The Per-Unit System and Impedance Diagrams

Chapter 3 ended with a single-line diagram and a difficulty. The generator's subtransient reactance was given as 0.20 per unit on its own 30 MVA rating; the first transformer as 0.10 on 35 MVA; the line as 100 ohms at 132 kV; the second transformer as 0.08 on 25 MVA. Every fault calculation, every load flow, every stability study needs those four elements in series. As they stand they cannot even be compared, let alone added.

One route is the classical one. Refer everything to a chosen voltage level using the transformer turns ratios: an impedance \(Z\) on the low-voltage side of a transformer of ratio \(a = N_1/N_2\) appears as \(a^2 Z\) on the high-voltage side. Do that consistently and the network becomes a single-voltage circuit that can be solved. The method is correct and it is what a student does by hand in a first course on transformers. It has three faults that matter at system scale: the \(a^2\) factors multiply up quickly and are easy to get upside down; the answers come out in ohms at some arbitrary reference level and must be translated back; and manufacturers do not quote impedances in ohms at all — the nameplate says "8%", which is already a normalised quantity on a base nobody has yet agreed to.

The per-unit system removes all three problems at once by choosing, in advance, a reference value for every kind of quantity and expressing everything as a ratio to it.

The advantages are worth listing once, because the rest of the chapter is an elaboration of them.

• Impedances on both sides of a transformer become equal, so the turns ratio disappears from the network and the whole system is one circuit at one voltage level.
• The factors of 3 and \(\sqrt3\) that Chapter 3 introduced are absorbed into the base definitions and never appear in a calculation again.
• Manufacturers already quote apparatus impedance in percent or per unit, so the data arrives in the right form.
• Per-unit impedances of machines of the same type fall in a narrow band regardless of size — a 10 MVA and a 500 MVA transformer both have reactances near 0.1 pu though their ohmic values differ by a factor of hundreds. That makes a missing data value guessable and a wrong one conspicuous.
• Voltages come out near 1.0, currents near 1.0 at full load, so a number far from unity is immediately suspicious. In a numerical load flow this scaling also keeps the Jacobian of Chapter 20 well conditioned.

Four quantities need bases: voltage, current, apparent power and impedance. They are not independent, because Ohm's law and the power relation tie them together. Fixing any two fixes the other two:

The boxed form is the one used in practice: with the voltage in kilovolts and the power in megavolt-amperes, the impedance comes out directly in ohms, because \((10^3)^2/10^6 = 1\).

Which two are chosen? Always base MVA and base kV, for the simple reason that those are the two quantities the plant is rated in. The base MVA is a single number for the entire system — usually a convenient round figure such as 10, 100 or 1000 MVA, or the rating of the largest machine. The base kV is chosen once for one part of the system and then propagated through the transformers by their turns ratios, which Section 4-6 describes; the aim is to make base kV close to the nominal operating voltage of each zone, so that per-unit voltages sit near 1.0.

Every derived base follows, and the derived per-unit quantities inherit the ordinary circuit relations unchanged:

Note what this achieves for the reactive quantity. Because \(P_B\), \(Q_B\) and \(S_B\) are the same number, the power triangle of Chapter 1 keeps its shape in per unit: \(S_{pu}^2 = P_{pu}^2 + Q_{pu}^2\) and the power factor is unchanged by normalisation. Similarly \(V_{pu} = I_{pu}Z_{pu}\) holds exactly, since dividing both sides of \(V = IZ\) by \(V_B = I_BZ_B\) leaves the equation intact. Every circuit law survives the change of units. That is the property that makes the system safe to use.

Recovering physical values at the end is a multiplication: \(V = V_{pu}V_B\), \(I = I_{pu}I_B\), \(Z = Z_{pu}Z_B\), \(S = S_{pu}S_B\). A per-unit answer is never a final answer to a practical question — a breaker is bought in kiloamperes, not per unit — but it is always the fastest route to one.

Chapter 3 established the conventions: an MVA quoted for a three-phase system is the three-phase total, and a kV is line-to-line. Adopt those as the bases and see what happens to the derived quantities. Per-phase analysis works with the line-to-neutral voltage and the line current, so start there:

This is the deeper reason the per-unit system is universal in power engineering rather than merely convenient. The per-phase equivalent circuit of Section 3-6 removed two of the three phases; per-unit normalisation removes the remaining bookkeeping that distinguished a per-phase quantity from a terminal quantity. What is left is a single scalar circuit whose equations look like elementary AC circuit theory, and every method in Parts 4 through 6 of this book is written in it.

One consequence deserves emphasis because it catches people out. In ohms a delta-connected load and its star equivalent differ by a factor of three, \(Z_\Delta = 3Z_Y\); in per unit they are the same number. The base impedance appropriate to a delta branch is not \(Z_B\) but \(3Z_B\), because the branch carries the base current divided by \(\sqrt3\) while standing across the full line-to-line base voltage: \(Z_{B\Delta} = \sqrt3\,V_{B(L-L)}/I_B = 3V_{B(L-L)}^2/S_B = 3Z_B\). Dividing an impedance three times larger by a base three times larger leaves the ratio untouched, so a per-unit impedance carries no statement of the connection.

Here is the property that justifies everything. Take a two-winding transformer of turns ratio \(a = N_1/N_2\), with primary leakage impedance \(Z_p\), secondary leakage impedance \(Z_s\), and the magnetising branch neglected for the moment. Choose the same base MVA on both sides — power is conserved through a transformer, so any other choice would be perverse — and choose base voltages in the ratio of transformation:

Now put an ideal transformer alone between those bases. It satisfies \(V_2 = V_1/a\) and \(I_2 = aI_1\). In per unit,

Equal voltage on both sides, equal current through: in per unit an ideal transformer is a piece of wire, and it can be deleted from the diagram. The leakage impedance is handled by the same argument. Refer \(Z_s\) to the primary in the usual way and divide by the primary impedance base:

Three qualifications belong with the result. A three-phase transformer is represented by its equivalent single-phase transformer, with the delta winding replaced by its star equivalent, so that the ratio used is always the line-to-line voltage ratio regardless of the vector group. The \(30^\circ\) phase shift of a star–delta connection, which Section 3-3 traced to the \(\sqrt3\) relation, is lost in this representation; it is irrelevant to balanced studies and is restored where it matters in Chapter 23. And when the base voltages are not in the ratio of transformation — which happens whenever a transformer's rated ratio differs from the ratio of the chosen bases, or when an off-nominal tap is in service — the ideal transformer does not vanish and must be retained as an off-nominal turns ratio in the bus admittance matrix of Chapter 16.

A manufacturer quotes impedance on the apparatus' own nameplate rating, because that is the only base the manufacturer knows. A system study needs everything on one common base. The conversion between the two is a two-line derivation, and it is worth doing rather than memorising, because the direction of each ratio is where errors happen.

The physical impedance in ohms is a property of the copper and iron; it does not change when the accountant changes base. Write it twice:

In the great majority of problems the base voltage in a zone is chosen equal to the rated voltage of the apparatus in it, the kV ratio is unity, and only the MVA ratio survives. The voltage term matters in exactly two situations: when a generator's rated voltage differs from the nominal system voltage of its bus (13.8 kV machine on a 13.2 kV base), and when a transformer's rated ratio differs from the ratio of the base voltages either side of it. Both appear in the worked examples.

The procedure is short and must be followed in order.

1. Choose one base MVA for the entire system. It applies to every zone at every voltage level, because power is common to all of them. Any round number will do; 100 MVA is the near-universal choice for transmission work.
2. Choose the base kV in one zone. A "zone" is a region of the network bounded by transformers and operating at one voltage. Normally the nominal line-to-line voltage of that zone is used.
3. Propagate the base kV through every transformer by its line-to-line voltage ratio. This is the step that makes the ideal transformers vanish, and it must be done by the rated ratio of the transformer, not by any convenient round number.
4. Convert every impedance onto the base of its own zone — ohmic values by dividing by \(Z_B = (kV_B)^2/\text{MVA}_B\), per-unit values by the change-of-base rule of Section 4-5.
5. Draw the resulting network. Every impedance is now a pure number and can be combined by ordinary series and parallel rules across the whole system.

Applied to the transformer ratios of Chapter 3's diagram, step three reads

A remark on the third step, because it is the one that is misunderstood. The base voltages must be in the ratio of transformation; they need not equal the rated voltages. If a 13.8/138 kV transformer sits between a zone whose base is 13.2 kV and a zone whose base is 132 kV, the base ratio \(132/13.2 = 10\) equals the rated ratio \(138/13.8 = 10\), so the ideal transformer still vanishes — but the transformer's own per-unit reactance, quoted on 13.8 kV, must be converted to the 13.2 kV base by the squared voltage term of Section 4-5. If instead the ratios disagree, an off-nominal ideal transformer remains in the circuit and cannot be deleted.

A transformer nameplate carries a figure such as "\(Z = 8\%\)". Its meaning is precise and testable: with the secondary short-circuited, 8% of rated voltage applied to the primary drives rated current. That is the short-circuit test, and it is how the number is measured. Reading it as a per-unit impedance, \(Z_{pu} = 0.08\) on the transformer's own rating, gives exactly the same statement — \(V_{pu} = I_{pu}Z_{pu} = 1.0 \times 0.08\).

Turn the statement around and it becomes a fault calculation. If rated voltage is applied instead of 8% of it, the current is \(1/0.08 = 12.5\) times rated. The apparent power that would flow into a bolted three-phase short circuit is therefore \(12.5\) times the transformer rating:

Typical per-unit values, on each machine's own rating, are worth committing to memory as a sanity check on any data set.

The last row explains the two diagrams that follow. On a transmission network the reactance dominates the resistance by an order of magnitude or more, so there are calculations in which the resistance may be dropped entirely and calculations in which it may not. The distinction produces two different reductions of the single-line diagram.

The impedance diagram is the per-phase equivalent circuit of the whole system, drawn in per unit, with each component replaced by its simplified equivalent circuit. It is what a load-flow study of Part 4 operates on, because a load flow must account for real power losses and therefore cannot discard resistance.

Three approximations are standard, and each is defensible.

• The neutral-earthing impedance is omitted. Under balanced conditions the neutral carries no current, by the identity of Section 3-2, so an impedance in the neutral drops no voltage and has no effect. It reappears the moment the system becomes unbalanced, which is why Chapter 23 puts it back — multiplied by three — in the zero-sequence network.
• The transformer magnetising branch is omitted. The magnetising current of a modern power transformer is well under 1% of rated current, so the shunt admittance carries a negligible share of the load current and its omission changes the terminal behaviour by a fraction of a percent.
• Resistance is omitted where the reactance is very much larger. For a generator or a large transformer, \(X/R\) may exceed 30, and retaining \(R\) changes the magnitude of the series impedance by less than one part in a thousand. For a distribution feeder, where \(X/R\) may be near 1, it may not be omitted at all.

What remains is a network of series impedances between buses, shunt admittances from buses to the reference, generators as voltage sources behind their impedances, and loads as impedances or as specified injections of \(P\) and \(Q\). The reference node — the neutral — becomes the common return rail of the diagram.

A symmetrical fault calculation asks a narrower question: how much current flows into a three-phase short circuit, and how much must the breaker interrupt? The answer is dominated by the reactances of the machines and transformers between the source and the fault, and the currents involved are many times load current. That justifies a further round of approximations, each of which makes the answer slightly pessimistic — which for a protection calculation is the safe direction.

• All resistances are omitted. With \(X/R\) large, \(|R+jX| \approx X\), and dropping \(R\) raises the computed fault current very slightly.
• All static loads are omitted. A load impedance is large compared with the fault path and its current is small compared with the fault current, so it is an open circuit for this purpose.
• Line charging capacitance is omitted for the same reason: its shunt admittance is small and carries a leading current irrelevant beside the fault current.
• Induction motors are omitted when the interrupting current a few cycles after the fault is required, since the motor's contribution decays within two or three cycles once its terminal voltage collapses. They are included when the first-cycle momentary current is wanted.
• The neutral-earthing impedance is omitted, as in the impedance diagram, because a symmetrical fault draws no neutral current.

What is left is a network of pure reactances and constant-voltage sources behind subtransient reactances — small enough to reduce by hand, which is exactly the point.

Unless a problem says otherwise, take base MVA as three-phase and base kV as line-to-line, and assume base voltages are propagated through the rated transformer ratios.