BE 45'? Homework #2 Solution
1. Consider the twobus system shown in Fig. l l. The two generators and
transformers are assumed of equal rating  soc MVA  which is the 3
phase base power for all pu unit data given in what follows.
" Line has series react
February 18, 2016
NAME fOLqu Ford
1. (4 pts) For a line to line fault analysis using symmetrical components( choose the
right answer) :
a) The positive. negative and zero sequence networks at the fault point are connected
in series
(b) The positive and ne
EDC3
1.0 Introduction
In the last set of notes (EDC2), we saw how
to use penalty factors in solving the EDC
problem with losses. In this set of notes, we
want to address two closely related issues.
What are, exactly, penalty factors?
How
to obtain the p
AGC 2
1.0 Introduction
In the last set of notes, we developed a model of
the speed governing mechanism, which is given
below:
xE
KG
1
(PC )
1 TG s
R
(1)
In these notes, we want to extend this model so
that it relates the actual mechanical power into
the
AGC 3
1.0 Introduction
The primary controller response to a
load/generation imbalance results in generation
adjustment so as to maintain load/generation
balance. However, due to droop, it also results
in a nonzero steadystate frequency deviation.
This f
AGC 4
1.0 Problem 11.2
For the isolated generating station with local
load shown in Fig. 1 below, it is observed that
PL=0.1pu brings about =0.2rad/sec in the
steadystate.
1
R
PL

PC +

1
s 1
+
10
1 10s
Fig. 1
(a) Find 1/R.
Solution:
We need the trans
Stability 1
1.0 Introduction
We now begin Chapter 14.1 in your text.
Our previous work in this course has focused on
analysis of currents during faulted conditions in
order to design protective systems necessary to
detect and clear faults. Now we turn our
Stability 2
1.0 Introduction
We ended our last set of notes, concluding that
the following equation characterizes the
electromechanical dynamics of a synchronous
machine.
(t ) P0 Ea V sin
M
M
(1)
Xd
Now I want to do an example of the most simple
power sy
Stability 3
1.0 Introduction
In our last set of notes (Stability 2), we
described in great detail the behavior of a
synchronous machine following a faulted
condition that is cleared by protective relaying.
A key figure for us was Fig. 1 (it was Fig. 15 in
Protection 4
1.0 Introduction
Recall there are five basic classes of relays:
Magnitude relays
Directional relays
Ratio (impedance) relays
Differential relays
Pilot relays
We study the impedance relay in these notes.
Other names for impedance relays a
Protection 5
1.0 Introduction
Recall there are five basic classes of relays:
Magnitude relays
Directional relays
Ratio (impedance) relays
Differential relays
Pilot relays
We provide only a very brief overview of
differential and pilot relays in these
Protection 1
(read section 13.0 of text)
1.0 Introduction
Faults do not occur that frequently 1/year/100
miles of transmission is typical. Distribution
systems may see more than this.
However, when they do occur, it is imperative
that they be removed as q
Protection 2
1.0 Introduction
There are five basic classes of protective relays:
Magnitude relays
Directional relays
Ratio (impedance) relays
Differential relays
Pilot relays
We will study each of these.
But the simplest is the magnitude relay. We
ad
Protection 3
1.0 Introduction
Recall there are five basic classes of relays:
Magnitude relays
Directional relays
Ratio (impedance) relays
Differential relays
Pilot relays
We study the directional relay in these notes.
This material addresses section
Examples
I will do 2 of your assigned problems
Problem 12.3
The problem requires us to find the three
currents Ia, Ib, and Ic, and also Vng, given that
Ea=1, Eb=1, and Ec=j1.
Ia
j1
Ec=j
Eb=1
j1
Ea
Ea=1
g
n
j1
Eb
Ec
j1
j1
Ib
j1
Ic
The network is symmetri
Examples
Example 1: Compute sequence components
of the following balanced abc sequence
linetoneutral voltages.
Van 2770
V abc Vbn 277 120
Vcn 277120
Solution:
1 Van
1 1
1
1
V S A V abc 1 2 Vbn
3
2
1
Vcn
1 2770
1 1
1
1 2 277 120
3
2
1
Symmetrical Components 2
Sequence impedances
Although the following focuses on loads, the
results apply equally well to lines, or lines
and loads. Read these notes together with
sections 12.6 and 12.9 of text.
Consider the Yconnected balanced load,
Fig.
EDC2
1.0 Introduction
In the previous set of note (EDC1), we
developed the EDC problem with losses
included and provided the corresponding
KKT conditions which lead to its solution
procedure. However, in the example that
was given, we neglected losses. No
EDC1
1.0 Introduction
In EE 303, we study the economic dispatch
calculation (EDC) problem. We review our
EE 303 work on EDC in this class, but we
solve it in a different way. In addition, we
extend the problem to account for losses.
Economic dispatch is t
1.0
Costs of Generating Electrical Energy
Overview
The shortrun costs of electrical energy generation
can be divided into two broad areas: fixed and
variable costs. These costs are illustrated in Fig. 1
below.
Fixed costs
Interest on bonds
Return to stoc
EE 457
Problem Set #3 Solution
Due by March 10, 2106
1. Consider the transmission system shown below
The relay parameters are:
Line to neutral voltage =79.7kV
Current transformer ratio = 500:5
Voltage Transformer ratio: 80,000:100
The relay setting guidel
EE457 Assignment 1 Due Tuesday, Feb. 2 2016
1. Problem 12.2 from the textbook
Find the symmetrical components of Ea = 110°, Eb : 14—90", EC 2 21135°
Convert to balanced. Find 1;, If, L9. See Eq. 12.5 and 12.6, pg. 447.
Eu 1 1 1 E3
E1, = 1 a2 a E;—
EC 1 a
Homework #5 Solution
Assignment: 13.1, 13.2, 13.3 ,13.4, and 13.5 Bergen & Vittal
Solutions:
13.1
Assuming a fault at bus 4.
Deriving the Thevenin equivalent circuits:
Z ag j5 j10 j10 j10 j35
Z ag j5 j10 j10 j10 j35
0
Z ag j5 j10 j10 j10 j35
SLG Fault
34.
Homework #6 Solution
Assignment: 14.1, 14.4, 14.5, 14.8, 14.9, and 14.10 Bergen & Vittal
Solutions:
14.1
Part A:
Wkinetic
3
H 3 Wkinetic H * Sb 5 *100M 500Megajoules
SB
Part B:
To find the kinetic energy of a 10ton truck (see image below which is
actuall
Homework #7 Solution
Assignment: 11.1 through 11.6 Bergen & Vittal
Solutions:
11.1
PM
1
Modified Equation 11.6 because gen. speed not fed back
R
R * PM (.01rad / MW sec)(100MW ) 1rad / sec
1
speed (1) (60) 9.55 r. p.m.
2
The new speed is therefore:
Homework #9 Solution
Assignment: 11.8  11.12 and 11.16 11.18 Bergen & Vittal
Solutions:
11.8
IC1 8.0 .003P 1
G
IC2 8.0 .001PG 2
IC3 7.5 .002PG 3
It helps to estimate the common IC graphically and then fine tune it by
iteration.
Incremental Cost vs. Power
EE 457, Exam 2, Spring 2015, Dr. McCalley, Time: 75 minutes,
Calculator allowed; closed book, closed notes, communication devices not allowed
1.
(25 pts) Consider the 138 kV transmission system.
Bus 1
Bus 2
3.2+j32.0
R12
R21
3.2+j32.0
R23
Bus 3
R32
Bus 4
Name: _
EE 457, Exam 1, Spring, 2015
Closed Book, Closed Notes, Closed Computer, No Communication Devices, Calculators allowed
1.
(35 pts) Symmetrical Fault Analysis: Consider a single generator supplying a balanced R+jX load as
shown in the diagram below
Homework #1 Solution
Problem #1:
1. Using the output from the Matlab code provided above, for =,
compute the ratio K()=imax ()/i1max , where the max indicates
the maximum absolute value of the waveform.
2. Repeat for the following values of :
=3, 2.5,