BL20A0710 Introduction to Electrical Power Systems

BL20A0710
Introduction to Electrical
Power Systems

Fault currents

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BL20A0710 Introduction to Electrical Power Systems

Causes of the faults

▪ Climatic overvoltage

▪ Fault in operation or malfunctioning of devices

▪ Human error

▪ Overload

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BL20A0710 Introduction to Electrical Power Systems

Short circuit
▪ Direct conductive connection between live parts (not through a load
or a power source)

▪ Harms
– outage
– rise of temperature
– forces
– electric arc (burns)
– high voltage
– pressure
– gases

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BL20A0710 Introduction to Electrical Power Systems

Earth fault
▪ Connection between a phase wire and earth (not through a load or a
power source)
▪ This phenomena essentially differs from a short circuit when the neutral
point of the network is not grounded. A harmful asymmetry enters the
network. Fault current is rather small.
▪ In Finland, medium voltage networks (10 ja 20 kV) are isolated (neutral
points are not directly grounded) or connected to ground throuh an arc-
suppression coil (Petersen Coil, high reactance between neutral point
and ground)

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BL20A0710 Introduction to Electrical Power Systems

Different types of faults

3-phase short circuit 2-phase short circuit

Single phase short circuit Combined earth fault and short circuit
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BL20A0710 Introduction to Electrical Power Systems

Failure density in power system

A division of disturbances by fault (%)
Amount of disturbances 1-phase 2-phase 3-phase 2- or 3-phase
(amount/1000 km/a) short circuit short circuit short circuit earth fault
Average 2004–2013*
400 kV 4.8 80 2 3 16
220 kV 9.8 78 2 3 17
110 kV 20.6 81 3 2 14
During disturbances the amount of energy not supplied 0.007%.

Un / kV Ik’’ / kA
420 6.9 ... 13.8
Magnitude of short-circuit currents
245 2.4 ... 12.0
123 2.4 .. 14.4
* https://www.entsoe.eu/Documents/Publications/SOC/Nordic/2013_ENTSOE_HVAC_2014_11_06.pdf

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BL20A0710 Introduction to Electrical Power Systems

Example: Failure in power system

https://www.fingrid.fi/en/electricity-market/power-system/operation-disturbances/
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BL20A0710 Introduction to Electrical Power Systems Ik = ?

R L
Stages of the short circuit U

1) Initial stage
– Ik” , initial short-circuit current, defines the maximum asymmetric
short-circuit current (sysäysoikosulkuvirran)
– Xd” , initial reactance of the synchronous machine
– Td” , time constant (~ 0.1 s)
– Protection of the grid works in 0.1–0.5 s
2) Transition stage
– Ik’ , transient short-circuit current
– Xd’ , transient reactance of the synchronous machine
– Td’ , time constant (~ 3–6 s)
3) Steady stage Ik’’ > Ik’ > Ik (X’’ < X’ < X) especially if
– Ik , steady stage short-circuit current generators locate near the fault place
– Xd , longitudinal synchronous reactance of the synch. machine
– Ik defines the heating of conductors and components

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BL20A0710 Introduction to Electrical Power Systems

Calculating the short-circuit current

Rs Xs Rtr Xtr
S T

~ Up
Ik,3p

Thévenin’s theorem
𝑈p
𝐼k,3p = ZT = Thévenin’s impedance
𝑍T + 𝑍F ZF = Fault impedance

𝑈p
if 𝑍F = 0 ⇒ 𝐼k,3p = s = supplying network
𝑅s + 𝑅tr 2 + 𝑋s + 𝑋tr 2
tr = transformer

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BL20A0710 Introduction to Electrical Power Systems

Example 1 400/110 kV R = 0.1  X = 1.1 

A

Ik’’ = 15 kA SN = 400 MVA

Ik’ = 12 kA ux = 10%
Ik = 10 kA
▪ A 3-phase short-circuit occurs at point A. Calculate the initial, transient and the steady stage
short-circuit currents
a) short-circuit reactances of the 400 kV network
𝑈
3 400 kV
𝑋k ′′ = ′′ = = 15.40 Ω :: 𝑋k ′ = 19.25 Ω :: 𝑋k = 23.09 Ω
𝐼k 3 ⋅ 15 kA

Values referred to the 110 kV voltage level

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110 kV
𝑋k ′′ = 15.40 Ω ⋅ = 1.16 Ω :: 𝑋k ′ = 1.46 Ω :: 𝑋k = 1.75 Ω
400 kV

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BL20A0710 Introduction to Electrical Power Systems

Example 1
b) Transformer
𝑈N 2 (110 kV)2
𝑋tr = 𝑢x = 0.1 ⋅ Ω = 3.03 Ω
𝑆N 400 MVA

▪ Thévenin’s impedance at the site of fault

𝑍T ′′ = 𝑅2 + (𝑋 + 𝑋k + 𝑋tr )2
= 0. 12 + 1.1 + 1.16 + 3.03 2 Ω = 5.29 Ω :: 𝑍T ′ = 5.59 Ω :: 𝑍T = 5.88 Ω

110 kV
⇒ 𝐼k ′′ = kA = 12.01 kA :: 𝐼k ′ = 11.36 kA :: 𝐼k = 10.80 kA
3 ⋅ 5.29 Ω
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BL20A0710 Introduction to Electrical Power Systems

Example 2
6/20 kV 20/6 kV
X = 3.0  A X = 5.0 
G1 G2

Un = 6 kV Sn = 10 MVA Sn = 20 MVA Un = 6 kV
Sn = 10 MVA uX = 10% uX = 10% Sn = 20 MVA
Xd = 20% Xd = 20%

▪ Define the short-circuit current for point A. Voltage is 20 kV in point A

before the fault occurs.
a) Generators
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6 kV 2 20 kV 62 20
2
𝑋G1 = 0.2 ⋅ ⋅ =8Ω 𝑋G2 = 0.2 ⋅ ⋅ = 4Ω
10 MVA 6 kV 20 6

20 kV 2 202
b) Transformers 𝑋tr1 = 0.1 ⋅ =4Ω 𝑋tr2 = 0.1 ⋅ =2Ω
10 MVA 20

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BL20A0710 Introduction to Electrical Power Systems

Example 2
8 4.0  3.0  5.0  2.0  4

 Up ~

15  11 

 Up ~

15 ⋅ 11
Z 𝑍T =
15 + 11
Ω = 6.35 Ω

 Up ~ 𝐼k =
20 kV
3 ⋅ 6.35 Ω
= 1.82 kA

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BL20A0710 Introduction to Electrical Power Systems

Example 3
20 kA ~ ~ 20 kA

400 kV 400 kV
100 km

800 MW 800 MW
20% 20%

115 kV 115 kV
100 km (0.3 /km)
Ik/kA

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10 kA ~ a ~ 10 kA
l

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Magnitude of a short-circuit current

10
Ik in a 100 km interconnecting line
0
0 20 40 60 80 100 l/km

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BL20A0710 Introduction to Electrical Power Systems

Example 3 Ik/kA

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~ 31.5 kA
20
20.0 kA

115 kV 10
Ik 0.4 /km
0
l 0 10 20 30 40 50 l/km

Short-circuit current Ik for a

110 kV radial conductor

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BL20A0710 Introduction to Electrical Power Systems

Short-circuit current in the electricity

distribution network
110/20 kV

uk, SN

▪ 3-phase short-circuit current

Ik,3p = Up/ZT
ZT is Thévenin’s impedance at the site of fault
(impendance of the network supplying fault current)

k = supply network (generators and transmission grid)

𝑍T = 𝑅k + 𝑅tr + 𝑅cond 2 + 𝑋k + 𝑋tr + 𝑋cond 2
tr = 110/20 kV transformer (uk, SN)
cond = 20 kV conductor (NB! Temperature rise)
Short-circuit current typically 0.3–10 kA
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BL20A0710 Introduction to Electrical Power Systems

Restriction of short circuit current

▪ Short-circuit currents are becoming higher as the demand for electricity rises, as
− more power plants are being built
– more 110/20 kV transformers are being installed
– the size of transformers are becoming bigger
– more lines are being built
– impedance is decreasing in existing feeders
▪ Tools of limiting a short-circuit current:
– utilization of an reactors
– choosing uk –values for the transformer
– fuses
– isolation of neutral point of a transformer from the ground
– choosing a higher voltage level
– dividing the network into smaller sections

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BL20A0710 Introduction to Electrical Power Systems

About grounding
▪ Equipment earthing: earthing a non-electric circuit component e.g.
earthing an iron cover or a transformer cover.
▪ System earthing: part of a circuit is directly connected to earth or
through a low impedance
− system earthing of a 3-phase system is done when one of its point
is directly, or through a small impedance, connected to earth
• Target is to stabilize voltage in the starpoint
– Circuit is not system earthed when it is isolated from the ground or
connected to it with high impedance e.g., with an arc-suppression
coil (resonant earthed system)

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BL20A0710 Introduction to Electrical Power Systems

Earthing of networks

1. 400 kV networks
− star points are earthed
2. 110 kV networks
– star points are earthed (from) here and there
3. 20 kV networks
– resonant earthed systems or network isolated from the ground
4. 0.4 kV networks
– system earthing done directly

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BL20A0710 Introduction to Electrical Power Systems

Earth fault in an earthed network

▪ Fault current is in the level of short-circuit current. During an earth fault,
overvoltages occur in healthy (undamaged) phases
ER X R
~
~
ES X S
~
US, UR and UT= phase voltages
US
ET X T
~ If
f = earth fault
Xtr UT
If

▪ An asymmetrical situation→ symmetrical components are needed in

calculations; positive-, negative- and a zero-sequence network
▪ Overvoltages are smaller than in an isolated neutral system (an effectively
earthed network, earthing factor K < 1.4)
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BL20A0710 Introduction to Electrical Power Systems

Calculating fault current in different cases

▪ 3-phase short circuit

𝑈p
𝐼f = If = IR = IS = IT
𝑍1 + 𝑍F

▪ 2-phase short circuit Symmetrical components

Z1= positive sequence impedance
𝑈p If = IS = -IT Z2= negative sequence impedance
𝐼f = −𝑗 3 Z0= zero sequence impedance
𝑍1 + 𝑍 2 + 𝑍F
IR = 0
▪ Single-phase earth fault
3𝑈p
𝐼f =
𝑍1 + 𝑍2 + 𝑍0 + 3𝑍F If = IR
▪ 2-phase earth fault IS = IT = 0
2
𝑗 3 𝑎𝑍2 − 𝑍0 − 3𝑍F 𝑈p −𝑗 3 𝑎 𝑍2 − 𝑍0 − 3𝑍F 𝑈p
𝐼R = 𝐼S =
𝑍1 𝑍2 + 𝑍0 + 3𝑍F 𝑍1 + 𝑍2 𝑍1 𝑍2 + 𝑍0 + 3𝑍F 𝑍1 + 𝑍2

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BL20A0710 Introduction to Electrical Power Systems

Earth fault in a medium voltage network

▪ In Finland, medium voltage network is either isolated from earth or it is
earthed through an arc-suppression coil
R R
S S
T T

Single-phase earth fault 2-phase short-circuit earth

R R
S S
T T

Double earth fault Conductor fault and single-phase earth fault on the load side
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BL20A0710 Introduction to Electrical Power Systems

Earth fault in a medium voltage network

▪ Single-phase earth fault in an isolated neutral system

Isolated neutral system

There is no direct connection which would short the circuit,
➔ fault current is very small
Only route for the fault current is through the
capacitances (between ground and conductors)

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BL20A0710 Introduction to Electrical Power Systems

Isolated neutral system

L1

L2 (Feeder 1)

L3
I
110/21 kV C1

L1

L2 (Feeder 2)
a
L3
I C2
If Rf
b
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BL20A0710 Introduction to Electrical Power Systems

Single-phase fault current, isolated neutral

system
▪ If resistance exists in the fault place
𝐼earth,f = 𝑈p ⋅ 3𝜔𝐶0 = 3𝑈 ⋅ 𝜔𝐶0

▪ Fault resistance reduces further the small fault current

𝜔𝐶0
𝐼earth,f = 𝑈p
1 + 3𝜔𝐶0 𝑅f 2

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BL20A0710 Introduction to Electrical Power Systems

Single-phase fault current, isolated neutral

system
▪ The earth fault current is dependant of the length of the conductors, which is a
part of an integrated network, and its earth capacitance
▪ Equations for an approximate value:
𝑈 [kV] ⋅ 𝑙 [km]
overhead line network: 𝐼f ≈ 𝑈 [kV] ⋅ 𝑙 [km] A underground cable network: 𝐼f ≈ A
300 5
▪ For example, fault current:
20 km

110/20 kV Total length 60 km (20 kV voltage level)

20 km - overhead line network → 4A
- underground cable network → 240 A
20 km

− NB! This is the total length (not only the length of trunk cables).
f = earth fault
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BL20A0710 Introduction to Electrical Power Systems

Voltages during an earth fault in an isolated

neutral system
▪ During an earth fault, voltages of the phases and neutral point change in the
network US
UR
U0 = zero potential = voltage
U’S between neutral point and ground
U0
U’R

UT
Us, UR and UT= phase voltages
If Rf

▪ Voltage of the faulted phase decreases during a fault. If Rf = 0 ➔ Voltage of the

faulted phase = 0
▪ Voltages of healthy phases increase. If the fault resistance = 0 ➔ Voltages of
the phases are equal to the main voltage 𝑈 = 3 ⋅ 𝑈p

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BL20A0710 Introduction to Electrical Power Systems

Voltages during an earth fault in an isolated

neutral system
▪ Voltage of the neutral point rises during a fault. If Rf = 0, U0 = Up
1 1
𝑈0 = ⋅𝐼 = ⋅ 𝑈p
3𝜔𝐶0 f 1 + 3𝜔𝐶0 𝑅f 2

▪ In other words, bigger the fault resistance, smaller the zero sequence
voltage
▪ When Rf =  → U0 = 0 (healthy network)

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BL20A0710 Introduction to Electrical Power Systems

Resonant earthed network

L1
L2

L3
110/21 kV I C1

L R L1

IL IR L2
a L3
I C2
If Rf
IC
b
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Fault currents, hazard voltages

▪ Earthing voltage: Uearth = If∙Rf
▪ Maximum values, for the earthing voltage, are set by the electrical safety
regulations
e.g. 𝑈earth = 500ൗ V, in which t is duration of the earth fault
𝑡

Utouch
▪ Utouch = touch voltage, Ustep = step voltage
Uearth
Ustep

Ustep

If
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