Broken Delta Applications

APPLICATION Published for the Basler Electric Power Systems Group 

 s   s  e   e  t   t  o  N  #PC-59N01 • August, 2004 

The 59N and Broken Delta Applications  On an impedance grounded system, phase to ground faults are detected by monitoring the zero sequence component, V0, of the line voltages and tripping with a 59N function (high V 0). Two ways relays determine V0 are: 1) Numerically calculated V 0 from phase-gnd  voltages. A relay monitors all phase-gnd voltages from a set of 3 VTs connected Wye-gnd/Wyegnd. The relay calculates V 0 using the equation V0 = (1/3)( VAG + VBG + VCG). 2) Measured 3V 0 from Wye-Gnd/Broken Delta VTs. A relay with a single voltage input monitors the voltage across the th e break in a set of 3 VTs connected Wye-gnd/broken Delta. The voltage across the broken delta is simply the t he sum of system phase-gnd voltages, or 3V 0. The Wye side of the

Wye-Gnd/Broken Wye-Gnd/Broken delta VT can either be directly connected to the high voltage terminals or to the secondary 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 best with) line-gnd voltage anyway, so the inclusion of the V0 calculation and 59N protection requires no additional inputs to the relay and, hence, is easily included. Method 2 is a more historic method that evolved before the era of numeric relays, but it still sees substantial use, likely due to users being comfortable with past practices, its ability to add a small zero sequence load to help stabilize the system neutral, and because it allows a simple relay to be used. The BE1-59N was designed for this application and is an easy relay to use and set.

Figure1:59NBrokenDeltaSchematic 

Besides the BE1-59N, Basler’s BE1-951, -1051, -IPS, and -GPS numeric relays can also monitor a broken delta  voltage via their VX input. Using numeric relays gives all the advantages thereof, such as detailed event reports and programmable logic. The BE1-951, -1051, and -IPS relays also have the ability to perform the 67N function and have the user selectable option of determining direction to a fault by a number of means, including comparing the phase relationship of V0 from the broken delta to I 0 or IG, so that the BE1-951, -1051, and -IPS relays can effectively act as a more modern way to do the work of the classic solid state BE1-67N. In method 2, it is common to place a resistor in the broken delta as shown in Figures 1 and 2. One rationale for the resistance is that the resistance stabilizes the measured  voltage. It does this by a) reducing the risk of  ferroresonance, and b) allowing the VT to act as a very small ground bank. The ground bank effect will help hold the system neutral voltage closer to ground, helping to prevent small leakage impedance to ground from causing  high neutral shifts. The ground bank effect also provides a means for bleeding off the capacitive voltage buildup associated with arcing ground faults on high impedance grounded systems. The ability of the VT and resistor to act as a ground bank and hence stabilize the neutral is fairly limited and will be covered further below. The issue of whether an unloaded VT is at risk of  entering into ferroresonance is difficult to answer. One circuit for ferroresonance can be seen if the delta presents a path for a phase to ground voltage on one phase to energize another phase via the delta, which in turn has a capacitance to ground, creating an LC network where the L is saturable. If the circuit is lossy due to the resistance in the circuit, then resonance of the LC network is less likely, but on the other hand, leaving the delta completely open removes the resonant path, so an argument could be made to leave the resistance out entirely if this is the circuit of  concern. Past practice by several generations of engineers seems to indicate that including the resistance is advisable. This application note will not try to analyze the matter any further than indicated below.

Voltages During a Ground Fault  Refer to the phasor diagrams in Figure 2, Page 3. During ideal normal operation with no ground fault or line to ground current:

During a ground fault, virtually all of the faulted phase's voltage is impressed upon the neutral impedance. For a phase A to ground fault, Van=0, and the voltage across the neutral resistor is essentially the negative of the Phase A to neutral voltage. Mathematically: which means:

which therefore means:

Since a broken delta sums the 3 phase voltages,

From this line of reasoning, one should be able to see that the worst case steady state fundamental frequency  voltage seen at the broke n delta will be:

The VTR in the above equation is best thought of in terms of the winding turns ratio, rather than voltage ratios. This is highlighted because the question of whether one should use VLL or VLN to calculate VTR is avoided if one thinks in terms of winding turns ratio. For an example, if normal system voltage was 13.8kV LL and 7.97kVLG, and a VTR of 115 were used, for a phase A  fault, the worst case broken delta voltage would be:

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

During a fault, secondary voltage on the two unfaulted phases rises to 120V (=13800/115). Note that 69.3 * 3 = 207.9V, which agrees with the earlier calculations for the maximum broken delta voltage during the fault. If the ground fault impedance is high or the source ground impedance is low, the voltage that will arise during a ground fault will be something less than 3 per unit. The calculation of what will occur can be analyzed using a set of  simultaneous equations. The circuit in Figure 2 is analyzed in a Mathcad™ (Revision 7) document, "59N_R#.MCD" found in the "Downloads" section of the Basler Electric web site, www.basler.com.

VT Voltage Rating  It is important to note that, during a ground fault, two of the VTs must withstand and reproduce full line-line  voltage. It would b e an error to use a VT rated only for line-ground voltage. For instance, in the wye-broken delta system described above, the normal system voltage is 13.8kV LL and 7.97kVLG, and secondary voltage is 69.3 under normal operating conditions but rising to 120V during a fault. The VT in this case needs to be rated for 13.8kV/120V.

Figure2:Phasordiagrams duringline-to-groundfault 

Resistance Selection  To obtain the maximum capability of the resistor to dampen system transients and dampen ferroresonant circuits, a typical approach to sizing the resistor is to put in one that can handle all the power that the VTs can supply during a full neutral offset. Fully loading the VT will cause some level of voltage drop and error in the secondary, but if the VT is dedicated to the broken delta configuration, the  voltage drop that will re sult will not cause any i ll effects. However, if the "aux VT" approach in Figure 1 is used, then one might need to investigate if the fully loaded aux VT will pull down the main VT secondary voltage and affect other relaying in the system. Ignoring the voltage drop issue, sizing of the resistor to bring the VTs to a high loading level results in two approaches to sizing the resistor: a) size the resistor so that the amps drawn are equal to the continuous current rating of the transformer bank, or b) if the fault will be cleared quickly, size the resistor so that power in the resistor is equal to the full 3 phase VA of  the transformer bank, which, as seen below, will overcurrent the bank by a factor of sqrt(3).

a) Base resistance on VT continuous current rating  This approach needs to be used if the fault might be in place for an extended period. Assume the previous example where VSEC is 69.3 normally but rises to 120V during a fault. Assume a VT rated at 500 VA per phase and 13.8kV/120V.

During a ground fault, ignoring voltage drop in the VT when fully loaded, the resistor will see 207.9V per the previous 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 an extended period, the resistor must be sized to handle this heat dissipation continuously. b) Base resistance on VT 3 phase VA rating  This approach will overload the VTs and is appropriate only if the fault will be cleared before the overload can affect the VTs. Assume, again, a VT rated at 500VA per phase, for a total of 1500VA for all three phases. Also ignore the voltage drop in the fully loaded VT and assume that the full 207.9V from the previous calculations is seen across the resistor. The resistance required to load the VT bank to 1500VA will be:

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

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

Ground Bank Effect of a VT 

The current drawn during the fault will be:

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

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

Resistor Short Time Power Rating  The power calculated in the above examples is rather large. However, if the fault will be cleared quickly, then the short time rating of the resistor can be used to allow a smaller resistor watt rating. A "rule of thumb" is that a wirewound power resistor can handle a short time overload of:

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The ability of a VT to act as a small grounding bank, and therefore limit neutral voltage shift, is fairly weak, but it might be seen in cases where ground impedances are very high. As a point of comparison, the 28.8 Ω resistor in the previous example reflected to the primary by VTR2 (1152 ) is 381k Ω. This might help stabilize a substation bus or a short transmission line with only minor phase to ground capacitance or leakage resistance. Directly related to this concept, the current flowing in the delta will flow through the faulted phase VT, which will tend to push some current from the VT back into the ground fault. If the ground fault impedance is large, there can be some tendency for the VT to sustain voltage on the faulted phase. The previously mentioned Mathcad document that analyzes Figure 2 may help you to determine if the ground bank effect might be noticable in a ground fault in your system.

For More Information  For further information, visit the download section of  our website at www.basler.com to access product documentation on the BE1-59N, -951, -1051, -IPS, and -GPS, the Mathcad file referenced on Page 2 of this Application Note, and Application Notes on other topics. To discuss your specific application, consult Basler at the factory at (618) 654-2341.

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