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Digital Technique of fault study of power system
Pralay Roy1, Shubham kumar Gupta2
1Assistant professor, Dept. of Electrical Engineering, Siliguri Institute of Technology, sukna, Darjeeling 2Third year student, Dept. of Electrical Engineering, Siliguri Institute of Technology, sukna, Darjeeling
------------------------------------------------------------------***------------------------------------------------------------
Abstract -In huge complex power system networks, faults are inevitable. Itβs very important to analyze the fault level for avoiding the severe consequences and proving the conditions thus quality power supply need not to face any kind of discontinuity. The objective of this paper is to study the common fault types which are balanced and unbalanced fault of transmission line in power system using a systematic procedure of digital techniques and to perform the analysis and obtain the result from simulation on that type of fault using MATLAB. Key Words:Power system fault analysis, Asymmetrical fault, Symmetrical components, MATLAB, Z-parameters.
1.INTRODUCTION Power system works in normal balanced condition but whenever some fault arises it becomes unbalanced. Any unexpected disturbance in the normal working condition is termed as fault. Fault may arise due to various factors such as Lightening stroke, high wind which leads to falling of trees on line, wind and ice loading resulting in failure of insulators. Fault can be symmetrical or unsymmetrical, when the fault occurs at all the three phases at a time, it is said to be symmetrical fault and where one or more phases may be involved is unsymmetrical fault. Though the symmetric fault is rare, but this type of faults shows most severe effects to our power systems. Unsymmetrical fault can be treated as a regular problem of the power system and has to be analyzed more preciously thus the system can remain in its stable condition and can lead its operation without breaking continuity. The point at which the fault occurs behaves as sink point and the voltage at that point tends to become zero. Thus all the points have potential higher than faulty point starts to send current to these faulty point and thus the fault level rises to very higher magnitude than normal operating level. Therefore, it is important to determine the values of system voltage and current during fault conditions so that the protective devices may be set to detect the fault and isolate the faulty portion of the system.
1.1 Faults in Three phase system 1. Symmetrical three-phase fault
2. Single line-to-ground fault
3. Line-to-line fault
4. Double line-to-ground fault
Symmetrical fault is defined as the simultaneous short circuit across all the three phases. It is most infrequent fault but the most severe type of fault encountered. Symmetrical fault is the rarest one and calculation of fault level is much simpler, but in case of unsymmetrical symmetry faulted power system does not have three phase symmetry, so it cannot be solved by per phase analysis. To find fault current and voltage, it is first transformed into their symmetrical component, use of which help to reduce the complexity of transmission line and components are by large and symmetrical, although the fault may be unsymmetrical .This can be done by replacing three phase fault current by the sum of three phases zero sequence sources, a three phase positive sequence source, a three phase negative sequence.
1.2 BUS IMPEDENCE CALCULATION
Fig.1 shows an n bus passive linear network. The voltage of the ith bus (with respect to reference) is ππ and the current entering the ith bus isπΌπ . The knowledge of network theory tells us that this network can be described by
Reference
1
i
n
NETWORK
πΌπ
ππ
Fig. 1 General n port network
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πππ’π πΓ1 = πππ’π πΓπ πΌππ’π πΓ1
Where πππ’π is the bus voltage matrix,πΌππ’π is the bus current matrix and πππ’π is the bus impedance matrix. The general entry πππ of πππ’π can be obtained
πππ = ππ
πΌπ
πΌ1=πΌ2 .......=πΌπ=0
πΌπ β 0
When an existingπππ’π is added with new element having self-impedance ππ , a new bus may be created or a new bus may not be created. This leads to the addition of ππ in the following ways:
1. Injection ofππ from a new bus to reference. 2. Injection ofππ from a new bus to old bus. 3. Injection ofππ from an old bus to reference 4. Injection ofππ between two old buses.
TYPE 1 MODIFICATION Fig. 2 shows a branch ππ connected between reference bus and a new bus q. For this new branch we can write
ππ = π§π πΌπ
Or πππ’π πππ€ =
S
bus
Z
oldZ
0...0
0
:
0
TYPE 2 MODIFICTION
ππ = π§π πΌπ + ππ
= ππ1πΌ1 + ππ2πΌ2 + ......... + πππ πΌπ + ........ + πππ πΌπ + (πππ +π§π )πΌπ
The equation for π1can be written by considering that new current at kth bus isπΌπ + πΌπ .
π1= π11πΌ1 + π12πΌ2 + ......... + π1ππΌπ + ........ + π1ππΌπ +π1ππΌπ
The equations for π2, π3 β¦β¦ . ππ also get modified in a similar manner and is given by
Reference Fig. 2 Type 1 modification
Network
ππ ππ
πΌπ q
1
n
1
n
Network
q
πΌπ
πΌπ + πΌπ k
πΌπ
ππ
Fig. 3 Type 2 modification Reference
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TYPE 3 MODIFICATION The new set of the equation is
ππ πππ’π πππ€ = πππ’π πππ - 1
πππ +π§π
π1π
π2π:
πππ
ππ1 ππ2 β¦ . πππ
TYPE 4 MODIFICATION
ππΎ = ππ +ππ
Or =π§π πΌπ+ππ1πΌ1 + ππ2πΌ2+.......+πππ(πΌπ + πΌπ ) + πππ πΌπ +πππ (πΌπ β
πΌπ )+.........
Rearranging
0 = (ππ1 β ππ1)πΌ1 + ... + (πππ β πππ )πΌπ + (πππ β πππ )πΌπ +
(πππ β πππ )πΌπ +....+(π§π + πππ β πππ β πππ + πππ )
Equations π1, π2, π3 β¦β¦β¦ππ can be written in the
following form:
Or πππ’π πππ€ = πππ’π πππ -
1
ππ +πππ+πππ β2πππ
π1π β π1π
π2π β π2π::
πππ β πππ
(ππ1 β ππ1) β¦β¦ (πππ β πππ )
1.3 SEQUENCE MATRICES The sequence impedance matrices required for the short
circuit studies are:
π0βππ’π = zero sequence bus impedance matrix (nΓ
π),generally entry πππ0
π1βππ’π = positive sequence bus impedance matrix
(nΓ π),generally entry πππ1
π2βππ’π = negative sequence bus impedance matrix
(nΓ π),generally entry πππ2
The equations relating the sequence quantities are:
π0βππ’π = - π0βππ’π πΌ0βππ’π
π1βππ’π = πΈππ’π - π1βππ’π πΌ1βππ’π
Network
Reference
ππ πΌπ
1
n
k
Fig. 4 Type 3 modification
n
Network
ππ
i k
Reference Fig. 5 Type 4 modification
1
πΌπ
πΌπ
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π2βππ’π = - π2βππ’π πΌ2βππ’π
The prefault currents are neglected, vector E contains 1β 0
in all entries. The currents are all zero till the network is
terminated externally . At a time only one bus( i.e. faulted
bus k ) is terminated. Thus, only πΌπ0 , πΌπ1 , πΌπ2 have non-zero
entry
1.4 SHORT CIRCUIT STUDIES
Symmetrical fault :-For a symmetrical fault the negative sequence and zero sequence are absent, i.e.,π0βππ’π ,π2βππ’π ,πΌ0βππ’π and πΌ2βππ’π are zero.
ππ1=E-(ππ11πΌ11 + ππ21πΌ21 + β―+ πππ1πΌπ1 + β― +πππ1πΌπ1) (1)
But all the currents except the current at the faulted bus, i.e., πΌπ1are zero. Therefore,
ππ1 = E-πππ1πΌπ1 (2)
If ππ is the fault impedance
ππ1 =πΌπ1ππ (3)
From (2 and 3)
πΌπ1 = πΈ
πππ 1+ππ (4)
The voltage at ith bus is
ππ1 = E β Zik 1πΌπ1 = E (1- Zik 1
πππ 1+ππ) (5)
For i = 1, 2 ......... n
Single Line to Ground Fault
Fig. 7: - Single Line-to-ground fault.
Assuming that fault current πΌπ occurred on the phase with
fault impedance ππ . The boundary conditions are:
n
General zero
sequence
network
modified for
fault analysis
1
k
Reference
πΌ0π
π0π
Fig. 6(a) system zero
sequence network
πΌ1π
General positive
sequence network
modified for fault
analysis
N
Reference
1
n
k
π1π
Fig. 6(b) system positive
sequence network
General negative
sequence
network
modified for
fault analysis
1
k
n
Reference
πΌ2π
π2π
Fig. 6(c) system negative
sequence network
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πππ = πππΌπ
πΌπ = 0, πΌπΆ = 0
The symmetrical component of current are, withπΌπ=πΌπΆ=0 :
πΌπ0
πΌπ1
πΌπ2
= 1
3 1 1 11 π π2
1 π2 π
πΌπ = πΌπ00
πΌπ0=πΌπ1
=πΌπ2=πΌπ
3
πΌπ = πΌπ = πΌπ0+πΌπ1
+πΌπ2=
3ππ
π0+ π1+π2+3ππ
Where, π0, π1 , π2 is a zero, positive, negative sequence impedance.
Using Z-parameters analysis (Digital approach)
We know that
π°ππ = π°ππ = π°ππ = π For i = 1, 2 ...... n (6a)
iβ K (6b)
And πΌπ0 = πΌπ1 = πΌπ2
ππ0 + ππ1 + ππ2 = 3πππΌπ1 (7)
Also ππ0 = -πππ0πΌπ0
ππ1 = E-πππ1πΌπ1 (8)
ππ2= -πππ2πΌπ2
Combining equation (6, 7 and 8)
πΌπ1 = πΈ
πππ 0+πππ 1 +πππ 2+3ππ (9)
The sequence components of bus voltages at ith bus (i = 1, 2, 3 ..... n) are
ππ0= -πππ0πΌπ0 =βπππ0πΌπ1 (10a)
=βπππ0πΈ
πππ 0+πππ 1 +πππ 2+3ππ (10b)
ππ1 = E-πππ1πΌπ1 (11a)
= E [1-πππ1
πππ 0+πππ 1 +πππ 2+3ππ ] (11b)
= E [πππ 0+πππ 1 +πππ 2+3ππβπππ1
πππ 0+πππ 1 +πππ 2+3ππ ] (11c)
ππ2= -πππ2πΌπ2= -πππ2πΌπ1 (12a)
=βπππ2πΈ
πππ 0+πππ 1 +πππ 2+3ππ (12b)
Line to Line Fault
Fig.8: Line-to-Line fault
Line-to-Line fault takes place on phase b and c. The boundary conditions are:
πΌπ = 0
πΌπ+ πΌπΆ = 0
ππ= ππ
The symmetrical component of current are:
πΌπ0
πΌπ1
πΌπ2
= 1
3 1 1 11 π π2
1 π2 π
0 πΌπβπΌπ
πΌπ0
πΌπ1
πΌπ2
=
01
3(π β π2)πΌπ
β1
3(π β π2)πΌπ
Therefore, πΌπ1= - πΌπ2
Also ππ1 = ππ2
Or πΌπ1=
ππ
π1+π2
Where, π1 , π2 is positive, negative sequence impedances.
Using Z-parameters analysis (Digital approach)
For a line to line fault
π0βππ’π = πΌ0βππ’π = 0 (13)
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(6a and 8) are still valid
ππ1 = ππ2+ πΌπ1ππ (14)
Substituting the values for ππ1 and ππ2from (7) into (13)
E β πππ1πΌπ1 = -πππ2πΌπ2+ πΌπ1ππ (15)
We know that for a line to line fault πΌπ2 = -πΌπ1 . Making this substitution in (14)
E β πππ1πΌπ1 = πππ2πΌπ1 + πΌπ1ππ
Or πΌπ1 = πΈ
πππ 1+πππ 2+ππ (16)
The positive and negative sequence voltages at ith bus (i = 1, 2 ...... n) are
ππ1 = E-πππ1πΌπ1 (17a)
= E-πππ1πΈ
πππ 1+πππ 2+ππ (17b)
= E[πππ 1+πππ 2+ππβπππ1
πππ 1+πππ 2+ππ] (17c)
ππ2 = -πππ2πΌπ2 (18a)
= +πππ2πΌπ1 (18b)
= πππ2πΈ
πππ 1+πππ 2+ππ (18c)
Double Line to Ground Fault
Fig. 9: - Double-line-to-ground fault
The boundary conditions for Double Line-to-Ground:
πΌπ = 0
ππ= ππ = (πΌπ + πΌπ)ππ
The symmetrical components of voltage are:
ππ0
ππ1
ππ2
= 1
3 1 1 11 π π2
1 π2 π
ππππ
ππ
Also ππ1=ππ2
=ππ0-3πππΌπ0
And Since πΌπ = 0
Therefore, πΌπ0+ πΌπ1
+ πΌπ2= 0
Or πΌπ1=
ππ
π1+π2(π0+3ππ )
π2+π0+3ππ
Where,π0, π1, π2 is a Zero, positive, negative sequence impedance.
Using Z-parameters analysis (Digital approach)
We know ππ1 = ππ2 (19a)
ππ0 -ππ1 = 3πΌπ0ππ (19b)
And πΌπ0 + πΌπ1 + πΌπ2 = 0 (20)
Equations (6a and 8) are still valid. From (8, 19 and 20)
πΌπ1 = πΈβππ1
πππ 1 (21a)
πΌπ2= βππ2
πππ 2 = β
ππ1
πππ 2 (21b)
πΌπ0 = βππ0
πππ 0 =
βππ1
πππ 0+3ππ (21c)
Substituting (21) into (20) we have
βππ1
πππ 0+3ππ+
πΈβππ1
πππ 1β
ππ1
πππ 2 = 0
Or ππ1 =E(πππ 0+3ππ)πππ 2
πππ 1 πππ 2+πππ 0+3ππ +πππ 2( πππ 0+3ππ) (22)
πππ1 πππ2 + πππ0 + 3ππ + πππ2( πππ0 + 3ππ) =β (23)
From (21)
πΌπ1 = πππ2 + πππ0 + 3ππ E/β (24a)
πΌπ2 = -E(πππ0 + 3ππ)/β (24b)
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πΌπ0 = -Eπππ2/β (24c)
The sequence components of voltages at ith bus (i = 1, 2 β¦ n) are
ππ0 = - πππ0πΌπ0= πΈπππ0πππ2/β (25a)
ππ1= E-πππ1πΌπ1= E [β-πππ1 πππ2πππ0 + 3ππ ]/β
(25b)
ππ2=-πππ2πΌπ2 = πππ2[πππ0 + 3ππ)] E/β (25c)
Test Problem
A 3-phase fault occurs at the bus 2 of the system shown in
fig. below. Find (a) sequence current and voltage (b) fault
current of each phases (c) fault voltage (line to neutral) of
each phases at bus 3. Assume that prefault voltages at all
buses are 1 p.u. Given:-
[π1βππ’π ] =
π0.1507937 π0.1269841 π0.1071429 π0.0793651
π0.1269841 π0.1574603 π0.1328571 π0.0984127
π0.1071429π0.0793651
π0.1328571π0.0984127
π0.1542857 π0.1142857π0.1142857 π0.1365079
Since all the negative sequence reactance are the same as
the positive sequence reactance,
[π1βππ’π ] = [π2βππ’π ] . and [π0βππ’π ] =
π0.05 0 0 00 π0.0514286 π0.03 π0.02
00
π0.03π0.02
π0.105π0.07
π0.07π0.0933333
RESULT
Using MATLAB tools, the result of the problem is
Param
eters
of
justific
ation
For
symmetri
cal fault
For
single
line-to-
ground
fault
For line-to-
line fault
For
double
line-to-
ground
fault
πΌπ0 0 amp. 0-j716.36
amp. 0 amp.
0+J3.841
5amp.
πΌπ1 0-j1666.7
amp.
0-j716.36
amp.
0-j3.1754
amp.
0-j5.0961
amp.
πΌπ2 0 amp. 0-j716.36
amp.
j3.1754
amp.
j1.2547
amp.
πΌπ 0-j1666.7
amp.
0-j2149.1
amp.
0 amp. 0 amp.
Param
eters
of
justific
ation
For
symmetri
cal fault
For
single
line-to-
ground
fault
For line-to-
line fault
For
double
line-to-
ground
fault
πΌπ (-
1443.3+j8
33.25)
amp.
0-j0.0315
amp.
(-5.4998-
j0.0001)
amp.
(-
5.4998+j
5.7620)
amp.
πΌπ 1443.3+j8
33.33
amp.
0-
j0.0315a
mp.
5.4998+j0.0
001amp.
5.4998+j
5.7622
amp.
ππ0 19.8464
KV
0-0.0819
KV
0 KV 0.1152
KV
ππ1 0 KV 0.6373
KV
0.4219 KV 1.0000-
j0.0262
KV
ππ2 0 KV -0.3627 0.4219 KV 1.0000-
j0.0262
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KV KV
ππ 19.8464
KV
24.4906
KV
0.8437 KV 2.1152-
j0.0525
KV
ππ (-9.9223-
j17.1870)
KV
(-27.844-
j110.00)
KV
-0.4219 KV 0 KV
ππ (-
9.9223+j1
7.1870)
KV
(-
27.848+j
110.00)
KV
-0.4219 KV 0 KV
RESULT ANALYSIS
It is evident from the result that a 3-phase fault is the most
dangerous one and it has larger magnitude of current than
other types of fault. Also among the unsymmetrical, line-
to-line fault because of absence of zero sequence network
as there is no path to ground ,it has higher magnitude of
current with respect to single line-to-ground fault and
double line-to-ground fault. This could be matched with
theoretical equation (16) as well, where zero sequence
networks is not there in denominator. The magnitude of
Double line-to-ground fault lies between single line-to-
ground fault and line βto-line fault i.e. less than line-to-line
fault and more than single line-to-ground fault. As far as
voltage is concerned line-to-line voltage has least voltage
among the unsymmetrical faults.
Also, data and analysis came same when the fault occurred
at bus 3 and parameters calculated at bus 2.
CONCLUSION
This paper is basically related with finding the fault that
occurs in power system using digital approach and
through MATLAB programming and it is concluded that
the evaluation of fault is very important part of power
system analysis for stable and economic operations of
power system network and also because it provide data
such as voltage and currents which are necessary in
designing the protective schemes of power system. The
calculations are done theoretically and also by MATLAB
programming. Both results are found equal, so this type of
digital fault calculation is very useful.
REFERENCES
[1] D P Kothari and I J Nagrath, βUnsymmetrical Fault Analysisβ,in Power System Engineering,2nd ed.,Tata McGraw Hill,2011,pp.525-530.
[2] C.L.Wadhwa, βSymmetrical Components and Fault Calculationββ,in Electrical Power System,6th ed., New Age International Limited, 2010, pp.312-329.
[3] J.B.Gupta, βUnsymmetrical Fault Analysisβ,in A course in Power Systems,11th ed.,S.K.Kataria & Sons,2013,pp.80-83.
[4] P.S.R. Murthy, βUnbalaced Fault Analysisβ,in Power System Analysis,BS publication,2007,pp. 241-243.
[5] Md. Fadli Bin Samsudin, βTransmissions Line Fault Analysis Using Bus Impedence Matrix,βElect. Res. Lab ,Universiti Technologi Malaysia,Final Rep.,May 2011.
[6] Sushma Ghimire,"Analysis of fault location methods on transmission line,βM.S. thesis,Dept. Elect. Eng.,University of New Orleans,New Orleans,Louisiana,United states,2011.
BIOGRAPHIES
Mr. PRALAY ROY Assistant Professor of Electrical Engineering Dept., Siliguri Institute of Technology, Sukna, Darjeeling. Area of Interest-Power system, Electrical Machine, Power Electronics, Analog Electronics.
Mr. Shubham Kumar Gupta 3rd year B.TECH student, 2016 of Dept. of Electrical Engineering, Siliguri Institute of Technology, Sukna,Darjeeling. Area of Interest- Analog Electronics, Electrical Machines, Power System, Basic Electrical Embedded System.