Adding a 2 kW Boiler to a 16 A Tailoring Workshop Branch
A tailoring-cum-embroidery unit in Ranchi draws all its power from a single 230 V, 50 Hz single-phase branch protected by a 16 A miniature circuit breaker. Connected to that branch are six sewing machines of 350 W each, two steam irons of 1000 W each, four fluorescent tube fittings of 40 W each, and one motor-driven overlock machine rated 1.5 kW input with a power factor of 0.75. The owner now wants to add a 2 kW steam boiler on the same branch. The local electrician approves the addition, arguing that '16 A at 230 V is 3680 W, and the boiler is only 2000 W, so plenty of capacity is still left.' The unit runs nine hours a day, and in the peak season almost every machine and both irons run together for long stretches. The wiring is 2.5 mm² copper concealed in PVC conduit.
1. Calculate the existing branch current and decide whether the 16 A breaker is adequate even before the boiler is connected. State your assumptions.
Read the worked answer
Treat the heaters, irons and lamps as resistive (unity power factor): Sewing machines: 6 × 350 = 2100 W Irons: 2 × 1000 = 2000 W Tube fittings: 4 × 40 = 160 W Resistive subtotal = 4260 W → I = 4260 / 230 = 18.5 A.
Overlock motor: 1500 W at pf 0.75 → apparent power = 1500 / 0.75 = 2000 VA → I = 2000 / 230 = 8.7 A.
Total ≈ 18.5 + 8.7 = 27.2 A. Adding currents arithmetically ignores the phase difference between motor and resistive loads, so it slightly overstates the true rms current - but the true value is still well above 16 A. Hence the breaker is already inadequate, and tripping is guaranteed whenever the machines and both irons run together. The 80% continuous-load rule makes it worse: a 16 A breaker should carry only about 12.8 A continuously, roughly half the present demand.
2. Identify the specific error in the electrician's reasoning and list, in priority order, what should be done.
Read the worked answer
The error is a category mistake: 16 A × 230 V is the maximum the breaker will pass, not spare capacity. The electrician checked the boiler against the breaker's rating while ignoring (i) the load already connected, (ii) the motor's power factor, (iii) the derating of a breaker under continuous load, and (iv) whether the 2.5 mm² conductor and conduit can carry the resulting current.
Priority of action:
- Do not connect the boiler to this branch. Give it a dedicated circuit with its own MCB, sized for 2000 W / 230 V = 8.7 A (typically a 16 A MCB on 2.5 mm²).
- Split the existing branch - irons on one circuit, machines on another - so no single circuit carries more than its cable rating.
- Verify earthing and residual-current protection, essential with water-heating equipment in a workshop.
- Only then consider extra machines. Watt-by-watt addition without a load schedule is how overload fires start.
3. The owner proposes correcting the motor power factor from 0.75 to 0.90 instead of rewiring, to 'free up capacity'. Evaluate this proposal quantitatively.
Read the worked answer
At pf 0.90 the motor current becomes 1500 / (230 × 0.90) = 7.25 A, a saving of about 1.45 A.
Existing motor current was 8.7 A, so the branch total falls from roughly 27.2 A to 25.8 A - under 5%. That is nowhere near the 11 A or more that would be needed to bring the branch under the 16 A rating, let alone create room for an extra 8.7 A boiler.
Power-factor correction is still worthwhile: it lowers cable and transformer losses, may reduce kVA demand charges, and marginally improves voltage regulation at the machines. But it reduces reactive current only; it does not reduce the real power that the resistive loads draw. The correct judgement is that the pf correction and the rewiring are not alternatives. The rewiring is mandatory; the correction is an efficiency measure that can follow.