Cable Derating, Grouping and a Contractor's 'Equivalent' Substitution
A 400 V submain in a basement plant room runs 90 m from the main switchboard to a mechanical services panel. The design current is 320 A. The contractor proposes one 4-core 185 mm² XLPE/SWA copper cable clipped to a perforated tray. The tray already carries five other loaded circuits, and the plant room ambient is 40 °C. The cable is protected at the main switchboard by a 400 A MCCB with Ir = 0.9 and a 10% tolerance. The client's specification limits voltage drop to 4% from the main switchboard to the load and requires a minimum 25 kA fault withstand for 1 s. The contractor argues that the cable's tabulated rating is 421 A, well above 320 A, and that grouping and ambient factors are already covered by the manufacturer's 'installation factor'. Your firm is the client's independent design reviewer. The contractor has ordered 185 mm² cable and it arrives in three days.
1. Check the 185 mm² proposal against the design current and the 400 A MCCB, showing the correction factors and the required tabulated rating.
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The tabulated 421 A is a reference value, not an installed rating. For XLPE at 40 °C, the ambient correction factor is about 0.87. For six loaded circuits bunched on a tray, the grouping factor is about 0.57; even with spaced cables and a more favourable factor of about 0.75, the result is still inadequate. Using 0.57: Iz = 421 × 0.87 × 0.57 ≈ 209 A. Using 0.75: Iz ≈ 275 A. The design current is 320 A and the MCCB's effective setting is Ir = 0.9 × 400 = 360 A, subject to tolerance. The fundamental rule Ib ≤ In ≤ Iz is therefore breached on both counts: the cable is overloaded at design current, and the MCCB will not protect it. The required tabulated rating is at least 360 / (0.87 × 0.57) ≈ 726 A, which no single 185 mm² cable can provide. A larger conductor or a parallel-cable solution must be recalculated using the actual cable table, not a manufacturer's vague installation factor. I would not accept the 185 mm² cable and would tell the contractor immediately so the order can be changed.
2. The contractor says voltage drop is only about 1.9%, so the cable is fine. Explain why this does not resolve the issue and what voltage-drop check you would still require.
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Voltage drop and thermal capacity are separate design checks and both must pass. Using a typical three-phase mV/A/m value of about 0.25 for 185 mm² gives 0.25 × 320 × 90 / 1000 ≈ 7.2 V, or about 1.8% of 400 V. That may be acceptable on the submain alone, but it does not cure a corrected current rating of about 209 A. The cable would overheat and age prematurely, and the MCCB would not disconnect it before damage. The voltage-drop check should still be done properly: use the tabulated mV/A/m for the actual cable construction and installation method, the design current of 320 A, the actual route length of 90 m, and the load power factor. The check must also consider the whole distribution path from the origin, not just this submain, because the 4% limit applies overall. In addition, check the voltage dip during chiller or motor starting, and check that the selected cable's fault withstand is adequate. If a larger cable is selected, the voltage drop will improve further, so voltage drop is not the driver here; thermal capacity is.
3. The contractor offers to install two parallel 95 mm² cables instead. Evaluate this proposal, including fault withstand for 25 kA for 1 s.
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Parallel cables are permissible only if they are the same material, cross-section and length, follow the same route, and have their current shared predictably. Each 95 mm² cable would carry about 160 A. Its tabulated rating might be around 250 A, but after ambient and grouping factors it falls to roughly 250 × 0.87 × 0.57 ≈ 124 A, below 160 A; the grouping factor applies to each parallel circuit and the bunched group is now larger. The proposal fails thermally unless the cables are well spaced and the calculation proves otherwise. For short-circuit withstand, check the adiabatic equation k²S² ≥ I²t. For 95 mm² copper XLPE, k is about 143, so k²S² ≈ (143 × 95)² ≈ 1.85 × 10⁸ A²s, while 25 kA for 1 s gives 6.25 × 10⁸ A²s. The 95 mm² cable fails the 25 kA 1 s requirement; the 185 mm² cable, at about 7.0 × 10⁸ A²s, just passes. The fault current sharing between parallel cables is also not necessarily equal. I would reject the two 95 mm² proposal and require a correctly sized single cable or parallel cables that pass both thermal and fault-withstand checks.