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

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Genghiz
Lt. Commander Lt. Commander
Lt. Commander

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

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

 

 

1 Reply 1
dkpElect
Crewman
Crewman

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

That's really informative answer. 

Thanks.