (kA) = FLI(J3 E) (7)Reactances in series are combined by addition, and in parallel by the usual formula: 1/X = 1/Xl + I/X2 + ... +etc. . " (8)In AC circuits, impedance is the vector sum of resistanceand reactance.

ExampleA typical distribution circuit is shown in Figure 3.

The above combined relative impedance is in parallelwiththat of Alternator 2.Let the combined Xpr at point A be XprA;1/XprA = 1/(200 + 50) + 1/(160)XprA = 97,6%From (3),with 100MVA substituted for N,FL at A = 100 100/97,6 = 102,5MVAFrom (7), the prospective fault current at point A is102,5/(1,73 6,6) = 9,0 kATheimpedance of the cable is 0,327 ohms/km, or 0,171ohms for the 500 m run.From (6), Xpr for the cable is 100 100 0,17116,6 2 =39,3%XprB = XprA + Xpr for the cable,= 97,6 + 39,3 = 136,9%From (3),FL at B = 100 100/136,9 = 73,1 MVAFrom (7), the prospective fault current at B is73,1/(1,73 6,6) = 6,4 kA.The cable brings about a substantial reduction in prospective fault current.

FIGURE 3 Typical power system

HRC fuses and current-limiting circuit breakersIf there is not a convenient cable run or transformer toreduce the prospective fault current, current-limiting highrupturing-capacity (HRC) fuses or circuit breakers can beused to perform this duty.One HRC fuse design (GEC 1990) comprises a copperstrip inside the fuse, surrounded by silica granules. The copper strip has a number of sections of reduced cross-sectionalong its length. If the fuse is suddenly subjected to greatlyincreased current, heat builds up rapidly at the sites of reduced cross-sectionwhichmelt simultaneously.Multiple arcsform in series, increasing the resistance to current flowandgenerating high temperatures. The heat fusesthe silica, forming glasswhich is an excellent insulator, and the high currentis interrupted. This interruption takes place very rapidly(less than one quarter of a cycle) and the fault current isinterrupted while it is still building up, and before it attainsits maximum (prospective) value.Circuit breakers with very short operating times are nowalso available to limit prospective fault currents. These circuit breakers are more expensive than fuses,but are quickerand easier to reset, thus reducing the time necessary to restore the supply after tripping. The easeof resetting can leadto abuse by plant operators who may repeatedly reset a circuit breaker which has tripped because they believe the tripping to have been caused bya relatively innocuous overload,rather than by a potentially catastrophic short circuit.

SizingHRCfusesOnce the appropriate fuse type has been chosen from themanufacturer's catalogue, the fuse size can be determinedfrom graphs provided by the manufacturer (see Figure 4).Alternating currents are usually expressed as 'root meansquare' (RMS) values. In the case of a sine wave, the peak

value is J2 (i.e. 1,41) times the RMS value. However underfault conditions the reactance of the circuit can impose adirect-current component on the alternating fault current,increasing the ratio of RMS to peak value of typically 2,26.The 'cut-off' linein Figure4 provides a ratio of2,26betweenthe X-axis values ('prospective current - RMS symmetrical')and the Y-axis ('cut-off current - peak'). This line thereforeprovides a direct relationship between the prospective RMS

No.2 alternator7.5 MVA12% reactance

500mXLPE Cable70mm20.343 ohmsImpedance/km

No.1 alternator10MVA20% reactance

3,3 kVTransformer10 MVA5%

6.6 kV

Fault B73.1 MVA--... ..- .....-6.4 kA

From (3), Xpr of Alternator I = 20 100/10 = 200%.Xpr of Alternator 2 = 12 10017.5 = 160%.Xpr of transformer = 5 100/10 = 50%The impedances of Alternator 1 and the transformer arein series. The combined relative impedance of these itemsis 200% + 50%.

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Proceedings of The SouthAfrican Sugar Technologists' Association - June 1991current and its corresponding peak value. The 'fuse characteristic' line in Figure 4 shows how the peak current varieswith the RMS fault current. .

Typical application ofHRC fusesFigure 5shows a typical sub-station layout with a 1MVA380 V transformer and main distribution board. Two utility

services sub-boards are tapped off the main board, to pro- .vide lighting and small power in the sub-station building.

Transformer1 MVA5% impedence380 V

IIIIII IL J

80 AHRC FUSE

Sub distributionboard

---,IIII1 Switched IILights socket IL _ _outlets..JSub stationutilities

Main distribution board

Prospectivefault20MVA30.4 kA

FIGURE 5 Reduction of fault levels by means of HRC fuses.

One sub-board is tapped directly offthe main board whilethe other is supplied via a fuse.The prospective fault level on the main board busbars is20MVA or 30,4kA. In the direct connection the prospectivefault level is not reduced between the main bars and thesub-board, (ignoring the reactance of the very short cablesto the sub-board). All the circuit breakers on the sub-boardwill therefore have to be able to clear 30,4 kA. This meansthat in practice the five and 20 amp miniature circuit breakers (MCBS) controlling the room lights and socket outletswould be required to clear fault currents of almost 30,4 kA.As the rupturing capacity of these MCBS is usually between two and five kA, they would probably not clear thefault current resulting from a short in the lighting circuit.The MCBS would probably explode causing an even largerfault, which would be transferred to the main board. Seriousplant damage and danger to personnel could result. The second sub-distribution board is supplied via a HRC fusewhichis a simple and cost-effectivewayof reducing the prospectivefault level.

ConclusionsSugar mills are usually expanded in capacity during theirlives, with the addition ofalternators, transformers and largercables. These items all increase the prospective fault currents, possibly to levels greater than the original switchgearcan handle.It is necessary for plant engineers to check the effectsofproposed modifications to the electrical distribution system,

to ensure that the prospective fault currents can be adequately controlled by existing or new switchgear.

500000.0 10

HRC FUSE LINKS

1,0Prospective current kA (R.M.S. symmetrical)

I,I I ,~ 8 III

I Values given apply at 415V I II 550V values are upto 10% greater I6 6 0 ~ values areupto 15% greater1000 _I -750670 -60 -C""" 1#', 450; . """ ...... 355 -250 -~ COi'..... j",. (j )160 Q. -~ V"" E....... 100 s63 ~ - I -40 . -0 32 tila: -~ - 20 Q)sly L..-"'" ~ "" I/)Vr ~ .... V""" Lt -......,. "'" 10V """"", ~ " ' "l / V

4 '::

' S ~tel l. lo" c ' ( l ' 3 ( ~ CV !.

~ j,.o""V.10.1

1 000

100

/..>/.tilQ)Q....c:e 10:;c:I:c;>...:::lU

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FIGURE 4 Fuse sizing diagram

To determine what fuse rating must be used to reduce theRMS prospective fault current from, say 30 kA to 5 kA,determine the peak current equivalent of 5 kA by drawinga vertical line from 5 kA on the X-axis to the cut-off line.This peak current is 11,3 kA.A fuse must be chosen which will not permit more thanthis peak fault current to pass when a prospective fault current of 30 kA is applied. Draw a horizontal line at 11,3 kApeak current, and a vertical line from 30 kA prospectiveRMS symmetrical current. The fuse line below the intersection of these two lines (100 A) is the largest fuse thatshould be used, but any smaller fuse of the same type willalso be suitable.

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1991_Boshoff_Faults in Sugar Mill