Electrical faults can occur in power systems for various reasons, such as equipment failure or lightning strikes. When a fault occurs, a large current flows through the system, and it is important to determine the maximum current that can flow to protect the system from damage. This is known as the fault current.

The fault current can be calculated using Ohm's law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. In the case of a fault current, the voltage is typically constant, and the resistance is low, so the current is very high.

The fault current can be calculated using the following formula:

$I_{sc}=\frac{V}{Z}$

I_sc         is the fault current, A
V         is the voltage, V
Z         is the impedance of the system, Ω

The system's impedance measures its resistance to the flow of current. It is determined by the resistance of the conductors, the reactance of the transformers, and the capacitance of the cables.

In power systems, the fault current is typically classified into three types: three-phase, phase-to-phase, and single-phase-to-ground. The fault current in each case can be calculated using different equations.

Three-Phase Fault

In a three-phase fault, all three phases of the system are shorted together. The fault current, in this case, can be calculated using the following formula:

$I_{sc}=\frac{V_{LL}}{\sqrt{3} Z_p}$

V _LL      is the line-line voltage, V
Z_p         is the phase impedance of the system seen from the point of fault, Ω

Phase-to-Phase Fault

In a phase-to-phase fault, two phases of the system are shorted together. The fault current, in this case, can be calculated using the following formula:

$I_{sc}=\frac{V_{LL}}{2Z_p}$

Single-Phase-to-Ground Fault

In a single-phase-to-ground fault, one phase of the system is shorted to ground. The fault current, in this case, can be calculated using the following formula:

$I_{sc}=\frac{V_{LL}}{\sqrt{3} Z_g}$

Zg        is the grounding impedance of the system, Ω 

In the case of a fault, it is essential to determine the maximum fault current that can flow to ensure that the system is protected from damage. Using the equations described above, it is possible to calculate the fault current for different types of faults in power systems.

In conclusion, the fault current is an essential parameter in power systems determining the maximum current that can flow during a fault. The fault current can be calculated using Ohm's law and the system's impedance. 

The circuit protective conductor (CPC) is used to provide protection from electric shock and to allow sufficient current to flow so the protective devices can trip.  Various national and international standards define the requirements for CPC.  Any CPC is also an important consideration in the calculation of .

Common types of allowable CPC include:

1. a single core cable
2. a conductor in a multicore cable
3. an insulated or bare conductor in a common enclosure with insulated live conductors
4. a fixed or bare insulated conductor
5. the sheath or armouring of a cable
6. a metal conduit, cable management system, or electrical continuous support system
7. an extraneous-conductive-part (subject to regulations)

When considering CPC, myCableEngineering allows the following:

1. cable armour by itself
2. a separate conductor by itself
3. armour combined with a separate conductor

Note: although (3) is quite common, it is not recommended due to the difficulties in evaluating divisions of current between the armour and separate conductor.

Earth Fault Loop

The illustration below shows a typical earth fault path.

Earth Fault Loop Path

The earth fault loop impedance Z_s, is given by:

$Z_s=Z_e+Z_1+Z_2$

Where:
Z_s =        earth fault loop impedance,  Ω
Z_e =        external earth fault impedance, Ω
Z₁  =       line conductor impedance, Ω
Z₂  =        impedance, Ω

The external impedance, Z_e depends on the upstream network.  In the illustration,  the external impedance would be Z₀ + Z_PEN.  In other arrangements, the external impedance may be derived differently.

The CPC impedance, Z₂  depends on the protective conductor used (armour, separate cable, trunking, etc.).

BS 7671 CPC Requirements

Earth fault loop impedance is important for regulation 411 "Protective Measure: Automatic Disconnection of Supply".   This regulation prescribes a minimum disconnection time for different types of circuit.  The disconnection time is related to the protective device and the time it takes the device to trip is dependant on the earth fault loop impedance. 

Regulation 411.3.2 gives the following maximum disconnection times:

1. For  TT circuits incorporating equipotential bonding in accordance with Regulation 411.3.1.2, the maximum tripping times for a TN system may be used. 
2. For TN distribution circuits and circuits not covered by the above table, a disconnection time not exceeding 5 s is allowed.
3. For TT distribution circuits and circuits not covered by the above table, a disconnection time not exceeding 1 s is allowed.

The characteristics of protective devices should be such that:

 $Z_s\times I_a\le U_0\times C_{\min }$        - TN, TT systems

$Z_s\times \left(2\times I_a\right)\le U_0\times C_{\min }$         - IT system (second fault, neutral/mid-point conductor not distributed)

Where:
I_a         - current causing operation of the protective device within a specified time, A
U₀       - nominal a.c. or d.c. line voltage to earth, V
C_min    - minimum voltage factor (= 0.95)

Where an RCD is used for fault protection, in addition to the above the following should be satisfied (further limiting the maximum Z_s ):

  $R_A\times I_{\Delta n}\le 50V$

Where:
R_A         - sum of resistances of earth electrode and protective conductor, Ω
I_Δn         - rated current of the RCD, A

myCableEngineering and Earth Fault Loop

External Fault Loop Impedance

The external earth loop fault impedance Z_e is calculated in the complex form using earth data entered by the user:

$I_{k2E}=I_E\times p{f}_E-j{I}_E\times \sin (\cos (p{f}_E))$

$U=U_0/\sqrt{3}$

$Z_e=U/I_{k2E}$

where:
I_E        - source earth fault level in A
pf_E      - source fault power factor
I_k2E     - source earth fault current complex form, A
U        - phase voltage, V
U₀       - line-line voltage, V
Z_e        - source (external) impedance, Ω

Cable Loop Impedance

myCableEngineering calculates positive ( Z_1(60909) ) and zero ( Z_0(60909) ) sequence impedance in accordance with IEC  60909 "Short-circuit currents in three-phase a.c. systems". Given the fault conditions (three-phase, and single-phase) at the load end of the cable, the resultant fault levels can be calculated at the remote end.

From IEC 60909, for a line to earth short circuit (with Z_2(60909) = Z_1(60909)), the fault current is given by:

$I_k=\frac{\sqrt{3}c{U}_n}{\left|2Z_{1\left(60909\right)}+Z_{2\left(60909\right)}\right|}$

giving the following conversions between IEC 60909 calculated values, and the common usage of Z1 and Z2 (or R1, R2) :

Earth Fault Loop Impedance

Having obtained the source impedance components, the total loop impedance ( Z_t ) and load end fault level ( I_f ) are given by:

$Z_t=Z_e+Z_1+Z_0$

$I_f=U/Z_t$

The earth fault loop impedance is simply the magnitude of Z_t.

For common configurations, the maximum earth fault loop impedance is calculated (but can be overridden by the user).  For other configurations, the user needs to enter the required maximum earth fault loop impedance.  Circuit conditions for which the maximum earth fault loop impedance is calculated are dependant on the system type and selected protective devices. 

Note: maximum earth fault loop impedance for devices listed in BS 7671 is available for 0.1, 0.2, 0.4, 1 and 5 s maximum disconnect times.  For MCCB, the impedance is available for 0.4 and 5 s disconnect times at maximum setting,

Circuit Protective Conductor (CPC)

The cable armour is used as the CPC in the calculation of Z₀ and the earth fault loop impedance. In addition, the user has the option to add an additional conductor, which will be used in parallel with any armour to form the CPC.  The additional conductor can be external to the cable or internal. 

Regulation 543 of BS 7671, specifies minimum sizes for protective conductors.  The user is recommended to verify that his cable design complies with this regulation. 

Protective Device

A check on the device setting is also carried out to ensure that the relevant requirements are met.  Similar if an RCD is used.

For an example calculation, please see Cable Fault Calculation

To calculate system faults, we can use the guidance given in IEC 609096 "Short-circuit currents in three-phase a.c. systems. For faults far from the generator faults in three phase systems, each type of fault the symmetrical short-circuit current I"_k is given by:

The Formulae

Note:

1. Z₁, Z₂ and Z₀ are the symmetrical components
2. for most components (except synchronous machines), we can take Z₂ as equal to Z₁.
3. for single-phase line to line is equivalent to line-neutral, and for d.c. positive to negative

For several series circuits in the fault loop, the final fault current is given by:

$I_k^{"}=\frac{c{U}_n}{\sqrt{3}{\displaystyle \sum _i}{Z}_i}$        - for a.c. circtuis

$I_k^{"}=\frac{c{U}_n}{\sum _i{Z}_i}$        - for d.c. circuits

Source Impedance

Source impedance Z_Q , is given by:

$Z_Q=\frac{c{U}_n}{\sqrt{3}{I}_k^{"}}$        - three phase and single phase systems

$Z_Q=\frac{c{U}_n}{I_k^{"}}$        - d.c. systems

Maximum & Minimum Fault Levels

Typically both a maximum fault level (used for rating equipment), and minimum level (used for protections settings) are calculated.  When evaluating the maximum and minimum, the following condtions are taken into account:

1. Use the c_max or c_min voltage factor as appropriate
2. When calculating external network impedances (Z_q), use the maximum or minimum value of short circuit current as appropriate

Note: IEC 60909 recommends resistance calculated at 20 °C for the maximum short circuit, and at the end of short circuit temperature for the minimum short circuit level.  Presently, in myCableEngineering we use the conductor operating temperature as the reference for both maximum and minimum fault levels. User specified (input) fault levels are taken as being those with the voltage factor c = 1.

Symbols

c                   - voltage factor
I"_k                  - initial symmetrical three phase short-circuit current (r.m.s.), A
                        - I"_k1        line-to-earth
                        - I"_k2        line-to-line
                        - I"_kE2E     line-to-line-earth
U_n             - nominal system voltage, line-to-line (r.m.s.), V
                        - positive-negative voltage for d.c. systems
Z₁             - positive sequence short -circuit impedance, Ω
Z₂             - negative sequence short circuit impedance, Ω
Z₀             - zero sequence short circuit impedance, Ω
Z_q        - impedance or any external network, Ω

The fault level at the source of a cable due to all the upstream elements is the network fault level (or external fault level).  Network fault level can be specified either as an apparent power, impedance or fault current and power factor.  The relationship between the various quantities is given by:

Three Phase a.c System

$I_k=\frac{VA}{\sqrt{3}{U}_n}$                or        $I_k=\frac{U_n}{\sqrt{3}{Z}_e}$

$VA = \sqrt{3}{I}_k{U}_n$        or        $VA=\frac{{U_n}^2}{Z_e}$

$Z_e=\frac{U_n}{\sqrt{3}{I}_k}$                or        $Z_e=\frac{{U_n}^2}{VA}$

Single Phase a.c. System

$I_k=\frac{\sqrt{3}VA}{U_n}$                or        $I_k=\frac{U_n}{\sqrt{3}{Z}_e}$        or         $I_k=\frac{VA}{U_{LN}}$

$VA = \frac{I_k{U}_n}{\sqrt{3}}$              or        $VA=\frac{{U_n}^2}{3Z_e}$        or        $VA=I_k{U}_{LN}$

$Z_e=\frac{U_n}{\sqrt{3}{I}_k}$                or        $Z_e=\frac{{U_n}^2}{3VA}$        or         $Z_e=\frac{U_{LN}}{I_k}$

d.c. System

$I_k=\frac{VA}{U_n}$                or        $I_k=\frac{U_n}{Z_e}$

$VA = I_k{U}_n$           or        $VA=\frac{{U_n}^2}{Z_e}$

$Z_e=\frac{U_n}{I_k}$                or        $Z_e=\frac{{U_n}^2}{VA}$

Note: for a.c. systems the values are phasor quantities and the use of absolute values will lead to some error.  The easiest way to take this into account phasor relationships is by carrying out the calculations with numbers in complex form (ensuring that both resistance are reactance are considered appropriately).  

Symbols

I_k               - network/external fault current, A
VA             - network/external fault apparent power, VA or MVA (x by 10^-6 to convert to VA)
pf              - power factor of the network/external fault
U_LN           - nominal voltage, line-neutral (single phase circuit), V
U_n             - nominal voltage, line-line (or absolute for d.c.), V
Z_e             - network/external fault impedance, Ω

How to Calculate the Fault Level on the Secondary of a Three-Phase Transformer Using Percentage Impedance

Transformers are essential components in electrical power systems, transferring power between different voltage levels. One critical parameter to consider when designing and operating these systems is the fault level on the secondary side of a transformer. Knowing the fault level is essential for proper protective device selection and coordination.

The page will guide you through the process of calculating the fault level on the secondary side of a three-phase transformer using percentage impedance.

Step 1: Gather the necessary data

To calculate the fault level, you will need the following information:

1. Transformer's rated capacity (kVA)
2. Primary line-to-line voltage (V1)
3. Secondary line-to-line voltage (V2)
4. Transformer's percentage impedance (Z%)

Note: we will not be using the primary voltage, as we are assuming an infinite fault level on the transformer primary (network impedance = zero).

The secondary side fault current (Isc) using this formula can be found using the following formula:

$I_b=\frac{\mathrm{Transformer}\mathrm{kVA}\times 1000}{V2\times \sqrt{3}\times \left({\displaystyle \frac{Z\%}{100}}\right)}$        Ampere

The fault level on the secondary side in MVA is given by:

$\mathrm{Fault}\mathrm{Level}\left(\mathrm{MVA}\right)=\frac{I{\mathrm{sc}}_{}\times V2\times \sqrt{3}}{1,000,000}$

Example Calculation

Let's go through an example calculation using the following data for a three-phase transformer:

• Transformer kVA: 1000 kVA
• Primary line-to-line voltage (V1): 11000 V
• Secondary line-to-line voltage (V2): 415 V
• Transformer percentage impedance (Z%): 6%

$I_{sc}=\frac{1000\times 1000}{400\times \sqrt{3}\times \left(6/100\right)}\approx 24056 A \approx 24.1 kA$

$\mathrm{Fault}\mathrm{Level}=\frac{24056\times 400\times \sqrt{3}}{1,000,000}\approx 16.6 \mathrm{MVA}$

In this example, the fault level on the secondary side of the three-phase transformer is approximately 70.87 MVA.

Online Calculator

If you need to calculate transformer secondary fault levels, you can use our online calculator. Using the calculator, you can determine the transformer's secondary three-phase fault current using an infinite or known fault level on the primary.