Balanced (Symmetrical) Faults - University of Waterlooraelshat/COURSES_dr/eleb7.old...– To adjust...
Transcript of Balanced (Symmetrical) Faults - University of Waterlooraelshat/COURSES_dr/eleb7.old...– To adjust...
ELE B7 Power Systems ELE B7 Power Systems Engineering Engineering
Balanced (Symmetrical) FaultsBalanced (Symmetrical) Faults
Fault Analysis• Fault?
When the insulation of the system breaks down or a conducting object comes in touch with a live point, a short circuit or a fault occurs.
• Sources?• This breakdown can be due to a variety of different factors
• lightning• wires blowing together in the wind• animals or plants coming in contact with the wires• salt spray or pollution on insulators
Fault Types• There are two classes of faults
– Balanced faults (symmetric faults): A Fault involving all the three phases
• system remains balanced; • these faults are relatively rare, but are the easiest to analyze so
we’ll consider them first.– Unbalanced faults (unsymmetric faults): A fault involving only one
or two phases• The majority of the faults are unsymmetrical. • system is no longer balanced; • very common, but more difficult to analyze
• The most common type of fault on a three phase system is the single line-to-ground (SLG), followed by the line-to- line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults
Fault Types
IFb
Transformer
IFa
IFc
IF
Transformer
IF
Single Line to Ground Fault
IF
IF
IF
Transformer IF
Line to Line Fault
IF
IFb
Transformer
IFIF
IF
IFc
Double line to ground Fault
Three Phases Fault (Balanced Fault)
Fault Analysis• Fault Analysis involve finding the voltage and current
distribution throughout the system during fault condition.• Fault currents cause equipment damage due to both
thermal and mechanical processes• Why do we need that?
– To adjust and set the protective devices so we can detect any fault and isolate the faulty portion of the system.
– To protect the human being and the equipment during the abnormal operating conditions.
– need to determine the maximum current to insure devices can survive the fault
Symmetrical Faults (Balanced Faults)
• It is the most severe fault but can be easily calculated
• It is an important type of fault
• The circuit breaker rated MVA breaking capacity is selected based on the three phase short circuit MVA.
A Fault involving all the three phases.
Transformer
Three Phase
Short Circuit
Symmetrical Faults (Balanced Faults)
• It is the most severe fault but can be easily calculated
• It is an important type of fault
• The circuit breaker rated MVA breaking capacity is selected based on the three phase short circuit MVA.
A Fault involving all the three phases.
Transformer
Three Phase
Short Circuit
1. Synchronous machines are represented by a constant voltage sources behind subtransient reactances. The EMF of all generators are 1 per unit making zero angle.
oV 01
The following assumptions are made in three phase fault calculation :
This means that the system voltage is at its nominal value and the system is operating at no-load conditions at the time of fault.
Therefore, All the generators can be replaced by a single generator since all EMFs are equal and are in phase.
2. Transformers are represented by their leakage reactances
Winding resistances, shunt admittances and Δ-Y phase shifts are neglected
3. The shunt capacitances of the transmission
Lines
are neglected.
4. The system resistances are neglected
and only
the inductive reactance of different elements are
taken into account.
Symmetrical
Fault Calculation
Transformer
Three Phase
Short Circuit
Simple CircuitsLoad Ignored
Use Thevenin's Equivalent
Simple CircuitsLoad not Ignored
Use Thevenin's EquivalentFind Pre-fault voltage
Large CircuitsConstruct
The Bus Impedance Matrix
Three Phase FaultCalculation
2
3
1
1. Simple Circuits and Load is ignored
1. Draw a single line diagram for the system.
The Calculations for the three phase fault are easy because the
circuit is completely symmetrical and calculations can be done
for only one phase.
T1
T2
G1
G2
1
2
3 phasefault e
Steps For Calculating Symmetrical Faults:
2. Select a common base and find out the per unit reactances of all generators, transformers, transmission lines, etc.
4. Reduce the single line reactance diagram by using series, parallel and Delta-Why transformations keeping the identity of the fault point intact. Find the total reactance of the system as seen from the fault point (Using Thevenin’s
Theorem)
3. From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram.
GG15.0j
20.0j 16.0j
10.0j23.0j
GG
01
01 23.0j
5. Find the fault current and the fault MVA in per unit.
Convert the per unit values to actual values.
6. Retrace the steps to calculate the voltages and the currents throughout different parts of the power system
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA, 11 kV10% Reactance
100 MVA, 11 kV15% Reactance
100 MVA, 11/132 kV10% Reactance
50 MVA, 11/132 kV8% Reactance
200 km
3 phasefault e
Example:
A three phase fault occurs in the system as shown in the Figure.Find the total fault current, the fault level and fault current supplied by each generator.
Step 1: Draw a single line diagram for the system.
The single line diagram for the system is given in the example as shown.
Solution:
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA, 11 kV10% Reactance
100 MVA, 11 kV15% Reactance
100 MVA, 11/132 kV10% Reactance
50 MVA, 11/132 kV8% Reactance
200 km
3 phasefault e
Step 2: Select a common base and find the per unit reactances of all generators, transformers, etc.
Select the common base as:
100 MVA (100,000 kVA)
11 kV for Transformer low voltage side (LV)
132 kV for Transformer high voltage side (HV)
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA, 11 kV10% Reactance
100 MVA, 11 kV15% Reactance
100 MVA, 11/132 kV10% Reactance
50 MVA, 11/132 kV8% Reactance
200 km
3 phasefault e
kVVBase 11
kVVBase 132
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA, 11 kV10% Reactance
100 MVA, 11 kV15% Reactance
100 MVA, 11/132 kV10% Reactance
50 MVA, 11/132 kV8% Reactance
200 km
3 phasefault e
kVVBase 11
kVVBase 132
G1
G2
15.01 jXG The per unit reactance
2.050
100*1.02 jjXG The per unit reactance
1.01 jXT The per unit reactance
16.050
100*08.02 jjXT The per unit reactance
23.0)132(
100)200*2.0( 2 jkVMVAj
ZX
XBase
LINELINE The per unit reactance
T1
T2
TL
From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram.
T1
T2
G1
G2
1
2
3 phasefault e
G15.0j
20.0j 16.0j
10.0j23.0j
G
01
01 23.0j
G25.0j
36.0jG
01
01115.0j
Find the total impedance
(reactance) of the system as seen from the fault side.
25.0j
36.0j115.0j
pujX
jjjjjX
Total
Total
2625.0
115.036.025.036.0*25.0
pujj
IF 8095.32625.0
01
AI sidekVBase 4.437132*31000*100)( 132
AjI oActualF 9027.16664.437*8095.3)(
01
2625.0j
LLBase
BaseLbase
V
SI
||3
|||| 3
BasepuFActualF III *)()(
The Fault Level is:
100)( BaseMVA
95.380100*8095.3)( LevelActualMVA
puMVA Level 8095.3)(
T1
T2
G1
G2
X=0.2 ohm/phase/km
1
2
50 MVA, 11 kV10% Reactance
100 MVA, 11 kV15% Reactance
100 MVA, 11/132 kV10% Reactance
50 MVA, 11/132 kV8% Reactance
200 km
3 phasefault e
1)( GFI
2)( GFI
FI
kVVBase 132kVVBase 11
The Fault Current at 11 kV side supplied by the two generators is:
AI sidekVBase 8.524811*3
1000*100)( 11
AjI oActualF 90199958.5248*8095.3)(
At 11 kV Side:
G15.0j
20.0j 16.0j
10.0j23.0j
G
01
01 23.0j
TotalFI )(
1)( GFI
2)( GFI
kVVBase 132kVVBase 11
111,2 )()()( GFkVTFGF III
Ajj
jI GF 903.1180025.036.0
36.0)9019995()( 1
AI GF 907.8194)( 2
G25.0j
36.0jG
01
01115.0j
T1
T2
G1
G2
1
2
1)( GFI
2)( GFI
1)( GFI
2)( GFI
TotalFI )(
Ajj
jII kVTFGF 903.1180025.036.0
36.0)()( 11,1
Example:Consider the single-line diagram of a power system shown below. The transient reactance of each part of the system is as shown and expressed in pu on a common 100 MVA base.
G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
Assuming that all generators are working on the rated voltages, when a three-phase fault with impedance of j0.16 pu occurs at bus 5. Find: The fault currents and buses voltages.
G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 516.0j
2.0j4.0j
4.0j
8.0j
4.0j
Bus 3 Bus 4
Bus 516.0j
5.016.034.0 jjjZTH
pu0.2j5.0j
1)I(0
5BusF
pujjj
jII FGF 2.14.06.0
6.0)( 1
pujjj
jII FGF 8.04.06.0
4.0)( 2
16.0j
Bus 5
34.0j
FI
16.0jBus 5
34.0j
FI
2.0j 4.0j
16.0j
Bus 3 Bus 4
Bus 5
2.0j 2.0j
1.0j
FI
1FGI 2FGI
16.0jBus 5
6.0j
1.0j
4.0j
FI
2FGI1FGI
0V&01V:VoltagesPrefault
Groundo
BUS
pu
VprefaultVVoltageFault
o 76.024.001
:
33
pujj
jIVisFaultDuringVariationVoltage
G
24.0)2.1()2.0(0
)2.0(0:
13
68.0)0.2)(16.0(01
:
5
jjV
FaultduringVariationVoltageo
pu
VprefaultVVoltageFault
o 32.068.001
:
55
NOTE: If the fault impedance is zero.
0.1)I)(0.0j(01V:FaultduringVariationVoltage
Fo
5 pu0.00.101VprefaultV o55 and
1GI 2GI
FI
o01
2.0j 4.0j
16.0j
Bus 3 Bus 4
Bus 5
2.0j 2.0j
1.0j+_
5V
3V
G1 G22.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
Problem:For the network shown in the figure, find the current flowing between buses 3 and 4 during symmetrical three phase fault at bus 5 as in the previous example.
16.0j
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 5
pu
jjprefaultVprefaultV
pu
VprefaultVVoltageFaultDuring
o
o
68.032.001
)8.0)(4.0(0
76.024.001
:
44
33
pujjj
jII FGF 2.14.06.0
6.0)( 1
pujjj
jII FGF 8.04.06.0
4.0)( 2
pujj
VVIF 1.0
8.0)( 4334
4.0)( 5335 j
VVIF
4.0)( 5445 j
VVIF
and
2.0j
2.0j
4.0j
8.0j
4.0j
1.0j
1.0j
Bus 1
Bus 3 Bus 4
Bus 2
Bus 516.0j
1GI2GI
34F )I(
45F )I(35F )I(
4V
5V
Symmetrical
Fault Calculation
Transformer
Three Phase
Short Circuit
Three Phase FaultCalculation
Simple CircuitsLoad Ignored
Use Thevenin's Equivalent
Simple CircuitsLoad not Ignored
Use Thevenin's EquivalentFind Pre-fault voltage
Large CircuitsConstruct
The Bus Impedance Matrix
2
2. Symmetrical Faults Considering the load current
Generally, the fault currents are much larger than the load currents. Therefore, the load current can be neglected during fault calculations.
The pre-fault voltage and the ZTH are used in calculating the fault current.
The fault current represents two components:1. The Load Current2. The Short Circuit Current
There are some cases where considering the load current is an essential factor in fault calculations. Superposition technique is proposed for such cases to compute the fault current.
For such case, it is necessary to compute the terminal voltage at fault location before fault takes place. This terminal voltage is known as the pre-fault voltage.
Connecting the load to the system
causes the current to flow in the network. Voltage drop due to system impedance cause the voltage magnitude at different buses to be deviated from 1.0 pu.
Consider a load (synchronous Motor) connected to a synchronous generator through a transmission line. The circuit model of the network could be represented as shown in the Figure
MIGI
jXG jXM
G M
jXL
FI
G M
FMIFGI jXL
jXG jXM
Normal Operation Condition:
LMG III
Fault condition at Load terminal:
LGFG III )(
LMFM III )(
MFGFF III )()(
During fault condition in the system, the generator as well as the load (synchronous motor) will supply the faulted terminals with power from the energy stored in their windings.
The Motor is drawing 40 MW at 0.8 pf. Leading with terminal
voltage of 10.95 kV.
Calculate the total current in the generator and motor during 3
phase s.c.
Example
100 MVA, 11 kV Generator0.25 pusub-transient reactance
50 MVA, Motor0.2 pusub-transient reactance
0.05 pu. Reactance
On 100 MVA Base
3 phase fault
G M
AkV
MVACurretBase
kVkVBaseMVAMVABase
88.5248)11(3
10011
100
Solution:The network could be represented as shown in the Fig.
A4.2636)8.0)(kV95.10(3
MW40CurretLoadThe
pujIjI
I
L
L
L
)301.0402.0()6.08.0(502.0
)8.0(cos502.0 1
puI puL 502.088.52484.2636)(
Under Normal Operation Condition (Prefault condition)
leading0.8isfactorpowerThe
25.0j 4.0j
G M
05.0j MIGI
Consider the motor terminal voltage as a reference.
puvoltage)terminal(Motor pu 995.011
95.10
pu)0201.0j9799.0(V995.0)050.0j)(301.0j402.0(V
V)050.0j(IV
G
G
MLG
pu)(V opuM 0995.0
Using KVL, calculate the pre-fault voltage at the generator terminal.
Find the equivalent reactance as seen from the fault terminals.
pujXjj
jjX
TH
TH
1607.0)45.0()25.0(
)45.0)(25.0(
25.0j 4.0j
05.0j
THX
25.0j 4.0j
G M
05.0j
MV
Calculate the fault current
puj)(Ij
j)(I
XVoltaegefaultpre)(I
puF
puF
THpuF
)09.6125.0(1607.0
0201.09799.0
pu(I)
puj(I)jj(I)
I)(I(I)
oG
G
G
LGFG
4.8264.3
)609.3482.0()301.0402.0()91.308.0(
Fault Current from Generator Side
puj)(Ipujj)(I
)(I)(I)(I
MF
MF
GFpuFMF
)18.2045.0()91.308.0()09.6125.0(
Fault Current from Motor Side
Current from Generator Side:
puj)(Ij0.25j0.45
j0.45)(I)(I
GF
puFGF
)91.308.0(
pu(I)
puj(I)jj(I)
I)(I(I)
oM
M
M
LMFM
8.261505.2
)48.2357.0()301.0402.0()18.2045.0(
Current from Motor Side:
25.0j 4.0j
G M
05.0j
FI
G(I) M(I)
Consider the system shown in the Figure.
The two generators are supplying the buses 1, 2, 3
and 4.
Let the pre-fault voltage between bus 2 and ref. Bus
be (Vf ).
3 - Symmetrical Fault Calculation Using Bus Impedance Matrix
3
T1
T2
G1
G2
1
2
4fV
If a 3 phase short circuit occurs a bus 2.
A huge current will run in the system
during the fault condition.
"fI
3
2
1
4"fI
3
T1
T2
G1
G2
1 2
4
fV
3 phasefault e
If the pre-fault voltage between bus 2 and ref. bus is Vf , then the voltage during
fault is zero.
Then, the system can be analyzed using the reactance
circuit.
During the three-phase fault, the voltage between bus 2 and the reference becomes zero. Therefore, the 3 phase fault at bus 2 can be simulated by making the voltage between bus 2 and Ref. is equal to zero during the fault condition.
3
2
1
4
"fI
fVfV
This is simulated by inserting a source with a magnitude
equal to
Therefore, the voltage between bus 2 and reference
is zero and fault current runs in the system.
)( fV
This means that, the Fault current during the three-phase short circuit results from the voltage source
"fI
)( fV
Therefore, to compute the short circuit current, we have to find the current entering
bus 2 due to only )( fV
4
3
2
1
VVVV
4
3
1
VVVfV
BUSZ
44434241
34333231
24232221
14131211
ZZZZZZZZZZZZZZZZ
=
00I0
''f=
The Nodal Impedance Equation during the fault is:
Note:is the voltage difference between node i and the reference node
due to the current .iV
"fI
Ignoring all other currents entering other buses (ignore sources).
3
2
1
4
"fI
fV
Or
4
3
2
1
VVVV
4
3
1
VVVfV
''f42
''f32
''f22
''f12
IZIZIZIZ
= =
And from the second row, the fault current is:
22
''ZV
I ff
Substituting in the previous equation:
4
3
2
1
VVVV
4
3
1
VVVfV
f22
42
f22
32
f
f22
12
VZZ
VZZ
V
VZZ
= =
Assuming no load is connected in the system, no current will flow before fault and all bus voltages are the same and equal to the pre- fault voltage )V( f
Using Superposition, then during the fault:
4
3
2
1
VVVV
fV
22
4222
32
22
12
ZZ1
ZZ10ZZ1
= +
i)voltagefaultPre(f)voltageiBus(i VVV
iV is the voltage change in bus i due to fault current in bus 2.
4
3
2
1
VVVV
f
f
f
f
V
V
V
V
=
Example:
A three-phase fault occurs at bus 2 of the network shown in the Figure.
Determine:1. the sub-transient current during the fault, 2.the voltages at all the buses3.The current flow during the fault from buses 3-2, 1-2 and 4-2.
Consider the pre-fault voltage at bus 2 equal to 1 pu. and neglect the pre-fault currents.
3
T1
T2
G1
G2
1 2
4
fV
3 phasefault e
Solution:
Construct the reactance circuit under normal operation condition.
pu2.0j pu1.0j
pu125.0j
pu25.0j
pu4.0j
pu25.0j
up2.0j
3
2
1
4pu2.0j pu1.0j
Construct the bus impedance Matrix (ZBUS )
1954.0j1046.0j1506.0j1456.0j1046.0j1954.0j1494.0j1544.0j1506.0j1494.0j2295.0j1938.0j1456.0j1544.0j1938.0j2436.0j
ZBUS
NOTE:Constructing ZBus will be discussed later.
Since there are no-load currents, the pre-fault voltage at all the buses is equal to 1.0 pu. When the fault occurs at bus 2,
pu357.4j2295.0j0.1
ZV
I22
f''f
fV
22
4222
32
22
12
ZZ
1
ZZ
10ZZ
1
=
4
3
2
1
VVVV
fV
22
4222
32
22
12
ZZ
1
ZZ
10ZZ
1
=
4
3
2
1
VVVV
=
2295.0j1506.0j1
2295.0j1494.0j1
02295.0j1938.0j1
343.0349.00
155.0
=
1954.0j1046.0j1506.0j1456.0j1046.0j1954.0j1494.0j1544.0j1506.0j1494.0j2295.0j1938.0j1456.0j1544.0j1938.0j2436.0j
ZBUS
3
T1
T2
G1
G2
1 2
4
fV
3 phasefault e
The current flows from bus 1 to bus 2 is:
125.0jVVI 21
12
pu2448.1jI12
125.0j1556.0I12
pu2.0j pu1.0j
pu125.0j
pu25.0j
pu4.0j
pu25.0j
pu2.0j
3
2
1
4"fI
pu2.0j pu1.0j
The current flows from bus 3 to bus 2 is:
25.0jV3VI 2
32
pu396.1jI32
pu719.1jI42
The current flows from bus 4 to bus 2 is:
The current flows from bus 3 to bus 1 is:
25.0jVV
I 1331
pu7736.0j