Illinois Institute of Technology
ECE Department
Analytical Methods in Power Systems (Spring 2017)
Problem Set 4
Issued: Tuesday, January 31, 2017
Due: Tuesday, February 07, 2017
Reading Assignment: Read chapters Two and Three.
Problems:
2.3
2.6
2.7
2.8
2.
Illinois Institute of Technology
ECE Department
Analytical Methods in Power Systems (Spring 2017)
Problem Set 1
Issued: Tuesday, January 10, 2017
Due: Tuesday, January 17, 2017
Reading Assignment: Read chapter one and glance at chapter two.
Problems:
1.1
Illinois Institute of Technology
ECE Department
Analytical Methods in Power Systems (Spring 2017)
Problem Set 2
Issued: Tuesday, January 17, 2017
Due: Tuesday, January 24, 2017
Reading Assignment: Read chapter one and glance at chapter two.
Problems:
1.2
Illinois Institute of Technology
ECE Department
Analytical Methods in Power Systems (Spring 2017)
Problem Set 5
Issued: Tuesday, February 07, 2017
Due: Tuesday, February 14, 2017
Reading Assignment: Read chapters Two and Three.
Problems:
2.11
2.12
2.13
2.
ECE 420 Chapter 1
Power System Basics
Grid Layout
Extra High Voltage
EHV
DC
765kV
500kV
345kV
Transmission Voltage
230kV
161kV
138kV
115kV
69kV
Distribution Voltage
46kV
34.5kV
24kV
13.8 kV
13.2kV
12.47kV
4.16kV
69kV
Low Voltage
480 / 277
208/120
240/120
CHAPTER 1
REVIEW OF POWER NETWORK FUNDAMENTALS
1.1 Introduction
The purpose of this course is to introduce a number of engineering concepts involved in
the planning, operation and control of large power systems that are affected by
economical, social and
PROBLEM 1
kVL
13.2 k
n
3
I cust1
kVA
3 kVL n
I cust1
65.6
I cust 2
kW / pf
3 kVL n
I cust 2
87.5
7.62 k
1,500
3 7.62
65.6 Amp
cos 1 0.9 65.6
1, 700 / .85
3 7.62
87.5 Amp
cos 1 0.85 87.5
65.6
25.8
31.8
I total
I cust1 I cust 2
25.8 87.5
31.8 153
29
Scust 2
Chapter 2
ECONOMIC DISPATCH
People use less electricity on Saturdays than on weekdays and at a lower rate between
midnight and 7:00 a.m. than during the daytime. Hence, electric utilities may have fewer
units in service during light load periods. We consi
ECE 420 Chapter 2 Rev 3
Economic Dispatch
Load Cycle
The system load has different cycles Daily, Weekly & annually. As a result different
generator types are required which vary in terms of running cost and starting ability
Generation Type
7,000
Base Load
Per Unit
Basic TRFM
Fault
IF
3-winder
Loads & Sources
For 3P fault, get single phase
model with voltage L-N. If working
in PU, all Z on same base & V = 1.
1P Power
3P Power
Delta Wye
Wye
B
C
A
Delta
B
ZBC
ZAB
A
ZAC
C
Wye Delta
Z AT
1000
IA
100
5'
TR
Time
ECE 420 Chapter 1
Power System Basics
Grid Layout
Extra High Voltage
EHV
DC
765kV
500kV
345kV
Transmission Voltage
230kV
161kV
138kV
115kV
69kV
Distribution Voltage
46kV
34.5kV
24kV
13.8 kV
13.2kV
12.47kV
4.16kV
69kV
Low Voltage
480 / 277
208/120
240/120
Illinois Institute of Technology
ECE Department
Analytical Methods in Power Systems (Spring 2017)
Problem Set 3
Issued: Tuesday, January 24, 2017
Due: Tuesday, January 31, 2017
Reading Assignment: Read chapters one and two.
Problems:
1.9
1-13
1-14
1-15
1-
ECE 420
Analytical Methods in Power Systems
Spring 2017
Instructor: Hassan M. Shanechi
Email: [email protected]
Phone: 312.567.3413
Fax: 312.567.8976
Office: Room 333 SH
Office Hours: My official office hours are 2:00 pm-4:00 pm Tuesdays and Thursdays.
How
CHAPTER 4
CONSTRAINED-ECONOMIC DISPATCH
USING LINEAR PROGRAMMING
This chapter consists of four sections:
1. Solution of Power Flow Equations
2. Loss calculation using Power Flow Equations
3. Real Power Optimization using linear programming
4. Loss minimiz
ECE 420
Spring 2014
Name:
Exam 2
ID #:
Problem 1:
a) Find the dc power flow solution (P and at all buses). Bus 1 is the slack bus.
b) Calculate the flow on line 12 (P12).
[
10
5
5
B= 5
10
5
5
5
10
]
Bus 1 is the slack bus
So,
[
B reduce= 10 5
5 10
]
[ ][
1. Z c = j0.2 , Z L= j0.1 , R=10
( j 0.2 ) ( j 0.1)
= j 0.2
j 0.2+ j 0.1
( j 0.2 ) ( 10 )
Z =R /(Z c Z L )=
=0.004+ j0.1999
j0.2+10
2
|V |
12
S= =
=0.1001+ j5.0005 VA
0.004 j 0.1999
Z
Z c Z L=
10. Z 1= j1 , Z 2=1+ j 1 ,|V |=200 V
Z =Z 1 / Z 2=1 j 1
2
7
8. a. For load is 150MW
If keep both of Pg1 and Pg2 running, we can get = $7.7, Pg1 = -62.5MW, and Pg2 =
212.5MW. So we set Pg1 to 0MW, and set Pg2 to 150MW, and the total cost is $1665 for
one hour.
C2 1.5 (6 150 0.004 1502 120) 1665
So Pg1=0MW and Pg2
15. ZY Z an Z bn Z cn 26 j15
Vac 250 30, Vbc 250 90
Vac I a ZY ( Ia Ib ) ZY Ia 4.808 30
Vbc I b ZY ( Ia Ib ) ZY Ib 4.808 150
Pa VL I L cos(30 (30) 1202.5W
Pa VL I L cos(90 (150) 600.25W
16. Z Z Z Z 12 j12
Y
AN
BN
CN
VL 130 3
set Vab 130 330, Vbc 130 3 90,
CHAPTER 3
LINEAR PROGRAMMING (LP)
The objective in using linear programming is to allocate the scarce resources to the
products in a manner such that profits are maximum, or alternately, costs are minimum.
Consider the following example:
max
Subject to
P
Chapter 2
ECONOMIC DISPATCH
People use less electricity on Saturdays than on weekdays and at a lower rate between
midnight and 7:00 a.m. than during the daytime. Hence, electric utilities may have fewer
units in service during light load periods. We consi
CHAPTER 1
REVIEW OF POWER NETWORK FUNDAMENTALS
1.1 Introduction
The purpose of this course is to introduce a number of engineering concepts involved in
the planning, operation and control of large power systems that are affected by
economical, social and