DOR-01-001-036v2
3/12/04
CHAPTER
1
12:54 PM
Page 1
Introduction to Control
Systems
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
Introduction 2
History of Automatic Control 4
Two Examples of the Use of Feedback 7
Control Engineering Practice 8
E
Chapter.3 Design of Commutator and Brushes
The Commutator is an assembly of Commutator segments or bars tapered in section. The
segments made of hard drawn copper are insulated from each other by mica or micanite, the
usual thickness of which is about 0.8
1
DESIGN OF TRANSFORMERS
Classification:
Based on the number of phases: single or three phase
Based on the shape of the magnetic media: core or shell type
Based on the loading condition: power or distribution type
Design features of power and distribution
C H A P T E R
Learning Objectives
Preference for Electricity
Comparison of Sources of
Power
Sources for Generation of
Electricity
Brief Aspects of Electrical
Energy Systems
Utility and Consumers
Why is the Three-phase a.c.
System Most Popular?
Cost
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UNIT 8
1. Illustrate the principle of force summing devices using suitable examples
and sketches?
Ans: Force summing devices serve as primary transducers and convert the pressu
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UNIT -6
1. Draw the Maxwells Bridge Circuit and derives the expression for the unknown element at
balance?
Ans:
Maxwell's bridge, shown in Fig. 1.1, measures an unknown inducta
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UNIT 7
1. What is transducer? Write the classifications of transducers?
Ans:
Transducer
A measuring device which measures and converts nonelectrical variable into electrical
va
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UNIT-5
1: Explain about storage oscilloscope with block diagram?
Ans:
Storage targets can be distinguished from standard phosphor targets by their ability to
retain a waveform
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UNIT-1
1. What are the basic performance characteristics of a system?
Ans:
STATIC CHARACTE RISTICS
The static characteristics of an instrument are, in general, considered for i
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UNIT-3
1. Draw the Block Schematic of AF Wave analyzer and explain its principle
and Working?
ANS: The wave analyzer consists of a very narrow pass-band filter section which ca
Electronic Measurements & Instrumentation
Question & Answers
UNIT 2
Q.1) Describe the functioning of standard signal generator
Ans.
STANDARD SIGNAL GENERATOR
A standard signal generator produces known and controllable voltages. It is used as power source
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UNIT-4
1. Explain briefly the Basic Features of a CRT?
Ans:
Electrostatic CRTs are available in a number of types and sizes to suit individual requirements.
The important featu
3.4.1
Method of LU decomposition
To solve for the matrices L and U for given matrix A, let us write the i, j th element of equation
(3.40). In general, the element aij can be represented as (from equation (3.40),
i1 1j + i2 2j + up to appropriate term = a
Module 3
Sparsity Technique
3.1
Sparse matrices
Sparse matrix is a matrix in which most (or, at least, signicant number) of the elements are zero.
In the context of power system analysis, the matrices associated with power ow solution are sparse.
For exam
3.2.1
Example of Gaussian elimination
We wish to solve the following matrix equation by Gaussian elimination:
11
23
22
12
17
27
32
15
18
25
34
41
16 x1 10
28 x2 20
=
36 x3 30
36 x4 40
(3.31)
Towards this goal, we proceed with the dierent ste
2.12.2
Example for A.C-D.C load ow
To illustrate the application of the above procedure, let us rst consider the 5-bus system. In this
system, it is now assumed that one bipolar HVDC link is connected between bus 4 and 5 (rectier
at bus 4 and inverter at
2.12
A.C.-D.C. LOAD FLOW
For solving the load ow problem of an A.C. system in which one or more HVDC links are present,
either of the following two approaches are followed;
a. Simultaneous solution technique
b. Sequential solution technique
In simultaneou
2.11.1
Example for Fast-decoupled load-ow technique
As an example, the 5 bus system described earlier is considered again. Starting from the at start,
the initial calculations are shown in Table 2.29. As the mismatch is more than the tolerance, the
algori
2.11
Fast-decoupled load-ow (FDLF) technique
An important and useful property of power system is that the change in real power is primarily
governed by the charges in the voltage angles, but not in voltage magnitudes. On the other hand,
the charges in the
2.10.1
Example for NRLF (rectangular) technique
Taking the 5 bus system as an example again, the NRLF algorithm starts with the at voltage prole
and subsequently dierent quantitative are calculated as shown in Table 2.19.
Table 2.19: Initial calculation w
2.10
NRLF in Rectangular co-ordinates [NRLF (Rect.)]
In rectangular co-ordinates, every complex quantity is expressed in terms of its real and imaginary
parts. Hence, let Vk = Vk ejk = ek + jfk ; Ii = ai + jci and Yik = gik + jbik .
n
n
k=1
k=1
Thus, fr
2.9.1
Example for NRLF (polar) technique
As an example, let us consider again the 5-bus system shown in Fig. 2.16. In this System, n = 5 and
m = 3 (as the number of generator in 3). Initially, let us assume that there is no violation of reactive
power lim
2.8
Basic Newton - Raphson (NR) Techniques
Before discussing the application of NR technique in load ow solution, let us rst review the basic
procedure of solving a set of non-linear algebraic equation by means of NR algorithm. Let there be
n equations in
2.7.1
Example of Gauss Seidel load ow technique
To illustrate the basic procedure of GSLF, let as consider a small 5-bus system as shown in Fig.
2.16. In this system, buses 1-3 are generator buses and buses 4-5 are load buses. Therefore, in this
system, n
Module 1
Introduction
Electricity is the most preferred used form of energy used in industry, homes, businesses and transportation. It can be easily and eciently transported from the production centers to the point of
use. It is highly exible in use as it
2.4
Formation of YBUS matrix in the presence of mutually
coupled elements
Let us consider Fig. 2.9. In this gure, the impedance Zc connected between nodes u and v is
mutually coupled with the impedance Zd connected between nodes x and y through a mutual
i
2.5
Basic power ow equation
From equation (2.7), for a n bus system,
I1 Y11
I Y
2 21
=
In Yn1
Or,
Y12 Y1n V1
Y22 Y2n V2
Yn2 Ynn Vn
(2.24)
n
Ii = Yij Vj
(2.25)
j=1
Complex power injected at bus i is given by,
Si = Pi + j
Module 2
Load Flow Analysis
AC power ow analysis is basically a steady-state analysis of the AC transmission and distribution
grid. Essentially, AC power ow method computes the steady state values of bus voltages and line
power ows from the knowledge of e
5.5.1
Example of contingency analysis for two line outages
As an example, let us consider outage of two lines in the IEEE-14 bus system. Towards this goal,
let us assume that line no. 3 (let it be considered as line e-f) and 17 (let it be considered as li
5.5
Analysis of multiple contingencies (contd.)
Let us now consider a case where two lines, having impedances Za and Zb are removed from the
terminals e-f and g-h respectively. Here it is to be noted that in the original system, impedance Za
and Zb are al