Faults And Short-circuit Study

Mustafa Bağrıyanık
Prof.Dr.
Istanbul Technical University
Department of Electrical Engineering
Maslak, 34422, TURKEY

“ These notes are only to be used in class presentations”

Distribution of Electrical Energy

 Outline

• Faults
• Short Circuit
• Symmetrical Components
• Calculation of Short-circuit currents
• Examples
 Short-circuit Duration
• Self-extinguishing
– fault too short to cause tripping of protections
– example: bird on a line

• Fugitive
– resolved after tripping of the protections and
reclosing
– example: lightning

• Permanent
– does not disappear after tripping of the protections
– example: breakage of the conductor
• Faults are either temporary or permanent
• A permanent fault is one where permanent
damage is done to the system.
• This includes insulator failures, broken wires,
or failed equipment such as transformers or
capacitors.
• Virtually all faults on underground equipment
are permanent.
• Most equipment fails to a short circuit.
permanent fault

• Permanent faults on distribution

circuits usually cause sustained
interruptions for some customers.
• To clear the fault, a fuse, recloser, or
circuit breaker must operate to
interrupt the circuit.
• The most critical location is the three-
phase mains, since a fault on the main
feeder will cause an interruption to all
customers on the circuit.
permanent fault

• A permanent fault also causes a voltage

sag to customers on the feeder and on
adjacent feeders.
• Permanent faults may cause momentary
interruptions for a customer.
• A common example is a fault on a fused
lateral (tap).
permanent fault

• With fuse saving (where an upstream circuit

breaker or recloser attempts to open before
the tap fuse blows), a permanent fault
causes a momentary interruption for
customers downstream of the circuit breaker
or recloser.
• After the first attempt to save the fuse, if the
fault is still there, the circuit breaker allows
the fuse to clear the fault.
• If a fault is permanent,
– all customers on the circuit experience a
momentary interruption, and
– the customers on the fused lateral experience a
sustained interruption.
temporary fault

• A temporary fault does not permanently

damage any system equipment.
• If the circuit is interrupted and then
reclosed after a delay, the system
operates normally.
• Temporary (non-damage) faults make
up 50 to 90% of faults on overhead
distribution systems.
temporary fault

• The causes of temporary faults include

– lightning,
– conductors slapping together in the wind,
– tree branches that fall across conductors and
then fall or burn off,
– animals that cause faults and fall off, and
– insulator flashovers caused by pollution.
• Temporary faults are the main reason that
– reclosing is used almost universally on
distribution circuit breakers and
– reclosers (on overhead circuits).
temporary fault

• Temporary faults will cause voltage

sags for customers on the circuit with
the fault and possibly for customers on
adjacent feeders.
• Temporary faults cause sustained
interruptions if the fault is downstream
of a fuse, and fuse saving is not used
or is not successful.
• For temporary faults on the feeder
backbone, all customers on the circuit
are momentarily interrupted.
temporary fault

• Faults that are normally temporary can turn

into permanent faults.
• If the fault is allowed to remain too long, the
fault arc can do permanent damage to
conductors, insulators, or other hardware.
• In addition, the fault current flowing through
equipment can do damage.
• The most common damage of this type is to
connectors or circuit interrupters such as
fuses.
• The majority of faults on overhead
distribution circuits are temporary.
• It is absolutely essential to have a precise
knowledge of the short-circuıt currents at
any point in a system

– Maximum short-circuit current r.m.s. value

• > circuit breaker and fuse breaking capacity
• > thermal stress
– Maximum short-circuit current peak value
• > making capacity
• > electrodynamic stress
– Minimum short-circuit current value
• > protection setting
• As the main purpose of short-circuit
calculations is
– to select and apply the devices properly, it
is meaningful for the calculations to be
related to current interruption phenomena
and the rating structures of interrupting
devices.
• The objectives of short-circuit calculations
can be summarized as follows:
– Determination of short-circuit duties on switching
devices, i.e., high-, medium- and low-voltage
circuit breakers and fuses.
– Calculation of short-circuit currents required for
protective relaying and coordination of protective
devices.
– Evaluations of adequacy of short-circuit
withstand ratings of static equipment like cables,
conductors, bus bars, reactors, and transformers.
– Calculations of fault voltage dips and their time-
dependent recovery profiles.
• A bolted fault means as if three phases were
connected together with links of zero
impedance prior to the fault,
– i.e., the fault impedance itself is zero and
– the fault is limited by the system and machine
impedances only.
• Such a fault is called a symmetrical three-
phase bolted fault, or a solid fault.
• Bolted three-phase faults are rather
uncommon.
• Generally, such faults give the maximum
short-circuit currents and form the basis of
calculations of short-circuit duties on switching
devices.
• Under certain conditions, the line-to-
ground fault or double line-to-ground
fault currents may exceed three-phase
symmetrical fault currents.
• Unsymmetrical faults are more
common as compared to three-phase
faults,
– i.e., a support insulator on one of the
phases on a transmission line may start
flashing to ground, ultimately resulting in a
single line-to-ground fault.
• Short-circuit calculations are, thus, the
primary study whenever a new power system
is designed or an expansion and upgrade of
an existing system are planned.
• In order to select and determine the
characteristics of equipment for electrical
networks it is necessary to know
– the magnitudes of the short-circuit currents
and
– short-circuit powers which may occur.
• The short-circuit
current at first runs
asymmetrically to
the zero line,
• It contains an
alternating-current
component and a
direct-current
component.

I k" initial symmetrical short-circuit

current,
IP peak short-circuit current,
Ik steady state short-circuit
current,
A initial value of direct current,
1 upper envelope,
Curve of short-circuit current:
2 lower envelope,
3 decaying direct current. a) near-to-generator fault,
b) far-from-generator fault
Establishment of short-circuit current
The appearance of a short-circuit (SC)
R X
in the system:
I sc A
Z sc • leads to sharp variation in load
impedance
• gives rise to a high short-circuit
E current, Isc, which is limited by the
B
SC impedance, Zsc .
The fault impedance is ignored.

Establishment of Isc :

• according to transient operation,

dependent on Zsc
• affected by the electrical proximity
of the fault and the system
generators
 Fault a long way from generator
• Most frequent case

• The short-circuit impedance, Zsc, remains constant during the fault.

• SC current in transient operation = 2 currents superimposed:

Ia : alternating sinusoidal component
Ic : aperiodic component
 Effect of the switching angle a
Switching angle a :
characterizes the shift between the initial moment of the
fault and the origin of the voltage wave
 a  j : Symmetrical operation
Ic = 0, no transient operation
Isc peak minimum
Isc _ peak  2  Irms
 a -j  p/2 : Asymmetrical operation
Ic maximum
Isc peak maximum
R
Isc _ peak  K  Irms with K  f( )
X
 Value of peak current

K
• Relationship between peak current and
2,8 rms current :
2,4

2,0
Isc _ peak  K  Irms
1,6
1,4
1,2

R
0,8 - p
0,4 where K  2 (1  e X
)
0 R
0,2 0,4 0,6 0,8 1,0 1,2 1,4 X

• Typical values:
– X=0: resistive circuit K 2
– R=0: non-damped system K  2 2
– for MV, R/X ~ 0.1 K 2.5
 Fault near a generator
Fault near a generator:
Zsc = generator impedance, variable
during the fault.

SC current near a generator = 2 currents

superimposed:
• Ia : damped current that can be
broken down into 3 states:
– subtransient (duration 10-20 ms)
– transient (duration 100-400 ms)
– synchronous (steady)
• Ic : aperiodic current
(duration 10-100 ms)

substransient transient steady state

 Fault near a generator

• The generator emf may be

considered as constant, its
internal reactance as variable
and its internal resistance as
negligible

• The 3 damped sinusoidal states

correspond to 3 reactances:
– X"d, subtransient reactance
– X'd, transient reactance
– Xd, synchronous reactance

substransient transient steady state

 Effect of the switching angle a

Switching angle a :
characterizes the shift between
the initial moment of the fault
and the origin of the voltage
wave

 a  j  p/2 : Symmetrical state

Ic = 0
no aperiodic state
Isc peak minimum

 a  0 : Asymmetrical state
Ic maximum
Isc peak maximum

substransient transient steady state

 Fault near a generator: breaking

 a  0 : Asymmetrical state
Ic maximum
Isc peak maximum
Isc does not go through 0 for
several periods !

=> BREAKING PROBLEM

substransient transient steady state

 Fault near a generator: detection

• Steady state established 0.5s to 1s

after the appearance of the fault

• Weak steady Isc current:

30% to 50% of rated current, In !

=> DETECTION AND

PROTECTION PROBLEM

• Solution :
– specific protection function
– classical protection function if
using generators designed so
substransient transient steady state that Isc = 3xIn for a few
seconds
 Contribution of asynchronous motors

During a short-circuit, asynchronous motors :

• operate like asynchronous generators

• feed the flow during the flow extinction time

(time constant from 10 to 50 ms)

• are considered as voltage sources

of transient impedance X'd (%) = In x 100
Istart
• In practice:
Motor contribution to SC current= 3 x S In
S In: sum of the rated currents of all the motors installed,
at the fault point
 Symmetrical components
of asymmetrical three-
phase systems
• A 3-phase fault affects the three-phase
network symmetrically.
• All three conductors are equally involved and
carry the same rms short-circuit current.
• Calculation need therefore be for only one
conductor.
• All other short-circuit conditions, on the other
hand, incur asymmetrical loadings.
• A suitable method for investigating such
events is to split the asymmetrical system
into its symmetrical components.
 Symmetrical components

• With a symmetrical voltage system the

currents produced by an asymmetrical
loading (I1, I2 and I3) can be
determined with the aid of the
symmetrical components (positive-,
negative- and zero-sequence system).
• The symmetrical components can be
found with the aid of complex
calculation or by graphical means.
 1
I
Current in positive-sequence system 1  ( I1  a I 2  a 2 I 3 )
3

1
Current in neg.-sequence system I 2  ( I1  a 2 I 2  a I 3 )
3
1
Current in zero-sequence system I 0  ( I1  I 2  I 3 )
3

For the rotational operators of value 1:

a  e j 120  a 2  e j 240  1 a  a2  0
• If the current vector leading the current in
the reference conductor is rotated 120°
backwards, and the lagging current vector
120 ° forwards, the resultant is equal to
three times the vector Im in the reference
conductor.
• The negative-sequence components are
apparent.
• If one turns in the other direction, the
positive-sequence system is evident
and the resultant is three times the
vector Ig in the reference conductor.
• Geometrical addition of all three
current vectors (I1, I2 and I3) yields
three times the vector I0 in the
reference conductor.
• If the neutral conductor is unaffected,
there is no zero-sequence system.
 Study of a three-phase system using symmetrical components

• The electrical faults can be

• The power supply source of a balanced
system is balanced – three-phase fault

– three-phase to earth fault
E3  
E(1)  E
  or unbalanced
E1 E(2)  0 – phase to phase fault


E2
E(0)  0 – two-phase to earth fault
– phase to earth fault

• The equipment and electrical

• Study of a SYSTEM
loads are balanced UNBALANCED BY A FAULT
– lines replaced by study of the 3
– transformers EQUIVALENT SYMMETRICAL
SYSTEMS, POSITIVE-
– capacitors SEQUENCE, NEGATIVE-
– motors SEQUENCE and ZERO-
SEQUENCE systems
 Currents and voltages at the fault point
System of equations to be resolved

• Breakdown of currents and voltages into symmetrical

components

          
V1  V(1)  V(2)  V(0) I 1  I (1)  I (2)  I (0) V(1)  E - Z(1) I (1)
         
V2  a V(1)  aV(2)  V(0) I 2  a I (1)  a I (2)  I (0) V(2)  - Z(2) I (2)
2 2

         
V3  aV(1)  a V(2)  V(0) I 3  a I (1)  a I (2)  I (0) V(0)  - Z(0) I (0)
2 2

• Equations governing the fault point, depending on the type of

fault.
 Currents and voltages at the fault point
Resolution of the equations

• Given ( complex values )

– E System power supply phase-to-neutral voltage
– Z Fault impedance
– Z(1) System positive-sequence impedance
– Z(2) System negative-sequence impedance
– Z(0) System zero sequence impedance

• Unknowns ( complex values )

– I1 , I2 , I3 Real phase currents at the fault point
– V1, V2, V3 Real phase-to-neutral voltages at the fault point
– I(1), I(2), I(0) Positive-sequence, negative-sequence and zero-sequence
current system at the fault point
– V(1), V(2), V(0) Positive-sequence, negative-sequence and zero-sequence
voltage system at the fault point
 Three-phase fault
Equations
• Three-phase faults are balanced
3-phase fault
no negative-sequence or zero-
Phase 3 sequence components
Phase 2 • Equations independent of Z
Phase 1 valid for isolated or earth fault

3-phase to earth fault E

I1 
Phase 3 Z(1)
Phase 2 E E
I2  a 2  I1 = I2 = I3 =
Phase 1 Z(1) Z(1)
E
V1  V 2  V 3  Z(I1  I2  I3 )
Z I3  a
Z(1)
V1  V2  V3  0
 Three-phase fault
Equivalent single phase layout

V(1)
I(1)
E Z(1)

E V(2)
I(1)  I(2)
Z(1) Z(2)

I(2)  I(0)  0
V(0)
V(1)  V(2)  V(0)  0 I(0)
Z(0)
 Phase-to-earth fault
Equations

Phase 3
Phase 2
Phase 1

I2  I3  0 Z
3E
V1  ZI1 I1 
Z(1)  Z(2)  Z(0)  3Z
3ZE
• Simplified equation : V1 =
– non-resist. fault Z=0 Z(1)  Z(2)  Z(0)  3Z
– Z(0) ~ 3ZN >> Z(1) + Z(2)
E
I1 
ZN
 Phase-to-earth fault
Equivalent single phase layout

V(1)
E I(1)
I(1)  I(2)  I(0)  E Z(1)
Z(1)  Z(2)  Z(0)  3Z
V(2)
E(Z(2)  Z(0)  3Z)
V(1)  I(2)
Z(1)  Z(2)  Z(0)  3Z Z(2)

- Z(2)E
V(2)  V(0)
Z(1)  Z(2)  Z(0)  3Z I(0)
Z(0)
- Z(0)E
V(0) 
Z(1)  Z(2)  Z(0)  3Z 3Z
 Phase-to-phase fault
Equations

Phase 3 3E
I2  - j
Phase 2 Z(1)  Z(2)  Z
Phase 1 3E
I3  j
I1 = 0, I2  - I3 Z Z(1)  Z(2)  Z
V3 - V2  ZI3 E(2Z(2)  Z)
V1 
Z(1)  Z(2)  Z
• Simplified equation :
E(a Z - Z(2))
2
– non-resist. fault Z=0
V2 
– In general Z(1) ~ Z(2) Z(1)  Z(2)  Z
E(aZ - Z(2))
3E V3 
I2  - I3  Z(1)  Z(2)  Z
2Z(1)
 Phase-to-phase fault
Equivalent single-phase layout

E
I(1)  V(1)
Z(1)  Z(2)  Z I(1)
E Z(1)
E
I(2)  -
Z(1)  Z(2)  Z V(2) Z
I(2)
I(0)  0 Z(2)
E(Z(2)  Z)
V(1)  V(0)
Z(1)  Z(2)  Z
I(0)
EZ(2) Z(0)
V(2) 
Z(1)  Z(2)  Z
V(0)  0
 2-phase-to-earth fault
Equations

3E(Z(0) - aZ(2)  3Z)

Phase 3 I2  - j
Phase 2 Z(1)Z(2)  (Z(1)  Z(2))(Z(0)  3Z)
Phase 1
3E(Z(0) - a Z(2)  3Z)
2

I3  j
I1 = 0 Z Z(1)Z(2)  (Z(1)  Z(2))(Z(0)  3Z)
V 2  V 3  Z(I2  I3 )
3Z(2)E(Z(0)  2Z)
V1 
Z(1)Z(2)  (Z(1)  Z(2))(Z(0)  3Z)
3Z(2)ZE
V2  -
Z(1)Z(2)  (Z(1)  Z(2))(Z(0)  3Z)
V3  V2
 2-phase-to-earth fault
Equivalent single-phase layout

V(1)
E(Z(2)  Z(0)  3Z) I(1)
I(1)  E Z(1)
Z(1)Z(2)  (3Z  Z(0))(Z(1)  Z(2))
V(2)
-E(Z(0)  3Z)
I(2)  - Z(2)
I(2)
Z(1)Z(2)  (3Z  Z(0))(Z(1)  Z(2))
- Z(2)E V(0) 3Z
I(0)  - I(0)
Z(1)Z(2)  (3Z  Z(0))(Z(1)  Z(2)) Z(0)

V(1), V(2), V(0) : very complex ...

 Fundamentals of calculation of
Short-circuit currents
• The calculation of short-circuit currents is
always based on the assumption of a dead
short circuit.
• Other influences, especially arc resistances,
contact resistances, conductor temperatures,
inductances of current transformers and the
like, can have the effect of lowering the short-
circuit currents.
• Since they are not amenable to calculation,
they are accounted for in Table 2 by the
factor c.
• Nominal system voltage (Un ) is the (line-to-
line) voltage by which a system is specified
and to which certain operating characteristics
are referred.
• Equivalent voltage source ( cU n / 3 ) is
the voltage of an ideal source applied at the
short-circuit location in the positive-sequence
system as the network’s only effective voltage
in order to calculate the short-circuit currents
by the equivalent voltage source method.
• Voltage factor (c) is the relationship between
the voltage of the equivalent voltage source
and U n / 3
Voltage factor (c)
Initial symmetrical short-circuit currents are calculated with
the equations in Table.
 Calculation of peak
short-circuit current Ip
• When calculating the peak short-circuit
current Ip, sequential faults are
disregarded.
• Three-phase short circuits are treated as
though the short circuit occurs in all three
conductors simultaneously.
• We have: "
ip  k 2 Ik

• The factor k takes into account the decay of

the d.c. component. It can be calculated as
-3 R / X
k  1.02  0.98 e
• The factor k can be taken from Figure
• The maximum value of k = 2 is attained
only in the theoretical limiting case with an
active resistance of R = 0 in the short-circuit
path.
• Experience shows that with a short-circuit
at the generator terminals a value of k = 1.8
is not exceeded, if machines’ power is less
than 100 MVA.
• With a unit-connected generator and high-
power transformer, however, a value of k =
1.9 can be reached in unfavourable
circumstances in the event of a short circuit
near the transformer on its high-voltage side,
owing to the transformer’s very small ratio R/X.
• The same applies to networks with a high fault
power if a short circuit occurs after a reactor.
• The impedances of equipment in the higher-
or lower-voltage networks have to be
recalculated with the square of the rated
transformer ratio (main tap).
• Synchronous motors and synchronous
condensers are treated as synchronous
generators.
• Induction motors contribute values to Ik”, ip
and Ia and in the case of a two-phase short
circuit, to Ik as well.
When calculating minimum short-circuit currents
one has to make the following changes:
• Reduced voltage factor c
• The network’s topology must be chosen so as to
yield the minimum short-circuit currents.
• Motors are to be disregarded
• The resistances RL of the lines must be
determined for the conductor temperature te at
the end of the short circuit (RL20 conductor
temperature at 20 °C).
• RL = [1 + 0.004 (te – 20 °C) / °C] · RL20
• For lines in low-voltage networks it is sufficient to
put te = 80°C.
 Short-circuit power

• Short-circuit power is the maximum

power that a network can supply to
equipment with a fault in it.
• It is expressed either in MVA or in
effective kA for a given service voltage.
• At a given point, the short-circuit power
depends on the network configuration
and on its components : generators,
lines, cables, transformers, motors…
• The rated short time withstand current
and its peak value are dependent on
the short-circuit current at a given point
of the panel.
• The breaking and closing capacities
required in circuit breakers are
determined on the basis of these
values.
• Short-circuit power is equivalent to an
apparent power.
• It is a conventional, practical value, without
any physical reality.
• Short-circuit power is defined by the
following expression :
Ssc  3 Us Isc
where, Ssc is short-circuit power,
Us is the service voltage, and
Isc is the short circuit current.

• Short-circuit power is expressed in MVA which

corresponds to kA for a given service voltage.
• Since we often do not have the necessary
information for calculation of the short-
circuit power value it is often specified by
the customer.
• Determination of short-circuit power
requires analysis of power flux that supply
the short-circuit in the worst case.
• Possible sources are :
– network supply via one or more power
transformers
– alternator supply
– power return due to revolving machines
(motors,…) or from MV/LV transformers

• We must calculate each of these flux.

 Example:

Figure 5. Calculation procedures for

industrial installations are defined in
IEC 909, published in 1988. [Merlin
Gerin]

Three sources are circulating in the panel (T1 - A - T2).

If transformer (T4) is supplied from another source, Isc5 is also possible return via LV.
Circuit breaker D1 (s/c in A): Isc2 + Isc3 + Isc4 + Isc5
Circuit breaker D2 (s/c in B): Isc1 + Isc3 + Isc4 + Isc5
Circuit breaker D3 (s/c in C): Isc1 + Isc2 + Isc4 + Isc5
• The rated short time withstand current
and its peak value are dependent on
the short-circuit power that is set up in
a short circuited installation.
• The short-circuit current that is set up will be
eliminated within a given time, and, as a result,
equipment should be chosen with a
corresponding short time thermal withstand (Ith).
• This short-circuit current has a peak value that
leads to electrodynamic forces that the
equipment should be capable of withstanding.
• This is why it will be necessary for the
electrodynamic withstand to correspond at least
to this peak value (Idyn).
The short-circuit power, thermal and electrodynamic currents are
related by the following equations :

Short-circuit current : Isc = (kA rms)

Thermal withstand : Ith = (kA rms/1 s or 3 s)
Electrodynamic withstand : Idyn = (kA peak).

Example :
Short-circuit current :
25 kA rms
Thermal withstand :
25 kA rms/1 s
Electrodynamic withstand :
2.5 x 25 = 62.5 kA peak.
• Knowing the short-circuit power of a network
allows us to choose various elements of the
panel so as to be able to withstand
considerable heating and electrodynamic
stresses.
• For example, isolating switches, switchboards
and circuit breakers.
• The breaking and closing capacities required
in circuit breakers are directly dependent on
the short-circuit power value at their terminals
and on the devices that are directly
connected to them.
• The breaking capacity, the maximum
amount of current that a circuit breaker
is capable of breaking on occurance of
a short-circuit under specific standard
conditions, must be greater than or
equal to the rated short time withstand
current that can be set up under these
circumstances.
 Short-circuit breaking current Ib
Calculation of the short-circuit breaking current Ib is required only when
the fault is near the generator and protection is ensured by timedelayed
circuit breakers. Note that this current is used to determine the breaking
capacity of these circuit breakers. This current may be calculated with a
fair degree of accuracy using the following equation:
Ib = μ . I"k
where μ = is a factor defined by the minimum time delay tmin and the
I"k / Ir ratio which expresses the influence of the subtransient and
transient reactances, with Ir as the rated current of the generator.
• Closing capacity, the maximum
amount of current that a circuit breaker
is capable of withstanding and of
maintaining on a short circuited
installation, must be greater than or
equal to the peak value of the rated
short time withstand circumstances.
• Calculation of the short-circuit power at
the terminals of a synchronous
generator is very complex since the
internal impedance of the latter varies as
a function of time.
• It increases progressively, with the
current decreasing correspondingly and
passing through three typical stages :
– subtransient
– transient
– steady-state
The transient
stages during
short-circuit
occurred at the
terminals of a
synchronous
generator

• subtransient (which enables determination of the closing

capacity of circuit breakers) ; average duration 10 ms
• transient (that fixes the breaking capacity of circuit
breakers) ; average duration 250 ms
• steady-state (short-circuit current value in the steady-
state stage).
• The complexity of the short-circuit current
calculation lies essentially in determination of
the equivalent impedance of the network
upstream of the point

• Each component of a network (supply

network, transformer, alternator, motors,
cables, busbars,…) is represented by an
impedance (Z) made up of a resistive
component (R) and an inductive component
(X) called the reactance. X, R and Z are
expressed in ohm.
The method consists of :
• splitting the network down into
sections,
• calculating values of R and X for each
section,
• calculating for the network :
• the equivalent value of R or X,
• the value of equivalent impedance,
• the short-circuit current.
 Association of impedances
in series and in parallel

Z1

Ze Z1 Z2 Ze

Z2

Z1Z 2
Ze  Z 1  Z 2 Ze 
Z1  Z 2
 Association of impedances
in star and delta layouts

Z2 Z21

Z1 Z32

Z3 Z13

Z 21Z13 Z 2 Z1
Z1  Z 21  Z 2  Z1 
Z 21  Z 32  Z13 Z3
Z 32 Z 21 Z3 Z2
Z2  Z 32  Z 3  Z 2 
Z 21  Z 32  Z13 Z1
Z13 Z 32 Z1Z 3
Z3  Z13  Z1  Z 3 
Z 21  Z 32  Z13 Z2
 Diagram reduction
Example

E E E
Source

Rs

Line Xs Ra
R
Rl Xa
Xl

R1 R2 Rb
X
X1 X2 Xb
Cable 1 Cable 2
 Diagram reduction example

network layout equivalent layout

equivalent layout equivalent layout

Calculation of the value of equivalent impedance by step by step transformation method

• More complex phase fault calculations are
made with computer programs.
• When calculating short-circuit currents in
high-voltage installations, it is often sufficient
to work with reactances because the
reactances are generally much greater in
magnitude than the effective resistances.
• Also, if one works only with reactances in the
following examples, the calculation is on the
safe side.
• Corrections to the reactances are
disregarded.
• The ratios of the nominal system voltages are
taken as the transformer ratios.
• Instead of the operating voltages of the faulty
network one works with the nominal system
voltage.
• It is assumed that the nominal voltages of the
various network components are the same as
the nominal system voltage at their respective
locations.
• Calculation is done with the aid of the %/MVA
system.
 Characteristics of the upstream network (distributor)

• Upstream network
– outside power supply source
– positive-sequence voltage symmetric three-phase system
– delivers the rated voltage under all circumstances

• Impedance of the upstream network calculated from its short-circuit power

3Vn 2 Un 2
Ssc   Zsc 
Zsc Ssc

Where Ssc(VA)= upstream network short-circuit power

Un(V) = upstream network rated phase-to-phase voltage
 Transformer characteristics
Short-circuit voltage

Short-circuit voltage of a transformer =

voltage to be applied to primary circuit to cause the rated current to flow in
the secondary circuit during a short-circuit

Usc(%) Usc(%) Vnsec

 Vnsec  Zsec  Insec  Zsec( )  
100 100 Insec
Usc(%) Unsec 2
Sn  3  Vnsec  Insec  Zsec( )  
100 Sn
Where Usc(%) = transformer short-circuit voltage
Insec = secondary circuit rated current
Vnsec = secondary circuit rated phase-to-neutral voltage
Unsec = secondary circuit no load rated phase-to-phase voltage
Sn = transformer apparent power
Zsec = transformer impedance seen from secondary circuit
 Transformer characteristics
Copper losses

• Copper losses of a transformer =

Losses due to Joule effect caused by transformer resistance
2
PC u  Un sec 
PCu  3 Rt sec  In  Rt sec  C u  
2
sec =P
3 In
2
 Sn 

• Where Pcu = transformer copper losses

Insec = secondary circuit rated current
Unsec = secondary circuit no load rated phase-to-phase
voltage
Sn = transformer apparent power
Rtsec = transformer resistance seen from secondary
Rt  0.1Xt to 0.3 Xt
• Generally speaking, Rt is ignored when compared to Xt
 Transformer characteristics
Impedance transformation

• Impedance transformation:
All impedances must be brought to the level of the rated voltage at the
fault point
Up, Ip

Zp 2
 Up 
Zp     Zs  Zp  k 2  Zs
 Us 
Up
k , transformation ratio
Zs Us

Us, Is
 Transformer characteristics
Typical impedances

Transformer coupling
(detected at R X(1) X(2) X(0)
secondary end)

without neutral 

Yyn or Zyn, free flow 

Zsc - R
2 2

PCu Un 2
Yyn or Zyn, forced flow Zsc  Usc  X(1) 10 to 15 X(1)
3  In 2 Sn
Dyn or YNyn 4%  Usc  20% X(1)

-- zn 0.1 to 0.2 X(1)

 Transformer characteristics
Zero sequence layout: principle
• Zero sequence layout :
Depends on transformer coupling of
windings. ? ? ? Zt(0) ?

• There is a path that enables the zero-

sequence current to circulate between the ? ? Zt(0)
system and the transformer (connection
between windings and earth)
? Zt(0) ?

? Zt(0) ?
• There is a path that enables the zero-
sequence current to circulate in the
? ? Zt(0)
transformer windings without circulating
through the system (delta coupling)
 Transformer characteristics
Zero sequence layouts: examples
Coupling of Zero-sequence Coupling of Zero-sequence
windings layout windings layout

Zt(0)
Zt(0)

Zt(0)
Zt(0)

Zt(0) Zt(0)
 Characteristics of rotating machines
Impedances expressed as %
• For rotating machines, the values of subtransient impedances, X", transient
impedances, X' and synchronous impedances, X, are expressed as
percentages rather than in ohms.

X(%) Vn X(%) U2n

X(  )    
100 In 100 Sn
Where In = rated current
Vn = rated phase-to-neutral voltage
Un = rated phase-to-phase voltage
Sn = rated apparent power

• Generally speaking, R is ignored when compared to X

R  X  Z  X
 Characteristics of rotating machines
Typical turbo-generator impedances

R X(1) X(2) X(0)

Subtransient 10 to 20 %

Transient << X 15 to 25 % 10 to 20 % 5 to 10 %

Synchronous 150 to 250 %

 Connection characteristics

R X(1) X(2) X(0)

Overhead lines l 0.3 to 0.4 X(1) ~3X(1)

R( )   
s /km
 in mm / m
2

at 20 C

MV cables Cu  0.018 0.1 to 0.2 X(1) ~3X(1)

Al  0.029 /km
 Calculation of 3-phase Isc, effect of R

63 kV system
PROBLEM:
Ssc = 800 MVA
R/X = 0.1
• Calculate the rms value of the
current circulating at the 6.3 kV
Transformer end during a three-phase non-
63 / 6.3 kV
Sn = 10 MVA resistive short-circuit at 6.3 kV
Usc = 10%
Pcu = 1%
3-phase SC • Draw conclusions concerning
the effect of the circuit
resistance
 Calculation of 3-phase Isc, effect of R

ANSWER:
R X
( 6.3  10 )
2 3 2
Uns
Network : Zsc =   0.05 
800  10
6
Ssc 0.005  0.05 
R = 0.1  X  X  Zsc
( 6.3  10 )
2 3 2
PCu Uns
Transformer : R =   0.01   0.04 
10  10
6
Sn Sn
0.04  0.4 
( 6.3  10 )
2 3 2
Usc Uns
Z=   0.1   0.4 
10  10
6
Uns Sn
: Z = 0.045  0.45  0.45 
2 2
Total 0.045  0.45 
6.3  10
3
Uns
Isc =   8083 A
3  Zsc 3  0.45
the resistance R is NEGLIGIBLE
Exercise: Calculation of 3-phase Isc, effect
of MV cables
63 kV system
PROBLEM:
Ssc = 800 MVA
R/X = 0.1
• Calculate the rms value of the
Transformer current circulating at the 6.3 kV
63 / 6.3 kV end during a three-phase non-
Sn = 10 MVA
Usc = 10%
resistive short-circuit 2 km
Pcu = 1% downstream in the cables

6 kV cable • Draw conclusions concerning

50 mm² Al
2 km (1.3 miles)
the effect of MV cables

3-phase SC
 Calculation of 3-phase Isc, effect of MV cables
ANSWER:

Cable : R X
1 1 2000
= mm / m  R =  = 1.18
2
Network
34 34 50 + trans. 0.045  0.45 
x  0.15 / km  X  0.15  2  0.3 
Cable 1.18  0.3 

: Z = 1.225  0.75  144

. 
2 2
Total Total 1.225  0.75 
6.3  10
3
Uns
Isc =   2532 A
3  Zsc 3  144
.

The impedance of the MV cables till 100 m is often NEGLIGIBLE

 Exercise:Calculation of 2-phase Isc

63 kV system
PROBLEM:
Ssc = 800 MVA
R/X = 0.1
• Establish the equivalent single-
phase layouts according to the
Transformer breakdown of the system into
63 / 6.3 kV
Sn = 10 MVA symmetrical components
Usc = 10%
Pcu = 1%
2-phase SC • Calculate the rms value of the
current circulating at the 6.3 kV
end during an isolated non-
resistive
2-phase short-circuit at 6.3 kV
 Calculation of 2-phase Isc

ANSWER:

I(1)
E Zr(1) Zt(1) I1  0 (SC between phases 2 & 3)
I(2) I2  a ²  I(1) + a  I( 2 ) + I( 0 ) = (a² - a)I(1)
Zr(2) Zt(2)
3 E
I2  - j 3  I(1) =
Zr(0) Zt(0) 2( Zr (1)  Zt (1))
with : R  0, Xr(1) = 0.05, Xt(1) = 0.4 
E
I(1) 
Zr (1)  Zr ( 2 )  Zt (1)  Zt ( 2 )
6.3  10
3
I( 2 )  I(1),I( 0 )  0  I2 = -I3 =  7000 A
Zr (1)  Zr ( 2 ), Zt (1)  Zt ( 2 ) 2  0.45
Exercise: Calculation of phase-to-earth Isc

63 kV system
PROBLEM:
Ssc = 800 MVA
R/X = 0.1
• Establish the equivalent single-
phase layouts according to the
Transformer breakdown of the system into
63 / 6.3 kV
Sn = 10 MVA symmetrical components
Usc = 10%
Pcu = 1%
phase-to-earth SC • Calculate the rms value of the
current circulating at the 6.3 kV
end during a non-resistive
phase-to-earth short-circuit at
6.3 kV
 Calculation of phase-to-earth Isc
ANSWER:

I(1) I2  I3  0 (SC between phase 1 & earth)

E Zr(1) Zt(1)
I1  I(1)  I( 2 )  I( 0 )  3  I( 0 )
I(2) 3E
Zr(2) Zt(2) I1 
2  Zr (1)  3  Zt (1)
I(0) avec : R  0, Xr(1) = 0.05, Xt(1) = 0.4
Zr(0) Zt(0)
3  6.3  10
3

 I1 =  8394 A
2  0.05  3  0.4
I(1)  I( 2 )  I( 0 )
 I1_ phase - to - earth > I1_3 - phase
E
I( 0 )  NB : In the case of a transformer with the neutral
Zr (1)  Zr ( 2 )  Zt (1)  Zt ( 2 )  Zt ( 0 )
earthed by impedance Z, it is necessary to
Zr (1)  Zr ( 2 ), Zt (1)  Zt ( 2 )  Zt ( 0 ) add 3Z to Zt(0)
(see chapter on earthing systems)
the reactances of the equivalent circuit
• The reactances entered in column 3 are added in the case of series
circuits, while the susceptances in column 4 are added for parallel
configurations.
• Columns 6 to 9 are for calculating the maximum short-circuit current and
the symmetrical breaking current.
• To determine the total reactance of the network at the fault location, one
first adds the reactances of the 220 kV network and of transformer 1.
• The sum 0.1438 % /MVA is in column 3, line 3.
• The reactance of the generator is then connected in parallel to this total.
• This is done by forming the susceptance relating to each reactance and
adding the susceptances (column 4, lines 3 and 4).
• The sum of the susceptances 15.1041 % /MVA is in column 4, line 5.
• Taking the reciprocal gives the corresponding reactance 0.0662 %/MVA,
entered in column 3, line 5.
• To this is added the reactance of transformer 2.
• The sum of 0.9412 %/MVA is in column 3, line 7.
• The reactances of the induction motor and of the induction motor group
must then be connected in parallel to this total reactance.
• Again this is done by finding the susceptances and adding them
together.
• The resultant reactance of the whole network at the site of the fault,
0.7225 %/MVA, is shown in column 3, line 10.
• This value gives
 References
• “ABB Switchgear Manual”.
• Schneider Electric, “Calculation of
short-circuit currents”, Cahier
technique.
• “Electrical Installations Handbook”,
Günter G. Seip, John Wiley & Sons.