Pool Pump Submain: Cable Selection and Verification
A 12 kW, 230 V single-phase pump set is to be installed in a pool plant room 42 m cable route from the origin. The cable will be buried in a 110 mm duct at 0.6 m depth, along with two other loaded circuits, and will pass through a plant room where the ambient temperature reaches 35 °C. The supply is TN-C-S with a measured Ze of 0.28 Ω and a 100 A main switch; the intended protective device is a 63 A BS EN 60898 Type B MCB. The client insists on a maximum 3% voltage drop for the pump circuit and on using the SWA armour as the circuit protective conductor. A 6 mm² four-core SWA is already on site and the client asks whether it can be used. The armour is to be the only CPC, and the installation is in Cyprus, where the ambient design temperature for buried ducts is taken as 20 °C.
1. Calculate the design current and the required tabulated current-carrying capacity after derating, and determine whether the 6 mm² SWA on site is suitable.
Read the worked answer
Design current: Ib = P/U = 12,000/230 = 52.2 A. The next standard device rating above Ib is In = 63 A, so the cable must have a derated capacity at least equal to In.
Derating factors: Ca = 0.94 for 35 °C (70 °C thermoplastic); Cg = 0.70 for three loaded circuits grouped together. Required tabulated capacity: It ≥ In/(Ca × Cg) = 63/(0.94 × 0.70) = 95.7 A.
A 6 mm² four-core SWA in a buried duct has a tabulated capacity of roughly 47–50 A. That is already below Ib = 52.2 A before derating, and after derating it is about 0.658 × 48 ≈ 32 A. The 6 mm² cable fails on current-carrying capacity.
It also fails on voltage drop: using about 7.3 mV/A/m, ΔU = 7.3 × 52.2 × 42/1000 ≈ 16 V, about 7% of 230 V, far above the 3% limit of 6.9 V.
Select a cable with tabulated It ≥ 96 A, for example 25 mm² SWA at about 110 A in duct, giving 110 × 0.94 × 0.70 ≈ 72 A > 63 A.
2. Check the voltage drop for the selected cable and explain what to do if it fails.
Read the worked answer
ΔU = mV/A/m × Ib × L/1000. For 25 mm² SWA, mV/A/m is about 1.75, so ΔU = 1.75 × 52.2 × 42/1000 = 3.84 V, which is 1.67% of 230 V. This complies with the client's 3% limit (6.9 V).
If it failed, do not simply increase the MCB. Options in order of preference:
- increase the conductor cross-sectional area;
- reduce the route length or change the route;
- split the load or use a three-phase pump and submain, which reduces current per phase;
- review the design current and the duty cycle of the pump.
The 3% figure is a client/design limit; BS 7671 permits 5% for power, but the stricter figure governs here. Show the calculation on the schedule of circuit details so the designer's intent is recorded.
3. Verify the armour as a CPC: disconnection and the adiabatic check.
Read the worked answer
Zs = Ze + R1 + R2. For 25 mm² copper over 42 m, R1 ≈ 18.1 × 42/25 = 30 mΩ at 20 °C, corrected to about 36 mΩ at operating temperature. The armour resistance is of similar order, so Zs is around 0.35 Ω. For a 63 A Type B MCB the 0.4 s maximum Zs is about 0.57 Ω, so disconnection is satisfied. Prospective fault current ≈ 230/0.35 ≈ 657 A, well above the instantaneous threshold.
Adiabatic check: k = 51 for steel armour. Compare k²S² with the let-through I²t from the device curve. With I ≈ 657 A and a short trip time of the order of 0.1 s, I²t ≈ 4.3 × 10^4 A²s. If the armour equivalent cross-section is, say, 40 mm², k²S² = 51² × 40² ≈ 4.2 × 10^6 A²s, so it passes comfortably. Confirm the armour cross-section from the manufacturer's data and verify continuity. Also provide equipotential bonding to the pool's metallic parts; if the armour is inadequate, run a separate CPC.