1. Fault analysis is important for safety, equipment protection, and power system stability. It allows engineers to design protective devices that can safely interrupt high fault currents. 2. When a fault occurs, system operators generate a fault report documenting the time, location, cause, affected equipment, actions taken, and recommendations. This helps guide future maintenance. 3. Faults can be caused by equipment breakdown, aging insulation, loose connections, overloading, or natural disasters. The main types are symmetrical, phase-to-phase, phase-to-ground, and double line-to-ground faults. Fault analysis methods include calculating current flows and voltages.

1. 1.4 Fault
Calculations
Module 1

2. Need for Fault Analysis
1.Safety: Fault currents can be very high and dangerous. By knowing the
magnitude of the fault current, we can design protective devices, such as
circuit breakers and fuses, to handle and interrupt this current safely. This
helps prevent electrical accidents and protects people and equipment from
harm.
2.Equipment Protection: Fault currents can cause damage to electrical
equipment, like transformers, generators, and cables. By determining the fault
current, we can select and install protective devices that can handle the
maximum expected fault current. This ensures that equipment is protected
from excessive current flow and prevents costly repairs or replacements.
3.System Stability: Fault currents can affect the stability of the power system.
High fault currents can cause voltage drops, frequency deviations, and even
blackouts. By analyzing and understanding the fault currents, engineers can
design and operate the power system in a way that maintains stability and
minimizes disruptions to the electricity supply.

3. What happens after a fault?
▪ Fault reports are generated by system operators or utility
companies responsible for managing the power grid. System
operators receive alarms or notifications from monitoring devices
and protective relays when a fault occurs.
▪ Fault reports document contains the
1.time, duration, location, and cause of the fault
2.information on the affected equipment
3.actions taken to address the fault
4.any findings or recommendations.
▪ Fault reports serve as a record of the incident and helps guide
future maintenance and improvement efforts.

4. Causes of fault
Internal Causes
▪ breakdown of transmission lines or equipment
▪ aging of insulation
▪ deterioration of insulation in generator,
transformer and other electrical equipment
▪ improper installations and inadequate design.
▪ Loose connections or failure of insulation that
leads to short circuit.
External Causes
▪ Overloading of equipment
▪ Insulation failure due to lighting surges
▪ Mechanical damage by public.
▪ Natural Disasters

5. Types of Faults
▪ Series fault (Open Circuit fault)
▪ Shunt fault (Short Circuit fault)
Symmetric fault Unsymmetric fault
Triple line fault, L-L-L Single line to ground fault, L-G
Triple line to ground fault, L-L-L-G Double line fault, L-L
Double line to ground fault, L-L-G

6. Symmetrical Fault

7. Symmetrical Fault:
▪ Also known as balanced fault.
▪ This is because fault current magnitudes are very high but are
equal in magnitude and has a phase shift of 120°
▪ Imposes heavy duty on the circuit breaker

8. Symmetrical Fault Analysis - Steps
1. Draw the single line diagram
2. Evaluate 𝑝. 𝑢 reactances of each element (generator, transformers,
transmission lines, motors)
3. Draw the single line reactance diagram
4. Evaluate the equivalent reactance 𝑋𝑒𝑞 in per unit.
5. Find 𝐼𝑓 by applying KVL, 𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) = 𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒
Where,
𝐼𝑓,𝑏𝑎𝑠𝑒 =
𝐾𝑉𝐴𝑏
3𝐾𝑉𝑏

9. Example 1: Symmetrical Fault Analysis
1. A 3-phase, 5 MVA, 6.6 KV alternator with a
reactance of 8% is connected to a feeder of
series impedance of 0.12+j0.48 ohms/phase per
km. The transformer is rated at 3 MVA, 6.6
kV/33KV and has a reactance of 5%. Determine
the fault current supplied by the generator
operating under no-load with a voltage of 6.9KV,
when a 3-phase symmetrical fault occurs at a
point 15 km along the feeder.
Solution:
Step 1: Draw the single line diagram
Steps:
1. Draw the single
line diagram
2. Evaluate
𝑝. 𝑢 reactances of
each element
(generator,
transformers,
transmission
lines, motors)
3. Draw the single
line reactance
diagram
4. Evaluate the
equivalent
reactance 𝑋𝑒𝑞 in
per unit.
5. Find 𝐼𝑓 by
applying KVL,
𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) =
𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒

10. Example 1: Symmetrical Fault Analysis
1. A 3-phase, 5 MVA, 6.6 KV alternator with a reactance of 8% is connected to a feeder of series
impedance of 0.12+j0.48 ohms/phase per km. The transformer is rated at 3 MVA, 6.6 kV/33KV and
has a reactance of 5%. Determine the fault current supplied by the generator operating under no-
load with a voltage of 6.9KV, when a 3-phase symmetrical fault occurs at a point 15 km along the
feeder.
Solution:
Step 1: Draw the single line diagram
Step 2: Evaluate 𝑝.𝑢 reactances of each element
Steps:
1. Draw the single
line diagram
2. Evaluate
𝑝. 𝑢 reactances of
each element
(generator,
transformers,
transmission
lines, motors)
3. Draw the single
line reactance
diagram
4. Evaluate the
equivalent
reactance 𝑋𝑒𝑞 in
per unit.
5. Find 𝐼𝑓 by
applying KVL,
𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) =
𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒

11. Example 1: Symmetrical Fault Analysis Steps:
1. Draw the single
line diagram
2. Evaluate
𝑝. 𝑢 reactances of
each element
(generator,
transformers,
transmission
lines, motors)
3. Draw the single
line reactance
diagram
4. Evaluate the
equivalent
reactance 𝑋𝑒𝑞 in
per unit.
5. Find 𝐼𝑓 by
applying KVL,
𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) =
𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒

12. Example 1: Symmetrical Fault Analysis Steps:
1. Draw the single
line diagram
2. Evaluate
𝑝. 𝑢 reactances of
each element
(generator,
transformers,
transmission
lines, motors)
3. Draw the single
line reactance
diagram
4. Evaluate the
equivalent
reactance 𝑋𝑒𝑞 in
per unit.
5. Find 𝐼𝑓 by
applying KVL,
𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) =
𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒

13. Example 1: Symmetrical Fault Analysis

14. Example 1: Symmetrical Fault Analysis

15. Example 1: Symmetrical Fault Analysis Steps:
1. Draw the single
line diagram
2. Evaluate
𝑝. 𝑢 reactances of
each element
(generator,
transformers,
transmission
lines, motors)
3. Draw the single
line reactance
diagram
4. Evaluate the
equivalent
reactance 𝑋𝑒𝑞 in
per unit.
5. Find 𝐼𝑓 by
applying KVL,
𝐼𝑓 =
𝐸𝑓
𝑗𝑋𝑒𝑞
𝑝. 𝑢
6. 𝐼𝑓(𝑎𝑐𝑡𝑢𝑎𝑙) =
𝐼𝑓 (𝑝.𝑢) ∗ 𝐼𝑓 𝑏𝑎𝑠𝑒

16. Unsymmetrical Fault

17. Unsymmetrical Fault:
▪ Also known as unbalanced fault.
▪ In an unbalanced fault, the fault currents or fault impedances in
different phases are unequal or unbalanced.
▪ This means that the fault currents or impedances do not have the
same magnitude or phase relationship as they would in a
balanced system.
▪ As a result, the phase shift between the currents in the different
phases can vary and may not be 120 degrees.

18. Types of Unsymmetrical Fault :
1.Phase-to-Phase Fault (line to line fault) L-L :
▪ A fault that occurs when two phases of a three-phase power system come into
contact with each other or experience a short circuit between them.
▪ This can happen due to equipment failure, insulation breakdown, or accidental
contact.
2.Phase-to-Ground Fault (Single Line to ground fault) L-G:
▪ A fault that occurs when one phase of a power system comes into contact with the
ground or an earthed object.
▪ This can be caused by insulation failure, conductors touching the ground, or
equipment malfunctions.
3.Double Line-to-Ground Fault, L-L-G:
▪ A fault that occurs when two phases of a power system simultaneously come into
contact with the ground or an earthed object.
▪ This can happen due to insulation breakdown, fallen conductors, or accidental
contact with grounded objects.

19. L-L-G fault:

22. Apply current division rule,

25. ▪ Case 2: If fault occurs through an impedance
Since 𝑉𝑏 = 𝑉
𝑐 ≠ 0, we must know if the sequence
components are equal or not. That is, in case 1, 𝑉𝑎0,
𝑉𝑎1 and 𝑉𝑎2 were equal.
To see how the sequences 𝑉𝑎0, 𝑉𝑎1 and 𝑉𝑎2 are related
in this case 2, we can use the phase matrix.

27. Now, we’ll see whether 𝑉𝑎0 is also equal to 𝑉𝑎1 and 𝑉𝑎2

28. From this equation, we understand that 𝑉𝑎0 is not equal to 𝑉𝑎1 and 𝑉𝑎2.

29. When you solve for 𝐼𝑎0 using the equivalent circuit of case 2, we’ll get,