What the adiabatic check actually asks

When to use this: sizing a circuit protective conductor that is smaller than the line conductors, deciding whether an SWA cable’s armour is adequate as the cpc, checking a reduced cpc in twin and earth, or justifying a conductor size on a design that somebody will inspect.

Every other cable calculation on this site is about normal service — will the cable carry the load without overheating, will the load get the voltage it needs. This one is about the few hundred milliseconds when everything has already gone wrong.

A fault puts thousands of amps through a conductor sized for tens. The protective device will clear it; that is not in question. The question is whether the conductor is still a conductor afterwards, or whether the copper has spent a fifth of a second at 400 °C and the insulation around it has quietly stopped being insulation.

S = √(I²t) ÷ k

S is the minimum cross-section in mm², I the fault current in amperes, t the disconnection time in seconds, and k a constant that carries both the conductor material and the temperature rise it is allowed. It is Regulation 543.1.3 in BS 7671.

What “adiabatic” buys, and what it costs

An adiabatic process is one where no heat enters or leaves the system. Applied here it means assuming that every joule of fault energy goes into raising the temperature of the conductor itself, and none of it escapes into the insulation, the armour, the terminations or the surrounding air.

That is not true. It is, however, conservative, and it is very nearly true for the short durations a fault lasts — which is what makes a one-line equation usable in place of a thermal model. The price is a hard validity limit: about 5 seconds. Past that, enough heat has genuinely left the conductor that the assumption stops being a safe simplification and starts being wrong in a way that depends on the installation, which the equation has no way to represent. The calculator says so when you cross it.

Where k comes from — and why this page derives it

Most calculators offer k as a dropdown of published values. This one computes it:

k = √[ Qc(β + 20) ÷ ρ₂₀ × ln((β + θf) ÷ (β + θi)) ]

Qc is the conductor’s volumetric heat capacity, β the reciprocal of its temperature coefficient of resistivity at 0 °C, and ρ₂₀ its resistivity at 20 °C. θi is the temperature the conductor is at when the fault starts; θf is the highest it may reach.

Read the equation in two halves and the whole subject becomes clear. The first bracket is a property of the metal and never changes — it works out at 226 for copper, 148 for aluminium and 78 for steel. The logarithm is the entire design decision: how hot the conductor already was, and how hot it is allowed to get.

Every published k table is those two temperatures and nothing else. That is why deriving k is both legitimate and more useful than looking it up: a construction the tables do not list still gets a defensible number. The reference page prints the derived value beside the published one for every row of the standard, so you can check the claim rather than take it.

The mistake that changes the answer by a quarter

There are three k tables, and the difference between them is only θi — the temperature the conductor starts at.

• A line conductor is carrying its load when the fault arrives, so it starts at the cable’s rated operating temperature: 70 °C for thermoplastic, 90 °C for thermosetting.
• A protective conductor inside a cable, or bunched with one, carries no load current itself but is heated by the conductors around it, so it starts at the same temperature they do.
• A protective conductor run separately carries nothing and is next to nothing, so it sits at ambient — taken as 30 °C.

Same copper, same insulation, same fault. Inside the cable, k = 115. Run on its own, k = 143. That is a 20% difference in the conductor you need, in both directions: read the separate-conductor row for a cpc that is actually inside the cable and you will undersize it.

The headroom is real but conditional. A cpc is only “separate” while it genuinely is — pull it through the same conduit as the loaded conductors and it is heated by them, and the bunched figure applies again.

Steel armour as the cpc

On an SWA run the armour is very often the circuit protective conductor, and this is the calculation that decides whether it may be. Steel’s k is roughly a third of copper’s — 52 against 143 for a 90 °C thermosetting cable — so the armour needs about three times the cross-sectional area to survive the same fault.

On larger cables it comfortably has that. On smaller ones it frequently does not, which is the reason a separate cpc gets pulled in alongside an SWA cable that appears to have a perfectly good earth path already. Take the armour’s cross-sectional area from the manufacturer’s data for the specific cable — it is published, and it is not proportional to the conductor size in any way you can guess.

Disconnection time, or let-through energy?

This is where most adiabatic calculations go wrong, and always in the expensive direction.

The obvious method is to read the prospective fault current at the point of the fault, look up the disconnection time on the device’s time/current characteristic, and multiply. For a slow device that is right. For a current-limiting device it is badly wrong: the device interrupts the fault before the current reaches its first peak, so the conductor never experiences the full prospective current, and never for the full time you read off the curve.

Manufacturers publish what actually gets through as a let-through energy, in A²s, per device and per prospective fault level. Use it whenever it exists. The two presets on this page show the same 1.5 mm² cpc failing on current-and-time and passing on let-through energy — a three-fold difference in required conductor, for one circuit, from choosing the right input.

If the conductor fails

Four options, roughly in order of cost:

1. A faster or current-limiting device. Usually the cheapest fix, and it attacks the actual problem — the energy reaching the conductor — rather than compensating for it.
2. Run a separate cpc rather than relying on the armour or a reduced conductor inside the cable. Worth a quarter on its own.
3. A larger conductor. Always works, always costs.
4. Reduce the prospective fault current. Effective, and almost always outside your control.

What passing this does not prove

Surviving the fault is one requirement out of several, and it is easy to treat it as the whole job because it produces a satisfying number. A protective conductor also has to give an earth fault loop impedance low enough that the device disconnects within the required time — a separate calculation with a separate failure mode — and where the Table 54.7 route is used it must meet that table’s minimum size regardless of what the adiabatic equation says.

The adiabatic equation is the alternative to Table 54.7 for sizing. It is not an alternative to the disconnection time requirement, and nothing on this page evaluates that.

How this calculator is checked

Technical reviewer

Salamot Hok

Electrician · 10+ years of installation work in Bangladesh and the wider South Asian region

He reads the result the way an installer would: are the defaults values people actually meet, does the warning fire where you would stop and think, and is the answer something you could buy and fit? The code figures themselves come from the published standards cited below, not from him — that boundary is set out on his profile.

• The maths lives in a pure function with its own test suite, asserted against worked examples from published references and standards. A calculator does not ship until those tests pass.
• 4 sources cited by name and linked, so any figure on the page can be traced back to the document it came from.
• Last reviewed 25 August 2026. Review dates are advanced only when the page is actually re-read, never to look fresh.
• Unusable input returns no answer. Where the inputs do not describe a real design, the calculator says so and withholds the number rather than printing a plausible-looking wrong one.

The full process is written up in the methodology and editorial policy. Results are engineering guidance, not a code sign-off — see the disclaimer. If a result looks wrong, tell us; corrections are answered before anything else.