Formulas for Calculating Grounding Resistance of Simple Forms with Ground Enhancement Material by Shirley Wang

Formulas for Calculating Grounding Resistance of Simple Forms with Ground Enhancement Material

www.ijape.org International Journal of Automation and Power Engineering (IJAPE) Volume 2 Issue 2, February 2013

C 

ab (19)

So, total resistance: RT = R1 + R2 +R3



and we will receive the formulas as in the case of the GEM layer of cylinder form.

RT 

 l2  1    1  ln C t  C 2 l  ( 2 )  

(25)

d  C(2t  C)    2 ln    3 ln  dt 2l   

Grounding Resistance of Single Horizontal Rod

Grounding Resistance of Equivalent Horizontal Rod

1) GEM Layer of Cylinder Form Single horizontal rod with diameter d , length l and resistivity 3 is fixed in the ground with the depth t . It is environed by the GEM environment with resistivity 2 and the soil environment with resistivity 1 (see FIG. 6)

We use the calculating method as at the above point and receive the results as follows: 1)

The Component in the Metal Environment

R 1 '  2)

 3   d   (26) ln   2  l   2 l  

The Component in the Equivalent Soil Environment

R ' 3 

FIG.6 SINGLE GROUNDING HORIZONTAL ROD WITH THE GEM LAYER

Applying (1) and after some transformations, we found the magnetic field that is created by the horizontal rod at the point (x, z)   ( x )  I  ln   l   1 2 l   2 x  

  l     1  (20)   2x   2

and the magnetic field is created by the image of the horizontal rod at the point (x, z):

 2 ( x) 

  I  l  ln  2 l  2(2t  x) x  

We use the calculating method as at the above point and receive the results as follows :



 l2 1    eq ln  2 l   dt

2) The Component in the GEM Environment R 2    2   2 ln  C ( 2 t  C )  (23) I 2 l  dt  3) The Component in the Soil Environment

  3 I 1  l2 R 3   (24)  ln  I 2 l  C ( 2 t  C ) 

  d   (28)    3 ln    2l  

Hence, In order to determine the equivalent resistivity we will equalize (25) and (28), from this



 l2  C ( 2t  C     2 ln   C t C dt  ( 2 )    

 1 ln 

l   eq ln   dt 2

   0 

The equivalent soil resistivity will be

 eq 

  l2  ln   C ( 2t  C ) 

1

 C ( 2t  C )    dt  

2

   (29)

l2  ln    dt 

1) The Component in the Metal Environment

  1  d  (22)  3 ln   I 2l  2l 

 (27)  

Total resistance: RT = R’1 + R’3

2    l (21)    1   2(2t  x)  

Grounding Resistance of Horizontal Rod with the GEM

R1 

 eq  l2 ln  2 l  dt

So, we can count as single grounding horizontal rod (with parameters d , l ,3) that is only buried in one single soil layer with equivalent resistivity ρeq. This equivalent resistivity is calculated by equation (29) (see FIG.7). In reality,  3

 0 , from (28) we have:

R T 

 eq 2 l

  l 2    (30)  ln  dt   