Sizing a Tailoring Room Supply Against What the Owner Already Bought
A single-phase 230 V supply must be wired for a small tailoring training room. The load schedule is: 12 LED fittings of 18 W, 6 ceiling fans of 75 W, 6 socket outlets counted at 100 W each, one 1.6 kW split AC drawing about 8.7 A at 0.8 power factor, and one 2 kW water heater. The distribution board is 22 m of cable away from the main switch, and the AC is a further 14 m from the board. The owner has already bought a 32 A MCB for the main switch, a 6 A MCB for the AC, and 2.5 sq mm copper PVC cable for both runs, and does not want to buy anything else. The supply authority fuse at the meter is 25 A. The earth electrode is a GI pipe measuring 6 ohm. Take copper resistivity as 0.0175 ohm sq mm per metre and 2.5 sq mm PVC copper as rated about 21-24 A depending on installation.
1. Work out the connected current, apply a diversity factor of 0.75, and decide whether the 32 A main MCB is acceptable for the 2.5 sq mm cable the owner has bought.
Read the worked answer
Add the currents branch by branch rather than only the watts, so that the AC power factor is not lost:
- LEDs: 216 W / 230 V = 0.94 A
- Fans: 450 W / 230 V = 1.96 A
- Sockets: 600 W / 230 V = 2.61 A
- AC: 8.7 A (given, at 0.8 pf)
- Heater: 2000 W / 230 V = 8.70 A Connected current = 22.9 A, connected load about 4.9 kW.
Apply diversity 0.75 (the heater and the AC will not normally run together, and not all sockets are loaded): design current Ib = 22.9 x 0.75 = 17.2 A.
The protective device rule is Ib <= In <= Iz. Here Ib = 17.2 A and Iz for 2.5 sq mm PVC copper is about 21-24 A. A 20 A MCB satisfies the rule. The 32 A MCB does not: it allows 32 A to flow in a cable rated at best 24 A for an indefinite period, so the cable, not the MCB, becomes the weak link and can overheat inside conduit before the MCB ever trips. The 32 A MCB is also larger than the 25 A meter fuse, so discrimination is lost and a fault would blow the authority fuse first.
Verdict: the 2.5 sq mm cable is acceptable for the present load, but the main MCB must be changed to 20 A; if the owner intends to add load later, go to 4 sq mm with a 25 A device.
2. Evaluate the 6 A MCB chosen for the AC. State the rating and curve you would use and check whether the 2.5 sq mm cable remains adequate for that branch.
Read the worked answer
The AC draws 8.7 A running. A 6 A MCB is below the running current, so Ib > In and the rule is broken: the breaker will trip in normal service, not just on fault. Worse, an AC compressor draws roughly 3 to 5 times running current for a fraction of a second on every start, that is about 26-43 A. A 6 A MCB on a B curve trips magnetically at 3-5 times rating, i.e. 18-30 A, so it will nuisance-trip on start.
Correct choice: a 16 A MCB with a C curve (magnetic trip 5-10 times rating, i.e. 80-160 A), which rides through the start surge and still clears a short circuit. A 20 A C-curve is also seen in practice but gives less margin for the cable.
Check the cable: 2.5 sq mm PVC copper, Iz about 21-24 A, is comfortably above In = 16 A, so the cable is protected and no change is needed.
Check the branch voltage drop: R = 0.0175 x 28 / 2.5 = 0.196 ohm for the 14 m two-wire run; drop = 8.7 x 0.196 = 1.7 V, that is 0.74 percent of 230 V, well inside 3 percent.
Also required: a local isolator next to the indoor unit, a 30 mA RCCB covering the socket and AC circuits, and an outdoor unit disconnect within sight for maintenance.
3. Check the voltage drop on the 22 m main run at the design current and at the full connected current, and say whether the owner's cable passes.
Read the worked answer
For a single-phase run the current goes out and comes back, so use twice the length.
Resistance of the run: R = 0.0175 x (2 x 22) / 2.5 = 0.0175 x 44 / 2.5 = 0.308 ohm.
At the design current of 17.2 A: drop = 17.2 x 0.308 = 5.3 V, which is 5.3 / 230 = 2.3 percent. This passes the usual 3 percent limit for a final circuit.
At the full connected current of 22.9 A (everything on together, which is unlikely but possible with a winter heater and the AC): drop = 22.9 x 0.308 = 7.05 V, i.e. 3.1 percent. That is marginally outside 3 percent and would leave very little margin for the service cable and the AC branch.
Judgement: the 2.5 sq mm cable is acceptable for today's load at 20 A, but it is at the edge. Two things follow. First, keep the run as short and as direct as possible rather than adding loops around the building. Second, if any load is added (a press, a second AC), the correct upgrade is 4 sq mm, which halves the resistance to about 0.154 ohm and brings the drop at 22.9 A down to about 3.5 V, or 1.5 percent.
Record the calculation with the cable size, length, design current and measured earth resistance so the installation can be inspected later.