# ChapterVIII solutions

```Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8. Short Circuit Analysis
8-1
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.1 Comparison of a short circuit in RL circuit and a generator
a. DC component of the short-circuit current
Solution of the differential equation:
2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ i + L
di
→
dt
Homogenous solution:
di
=0
dt
di
= λ ⋅ e λt
i = e λt ;
dt
R ⋅i + L
i = C ⋅e
−
t
τ
→ R ⋅ e λt + L λ ⋅ e λt = 0 → λ = −
( C :int egration
R
1
=−
L
τ
cons tan t )
Non-homogenous solution:
i = A. cos( ωt + α ) + B sin( ωt + α ) →
di
= − A.ω ⋅ sin( ωt + α ) + B ⋅ ω ⋅ cos( ωt + α ) →
dt
di
→
dt
2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ ( A. cos( ωt + α ) + B sin( ωt + α )) + L ⋅ (− A.ω ⋅ sin( ωt + α ) + B ⋅ ω ⋅ cos( ωt + α )) →
2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ i + L
2 ⋅U = R ⋅ B − ω ⋅ L ⋅ A → B =
0 = R ⋅ A + X ⋅ B → B = −A ⋅
2 ⋅ U + A ⋅ ωL
R
R
=
X
R
2 ⋅ U + A ⋅ ωL
= −A ⋅ →
R
X
2 ⋅U ⋅ X
A=−
with Z = R 2 + X 2
2
Z
2 ⋅U ⋅ R
B=
→
Z2
(
i=−
2 ⋅U ⋅ X
)
2 ⋅U ⋅ R
⋅ sin( ωt + α ) =
Z
Z2
= 2 ⋅ I ⋅ (− sin θ . cos( ωt + α ) + cos θ ⋅ sin( ωt + α ))
2
. cos( ωt + α ) +
( ωL = X )
X
R

and cos θ = 
 with sin θ =
Z
Z

8-2
Exercises
Electric Power Systems I - III
Trigonometric identitiy : cosu ⋅ sin v =
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1
⋅ (sin( u + v ) − sin( u − v )) →
2
 1

i = 2 ⋅ I ⋅  − ((sin( ωt + α + θ ) − sin( ωt + α − θ ) + (sin( ωt + α + θ ) + sin( ωt + α − θ ))
 2

= 2 ⋅ I ⋅ sin( ωt + α − θ )
Overall solution:
i( t ) = C ⋅ e
−
t
τ
+ 2 ⋅ I ⋅ sin( ωt + α − θ )
Determination of the integration constant C and the final solution:
i( 0 ) = 0 = C + 2 ⋅ I ⋅ sin( α − θ ) → C = − 2 ⋅ I ⋅ sin( α − θ ) ⇒
t

−
τ

i( t ) = 2 ⋅ I ⋅ sin( ωt + α − θ ) − sin( α − θ ) ⋅ e






8-3
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8-4
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
t

−
i( t ) = 2 ⋅ I ⋅  sin( ωt + α − θ ) − sin( α − θ ).e τ


with

,


U = 215V
L = 0.27 H
R = 15 Ω
I=
U
R 2 + ( ωL )2
= 2 .5 A
 ωL 
0
θ = tan −1 
 = 80
R


L
τ = = 18 ms
R
The DC component:
u( t ) = 2 ⋅ 215 V ⋅ sin α = 100 V → α = 19.2 0
i d.c. = − 2 ⋅ I ⋅ sin( α − θ) ⋅ e
−
t
τ
= 3.08 A
b. Instantaneous value of the voltage that produces maximum DC
8-5
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
sin(α − θ) = −1 → α − θ = −90 0
α = −10 0 → i d.c. = 3.54 A
c. Instantaneous value of the voltage that results in the absence of any
DC component
sin(α − θ) = 0 → α − θ = 0 0
α = 80 0 → i d.c. = 0
d. Current values for switching at instantaneous voltage value of zero
α=0
i( t ) = 2 ⋅ I ⋅ (sin( ωt − θ) + sin(θ).e
−
t
τ
0
) = 3.5 A ⋅ (sin( ωt − 80 ) + 0.985.e
−
t
0.018
)
t = 0.5 ⋅ 0.02 = 0.01s → i = 5.42 A
t = 1.5 ⋅ 0.02 = 0.03 s → i = 4.1A
t = 5.5 ⋅ 0.02 = 0.11s → i = 3.5 A
e. Difference between short circuit currents in an R - L circuit and a
synchronous generator
While the short-circuit current in an R – L circuit maintains a constant amplitude for the entire
duration of the short-circuit (assuming that the DC component is neglected), the short-circuit
current in a synchronous generator does not. The flux across the air gap of a synchronous
machine is determined by the combined action of the field, the armature, and the damper
windings (or iron parts of the cylindrical rotor). Depending on the level of the transient
processes in these windings and their interaction with one another, the post-fault period is
categorized into sub-transient, transient and steady-state phases, each with its own
characteristic reactance (Xd”, Xd’ and Xd) and the corresponding components of the shortcircuit current have decreasing magnitudes, i.e. Ik” > Ik’ > Ik.
8-6
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.2 Symmetrical short-circuit in a generator – transformer unit
Single-line diagram (with the fault on the low voltage side of the transformer):
∼
a. the sustained short-circuit current in the breaker
"
E =1
"
Ik =
E
1
S
100
=
= −0.769 p.u. ⇒ Ik = 0.769 ⋅
= 0.769 ⋅
= 2.467 kA
jx d j1.3
3 ⋅U
3 ⋅ 18
b. sub-transient, transient and the initial symmetrical RMS currents
Sub-transient short-circuit current:
"
Ik =
E
"
jx "d
=
1
S
100
"
= − j5.263 p.u. ⇒ Ik = 5.263 ⋅
= 5.263 ⋅
= 16.88 kA
j0.19
3 ⋅U
3 ⋅ 18
Transient short-circuit current
'
Ik =
E
"
jx 'd
=
1
100
'
= − j3.846 p.u. → Ik = 3.846 ⋅
= 12.34 kA
j0.26
3 ⋅ 18
The initial symmetrical RMS current is the same as the sub-transient short-circuit
current.
c. the maximum possible dc component in the breaker
i d.c. = 2 ⋅ Ik" = 23.874 kA
Single-line diagram (with the fault on the high voltage side):
∼
d. the initial symmetrical RMS current on the high voltage side
8-7
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
x ps = 0.1 ⋅
10.5 2
E
"
Ik =
j(
"
+ x ps )
x"d
S
"
I k = 4.464 ⋅
= 4.464 ⋅
= 0.034
18 2
3 ⋅U
100
=
1
= − j 4.464 p .u . ⇒
j ( 0.19 + 0.034 )
=
= 0.683kA
18
3 ⋅ 220 ⋅
10.5
e. the initial symmetrical RMS current on the low voltage side.
"
I k = − j 0.683 kA ⋅
220
= − j14.32kA
10.5
f. the breaker interrupting current.
For the time delay specified (t = 0.1 s):
"
μ = 0.62 + 0.72 ⋅ e −0.32⋅IkG / IrG
(DIN EN 60909-0)
where:
I "kG = 14.32kA
IrG =
100 MVA
3 ⋅ 18 kV
(initial symmetrical RMS current)
= 3.2075 kA →
(rated generator current)
μ = 0.62 + 0.72 ⋅ e −0.32⋅14.32 / 3.2075 = 0.62
Breaker interrupting current:
I b = μ ⋅ I "k = 0.62 ⋅ 14.32kA = 8.9 kA
8-8
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.3
Short-circuit current computation: internal emf method vs.
Thévenin theorem I
a. Short-circuit current computation using the internal emf method
I=−
400 MW
3 ⋅ 20 kV
X "d = x "d ⋅
"
E =
Un2
20 2
= 0. 2 ⋅
Ω = 0.16 Ω
500
Sn
20 kV
3
= −11.547 kA
∠0 0 − I ⋅ jX "d = 11.547 + j1.8475 kV = 11.69 kV∠9.09 0
"
E
11.69 kV∠9.09 0
= 73.06 kA∠ − 80.910
Ik = " =
j0.16 Ω
jX d
"
b. Short-circuit current computation using the Thévenin theorem.
Xd“
ZL
Pre-fault voltage at the fault location:
U th =
20
3
kV = 11.547 kV
Equivalent impedance at the fault location:
ZL =
20 kV 2
= 1Ω
400 MW
Z th = jX "d // Z L = j0.16 // 1 = 0.158 Ω∠80.9 0
8-9
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Fault current:
"
Ik =
U th
11.547 kV
=
= 73.08 kA∠ − 80.9 0
0
Z th 0.158 Ω∠80.9
Current flowing through the generator:
(part of the current flowing through the generator + pre-fault current )
"
"
IkG = Ik ⋅
1
"
+ j11.547 kA = − j72.162 + 11.547 kA = 73.08 kA∠ − 80.8 0 = Ik
1 + j0.16
8-10
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.4
Short-circuit current computation: internal emf method vs.
Thévenin theorem II
a. The sub-transient short-circuit currents using the internal EMF method
Internal voltage of the generator:
IM =
20 MW
0.8 ⋅ 12.8 kV ⋅ 3
2
(
13.2kV )
= 0. 1⋅
X line
30 MVA
X "d = X M" = 0.2 ⋅
E "G =
I
"
kG
13.2
3
30 MVA
j( X + X line )
13.2
3
= 1.1616 Ω
kV∠0 0 + IM ⋅ ( jX line + jX "d ) = 6.41kV∠14.2 0
"
d
E M" =
= 0.58 Ω
(13.2kV )2
E "g
=
∠36.9 0 = 1127.64 A ∠36.9 0
= 3.679 kA∠ − 75.8 0
kV∠0 0 − IM ⋅ jX M" = 8.24 kV∠ − 7.3 0
"
M
"
M
E
= 7.1kA∠ − 97.3 0
jX
"
=
IkM
Ik" = I"fG + I"fM = 3.679 kA∠ − 75.8 0 + 7.1kA∠ − 97.3 0 = − j10.6 kA
b. Solution using the Thévenin theorem.
Pr efualt voltage :
U th =
12.8
kV∠0 0
3
Thevenin Equivalent impedance :
X th = ( X "d + X line ) // X M" =
Ik" =
1.74 ⋅ 1.1616
Ω = 0 .7 Ω
2 .9
U th 12.8 / 3
=
kA = −10.6 kA
jX th
j0.7
Contribution of the motor and the generator to the short-circuit current:
8-11
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Pr e − fault currents :
IM =
S
3 ⋅U
IG = −IM
=
20 MW
∠36.9 0 = 1127.64 A∠36.9 0
0.8 ⋅ 12.8 kV
Thevénin current (as a result of fault):
IGth
X M"
1.16
= Ik ⋅ "
= −10.6 kA ⋅
= − j4.24 kA
"
2 .9
X M + X d + X line
IMth = Ik ⋅
X "d + X line
1.7424
= −10.6 kA ⋅
= −6.37 kA
"
"
2 .9
X M + X d + X line
"
IkG
= 1127.64∠36.9 0 A
"
IkM
= IMth − IM = − j6.37 kA − 1127.64 A∠36.9 0 = −0.9 − j7.047 kA = 7.1kA∠ − 97.3 0
"
IkG
= IGth − IG = − j4.24 kA + 1127.64 A∠36.9 0 = 0.9 − j3.563 kA = 3.675 kA∠ − 75.82 0
How does the pre-fault load current affect the current flowing into the fault location?
The current flowing into the fault location is the same whether the pre-fault load current is
considered or neglected. But in the equipments (generator, transformer, lines, etc.) the
current during fault is the resulting current when the pre-fault current is superimposed on the
fault current. Additionally, the pre-fault current indirectly affects the fault current in the sense
that the Thevénin voltage at the fault location is determined by the load flow situation in the
network.
8-12
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.5 Short-circuit in a generator connected to motors I
a. The impedance diagram
xd“
XM“
x "d = 0.2 p.u.
x M" = 0.2 ⋅
625
= 0.67 p.u.
187.5
b. Symmetrical short-circuit current (interrupted by breakers A and B) for
a fault at P.
Brea ker A :
"
IkA =
1
1
=
= − j5 p.u.
"
jx d j0.2
Brea ker B :
"
IkB =
1
1
1
1
1
1
+ " + " =
+
+
= − j8 p.u
"
jx d jx M jx M j0.2 j0.67 j0.67
c. For a fault at point Q.
Brea ker A :
"
IkA =
1
1
=
= − j5 p.u.
"
jx d j0.2
Brea ker B :
"
IkB =
1
1
=
= − j1.5 p.u
"
jx M j0.67
8-13
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
d. For a fault at point R.
Brea ker A :
1
1
1
1
1
1
"
IkA = " + " + " =
+
+
= − j4.5 p.u
jx M jx M jx M j0.67 j0.67 j0.67
Brea ker B :
"
IkB =
1
= − j1.5 p.u
jx M"
R
A
Q
P
B
8-14
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.6 Short-circuit in a generator connected to motors II
a. the sub-transient current into the fault location
All reactances on a base of 5 MVA:
Generator :
x "d = 0.2 p.u.
Transforme r :
x ps = 0.1p.u.
Each Motor
x "d = 0.2 ⋅
25 MVA
= 1p.u.
5 MVA
1
X th = 1//( // (0.1 + 0.15 ) = 0.125
3
1
= − j8.0 p.u.
I"k =
j0.125
b. the sub-transient current in breaker A
"
=
IkA
1
1
j //(0.1 + 0.15)
3
= − j7.0 p.u.
8-15
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.7
Symmetrical short-circuit current computation
110kV
G1
T3
220kV
T6 110kV
220kV
T1
T4
T5
T2
G2
T7
Q
2X240/40-Al/St
T9
G9
T8
50km,
double line
SS2
SS3
240/40-Al/St
50km,
double line
SS1
SS4
Data of the network components:
Xb’=0.298 Ω/km (220-kV overhead line; for the bundle in each system)
Xb’=0.393 Ω/km (110-kV overhead line; for each system)
Overhead line 220 kV: R/X = 0.26
Overhead line 110 kV: R/X = 0.3
G1, G2:
10.5 kV; 100 MVA; xd“ = 0.16; RsG/Xd“ = 0.05;
G9:
21 kV; 225 MVA; xd“ = 0.19; RsG/Xd“ = 0.05;
T1, T2:
120 MVA; uk =10%; ür = 115 kV/10.5 kV
T3 …T8:
200 MVA; uk =12%; ür = 240 kV/110 kV
T9:
250 MVA; uk =10%; ür = 112 kV/21 kV; R/X = 0.03
Q:
SkQ“ = 20 GVA
Calculation of reactances (all referred to the 110-kV side):
10.5 2
100
2
 115 
⋅
 Ω = 21.16 Ω
 10.5 
G1, G2:
X "d = 0.16 ⋅
G9:
212  112 
⋅
X = 0.19 ⋅
 Ω = 10.6 Ω
225  21 
2
"
d
8-16
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
115 2
= 0. 1⋅
Ω = 11.02 Ω
120
T1, T2:
X ps
T3…T8:
X ps = 0.12 ⋅
T9:
X ps = 0.1⋅
Q:
1.1⋅ 220 2  110 
⋅
XN =
 Ω = 0.559 Ω
20000  240 
Xb220:
1  110 
X b 220 = 0.298 ⋅ 50 ⋅ ⋅ 
 Ω = 1.565 Ω
2  240 
Xb110:
1
X b110 = 0.393 ⋅ 50 ⋅ Ω = 9.825 Ω
2
110 2
Ω = 7.26 Ω
200
112 2
Ω = 5.02 Ω
250
2
2
Calculation of impedances:
G1, G2:
Z G1 = 21.16 ⋅ (0.05 + j)Ω = 1.058 + j21.16 Ω = 21.186 Ω∠87.10
G9:
Z G 9 = 10.6 ⋅ (0.05 + j)Ω = 0.53 + j1.06 Ω = 10.61Ω∠87.10
T1, T2:
Z psT1 = j11.02 Ω
T3…T8:
Z psT 3 = j7.26 Ω
T9:
Z psT 9 = 5.02 ⋅ (0.03 + j)Ω = 0.1506 + j5.02 Ω = 5.022 Ω∠88.28 0
Q:
Z N = j0.559 Ω
"
"
220-kVLine: Z L 220 = 1.565 ⋅ (0.26 + j)Ω = 0.4069 + j1.565 Ω = 1.617 Ω∠75.4 0
110-kVLine: Z L110 = 9.825 ⋅ (0.3 + j)Ω = 2.9475 + j9.825 = 10.26 Ω∠73.3 0
Calculation of equivalent impedance at the fault location:
The impedance diagram of the network is given below.
8-17
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FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
SS3
SS2
SS1
Z3
Z2
SS4
Z1
After Delta → Star conversion:
Z 1 = 0.4688 + j2.4349 Ω
Z 2 = 0.1311 + j1.4921Ω
Z 3 = −0.0596 + j0.5819 Ω
Z eq1
"

Z G1 + Z psT1 

= Z1 + Z 2 +
// (Z 3 + Z Q ) = 0.4147 + j3.5065 Ω


2


"
Z eq2 = Z psT 9 + Z G 9 = 0.6806 + j15.62 Ω
Z eq = Z eq 1 // Z eq 2 = 0.2993 + j2.866 Ω = 2.882 Ω∠84.04 0
a. The initial short-circuit current ( Ik" ) for a fault at SS4
"
Ik =
1.1⋅ 110 kV
3 ⋅ Z eq
= 24.24 kA∠ − 84.04 0
b. The first amplitude of the fault current (peak short-circuit current)
κ = 1.02 + 0.98 ⋅ e −3R / X = 1.7364
ip = κ ⋅ 2 ⋅ Ik" = 59.525 kA
c. The fault interrupting current for the maximum interrupting time delay of
0.25 s
Distribution of the fault current:
8-18
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
"
Q:
IQ = 2.1151 − j18.4589 kA = 18.58 kA∠ − 83.46 0
Generator 9 :
IG9 = 0.1945 − j4.464 kA = 4.468 kA∠ − 87.5 0
"
Generators 1& 2 : I"G = 0.2087 − j1.189 kA = 1.2072 kA∠ − 80 0
Ik" = 2.5183 − j24.112 kA = 24.24 kA∠ − 84.04 0
Determination of µ:
µ=1
Q:
(Distance fault)
IbQ = 18.58kA
2
IrG
G1, G2:
2 ⋅ 100 MVA  10.5 
=
⋅
 = 91.68 A
3 ⋅ 10.5 kV  115 
I"G 1207.2
=
= 13.17
IrG
91.68
μ = 0.56 + 0.94 ⋅ e −0.38⋅13.17 = 0.566
IbG = 0.566 ⋅ 1207.2 A = 683.65 A
2
IrG9
225 MVA  21 
=
⋅
 = 217.47 A
3 ⋅ 21kV  112 
I"G
4468
=
= 20.55
IrG 217.47
G9:
μ = 0.56 + 0.94 ⋅ e −0.38⋅20.55 = 0.56
IbG9 = 0.56 ⋅ 4.468 kA = 2.5 kA
Total interrupting current at the fault location:
Ib = 18.58 kA+0.68365 kA + 2.5 kA = 21.76 kA
d. The maximum dc component of the short-circuit current
i d.c. = 2 ⋅ Ik" = 34.28 kA
e. The influence of the resistive components of the impedances on the fault
current
Neglecting all the resistive components results in the following values:
After Delta → Star conversion:
Z1 = j2.41Ω
Z 2 = j1.469 Ω
Z 3 = j0.594 Ω
8-19
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Z eq1
"


Z
G + Z psT1 
= Z1 +  Z 2 + 1
// (Z 3 + Z Q ) = j3.492 Ω


2


"
Z eq2 = Z psT 9 + Z G 9 = j15.62 Ω
Z eq = Z eq1 // Z eq 2 = j2.854 Ω
"
Ik =
1.1⋅ 110 kV
3 ⋅ Z eq
= 24.48 kA∠ − 90 0
i d.c. = 2 ⋅ Ik" = 34.62 kA
κ = 1.02 + 0.98 ⋅ e −3R / X = 2
ip = κ ⋅ 2 ⋅ Ik" = 69.23 kA
By neglecting the resistive components in the impedances of the system elements, the
computational effort can be immensely reduced without any significant loss of accuracy.
8-20
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.8
Mechanical and thermal stress caused by the short-circuit
current
21 kV
G
110 kV
T
G:
UrG = 21 kV; SrG = 225 MVA; xd“ = 0.18; xd = 2; R/X= 0.05
T:
SrT = 250 MVA; uk = 10%
X "d =
0.18 ⋅ 21kV 2
= 0.3528 Ω
225 MVA
Ik" =
R
= 0.05 → κ = 1.02 + 0.98 ⋅ e −3⋅0.05 = 1.86
X
→
1.1⋅ 21kV
3 ⋅ 0.3528 Ω
= 37.8 kA
ip = κ ⋅ 2 ⋅ Ik" = 99.62 kA
Arrangement of the Al – bars:
Phase A
1
2
Phase C
Phase B
3
1
2
1
3
2
3
b
am
am
am = 350mm
b = 200 mm
as = 15mm
d = 15 mm
d
as
a11
a12
a13
8-21
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
a. The type of fault that causes maximum stress
The symmetrical fault calls for the maximum short-circuit current and therefore causes
the maximum mechanical (as well as thermal) stress on the aluminium bars. The bars of
the middle phase experience the largest force.
b. The total force on the middle phase bars by the fault current flowing in
the other two phases
General:
Force on conductor 1 due to a current flowing in conductor 2 (placed at a distance a):
(
)
F = l 1 X B 2 ⋅ I1 =
μ0 ⋅ I1 ⋅ I2 ⋅ l
2π ⋅ a
Three-phase system:
In a three-phase system, the force on the middle conductor (phase b) assuming phases a
and c are placed at a distance a on either side of b:
F=
μ0 ⋅ Ib ⋅ (Ia − Ic ) ⋅ l
2π ⋅ a
The maximum force (Fmax) for the following current relation:
i a = im ⋅ cos θ
F( t ) =
ib = im ⋅ cos(θ − 2π / 3)
ic = im ⋅ cos(θ + 2π / 3)
( θ = ωt ) →
μ0 ⋅ im2 ⋅ (cos(θ − 2π / 3) ⋅ (cos θ − cos(θ + 2π / 3)))
⋅l
2π ⋅ a
dF( t ) d  μ0 ⋅ im2 ⋅ (cos(θ − 2π / 3) ⋅ (cos θ − cos(θ + 2π / 3))) 
= 
⋅ l  = 0
dt
dt 
2π ⋅ a

− sin(θ − 2π / 3) ⋅ ((cos θ − cos(θ + 2π / 3))) + cos(θ − 2π / 3) ⋅ (− sin θ + sin(θ + 2π / 3)) = 0 ⇒
− sin( 2θ − 2π / 3) + sin 2θ = 0
[Follows from the trigonomet ric identity ]
sin( α + β) = sin α ⋅ cos β + cos α ⋅ sin β)
The solution of the trigonometric equation:
− sin( 2θ − 2π / 3) + sin 2θ = 0 →
sin 2θ = sin( 2θ − 2π / 3) →
2θ − 2π / 3 = π − 2θ
θ=
[Follows from
sin δ = sin( π − δ) ]
5
⋅ π = 75 0 ⇒
12
The peak force, therefore, is:
Fmax =
(
(
))
μ0 ⋅ im2 ⋅ cos(75 0 − 120 0 ) ⋅ cos 75 0 − cos(75 0 + 120 )
3 μ0 ⋅ im2
⋅l =
⋅
⋅l
2π ⋅ a
2 2π ⋅ a
8-22
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
This relationship, however, is only valid under the assumption that the current is
concentrated at the centre of the conductor and the mutual distance between the
conductors is much larger than the conductor dimensions.
To account for the deviation from this idealised assumption, the mutual distance should
be replaced by an effective distance using a correction factor in the form:
a' =
a
, where a = the actual distance and k = the correction factor.
k
The above peak force (including the correction factors) thus becomes:
Fmax
3 μ0 ⋅ im2 ⋅ l  k 11 k 12 k 13 k 21 k 22 k 23 k 31 k 32 k 33
=
⋅
⋅ 
+
+
+
+
+
+
+
+
2
2π
 a11 a12 a13 a 21 a 22 a 23 a 31 a 32 a 33



where kij are the correction factors. The mutual distances aij are shown in the figure
above.
The correction factors for rectangular bars can be obtained from Figure 1 (given in the
problem) as follows:
height
b 200
= =
= 13.33 → identify (or estimate) the applicable curve in Figure 1.
Then,
a11 a m 350
=
=
= 23.33 → from curve : k 11 = 0.95 = k 22 = k 33
d
d
15
a12 a m + 2d 380
=
=
= 25.33 → from curve : k 12 = 0.96 = k 23
d
d
15
a13 a m + 4d 410
=
=
= 27.33 → from curve : k 13 = 0.97
d
d
15
a 21 a m − 2d 320
=
=
= 21.33 → from curve : k 21 = 0.94 = k 32
d
d
15
a 31 a m − 2d 290
=
=
= 19.33 → from curve : k 31 = 0.93
d
d
15
'
max
F
Fmax
3 μ0 ⋅ im2  0.95 0.96 0.97 0.94 0.95 0.96 0.93 0.94 0.95 
=
=
⋅
⋅
+
+
+
+
+
+
+
+

2
2π  350 380 410 320 350 380 290 320 350 
l
=
3 μ0 ⋅ im2
⋅
⋅ 24.64 m −1
2
2π
According to DIN VDE 0103, the peak current is to be used for the calculation of the
maximum stress. Thus,
im =
ip
3
=
99.62
kA = 33.2 kA
3
(Division by 3 is necessary since the total current divides
between the three bars in each of the phases)
8-23
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
3 4π ⋅ 10 −7 H ⋅ m −1 ⋅ (33.2 kA )
F =
⋅
⋅ 24.64 m −1 = 4.7 kN / m
2
2π
V ⋅ s ⋅ A ⋅ A V ⋅ s ⋅ A W ⋅ s N⋅m
Unit check :
=
=
=
= N/m
A ⋅m⋅m
m⋅m
m⋅m m⋅m
2
'
m
c. The force exerted on the outer bar of the middle phase due to the
currents flowing in the middle phase itself
In part (a) it was stated that the middle phase experiences the biggest mechanical stress.
Within the conductors of the middle phase, however, the middle bar experiences the least
force (due to the currents flowing in the other two bars of the same phase) since the two
outer bars apply antipodal forces on it with the resultant being zero).
The force on one of the outer bars:
Fs' =
μ0 ⋅ im2  k 11 k 12 
⋅
+

2π  2d 4d 
Please note that the factor
3
is not necessary here since the two currents are in phase
2
and the force has its peak value when the current has its maximum value.
2d
= 2 → k 11 = 0.35
d
4d
= 4 → k 12 = 0.55
d
Fs' =
μ0 ⋅ im2  k 11 k 12 
 0.35
0.55 
2
 = 4.59 kN / m
⋅
+
+
 = 2 ⋅ (33.2 kA ) ⋅ 10 −7 ⋅ 
2π  2d 4d 
 30 mm 60 mm 
d. The combined force exerted on the mounting and its direction
Although the two forces (one arising from the currents in the other two phases and the
other from currents of the same phase) obviously have their maxima at two different
instants. In spite of this they should be added directly (as an additional safety margin) in
the dimensioning of the setup (DIN VDE 0103).
Thus, the combined force:
F = (Fs' + Fm' ) ⋅ l = ( 4.59 + 4.71) kN = 9.3 kN
Each of the mountings absorbs half of this force.
The force is horizontally directed.
e. Capability of the Al bars to withstand the thermal stress during fault
8-24
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Determination of m:
Fault current: Ik" = 37.8 kA
(Calculated at the beginning).
RG
= 0.05 → κ = 1.86
X "d
Tk = 0.2 s → Tk ⋅ f = 10
Using these values, we obtain from the curve → m = 0.35
Determination of n:
Rated generator current: IrG =
225 MVA
3 ⋅ 21kV
= 6185 .9 A
Ik = 1.76 ⋅ IrG = 1.76 ⋅ 6185.9 A = 10887 .18 A
Ik"
37.8 kA
=
= 3.47
Ik 10887 .18 A
With Tk = 0.2 s from curve → n = 0.69
The thermal equivalent of the short circuit current:
I th = Ik" ⋅ m + n = 38.55 kA
Short-time current density:
S th =
Ith
38.55 kA
=
= 4.28 A / mm 2
A 9000 mm 2
( A = 3 ⋅ 200 mm ⋅ 15 mm = 9000 mm 2 )
The Al – bar is designed for the allowable short-time current density of:
S th _ max = 87 A / mm 2 (given in the problem), which leads to a temperature rise of 650C
(operating temperature) to 2000C (maximum allowable temperature) in 1 s.
The fault duration in this case is Tk = 0.2 s. This means, even a larger current density is
tolerable, i.e.
(
S 2th _ max ⋅ Tk = S 'th _ max
)
2
⋅ 1s → S 'th _ max = S th _ max ⋅
1s
= 194.5 A / mm 2
Tk
While the Al – bars are designed to withstand a short-time short-circuit current density of
up to 194.5 A/mm2 for the duration of the short circuit considered here, the current density
occurring here is only a tiny proportion of that, namely 4.28 A/mm2
→ Capable of withstanding the thermal stress caused by the short circuit.
8-25
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.9
Calculation of short circuit currents in a meshed network
Reactances (referred to the 110-kV level):
1.1⋅ Un2 1.1⋅ (110kV ) 2
XN =
=
= 3.33Ω
4GVA
S k''
X G1 = X G2 = X T1 = X T 2 = X T 3 = X T 4
X T5 = X T6
XM =
0.12 ⋅ Un2 0.12 ⋅ (110kV ) 2
=
=
= 46.1Ω
Sn
31.5MVA
0.11⋅ Un2 0.11⋅ (110kV ) 2
=
=
= 83.19Ω
Sn
16MVA
x M' ⋅ Un2
Pn / cos
=
n
0.3 ⋅ (110kV ) 2
= 242Ω
12MW / 0.8
X L1 = X b' ⋅ l = 0.4Ω / km ⋅ 15km = 6Ω
X L 2 = X b' ⋅ l ⋅
110 2
= 0.4Ω / km ⋅ 2.5km ⋅ 121 = 121Ω
10 2
Currents at fault locations:
Fault location F1:
U" =
''
IkN
=
1.1 ⋅ 110kV
3
= 69.859 kV
U"
= 20.98kA
XN
''
''
IkG
1 + IkG2 =
U"
= 0.93kA
( X G1 + X T1 + X T 3 ) / 2 + X L1
''
IkM
≈0
''
''
''
''
Ik'' = IkN
+ IkG
1 + IkG2 + IkM = 21.91kA
Fault location F2:
''
IkN
=
U"
= 7.49kA
X N + X L1
''
''
IkG
1 + IkG2 =
U"
( X G1 + X T1 + X T3 ) / 2
= 1.01kA
''
IkM
≈0
''
''
''
''
Ik'' = IkN
+ IkG
1 + IkG 2 + IkM = 8.5kA
8-26
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Fault location F3:
U"
= 2.16kA
X N + X L1 + X T 3 // X T 4
''
=
IkN
''
''
IkG
1 = IkG 2 =
U"
= 0.76kA
X G1 + X T1
U"
= 0.215kA
XM + X T5
''
=
IkM
''
''
''
Ik'' = IkN
+ 2 ⋅ IkG
1 + IkM = 3.895kA
Referred to the 30-kV level:
Ik'' 30 = Ik'' ⋅
110
= 14.28kA
30
Fault location F4:
''
''
+ IkG
=
IkN
U"
= 683.5 A
( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5
''
''
''
= (IkN
+ IkG
IkN
)⋅
( X G1 + X T1 ) / 2
= 401.5 A
X N + X L1 + X T 3 / 2 + ( X G1 + X T1 ) / 2
''
''
''
''
''
IkG
1 = IkG 2 = 0.5 ⋅ (IkN + IkG ) − IkN = 141A
1.1⋅ 110kV
''
=
IkM
3
I =I
''
k
''
kN
⋅
1
= 289 A
XM
''
''
+ 2 ⋅ IkG
1 + IkM = 972 A
Referred to the 10-kV level:
Ik'' 10 = Ik'' ⋅
110
= 10.7kA
10
Fault location F5:
Ik'' =
U"
= 0.3622kA
[( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] // X M + X L 2
''
''
(IkN
) = Ik'' ⋅
+ IkG
XM
= 254.6 A
[( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] + X M
''
''
(IkN
) = Ik'' ⋅
+ IkG
XM
= 254.6 A
[( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] + X M
''
''
''
IkM
= Ik'' − (IkN
+ IkG
) = 107.6 A
8-27
Exercises
Electric Power Systems I - III
''
''
''
IkN
= (IkN
+ IkG
)⋅
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
( X G1 + X T1 ) / 2
= 149.6 A
X N + X L1 + X T 3 / 2 + ( X G1 + X T1 ) / 2
''
''
''
''
''
IkG
1 = IkG2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 52.5 A
Referred to the 10-kV side:
Ik'' 10 = Ik'' ⋅
110
= 3.984kA
10
8-28
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
8.10 Short Circuit Current Limitation
a. Possibilities for short circuit current limitation
◊
Measure 1: opening the bus bar coupling A – B (for fault location F3)
◊
Measure 2: installation of an inductance in series with L2 (for fault location F5)
b. Effectiveness of the alternative options
Measure 1:
No effect for fault locations F1 and F2.
Fault location F3:
''
''
+ IkG
IkN
2 =
U"
= 1.274kA
( X N + X L1 ) //( X G2 + X T 2 + X T 4 ) + X T 3
''
''
''
IkG
2 = (IkN + IkG 2 ) ⋅
X N + X L1
= 0.0805kA
X N + X L1 + X G2 + X T 2 + X T 4
''
''
''
''
= (IkN
+ IkG
IkN
2 ) − IkG 2 = 1.1935kA
I
''
kG1
U"
=
= 0.76kA
X G1 + X T1
I
''
kM
U"
=
= 0.215kA
XM + X T5
''
''
''
''
Ik'' = IkN
+ IkG
2 + IkG1 + IkM = 2.249kA
F3 (30 kV): Ik'' 30 = Ik'' ⋅
110
= 8.246kA
30
→ Only 57.7 % of the short-circuit current without the countermeasure flows.
Fault location F4:
8-29
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
''
''
+ IkG
=
IkN
[( X N + X L1 ) //( X G2 + X T 2
''
''
+ IkG
IkN + IkG 2 = (IkN
)⋅
I
= (I
''
kG1
''
kN
+ I ) − (I
''
kG
''
kN
''
''
''
IkG
2 = (IkN + IkG 2 ) ⋅
U"
= 594.2A
+ X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5
( X N + X L1 ) //( X G2
X G1 + X T1
= 372.6 A
+ X T 2 + X T 4 ) + X T 3 + X G1 + X T1
''
+ IkG
2 ) = 221 .6 A
X N + X L1
= 23.5 A
X N + X L1 + X G2 + X T 2 + X T 4
''
''
''
''
= (IkN
+ IkG
IkN
2 ) − IkG 2 = 349 .1A
I
''
kM
U"
=
= 288.7 A
XM
''
''
''
''
Ik'' = IkN
+ IkG
2 + IkG1 + IkM = 882 .9 A
F4 (10 kV): Ik'' 30 = Ik'' ⋅
110
= 9.71kA
10
→ The fault current now is reduced to 90% of the value without the reactance.
Fault location F5:
I =
''
k
({[( X N + X L1 ) //( X G2 + X T 2
''
''
IkN
+ IkG
= Ik'' ⋅
U"
= 349.1A
+ X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5 } // X M ) + X L 2
({[( X N + X L1 ) //( X G2 + X T 2
XM
= 234.9 A
+ X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5 } + X M )
''
''
''
IkM
= Ik'' − (IkN
+ IkG
) = 114.2A
''
''
''
''
(IkN
+ IkG
2 ) = (IkN + IkG ) ⋅
( X N + X L1 ) //( X G2
X G1 + X T1
= 147.3 A
+ X T 2 + X T 4 ) + X T 3 + X G1 + X T1
''
''
''
''
''
IkG
1 = (IkN + IkG ) − (IkN + IkG 2 ) = 87.6 A
''
''
''
IkG
2 = (IkN + IkG 2 ) ⋅
X N + X L1
= 9 .3 A
X N + X L1 + X G2 + X T 2 + X T 4
''
''
''
''
IkN
= (IkN
+ IkG
2 ) − IkG 2 = 138 A
F5 (10 kV): Ik'' 30 = Ik'' ⋅
110
= 3.84kA
10
Measure 2:
XD =
x D ⋅ Un2
3 ⋅ U n ⋅ In
= 36.963Ω (referred to 110-kV level)
Calculations similar to Measure 1, only transformer reactances XT3 and XT4 need to be
increased by 36.936 Ω each
Fault currents:
Fault location F1:
8-30
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
I
''
kN
U"
=
= 20.98kA
XN
''
''
IkG
1 + IkG 2 =
U"
= 0.746kA
( X G1 + X T1 + X T 3 + X D ) / 2 + X L1
''
IkM
≈0
''
''
''
''
Ik'' = IkN
+ IkG
1 + IkG 2 + IkM = 21 .726kA
Fault location F2:
U"
= 7.49kA
X N + X L1
''
=
IkN
''
''
IkG
1 = IkG 2 =
1.1⋅ 110kV
3
⋅
1
= 0.399kA
X G1 + X T1 + X T 3 + X D
''
IkM
≈0
''
''
''
Ik'' = IkN
+ 2 ⋅ IkG
1 + IkM = 8.29kA
Fault location F3:
''
=
IkN
X N + X L1 + ( X T 3
''
''
IkG
1 = IkG 2 =
U"
= 1.374kA
+ X D ) || ( X T 4 + X D )
1.1⋅ 110kV
3
⋅
1
= 0.76kA
X G1 + X T1
"
U
= 0.215kA
XM + X T5
''
=
IkM
''
''
''
Ik'' = IkN
+ 2 ⋅ IkG
1 + IkM = 3.109kA
F3 (30 kV): Ik'' 30 = Ik'' ⋅
110
= 11.4kA
30
Fault location F4:
I
''
kN
+I
''
kG
U"
=
= 650.6 A
( X N + X L1 + X T 3 / 2 + X D / 2) || ( X G1 + X T1 ) / 2 + X T 5
''
''
''
= (IkN
+ IkG
IkN
)⋅
( X G1 + X T1 ) / 2
= 309.3 A
X N + X L1 + ( X T 3 + X D ) / 2 + ( X G1 + X T1 ) / 2
''
''
''
''
''
IkG
1 = IkG 2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 171 .15 A
''
=
IkM
U"
= 288.7 A
XM
''
''
''
Ik'' = IkN
+ 2 ⋅ IkG
1 + IkM = 939 .3 A
F4 (10 kV): Ik'' 10 = Ik'' ⋅
110
= 10.3kA
10
8-31
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Fault location F5:
Ik'' =
[( X N + X L1 + X D / 2 + X T 3
''
''
(IkN
) = Ik'' ⋅
+ IkG
U"
= 357.6 A
/ 2) || ( X G1 + X T1 ) / 2 + X T5 ] || X M + X L 2
[( X N + X L1 + X D / 2 + X T 3
XM
= 247.7 A
/ 2) || ( X G1 + X T1 ) / 2 + X T 5 ] + X M
''
''
''
IkM
= Ik'' − (IkN
+ IkG
) = 109.9 A
''
''
''
IkN
= (IkN
+ IkG
)⋅
( X G1 + X T1 ) / 2
= 117.8 A
X N + X L1 + X D / 2 + X T 3 / 2 + ( X G1 + X T1 ) / 2
''
''
''
''
''
IkG
1 = IkG2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 65 A
F5 (10 kV) : Ik'' 10 = Ik'' ⋅
110
= 3.934kA
10
8-32
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.11
Analysis of unsymmetrical faults
a. Fault conditions
(a)
I A + IB + IC = 0
U A =U B =UC
(b)
IB = IC = 0
UA =0
(c)
IA =0
(d )
I A = 0 ; I B = −I C
U B =UC = 0
U B =UC
b. Fault conditions in symmetrical components and sequence networks
(a)
I A + I B + I C = 0 → I A0 = 0
U A = U B = U C → U A0 = U A1 = U A2 = 0
UG“
UA1
UA2
IA0 =
0
(b)
UA0
I B = I C = 0 → I A0 = I A1 = I A2
U A = 0→ U A0 + U A1 + U A2 = 0
8-33
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
IA1
UG“
UA1
IA2
UA2
IA0
UA0
(c)
I A = 0 → I A0 + I A1 + I A2 = 0
U B = U C = 0 → U A0 = U A1 = U A 2
IA1
UG“
UA1
IA2
UA2
IA0
UA0
(d )
I A = 0 → I A0 + I A1 + I A2 = 0 ;
I B = − I C → I A0 = 0
U B = U C → U A1 = U A 2
8-34
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
IA1
UG“
UA1
IA2
UA2
Positive sequence network:
01
U“
U“
Xd“
Xps
2
1
XN
XL
F
3
Negative sequence network:
01
Xd“
Xps
2
1
XN
XL
F
3
Zero sequence network:
00
XG0+3Xn
Xps
1
XG0→∞
XL
2
F
3
Reactances:
Generators G1 external network:
8-35
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
X "d = X 1 = X 2 = 0.19 ⋅
X G 0 = 0.067 ⋅
380 2
Ω = 144.07Ω
13.8 2
380 2
Ω = 50.8Ω
13.8 2
380 2
X n = 0.095 ⋅
Ω = 72.03Ω.
13.8 2
Transformers T1:
X ps = 115.5Ω
Transmission line:
X L1 = X L 2 = 21.66Ω
X L 0 = 72.2Ω
X N = 1.1 ⋅
U " = 1. 1
(380kV )2
10000MVA
380
3
= 15.88Ω
kV = 241.33kV
c. Single line-to-ground fault at bus 3
Z 1 = Z 2 = j ( X d" + X ps + X L1 ) // jX N = j (144.07 + 115.5 + 21.66) // 15.88Ω = j15.03Ω
Z 0 = j ( X L 0 + X ps ) = j187.7Ω
i.
"
The sub-transient fault current for a single line to ground fault
I k (1) = 3 ⋅
U"
241.33kV
= 3⋅
= − j 3.325kA
Z 0 + Z1 + Z 2
j (15.03 + 15.03 + 187.7)
ii. voltage at the fault location and at the terminals of G1
• voltage at the fault location
8-36
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
"
− j 3.325kA
= − j1.108kA
3
3
= U "− I a1−G1 ⋅ jX 1 = 241.33kV − (− j1.108kA ⋅ j15.03Ω ) = 224.68kV
I a 0 = I a1 = I a 2 =
U a1
I k (1)
=
U a 2 = − I a 2−G 2 ⋅ jX 2 = −(− j1.108kA ⋅ j15.03Ω ) = −16.65kV
U a 0 = − I a 0−G1 ⋅ jX 0 = − − j1.108kA ⋅ j187.7Ω = −207.97kV
U a = U a 0 + U a1 + U a 2 = −207.97kV + 224.68kV − 16.65kV ≈ 0
U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = −207.97kV + 224.68kV∠240 0 − 16.65kV∠120 0 =
2
− 311.99 − j 209 = 375.52∠213.82 0
U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = −207.97kV + 224.68kV∠120 0 − 16.65kV∠240 0 =
2
− 311.99 + j 209 = 375.52∠146.18 0
• voltage at the terminals of generator 1
Current out of generator 1 :
I a1−G1 = I a 2−G1 = − j1.108kA ⋅
15.88
= − j 0.0592kA
15.88 + 21.66 + 115.5 + 144.07
I a 0−G1 = − j1.108kA
• Voltages Ua, Ub, Uc:
U a1 = U "− I a1−G1 ⋅ jX d" = 241.33kV − (− j 0.0592kA ⋅ j144.07Ω ) = 232.8kV
U a 2 = − I a 2−G 2 ⋅ jX d" = −(− j 0.0592kA ⋅ j144.07Ω ) = −8.53kV
U a0 = 0
U a = U a 0 + U a1 + U a 2 = −8.53kV + 232.8kV = 224.27kV
2
U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 232.8kV∠240 0 − 8.53kV∠120 0 =
→ U b = −112.14 − j 209 = 237.18kV∠241.78 0
U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 232.8kV∠120 0 − 8.53kV∠240 0 =
2
→ U c = −112.14 + j 209 = 237.18kV∠92.910
iii. Phase c current out of generator 1
(
)
I c = a ⋅ I a1−G1 + a ⋅ I a 2−G1 = − j 0.0592kA ⋅ a ⋅ + a = j 0.0592kA
2
2
d. Line-to-line fault at bus 3
i.
fault current
I a1 = − I a 2 =
1
241.33kV
=
= − j8.03kA
Z1 + Z 2
j (15.03 + 15.03)
I a0 = 0
2
I b = − I c = (a − a ) ⋅ (− j8.03kA) = − j 3 (− j8.03kA) = −13.9kA
8-37
Exercises
Electric Power Systems I - III
ii.
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Line-to-line voltage at the fault location
U a1 = 241.33kV − I a1 ⋅ Z 1 = 241.33kV − (− j8.03kA) ⋅ j15.03Ω = 120.64kV
U a 2 = −( j8.03kA) ⋅ j15.03Ω = 120.69kV
U a = 120.64kV + 120.69kV = 241.33kV
2
U b = a ⋅ U a1 + a ⋅ U a 2 ≈ −U a1 = −120.64kV
2
U c = a ⋅ U a1 + a ⋅ U a 2 = U b
U ab = 362kV
U bc = 0
e. Double line-to-ground fault at bus 3
Fault current
I a1 =
241.33kV
241.33kV
=
= − j8.34 kA
Z 1 + Z 2 // Z 0
j (15.03 + 15.03 // 187.7)Ω
I a 2 = − I a1 ⋅
Z0
= j 7.69kA
Z0 + Z2
I a 0 = − I a1 ⋅
Z2
= j 0.62kA
Z0 + Z2
"
I k ( 2) = 3 ⋅I a 0 = j1.85kA
Voltage at the fault location
U a1 = 241.33kV − I a1 ⋅ Z 1 = 115.98kV
U a 2 = − I a 2 ⋅ Z 2 = 115.58kV
U a 0 = − I a 0 ⋅ Z 2 = 116.37 kV
U a 0 ≈ U a1 ≈ U a 2
U a = U a 0 + U a1 + U a 2 = 348kV
2
U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 0
2
U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 0
8-38
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.12
Effect of transformer neutral point connection on system
performance during fault
Network parameters:
Overhead line B – C:
Positive and negative sequence:
j0.32 ⋅ 100
= j16 Ω
2
Zero-sequence:
j1.12 ⋅ 100
= j56 Ω
2
Overhead line B – A:
Positive and negative sequence:
j0.32 ⋅ 50
= j8 Ω
2
Zero-sequence:
j1.12 ⋅ 50
= j28 Ω
2
Overhead line C – A:
Positive and negative sequence:
j0.32 ⋅ 50
= j8 Ω
2
Zero-sequence:
j1.12 ⋅ 50
= j28 Ω
2
220-kV Cable
Positive and negative sequence:
j 0.16 ⋅ 30
= j 2 .4 Ω
2
Zero-sequence:
j 0.56 ⋅ 30
= j 8 .4 Ω
2
110-kV Network:
Substation A:
Positive and negative sequence:
Z N = 1 .1 ⋅
Zero-sequence:
Z0 →
220 2
= 17.75 Ω
3000
∞
8-39
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Substation B
Positive and negative sequence:
Z N = 1 .1 ⋅
Zero-sequence:
Z0 →
220 2
= 11.83 Ω
4500
∞
220-kV Network:
Positive and negative sequence:
Z N = 1.1 ⋅
Zero-sequence:
Z0 →
220 2
= 21.3 Ω
2500
∞
380-kV Network:
220 2
= 5.32 Ω
10000
Positive and negative sequence:
Z N = 1 .1 ⋅
Zero-sequence:
 231 
Z 0 = 50 ⋅ 
 = 16.68 Ω
 400 
2
Transformer A
Positive and negative sequence:
Zero-sequence:
Xp =
2312
2312
2312
1 
+ 0.148 ⋅
− 0.09 ⋅
0.155 ⋅
83
83
250
4 

 = 17.59Ω


Xs =
2312
2312
2312
1 
− 0.148 ⋅
+ 0.09 ⋅
0.155 ⋅
83
83
250
4 

 = −1.05 Ω


Xt =
2312
2312
2312
1 
− 0.155 ⋅
+ 0.148 ⋅
+ 0.09 ⋅
83
83
250
4 

 = 29.98 Ω


X 0 = X pt = 47.57 Ω
8-40
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Transformer B
Positive and negative sequence:
X p = 17.59 / 2 = 8.8 Ω
X s = - 1.05 / 2 = -0.53Ω
X t = 29.98 / 2 = 15.0 Ω
X 0 = X pt = 23.79 Ω
Zero-sequence
Transformer C
1 0.145 ⋅ 2312
⋅
= 8.23 Ω
4
235
Positive and negative sequence:
X ps =
Zero-sequence
X 0 = X ps = 8.23 Ω
Transformer D
Positive and negative sequence:
2312
2312
2312
1 
+ 0.166 ⋅
− 0.113 ⋅
X p = j ⋅  0.1 ⋅
210
210
630
2 

 = j10.97 Ω


2312
2312
2312
1 
− 0.166 ⋅
+ 0.113 ⋅
X s = j ⋅  0.1 ⋅
210
210
630
2 

 = -j 2.5 Ω


2312
2312
2312
1 
+ 0.166 ⋅
+ 0.113 ⋅
X t = j ⋅  − 0.1 ⋅
210
210
630
2 

 = j31.21Ω


Generator C
Positive and negative sequence
X "d =
1 0.19 ⋅ 2312
⋅
= 10.79 Ω
4
235
X 2 = 10.79 Ω
8-41
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Positive-sequence network:
01
B110
C220
B220
C015
A220
A110
D220
Branch
B110 - G
A110 - G
B220 - G
C015 - G
D380 - G
A220 - A110
D220 - D380
A220 - D220
B220 - B110
C220 - C015
B220 - A220
B220 - C220
A220 - C220
D380
Reactance [Ω]
11.83
17.75
21.3
10.79
5.32
16.54
8.47
2.40
8.27
8.23
8.00
16.00
8.00
8-42
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Negative sequence:
The negative-sequence network is similar to the positive sequence network without the
voltage sources. For generators: X 2G =
X "d + X "q
2
≈ X1G (during the sub-transient phase);
otherwise X1G ≠X2G
Zero-sequence network:
00
B110
A110
C220
B220
A220
Branch
B220 - 00 (Xpt) (4 trafos in parallel)
A220 - 00(Xpt) (2 trafos in parallel)
C220 - 00 (Xps) (4 trafos in parallel)
A220 - D220 (cable)
D220 - D0 (Xs)
D0 - 00 (Xt)
D0 - D380 – 00 (ZN0+Xp)
D0 - D380 - 00 (380-kV side isolated)
A220 - C220 (line)
A220 - B220 (line)
C220 - B220 (line)
D220
D380
D0
Reactance
[Ω]
23.79
47.57
8.23
8.40
-2.50
31.21
27.65
→∋
28.00
28.00
56.00
8-43
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
The fault location is assumed to be the bus bar A 220.
a. The initial single-phase (I‘‘k(1)) and three-phase (I‘‘k(3)) short-circuit
currents
Delta → Star conversion of the positive-sequence line reactance:
8⋅8
= 2Ω
32
16 ⋅ 8
C220 − 0 : X =
= 4Ω
32
16 ⋅ 8
B220 − 0 : X =
= 4Ω
32
A 220 − 0 : X =
U" =
1.1 ⋅ 220
3
= 139.72kV
Equivalent reactance of the positive (negative)-sequence network:
X B220 = ((11.83 + 8.27 ) // 21.3) + 4 = 14.34 Ω
X C220 = ( 4 + 8.23 + 10.79) = 23.02 Ω
X A220 = (2.4 + 8.47 + 5.32) // (16.54 + 17.75) = 11.0 Ω
X1 = X 2 = X A220 //(2 + X B220 //X C220 ) = 5.46Ω
I"k(3) =
U"
= − j25.59kA
jX1
Delta → Star conversion of the zero-sequence line reactance:
28 ⋅ 28
= 7Ω
112
28 ⋅ 56
C220 − 0 : X =
= 14 Ω
112
28 ⋅ 56
B220 − 0 : X =
= 14 Ω
112
A 220 − 0 : X =
X 0 = ((23.79 + 14 ) //(14 + 8.23 ) + 7) // 47.57 //(8.4 − 2.50 + 31.21 // 27.65 ) = 8.527 Ω
I"k(1) = 3 ⋅
U"
= − j21.55kA
2 ⋅ X1 + X 0
b. Earth fault factor
U a0 =
− Ia0 ⋅ jX 0
"
Ua1 = U − Ia1 ⋅ jX1
Ua2 =
− Ia 2 ⋅ jX 2
8-44
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
 Ua  1 1
  
2
 Ub  =  1 a
 U  1 a
 c 
1   U a0 

 
a  ⋅  Ua1 


2
a   Ua2 
Ua = 0
"
Ub = −Ia0 ⋅ jX 0 − a 2 ⋅ (U − Ia1 ⋅ jX1 ) − a ⋅ Ia 2 ⋅ jX 2
= U" ⋅
X 0 ⋅ (a 2 − 1) + X1 ⋅ (2a 2 + 1)
X 0 + 2 X1
−
"
=U ⋅

3 X0
3  X0
⋅
−j
⋅ 
+ 2 
2 X1
2  X1

X0
+2
X1
"
Uc = −Ia0 ⋅ jX 0 − a ⋅ (U − Ia1 ⋅ jX1 ) − a 2 ⋅ Ia 2 ⋅ jX 2
"
=U ⋅
X 0 ⋅ (a − 1) + X1 ⋅ (2a + 1)
X 0 + 2 X1
−
"
=U ⋅

3 X0
3  X0
⋅
+j
⋅ 
+ 2 
2 X1
2  X1

X0
+2
X1
2
 X0 
X 

 +  0  + 1
 X1 
 X1 
Ub = Uc = U" ⋅ 3 ⋅
X0
+2
X1
=
UN
3
⋅ cf
2
 X0 
X 

 +  0  + 1
 X1 
 X1 
c f (Earth fault factor ) = 1.1 ⋅ 3 ⋅
X0
+2
X1
c. Voltage of the phases that are not directly affected by the fault
i.
For the given case
8-45
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
X 0 8.53
=
= 1.56 → c f = 1.196
X1 5.46
Ub = Uc = 1.196 ⋅
220
3
= 151.91 kV
ii. only one 220-kV neutral point grounded
X1 = 5.46 Ω
X 0 = (( 4 ⋅ 23.79 + 14 ) //(14 + 4 ⋅ 8.23 ) + 7) //(2 ⋅ 47.57 ) //
(8.4 − 2.50 + 31.21 // 27.65 ) = 11.51Ω
X 0 11.51
=
= 2.11 → c f = 1.275
X1
5.46
Ub = Uc = 1.275 ⋅
220
3
= 161.95 kV
iii. transformer with isolated neutral point in substation D
X1 = 5.46 Ω
X 0 = (( 4 ⋅ 23.79 + 14 ) //(14 + 4 ⋅ 8.23) + 7) //( 2 ⋅ 47.57) //(8.4 − 2.50 + 31.21) = 15.98 Ω
X 0 15.98
=
= 2.93 → c f = 1.367
X1
5.46
Ub = Uc = 1.367 ⋅
220
3
= 173.63 kV
d. The neutral point to ground voltage (UEM) of the isolated transformers
Iao =
U"
139.72 kV
=
= − j5.194 kA
j(2 ⋅ X1 + X 0 ) j(2 ⋅ 5.46 + 15.98 ) Ω
Distribution of the zero-sequence current:
Substation
Zero-sequence
current
A
-j0.872kA
B
-j0.62678kA
C
-j1.458kA
D
2.237kA
8-46
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
0
UEM
C220
A110
A220
D220
D0
D380
The neutral to ground voltage (UEM) for substation A is illustrated in the figure above. As can
be seen in the figure, this voltage is equal to the zero-sequence voltage at the corresponding
bus bar (in this case A).
Thus,
A
UEM
= 0.872 kA ⋅ 95.14Ω = 83kV
UBEM = 0.62678 kA ⋅ 95.16Ω = 59.64kV
C
UEM
= 1.458 kA ⋅ 32.92Ω = 48kV
8-47
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.13
Neutral point grounding in medium voltage networks
a. sequence network
Upa1
XN
Xps
XN
Xps
3ZME
Xst
CE
X ps = 1.26 Ω
X st = 0.51Ω
S k" = 2500 MVA → X N =
1.1 ⋅ 20 2
Ω = 0.176 Ω
2500
IC = 0.02 ⋅ 100 + 1.47 ⋅ 30 = 46.1 A
IC =
X CE
20 kV
⋅ ωCE → CE = 12.708 μF
3
≥ 3Z ME ≥ 0
→ X CE =
1
= 250.48 Ω
ωCE
8-48
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
b. Single line-to-ground fault current for isolated neutral point (3ZME→∞)
Upa1
"
Ik (1) = 3 ⋅
2 ⋅ j( X ps + X N ) + j( X st + 3Z ME ) //
Z ME → ∞ and ( X ps + X N ) <<
1
jωC E
1
⇒
ωC E
"
Ik (1) ≈ 3 ⋅ ωC E ⋅ Upa1 = 3 ⋅ IC = 3 ⋅ 46.1A = 138.3 A
This same result can be obtained directly from the network without using the symmetrical
components, as illustrated below.
Uac.jωCE
c
Uab.jωCE
b
a
CE
ICE
Ik" (1) ≈ ICE = Uab ⋅ jωC E + Uac ⋅ jωC E = jωCE ⋅ (Ua − Ub + Ua − Uc )
= 3 ⋅ jωC E ⋅ Ua = 3 ⋅ jωCE ⋅ Upa1 = 3 ⋅ IC
with:
I“k(1) = single line-to-ground fault current;
ICE = capacitive earth leakage current
IC = charging current
And the single line-to-ground fault current considering the positive and negative sequence
reactances is:
Ik" (1) = 3.
Upa1
2 jX1 + ( − jX C )
=
3 ⋅ 20 kV
= j139.90 A
j(2 ⋅ (0.176 + 1.26 ) − 250.48 ) Ω
c. Resonant grounding (3ZME=3XME=XCE)
i. reactance for resonant tuning
8-49
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
For resonance tuning:
IL (coil current) = ICE (capacitive earth leakage current) = 138.3 A→
ω ⋅ L ⋅ Upa1 = 138.3 A → ω ⋅ L = X L =
20 kV
3 ⋅ 138.3 A
= 83.5 Ω = X ME
ii. Equivalent circuit of the zero-sequence network for a damping factor of 4% and a
detuning (off-resonance factor) of -3% (XME ≠ XL)
Y0 =
1
1
1
1
+
+
=
j3 ⋅ X ME − jX CE R D X CE
 3 ⋅ X ME − X CE X CE
⋅  j
+
3 ⋅ X ME
RD




X st << 3 X ME → 3 X ME + X st ≈ 3 X ME
Definition:
v (off - resonance factor) =
d (damping factor) =
Y0 =
1
X CE
3 ⋅ X ME − X CE ICE − IL
=
3 ⋅ X ME
ICE
X CE
RD
⋅ (d + jv )
X CE
= 0.04 → R D = 6.262kΩ
RD
3 ⋅ X ME − X CE
= −0.03 → 3 ⋅ X ME = 243.18Ω
3 ⋅ X ME
iii. Single line-to-ground fault current
1
"
Ik(1) = Ir = 3 ⋅ Upa1 ⋅ Y0 = 3 ⋅ Upa1 ⋅
X CE
⋅ (d + jv ) = ICE ⋅ (d + jv )
= 0.05 ⋅ ICE ∠ − 36,87 0 = 6.915 A∠ − 36,87 0
Ir = residual current
Irw = ICE ⋅ d= 138.3 ⋅0.04 = 5.532 A
(real part of the residual current )
Irb = ICE ⋅ v = 138.3 ⋅ 0.03 = 4.149 A
(imaginary part of the residual current )
d. Impedance grounding (0<3XME<3XCE)
i. Grounding reactance for a maximum fault current of 1000A
8-50
Exercises
Electric Power Systems I - III
1
= 1000 A →
3 2 ⋅ X1 + X 0
2 ⋅ X1 + X 0 = 36.37 Ω
Ik" (1) = 3 ⋅ 1.05 ⋅
20
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
⋅
X1 = X 2 = X ps + X N = 1.26 Ω + 0.176 = 1.436 Ω
X st = 0.51Ω
X 0 = 3 ⋅ X ME + X st = 36.37Ω − 2.52 Ω = 33.5 Ω → X ME =
X 0 − X st
= 11 Ω
3
ii. The effect of neglecting the zero-sequence capacitance
X 0 = ( j3 ⋅ X ME + jX st ) // − jX CE = ( j33 + j0.51) // − j250.48 = j38.69Ω
Ik" (1) = 3 ⋅ 1.05 ⋅
20
1
= − j875.15 A
3 2 ⋅ X1 + X 0
⋅
The earth capacitance compensates part of the grounding reactance, and, as a result, the
actual fault current is significantly smaller. Neglecting the zero-sequence capacitance,
therefore, would lead to a significant error (the error in this example = 1000 A – 875.15 A=
124.85 A).
e. Transformer neutral point solidly grounded (3ZME = 0)
i. Single line-to-ground fault
Ik" (1) = 3 ⋅ 1.1 ⋅
20
1
= 11.267kA
3 2 ⋅ X 1 + X st
⋅
ii. Effect of the zero-sequence capacitance on the fault current
X 0 = j051 // − 250.58 = j0.511
Ik" (1) = 3 ⋅ 1.1⋅
◊
20
1
= 11.264kA →
3 2 ⋅ X1 + X0
⋅
The effect of the zero-sequence capacitance on the fault current is negligible. As a
result of the small zero-sequence transformer reactance (Xst), the capacitance is
practically parallel short-circuited, and can thus be neglected.
f. Discussion of the results
Basically, all methods of neutral point grounding can be viewed as alternative options.
However, for this network practical considerations reduce the options to only resonant and
impedance grounding because of the following reasons:
◊
Isolating the neutral point is not an option since the capacitive earth leakage current
of ICE ≈ 138A does not guarantee a self-extinguishing arc.
8-51
Exercises
Electric Power Systems I - III
◊
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
In case of solid grounding, the high short-circuit current (I“k(1) ≈ 11.3kA) would cause
an unacceptably high stress on equipment along the short-circuit path.
Of the two remaining options, the resonant grounding offers a clear advantage over the (low)
impedance grounding in this network. Because, although the large overhead line contributes
only a small amount to the capacitive earth leakage current, it gives rise to a much larger
probability of earth fault. In case of impedance grounding, this would lead to tripping of the
protective relays to isolate the faulted section of the distribution network, and, therefore, to
service interruption. In the event of resonant grounding, a fault does not necessarily lead to
immediate tripping, resulting in improved supply continuity.
8-52
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.14
Problem of voltage un-symmetry in three phase transmission
systems
c
M
b
a
Note: only one of the 3 impedances
YaE=GaE+jωCaE is drawn for clarity.
ZEM
E
Applying Kirchoff’s current law:
I EM = I aE + I bE + I cE →
I EM = U aE ⋅ Y aE + U bE ⋅ Y bE + U cE ⋅ Y cE
U aE = U aM − U EM because (Z LE >> Z L )
2
U bM = a ⋅ U aM
U cM = a ⋅ U aM →
ZL: line impedance; ZLE: line – earth impedance
U EM
(Y
=
U EM =
=
=
d=
)
+ a 2 ⋅ Y bE + a ⋅ Y cE ⋅ U aM
3 ⋅ Y 01
=
⋅ U aM
(Y aE + Y bE + Y cE ) + Y EM
3 ⋅ Y 00 + Y EM
aE
G EM
G EM
3Y 01
⋅ U aM
+ jB EM + 3(G 00 + jB00 )
3Y 01
⋅ U aM
+ 3G 00 + j (B EM + 3B00 )
3Y 01
G
+ 3G 00
+ 3B00
B
3B00  EM
+ j EM
3B00
3B00

G EM + 3G 00
I
= wr
3B00
I CE
v=



⋅ U aM =
B EM + 3B00
=
3B00
Y 01
⋅ U aM
B00 (d + jv )
3ωC 00 −
1
X EM
3ωC 00
=
I CE − I L
I CE
8-53
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
I wr = (G ME + 3G 00 ) ⋅ U EM
I CE = 3ωC 00 ⋅ U EM
IL =
1
X ME
⋅ U EM
a. Ohmic symmetry, capacitive un-symmetry
G aE = GbE = G cE
C aE ≠ C bE = C cE →
C aE + ∆C = C bE = C cE ( for example if C aE > C bE )
UEM =
j
(Y
2
aE
)
+ a ⋅ Y bE + a ⋅ Y cE ⋅ UaM
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
=
(C
2
aE
)
+ ∆C + a ⋅ C bE + a ⋅ C cE
⋅ jω ⋅ UaM =
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
∆C ⋅ ( v + jd)
∆C
⋅ UaM =
⋅ UaM
3 ⋅ C 00 ⋅ (d + jv )
3 ⋅ C 00 ⋅ d 2 + v 2
v = 0 → UEM = j
∆C
⋅U
3 ⋅ C 00 ⋅ d aM
v = ±∞ → UEM =
∆C
⋅ UaM
3 ⋅ C 00
b. Capacitive symmetry, Ohmic un-symmetry
G aE ≠ GbE = G cE
C aE = C bE = C cE →
G aE + ∆G = GbE = G cE ( for example if G aE > G bE )
UEM
(Y
=
2
aE
)
+ a ⋅ Y bE + a ⋅ Y cE ⋅ UaM
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
(G
=
2
aE
)
+ ∆C + a ⋅ GbE + a ⋅ G cE
⋅ ω ⋅ UaM =
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
∆G ⋅ (d − jv )
∆G
⋅ UaM =
⋅ UaM
3 ⋅ C 00 ⋅ (d + jv )
3 ⋅ C 00 ⋅ d 2 + v 2
v = 0 → UEM =
∆G
⋅U
3 ⋅ C 00 ⋅ d aM
v = +∞ → UEM = − j
v = −∞ → UEM = j
∆G
⋅ UaM
3 ⋅ C 00
∆G
⋅ UaM
3 ⋅ C 00
c. Ohmic symmetry, capacitive un-symmetry (general)
8-54
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
G aE = GbE = G cE
C aE ≠ C bE ≠ C cE →
C aE = C aE
C bE = C aE + ∆C bE
C cE = C aE + ∆C cE
UEM =
(Y
2
aE
)
+ a ⋅ Y bE + a ⋅ Y cE ⋅ UaM
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
(a
)
2
⋅ ∆C bE + a ⋅ ∆C cE
⋅ jω ⋅ UaM
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
=
with :
Δc bE =
ΔC bE
3 ⋅ ω ⋅ C 00
Δc cE =
ΔC cE
→
3 ⋅ ω ⋅ C 00
( per unit )
(
)
(
2
2
U EM
a ⋅ ΔC bE + a ⋅ ΔC cE
j a ⋅ Δc bE + a ⋅ Δc cE
=
⋅ jω =
3 ⋅ ω ⋅ C 00 ⋅ (d + jv )
U aM
(d + jv )
)
d. Conductor-to-earth capacitances
UEM
UaM
=
(
)
2
j a ⋅ ∆c bE + a ⋅ ∆c cE
→
(d + jv )
(
2
)
j a ⋅ ∆c bE + a ⋅ ∆c cE
11.6 kV∠245 0
=
→
110
0.025
kV
3
4.5663 ⋅ 10 −3 ∠245 0 = ∆c bE (cos( −30 0 ) − j sin 30 0 ) + ∆c cE (cos 210 0 + j sin 210 0 ) →
∆c bE = 3.0242 ⋅ 10 −3
∆c cE = 5.2528 ⋅ 10 −3
ICE = 3 ⋅ ω ⋅ C 00 ⋅
Un
3
= 3 ⋅ C 00 ⋅ 100 ⋅ π ⋅
110
3
kV = 395 A → 3 ⋅ C 00 = 19.8µF
∆C bE = ∆c bE ⋅ 3 ⋅ C 00 = 3.0242 ⋅ 10 −3 ⋅ 19.8 µF = 0.059879 µF
∆C cE = ∆c cE ⋅ 3 ⋅ C 00 = 5.2528 ⋅ 10 −3 ⋅ 19.8 µF = 0.104 µF
C aE + C bE + C cE = 3 ⋅ C aE + ∆C bE + ∆C cE = 3 ⋅ C 00 →
CaE = 6.5454 µF
CbE = 6.5454 µF + 0.059879 µF = 6.6 µF
CcE = 6.5454 µF + 0.104 µF = 6.65 µF
8-55
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.15
Low impedance grounding in cable networks
Sequence networks:
01
1.05
UN
3
j0.176Ω
A110
j1.26Ω
A20-2
0.45+j0.3 Ω
B2
j1.26Ω
A20-1
0.45+j0.3Ω
B1
0.45+j0.3Ω
0.6+j0.4Ω
C2
0.45+j0.3Ω
D2
C1
0.3+j0.2Ω
D1
Negative sequence equivalent circuit is identical to the positive sequence network without the
voltage sources.
8-56
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Zero-sequence network:
00
3ZME
3ZME
j0.51Ω
j0.51
XCE
A20-2
0.3+j1.92Ω
B2
0.3+j1.92Ω
C2
0.3+j1.92Ω
D2
X CE =
3 ⋅ Un
=
ICE
A20-1
0.3+j1.92Ω
B1
0.4+j2.56Ω
C1
0.2+j1.28Ω
D1
3 ⋅ 20 ⋅ 10 3
Ω;
ICE [ A ]
X CE = 679.24Ω
(ICE = 51A )
X CE = 407.54Ω
(ICE = 85 A )
X ME = R ME = 12 Ω
Upa1 = 1.05 ⋅
20
3
= 12.124kV
XN = 0.176 Ω
Xps = 1.26 Ω
a.
Single line-to-ground fault at bus A
i.
Coupling switch in bus bar A open, loop opened at bus D, transformer neutral
points not interconnected
8-57
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Z 0 = ( j3 ⋅ X ME + jX st ) // − jX CE = j38.58Ω
reac tan ce grounding
Z 0 = (R ME + jX st ) // − jX CE = 36Ω∠ − 2.23 0
resis tan ce grounding
Z1 = Z 2 = 0.176 Ω + 1.26 Ω = 1.436 Ω
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 1011 .1∠ − 2.2 0 A
⇔ for grounding through R ME
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 877.47∠ − 90 0 A
⇔ for grounding through X ME
ii. Coupling switch in bus bar A open, loop opened at bus D, transformer neutral
points not interconnected
Z 0 = − jX CE //( jX st + (3Z ME // 3Z ME //( jX st − jX CE )))
= j19.55Ω ( X ME )
= 17.97Ω∠ − 1.5 0 (R ME )
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 2007.2∠ − 7.7 0 A
⇔ for grounding through R ME
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 1620.8∠ − 90 0 A
⇔ for grounding through X ME
iii. Coupling switch in bus bar A closed, loop opened at bus D, transformer neutral
points not interconnected
Z0 =
− jX CE //( jX st + 3Z ME ) − j679.24 //( j0.51Ω + 3Z ME )
=
2
2
= j19.29Ω ( Z ME = j3 X ME = j36 Ω)
= 17.99Ω∠ − 2.23 0 ( Z ME = 3R ME = 36 Ω)
Z1 = Z 2 = j1.436Ω
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 2007.73∠ − 6.88 0 A
⇔ for grounding through R ME
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
= 1640.65∠ − 90 0 A
⇔ for grounding through X ME
iv. Coupling switch in bus bar A closed, loop opened at bus D, transformer neutral
points interconnected
- same as (iii) -
b. Fault at Bus B2, CB to the left of B2 open
8-58
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Z 0 = − jX CE //( jX st + 3Z ME ) + (1.5 + j9.6) Ω
X CE = 407.54 Ω
Z 0 = 1.5 + j49.7 = 49.73Ω∠88.27 0 ( Z ME = jX ME )
Z 0 = 37.95Ω∠10.54 0 ( Z ME = R ME )
Z1 = Z 2 = 2.25 + j2.936 Ω = 3.7∠52.5 Ω
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
=
36.372kV
2 ⋅ 3.7Ω∠52.5 0 + 37.95Ω∠10.54 0
= 831.74 A∠ − 17.03 0 ⇔ (grounding through R ME )
Ik" (1) = 3 ⋅
Upa1
2 ⋅ Z1 + Z 0
=
36.372kV
2 ⋅ 3.7Ω∠52.5 0 + 49.72Ω∠88.27 0
= 650.78∠ − 83.84 0 A ⇔ ( for grounding through X ME )
c. Discussion of the results
•
In parts (a) and (b) the network is to be considered as not effectively
grounded.
•
The cable impedances have significant bearing on the magnitude of the fault
current. This fact needs to be taken into consideration during the setting of
protective relays.
•
As the grounding impedance becomes smaller, the effect of the earth
capacitance on the fault current becomes also smaller.
•
As a result of the compensating effect of the grounding reactance, grounding
of the neutral point through a resistance will always cause a larger fault
current during a single line-to-ground fault.
8-59
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.16
Transient stability analysis I
An elementary principle of dynamics states that:
d 2 δm
J ⋅ 2 = Tm − Te = Ta
dt
(1)
where:
J = the total moment of inertia of the rotating masses in kg.m2
δm = the angular displacement of the rotor with respect to a synchronously
rotating reference in mechanical radians
Tm = the mechanical torque in N.m.
T2 = the electrical torque in N.m.
Ta = accelerating torque in N.m.
Multiplying both sides of the equation by ωm
ωm ⋅ J ⋅
d 2 δm
= ωm ⋅ (Tm − Te ) = ( Pm − Pe )
dt 2
(2)
Definition:
1
⋅ J ⋅ ωm2
H= 2
Sr
where:
H = inertia constant in seconds
Sr = machine rated power in MVA
ωm = angular speed in mechanical radians/second
Dividing both sides of equation (2) by Sr/2ωm and re-arranging, we have:
d 2δ m
dt 2
=
ωm
2H
⋅ ( Pm − Pe ) , Pm, Pe: in per unit.
Introducing δm = δ/p (δ = angle in electrical radians; p = number of pole pairs), we have
the swing equation in final form:
d 2δ π ⋅ f
=
⋅ (Pm − Pe )
H
dt 2
(3)
where:
δ
= the angular displacement of the rotor (in electrical radians) with respect to a
reference frame rotating at synchronous speed
f = the frequency
Pm = shaft mechanical power in per unit
Pe = electrical output power in per unit
8-60
Exercises
Electric Power Systems I - III
8.17
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Transient stability analysis I
a. Kinetic energy, inertia constant
S = P / cos φ = 150 / 0.85 MVA = 176.47 MVA
Un = 15 kV, n = 1500 rpm, J =75 ·103 kgm2
i. the stored kinetic energy of the rotating masses at nominal speed
1
⋅ J ⋅ ωm2
2
2⋅π ⋅n
ωm =
= 157.08 mech.rad / s →
60
75 ⋅ 10 3 ⋅ 157.08 2
Wk =
kW ⋅ s = 925.279 MW ⋅ s
2
Wk =
ii. the inertia constant H (in seconds)
H=
1 J ⋅ ωm2 Wk
⋅
=
= 5.243 s
2
S
S
Mechanical time constant Tm is defined as follows:
Tm =
J ⋅ ωm2
= 2H
S
iii. the inertia constant M (in joule-seconds/mechanical radian).
M = J ⋅ ωm = 11.777 MJ / rad
J ⋅ ωm
1 J ⋅ ωm2
H= ⋅
→ 2⋅ H =
= M [ p .u ]
S
2
S / ωm
b. Stored kinetic energy of a 100-MVA generator having H = 8.0 s.
i. Kinetic energy at synchronous speed
H=
1
1 J ⋅ ωm2
⋅
= 8s → ⋅ J ⋅ ωm2 = H ⋅ S = 800 MWs
2
2 100 MVA
ii. Speed of a 40-tonne lorry to have the same amount of kinetic energy
1600⋅ 10 6 Ws
1
⋅ m ⋅ v 2 = 800MWs → v =
= 200m / s
2
40 ⋅ 10 3 kg
Unit check :1W ⋅ s = 1N ⋅ m;1 N = 1kg ⋅ m / s 2 → W ⋅ s / kg = (m / s )2
iii. Total mass of a train travelling at 100 km/h to have the same kinetic energy
2
 10 5

1
⋅ m ⋅ 
m / s  = 800 ⋅ 10 6 Ws → m = 2073.6 tonne
2
 3600

c. Angular acceleration, angular speed, displacement angle
i.
Angular acceleration for a mechanical power increase from 50 MW to 80 MW
8-61
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
π ⋅ 50 (80 − 50)
d 2δ π ⋅ f
=
⋅ (Pm − Pe ) =
⋅
= 5.89 ele. rad / s 2
2
8
100
H
dt
ii.
The rotor angle and rotor speed for a 10-cycle constant acceleration
d 2δ
1
dδ
= k = 5.89 rad / s 2 →
= k ⋅ t + ω0 = Δω + ω0 → Δω = 5.89 ⋅ 10 ⋅ ⋅ rad / s
2
50
dt
dt
= 1.178rad / s ( all in ele. radians )
δ( t ) =
1
⋅ k ⋅ t 2 + ω0 ⋅ t + δ0 →
2
2
1 
1
1

Δδ = δ( t ) − ω0 ⋅ t = ⋅ k ⋅ t 2 + δ0 = ⋅ 5.89 s − 2 ⋅ 10 ⋅ s  = 0.1178 el . rad . = 6.75 0 el .
50 
2
2

(In the last expression, subtraction of ω0 ⋅ t is necessary since the angle is
measured with respect to a reference frame rotating at the constant speed of ω0 .)
Brief explanatory note on the relationship between mechanical and electrical radians:
Assume the generator is running at the speed n rpm→n/60 revolutions per second (rps).
That means the machine is traversing in 1 s :
rotational speed of
ωm =
n
⋅ 2π
60
n
⋅ 2π ( mech.) rad / s . Whereas one complete revolution is
60
always 2π rad mechanically, electrically it is 2π.p (p = number of pole pairs). Thus, the
speed
in
electrical
is: ωe
= p⋅
n
⋅ 2π ( el .)rad / s = p ⋅ ωm . Generally:
60
ωe
δ
= e = p.
ωm δ m
d.
The critical fault clearing angle
2,5
2
1,5
P [p.u.]
pre-fault
during fault
1
post fault
mech. power
0,5
0
0
0,5
1
1,5
2
2,5
3
3,5
-0,5
8-62
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
The accelerating and decelerating areas are indicated above. Assuming the following
general values:
Pm
= Mechanical power
P0max
= Pre-fault amplitude of the power vs. power angle curve
Pfmax
= Amplitude during fault
Ppfmax
= Post-fault amplitude
The critical fault clearing angle can be obtained by equating the accelerating and
decelerating areas:
δc
∫ (Pm −
δ0
f
Pmax
)
⋅ sin δ ⋅ dδ =
δ max
∫ (Pmax ⋅ sin δ − Pm )⋅ dδ →
pf
δc0
f
pf
Pm ⋅ (δc − δ0 ) + Pmax
⋅ (cos δc − cos δ0 ) = Pmax
⋅ (cos δc − cos δ max ) + Pm ⋅ (δc − δmax ) →
A general formula for calculating critical fault clearing angle:
pf
f
 P ⋅ (δ − δ ) + Pmax
⋅ cos δmax − Pmax
⋅ cos δ0 

δc = cos −1  m max 0
pf
f

Pmax − Pmax


With:
δ0 = sin −1 ( 1 / 2 ) = 0.523 rad = 30 0
δmax = π − δ0 = 2.618rad = 150 0
We have for the critical fault clearing angle:
δc = cos −1 (δmax − δ0 + 1.5 ⋅ cos δmax − 0.5 ⋅ cos δ0 + ) = 68.710 = 1.2rad
8-63
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.18
Transient stability analysis III
a. Assuming a pre-fault generator load of 1 p.u, calculate the critical fault clearing
angle for a solid three-phase fault occurring at point P in the system given below.
j0.15
j0.1
j0.15
j0.15
j0.28
UN=1.0
j0.15
∼
P
E=1.2
j0.15
j0.14
j0.14
j0.15
Infinite bus
Before fault:
P=
E ⋅U N
0
⋅ sin δ = Pmax
⋅ sin δ
XΣ
X Σ = 0.15 + 0.1 +
0.15 + 0.28 + 0.15
+ 0.15 = 0.69 →
2
1 .2
= 1.74
0.69
E ⋅U N
1= P =
⋅ sin δ0 → δ0 = 35.10
XΣ
0
=
Pmax
The equivalent circuit of the network during fault:
j0.15
j0.1
j0.15
j0.15
j0.28
UN=1.0
j0.15
∼
P
E=1.2
j0.15
j0.14
j0.14
j0.15
Infinite bus
j0.58
j0.15
E=1.2
∼
Generator - Trafo
j0.15
j0.1
j0.29
j0.29
∼
External network
The Thevenin equivalent circuit of the external network:
8-64
UN=1
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
X Th = 0.29 // (0.58 + 0.15 // 0.29 ) = 0.203
U Th = 0.29 ⋅
1
0.29
⋅
= 0.197
0.15 + (0.58 + 0.29 ) // 0.29 0.58 + 0.29 + 0.29
The equivalent circuit during fault:
j0.15
E=1.2
j0.1
j0.203
∼
∼
UTh=0.197
The power equation for the duration of the fault:
Pf =
1.2 ⋅ 0.197
⋅ sin δ = 0.522 ⋅ sin δ
0.25 + 0.203
The power equation after the fault is cleared:
pf
=
Pmax
1 .2 ⋅ 1
⋅ sin δ = 1.22 ⋅ sin δ
0.98
With
Pfmax = 0.522,
δ0 = 35.10 =0.61rad
δmax = 124.950 = 2.18rad
and using the formula derived in 8.17 (d), we have for the critical clearing angle.
pf
f
 P ⋅ (δ − δ0 ) + Pmax
⋅ cos δmax − Pmax
⋅ cos δ0
δc = cos −1  m max
pf
f
− Pmax
Pmax


→


 2.18 − 0.61 − 1.22 ⋅ 0.573 − 0.522 ⋅ 0.818 
0
δc = cos −1 
 = 50.5
0.698


b. A solid three- phase fault is assumed to occur at point P in the system given below.
All the necessary data are given in the figure and the system frequency is 50 Hz.
The fault is cleared by disconnecting the faulted line in both ends. What are the
critical fault clearing angle and critical fault clearing time for a pre-fault generator
load of 1 p.u.
8-65
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
j0.15
j0.1
UN=1.0
j0.5
j0.05
∼
E =1.2 p.u
H=4s
j0.4
P
Infinite bus
The approach as the same as (a).
Pre-fault:
0
=
Pmax
1 .2
= 2 .3
0.15 + 0.1 + 0.4 // 0.5 + 0.05
During fault:
The external network incorporates the rest of the network except generator and
transformer. Parameters of the Thevenin equivalent for this part of the network are:
f
Pmax
=0
After fault:
pf
Pmax
=
1 .2
= 1 .5
0.15 + 0.1 + 0.5 + 0.05
With
δmax = 138.190 = 2.41rad
δ0 = 25.770 =0.45rad
and again using the formula derived in 8.17 (d), we have for the critical clearing angle.
pf
f
 P ⋅ (δ − δ0 ) + Pmax
⋅ cos δmax − Pmax
⋅ cos δ0
δc = cos −1  m max
pf
f
Pmax − Pmax

−1 
δc = cos 

(2.41 − 0.45) + 1.5 ⋅ cos 138.19 0  = 55.85 0
1.5



→


c. The generator in the figure below has an inertia constant H = 4 s and feeds 1 p.u.
power into the infinite bus. Data for the system are: UN = Ut = 1.0 p.u., xd’ = 0.25 p.u.,
xt = 0.10 p.u., xl = 0.5 p.u., f = 50 Hz.
xd’
xt
xl
UN
∼
Up’
Ut
xl
Infinite bus
8-66
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
i. What is the value of the emf (Up’) behind the transient reactance?
The magnitude of the generator terminal voltage (Ut) is given, but its phase angle is
unknown. It can be determined from the power relationship:
P =1=
U t ⋅U N
1
⋅ sin δG =
⋅ sin δG → δG = 20.48 0 → U t = 1∠20.48 0
xt + xl / 2
0.1 + 0.25
The current flowing out of the generator:
IG =
Ut − U N
1∠20.48 0 − 1 − 0.063 + j 0.35
=
=
= 1 + j 0.18 = 1.016∠10.2 0
j ( xt + xl / 2 )
j 0.35
j 0.35
The transient voltage of the generator can now be calculated using the relationship:
'
U p = U t + jx'd ⋅ I G = 1∠20.48 0 + j 0.25 ⋅ 1.016∠10.2 0 = 0.89 + j 0.6 = 1.07∠34.0 0
ii. Calculate the maximum power that can be transmitted if:
− the topology of the system is as shown in the figure.
0
Pmax
=
U 'p ⋅ U N
x'd
+ xt + xl / 2
=
1.07
= 1.78
0.25 + 0.1 + 0.25
− When one of the lines experiences a three-phase fault midway along the line.
X Th = xl // xl / 2 = 0.167
U Th =
f
Pmax
=
1
3
1.07 ⋅ 1 / 3
= 0.69
0.25 + 0.1 + 0.167
iii. What is the accelerating power immediately after the fault described in (ii)?
d 2δ π ⋅ f
=
⋅ ΔP
H
dt 2
At the instant of fault: δ = δ0 and the electrical output
f
⋅ sin δ0 = 0.69 ⋅ sin( 34 0 ) = 0.3858 → ΔP = 1 − 0.3858 = 0.61 →
power: P = Pmax
π ⋅ 50
d 2δ π ⋅ f
=
⋅ ΔP =
⋅ 0.61 = 23.95 rad / s 2
2
4
H
dt
iv. What is the value of the rotor angle if the acceleration is maintained constant at
the value calculated in (iii) for 0.05 s?
Δδ =
1 π ⋅ f ⋅ ΔP
⋅
⋅ 0.05 2 = 1.72 0
2
4
v. What is the accelerating power at the rotor angle calculated in (iv)?
f
δ = δ0 + Δδ = 34 0 + 1.72 0 = 35.72 0 → P = Pmax
⋅ sin 35.72 0 = 0.4 →
Pa = Pm − P = 1 − 0.4 = 0.6
vi. Discuss the difference between the values obtained in (iii) and (v)!
8-67
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
The value obtained in (v) is smaller since the output electrical power increases as
the power angle increases.
8-68
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
8.19 Transient stability analysis
Derivation of the Equal Area Criterion:
Equation of motion:
J
1
dωL
= MT − M E =
⋅ ( PT − P )
ωL
dt
dδGN
→
dt
d 2δGN
1
J
= MT − M E =
⋅ ( PT − PE )
dt
ωL
ωL =
J=
Tm ⋅ S n
→ from the definition of Tm
ωL2 0
d 2δGN
= k ⋅ ( PT − PE )
dt
with
k=
ω0
S nTm
Equal area criterion:
d 2δGN
dδ
dδ
⋅ 2 GN = k ⋅ ( PT − PE ) ⋅ 2 GN →
dt
dt
dt
d 2δGN
d
δ
⋅ 2 GN =
dt
dt
dt
 dδ 
⋅  GN 
 dt 
2
2
dδGN
d  dδGN 
d
dx
dδGN d 2δGN
2
(Check: x =
→ ⋅
=2
⋅
)
 = ⋅ x = 2x ⋅
dt
dt  dt 
dt
dt
dt
dt
Thus,
2
 dδ 
d  GN  = k ⋅ ( PT − PE ) ⋅ 2 ⋅ dδGN →
 dt 
δ
 dδ  2  GN ( max )
=
 GN  
dt


 δGN

(0)
δGN ( max )
∫ 2k ⋅ ( PT − PE ) ⋅dδGN
δGN ( 0 )
dδGN
= 0 at δGN ( 0 ) and δGN (max)
dt
8-69
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
δGN ( max )
∫ ( PT − PE )⋅dδGN = 0
δGN ( 0 )
Prefault power output of the generator:
Us ⋅ U∞
⋅ sin δ = 1.4
x
0.1 + 0.2 + 0.1
x=
= 0.2 p.u.
2
P=
δ s∞ = 16.26 0
I=
Us − U∞
E "G
=
jx
1∠16.26 0 − 1
= 1.414∠8.13 0
j0.2
= U s + I ⋅ jx "d = 1∠16.26 0 + j0.12 ⋅ 1.414∠8.13 0 = 0.936 + j0.448 = 1.04∠25.58 0
→ δ 0 = 25.58 0
Pre-fault power vs. power angle curve:
Pe =
E "G ⋅ U ∞
x+
x "d
⋅ sin δ =
1.04 ⋅ 1
⋅ sin δ = 3.25 ⋅ sin δ
0.32
a. Symmetrical fault
Equivalent circuit during the symmetrical fault:
xps
xd
xL
xps
“
EG“
U∞
Generator
External system
Output power during fault:
E th =
x ps
x ps + x L + x ps
0.1
=
= 0. 2
0. 5
xTh
ETh
x Th = x ps //( x ps + x L + x ps ) = 0.1// 0.4 = 0.08
P=
E "G ⋅ E Th
x "d
+ x Th
⋅ sin δ =
1.04 ⋅ 0.2
⋅ sin δ = 1.04 ⋅ sin δ
0 .2
8-70
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Output power, when one of the parallel lines is disconnected:
P=
E "G ⋅ U ∞
x "d
+ x ps + x L + x ps
⋅ sin δ =
1.04 ⋅ 1
⋅ sin δ = 2 ⋅ sin δ
0.12 + 0.1 + 0.2 + 0.1
Power vs. power angle curve
3,5
3
power [p.u.]
2,5
2
1,5
1
0,5
0
0
0,5
1
1,5
2
2,5
3
3,5
∆t = 0.05 s
k=
π⋅f
(∆t )2 = 0.0785398
H
δ n = δ n−1 + ∆δ n (in electrical radians )
∆δ n = ∆δ n−1 + k ⋅ Pa,n−1
Pa, n = Pm, n − Pel, n
(Pm,n, Pel,n: mechanical input and output electrical power, respectively)
t[s]
δ
Pa
∆δ
t[s]
δ
Pa
0
0.05
0.4464
0.4755
0.0373
0.1
0.4838
0.9162
0.1093
0.15
0.5931
0.8187
0.1736
0.2
0.7667
0.6785
0.2269
0.25
0.9936
0.5285
0.2684
1.2620
-0.04811
0.2646
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
2.528
1.5266 1.7443 1.9172 2.0523 2.1581 2.2431 2.3152 2.3817 2.45
-0.5980 -0.5700 -0.4812 -0.3726 -0.2649 -0.1648 -0.071 0.0223 0.12437 0.2483
8-71
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
∆δ
0.2177
0.1729
0.1351
0.1058
0.0850
0.0721 0.0665
0.0683
0.078
0.0975
The angles corresponding to the fault clearing and switching times can be obtained from the
table:
t =0 s
→ δ =0.446 rad
t =0.25 s
→ δ =1.262 rad
t =0.75 s
→ δ = 2.528 rad
Application of the equal area criterion and determining whether or not the generator remains
in synchronism with the rest of the network:
Area corresponding to 0 ≤ t ≤0.25 s:
1.262
∫ (1.4 − 1.04 ⋅ sin δ) ⋅ dδ = 0.52
A1 =
0.446
Area corresponding to 0.25 < t ≤0.75 s:
2.528
∫ (1.4 − 2 ⋅ sin δ) ⋅ dδ = −0.47
A2 =
1.262
Accelerating area: A1 + A2 = 0.52 – 0.47 = 0.05
Reserve area (t>0.75 s):
2.6958
∫ (3.25 ⋅ sin δ − 1.4) ⋅ dδ = 0.04
A3 =
2.528
Accelerating area (0.05) > Reserve area (0.04) → the generator will go out of synchronism!
This is demonstrated by the following plot of the power angle against time.
Swing curve
4,5
4
3,5
3
2,5
2
1,5
1
0,5
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t [s]
8-72
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
In this example, the generator loses synchronism just marginally. If the fault is cleared, for
example, at t = 0.2 s (instead of t=0.25) the generator will retain synchronism, as shown in
the following curve.
Swing curve
2
1,5
1
0,5
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
-0,5
t [s]
b. Double line-to-ground fault
Equivalent circuits of the sequence networks:
01
EG“
xd“
xps
xL
xps
U∞
02
xd“
xps
xL
xps
x2=0.1172
00
xps
xL
xps
x0=0.1
8-73
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Double line-to-ground fault calls for the parallel connection of the three networks, with the
fault location as the terminal.
01
EG“
xd“
xps
xL
xps
U∞
xeq= x2//x0=0.054
External network
xd“
xTh=0.1066
ETh=0.3776
EG“
Generator
Output power during the double line-to-ground fault:
P=
E "G ⋅ E Th
x "d + x Th
⋅ sin δ = 1.73 ⋅ sin δ
8-74
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Exercises
Electric Power Systems I - III
Power vs. power angle curve
Double line-to-ground fault
2
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
0
0,5
1
1,5
2
2,5
3
3,5
Accelerating area:
A+ =
0.943
∫ (1.4 − 1.73 ⋅ sin δ) ⋅ dδ = 0.151
0.446
Reserve (decelerating) area:
A− =
π−0.943
∫ (1.73 ⋅ sin δ − 1.4 ) ⋅ dδ = 0.2744
0.943
A- > A+→ Even for a sustained fault (fault not cleared), the generator would remain in
synchronism, as demonstrated by the following swing curve.
Sustained double line-to-ground fault
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t [s]
Maximum swing angle:
8-75
Exercises
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
δx
∫ (1.73 ⋅ sin δ − 1.4) ⋅ dδ = 1.4 ⋅ (0.943 − δ x ) + 1.73 ⋅ (cos 0.943 − cos δ x ) = 0.151 →
0.943
cos δ x + 0.809 ⋅ δ x = 1.2632 → δ x = 1.61 = 92.25 0
→ Note: This value corresponds to the maximum angle in the curve above.
8-76
```