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Worked Example: Motor Circuit Cable Sizing for a 75 kW Escalator Drive — The King's Cross Fire Investigation

Full motor circuit design for a 75 kW escalator drive: cable sizing, starter selection, protection coordination, and voltage drop at starting current. Inspired by the 1987 King's Cross Underground fire investigation.

BS 767122 min readUpdated February 24, 2026
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The Incident: When Maintenance Failures Meet Thermal Degradation

On 18 November 1987, a fire at King’s Cross St Pancras underground station killed 31 people. The fire started in accumulated grease and debris beneath a wooden escalator — ignited by a discarded match that fell through the treads. The subsequent investigation by Desmond Fennell QC revealed systemic maintenance failures throughout London Underground’s escalator systems.

Among the investigation findings was that the escalator motor electrical installations had not been properly maintained for years. Cable insulation degradation from thermal cycling, combined with accumulation of flammable debris near cable terminations, created conditions where a minor ignition source could escalate into a major fire. The motors started and stopped frequently throughout the day (escalator reversals, traffic-controlled starts), and each start imposed a brief but intense thermal pulse on the feeder cable.

A properly designed motor circuit considers not just the cable size for running current, but the thermal environment around the motor, the starting current that heats the cable cyclically, and the short-circuit protection that must clear faults before cable damage occurs. The King’s Cross fire was a failure of maintenance, but the engineering design must account for the reality that maintenance will sometimes be inadequate — margins exist precisely for this reason.

Scenario: 75 kW Escalator Drive Motor

Design the motor circuit for a 75 kW escalator drive in an underground railway station.

ParameterValue
Supply415 V three-phase, 50 Hz
Motor75 kW, 4-pole induction motor, efficiency 0.93, PF 0.86
Starting methodDirect on line (DOL)
Starting current ratioIst/In = 6.5
Starting duration8 seconds to full speed
Starting frequency10 starts per hour (traffic-controlled)
Cable route length60 m (switchroom to escalator pit)
Installation methodMulticore in enclosed trunking, 3 other circuits (Method B1)
Ambient temperature35°C (underground station, above suspended ceiling)
Cable typeXLPE insulated, copper, 90°C rated
Primary standardBS 7671:2018+A2

Step 1: Calculate Full-Load Current (I_FL)

For a three-phase induction motor:

IFL = P / (√3 × V × PF × η) — (Eq. 1)

IFL = 75,000 / (√3 × 415 × 0.86 × 0.93)

IFL = 75,000 / 574.7

IFL = 130.5 A

Step 2: Calculate Starting Current

For DOL starting with Ist/In = 6.5:

Istart = 6.5 × IFL = 6.5 × 130.5 — (Eq. 2)

Istart = 848 A

This starting current lasts for approximately 8 seconds until the motor reaches full speed. During this period, the motor power factor is approximately 0.3 (heavily inductive at locked rotor), rising to 0.86 at full speed.

Step 3: Size Cable for Continuous Running Current

Per BS 7671, Section 552.1, the cable for a motor circuit must be rated for the motor’s full-load current with appropriate derating factors.

Ambient temperature correction (Ca):

From BS 7671 Table C.1, Row 35°C, Column 90°C XLPE:

Ca = 0.96

Grouping correction (Cg):

4 circuits in enclosed trunking (Method B1). From BS 7671 Table C.3, Row: 4 circuits, enclosed:

Cg = 0.65

Combined derating factor:

Ctotal = Ca × Cg = 0.96 × 0.65 = 0.624 — (Eq. 3)

Required tabulated current rating:

It ≥ IFL / Ctotal = 130.5 / 0.624 = 209.1 A — (Eq. 4)

From BS 7671 Table 4D2A, Method B1, XLPE copper multicore:

Cable Size (mm²)Rating (A)Result
50153✗ Too low
70196✗ Too low (196 < 209.1)
95238✓ Passes (238 ≥ 209.1)
120276✓ Passes with margin

Selected on current capacity: 95 mm². But we must verify voltage drop at both running and starting current.

Step 4: Voltage Drop at Running Current

From BS 7671 Table 4E2B for 95 mm² three-phase XLPE copper cable:

r = 0.470 mV/A/m, x = 0.135 mV/A/m

Using the full impedance method at PF 0.86 (cosφ = 0.86, sinφ = 0.51):

ΔV = √3 × IFL × L × (r cosφ + x sinφ) / 1000 — (Eq. 5)

ΔV = 1.732 × 130.5 × 60 × (0.470 × 0.86 + 0.135 × 0.51) / 1000

ΔV = 13,562 × (0.404 + 0.069) / 1000

ΔV = 13,562 × 0.473 / 1000

ΔV = 6.41 V

ΔV% = 6.41 / 415 × 100 = 1.55%

Within the BS 7671 Appendix 12 limit of 5%. PASS.

Step 5: Voltage Drop at Starting Current — Motor Start Viability

At starting current (848 A), the power factor is approximately 0.3 (locked rotor). This is the critical check for motor startability:

ΔVstart = √3 × Istart × L × (r cosφstart + x sinφstart) / 1000 — (Eq. 6)

ΔVstart = 1.732 × 848 × 60 × (0.470 × 0.30 + 0.135 × 0.954) / 1000

ΔVstart = 88,122 × (0.141 + 0.129) / 1000

ΔVstart = 88,122 × 0.270 / 1000

ΔVstart = 23.8 V

ΔVstart% = 23.8 / 415 × 100 = 5.73%

Motor terminal voltage during starting: 415 − 23.8 = 391.2 V

Motor starting torque as a fraction of rated starting torque (IEEE 141, Red Book):

Tstart / Trated-start = (Vterminal / Vrated)² = (391.2 / 415)² = 0.889 — (Eq. 7)

The motor retains 88.9% of rated starting torque. IEEE 141 recommends a minimum of 80% terminal voltage (64% torque). At 94.3% terminal voltage, this motor will start reliably. PASS.

Step 6: Protection Device Selection

Motor circuit protection requires coordination between overload and short-circuit elements per BS 7671, Section 552.1.

MCCB with adjustable thermal-magnetic trip:

SettingValuePurpose
Thermal (overload) trip1.05 × IFL = 137 AOverload protection with Class 10 trip curve
Magnetic (instantaneous) trip10 × In = 1,600 AShort-circuit protection, above starting current peak
MCCB frame rating160 A (adjustable 128–160 A thermal, 1,280–1,600 A magnetic)Standard frame size

Time-current coordination check:

  • At starting current (848 A = 5.3× In): the thermal trip has a 30–40 second delay at this multiple for Class 10. Starting duration is 8 s. ✓ Will not nuisance trip during starting.
  • At stalled rotor (848 A sustained beyond 12 s): thermal trip operates in 10–15 s for Class 10. ✓ Will trip before motor winding damage occurs.
  • At magnetic trip point (1,600 A): instantaneous operation (< 20 ms). This is above any starting transient peak. ✓ Only activates on genuine short circuit.

Step 7: Earth Fault Loop Impedance

For the 160 A MCCB to disconnect under earth fault within the required time per BS 7671, Regulation 411.3.2.3 (5 s for fixed equipment):

Zs(max) from Table 41.3 for 160 A MCCB: 1.5 Ω

Calculated Zs for 95 mm² cable with 25 mm² CPC, 60 m route:

Zs = Ze + (R1 + R2) × L × 1.2 — (Eq. 8)

Where Ze = 0.35 Ω (TN-S supply), R1 = 0.235 mΩ/m (95 mm² at 20°C), R2 = 0.895 mΩ/m (25 mm² CPC at 20°C):

Zs = 0.35 + (0.235 + 0.895) × 0.060 × 1.2

Zs = 0.35 + 0.0813

Zs = 0.431 Ω

0.431 Ω < 1.5 Ω — PASS with large margin.

Step 8: Short-Circuit Withstand (Adiabatic Equation)

Verify the cable and CPC can withstand the prospective short-circuit current:

If = U0 / Zs = 240 / 0.431 = 557 A

For the phase conductor (95 mm², XLPE copper, k = 143):

k²S² = 143² × 95² = 20,449 × 9,025 = 184,552,225 A²s — (Eq. 9)

MCCB clears the 557 A fault in approximately 0.5 s (thermal trip region):

I²t = 557² × 0.5 = 155,125 A²s

155,125 < 184,552,225 — PASS.

For the CPC (25 mm² copper, k = 143):

k²S² = 143² × 25² = 20,449 × 625 = 12,780,625 A²s

155,125 < 12,780,625 — PASS.

Step 9: Cyclic Starting — Accumulated Thermal Stress

This is the check that relates directly to the King’s Cross scenario. With 10 starts per hour, each imposing 848 A for 8 seconds on the cable, the accumulated I²t per hour from starting alone is:

I²tstart-per-hour = n × Istart² × tstart — (Eq. 10)

I²tstart-per-hour = 10 × 848² × 8

I²tstart-per-hour = 10 × 719,104 × 8 = 57,528,320 A²s per hour

The cable’s total adiabatic withstand (from Step 8) is 184,552,225 A²s. In a single hour of operation, the starting events consume:

57,528,320 / 184,552,225 = 31.2% of the cable’s adiabatic capacity

The aha moment: This doesn’t mean the cable fails after 3 hours. The adiabatic equation assumes all energy is absorbed before any heat can dissipate. In reality, the cable cools between starts. But the calculation reveals that the 95 mm² cable is working hard during frequent starting duty. Over years, this cyclic thermal stress accelerates insulation ageing — exactly the degradation observed at King’s Cross. A cable that passes every static check can still degrade prematurely under frequent starting if there is no maintenance regime to detect early insulation deterioration.

Engineering mitigation: For escalators with ≥ 10 starts/hour, consider:

  • Upsizing to 120 mm² (provides 46% more thermal mass)
  • Using a soft starter or VFD to reduce starting current from 848 A to ~200 A (eliminates the cyclic thermal stress almost entirely)
  • Implementing thermal monitoring on the cable route

Result Summary

CheckRequirementActualStatus
Current capacity (derated)≥ 130.5 A148.5 A (95 mm², derated by 0.624)✓ PASS
Voltage drop (running)≤ 5%1.55%✓ PASS
Voltage drop (starting)≤ 20% (motor start)5.73%✓ PASS
Motor starting torque≥ 64% (IEEE 141)88.9%✓ PASS
Earth fault loop≤ 1.5 Ω0.431 Ω✓ PASS
Short circuit (phase)k²S² ≥ I²t184.6M >> 155k A²s✓ PASS
Short circuit (CPC)k²S² ≥ I²t12.8M >> 155k A²s✓ PASS
Cyclic starting stressWithin cable thermal life31.2% adiabatic/hour — marginal▵ REVIEW

Selected cable: 95 mm² XLPE copper, 4-core + 25 mm² CPC, protected by 160 A MCCB (Class 10, Imag = 10× In).

Recommendation: For the high starting frequency (10 starts/hour), strongly consider upsizing to 120 mm² or installing a soft starter to reduce cyclic thermal stress. The 95 mm² cable passes all static checks but the cumulative starting stress warrants engineering review of the motor starting method.

What Would Have Prevented This?

The King’s Cross fire was caused by accumulated debris ignited by a match, not by cable failure. But the investigation revealed that cable insulation degradation from thermal cycling was a contributing factor to the overall risk profile of the underground’s electrical installations:

  • Account for starting frequency in cable selection — the standard current rating tables assume steady-state operation; frequent motor starting imposes additional thermal cycling that accelerates insulation ageing
  • Consider VFDs or soft starters for frequent-start applications — reducing starting current from 6.5× to 1.5× virtually eliminates cyclic thermal stress on cables
  • Implement thermographic inspection schedules — annual infrared surveys of motor terminations and cable routes detect hotspots before insulation failure occurs
  • Design cable routes away from combustible accumulations — escalator pits and trunking near machinery collect debris; cable routes should be cleanable and inspectable
  • Size with margin for maintenance reality — a cable that passes with 5% margin on paper has zero margin after 20 years of thermal cycling; designing with 20–30% margin extends the service life and compensates for real-world conditions

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Frequently Asked Questions

Direct on line (DOL) starting draws 6-8× full load current. Star-delta reduces starting current to approximately 33% of DOL (about 2-2.5× FLC) but produces only 33% of DOL starting torque. For cable sizing, DOL requires a cable that tolerates the full starting current inrush; star-delta allows a smaller cable because the transient current is much lower. However, star-delta introduces a second current peak when the motor transitions from star to delta connection, which the protection must accommodate.
A soft starter (thyristor-based) gradually increases motor terminal voltage over a preset ramp time (typically 5-15 seconds), limiting starting current to 2-3× FLC instead of 6-8× FLC. This reduces the I²t energy during each start by a factor of 4-8, dramatically reducing cumulative thermal stress on the cable. For high-frequency starting applications (escalators, conveyors, pumps with frequent cycling), soft starters or VFDs can extend cable insulation life by decades.
The circuit protective conductor (CPC) only carries fault current for the brief duration until the protective device trips — typically less than 0.5 seconds. It does not carry continuous load current. BS 7671 Table 54.7 allows the CPC to be smaller than the phase conductor: for phase conductors up to 16 mm², the CPC should be the same size; for 25-35 mm², the CPC is 16 mm²; and for phase conductors above 35 mm², the CPC should be at least half the phase conductor cross-section. In this example, half of 95 mm² = 47.5 mm², so 25 mm² appears undersized. However, the adiabatic calculation in Step 8 confirms that 25 mm² is adequate for the specific fault current and clearing time.

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