Short Circuit Studies Indonesia Clean Energy Development (ICED) project Indonesia Wind Sector Impact Assessment Presented by: Dr. K. Balaraman, Makassar, February 17 to 21, 2014
Feb 25, 2016
Short Circuit Studies
Indonesia Clean Energy Development (ICED) project
Indonesia Wind Sector Impact AssessmentPresented by:
Dr. K. Balaraman,
Makassar, February 17 to 21, 2014
Short circuit study requirements:
• Determining the fault level at buses
• Selection of breaker ratings
• Protective device co-ordination
Symmetrical components
• Unbalanced system of ‘n’ related phasors can be resolved to ‘n’ system of balanced phasors
• In each balanced phasor, angle between two phasors and magnitude of each phasor are equal
• Phase Quantities: Ia , Ib & Ic and• Sequence components are:
– Ia1 stands for Positive sequence current– Ia2 stands for Negative sequence current– Ia0 stands for Zero sequence current.
021
021
021
ccc
bbb
aaa
IIIIIIIII
c
b
a
III
=
Sequence components
• Ia1, Ib1 & Ic1 : Same phase sequence as Ia, Ib & Ic
• Ia2, Ib2, & Ic2: Opposite phase sequence as Ia, Ib and Ic
• Ia0, Ib0, & Ic0 : All in-phase
Ia
Ic
Ib
Ia1
Ic1
Ib1
120
120 120 Ia2
Ib2
Ic2
120
120 120
Ia0 Ib0 Ic0
Unbalanced current phasor
Positive sequence current
phasors
Negative sequence current
phasors
Zero sequence current phasor
c0c2c1c
b0b2b1b
a0a2a1a
VVVV
VVVV
VVVV
a0
a2
a1
2
2
c
b
a
VVV
1aa
1aa
111
VVV
Similarly for voltages
sequencetoPhaseps
aa
aa
T
phasetoSequencespand
aa
aa
aaT
ps
sp
1111
1
31
12012401,1201
1
1
2
2
2
2
2
111
alwaysVVV
VVVV
VVV
aa
aa
VVV
cabcab
cbaa
c
b
a
a
a
a
0
31
1111
1
31
0
2
2
0
2
1
Therefore no zero sequence exists in line voltage phasor. Since phase voltage sum is not always zero, in the phase voltage phasor, the zero sequence voltage exists.
III
a aa a
III
I I I I
a
a
a
a
b
c
a a b c
1
2
0
2
2
0
13
111 1 1
13
Delta Connection
Ia
a
b
Ib
Ic
c
Ia+ Ib + Ic= 0
Therefore no zero sequence current flows into delta connection.
Sequence Currents
Ia a
Ib
Ic
b
c
Star connection Star grounded
Ia
I0a
3I0
Ib
Ic
I0
I0
b
c
Ia+Ib+Ic = 0, There fore no zero sequence current flows into star connection
Ia+Ib+Ic may not be zero. Hence path always exists for zero sequence currents.
Star - Delta +ve and -ve Zero
Transformer sequence impedance diagram
Star grounded-Delta +ve and -ve Zero
Transformer sequence impedance diagram
Three winding Transformer
a
b
c
a’
b’
c’
Zaa
Zbb
Zcc
c
b
a
cccbca
bcbbba
acabaa
'cc
'bb
'aa
III
ZZZZZZZZZ
VVV
Let Zaa = Zbb = Zcc = Zs and Zab = Zac = Zba = Zbc = Zca
= Zcb = ZmTsp Vs = [Z] . Tsp.Is
Sequence Impedance
1 1 111
1 1 111
0 00 00 0 2
2
2
1
2
0
2
2
1
2
0
1
2
0
1
2
0
1
a aa a
VVV
Z Z ZZ Z ZZ Z Z
a aa a
III
VVV
Z ZZ Z
Zs Z
III
Z
a
a
a
s m m
m s m
m m s
a
a
a
a
a
a
s m
s m
m
a
a
a
, ,2 00 0
0 00 0 2
ZZ Z
Z ZZ Z
s
s m
s m
s m
Rotating Machines
• Za,b,c is not symmetric. Even then, the Z1,2.0 is diagonalized
Exercise:• Find the expression for Z1,2,0 • and Prove that it is a diagonal matrix:
VVV
Z Z ZZ Z ZZ Z Z
III
a
b
c
s m m
m s m
m m s
a
b
c
1 2
2 1
1 2
a
b
c
Ia
Ib
Ic
Ea
Ec Eb
ZnIn
+
+ +
Steady State
c
a
b
Ia1
Ib1
Ic1
Ea
Ec Eb
Z1
Z1
Z1
Z1
Ea Va1
Ia1
Reference Bus
Va1 = Ea - Ia1 Z1
a
+
-
Positive Sequence Network
c
Ia2
Ib2
Ic2
Z2
Z2
Z2
Z2 Va2
Reference Bus
Va2 = - Ia2 Z2
a b
a
Negative Sequence Network
c
a
b
Ia0
Ib0
Ic0
Zg0
3Zn Va0
Reference Bus
Va3 = - Ia0 Z0
a Zg0
Zg0
Zn 3Ia0 Zg0 Z0
Zero Sequence Network
Generator impedance for fault study :
• Transient (xd’) or sub-transient ( xd”) is considered for positive sequence
• X2 i.e. Negative sequence which is close to xd”. (Approximately)
• X0 is small 0.1 to 0.7 times xd”
Typical values on own rating: Xd 100 to 200 %Xq 60 to 200 %
Xd’ 21 to 41 %
Xd” 13 to 30 %
X2 Xd”
Transmission line:
• Positive sequence impedance = Negative sequence impedance• Zero sequence impedance depends upon: Return path, Ground
wires and Earth resistively• Zero sequence reactance is approximately 2 - 2.5 times
positive sequence impedanceR0 is usually large. May be 5 to 10 times also
• B0 is 65 to 80 % of B positive sequence
Transformers :• All are equal i.e. Zpt = Znt = Zzt for transformer
Va1 Vf
Z1
Ia1
Va2
Z2
Ia2
Va0 Z0
Ia0
Va1 = Vf - Z1.Ia1 Va2 = -Z2 . Ia2 Va0 = - Z0 . Ia0
a0
a2
a1
0
2
1f
a0
a2
a1
III
Z000Z000Z
00V
VVV
Fault Representation
a
b
c
a
b
c
Vab = 0 Va = 0 Va1 = 0Vbc = 0 Vb = 0 Va2 = 0Vca = 0 Vc = 0 Va0 = 0
a0
a2
a1
0
2
1f
III
Z000Z000Z
00V
000
Three phase fault representation
Ia1 = Vf Z1 , Ia2 = 0 , Ia0 = 0
a0
a2
a1
2
2
c
b
a
III
1aa
1aa
111
III
Ia = Ia1
Ib = a2Ia1
Ic = aIa1
Ia = Ib = Ic
Zf
a
Zf
a
Zf
a
Z0
Zf
a
Zf
a
Zf
a
0I0IZZ
VI a0a2f1
fa1
, ,
Fault through impedance :
ab
c
IcIb Ia
Va = 0
Ib = 0
Ic = 0
Single line to ground fault representation
021
3
ZZZV
If
a 021
021 ZZZV
IIIf
aaa
IZIZIZVVVV 002211021 aaafaaa 000 IZV aa 222 IZV aa
111 IZVV afa
021
02
1
021
000000
00
III
ZZ
ZV
VVV
aaaf
aaa
021 III aaa
22
11111
31 aa
aa
III
cba
021
III
aaa
021 0VVVV aaaa Q
a
b
c
Ia
Ic
Ib
Zf
f021
fa0a2a1 3zZZZ
VIII
Fault through Impedance
a
b
c
Ic
Ib
Ia
V b = V c
Ia = 0
Ib = -I c
c
b
a2
2
a0
a2
a1
VVV
111aa1
aa1
31
VVV
Va1 = 1/3 [Va+aVb+a2Vc] = 1/3 [Va+aVb +a2Vb]Va2 = 1/3 [Va+a2Vb+aVc] = 1/3 [Va+a2Vb+aVb]Va1 = Va2
Line to line fault representation
III
a a
a aIII
a
a
a
a
b
c
1
2
0
2
213
1
11 1 1
Ia0 = 1/3 (Ia +Ib +Ic) = 1/3 (Ib - Ib) = 0
Ia1 = 1/3 (aIb+a2Ic) = 1/3(aIb-a2Ib)
Ia2 = 1/3 (a2Ib + aIc) = 1/3 (a2Ib-aIb) Ia1 = -Ia2
0
000000
00 1
1
0
2
1
0
2
1
a
af
a
a
a
II
ZZZV
VVVVa1 = Vf - Z1Ia1
Va2 = Z2Ia2 = Va1 = Vf - Z1Ia1
Z2Ia1 = Vf - Z1Ia1 I VZ Z and I I I Va
fa a a a1
1 22 1 0 00 0, ,
a
b
cIcIb
Ia
Zf
I IV
Z Z Za af
f1 2
1 2
Fault through impedance
a
b
cIcIb
IaVb = 0
Vc = 0
Ia = 0
VVV
a aa a
Vaaa
a120
2
213
111 1 1
00
V V Va a a1 2 0
1 1 1V V V V V Va a a a a a1 2 03 3 3
, ,
Double line to ground fault representation
III
a a
a a II
a
a
a
b
c
1
2
0
2
213
1
11 1 1
0
Ia1 + Ia2 + Ia0 = IA =0Ia1 = 1/3 (aIb + a2Ic)Ia2 = 1/3 (a2Ib + aIc)Ia0 = 1/3 (Ib + Ic)
1/3(aIb + a2Ic) + 1/3(a2Ib + aIc) + 1/3(Ib +Ic) = 0aIb + a2Ic + a2Ib + aIc + Ib + Ic = 0
VVV
V ZZ
Z
III
a
a
a
f a
a
a
1
2
0
1
2
0
1
2
0
00
0 00 00 0
Va1 = Vf - Z1Ia1
Va2 = - Z2Ia2
Va0 = - Z0Ia0
Va1 = Va2 = Va0
\ Vf - Z1Ia1 = - Z2Ia2 - Z0Ia0
\ I V
Z Z ZZ Z
af
1
12 0
2 0
æèç
öø÷
V a0
Z 1 Z 2 Z 0
V f
I ao
V a2V a1
I a2I a1
Fault through impedance:a
b
c
ZfZf
Zg
I V
Z Z ZZ Z
Z Z ZZ Z ZZ Z Z Z
af
f
f
f g
1
12 0
2 0
1 1
2 2
0 0 3
æèç
öø÷
'' '
' '
'
'
'
bc
b’c’
a a’Ib = 0 and Ic = 0
V V Z I Z Iia
ka
aa a aa f ' '
III
a aa a
I
I I I I I
a
a
a
a
a a a a f
1
2
0
2
2
1 2 0
13
111 1 1
00
13
13
Open in phase B & C
VVV
V ZZ
Z
III
a
a
a
f a
a
a
1
2
0
1
2
0
1
2
0
00
0 00 00 0
V V Z IV Z IV Z I
V V V V V Z I Z IV V
Z I V Z I Z I Z I
I I IV
Z Z Z Z
a f a
a a
a a
ka
la
kl kl kl zz f aa a
f kl
aa a kl a a a
a a aokl
aa
1 1 1
2 2 2
0 0 0
1 2 01
1 1 1 2 2 0 0
1 21 2 0
3
3
3
' '
'
'
f0
f2
f1f
'aa
0kl
0lk
0ll
0kk0
2kl
2lk
2ll
2kk2
1kl
1lk
1ll
1kk1
ZZZZ3Z
ZZZZZ
ZZZZZ
ZZZZZ
• Vkl is the Thevenin’s equivalent voltage, once the line is removed between buses k & l
Solution Methodology:• 1. Do the load flow with the line isolated• 2. Then insert the line with two phase open and
perform short circuit study
Open in phase A• Ia = 0• Similar type of analysis like double line fault• Voltage is between two nodes, rather than between
node and ground• Impedance is the Thevenin’s equivalent impedance
between nodes.
IV V
ZZ Z
Z Z
fl k l
1
2 0
2 0
'' '
' '.
Z Z Z Z Z Z
Z Z Z Z Z Z
Z Z Z Z Z Z
lkk ll lk kl f
lkk ll lk kl f
lkk ll lk kl f
11 1 1 1 1
22 2 2 2 2
00 0 0 0 0
I IZ
Z Zf f2 1 0
0 2
.'
' ' II Z
Z Zff01
2
0 2
. '
' '
Solution methodology
• Form the Y bus for +ve, -ve and zero sequence• Do the LU factorisation for +ve, -ve and zero
sequence• To find the driving point impedance in fault study,
particular row or column of Z bus is requiredConsider, Yx = zx=Y-1zx= Z.z
Let z have 1 in pth location and 0, at all other location
Solution methodology
• Pass in z, 1.0 for the desired bus and 0 else where.• Call LU solution - x will be the desired column of Z bus.• Find the sequence fault current I1
f, I2f and I0
f.• Determine the post fault bus voltage
x
Z Z Z ZZ Z Z Z
Z Z Z Z
ZZ
Z
Yx z
p n
p p pp pn
n n np nn
p
pp
np
11 12 1 1
1 2
1 2
101
0
LL
ML
M M
[Vf] = [Z] [If1]
= [Z] {[I0] - [If]}
= [Z] [I0] + [Z][-If]
= [V0] + [V]
V = [Z] [-If]
Y[V] = [-If]
Solve V using LU solution.
Above can be written as :
[Y]p[V]p = [-If]p , p: positive sequence and similarly for other sequences.
For open fault, between k & l,
I k II
f p f
f
00
0
MMMM
For faults other than open fault,
I If p f
p
00
0
:
:
A
BC
DE
F
G
20% on 100 MVA
Xd”= 9%
100 MVA ,138/13.8 kV11%
40 Mile, 0.77 Ohm/mile X.X0 = 3X
50 MVA ,13.8/138 kV11%
Xd”=24% High resistanceGrounding
Example:
Grounding practice in Power System
Advantages of ungrounded system : • Ground connection normally doesn’t carry current.
Hence elimination saves the cost.• Current can be carried in other phases, with fewer
interruptions.
Limitations of ungrounded system
• With the increase in voltage and line length current has increased and self clearing nature advantage couldn’t be seen
• Arcing ground : Phenomena of alternate clearing and re-striking of the arc, which cause high voltage (surge and transient)
• If grounded, the insulation can be graded in transformer from line to neutral, there by reducing the cost
• In case of ungrounded system, influence on communication lines is more.
• Ground current can’t be limited
a
b
c
3.01.73 1.73
0.0
Perfectly transposed line : Neutral of transformer is at zero potential.Un-transposed line : Neutral is shifted.Capacitance grounded : Perfectly transposed line.
Ungrounded system
Resistance grounded system :
• The resistance to neutral limits ground current. Selection of resistance value:• Amount of ground fault currentPower loss in resistor during ground fault• Power loss: usually expressed as a percentage of
system rating
R
Xg = 16 %Xg = 8%
Z1 = 24% , Z2=24% , Z0=3R+j8
IZ Z Z j j R j R j
Power Loss I RR j
R
f
f
´
æ
èç
ö
ø÷
3 3 100
24 24 3 8
300
3 56
300
3 56
1 2 0
2
2
.
If : in pu, R : in pu For three phase system, power loss
I R
percentage of phasesystem kVA ratingf
2
33
R is in percent p.u.Maximum power loss = ? Selection : How much be the value of ground fault current ?What should be the percent power loss in the ground resistance ?
X system
X systemR
X systemo0
1 130 1 0 . .
X1 =X2 =25%
X1 =7% =X2=X0
X1 =34%X2 =120%
69 kV25 MVA
69 kV
B
A
Effectively grounded system
• Solidly grounded : No impedance between neutral and earth
• Effectively grounded : As per AIEE standard No.32, Section 32 - 1.05, May 1947 :
For fault at AXX
For fault at BXX
ac ce Grounded System
XX
but no resonance resonance grounding
0
1
0
1
0
1
732
12732 34
12766
3 0
³
Re tan :
. ;
If Xg = 50 at A , find the nature of grounding at A & B.
0
0
I
I
I
0
I
I 2
Resonant - Grounded System :
• Capacitance current is tuned or neutralised by a neutral reactor or similar device
• Fault current is made zero by selecting the suitable tap on the ground reactor
Selection of Breaker
• Determine the symmetrical current (maximum) for any type of fault
• Multiply the current by factor obtained from standard tables depending on the duty cycle (2 cycles clearing, 5 cycles clearing etc.)
• Use the above current to specify the interrupting current
References
1. Stagg and A.H. El-Abiad, "Computer Methods in Power System Analysis", McGraw-Hill, 13th print, New Delhi,1988.
2. William D. Stevenson "Elements Of Power System Analysis", McGraw-Hill, Fourth edition, New Delhi, 1982.
3. George L. Kusic, "Computer Aided Power System Analysis", Prentice-Hall, International, N.J.,New Delhi, 1989.
4. J. Arrillaga, C.P. Arnold and B.J.Harker ,"Computer Modeling of Electrical Power Systems", John Wiley and Sons,1983.