#12317
Spir Georges GHALI
Participant

Dear ;

As there's no answer till now, please find bellow the calculation of the Cables' Sections.

Assume that :

• ” Ku ≈ 70 % ” that means  50  of  70 AC  are working in tha same time.
• The Methods of Installation :

– ” Horizontaly Perforated Cable Tray “

– ” Touching – Flat “

• ” Only 1 Cable Tray “
• The kind of Cable : ” Single Core Cable – Copper – PVC “

The total load current :  Ib = 50 x 23.5 = 1175 A

The Reduction Factors will be defined accordingly to the following points :

• 3  or  4  cables per Phase  →  The Reduction Factor : 0.87
• The Ambiant Temperature is ” 55 °C ” in the air  →  The Reduction Factor : 0.61

So, the total Reduction Factors is : 0.87 x 0.61 = 0.5307

And the nominal current for needed cables is : 1175 / 0.5307 = 2214 A

Suppose that we will use ” Single Core Cable – 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 “, the number of cable per phase is : 2214 / 587 = 3.77

So, we will use :

• For each Phase        : ” 4 Single Core Cables of 300 mm2
• For Neutral              : ” 2 Single Core Cables of 300 mm2

Noting that :

• For Neutral : as all loads are ” 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 ” Drop Voltage “ to know if these cables are suitable or not.

Regards.