à during short circuit studies, power systems faults at ... · short circuit studies àduring...
TRANSCRIPT
@ Salvador Acevedo, 2000
Short Circuit Studies
à During Short Circuit Studies, power systems are solved to obtain current magnitudes during faults at different points in the network.MFault: “Failure in a circuit which interferes with
the normal flow of current”
à Purposes of a Short Circuit StudyTo design a PROTECTION scheme to prevent damage to the electric equipment in case of the occurrence of a fault.ð Location of breakersð Selection of breakers
o Ratings of breakers
ð Proper adjustment of breakersð Coordination of the Protection
o Interruption of the currento Isolation of the faulto Sequence of operationo Protection backup
à When is a Short Circuit Study performed?ð When designing the electrical installationð When changing operating conditions ð When installing or removing equipmentð When planning expansion
@ Salvador Acevedo, 2000
Types of Faults
à Symmetrical Faultsð Faults involving the three-phases
o about only 5% of the cases
ð Easiest to evaluateð Required in a Short Circuit Study because they
are commonly the worst case
à Unsymmetrical Faultsð Faults involving some unbalance
o Line to ground faults (one phase to ground)à about 70% of the faults
o Line-to-line faults (between two phases)à about 25% of the faults are line-to-line faults
ð To solve for these faults, we require the use of symmetrical components and sequence networks
@ Salvador Acevedo, 2000
Faults in a Three-Phase Line
abc
abc
Solid three-phase fault Three-phase to ground fault
Fault impedance
abc
Line-to-line fault Line-to-line to ground fault
abc
Fault impedance
Line to ground fault Line to ground fault through impedance
abc
Fault impedance
abc
@ Salvador Acevedo, 2000
Transients in RL Circuits
+
Vs
-
R
Li(t)
( )
( )
The so lu t i o n f o r t h e cur ren t c o n t a i n s a fo r ced r e sponse ( s t e ady s t a t e ) ,and a t r an s ien t resp onse (na tu ral) :i(t) = i i
T h e s t e a d y - s ta te or f o rced re sp o n s e c a n b e o b t a i n e d u s i n g p h aso r s :
I =V sZ
i
T h e t r a n s ie n t r e s p o n se i s the na tu ra l r e s p o n s e o f t he c i r cu it , w h ich i s t h e
so lu t ion t o t h e h o m o g e n e o u s d if fe ren t ia l e q u a t i o n :
s t eady -state t r ans i en t
s t e a d y -s ta te
s t e a d y -s ta te
+
=∠
∠=
∠ − = ∠
= + −
= + =
+ =
=
−
−
VZ
VZ
I
I w t
w h e r e Z R w L a n dw LR
R i Ld id t
i K et r a n s i e n t
R
m a x m a xm a x
m a x sin
: tan
αθ
α θ θ
α θ
θ2 2 2 1
0
( )
( )( ) ( )
( ) ( )
Lt
RL
t
RL
t
K e I w t
I K e
K I I
I e w t
T h e r e f o r e , the to ta l r e s p o n s e i s :
i(t) =
if i (0) = 0 , then 0 =
a n d
i(t) =
−
−
+ + −
− +
= − − = −
− + + −
m a x
m a x
m a x m a x
m a x
sin
sin
sin sin
sin sin
α θ
α θ
α θ θ α
θ α α θ
0
vs Ri Ldidt
i
V wt Ri Ldidt
= + =
+ = +
( )
sin( )max
0 0
α
@ Salvador Acevedo, 2000
RL circuit
( ) ( )
Plot for the current
i(t) = I e I wtR
Lt
max maxsin sin−
− + + −θ α α θ
+
=
total current
natural
0 0.05 0.10
20
40
60
forced
0 0.05 0.1-100
-50
0
50
100
0 0.1-100
-50
0
50
100
@ Salvador Acevedo, 2000
Short Circuit Current in a Synchronous Generator
-600
-400
-200
0
200
400
600
transient steady-statesubtransient
Subtransient generator equivalent
+Ed”
-
jXd” ra +Vt-
Ia
Ed” : Internal subtransient voltageVt : Terminal voltageXd ”: Subtransient reactance
Transient generator equivalent
jXd’ ra+Ed’
-
+Vt-
Ia
Ed’ : Internal transient voltageVt : Terminal voltageXd’ : Transient reactance
Steady-state generator equivalent
jXd ra+Ed-
+Vt-
Ia
Ed : Internal steady-state voltageVt : Terminal voltageXd: Steady-state reactance
Line current during a three-phase short circuit
Xd > Xd’ > Xd”
@ Salvador Acevedo, 2000
Short-Circuit Example: Unloaded Generator
Three-phaseshort circuit
GENERATOR50 MVA, 13.8 KV
xd"=15%xd'=25%xd=80%
TRANSFORMER50 MVA, 13.8-69 KV
x=10%
Find the subtransient, transient, and steady-state generator current when a three-phase short circuit occurs at the high-voltage transformer terminals. Before the fault, there is no load connected and the open circuit voltage at the line terminals is 69KV. Neglect all resistances.
@ Salvador Acevedo, 2000
Solution
The equivalent circuits for subtransient, transient, and steady-state periods are shown below. Solving each one will give the magnitude of the fault current during its corresponding stage (subtransient, transient, and steady-state).
+1 p.u.-
+1 p.u.-
+1 p.u.-
j0.15 j0.10
j0.25 j0.10
j0.80 j0.10
If ”
If ’
If
Subtransient solution:
Transient solution:
Steady-state solution:
@ Salvador Acevedo, 2000
Short-Circuit Currents for the Example
Since the generator is not supplying any current, we assume Ed=Ed’=Ed”=100%
( )
For each circuit, we find the short - circuit current as:
Subtransient: If ' '=1
j0.25 p.u.
Transient: If '=1
j0.35 p.u.
Steady - state: If =1
j0.90 p.u.
Generator currents are found using the base current:
Ibase1=50000
2091.85 amperes
Generator current magnitudes are:Subtransient: If ' '= 4.000 x Ibase1 = 8367 amperesTransient: If '= 2.857 x Ibase1 = 5976 amperesSteady - state: If = 1.111 x Ibase1 = 2324 amperes
= −
= −
= −
=
j
j
j
4
2 857
1111
3 138
.
.
.
@ Salvador Acevedo, 2000
Loaded Machine under Fault Conditions
Three-phaseshort-circuit
GENERATOR50 MVA, 13.8 KV
xd"=15%xd'=25%xd=80%ra=2%
TRANSFORMER50 MVA, 13.8-69 KV
x=10%r=1%
P
Load50MVA69KV
pf=0.9 (-)
Assume a three-phase short-circuit occurs at point ‘P’. To evaluate a fault during the subtransient or transient period, we need to know the pre-fault current value IL.
+Ed-
ra jXd rt jXt
Z
P
IL Switch'S'
Switch ‘S’ is normally open. Close it to simulate a fault at point ‘P’.
Impedance diagram for the circuit before fault:
@ Salvador Acevedo, 2000
Pre-Fault Conditions (subtransient)
The load current (pre-fault current) will help us determine the internal voltage for the subtransient, before the fault:
+Ed"
-
ra jXd" rt jXt
Z
P
IL +Vf-
+Vt-
E d V t jXd I V f j X t X d IL L' ' ' ' ( ' ' )≅ + = + +
+Ed"
-
jXd" jXt
Z
+Vt-
P
IL If "+Vf-
Ed Vt ra jXd I V f ra rt j X t Xd IL L' ' ( ' ' ) [( ) ( ' ' )]= + + = + + + +
Neglecting transformer and generator resistances:
Before the fault, If ”=0.
@ Salvador Acevedo, 2000
Pre-Fault Conditions (transient)
E d V t j X d I V f j X t X d IL L' ' ( ' )≅ + = + +
+Ed'-
jXd' jXt
Z
P
IL If '+Vf-
+Vt-
The corresponding internal voltage Ed’ for the transient is:
Fault Current
Before the fault, If ’=0.
To simulate the fault, switch ‘S’ is now closed.
Subtransient short - circuit current:
Transient short - circuit current:
Steady - state short - circuit current:
IfEd
j Xd Xt
IfEd
j Xd Xt
IfEd
j Xd Xt
""
( " )
''
( ' )
( )
=+
=+
=+
@ Salvador Acevedo, 2000
Multi-Machine System
Three-phaseshort-circuit
GENERATOR SynchronousMOTOR
P
TRANSFORMER
+Eg-
jXg jXt P
IL IL+
Em-
jXm
+Vf-
Equivalent circuit before the fault:
+Eg'-
jXg' jXt P
IL IL+
Em'-
jXm'
+Vf-
+Eg"
-
jXg" jXt P
IL IL+
Em"-
jXm"
+Vf-
Steady-state
Transient
Subtransient
@ Salvador Acevedo, 2000
Subtransient Short-Circuit Solution
1. Evaluate subtransient internal voltages for generator and motor under the operating conditions (switch ‘S’ open).
Eg Vf j Xt Xg ILEm Vf jXm IL
" ( " )" "= + += −
2. Close switch ‘S’, and find the current contribution from generator and motor to the fault.
+Eg"
-
jXg" jXt P
Ig" Im" +Em"
-
jXm"
If "
+Eg"
-
jXg" jXt P
IL IL+
Em"-
jXm"
+Vf-S
If "=0
IgEg
j Xt Xg"
EmjXm
If Ig
""
( " )Im
""
" " Im"
=+
=
= +
@ Salvador Acevedo, 2000
Thévenin Equivalent Method
Same multi-machine example (2 machines).Combining steps 1 and 2.
IgEg
j Xt XgVf j Xt Xg IL
j Xt Xg
IgVf
j Xt XgIL
"EmjXm
Vf jXm ILjXm
VfjXm
IL
If IgVf
j Xt XgIL
VfjXm
IL
IfVf
j Xt XgVf
jXm
""
( " )( " )
( " )
"( " )
Im""
""
Im""
" " Im"( " ) "
"( " ) "
=+
=+ +
+
=+
+
= =−
= −
= + =+
+ + −
=+
+
Igf ’’ (lets name this term: Igf ”)
{{
Imf ’’ (lets name this term: Imf ”)
Igf ’’
{ {
Imf ’’
@ Salvador Acevedo, 2000
Thévenin Equivalent (continued)
IfVf
j Xt XgVf
jXm"
( " ) "=
++
Igf ’’ Imf ’’
{ {
This expression can be represented in the following circuit:
IfVf
j Xt XgVf
jXmVf
j Xt Xg jXm
If VfjXtg jXm
IfVfZth
Zth jXtg XmXtg Xm
"( " ) " ( ' ' ) ' '
' '"
"( )( ' ' )
' '
=+
+ =+
+
= +
= =+
1 1
1 1where: X tg = Xt + Xg"
where:
Vf is the pre-fault voltage at the fault point (Thévenin voltage)Zth is the Thévenin Impedance seen from the fault point.If ’’ is the subtransient fault current.
jXg" jXt P
Igf " Imf "
jXm"
If "
-Vf+
+Vf-
If "
j(Xg"+Xt)
jXm"
Igf "
Imf "
@ Salvador Acevedo, 2000
Thévenin Equivalent (continued)
Remember that the total subtransient generator current is
Ig ” = Igf ”+IL
and the subtransient motor current is
Im” = Imf ” - IL
The same problem can be solved applying superposition. The Pre-fault Solution plus the Thévenin Equivalent Solution
+Eg"
-
jXg" jXt P
Ig" Im" +Em"
-
jXm"
If "+Vf-+
-Vf-
Vf is the pre-fault voltage at ‘P’ and the short-circuit is represented by two opposed Vf sources connected in series.
@ Salvador Acevedo, 2000
Superposition and Thévenin
2. The contribution to the fault is obtained with -Vf only.This will make
Ig”2=Igf ”, Im”2=Imf ”, If ”=Igf ”+Imf ”.(Here is where we use the Thévenin equivalent)
+Eg"
-
jXg" jXt
Ig"1=IL Im"1=-IL +Em"
-
jXm"
If "=0
+Vf-
IL
To obtain the total solution, we apply superposition: 1. The pre-fault solution is obtained with Eg”, Em” and Vf.
This will makeIg”1=IL, Im”1=-IL, If ”=0.
3. Add steps 1 and 2. This will give the total fault currentsIg”=Igf ”+IL, Im”=Imf ”-IL, If”=Igf ”+Imf ”.
+Eg"=0
-
jXg" jXt
Ig"2=Igf" Im"2=Imf " +Em"=0
-
jXm"
If "=Igf "+Imf "
+-Vf-
-Vf+
@ Salvador Acevedo, 2000
Summary of Fault Analysis Using Thévenin Method
à Locate the fault point ‘P’.à Represent the system in admittance form.
ð Convert synchronous machines to their Norton equivalents.
ð Build admittance matrix for nodal analysis [YBUS].
G1
G2
G3
Gi
Generators,Transformers,
Loads,Transmission
Lines, etc.
...
Load 1Load 2Load 3..Load j
FAULTED BUS ‘P’
If+Vf-
If switch open If = 0If switch closed Vf = 0
@ Salvador Acevedo, 2000
Step 1
à Find the pre-fault operating conditions.ð Use steady-state values.ð Name Vf the pre-fault voltage at point ‘P’.ð With nodal analysis or Load Flow analysis obtain
prefault voltages V1o, V2
o, V3o….Vf
ð Calculate currents Ig1, Ig2,… Iline1, Iline2, …...
+Vf = Vpre-fault-
→
→
→
→
System matrix[Ybus]
According to the method used todetermine the
pre-fault operatingconditions
...
Load 1Load 2Load 3..Load m
‘P’
If = 0
J1
J2
J3
Ji
If = Fault current
@ Salvador Acevedo, 2000
Step 1 (continued)
+Vf = Vpre-fault voltage-
→
→
→
→
System matrix[Ybus]...
Load 1Load 2Load 3..Load m
‘P’
If = 0
J1
J2
J3
Ji
[ ]Y V J V Y J Z J
V
V
V f
V
Y Y Y YY Y Y Y
Y Y Y Y
Y Y Y Y
JJ
Jp
Jn
B U S B U S B U S
o
o
n
p n
p n
p p p p p n
n n n p n n
= = =
=
=
−
−
1
1
2
0
1 1 1 2 1 1
2 1 2 2 2 2
1 2
1 2
12
0.
.
. . . . . .
. . . . . ..
. . . . . ..
. . . . . .
.
.
1
=
=
=
V
V
V f
V
Z Z Z ZZ Z Z Z
Z Z Z Z
Z Z Z Z
JJ
Jp
Jn
o
o
n
p n
p n
p p p p p n
n n n p n n
1
2
0
1 1 1 2 1 1
2 1 2 2 2 2
1 2
1 2
12
0.
.
. . . . . .
. . . . . ..
. . . . . ..
. . . . . .
.
.
This term is zerobecause before the faultthere is no fault current(switch is open)
@ Salvador Acevedo, 2000
Step 2
à Find the Thévenin contribution to the fault.ð Set all sources to zero (including synchronous motors
internal sources).ð Use the subtransient, transient or steady-state
impedances depending on the solution desired.ð Apply a voltage source ‘-Vf ’ at point ‘P’ and solve
the network with this source only. This source injects current ‘-If ’ into faulted node.o This will give the fault current If and all changes
in voltages and currents needed.àName voltage changes ∆V1, ∆V2, ∆V3…àName current changes ∆Ig1, ∆Ig2,….,
∆Iline1, ∆Iline2…
Machine Impedances(for Subtransient, Transient,
or Steady-state),Transformers,
and TransmissionLines
...
Load 1Load 2Load 3..Load m
If ’’If ’If
-Vf+
‘P’
J1=0
J2=0
J3=0
Ji=0
@ Salvador Acevedo, 2000
Step 2 (continued)
Machine Impedances (for Subtransient, Transient,
or Steady-state),Transformers,
and TransmissionLines
...
Load 1Load 2Load 3..Load m
If ’’If ’If
-Vf+
‘P’
J1 = 0
J2 = 0
J3 = 0
Ji = 0
∆∆
∆
∆
∆∆
∆
VV
Vp
Vn
VV
Vf
Vn
Z Z Z ZZ Z Z Z
Z Z Z Z
Z Z Z Z
p n
p n
p p pp pn
n n np nn
12
12
00
11 12 1 1
21 22 2 2
1 2
1 2
.
.
.
.
... ...
... ....
... ....
... ...
.
=−
=
− If.0
This matrix ZBUS is formed using the appropriate impedances (subtransient, transient or steady-state) to form YBUSbefore inverting.
This term equals -Vf because -Vf is the voltagethat we need to add tothe prefault voltage Vfto have a zero voltageat point ‘P’
@ Salvador Acevedo, 2000
Step 2 (continued)
[ ]
The last equations can be sim plified to:
or
from w h ich
where is the fault current
is the pre - fault voltage at poi
∆∆
∆
∆
∆
∆
∆
VV
Vf
Vn
ZZ
Z
Z
If
V Z If
V Z If
Vp Vf Z If
Vn Z If
V Z I IV
Z
I
V
p
p
pp
np
p
p
pp
np
f pp f ff
pp
f
f
12
1
2
1
2
1
2
.
.
.
.
,
,
. . . . .,
. . . .
−
=
−
= −
= −
= − = −
= −
= =
nt ' P '
is the Thévenin Im pedanceZ pp
@ Salvador Acevedo, 2000
Step 3
à Add solutions for steps 1 and 2.ð This is equivalent to solving the network with
the switch closed.
à Voltages during the fault areV1
f=V10+∆V1, V2
f=V2+∆V2, V3f=V3+∆V3, …
à at the fault Vp=Vf +(-Vf)=0à Currents during the fault are
Ig1+∆Ig1, Ig2+∆Ig2…Iline1+∆Iline1, Iline2+∆Iline2….
G1
G2
G3
Gi
System matrix[Ybus]...
Load 1Load 2Load 3..Load m
If+Vf = 0-
‘P’
@ Salvador Acevedo, 2000
Study Case: General Solution for Symmetrical Faults
The power system shown operates under steady-state
conditions with Eg1=1 ∠0° p.u. and Eg2=0.9 ∠30° p.u. when a solid three-phase fault occurs at node 2. Obtain the transient short-circuit currents in lines, generators and transformers. Evaluate the transient node voltages V1
f, V2f and V3
f during the fault (transient period).
GENERATOR-1Xd=85%Xd'=25%Xd"=10%
ra=1%
TRANSFORMER-1Y-Y
Zt=0.01+j0.15 p.u.
TRANSFORMER-2Y-Y
Zt=0.01+j0.20 p.u.
GENERATOR-2Xd=120%Xd'=40%Xd"=20%
ra=2%
Transmission line 1-2Z=0.03+j0.4p.u.
Transmission line 1-3Z=0.05+j0.5 p.u.
LOADR=10 p.u.
Transmission line 2-3Z=0.05+j0.5 p.u.
1 2
3
Fault
@ Salvador Acevedo, 2000
Step 1. Pre-Fault Solution
+E1-
zLoad
+E2-
zgt1
z12
z13
zgt2
z23
i12
i13 i23
iG1 iG2
+
V1
-
+
V2
-
+V3-
I1 I2ygt1
y12
y13
ygt2
y23
zgt1=zg1+zt1=0.02+j1z12=0.03+j0.4z13=z23=0.05+j0.5zgt2=zg2+zt2=0.03+j1.4zLoad=10E1=1
E2=0.9∠30°
ygt1=0.02-j0.9996y12=0.187-j2.486y13=y23=0.198-j1.980ygt2=0.0153-j0.714yload=0.1I1=0.02-j0.9996I2=0.333-j0.549
Impedance diagram (all values in p.u.)
Admittance diagram (all values in p.u.)
To solve it, we use nodal analysis.
Y Y YY Y YY Y Y
V1V2V3
J1J2J3
-0.187 + j2.486 -0.198 + j1.980-0.187 + j2.486 0.399 - j5.180 -0.198 + j1.980
- 0.198 + j1.980 - 0.198 + j1.980
V1V2V3
11 12 13
21 22 23
31 32 33
=
−
−
=−−
0 405 5466
0 496 3960
0 02 0 99960 333 0549
0
. .
. .
. .
. .j
j
jj
=∠ °∠ °∠ °
V1V2V3
0 929 710 916 10 20 920 7 2
. .
. .
. .
@ Salvador Acevedo, 2000
Pre-Fault Currents
Line Currents:Line 1-2I12=y12 (V1-V2)
I12 =(0.187-j2.486)(0.929∠ 7.1°- 0.916∠ 10.2°)
I12 = 0.129∠ −151.9° p.u.
Line 1-3I13=y13 (V1-V3)
I13 =(0.198-j1.980)(0.929∠ 7.1°- 0.920∠ 7.2°)
I13 = 0.019∠ −87.4° p.u.
Line 2-3I23=y23 (V2-V3)
I23 =(0.198-j1.980)(0.916∠ 10.2°- 0.920∠ 7.2°)
I23 = 0.096∠ −18.9° p.u.
Generator Currents:Generator 1
IG1= I12 + I13 = 0.129∠ −151.9° + 0.019∠ −87.4°
IG1= 0.138 ∠ −144.6°
Generator 2
IG2= -I12 + I23 = -0.129∠ −151.9° + 0.096∠ −18.9°
IG2= 0.224 ∠ −24.2°
@ Salvador Acevedo, 2000
Power Balance (Pre-Fault)
To verify the solution, a real Power Balance is now calculated (as an exercise):
Generated PowerGenerator 1 + Transformer 1:
SG1=V1 IG1*= (0.929∠ 7.1° )(0.138 ∠ 144.6° )SG1= -0.113 + j 0.061PG1= - 0.113 (where the minus sign means this generator
absorbs P=0.113 p.u. and therefore is acting as a motor)
QG1=0.061
Generator 2 + Transformer 2:
SG2=V2 IG2*= (0.916∠ 10.2°)(0.224 ∠ 24.2° )SG2=0.1986 - 0.0495iPG2= 0.1986QG2= - 0.0495 (this generator absorbs Q = 0.0495 p.u. and
still generates P = 0.1987 p.u., therefore this machine acts as a generator)
@ Salvador Acevedo, 2000
Power Balance (continued)
Absorbed PowerLoad:PLoad=(V3)2/Rload=(0.920) 2*0.1=0.0846 p.u.
Real Power dissipated in lines:line 1-2: P=I12
2 * Rline12=(0.129) 2(0.03)=0.0005
line 1-3: P=I132 * Rline13=(0.019) 2(0.05)=0.00002
line 2-3: P=I232 * Rline23=(0.096) 2(0.05)=0.00046
The power balance is:
Pgenerated=Pabsorbed
PG2=PLoad+Pline12+Pline13+Pline23+Pabsorbed-G1
0.1986=0.0846+0.0005+0.00002+0.0004+0.113=0.1986
Note: Nodal analysis has been used to find the operating conditions of the system before the fault. In practice, a Load-flow solution would have been used instead.
@ Salvador Acevedo, 2000
Step 2. Fault at Bus 2 (Thévenin Contribution)
To simulate a Fault at Bus 2, we will add the pre-fault response to the Thévenin Contribution.
We use a source equal to the pre-fault voltage at point 2 and set all the original sources to zero. To solve this network for the transient period, we require the use of transient values for the machines impedances.
The machine impedances for the transient period are:zg1’=0.01 + j 0.25, zg2’=0.03 + j0.4
Including transformers:zgt1’=(0.01+0.01)+j(0.25+0.15)=0.02 + j 0.40zgt2’=(0.02+0.01)+j(0.40+0.20)=0.03 + j 0.60
The matrix [YBUS] and its inverse [ZBUS] become:
[ ]
Yj
j
Z Yj
j
BUS
BUS BUS
=−
−
= =+
+
−
0 509 6 960
0 496 3 960
0 020 0 275
0 061 0 487
1
. .
. .
. .
. .
- 0.187 + j2.486 - 0.198 + j1.980- 0.187 + j2.486 0.468 - j6.129 - 0.198 + j1.980
- 0.198 + j1.980 - 0.198 + j1.980
0.014 + j0.186 0.023 + j0.2290.014 + j0.186 0.024 + j0.319 0.025 + j0.2510.023 + j0.229 0.025 + j0.251
@ Salvador Acevedo, 2000
Thévenin Contribution (step 2)
∆∆∆
∆
∆
∆
∆
∆∆
VVV
VVfV
Z If
VVfV
Z Z ZZ Z ZZ Z Z
If
V Z Z If Z Z IfV Vf Z
BUS
123
1
3
0
0
1
3
0
0
1 0 02
11 12 13
21 22 23
31 32 33
11 12 13 12
= −
= −
−
=
−
= ⋅ + − + ⋅ = −= − =
from where:( )
12 22 23 22
31 32 33 32
22 22
22
0 03 0 0
⋅ + − + ⋅ = −= ⋅ + − + ⋅ = −
=−−
= =
=
( )( )
Z If Z Z IfV Z Z If Z Z If
IfVfZ
VfZ
VfZ
Z Z
Thev
Thev
∆
* Note that only elements from column ' P' are needed.
From the second equation:
where is the Thevenin impedance for a fault at node 2.
Fault current and changes in voltages are now obtained inin the following way:
@ Salvador Acevedo, 2000
Thévenin Contribution (step 2)
Fault current and changes in voltages are now calculated.The Thévenin Impedance for a fault at bus 2 is:
Z = Z = 0.024 + j0.319
Using the pre - fault voltage at node 2:V = V = 0.916 10.2
we find the fault current:
I'VZ
0.916 10.2 0.024 + j0.319
0.916 10.2 0.320 85.6
The voltage changes at the other nodes are found from:V = Z (-I' ) = (0.014 + j0.186)(V = Z (-I' ) = (0.024 + j0.319)(
22 Thev
f 2
ff
22
1 12 f
2 22 f
∠ °
= =∠ °
=∠ °∠ °
= ∠ − °
∠ − °+ °∠ − °+ °
2 862 754
2 862 754 1802 862 754 180
. .
. . ). . )
∆∆∆V = Z (-I' ) = (0.025+ j0.251)(
V = Z (-I' )V = Z (-I' ) =V = Z (-I' ) =
3 32 f
1 12 f
2 22 f
3 32 f
2 862 754 180
0533 169 70 916 169 80 723 1712
. . )
. .. .. .
∠ − °+ °
= ∠ − °∠ − °= −∠ − °
∆∆∆
Vf
@ Salvador Acevedo, 2000
Step 3. Fault Conditions
Adding results from steps 1 and 2, we obtain the faulted voltagesat each node:
V V + V = 0.929 7.1 +0.533 -169.7 = 0.398 2.7
V V + V = 0.916 10.2 +0.916 -169.8 = 0
V V + V = 0.920 7.2 +0.723 -171.2 = 0.199 1.3
The current contributions from the lines during the fault are:I' V V 0.398 2.7I' V
f 01
f 02
3f 0
3
12f 1f 2f
32f
1
2
3
1
2
12
32
0187 2 486 0 0 992 82 9
= ∠ ° ∠ ° ∠ °
= ∠ ° ∠ °
= ∠ ° ∠ ° ∠ °
= − = − ∠ °− = ∠ − °=
∆
∆
∆
y jy
( ) ( . . )( ) . .( 3f 2f
g2f f 12f 32f
V 0.398 2.7
The generator contribution is found by Kirchhoff Currents Lawat the faulted node:I' I' I' I'
− = − ∠ °− = ∠ − °
= − − = ∠ − °
) ( . . )( ) . .
. .
0198 1980 0 0 395 83 0
1497 68 5
j
1
2
3
I12f '
I32f 'Ig2f '
If '=2.862∠−75.4°
All quantities have been calculated in per unit.Results are for phase ‘a’.