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.