The maximum grid current (IG) is the design current that passes from the substation grid into the soil during a ground fault. It is not the whole fault current: part returns through shield wires, neutrals and buried elements. The split factor gives the fraction that enters the soil, and the decrement factor adds the effect of asymmetry.
Short answer
The maximum grid current (IG) is the design value of the current that flows between the grounding grid and the surrounding soil during a ground fault. It is obtained in three steps: take the symmetrical ground fault current, apply the split factor (Sf) to keep the fraction that actually enters the soil, and multiply by the decrement factor (Df), which includes the effect of the dc component of the fault. IG is used to calculate the ground potential rise of the grid and the step and touch voltages. Which fault produces the largest current is explained in ground fault type and most severe fault.
What IG is and how it relates to the fault
IEEE Std 80-2013 separates three currents, which must not be confused:
| Current | What it is |
|---|---|
| Symmetrical ground fault current | The rms value of the ground fault, equal to three times the zero-sequence current |
| Symmetrical grid current (Ig) | The part of that current that passes between the grid and the soil; it is the fault current multiplied by the split factor |
| Maximum grid current (IG) | Design value: the symmetrical grid current multiplied by the decrement factor |
The split factor is the ratio between the current entering the soil through the grid and three times the zero-sequence current. The standard treats it as constant throughout the fault, although in reality it changes with time depending on how the contributions decay and the breakers operate.
In most cases, the largest grid current produces the most dangerous condition. For that reason the standard procedure first evaluates which fault type and location give the greatest current flow between the grid and the soil, calculates Sf and the symmetrical grid current for those faults, applies the decrement according to the duration of each fault, picks the largest result and considers future increases of fault current.
- Most severe fault. Type and location that give the most current between grid and soil
- Split factor. Fraction of the fault that enters the soil through the grid
- Symmetrical grid current. Ground fault current times the split factor
- Decrement factor. According to fault duration and waveform asymmetry
- Maximum grid current IG. Checked against future increases of the fault level
How much current leaves through other paths
Shield wires and neutrals
When the shield wires of transmission lines or the neutral conductors are bonded to the substation ground, a significant part of the fault current is diverted away from the grid. That connection usually increases the potential rise at the tower bases and decreases it at the substation: each nearby tower shares the voltage rise of the grid, whatever the cause, instead of being affected only by a local insulation fault.
Buried cables and pipes
Buried cables with a shield or armor in good contact with the soil, and metallic pipes bonded to the grid that extend beyond its perimeter, have a similar effect. They divert part of the current and slightly reduce the potential rise and the local gradients.
Because the current pattern is complex, this effect is hard to calculate. In addition, the standard warns that these elements can introduce hazards outside the substation, which are treated in the clause on special areas.
What the split factor depends on
The standard lists four factors:
- the location of the fault,
- the impedance of the grid,
- the buried cables and pipes near or connected to the grid,
- the shield wires, neutrals or other return paths.
If the additional paths in the last two items are ignored, the split can be calculated with traditional symmetrical components, but the result may be too conservative.
How the split is calculated
The calculation consists of building an equivalent representation of the shield wires and neutrals connected to the grid and solving it to find what fraction of the total current flows between the grid and the soil and what fraction through the shield wires or neutrals.
The standard describes several published methods, from span-by-span models to algorithms with multilayer soil. In a simplified span-by-span method, each shield wire is modeled with the impedance of each span and the grounding resistance of each tower, forming a ladder network that is reduced to an input impedance seen from the fault point; the split factor comes from dividing the current between the grid resistance and those impedances. Detailed models of wires, phase conductors and towers give the best evaluation, and approximate methods give comparable results in many simple cases.
Annex C of the standard, informative, offers curves of Sf against grid resistance, based on an EPRI computer program. The curves are grouped in four categories according to how much of the fault current comes from remote sources and how much from local sources:
| Category | Remote source | Local source | Typical example |
|---|---|---|---|
| A | 100% | 0% | Distribution substation with a delta-wye transformer |
| B | 75% | 25% | Transmission substation or plant with local zero-sequence sources |
| C | 50% | 50% | Same |
| D | 25% | 75% | Same |
Category A works well in practical cases; B to D are rough approximations and their accuracy depends on the local source of fault current.
Effect of resistances
- Grid resistance. In most cases it is enough to calculate IG without considering the system, grid or fault resistance. The error is small and on the safe side. Only in unusual cases, when the estimated grid resistance is large compared with the system reactance, is it advisable to include it.
- Fault resistance. If the insulation failure occurs inside the substation, it must be assumed to be zero. If it occurs outside, on a line connected to the bus, a conservative minimum value of fault resistance can be used, multiplied by three and added to the other zero-sequence resistances. If the actual resistance does not reach that value, it is neglected.
Future changes
IG must reflect the expected growth of the system. If growth is considered and current division is ignored, the grid ends up oversized. Disconnecting the shield wires of the lines reaching the substation can change the current through the grid and leave the system inadequate. Adding shield wires can reduce the part that enters the soil and leave the existing grid oversized.
Where IG enters the design
IG determines the potential rise of the grid and is the input that multiplies the soil resistivity in the step and touch voltages. The IEEE 80 design procedure places it at step 6. The starting fault current comes from the short-circuit study, and asymmetry is explained in short-circuit peak factor and X/R ratio. The free Grounding resistance tool lets you see the effect of a change in the current.
How we do it at Tiza Group
Determining IG with real data needs the short-circuit study, the data of the lines with shield wires and the estimated grid resistance. At Tiza Group we do that calculation within the grounding study of medium-, high- and extra-high-voltage substations, remotely and worldwide. See our engineering design and studies service or write to us.
Related guides
- IEEE 80: Substation Grounding Safety
- Substation Grounding Design: IEEE 80 Method
- Decrement Factor Df and Fault Asymmetry
- Which Ground Fault Sizes the Substation Grid
- Short-Circuit Study: Sources, IEC 60909 and ANSI/IEEE
References
- IEEE Std 80-2013, IEEE Guide for Safety in AC Substation Grounding, clauses 15.1, 15.2, 15.4 to 15.7 and 15.9; Annex C.
- IEEE Std 80-2013, clause 15.11, effect of future changes.
Frequently asked questions
Why is the grid not designed with the whole ground fault current?
Because part of that current returns to the source through the shield wires of the lines, through connected neutrals and through buried cables or pipes bonded to the grid, without passing through the substation soil. Designing with the total current oversizes the grid; the standard says to use only the part that flows between the grid and remote earth.
What happens to IG if the shield wires of a line reaching the substation are removed?
The split factor increases, because the current no longer has that return path and a larger fraction enters through the soil. A grid that complied may end up too small. Future changes that add shield wires do the opposite: they reduce the part that enters the grid and leave it oversized.
Can the fault resistance be included to reduce IG?
Only if the fault occurs outside the substation, on a line connected to the bus, and a safe minimum value of that resistance can be assigned. If the fault is an insulation failure inside the substation, the only safe assumption is zero resistance.