LibraryReferenceShort-circuit currents

Short-circuit currents by IEC 60909-0

The short-circuit currents of a substation are checked against the ratings of its switchgear and the design current of its earth mat. IEC 60909-0:2016 calculates them from one equivalent voltage source at the fault location and the impedances of the network. This topic follows the method as the design applies it to a pooling substation: the grid, the lines, the interconnecting transformers (ICTs) and the plant.

1The equivalent voltage source

The method replaces the driving voltages of the network by one source at the fault location, $c U_{n} / 3$ , with $U_{n}$ the nominal voltage of the faulted bus. The other elements of the network enter by their impedances. The voltage factor $c$ sets the source for the case: $c_{m a x}$ = 1.10 for the maximum currents and $c_{m i n}$ = 1.00 for the minimum currents, at nominal voltages above $1 kV$ (IEC 60909-0:2016 Table 1).

An unbalanced fault is calculated with the symmetrical components of the network at the fault location: the positive-sequence impedance $Z_{1}$ , the negative-sequence impedance $Z_{2}$ and the zero-sequence impedance $Z_{0}$ . The design takes $Z_{2} = Z_{1}$ , as for a network far from a generator. The initial symmetrical current $I_{k}^{''}$ is the symmetrical short-circuit current at the first instant of the fault.

2The impedances of the network

The grid behind the substation is a network feeder. Its impedance follows from the three-phase fault level $I_{k Q}^{''}$ at the remote grid bus, and its resistance from the grid's X/R ratio:

Z_{Q} = \frac{c U _{n Q}}{3 I _{k Q}^{''}}, R_{Q} = \frac{X _{Q}}{X / R}

(1)

IEC 60909-0:2016 §6.2, Eq. 4–5

The design takes the feeder with $c_{m a x}$ for both cases, and applies the case's $c$ at the fault. The zero-sequence feeder impedance follows from the single-phase fault level: the design solves Eq. (9) for $∣ Z_{Q 0} ∣$ , with the grid's X/R ratio. Unless the grid operator's levels are entered, the design takes 80 % of the level's standard fault rating for both levels.

An overhead line enters by its impedance per kilometre and its length. Parallel line bays are parallel circuits, without their mutual zero-sequence coupling:

Z_{L} = (R_{1} + j X_{1}) \frac{l}{n}

(2)

IEC 60909-0:2016 §6.4

A transformer enters by its short-circuit voltage and rating. A network transformer's impedance is multiplied by the correction factor $K_{T}$ :

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

(3)

IEC 60909-0:2016 §6.3.1, Eq. 7

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

(4)

IEC 60909-0:2016 §6.3.3, Eq. 12a

The ICT's positive-sequence impedance is $Z_{T 1} = K_{T} Z_{T}$ . Its zero-sequence impedance is $Z_{T 0} = k_{0} K_{T} Z_{T}$ , with $k_{0}$ the transformer's Z0 / Z1 ratio, 0.9 by default. Seen from the high-voltage (HV) bus, the earthed star with its closed delta tertiary is an earth path $Z_{T 0} / n$ for $n$ ICTs, in parallel with the grid's. A transformer with a delta HV winding (Dyn11) has no HV neutral, and adds no earth path at the HV bus.

where
$U_{n}$ , $U_{n Q}$	nominal voltage of the faulted bus, of the grid bus	kV
$c$	voltage factor, $c_{m a x}$ or $c_{m i n}$
$I_{k Q}^{''}$	three-phase fault level of the grid at the remote bus	kA
$Z_{Q}$ , $R_{Q}$ , $X_{Q}$	impedance, resistance and reactance of the network feeder	Ω
$R_{1}$ , $X_{1}$	positive-sequence resistance and reactance of a line circuit	Ω/km
$l$ , $n$	length of the line, number of parallel circuits	km,
$u_{k r}$	rated short-circuit voltage of the transformer	%
$U_{r T}$ , $S_{r T}$	rated voltage on the HV side, rated apparent power	kV, MVA
$x_{T}$	relative reactance of the transformer, $X_{T}$ per unit of $U_{r T}^{2} / S_{r T}$
$K_{T}$	correction factor of a network transformer
$k_{0}$	ratio of the transformer's zero-sequence to positive-sequence impedance

The medium-voltage bus

The medium-voltage (MV) bus is sectionalised, with the bus-section circuit breakers normally open. A fault on one section sees one ICT, or one low-voltage winding of a split-winding unit, and the section's share of the plant. The HV impedances are referred to the MV bus by $t = U_{n, MV} / U_{n, HV}$ : $Z_{1, MV} = (Z_{Q 1} + Z_{L 1}) t^{2} + Z_{T 1} t^{2}$ . Each low-voltage winding is earthed through a neutral earthing resistor (NGR) of resistance $R_{NGR} = (U_{n} / 3) / I_{NGR}$ . The zero-sequence impedance of the section is then $Z_{0, MV} = 3 R_{NGR} + Z_{T 0} t^{2}$ , and the NGR limits the earth-fault current.

The plant

A plant of full-size converter units feeds the fault as a current source, a multiple $k$ of its rated current $I_{r}$ :

I_{plant} = k I_{r}

(5)

IEC 60909-0:2016 §6.9

The design takes $k$ = 1.2 by default, and the plant's apparent power as its rated power over a power factor of 0.95. The infeed acts in the positive sequence only. It is added to the three-phase current at the faulted bus, referred to the HV bus by the ICT ratio. A generating plant's units enter instead as impedances, each a generator and its transformer with an on-load tap changer, corrected by $K_{S}$ (IEC 60909-0:2016 §6.7.1).

3The initial symmetrical currents

With the impedances at the fault location, the three-phase, line-to-line, line-to-line-to-earth and line-to-earth currents follow from the equivalent voltage source. The three-phase current involves the positive sequence only:

I_{k 3}^{''} = \frac{c U _{n}}{3 ∣ Z _{1} ∣}

(6)

IEC 60909-0:2016 §7.2, Eq. 29

I_{k 2}^{''} = \frac{c U _{n}}{∣ Z _{1} + Z _{2} ∣}

(7)

IEC 60909-0:2016 §7.3, Eq. 49

I_{k 2 E L 2}^{''} I_{k E 2 E}^{''} = \frac{∣ - j c U _{n} ( Z _{0} - a Z _{2} ) ∣}{∣ Z _{1} Z _{2} + Z _{1} Z _{0} + Z _{2} Z _{0} ∣} = \frac{3 c U _{n} ∣ Z _{2} ∣}{∣ Z _{1} Z _{2} + Z _{1} Z _{0} + Z _{2} Z _{0} ∣}

(8)

IEC 60909-0:2016 §7.4, Eq. 50 and 51

I_{k 1}^{''} = \frac{3 c U _{n}}{∣ Z _{1} + Z _{2} + Z _{0} ∣}

(9)

IEC 60909-0:2016 §7.5, Eq. 52

where
$I_{k 3}^{''}$	initial symmetrical current of a three-phase fault	kA
$I_{k 2}^{''}$	initial symmetrical current of a line-to-line fault	kA
$I_{k 2 E L 2}^{''}$	current in phase L2 of a line-to-line-to-earth fault (phase L3 with $a^{2}$ )	kA
$I_{k E 2 E}^{''}$	current to earth of a line-to-line-to-earth fault	kA
$I_{k 1}^{''}$	initial symmetrical current of a line-to-earth fault	kA
$Z_{1}$ , $Z_{2}$ , $Z_{0}$	positive-, negative- and zero-sequence impedance at the fault location	Ω
$a$	rotation operator, $e^{j 2 π /3}$

The design adds the plant's infeed (Eq. (5)) to the three-phase current of the maximum case. The maximum case takes $c_{m a x}$ with the plant feeding in. The minimum case takes $c_{m i n}$ without the plant. The larger of the line-to-line-to-earth phase currents is reported, with the current to earth. The MV values are per section.

Example 1Initial symmetrical currents of the default 220 kV project

Given the default 220 kV / 33 kV project at 50 Hz, for a 250 MW plant.

Quantity	Value
Grid fault levels $I_{k Q}^{''}$ , three-phase and single-phase	40 kA, 40 kA
Grid X/R ratio	15
Line: circuits × length, conductor	1 × 10 km, Twin ACSR Moose
Line constants $R_{1}$ , $X_{1}$ , $R_{0}$ , $X_{0}$	0.028 Ω/km, 0.31 Ω/km, 0.20 Ω/km, 1.00 Ω/km
ICTs: number × rating, $u_{k r}$ , X/R ratio	2 × 160 MVA, 12.5 %, 40
MV sections, NGR per winding	2, 1000 A (19.05 Ω)
Plant: apparent power, infeed factor $k$	263 MVA, 1.2

Network feeder by Eq. (1):

Z_{Q} = \frac{1.10 \times 220}{3 \times 40} = 3.493 Ω, Z_{Q 1} = 0.232 + j 3.485 Ω

Line by Eq. (2):

Z_{L 1} = (0.028 + j 0.31) \times \frac{10}{1} = 0.28 + j 3.10 Ω

Transformer by Eq. (3) and Eq. (4):

Z_{T} = \frac{12.5}{100} \times \frac{220 ^{2}}{160} = 37.81 Ω K_{T} = \frac{0.95 \times 1.10}{1 + 0.6 \times 0.1250} = 0.9721, ∣ Z_{T 1} ∣ = 36.76 Ω

HV bus by Eq. (6), with the plant's 829 A referred to the HV bus:

Z_{1} = Z_{Q 1} + Z_{L 1} = 0.512 + j 6.585 Ω I_{k 3}^{''} = \frac{1.10 \times 220}{3 \times 6.605} + 0.83 = 21.98 kA

MV section by Eq. (6), referred by $t$ = 0.1500, with the section's 50 % of the plant, 2762 A:

Z_{1, MV} = 0.032 + j 0.975 Ω, I_{k 3}^{''} = \frac{1.10 \times 33}{3 \times 0.975} + 2.76 = 24.25 kA

Fault types by Eq. (7), Eq. (8) and Eq. (9), with the zero-sequence impedances $Z_{0}$ of 0.758 Ω + j7.464 Ω at the HV bus and 57.176 Ω + j0.744 Ω on a MV section ( $3 R_{NGR}$ = 57.16 Ω):

Current	HV maximum	HV minimum	MV maximum	MV minimum
Three-phase $I_{k 3}^{''}$	21.98 kA	19.23 kA	24.25 kA	19.53 kA
Line-to-line $I_{k 2}^{''}$	18.32 kA	16.65 kA	18.61 kA	16.91 kA
Line-to-line-to-earth, phase	20.87 kA	18.97 kA	18.88 kA	17.16 kA
Line-to-line-to-earth, to earth $I_{k E 2 E}^{''}$	19.40 kA	17.63 kA	0.55 kA	0.50 kA
Line-to-earth $I_{k 1}^{''}$	20.24 kA	18.40 kA	1.10 kA	1.00 kA

Result The largest current at the HV bus is the three-phase current of 21.98 kA, and on a MV section the three-phase current of 24.25 kA. The NGR limits the MV line-to-earth current to 1.10 kA. The HV earth-fault current of 20.24 kA is within the earth mat's design current of 50 kA.

Sources

[1]IEC 60909-0:2016, Short-circuit currents in three-phase AC systems, Part 0: Calculation of currents. §6.2, Eq. 4–5; §6.4; §6.3.1, Eq. 7; §6.3.3, Eq. 12a; §6.9; §7.2, Eq. 29; §7.3, Eq. 49; §7.4, Eq. 50 and 51; §7.5, Eq. 52.

CitationSubstationLayout Reference, Short-circuit currents by IEC 60909-0, revision 1, 8 Nov 2026, https://substation.solarlayout.app/library/reference/short-circuit-currents