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Driving ground rods

0

0_Qed

Jan 1, 1970
0
Readership,

Is an 'almost' horizontally driven 8' 5/8" rnd ground rod as effective
as a vertically driven ground rod ???

Clay/Shale at about 4' depth ,
thus the query about 'near' horizontal.

Ed
 
S

SQLit

Jan 1, 1970
0
0_Qed said:
Readership,

Is an 'almost' horizontally driven 8' 5/8" rnd ground rod as effective
as a vertically driven ground rod ???

Clay/Shale at about 4' depth ,
thus the query about 'near' horizontal.

Ed

ground rods can be installed in a varity of ways. However, the differences
in the installation can and usually do effect the usefulness of the rod. I
personally do not ever install one. I install them in pairs spaced a minimum
of the distance that they are long. It is possible to drive one or even two
rods in poor grounding conditions and not get the <25 ohms for a safety
ground (NEC).

Most of my grounding installations are considered "performance" < 5 ohms.
All of my installed ground systems are tested. Several ground systems are
tested several times a year. I live in the desert and grounding can be
problem.
 
Readership,

Is an 'almost' horizontally driven 8' 5/8" rnd ground rod as effective
as a vertically driven ground rod ???

Clay/Shale at about 4' depth ,
thus the query about 'near' horizontal.
The code says up to 45 degrees is OK. If you have a question, drive
another rod at least 6 feet away and connect them together
 
D

Don Kelly

Jan 1, 1970
0
----------------------------
0_Qed said:
Readership,

Is an 'almost' horizontally driven 8' 5/8" rnd ground rod as effective
as a vertically driven ground rod ???

Clay/Shale at about 4' depth ,
thus the query about 'near' horizontal.

Ed

Dwight gives :
vertical rod Resistance =(rho/2*pi*L)(ln(4L/a)-1)
where rho is the soil resistivity L is the rod length and a is its radius

horizontal rod: Resistance =(rho/4*pi*L){ln(4L/a)+ln(4L/s) -2 +s/2L -smaller
terms)
where the length is 2L depth s/2 and the rest as above

for L=3m a=0.01m and s =0.3m
R vert/R horiz =1.07/1.24 =86%
This is assuming uniform earth which you don't have. Use a longer horizontal
rod (say by a factor of at least 2 and 3 would be better); or a grid such as
a 3 point horizontal star with arms of length L
(this gives, for the same L, less than 60% of the resistance of a single
vertical rod.)
Hey, a star or grid uses more conductor but it is still cheaper than trying
to get a good ground with a vertical rod.
 
0

0_Qed

Jan 1, 1970
0
Don Kelly wrote:

....snip...

Don,

Thanks for the detailed reply ...
now have a 'PhD' in horiz driven grnd rods.

Ed
 
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