Electronics Guide

Why Power Engineers Work in Per-Unit

The Trouble with Ohms

An impedance on the 12.47 kV side of a substation transformer cannot simply be added to the impedance of the 138 kV line that feeds it. One must first be referred across the transformer by the square of its turns ratio, and every further transformer adds another referral and another chance for error. The ohmic values also span a very wide range. The 100 MVA, 10 percent step-up transformer in the worked example below has a leakage reactance of 0.190 Ω referred to its 13.8 kV winding and 19.0 Ω referred to its 138 kV winding.

What Per-Unit Changes

A per-unit value is an actual value divided by a base value of the same dimension:

quantity in per unit = actual quantity / base quantity

Percent is the same ratio times 100: a nameplate impedance of 10 percent is 0.10 per unit on the transformer's own rating. Four practical gains follow:

• Transformers become series impedances. With base voltages in the ratio of a transformer's rated voltages, its ideal transformer becomes a 1:1 connection, and a multi-voltage network becomes one circuit.
• The three-phase factors cancel. With the conventional three-phase bases, the √3 and 3 appear only when a result converts back to physical units.
• Equipment data fall in narrow ranges. Machines and transformers of very different ratings have similar impedances on their own ratings, so a data error stands out.
• Normal operation sits near 1. A bus at 0.88 per unit or a transformer loaded to 1.3 per unit of its rating stands out immediately.

The price is bookkeeping: every per-unit number carries an implied base, data arrive on many different bases, and phase shifts, taps, and off-nominal nameplate voltages need explicit handling.

Per-Unit Beyond Impedance

Standards and settings use the same language. IEEE 1547-2018 states the reactive-power capability of distributed energy resources as a fraction of nameplate apparent power, and utilities publish inverter volt-var curves with voltage breakpoints in per unit, as Hosting Capacity and Distribution Impacts describes. Many protective relays set pickup as a multiple of nominal current, which is per-unit current under another name.

Base Quantities and Their Relationships

Choose Two, Derive the Rest

Voltage, current, apparent power, and impedance are linked by two equations, so only two bases can be chosen freely. Power engineers choose a base apparent power, which is the same everywhere in the network, and a base voltage, which changes from zone to zone at each transformer. The base current and base impedance follow. Bases are real numbers with no phase angle, so dividing a phasor by its base scales the magnitude and leaves the angle unchanged.

Single-Phase Relationships

For a single-phase circuit with base voltage V_base and base apparent power S_base:

I_base = S_base / V_base

Z_base = V_base / I_base = V_base² / S_base

Y_base = 1 / Z_base

Resistance, reactance, and impedance share Z_base; conductance, susceptance, and admittance share Y_base; real, reactive, and apparent power share S_base. A complex impedance converts term by term, so Z_pu = R/Z_base + jX/Z_base. Because the bases themselves obey Ohm's law and the power equation, both hold unchanged in per unit:

V_pu = Z_pu I_pu and S_pu = V_pu I_pu^*

where the asterisk marks the complex conjugate.

Three-Phase Relationships

Three-phase work takes S_base as a three-phase apparent power, usually in MVA, and V_base as a line-to-line voltage, usually in kV. The derived bases are

I_base = S_base,3φ / (√3 V_base,LL)

Z_base = V_base,LL² / S_base,3φ = kV_LL² / MVA_3φ

The last form gives ohms directly when voltage is in kilovolts and power in megavolt-amperes. With kilovolt-amperes a factor of 1,000 appears: Z_base = 1,000 × kV_LL² / kVA_3φ ohms. The base current in amperes is 1,000 × MVA_3φ / (√3 × kV_LL).

These forms agree exactly with the single-phase formulas applied to one phase of the wye equivalent, whose base power is S_base,3φ/3 and whose base voltage is V_base,LL/√3:

I_base = (S_base,3φ/3) / (V_base,LL/√3) = S_base,3φ / (√3 V_base,LL)

Z_base = (V_base,LL/√3)² / (S_base,3φ/3) = V_base,LL² / S_base,3φ

For a 100 MVA, 138 kV base, both routes give Z_base = 190.44 Ω and I_base = 418.37 A. In either form, the impedance being converted is the per-phase impedance of the wye equivalent.

Two consequences justify the convention. First, in a balanced system a line-to-line voltage and its line-to-neutral counterpart have the same per-unit magnitude, because both the actual value and the base carry the same factor of √3; only the 30° angle between them remains. Second, three-phase power in per unit equals per-phase power in per unit, so S_pu = V_pu I_pu^* needs no factor of 3 or √3.

Delta-Connected Elements

A delta-connected impedance can be replaced by its wye equivalent, Z_Y = Z_Δ/3, and divided by the base impedance above, or divided directly by a delta base, the line-to-line voltage squared over the per-phase power, which is three times the wye base. The factor of 3 cancels, so both routes give the same per-unit value, and a per-unit load impedance carries no record of how the load is connected.

Changing the Base of an Impedance

Generator, transformer, and motor data arrive on each unit's own rating, but a study needs every element on one system base. A round figure such as 100 MVA is common; the IEEE 14-bus test case distributed with the MATPOWER toolbox, for example, uses a 100 MVA base.

The Change-of-Base Formula

An impedance has one value in ohms, whatever base describes it. Writing that value on both bases gives

Z_Ω = Z_pu,old × V_old² / S_old = Z_pu,new × V_new² / S_new

and therefore

Z_pu,new = Z_pu,old × (S_new / S_old) × (V_old / V_new)²

The formula holds under three conditions:

• Both voltage bases apply at the same point in the network, on the same side of any transformer. Carrying an impedance across a transformer is the job of the base-voltage zones described below, not of this formula.
• The two power bases are of the same kind, both three-phase or both per phase, and the two voltage bases are of the same kind, both line-to-line or both line-to-neutral.
• Both per-unit values describe the same physical impedance, such as the same transformer at the same tap position.

The formula is linear in Z, so it converts resistance, reactance, and complex impedance alike; admittance converts by the reciprocal factor. When the voltage bases match, only the power ratio remains. A 90 MVA transformer with 10 percent impedance has 0.10 × 100/90 = 0.1111 per unit on a 100 MVA base. The number grows because the same ohms are now compared with a smaller base impedance: at 138 kV, 190.44 Ω on 100 MVA rather than 211.60 Ω on 90 MVA.

When Nameplate Voltage Differs from System Voltage

The voltage term matters whenever equipment is rated at a voltage other than the base of its zone. Motors are the common case. The U.S. Department of Energy's Premium Efficiency Motor Selection and Application Guide (2014) notes that NEMA members customarily rate motors at about 95.8 percent of nominal system voltage, so a motor for a 480 V system carries a 460 V nameplate. Consider a motor with a rated input of 450 kVA at 460 V that draws six times rated current with its rotor locked at rated voltage. Its locked-rotor impedance is the reciprocal of that current, 1/6 = 0.1667 per unit on its own rating. On a 1 MVA study base with 480 V as the base voltage,

Z_pu,new = 0.1667 × (1,000 / 450) × (460 / 480)² = 0.3401 per unit

Both values describe 0.0784 Ω. Omitting the voltage term gives 0.3704 per unit, 8.9 percent too high, and every motor-starting voltage dip or fault contribution computed from it inherits the error.

Percent Impedance on Transformer Nameplates

A transformer's percent impedance comes from a short-circuit test. Schneider Electric's Cahier Technique no. 158 defines it as the voltage, in percent of rated voltage, that must be applied to the primary winding to drive rated current through the secondary winding with the secondary terminals short-circuited. Divided by 100, it is the per-unit impedance on the transformer's rated power and rated voltages, and the same number applies from either winding, for the reason given in the next section.

Other Quantities

Voltage, current, and power change base by simpler ratios: V_pu,new = V_pu,old × V_old/V_new, I_pu,new = I_pu,old × I_old/I_new, and S_pu,new = S_pu,old × S_old/S_new. A machine's inertia constant H, its stored kinetic energy at rated speed divided by its rated apparent power, converts by the power ratio alone: H on the system base equals H on the machine base times S_machine/S_system.

Transformers in the Per-Unit System

Why the Turns Ratio Drops Out

Magnetically Coupled Circuits develops the practical transformer model: an ideal transformer of ratio a = N₁/N₂, series winding resistance and leakage reactance, and a shunt magnetizing branch. Choose base voltages on the two sides in the transformer's rated voltage ratio, V_base,1/V_base,2 = a, and use one S_base for both sides. Three results follow.

1. Voltage. The ideal transformer sets V₁ = aV₂. Dividing each side by its base gives V₁/V_base,1 = aV₂/(aV_base,2), so V_1,pu = V_2,pu.
2. Current. The ideal transformer sets the current magnitudes in the inverse ratio, I₁ = I₂/a. The base currents, S_base/V_base, stand in the same inverse ratio, so I_1,pu = I_2,pu.
3. Impedance. The base impedances satisfy Z_base,1 = a²Z_base,2. An impedance Z₂ on side 2, referred to side 1, becomes a²Z₂, and a²Z₂/Z_base,1 = Z₂/Z_base,2. Its per-unit value is the same on either side.

In per unit the ideal transformer has a ratio of 1:1, which is a plain connection, so it drops out of the circuit. What remains is the per-unit leakage impedance in series, plus the magnetizing branch in shunt where a study needs it. A power transformer's magnetizing current is a small fraction of its rated current, so load-flow and short-circuit models often omit that branch, while energization and no-load loss studies keep it.

A 50 kVA, 7,200/240 V single-phase distribution transformer has a turns ratio of 30 and base impedances of 1,036.8 Ω on its high-voltage side and 1.152 Ω on its low-voltage side, a ratio of 900, or 30². An illustrative leakage impedance of 2 percent is 20.736 Ω referred to the 7,200 V winding and 23.04 mΩ referred to the 240 V winding: two ohmic values, one per-unit value.

Base-Voltage Zones

Transformers divide a network into zones, each with one base voltage. Assigning the bases is mechanical:

1. Choose one base power for the entire network.
2. Choose the base voltage in one zone, usually the nominal voltage there or the rated voltage of a major machine or transformer.
3. Cross each transformer, multiplying the base voltage by the ratio of the transformer's rated voltages, until every zone has a base.
4. Compute the base impedance and base current of each zone, and convert every element with the bases of the zone in which it sits.

For a three-phase transformer, the ratio is that of its rated line-to-line voltages, whatever its connections. A network with loops supplies a check: going around any loop through its transformers must return to the starting base. When two paths between the same zones pass through transformers of different rated ratios, no single set of bases satisfies both, and at least one transformer must carry an off-nominal ratio in the model.

Phase Shift in Delta–Wye Transformers

The magnitude of the ratio drops out; its angle does not. A delta–wye or wye–delta transformer shifts positive-sequence voltages and currents on one side relative to the other by an odd multiple of 30°, and the shift reverses in negative sequence. For North American units, IEEE C57.12.00 fixes the direction: in positive sequence, with standard terminal markings, the line-to-neutral voltage at low-voltage terminal X1 lags that at high-voltage terminal H1 by 30°, whichever winding is the delta, and delta–delta and wye–wye units have no shift. IEC vector groups record either direction: Dyn1 places the low-voltage side 30° behind, and Dyn11 places it 30° ahead. Three-Phase Circuits and Power explains the shift and the clock-number notation.

In per unit such a transformer is an ideal phase shifter of unit magnitude, modeled below. In a radial network, or wherever every loop passes through matching shifts, a positive-sequence study can omit the shift; magnitudes and powers come out unchanged, and the true phasors in each zone are rotated by a known angle. Loops that close through unequal shifts, and all unbalanced studies, must keep it.

Off-Nominal Turns Ratios and Taps

The turns ratio drops out only when the ratio of the base voltages equals the actual turns ratio. A tap changer moves the actual ratio away from the rated one, and a transformer whose rated voltages differ from the bases of its zones does the same. The per-unit model then keeps an ideal transformer of ratio t:1, where t is the actual turns ratio divided by the ratio of the base voltages. Place the ideal transformer at bus i, so that V_i = tV′, where V′ is the voltage at the internal node between the ideal transformer and the series admittance y = 1/Z that joins that node to bus j. With both currents taken as flowing from their buses into the branch, a real t gives

I_i = (y / t²) V_i − (y / t) V_j

I_j = −(y / t) V_i + y V_j

The same equations describe a π network with a series admittance of y/t between the buses, a shunt admittance of y(1 − t)/t² at bus i, and a shunt admittance of y(t − 1)/t at bus j. With t = 1 both shunts vanish and the ordinary series impedance returns.

For a transformer with x = 0.10 per unit and t = 1.05, y = −j10. The series admittance is −j9.5238 per unit, a reactance of 0.1050; the shunt at the tap bus is +j0.4535, a capacitive susceptance; and the shunt at bus j is −j0.4762, an inductive one. With bus j open, the network gives V_j = V_i/1.05 = 0.9524 V_i, as the ratio requires.

A phase-shifting transformer has a complex ratio, t = |t|e^jθ, so V_i leads V′ by θ. The first equation becomes I_i = (y/|t|²) V_i − (y/t^*) V_j, while the coefficient of V_i in I_j stays −y/t. The admittance matrix is then no longer symmetric, so no passive π network represents the branch. The delta–wye shift is the special case |t| = 1: a C57.12.00 unit with its ideal transformer at the high-voltage bus has θ = +30° in positive sequence and −30° in negative sequence. Study programs build these terms directly into the bus admittance matrix. The open-source MATPOWER toolbox uses exactly these expressions, places the tap at a branch's from bus and the impedance at its to bus, and treats a positive shift angle as a delay of the to side. Other programs choose differently, so tap and shift data must follow the convention of the program that reads them.

Three-Winding Transformers

A three-winding transformer is modeled as a star of three impedances joined at a fictitious internal node. Its test report gives three pairwise impedances, Z_ps, Z_pt, and Z_st, each measured between two windings with the third open. The windings often have different ratings, and a report may state each pairwise value on a different MVA base, so all three must be converted to one base before the star is computed:

Z_p = ½(Z_ps + Z_pt − Z_st)

Z_s = ½(Z_ps + Z_st − Z_pt)

Z_t = ½(Z_pt + Z_st − Z_ps)

Suppose a 138/13.8/4.16 kV unit reports Z_ps = 0.08 per unit on 40 MVA, and Z_pt = 0.10 and Z_st = 0.07 per unit on 20 MVA. On a 100 MVA base these become 0.2000, 0.5000, and 0.3500, and the star impedances are Z_p = 0.1750, Z_s = 0.0250, and Z_t = 0.3250 per unit; each pair of branches sums back to its converted test value. Applied to the unconverted values, which mix two MVA bases, the same formulas give a meaningless 0.0550, 0.0250, and 0.0450. A star branch can come out very small, as Z_s does here, or even slightly negative, which is a property of the equivalent circuit rather than an error.

Typical Per-Unit Impedances

The practical payoff of per unit is that similar equipment has similar impedance on its own rating, whatever its size. Schneider Electric's Cahier Technique no. 158, Calculation of Short-Circuit Currents (updated June 2000), tabulates typical values for short-circuit work, summarized below. They suit data checks and preliminary studies; nameplate and test data govern a real study.

CT158 takes the distribution transformer values from the European harmonization document HD 428-1 S1 of October 1992, and it notes that rectifier transformers reach their higher impedance deliberately, to limit short-circuit current. Its line and cable figures are 50 Hz values, as the document's reactance formula shows; reactance scales with frequency, so they rise by a factor of 1.2 on a 60 Hz system.

Narrow Ranges, Wide Ohms

A subtransient reactance of 0.15 per unit, near the middle of the turbo-generator range, is 0.317 Ω for a 90 MVA, 13.8 kV machine and 3.63 Ω for a 5 MVA, 11 kV machine, a factor of more than 11 in ohms for one per-unit value. Across voltage levels the contrast grows. The 100 MVA step-up transformer of the worked example, at an illustrative 10 percent, presents 19.0 Ω referred to 138 kV, while a 630 kVA distribution transformer at 4 percent presents 10.7 mΩ referred to a 410 V winding. In ohms the two differ by more than three orders of magnitude; in per unit, by a factor of 2.5.

Resistance is usually much smaller than reactance in this equipment. CT158 puts generator R/X ratios at about 0.05 to 0.1 for medium-voltage machines and 0.1 to 0.2 for low-voltage machines, and it puts transformer resistance at about 0.2 times the reactance in general, with a higher ratio in small units. A large power transformer's impedance is a design choice, trading fault-current limitation against voltage regulation, and comes from its test report.

Ranges as a Data Check

A value far outside its range usually signals a base error rather than unusual equipment. A generator subtransient reactance entered as 15 was probably meant as 15 percent. A transformer impedance of 0.0010 has probably been divided by 100 twice. An induction motor impedance of 34 per unit is almost certainly on a system base: the 450 kVA motor described earlier has 34.01 per unit on a 100 MVA, 480 V base.

One-Line Diagrams

One Line for Three Phases

A one-line diagram draws a three-phase power system with a single line for each three-phase circuit, whether that circuit is an overhead line, a cable, or a bus. The simplification rests on the same fact as per-phase analysis: in a balanced system the three phases carry identical waveforms displaced in time, so one phase describes all three. The diagram shows how equipment is connected, not where it sits. It omits control wiring and shows neutral paths only where they matter, such as a neutral grounded through a resistor. Schematic Diagrams and Symbols places the one-line diagram among the other kinds of electrical diagram.

What a One-Line Diagram Records

Drawn for analysis, the diagram doubles as a data sheet, recording beside each symbol:

• Sources: utility connections with their available short-circuit power and X/R ratio; generators with rated power, voltage, power factor, reactances, and neutral grounding.
• Transformers: rated power and voltages, percent impedance, winding connections and grounding, tap range and setting, and phase displacement.
• Buses: a name or number and the nominal voltage.
• Lines and cables: length, conductor or cable type, and impedance.
• Switching devices: circuit breakers, fuses, and switches, with their ratings and normal positions.
• Loads and motors: rated power, power factor, and starting data for large motors.
• Compensation: capacitor banks and reactors with their ratings in Mvar.
• Protection and metering: current and voltage transformers with their ratios, and the relay functions each one supplies.

Kinds of One-Line Diagram

One system usually has several diagrams, each drawn for a purpose:

• Study diagrams reduce each substation to buses and the branches between them. Load-flow and short-circuit programs solve this bus-branch form.
• Operating diagrams keep every breaker and disconnect switch in node-breaker form, because operators must know which devices energize which equipment.
• Protection diagrams add the locations of current and voltage transformers, the relay functions they supply, and the breakers each relay trips.
• Three-line diagrams draw every phase and the neutral where one line is not enough, as for current transformer polarity and metering.

Keeping the Diagram True

A model built from an outdated diagram gives confident wrong answers. Short-circuit, coordination, and arc flash studies all start from the one-line diagram, and arc flash study documentation typically includes it. In IEC 61850 substations the single-line description has also become machine-readable: the Substation Configuration Language of IEC 61850-6 can carry the single-line diagram and its equipment, as Substation Automation and IEC 61850 describes.

Symbols and Annotation in IEEE and IEC Practice

Symbol Standards and Power Conventions

One-line symbols come from the same standards as schematic symbols. North American drawings follow IEEE 315, which IEEE has listed as inactive-reserved since November 7, 2019, but which remains in wide use; international drawings follow the IEC 60617 database. Power drawings add conventions of their own, and a drawing's legend governs any doubtful symbol:

• Buses are commonly drawn as heavy lines, with the circuits that leave them drawn lighter.
• Rotating machines appear as circles marked with a letter, such as G for a generator or M for a motor.
• A two-winding transformer is commonly drawn as two overlapping circles or, in many North American drawings, as two facing winding symbols. Small delta and wye marks beside the windings, with a ground symbol on a grounded neutral, record the connections.
• IEC-style diagrams commonly mark the function of a switching device at its contact, with a cross for a circuit breaker and a short bar for a disconnector. North American one-line diagrams commonly draw a power circuit breaker as a small square.
• Many operating diagrams show each switching device in its normal state, so a normally open tie between two buses stands out at a glance.

Device Function Numbers

North American protection diagrams identify relay functions with the device function numbers of IEEE C37.2. The current edition, IEEE C37.2-2022, assigns 95 numbers and 22 acronyms. On a one-line diagram the numbers commonly appear in small circles beside the current transformers that supply the relays, such as 50 for instantaneous overcurrent and 51 for time overcurrent. Suffix letters narrow the application, as in 87T for transformer differential protection, and the circuit breaker itself is device 52. Power System Protection lists the numbers most often seen.

IEC Designations

IEC practice labels the same functions differently. IEC 60617 supplies quantity-and-comparison symbols, such as I> for overcurrent, and IEC 61850 names each function as an instance of a logical node class, such as PTOC for time overcurrent. Relay makers publish the schemes side by side. The product guide for ABB's REF615 feeder relay (document 1MRS756379, revision U, December 2021) lists the low stage of its three-phase non-directional overcurrent protection as PHLPTOC in IEC 61850 terms, 3I> in IEC 60617 notation, and 51P-1 in its IEC-ANSI column. The rules for preparing the diagrams themselves come from IEC 61082-1, Preparation of Documents Used in Electrotechnology, Part 1: Rules, and equipment designations follow IEC 81346.

Impedance and Reactance Diagrams

An impedance diagram is the one-line diagram redrawn as a per-phase circuit, with every element replaced by its per-unit equivalent on the system base. It is the circuit a study actually solves.

The Reactance Diagram

A reactance diagram simplifies further by dropping resistances, magnetizing branches, line charging, and often static loads, which leaves a network of reactances alone. The simplification suits hand calculations in high-voltage networks, where reactance dominates. CT158 gives the resistance of an upstream network as about 0.1 times its impedance at 150 kV, which makes the reactance 99.5 percent of the impedance. The same source gives a ratio of about 0.3 at 6 kV and treats low-voltage conductors smaller than 150 mm² as purely resistive, so a reactance diagram of a low-voltage system misleads. Losses and feeder voltage drop always need resistance.

Sequence Networks

Everything above describes the positive-sequence network, which is all a balanced study needs. Unbalanced faults and unbalanced loads also need negative- and zero-sequence networks, built from the same one-line diagram and the same bases. Transformer data differ most, because winding connections and grounding decide whether zero-sequence current can pass. Building and connecting those networks is the subject of fault analysis and symmetrical components.

Building a Network Model for a Study

The two tools meet when an engineer builds a model. The steps are much the same for any balanced study:

1. Define the study and its boundary. Load-flow, short-circuit, and motor-starting studies need different data and machine models, and the system beyond the boundary becomes an equivalent source.
2. Collect and verify the data. Take ratings and impedances from nameplates and test reports, line and cable data from construction records, and the utility's short-circuit power and X/R ratio at the point of connection. Check taps, cable lengths, and normally open points in the field.
3. Choose the bases by the zone procedure above, and record each zone's base impedance and base current.
4. Convert every element to the system base with the change-of-base formula, and convert line and cable ohms with the base impedance of their zone.
5. Represent what does not drop out: off-nominal taps, phase shifts where loops or unbalanced studies need them, and three-winding transformers as stars.
6. Assemble the impedance diagram, number the buses, and enter the data in the study program in the units its format expects.
7. Validate before trusting the results. Compare computed voltages and flows with measurements at a known operating point, check each per-unit impedance against its typical range, and confirm that the short-circuit level at the boundary matches the utility's figure.

The External System as an Equivalent Source

A utility usually describes its system at the point of connection by the three-phase short-circuit power available there, S_sc. At a prefault voltage of 1.0 per unit, the per-unit short-circuit current equals the per-unit short-circuit power, and both equal the reciprocal of the source impedance. When the base voltage equals the voltage at which S_sc is stated,

Z_source,pu = S_base / S_sc

which in ohms is CT158's Z_up = U²/S_sc, with U the no-load line-to-line voltage. For 2,000 MVA at 138 kV on a 100 MVA base, Z_source = 0.0500 per unit, or 9.522 Ω. When only S_sc is known, CT158 suggests estimating resistance from an R/Z ratio of about 0.1 for a 150 kV network. At that ratio, R = 0.0050 and X = 0.0497 per unit, an X/R ratio of 9.95. A utility-supplied X/R ratio replaces the estimate.

Loads and Study Programs

Loads convert from MW and Mvar by dividing by the base power, and the study must also decide how each load responds to voltage. Constant-power loads draw the same power at any voltage and are the usual starting point in load flows. A constant-impedance load, Z_pu = |V_pu|²/S_pu^*, draws power in proportion to the square of its voltage. MATPOWER's case format stores one system base in a baseMVA field and gives branch resistance, reactance, and charging susceptance in per unit, with a transformer's off-nominal ratio and phase-shift angle in separate columns. Whatever the program, the engineer answers for the base on which each number was entered.

Worked Example: Generator, Transformers, Line, and Load

A 60 Hz generating station supplies an industrial substation over a 138 kV line. The task is to build the per-unit model and find the generator voltage needed to hold the substation's 12.47 kV bus at nominal voltage. The illustrative data, chosen inside the typical ranges above, are:

• Generator G: 90 MVA, 13.8 kV, with synchronous reactance X_d = 1.80 per unit and subtransient reactance X″_d = 0.15 per unit on its rating. Resistance is neglected.
• Step-up transformer T1: 100 MVA, 13.8 kV delta to 138 kV grounded wye, leakage reactance 10 percent.
• Line: 40 km at 138 kV with a series impedance of 0.10 + j0.48 Ω/km. The reactance is CT158's 0.4 Ω/km average for medium- and high-voltage lines scaled from 50 Hz to 60 Hz; the resistance is assumed. Shunt capacitance is neglected, the usual short-line approximation.
• Step-down transformer T2: 60 MVA, 138 kV delta to 12.47 kV grounded wye, leakage reactance 8 percent.
• Load: 30 MW at a power factor of 0.95 lagging, to be held at 12.47 kV.

Bases and Conversions

1. Choose the bases. Take S_base = 100 MVA and 13.8 kV in the generator zone. T1's rated ratio carries the base to 138 kV on the line, and T2's carries it to 138 × 12.47/138 = 12.47 kV at the load. The first table below lists the derived bases.
2. Convert the generator. Its rated voltage matches its zone base, so only the power ratio applies: X_d = 1.80 × 100/90 = 2.0000 per unit and X″_d = 0.15 × 100/90 = 0.1667 per unit.
3. Convert the transformers. T1 is already on a 100 MVA base at its rated voltages, so X_T1 = 0.1000 per unit. T2 converts by the power ratio: X_T2 = 0.08 × 100/60 = 0.1333 per unit, the same from either winding.
4. Convert the line. Its impedance is 40 × (0.10 + j0.48) = 4.0 + j19.2 Ω. Dividing by 190.44 Ω gives Z_line = 0.0210 + j0.1008 per unit.
5. Convert the load. The reactive power is 30 × tan(arccos 0.95) = 9.861 Mvar, so S_load = 0.3000 + j0.0986 per unit. As a constant impedance at 1.0 per unit, the load would be 3.0083 + j0.9888 per unit, or 4.678 + j1.538 Ω per phase.

The Impedance Diagram

No ideal transformer remains, because both transformers have rated ratios equal to the ratios of their zone bases, and every series element carries the same per-unit current. A fault study would use the same diagram with the subtransient reactance, j0.1667, in place of j2.0000.

Solving for the Source Voltage

1. Take the load voltage as the reference, V₄ = 1.0000 + j0 per unit. The current is I = (S_load/V₄)^* = 0.3000 − j0.0986 per unit, a magnitude of 0.3158 per unit lagging the reference by 18.19°.
2. Add the drop across T2: V₃ = V₄ + jX_T2I = 1.0131 + j0.0400, or 1.0139 per unit (139.92 kV).
3. Add the drop across the line: V₂ = V₃ + Z_lineI = 1.0294 + j0.0682, or 1.0316 per unit (142.37 kV).
4. Add the drop across T1: V₁ = V₂ + jX_T1I = 1.0393 + j0.0982, or 1.0439 per unit at an angle of 5.40° ahead of the load voltage. The generator must hold its terminals at 1.0439 × 13.8 = 14.41 kV.
5. Find the generator output: S_G = V₁I^* = 0.3021 + j0.1319 per unit, or 30.21 MW and 13.19 Mvar. The line consumes 0.209 MW, and the series reactances absorb 3.332 Mvar. The machine delivers 32.96 MVA, 0.366 per unit of its own rating, at a power factor of 0.916 lagging.
6. Find the internal voltage behind synchronous reactance, using an unsaturated round-rotor model: E = V₁ + jX_dI = 1.2365 + j0.6982, or 1.4200 per unit, leading the terminal voltage by 24.05°.
7. Convert the current to amperes with each zone's base: 0.3158 per unit is 1,462 A at 12.47 kV, 132.1 A at 138 kV, and 1,321 A at 13.8 kV.

Checking in Ohms

The same answer follows without per unit, at the cost of referring every element to one voltage. Referred to 138 kV, T1 is j19.044 Ω, T2 is j25.392 Ω, and the line is 4.0 + j19.2 Ω. The load takes 10 MW and 3.287 Mvar per phase at 79,674 V line-to-neutral, a current of 125.51 − j41.25 A, or 132.12 A. The generator voltage referred to 138 kV comes to 83,170 V line-to-neutral, 144.06 kV line-to-line, at the same angle of 5.40°. Dividing by T1's 10:1 ratio gives 14.41 kV, the per-unit result.

Reading the Result

Holding the load bus at 1.00 per unit takes 1.044 per unit at the generator, a rise of 4.39 percent. Most of it comes from reactive power flowing through 0.3342 per unit of series reactance: in the in-phase part of the drop, the reactive term contributes about 0.033 per unit and the resistive term about 0.006. A tap change on T2 or reactive support at the load would reduce it. The angles computed here omit the transformer shifts. If T1 and T2 follow IEEE C57.12.00, the true 138 kV phasors lead the computed ones by 30°, while the two shifts cancel between the load and generator zones, so the generator's 5.40° lead over the load bus stands as computed. No magnitude or power changes.

Common Errors and Troubleshooting

• Mixed kinds of base. A per-phase MVA base with a line-to-line kV base makes base impedances three times too large and per-unit impedances one-third of their true values. A three-phase MVA base with a line-to-neutral kV base does the reverse.
• Unconverted nameplate data. A transformer's percent impedance on its own rating is entered as if it were on the system base, a 460 V motor on a 480 V bus loses the voltage term of the change-of-base formula, or three-winding test values on different bases are combined without conversion.
• Inconsistent zone bases. Taking one zone's base from its nominal voltage and the next from a transformer rating whose ratio does not match creates a spurious off-nominal ratio and, in loops, false circulating current.
• Wrong zone for amperes. A per-unit current converts to amperes only with the base current of the zone in which it flows.
• Forgotten phase shift. A loop that closes through delta–wye and wye–wye transformers, or any unbalanced study, needs the 30° shift that a radial balanced study may omit.

Summary

The per-unit system divides every quantity by a base. One base power and a base voltage in each zone fix the base current and base impedance, Z_base = kV²/MVA in three-phase terms, and the change-of-base formula moves an impedance between bases at one point in the network. When base voltages follow the rated voltage ratios of the transformers, every ideal transformer becomes a 1:1 connection, and a network of many voltage levels becomes one circuit.

What does not drop out must be modeled: off-nominal taps, delta–wye shifts where loops or unbalanced studies need them, and three-winding transformers as stars on one base. The one-line diagram supplies the connections and ratings, and redrawn with per-unit values it becomes the impedance diagram that load-flow, short-circuit, and stability studies solve.

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