## Few words about Sub-main circuits

**A sub-main electrical circuit** can be defined as a circuit connected directly from the main LV switchboard to a sub-main distribution panel or a rising main for final connection of the minor current-using equipment.

The Code requires that the maximum copper loss in every sub-main circuit **should not exceed 1.5% of the total active power transmitted** along the circuit conductors at rated circuit current.

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

Alternatively, an energy efficiency method introduced by the Code could also be used for preliminary cable sizing. This energy efficiency method for cable sizing requires the calculation of the **maximum allowable conductor resistance** based on the maximum copper loss requirement as stipulated in the code.

**For a 3-phase 4-wire circuit (assuming balanced, linear or non-linear):**

Active power transmitted via the circuit conductors //

P = √3 · U_{L} · I_{1 }· cosθ

Total copper losses in conductors //

P_{copper} = (3 · I_{b}^{2} + I_{N}^{2}) · r · L

**where:**

**U**– Line to line voltage, 380V_{L}**I**– Design current of the circuit in ampere_{b}**I**– Fundamental current of the circuit in ampere_{1}**I**– Neutral current of the circuit in ampere_{N}**cosθ**– Displacement power factor of the circuit**r**– a.c. resistance / conductor/ metre at the conductor operating temperature**L**– 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 the common 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 the correction factors accordingly. **The effective-current 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 protective device.

## Calculate an appropriate cable size

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

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

### Case #1

#### Undistorted balanced condition using conventional method:

- I
_{b}= 100A - I
_{n}= 100A - I
_{t(min)}= 100A

Assume the correction factors **C _{a}**,

**C**,

_{p}**C**and

_{g}**C**are all unity.

_{i}Refer to **BS7671:2008**, The Requirements for Electrical Installations,

Table 4D4A for 25mm^{2} – 4/C PVC/SWA/PVC cable – It = **110A**

Table 4D4B for r = 1.5mV/A/m x = **0.145mV/A/m** (negligible)

Conductor operating temperature | t= 30 + 100_{1 }^{2} / 110^{2} · (70 – 30) = 63°C |

Ratio of conductor resistance 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 in conductors (P_{cu}) | = 3 · 100^{2} A^{2} · 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 an approximate conductor size required by calculating the max. conductor resistance at **1.5% power loss**:

**From Table 4.2A 35 mm ^{2}** – 4/C PVC/SWA/PVC cable having a

**conductor resistance of 0.625mΩ/m**is required. Refer to BS7671:2008, The Requirements for Electrical Installations:

Table 4D4A for 35mm^{2} 4/C PVC/SWA/PVC cable It = **135A**

Table 4D4B for r = 1.1mV/A/m x = **0.145mV/A/m**

Conductor operating temperature | t= 30 + 100_{1 }^{2} / 135^{2} · (70 – 30) = 52°C |

Ratio of conductor resistance at 52°C to 70°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 in conductors (P_{cu}) | = 3 · 100^{2} A^{2} · 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 4.2A

**Multicore Armoured and Non-armoured Cables (Copper Conductor), Conductor Resistance at 50 Hz Single-phase or Three-phase a.c.**

#### TABLE 4.2B

**Single-core PVC/XLPE Non-armoured Cables, with or without sheath (Copper Conductor), Conductor Resistance at 50 Hz Single-phase or Three-phase a.c.**

#### TABLE 4D4A

**Multicore armoured 70C thermoplastic insulated cable**

#### TABLE 4D4B

**Voltage drop (per ampere per-metre)**

**Reference //** Code of Practice for Energy Efficiency of Electrical Installations – Electrical and Mechanical Services Department – The Government of the Hong KongSpecial Administrative Region

Hello well calculation, how about distribution transformed cable in 100Kva in secondery side and distance?

Thank you

Dear Sir,

How can I upload data/article at this page.

Best Regards

Md Fazlul Haque

Bangladesh

how do u get the vale of 230 in case of ratio of resistance calculation, please. overall nice article..