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ELEC9713 Industrial and Commercial Power Systems

Fault Calculation Methods

There are two major problems that can occur in electrical systems: these are open circuits and short circuits. Of the two, the latter is the most dangerous because it can lead to very high fault currents and these currents can have very substantial effects (thermal heating and electromechanical forces) on equipment that may require replacement of equipment and may even cause fires and other similar ensuing effects in the electrical system. Building systems are particularly at risk. To prevent problems from short circuits, it is necessary to design electrical protection systems that will be able to detect abnormal fault currents that may occur and then take remedial action to isolate the faulty section of the system in as short a time as is consistent with the magnitude of the short circuit fault current level. This requires that the fault current be predicted for a fault in any particular location of the circuit system. We thus need to establish methods of fault calculation. Fault calculation is not simple for a number of reasons:

There are many different types of fault in three phase systems

ELEC9713: Industrial and Commercial Power Systems p. 1

The impedance characteristics of all electrical items in the system must be known

The fault impedance itself may be non-zero and difficult to estimate

There may be substantial fault current contribution from rotating machines etc.

The initial cycles of fault current may be asymmetric with substantial DC offset

The earth impedance in earth faults can be difficult to estimate accurately

DC system faults also include inductance effects in fault current growth

For example, the possible fault types that may occur in a three-phase system are:

Three phase (symmetrical) faults (the most severe in terms of current)

Phase to phase fault

Single phase to earth fault

Three phase to earth fault

Phase to phase to earth fault Each of these fault types will have different fault current when they occur at the same location and the electrical protection system will need to take this into account when operating time is determined.


In the very simplified coverage of fault calculations that follows we will look only at symmetrical three phase faults. We do not cover any asymmetrical faults (phase to phase and single phase) except for some general comments on their behaviour. In general, three phase symmetrical faults will give the maximum fault current level at any location and thus such calculations represent worst case situations in general. Because they have low impedance systems, low voltage electrical systems, such as those in buildings, generate very high levels of fault currents. The prospective short-circuit current and the fault level (power) at the connection point to the utility supply are important parameters that the designer of an electrical installation needs to know and can be obtained from the electricity utility distributor. The prospective short-circuit current is defined as the current which would flow as a result of a bolted 3-phase fault. Typical value at the point of supply for 230/400V NSW distribution systems:

Suburban residential areas: 10 kA Commercial and industrial areas: 25 kA

(the bolted terminology means a zero impedance fault that has no effect on the fault current, as opposed to a high impedance fault where the fault impedance will have some effect on the fault current (by diminishing it)). ELEC9713: Industrial and Commercial Power Systems p. 3

Knowing the fault level, the impedance of the upstream circuit and devices (e.g. transformers, conductors) can be derived. The prospective fault current varies at different points in the supply:

At the supply transformer terminals, it is limited by the impedance of the distribution transformer and feeder conductors.

At the main switchboard, the fault current is further reduced because of the additional cable impedance of the consumer’s mains.

At the distribution board, the fault current is further decreased by the cable impedance of the submains.

Y

MSB DBSupply

Transformer

Consumersmains cables

Submainscables

Utilitycables / lines

YY

MSB DBSupply

Transformer

Consumersmains cables

Submainscables

Utilitycables / lines

Example: An 11kV to 400/230V transformer has a prospective fault current of 32kA at the secondary terminals. The consumers mains circuit has a route length of 25m, using single-core 120mm2 active conductors and 70mm2 neutral. The submains circuit has a route length of 35m, using 16mm2 multi-core cables.



We want to determine the prospective fault current at the main switchboard and the distribution board (for the purpose of selecting appropriately rated protection devices and switchgear). Transformer impedance:

230 0.00718 32000TX

VZI

= = = Ω

Assume cable temperature is 45oC, from Table 34 of AS3008.1, impedance of consumers mains (1 phase):

0.170 25 0.00425 1000CMZ = × = Ω

Prospective fault current at main switchboard:

230 20.2 kA0.00718 0.00425SCI = =

+

Assume cable temperature is 45oC, from Table 35 of AS3008.1, impedance of the submains (1 phase):

1.26 35 0.0441 1000SMZ = × = Ω

Prospective fault current at the distribution board:

230 4.158 kA0.00718 0.00425 0.0441SCI = =

+ +

Note that the above calculations are for a balanced three phase short-circuit fault across all three phases with the result that there is no neutral impedance to consider. A single phase short-circuit from a single phase conductor to neutral will produce a lower fault current. Here, we need to include the impedance of the neutral cable because the (unbalanced) fault current will flow in the neutral.

1. The Per Unit System Fault calculations are simplified very substantially if they are performed using the per-unit system and normalising all electrical quantities relating to the fault in per unit values for the fault analysis. This allows the removal of the complexity of transformer ratios in the fault calculations. The transformer can be included as a simple impedance. In the per unit system we express voltage, current, kVA and impedance as per unit values of selected base values of those quantities. Thus

puB

VVV

= VB is the voltage base value

puB

III

= IB is the current base value

puB

SSS

= SB is the kVA base

puB

ZZZ

= ZB is the impedance base

It is usual to specify the two base values VB and SB and then the other two base values IB and ZB are able to be determined from the specified VB and SB values by normal (Ohm’s Law) electrical relationships:


BB

B

SIV

=

2

B BB

B B

V VZI S

= =

Normally, the voltage base VB is taken as the rated system voltage and SB is arbitrarily specified (often 100, 10 or 1 MVA is chosen), although a common method is to use the rating of a major element in the system such as a transformer or generator as the base SB. For balanced symmetrical three phase faults the fault calculation is able to be done on a single phase basis using the per unit phase impedances in the one-line diagram of the fault circuit. Some care must be taken to use the proper phase kVA and voltage levels in the single-phase circuit to calculate the appropriate base values of current and impedance.

3

BB

B

SIV

=

2

BB

B

VZS

=

where VB and SB are the line voltage and three phase kVA values.


In the fault calculation the impedances in the fault circuit must include all significant components and all of these must have their impedance expressed in per unit terms using the appropriate base value. This requires changes in some per unit values if they are already expressed (for example on the name plate) using different base values. This may commonly occur with transformer impedances. To change per unit impedances from one base value to another we have to use the following equation as the basis for change:

ohmspu ohms 2

B

B B

Z SZ ZZ V

= =

Thus: (i) For change of kVA base (SB), the new Zpu is given by:

( ) ( )( )

( )

newpu new pu old

old

B

B

SZ Z

S= ×

(ii) For change of voltage base (VB)

( ) ( )( )

( )

2old

pu new pu old 2new

B

B

VZ Z

V= ×

(iii) For change of both kVA and voltage bases at the same

time:

( ) ( )( )

( )

( )

( )

2new old

pu new pu old 2old new

B B

B B

S VZ Z

S V= × ×


In most cases the impedances of items such as transformers, generators, motors etc, will be given on name plates in per unit or percentage terms based on the equipment’s rated voltage and power levels. These given values must be adjusted to the base values chosen for fault calculations if these are different from the nameplate values. For cables, overhead lines, busbars, etc, the impedances will most likely be given or obtained in ohmic values. These must then be used with the appropriate base values to get their per unit values referred to the common bases. Thus the appropriate operating voltage and chosen SB must be used to get ZB and IB. The base impedances and currents for a 1 MVA (1000 kVA) base and typical common voltage levels are shown below, using VB equal to rated voltage: [Note that 1000 kVA is the 3-phase base value]

Line Voltage

(V)

Phase Voltage

(V)

Base Current

(A)

Base Impedance

(Ω) 415

3300 6600 11000 33000 66000

240 1905 3810 6351 19053 38105

1391 175 87.5 52.5 17.5 8.75

0.1722 10.89 43.56 121.0 1089 4356



Example: A 3-phase radial transmission system is shown below. Calculate terminal voltage of the generator. Use a base of 100MVA for all circuits.

50MW0.8pflagging

30kV

132kV 33kV11kV 132kV

50MVAX=10%

50MVAX=12%

j100Ω

Line

VS VS

50MW0.8pflagging

30kV

132kV 33kV11kV 132kV

50MVAX=10%

50MVAX=12%

j100Ω

Line

VS VS

Base impedance of the line:

( )232

6

132 10174

100 10B

BB

VZS

×= = =

×Ω

Per-unit reactance of the line:

100 0.575174B

Z j jZ

= = =

Per-unit reactance of sending-end transformer:

(new)pu(old)

(old)

1000.1 0.250

B

B

SZ j j

S= = =

Per-unit reactance of receiving-end transformer:

(new)pu(old)

(old)

1000.12 0.2450

B

B

SZ j

S= = = j

Load current (using formula 3 cosL LP V I φ= ):

( )

6

3

50 10 12033 30 10 0.8

×= =

× × × A

Base current for the 33kV line:


6

3

100 10 17503 3 33 10

B

B

SV

×= = =

× × × A

Hence, per-unit load current is:

1203 0.6871750B

II

= = = pu

Per-unit voltage of load busbar:

30 0.9133B

VV

= = = pu

The equivalent circuit is shown below:

j0.2pu j0.575pu j0.24pu0.687pu0.8pflagging

ESVS VR=

0.91pu Load

j0.2pu j0.575pu j0.24pu0.687pu0.8pflagging

ESVS VR=

0.91pu Load

Hence,

( )( ) (0.687 0.8 0.6 0.2 0.575 0.24 0.91 0.0SV j j j j= − + + + + )j

1.328 0.558SV j= + pu

1.44SV = pu or 1.44 11 kV 15.84 kV× =

2. Fault Calculation Effects and Requirements Fault levels in a power system are required to be determined at the design stage to allow determination of the following parameters:

(i) overcurrent protection requirements (ii) peak electromagnetic forces (iii) thermal heating effects (iv) the maximum fault current (and the minimum fault

current) (v) the (time) discrimination requirements of protection

operation (vi) the touch voltages on earthed objects (personnel

safety) 2.1 Sources of fault currents In a complex electrical system, there are a number of potential sources of fault current when a short circuit occurs in the system. These are:

(i) the electrical utility supply grid system (ii) any in-house generation systems operating at the

time of the fault (iii) any motors operating within the system at the time of

the fault (iv) any electrical storage elements in the system (e.g

capacitors)


Static equipment such as power electronic inverters and converters, transformers, induction heaters are not sources of fault current. Capacitors in power factor correction systems and battery operated uninterruptible power supplies may be fault current sources however, although generally the contribution of fault current is low and of very short duration. The supply utility contribution to the fault provides a constant fault current (see the diagram below) as will the in-house synchronous generation for a short period, but operating motors will provide decaying fault current contributions as their magnetic excitation fields collapse. Synchronous motors will sustain their fault current level much longer than induction motors. 2.2 Fault impedance variation In calculating fault currents, all components, including the source impedances, must be represented in the one line diagram by an effective impedance in per unit values. For the utility supply this is constant (a stiff power source) but for the motors there is a time-varying impedance depending on the time after the short circuit. Depending on when the fault current needs to be calculated, any of three rotating impedances may need to be used:

(i) sub-transient reactance (Xd”) (ii) transient reactance (Xd’) (iii) synchronous reactance (Xs)


We must use the sub-transient reactance for the fault current during the first few cycles, the transient reactance for the fault current up to a fraction of a second and the synchronous reactance for very long duration faults (usually synchronous reactance is not necessary as the protection should operate before it comes into effect). For synchronous motors only the sub-transient and transient reactance are normally used before the exciting field dies away and the fault current contribution is then effectively reduced to zero. For induction motors, only the sub-transient reactance is used before the fault current contribution dies to zero. The combined overall fault current will thus decrease in the manner shown below.

Examples of fault current waveforms


2.3 DC Offset (See figure below) This must be included in fault calculations, particularly in low voltage systems as the offset can increase the initial current levels substantially and this is an important consideration for the circuit breaker (the momentary current rating): a very high initial current even for less than a cycle can open breaker contacts prematurely and result in contact welding. The magnitude and duration of the DC offset level is governed primarily by the X/R ratio of the faulted circuit. [The offset magnitude is also dependent on the angle on the voltage waveform at which the fault occurs. However the worst-case situation is always assumed in the fault calculation].


2.4 Types of AC faults

The classes of faults that can occur in AC power systems are:

Three phase fault Three phase to earth fault Phase to phase fault Phase to phase to earth fault Single phase to earth fault.

The first of these gives the highest fault current and is the one which will be used in the following examples. However the most common fault is the last type, the single phase to earth fault and at low voltages the fault impedance becomes an important factor in that type, particularly at low voltages. The estimation of fault impedance in such cases is very difficult. Whereas in high voltage faults (11kV and above) it is usual to assume zero or negligible fault impedance, this cannot be done in LV faults where the voltage drop over the fault itself is significant compared to the system voltage and thus there is substantial impact on the fault current. I single phase faults it is normal to assign some multiplying factors (less than unity) when the bolted short circuit levels are calculated to give an estimate of the high impedance arcing fault current.


The following diagrams show some of the above effects of fault currents.

Fault type Magnitude 3-phase (most severe)

Line-to-line Line-to-ground

(usually least severe)

(E/Z) x multiplier About 0.87 x 3-phase fault

Depends on system grounding


3. Fault Calculation Methods For the simple fault calculations that we will cover here, we assume the following: (i) The fault is balanced 3-phase symmetrical.

(ii) The per unit impedances are pure reactances: any resistance is neglected, i.e. it is effectively a DC analysis. This is not a very good approximation for LV systems where the resistance can be significant (see item (vii) below).

(iii) All significant component impedances are included

(iv) The fault itself has zero impedance [that is, it is a “bolted” short circuit]

(v) Earth circuit impedance is neglected because of the balanced 3-phase nature which eliminates the earth impedance.

(vi) The appropriate rated voltage is used as the voltage base value.

(vii) For LV systems where resistance is important, we use the impedance determined by 2 2Z R X= + and a DC analysis.

(viii) Record X R for all equipment, if necessary, to calculate the level of the DC offset multiplier after the symmetrical fault current has been calculated. It is necessary to know R and L separately.



The first step in the process is to convert all impedances to per unit values and to then use these to draw a single line diagram of the fault circuit, including all possible sources modelled as an ideal voltage source with their appropriate source impedance value connected. Then, by a process of circuit simplification the impedance diagram is reduced to a single per unit impedance ZF connected to true earth and to an ideal voltage source.

( )puFZ

pu 1V =

( )puFZ

pu 1V =

Then the fault current and fault power in per unit value are:

( )pu

pu(pu)

FF

VI

Z= and

( )

2pu

(pu)pu

FF

VS

Z=

Thus: ( )

(pu) (pu)pu

1F F

F

I SZ

= = when we define pu 1V =

The actual fault current is ( )puF FI I IB= × amps and fault power is . ( )pu VAF BFS S S= ×

The advantage of using Vpu 1= is evident from the above.


4. Faults in DC Systems

C systems are becoming increasingly common with the

n DC systems the impedance elements which determine

he sources of DC fault currents are, typically, any of the

erators nverters

tems

nother factor that must be considered in the design of the protection system is that DC arc currents are more difficult

Duse of power electronics and the calculation of fault currents in such systems is also necessary to consider in modern commercial and industrial systems. Ithe steady state fault current level are only resistance elements. However in most cases the system inductance will also have a significant effect in that it will determine the rate of increase of the fault current level in DC system faults. The L/R time constants of such systems are usually long enough that the steady state fault current will not be reached before protection operates and the protection will thus be interrupting current when that current is still rising. Thus DC fault calculations are not necessarily simple to perform. Tfollowing:

DC gen Synchronous co DC motors Rectifier sys Battery banks UPS systems

A


to interrupt than AC arc currents. An AC circuit breaker has 100 current zeroes per second to interrupt the fault current, while a DC breaker has none. Thus the arc interruption is much more difficult for DC than for AC. In a DC breaker the arc voltage developed is an important factor in the protection design and in determining fault current levels. As a result of the difference between AC and DC faults, either specialised DC breakers or fuses must be used or, more commonly, if AC breakers are used they must be de-rated for use on a DC system. The fault calculation procedure must involve the

etermination of the time constant and thus the initial

in series with a ixed circuit resistance, a fixed circuit inductance and a

sient is:

dexponential rate of rise of current as it is most likely that interruption will occur during this period. A DC fault is modelled by a DC supply fvariable resistance in the form of the circuit breaker arc when its contact open (see figure over page). The governing equation during the initial tran

S R adIV V V Ldt

= + +

or ( )S RdIL V Vdt

= − − aV

Va is small or zero, Initially, when ( )S RV V Va− > and dI dt is positive and current increases, but later as the arc


develops and lengthens, ( )R VSV V a− < and dI dtnegative and current decreases. The typical behaviour is shown below.

is

+

_

+

_

IaVSV

LR

C.B.

+ +RVdILdt

DC fault circuit and C.B.

+

_

+

_

IaVSV

LR

C.B.

+ +RVdILdt

DC fault circuit and C.B.


5. Fault calculation data and calculation example The following tables give details relating to various parameters required for fault calculations and an example of a typical fault calculation procedure.


5.1 Example of a Simple Fault Current Calculation

The system has supply from a utility connection at top left and an in-house generator at top right. In addition there will be a fault current contribution from the 4.8kV, 200kVA motor which we take to be an induction motor because only one impedance value is given. We have to use an impedance diagram as below and we have to use per unit impedances so that we can remove the transformer ratio complications.


We are required to find the three phase fault current at location A: the voltage there is 480V.

(1) Utilitysupply

(2) Transformers

(3) Generator

(4) Cable

(4’)

(7)

(8)

(9)

(10)

(6) (5)

MotorCable

Cable

Cable

Cable

Power transformer

Current transformer

(1) Utilitysupply

(2) Transformers

(3) Generator

(4) Cable

(4’)

(7)

(8)

(9)

(10)

(6) (5)

MotorCable

Cable

Cable

Cable

Power transformer

Current transformer

Impedance circuit Use base of 20 MVA for p.u calculation, i.e.

20 MVABS =

At 4.8 kV:

pu 1V = 2

1.152 BB

B

VZS

= = Ω

2406 A3

BB

SI = =×

BV

t 480 V: A

pu 1V = 0.01152 BZ = Ω 24056 ABI =


(1) Source impedance:

500 MVA fault level 25≡ p.u pu1 0.0425

Z = = p.u

6X R = ⇒ pu 0.0066 0.0395Z j= + (2) Transformer:

3000 kVA, 6% pu200.06 0.4 p.u3

Z = × =

8X R = ⇒ pu 0.05 0.4Z j= + (3) Generator:

1000 kVA, 15% pu200.15 31

Z = × = p.u.

10X R = (negligible R) ⇒ pu 3.0Z j= 4) Cable (4.8kV): (

0.06 0.03j+ Ω 1.152 BZ = Ω

⇒ pu0.06 0.03 0.052 0.026

1.152jZ j+

= = + p.u.

4’) Cable (4 8kV):

( .

0.05 0.02j+ Ω 1.152 BZ = Ω

⇒ pu0.05 0.02 0.043 0.017

1.152jZ j+

= = + p.u.

5) Motor: (


200 kVA, 10% pu200.1 100.2

Z = × = p.u.

10X R = (negligible R) ⇒ pu 10Z j= (6) Cable (4.8 kV):

Αs (4) p.u.

):

⇒ =pu 0.052 0.026Z j+ 7) Cable (4.8 kV(

0.1 0.04 j+ Ω 2 1.15BZ = Ω

⇒ pu0.1 0.04 0.087 0.035

1.152jZ j+

= = + p.u

1000 kVA

(8) Power transformer:

, 4% pu .04Z = 200 0.81

× = p.u.

4X R = ⇒ pu 0.194 0.776Z j= + p.u.

9) Current transfo (480V): ( rmer

00. 001 0.0005Z j+= 0.01152 BZ = Ω

⇒ pu0.0001 0.0005 0.0087 0.0043

.011520jZ j+

= = + p.u.

e 80V): (10) Cabl (4

0.002 0.002j+ Ω 0.01152BZ = Ω ⇒ pu 0.174 0.174Z j= + p.u.


T we have the imped ed to

hus ance diagram simplifi :

(1)+(2)

(4’)

(7)+(8)+(9)+(10)

(5)+(6)

(3)+(4)

F

(1)+(2)

(4’)

(7)+(8)+(9)+(10)

(5)+(6)

(3)+(4)

F ( ) ( )1 2 0.0566 0.4j+ = + 395 0.443 p.u=

( ) ( )3 3.03 p.u+ = 4 0.052 3.03j= +

( ) ( )5 6 0.05 10.03 10.03 p.uj+ = = 2 +

( )4' 0.043 0.017 0.046 p.uj= + =

( ) ( ) ( ) ( )7 8 9 10 0.4637 0.9893 1.093 p.uj+ + + = + = Approxim tia on :

( ) ( ) ( ) ( )1 2 3 4 0.443 3.03 0.384⎡ ⎤ ⎡ ⎤+ + = =⎦ ⎣ ⎦ ⎣

( ).384 4' 0.430+ = 0

( ) ( )0.430 5 6 0.430 0.03 0.412⎡ ⎤+ = =⎣ ⎦ 1


( ) ( ) ( ) ( ).412 7 8 9 10 0.412 1.093 1.505⎡ ⎤+ + + + = + =0 ⎣ ⎦

otal p.u.

Fault

T pu 1.505Z =

pupu

1 0.664IZ

= =

At 480V: 0.664 0.664 24056 15970 AF BI I= × = × =

At 4.8 kV: 0.664 0.664 2406 1597 AF BI I= × = × = Alternatively :

( ) ( ) ( ) ( )1 2 3 4 0.0440 0.3843j⎡⎣ ⎤ ⎡ ⎤+ + = +⎦ ⎣ ⎦

( ) ( )0.0440 0.3843 4' 0.0870 0.4013j j+ + = +

( ) ( ) ( )0.0870 0.4013 5 6 0.0805 0.3865j j⎡ ⎤+ + = +⎣ ⎦

( ) ( ) ( ) ( ) ( )0.0805 0.3865 7 8 9 10j ⎡ ⎤+ + + + +⎣ ⎦ 0.5442 3758j= +

otalT pu 0.5442 1.3758 1.4795Z j= + = p.u.

Fault pu1I = =pu

0.676Z

At 480V: 0.676 0.676 24056 6262 AF BI I 1= × = × =

At 4.8 kV: 0.676 0.676 2406 1626 AF BI I= × = × =

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