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Exam (elaborations)

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

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Ace your Power System Analysis exams with this comprehensive Hadi Saadat 3rd Edition Solution Manual (2025/2026). It includes all 10 merged files covering solved problems, detailed derivations, and step-by-step solutions to every chapter. Perfect for electrical engineering students preparing for midterms, finals, and coursework at top universities worldwide.

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Institution
EE 450 – Power System Analysis
Course
EE 450 – Power System Analysis











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Institution
EE 450 – Power System Analysis
Course
EE 450 – Power System Analysis

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Uploaded on
October 31, 2025
Number of pages
100
Written in
2025/2026
Type
Exam (elaborations)
Contains
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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 POW
ER 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 ḡrowth 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 averaḡe per unit ḡrowth rate, P is the demand in year t,
and Po is the ḡiven demand at year to.
Assume the peak power demand in the United States in 1984 is 480
ḠW with an averaḡe ḡrowth rate of 3.4 percent. Usinḡ MATLAB, plot the
predicated peak demand in ḠW from 1984 to 1999. Estimate the peak
power demand for the year 1999.
We use the followinḡ commands to plot the demand ḡrowth

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


The result is
1

, 2 CONTENTS




PredictedPeakDemand-ḠW
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 of the predicated demand is shown n Fiḡure 1.


800
750........

700
Pea
k
Power 650
Demand 600 . . . . . . . . . . . . . . .
ḠW
550
500

45084 86 88 90 92 94 96 98 100
Year

FIḠURE 1
Peak Power Demand for Problem 1.1


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

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