Published for the Basler Electric Power Systems Group #PC-59N01 • August, 2004

APPLICATION

Notes

On an impedance grounded system, phase to groundfaults are detected by monitoring the zero sequencecomponent, V

0, of the line voltages and tripping with

a 59N function (high V0). Two ways relays determine

V0 are:

1) Numerically calculated V0 from phase-gnd

voltages. A relay monitors all phase-gnd voltagesfrom a set of 3 VTs connected Wye-gnd/Wye-gnd. The relay calculates V

0 using the equation

V0 = (1/3)( V

AG + V

BG + V

CG).

2) Measured 3V0 from Wye-Gnd/Broken Delta

VTs. A relay with a single voltage input monitorsthe voltage across the break in a set of 3 VTsconnected Wye-gnd/broken Delta. The voltageacross the broken delta is simply the sum of systemphase-gnd voltages, or 3V

0. The Wye side of the

The 59N and Broken Delta Applications

Figure 1: 59N Broken Delta Schematic

Wye-Gnd/Broken delta VT can either be directlyconnected to the high voltage terminals or to thesecondary of a main stepdown VT.

Method 1 is used by numeric multifunction relays,such as the BE1-951, -1051, -IPS, and -GPS relays.Most applications of such relays need (or work bestwith) line-gnd voltage anyway, so the inclusion of theV

0 calculation and 59N protection requires no

additional inputs to the relay and, hence, is easilyincluded. Method 2 is a more historic method thatevolved before the era of numeric relays, but it stillsees substantial use, likely due to users beingcomfortable with past practices, its ability to add asmall zero sequence load to help stabilize the systemneutral, and because it allows a simple relay to beused. The BE1-59N was designed for thisapplication and is an easy relay to use and set.

which therefore means:

Since a broken delta sums the 3 phase voltages,

From this line of reasoning, one should be able to seethat the worst case steady state fundamental frequencyvoltage seen at the broken delta will be:

The VTR in the above equation is best thought of interms of the winding turns ratio, rather than voltage ratios.This is highlighted because the question of whether oneshould use VLL or VLN to calculate VTR is avoided if onethinks in terms of winding turns ratio.

For an example, if normal system voltage was 13.8kVLLand 7.97kVLG, and a VTR of 115 were used, for a phase Afault, the worst case broken delta voltage would be:

With a VTR of 115, the normal secondary voltageduring unfaulted conditions on each leg of the delta will be:

During a fault, secondary voltage on the two unfaultedphases rises to 120V (=13800/115). Note that 69.3 * 3 =207.9V, which agrees with the earlier calculations for themaximum broken delta voltage during the fault.

If the ground fault impedance is high or the sourceground impedance is low, the voltage that will arise during aground fault will be something less than 3 per unit. Thecalculation of what will occur can be analyzed using a set ofsimultaneous equations. The circuit in Figure 2 is analyzedin a Mathcad (Revision 7) document, "59N_R#.MCD"found in the "Downloads" section of the Basler Electricweb site, www.basler.com.

VT Voltage RatingIt is important to note that, during a ground fault, two

of the VTs must withstand and reproduce full line-linevoltage. It would be an error to use a VT rated only forline-ground voltage. For instance, in the wye-broken deltasystem described above, the normal system voltage is13.8kVLL and 7.97kVLG, and secondary voltage is 69.3under normal operating conditions but rising to 120Vduring a fault. The VT in this case needs to be rated for13.8kV/120V.

Besides the BE1-59N, Baslers BE1-951, -1051, -IPS,and -GPS numeric relays can also monitor a broken deltavoltage via their VX input. Using numeric relays gives all theadvantages thereof, such as detailed event reports andprogrammable logic. The BE1-951, -1051, and -IPS relaysalso have the ability to perform the 67N function and havethe user selectable option of determining direction to a faultby a number of means, including comparing the phaserelationship of V0 from the broken delta to I0 or IG, so thatthe BE1-951, -1051, and -IPS relays can effectively act as amore modern way to do the work of the classic solid stateBE1-67N.

In method 2, it is common to place a resistor in thebroken delta as shown in Figures 1 and 2. One rationale forthe resistance is that the resistance stabilizes the measuredvoltage. It does this by a) reducing the risk offerroresonance, and b) allowing the VT to act as a verysmall ground bank. The ground bank effect will help holdthe system neutral voltage closer to ground, helping toprevent small leakage impedance to ground from causinghigh neutral shifts. The ground bank effect also provides ameans for bleeding off the capacitive voltage buildupassociated with arcing ground faults on high impedancegrounded systems. The ability of the VT and resistor to actas a ground bank and hence stabilize the neutral is fairlylimited and will be covered further below.

The issue of whether an unloaded VT is at risk ofentering into ferroresonance is difficult to answer. Onecircuit for ferroresonance can be seen if the delta presentsa path for a phase to ground voltage on one phase toenergize another phase via the delta, which in turn has acapacitance to ground, creating an LC network where the Lis saturable. If the circuit is lossy due to the resistance in thecircuit, then resonance of the LC network is less likely, buton the other hand, leaving the delta completely openremoves the resonant path, so an argument could be madeto leave the resistance out entirely if this is the circuit ofconcern. Past practice by several generations of engineersseems to indicate that including the resistance is advisable.This application note will not try to analyze the matter anyfurther than indicated below.

Voltages During a Ground FaultRefer to the phasor diagrams in Figure 2, Page 3.

During ideal normal operation with no ground fault or lineto ground current:

During a ground fault, virtually all of the faultedphase's voltage is impressed upon the neutral impedance.For a phase A to ground fault, Van=0, and the voltageacross the neutral resistor is essentially the negative of thePhase A to neutral voltage. Mathematically:

which means:

Figure 2: Phasor diagrams during line-to-ground fault

Resistance SelectionTo obtain the maximum capability of the resistor to

dampen system transients and dampen ferroresonantcircuits, a typical approach to sizing the resistor is to put inone that can handle all the power that the VTs can supplyduring a full neutral offset. Fully loading the VT will causesome level of voltage drop and error in the secondary, butif the VT is dedicated to the broken delta configuration, thevoltage drop that will result will not cause any ill effects.However, if the "aux VT" approach in Figure 1 is used,then one might need to investigate if the fully loaded auxVT will pull down the main VT secondary voltage andaffect other relaying in the system. Ignoring the voltagedrop issue, sizing of the resistor to bring the VTs to a highloading level results in two approaches to sizing theresistor:

a) size the resistor so that the amps drawn are equal tothe continuous current rating of the transformer bank, or

b) if the fault will be cleared quickly, size the resistor sothat power in the resistor is equal to the full 3 phase VA ofthe transformer bank, which, as seen below, willovercurrent the bank by a factor of sqrt(3).

a) Base resistance on VT continuous current ratingThis approach needs to be used if the fault might be in

place for an extended period. Assume the previousexample where VSEC is 69.3 normally but rises to 120Vduring a fault. Assume a VT rated at 500 VA per phase and13.8kV/120V.

During a ground fault, ignoring voltage drop in the VTwhen fully loaded, the resistor will see 207.9V per theprevious example. To limit current to 4.167A,

The power in the resistor will be:

Since it is anticipated that the fault might be held for anextended period, the resistor must be sized to handle thisheat dissipation continuously.

b) Base resistance on VT 3 phase VA ratingThis approach will overload the VTs and is appropri-

ate only if the fault will be cleared before the overload canaffect the VTs. Assume, again, a VT rated at 500VA perphase, for a total of 1500VA for all three phases. Alsoignore the voltage drop in the fully loaded VT and assumethat the full 207.9V from the previous calculations is seenacross the resistor. The resistance required to load the VTbank to 1500VA will be:

The current drawn during the fault will be:

This current is sqrt(3) times the continuous currentrating of the VT, calculated previously. The power in theresistor will be:

In this case, since the VT is overloaded, it is likely thatthe intent is to clear the fault relatively quickly. Since thefault will be cleared quickly, the short time rating of theresistor might be considered, allowing a smaller resistor tobe used.

Resistor Short Time Power RatingThe power calculated in the above examples is rather

large. However, if the fault will be cleared quickly, then theshort time rating of the resistor can be used to allow asmaller resistor watt rating. A "rule of thumb" is that a wire-wound power resistor can handle a short time overload of:

PC-59N01 (08/04)

where ts = time in seconds. Solving for the requiredcontinuous rating:

In the above equation, if t is less than 1s, use t=1, andif t>25s, then it may be best to consider using a fully ratedresistor. If a 28.8Ω resistor is to be used as in the aboveexample, but all faults will be cleared in 5 seconds, then bythe above equation and using a x2 safety margin, a 300Wresistor might be used (1500*(5/50)* (2) = 300).

Ground Bank Effect of a VTThe ability of a VT to act as a small grounding bank,

and therefore limit neutral voltage shift, is fairly weak, but itmight be seen in cases where ground impedances are veryhigh. As a point of comparison, the 28.8Ω resistor in theprevious example reflected to the primary by VTR2 (1152) is381kΩ. This might help stabilize a substation bus or a shorttransmission line with only minor phase to ground capaci-tance or leakage resistance. Directly related to this concept,the current flowing in the delta will flow through the faultedphase VT, which will tend to push some current from theVT back into the ground fault. If the ground fault imped-ance is large, there can be some tendency for the VT tosustain voltage on the faulted phase. The previouslymentioned Mathcad document that analyzes Figure 2 mayhelp you to determine if the ground bank effect might benoticable in a ground fault in your system.

For More InformationFor further information, visit the download section of

our website at www.basler.com to access product docu-mentation on the BE1-59N, -951, -1051, -IPS, and -GPS,the Mathcad file referenced on Page 2 of this ApplicationNote, and Application Notes on other topics.

To discuss your specific application, consult Basler atthe factory at (618) 654-2341.

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