Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Earthing 2

Status
Not open for further replies.

IrVen

Electrical
Apr 18, 2009
4
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.
 
Replies continue below

Recommended for you

Soil is a relatively good conductor when moist. Rock is conductive too, but the resistance is typically much higher.

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
 
I agree with alehman. I'll try to explain this with an example.
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
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor