These answers are given by our panel of experts. I am chairing this panel, and the other panel members include Bill Summers, James Stallcup, and Dan Leaf. The opinion expressed is that of the panel. If a panelist disagrees with the majority opinion, his explanation is printed following the answer, Although authoritative, the answers printed here are not, and cannot be relied on as formal interpretations of the National Electrical Code.
A long circuit length raises questions abut fault-current return capabilities when a standby system, not separately derived, is running on its generator.
More often than not, grounding discussions concerning generators involve separately derived systems. From the definition in Art. 100, a separately derived system "has no direct electrical connection, including a solidly connected grounded circuit conductor, to supply conductors originating in another system." The drawing shows a standby system with a solid neutral bar in the transfer switch, and which, therefore, is not separately derived.
These feeders are also quite long, running about 300 ft between the transfer switch and both the normal (utility) and standby sources. We were asked whether the generator neutral could be grounded to its frame. This would take 600 ft out of the ground-fault circuit length since the normal switchboard would no longer be part of the circuit path while the generator was running. The suggested arrangement does represent a reduction in the impedance in the ground-fault return path, a primary objective of Sec. 250-51:
250-51. Effective Grounding Path. The path to ground from circuits, equipment, and metal enclosures for conductors shall (1) be permanent and continuous; (2) have capacity to conduct safely any fault current likely to be imposed on it; and (3) have sufficiently low impedance to limit the voltage to ground and to facilitate the operation of the circuit protective devices.
On the other hand, such a connection clearly violates the general rule against making equipment grounding connections to grounded circuit conductors downstream of the service disconnect, as Sec. 250-23(a) provides in its final sentence:
A grounding connection shall not be made to any grounded circuit conductor on the load side of the service disconnecting means.
One reason for the 3-pole transfer switch in this case is that the local electric utility will not allow the facility to use a 4-pole switch because of bad experiences with transient disturbances during some transfer operations. What should be done?
The EC&M panel's response
Although we applaud the concern for reducing the impedance in the grounding return path, the requirement in Sec. 250-23(a) must be strictly observed. If the generator neutral were grounded to the frame, neutral return current from the load during normal operation would routinely divide in the transfer switch. Most of it would stay on the normal neutral, but some would travel to the generator frame over the standby neutral conductor. At that point these currents would return over the equipment grounding conductors to the main bonding jumper in the service.
Due to voltage drop from current passing over the equipment grounding systems, the result would be measurable voltages throughout the building on every conductive surface of electrical equipment. Equipment grounding conductors must never routinely carry load currents.
Nevertheless, the requirement in Sec. 250-51 must also be observed. In the event of a fault while the generator is running, that fault must be cleared promptly. In this specific example, the 100A feeder circuits used No. 2 THHN installed in 1 1/4-in. galvanized rigid conduit (GRC). The return path therefore consists of 300 ft of GRC and 600 ft of No. 2 in a steel raceway.
Calculating impedance
Figuring the impedance of this return circuit is surprisingly complicated, and not commonly understood. In order to have full confidence in this procedure, we went back to the original research based on lengths of 3-in. and 4-in. rigid conduit that was published in 1954.
The first problem is that the impedance of a magnetic raceway decreases with increasing current. The reactance decreases due to increasing magnetic saturation of the metallic conduit; the resistance decreases because high currents increasingly utilize the entire thickness of the conduit instead of only the outer skin. For example, the DC resistance of 1 1/4-in. GRC is 0.012 ohms per hundred feet. If the conduit is carrying 200A, its impedance is 0.0528 ohms per hundred feet; at 500A and 1000A, the equivalent impedances fall to 0.0329 and 0.0204 ohms respectively. These values are from industry sources largely based on testing, and cannot be reliably calculated in the field.
The other problem is that the circuit cannot be analyzed as a series path from the load panel to the main bonding jumper over the raceway, and then back to the transfer switch over the neutral conductor. In this circuit, due to magnetic coupling between the conduit and the neutral, the total impedance is essentially the one-way impedance of the conduit (for which the quality of workmanship is crucial), and the size of the conductor matters little. Using this principle and 500A impedance values, chosen by the rule of thumb that a current of five times the overcurrent device setting will trip reasonably quickly (in the 10-see range), the total impedance from transfer switch to service and back is 3 x 0.0329 = 0.0987 ohms.
The impedance of the 300-ft neutral run from the transfer switch to the generator is much easier. Using Chapter 9, Table 8 (recalculated to 25 [degrees] C) the resistance for the No. 2 will be 0.0163 ohms per 100 ft. From Table 9 the reactance in a magnetic raceway is 0.0057 ohms/100 ft. The total impedance is then [square root of ([R.sup.2] + [X.sup.2])] = 0.0173 ohms/100 ft or 0.0519 ohms. This brings the overall impedance to (0.0519 + 0.0987) = 0.151 ohms.
Using Ohms law, and allowing for a 50V drop in an arcing fault on a 208Y/120V system (again, a rule of thumb), I = E/Z = 70/0.151 = 464A. This is somewhat low and needs attention, especially considering generator impedance that will further reduce current. If the neutral were increased to No. 1/0 on the generator side of the transfer switch (as noted, there is little to be gained on the other side), then the impedance in this segment falls to 0.0348 ohms, for a total of 0.0348 + 0.0987 = 0.134 ohms. This would produce a current of 522A, which is probably OK. Note that the 50V drop in the arc is probably unrealistically high; at 120V the arcing circuit is unlikely to sustain itself without welding. The resulting 900A fault (120V/0.134 ohms) would trip almost immediately.