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 Lackenby, 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.

Three earthing systems in a line. COAM, the CCGT and PCC site at 0.17 ohms, connects by a 275 kV double-circuit underground cable route to DOWN at 1.24 ohms, which continues by overhead line with a 20-tower chain to Lackenby, the National Grid connection, at 0.07 ohms. The joint bays sit on the cable route about 550 m from COAM, with their own earth grid at 0.71 ohms, and that is where the fault is applied. Two earth continuity conductors of 300 square millimetres run the full cable route and are bonded at every termination; the overhead section has no ECC. The CCGT contributes 9.5 kA at COAM and the National Grid 22 kA arriving via DOWN, 31.5 kA in total, against a design case of 40 kA for 1 second. — Each impedance is a lumped value to true remote earth, not to the neighbouring site. Lackenby’s 0.07 Ω is a combined 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.
Lackenby TOD Point, grid connection	0.07 Ω	A combined figure — it already includes the OHL earth wires and the drainage of the tower chain. Roughly 94 % of it is that drainage, not the substation grid.

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 Lackenby; 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. At COAM the two sources oppose one another, which a single net arrow would hide entirely.

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 Lackenby 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 on its own, 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)

Do it in complex arithmetic, not on magnitudes: an ECC's earth-return impedance is strongly reactive while an earth grid's is predominantly resistive, so adding magnitudes overstates the metallic branch — here, 29.5 % against 26.4 %.

A divider also 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 %: the CCGT's return along the ECC lands on the COAM grid, which is the very grid its star point is earthed into, so it never crosses soil at all. The grid source, earthed at Lackenby, has no such shortcut on this route — and that qualifier is doing a great deal of work. §9 is about what happens when it does.

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 the 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.

There is a second mutual effect in the computed model, and it pulls the other way. Two earthing systems in the same ground are not independent: the model solves them in a shared soil, where this one treats each as a separate impedance to a perfect remote earth. The correction is small and it is worth knowing which way it goes.

Z_soil(a → b) = Z_a + Z_b − 2ρ / 2πD

The soil between two grids adds nothing. An electrode's resistance integral converges — almost all of it sits in the first few metres of ground around it — so by the time you are a few hundred metres away there is nothing left to add. What is left is that mutual term, and ρ/2πD is exactly the self-resistance of a hemisphere of radius D: the whole correction is worth no more than an electrode the size of the separation. Over the 550 m from the bay to COAM it is 0.058 Ω against 0.88 Ω of electrode — and it subtracts, so ignoring it over-states that soil path by about 7 %. The 20.5 % is therefore the net of two mutual effects, not just the conductor coupling above.

That expression is exact for point electrodes in uniform soil, from the same derivation as the hemisphere formula, and it is an approximation for real grids in the two-layer ground a survey actually finds. It is derived here rather than taken from a source, and it has not yet been cross-checked against a computed model — which is the right place to settle it.

So use the hand method for what it is good at. It is the same mechanism as a cable screening factor and it always errs the safe way — less current in the soil than predicted, not more. It gets every sensitivity right and tells you which lever to pull. What it cannot do is produce the number that goes in the report. That needs a model solving all the conductors together along a distributed route, which is what software like CDEGS is for.

5. Building it in the software

A conductor-based solver models electrodes in soil. It has no natural way to say “and over there is a grounding system worth 0.17 Ω that I am not going to model” — which is exactly what you have, three times over, the moment the terminal impedances arrive from the network operator. The standard arrangement, which SES document in SES FAQ 290, uses two conductors and neither of them has to be buried:

Two conductors, and above ground is fine. They are a circuit device, not a physical electrode. Burying them would give them a real earth impedance of their own, which is precisely what you are trying to avoid.
Give the first a GPR energization of 0 V. That pins it at zero potential, which is the definition of remote earth. It becomes the reference the lumped impedance is measured against.
Leave the origin node of that first conductor floating. Nothing connects there — the energization does the work, and a connection would give the current somewhere else to go.
Give the second conductor a lumped impedance. The value you want: 0.17 Ω for COAM, 0.07 Ω for Lackenby, 0.05 Ω for Hartlepool.
Connect one end of the impedance conductor to the END node of the first conductor's first segment — and the other end into your network. Your network now sees a path to true remote earth through exactly the impedance you specified, and nothing else.

The arrangement drawn as a circuit. On the left, conductor 1: a floating origin node, then a conductor carrying a GPR energization set to zero volts, which makes it remote earth. Its far end is the end node of segment 1. From there, conductor 2 carries a lumped impedance of the value wanted, here 0.17 ohms, and its other end goes into a box labelled your network, at the node you want earthed. Neither conductor has to be buried — they are a circuit device, not an electrode. Three checks are set out beneath: whether the figure is a combined one, in which case the earth-wire and tower-chain drainage is already inside it and no chain should be modelled beside it; where the number came from, since an impedance from a one amp injection test is a derived result rather than a measurement; and what it is measured to, since every value is to true remote earth rather than to the neighbouring site. SES document the arrangement in FAQ 290. — How a number from the DNO actually gets into a model that otherwise only knows about buried conductors. Check which convention the number follows before you type it in.

Two things to be careful about before you type a number in. If it is a combined figure, the earth-wire and tower-chain drainage is already inside it — the next section is entirely about that. And a grid impedance obtained from a 1 A injection test, which is what the 0.71 Ω at the bay is, is a derived result rather than a measurement of anything installed: it is what the model says its own electrodes are worth. Say which it is when you report it.

6. A published terminal impedance already contains the chain

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 and the network beyond. Published figures are normally combined — both together. Lackenby is quoted at 0.07 Ω and Hartlepool at 0.05 Ω, and in both cases the substation grid on its own is around 1.21 Ω.

Three ways to represent a terminal that also has an overhead line leaving it. A COMBINED 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, and the same network written out. A combined impedance with a separate tower chain added alongside it: wrong, because the same drainage path is counted twice. Here Lackenby is 0.07 ohms combined against 1.21 ohms of substation grid, which puts about 94 per cent of the drainage through the chain, so counting it twice roughly doubles the path that was already carrying almost all of it. A tower chain on its own is worth about 1.9 ohms, which is 27 times the combined figure, so what dominates a combined number is the wider earthed network rather than the towers. — Both correct forms describe the same physical network. Check which convention a quoted terminal impedance follows before putting it beside anything else.

Work backwards from those two numbers and the second path is worth about 0.074 Ω at Lackenby — which means roughly 94 % of the drainage goes that way rather than through the grid. Hartlepool comes out at 96 %. So modelling a tower chain beside a combined figure does not add a small error. It roughly doubles the path that was already carrying almost all of it, the terminal looks better earthed than it is, more current is drawn away from the fault, and the EPR comes out optimistic.

One thing this rule does not forbid, because it comes up immediately: putting a conductor between two modelled sites. A tie is a series branch from one node to another; a terminal impedance is a shunt from one node to remote earth. They are different circuit elements, and no shunt can stand in for a tie. What would double count is giving that tie its own drainage into tower footings — so it gets none, because that leakage is already inside the combined figure. §9 puts one in.

It is also worth knowing what that 0.074 Ω is not. A chain of spans and tower footings is self-similar, so its impedance is a fixed point:

Z_chain = [ Z_s + √( Z_s² + 4 Z_sR_f ) ] / 2

For a plausible span impedance of 0.3 Ω on 10 Ω footings that converges to about 1.9 Ω — some twenty-five times the figure implied above. What dominates a combined number is not the twenty towers you can see from the fence; it is the wider earthed network they connect to. Two conventions are correct: a combined figure with nothing else modelled, or a grid-only figure with the chain modelled explicitly. Never both. Ask which one a quoted number is before it goes anywhere near a model.

7. 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 — and 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: its clamping voltage is set for lightning and switching transients, far above anything a sustained power-frequency fault produces along a sheath. It never conducts, so there is no closed circuit at 50 Hz and the ECC is the sole metallic return path. Worth stating plainly, because the intuition that a big conductor beside a fault must be carrying some of it is exactly wrong here — and because if those SVLs did conduct, the whole division would change.

8. What actually moves the answer

Five cases compared. As modelled the soil fraction is about 26 per cent. Unbonding the ECC at DOWN raises it slightly to about 31 per cent. Making both terminals as weak as DOWN at 1.24 ohms raises it to about 48 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.
Whether the earth path stops at DOWN — the only one of these that can make the answer better, and it is worth more than most of the others make it worse. It gets its own section.

9. Does the earth path stop at DOWN?

Everything above models the eastern ECC as terminating at DOWN, on DOWN's 1.24 Ω. That is what this route does, and it is the conservative case. But it is an input the study has to establish, not a fact of the world — and it turns out to be worth more than almost anything else on the page.

Carry the ECC on past DOWN and bond it into the Lackenby earthing system, and something changes that no amount of tuning the terminal impedances can achieve: the National Grid's 22 kA — much the larger of the two sources — can reach its own neutral without crossing soil at all, exactly as the CCGT's share already does at COAM. The asymmetry that §3 turned on disappears.

Taking the same 2 × 300 mm² construction as the ECC itself, at 1 km the soil share falls from 23.3 % to 19.9 % and the EPR from 6,613 V to 5,643 V. At a quarter of a kilometre it is 16.3 % and 4,620 V. Push the run out to 20 km and it is back to 23.1 % — the tie is then a worse path than DOWN's own grid, and the answer returns to where it started. Every step of that is in the widget above.

And a second result, which is the more useful one on site. With the run continued, whether the ECC is bonded at DOWN barely matters — 16.3 % against 16.4 % at a quarter of a kilometre. A continuous metallic path to a 0.07 Ω terminal makes the weak terminal's own bond close to irrelevant. Without it, losing that same bond costs you 3.6 points of soil fraction and a thousand volts.

This is not the double count of §6: a tie is a series branch between two modelled nodes, where a terminal impedance is a shunt to remote earth. They are different circuit elements. The tie is given no drainage of its own, because that leakage is already inside Lackenby's combined figure — which is exactly the discipline §6 asks for, applied rather than repeated.

10. What comes out of it

The governing output is the current to earth at the joint bay: the EPR is that current times the bay grid impedance, and the touch and step voltages follow from it. The link box, where the ECC and sheaths terminate, is the only above-ground metalwork at the bay — the perimeter fencing is GRP — so touch voltage is assessed from the 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.

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.

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.

Does it matter whether the earth path continues past DOWN?

A great deal, and it is the one input that can make the answer better rather than worse. The route modelled here stops at DOWN, on its 1.24 \u03a9 grid, and that is the conservative case. Carry the ECC on to Lackenby and the National Grid\u2019s 22 kA \u2014 much the larger source \u2014 reaches its own neutral without crossing soil at all, exactly as the CCGT\u2019s share already does at COAM. On the same 2 x 300 mm\u00b2 construction that is 23.3 per cent to soil falling to 19.9 per cent at 1 km, and 16.3 per cent at a quarter of a kilometre. It is also not a double count of a combined terminal impedance: a tie is a series branch between two nodes, where a terminal impedance is a shunt to remote earth.

How do you put a grounding system of known impedance into a model?

You represent it rather than model its electrodes, using two conductors that do not have to be buried. Give the first a GPR energization set to 0 V, which makes it remote earth, and leave its origin node floating. Give the second a lumped impedance of the value you want. Connect one end of the impedance conductor to the end node of the first conductor\u2019s first segment and the other end into your network. SES document the arrangement in SES FAQ 290. It is how a figure that arrives from the network operator \u2014 0.17 \u03a9, 0.07 \u03a9, 0.05 \u03a9 \u2014 gets into a model that otherwise only knows about buried conductors.

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 one combined impedance which already contains both, or as a grid-only impedance with the chain modelled explicitly beside it. Both are correct. Mixing them counts the same drainage twice, and it is not a small error: Lackenby is published as 0.07 Ω combined against about 1.21 Ω of substation grid, which puts roughly 94 per cent of the drainage through the chain rather than the grid. Add a chain beside that figure and you have doubled the path that was already carrying almost all of it.

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.

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.

Need a joint bay, substation or cable route earthing model built or checked before it goes to the DNO or to National Grid? Get in touch.

Fault Current Division at a Cable Joint Bay