I want formula to calculate cable size as per load given in
kw & amp.I searched many sites but didn't right answer.Plz
reply me asap.
Answer Posted / er. naushad ashraf
Dear See 1 eg. of Cable Designing
From - 5 MVA power T/f LV side
To - 11 kV LA’s
• RMS Symmetrical Short Ckt current = 17.39 kA (As per Tender spec.)
• Fault Clearing Time = 1 sec
• Min. Cable Size = (17.39x1000xsq root 1)/94= 185 sqmm.
• Min. Cable Size = 1-3C 185 sqmm.
• Cable Length = 0.06 Km. (Assumed).
• Ifl = (5000x1000)/(1.734x11000x0.9x0.95) = 307 A.
• Derating Factor = Ca X Cg = .88x1 = 0.88
• Cable sized considered = 3C X 185 sqmm. (XLPE+Al).
• Cable Resistance (Ohm/Km) = 0.21
• Cable Reactance (Ohm/Km) = 0.087
• Cable Ampacity = 330 A.
• Cable Derated Current = 330x0.88 = 290.4 A.
• Min. No. of cable required = 307/290 = 1.0586 = 1 No.
• No. of Cable Runs Selected = 1 No.
• % Voltage Drop = (sqrt 3x307x0.06(0.21x0.9+0.087x0.43)x100)/(1x11000)
= 0.066 % (Acceptable)
Is This Answer Correct ? | 2 Yes | 2 No |
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