Per-unit analysis

A power system contains many voltage levels linked by transformers: a generator at 13.8 kV, transmission at 230 kV, subtransmission at 69 kV, distribution at 25 kV or 12.47 kV, and buildings at a few hundred volts. Analysing such a network in volts, amperes and ohms means referring every impedance across every transformer with the square of its turns ratio, a tedious and error-prone process. The per-unit system removes this burden. Every quantity is expressed as a fraction of a chosen base value, and with bases chosen consistently, ideal transformers disappear from the circuit entirely. Per-unit values also have the useful property that similar equipment has similar per-unit impedances regardless of its size, which makes data easy to check and estimate. This lesson defines the per-unit system, shows how to change bases and handle transformers, and builds the per-unit impedance diagram of a small system.

Base quantities

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

In a three-phase system, two base quantities are chosen and the others follow from circuit laws. The usual choice is a three-phase power base \(S_b\) (for example 100 MVA), common to the whole system, and a line-to-line voltage base \(V_b\) for each voltage zone, normally the nominal voltage. Then

These formulas use three-phase power and line-to-line voltage, and they give exactly the same per-unit values as a per-phase calculation with single-phase power \(S_b/3\) and line-to-neutral voltage \(V_b/\sqrt3\). In per unit, the factor \(\sqrt3\) vanishes from the power formula: \(S_{\mathrm{pu}} = V_{\mathrm{pu}}I_{\mathrm{pu}}^*\), whether the system is single-phase or three-phase.

Changing base

Equipment impedances are given on the equipment's own rating as base: a transformer nameplate might state "7.5% impedance" on its 2.5 MVA rating and rated voltages. To combine them in a system study, every impedance must be converted to the common system base. Since the ohmic value is the same,

When the voltage bases agree, as is usually arranged, only the power ratio matters: a 0.075 pu impedance on 2.5 MVA becomes \(0.075 \times 100/2.5 = 3.0\) pu on 100 MVA. Small equipment has large per-unit impedances on a large system base, reflecting its small capacity.

Transformers in per unit

Consider a single-phase transformer with turns ratio \(N_1 : N_2\) and leakage impedance \(Z\) referred to the primary. Referred to the secondary, the impedance is \(Z(N_2/N_1)^2\). If the voltage bases on the two sides are chosen in the ratio of the turns, \(V_{b2}/V_{b1} = N_2/N_1\), then the base impedances are in the ratio \((N_2/N_1)^2\) too, so the per-unit impedance is the same on both sides:

The ideal transformer becomes a 1:1 ratio in per unit and can be removed from the circuit, leaving only its series impedance (and, if needed, a shunt magnetizing branch, usually neglected in system studies). This is the main reason for using per unit. The procedure is to pick the voltage base in one zone and carry it through each transformer by its rated voltage ratio, so that every zone has a consistent base.

Three-phase transformer connections

For three-phase transformers, the voltage base ratio is the ratio of rated line-to-line voltages, whatever the winding connection, and the per-unit impedance on the transformer's own three-phase rating is the nameplate percentage impedance. Wye-delta transformers also introduce a 30° phase shift between primary and secondary line voltages (the North American convention is that the high-voltage side leads). In balanced load flow and three-phase fault studies the shift can be ignored, since it shifts all quantities on one side equally, but it matters for paralleling transformers, which must have the same phase shift, and in the unbalanced fault analysis of Lesson 9.

Off-nominal taps

Most power transformers have taps that change the ratio by up to about ±10% in steps of 0.625% to 2.5%, to regulate voltage. If the actual ratio differs from the ratio of the voltage bases, the transformer is represented by its per-unit impedance in series with an ideal transformer of ratio \(a\):1, where \(a\) is the off-nominal ratio (for example 1.025 for a +2.5% tap). For load flow, this combination is converted to an equivalent π circuit whose elements depend on \(a\), and the tap ratio can be adjusted automatically to hold a bus voltage, as Lesson 6 describes.

Three-winding transformers

A three-winding transformer (for example 230/69/13.8 kV) is represented by a wye of three impedances \(Z_p\), \(Z_s\) and \(Z_t\). Test data give the impedances between pairs of windings, \(Z_{ps}\), \(Z_{pt}\) and \(Z_{st}\), each measured with the third winding open and all converted to a common base, from which

One of the three can come out slightly negative; it is a mathematical equivalent, not a physical component.

Typical per-unit values

On their own ratings, equipment impedances fall into narrow ranges, which makes per unit a powerful check on data: a transformer with "75% impedance" is clearly a typing error.

Building a per-unit impedance diagram

The procedure for a system study is:

• Choose a system power base, commonly 100 MVA.
• Choose a voltage base in one zone, usually the nominal voltage, and carry it through each transformer by its rated voltage ratio.
• Convert every impedance to the system base, using the change-of-base formula for equipment data and \(Z/Z_b\) for impedances given in ohms.
• Convert loads to per-unit power (or impedance) and sources to per-unit voltages.
• Draw the per-unit impedance diagram, which is now a single circuit with no transformers, and solve it.
• Convert results back to actual units with the base values of each zone.

With the impedance diagram in hand, the load voltage can be found. A constant-power load makes the circuit non-linear, since the current depends on the voltage, so a simple iteration is used: guess the load voltage, compute the current \(I = (S/V)^*\), recompute the voltage drop, and repeat. This is a miniature version of the load flow calculations of Lesson 6.

import numpy as np

S_load = 0.8 + 0.3j          # pu on 100 MVA
X = 0.10 + 0.0473 + 0.1067   # T1 + two lines in parallel + T2, from the generator terminals
V_gen = 1.0                  # generator terminal voltage held at 1.0 pu by its AVR

V = 1.0 + 0j
for it in range(1, 9):
    I = np.conj(S_load / V)
    V = V_gen - 1j * X * I
    print(f"iteration {it}: |V| = {abs(V):.5f} pu, angle = {np.degrees(np.angle(V)):7.3f} deg")

base_kV, base_A = 34.5, 100e6 / (np.sqrt(3) * 34.5e3)
print(f"load voltage = {abs(V) * base_kV:.2f} kV, load current = {abs(I) * base_A:.0f} A")
print(f"reactive power supplied by the generator = {(abs(I) ** 2 * X + S_load.imag) * 100:.1f} Mvar")

iteration 1: |V| = 0.94588 pu, angle = -12.405 deg
iteration 2: |V| = 0.89609 pu, angle = -12.405 deg
iteration 3: |V| = 0.89170 pu, angle = -13.172 deg
iteration 4: |V| = 0.88823 pu, angle = -13.172 deg
iteration 5: |V| = 0.88790 pu, angle = -13.230 deg
iteration 6: |V| = 0.88764 pu, angle = -13.230 deg
iteration 7: |V| = 0.88762 pu, angle = -13.234 deg
iteration 8: |V| = 0.88760 pu, angle = -13.234 deg
load voltage = 30.62 kV, load current = 1611 A
reactive power supplied by the generator = 53.5 Mvar

The iteration converges in a few steps, and the voltage at the load settles at about 0.89 pu, an 11% drop that would be unacceptable in practice (service voltages are normally kept within about ±5%). It is driven mainly by the reactive power flowing through the series reactance: the generator must supply 53.5 Mvar: the 30 Mvar of the load plus 23.5 Mvar consumed by the transformers and lines themselves. Adding a capacitor bank at the load bus, or raising the T2 tap, would correct it. Being able to see such problems quickly, without wading through turns ratios, is the payoff of the per-unit system.

Advantages and cautions

Per-unit analysis offers several advantages beyond eliminating transformers: per-unit voltages near 1.0 make abnormal values obvious (a bus at 0.88 pu is clearly low), equipment data can be checked against typical ranges, the \(\sqrt3\) factors disappear, and computer programs work naturally in per unit. The cautions are to choose voltage bases consistently through every transformer (a mismatch silently corrupts results, unless it is represented as an off-nominal tap), to convert all equipment data to the system base before combining them, and to remember which zone's base to use when converting results back.

Lesson 3 uses per-unit values to describe generator reactances and capability, and Lesson 10 extends the transformer models of this lesson to zero-sequence networks and phase shifts.