Earth pit?

[farale]

Hi everybody,
Assume that there are two earth pits. Resistance of each of them is 20 Ohms (Calculated and measured).

We connect an Ohmmeter to these two pits. Assuming negligible resistance for wires which connect Ohmmeter to pits, what will be measured value?

Does it depend on distance between two earth pits?

Please give me a reference.
Thanks.

 

[jghrist]

Yes. IEEE Std 142-1991 (Green Book) Table 13

Resistance of two ground rods, resistivity rho, length L, spacing s, radius a, s>L

R=(rho/4/pi/L)*(ln(4*L/a)-1) + (rho/4/pi/s)*ln(1-L²/3/s²+(2*L^4)/5/(s^4)...)

 

[HamburgerHelper]

The IEEE Color books got transferred over to the DOT system and collected. One of the IEEE 3000 to 3007 books probably has what jghrist has posted, too. I would guess it would be IEEE 3003, Power Systems Grounding.

------------------------------------------------------------------------------------------
If you can't explain it to a six year old, you don't understand it yourself.

 

[farale]

Thanks for all replies.
I was asking about resistance between two rods (not resistance of combination of rods to the earth).

In other words: resistance of each of these rods to the earth is 20 Ohms. What is resistance between first rod to the second rod? 40 Ohms? More? Less?

 

[jghrist]

That's not something normally measured or calculated. I don't know of what use the value would be. It certainly would depend on the distance between rods. It wouldn't be directly related to the resistance to remote earth of either rod, which depends on current flowing through a large area of earth, not just the earth between the rods. I would guess a much lower value. You could calculate it with a grounding analysis program like SES CDEGS.

 

[aussiejohn2]

Hi Farale,
Consider two limiting cases:
1. If the rods are very (infinitely) far apart then you will measure the sum of the two rods i.e. 40 ohms. Neglecting the lead resistance is not entirely unreasonable, since you could use superconducting leads or (easier) use a four-terminal measurement to eliminate the effect of the lead resistance from the measurement.
2. If the rods are negligible ((zero) distance apart, then the current can all take the short path, bypassing most of the earth resistance. So the resistance approaches zero.

I do agree though that there are not many cases where the value is of any practical interest.
John.

 

[farale]

Thank you all

Practical usage for this question is to calculate fault current in TT systems.
(as well as better understanding the concepts behind any formula about resistivity of a rod to ground)

 

[erdep]

     A program can calculate the resistance to ground of any electrode,
     and the mutual resistance between any set of electrodes.

     The electrode to electrode resistance can be calculated using the
     same program result.



     In the shunt circuit with R1, R2 and Rm

     Req= R1*R2-Rm**2)/(R1+R2-2*Rm)

     o    V   o
     !        !
     R1  Rm   R2
     !        !
     ---------- ground


     o    V   o
     !        !
     R1-Rm    R2-Rm
     !        !
     -----!-----
          Rm
          !
          !
     ----------  ground


     I = V /[(R1-Rm)//(R2-Rm)]+Rm = V /Req

     Req = ((R1-Rm)*(R2-Rm)/(R1-Rm+R2-Rm)) + Rm

       = (R1R2-R1Rm-R2Rm+RmRm)/(R1+R2-2Rm) + Rm

       = [R1R2-Rm(R1+R2)+RmRm)+Rm(R1+R2-2Rm)]/(R1+R2-2Rm)

       = [R1R2-Rm(R1+R2)+RmRm+Rm(R1+R2)-2RmRm]/(R1+R2-2Rm)

       = (R1R2-Rm**2)/(R1+R2-2Rm)




       In the series circuit

       I        I
       o   V    o
       !        !
       R1  Rm   R2
       !        !
       ----------

       o        o
       !        !
       R1-Rm    R2-Rm
       !        !
       ----------

       or

       o        o
       !        !
       R1       R2
       !        !
       -- -2Rm---

       I= V/(R1+R2-2Rm)

        =V/Req

      Req= R1+R2-2Rm

      For example, two rods of 10 m, 0.02 m diameter, in 100 Ohm.m resistivity
      separation of 10 m.

      One rod has R1= 11.57308 Ohm (= R2)

      Rmutual     Rm= 1.3027 Ohm

      The resistance of the two rods combined results

      Rc= (R1*R2-Rm**2)/(R1+R2-2*Rm)

        =  6.43782 Ohm

      The same set of two rods, as calculated by the same program is

      Rc = 6.43789 Ohm

      The resistance of rod to rod is

      Rr = R1+R2-2*Rm

      Rr = 21.84346 Ohm

      a bit smaller than simply R1+R2

      Rm is big for small separation, reducing the electrode to electrode resistance,

      If the separation is big, the mutual resistance is very small, and the electrode
      to electrode resistance tends to R1+R2.

      D12        Rm         R1-2
      0.05       7.7838     15.36236
      0.10       7.0517     16.09446
      0.20       6.2883     16.85786
      0.50       5.3530     17.79316
      1.00       4.2929     18.85326
      2.00       3.2794     19.86676
      5.00       2.0737     21.08246
      10.0       1.3027     21.84346
      100.0      0.1586     22.98630

OPH.2019-07-06 17:34