Cable Derating and Voltage Drop for a Long Pump Feeder
A 90 kW, 400 V, 0.86 pf, 92% efficient pump is fed by a 240 m run of four-core XLPE/SWA cable installed in a buried duct bank with three other loaded cables. The designer proposes 95 mm2 copper conductors. The cable is grouped with three other circuits (grouping factor 0.75), buried at 0.8 m in soil at 25 C (soil correction 0.95), and the pump runs continuously. The motor starter limits starting current to 3x full-load current. The permitted voltage drop for the installation is 5% at the motor terminals, and the supply authority allows a 2% drop to the origin of the installation, so the cable may drop 3%. The motor is not inverter-driven.
1. Calculate the full-load current and check the derated cable current rating.
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Motor output Pout = 90 kW. Input power Pin = 90 / 0.92 = 97.83 kW.\nFull-load current I = Pin / (sqrt(3) x 400 x 0.86) = 97,830 / (1.732 x 400 x 0.86) = 97,830 / 595.8 = 164.2 A.\n\nA 95 mm2 four-core copper XLPE cable in free air is rated around 250-260 A, but in a buried duct the base rating is lower, typically around 210-230 A depending on installation method. Taking a base tabulated rating of 220 A for the buried ducted case (the designer must use the actual table value):\n\nDerated rating = 220 x 0.75 (grouping) x 0.95 (soil) = 156.8 A.\n\nThis is below the 164.2 A full-load current, so 95 mm2 fails the current-carrying capacity check before voltage drop is even considered. The next size up (120 mm2) should be checked. Note: the exact base rating must come from the applicable cable table for the specific installation method - the principle is that the derated rating must be at least the design current, and here it is not.
2. Check the voltage drop on the proposed 95 mm2 cable and comment on starting.
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Voltage drop for a three-phase run: dV = sqrt(3) x I x L x (R cos(phi) + X sin(phi)).\n\nFor 95 mm2 copper, R at operating temperature is approximately 0.247 ohm/km at 90 C (or about 0.193 ohm/km at 20 C); use the hot value for a continuously loaded cable. X is about 0.08 ohm/km. L = 0.24 km.\n\nR cos(phi) = 0.247 x 0.86 = 0.2124; X sin(phi) = 0.08 x 0.51 = 0.0408. Sum = 0.2532 ohm/km.\n\ndV = 1.732 x 164.2 x 0.24 x 0.2532 = 1.732 x 164.2 = 284.4; x 0.24 = 68.3; x 0.2532 = 17.3 V.\n\n17.3 V / 400 V = 4.3% - this exceeds the 3% allowance for the cable, so 95 mm2 fails voltage drop as well.\n\nStarting: at 3x FLC (492 A) the starting drop scales roughly with current, so about 3 x 4.3% = 12.9% at the motor during start, ignoring motor impedance. That is a deep dip that may affect contactor hold-in and other loads on the same board. The cable must be upsized; a 150 mm2 or 185 mm2 conductor should be evaluated for both drop and derated rating, and the starting dip checked against the contactor's drop-out voltage.
3. Recommend a cable size and state the checks that must accompany it.
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Recommendation: select the smallest standard size that satisfies both the derated current rating and the 3% running drop, with margin for the starting dip. From the calculation, 95 mm2 fails both checks; 150 mm2 (R approx 0.16 ohm/km hot) gives dV = 1.732 x 164.2 x 0.24 x (0.16 x 0.86 + 0.08 x 0.51) = 68.3 x (0.1376 + 0.0408) = 68.3 x 0.1784 = 12.2 V = 3.05%, marginally over; 185 mm2 brings it comfortably under. The designer should tabulate 120, 150 and 185 mm2 against both criteria and choose the first that passes with margin.\n\nAccompanying checks: (1) use the actual tabulated base rating for the ducted, grouped, buried installation, not a free-air value; (2) verify the derated rating at the chosen size exceeds 164.2 A; (3) check the starting dip against contactor drop-out and other sensitive loads; (4) check the earth fault loop impedance and disconnection time for the chosen size and length; (5) confirm the cable's short-circuit withstand against the board's fault level; (6) confirm the pump duty is continuous and no cyclic overload applies. The final choice is governed by the worst of these, not by current alone.