Symmetrical Faults - Spyros Chatzivasileiadis

Spyros Chatzivasileiadis

31730: Electric Power Engineering, fundamentals

Symmetrical Faults

DTU Electrical Engineering, Technical University of Denmark

Exercises are important

By induction… exercise sessions are important

2


Empires of Light

• The war of currents

• AC vs DC

• Tesla and Westinghouse vs Edison

• The birth of power systems

3


Goals for Today• Fault current components during

transients of R-L circuits

• Transient and subtransientreactances: Xd’, Xd’’

4 1 November 2016

• How to calculate the fault currents during three-phase short-circuits


Have you ever experienced a blackout?

5


Blackouts

6


Blackouts

• US blackout in 2003: affected 55 million people• India blackout in 2012: affected 700 million people

• Blackouts are rare events:à Frequency of interruptions: ~1 hr/year• High cost à Economic damage (US only, 2005): ~80 billion US dollars/year

– Total electric energy cost in the US: ~370 billion US dollars/year7


Faults• Cause: insulation breakdown

– Lightning– Wires blowing together in the wind– Animals or plants coming in contact with the wires– Salt spray or pollution of insulators

• Two main types of faults– Symmetric faults: system remains balanced; relatively rare, but the easiest to analyze– Asymmetric faults: system is no longer balanced; much more common, but more difficult

to analyze

• Most common type of fault: single-phase-to-ground fault

8


Three-phase faults

• 3-phase faults are the most severe type of fault

• Engineers carry out a lot of studies and simulations, studying blackouts

• During both power system design and operation we study the impact of such faults, in order to:

– Calculate the fault currents– Calculate the voltages during the fault

so that:– We can dimension our equipment to survive the fault, e.g. conductors/cables,

transformers, etc.– Properly size and tune all our protection devices

https://www.youtube.com/watch?v=BFHtVqoyj8g

9


Benchmarking on Continuity of Supply I/II

16

Source: http://www.ceer.eu

Average annual interruption time in European countries

10

Source: M. Bollen, W. Friedl, PES GM 2014

http://grouper.ieee.org/groups/td/dist/sd/doc/2014-08%20European%20Benchmarking%20Report%20on%20Quality%20of%20Supply-%20Mathew%20Bollen+Werner%20Friedl.pdf


Series R-L circuit

11

What equation described the relationship between v(t) and i(t)?

Figure taken from:J. Glover, T. Overbye, M. Sarma, Power System Analysis and Design,

Cengage Learning, Sixth Edition, 2016


Series R-L circuit transients

12

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i =V

R(1� e�t/T ) (81)

T =L

R(82)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i =V

R(1� e�t/T ) (81)

T =L

R(82)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i(t) =V

R(1� e�t/T ) (81)

T =L

R(82)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i(t) =V

R(1� e�t/T ) (81)

T =L

R(82)

e = V (83)

e(t) =p2V sin(!t+ ↵) (84)

4

Constant voltage

(DC, e.g. battery)




Series R-L circuit transients

13

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i =V

R(1� e�t/T ) (81)

T =L

R(82)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i =V

R(1� e�t/T ) (81)

T =L

R(82)

4

i(t) = iac(t) + idc(t) (83)

i(t) =

p2V

Z[sin(!t+ ↵� ✓)� sin(↵� ✓)e�t/T ] (84)

iac(t) =

p2V

Z[sin(!t+ ↵� ✓)] (85)

idc(t) = �p2V

Z[sin(↵� ✓)e�t/T ] (86)

5

i(t) = iac(t) + idc(t) (83)

i(t) =

p2V

Z[sin(!t+ ↵� ✓)� sin(↵� ✓)e�t/T ] (84)

iac(t) =

p2V

Z[sin(!t+ ↵� ✓)] (85)

idc(t) = �p2V

Z[sin(↵� ✓)e�t/T ] (86)

5

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i(t) =V

R(1� e�t/T ) (81)

T =L

R(82)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i(t) =V

R(1� e�t/T ) (81)

T =L

R(82)

e = V (83)

e(t) =p2V sin(!t+ ↵) (84)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

V = Ri+ Ldi

dt(80)

i(t) =V

R(1� e�t/T ) (81)

T =L

R(82)

e = V (83)

e(t) =p2V sin(!t+ ↵) (84)

4

Constant voltage

(DC, e.g. battery) ac voltage

Steady-state fault current

dc offset currentasymmetrical

fault current

DTU Electrical Engineering, Technical University of Denmark14

i(t) = iac(t) + idc(t) (83)

i(t) =

p2V

Z[sin(!t+ ↵� ✓)� sin(↵� ✓)e�t/T ] (84)

iac(t) =

p2V

Z[sin(!t+ ↵� ✓)] (85)

idc(t) = �p2V

Z[sin(↵� ✓)e�t/T ] (86)

5


dc offset current




i(t) = iac(t) + idc(t) (83)

i(t) =

p2V

Z[sin(!t+ ↵� ✓)� sin(↵� ✓)e�t/T ] (84)

iac(t) =

p2V

Z[sin(!t+ ↵� ✓)] (85)

idc(t) = �p2V

Z[sin(↵� ✓)e�t/T ] (86)

5


dc offset current

For which values of αdo we get

the minimum, and the maximumdc fault current?

How much is the faultcurrent at t=0?


Unloaded synchronous machine

16

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

Direct axis synchronous reactance

Direct axis transient reactance

Direct axis subtransient reactance

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

4

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

T 0d (75)

T 00d (76)

X 00d < X 0

d < Xd (77)

(78)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(79)

4

Steady-state

Instantaneous AC Fault Current


Unloaded synchronous machine

17

�a1 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D12+ Ic ln

1

D13

�(63)

�a2 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D23+ Ic ln

1

D12

�(64)

�a3 =µ0

2⇡

Ia ln

1

r0+ Ib ln

1

D31+ Ic ln

1

D23

�(65)

�a =�a1

l3 + �a2

l3 + �a3

l3

l=

�a1 + �a2 + �a3

3(66)

�a =1

3

µ0

2⇡

3Ia ln

1

r0+ Ib ln

1

D12D23D31+ Ic ln

1

D13D12D23

�(67)

�a =1

3

µ0

2⇡

3Ia ln

1

r0� Ia ln

1

D12D23D31

�(68)

�a =1

3

µ0

2⇡

3Ia ln

3pD12D23D31

r0

�(69)

Deq = 3p

D12D23D31 (70)

La =µ0

2⇡ln

Deq

DSL(71)

Xd (72)

X 0d (73)

X 00d (74)

X 00d < X 0

d < Xd (75)

(76)

iac(t) =p2Eg

✓1

X 00d

� 1

X 0d

◆e�t/T 00

d +

✓1

X 0d

� 1

Xd

◆e�t/T 0

d +1

Xd

�sin

⇣!t+ ↵� ⇡

2

⌘

(77)

4

How much is iac for t=0 ?

How much is iac for t →∞ ?

*This is just the symmetrical (ac) component. The dc offset has been removed.


Assumptions for 3-phase short-circuit studies1. Transformers represented by their leakage reactances

2. Lines represented by their equivalent series reactance

3. Synchronous machines represented by constant voltage-sources behind subtransientreactances

4. Neglect non-rotating impedance loads

5. For small motors under 40 kW: either neglect induction motors or represent them as synchronous machines

18

These assumptions are made to simplify our calculations.

In everyday practice, there are cases where we have to avoid such simplifications.

Can you give an example where such simplifications should be avoided?


3-phase short circuit

19

How will you calculate the fault current?





20

• Step 1: This is a balanced 3-phase fault: Design the single-phase equivalent. – Follow the simplifying assumptions for 3-phase short-circuits– calculate all elements in p.u.



21

1. Transformers represented by their leakage reactances

2. Lines represented by their equivalent series reactance

3. Synchronous machines represented by constant voltage-sources behind subtransient reactances

4. Neglect non-rotating impedance loads5. For small motors under 40 kW: either

neglect induction motors or represent them as synchronous machines



22

Figure taken from:J. Glover, T. Overbye, M. Sarma, Power System Analysis

and Design, Cengage Learning, Sixth Edition, 2016


Short-circuit = SW closes

=Then the two circuits are

equivalent

+VF

-VF

3-phase short-circuit: superposition technique




3-phase short-circuit: superposition technique

24

= = +

+


3-phase short circuit: alternative method

25

What is a more traditional method to calculate IF’’ in this circuit?

What values do you know?

What values you don’t know and need to find?



26

What is a more traditional method to calculate IF’’ in this circuit?

1. You know VF at point 12. You know that IF’’ = Ig’’ + Im’’3. You know all impedances

4. You don’t know Eg’’ and Em’’5. You don’t know Ig’’ and Im’’



27

1. Calculate the pre-fault current at point 1

2. Calculate Eg’’ and Em’’

3. Calculate the fault current Ig’’ at point 1, due to Eg’’

4. Calculate the fault current Im’’ at point 1, due to Em’’

5. IF’’ = Ig’’ + Im’’


Positive-sequence?• The term “positive-sequence” refers to the modeling approach we use for unbalanced

conditions: the symmetrical components

• Symmetrical components consist of the positive-sequence, negative-sequence, and zero-sequence

• In this course we deal only with balanced systems

• In balanced systems, the positive-sequence component corresponds to the single-phase equivalent of our three-phase system

28


Per unit calculations

29

Why do we calculate in p.u.?


Per unit calculations

30

Why do we calculate in p.u.?

•Proper selection of base values eliminates the need for transforming from one voltage level to the other

•Then: voltage and current values in p.u. remain the same from both sides of the transformer


Per unit calculations1. Divide our system in as many areas as NTRAFO+1

2. Select a base voltage for each area– Make sure that the voltage ratio between two neighboring areas is the same as the

transformer ratio between the two areas it connects

3. Select a single base MVA that will be the same for the whole system

4. Calculate the base impedance in each area

5. Calculate the impedances in p.u.

6. Calculate the voltages in p.u.

7. Calculate the power in p.u. (i.e. usually of loads or generators)

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31xxx : Optimization Methods for Power Systems

• Special 3-week course in January 2017 on power system optimization• 5 ECTS

Questions that we will try to answer:

• How do you plan the operation of our electricity system in order to:– Minimize costs?– Ensure a safe operation (no blackouts) ?

• What methods do power system operators currently use?

• What are the state-of-the-art methods?

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Symmetrical Faults - Spyros Chatzivasileiadis

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Transcript of Symmetrical Faults - Spyros Chatzivasileiadis