Substation Grounding IEEE80 Current Split Factor
Substation Grounding IEEE80 Current Split Factor
(OP)
IEEE80(2000) has Sf current split factors shown on page 153 for a Dist gnd of 25 ohms vs one for 200 ohms. The table on page 151 line 4 from the bottom says R=.322 and X=.242 for 25 ohms and then R=1.65 and X=.291 for 200 ohms (the far right column heading is wrong and should say Rdg=200).
If I am at 30 ohms for my Dist gnd, am I forced to use the R=1.65/X=.291 and not the R=.322/X=.242 because I am over 25 ohms? Can I interpolate a value for my 30 ohm actual condition? If so, how?
At R=.322/X=.242, I easily pass my Touch and Step potentials, but at R=1.65/X=.291 I have a hard time passing as the R value is five times larger. I just barely missed the 25 ohm limit at 30 ohms and want to know if there is a way to prorate the R and X values for a Dist gnd of 30 ohms.
If I am at 30 ohms for my Dist gnd, am I forced to use the R=1.65/X=.291 and not the R=.322/X=.242 because I am over 25 ohms? Can I interpolate a value for my 30 ohm actual condition? If so, how?
At R=.322/X=.242, I easily pass my Touch and Step potentials, but at R=1.65/X=.291 I have a hard time passing as the R value is five times larger. I just barely missed the 25 ohm limit at 30 ohms and want to know if there is a way to prorate the R and X values for a Dist gnd of 30 ohms.






RE: Substation Grounding IEEE80 Current Split Factor
a) Transmission line length of 37.82 km and a distance between grounds of 152 m.
b) Transmission tower footing resistance of 15 or 100ohm.
c) Transmission line structure single pole with 1–7#10 alumoweld shield wire and 336.4 kcmil, 26/7 ACSR conductor.
d) Distribution line length of 4 km and a distance between grounds of 122 m.
e) Distribution pole footing resistance of 25 ohm or 200 ohm.
f) Distribution pole three-phase triangular layout, with one 336.4 kcmil, 26/7 ACSR phase and 1/0 ACSR neutral conductor.
g) Soil resistivity of 100ohm•m.
h) Substation grounding system resistances of 0.1ohm, 0.5 ohm , 1.0 ohm, 5.0ohm, 10.0 ohm, and 25.0 ohm.
i) Number of transmission lines varied from 0, 1, 2, 4, 8, 12, and 16.
j) Number of distribution lines varied from 0, 1, 2, 4, 8, 12, and 16.
k) One remote source for each two transmission lines.
As you can see, it is an average system and a difference from 25ohm to 30ohm it is entirely not important. In my opinion this calculation is for information only not an exact one.
For a more accurate solution, you have to proceed as indicated in section 15.9 Computation of current division.