Analysis of Fault

1. BEF43303
POWER SYSTEM ANALYSIS AND PROTECTION
ANALYSIS OF BALANCED AND UNBALANCED FAULTS
WEEK 6

2. 6.0 CONTENTS
6.1 SINGLE-LINE-TO-GROUND FAULT
6.2 LINE-TO-LINE FAULT
6.3 DOUBLE-LINE-TO-GROUND FAULT
6.4 UNBALANCED FAULT ANALYSIS

3. 6.1 SINGLE-LINE-TO-GROUND FAULT
• Figure 1 shows a three-phase
generator with neutral
grounded through impedance
𝑍𝑛.
• Suppose a line-to-ground fault
occurs on phase 𝑎 through
impedance 𝑍𝑓.
Figure 1

4. 6.1 SINGLE-LINE-TO-GROUND FAULT
• Assume that the generator is initially on no-load, the conditions at
the point of fault are:
𝑉
𝑎 = 𝑍𝑓𝐼𝑎
𝐼𝑏 = 𝐼𝑐 = 0
• Substituting the above conditions into the symmetrical components
of currents:
𝐼𝑎
0
𝐼𝑎
1
𝐼𝑎
2
=
1
3
1 1 1
1 𝑎 𝑎2
1 𝑎2 𝑎
𝐼𝑎
0
0

5. 6.1 SINGLE-LINE-TO-GROUND FAULT
• From the previous equation:
𝐼𝑎
0 = 𝐼𝑎
1 = 𝐼𝑎
2 =
1
3
𝐼𝑎
• Phase 𝑎 voltage in terms of symmetrical component is given as:
𝑉
𝑎 = 𝑉
𝑎
0
+ 𝑉
𝑎
1
+ 𝑉
𝑎
2
• Consequently:
𝑉
𝑎 = 𝐸𝑎 − 𝑍0
+ 𝑍1
+ 𝑍2
𝐼𝑎
0
3𝑍𝑓𝐼𝑎
0 = 𝐸𝑎 − 𝑍0 + 𝑍1 + 𝑍2 𝐼𝑎
0
𝐼𝑎
0 =
𝐸𝑎
𝑍0 + 𝑍1 + 𝑍2 + 3𝑍𝑓

6. 6.1 SINGLE-LINE-TO-GROUND FAULT
• Hence, the fault current is given by:
𝐼𝑎 = 3𝐼𝑎
0 =
3𝐸𝑎
𝑍0 + 𝑍1 + 𝑍2 + 3𝑍𝑓
• These equations can be represented in the equivalent circuit of
Figure 2.
Figure 2

7. 6.2 LINE-TO-LINE FAULT
• Figure 3 shows a three phase
generator with a fault through
an impedance 𝑍𝑓 between
phases 𝑏 and 𝑐.
• Assuming the generator is
initially no-load, the condition
at the fault point are:
𝑉𝑏 − 𝑉
𝑐 = 𝑍𝑓𝐼𝑏
𝐼𝑎 = 𝐼𝑏 + 𝐼𝑐 = 0
Figure 3

8. 6.2 LINE-TO-LINE FAULT
• Considering the previous conditions, the symmetrical components
of currents are:
𝐼𝑎
0
𝐼𝑎
1
𝐼𝑎
2
=
1
3
1 1 1
1 𝑎 𝑎2
1 𝑎2
𝑎
0
𝐼𝑏
−𝐼𝑏
• In component form:
𝐼𝑎
0 = 0
𝐼𝑎
1 =
1
3
𝑎 − 𝑎2 𝐼𝑏
𝐼𝑎
2 =
1
3
𝑎2 − 𝑎 𝐼𝑏

9. 6.2 LINE-TO-LINE FAULT
• We know that:
𝑉𝑏 − 𝑉
𝑐 = 𝑎2 − 𝑎 𝑉
𝑎
1 − 𝑉
𝑎
2 = 𝑍𝑓𝐼𝑏
𝑎2 − 𝑎 𝐸𝑎 − 𝑍1 + 𝑍2 𝐼𝑎
1 = 𝑍𝑓𝐼𝑏
• Replacing 𝐼𝑏 from equation:
𝐸𝑎 − 𝑍1
+ 𝑍2
𝐼𝑎
1
= 𝑍𝑓
3𝐼𝑎
1
(𝑎 − 𝑎2)(𝑎2 − 𝑎)
• Solving for 𝐼𝑎
1 results in:
𝐼𝑎
1
=
𝐸𝑎
𝑍1 + 𝑍2 + 𝑍𝑓

10. 6.2 LINE-TO-LINE FAULT
• The phase currents are:
𝐼𝑎
𝐼𝑏
𝐼𝑐
=
1 1 1
1 𝑎2
𝑎
1 𝑎 𝑎2
0
𝐼𝑎
1
−𝐼𝑎
1
• The fault current is:
𝐼𝑏 = −𝐼𝑐 = (𝑎2 − 𝑎)𝐼𝑎
1
• These equations can be
represented in the equivalent
circuit shown in Figure 3.
Figure 3

11. 6.3 DOUBLE-LINE-TO-GROUND
FAULT
• Figure 4 shows a three-phase
generator with a double-line-
to-ground fault (phase 𝑏 − 𝑐).
• Assuming the generator is
initially on no-load, the
conditions at the fault point
are:
𝑉𝑏 = 𝑉
𝑐 = 𝑍𝑓 𝐼𝑏 + 𝐼𝑐
𝐼𝑎 = 𝐼𝑎
0
+ 𝐼𝑎
1
+ 𝐼𝑎
2
= 0
Figure 4

12. 6.3 DOUBLE-LINE-TO-GROUND
FAULT
• The phase voltage 𝑉𝑏 and 𝑉
𝑐 are given by:
𝑉𝑏 = 𝑉
𝑎
0 + 𝑎2𝑉
𝑎
1 + 𝑎𝑉
𝑎
2
𝑉
𝑐 = 𝑉
𝑎
0
+ 𝑎𝑉
𝑎
1
+ 𝑎2
𝑉
𝑎
2
• Since 𝑉𝑏 = 𝑉
𝑐, the previous equations become:
𝑉
𝑎
1
= 𝑉
𝑎
2
• The symmetrical components of currents are rewritten as:
𝑉𝑏 = 𝑍𝑓(𝐼𝑎
0 + 𝑎2𝐼𝑎
1 + 𝑎𝐼𝑎
2 + 𝐼𝑎
0 + 𝑎𝐼𝑎
1 + 𝑎2𝐼𝑎
2)
𝑉𝑏 = 3𝑍𝑓𝐼𝑎
0

13. 6.3 DOUBLE-LINE-TO-GROUND
FAULT
• Hence,
3𝑍𝑓𝐼𝑎
0 = 𝑉
𝑎
0 + 𝑎2 + 𝑎 𝑉
𝑎
1 = 𝑉
𝑎
0 − 𝑉
𝑎
1
• Substituting the symmetrical components of voltage and solving for
𝐼𝑎
0:
𝐼𝑎
0 = −
𝐸𝑎 − 𝑍1𝐼𝑎
1
𝑍0 + 3𝑍𝑓
• Substituting for the symmetrical components of voltage:
𝐼𝑎
2 = −
𝐸𝑎 − 𝑍1𝐼𝑎
1
𝑍2

14. 6.3 DOUBLE-LINE-TO-GROUND
FAULT
• Substituting 𝐼𝑎
0 and 𝐼𝑎
2 into the condition at the point of fault and
solving for 𝐼𝑎
1.
𝐼𝑎
1 =
𝐸𝑎
𝑍1 +
𝑍2(𝑍0 + 3𝑍𝑓)
𝑍2 + 𝑍0 + 3𝑍𝑓
• The fault current is obtained from:
𝐼𝑓 = 𝐼𝑏 + 𝐼𝑐 = 3𝐼𝑎
0
• These equations can be represented in the equivalent circuit shown
in Figure 5.

15. 6.3 DOUBLE-LINE-TO-GROUND
FAULT
Figure 5

16. 6.4 UNBALANCED FAULT ANALYSIS
Example 1
The one-line diagram of a simple power system is shown in Figure 6. The neutral of each generator
is grounded through a current-limiting reactor of 0.25/3 per unit on a l00-MVA base. The system
data expressed in per unit on a common l00-MVA base is tabulated in Table 1. The generators are
running on no-load at their rated voltage and rated frequency with their emfs in phase. Determine
the fault current for the following faults.
a) A balanced three-phase fault at bus 3 through a fault impedance 𝑍𝑓 = 0.1𝑖 per unit.
b) A single line-to-ground fault at bus 3 through a fault impedance 𝑍𝑓 = 0.1𝑖 per unit.
c) A line-to-line fault at bus 3 through a fault impedance 𝑍𝑓 = 0.1𝑖 per unit.
d) A double line-to-ground fault at bus 3 through a fault impedance 𝑍𝑓 = 0.1𝑖 per unit.

17. 6.4 UNBALANCED FAULT ANALYSIS
Table 1
Figure 6

18. 6.4 UNBALANCED FAULT ANALYSIS
• For further exercises, check out Problems 10.11-10.15 of Power
System Analysis by Hadi Saadat.