Cable Derating and Device Selection for a Roof-Void Pump Feeder
A 400 V, 50 Hz TN-S warehouse extension feeds a 45 kW, 4-pole pump motor (efficiency 0.93, power factor 0.86) through a 65 m run of 4-core XLPE/SWA copper cable. The cable is clipped to a steel tray in an unventilated roof void where the measured air temperature reaches 45 °C, and it shares the tray with three other loaded circuits, all bunched and touching. A 25 mm² cable protected by an 80 A MCCB was installed by the previous contractor. The tabulated current-carrying capacity for 25 mm² under this reference method is 114 A, and the manufacturer's derating factors are 0.87 for 45 °C ambient and 0.75 for four bunched circuits. The motor has tripped on the MCCB three times in the past month, always mid-afternoon. The pump is started direct-on-line with a locked-rotor current of 6.5 times full-load current. The client wants to know whether the cable is adequate and what must change before a second identical pump is added on the same tray.
1. Calculate the motor full-load current and the derated cable capacity, and give a verdict on the existing installation. Explain the afternoon trips.
Read the worked answer
Motor full-load current. I_b = P / (√3 × U × η × cos φ) = 45,000 / (1.732 × 400 × 0.93 × 0.86) = 45,000 / 554.6 = 81.1 A
Derated cable capacity. I_z' = I_tab × k_ambient × k_group = 114 × 0.87 × 0.75 = 74.4 A
Verdict. Two separate non-compliances:
- The cable is overloaded: 81.1 A load on 74.4 A derated capacity, about 9% over. The insulation will age faster; the rule of thumb that insulation life halves for every 8–10 °C of extra conductor temperature applies, and the roof void is already the hottest location in the building.
- The protective device does not protect the cable: the 80 A MCCB rating exceeds the 74.4 A derated capacity, so the condition I_b ≤ I_n ≤ I_z' is not met on either side.
The afternoon trips. The MCCB thermal element is carrying 81.1 A continuously, about 101% of its rating, in a panel whose internal temperature is also elevated. A thermal-magnetic breaker in that condition does not trip immediately; it trips after hours of accumulated heating, which is exactly why the trips occur in the afternoon and not at start-up. This is a genuine overload protection operation, not a fault, and resetting it repeatedly is masking an overloaded cable.
Note on start-up. The magnetic element of an 80 A MCCB is typically fixed around 10 × I_n = 800 A, comfortably above the 6.5 × 81.1 = 527 A starting current, which is consistent with the trips occurring in steady running rather than at start.
2. Select the replacement cable and protective device, and state the additional check required when the second pump is added to the same tray.
Read the worked answer
Cable selection. Try 35 mm², tabulated capacity 141 A under the same reference method: I_z' = 141 × 0.87 × 0.75 = 92.0 A ≥ 81.1 A → acceptable. 25 mm² cannot be made to work by any realistic improvement in installation method short of separating the circuits and ventilating the void.
Protective device. The condition I_b ≤ I_n ≤ I_z' requires: I_n ≤ 92 A, so a 100 A frame MCCB with an adjustable thermal setting (Ir) set to 90 A is suitable; a fixed 100 A device is not, because 100 A > 92 A. Also apply the enclosure derating: a breaker in a 45–50 °C enclosure typically carries only about 0.9 of its rated current, so an Ir of 90 A in a hot panel is marginal. Either improve panel ventilation or step up to a 125 A frame with Ir set at 90 A, which gives thermal headroom.
Motor protection. Set the overload relay to the motor's rated current (about 81 A, or to the motor nameplate service factor), not to the cable capacity, and confirm the relay is ambient-compensated for a 45 °C panel.
Second pump on the same tray. Adding one more loaded circuit changes the grouping factor from four to five bunched circuits. Using a typical five-circuit factor of 0.73: I_z' = 141 × 0.87 × 0.73 = 89.5 A ≥ 81.1 A - still acceptable, but the margin is now only about 10%, so recalculate with the actual tabulated factor for the final arrangement and confirm the tray fill does not exceed its permitted ratio.
Other checks: terminal capacity at the motor and at the MCC, gland size for the larger SWA, bending radius in the roof void, and whether the tray now needs additional supports for the extra cable weight.
3. Calculate the running voltage drop and the starting voltage dip for the 35 mm² cable, and state what should be done if the pump is a positive-displacement type.
Read the worked answer
Cable impedance at operating temperature. For 35 mm² copper: R_20 ≈ 0.524 Ω/km; at 90 °C conductor temperature: R = 0.524 × [1 + 0.00393 × (90 − 20)] = 0.524 × 1.275 ≈ 0.668 Ω/km X ≈ 0.08 Ω/km For 65 m: R = 0.0434 Ω, X = 0.0052 Ω.
Running voltage drop. ΔU = √3 × I_b × (R cos φ + X sin φ) = 1.732 × 81.1 × (0.0434 × 0.86 + 0.0052 × 0.512) = 1.732 × 81.1 × (0.03732 + 0.00266) = 1.732 × 81.1 × 0.03998 ≈ 5.6 V = 1.4% of 400 V - well inside the usual 5% running limit.
Starting dip from this cable alone. I_start = 6.5 × 81.1 = 527 A ΔU_start ≈ 5.6 × 6.5 ≈ 36 V = 9.1%
That is only the cable contribution. The transformer and upstream feeder add their own impedance at 527 A, so the terminal voltage at the motor during starting will be lower than 91% of nominal. Motor torque falls with the square of voltage: at 91% terminal voltage the available torque is about 83% of nominal; at 85% it is about 72%.
What this means by pump type.
- Centrifugal pump: load torque rises with speed squared, breakaway torque is low, and a 9–15% dip is normally tolerated. DOL starting is acceptable, but the transformer dip should still be calculated and the utility's flicker limit checked.
- Positive-displacement pump: high breakaway torque, often 100% or more of rated torque at zero speed. A 15% terminal dip takes away nearly 28% of starting torque and the motor may stall, draw locked-rotor current, and trip on overload - or damage the coupling.
Action if it is positive-displacement: calculate the total starting dip from the transformer to the motor terminals, and if the terminal voltage falls below about 85%, fit a soft starter (or a star-delta starter if the motor has six terminals and the duty allows it) and re-check the starting current, starting torque and acceleration time against the pump curve.