Electrical Engineering - Grounding(Earthing)Resistance Calc.

Author

Grounding(Earthing)Resistance Calc.

dsky

2006-02-20, 6:21 am

Dear Experts,
I'm in the mountains doing installation of a substation grounding in
Thailand, after 15 years of labor suprvision on construction works and
rote following of plans has really blunted my engineering/math analysis
skills,forgotten most of my college elect. theories, dont have
university access now so pls bear with me coz im not planning to be a
design engineer but only want to understand.
I was wondering why in most calculations does the nat.logarithm terms
always appears, for example :
(1)Basic Resistance Formula : R = rho*Lenght/Area

(2)Resistance of ground rod bed R2=
(rho/(2pi*N*L))*(ln(4L/a)-1+((2kL/A^0.5)*(N^0.5-1)^2

(3)Resistance of grid R1=
(rho/pi*Lc)*(ln(2Lc/(b*2h)^0.5)+(k*Lc/A^0.5)-k')

(4)Mutual Resistance Rm = (rho/pi*Lc)*(ln(2Lc/L)+(k*Lc/A^0.5)-k'+1)

(5)Final ground reisitance Rg = R1*R2-Rm^2/ (R1+R2-2Rm)
whre :
rho = soil resistivity in ohm*m , linear units in meters
a = diameter of ground rod
b = diametrer of ground grid or mesh & connecting conductor
L = lenght of rod
Lc = total lenght of grid conductors
A = land area occupied by ground grid with rod
k,k' are experimental constants to be sourced from IEEE table
Lt = Lc + L
Also why is the above are different from recommended formula :
(5) Rg = (rho/4)*(pi/A)^0.5 + rho/ Lt
which is taken basically from
(6) Rg = (rho/4)*(pi/A)^0.5
Can remove the terms with "A" on formulas (2) to (4) coz the result
will give me lower resistance result as compared to (5). ( just
theoretically as I would not do lest I compromise the safety of the
installation.)
also how is it different from the Dwight Equation.
all formulas are taken from IEEEstd 80 except for (1), that my friend
lend to me but would lessen my professional leverage if I further ask.

thanks for any expalination on this matters, laymans terms are also
welcome.

daestrom

2006-02-20, 3:21 pm

"dsky" <dsky_max@yahoo.com> wrote in message
news:1140428244.345970.183430@g43g2000cwa.googlegroups.com...
> Dear Experts,
> I'm in the mountains doing installation of a substation grounding in
> Thailand, after 15 years of labor suprvision on construction works and
> rote following of plans has really blunted my engineering/math analysis
> skills,forgotten most of my college elect. theories, dont have
> university access now so pls bear with me coz im not planning to be a
> design engineer but only want to understand.
> I was wondering why in most calculations does the nat.logarithm terms
> always appears, for example :
> (1)Basic Resistance Formula : R = rho*Lenght/Area
>
> (2)Resistance of ground rod bed R2=
> (rho/(2pi*N*L))*(ln(4L/a)-1+((2kL/A^0.5)*(N^0.5-1)^2
>
> (3)Resistance of grid R1=
> (rho/pi*Lc)*(ln(2Lc/(b*2h)^0.5)+(k*Lc/A^0.5)-k')
>
> (4)Mutual Resistance Rm = (rho/pi*Lc)*(ln(2Lc/L)+(k*Lc/A^0.5)-k'+1)
>
> (5)Final ground reisitance Rg = R1*R2-Rm^2/ (R1+R2-2Rm)
> whre :
> rho = soil resistivity in ohm*m , linear units in meters
> a = diameter of ground rod
> b = diametrer of ground grid or mesh & connecting conductor
> L = lenght of rod
> Lc = total lenght of grid conductors
> A = land area occupied by ground grid with rod
> k,k' are experimental constants to be sourced from IEEE table
> Lt = Lc + L
> Also why is the above are different from recommended formula :
> (5) Rg = (rho/4)*(pi/A)^0.5 + rho/ Lt
> which is taken basically from
> (6) Rg = (rho/4)*(pi/A)^0.5
> Can remove the terms with "A" on formulas (2) to (4) coz the result
> will give me lower resistance result as compared to (5). ( just
> theoretically as I would not do lest I compromise the safety of the
> installation.)
> also how is it different from the Dwight Equation.
> all formulas are taken from IEEEstd 80 except for (1), that my friend
> lend to me but would lessen my professional leverage if I further ask.
>
> thanks for any expalination on this matters, laymans terms are also
> welcome.

Well, I won't go into the details of this *particular* equation, but in
general, you'll find the natural log in a lot of places where the solutions
involve calculus and where the property of something varies with 1/R.

For a lot of things that involve adding up the 'bits' of each section spread
out over a large area, the whole thing is done by integration. If you can
come up with a formula for how the property of something changes with
distance from some point, you can usually come up with a formula that
describes how you would add up this constantly changing property from 'A' to
'B'.

One common set of problems involves moving in ever increasing circles around
some central point (as your problem does). When transferring heat, or
conducting electricity through a material from the center outward, you can
break it down into a series of concentric rings. Each ring adds another
layer of thickness, but each ring has a larger diameter so the surface area
of each ring also grows by 1/R.

And whenever you start integrating something that varies by 1/R, (or more
generally, integrating a term 1/'x' with respect to 'x'), you'll get an
e^'x' term (by definition in calculus).

daestrom

daestrom

2006-02-20, 6:21 pm

"daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in message
news:VnoKf.15280$z%5.1134@twister.nyroc.rr.com...
>
> "dsky" <dsky_max@yahoo.com> wrote in message
> news:1140428244.345970.183430@g43g2000cwa.googlegroups.com...
>
> Well, I won't go into the details of this *particular* equation, but in
> general, you'll find the natural log in a lot of places where the
> solutions involve calculus and where the property of something varies with
> 1/R.
>
> For a lot of things that involve adding up the 'bits' of each section
> spread out over a large area, the whole thing is done by integration. If
> you can come up with a formula for how the property of something changes
> with distance from some point, you can usually come up with a formula that
> describes how you would add up this constantly changing property from 'A'
> to 'B'.
>
> One common set of problems involves moving in ever increasing circles
> around some central point (as your problem does). When transferring heat,
> or conducting electricity through a material from the center outward, you
> can break it down into a series of concentric rings. Each ring adds
> another layer of thickness, but each ring has a larger diameter so the
> surface area of each ring also grows by 1/R.
>
> And whenever you start integrating something that varies by 1/R, (or more
> generally, integrating a term 1/'x' with respect to 'x'), you'll get an
> e^'x' term (by definition in calculus).
>

OOPS, I mean you'll get an ln(x) term.

daestrom

dsky

2006-02-26, 3:21 am

Thanks Mr daestrom, it is clear explanation
till next time
dsky

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