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Cable Sizing — Ampacity, Volt Drop, Short-Circuit: Three Checks, and the Biggest Wins

The MCC schedule that nearly got issued

A 75 kW water-injection pump sits 500 m from its 415 V MCC — out at the far edge of the plot, fed along a hot outdoor tray with five other cables touching it. The draughtsman fills in the cable schedule: full-load current 129 A, so a 70 mm² three-core will do, add a margin, call it 95 mm². It goes out for procurement.

Guess first: how far wrong is 95 mm²? One size? Two?

The cable that actually works is 400 mm² — nearly six standard sizes up, and about four times the copper. Nothing in the process data changed. What changed is which check governs.

The misconception, stated plainly

"Size the cable for the running current. The ampacity table gives you the answer; everything else is a sanity check."

This is true often enough to be dangerous. On a short run inside a substation, fed from a modest fault level, the thermal check really does govern and the table really is the answer. So the habit forms. Then it silently produces the wrong cable in exactly the three situations where it matters:

There are three independent checks, each with its own physics, and the governing one varies from circuit to circuit. The correct size is simply the largest of the three — and on motor circuits there is a fourth, hidden one that beats all of them more often than anyone expects.

S_selected  =  max ( S_ampacity ,  S_voltdrop ,  S_shortcircuit )

Symbol key — every symbol on this sheet

Check 1 — ampacity: the table is not the rating

A cable's thermal limit is the temperature its insulation can sit at forever: 90 °C for XLPE/EPR, 70 °C for PVC. The tabulated current It is whatever current produces that temperature under a specific reference installation. Change the installation and you change the rating:

Iz = It × ca × cg × ci × cs        must satisfy      Iz ≥ Ib
equivalently, required table value:   It ≥ Ib / (ca · cg · ci · cs)

Typical factors (IEC 60364-5-52 family — illustrative values, always use the project's own table):

Effect Condition Factor
Ambient air, XLPE 90 °C (30 °C base) 35 / 40 / 45 / 50 / 55 / 60 °C 0.96 / 0.91 / 0.87 / 0.82 / 0.76 / 0.71
Grouping, single layer touching on perforated tray 2 / 3 / 4 / 6 / 9 cables 0.88 / 0.82 / 0.79 / 0.76 / 0.73
Grouping, bunched/enclosed in conduit 2 / 3 / 4 / 6 / 9 cables 0.80 / 0.70 / 0.65 / 0.57 / 0.50
Soil thermal resistivity, direct buried (2.5 K·m/W base) 1.0 / 1.5 / 2.0 / 3.0 K·m/W 1.50 / 1.28 / 1.12 / 0.90

Two things to notice. The factors multiply, so a hot bunched tray easily lands at 0.6–0.65 — a 35–40 % loss of rating before the cable has carried a single amp. And soil resistivity moves the answer by ±50 %: dry sand around a buried duct bank is a thermal blanket, and the soil thermal resistivity assumed at design stage is one of the least-verified numbers in the whole calculation. Spacing cables apart on the tray, rather than up-sizing them, is very often the cheaper fix — and it is a layout decision, taken early or not at all.

Watch the derating happen — add cables to the tray, raise the ambient, and see the conductors heat up towards their 90 °C limit while the spare capacity bar empties: ▶ open the interactive: electrical cable sizing 3d

Check 2 — volt drop: and the reactance floor nobody expects

For a three-phase circuit, with R and X in Ω/km and L in km:

ΔV = √3 · I · L · (R·cos φ + X·sin φ)            three-phase
ΔV = 2 · I · L · (R·cos φ + X·sin φ)             single-phase (out and back)
ΔV % = 100 · ΔV / Un

Typical limits — check the project electrical design basis, they vary:

Running, feeders + final circuit combined :  5 % of Un    (often 3 % for the feeder alone)
Lighting circuits                          :  3 %
Motor terminals during DOL starting        : 15 % (some specs 10 %); contactors drop out
                                              below roughly 80 % of coil voltage

Now the part that makes motor circuits behave differently from everything else. In the bracket, R and X are weighted by cos φ and sin φ:

That is the reactance floor: you cannot copper your way out of a starting dip. Past a certain size the drop stops falling and the money stops buying anything. The real fixes are a soft starter or VFD (starting current 2–3× instead of 6–7×), a star-delta starter (≈ 2.2×), a higher distribution voltage, or moving the substation. Cable is the last of those options, not the first.

Slide the length, the starter type and the size, and watch which of the three checks wins — the calculator shows all three answers side by side with the governing one flagged, plus the starting-dip indicator: ▶ open the interactive: electrical cable sizing calc

Why X is nearly constant, and why the R term shrinks with temperature too

Reactance per unit length for a three-phase circuit is X = 2πf · (µ₀/2π)·[ln(GMD/GMR) + ¼]·10³ Ω/km, with µ₀/2π = 2 × 10⁻⁷ H/m — the bracket times 2 × 10⁻⁷ is henries per metre, so the 10³ is what converts it to per kilometre (2π·50 · 2×10⁻⁷ · 1.27 · 10³ ≈ 0.08 Ω/km). It is a logarithm of the ratio of the spacing between conductors to the conductor's own geometric mean radius. Inside a multicore cable the conductors are pressed together, so both GMD and GMR grow when the conductor grows, and their ratio hardly changes. A logarithm of a near-constant is a constant: 0.08 Ω/km, essentially forever. Resistance has no such defence — it is ρ·L/S, straight inverse proportionality.

The other half of the story is temperature. Resistance is quoted at 20 °C but the conductor runs hot: R_θ = R₂₀ · [1 + α(θ − 20)] with α ≈ 0.00393 /K for copper. At the 90 °C XLPE limit that is a 27.5 % increase — a real and often-forgotten penalty. Some engineers size volt drop at the 90 °C value (conservative), others at the actual operating temperature estimated from θ = θ_amb + (θ_max − θ_amb)·(Ib/Iz)² (realistic, and can save a size on a lightly loaded feeder). State which one you used; a reviewer cannot tell from the answer alone.

Check 3 — short-circuit withstand: the adiabatic equation

During a fault the conductor heats so fast that no heat escapes into the insulation. All the I²R energy stays in the copper — adiabatic. Equate the energy let through by the protective device to the energy the conductor can absorb before the insulation is destroyed:

I² · t  ≤  k² · S²                   the withstand condition
S_min   =  I · √t / k                the size it demands

k bundles the conductor's heat capacity, resistivity and the allowed temperature excursion:

Conductor / insulation Normal θ Short-circuit θ k
Copper / XLPE, EPR 90 °C 250 °C 143
Copper / PVC (≤ 300 mm²) 70 °C 160 °C 115
Aluminium / XLPE, EPR 90 °C 250 °C 94
Aluminium / PVC (≤ 300 mm²) 70 °C 160 °C 76
Steel wire armour (used as the earth path) — — ≈ 50

Read I·√t as the real driver: halving the clearing time buys the same relief as reducing the fault current by 30 %. Protection settings and cable size are the same conversation. A 0.2 s clearance and a 1.0 s clearance differ by a factor √5 = 2.24 in required area — two or three standard sizes on a large feeder.

Where k actually comes from — and why the fault current is not the same at both ends

k is not a table constant handed down from nowhere. It falls straight out of equating the resistive energy to the conductor's heat capacity:

k = √[ (Qc(β + 20) / ρ20) · ln( (β + θf) / (β + θi) ) ]

with Qc the volumetric heat capacity in the units the formula wants — 3.45 × 10⁻³ J/(K·mm³) for copper, 2.5 × 10⁻³ for aluminium (that is 3.45 and 2.5 J/K·cm³; the 10³ between cm³ and mm³ is the classic way to get k out by a factor of 32) — ρ20 the resistivity at 20 °C in Ω·mm (17.24 × 10⁻⁶ for copper), β the reciprocal temperature coefficient (234.5 for copper, 228 for aluminium), θi the starting temperature (the insulation's normal limit) and θf the permitted short-circuit temperature. Put in copper, 90 °C → 250 °C, and you get 143. Put in aluminium and the lower heat capacity and higher resistivity together give 94. Everything in that formula is material physics; the only engineering decision is θf, which is an insulation property — which is why the same copper has k = 143 with XLPE and k = 115 with PVC.

The second half of the story is where you evaluate the fault. The prospective current at the supply end is set by the source impedance; by the far end of a long cable the cable's own impedance has reduced it substantially — a 500 m 240 mm² run adds roughly 0.05 Ω, which on a 415 V system caps the far-end fault at a few kA regardless of what the MCC could deliver. The withstand check is done at the supply end (highest current). But the disconnection check — will the protection actually see the fault and clear in time — must be done at the far end, where the current is lowest. Two checks, two ends, opposite directions of conservatism.

The adiabatic assumption is valid for roughly 0.1 s ≤ t ≤ 5 s. Faster than that, a current-limiting fuse never lets the full prospective current flow and you must use its published let-through I²t instead. Slower than 5 s, heat genuinely does leave the conductor and the adiabatic answer is conservative (but nobody complains about that).

Worked example — the 500 m pump feeder, all four numbers

Motor 75 kW, 415 V, 3-phase, η = 0.94, cos φ = 0.86, DOL (LRC = 6×, cos φ_start = 0.30)
Route 500 m, 3-core XLPE/SWA copper on perforated tray, 6 cables touching, ambient 45 °C
Fault level 31.5 kA at the MCC, breaker total clearing time 0.20 s
Limits: 3 % running volt drop, 15 % starting volt drop at the motor terminals

Full-load current

Ib = P / (√3 · Un · cos φ · η) = 75 000 / (1.732 × 415 × 0.86 × 0.94) = 129.1 A
I_start = 6 × 129.1 = 774 A

Check 1 — ampacity. ca = 0.87 (45 °C), cg = 0.76 (6 touching on tray) → product 0.661.

It required = 129.1 / 0.661 = 195 A    →   50 mm² (187 A) fails, 70 mm² (238 A) passes
S_ampacity = 70 mm²

Check 3 — short-circuit.

S_min = 31 500 × √0.20 / 143 = 31 500 × 0.4472 / 143 = 98.5 mm²   →   S_sc = 120 mm²

Check 2a — running volt drop. R at 90 °C ≈ 23.4/S Ω/km, X ≈ 0.079–0.082 Ω/km, cos φ = 0.86 → sin φ = 0.510. Limit 3 % of 415 V = 12.45 V.

120 mm²:  ΔV = 1.732 × 129.1 × 0.5 × (0.195×0.86 + 0.080×0.510) = 23.3 V = 5.62 %   ✗
240 mm²:  ΔV = 1.732 × 129.1 × 0.5 × (0.097×0.86 + 0.079×0.510) = 13.9 V = 3.34 %   ✗
300 mm²:  ΔV = 1.732 × 129.1 × 0.5 × (0.078×0.86 + 0.079×0.510) = 12.0 V = 2.89 %   ✓
S_voltdrop(run) = 300 mm²

Check 2b — starting volt drop. I = 774 A, cos φ = 0.30 → sin φ = 0.954. Limit 15 % = 62.3 V.

300 mm²:  ΔV = 1.732 × 774 × 0.5 × (0.078×0.30 + 0.079×0.954) = 66.2 V = 15.96 %   ✗
400 mm²:  ΔV = 1.732 × 774 × 0.5 × (0.058×0.30 + 0.078×0.954) = 61.7 V = 14.86 %   ✓
S_start = 400 mm²   ←  GOVERNS

Answer: 400 mm². And look at what the last step bought: +33 % copper for 1.1 percentage points of starting dip, because at cos φ = 0.30 the reactance term is 81 % of the bracket and refuses to shrink. That is the signal to stop sizing and start re-engineering. Fit a star-delta starter (2.2× → 284 A) and the starting drop at 300 mm² falls to 5.9 %: one size smaller, 1.3 tonnes less copper over 500 m (100 mm² × 3 cores × 500 m × 8 960 kg/m³), and a better answer for the switchgear too.

Final comparison — the four numbers from one circuit:

Check Required size Comment
Running current (the habit) 70 mm² the answer the misconception gives
Short-circuit, 31.5 kA / 0.2 s 120 mm² protection setting is half of this answer
Running volt drop, 3 % 300 mm² length is the driver
Starting volt drop, 15 % 400 mm² the hidden governor — reactance-limited

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