Earth rods in triangle configuration BS 7430:2011

[josealjim]

Hello,

Id really appreciate if you could help me to solve some doubts from the BS 7430:2011 ( the part that talks about the electrodes resistance calculation)
It is not quite clear for me if, in order to calculate the resistance to earth of a 3 rods triangle system, we ONLY use the equation given on 9.5.8.1 "three rods at the vertices of an equilateral triangle", or we ALSO have to calculate the resistance of the round conductor and then the total final resistance of the paralleled system.
In the section "9.5.8. Miscellaneous electrodes", there are several configurations like Three rods at the vertices of an equilateral triangle, two strips set at right angles to each other meeting at one corner...

- The equation given on 9.5.8.1, apart from the errors presented on the edition 2011 which seem to be amended on the a1:2015, doesn't include the diameter of the conductor that connect them all together.

-How is it possible that if a parallel connection is needed between the three rods and the ground electrode, it is not mentioned on the code?

-Assuming we have to calculate resistance of the round conductor too, Do we use the straight run round conductor electrode?


Thanks a lot in advance

 

[Julius Right]

I have not the 2011 edition of BS7430. I have with me the 1998 edition only.
Here the total resistance of n embedded rods will be:
Rn=R*(1+l*a)/n where l=1.66 for 3 embedded rods
a=ro/(2*pi()*R*s)
s is the distance between adjacent rods, in m
R=is the resistance of one rod in isolation, in ohm
s has to be equal or more then rod length.
ro is the resistivity of soil, ohm·m
You have at first to calculate R for a single embedded rod.
R=ro/(2*pi()*L)*(ln(8*L/d)-1)
where L is the rod length[m] and d is the rod diameter[m].

 

[Julius Right]

By the way, the metallic resistance is not taken into consideration since is negligible.
Let's say a stainless steel [ro=0.77 ohm.mm^2/m] rod of 3/4"[19 mm diameter] of 10 feet long presents the following resistance:
ro=100 ohm.m
R=100/(2*PI()*3.048)*(LN(8*3048/19)-1)=32.14 ohm
The same rod metallic
Rrod=0.77*3.048/(19^2/4*pi())=0.00823 ohm

 

[josealjim]

By the way, the metallic resistance is not taken into consideration since is negligible.
Let's say a stainless steel [ro=0.77 ohm.mm^2/m] rod of 3/4"[19 mm diameter] of 10 feet long presents the following resistance:
ro=100 ohm.m
R=100/(2*PI()*3.048)*(LN(8*3048/19)-1)=32.14 ohm
The same rod metallic
Rrod=0.77*3.048/(19^2/4*pi())=0.00823 ohm

Thanks a lot