EE369
POWER SYSTEM ANALYSIS
Lecture 3
Three Phase, Power System Operation
Tom Overbye and Ross Baldick
1
Reading and Homework
For lecture 3 read Chapters 1 and 2
For lectures 4 through 6 read Chapter
EE 369, Fall 2017
Homework 3
Assigned:
September 19, 2017
Due:
September 26, 2017
Bonus:
Surge arresters are often used to protect transformers.
1. For protection against switching surges and lightnin
Homework 4 Solution
EE369 Fall 2017
1. Problem 4.8
(a)
Lint = ( 12 107 )(1000)(1000) = 0.05 mH/km
(b)
The inductance of each conductor due to both internal and external flux leakages is expressed as,
EE 369, Fall 2017
Homework 4
Assigned:
September 26, 2017
Due:
October 5, 2017
Bonus:
From your previous homework, you learned about surge arresters. When a surge arrester fails, does it
act as an ope
EE 369, Fall 2017
Homework 1
Assigned:
September 5, 2017
Due:
September 12, 2017
Each problem is worth 2 points
1.
2.
3.
4.
5.
Page 45, Example 2.1 with the following changes: The resistor and the ind
Name:
EE 369, Fall 2017
1. You started your new job as a power systems engineer and your manager hands you a
customers utility bill and a printed e-mail from the customer. The bill shows they were cha
5.1 A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance 2 = 0.19 +
10.34 ohm/km. The load at the receiving end absorbs 10 MVA at 33 kV. Assuming a short line,
calculate:
EE 369, Fall 2017
Homework 7
Assigned:
October 24, 2017
Due:
November 2, 2017
Bonus:
In class, we talked about relay protection utilizing ANSI numbers. For example, 50/51 is instantaneous
overcurrent
Homework 2 Solution
EE369 Fall 2017
1. Problem 2.24
The complex power supplied to lighting load,
S1 = 10 + j0 kVA
The complex power supplied to induction motor,
S2 = S cos1 (p.f) (lagging)
= 10 cos1 (
Name:
EE 369, Fall 2017
/20
1. A 3-phase, 60 Hz, 80 mile completely transposed transmission line is built
composed of Parrot conductors. The outside diameter is 1.506 inches; stranding
of 54/19 (Al/St
Homework 5 Solution
EE369 Fall 2017
1. Problem 5.1
(a)
The relation between sending-end and receiving-end is given by,
VS = AVR + BIR
IS = CVR + DIR
As Kirchhoffs voltage law,
VS = VR + zlIR
= VR + (3
Homework 7 Solution
EE369 Fall 2017
1. Problem 6.9
To solve using the Gauss method, rearrange the equations to solve for x1 and x2 as follows:
x1 = 13 x2 + 0.6333
x2 = 3.0 x1 2
Substitute the initial
R
C
L
7.28 ohm
3.22331E-011 C/m
3.57681E-007 H/m
radius
1.41 cm
conductor r
0.0728 ohm/km
length
200 km
GMR_L
22.9 cm
GMR_C 0.2594909958 m
GMD
8.1888937121
r'd
rd
L_total
XL
0.052441
0.0673355769
0.07
EE 369, Fall 2017
Homework 5
Assigned:
October 10, 2017
Due:
October 17, 2017
Note j4.4e-6 S/km = j4.4*10-6 S/km
345kV Double-Circuit Transmission Line
Scale: 1 cm = 2 m
5.7 m
7.8 m
8.5 m
7.6 m
7.6 m
EE369
POWER SYSTEM ANALYSIS
Lecture 4
Power System Operation, Transmission
Line Modeling
Tom Overbye and Ross Baldick
1
Reading and Homework
For lectures 4 through 6 read Chapter 4
we will not be co
EE369
POWER SYSTEM ANALYSIS
Lecture 6
Development of Transmission Line Models
Tom Overbye and Ross Baldick
1
Homework
HW 5 is problems 4.26, 4.32, 4.33, 4.36,
4.38, 4.49, 5.1, 5.7, 5.8, 5.10, 5.16, 5
EE 369
POWER SYSTEM ANALYSIS
Lecture 5
Development of Transmission Line Models
Tom Overbye and Ross Baldick
1
Reading
For lectures 5 through 7 read Chapter 4
we will not be covering sections 4.7, 4.1
EE369
POWER SYSTEM ANALYSIS
Lecture 2
Complex Power, Reactive Compensation, Three
Phase
Tom Overbye and Ross Baldick
1
Reading and Homework
Read Chapters 1 and 2 of the text.
HW 1 is Problems 2.2, 2.3
EE 369
POWER SYSTEM ANALYSIS
Lecture 13
Newton-Raphson Power Flow
Tom Overbye and Ross Baldick
1
Announcements
Read Chapter 12, concentrating on
sections 12.4 and 12.5.
Homework 11 is 6.43, 6.48, 6.
EE 369
POWER SYSTEM ANALYSIS
Lecture 15
Economic Dispatch
Tom Overbye and Ross Baldick
1
Announcements
Read Chapter 12, concentrating on
sections 12.4 and 12.5.
Read Chapter 7.
Homework 11 is 6.43, 6.
EE 369
POWER SYSTEM ANALYSIS
Lecture 14
Power Flow
Tom Overbye and Ross Baldick
1
Announcements
Read Chapter 12, concentrating on
sections 12.4 and 12.5.
Homework 11 is 6.43, 6.48, 6.59, 6.61,
12.19,
EE 369
POWER SYSTEM ANALYSIS
Lecture 12
Power Flow
Tom Overbye and Ross Baldick
1
Announcements
Homework 9 is: 3.47, 3.49, 3.53, 3.57, 3.61,
6.2, 6.9, 6.13, 6.14, 6.18, 6.19, 6.20; due
November 7. (U
EE 369
POWER SYSTEM ANALYSIS
Lecture 8
Transformers, Per Unit
Tom Overbye and Ross Baldick
1
Announcements
For lectures 8 to 10 read Chapter 3
Homework 6 is 5.9, 5.14, 5.24, 5.26, 5.27,
5.28, 5.29, 5.
EE 369
POWER SYSTEM ANALYSIS
Lecture 17
Optimal Power Flow, LMPs
Tom Overbye and Ross Baldick
1
Announcements
Read Chapter 7.
Homework 11 is 6.43, 6.48, 6.59, 6.61,
12.19, 12.22; due November 21.
Home
EE 369, Fall 2017
Homework 2
Assigned:
September 12, 2017
Due:
September 19, 2017
Bonus
Given the simplified system below, assume a line is lossless with impedance jX and derive the real power
transfe
Homework 3 Solution
EE369 Fall 2017
1. Problem 3.4
(a)
The turns ratio is,
at =
N1
N2
=
E1rated
E2rated
=
2400
240
= 10
Thus, the primary voltage is,
E1 = at E2 = 10 230 0 = 2300 0 V
(b)
The current t