Lathe Circuit: Voltage Drop and Derating in a Shared Conduit
A vocational institute's machine shop has just added a lathe driven by a 3 kW, 230 V, single-phase capacitor-start motor. At full load the motor draws 16.3 A at a power factor of 0.80. The motor is fed from the workshop distribution board through 32 m of 2.5 mm² copper PVC single-core cable pulled into an existing 25 mm conduit that already carries three other single-phase circuits; the contractor fitted a 20 A MCB at the DB. Copper resistivity is to be taken as 0.0175 Ω·mm²/m. The tabulated current rating of 2.5 mm² cable in conduit is 24 A before derating; the grouping factor for four circuits in one conduit is 0.80 and the ambient correction factor at 40 °C is 0.87. The board's no-load voltage is 228 V and the institute's internal standard for a final circuit is a maximum voltage drop of 3 %. During trial runs the motor accelerates slowly and the fluorescent fittings in the same conduit flicker at every start.
1. Calculate the voltage drop on the 2.5 mm² cable at full-load current and state, with figures, whether the circuit meets the institute's 3 % limit.
Read the worked answer
Voltage drop calculation
For a single-phase two-wire circuit the current flows out and back, so the loop resistance is twice the one-way resistance:
Vd = 2 × ρ × L × I / A = 2 × 0.0175 × 32 × 16.3 / 2.5 = 7.30 V
As a percentage of 230 V: 7.30 / 230 × 100 = 3.17 %; measured against the actual 228 V at the board it is 3.20 %.
Judgement: the circuit is marginally non-compliant. The margin is small, so the conclusion must not rest on the drop alone - the cable's derated current capacity and the motor starting current still have to be checked before deciding the remedy. Note that the motor terminal voltage at full load would be about 228 − 7.3 ≈ 221 V, i.e. 4 % below nominal, which is just inside the ±6 % band usually allowed at the terminals but leaves nothing for supply dips.
2. Check the 20 A MCB against the derated current capacity of the cable using Ib ≤ In ≤ Iz, and state the minimum practical change needed.
Read the worked answer
Derated capacity
Iz = 24 × 0.80 × 0.87 = 16.70 A
Coordination check (Ib ≤ In ≤ Iz):
- Ib = 16.3 A
- In = 20 A (MCB fitted)
- Iz = 16.70 A
The condition In ≤ Iz fails: the MCB allows 20 A to flow indefinitely while the cable in that hot, crowded conduit can only carry 16.7 A. The cable would run hot and its insulation would age before the MCB ever operated. Even the margin between Ib (16.3 A) and Iz (16.7 A) is only 0.4 A, so there is no allowance for the motor's service factor or a hot afternoon.
Minimum change: replace the 2.5 mm² run with 4 mm² copper in the same route. The tabulated rating (about 32 A) derated becomes 32 × 0.80 × 0.87 ≈ 22.3 A, which is above the 20 A MCB, and the voltage drop falls to 7.30 × 2.5/4 = 4.56 V (1.98 %). A 20 A type C MCB suits the motor's inrush.
3. Explain the flicker of the fluorescent fittings and the slow acceleration, support your explanation with a starting-current calculation, and list the actions you would take.
Read the worked answer
Why the lights flicker
A capacitor-start induction motor draws roughly 5–7 times full-load current at switch-on. Taking 6 ×:
Ist ≈ 6 × 16.3 = 98 A
Vd(start) = 2 × 0.0175 × 32 × 98 / 2.5 = 43.9 V, i.e. about a 19 % dip at the motor terminals and a smaller but very visible dip at the fittings that share the same conduit and board. The motor therefore accelerates on a depressed terminal voltage, which reduces starting torque (torque falls roughly with the square of voltage) and lengthens acceleration - exactly the symptom described.
Actions
- Upsize the motor circuit to 4 mm² (starting dip ≈ 27.4 V, about 12 %) and preferably give the motor its own conduit and its own way in the DB.
- Keep lighting on a separate circuit, and balance the lighting phase away from the motor phase.
- Fit a type C or D MCB so the inrush does not trip a magnetic release.
- Check the start capacitor and centrifugal switch, since a failed capacitor alone causes slow acceleration and high current.
- Re-measure the board voltage with the motor running; if the dip persists on a 4 mm² circuit, the cause is upstream supply impedance and must be raised with the utility.