Answer / chandan tripathy

As per IS 3043, Cross Section Area of Earthing Conductor
(A) =(If x√t) / K
Where:
t = Fault current Time (Second).
K = Material Constant.
Example: Calculate Cross Section Area of GI Earthing Conductor for System has 50KA Fault Current for 1 second. Corrosion will be 1.0 % Per Year and No of Year for Replacement is 20 Years.
Cross Section Area of Earthing Conductor (A) =(If x√t) / K
Here:
If = 50000 Amp
T = 1 Second
K = 80 (Material Constant, For GI=80, copper K=205, Aluminium K=126).
• Cross Section Area of Earthing Conductor (A) = (50000×1)/80
• Cross Section Area of GI Earthing Conductor (A) = 625 Sq.mm
• Allowance for Corrosion = 1.0 % Per Year & Number of Year before replacement say = 20 Years
• Total allowance = 20 x 1.0% = 20%
• Safety factor = 1.5
• Required Earthing Conductor size = Cross sectional area x Total allowance x Safety factor
• Required Earthing Conductor size = 1125 Sq.mm say 1200 Sq.mm
• Hence, Considered 1Nox12x100 mm GI Strip or 2Nox6 x 100 mm GI Strips

Answer / bala.a

you will follow as per State electricity board norms - 50
x 6 mm Cu for Nu and 50 x 6 mm GI FOR body 0r
you will calculate as per short circuit calculation
Formula 0.0054 x root t x I
Where
t =1 second
I = SHORT CIRCUIT CURRENT
example
Fault current- 33kA( assumed )
0.0054 x root 1 x 33000= 178.2 mm
so you will recommended 50 x 4 mm Cu

Answer / k.prakashchandra

For above 750 KVA ,earth lead is determined as per I.S.1886

Answer / nitin

Mr. Bala,
what is 0.0054factor? Is it root3 divided by current rating
in A/sq.mm? Then for Cu it is 205 A/sq.mm and so root 3
divided by 205 comes to 0.0084. OK?

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