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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 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 “, the number of cable per phase is : 2214 / 587 = **3.77**

So, we will use :

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

Noting that :

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

Regards.