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Re: [TowerTalk] Fwd: Tower base/Ufer ground

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Fwd: Tower base/Ufer ground
From: "Roger (K8RI) on TT" <K8RI-on-TowerTalk@tm.net>
Date: Mon, 19 May 2014 02:21:35 -0400
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
On 5/18/2014 7:44 PM, Jim Lux wrote:
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?

Well...not quite, the S/I unit Om/meter is not the same as ohms per cubic meter./ One is squared and the other cubed. It's an unfortunate choice of units. They "can" be interchangeable "AT TIMES" and not at others.

Given a reistor which is for all intensive purposes, dimensionless. It can be a 1/2" long and a 1/4" in diameter, ot it can be a f18" long ans an inch in diameter. It's value is ohms. You could add wattage, but that's not relevant in this discussion and The resistance and resistivity are the same. It was the use of resistivity in the case that causes learning problems.

But instead of the resistor, let's look at the material (other than wire). A carbon composition resistor depends not only the material, but the heat and pressure to get the desired "Resistivity" of the material. Another is concrete. In both cases we are using the resistivity of a unit volume. We have added a dimension and the equations use for oms/meter no longer work because we have a third dimension. You can not get from a linear resistance to a unit volume because we are short one dimension. It's easy going the other way


 *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


If you need the resistivity of a material such as DI cooling water, the easiest way is to measure the resistance between two 1 cm square plates, with 1 cm spacing. This happens to be a spot where the two systems are the same and I could give a tech the probe and just take the meter reading directly in ohms per cubic cm. Just sticking two meter probes in will not directly give the resistivity needed to know if the water is still good although this is often referred to as resistivity.

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"

Actually there is when purchasing wire. CAT5 and 6 have a resistance per unit length, but that's straying from the discussion.


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

I can get a wire wound twice as long, with the same wire and the same resistance. The resistor is longer, but the wire is the But that is of concern in making the resistors, not using them, or dimensions are a second order of importance. We ned resistance, we need to dissipate power and we may be limited at to available weight and volume.

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.

But formal resistivity is in ohms per cubic centimeter, or meter. If you sened a request to a lab for material resistivity it will come back, typically as ohms per cubic unit, not in 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 did at he beginning of the previous post,

"The SI unit of electrical resistivity is the ohm⋅metre although other units like ohm⋅centimetre are also in use. 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." They calculated a 3 dimensional case and presented the results in two. They are adding the third dimension back by multiplying Ohms (two dimensionally derived)by m... It would be a whole lot simpler to just say M^3
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.

If the bar dimensions are known, it can be treated as a simple resistor. Again, this is a case where the two systems coincide.



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?

Are you defining a bar 20 X 20 cm sq?  4M^2?

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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