This document is a lecture on sequence networks of impedance loads and fault analysis. It discusses sequence networks diagrams and their properties. It provides examples of drawing sequence networks for different systems and analyzing different types of faults, including three-phase, single line-to-ground, line-to-line, and double line-to-ground faults. The document contains examples and solutions for determining fault currents for different faults in a sample power system.

1. EE 8304
Electrical Power Systems Analysis II
Time: Tuesday & Friday (1540-1755)Hours
1
Course facilitator: Mr. Malamla
Title: Assistant Lecturer
Email: smalamla@gmail.com
Phone No. +255716771718
MBEYA UNIVERSITY OF SCIENCE AND TECHNOLOGY
(MUST)

2. Lecture 7.2

3. SEQUENCE NETWORKS OF IMPEDANCE LOADS
Fig 1: Balanced-Y impedance load

4. SEQUENCE NETWORKS OF IMPEDANCE LOADS

5. SEQUENCE NETWORKS OF IMPEDANCE LOADS

6. SEQUENCE NETWORKS OF IMPEDANCE LOADS

7. SEQUENCE NETWORKS OF IMPEDANCE LOADS

8. SEQUENCE NETWORKS DIAGRAMS
Figure 2

9. SEQUENCE NETWORKS DIAGRAMS
Observations:
 The three sequences are independent.
 The positive sequence network is the same as the one-
line diagram used in studying balanced three-phase
currents and voltages.
 Only positive sequence has a voltage source.
Therefore the positive sequence current causes only
positive sequence voltage drops.
 There is no voltage source in the negative or sequence
networks
 Negative and zero sequence currents cause only
negative and zero sequence voltage drop only.

10. SEQUENCE NETWORKS DIAGRAMS
 The neutral of the system is the reference for positive
and negative sequence networks, but ground is the
reference for zero sequence networks.
 The zero sequence current flow only if the circuit
from the system neutral to ground is complete.
 The ground impedance is reflected in zero sequence
network as 3Zn.
 The three sequence system can be solved separately
on per phase basis.
 The phase currents and voltage can be determined by
superposing their symmetrical components of current
and voltage respectively.

11. Example 1: Draw the sequence networks for the circuit
of figure 3 below.
Figure 3: One-line diagram of power system
SEQUENCE NETWORKS DIAGRAMS

12. Solution 1
Figure 4: Positive sequence network diagram

13. Solution 1
Figure 5: Negative sequence network diagram

14. Solution 1
Figure 6: Negative sequence network diagram

15. ∆-∆ TRANSFORMER BANK AND ITS ZERO-SEQUENCE
NETWORK

16. SINGLE LINE TO GROUND FAULT
The figure below illustrates a three-phase generator with
neutral grounded through impedance Zn.
Fig 7.

17. SINGLE LINE TO GROUND FAULT

18. SINGLE LINE TO GROUND FAULT

19. Thus, for the line-to-ground fault the Thevenin
impedance to the point of fault is obtained for each
sequence network, and the three sequence networks are
connected in series.
SINGLE LINE TO GROUND FAULT

20. Fig. 8:
 In many applications cases, the positive and negative
sequence impedance are found to be equal.
 If the generator neutral is solidly ground Zn = 0 and for
the bolted faults Zf = 0.
SINGLE LINE TO GROUND FAULT

21. LINE TO LINE FAULT
The figure below shows a three phase generator with a
fault through an impendence Zf between phases a
and b.
Fig. 9:

22. Assuming the generator is initially on no-load, the boundary
conditions at the fault point are:
LINE TO LINE FAULT

23. LINE TO LINE FAULT

24. LINE TO LINE FAULT

25. LINE TO LINE FAULT

26. The symmetrical components of voltage and phase voltages
at the point of fault can be obtained by substituting the
symmetrical components of currents in equation (10.54)
LINE TO LINE FAULT

27.  Equations (10.71) and (10.75) can be represented by
connecting the positive and negative sequence
networks in (opposition) parallel.
 In many applications cases, the positive and negative
sequence impedance are found to be equal.
 For the bolted fault, Zf = 0.
LINE TO LINE FAULT

28. Fig. 10.
LINE TO LINE FAULT

29. DOUBLE LINE TO GROUND FAULT
The figure below shows a three phase generator with a
fault phase a and b. an through an impendence Zf to
ground.
Fig. 11:

30. Assuming the generator is initially on no-load, the
boundary conditions at the fault point are:
DOUBLE LINE TO GROUND FAULT

31. DOUBLE LINE TO GROUND FAULT

32. DOUBLE LINE TO GROUND FAULT

33.  Equations (10.86)-(10.88) can be represented by
connecting the positive sequence in series with the
parallel combination of the negative sequence and the
zero sequence networks.
 The value of found from (10.88) is substituted to
in (10.86) and (10.87) and are found.
 The phase currents are found from the symmetrical
components of current.
 And finally the fault currents are found.
1
a
I
0 2
and
a a
I I
DOUBLE LINE TO GROUND FAULT

34. Fig 12.
DOUBLE LINE TO GROUND FAULT

35. Example 2
The one-line diagram of a simple power system shown
in fig. 13. The neutral of each generator is grounded
through a current limiting reactor of 0.25/3 per unit
on a 100-MVA base. The system data expressed in per
unit on a common 100-MVA base is tabulated in
table1.
The generators are running on no-load at their rated
voltage and rated frequency with their emfs in phase.

36. Example 2
Determine the fault current for the following faults.
(a) A balanced three phase fault at bus 3 through a fault
impedance Zf = j0.1 per unit.
(b) A single line to ground fault at bus 3 through a fault
impedance Zf = j0.1 per unit.
(c) A line to line fault at bus 3 through a fault
impedance Zf = j0.1 per unit.
(d) A double line to to ground fault at bus 3 through a
fault impedance Zf = j0.1 per unit.

37. Example 2
Table 1

38. Example 2
Figure13:

39. Solution 2
Figure 14:

40. Solution 2

41. Solution 2

42. The equivalent circuit for zero sequence network is
constructed according to the transformer winding
connections in as shown in figure 16.
Solution 2

43. Solution 2

44. Solution 2

45. Solution 2

46. Solution 2

47. Solution 2

48. Solution 2

49. Solution 2

50. Solution 2

51. End of Lecture 7.2