# According to the IEC 60909

## Introduction

This article looks at the calculation of short circuit currents for bolted three-phase and single-phase to earth faults in a power system. A short circuit in a power system can cause very high currents to flow to the fault location. The magnitude of the short circuit current depends on the impedance of system under short circuit conditions. In this calculation, the short circuit current is estimated using the guidelines presented in IEC 60909.

### Why do the calculation?

Calculating the prospective short circuit levels in a power system is important for a number of reasons, including:

• To specify fault ratings for electrical equipment (e.g. short circuit withstand ratings)
• To help identify potential problems and weaknesses in the system and assist in system planning
• To form the basis for protection coordination studies

### When to do the calculation?

The calculation can be done after preliminary system design, with the following pre-requisite documents and design tasks completed:

• Key single line diagrams
• Major electrical equipment sized (e.g. generators, transformers, etc)
• Cable sizing (not absolutely necessary, but would be useful)

## Calculation Methodology

This calculation is based on IEC 60909-0 (2001, c2002), "Short-circuit currents in three-phase a.c. systems - Part 0: Calculation of currents" and uses the impedance method (as opposed to the per-unit method). In this method, it is assumed that all short circuits are of negligible impedance (i.e. no arc impedance is allowed for).

There are six general steps in the calculation:

• Step 1: Construct the system model and collect the relevant equipment parameters
• Step 2: Calculate the short circuit impedances for all of the relevant equipment
• Step 3: Refer all impedances to the reference voltage
• Step 4: Determine the Thévenin equivalent circuit at the fault location
• Step 5: Calculate balanced three-phase short circuit currents
• Step 6: Calculate single-phase to earth short circuit currents

### Step 1: Construct the System Model and Collect Equipment Parameters

The first step is to construct a model of the system single line diagram, and then collect the relevant equipment parameters. The model of the single line diagram should show all of the major system buses, generation or network connection, transformers, fault limiters (e.g. reactors), large cable interconnections and large rotating loads (e.g. synchronous and asynchronous motors).

The relevant equipment parameters to be collected are as follows:

• Network feeders: fault capacity of the network (VA), X/R ratio of the network
• Synchronous generators and motors: per-unit sub-transient reactance, rated generator capacity (VA), rated power factor (pu)
• Transformers: transformer impedance voltage (%), rated transformer capacity (VA), rated current (A), total copper loss (W)
• Cables: length of cable (m), resistance and reactance of cable ($\Omega \ km$)
• Asynchronous motors: full load current (A), locked rotor current (A), rated power (W), full load power factor (pu), starting power factor (pu)
• Fault limiting reactors: reactor impedance voltage (%), rated current (A)

### Step 2: Calculate Equipment Short Circuit Impedances

Using the collected parameters, each of the equipment item impedances can be calculated for later use in the motor starting calculations.

#### Network Feeders

Given the approximate fault level of the network feeder at the connection point (or point of common coupling), the impedance, resistance and reactance of the network feeder is calculated as follows:

$Z_{f} = \frac{c V_{n}^{2}}{S_{f}} \,$
$R_{f} = \frac{Z_{f}}{\sqrt{1 + \left( \frac{X}{R} \right)^{2}}} \,$
$X_{f} = \frac{X}{R} \times R_{f} \,$

Where $Z_{f} \,$ is impedance of the network feeder ($\Omega$)

$R_{f} \,$ is resistance of the network feeder ($\Omega$)
$X_{f} \,$ is reactance of the network feeder ($\Omega$)
$V_{n} \,$ is the nominal voltage at the connection point (Vac)
$S_{f} \,$ is the fault level of the network feeder (VA)
$c \,$ is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)
$\frac{X}{R} \,$ is X/R ratio of the network feeder (pu)

#### Synchronous Generators and Motors

The sub-transient reactance and resistance of a synchronous generator or motor (with voltage regulation) can be estimated by the following:

$X_{d}^{''} = \chi_{d}{''} \times K_{g} \times \frac{V_{g}^{2}}{S_{g}} \,$
$R_{g} = \frac{X_{d}^{''}}{\frac{X}{R}} \,$
$K_{g} = \frac{V_{n}}{V_{g}} \frac{c}{1 + \chi_{d}{''} \sin \phi_{g}} \,$

Where $X_{d}^{''} \,$ is the sub-transient reactance of the generator ($\Omega$)

$R_{g} \,$ is the resistance of the generator ($\Omega$)
$K_{g} \,$ is a voltage correction factor - see IEC 60909-0 Clause 3.6.1 for more details (pu)
$\chi_{d}{''} \,$ is the per-unit sub-transient reactance of the generator (pu)
$V_{g} \,$ is the nominal generator voltage (Vac)
$V_{n} \,$ is the nominal system voltage (Vac)
$S_{g} \,$ is the rated generator capacity (VA)
$\frac{X}{R} \,$ is the X/R ratio, typically 20 for $S_{g} \geq \,$ 100MVA, 14.29 for $S_{g} \lt \,$ 100MVA, and 6.67 for all generators with nominal voltage $V_{g} \leq \,$ 1kV
$c \,$ is a voltage factor which accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)
$\cos \phi_{g} \,$ is the power factor of the generator (pu)

For the negative sequence impedance, the quadrature axis sub-transient reactance $\chi_{q}{''} \,$ can be applied in the above equation in place of the direct axis sub-transient reactance $\chi_{d}{''} \,$.

The zero-sequence impedances need to be derived from manufacturer data, though the voltage correction factor $K_{g} \,$ also applies for solid neutral earthing systems (refer to IEC 60909-0 Clause 3.6.1).

#### Transformers

The positive sequence impedance, resistance and reactance of two-winding distribution transformers can be calculated as follows:

$Z_{t} = u_{k} \frac{V_{t}^{2}}{S_{t}} \,$
$R_{t} = \frac{P_{kt}}{3 I_{t}^{2}} \,$
$X_{t} = \sqrt{Z_{t}^{2} - R_{t}^{2}} \,$

Where $Z_{t} \,$ is the positive sequence impedance of the transformer ($\Omega$)

$R_{t} \,$ is the resistance of the transformer ($\Omega$)
$X_{t} \,$ is the reactance of the transformer ($\Omega$)
$u_{k} \,$ is the impedance voltage of the transformer (pu)
$S_{t} \,$ is the rated capacity of the transformer (VA)
$V_{t} \,$ is the nominal voltage of the transformer at the high or low voltage side (Vac)
$I_{t} \,$ is the rated current of the transformer at the high or low voltage side (I)
$P_{kt} \,$ is the total copper loss in the transformer windings (W)

For the calculation of impedances for three-winding transformers, refer to IEC 60909-0 Clause 3.3.2. For network transformers (those that connect two separate networks at different voltages), an impedance correction factor must be applied (see IEC 60909-0 Clause 3.3.3).

The negative sequence impedance is equal to positive sequence impedance calculated above. The zero sequence impedance needs to be derived from manufacturer data, but also depends on the winding connections and fault path available for zero-sequence current flow (e.g. different neutral earthing systems will affect zero-sequence impedance).

#### Cables

Cable impedances are usually quoted by manufacturers in terms of Ohms per km. These need to be converted to Ohms based on the length of the cables:

$R_{c} = R \times \frac{L_{c}}{1000} \,$
$X_{c} = X \times \frac{L_{c}}{1000} \,$

Where $R_{c} \,$ is the resistance of the cable {$\Omega$)

$X_{c} \,$ is the reactance of the cable {$\Omega$)
$R \,$ is the quoted resistance of the cable {$\Omega /km$)
$X \,$ is the quoted reactance of the cable {$\Omega /km$)
$L_{c} \,$ is the length of the cable {m)

The negative sequence impedance is equal to positive sequence impedance calculated above. The zero sequence impedance needs to be derived from manufacturer data. In the absence of manufacturer data, zero sequence impedances can be derived from positive sequence impedances via a multiplication factor (as suggested by SKM Systems Analysis Inc) for magnetic cables:

$R_{c(0)} = R_{c} \times 3.15155 \,$
$X_{c(0)} = R_{c} \times 2.46274 \,$

#### Asynchronous Motors

An asynchronous motor's impedance, resistance and reactance is calculated as follows:

$Z_{m} = \frac{1}{I_{LRC}/I_{FLC}} \times \frac{V_{m}^{2} \cos \phi_{m}}{P_{m}} \,$
$R_{m} = \frac{P_{m} \times I_{LRC}/I_{FLC} \times \cos \phi_{s}}{3 I_{LRC}^{2} \times \cos \phi_{m}} \,$
$X_{m} = \sqrt{Z_{m}^{2} - R_{m}^{2}} \,$

Where $Z_{m} \,$ is impedance of the motor ($\Omega$)

$R_{m} \,$ is resistance of the motor ($\Omega$)
$X_{m} \,$ is reactance of the motor ($\Omega$)
$I_{LRC}/I_{FLC} \,$ is ratio of the locked rotor to full load current
$I_{LRC} \,$ is the motor locked rotor current (A)
$V_{m} \,$ is the motor nominal voltage (Vac)
$P_{m} \,$ is the motor rated power (W)
$\cos \phi_{m} \,$ is the motor full load power factor (pu)
$\cos \phi_{s} \,$ is the motor starting power factor (pu)

The negative sequence impedance is equal to positive sequence impedance calculated above. The zero sequence impedance needs to be derived from manufacturer data.

#### Fault Limiting Reactors

The impedance of fault limiting reactors is as follows (note that the resistance is neglected):

$Z_{r} = X_{r} = \frac{u_{k} \times V_{n}}{|sqrt{3} I_{r}} \,$

Where $Z_{r} \,$ is impedance of the reactor ($\Omega$)

$X_{r} \,$ is reactance of the reactor($\Omega$)
$u_{k} \,$ is the impedance voltage of the reactor (pu)
$V_{n} \,$ is the nominal voltage of the reactor (Vac)
$I_{r} \,$ is the rated current of the reactor (A)

Positive, negative and zero sequence impedances are all equal (assuming geometric symmetry).

#### Static Converters

Static converters and converter-fed drivers (i.e. feeding rotating loads) should be considered for balanced three-phase short circuits. Per IEC 60909-0 Clause 3.9, static converters contribute to the initial and peak short circuit currents only, and contribute 3 times the rated current of the converter. An R/X ratio of 0.1 should be used for the short circuit impedance.

#### Other Equipment

Line capacitances, parallel admittances and non-rotating loads are generally neglected as per IEC 60909-0 Clause 3.10. Effects from series capacitors can also be neglected if voltage-limiting devices are connected in parallel.

### Step 3: Referring Impedances

Where there are multiple voltage levels, the equipment impedances calculated earlier need to be converted to a reference voltage (typically the voltage at the fault location) in order for them to be used in a single equivalent circuit.

The winding ratio of a transformer can be calculated as follows:

$n = \frac{V_{t2} \left( 1 + t_{p} \right)}{V_{t1}} \,$

Where $n \,$ is the transformer winding ratio

$V_{t2} \,$ is the transformer nominal secondary voltage at the principal tap (Vac)
$V_{t1} \,$ is the transformer nominal primary voltage (Vac)
$t_{p} \,$ is the specified tap setting (%)

Using the winding ratio, impedances (as well as resistances and reactances) can be referred to the primary (HV) side of the transformer by the following relation:

$Z_{HV} = \frac{Z_{LV}}{n^{2}} \,$

Where $Z_{HV} \,$ is the impedance referred to the primary (HV) side ($\Omega$)

$Z_{LV} \,$ is the impedance at the secondary (LV) side ($\Omega$)
$n \,$ is the transformer winding ratio (pu)

Conversely, by re-arranging the equation above, impedances can be referred to the LV side:

$Z_{LV} = Z_{HV} \times n^{2} \,$

### Step 4: Determine Thévenin Equivalent Circuit at the Fault Location

The system model must first be simplified into an equivalent circuit as seen from the fault location, showing a voltage source and a set of complex impedances representing the power system equipment and load impedances (connected in series or parallel).

The next step is to simplify the circuit into a Thévenin equivalent circuit, which is a circuit containing only a voltage source ($V_{s} \,$) and an equivalent short circuit impedance ($Z_{k} \,$).

This can be done using the standard formulae for series and parallel impedances, keeping in mind that the rules of complex arithmetic must be used throughout.

If unbalanced short circuits (e.g. single phase to earth fault) will be analysed, then a separate Thévenin equivalent circuit should be constructed for each of the positive, negative and zero sequence networks (i.e. finding ($Z_{k(1)} \,$, $Z_{k(2)} \,$ and $Z_{k(0)} \,$).

### Step 5: Calculate Balanced Three-Phase Short Circuit Currents

The positive sequence impedance calculated in Step 4 represents the equivalent source impedance seen by a balanced three-phase short circuit at the fault location. Using this impedance, the following currents at different stages of the short circuit cycle can be computed:

#### Initial Short Circuit Current

The initial symmetrical short circuit current is calculated from IEC 60909-0 Equation 29, as follows:

$I_{k}^{''} = \frac{cV_{n}}{\sqrt{3}Z_{k}} \,$

Where $I_{k}^{''} \,$ is the initial symmetrical short circuit current (A)

$c \,$ is the voltage factor that accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)
$V_{n} \,$ is the nominal system voltage at the fault location (V)
$Z_{k} \,$ is the equivalent positive sequence short circuit impedance ($\Omega$)

#### Peak Short Circuit Current

IEC 60909-0 Section 4.3 offers three methods for calculating peak short circuit currents, but for the sake of simplicity, we will only focus on the X/R ratio at the fault location method. Using the real (R) and reactive (X) components of the equivalent positive sequence impedance $Z_{k} \,$, we can calculate the X/R ratio at the fault location, i.e.

$X/R = \frac{X_{k}}{R_{k}} \,$

The peak short circuit current is then calculated as follows:

$I_{p} = \kappa \times \sqrt{2} I_{k}^{''} \,$ (for non-meshed networks)

or

$I_{p} = 1.15 \kappa \times \sqrt{2} I_{k}^{''} \,$ (for meshed networks - see clause 4.3.12b)

Where $I_{p} \,$ is the peak short circuit current (A)

$I_{k}^{''} \,$ is the initial symmetrical short circuit current (A)
$\kappa \,$ is a constant factor, $\kappa = 1.02 + 0.98e^{-\frac{3}{X/R}} \,$

#### Symmetrical Breaking Current

The symmetrical breaking current is the short circuit current at the point of circuit breaker opening (usually somewhere between 20ms to 300ms). This is the current that the circuit breaker must be rated to interrupt and is typically used for breaker sizing. IEC 60909-0 Equation 74 suggests that the symmetrical breaking current for meshed networks can be conservatively estimated as follows:

$I_{b} = I_{k}{''} \,$

Where $I_{b} \,$ is the symmetrical breaking current (A)

$I_{k}{''} \,$ is the initial symmetrical short circuit current (A)

For close to generator faults, the symmetrical breaking current will be higher. More detailed calculations can be made for increased accuracy in IEC 60909, but this is left to the reader to explore.

#### DC Short Circuit Component

The dc component of a short circuit can be calculated according to IEC 60909-0 Equation 64:

$I_{dc} = \sqrt{2} I_{k}^{''} e^{\frac{-2 \pi f t}{X/R}} \,$

Where $I_{dc} \,$ is the dc component of the short circuit current (A)

$I_{k}{''} \,$ is the initial symmetrical short circuit current (A)
$f \,$ is the nominal system frequency (Hz)
$t \,$ is the time (s)
$X/R \,$ is the X/R ratio - see more below

The X/R ratio is calculated as follows:

$X/R = \frac{X_{k}}{R_{k}} \frac{f}{f_{c}} \,$

Where $X_{k} \,$ and $R_{k} \,$ are the reactance and resistance, respectively, of the equivalent source impedance at the fault location ($\Omega$)

$\frac{f}{f_{c}} \,$ is a factor to account for the equivalent frequency of the fault. Per IEC 60909-0 Section 4.4, the following factors should be used based on the product of frequency and time ($f . t \,$ ):
$f . t \,$ $\frac{f}{f_{c}} \,$
<1 0.27
<2.5 0.15
<5 0.092
<12.5 0.055

### Step 6: Calculate Single-Phase to Earth Short Circuit Currents

For balanced short circuit calculations, the positive-sequence impedance is the only relevant impedance. However, for unbalanced short circuits (e.g. single phase to earth fault), symmetrical components come into play.

The initial short circuit current for a single phase to earth fault is as per IEC 60909-0 Equation 52:

$I_{kl}^{''} = \frac{\sqrt{3} c V_{n}}{Z_{k(1)} + Z_{k(2)} + Z_{k(0)}} \,$

Where $I_{kl}^{''} \,$ is the initial single phase to earth short circuit current (A)

$c \,$ is the voltage factor that accounts for the maximum system voltage (1.05 for voltages <1kV, 1.1 for voltages >1kV)
$V_{n} \,$ is the nominal voltage at the fault location (Vac)
$Z_{k(1)} \,$ is the equivalent positive sequence short circuit impedance ($\Omega$)
$Z_{k(2)} \,$ is the equivalent negative sequence short circuit impedance ($\Omega$)
$Z_{k(0)} \,$ is the equivalent zero sequence short circuit impedance ($\Omega$)

## Worked Example Figure 3. System model for short circuit example

In this example, short circuit currents will be calculated for a balanced three-phase fault at the main 11kV bus of a simple radial system. Note that the single phase to earth fault currents will not be calculated in this example.

### Step 1: Construct the System Model and Collect Equipment Parameters

The system to be modelled is a simple radial network with two voltage levels (11kV and 415V), and supplied by a single generator. The system model is shown in the figure to the right. The equipment and cable parameters were collected as follows:

Equipment Parameters
Generator G1
• $S_{g1} \,$ = 24,150 kVA
• $V_{g1} \,$ = 11,000 V
• $\chi_{d}^{''} \,$ = 0.255 pu
• $\cos \phi \,$ = 0.85 pu
Generator Cable C1
• Length = 30m
• Size = 2 parallel circuits of 3 x 1C x 500 $mm^{2}$

(R = 0.0506 $\Omega$\km, X = 0.0997 $\Omega$\km)

Motor M1
• $P_{m1} \,$ = 500 kW
• $V_{m1} \,$ = 11,000 V
• $I_{LRC} \,$ = 200.7 A
• $I_{LRC}/I_{FLC} \,$ = 6.5 pu
• $\cos \phi_{m} \,$ = 0.85 pu
• $\cos \phi_{s} \,$ = 0.30 pu
Motor Cable C2
• Length = 150m
• Size = 3C+E 35 $mm^{2}$

(R = 0.668 $\Omega$\km, X = 0.115 $\Omega$\km)

Transformer TX1
• $S_{tx1} \,$ = 2,500 kVA
• $V_{t1} \,$ = 11,000 V
• $V_{t2} \,$ = 415 V
• $u_{k} \,$ = 0.0625 pu
• $P_{kt} \,$ = 19,000 W
• $t_{p} \,$ = 0%
Transformer Cable C3
• Length = 100m
• Size = 3C+E 95 $mm^{2}$

(R = 0.247 $\Omega$\km, X = 0.0993 $\Omega$\km)

Motor M2
• $P_{m1} \,$ = 90 kW
• $V_{m1} \,$ = 415 V
• $I_{LRC} \,$ = 1,217.3 A
• $I_{LRC}/I_{FLC} \,$ = 7 pu
• $\cos \phi_{m} \,$ = 0.8 pu
• $\cos \phi_{s} \,$ = 0.30 pu
Motor M3
• $P_{m1} \,$ = 150 kW
• $V_{m1} \,$ = 415 V
• $I_{LRC} \,$ = 1,595.8 A
• $I_{LRC}/I_{FLC} \,$ = 6.5 pu
• $\cos \phi_{m} \,$ = 0.85 pu
• $\cos \phi_{s} \,$ = 0.30 pu

### Step 2: Calculate Equipment Short Circuit Impedances

Using the patameters above and the equations outlined earlier in the methodology, the following impedances were calculated:

Equipment Resistance ($\Omega$) Reactance ($\Omega$)
Generator G1 0.08672 1.2390
Generator Cable C1 0.000759 0.001496
11kV Motor M1 9.4938 30.1885
Motor Cable C2 0.1002 0.01725
Transformer TX1 (Primary Side) 0.36784 3.0026
Transformer Cable C3 0.0247 0.00993
415V Motor M2 0.0656 0.2086
415V Motor M3 0.0450 0.1432

### Step 3: Referring Impedances

We will model a fault on the main 11kV bus, so all impedances must be referred to 11kV. The two low voltage motors need to be referred to this reference voltage. Knowing that the transformer is set at principal tap, we can calculate the winding ratio and apply it to refer the 415V motors to the 11kV side:

$n = \frac{415 \left( 1 + 0% \right)}{11,000} = 0.03773 \,$

The 415V motor impedances referred to the 11kV side is therefore:

Equipment Resistance ($\Omega$) Reactance ($\Omega$)
415V Motor M2 46.0952 146.5735
415V Motor M3 31.6462 100.6284

### Step 4: Determine Thévenin Equivalent Circuit at the Fault Location

Using standard network reduction techniques, the equivalent Thévenin circuit at the fault location (main 11kV bus) can be derived. The equivalent source impedance is:

$Z_{k} = \left[ Z_{G1} + Z_{C1} \right] || \left[ Z_{M1} + Z_{C2} \right] || \left[ Z_{C3} + Z_{TX1} + \left( Z_{M2} || Z_{M3} \right) \right] \,$
$= 0.08618 + j 1.16531 \,$

### Step 5: Calculate Balanced Three-Phase Short Circuit Currents

#### Initial Short Circuit Current

The symmetrical initial short circuit current is:

$I_{k}^{''} = \frac{cV_{n}}{\sqrt{3}Z_{k}} \,$
$= 5.9786 \,$ kA

#### Peak Short Circuit Current

The constant factor $\kappa \,$ at the fault location is:

$\kappa = 1.02 + 0.98e^{-\frac{3}{X/R}} \,$
$= 1.805 \,$

Therefore as it is a simple radial system (non-meshed), the symmetrical peak short circuit current is:

$I_{p} = \kappa \times \sqrt{2} I_{k}^{''} \,$
$= 15.2614 \,$ kA

## Computer Software

Short circuit calculations are a standard component of power systems analysis software (for example, using commercial software such ETAP, PTW, DIgSILENT, etc, or alternatively an open source package such as pandapower) and the calculations are far easier to perform with software than by hand. However manual calculations could be done as a form of verification to confirm that the software results are reasonable.

## What Next?

The results from the short circuit calculations can be used to specify the fault ratings on electrical equipment (e.g. switchgear, protective devices, etc) and also for protection coordination studies.