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Exam (elaborations)

Power System Analysis by Hadi Saadat (3rd Edition) – Complete Solution Manual [10 Files Merged]

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Ace your Power System Analysis (EE/EEE 432 or EE 451) exams with this comprehensive 2025/2026 solution manual for Hadi Saadat’s Power System Analysis, 3rd Edition. This merged edition (10 files combined) includes step-by-step solutions, illustrative examples, and solved numerical problems from all major chapters—perfect for electrical engineering students focusing on power flow, fault analysis, stability, and load modeling. Ideal for study, homework help, or final exam preparation at top engineering universities worldwide.

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EE 432 – Power System Analysis
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EE 432 – Power System Analysis











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Institution
EE 432 – Power System Analysis
Course
EE 432 – Power System Analysis

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Uploaded on
October 31, 2025
Number of pages
100
Written in
2025/2026
Type
Exam (elaborations)
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All Chapters Covered




P OW E R SY ST E M
· A NA LY S I S




H a d i Saa d at
O0 ( 00 IOI
':
'
100
.
100 l!O

, CONTENTS




1 THE POWER SYSTEM: AN OVERVIEW 1

2 BASIC PRINCIPLES 5

3 GENERATOR AND TRANSFORMER MODELS;
THE PER-UNIT SYSTEM 25

4 TRANSMISSION LINE PARAMETERS 52

5 LINE MODEL AND PERFORMANCE 68

6 POWER FLOW ANALYSIS 107

7 OPTIMAL DISPATCH OF GENERATION 147

8 SYNCHRONOUS MACHINE TRANSIENT ANALYSIS 170

9 BALANCED FAULT 181

10SYMMETRICAL COMPONENTS AND UNBALANCED FAULT 208

11STABILITY 244

12 PO
WER SYSTEM CONTROL 263

, CHAPTER 1 PROBLEMS




1.1 The demand estimation is the starting point for planning the future electric
power supply. The consistency of demand growth over the years has led to numer-
ous attempts to fit mathematical curves to this trend. One of the simplest curves is

p = Poea( t-to)

where a is the average per unit growth rate, P is the demand in year t, and Po
is the given demand at year to.
Assume the peak power demand in the United States in 1984 is 480 GW with
an average growth rate of 3.4 percent. Using MATLAB, plot the predicated peak
demand in GW from 1984 to 1999. Estimate the peak power demand ḟor the
year 1999.
We use the ḟollowing commands to plot the demand growth

tO = 84; PO = 480;
a=.034;
t =(84:1:99)';
P=PO*exp(a*(t-tO));
disp('Predicted PeakDemand -GW') disp([t, P])
plot(t,P),grid
xlabel('Year'),ylabel('Peakpower demand GW') P99=PO*exp(a*(99 -
tO))


The result is
1

, 2 CONTENTS




PredictedPeakDemand-GW
84.0000 480.0000
85.0000 496.6006
86.0000 513.7753
87.0000 531.5441
88.0000 549.9273
89.0000 568.9463
90.0000 588.6231
91.0000 608.9804
92.0000 630.0418
93.0000 651.8315
94.0000 674.3740
95.0000 697.6978
96.0000 721.8274
97.0000 746.7916
98.0000 772.6190
99.0000 799.3398

P99 =

799.3398

The plot oḟ the predicated demand is shown n Ḟigure 1.

800
750........

700
Pea
k
Power 650
Demand 600 . . . . . . . . . . . . . . .
GW
550
500
45084 86 88 90 92 94 96 98 100
Year

ḞIGURE 1
Peak Power Demand ḟor Problem 1.1

1.2 In a certain country, the energy consumption is expected to double in 10 years.

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