this chapter deals with fault analysis of a power system. under this topic, only symmetrical fault analysis is given. it will describe the methods used to determine fault current and voltage values.

1. ASTU
SCHOOL OF ELECTRICAL ENGINEERING AND
COMPUTING
DEPT. OF POWER AND CONTROL ENGINEERING
COMPUTER APPLICATION IN POWER SYSTEM
(PCE5307)
CHAPTER THREE
FAULT STUDIES
BY: MESFIN M.

2. OUTLINE
 Three Phase Fault Analysis
Admittance Matrix equations
Impedance Matrix equations
Fault Calculation
Analysis of Unbalanced Fault
Short Circuit Faults
Open Circuit Faults

3. INTRODUCTION
 The main objective of fault analysis is to calculate fault
currents and voltages for the determination of circuit
breaker capacity and protective relay performance.
Early methods used in the calculation of fault levels
involved the following approximations.
All voltage sources assumed a one pu magnitude and zero relative phase,
which is equivalent to neglecting the pre-fault load current contribution.
Transmission plant components included only inductive parameters.
Transmission line shunt capacitance and transformer magnetizing
impedance were ignored.

4. CONT.…
Based on the above assumptions, simple equivalent
sequence impedance networks were calculated and these
were interconnected according to the fault specification.
Conventional circuit analysis was then used to calculate the
sequence voltage and currents
Although the basic procedure of the computer solution is
still the same, the need for the various approximations has
disappeared.

5. ANALYSIS OF THREE-PHASE FAULTS
 A preliminary stage to the analysis is the collection of
appropriate data specifying the system to be analyzed in
terms of:
Pre-fault voltage,
loading and
Generating conditions.
 Such data is then processed to form a nodal equivalent
network constituted by admittances and injected currents.

6. CONT.…
Generator model and its converted nodal equivalent
representation.

7. CONT.….
The injected nodal current for the above circuit is given by:
𝑰𝒋 = 𝑬𝒋
𝑴
𝒀𝒋
𝑴
But using KVL
𝑬𝒋
𝑴
= 𝑽𝒋 +
𝑰𝒋
𝑴
𝒀𝒋
𝑴
Thus,
𝑰𝒋 = 𝑽𝒋 𝒀𝒋
𝑴
+ 𝑰𝒋
𝑴
𝑰𝒋
𝑴
is the current required at the voltage 𝑽𝒋 to produce the
machine power 𝑷𝒋
𝑴
+ 𝒋𝑸𝒋
𝑴
, so

8. CONT.…
(𝐼𝑗
𝑀
)∗
𝑉𝑗 = 𝑃𝑗
𝑀
+ 𝑗𝑄𝑗
𝑀
Thus, from load flow data for P, Q and V we can calculate the
injected nodal current 𝐼𝑗.
𝐼𝑗 = 𝑽𝒋 𝒀𝒋
𝑴
+
𝑃𝑗
𝑀
− 𝑗𝑄𝑗
𝑀
𝑽𝒋
∗

9. ADMITTANCE MATRIX EQUATION
 Let us take as a reference the small system given below,

10. CONT.…
Substituting the above network with equivalent model

11. CONT.….
The final equivalent circuit of the network with branch elements

12. CONT.…
The following equations may then be written for the final
equivalent network.
𝐼1 = 𝑦11 𝑣1 + 𝑦12 𝑣1 − 𝑣2
𝐼2 = 𝑦22 𝑣2 + 𝑦21 𝑣2 − 𝑣1 + 𝑦23 𝑣2 − 𝑣3 + 𝑦24 𝑣2 − 𝑣4
𝐼3 = 𝑦33 𝑣3 + 𝑦32 𝑣3 − 𝑣2 + 𝑦34 𝑣3 − 𝑣4
𝐼4 = 𝑦44 𝑣4 + 𝑦42 𝑣4 − 𝑣2 + 𝑦43 𝑣4 − 𝑣3 + 𝑦45 𝑣4 − 𝑣5
𝐼3 = 𝑦55 𝑣5 + 𝑦54 𝑣5 − 𝑣4

13. CONT.…
 or in matrix form after grouping together the terms
common to each voltage
𝐼1
𝐼2
𝐼3
𝐼4
𝐼5
=
𝑌11 𝑌12 𝑌13 𝑌14 𝑌15
𝑌21 𝑌22 𝑌23 𝑌24 𝑌25
𝑌31
𝑌41
𝑌51
𝑌32
𝑌42
𝑌52
𝑌33
𝑌43
𝑌53
𝑌34
𝑌44
𝑌54
𝑌35
𝑌45
𝑌55
∙
𝑉1
𝑉2
𝑉3
𝑉4
𝑉5
Where
𝑌𝑖𝑖 =
𝑗
𝑦𝑖𝑗 and 𝑌𝑖𝑗 = −𝑦𝑖𝑗 𝑖 ≠ 𝑗

14. CONT.…
 The matrix is usually written as;
𝐼 = 𝑌 ∙ 𝑉
where [I] and [V] are the current and voltage vectors and [Y] is
the nodal admittance matrix of the system.
In the admittance matrix nonzero elements only occur where
branches exist between nodes.
Such sparsity is exploited by only storing and processing the
nonzero elements.
Moreover, the symmetry of the matrix ( Yij = Yji) permits using
only the upper right-hand terms in the calculations.

15. IMPEDANCE MATRIX EQUATION
 The nodal admittance equation is inefficient as it requires a
complete iterative solution for each fault type and location. Thus,
𝑉 = 𝑌 −1
∙ 𝐼
𝑉 = 𝑍 ∙ 𝐼

16. FAULT CALCULATIONS
 From the initial machine data, the values of [I] are first
calculated from equation for injected nodal current using one
per unit voltages.
These may now be used to obtain a better estimate of [V], the
pre-fault voltage at every node from equation 𝑉 = 𝑍 ∙ 𝐼 .
The program now has sufficient information to calculate the
voltages and currents during a fault.
𝑉𝑘
𝑓
= 𝑍 𝑓
𝐼 𝑓
where k is the bus to be faulted, Zf is the fault impedance and
𝐼 𝑓
is the fault current.

17. CONT.…
The equation 𝑉 = 𝑍 ∙ 𝐼 for a fault at K bus can be expanded
as,
𝑉1
𝑉2
∙
𝑉𝑘
∙
𝑉𝑛
=
𝑍11 𝑍12 ∙ 𝑍1𝑘 ∙ 𝑍1𝑛
𝑍21 𝑍22 ∙ 𝑍2𝑘 ∙ 𝑍2𝑛
∙
𝑍 𝑘1
∙
𝑍 𝑛1
∙
𝑍 𝑘2
∙
𝑍 𝑛2
∙
∙
∙
∙
∙
𝑍 𝑘𝑘
∙
𝑍 𝑛𝑘
∙ ∙
∙ 𝑍 𝑘𝑛
∙ ∙
∙ 𝑍 𝑛𝑛
∙
𝐼1
𝐼2
∙
𝐼 𝑘
∙
𝐼 𝑛
Selecting row k and expanding gives
𝑉𝑘 = 𝑍 𝑘1 𝐼1 + 𝑍 𝑘2 𝐼2 + ⋯ + 𝑍 𝑘𝑘 𝐼 𝑘 + ⋯ + 𝑍 𝑘𝑛 𝐼 𝑛
This equation describes the voltage at bus k prior to the fault.

18. CONT.…
During a fault a large fault current If flows out of bus k thus,
𝑉𝑘
𝑓
= 𝑍 𝑓 𝐼 𝑓 = 𝑍 𝑘1 𝐼1 + 𝑍 𝑘2 𝐼2 + ⋯ + 𝑍 𝑘𝑘 𝐼 𝑘 + ⋯ + 𝑍 𝑘𝑛 𝐼 𝑛 − 𝑍 𝑘𝑘 𝐼 𝑓
Or
𝑍 𝑓 𝐼 𝑓 = 𝑉𝑘 − 𝑍 𝑘𝑘 𝐼 𝑓
Thus, the fault current can be found as,
𝐼 𝑓 =
𝑉𝑘
𝑍 𝑓 + 𝑍 𝑘𝑘
Also during fault voltage at any other bus j is.
𝑉𝑗
𝑓
= 𝑉𝑗 − 𝑍𝑗𝑘 𝐼 𝑓

19. ANALYSIS OF UNBALANCED FAULT
 Reading Assignment
Short circuit
Open circuit