34.5 kV Grounding Banks, X₀/X₁ & TOV

Look at every transformer winding facing the 34.5 kV collector. The pad-mounts are delta on the 34.5 kV side. In a transmission-connected plant the main power transformer is usually grounded-wye on the HV side and delta on 34.5 kV, which keeps collector ground faults out of the transmission system. Either way, not one plant winding ties the 34.5 kV neutral to earth.

An ungrounded three-phase system still runs balanced, because the neutral "floats" at the centroid of the three voltages. When one phase touches ground, that phase is pinned to 0 V and the whole triangle slides. The two healthy phases now sit a full line-to-line voltage above ground: $\sqrt3 \approx 1.73$ pu, or about 34.5 kV on insulation and arresters chosen for a nominal 19.9 kV phase-to-ground voltage.

This module assumes the configuration of the example study: HV-delta pad-mounts (so no pad-mount is a 34.5 kV ground source) and a distribution-voltage POI at 34.5 kV, so this plant has no MPT; its only 34.5 kV ground reference is on the utility side. Module 1.2 compares this with HV-grounded-wye pad-mounts, which ground every feeder on their own.

Grid-connected, the utility may provide some ground reference through its own station. The dangerous moment comes when the POI breaker opens during a ground fault. For a few hundred milliseconds the inverters keep the collector energized as an island, with no ground source at all. That is the temporary overvoltage (TOV) that destroys surge arresters.

A system is effectively grounded when its coefficient of grounding (the highest sound-phase voltage to ground during an SLG fault, divided by the line-to-line voltage) stays at about 80% or below, at every location and in every credible operating state. IEEE C62.92.1 (and IEEE 142) give the impedance test that ensures it: the driving-point sequence impedances satisfy

The ratios are a sufficient design test; the COG is the quantity they protect (check the exact wording of the edition you cite). The worked example below meets the 80% COG yet fails the ratio test, which is why designers check both:

With an SLG fault on phase A, $I_1 = I_2 = I_0 = E/(Z_1+Z_2+Z_0)$, and with $Z_2 = Z_1$:

If $Z_0 = Z_1$ (solidly grounded and "stiff"), the subtracted term vanishes and the sound phases stay at 1.0 pu. As $Z_0 \to \infty$ (ungrounded) the term tends to 1 and $|a^2 - 1| = \sqrt3$. With a capacitive Z₀ from cable charging, the voltage can even exceed √3.

Grid-connected, zero-sequence current from a collector ground fault can return through the utility's grounded sources: in this example a legacy zig-zag at the 138 kV station, reached through 2.5 miles of 34.5 kV line and cable. When the POI breaker opens, that return path disappears, but the inverters do not stop instantly. Ride-through rules expect them to stay online through grid faults, and anti-islanding detection takes time. For that interval the plant side is:

• energized by the inverters (current-limited sources),
• connected to ground only through the faulted phase and cable charging capacitance,
• so effectively ungrounded, with the sound phases near or above 1.73 pu.

Start with the as-found design (no plant bank) and the 22 kV MCOV arresters: the island destroys them. Enable the bank and sweep its Z₀. Your goal is the highest Z₀ that keeps X₀/X₁ ≤ 3 and R₀/X₁ ≤ 1 in both grid-connected states, while the arresters survive the island and the cable shield, the utility's 51N and the ground grid all stay inside their limits. The islanded ratio is shown for information: it rests on an assumed inverter Z₁ (see the C62.92.6 callout above).

It is tempting to install the stiffest bank you can buy. Once the COG is under about 80%, lowering Z₀ further buys little that the arresters need: in the model, islanded TOV falls from 1.26 pu at 13.8 Ω to 1.16 pu at 9.2 Ω, both well inside the 22 kV MCOV capability. And it costs you elsewhere:

• Fault duty rises. 3I₀ climbs toward the cable concentric-neutral and shield thermal limits, and the bank's own short-time rating (and price) grows with it.
• Ground potential rise moves. IEEE 80 GPR comes from the share of fault current that returns through earth to remote sources (the split factor). For a fault in the substation a stiffer local bank keeps more current local, so the substation's GPR actually falls. For faults out on the collector, at pad-mount ground grids, the bank's current returns through earth and cable neutrals, so a stiffer bank raises earth current and touch voltages there.
• You steal the utility's current. A stiff plant bank becomes the preferred zero-sequence path, so the utility's residual 51N sees less current for resistive faults and loses sensitivity and coordination margin.
• SLG can overtake three-phase duty. If the bank is so stiff that X₀ falls below roughly X₁ (strictly, once $|2Z_1 + Z_0| < 3|Z_1|$), SLG current exceeds the three-phase fault current, and breaker interrupting duty must be checked for the SLG case. IEEE C37.010 permits SLG duty above the rated short-circuit current (up to about 1.15×) only under stated conditions; check the edition you use and the breaker's own ratings. In this example even a 4.6 Ω bank keeps max-source SLG below the three-phase 6,780 A, but check it in your own study.

In the sandbox the grid-connected ratios allow up to about 18.4 Ω/phase, and the utility's 51N needs at least about 12.2 Ω for a 40 Ω fault, so the model's window is roughly 12–18 Ω. The study selected 13.8 Ω/phase, inside that window, on the strength of its EMT results for the island: X₀/X₁ ≈ 2.60 at maximum source and ≈ 2.9 islanded (proxy), with a maximum-source 3I₀ ≈ 4383 A.

Specifying the bank (IEEE C57.32)

• Zero-sequence impedance: 13.8 Ω per phase, the impedance each phase presents to I₀. The full 3I₀ flows in the bank neutral, so specify the impedance per phase (not in neutral ohms) to avoid a factor-of-3 error on the datasheet.
• Short-time rating: the bank's own current is less than the total bus 3I₀. Grid-connected, part of the fault current returns through the utility, so the bank carries about 2,460 A at maximum source. Islanded, the bank is inductive and the cable charging capacitive, so the capacitance circulates extra current through the bank: it carries about 2,570 A, slightly more than the 2,510 A islanded fault current. The larger, about 2,570 A, is the bank's own duty in the model. The example study specified on the total bus 3I₀ instead: the max-source 4,380 A, or the bolted "terminal" bound $3E/|Z_0|$ ≈ 4,320 A, which is the same plant-bus fault with Z₁ taken as zero. Specifying on the total is conservative, and a reasonable choice when future source changes are uncertain. It gave 4,400 A for 10 s, enveloping a 0.75 s backup clearing time with margin for repeated operations.
• Continuous neutral rating: sized for standing neutral current from voltage unbalance and triplen harmonics. IEEE C57.32 tabulates the continuous rating of a 10 s-rated grounding transformer at only a few percent (about 3%) of its short-time rating; confirm the figure in the edition you use. The example project chose to specify 400 A (≈ 9% of the short-time rating) against about 11 A measured in the field. That is a special, more expensive specification, not the standard relationship.

Protecting the bank, and losing it

• Bank protection: phase overcurrent (50/51) on the bank feed, a neutral 51N set below the bank's short-time capability and coordinated with the feeder ground elements, and inclusion of the bank in the 34.5 kV bus differential (87B) or, if it hangs on the MPT LV leads, in the 87T zone (see Module 2.4).
• Loss of the bank: if the bank trips or is taken out of service, the collector is ungrounded again, which is exactly what the bank exists to prevent. Interlock it so that losing the bank blocks or trips the plant (or at least the inverters) rather than leaving it running ungrounded.

Wherever part of the system could end up isolated from its ground bank (a bus section, a feeder with the bank out of service), add zero-sequence overvoltage protection (59G). It measures $3V_0$ from a broken-delta auxiliary VT or from the residual sum of wye-connected bus VTs. On an ungrounded system a ground fault shifts the neutral by a full phase voltage, so 3V₀ approaches 3 pu. That is an unmistakable signal to trip the generation off. On an inverter-fed island that means tripping (or transfer-tripping) the inverters: opening a breaker on the bus side doesn't collapse an island the inverters are still feeding.

The candidate table below is reproduced from the source study as reported. Our simplified sandbox follows the same trends and picks the same bank, but it does not reproduce every value. Applying the sequence equation exactly to the reported impedances gives about 1.37 pu for the as-found minimum-source case, not the 1.52 pu (COG 0.88) quoted. Treat the figures as illustrative and recompute them with your own short-circuit and EMT models.

The gaps are not limited to that row. For the reported islanded X₀/X₁ ratios, the sequence equation gives COG values about 5–8 points lower than the table's (2.94 → about 73%, not 79%). Away from 13.8 Ω the table's islanded ratios and POI 3I₀ also differ substantially from the sandbox (at 27.6 Ω: 4.42 vs about 6.0, and 2,640 A vs about 3,600 A max source). The as-found row mixes states: it quotes the grid-connected X₀/X₁ in the islanded column. Both sources agree that 13.8 Ω gives an islanded X₀/X₁ of about 2.94.