Per-Unit Calculator for Three-Phase Fault Current

This per-unit calculator puts the utility, transformers, cables, reactors and motors of a three-phase system on one base and gives the bolted fault current at the bus you choose, with every step shown. It also converts an impedance between ohms, percent and per unit and moves it to a new base. In the worked example, a 500 MVA utility at 13.8 kV, a 1,500 kVA transformer at 5.75%, 100 ft of four sets of 500 kcmil copper and 500 hp of motors give 28.72 kA at 480 V, with an X/R of 4.79.

How to use the calculator

Choose the one-line or the base converter. For the one-line, keep 100 MVA as the system base, enter the utility voltage and its available fault level in MVA or kA with its X/R, then list up to six elements in order from the source to the fault: transformers from their nameplate (kVA, %Z, X/R and the two voltage ratings), cables from their ohms per length and the number of parallel sets, and reactors from their ohms and X/R. Pick the bus where the fault is and, if motors run at that bus, enter their total size. The result shows the fault current, the X/R at the fault and the fault MVA, and the full table lists every element in ohms and per unit so you can check it by hand.

How to calculate fault current with the per-unit method

1. Pick a base. Choose one MVA base for the whole system, usually 100 MVA, and use the source voltage as the base voltage of the first zone.
2. Set the base voltage of each zone. Every transformer starts a new zone. Its base voltage is the base of the zone before it times the transformer’s secondary rating divided by its primary rating.
3. Put each element on the system base. Divide ohms by the zone base impedance, kV² ÷ MVA. Move a transformer’s %Z from its own kVA and voltage to the system base with Zpu new = Zpu old × (kV old ÷ kV new)² × (MVA new ÷ MVA old). Split each impedance into R and X with its X/R ratio.
4. Add the series path. Add the R values and the X values of the source and every element between the source and the fault.
5. Add the motors in parallel. Convert the motors’ subtransient reactance to the system base and combine it with the path: Z = Z1 × Z2 ÷ (Z1 + Z2).
6. Find the current. With a prefault voltage of 1.0 per unit, the fault current is 1 ÷ |Z| per unit. Multiply by the base current at the fault bus, MVA ÷ (1.732 × kV), to get kA.

Formulas

• Base impedance: Zbase = kV² ÷ MVA, in ohms.
• Base current: Ibase = MVA ÷ (1.732 × kV), in kA.
• Per unit from ohms: Zpu = Z ÷ Zbase. Per unit from percent: Zpu = %Z ÷ 100.
• Change of base: Zpu new = Zpu old × (kV old ÷ kV new)² × (MVA new ÷ MVA old).
• R and X from Z and X/R: R = Z ÷ √(1 + (X/R)²), X = R × X/R.
• Utility source: Zpu = MVA base ÷ short-circuit MVA, with short-circuit MVA = 1.732 × kV × kA.
• Fault current: I = Ibase ÷ |Z|. Fault MVA = MVA base ÷ |Z|. X/R at the fault = X ÷ R of the total impedance.

Worked example

A 13.8 kV utility service with 500 MVA available at an X/R of 15 feeds a 1,500 kVA transformer, 13.8 kV to 480 V, 5.75% impedance and X/R 6. The secondary runs 100 ft through four sets of 500 kcmil copper in steel conduit (0.029 + j0.048 Ω per 1,000 ft, from NEC Chapter 9, Table 9) to a switchboard with 500 hp of 460 V motors. Base: 100 MVA, 13.8 kV on the primary and 0.48 kV on the secondary, where Zbase is 0.002304 Ω and Ibase is 120.28 kA.

|Z| = √(0.8556² + 4.100²) = 4.189 pu. Fault current = 120.28 ÷ 4.189 = 28.72 kA symmetrical. X/R = 4.100 ÷ 0.8556 = 4.79. Fault level = 100 ÷ 4.189 = 23.9 MVA. The utility path gives 26.13 kA and the motors 2.58 kA.

Assumptions

• The fault is a balanced three-phase fault, so only the positive-sequence network is used. Line-to-ground and line-to-line faults need the negative and zero sequence networks too.
• The voltage before the fault is 1.0 per unit of the bus base voltage.
• The fault is bolted: no arc resistance and no fault resistance.
• Load current before the fault is ignored.
• Motors are represented by their subtransient reactance only, so the result is the first-cycle symmetrical current. Their contribution decays over the next few cycles, which the calculator does not model.
• The system is radial: one source and one series path to the fault, with motors only at the fault bus.
• Results are symmetrical rms currents. The DC offset and the asymmetrical peak are not included.

Frequently asked questions

Why use per unit for fault calculations?

Per unit turns every impedance into a fraction of a base, so a transformer looks the same from either side and a whole system with several voltages becomes one simple circuit. Equipment impedances already come in percent of their own rating, which is per unit times 100, so most of the data needs only a change of base.

How do I change the base of an impedance?

Multiply the per-unit value by the square of the old base voltage over the new one, and by the new base MVA over the old one. A transformer of 1,500 kVA at 5.75% has 0.0575 × 100 ÷ 1.5 = 3.833 per unit on a 100 MVA base at its rated voltage.

Is a transformer’s %Z on its own base?

Yes. The nameplate %Z is on the transformer’s own kVA rating and rated voltages. The calculator converts it to the system base, and when a rated voltage differs from the zone base voltage it applies the (rated kV ÷ base kV)² correction and says so.

How does the calculator handle motor contribution?

Running motors act as small generators for a few cycles after a fault. The calculator takes the total motor load at the fault bus, one hp as about one kVA, with its subtransient reactance and X/R, converts it to the system base and places it in parallel with the source path. Because the currents are added with their angles, the total can differ slightly from the plain sum of the two contributions.

Does the choice of base MVA change the answer?

No. The per-unit impedances and the base current change together, so the fault current in kA is the same on any base. 100 MVA is common because it keeps utility and plant values in a familiar range.

Why can the result differ from the available fault current calculator?

The point-to-point result of the available fault current calculator adds impedance magnitudes, while this calculator adds R and X separately: the two agree when every element has the same X/R, and the per-unit result is slightly higher when the X/R values differ. The sequence network section on that page works in per unit like this one and also covers line-to-ground and line-to-line faults. Both of its methods can reduce the transformer %Z by a tolerance factor such as 0.9; to compare, enter the same reduced %Z here.

When to bring in an engineer

A radial sketch is a good check, but a real system has more sources, more branches and equipment ratings to verify. Dreiym’s licensed Professional Engineers build the full model in a short circuit study, check each device’s interrupting and withstand rating, and carry the results into arc flash and coordination work, PE-stamped where required. Related tools: the available fault current calculator and, to go from the fault current to incident energy, the arc flash calculator. See our short circuit analysis and electrical power studies services, or contact us for a free quote. We typically respond within one business day.