For the design described, use the customer-required Sf = 1.0 for the system grounding calculation, and do not apply a second, assumed 70% reduction to buried-grid conductors without demonstrating the current in each conductor segment. System current division and local current sharing in a mesh are different calculations.
Check which current the design is calculating
- Prerequisite: Identify the fault location and type, the grounding conductor or mesh segment being sized, and the fault-clearing time. Record the maximum applicable fault current and the zero-sequence current at that location. Confirm these values from the system study before proceeding.
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Read the design basis: Determine whether
Sfrepresents the fraction of system fault current injected into the grounding electrode, or whether a proposed percentage represents current in a particular buried-mesh conductor. If the basis says system division factor, use it for the system injection calculation; it does not by itself establish current in each mesh segment. -
Apply the customer requirement: Where the approved design basis requires
Sf = 1.0, calculate system grounding duty using the full specified fault current. Proceed to a mesh-current calculation only after identifying the path currents and junctions within the grid.
Keep a calculation sheet with separate entries for the system fault current, the system division factor, the resulting ground-injected current, and the current assigned to each conductor segment. This prevents a percentage from being applied twice simply because both calculations concern grounding.
Check the system split-factor basis
Read the fault study and the utility-provided grounding data for the fault location. The split factor estimates how much fault current enters the station grounding system rather than returning through other parallel paths. The answer can vary substantially between substations because grounding connections, shield-wire terminations, cable shields, buried metallic services, and network configuration change the available paths.
| Reading or input | What it determines | Next check |
|---|---|---|
| Fault type and maximum fault current | Which system fault case governs the grounding study | Confirm the applicable zero-sequence current |
Zero-sequence current, I0
|
Input to the current-division calculation | Confirm the adopted factor or calculate it |
Utility-provided Sf and its fault location |
Whether a project-specific value applies to this case | Check its assumptions and calculation basis |
| Ground-grid resistance and network configuration | Inputs to the graphical or impedance method | Compare with a detailed study if needed |
A cited example in the supplied design discussion uses a 138 kV project with a 30 kA single-line-to-ground fault and a utility-provided division factor of 20%. The resulting ground-injected current is 30 kA × 0.20 = 6 kA. The stated explanation is the presence of an underground pipe-type cable feeder. Treat that as a project example, not a default factor for other stations.
Choose a defensible way to calculate system division
Use one of the stated methods, with inputs matched to the fault location and return network. IEEE Std 80-2000 Annex C is identified as a reference for the simplified graphical and impedance approaches. Verify the governing edition and project requirements before using it.
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Graphical method: Select the applicable graph for the fault category and relevant tower-footage resistance. Use the calculated grid resistance and the curve corresponding to
A/B, whereAis the number of transmission lines andBis the number of secondary feeders, as described in the method. ReadSffrom the graph’s vertical axis. Confirm that the graph’s configuration matches the station being studied before adopting the result. -
Impedance calculation: Calculate the maximum applicable fault current to ground and the zero-sequence current at the location. The supplied reference gives the relation
Sf = Ig / (3 × I0). Confirm howIgandI0are defined in the governing calculation method, and use consistent fault-current bases. A different fault type or a different shield-wire termination can change the current return paths. - Specialized grounding software: Use a suitable program when the network and geometry require detailed modeling. The cited examples are EPRI substation grounding programs and CDEGS from SES Technologies. These tools require training and accurate system parameters; compare inputs and assumptions with the fault study rather than treating software output as self-validating.
If a utility supplies a factor, ask for the fault location, fault case, return-path model, and calculation basis. If those do not match the design case, resolve the mismatch before substituting the supplied value.
Check where current enters and divides in the buried grid
Read the physical grounding layout at the faulted equipment, including grid conductors, bus supports, take-offs, tower connections, shield wires, and equipment grounding leads. Then determine the current entering the grid and the paths available from that point. Current divides according to the impedance and connectivity of the parallel paths; it does not necessarily split equally between two conductors.
At a bus support, for example, fault current may return through both a ground riser and the connected steel structure, anchor bolts, foundation, and surrounding soil. Close proximity between anchors and reinforcing steel, together with concrete’s semiconductive behavior, can provide a path even when the bolts are not directly bonded to the rebar. That establishes a possible parallel path, not a verified percentage. A fault at a bus support may also send current in both directions along a grid conductor; the path toward a take-off or shield-wire connection can carry a different share than a distant branch.
| Observed condition | What it means | Design action |
|---|---|---|
| Parallel metallic or foundation paths exist | Current may divide outside the intended riser or mesh branch | Include credible paths in the current model; do not assign an unmeasured split |
| Fault point is near a take-off or shield-wire connection | Return current can favor the shorter, lower-impedance route | Check conductor segment currents at the junctions |
| Proposed mesh factor is 70%, with no path calculation | The percentage does not establish a maximum conductor duty | Use full assigned current or calculate the segment current |
| Equipment grounding lead terminates on a smaller mesh conductor | The joint may see the lead’s full fault current even if downstream mesh current divides | Check the lead and joint for the applicable duty |
Decide whether a mesh reduction is justified
Read the current entering each mesh junction and the modeled or calculated current through the conductor segment under review. A general assumption that at least two paths exist is insufficient to establish a 70% maximum in any one path. The actual division depends on conductor layout, path impedances, connection locations, and the fault position. If the design cannot demonstrate those currents, do not use an assumed mesh reduction to reduce conductor size.
Do not multiply the system factor by the mesh percentage merely to produce a smaller conductor duty. If Sf = 0.70 is a system division factor and 70% is also proposed as a mesh-segment share, applying both yields 49% of the original current—but that result is valid only if each factor is independently defined and supported for the same fault case. The second factor needs its own current-path analysis; it is not a consequence of the first.
Where a customer requires full system fault current for the earthing calculation, retain Sf = 1.0 for that calculation. Size each mesh segment from its demonstrated current duty, or conservatively use the full applicable current where no defensible segment-current study exists. Keep equipment grounding leads full size for their assigned fault duty unless a separate calculation establishes otherwise. Check the joints where leads connect to smaller conductors for the current and thermal duty at that specific connection.
Close the thermal and connection checks
Read the fault duration, conductor material and size, initial and permissible final temperatures, and connection type used in the sizing method. Confirm that the conductor withstand calculation and the joint’s thermal capability use compatible assumptions. A conductor can survive a short fault while a connection overheats; conductor capacity alone does not clear the joint.
It does not provide the assumptions behind those figures. Do not use them as project ratings; calculate against the actual conductor data, fault duration, and design criteria.
The same discussion lists maximum allowable temperatures of 1083°C for a copper exothermic weld, 450°C for a brazed joint, and 250°C for a mechanical joint. Treat these as cited examples, not universal connector ratings. Check the selected connector’s manufacturer data and applicable design criteria. A design that relies on current division must also verify every relevant connection for the current it could carry under the modeled fault, including a case where another path is unavailable.
Grounding redundancy also matters. The discussion describes substation steel members commonly connected to the station electrode with at least two copper conductors of at least No. 2/0 AWG, arranged so damage to one conductor does not isolate a structure or equipment frame. Treat this as a described practice, not a universal minimum; check the project specification and grounding design basis for required conductor quantity and size.
Record the resolving calculation and verification
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Prerequisite: Approve the governing fault case, fault current, clearing time, and system split-factor basis. Confirm whether the customer requires
Sf = 1.0. If those values conflict, resolve the basis with the responsible design authority before final sizing. - Calculate system duty: Apply the approved system factor to the applicable fault current to determine ground-injected current. Record the factor, source, fault location, and calculation method. Verify the arithmetic before calculating mesh currents.
- Calculate segment duty: Trace the current from each injection point through the grid and determine the current in each conductor section and connection. Include credible parallel paths and the effect of the fault location. If a claimed 70% share has no calculation or defensible model, remove it from the sizing basis.
- Check thermal ratings: Compare each conductor’s assigned current and fault duration with the selected sizing method. Check each lead, joint, and connection for its own current duty and temperature limit. Confirm redundancy against project requirements.
- Verify the installed system: Compare as-built conductor routing, bonding, terminations, and connector types with the analyzed paths; confirm that no modeled parallel path or redundant connection is missing. Final verification is the documented match between the approved fault case, calculated current in every sized segment and joint, and the installed grounding layout.
Frequently asked questions
Why does the system split factor differ from a mesh split?
The system factor estimates the share of fault current entering the grounding electrode rather than returning through other system paths. A mesh split concerns current in a specific buried conductor or junction and requires its own path-current calculation.
Why use a split factor of 1.0 when sizing the grounding system?
A customer or project design basis may require full fault current for the system calculation. In that case, use Sf = 1.0 for system duty; do not treat it as proof that a separate mesh segment carries only a fraction.
Can I size mesh conductors at 70% of fault current?
Only when a calculation or suitable model demonstrates that no segment exceeds that share for the governing fault cases and locations. Otherwise, size for the full assigned duty and check equipment leads and joints independently.
How do I verify a 20% system split factor?
Confirm the fault location, applicable fault current, zero-sequence current, return-path configuration, and method behind the factor. For the cited 30 kA example, 20% gives 6 kA injected current; the project value must match its own network conditions.