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Dears ;

please find below an exsemple to define the section of a cable supplying a load.

Assume that we have the following load :

– Power : 450 kW

– Cos φ : 0.82

– Supply : 3Ph + N

– Voltage : 400/231V-50Hz

– Length from the panel : 125 m

– There's no Harmonics

– ku = 100%

So, the nominal current for this load is :

Ib = 450 x 10^{3} / 1.73 x 400 x 0.82

**I**b = **793 A**

The Correction Factors ( or Reduction Factors ) will be defined accordingly to the Method of Installation and the kind of cable, assume that :

– Method of installation ” F ” :

- ” Horizontaly Perforated Cable Tray “
- ” Touching – Flat “

– The kind of Cable : ” Single Core Cable – Copper – PVC “

– 2 cables per Phase.

**So, the Reduction Factor : 0.91**

– The Ambiant Temperature is ” 45 °C ” in the air

**So, the Reduction Factor : 0.79**

And the total Reduction Factors is : 0.91 x 0.79 = **0.718**

And the nominal current for needed cables is : 793 / 0.718 = 1105 A

We assume that we will use ” Single Core Cable – Cu – PVC – 300 mm^{2} “, and from table

” A.52-10 ” of ” IEC 364-5-52 “, we find that the Current Carrying Capcity for ” 300mm2 ” for ” Touching – Flat / Methode F ” is ” 587 A “, and the number of cable per phase is :

1105 / 587 = 1.88

So, we will use :

– For each Phase : **” 2 Single Core Cables of 300 mm ^{2} “**

**– For Neutral : ” 1 Single Core Cable of 300 mm ^{2} “**

** **

Noting that :

– for Neutral : the load is ” 3 phases “, and we haven't the 3^{rd} Harmonic.

– If ” Ku ” is different, the calculation should be re-done.

After that, you should calculate the **” total Voltage Drop “** from the source till the load to decide if that cables are suitable or not.

To calculate the ” Voltage Drop “, we use the following formula :

**Δ**** U = 1.73 . I ( R . Cos ****φ**** + X . Sin ****φ )**** **

**1 – For section : from the panel to the load :**

As the cable is ” Singl core, 300 mm^{2}, Touching-Flate “, So :

R = ρ . L / S → 18.51 x 1.23 x 125 / 300 x 2 → 4.74 mΩ

X = λ . L → 0.13 x 125 / 2 → 8.12 mΩ

Cos φ = 0.82 → Sin φ = 0.57

Δ U = 1.73 x 793 ( 0.00474 x 0.82 + 0.00812 x 0.57 ) → 11.68 V

**Δ**** ****U = ****2.92%**

**2 – For section : from the source till the load :**

To be able to calculate this section we should have the following information :

– The section of cable if the panel is supplying directly from the source, or the sections of cables if there's many sections.

– The length of cable or lenghts of cables.

– The kind of cable or cables.

– The nominal for this section or the nominal currents for these sections.

– The Cos φ for this section or for these sections.

– The Method of installation for each section.

By using the same formula, we caculate the percentage of Voltage Drop for each section, then we sum these perecentages to have the ” Total Voltage Drop ” that should be compliance with the standard percentage, if not, the section of cable should be oversized.

By the way, if the load is a motor, we should calculate the Total Voltage Drop 2 times that :

– After running ( nominal current for the motor ) ( as before )

– During running ( current & Cos φ for the motor during running )

And the Total Drop Voltage during running shouldn't be exceeded 10-12%