Home › Electrical Engineering Forum › General Discussion › CABLE SIZE
 This topic has 18 replies, 5 voices, and was last updated 12 years, 10 months ago by ahmadalsayed20.

AuthorPosts

2011/09/20 at 6:59 am #12511adminKeymaster
Did any one has selection table of cables, I mean for every size what is the current carrying capacity and the derating tables.
2011/09/21 at 12:04 pm #11315Spir Georges GHALIParticipantDears ;
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 : 400/231V50Hz
– 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
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 mm^{2} “, and from table
” A.5210 ” of ” IEC 364552 “, 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 = 1.88So, 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 redone.
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}, TouchingFlate “, 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 1012%
2011/09/21 at 12:07 pm #12506Spir Georges GHALIParticipantSoufi 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 email to send them. noting that my email is :
regards.
2011/09/22 at 9:07 am #12520 
AuthorPosts
 You must be logged in to reply to this topic.