An example for 'determining the smallest allowable cross-sectional area of circuit conductors.'

[Genghiz]

From the available inputs;

let us see how to determine the 'smallest allowable cross sectional area' of the circuit conductors.

Available inputs;

Load Current	200 A
THDi	48 %
Conductor Insulation*	XLPE
Conductor Type	Multi-Core Cable
Conductor Material*	Copper
Installation	Directly in Ground
Soil Resistivity*	2 Km/w
Soil Temperature*	40*C

*Some of the inputs are assumed to provide in this example; and they were neither received from the user, nor from the environment.

Procedure:

• Determinant conductor : Neutral

          As the THDi is > 33 % (that is THDI = 48 %). So, the Size-determining" live conductor is Neutral

• Neutral Current               :  3 × 0.48 × 200 = 288 A

• Correction factor K3       : The factor based on the overload protection device.

           This circuit is assumed to be protected by the circuit-breaker.

           So that the K3 =1.0

• F1 Temperature              : Table B.52.15

          For 40°C and with the insulation XLPE. F1 = 0.85

• Installation method       : 72                                                                                                           (Sheathed single-core or multi-core cables direct in the ground - without added mechanical protection) Reference method          : D2

• F2 Group                          : Table B.52.18

          There is no additional touching circuits i.e. total no. of circuits is 1

          F2group = 1.00

• Correction factor F3       : Table (E.52.1)

          THDI >45 % = THDI = 48 %  à f3NeutralTHDI = 1.00

• F3RSoil                              : Table B.52.16

           Installation method = in Ground -> Rth = 2.0 K.m/w

           f3RthSoil = 1.12  

• K user = 1.0

Correction factor f

f = 0.85 × 1.00 × 1.00 × 1.05 × 1.00 = 0.952

The overall correction factor is determined by multiplying all 'the available correction factors."

Theoretical Iz’                 = (K3 x IB) / (F x N live) = (1.0 x 288) / (0.952 x1) = 302.52 A = 303 A

The conductor selection is done from the given Table B.52.5 Column 8 

185 mm2 will be able to carry 324 Amps

240 mm2 will be able to carry 375 Amps

Considering the +5% of tolerance, It would be recommended to consider the size as 185 mm²

K3(Correction Factor based on Protection Device)	1
F1(Correction Factor based on Temperature)	0.85
F2(Correction Factor based on Group of conductors)	1
F3(Correction Factor based on THDi)	1
F3(Correction Factor based on Soil Resistivity)	1.05
K user(Correction Factor defined by the user)	1
Iz’(A) Theoretical	303 A
Iz (A) considered from the table	324 A
Theoretical cross sectional area  Ph (mm²)	1 x 185   mm²
Theoretical cross sectional area Neutral (mm²)	1 x 185   mm²
Theoretical cross sectional area PE (mm²)	1 x 95   mm²

[dkpElect · Accepted solution]

That's really informative answer. 

Thanks.