LibraryReferencePer-unit quantities

Per-unit quantities

A transformer's nameplate gives its impedance as a percentage, and a generator's datasheet gives its reactance per unit. Neither is an impedance in ohms. Each is the impedance as a fraction of a base impedance that the equipment's own rated voltage and rated power define. This topic explains these quantities, how the design converts them to ohms for the short-circuit study, and where the study keeps a value per unit.

1Per unit and percent

A quantity per unit (pu) is the quantity divided by a base of the same kind. In percent it is 100 times that. For equipment, the bases are its rated voltage $U_{r}$ and its rated apparent power $S_{r}$ . They fix a base impedance, $Z_{b} = U_{r}^{2} / S_{r}$ , the impedance that would draw the rated power at the rated voltage.

The design takes one percent impedance per voltage level for its interconnecting transformers (ICTs), whatever their rating (Power transformers, Impedance): 12.5 % at 220 kV.

The design's short-circuit study does not set up a common MVA base. It converts each percent or per-unit impedance to ohms at the nominal voltage of the bus it is connected to. It then combines the impedances in ohms, referring those of the other voltage level through the transformer ratio. Two quantities stay per unit: the relative reactances in the correction factors, and the driving voltage of the ANSI / IEEE study.

where
$U_{r}$ , $S_{r}$	rated voltage and rated apparent power of the equipment	kV, MVA
$Z_{b}$	base impedance, $U_{r}^{2} / S_{r}$	Ω

2The transformer

A transformer's rated short-circuit voltage $u_{k r}$ , in percent, is its impedance in percent of its base impedance. The impedance in ohms follows:

Z_{T} = \frac{u _{k r}}{100} \frac{U _{r T}^{2}}{S _{r T}}

(1)

IEC 60909-0:2016 §6.3.1, Eq. 7

$U_{r T}$ is the rated voltage on the side to which the impedance is referred. The design takes the high-voltage (HV) side, and refers the result to the medium-voltage (MV) side by $t^{2}$ , with $t$ the ratio of the MV to the HV nominal voltage. The same per-unit value on the MV base gives the same result. A per-unit impedance is the same on both sides of a transformer, and only the base changes.

The correction factor $K_{T}$ of a network transformer uses its relative reactance $x_{T}$ , the reactance $X_{T}$ per unit of $U_{r T}^{2} / S_{r T}$ :

K_{T} = \frac{0.95 c _{m a x}}{1 + 0.6 x _{T}}

(2)

IEC 60909-0:2016 §6.3.3, Eq. 12a

The design's transformers have an X/R ratio of 40 by default, so $x_{T}$ is close to $u_{k r} /100$ .

where
$u_{k r}$	rated short-circuit voltage of the transformer	%
$U_{r T}$ , $S_{r T}$	rated voltage of the transformer on the side considered, rated apparent power	kV, MVA
$Z_{T}$ , $X_{T}$	impedance and reactance of the transformer	Ω
$x_{T}$	relative reactance of the transformer
$t$	ratio of the MV to the HV nominal voltage
$K_{T}$	correction factor of a network transformer

Example 1Impedance of the default ICT in ohms and per unit

Given the default 220 kV / 33 kV project: 2 ICTs of 160 MVA, $u_{k r}$ = 12.5 %, X/R 40.

Base impedance on the HV side:

Z_{b, HV} = \frac{220 ^{2}}{160} = 302.5 Ω

Impedance by Eq. (1), 0.125 pu of the base:

Z_{T} = 0.125 \times 302.5 = 37.81 Ω, Z_{T} = 0.945 + j 37.80 Ω

Relative reactance and correction factor by Eq. (2):

x_{T} = \frac{37.80}{302.5} = 0.1250, K_{T} = \frac{0.95 \times 1.10}{1 + 0.6 \times 0.1250} = 0.9721

Referred to the MV side by $t$ = 0.1500:

Z_{T} t^{2} = 37.81 \times 0.02250 = 0.8508 Ω, Z_{b, MV} = \frac{33 ^{2}}{160} = 6.8063 Ω

Result The ICT's 12.5 % is 37.81 Ω on the HV side and 0.8508 Ω on the MV side, 0.125 pu of the MV base in both cases. With $K_{T}$ the study takes 36.76 Ω on the HV side.

3The generator

A synchronous generator's datasheet gives its subtransient reactance $x_{d}^{''}$ per unit of its own base, its rated voltage $U_{r G}$ and its rated apparent power $S_{r G}$ . The rated apparent power is the unit's rated active power over its rated power factor. The reactance in ohms is $x_{d}^{''} U_{r G}^{2} / S_{r G}$ .

The design models each generator with its generator transformer as a power station unit with an on-load tap changer, referred to the HV bus:

Z_{S} = K_{S} (t_{r}^{2} Z_{G} + Z_{T H V})

(3)

IEC 60909-0:2016 §6.7.1, Eq. 21

K_{S} = \frac{U _{n Q}^{2}}{U _{r G}^{2}} \frac{U _{r T L V}^{2}}{U _{r T H V}^{2}} \frac{c _{m a x}}{1 + ∣ x _{d}^{''} - x _{T} ∣ sin φ _{r G}}

(4)

IEC 60909-0:2016 §6.7.1, Eq. 22

The design takes the transformer's HV rating equal to the nominal voltage of the HV bus, and its LV rating equal to the generator's. The voltage ratios of Eq. (4) are then 1. The term $t_{r}^{2} Z_{G}$ is the generator's reactance per unit times the HV base, $x_{d}^{''} U_{n Q}^{2} / S_{r G}$ . The factor $K_{S}$ compares two per-unit reactances on different bases, the generator's $x_{d}^{''}$ on $S_{r G}$ and the transformer's $x_{T}$ on its own rating.

where
$x_{d}^{''}$	subtransient reactance of the generator, per unit of its rating
$U_{r G}$ , $S_{r G}$	rated voltage and rated apparent power of the generator	kV, MVA
$Z_{G}$	impedance of the generator	Ω
$Z_{T H V}$	impedance of the generator transformer, referred to its HV side	Ω
$t_{r}$	rated ratio of the generator transformer, HV over LV
$U_{n Q}$	nominal voltage of the HV bus	kV
$U_{r T H V}$ , $U_{r T L V}$	rated voltages of the generator transformer, HV and LV side	kV
$φ_{r G}$	phase angle of the generator at its rated power factor
$K_{S}$	correction factor of a power station unit with an on-load tap changer
$Z_{S}$	impedance of the power station unit, referred to the HV bus	Ω

Example 2Generator reactance of a thermal plant at 220 kV

Given a thermal plant at 220 kV with the level's defaults: 2 units of 250 MW at power factor 0.85, $x_{d}^{''}$ = 0.20 pu, each on a generator transformer of 315 MVA at 12.5 %.

Rated apparent power of a unit:

S_{r G} = \frac{250}{0.85} = 294.1 MVA

Generator reactance referred to the HV bus, on the base $U_{n Q}^{2} / S_{r G}$ :

\frac{220 ^{2}}{294.1} = 164.6 Ω, X_{d}^{''} = 0.20 \times 164.6 = 32.91 Ω

Power station unit by Eq. (3), with the transformer's 19.21 Ω by Eq. (1) and $K_{S}$ = 1.0582 by Eq. (4): $∣ Z_{S} ∣$ = 55.19 Ω.

Result The generator's 0.20 pu on its own 294.1 MVA is 32.91 Ω at 220 kV, and the unit enters the HV bus with 55.19 Ω.

4Voltage per unit

The IEC 60909-0 method sets the driving voltage as a multiple of the nominal voltage, $c U_{n} / 3$ . The voltage factor $c$ is that multiple per unit: $c_{m a x}$ = 1.10 for the maximum currents (Short-circuit currents by IEC 60909-0, The equivalent voltage source).

The ANSI / IEEE set's study by IEEE C37.010-2016 takes the driving voltage $E$ at the maximum system voltage of each bus, per unit of its nominal voltage (Short-circuit duty by IEEE C37.010, The E/X method). In the 230/34.5 kV example, $E$ is 242 kV over 230 kV, 1.052 pu, at the HV bus. It is 36.5 kV over 34.5 kV, 1.058 pu, at a MV section. The minimum case takes 1.0 pu.

Sources

[1]IEC 60909-0:2016, Short-circuit currents in three-phase AC systems, Part 0: Calculation of currents. §6.3.1, Eq. 7; §6.3.3, Eq. 12a; §6.7.1, Eq. 21; §6.7.1, Eq. 22.

[2]IEEE C37.010-2016, IEEE Application Guide for AC High-Voltage Circuit Breakers.

CitationSubstationLayout Reference, Per-unit quantities, revision 1, 8 Nov 2026, https://substation.solarlayout.app/library/reference/per-unit-quantities

Per-unit quantities

1Per unit and percentCopy link to Per unit and percent

2The transformerCopy link to The transformer

3The generatorCopy link to The generator

4Voltage per unitCopy link to Voltage per unit

1Per unit and percent

2The transformer

3The generator

4Voltage per unit