Long 230 V Submain: Voltage Drop and Added Load
A 230 V single-phase submain feeds a compressor in a workshop. The route from the main intake to the compressor isolator is 60 m, run as 4 mm² two-core PVC-insulated cable clipped direct. The compressor draws a measured 31 A continuously. At the intake, the voltage is 228 V; at the compressor terminals it is 206 V while running. The cable's tabulated voltage drop is 11.0 mV/A/m. The circuit is protected by a 32 A type C MCB. The workshop owner wants to add a 3 kW fan heater on the same circuit and argues that because the cable is 'rated at 32 A' and the MCB has never tripped, the existing arrangement is fine. The supply is TN-S with Ze 0.4 Ω and the circuit's R1+R2 is 0.28 Ω. The compressor manufacturer requires the terminal voltage to remain within ±5% of nominal.
1. Calculate the voltage drop at the compressor using both the measured voltages and the tabulated mV/A/m figure. Does the circuit meet the 5% limit?
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Measured drop: 228 V − 206 V = 22 V. As a percentage: 22/228 × 100 = 9.6%. Tabulated drop: 11.0 mV/A/m × 31 A × 60 m / 1000 = 20.5 V, which is 20.5/228 × 100 = 9.0%. The 5% limit is 0.05 × 228 = 11.4 V. Both the measured and calculated values exceed 11.4 V, so the circuit fails the voltage-drop requirement. The measured drop is slightly higher than the tabulated value because of supply variation, connection resistance and the motor's starting/load characteristic; it does not indicate a fault, but it confirms the design is marginal. Note that 4 mm² clipped direct is thermally adequate for 31 A; voltage drop, not current-carrying capacity, is the governing constraint here.
2. The owner adds a 3 kW fan heater. Determine the new design current and state whether the existing 4 mm² cable and 32 A MCB can remain. What minimum cable size would comply on voltage drop?
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Heater current = 3000 W / 230 V = 13.0 A. New design current = 31 + 13 = 44 A. The 32 A MCB is now under-rated for the load, and 4 mm² is not adequate. Required mV/A/m = (11.4 V × 1000) / (44 A × 60 m) = 4.32 mV/A/m. A 10 mm² cable at 4.4 mV/A/m gives 4.4 × 44 × 60 / 1000 = 11.6 V, which is still slightly above 11.4 V; 16 mm² at 2.8 mV/A/m gives 7.4 V (3.2%), which complies. The protective device would also need to be increased to suit 44 A, with the cable's tabulated capacity checked after any derating. The better practical solution is a separate heater circuit or a separately protected 16 mm² submain, not simply a larger MCB on the existing 4 mm² cable.
3. Why is 'the MCB has never tripped' not evidence that the circuit is compliant?
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An MCB responds to overload and short-circuit current, not to voltage drop. A 31 A continuous load on a 32 A device leaves almost no margin, and the thermal element has tolerances and ambient-temperature effects; the fact that it has not tripped may simply mean the load is intermittent or the device is running hot without reaching its trip threshold. Voltage drop causes the motor to draw more current for the same output, increases winding temperature, reduces starting torque and can shorten motor life. Compliance is established by design calculation and verified by test, not by absence of tripping. The correct judgement is to calculate the drop, check the protective device rating and cable capacity, and correct the design rather than relying on operational luck.