CABLE SIZE

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• #12511

Did any one has selection table of cables, I mean for every size what is the current carrying capacity and the derating tables.

#11315

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 103 / 1.73 x 400 x 0.82

Ib = 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 mm2 “, 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 mm2

– For Neutral              : ” 1 Single Core Cable of 300 mm2

Noting that :

–          for Neutral : the load is ” 3 phases “, and we haven't the 3rd 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 mm2, 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%

#12506

Soufi said:

Did any one has selection table of cables, I mean for every size what is the current carrying capacity and the derating tables.

Dear Soufi ;

As there's many tables, please send me your e-mail to send them. noting that my e-mail is :

hagersg@aloola.sy

regards.

#12520
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