Earthing
Earthing
(OP)
I would really like to know more about electrical earthing because i understood it is one of the conductor during faults. But just wondering how soils and rocks could conduct electricity.
When was the last time you drove down the highway without seeing a commercial truck hauling goods?
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RE: Earthing
When designing a grounding system, soil resistivity is measured to determine the extent of the grounding electrodes needed to achieve a specified resistance to earth for the system. Also it's important to calculate the voltage gradients in the soil/rock surrounding the grounding electrodes in a fault condition. Too much gradient is hazardous to someone standing in the vicinity.
IEEE 80 is a good reference.
Alan
----
"It's always fun to do the impossible." - Walt Disney
RE: Earthing
If you compare the resistivity of copper [ro=1.72 x 10^-8 ohm.m] with the good conductive earth ro= 30 ohm.m [agricultural plains, streams, richest loam soil] or sea water [away from river estuaries] ro=0.22 ohm.m the difference is enormous, indeed.
But, if we compare the resistance of the current conducting objects: the wire and the remote earth resistance it seems they are close.
The wire resistance is R=ro*length/cross section .If it is a long wire and a small cross section the resistance may be a couple of hundred ohm. In the contrary, even in moderate high soil resistance one can get a few ohms remote earth resistance due to huge cross section of the ground.
for example:
A conductor of 18 awg[0.81 sqr.mm] copper at 20 degree C and a length of 1000 ft [304.8 m]
Since 1.724/10^8 ohm.m =1/58 ohm.sqr.mm/m then: R=1/58*304.8/.81= 6.488 ohm
For a single electrode:
R = ro*[ln(4*L/r)-1]/2/pi()/L where:
R = resistance in ohms of the ground rod to the earth (or soil) [ohm]
L = grounding electrode length[m]
r = grounding electrode radius[m]
ro = average resistivity in ohms-cm.
If ro=100 [average soil resistivity] L=50 ft and r=3/8"[L=15.24 m; r=0.009525 m]
R= 8.108 ohm