testing numarical relay

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1 www.netaworld.org Fall 201 0 NETA WORLD Feature by Steve Turner,Senior Applications Engineer, Beckwith Electric Co., Inc Dynamic Testing Numerical Protection Relays A numerical protection relay provides multiple functions and typically uses algorithms that do not work on the same principles as older electro- mechanical relays. Often, numerical relays are still tested using outdated techniques developed long ago for electro- mechanical relays. One such common technique is referred to as a static test. Fault voltage or fault current is injected to see if the selected protection function operates at the set point and is in tolerance. Some protection functions have dynamic characteristics, and if a static test is applied, then the true performance of the selected function may not be properly tested. A good example is a mho distance element that is polarized using memorized positive-sequence voltage. Te mho circle expands during an actual fault and has more resistive coverage which is dependent on the apparent source strength. Static testing is only useful to measure the reach point setting along the line angle and the line angle setting itself. A dynamic test usually consists o f a pre-fault condition and a fault condition. able 1 illust rates a simple two-stage dynamic test for a three-phase fault on a distribution feeder. est Signal Pre Fault Fault V A 67 0° volts 53.6 0° volts V B 67 240° volts 53.6 240° volts V C 67 120° volts 53.6 120° volts I A 1 -3° amps 13.4 -80° amps I B 1 237° amps 13.4 160° amps I C 1 -3° amps 13.4 40° amps able 1 D ynamic Tree-Phase Feeder Fault Tis article demonstrates the benets of dynamic testing and shows some examples of how to apply them. Assume that all values shown are secondary unless otherwise noted. Power System Models o perform a dynamic test, you must inject signals the relay would measure during an actual power system fault. Terefore, you must use a simple power system model to dynamically test a numerical protection function. Te following parameters will generate the three-phase fault signals shown in able 1 above. E S is the apparent source voltage. Z S is the apparent source impedance. Z L is the feeder impedance. E S = 1 0° volts line-to-neutral Z S = 1 0° Z L = 4 0° E S I F = Z S + Z L V F = I F Z L Figure 1 Dynamic Mho Characteristic versus Static Mho Characteristic

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Feature

by Steve Turner,Senior Applications Engineer, Beckwith Electric Co

Dynamic Testing NumericalProtection Relays

A numerical protection relay provides multiplefunctions and typically uses algorithms that do

not work on the same principles as older electro-mechanical relays. Often, numerical relays are still testedusing outdated techniques developed long ago for electro-mechanical relays. One such common technique is referredto as a static test. Fault voltage or fault current is injectedto see if the selected protection function operates at the setpoint and is in tolerance. Some protection functions havedynamic characteristics, and if a static test is applied, thenthe true performance of the selected function may not be

properly tested. A good example is a mho distance elementthat is polarized using memorized positive-sequence voltage.Te mho circle expands during an actual fault and has moreresistive coverage which is dependent on the apparent sourcestrength. Static testing is only useful to measure the reachpoint setting along the line angle and the line angle settingitself. A dynamic test usually consists of a pre-fault conditionand a fault condition. able 1 illustrates a simple two-stagedynamic test for a three-phase fault on a distribution feeder.

est Signal Pre Fault FaultV A 67∠0° volts 53.6∠0° voltsV B 67∠240° volts 53.6∠240° voltsV C 67∠120° volts 53.6∠120° voltsIA 1∠-3° amps 13.4∠-80° ampsIB 1∠237° amps 13.4∠160° ampsIC 1∠-3° amps 13.4∠40° amps

able 1 Dynamic Tree-Phase Feeder Fault

Tis article demonstrates the bene ts of dynamic testingand shows some examples of how to apply them. Assumthat all values shown are secondary unless otherwise note

Power System Modelso perform a dynamic test, you must inject signals the

relay would measure during an actual power system faulTerefore, you must use a simple power system modelto dynamically test a numerical protection function. Tefollowing parameters will generate the three-phase faulsignals shown in able 1 above. ES is the apparent sourcevoltage. ZS is the apparent source impedance. ZL is thefeeder impedance.

ES = 1∠0° volts line-to-neutralZS = 1∠0° ZL = 4∠0°

ESIF = ZS + ZL

V F = IF • ZL

Figure 1 Dynamic Mho Characteristic versusStatic Mho Characteristic

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♦ - Static characteristic • - Dynamic characteristic

Figure 1 illustrates the dynamic mho characteristic. Tediameter of the dynamic mho circle is the actual reach plusthe apparent source impedance. Te red shaded area showsthe expanded resistive reach of the expanded characteristic.Te weaker the apparent source, the greater the additionalresistive reach.Figure 1 demonstrates that per orming a test by

checking the static characteristic actually serves no use ul purpose or the case o a mho circle.

Stator Ground Fault Protection UsingSubharmonic Voltage Injection

One hundred percent stator ground fault protectionis provided by injecting a 20 Hz voltage signal into thesecondary of the generator neutral grounding transformerthrough a band-pass lter. Te band-pass lter passes only the 20 Hz signal and rejects out-of-band signals. Te mainadvantage of this protection is 100 percent protection of the stator windings for ground faults, including when the

machine is o -line (provided that the 20 Hz signal is pres-ent). Conventional test methods do not apply to this specialprotection. Tis section shows how to use the power systemmodel to calculate test signals for this protection function.

Figure 2 illustrates a typical application where a 20 Hz voltage signal is impressed across the grounding resistor(R N) by the 20 Hz signal generator. Te band-pass lter only passes the 20 Hz signal and rejects out-of-band signals. Tevoltage across the grounding resistor is also connected acrossthe voltage input (V N) of the 64S relay. Te current input(IN) of the 64S relay measures the 20 Hz current owingon the grounded side of the grounding transformer and isstepped down through a C . It is important to note that therelay does not measure the 20 Hz current owing throughthe grounding resistor. Te 20 Hz current increases duringground faults on the stator winding and an overcurrent ele-ment that operates on this current provides the protection.

Figure 2 20 Hz Injection Grounding Network

Te following series of equations shows how to modand calculate the 20 Hz voltage and current measured the 64S relay.

Grounding Transformer Turns Ratio (N)Assume that the turns ratio of the grounding transform

is equal to:

8,000N = (1)

240Capacitive Reactance

Te total capacitance to ground of the generator statowindings, bus work and delta-connected transformer winings of the unit transformer is expressed as C0. Generatorstep-up transformers have delta-connected windings facithe generator so capacitance on the high side is ignored. Tcorresponding capacitive reactance is calculated as follo

1X c0 = (2)

2 π 0C0

Te capacitive reactance for 1 microfarad is equal to: 1X c0 =

2 • π • (20 HZ) • (10-6 F)= 7,958 primary

Re ect the capacitive reactance to the secondary of grounding transformer:

7,958 Ω 7,958 ΩX c0 = =

N2

= 7.162 secondary

Grounding Resistor (RN )o avoid high transient overvoltage due to ferrores

nance, the ohmic value of the grounding resistor can sized as follows:

X c0R N = (3)3

7,162 ΩsecR N = = 2.387 Ωsecondary 3

A value of 2.5 ohms secondary is used for this example

20 Hz Signal Generator and Band-Pass Filter CharacterAssume that the 20 Hz signal generator outputs 25 vol

Te band-pass lter has a resistance equal to eight ohms

V = 25∠0° volts (4R BPF = 8Ωsecondary (5

8,000 ⁄ 2 ⁄ 240

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Stator Insulation Resistance (RS)R S is the insulation resistance from the stator windings

to ground. A typical value or non ault conditions is 50,000 ohms primary.

50,000Ωpri 50,000ΩpriR S = = N2

= 45 secondary

Current Transformer Te current input (I N) of the 64S relay measures the 20

Hz current owing on the grounded side of the groundingtransformer and is stepped down through a C .

C R = 80/1 (6)

Grounding NetworkNow there are all of the elements needed to mathemati-

cally represent the grounding network and determine the20 Hz signals measured by the 64S relay. Figure 2A showsthe insulation resistance and the stator windings referred tothe primary of the grounding transformer. Figure 2B showsthe insulation resistance and the stator windings referred tothe secondary of the grounding transformer.

8,000 ⁄ 2 ⁄ 240

Figure 2A 20 Hz Grounding Network –Referred to Primary of Grounding ransformer

Figure 2B 20 Hz Grounding Network –Referred to Secondary of Grounding ransformer

20 Hz Current (I N ) Measured by 64S RelayTe current input (I N) of the 64S relay measures the 20

Hz current owing on the grounded side of the groundingtransformer and is stepped down through a C . As notedpreviously, the relay does not measure the 20 Hz curren

owing through the grounding resistor.

Total 20 Hz Current Supplied by Signal Generator Te 20 Hz signal generator looks into the band-pass

lter resistance (R BPF ) which is in series with the parallelcombination of the following:• Zc0• R S• R N

Terefore, the total loop impedance of the 20 Hz ground-ing network can be expressed as follows:

Z = R BPF + R N//R S//Z c0 (7)

Z = 8 + (2.5)//(45)//(- 7.162 j) = 10.135 – 0.706jΩsecondary

Te total 20 Hz current supplied by the signal generator isdetermined as follows:

| V ||I | = | | (8)

| C R • Z |

| 25 ∠0°V ||I | = | | = 30.759 mA

| 80 | • (10.135-0.706jΩ) |

1 |

20 Hz Current Measured by 64S Relay (I N ) during Non-faulted Conditions

Te 20 Hz current measured by the 64S relay is the por-tion of the total current supplied by the signal generator tha

ows into the primary side of the grounding network (Zc0//RS). Calculate the 20 Hz current measured by the 64S relayas follows using the current divider method:

| R N ||I N| = |I • | (9)

| R N + Zc0 // R S |

| 2.5 ||I N| = |30.579 • || 2.5+(–7.162j)//(45) |

= 9.779 mA (Non aulted )

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Figure 3. Black Start at Hydroelectric Plant

do not have the same saturation voltage characteristics, athere is a large DC o set present in the inrush current, ththe current transformers can saturate with restraint curresigni cantly less than two times the nominal relay curr

Generator di erential protection must have special mesures built in to account for C saturation, thus preventiunwanted tripping. An easy method to test C saturatiodetectors is via COM RADE playback. COM RADEplayback is a form of dynamic testing since you are injecactual fault signals recorded during a system event. Tis ivery practical method to check that any protection functipotentially a ected by transformer inrush is properly se

Figure 3 shows the switching con guration for an actblack start. Te step-up transformer is normally energizefrom the 230 kV transmission grid. Tis particular test wperformed to prove that the generating company can piup the station service transformer for essential loads, suas the sluice gate spillway supply, should a system blackoccur. Tis is considered to be a dead bus operation. Tetransmission company opened the step-up transformhigh-side breaker and the generators continued to turn speed with no load. Te plant was attempting to only picup station service. However, the step-up bank had to picked up as well since the plant cannot remotely isolthe step-up transformer.

20 Hz Current Measured by 64S Relay (I N ) during Ground Fault on Stator Windings

A typical value to represent the insulation resistance of the stator windings breaking down during a ground faultis 5,000 ohms primary. If the calculations for equations 7through 9 are repeated for a fault resistance equal to 5,000ohms primary (4.5 ohms secondary), then the 20 Hz currentmeasured by the relay is as follows:

|I N| = 13.486 mA (5,000 ohm primary ground ault )If the calculations for (7) through (9) are repeated for

a fault resistance equal to 1,000 ohms primary (0.9 ohmsecondary), then the 20 Hz current measured by the relay is as follows:

|I N| = 26.640 mA (1,000 ohm primary ground ault )

able 2 summarizes the 20 Hz current measured by therelay for nonfaulted and faulted conditions. Use these valuesto test the function.

RS (primary) |I N| (secondary)

50,000Ω 09.779 mA5,000 Ω 13.486 mA1,000 Ω 26.640 mA

TABLE 2 20 H z C urrEnT M EAsurEMEnTs

Hidden Features and COMTRADEPlayback

Black Start and Generator Diferential ProtectionOften, numerical protection functions have hidden fea-

tures of which the user may not be aware when testing therelay.A good example is how a generator diferential protection

unction provides proper restraint to prevent a misoperationduring a black start . Tis section provides such an examplewhere a backup numerical generator protection relay misop-erated during a black start. Te primary numerical generatorprotection relay properly restrained due to a built-in Csaturation detector. Te current waveforms captured by thenumerical relays can be played back to test that the backuprelay works properly after modifying the protection settings.

Tis playback can also be used for generator relays at otherlocations that also have black start capability.

What is Black Start? A black start is the process of restoring a power station

to operation without relying on external power sources.O -site power source is not available during a wide-areaoutage. Terefore, a black start is required in the absence of grid power. Generator relay current inputs are subjected tohigh levels of dc o set and harmonic current due to ener-gizing the step-up transformer during a black start. If thesystem-side and neutral-side sets of current transformers

Review of the oscillography captured by the relay tmisoperated revealed that there was signi cant mismapresent in A-Phase and C-Phase C secondary currenfrom both the neutral (Iabc) and system (IABC) sides of thgenerator windings. Te mismatch was due to the di erenvoltage class ratings for the neutral and system side CFigure 4A and Figure 4B show the individual di erentcurrents for A-Phase and C-Phase. It can be seen that Ia i

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Figure 4A A-Phase Di erential Current

Figure 4B C-Phase Di erential Current

Figure 4C C-Phase Di erential Current (superimposed)

0.5Minimum

Pickup

1.0

1.5

Differential(Per Unit)

1.0 2.0 3.0 4.0 Bias(Per Unit)

A Phase

C Phase

Slope = 40%

S lope = 10 %

more saturated in comparison to IA if the 9th positive peak inFigure 3A is examined. Figure 4C is the same as Figure 4B,but all three signals (IC, Ic, C_Ph_DIFF) are superimposedonto the same channel.

Te primary numerical generator protection relay phasedi erential protection used for this particular applicationhas a C saturation detection algorithm to improve security.If there is dc o set present, C s can saturate with restraintcurrent signi cantly less than two times nominal.e pri-

mary numerical generator diferential protection automaticallymultiples the slope by our times (see Figure 5) i the total rmsdiferential current is greater than the undamental diferential current. e total rms diferential current calculated by the relaycontains dc and the 1st through 5 th harmonic components.

Figure 5 Generator di erential protection characteristics

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Figure 6 High Impedance Grounded Stator Windings

Figure 7 Arc Detector Logic

Arc-Fault ProtectionIt is extremely di cult to model an arcing fault even with the use of advanced

transient simulation tools such as the Alternate ransients Program (A P).COM RADE playback is an extremely valuable tool for testing arc-fault pro-tection. One customer captured a long event where an undetected intermittentarcing fault eventually evolved into a three-phase fault and completely destroyedthe machine.

Arcing faults can occur due to dirty insulators or broken strands in the gen-erator stator windings. Such faults, if undetected, can lead to overheating alongwith catastrophic electrical failure.ese events typically require extensive repairswith an extended shutdown o the machine . It is highly desirable to detect such aultsat an early or incipient stage so that remedial action can be taken be ore a complete

ailure occurs. A common practice for large synchronous machines is to limit theground fault current through the generator stator windings by grounding theneutral through a distribution transformer with a resistor connected across thesecondary winding. Te neutral resistor is re ected to the primary and provideshigh resistance grounding during single phase-to-ground faults and is typically sized to limit the ground fault current from 3 to 25 amperes primary. Te ohmicvalue of the grounding resistor is selected to avoid high transient voltage due toferroresonance. Figure 6 illustrates how the stator windings are grounded throughhigh impedance using the distribution transformer and grounding resistor acrossthe secondary.

An arcing fault is intermittent, and if the duration of each arc is shorter thanthe conventional 100 percent stator ground fault protection time delay on pickup,then no trip occurs. Figure 7 illustrates the logic for a new method of arc protec-tion. 59N is the neutral overvoltage protection.

Figure 8 shows a typical timinsequence for a trip during an intermitent arcing ground fault.

Te reset timer ( R ) has memor

and stalls the delay on pickup timer (pwhen the initiating function pickudrops out intermittently as is alwaythe case for an arcing fault. Te initiating function that drives this logic 59N. For the purpose of this test, thtime delay on pickup was set equto 18 cycles, and the reset timer wequal to 30 cycles. Set the reset timgreater than the period when the arcing fault is o ; otherwise, the pickutimer will reset prior to a trip. Figure

shows the arcing fault captured by thcustomer. Te relay operated twice andthe variable operating time indicateis a function of both the arc and restimer.

ConclusionsA numerical protection relay pr

vides multiple functions and typicauses algorithms that do not work othe same principles as older electrmechanical relays. Often, relay tepersonnel still want to use outdatetest methods developed long ago felectromechanical relays. est resuobtained from these old test methods may not be a good indication whether or not the numerical relafunctions properly. One such obsoletechnique still often used is referrto as astatic test. Fault voltage or fau

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