Infrastructure Engineering · Lesson 15 of 24 · 25 minutes

Declare transformer and fault-current boundaries

A current value without its voltage, power base and network boundary is easy to misuse. This lesson builds a small equivalent circuit in layers, then shows how changing an assumption changes a fault-current screen. The purpose is to expose model limits before equipment selection, not to turn an attractive fault animation into a completed short-circuit study.

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Learning objectives

Calculate consistent three-phase current and impedance bases.
Compare infinite- and finite-source fault estimates.
Keep continuous, peak, interrupting and thermal duties distinct.

Editorial derivation

Choose consistent bases

For total three-phase apparent power S and line-to-line voltage V, base impedance is V²/S and full-load current is S/(√3 V). Percentage impedance must be divided by 100 on its stated base. The source contribution and transformer contribution cannot be added as unrelated percentages. Complex impedances normally require separate resistance and reactance; the teaching comparison explicitly adopts purely reactive values.

Illustrative assumption

Build the two-megavolt-ampere example

At 2 MVA, 400 V and 6% impedance, full-load current is 2,886.75 A, base impedance is 0.08 Ω and transformer impedance is 0.0048 Ω. With an infinite source, the simplified three-phase fault is about 48.11 kA. Adding an illustrative 250 MVA source strength contributes 0.00064 Ω and reduces the screen to about 42.45 kA. Both calculations omit the actual cable route and equipment tolerances.

Unresolved

Read the missing-duty list

The initial symmetrical RMS fault is not the asymmetric peak or the current at interruption. Earth faults require the actual grounding and zero-sequence paths. Minimum faults under weak-source or inverter operation can be harder to detect even when maximum duty is lower. The manuscript also preserves a conflict between a 25 kA note, a 26.2 kA duty and a 31.5 kA schedule; one favourable schedule entry does not reconcile the whole design.

Predict → run → explain

Read the declared boundary and units. Predict the selected result before calculating.
Run the starting case and inspect the ledger, assumptions and model limits.
Apply the challenge, run again, and explain the difference. Export only the last completed inputs and results.

Zbase = V²/(S × 10⁶); Ztransformer = (z%/100) × Zbase; Zsource = V²/(source MVA × 10⁶); Iload = S × 10⁶/(√3 V); Ifault = V/[√3 (Ztransformer + Zsource)]. Purely reactive, bolted three-phase screen.

Declared starting inputs

{
  "mva": 2,
  "voltageV": 400,
  "zPercent": 6,
  "sourceMVA": 250
}

Worked default — unrounded educational outputs
Quantity	Value	Unit
fullLoadA	2886.751345948129	A
baseOhm	0.08	Ω
transformerOhm	0.0048	Ω
sourceOhm	0.00064	Ω
faultKA	42.45222567570779	kA

Challenge

What changes when the upstream source is idealized as infinite?

{
  "sourceMVA": null
}

The source impedance becomes zero and the fault estimate rises from approximately 42.45 kA to 48.11 kA. This is a changed assumption, not a better measurement.

Worked challenge
Quantity	Value	Unit
fullLoadA	2886.751345948129	A
baseOhm	0.08	Ω
transformerOhm	0.0048	Ω
sourceOhm	0	Ω
faultKA	48.112522432468815	kA

The layered equivalent circuit — static equivalent

Reveal the source, transformer and route as separate blocks.
Expose each impedance with units and base.
Trigger a user-selected three-phase fault marker; update the model-derived RMS value.
Open distinct duty cards for RMS, peak, interruption and minimum detection.

Equivalent-circuit description and a table of assumptions, impedances and duty categories.

Toggle finite/infinite source and inspect missing model contributions; do not animate arcing or imply safe working distances.

Narrative steps are not physical seconds. Essential states and quantities remain available without animation.

Assessment and answer criteria

3.1 · Find full-load and infinite-source fault current for 1.6 MVA, 400 V and 6%.

Read answer and marking criterion

2,309.40 A full load; 38.49 kA fault under the simplified model.

Criterion: Correct total-three-phase base and percentage conversion.

Source: DCIE-R1 Appendix A, PDF pp.59-61

3.2 · Convert per-unit impedance to a new apparent-power base at unchanged voltage.

Read answer and marking criterion

z_new = z_old × S_base,new / S_base,old.

Criterion: Keep the physical ohmic impedance unchanged.

Source: DCIE-R1 Appendix A, PDF pp.59-61

3.3 · Why can a larger transformer improve capacity but worsen fault duty?

Read answer and marking criterion

At fixed voltage and percentage impedance, higher MVA raises rated current and lowers ohmic impedance.

Criterion: Identify which quantities were held fixed.

Source: DCIE-R1 Appendix A, PDF pp.59-61

Source and model boundaries

Data Centre Infrastructure Engineering, reference manuscript revision 1 (DCIE-R1), 11 October 2026, chapter 3, PDF pp.18–21. PDF page 1 includes the cover. manuscript-derived; not independent source verification.

Purely reactive, bolted three-phase model; voltage factor is 1.
No zero-sequence, motor, cable, tolerance, protection-setting or breaker-selection result.

Original educational adaptation. The private PDF and missing archive are not public downloads. Source statement, Editorial derivation, Illustrative assumption and Unresolved remain distinct.

Result schema: infrastructure-screen-1; content 2.0.0; model 1.0.0. Raw inputs, outputs, units, assumptions, source locators and canonical hashes. No supported in-browser Python verification protocol exists for this result family.