[TowerTalk] Mot R36 related to NUMBER of installed ground rods

[jimlux]

On 11/2/17 10:10 PM, Jeff wrote:
> I’m laying out the ground rod scheme for the towers.
> 
> As I look at Mot R56, it seems like about 45’ of rods gets you to the 5 ohm target based on one of the graphics they provide  That’s about 9 ground rods spread out in 3 legs each with 3 rods separated by 16’ assuming 8’ rods and average soil.
> 
> While I realize more is better, how do you determine “enough” in the context of ground rod count?
> 


It's arbitrary - in the sense that most of these requirements aren't 
based on some sort of analysis, rather they are representative of what 
worked in the past, and are suitable for putting in a contractual document.

"The contractor shall install sufficient ground rods in accordance with 
document XYZ that the measured ground resistance is less than 5 ohms."

At the end of the job, the customer goes out, measures the resistance, 
finds it's 4.2 ohms and says "you get paid"

In most cases, the performance requirement (and the calculations such as 
in R56) harken back to some IEEE standard:
IEEE 1100 Powering and Grounding Electronic Equipment
IEEE 142 IEEE Recommended Practice for Grounding of Industrial and 
Commercial Power Systems
IEEE 80 IEEE Guide for Safety in AC Substation Grounding
IEEE 81 IEEE Guide for Measuring Earth Resistivity, Ground Impedance, 
and Earth Surface Potentials of a Ground System

Those documents tend to have a fair amount of analysis and test data to 
back up the specific recommendation.


Noting that lightning is a impulse and the voltage rises are mostly due 
to inductance, the actual "resistance" is almost insignificant - on the 
other hand, it's easy to measure (back to the "did the contractor 
complete the job" question).  Measuring the impulse impedance of a 
lightning protection system is well nigh impossible.



There is a "integrated I^2*R*t for a stroke" (called the "action") - 
basically related to how much heat will be dissipated in the ground 
system. 20 kA for 50 microseconds into 5 ohms is 100 kilojoules.  That 
will raise 25 kilos of water about 1 degree C.  How much is 25 kg? Call 
it 25 liters, an 8 foot rod is about 250 cm long - so a 100 square cm 
area around the rod  - a radius of 6cm - less than 4"

A lightning strike has multiple strokes, each dumping a shot of energy 
into the system - so the actual temperature rise will be a bit more.

But, you can see that a 5 ohm resistive loss is going to keep the 
temperature rise to a fairly low number - no "steaming rods" after a strike.