Residential Tower Main Feeder: Demand Factor Versus Derated Ampacity
A 12-storey residential tower in Karaj has 48 identical units, each with an 8 kW connected load (including a 3 kW instantaneous water heater and a 2.5 kW split AC). The consultant's panel schedule totals 384 kW and applies a 0.45 demand factor, giving 172.8 kW. At 400 V, 3-phase, 0.90 power factor, the calculated current is 277 A, so a 300 A MCCB and a 3×120+70 mm² Cu XLPE feeder are proposed. The feeder route is 120 m in a dedicated duct bank, ambient 35 °C, grouped with three other loaded feeders, and the main earthing is TN-C-S. The client wants to value-engineer the feeder to 3×95+50 mm² to save cost, citing that the measured maximum demand in a similar tower was only 190 A. The contractor has already ordered the 95 mm² cable. You are the licensed electrical supervisor asked to approve or reject the substitution before installation.
1. Check the design current and decide whether the 300 A MCCB and the proposed 120 mm² cable are adequate after installation conditions are applied.
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Design current: I = 172.8 kW / (√3 × 0.40 kV × 0.90) = 277 A. A 300 A MCCB is acceptable only if the cable's corrected ampacity is at least 277 A and the breaker setting does not exceed that ampacity. Using IEC 60364-5-52 method D for Cu XLPE, 120 mm² has a base 3-loaded ampacity of roughly 270–300 A. Apply about 0.96 for 35 °C and about 0.75 for four loaded circuits: 270 × 0.72 ≈ 195 A; even an optimistic 300 A base gives about 216 A. The 120 mm² cable is therefore not adequate, and the 300 A breaker is larger than the corrected cable capacity. The 95 mm² cable is worse: its corrected ampacity is roughly 160–180 A. A 240 mm² conductor (base about 420 A × 0.72 ≈ 300 A) is more credible, subject to the actual tabulated values and grouping arrangement. Do not approve the 95 mm² substitution.
2. If the client insists on 95 mm², estimate the voltage drop and explain why voltage drop alone is not the decisive test.
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For 95 mm² Cu, take R at operating temperature as approximately 0.25 Ω/km and X as approximately 0.08 Ω/km. Vd = √3 × 277 × (0.25 × 0.90 + 0.08 × 0.436) × 0.12 ≈ 14.9 V, which is about 3.7% of 400 V. That exceeds a typical 2–3% feeder allowance and leaves little margin for final circuits. For 240 mm² the drop is roughly 7 V, about 1.7%. However, the decisive failure for 95 mm² is ampacity: the corrected current-carrying capacity is far below 277 A, so the cable would overheat even if the voltage-drop result were acceptable. Voltage drop and ampacity are separate checks; passing one does not excuse failing the other.
3. State what you would require before signing any approval, including how to treat the measured 190 A demand in the similar tower.
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Require a signed load calculation showing the demand factor source, unit load basis, and any diversity assumptions. The measured 190 A in another building is not a design basis: it may reflect different occupancy, weather, tariffs, or incomplete fit-out, and it provides no margin for future loads such as EV charging or heat pumps. Require verification of the actual grouped duct-bank arrangement and ambient temperature, a short-circuit withstand check, neutral sizing for harmonic loads, and selectivity between the 300 A MCCB and the upstream utility fuse. Issue a non-conformance for the ordered 95 mm² cable, do not energize, and require either 240 mm² or a documented redesign with a lower verified demand and suitable protection. Update the panel schedule and as-built drawings before approval.