Fault Current Division at a Cable Joint Bay: EPR on a 275 kV Route

A joint bay is an unremarkable thing — a concrete chamber on a cable route where the cable lengths are jointed. It has an earth grid, a link box, and no reason to be interesting. But put an earth fault on it and you have to answer a question that decides whether the design passes: how much of that fault current actually goes into the ground here?

The short version: fault current injected into a joint bay earth grid has two ways home — metallically along the earth continuity conductors, or through the soil. Only the soil part raises the earth potential. On this route roughly a fifth goes to soil and four fifths returns in the metal, and getting that fraction right matters far more than getting the fault current right.

1. The route

Three earthing systems in a line. COAM, a CCGT and PCC site, connects by a 275 kV double-circuit underground cable route to DOWN, which continues by overhead line to TOD Point, the National Grid connection. The two joint bays — North and South, sharing a common wall — sit underground on the cable route about 550 m from COAM.

The route drawn with the National Grid infeed on the left. TOD Point, the grid connection at 0.07 ohms, sits beside DOWN at 1.24 ohms, with a 20-tower overhead line running away from TOD into the wider network and no earth continuity conductor on that section. From DOWN the 275 kV double-circuit cable runs to COAM, the CCGT and PCC site at 0.17 ohms, passing the joint bays about 550 m from COAM where the earth grid is 0.71 ohms. Three runs are drawn separately: the 275 kV cores, the two 300 square millimetre earth continuity conductors bonded at every termination over the cable section only, and the overhead line earth wire. The fault is applied from the core to the joint bay earth. — Each impedance is a lumped value to true remote earth, not to the neighbouring site. TOD Point’s 0.07 Ω is a global figure that already contains the tower chain.

Node	Impedance	Basis
COAM CCGT and PCC site	0.17 Ω	Grid impedance. The CCGT generator neutral is earthed here.
South joint bay on the route, ~550 m from COAM	0.71 Ω	Derived from a 1 A injection test in the computed model.
DOWN cable route to overhead line	1.24 Ω	Grid impedance. The weakest terminal earth on the route.
TOD Point NGTP, grid connection	0.07 Ω	A global effective impedance — it already includes the OHL earth wires and the drainage of the tower chain.

Every one of these is a lumped impedance from that node to true remote earth — not to its neighbour. That distinction matters as soon as you start adding them together.

Two sources feed it: the National Grid at 22 kA arriving via DOWN, and the CCGT generator at 9.5 kA whose star point is earthed in the COAM grid. That is 31.5 kA in total, and the design assessment adopts 40 kA for 1 s against a real clearance time of 0.16 s — conservative on both magnitude and duration.

The cable route carries two 300 mm² copper earth continuity conductors over its full length, bonded to the earth grid at COAM, at every joint bay and at DOWN. That makes one continuous low-impedance metallic path spanning the whole route. The overhead section has no ECC at all.

2. Where the fault current actually goes

Apply a single-phase-to-earth fault at the South joint bay. Current arrives along the faulted core from both sources, transfers to the bay's earth grid, and has to get back to the neutral of whichever source drove it — and those neutrals are in different places. The National Grid's is earthed at TOD Point; the CCGT's star point is earthed in the COAM grid.

The ECC does not keep current out of the ground — it chooses where the ground is entered. More enters at COAM than at the joint bay, which is the ECC doing its job.

Metallically, 29 kA returns along the ECC toward COAM and 5.1 kA toward DOWN. The asymmetry is not about ECC length — the two runs here are the same — it is that COAM at 0.17 Ω lets current back into the earth far more readily than DOWN at 1.24 Ω.

Through the ground: 9.3 kA enters the soil at the joint bay, 18.0 kA at COAM, 5.2 kA at DOWN, and 27.9 kA comes back up at TOD Point into the neutral that drove it. As phasors those four balance exactly, which is the check that the picture is complete — though, as below, their magnitudes do not add up and are not meant to.

Look at COAM. More current enters the soil there than at the joint bay — 18 kA against 9.3 kA. That is not a failure; it is the ECC doing exactly what it is for. An ECC does not keep fault current out of the ground. It chooses where the ground is entered — off a chamber in a field and onto a substation with a proper grid, fencing and controlled access. Judge it on where the current ends up, not on whether it stays out of the soil, because it never does.

EPR = I_soil × Z_grid. Not the fault current times the grid impedance. Only the 9.3 kA that enters the soil at the bay raises the potential there — 9.3 kA × 0.71 Ω = 6,613 V — and everything that follows from it, the touch and step voltages, follows from that fraction alone.

One thing worth noticing before you check the arithmetic: 29 + 5.1 + 9.3 does not make 40. Branch currents are phasors. The soil path is essentially resistive and the ECC paths strongly reactive, so their magnitudes do not add. It is the phasor sum that balances — for the ground currents at the four grids just as much as for these three.

3. Reduce it, and it is a current divider

Strip the geography away and the network is three branches from one node to remote earth:

The same network drawn as three branches between the injection node and remote earth: the western ECC plus the COAM terminal impedance, the bay grid impedance alone, and the eastern ECC plus the DOWN terminal impedance. The soil fraction is the return impedance divided by the sum of the return impedance and the bay grid impedance. The calculation is done in complex arithmetic rather than on magnitudes, because the ECC is strongly reactive while the earth grids are predominantly resistive. — Everything the study varies — terminal impedance, ECC bonding, grid impedance, source location — moves one of these three branches.

Z_return = (Z_{ecc W} + Z_COAM) ∥ (Z_{ecc E} + Z_DOWN) soil fraction = Z_return / (Z_return + Z_bay)

One expression, and every sensitivity the study varies moves one of its terms. It is worth doing in complex arithmetic rather than on magnitudes: an ECC's earth-return impedance is strongly reactive while an earth grid's impedance is predominantly resistive, and adding magnitudes overstates the metallic branch. On this route that is the difference between 29.5 % and 26.4 %.

But a divider assumes every return path ends in the same place, and they do not. Solve each source to its own neutral and the answer drops to 23.3 %, because the CCGT's share behaves completely differently: the ECC lands on the COAM grid, which is the very grid its star point is earthed into, so that current returns metallically and never crosses soil at all. The grid source, earthed at TOD, has no such shortcut.

4. What the hand calculation misses

Three numbers, each closer than the last. The three-branch divider gives 26.4 % to soil. Solving each source to its own neutral gives 23.3 %. The computed model gives 20.5 %. None of those is an error — each step adds a mechanism the one before it left out, and the last one is the biggest.

The earth continuity conductor runs about half a metre from the faulted core along the whole route, so flux from the fault current induces a voltage along the ECC that drives current into it. Carson self impedance for a 300 square millimetre conductor with earth return is 0.107 plus j0.736 ohms per kilometre, the mutual impedance at half a metre spacing is 0.049 plus j0.473, and the equivalent earth return depth is 932 metres. The resulting reduction factor of 0.36 means coupling alone captures most of the metallic return before the impedance divider is considered. The hand divider gives 26.4 per cent to soil where the computed model gives 20.5 per cent. — The same mechanism as a cable screening factor, and it always moves the answer the safe way. The hand method gives the shape; the number needs the computed model.

The ECC does not merely offer a low-impedance path. It lies a few hundred millimetres from the faulted conductor for the entire route, so the flux from the fault current induces a voltage along it that drives current into it. Working the earth-return impedances through, the self impedance of a 300 mm² conductor is about 0.107 + j0.736 Ω/km and the mutual impedance at that spacing about 0.049 + j0.473 Ω/km — comparable quantities. The resulting reduction factor of about 0.36 means coupling alone captures most of the return before the impedance divider gets a say.

It is the same mechanism as a cable screening factor, and it always moves the answer the safe way: less current in the soil than the hand method predicts, not more.

So use the hand method for what it is good at. It gets every sensitivity right, it tells you which lever to pull, and it errs pessimistically. What it cannot do is produce the number that goes in the report — that needs a model that solves all the conductors together along a distributed route, which is what software like CDEGS exists for.

5. Why the sheaths carry none of it

There is a great deal of metal in that trench: six single-core 275 kV cables, each with a metallic sheath. None of it carries any of this fault.

The sheaths are single-point bonded — earthed at one end only, with sheath voltage limiters at the free end. An SVL is a surge arrester. It clamps transient overvoltage from lightning and switching, and its clamping voltage is set far above anything a sustained power-frequency fault produces along a sheath. So it never conducts, the sheath has no closed circuit at 50 Hz, and it carries nothing.

The ECC is therefore the sole metallic return path. Worth stating plainly, because the intuition that a big conductor lying next to a fault must be carrying some of it is exactly wrong here — and because if those SVLs did conduct, the whole division would change.

6. What actually moves the answer

Five cases compared. As modelled the soil fraction is about 23 per cent. Unbonding the ECC at DOWN raises it slightly to about 27 per cent. Making both terminals as weak as DOWN at 1.24 ohms raises it to about 34 per cent. Unbonding the ECC at COAM raises it to about 65 per cent, far more damaging than losing the DOWN bond, because COAM at 0.17 ohms was carrying most of the metallic return. Unbonding both ends sends all of the fault current into the soil. — Losing the bond at the STRONG terminal costs far more than losing it at the weak one — which is the whole argument for terminal impedance mattering.

Where the current is sourced — the largest single effect. Sourcing everything beyond DOWN is the bounding case, because the metallic return then has to re-enter the earth through the weakest terminal at 1.24 Ω. Splitting it realistically gives lower voltages, because part of it drains through COAM instead.
The ECC bonds — and asymmetrically. Losing the bond at COAM is far more damaging than losing it at DOWN, because COAM was carrying most of the return. Losing both sends every ampere into the soil.
The terminal earth impedances — a low-impedance terminal draws the return toward itself. This is the whole reason those numbers are worth establishing carefully rather than assuming.
Soil resistivity — it sets the bay grid impedance, and therefore both the soil branch of the divider and the EPR that the soil current produces.
Fault magnitude and duration — magnitude scales the EPR directly; duration does not change the EPR at all, but it changes what is permissible.

7. A modelling trap worth knowing

A terminal with an overhead line leaving it drains two ways: through its own earth grid, and out along the earth wires into the tower chain. There are two correct ways to write that down and one wrong one.

Three ways to represent a terminal that also has an overhead line leaving it. A global effective impedance already contains the earth wires and tower chain in parallel with the grid and nothing else is modelled: correct. A grid-only impedance with the earth wires and tower chain modelled as their own explicit branch beside it: also correct, the same network written out. A global impedance with a separate tower chain added alongside it: wrong, because the same drainage path is counted twice, the terminal looks better earthed than it is, and the earth potential rise comes out optimistic. — Both correct forms describe the same physical network. Check which convention a quoted terminal impedance follows before putting it beside anything else.

Here, TOD Point's 0.07 Ω is a global effective impedance: the earth wires and the 20-tower chain are already inside it. The alternative representation is a grid-only figure of about 1.21 Ω with the tower chain modelled explicitly beside it. Both describe the same physical network and both give the same answer.

Mixing them does not. Take the global figure and then add a tower chain alongside it and the same drainage path is counted twice; the terminal looks better earthed than it is, more current is drawn away from the fault, and the EPR comes out optimistic. It is an easy mistake to make when a number arrives from somewhere else, which is why it is worth asking which convention a quoted terminal impedance follows before putting it in a model beside anything else.

8. What comes out of it

The governing output is the current to earth at the joint bay. Everything else follows: the EPR is that current times the bay grid impedance, and the touch voltage at the link box and the step voltage across the bay area are what a person could actually be exposed to.

The link box, where the ECC and sheaths terminate, is the only above-ground metalwork at the bay. The perimeter fencing is GRP and non-conductive, so there are no other extraneous-conductive-parts to bring into it. Touch voltage is assessed from the link box to a point 1 m away.

The pass criteria are the permissible touch and step voltages for the adopted fault duration, from ENA TS 41-24 and BS EN 50522. Note what is not a criterion: the EPR itself. A high EPR on a site where nobody can be exposed to a dangerous fraction of it is perfectly acceptable — which is the same point the earth potential rise explainer makes from the low-voltage side.

The soil model underneath all of it comes from a resistivity survey — here a two-layer model fitted to 9.678 % RMS error. The bay's own grid is copper tape 25 × 3 mm with 11 earth rods of 16 mm diameter, each 2.4 m long, at 2.827 m depth, 367.73 m of conductor in total — which reaches 0.71 Ω only because the ground there is low-resistivity.

Frequently asked questions

Why does only part of the fault current raise the earth potential?

Because earth potential rise is caused by current passing through the resistance of the soil. Current that returns to its source along a metallic conductor never enters the ground at the fault location, so it contributes nothing to the local rise. EPR is the soil current multiplied by the earthing system impedance, not the fault current multiplied by it.

What decides how the current splits?

The relative impedance of the competing paths. From the injection point the soil path is the bay earth grid impedance; each metallic path is the ECC run plus the terminal earth grid it lands on. The soil fraction is the parallel combination of the metallic paths divided by that combination plus the bay grid impedance. It is an ordinary current divider.

Why do the terminal earth impedances matter so much?

Because the metallic return has to re-enter the earth somewhere. A low-impedance terminal lets it back into the ground easily and therefore draws current toward itself; a high-impedance terminal resists it. On this route COAM at 0.17 Ω takes roughly five times the current DOWN does at 1.24 Ω, even though the ECC runs to each are similar.

Do the cable sheaths carry any of the fault return?

No, and this surprises people. The sheaths are single-point bonded with sheath voltage limiters at the free end. An SVL is a surge arrester: it clamps transient overvoltage, and its clamping voltage is never reached by a sustained power-frequency fault. With no closed circuit at 50 Hz the sheaths stay non-conducting, so the ECC is the sole metallic return path even though there is a great deal of other metal in the trench.

More current enters the soil at COAM than at the joint bay. Is that wrong?

No — it is the ECC working. The ECC carries the grid-sourced share along the route to COAM, and COAM at 0.17 ohms is an easy way back into the earth, so that is where it goes: about 18 kA at COAM against 9.3 kA at the bay. An ECC does not keep fault current out of the ground, it chooses where the ground is entered. Moving it off a chamber in a field and onto a substation with a proper grid, fencing and controlled access is the point of installing one.

The three currents do not add up to the fault current. Why?

Because they are phasors. The soil path is essentially resistive and the ECC paths are strongly reactive, so their magnitudes do not sum arithmetically — 29 plus 5.1 plus 9.3 is not 40. Only the phasor sum balances, and that applies to the ground currents at the four grids too: they cancel as phasors, not as the numbers printed beside them.

Why does a hand calculation not match the computed model?

Because a hand divider treats the ECC only as a low-impedance path, and it is more than that. It runs a few hundred millimetres from the faulted conductor for the whole route, so flux from the fault induces a voltage along it that actively drives current into it. That mutual coupling — the same mechanism as a cable screening factor — captures much of the return before the impedance divider is considered. Here the divider gives about 26 per cent to soil where the computed model gives 20.5 per cent.

So is the hand calculation useless?

Far from it. It gets every sensitivity right and the magnitude approximately, and it tells you which lever to pull. It also errs the safe way: it over-states the current going into the soil, so an EPR estimated this way is pessimistic. What it cannot do is produce the number you put in a report.

What is the worst realistic case?

Sourcing all of the current beyond the weakest terminal earth, because the metallic return then has nowhere easy to re-enter the ground. Splitting it realistically — 22 kA arriving via DOWN and 9.5 kA at COAM — gives lower voltages, because part of the return drains through COAM at 0.17 Ω. Losing an ECC bond at the strong terminal is worse than losing it at the weak one, by a wide margin.

What is the double-counting trap with terminal impedances?

A terminal with an overhead line leaving it drains through its own grid and out along the earth wires into the tower chain. You can model that as a single global effective impedance which already contains both, or as a grid-only impedance with the tower chain as an explicit separate branch. Both are correct and equivalent. Putting a global figure in a model AND adding a tower chain beside it counts the same drainage twice, so the terminal looks better earthed than it is and the EPR comes out optimistic.

What fault current and duration should be assessed?

The design assessment here adopts 40 kA for 1 s against a real total of 31.5 kA and a real clearance time of 0.16 s — conservative on both magnitude and duration. Duration does not change the EPR, but it changes the permissible touch and step voltages, because the body tolerates more for less time.

What is the pass criterion?

The permissible touch and step voltages for the adopted fault duration, from ENA TS 41-24 and BS EN 50522. Those documents set the limits; the model produces the exposures to compare against them. Note that EPR itself has no pass or fail — a high EPR on a site where nobody can be exposed to a dangerous fraction of it can be perfectly acceptable.

What is actually assessed at the joint bay?

The current entering the soil, the EPR that follows from it, the touch voltage at the link box measured from the box to a point 1 m away, and the step voltage across the bay area. The link box is the only above-ground metalwork; the perimeter fencing is GRP and non-conductive, so there are no other extraneous-conductive-parts to consider.

Verify before you rely on it. The current divider on this page is exact given its impedances, but it is a lumped model of a distributed route, and it does not contain the mutual coupling between the faulted conductor and the ECC — so it over-states the soil current and the EPR that follows. The earth-return impedances assume uniform soil of a single resistivity. And nothing here establishes compliance: the permissible touch and step voltages live in ENA TS 41-24 and BS EN 50522, neither of which is reproduced or summarised on this page. This computes the exposures; those documents judge them.

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Fault Current Division at a Cable Joint Bay