Cable Sizing of Sub­Main Circuits – WorkingExamples

Cable Sizing of Sub­Main Circuits – Working Examples (photo credit: topcable.com)

Few words about Sub­main circuits…A sub­main circuit can be defined as a circuit connected directly from the main LV switchboard toa sub­main distribution panel or a rising main for final connection of the minor current­usingequipment. The Code requires that the maximum copper loss in every sub­main circuit should notexceed 1.5% of the total active power transmitted along the circuit conductors at ratedcircuit current.

Similar approach could be followed for sizing conductor as for feeder circuit. However, assumptionhas to be made in the design for various characteristics of the sub­main circuit including designcurrent,expected harmonic current (THD) in the circuit, degree of unbalance, etc.

Alternatively, an energy efficiency method introduced by the Code could also be used forpreliminary cable sizing. This energy efficiency method for cable sizing requires the calculation of

the maximum allowable conductor resistance based on the maximum copper loss requirementas stipulated in the code.

For a 3­phase 4­wire circuit (assuming balanced, linear or non­linear):

Active power transmitted via the circuit conductors //

P = √3 · UL · I1 · cosθ

Total copper losses in conductors //

Pcopper = (3 · Ib2 + IN2) · r · L

where:

UL – Line to line voltage, 380VIb ­ Design current of the circuit in ampereI1 ­ Fundamental current of the circuit in ampereIN ­ Neutral current of the circuit in amperecosθ – Displacement power factor of the circuitr – a.c. resistance / conductor/ metre at the conductor operating temperatureL – Length of the cable in metre

Percentage copper loss with respect to the total active power transmitted:

Therefore,

Table 4.2A and 4.2B in the Code provide a quick initial assessment of cable size required for thecommon cable types and installation methods used in Hong Kong for this example.

The tabulated current rating of the selected cable could then be corrected by applying thecorrection factors accordingly. The effective­current carrying capacity of the selected cable must

be checked so that its value is larger than or equal to the nominal rating of the circuit protectivedevice.

Calculate an appropriate cable sizeA 3­phase sub­main circuit having a design fundamental current of 100A is to be wired with 4/CPVC/SWA/PVC cable on a dedicated cable tray. Assuming an ambient temperature of 30°C and acircuit length of 40m, calculate an appropriate cable size for the following conditions:

1. CASE 1 // Undistorted balanced condition using traditional method (cosθ = 0.85)2. CASE 2 // Undistorted balanced condition with a max. copper loss of 1.5% (cosθ = 0.85)

Case #1

Undistorted balanced condition using conventional method:Ib = 100A

In = 100A

It(min) = 100A

Assume the correction factors Ca, Cp, Cg and Ci are all unity.

Refer to BS7671:2008, The Requirements for Electrical Installations,

Table 4D4A for 25mm2 ­ 4/C PVC/SWA/PVC cable ­ It = 110ATable 4D4B for r = 1.5mV/A/m x = 0.145mV/A/m (negligible)

Conductor operatingtemperature

t1 = 30 + 1002 / 1102 · (70 – 30) = 63°C

Ratio of conductorresistance at 63°C to 70°C

r = (230 + 63) / (230 + 70) = 0.98

Voltage drop u = 1.5mV/A/m · 0.85 · 0.98 · 100A · 40m = 5V(1.3%)

Active power transferred (P) P = √3 · 380V · 100A · 0.85 = 56kW

Total copper losses inconductors (Pcu)

= 3 · 1002 A2 · 0.0015Ω/m / √3 · 0.98 · 40m

= 1.02kW (1.82%)

(Cable size selected is not acceptable if the

maximum allowable copper loss is 1.5%)

Case #2

Undistorted balanced condition with a maximum copper loss of 1.5%(cosθ = 0.85)Maximum copper loss method using Table 4.2A in the Code for initial assessment of anapproximate conductor size required by calculating the max. conductor resistance at 1.5% powerloss:

From Table 4.2A 35 mm2 ­ 4/C PVC/SWA/PVC cable having aconductor resistance of0.625mΩ/m is required. Refer to BS7671:2008, The Requirements for Electrical Installations:

Table 4D4A for 35mm2 4/C PVC/SWA/PVC cable It = 135ATable 4D4B for r = 1.1mV/A/m x = 0.145mV/A/m

Conductor operatingtemperature

t1 = 30 + 1002 / 1352 · (70 – 30) = 52°C

Ratio of conductorresistance at 52°C to70°C

r = (230 + 52) / (230 + 70) = 0.94

Voltage drop u = 1.1mV/A/m · 0.85 · 0.94 · 100A · 40m = 3.5V (0.92%)

Active power transferred(P)

P = √3 · 380V · 100A · 0.85 = 56kW

Total copper losses inconductors (Pcu)

= 3 · 1002 A2 · 0.0011Ω/m / √3 · 0.94 · 40m

= 716kW (1.28%)

(Cable size selected is acceptable, i.e. power loss <

1.5%, under undistorted and balanced conditions)

RELATED TABLES //

TABLE 4.2AMulticore Armoured and Non­armoured Cables (Copper Conductor), Conductor Resistance at50 Hz Single­phase or Three­phase a.c.

Multicore Armoured and Non­armoured Cables (Copper Conductor), ConductorResistance at 50 Hz Single­phase or Three­phase a.c.

Go back to Cases ↑

TABLE 4.2BSingle­core PVC/XLPE Non­armoured Cables, with or without sheath (Copper Conductor),Conductor Resistance at 50 Hz Single­phase or Three­phase a.c.

TABLE 4.2BSingle­core PVC/XLPE Non­armoured Cables, with or without sheath (CopperConductor), Conductor Resistance at 50 Hz Single­phase or Three­phase a.c.

Go back to Cases ↑

TABLE 4D4AMulticore armoured 70C thermoplastic insulated cable

TABLE 4D4A – Multicore armoured 70C thermoplastic insulated cable

Go back to Cases ↑

TABLE 4D4BVoltage drop (per ampere per­metre)

TABLE 4D4B – Voltage drop (per ampere per­metre)

Go back to Cases ↑

Reference // Code of Practice for Energy Efficiency of Electrical Installations – Electrical andMechanical Services Department – The Government of the Hong KongSpecial AdministrativeRegion