Question

What size Grounding Electrode Conductor would be required for 1000A, 277/480V., 3-Phase, 4-Wire Service with (4) sets of #250 kCMIL copper conductors per phase?

Answer 1

We need to find the size of Grounding electrode Conductor

I=1000A

Primary / Secondary Voltage = 277/480 V

3 phase 4 wire service .

250 kCMIL copper conductor per phase .

cMil is unit of area

Given as :

1 cMil =5.067 * 10 ^-4 mm²

250 kCMIl = 250 * 1000 CMil

=250* 1000* 5.067 * 10 ^-4 mm²

=126.675 mm²

It is given that

3 phase 4 wire

Hence, current in 1 conductor = 1000/4 =250 A

Since it is 3 phase hence voltage across 1 phase given by :

= 480/?3 V

Now , using

V=IR

R=V/I

R= (480/?3 )/ 250

=1.1 ohms

We know that

R=( Re*L) /A

Re is resistivity

L is Length of wire

A is area

Hence ,

L=R*A/Re

Asuuming ideal conductor Re = 1

L=1.01 *126.67 mm

L=140.4 mm