Here are the steps to solve this problem: 1) Choose base voltage VB1 = 20kV, VB2 = 200kV 2) Change base of generator and transformer impedances to 100MVA base 3) Determine line and load impedances in per unit 4) Draw the single line diagram showing all impedances in per unit The solution is left as an exercise for the student.

1. BEF43303
POWER SYSTEM ANALYSIS AND PROTECTION
POWER FLOW ANALYSIS
WEEK 1

2. LECTURER INFO
1) Mohd Aifaa bin Mohd Ariff
2) B. Eng. (UTM), M. Eng. (UTM), PhD (ICL), DIC
3) Room: D1-305 (By appointment only)
4) Contact Number: 017-7326074
5) Email: aifaa@uthm.edu.my
6) Notes: https://www.edmodo.com/ (Code: ji8fq4)
7) Main reference: “Power System Analysis”, Hadi Saadat, Third
Edition, PSA Publishing.

3. COURSE GOALS
The course discusses the topics related for power system analysis and
protection. The topics covered in this course are per-unit system,
power flow analysis, analysis of balanced and unbalanced faults, power
system stability, overcurrent protection, differential protection, and
distance protection. This course focuses on the analytical techniques
for power system analysis and the application of protection schemes
implemented in practical power system network.

4. COURSE SYNOPSIS
This course is designed to educate students with the understanding of
electric power system analysis and the protection schemes. The
information gained from power system analysis is important for wide
field of power system engineering especially for power system
protection. Power system protection is vital in power system operation
to ensure the security and reliability power delivery to consumers. This
highlights the important of this course to be delivered to the future
power system engineers

5. COURSE LEARNING OUTCOMES
At the end of this course the student will be able to:
1) Analyze power system faults based on balance and unbalanced
faults techniques. (PLO4-CTPS-C4).
2) Express power flow analysis using analytical software. (PLO3-CS-
P4).
3) Explain the suitable protection schemes to ensure power system
delivery and sustainability for given case study. (PLO11-EVTS-A3).

6. LECTURE PLAN
WEEK SUBJECT ASSESSMENT
1-3 POWER FLOW ANALYSIS TEST 1, FINAL EXAM
4-5 ANALYSIS OF BALANCED AND UNBALANCED FAULTS TEST 1, FINAL EXAM
6 TEST 1
6-8 POWER SYSTEM STABILITY PROJECT
9 POWER SYSTEM PROTECTION FINAL EXAM
10 DISTANCE PROTECTION TEST 2, FINAL EXAM
11-12 OVERCURRENT AND EARTH FAULT PROTECTION TEST 2, FINAL EXAM
13 TEST 2
13-14 DIFFERENTIAL PROTECTION SCHEME FINAL EXAM

7. COURSE INFO
Project : 20%
Test 1 : 15%
Test 2 : 15%
Final : 50%
Total : 100%

8. 1.1 INTRODUCTION TO POWER
SYSTEM ANALYSIS
• Power flow studies (AKA load flow) are the backbone of power
system analysis and design.
• Necessary for planning, operation, economic scheduling, and
exchange of power between utilities.
• It is required for many other analysis such as transient stability and
contingency studies.

9. 1.1 INTRODUCTION TO POWER
SYSTEM ANALYSIS
• Deals with the steady-state analysis of an interconnected power
system during normal operation.
• The system is assumed under balanced condition, hence it is
represented by single-phase network.
• The network contains hundred nodes and branches.
• Impedances in the network are specified in PER UNIT on a common
MVA base.

10. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Per-unit system
• An interconnected power system has several different voltage levels.
• Requires transformations of all impedances to a single voltage level
in order to analyze the system.
• The per-unit system has been devised such that the various physical
quantities such as power, voltage, current, and impedance are
expressed as a decimal fraction of base quantities.
• The voltage levels disappear.
• Reduced a power network to a system of simple impedances.

11. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
The advantages of per-unit system:
• Clear idea of relative magnitudes of various quantities.
• The per-unit impedance of equipment based on their own ratings
fall in a narrow range regardless of the rating of the equipment.
• The per-unit variables of a transformer are the same regardless of
whether they are referred to the primary or the secondary side.
• Ideal for the computerized analysis and simulation of complex
power system problems.
• The circuit laws are valid in per-unit system.

12. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
• The per-unit value of any quantity is defined as:
Quantity in per−unit =
Actual quantity
Base value of quantity
• A minimum of four base quantities are required to completely
define a per-unit system; S, V, I and Z.
• Usually, base apparent power 𝑆𝐵 and base voltage 𝑉𝐵 are selected.
• Base current 𝐼𝐵 and base impedance 𝑍𝐵 are dependent on 𝑆𝐵 and 𝑉𝐵.
𝐼𝐵 =
𝑆𝐵
3𝑉𝐵
, 𝑍𝐵 =
𝑉𝐵
3
𝐼𝐵

13. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Change of base
• The impedance of individual generators and transformers are
generally in terms of per-unit quantities based on their own rating.
• The impedance of transmission lines are usually expressed by their
ohmic values.
• For power system analysis, all impedances must be expressed in per
unit on a common system base.
• To accomplish this, an arbitrary base for apparent power is selected.
• Then, the voltage bases must be selected.

14. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
• Once a voltage base has been selected for a point in a system, the
remaining voltage bases are no longer independent.
• They are determined by the various transformer turns ratios.
• The relationship between the old and the new per-unit values is:
𝑍𝑝𝑢
𝑛𝑒𝑤
= 𝑍𝑝𝑢
𝑜𝑙𝑑
𝑆𝐵
𝑛𝑒𝑤
𝑆𝐵
𝑜𝑙𝑑
𝑉𝐵
𝑜𝑙𝑑
𝑉𝐵
𝑛𝑒𝑤
2

15. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Example 1
The one-line diagram of a three-phase power system is shown in Figure 1. Select a common base of
100MVA and 22kV on the generator side. Draw an impedance diagram with all impedances
including the load impedance marked in per-unit. The manufacturer's data for each device is given
as follow:
G: 90.0MVA 22kV X = 18%
T1: 50.0MVA 22/220kV X = 10%
T2: 40.0MVA 220/11kV X = 6.0%
T3: 40.0MVA 22/110kV X = 6.4%
T4: 40.0MVA 110/11kV X = 8.0%
M: 66.5MVA 10.45kV X = 18.5%
The three-phase load at bus 4 absorbs 57MVA, 0.6 power factor lagging at 10.45kV. Line 1 and line 2
have reactances of 48.4 and 65.43Ω , respectively.

16. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Figure 1 – Single-line diagram for Example 1

17. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
1) Determine the base voltage for all sections. Choose generator
voltage as base voltage. Hence,
𝑉𝐵1 = 22𝑘𝑉
𝑉𝐵2 = 𝑉𝐵3 = 22 ×
220
22
= 220𝑘𝑉
𝑉𝐵4 = 220 ×
11
220
= 11𝑘𝑉
𝑉𝐵5 = 𝑉𝐵6 = 22 ×
110
22
= 110𝑘𝑉

18. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
2) Change the base for per-unit reactances of generator and
transformers to 100 MVA
𝑋𝐺 = 0.18 ×
100
90
= 0.2𝑝𝑢
𝑋𝑇1 = 0.1 ×
100
50
= 0.2𝑝𝑢
𝑋𝑇2 = 0.15𝑝𝑢, 𝑋𝑇3 = 0.16𝑝𝑢, 𝑋𝑇4 = 0.2𝑝𝑢

19. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
3) Change the base for per-unit reactance of motor 100 MVA, 11kV
𝑋𝑀 = 0.185 ×
100
66.5
×
10.45
11
2
= 0.25𝑝𝑢
4) Determine the impedance bases of Line 1 and Line 2.
𝑍𝐵1 =
220 2
100
= 484Ω
𝑍𝐵2 = 121Ω

20. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
5) Determine Line 1 and 2 per-unit reactances:
𝑋1 =
48.4
484
= 0.1𝑝𝑢
𝑋2 = 0.54𝑝𝑢
6) The load apparent power at 0.6PF lagging.
𝑆𝐿(3𝜙) = 57∠53.13°

21. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
5) Determine Line 1 and 2 per-unit reactances:
𝑋1 =
48.4
484
= 0.1𝑝𝑢
𝑋2 = 0.54𝑝𝑢
6) The load impedance in ohm at 0.6PF lagging.
𝑆𝐿(3𝜙) = 57∠53.13°
𝑍𝐿(Ω) =
𝑉𝐿−𝐿
2
𝑆𝐿(3𝜙)
∗ =
10.45 2
57∠ −53.13°
= 1.1495 + 𝑗1.53267 Ω

22. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
7) The load impedance in per-unit
𝑍𝐵4 =
11 2
100
= 1.21Ω
𝑍𝐿 =
1.1495 + 𝑗1.53267
1.21
= 0.95 + 𝑗1.2667 𝑝𝑢
8) The per-unit equivalent circuit is shown in Figure 2

23. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
7) The load impedance in per-unit
𝑍𝐵4 =
11 2
100
= 1.21Ω
𝑍𝐿 =
1.1495 + 𝑗1.53267
1.21
= 0.95 + 𝑗1.2667 𝑝𝑢
8) The per-unit equivalent circuit is shown in Figure 2

24. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 1
Figure 2 – Per-unit impedance diagram for Example 1

25. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Example 2
The motor of Example 1 operates at full-load 0.8 power factor leading
at a terminal voltage of 10.45kV.
a) Determine the voltage at the generator busbar (bus 1).
b) Determine the generator and the motor internal emfs.

26. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 2(a)
1) The per-unit voltage at bus 4, taken as reference is:
𝑉4 =
10.45
11
= 0.95∠0°
𝑝𝑢
2) The motor apparent power 0.8 leading power factor is given by:
𝑆𝑀 =
66.5
100
∠ −36.87°
𝑝𝑢
3) Current drawn by the motor is:
𝐼𝑀 =
𝑆𝑀
∗
𝑉𝑀
∗ =
0.665∠36.87°
0.95∠0°
= 0.56 + 𝑗0.42 𝑝𝑢

27. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 2(a)
4) Current drawn by the load is:
𝐼𝐿 =
𝑉𝐿
𝑍𝐿
=
0.95∠0°
0.95 + 𝑗1.2667
= 0.36 − 𝑗0.48 𝑝𝑢
5) Total current drawn from bus 4 is:
𝐼 = 𝐼𝐿 + 𝐼𝑀 = 0.92 − 𝑗0.06 𝑝𝑢
6) The equivalent reactance of the parallel branches is:
𝑋𝐿−𝐿 =
0.45 × 0.9
0.45 + 0.9
= 0.3𝑝𝑢

28. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 2(a)
7) The generator terminal voltage is:
𝑉1 = 𝑉4 + 𝑍𝐿−𝐿𝐼 = 1.0∠15.91°
𝑝𝑢 = 22∠15.91°
kV

29. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Solution for Example 2(b)
1) The generator internal emf is:
𝐸𝐺 = 𝑉1 + 𝑍𝐺𝐼 = 1.0826∠25.14°
𝑝𝑢 = 23.84∠25.14°
𝑘𝑉
2) The motor internal emf is:
𝐸𝑀 = 𝑉4 − 𝑍𝑀𝐼𝑀 = 1.064∠ −7.56°
𝑝𝑢 = 11.71∠ −7.56°
𝑘𝑉

30. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 1
A three-phase Y connected 75MVA, 27kV synchronous generator has a
synchronous reactance of 9.0 per phase. Using rated MVA and voltage
as base values, determine the per-unit reactance. Then refer this per-
unit value to a 100MVA, 30kV base. (Answer: 0.926pu, 1.0pu).

31. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 2
A 40MVA, 20kV/400kV, single-phase transformer has the following
series impedances:
𝑍1 = 0.9 + 𝑗1.8 Ω, 𝑍2 = 128 + 𝑗288 Ω
Using the transformer rating as base, determine the per-unit
impedance of the transformer from the ohmic value referred to the
low-voltage side. Compute the per unit impedance using the ohmic
value referred to the high-voltage side. (Answer: 𝑍𝑅1 = 𝑍𝑅2 = 0.122 +
j0.252 pu).

32. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 3
Draw an impedance diagram for the electric power system shown in Figure 3 showing
all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for
generator. The three-phase power and line-line ratings are given below.
G1: 90MVA 20kV X = 9%
T1: 80MVA 20/200kV X = 16%
T2: 80MVA 200/20kV X = 20%
G2: 90MVA 18kV X = 9%
Line: 200kV X = 120 Ω
Load: 200kV S = 48MW + j64MVar

33. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 3
Answer:
Figure 3 – Per-unit impedance diagram for Tutorial 3

34. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 4
The one-line diagram of a power system is shown in Figure 4. The three-phase power and line-line
ratings are given below.
G: 80MVA 22kV X = 24%
T1: 50MVA 22/220kV X = 10%
T2: 40MVA 220/22kV X = 6.0%
T3: 40MVA 22/110kV X = 6.4%
Line 1: 220kV X = 121Ω
Line 2: 110kV X = 42.35Ω
M: 68.85MVA 20kV X = 22.5%
Load: 10MVar 4kV Δ-connected capacitors

35. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 4
The three-phase ratings of the three-phase transformer are:
Primary: Y-connected 40MVA, 110 kV
Secondary: Y-connected 40 MVA, 22 kV
Tertiary: Δ-connected 15 MVA, 4 kV
The per phase measured reactances at the terminal of a winding with the second one short-
circuited and the third open-circuited are:
ZPS = 9:6% 40 MVA, 110kV/22 kV
ZPT = 7:2% 40 MVA, 110kV/4 kV
ZST = 12% 40 MVA, 22kV/4 kV
Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 22 kV
as the voltage base for generator.

36. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 4
Answer:
Figure 4 – Per-unit impedance diagram for Tutorial 4

37. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 5
The three-phase power and line-line ratings of the electric power system shown in Figure 5 are
given below.
G1: 60MVA 20kV X = 9%
T1: 50MVA 20/200kV X = 10%
T2: 50MVA 200/20kV X = 10%
M: 43.2MVA 18kV X = 8%
Line: 200kV Z = 120 + j200 Ω
Figure 5 - One-line diagram for Tutorial 5

38. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 5
a) Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 20
kV as the voltage base for generator. (Answer: Figure 6)
b) The motor is drawing 45 MVA, 0.80 power factor lagging at a line-to-line terminal voltage of 18
kV. Determine the terminal voltage and the internal emf of the generator in per unit and in kV.
(Answer: 1.318∠11.82°
𝑝𝑢, 26.359∠11.82°
𝑘𝑉,1.3756∠13.88°
𝑝𝑢, 27.5∠13.88°
𝑘𝑉)
Figure 6 – Per-unit impedance diagram for Tutorial 5

39. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 6
The one-line diagram of a three-phase power system is as shown in Figure 7. Impedances are
marked in per unit on a 100MVA, 400kV base. The load at bus 2 is S2 = 15.93 MW- j33.4 MVar, and at
bus 3 is S3 = 77 MW +j14 MVar. It is required to hold the voltage at bus 3 at 400kV. Working in per
unit, determine the voltage at buses 2 and 1.
(Answer: 1.1∠16.26°
𝑝𝑢, 440∠16.26°
𝑘𝑉, 1.2∠36.87°
𝑝𝑢, 480∠36.87°
𝑘𝑉)
Figure 7 – One line diagram for Tutorial 6

40. 1.2 REVIEW OF TRANSMISSION LINE
MODELS
Tutorial 7
The one-line diagram of a three-phase power system is as shown in Figure 8. The transformer
reactance is 20 percent on a base of 100MVA, 23/115kV and the line impedance is Z = j66.125Ω.
The load at bus 2 is S2 = 184.8MW + j6.6MVar, and at bus 3 is S3 = 0MW + j20MVar. It is required to
hold the voltage at bus 3 at 115kV. Working in per unit, determine the voltage at buses 2 and 1.
(Answer: 1.1∠0°
𝑝𝑢, 126.5∠0°
𝑘𝑉, 1.2∠16.26°
𝑝𝑢, 27.6∠16.26°
𝑘𝑉)
Figure 8 – One line diagram for Tutorial 7