[TowerTalk] Fwd: Tower base/Ufer ground

[Jim Lux]

On 5/18/14, 7:39 AM, Roger (K8RI) on TT wrote:
>
>  From the dictionary:  There is no need to go any farther.  Forget your
> calculations, this is the definition.
>
> Electrical resistivity quantifies how strongly a given material opposes
> the flow of electric current. A low resistivity indicates a material
> that readily allows the movement of electric charge. Resistivity is
> commonly represented by the Greek letter ρ.


The SI unit of electrical
> resistivity is the ohm⋅metre although other units like ohm⋅centimetre
> are also in use.

Seems like what I was saying, no?

  *As an example, if a 1 m × 1 m × 1 m solid cube of
> material has sheet contacts on two opposite faces, and the resistance
> between these contacts is 1 Ω, then the resistivity of the material is 1
> Ω⋅m.* (that is misleading as the formula actually  gives ohms per cubic
> meter, not ohms per meter.

Actually not.. if the resistance between the two faces is R ohms, then 
the resistivity (rho) is rho=R * Area/length which has units of Ohms * 
meters

>
> Please note Resistivity is in ohms per cubic Centimeter or ohms per
> cubic meter not ohms per meter, or centimeter as is often stated which
> is incorrect.

Yes, I agree, there's no "ohms/meter"


The resistance has to be proportional to length (distance between 
electrodes), because if you double the length, the resistance doubles.

Likewise, the resistance is inversely proportional to the cross 
sectional area (put two identical bars in parallel, and the resistance 
is half the resistance of a single bar).

Therefore, so if you have a property called "resistivity" which turns 
into "resistance" by factoring in the geometry, it has to be

R = Length/Area * rho.

Since Length/Area has units of 1/length (1/meters), and R is in ohms, 
then rho has to be ohms * meters, so that when you do ohms * meters * 
(1/meters) it comes out as ohms.


I note that a lot of software and tables mis-identify conductivity (not 
conductance) and resistivity.  More than one table shows resistivity as 
ohms/meter, which is, of course, in correct.  Ditto for conductivty, 
you'll see Siemen-meter or mho-cm, when it should be Siemen/meter or mho/cm

And that doesn't even get into when you look at older texts before units 
were systematized. Take a look at NBS pubs from the early 20th century 
or handbooks of the era and you see a lot of weird units (for instance, 
inductance specified in terms of length).  For any consistent system of 
units you get 3 that are nice easy numbers, and one oddball one with 
weird constants (usually involving 4*pi or its inverse).

> When calculating you must use ohms cc^3 or M^3 for the units to properly
> cancel.
>
I don't think so.. can you show an example using cubic units?

I can give one using ohm-meters as resistivity.

Calculate the current through a concrete bar that is 10 ohm-meter 
resistivity. The bar is 1 meter long and 10x10 cm in cross section. A 
voltage of 1000 Volts is applied to the long axis of the bar.

R = 10 ohm-meter * 1 meter / (0.1 * 0.1 meter^2) = 10/0.01 = 1000 ohms.
I = V/R, so current is 1 Amp.
The current density is 100 Amps/square meter = 1 Amp/ 0.01 square meters

If one makes the bar 20x20 cm, what is the current?
R = 10*1 /0.04 = 250 ohms
The current density is 4 amps / 0.04 square meters or the same 100 
Amps/square meter, which makes sense, because the voltage between the 
ends is the same (1000V) and the current is flowing "along" the bar, not 
crossways


A more complex question.. Given that the soil has a conductivity (sigma) 
of 5 mS/m, what is the current flow through a 2 meter trough which is 
30x30cm with 10,000 volts along the length it?

The conductance G is sigma * Area/Length  = 0.005 * 0.09/2 Siemens = 
0.000225 Siemens (or mhos).

I = G * V

So, I = 2.25E-4 * 1E4 = 2.25 Amps

so the current is 4 Amps



> Too many texts use the simplified versions which are misleading.


>
> 73
>
> Roger (K8RI)
>
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