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Theoretically would 24 ohms to ground be the correct resistance value for driven ground rod in drawing below: http://www.geocities.com/cinkerf/misc.html [This message has been edited by Frank Cinker (edited 08202002).]




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Frank,
The 5 Amp current is flowing in a circuit thru the resistance of the ground rod plus the resistance of the wires, connections, fuse, the source impedance and, most importantly, the connection back out of the earth to the transformer.
The circuit resistances (other than the ground connections) will get larger as the transformer is smaller and further away, and as the conductors in the circuit are smaller.
A residential setting with a pole mounted transformer two spans away and with the ground rod tied at the service disconnect will likely have between a quarter and a half an Ohm in the path through the wire and connections.
Then, the current goes into earth at the ground rod, and, in the real world with several services connected to the same transformer, the current returns out of the earth at the pole ground below the transformer, perhaps a butt plate or a ground ball, as well as at any other ground rods that have parallel return paths to the transformer neutral terminal.
I'd hazard a guess that your theoretical ground rod has 12 to 22 Ohms in its connection to earth. The fewer returns out of the ground, the closer the resistance will get to 12 Ohms.
Al
Al Hildenbrand




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Al, I don't understand. The only impedance to the flow of current of any real importance in that circuit is that caused by the earth. The others will be orders of magnitude smaller. With Frank's circuit, the resistance of the earth between the rod and the transformer grounded conductor will be very nearly 24 ohms. Don
Don(resqcapt19)




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Don,
The idea of the earth as a conductor tricks my mind into thinking of a really fat wire, with a cross section of a couple square feet runing from the service to the xfmr pole, and since dirt intuitively doesn't conduct well, I think it must be the source of the resistance. Well, I turned my minds perspective to a half sphere, with the ground rod being at the center of the radius (for simplicity, I'm treating the ground rod as a point). At 10 feet distance from the rod, the conductor formed by the earth has the surface, or cross section, of a half sphere, or ½(4Pir²) where: Pi = 3.14... r = 10 And the cross section of the earth conductor is about 628 ft². At 20 feet from the rod the earth conductor cross section becomes 2,512 ft². The conductor becomes huge quickly, and is thought of as perfect. What we, as electricians, measure is the resistance of the connection to that huge earth conductor.
The earth is simultaneously an infinite source and an infinite sink of electrons. The electron that leaves the ground rod will almost certainly not be the electron that goes up the pole to the xfmr, but the effect is the same as if it is.
The resistance that we measure with a ground resistance test set tells us how good (or bad) the connection is between the GEC and the infinite conductivity of the Earth itself. If we test the ground rod we will get one measurement. Then if we take the test set to the xfmr ground and test it, we will get another, different, reading depending on how well that ground is made.
Since what goes in, must come out, there are two contact points, at the minimum, for Frank's circuit to be complete. Each of the contacts to earth (ground rod and xfmr ground) have their own resistance. 5 Amps flowing says the total resistance in the circuit is 24 Ohms. The two big pieces are the two ground connections, not the earth itself.
Al
[This message has been edited by ElectricAL (edited 08202002).]
Al Hildenbrand




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Practically, replace the fuse with a carbon pile {~0.5kW variable resistor} and you may have something. Don {no pun…} the right PPE and have at it. The 128135Hz ‘namebrand’ test sets minimize powerfrequency and galvanicDC current effects.




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As the others have pointed out, the 24 ohms you calculate from 120/5 is the total loop impedance of the circuit, including copper conductors, xfmr winding, and the resistance of the two ground connections.
Assuming a relatively short copper path, the bulk of the loop impedance will consist of the resistance areas or voltagegradient areas of the two ground rods. So long as the two rods are far enough apart that their gradients do not overlap, increasing the distance between them makes very little difference.




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Al, The IEEE Green Book says: The grounding resistance of an electrode is made up of: 1) Resistance of the (metal) electrode 2) Contact resistance between the electrode and the soil 3) Resistance of the soil, from the electrode surface outward, in the geometry set up for the flow of current outward from the electrode to infinite earth. The first two resistances are or can be made small with respect to the third (fraction of an ohm) and can be neglected for all practical purposes. As the earth is (relatively) infinite in its size compared to the grounding systems as we know them, so too is its capacity to absorb a virtually unlimited supply of current. Practically, however, this unlimited current to the earth is transmitted across the electrodeearth soil interface in a way that is best described as follows: Around a grounding electrode, the resistance of the soil is the sum of the series resistances of virtual shells of earth, located progressively outward from the rod. The shell nearest to the rod has the smallest circumferential area of cross section, so it has the highest resistance. Successive shells outside this one have progressively larger areas, and thus progressively lower resistances. As the radius from the rod increases, the incremental resistance per unit of radius decreases effectively to zero. They go on to show a table that if you measure the resistance between a rod and a point in the earth 25' away, 25% of that total resistance will occur in the first 0.1' and 52% in the first 0.5'.
Don(resqcapt19)




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Would my method of testing satisfy NEC 250.56 ie. No additional ground rod would been needed. I know this isn't Industry standard. I simply used Ohms Law and wandered if it would work.




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Frank, Your test current of 5 amps, comes out of the ground at a second ground, the ground at the transformer. The 5 amp current is running thru two ground rods. To get the resistance of the ground rod being tested as you suggest, one has to already know the resistance of the other ground connection at the transformer. Use Ohm's law to calculate the resistance of the circuit 120 V ÷ 5 A = 24 Ohms. Then subtract the resistance of the ground connection at the transformer and you will be left with the resistance of the ground rod you are testing. Don, The IEEE Green Book quote is a much better summation than the way I say it with my words. But it says the same thing I'm trying to say. Over in General Discussion, in Frank's thread \"Gnd. rod resistance tester\"[/b] BJarney references [b]http://www.tradeport.on.ca/Appli...Ground Resistance Testing\"[/i] . In AEMC's document, on page 7, section (C) and Figure 9 show what the Green Book describes when saying: IEEE Green Book Around a grounding electrode, the resistance of the soil is the sum of the series resistances of virtual shells of earth, located progressively outward from the rod. . .As the radius from the rod increases, the incremental resistance per unit of radius decreases effectively to zero. "Effectively to zero" says the Earth has zero resistance once the current gets outside of the "effective resistance area" of the ground rod. AEMC states: Understanding Ground Resistance Testing Eventually, adding shells at a distance from the grounding electrode will no longer noticeabley affect the overall earth resistance surrounding the electrode. The distance at which this effect occures is referred to as the effective resistance area and is directly dependent on the depth of the grounding electrode. In AEMC's document, Figure 14 & 15, on page 12, show graphs of the resistance between two points when close together and far apart. In Figure 15, two ground connections have their own seperate resistances. Figure 15 shows the simplest state to consider with respect to Frank's test procedure. Al
Al Hildenbrand



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