Single Line-to-Ground Fault Data Analysis

Have you ever tried to coordinate feeder relays with the substation transformer overcurrent elements and felt the math didn’t quite line up? It happens because the current seen on the transformer high side is not the same as what the feeder relays measure on the low side. The transformer’s turns ratio and winding configuration reshape the fault current before it reaches the high-side device. Here’s the step-by-step logic I personally use when checking coordination: 1) Understand the transformer connection A common North American distribution substation transformer is high side Delta / low side Yg. Don't forget: the Delta blocks zero sequence current from passing to the high side. 2) Know what each relay is measuring • Low-side feeder relays (phase/ground) measure positive, negative, and zero sequence current on the low-voltage base. • High-side phase overcurrent sees only positive and negative sequence current for a low-side line-to-ground fault because the delta traps I0. 3) Compare currents for the same fault For a single-line-to-ground fault on the feeder: • Feeder current: I(feeder) = I1 + I2 + I0 • High-side current: I(high side) = I1 + I2 • The feeder device responds to the full residual current, while the transformer protection is blind to I0. 4) Identify the tightest point of coordination Surprisingly, it’s not the LG fault. The toughest case is a LL fault near the substation: • Feeder side 50/51P sees about 87 % of the current it would see for a 3ϕ fault. • High-side transformer 50/51P sees nearly the full 3ϕ current because the delta winding passes positive and negative sequence unchanged. If you coordinate the feeder phase time-overcurrent 50/51P pickup and curve to clear before the high-side 50/51P for this LL case, you’ll generally maintain margin for all other fault types (including LG and 3ϕ faults). 5) Verify with actual curves Time-current curves on the low-side feeder relays and the high-side transformer protection must be compared using the converted current magnitudes each will experience. Only then can you be sure the feeder clears before the transformer trips for downstream faults. Real systems complicate this: zero-sequence compensation on feeder relays, different CT ratios, and relay curve shapes can all shift coordination. Questions for the community: • Have you seen feeders miscoordinate because someone forgot the delta blocks zero sequence? • Any lessons from real faults where the high-side transformer protection tripped first? I’d like to hear how others are refining these checks with today’s digital relays and modeling tools (ASPEN Inc., CYME, ETAP Software, EasyPower Software, SKM, etc). Comment or share your experience (or share this post if you found it valuable)!

A 138 kV system supplies a load connected to the 12.47 kV via two step down transformers (xfmrs): (138/34.5 kV & 34.5/12.47 kV, each Dyn1). Connecting them in ‘series’ creates a zero phase shift between the 138 & 12.47 kV systems. An alternative could be to use a 138/12.47 kV (YNyn0). A single-line-to-ground (SLG) bus fault occurs on the 12.47 kV side of xfmr 2. Right-Hand Rule: current into the dot on one side of the xfmr will flow out of the dot on the other side of the same xfmr when comparing the windings that are wound around the same core leg. Ampere-turn balance needs to be maintained. Typically all winding phases (high, low, tertiary, etc.) that are wound around the same core leg are shown “phasiorally” in parallel due to same flux linkage. If fault current (x) flows through the faulted phase (W1), then (x) will be reflected to the 34.5 kV side of xfmr 2 using its ratio. Xfmr 1 ratio=TR1=138/[34.5/sqrt(3)]=6.93 Xfmr 2 ratio=TR2=34.5/[12.47/sqrt(3]=4.79. Let's assume the fault current: x=6000 A. Windings (W1) & (W4) of xfmr 2 are wound around the same core leg. x=6000 A flows out of the dot of (W1) is reflected to (W4) as y=x/TR2=6000/4.79=1252.6 A, which flows into the dot of (W4). y=1252.6 A. 0 A flows through (W2) & (W3) since current flows through the faulted phase only. (W2) & (W5) are wound around the same core leg. 0 A flows through (W2), so 0 A flows through (W5). (W3) & (W6) are wound around the same core leg. 0 A flows through (W3), so 0 A flows through (W6). Using KCL at the node joining (W5) & (W6), 0 A will flow through the line connected to this node as well as through (W9). Since 0 A flows through (W5), (W6), & (W9), then y=1252.6 A flows through (W8) & (W7). (W9) & (W12) are wound around the same core leg. Since 0 A flows through (W9), then 0 A flows through (W12). (W8) & (W11) are wound around the same core leg. y=1252.6 A flows into the dot of (W8), which flows out of the dot of (W11) as (z) based on xfmr 1 ratio. z=y/TR1=1252.6/6.93=180.75 A. z=180.75 A. 2z=2*180.75=361.5 A enters the node joining (W10) & (W11) per KCL. Note: while no zero-sequence current flows on the lines connected to the delta of xfmr 1 since the delta traps zero-sequence current, z=180.75 A flows in each of the two lines & 2z=361.5 A flows through the other line. Similarly, no zero-sequence current flows through the lines connected to the delta of xfmr 2. However, y=1252.6 A flows through two of the lines. The phase/line quantity should not be confused with sequence quantity. In this example fault current (x) is given. However, if fault current is not known, you need positive-, negative-, & zero-phase sequence impedance of all systems as well as xfmrs (T-model) to compute fault current. Pre-fault voltage is also required, which can typically be assumed 1-1.05 per unit. Symmetrical components & sequence networks can then be used to calculate sequence & phase currents, etc. ———————————— Please comment (add/correct) as needed.

𝗠𝗮𝗻𝘆 𝗲𝗻𝗴𝗶𝗻𝗲𝗲𝗿𝘀 𝗿𝘂𝗻 𝗮 𝗳𝗮𝘂𝗹𝘁 𝗶𝗻 𝗘𝗧𝗔𝗣, 𝘀𝗲𝗲 𝗮 𝗻𝘂𝗺𝗯𝗲𝗿, 𝗮𝗻𝗱 𝗺𝗼𝘃𝗲 𝗼𝗻. 𝗬𝗼𝘂 𝗰𝗮𝗿𝗲 𝗮𝗯𝗼𝘂𝘁 𝘄𝗵𝗮𝘁 𝘁𝗵𝗮𝘁 𝗻𝘂𝗺𝗯𝗲𝗿 𝗺𝗲𝗮𝗻𝘀 𝗼𝗻 𝗲𝗮𝗰𝗵 𝘀𝗶𝗱𝗲 𝗼𝗳 𝘁𝗵𝗲 𝘁𝗿𝗮𝗻𝘀𝗳𝗼𝗿𝗺𝗲𝗿. 𝗧𝗵𝗿𝗲𝗲-𝗽𝗵𝗮𝘀𝗲 𝗳𝗶𝗿𝘀𝘁. Here the rule is simple.  • Reflected HV current = LV fault current × (LV/HV ratio).  • Δ–Y, Y–Y, Y–Δ. It does not change the value.  • Positive sequence drives the result. 𝗦𝗶𝗻𝗴𝗹𝗲-𝗹𝗶𝗻𝗲-𝘁𝗼-𝗴𝗿𝗼𝘂𝗻𝗱 𝗶𝘀 𝘄𝗵𝗲𝗿𝗲 𝗳𝗲𝗮𝗿 𝗰𝗿𝗲𝗲𝗽𝘀 𝗶𝗻.  • In Δ–Y, zero sequence loops inside the delta.  • Your HV side does not “see” that earth fault since zero sequence will not cross the delta.  • It appears as a line-to-line fault across two phases.  • Your HV earth-fault relay stay silent while the LV burns due to SLG fault. 𝗧𝗵𝗮𝘁 𝗶𝘀 𝗮 𝘀𝗰𝗮𝗿𝘆 𝗴𝗮𝗽.  • You want clarity, not surprises during utility review.  • You want settings that stand when someone asks “why.”  • You want to stop over-protecting one side and leaving the other exposed.  • Do this in your next study.  • Model grounding honestly.  • Solid, resistance, reactance, NGR.  • Feed the right zero-sequence impedances.  • Create one fault at a time. Single bus. Then read both sides.  • If you fault every LV bus together, you will hide the reflection.  • Track what you are plotting.  • For SLG fault in LV Side, I0 at HV is zero in Δ–Y transformer.  • Switch to phase currents and you will catch the two-phase reflection.  • Watch the healthy-phase voltage.  • With C-factor at 1.1, it lifts toward line value. 𝗗𝗼 𝗻𝗼𝘁 𝗽𝗮𝗻𝗶𝗰.  • It is expected behaviour.  • Use load flow to set OLTC range and PF correction.  • Use short circuit to set short-time duties for cables, CTs, busbars, and breakers.  • Map these to relays.  • Then write it in your report in plain language. Want to learn the basics of etap check this out https://lnkd.in/gCmHu8QB #powerprojects #etap #electricalengineering