Three Phase Synchonours Generator and AVR

Three Phase Synchonours Generator
&
Automatic Voltage Regulator
(AVR)
EMRE ÖZTOKLU
2014
International University of Sarajevo
Faculty of Engineering and Natural Sciences

Dean
Professor Dr.Fuat GÜRCAN
Program Coordinator
Professor Dr. Migdat Hodzic
Referees
Assoc. Professor Dr. Izudin Džaﬁć
Assist. Professor Dr. Emir Karamehmedovic
Date of the graduation
2014

Contents
Abstract 1
1 Introduction 3
1.1 History of Electric Generator . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Method of Regulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.1 Conventional Controllers . . . . . . . . . . . . . . . . . . . . . 8
2 Synchronous Generator 13
2.1 Main Components of a Synchronous Generator . . . . . . . . . . . . . 16
2.1.1 The Stator: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.1.2 The Rotor: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.1.3 Slip Rings: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.2 The Generators Synchronous Rotation Speed . . . . . . . . . . . . . . 28
2.3 Field Excitation & Exciters . . . . . . . . . . . . . . . . . . . . . . . 31
2.4 Brushless Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3 Control of Generator (Automatic Voltage Regulator -AVR ) 35
3.1 Principals of Automatic Voltage Control . . . . . . . . . . . . . . . . 35
3.2 PID Theory Explained . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2.1 PI and PID Theory . . . . . . . . . . . . . . . . . . . . . . . . 40
3.3 Three Phase Diode Bridge Rectiﬁers . . . . . . . . . . . . . . . . . . 42
4 Work and Result 47
4.1 Bulding of Three Phase Synchronous Generator . . . . . . . . . . . . 47
4.1.1 The Stator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4.1.2 The Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.1.3 Slip rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.2 Diameters of The Three Phase Synchronous Generator . . . . . . . . 49
4.3 Equivalent Circuit of Synchronous Generator . . . . . . . . . . . . . . 50
4.4 Phasor diagram of a synchronous generator . . . . . . . . . . . . . . . 55
4.5 Calculating of Power, Torque and Eﬃciency . . . . . . . . . . . . . . 56
4.6 Open Circuit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.7 Short Circuit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.8 Testing of PI controller . . . . . . . . . . . . . . . . . . . . . . . . . . 62
i

Contents Contents
Acknowledgments 71
.1 Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Bibliography 81
ii

Abstract
In this project, it will explain prensible that how to build three phase synchronous
generator with DC exitation. In a synchronous generator, a DC current is applied to
the rotor winding, which produces a rotor magnetic field. The rotor of the generator
is then turned by a prime mover, producing a rotating magnetic field within the
machine. This rotating magnetic field induces a three-phase set of voltages within
the stator windings of the generator. Moreover, I will explain and/or implement an
Automatic Voltage Regulator (AVR) for the generator. For that reason, the part
of generator is more important for all generator to protect itself and our electrical
staffs. Although manufacturers often use the analog AVR’s, advanced features of
microcontrollers that can be solved various control algorithms and protection designs
with simple softwares will cause the system to gain more performance. AVR designed
to keep the generator terminal voltage on nominal value.
1

1 Introduction
Classification of AC Rotating Machines
Synchronous Machines
• Synchronous Generators : A DC current is applied to the rotor winding
producing a rotor magnetic field. The rotor is then turned by external means
producing a rotating magnetic field, which induces a 3-phase voltage within
the stator winding.
• Synchronous Motors : A 3-phase set of stator currents produces a rotating
magnetic field causing the rotor magnetic field to align with it. The rotor
magnetic field is produced by a DC current applied to the rotor winding. It
used as motors as well as power factor compensators (synchronous condensers).
Asynchronous (Induction) Machines:
• Induction Motors : Most widely used electrical motors in both domestic
and industrial applications.
• Induction Generators : Due to lack of a separate field excitation, these
machines are rarely used as generators.
Energy Conversion
• Generators convert mechanical energy to electric energy.
• Motors convert electric energy to mechanical energy.
• The construction of motors and generators are similar
• Every generator can operate as a motor and vice versa.
• The energy or power balance is :
–Generator: Mechanical power = electric power + losses
–Motor: Electric Power = Mechanical Power + losses
In electricity generation, an electric generator is a device that converts mechanical
energy to electrical energy in the form of alternating current. A generator forces
electric current to flow through an external circuit. The source of mechanical energy
3

Chapter 1 Introduction
may be a reciprocating or turbine steam engine, water falling through a turbine
or waterwheel, an internal combustion engine, a wind turbine,[1] a hand crank,
compressed air, or any other source of mechanical energy. Generators provide nearly
all of the power for electric power grids. The reverse conversion of electrical energy
into mechanical energy is done by an electric motor, and motors and generators have
many similarities. Many motors can be mechanically driven to generate electricity
and frequently make acceptable generators.
In a synchronous generator, a dc current is applied to the rotor winding, which
produces a rotor magnetic field. The rotor of the generator is then turned by a
prime mover, producing a rotating magnetic field within the machine. This rotating
magnetic field induces a three-phase set of voltages within the stator windings of
the generator.
1.1 History of Electric Generator
Before the connection between magnetism and electricity was discovered, electro-
static generators were used. They operated on electrostatic principles. Such gener-
ators generated very high voltage and low current. They operated by using moving
electrically charged belts, plates, and disks that carried charge to a high potential
electrode. The charge was generated using either of two mechanisms:
• Electrostatic induction
• The triboelectric effect, where the contact between two insulators leaves them
charged.
Because of their inefficiency and the difficulty of insulating machines that produced
very high voltages, electrostatic generators had low power ratings, and were never
used for generation of commercially significant quantities of electric power. The
Wimshurst machine and Van de Graaff generator are examples of these machines
that have survived.
In 1827, Hungarian Anyos Jedlik started experimenting with the electromagnetic
rotating devices which he called electromagnetic self-rotors, now called the Jedlik’s
dynamo. In the prototype of the single-pole electric starter (finished between 1852
and 1854) both the stationary and the revolving parts were electromagnetic. He for-
mulated the concept of the dynamo at least 6 years before Siemens and Wheatstone
but didn’t patent it as he thought he wasn’t the first to realize this. In essence the
concept is that instead of permanent magnets, two electromagnets opposite to each
other induce the magnetic field around the rotor. It was also the discovery of the
principle of self-excitation.[2]
In the years of 1831–1832, Michael Faraday discovered the operating principle of
electromagnetic generators. The principle, later called Faraday’s law, is that an
electromotive force is generated in an electrical conductor which encircles a varying
4

1.1 History of Electric Generator
magnetic flux. He also built the first electromagnetic generator, called the Faraday
disk, a type of homopolar generator, using a copper disc rotating between the poles
of a horseshoe magnet. It produced a small DC voltage.
The horseshoe-shaped magnet (A) created a magnetic field through the disk (D).
When the disk was turned, this induced an electric current radially outward from
the center toward the rim. The current flowed out through the sliding spring contact
m, through the external circuit, and back into the center of the disk through the
axle.
Figure 1.1: The first electric generator ( Faraday disk )
This design was inefficient, due to self-cancelling counterflows of current in regions
that were not under the influence of the magnetic field. While current was induced
directly underneath the magnet, the current would circulate backwards in regions
that were outside the influence of the magnetic field. This counterflow limited the
power output to the pickup wires, and induced waste heating of the copper disc.
Later homopolar generators would solve this problem by using an array of magnets
arranged around the disc perimeter to maintain a steady field effect in one current-
flow direction.
Another disadvantage was that the output voltage was very low, due to the single
current path through the magnetic flux. Experimenters found that using multiple
turns of wire in a coil could produce higher, more useful voltages. Since the output
voltage is proportional to the number of turns, generators could be easily designed to
produce any desired voltage by varying the number of turns. Wire windings became
a basic feature of all subsequent generator designs.
The dynamo was the first electrical generator capable of delivering power for in-
dustry. The dynamo uses electromagnetic induction to convert mechanical rotation
5

into direct current through the use of a commutator. The first dynamo was built by
Hippolyte Pixii in 1832.
A dynamo machine consists of a stationary structure, which provides a constant
magnetic field, and a set of rotating windings which turn within that field. On small
machines the constant magnetic field may be provided by one or more permanent
magnets; larger machines have the constant magnetic field provided by one or more
electromagnets, which are usually called field coils.
Through a series of accidental discoveries, the dynamo became the source of many
later inventions, including the DC electric motor, the AC alternator, the AC syn-
chronous motor, and the rotary converter.
Alternating current generating systems were known in simple forms from the discov-
ery of the magnetic induction of electric current. The early machines were developed
by pioneers such as Michael Faraday and Hippolyte Pixii.
Faraday developed the "rotating rectangle", whose operation was heteropolar - each
active conductor passed successively through regions where the magnetic field was
in opposite directions.[3]The first public demonstration of a more robust "alterna-
tor system" took place in 1886.[4] Large two-phase alternating current generators
were built by a British electrician, J.E.H. Gordon, in 1882. Lord Kelvin and Se-
bastian Ferranti also developed early alternators, producing frequencies between
100 and 300 Hz. In 1891, Nikola Tesla patented a practical "high-frequency" al-
ternator (which operated around 15 kHz).[5]After 1891, polyphase alternators were
introduced to supply currents of multiple differing phases.[6] Later alternators were
designed for varying alternating-current frequencies between sixteen and about one
hundred hertz, for use with arc lighting, incandescent lighting and electric motors.[7]
Large power generation dynamos are now rarely seen due to the now nearly universal
use of alternating current for power distribution. Before the adoption of AC, very
large direct-current dynamos were the only means of power generation and distri-
bution. AC has come to dominate due to the ability of AC to be easily transformed
to and from very high voltages to permit low losses over large distances.
6

1.2 Method of Regulation
Definitions of Automatic Voltage Control
The process of maintaining voltage at the junction points of an electrical system
within given limits to ensure technically feasible conditions of operation for users of
electrical energy and of separate systems and to increase the economic efficiency of
their operation.
Most users of electrical energy may tolerate long-term voltage deviations from the
nominal value of not more than ±5 percent. An increase of the voltage above its
nominal value leads to a shorter service life for electrical equipment, and a decrease
lowers the productivity and economically efficient operation of the machinery and
the capacity of the transmission lines; it may also interfere with the stable operation
of synchronous machines and induction motors.
Automatic voltage control is needed because of variable operating conditions at the
user’s plant and at electrical energy sources. Because an increase in load leads to an
increase in current strength and consequently also to an increase in voltage losses
in various sectors of the grid, the voltage at the machine may be out of permissible
limits. Therefore, the buses of electric power stations and the secondary voltage
buses of regional substations are usually regulated by reverse (balancing) means.
In this case, the voltage is held above its nominal value with increased loads and
lowered with decreased loads. This lessens the variation of voltage fluctuation for the
user. However, such regulation does not generally exclude the need for automatic
voltage control for each user.
Automatic voltage control at electric power stations is obtained by regulating the
excitation of the synchronous generators. In substations automatic voltage control
is obtained by regulation of the excitation of synchronous compensators, if such have
been installed in these substations, by automatic change under load of the ratio of
transformers, and by control of the power of static capacitor banks. At the users
of electrical energy, automatic voltage regulation is applied by means of control of
high-power synchronous motors and of the power of static capacitor banks. The
problem of specific selection of regulating devices is usually solved on the basis of
technical and economic analysis.[8, 9, 10, 11]
7

Some examples of AVR :
Figure 1.2: Conventional and Static Exciter with AVR
Figure 1.3: via AVR , Brusless Excitations with exciter & without exciter
1.2.1 Conventional Controllers
Today, a number of different controllers are used in industry [12] and in many other
fields. In quite general way those controllers can be divided into two main groups:
• conventional controllers
• unconventional controllers
As conventional controllers we can count a controllers known for years now, such
as P, PI, PD, PID, Otto-Smith, all their different types and realizations, and other
controller types [13]. It is a characteristic of all conventional controllers that one has
to know a mathematical model of the process in order to design a controller. Un-
conventional controllers utilize a new approaches to the controller design in which
knowledge of a mathematical model of a process generally is not required. Ex-
amples of unconventional controller are a fuzzy controller and neuro or neuro-fuzzy
controllers. Many industrial processes are nonlinear and thus complicate to describe
mathematically. However, it is known that a good many nonlinear processes can sat-
isfactory controlled using PID controllers providing that controller parameters are
tuned well. Practical experience shows that this type of control has a lot of sense
since it is simple and based on 3 basic behavior types: proportional (P), integrative
8

(I) and derivative (D). Instead of using a small number of complex controllers, a
larger number of simple PID controllers is used to control simpler processes in an
industrial assembly in order to automates the certain more complex process. PID
controller and its different types such as P, PI and PD controllers are today a basic
building blocks in control of various processes. In spite their simplicity, they can be
used to solve even a very complex control problems, especially when combined with
different functional blocks, filters (compensators or correction blocks), selectors etc.
A continuous development of new control algorithms insure that the time of PID
controller has not past and that this basic algorithm will have its part to play in
process control in foreseeable future. It can be expected that it will be a backbone
of many complex control systems.
1.2.1.1 Basic Controller Types
PID controllers use a 3 basic behavior types or modes:
• P - proportional
• I - integrative
• D - derivative.
While proportional and integrative modes are also used as single control modes, a
derivative mode is rarely used on it’s own in control systems. Combinations such
as PI and PD control are very often in practical systems. It can be also shown that
PID controller is a natural generalization of a simplest possible controller - On-off
controller.
P controller A proportional control system is a type of linear feedback control
system. Proportional control is how most drivers control the speed of a car. If the
car is at target speed and the speed increases slightly, the power is reduced slightly,
or in proportion to the error (the actual versus target speed), so that the car reduces
speed gradually and reaches the target point with very little, if any, "overshoot", so
the result is much smoother control than on-off control [14]. In the proportional
control algorithm, the controller output is proportional to the error signal, which is
the difference between the set point and the process variable. In other words, the
output of a proportional controller is the multiplication product of the error signal
and the proportional gain. This can be mathematically expressed as
Pout = Kp ∗ e(t) (1.1)
Where ;
Pout: Output of the proportional controller
9

Kp: Proportional gain
e(t): Instantaneous process error at time ’t’ [e(t) = SP−PV ]
SP: Set point
PV : Process variable
With increase in Kp :
• Response speed of the system increases.
• Overshoot of the closed-loop system increases.
• Steady-state error decreases.
But with high Kp value, closed-loop system becomes unstable.
I controller In a proportional control of a plant whose transfer function does not
possess an integrator1/s, there is a steady-state error, or offset, in the response to a
step input. Such an offset can be eliminated if integral controller is included in the
system. In the integral control of a plant, the control signal, the output signal from
the controller, at any instant is the area under the actuating error signal curve up to
that instant. But while removing the steady-state error, it may lead to oscillatory
response of slowly decreasing amplitude or even increasing amplitude, both of which
is usually undesirable [14].
PI controller In control engineering, a PI Controller (proportional-integral con-
troller) is a feedback controller which drives the plant to be controlled by a weighted
sum of the error (difference between the output and desired set-point) and the inte-
gral of that value. It is a special case of the PID controller in which the derivative
(D) part of the error is not used. The PI controller is mathematically denoted as:
Gc = Kp +
KI
s
(1.2)
Gc = K ∗ (1 +
1
sTi
) (1.3)
10

Figure 1.4: courtesy
Integral control action added to the proportional controller converts the original
system into high order. Hence the control system may become unstable for a large
value of Kp since roots of the characteristic eqn. may have positive real part. In this
control, proportional control action tends to stabilize the system, while the integral
control action tends to eliminate or reduce steady-state error in response to various
inputs. As the value of Ti is increased;
• Overshoot tends to be smaller
• Speed of the response tends to be slower.
PI controllers are the most often type used today in industry. A control without D
mode is used when:
• fast response of the system is not required
• large disturbances and noise are present during operation of the process
• there is only one energy storage in process (capacitive or inductive)
• there are large transport delays in the system
If there are large transport delays present in the controlled process, error prediction is
required. However, D mode cannot be used for prediction because every information
is delayed till the moment when a change in controlled variable is recorded. In such
cases it is better to predict the output signal using mathematical model of the
process in broader sense (process + actuator).
11

2 Synchronous Generator
Synchronous generators are built with either a stationary or a rotating dc magnetic
field. A stationary-field synchronous generator has the same outward appearance
as a dc generator. The salient poles create the dc field, which is cut by a revolving
armature. The armature possesses a 3-phase winding whose terminals are connected
to three slip-rings mounted on the shaft. A set of brushes, sliding on the slip-rings,
enables the armature to be connected to an external 3-phase load. The armature
is driven by a gasoline engine, or some other source of motive power. As it rotates,
a 3-phase voltage is induced, whose value depends upon the speed of rotation and
upon the dc exciting current in the stationary poles. The frequency of the voltage
depends upon the speed and the number of poles on the field. However, for greater
outputs, it is cheaper, safer, and more practical to employ a revolving dc field.
A revolving-field synchronous generator has a stationary armature called a stator.
The 3-phase stator winding is directly connected to the load, without going through
large, unreliable slip-rings and brushes. A stationary stator also makes it easier
to insulate the windings because they are not subjected to centrifugal forces. Fig.
Fig. 2.16 is a schematic diagram of such a generator, sometimes called an alternator.
The field is excited by a dc generator, usually mounted on the same shaft. Note
that the brushes on the commutator have to be connected to another set of brushes
riding on slip-rings to feed the dc current Ix into the revolving field.
Like the previous DC generator, the operation of the Synchronous Generator is
also based on Faraday’s law of electromagnetic induction and works in a similar
way to the automotive alternator. The difference this time is that the synchronous
generator produces a three-phase AC voltage output from its stator windings, unlike
the DC generator which produces a single DC or direct current output. Single-
phase synchronous generators are also available for low power synchronous generator
systems.
13

Chapter 2 Synchronous Generator
Figure 2.1: Section view of Synchronous Generator
Basically, the synchronous generator is a synchronous electro-mechanical machine
used as a generator and consists of a magnetic field on the rotor that rotates and
a stationary stator containing multiple windings that supplies the generated power.
The rotors magnetic field system (excitation) is created by using either permanent
magnets mounted directly onto the rotor or energised electromagnetically by an
external DC current flowing in the rotor field windings. This DC field current is
transmitted to the synchronous machine’s rotor via slip rings and carbon or graphite
brushes. Unlike the previous DC generator, synchronous generators do not require
complex commutation allowing for a simpler construction.
These machines are used for nearly all electrical-energy generation by utility com-
panies. As motors, they tend to be used in higher-power, lower-speed applications
than those forwhich inductionmotors are used. Unlike other types of ac and dc mo-
tors that we have studied to this point, the speed of a synchronous motor does not
vary with mechanical load (assuming a constant-frequency ac source). Instead, we
will see that they run at synchronous speed Ns, which is given by Formula (2.1) [15]
When talking about the "synchronous generator", the terminology used for the de-
scription of the machines parts is the reverse to that for the description of the DC
generator. The field windings are the windings producing the main magnetic field
which are the rotor windings for a synchronous machine, and the armature wind-
ings are the windings where the main voltage is induced usually called the stator
windings. In other words, for a synchronous machine, the rotor windings are the
field windings and the stator windings are the armature windings as shown.
14

Synchronous Generator
Figure 2.2: Synchronous Generator Construction
The example above shows the basic construction of a synchronous generator which
has a wound salient two-pole rotor. This rotor winding is connected to a DC supply
voltage producing a field current, If .The external DC excitation voltage which
can be as high as 250 volts DC, produces an electromagnetic field around the coil
with static North and South poles. When the generators rotor shaft is turned by
the turbines blades (the prime mover), the rotor poles will also move producing a
rotating magnetic field as the North and South poles rotate at the same angular
velocity as the turbine blades, (assuming direct drive). As the rotor rotates, its
magnetic flux cuts the individual stator coils one by one and by Faraday’s law, an
emf and therefore a current is induced in each stator coil.
The magnitude of the voltage induced in the stator winding is, as shown above, a
function of the magnetic field intensity which is determined by the field current, the
rotating speed of the rotor, and the number of turns in the stator winding. As the
synchronous machine has three stator coils, a 3-phase voltage supply corresponding
to the windings, A, B and C which are electrically 120◦
apart is generated in the
stator windings and this is shown above.
This 3-phase stator winding is connected directly to the load, and as these coils are
stationary they do not need to go through large unreliable slip-rings, commutator
or carbon brushes. Also because the main current generating coils are stationary, it
makes it easier to wind and insulate the windings because they are not subjected to
rotational and centrifugal forces allowing for greater voltages to be generated.[16]
Then the synchronous generator operates in a similar way to the automotive car
alternator and consists of the two following common parts:
15

2.1 Main Components of a Synchronous Generator
2.1.1 The Stator:
Main Components of a Synchronous Generator
The Stator(Fig. 2.3) carries the three separate ( 3-phase ) armature windings
physically and electrically displaced from each other by 120 degrees producing an
AC voltage output. The stator of a synchronous machine has the same construction
as the stator of a three-phase induction motor. In review, the stator contains a
set of three-phase windings that establish the stator field. This field consists of P
magnetic poles, alternating between north and south around the circumference of
the stator and rotating at synchronous speed. In a synchronous machine, the set of
stator windings is called the armature. [15]
Figure 2.3: Stator
The stator produces a rotating magnetic field that is proportional to the frequency
supplied. This motor rotates at a synchronous speed, which is given by the following
equation:
Ns =
120 ∗ f
P
(2.1)
where
f = frequency of the supply frequency [Hz]
p = number of poles
Ns = Synchronous speed
16

The winding is always connected in wye and the neutral is connected to
ground. A wye connection is preferred to a delta connection because:
1. The voltage per phase is only 1/
√
3 or 58%of the voltage between the lines.
This means that the highest voltage between a stator conductor and the
grounded stator core is only 58% of the line voltage. We can therefore reduce
the amount of insulation in the slots which, in turn, enables us to increase the
cross section of the conductors. A larger conductor permits us-to increase the
current and, hence, the power output of the machine.
2. When a synchronous generator is under load, the voltage induced in each
phase becomes distorted, and the waveform is no longer sinusoidal. The dis-
tortion is mainly due to an undesired third harmonic voltage whose frequency
is three times that of the fundamental frequency. With a wye connection, the
distorting line-to-neutral harmonics do not appear between the lines because
they eﬀectively cancel each other. Consequently, the line voltages remain si-
nusoidal under all load conditions. Unfortunately, when a delta connection
is used, the harmonic voltages do not cancel, but add up. Because the delta
is closed on itself, they produce a third-harmonic circulating current, which
increases the I2
R losses.
2.1.1.1 The Wiring of Stator:
1. Single- layer winding
2. Double layer winding
Single- layer winding: In this type of winding, as shown in Fig. 2.4, each slot
contains only one coil side. It means a coil occupies two complete slots. The number
of coils in the machine is equal to half the number of slots in the stator, or rotor
and armature[17]
• Per slot only one coil side is placed
• Used only in small ac machines
North and South pole are generated by one coil group per phase.
Figure 2.4: Shematic of Single Wiring
17

Double -Layer Winding: In this type, as shown in Fig. 2.5, each slot contains two
coil sides, housed one over the other. The number of coils is equal to the number of
slots in the stator and armature.[17]
• Slot contains even number (may be 2,4,6 etc.) of coil-sides in two layers
• Double-layer winding is more common above about 5kW machines
North and South pole are generated by two coil groups and direction of current
ﬂow in N-pole and S-pole coils opposite.[18]
Figure 2.5: Double- layer wiring
The advantages of double-layer winding over single layer winding are as follows:
• Easier to manufacture and lower cost of the coils
• Fractional-slot winding can be used
• Chorded-winding is possible
• Lower-leakage reactance and therefore , better performance of the machine
• Better emf waveform in case of generators
Manufacturing much more expensive than single-layer winding, therefore used usu-
ally only in bigger machines. For example: High voltage machines up to 30 kV ( “
High Voltage” : U > 1000VRMS )[18]
18

2.1.1.2 The Connection of Stator:
Almost all electric power generation and most of the power transmission in the world
is in the form of three-phase AC circuits. A three-phase AC system consists of three-
phase generators, transmission lines, and loads.There are two major advantages of
three-phase systems over a single-phase system:
1. More power per kilogram of metal form a three-phase machine
2. Power delivered to a three-phase load is constant at all time, instead of pulsing
as it does in a single-phase system
A three-phase generator consists of three single-phase generators with voltages of
equal amplitudes and phase diﬀerences of 120◦
Figure 2.6: Relationship between three phase Voltages
VA(t) =
√
2 ∗ sin (ωt) [V olt] VA = V ∠0◦
[V olt]
VB(t) =
√
2 ∗ sin (ωt − 120◦
) [V olt] VB = V ∠ − 120◦
[V olt]
VC(t) =
√
2 ∗ sin (ωt − 240◦
) [V olt] VC = V ∠ − 240◦
[V olt]
Z = R + jX
θ = tan−1 jX
R
|Z| = Z∠θ◦
=⇒
VA/|Z| = I∠ − θ◦
VB/|Z| = I∠ − 120◦
− θ◦
VC/|Z| = I∠ − 240◦
− θ◦
There are two types of connections in three-phase circuits: Y and :
19

Table 2.1: Shematic of wye and delta connection
Wye (Y) Connection: Assuming a resistive load is that:
Figure 2.7: Wye connection resistive load
Van = Vφ∠0◦
Ia = Iφ∠0◦
Vbn = Vφ∠ − 120◦
Ib = Iφ∠ − 120◦
Vcn = Vφ∠ − 240◦
Ic = Iφ∠ − 240◦
The current in any line is the same as the current in the corresponding phase:
ILL = Iφ (2.2)
Magnitudes of the line-to-line voltages and the line-to-neutral voltages are related
as:
20

VLL = Vφ ∗
√
3 (2.3)
In addition, the line voltages are shifted by 30◦
with respect to the phase voltages.
In a connection with abc sequence, the voltage of a line leads the phase voltage.
Delta ( ) Connection: Assuming a resistive load is that:
Figure 2.8: Delta connection resistive load
Vab = Vφ∠0◦
Iab = Iφ∠0◦
Vbc = Vφ∠ − 120◦
Ibc = Iφ∠ − 120◦
Vca = Vφ∠ − 240◦
Ica = Iφ∠ − 240◦
21

Voltages are related:
VLL = Vφ (2.4)
The magnitudes are related:
IL = Iφ ∗
√
3 (2.5)
For the connections with the abcphase sequences, the current of a line lags the
corresponding phase current by 30◦
2.1.2 The Rotor:
The rotor carries the magnetic field either as permanent magnets or wound field
coils connected to an external DC power source via slip rings and carbon brushes.
The rotor of a synchronous machine is usually a P-pole electromagnet with field
windings that carry dc currents. (In smaller machines, the rotor can be a per-
manent magnet, but we will concentrate on machines with eld windings.) The eld
current can be supplied from an external dc source through stationary brushes to
slip rings mounted on the shaft. The slip rings are insulated from one another
and from the shaft. Another method is to place a small ac generator, known as
an exciter, on the same shaft and use diodes mounted on the shaft to rectify the
ac. This avoids the maintenance associated with brushes and slip rings. Two-pole
and four-pole synchronous machines are illustrated in Figure Fig. 2.9 The rotor can
either be cylindrical, as shown for the two-pole machine, or it can have salient poles
22

as illustrated for the four-polemachine. Generally, salient-pole construction is less
costly but is limited to low-speedmachines havingmany poles. High-speed machines
usually have cylindrical rotors. Salient-pole machines are common in hydroelectric
power generation, whereas cylindrical-rotor machines are common in thermal (coal,
nuclear, etc.) power plants.[15]
Figure 2.9: Cross sections of two synchronous machines. The relative positions of
the stator and rotor poles are shown for motor action. Torque is developed in
the direction of rotation because the rotor poles try to align themselves with the
opposite stator poles
Therefore, synchronous generators are built with two types of rotors:
2.1.2.1 Salient-Poles rotor
Salient-Poles rotor: Salient-Pole rotors are usually driven by low-speed hydraulic
turbines.
• The stator has a laminated iron-core with slots and three phase windings
placed in the slots.
• They are usually four and more poles.
for example: to produce 50 Hz electricity
p=12, n=500 rpm
p=24, n=250 rpm
• The rotor has salient poles excited by dc current.
• DC current is supplied to the rotor through slip-rings and brushes.[19]
23

Most hydraulic turbines have to turn at low speeds ( between 50 and 300 r/min) in
order to extract the maximum power from a waterwheel, Because the rotor is directly
coupled to the waterweel, and because a frequency of 50 Hz or 60 Hz is required,
a large number of poles are required on the rotor. Low-speed rotors always possess
a large diameter to provide the necessary space for the poles. The salient poles
are mounted on a large circular stell frame which is fixed to a revolving vertical
shaft. To ensure good cooling, the field coils are made of bare copper bars, with
the turns insulated from each other by strips of mica. The coils are connected in
series, with adjacent poles having opposite polarities. In addition to the DC field
winding, we often add a squrriel-cage winding,embedded in the polefaces. Under
normal conditions, this winding does not carry any current because the rotor turns
ay synchronous speed. However, when the load on the generator changes suddenly,
the rotor speed begins to fluctuate, producing momentary speed variations above
and below synchronous speed. This induces a voltage in the squrirel-cage winding,
causing a large current to flow therein. The current reacts with magnetic field of the
stator, producing forces which dampen the oscillation of the rotor. For this reason,
the squirrel-cage winding is sometimes called a damper-winding. [20]
The damper winding also tends to maintain balanced 3-phase voltages between the
lines, even when the line current are unequal due to unbalanced load conditions.
The Wiring of Salient Rotors:
Figure 2.10: winding of rotor (8 poles, 4 poles 1 slot and 2 slot )
24

Figure 2.11: 4 ploes 1 slot rotor winding
Figure 2.12: 4 ploes 2 slot rotor winding
Figure 2.13: 4 poles 3 slot rotor winding
25

2.1.2.2 Cylindrical Rotor(Non-Salient)
Cylindrical rotor (Round rotor or Non-Salient rotor): Round rotor structure
is used for high speed synchronous machines, such as steam turbine generators.The
stator is a ring shaped laminated iron-core with slots.
• Three phase windings are placed in the slots.
• They are usually two- and four-pole rotors ( it can not be so many poles.Because,
the magnetic field will be reduced )
for example: to produce 50 Hz electricity ;
p=2, n=3000 rpm
p=4, n=1500 rpm
• Round solid iron rotor with slots.
• A single winding is placed in the slots. Dc current is supplied through slip
rings.[19]
It is well known that high-speed steam turbines are smaller and more efficient than
low-speed turbines.The same is true of high-speed synchronous generators. However,
to generate the required frequency we can not use less than 2 poles, and this fixes the
highest possible speed. On a 60Hz system it is 3600 [r/min]. The next lower speed
is , corresponding to a 4 pole machine. Therefore, these steam-turbine generators
possess either 2 or 4 poles. The high speed of rotation produces strong centrifugal
forces, which impose an upper limit on the diameter of the rotor. In the case of a
rotor turning at 3600 [r/min], the elastic limit of steel requires the manufacturer to
limit the diameter to a maximum of 1-2 m. On the other hand, to build the powerful
1000 [MV A] to 1500 [MV A] generators the volume of the rotors has to be large.
It follows that high-power, high speed rotors have to be very long.[20]
- Rotors are made laminated to reduce eddy current losses
26

Figure 2.14: Salient pole rotor (left side) & Non-Salient pole rotor (right side)
2.1.3 Slip Rings:
Slip rings are metal rings completely encircling the shaft of a machine but insulated
from it. One end of a DC rotor winding is connected to each of the two slip rings on
the machine’s shaft. Graphite-like carbon brushes connected to DC terminals ride
on each slip ring supplying DC voltage to ﬁeld windings regardless the position or
speed of the rotor.
Slip rings and brushes have certain disadvantages: increased friction and wear (there-
fore, needed maintenance), brush voltage drop can introduce signiﬁcant power losses.
Still this approach is used in most small synchronous machines.
How many slip rings are there?
27

Figure 2.15: view of two slip rings of rotor and three slip rings of rotor
In conclusion, If the synchronous motor is single phase then there are two slip rings
& if this motor is three phase so the slip rings are three in number. A synchronous
motor has a separately excited ﬁeld. If the excitation comes from a stationary DC
source it has 2 slip (collector) rings. A brushless induction motor has no slip rings
because the exciter armature rotates and so do the rectiﬁers. A permanent magnet
motor, used with variable frequency drives, is another type of synchronous motor
that has no slip rings. A three phase motor with 3 slip rings is a Wound Rotor motor.
Wound rotor motors are variable speed motors that were used for such applications
as bridges and cranes before variable speed drives.
2.2 The Generators Synchronous Rotation Speed
The frequency of the output voltage depends upon the speed of rotation of the rotor,
in other words its "angular velocity", as well as the number of individual magnetic
poles on the rotor. In our simple example above, the synchronous machine has
two-poles, one North pole and one South pole. In other words, the machine has two
individual poles or one pair of poles, (North-South) also known as pole pairs. As the
rotor rotates one complete revolution, 360◦
, one cycle of induced emf is generated, so
the frequency will be one-cycle every full rotation or 360◦
. If we double the number
of magnetic poles to four, (two pairs of poles), then for every revolution of the rotor,
two cycles of induced emf will be generated and so on.
Since one cycle of induced emf is produced with a single pair of poles, the number of
cycles of emf produced in one revolution of the rotor will therefore be equal to the
number of pole pairs, P. So if the number of cycles per revolution is given as: P/2
28

2.2 The Generators Synchronous Rotation Speed
relative to the number of poles and the number of rotor revolutions Nm per second
is given as: Nm/60, then the frequency, ( ƒe ) of the induced emf will be defined as:
fe =
Nm ∗ P
120
(2.6)
Nm = NS (2.7)
ωm = ωS (2.8)
Where
fe : is the electrical frequency,[Hz]
NM : is mechanical speed of magnetic field (rotor speed for syn-
chronous machine), [rpm]
NS : is speed of stator
P : is the number of poles.
ω : omega [N.m]
In a synchronous motor, its angular velocity is fixed by the frequency of the supply
voltage so N is commonly known as the synchronous speed. Then for a "P"-pole
synchronous generator the speed of rotation of the prime mover (the turbine blades)
in order to produce the required frequency output of either 50Hz or 60Hz of the
induced emf will be:
So for a given synchronous generator designed with a fixed number of poles, the
generator must be driven at a fixed synchronous speed to keep the frequency of the
induced emf constant at the required value, either 50Hz or 60Hz to power mains
appliances. In other words, the frequency of the emf produced is synchronised
with the mechanical rotation of the rotor. Then from above, we can see that to
generate 60 Hz using a 2-pole machine, the rotor must rotate at 3600 [revs/min],
or to generate 50 Hz using a 4-pole machine, rotor must rotate at1500[revs/min].
For a synchronous generator that is being driven by an electrical motor or steam
generator, this synchronous speed may be easy to achieve however, when used as a
wind turbine synchronous generator, this may not be possible as the velocity and
power of the wind is constantly changing.
However, for a synchronous machine, the more poles it has the larger, heavier and
more expensive becomes the machine which may or may not be acceptable.
29

Table 2.2: Number of poles and their rpm
One solution is to use a synchronous machine with a low number of poles which can
rotate at a higher speed of 1500 to 3600 rpm driven through a gearbox. The low
rotational speed of the wind turbines rotor blades is increased through a gearbox
which allows the generator speed to remain more constant when the turbines blade
speed changes as a 10% change at 1500rpm is less of a problem than a 10% change at
100rpm. This gearbox can match the generators speed to variable rotational speeds
of the blades allowing for variable speed operation over a wider range.
However, the use of a gearbox or pulley system requires regular maintenance, in-
creases the weight of the wind turbine, generates noise, increases power losses and
reduces system eﬃciency as extra energy is required to drive the gearboxes cogs and
internal components.
There are many advantages to using a direct drive system without a mechanical
gearbox, but the omission of a gearbox means a larger synchronous machine with
an increase in both size and cost of the generator, which then has to operate at a
low speeds. So how can we operate a synchronous generator in a low speed wind
turbine system whose rotor blade speed is determined only by the winds power.
By rectifying the generated 3-phase supply into a constant DC or direct current
supply.[16]
30

2.3 Field Excitation & Exciters
2.3 Field Excitation & Exciters
The dc field excitation of a large synchronous generator is an important part of
its overall design.The reason is that the field must ensure not only a stable ac
terminal voltage, but must also respond to sudden load changes in order to maintain
system stability. Quickness of response is one of the important features of the field
excitation. In order to attain it. two dc generators are used: a main exciter and a
pilot exciter. Static exciters that involve no rotating parts at all are also employed.
The main exciter feeds the exciting current to the field of the synchronous generator
by way of brushes and slip-rings. For instance, under normal conditions the exciter
voltage lies between 125 V and 600 V. It is regulated manually or automatically
by control signals that vary the current IC, produced by the pilot exciter in Figure
(Fig. 2.16)
Figure 2.16: Schematic diagram and cross-section view of typical synchronous
generator and DC exciter
The power rating of the main exciter depends upon the capacity of the synchronous
generator. For Example,typically a 25 kW exciter is needed to excite a 1000 kVA
alternator (2.5% of its rating) whereas a 2500 kW exciter suffices for an alternator
of 500 MW (only 0.5% of its rating). Under normal conditions the excitation is
varied automatically. It responds to the load changes so as to maintain a constant
ac line voltage or to control the re¬active power delivered to the electric utility
system. A serious disturbance on the system may produce a sudden voltage drop
across the terminals of the alternator. The exciter must then react very quickly to
keep the ac voltage from falling. For example, the exciter voltage may have to rise to
twice its normal value in as little as 300 to 400 milliseconds. This represents a very
quick response, considering that the power of the exciter may be several thousand
kilowatts.
• For a constant load, the power factor of a synchronous motor can be varied
from a leading valueto a lagging value by adjusting the DC field excitation in
Figure (Fig. 2.17 ). Field excitation can beadjusted so that PF = 1 in Figure
31

(Fig. 2.17a). With a constant load on the motor, when the field excitationis
increased, the counter EMF (Vg) increases. The result is a change in phase
between statorcurrent (I ) and terminal voltage (Vt), so that the motor oper-
ates at a leading power factorin Figure (Fig. 2.17b). Vp in Figure (Fig. 2.17) is
the voltage drop in the stator winding’s due to the impedance of thewindings
and is 90◦
out of phase with the stator current. If we reduce field excitation,
the motorwill operate at a lagging power factor in Figure (Fig. 2.17c). Note
that torque angle,α also varies as fieldexcitation is adjusted to change power
factor.
Figure 2.17: Synchronous Motor Field Excitation
Synchronous motors are used to accommodate large loads and to improve the power
factor oftransformers in large industrial complexes.
2.4 Brushless Excitation
Due to brush wear and carbon dust, we constantly have to clean, repair, and replace
brushes, slip-rings, and commutators on conventional dc excitation systems. To
eliminate the problem, brushless excitation systems have been developed. Such a
system consists of a 3-phase stationary-field generator whose ac output is rectified
by a group of rectifiers. The dc output from the rectifiers is fed directly into the
field of the synchronous generator in Figure (Fig. 2.18)
32

2.4 Brushless Excitation
Figure 2.18: Typical brushless exciter system
The armature of the ac exciter and the rectifiers are mounted on the main shaft
and turn together with the synchronous generator. In comparing the excitation
system of Fig ( Fig. 2.18) with that of Fig (Fig. 2.16), we can see they are identical,
except that the 3-phase rectifier replaces the commutator, slip-rings, and brushes.
In other words, the commutator (which is really a mechanical rectifier) is replaced
by an electronic rectifier. The result is that the brushes and slip-rings are no longer
needed. The dc control current IC from the pilot exciter regulates the main exciter
output lx as in the case of a conventional dc exciter. The frequency of the main
exciter is generally two to three limes the synchronous generator frequency (60 [Hz]).
The increase in frequency is obtained by using more poles on the exciter than on the
synchronous generator. The Figure shows the rotating portion of a typical brushless
exciter. Sialic exciters that involve no rotating parts at all are also employed.
33

3 Control of Generator (Automatic
Voltage Regulator -AVR )
3.1 Principals of Automatic Voltage Control
Voltage transformers provide signals proportional to line voltage to the avr where
it is compared to a stable reference voltage. The difference (error) signal is used
to control the output of the exciter field. For example, if load on the generator
increases, the reduction in output voltage produces an error signal which increases
the exciter field current resulting in a corresponding increase in rotor current and
thus generator output voltage. Due to the high inductance of the generator field
windings, it is difficult to make rapid changes in field current. This introduces
a considerable "lag" in the control system which makes it necessary to include a
stabilizing control to prevent instability and optimize the generator voltage response
to load changes. Without stabilizing control, the regulator would keep increasing
and reducing excitation and the line voltage would continually fluctuate above and
below the required value. Modern voltage regulators are designed to maintain the
generator line voltage within better than +/- 1% of nominal for wide variations of
machine load.
35

Chapter 3 Control of Generator (Automatic Voltage Regulator -AVR )
Figure 3.1: Close loop PI controller
3.2 PID Theory Explained
Overview
Proportional-Integral-Derivative (PID) control is the most common control algo-
rithm used in industry and has been universally accepted in industrial control. The
popularity of PID controllers can be attributed partly to their robust performance
in a wide range of operating conditions and partly to their functional simplicity,
which allows engineers to operate them in a simple, straightforward manner.
As the name suggests, PID algorithm consists of three basic coefficients; propor-
tional, integral and derivative which are varied to get optimal response. Closed loop
systems, the theory of classical PID and the effects of tuning a closed loop control
system are discussed in this paper.
Three different types of error processing are commonly used in control systems, P,
I and D, named after three basic ways of manipulating the error information.
Proportional - Proportional error correction multiplies the error by a (negative)
constant P, and adds it to the controlled quantity.
Integral - Integral error correction incorporates past experience. It integrates the
error over a period of time, and then multiplies it by a (negative) constant I and
36

adds it to the controlled quantity. Equilibrium is based on the average error and
avoids oscillation and overshoot providing a more stable system.
Derivative - Derivative error correction is based on the rate of change of the er-
ror and takes into account future expectations. It is used in so called "Predictive
Controllers". The first derivative of the error over time is calculated, and multiplied
by another (negative) constant D , and also added to the controlled quantity. The
derivative term provides a rapid response to a change in the system.
Combinations of all three methods of error processing are often used simultaneously
in "PID" controllers to address different system performance priorities. Where noise
may be a problem, the derivative term is not used.
1. Control System
The basic idea behind a PID controller is to read a sensor, then compute the desired
actuator output by calculating proportional, integral, and derivative responses and
summing those three components to compute the output. Before we start to define
the parameters of a PID controller, we shall see what a closed loop system is and
some of the terminologies associated with it.
Closed Loop Systems (Automatic Control Fig. 3.1):Once the initial operat-
ing parameters have been set, an open loop system is not responsive to subsequent
changes or disturbances in the system operating environment such as temperature
and pressure, or to varying demands on the system such as power delivery or load
conditions. For continual monitoring and control over the operating state of a sys-
tem without operator intervention, for more precision or faster response, automatic
control systems are needed.[21]
Closed Loop System In a typical control system, the process variable is the system
parameter that needs to be controlled, such as temperature(ºC) , pressure(psi) , or
flow rate (liters/minute) . A sensor is used to measure the process variable and pro-
vide feedback to the control system. The set point is the desired or command value
for the process variable, such as 100 degrees Celsius in the case of a temperature
control system. At any given moment, the difference between the process variable
and the set point is used by the control system algorithm (compensator), to deter-
mine the desired actuator output to drive the system (plant). For instance, if the
measured temperature process variable is 100 ºC and the desired temperature set
point is 120 ºC, then the actuator output specified by the control algorithm might
be to drive a heater. Driving an actuator to turn on a heater causes the system
to become warmer, and results in an increase in the temperature process variable.
This is called a closed loop control system, because the process of reading sensors to
provide constant feedback and calculating the desired actuator output is repeated
continuously and at a fixed loop rate as illustrated in Fig. 3.2
37

In many cases, the actuator output is not the only signal that has an effect on the
system. For instance, in a temperature chamber there might be a source of cool
air that sometimes blows into the chamber and disturbs the temperature. Such a
term is referred to as disturbance. We usually try to design the control system to
minimize the effect of disturbances on the process variable.[22]
Figure 3.2: Block diagram of a typical closed loop system
Defintion of Terminlogies The control design process begins by defining the perfor-
mance requirements. Control system performance is often measured by applying a
step function as the set point command variable, and then measuring the response
of the process variable. Commonly, the response is quantified by measuring defined
waveform characteristics. Rise Time is the amount of time the system takes to
go from 10% to 90% of the steady-state, or final, value. Percent Overshoot is the
amount that the process variable overshoots the final value, expressed as a percent-
age of the final value. Settling time is the time required for the process variable to
settle to within a certain percentage (commonly 5%) of the final value. Steady-State
Error is the final difference between the process variable and set point. Note that
the exact definition of these quantities will vary in industry and academia.
Figure 3.3: Response of a typical PID closed loop system.
After using one or all of these quantities to define the performance requirements for
a control system, it is useful to define the worst case conditions in which the control
38

system will be expected to meet these design requirements. Often times, there is
a disturbance in the system that affects the process variable or the measurement
of the process variable. It is important to design a control system that performs
satisfactorily during worst case conditions. The measure of how well the control
system is able to overcome the effects of disturbances is referred to as the disturbance
rejection of the control system.
In some cases, the response of the system to a given control output may change over
time or in relation to some variable. A nonlinear system is a system in which the
control parameters that produce a desired response at one operating point might not
produce a satisfactory response at another operating point. For instance, a chamber
partially filled with fluid will exhibit a much faster response to heater output when
nearly empty than it will when nearly full of fluid. The measure of how well the
control system will tolerate disturbances and nonlinearities is referred to as the
robustness of the control system.
Some systems exhibit an undesirable behavior called deadtime. Deadtime is a delay
between when a process variable changes, and when that change can be observed.
For instance, if a temperature sensor is placed far away from a cold water fluid inlet
valve, it will not measure a change in temperature immediately if the valve is opened
or closed. Deadtime can also be caused by a system or output actuator that is slow
to respond to the control command, for instance, a valve that is slow to open or
close. A common source of deadtime in chemical plants is the delay caused by the
flow of fluid through pipes.
Loop cycle is also an important parameter of a closed loop system. The interval
of time between calls to a control algorithm is the loop cycle time. Systems that
change quickly or have complex behavior require faster control loop rates. [22]
Once the performance requirements have been specified, it is time to examine the
system and select an appropriate control scheme. In the vast majority of applica-
tions, a PID control will provide the required results
Figure 3.4: Response of a closed loop system with deadtime.
39

3.2.1 PI and PID Theory
PI controller : In control engineering, a PI Controller (proportional-integral con-
troller) is a feedback controller which drives the plant to be controlled by a weighted
sum of the error (difference between the output and desired set-point) and the inte-
gral of that value. It is a special case of the PID controller in which the derivative
(D) part of the error is not used.
u(t) = KP e(t) + K ˙I
ˆ
e(τ)dτ + Kd
d
dt
e(t) (3.1)
U(s) = KP +
K ˙I
s
+ Kds = KP (1 +
1
s Ti
+ sTd) (3.2)
where;
Kp: Proportional gain, a tuning parameter
Ki: Integral gain, a tuning parameter
Kd: Derivative gain, a tuning parameter
e(t) : is the error or deviation of actual measured value (PV) from the
setpoint (SP). (Error = SP − PV )
t : Time or instantaneous time (the present)
τ : Variable of integration; takes on values from time 0 to the present t.
Ti: is the integral time
Td: is the derivative time
U(s): The transfer function of the PID controller
Figure 3.5: Block diagram of a basic PID control algorithm
Proportional Response
The proportional component depends only on the difference between the set point
and the process variable. This difference is referred to as the Error term. The
40

proportional gain (Kp)determines the ratio of output response to the error signal.
For instance, if the error term has a magnitude of 10, a proportional gain of 5 would
produce a proportional response of 50. In general, increasing the proportional gain
will increase the speed of the control system response. However, if the proportional
gain is too large, the process variable will begin to oscillate. If Kp is increased
further, the oscillations will become larger and the system will become unstable and
may even oscillate out of control.[23]
The proportional term is given by:
Pout = Kp e(t) (3.3)
Integral Response
The integral component sums the error term over time. The result is that even a
small error term will cause the integral component to increase slowly. The integral
response will continually increase over time unless the error is zero, so the effect
is to drive the Steady-State error to zero. Steady-State error is the final difference
between the process variable and set point. A phenomenon called integral windup
results when integral action saturates a controller without the controller driving the
error signal toward zero.[23]
The integral term is given by:
Iout = Ki
ˆ t
0
e(τ) dτ (3.4)
Derivative Response
The derivative component causes the output to decrease if the process variable is
increasing rapidly. The derivative response is proportional to the rate of change of
the process variable. Increasing the derivative time (Td) parameter will cause the
control system to react more strongly to changes in the error term and will increase
the speed of the overall control system response. Most practical control systems use
very small derivative time (Td), because the Derivative Response is highly sensitive
to noise in the process variable signal. If the sensor feedback signal is noisy or if the
control loop rate is too slow, the derivative response can make the control system
unstable. [23]
The derivative term is given by:
Dout = Kd
d
dt
e(t) (3.5)
41

Figure 3.6: Block diagram of a basic PI control algorithm
The controller output is given by
u(t) = KP e(t) + KI
ˆ
e(τ)dτ (3.6)
s domain:
G(s) = Kp +
KI
s
=
KP ∗ s + KI
s
(3.7)
or
G(s) = Kp(1 +
1
s Ti
) (3.8)
where;
e(t)=is the error or deviation of actual measured value (PV) from the setpoint (SP).
e(t) = SP − PV (3.9)
3.3 Three Phase Diode Bridge Rectifiers
Rectifiers or AC/DC converts are used for conversion AC (alternating current) into
DC (direct current).
In general, rectifiers can be classified as controlled and uncontrolled devices depend-
ing whether they are built on diodes or thyristors. Rectifiers built on thyristors
42

Figure 3.7: Six Pulse Rectifiers
have the ability to control DC output. Based on their design and output, recti-
fiers can be bridge or midpoint rectifiers, single or three phase rectifiers, half or full
wave rectifiers. Six pulse three-phase rectifiers shown inFig. 3.7 is considered in the
following.[24]
Im most cases in each phase there is a single phase transformer before the rectifier
used to isolate rectifier from AC supply and to adjust the desired voltage level. Also,
different three phase transformers can be used to shift voltages and currents (de-
pending on transformer connection Wye or Delta ).This way it is possible to remove
some low order harmonics. Six pulse rectifier is built of six diodes or thyristors.
Cathodes of first three resistors are connected to one point as well as the anodes
of the remaining three. At any instant, only two diodes conduct: one from the
cathode-connected group, and one from the anode-connected group. Resulting out-
put voltage consists of six pulses per period. Combinations of open diodes from
Fig. 3.7 in one period are: D1 and D6, D6 and D2, D2 and D4, D4 and D3, D3 and
D5, D5 and D1. When poly-phase AC is rectified, the phase-shifted pulses overlap
each other to produce a DC output that is much “smoother” (has less AC con-
tent) than that produced by the rectification of single-phase AC. This is a decided
advantage in high-power rectifier circuits, where the sheer physical size of filtering
components would be prohibitive but low-noise DC power must be obtained. The
diagram in (Fig. 3.8 ) shows the full-wave rectification of three-phase AC.
43

Figure 3.8: Three-phase AC and 3-phase full-wave rectiﬁer output
DC output voltage can be calculated by this Formula ( 3.10 ):
Vout =
3
√
2 ∗ VL
π
(3.10)
Figure 3.9: Synchronous Generator Rectiﬁer Circuit
VL= line to line voltage ( VLL)
our generator connected as a Wye connection, magnitudes of the line to line voltages
and the line to neutral voltages are related as in Formula (3.11) ;
44

VL = VLL =
√
3 ∗ Vφ (3.11)
and current is related in this Formula ( 3.12 );
IL = Iφ (3.12)
In general;
Vout =
3 ∗
√
3.
√
2 ∗ Vφ
π
(3.13)
Vφ =line to neutral phase voltages ( in the RMS )
45

4 Work and Result
4.1 Bulding of Three Phase Synchronous Generator
4.1.1 The Stator
We said that synchronous generator carries the three separate ( 3-phase ) armature
windings physically and electrically displaced from each other by 120 degrees pro-
ducing an AC voltage output. Because of that, I had to create one stator and each
phases have to sperate from each other 1200
and each slots should be sperated from
each other 600
.the stator is as shown under.
Figure 4.1: The Stator
47

Chapter 4 Work and Result
4.1.2 The Rotor
For rotor, I decided to make two poles salient rotor which shown in Figure(Fig. 2.2)
and when we start to turn this rotor is that will eﬀect each stator coils and we will
creat magnetic ﬁeld with our excitation current and we will have three phase ac
voltage from our wye connected output of stator.
Figure 4.2: The Salient Rotor
Ns = 120∗f
P
= 120∗50
2
= 3000 [rpm]
number of poles are two in our generator, becuse of that, the rotor should turn with
3000 rpm.( frequency is 50 Hz). if our frequency would be 60 Hz , our rotation will
be 3600 rpm.
4.1.3 Slip rings
The synchronous motor is single phase then there are two slip rings & if this motor
is three phase so the slip rings are three due to our generator is single phase and it
has two slip rings.
48

4.2 Diameters of The Three Phase Synchronous Generator
4.2 Diameters of The Three Phase Synchronous
Generator
Figure 4.3: Diameters
49

Figure 4.4: 3D view of Sator and Rotor in Autocad
Figure 4.5: Demonstration of Our Generator
4.3 Equivalent Circuit of Synchronous Generator
The internally generated voltage in a single phase of a synchronous machineEA is
not usually the voltage appearing at its terminals. It equals to the output voltage
50

VØ only when there is no armature current in the machine. The reasons that the
armature voltage EA is not equal to the output voltage VØ are:
1. Distortion of the air-gap magnetic field caused by the current flowing in the
stator (armature reaction);
2. Self-inductance of the armature coils;
3. Resistance of the armature coils;
4. Effect of salient-pole rotor shapes.
Equivalent circuit of a synchronous generator:
• Each phase has resistance R and inductance L
• Synchronous reactance XS = 2 ∗ π ∗ f ∗ L = ωL
• R is typically << XS, therefore neglected unless interested in efficiency or
heating effects
The generator is connected to a lagging load, the load current If in figure ( Fig. 2.2)
will create a stator magnetic field Bs, which will produce the armature reaction
voltageEstat .Therefore, the phase voltage will be:
VØ = EA + Estat (4.1)
Assuming that the load reactance is X, the armature reaction voltage is:
Estat = −j ∗ XS ∗ IA − RA ∗ IA (4.2)
General, the phase voltage is:
VØ = EA − j ∗ XS ∗ IA − RA ∗ IA (4.3)
The net magnetic flux will be:
Bnet = BR + BS (4.4)
where
BR =Rotor Field
BS =Stator Field
51

Figure 4.6: Equvalent Circuit of Synchronous Generator
The equivalent circuit of a 3-phase synchronous generator is shown. The adjustable
resistor Radj controls the field current and, therefore, the rotor magnetic field.
when we apply 1.6 amper excitaton current (IF ) , our output voltage (VØ) shows
10.2 volt,
so when we put in formula 4.1 VØ = EA + Estat =⇒ 10, 2 = EA + Estat and now, we
have to know that Estat to find that our EA
how can we find our EA?
Example : A 200 kVA, 480 V, 50 Hz, Y-connected synchronous generator with a
rated field current of 5 A was tested and the following data were obtained:
1. VT,OC = 540 V at the rated IF .
2. IL,SC = 300 A at the rated IF .
3. When a DC voltage of 10 V was applied to two of the terminals, a current of
25 A was measured.
Find the generator’s model at the rated conditions (the armature resistance and the
approximate synchronous reactance).
Since the generator is Y-connected, a DC voltage was applied between its two phases.
52

Figure 4.7: when we appyl an external dc source to output of our generator we can
find our values of Ra , Rf, Xa and Xf with this methode
Therefore:
2 ∗ RA =
Vdc
Idc
=
10
2 ∗ 25
=⇒ RA = 0.2 [Ω] (4.5)
The internal generated voltage at the rated field current is
EA = VØ,OC =
VT
√
3
=
540
√
3
= 311, 76 [V olt] (4.6)
The synchronous reactance at the rated field current is precisely
XS = (Z)2 − (RA)2 =
311.8
300
2
− (0.2)2 = 1, 02 [Ω] (4.7)
We observe that if XSwas estimated via the approximate formula, the result would
be:
XS =
VØ
ISC
=
311, 8
300
= 1, 04 [Ω] (4.8)
Which is close to the previous result. The error ignoring RA is much smaller than
the error due to core saturation.
Therefore; when we apply this example to our generator;
Vdc= 4.5Volt & Idc = 2, 4Amper,
Then;
2 ∗ RA =
Vdc
Idc
=
4.5
2 ∗ 2.4
= 0, 9375 [Ω]
53

we divide by two becaus, our measuring between two coils.
Then;
EA
Iscc
=
VØ
Iscc
=⇒ Z = (R)2 + (jX)2 =⇒
10, 2
1, 35
= 7.55 [Ω]
XS = (Z)2 − (RA)2 = (7.55)2 − (0, 9375)2 = 7.49 [Ω]
We observe that if XSwas estimated via the approximate formula is that:
XS =
VØ
Iscc
= 7, 55 [Ω]
it means that:
XS RA
Figure 4.8: Equvalent Circuit of The generator
Then, we can calculate only with XSfrom in this equation 4.3 and IA = 1, 35.
10, 2 = EA − j7.55 ∗ IA (4.9)
Also EAis :
EA = KΦw (4.10)
54

4.4 Phasor diagram of a synchronous generator
where
K= is a constant representing the construction of the machine;
Φ= is flux in it
w= is its rotation speed
Since flux in the machine depends on the field current through it, the internal
generated voltage is a function of the rotor field current.
Figure 4.9: flux and ideal open-circuit characteristic
4.4 Phasor diagram of a synchronous generator
Since the voltages in a synchronous generator are AC voltages, they are usually
expressed as phasors. A vector plot of voltages and currents within one phase is
called a phasor diagram.
1. A phasor diagram of a synchronous generator with a unity power factor (re-
sistive load)
2. Lagging power factor (inductive load): a larger than for leading PF inter-
nal generated voltage EA is needed to form the same phase voltage. ( Our
generator )
3. Leading power factor (capacitive load).
55

Figure 4.10: Phasor diagrams
Eﬀects of adding loads can be described by the voltage regulation:
V R =
Vnl − Vfl
Vfl
∗ 100% (4.11)
Where;
• Vnl: is the no-load voltage of the generator
• Vfl: is full-load voltage
4.5 Calculating of Power, Torque and Eﬃciency
A synchronous generator needs to be connected to a prime mover whose speed is
reasonably constant (to ensure constant frequency of the generated voltage) for
various loads.
The applied mechanical power:
Pin = τapp ∗ ωm (4.12)
56

4.5 Calculating of Power, Torque and Eﬃciency
is partially converted to electricity
Pconv = τind ∗ ωm =
√
3 ∗ VT ∗ IL ∗ cos(θ) = 3VφIφ ∗ cos(θ) (4.13)
Pconv ≈ Pout
θ = cos−1
(PF)
where;
PF: power factor
θ: is the angle between Vφ and Iφ(Recall that the power factor angle θ is the angle between
Vφand Iφ and not the angle between VT and IL)
Figure 4.11: The power-ﬂow diagram of a synchronous generator
The real output power of the synchronous generator is:
Pout =
√
3 ∗ VT ∗ IL ∗ cos(θ) (4.14)
The reactive output power of the synchronous generator is:
Qout =
√
3 ∗ VT ∗ IL ∗ sin(θ) (4.15)
57

Example : A 480V , 50Hz, Y-connected four-poles synchronous generator has a per-phase
synchronous reactance of 1.0Ω. Its full-load armature current is 60A at 0.8PF lagging.
Its friction and windage losses are 2kW and core losses are 0.9kW at 50Hz at full load.
Assume that the armature resistance (and, therefore, the I2R losses) can be ignored. The
ﬁeld current has been adjusted such that the no-load terminal voltage is 480V
a.What is the speed of rotation of this generator?
b.What is the terminal voltage of the generator if
• It is loaded with the rated current at 0.8 PF lagging;
• It is loaded with the rated current at 1.0 PF;
• It is loaded with the rated current at 0.8 PF leading. and (What is the voltage
regulation of this generator at 0.8 PF lagging? at 1.0 PF? at 0.8 PF leading?)
c. What is the eﬃciency of this generator (ignoring the unknown electrical losses) when
it is operating at the rated current and 0.8 PF lagging?
d. How much shaft torque must be applied by the prime mover at the full load? how large
is the induced counter torque?
Answer:
Since the generator is Y-connected, its phase voltage is: Vφ = VT /
√
3 = 480/
√
3 =
277 [V ]
a. The speed of rotation of a synchronous generator is: Nm = 120 ∗ f/P = 120 ∗
50/4 = 1500 [rpm]
which is: ωm = (1500/50) ∗ 2π = 188, 5 [rad/s]
b.1 For the generator at the rated current and the 0.8 PF lagging. The phase voltage
is at 00
, the magnitude of EA is 277 V
therefore ;θ = cos−1
(−0, 8) = −36.83
E2
A = (Vφ +jXSIAsinθ)2
+(jXSIAcosθ)2
=⇒ 2772
= (V 2
φ +72Vφ +7488)= 236.58 [V ]
Since the generator is Y-connected: VT =
√
3Vφ = 410 [V ]
The voltage regulation of the generator is for 0.8 lagging:
V R = VLL−VT
VT
100% = 480−410
410
100% = 17%
b.2. For the generator at the rated current and the 1.0 PF ;
therefore; θ = cos−1
(1) = 0
E2
A = (Vφ + jXSIAsinθ)2
+ (jXSIAcosθ)2
=⇒ 2772
= (V 2
φ + 72Vφ + 7488)= 270 [V ]
VT =
√
3Vφ = 468 [V ]
The voltage regulation of the generator is for 1.0 PF:
58

4.6 Open Circuit Test
V R = VLL−VT
VT
100% = 480−468
468
100% = 2.5%
b.3. For the generator at the rated current and the 0.8 PF leading :
therefore= θ = cos−1
(0, 8) = 36.83
E2
A = (Vφ + jXSIAsinθ)2
+ (jXSIAcosθ)2
=⇒ 2772
= (V 2
φ + 72Vφ + 7488)= 308 [V ]
VT =
√
3Vφ = 535 [V ]
The voltage regulation of the generator is for 0.8 leading:
V R = VLL−VT
VT
100% = 480−535
535
100% = 17%
c. The output power of the generator at 60 A and 0.8 PF lagging,then:
Pout = 3VφIAcosθ = 3 ∗ 236 ∗ 60 ∗ cos(36.830
) = 34.1 [kW]
The mechanical input power is given by:
Pin = Pout + Pelec loss + Pcore loss + Pmech loss = 34.1 + 0 + 0.9 + 2 = 37 [kW]
The efficiency:
η = Pout
Pin
%100 = 34.1
37
∗ 100 = %92
d. The input torque of the generator :
τapp = Pin
ωm
= 37
188.5
= 196.2 [N.m]
The induced countertorque of the generator is (Pconverted ≈ Pout)(ωm = ωs)
τapp = Pconv
ωm
= 34.1
188.5
= 180.9 [N.m]
4.6 Open Circuit Test
We conduct first the open-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are disconnected from loads, the field
current is set to zero first. Next, the field current is increased in steps and the phase
voltage (whish is equal to the internal generated voltage EA since the armature
current is zero) is measured.
Therefore, it is possible to plot the dependence of the internal generated voltage on
the field current – the open-circuit characteristic (OCC) of the generator.
• Generator run at rated speed
59

• Exciting current is raised until rated voltage generated
• Exciting current (If ) and line-to-neutral voltage(VPhase = VØ) are recorded.
0 0.5 1 1.5 2
0
2
4
6
8
10
12
Excitation
Vphase
Open Circuit Test
Figure 4.12: The graphic of open circuit test of the generator
4.7 Short Circuit Test
We conduct next the short-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are short-circuited through ammeters,
the field current is set to zero first. Next, the field current is increased in steps and
the armature current IA is measured as the field current is increased.
The plot of armature current (or line current) vs. the field current is the short-circuit
characteristic (SCC) of the generator.
• Excitation is reduced to zero and armature is short-circuited
• Generator run at rated speed
• Excitation returned to value If
• Short-circuit ISC in the stator is measured
The SCC is a straight line since, for the short-circuited terminals, the magnitude of
the armature current is:
IA = ISC =
EA
R2
A + X2
S
(4.16)
60

4.7 Short Circuit Test
0 0.5 1 1.5 2
0
0.5
1
1.5
2
Value of Excitation Current ( If )
Isc
Short Circuit Test
Ia vs. If
I scc
Ia vs. If (smooth)
Figure 4.13: The graphic of short circuit test of the generator
Measuring of Test:
Excitation Current ( If ) Average of Open Circuit ( VØ) Excitation Current ( If ) Average of Open Circuit ( ISC )
0 0 0 0
0,2≈0,15 0,6 0,2≈0,15 0,16
0,4≈0,35 2,1 0,4≈0,35 0,32
0,6≈0,57 4,2 0,6≈0,53 0,48
0,8≈0,71 5,2 0,8≈0,78 0,71
1,0≈0,99 7,2 1,0≈0,92 0,84
1,2≈1,08 7,8 1,2≈1,14 1,01
1,4≈1,42 9,6 1,4≈1,40 1,17
1,6≈1,52 10,2 1,6≈1,52 1,35
1,8≈1,71 10,8 1,8≈1,72 1,53
2,0≈1,90 11,5 2,0≈1,99 1,75
2.2≈2,10 12,2 2.2≈2,22 1,95
2.4≈2,29 12,7 2.4≈2,37 2,16
Table 4.1: Results of Open Circuit and Short Circuit
61

0 0.5 1 1.5 2
0
2
4
6
8
10
12
Excitation Current
VphaseandIscc
Open Circuit and Short Circuit Test
Vphase vs. If (smooth) OCC
Ia vs. If (smooth) Iscc
Figure 4.14: VØand ISCCTest
4.8 Testing of PI controller
to understand and how to work a PI controller , we can use some special electronic
simulink program such that Matlab, Multisim ,Proteus to simulate a hardware of
PI controller. İn the industry, some manufacturing companies use hardware PI
controller or software PI controller , both of two types have some advantages. I will
show in this part that, how to simulate a PI controller and which kind of equipments
it has in a PI controller. İn this project, I will use a LM358 opamp which has two
operational ampliﬁer.
62

Figure 4.15: inside of LM358
The Differential Amplifier
A differential amplifier produces an output that is proportional to the difference
between two inputs.
the output of this opamp is that:
Vout =
R9
R7
(SetPoint − Measured V alue) (4.17)
R9 = R7 Then; =⇒ Vout = SetPoint − Measured V alue
This opearation is calculating difference of between Setpoint and Measured Value.
63

Figure 4.16: Desing of PI controller in Multisim
PI controller Ampliﬁer:
The proportional-integrating controller is commonly referred to as the PI controller.
A PI controller provides a transfer function of the form below:
U(s) =
C (s + a)
s
= Kp +
Ki
s
(4.18)
u(t) U(s)
Proportional R5
R3
∗ e(t) Kp ∗ e(t) KP
İntegrating 1
R3∗C1
´
e(t) ∗ dt Ki ∗
´
e(t)dt
K ˙I
s
The transfer function of a PI controller may be implemented with the circuit shown
inFig. 4.16.
İnverting Opamp:
In this opamp, it changes only sign of PI output signal for the input of voltage
controlled current source.
˙Inverter output = −
R13
R11
Vin (4.19)
Voltage Controlled Current Source:
The above circuit can be modiﬁed to produce a current source.
64

Figure 4.17: Voltage Controlled Current Source circuit
Assume that the op-amp drives a load of unknown resistance. A small resistor R
is placed in series with the load. Then the inverting input is v− = i ∗ R . At the
output of the ampliﬁer
Vout = A(v+−v−) =⇒ A(vin − iR) (4.20)
i =
1
R
Vin −
1
A
Vout (4.21)
if A 1 then;
i =
1
R
Vin (4.22)
which is independent of the load.
Test of the Circuit:
Setpoint Voltage = 5V , it means that there should appear same voltage on the load,
only current will change on the load.
Per Cent of Load 20Ω The load Voltage (xmm2) The load Current (xmm1)
%10 5, V 2.5A
%20 5V 1.26A
%30 5V 842mA
%40 5V 632mA
%50 5V 505mA
%60 5V 421mA
%70 5V 361mA
%80 5V 316mA
%90 5V 280mA
%100 5V 252mA
65

MATLAB EXAMPLE: TO UNDERSTAND PI CONTROLLER BEHAV-
IOR
The best way to understand something is by simulating it. So I simulated a PID
controller in matlab
Let me assume a suitable mathematical model for the plant and then go ahead with
designing the controller.
Let the transfer function of the plant be 1/(s2
+ 20s + 30)
The step response of a system is the output of the system when the input to the sys-
tem is a unit step. The open loop step response of the above plant is shownFig. 4.18
Step Response
Time (seconds)
Amplitude
0 1 2 3 4 5 6 7
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
System: Mc
Settling time (seconds): 4.6
Figure 4.18: The step response of a system
. It can be seen that the step response output is close to 0.035. The steady state
error = 1-0.035 = 0.965. That’s quite high! Also observe that the settling time is
around 3 sec. As we can see the step response of system is same like our open circuit
test characteristic.
Now lets see what is the eﬀect of PI controller on the system response. Lets see the
eﬀect of proportional element on the system output.
Keeping Kp = 10, Ki = 0 the step response of the system is shownFig. 4.19
66

0 0.5 1 1.5 2 2.5 3 3.5
0
0.05
0.1
0.15
0.2
0.25
Step Response
Time (seconds)
Amplitude
Figure 4.19: PI controller on the system response
The output is now 0.25 (The curve in red shows the open loop step response of the
plant). Now let me increase the Kp further and observe the response. Keeping Kp
= 100, Ki = 0 the step response of the system is
0 0.5 1 1.5 2 2.5 3 3.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Step Response
Time (seconds)
Amplitude
Figure 4.20: Kp = 100, Ki = 0 the step response
the output is now 0.77. So it’s clear now that increasing Kp will reduce the steady
state error.
Keeping Kp = 200, Ki = 200 the step response of the system is
67

Step Response
Time (seconds)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: Mc
Final value: 1
System: Mc
System: M_noController
Final value: 0.0323
Figure 4.21: Kp = 200, Ki = 200 the step response of the system
Observe that rise time has now reduced and steady state error is very small. İf we
can increase coeﬃency of Ki it may be much better
Keeping Kp = 200, Ki = 300 the step response of the system is
Step Response
Time (seconds)
Amplitude
0 0.5 1 1.5 2 2.5 3 3.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Rise time (seconds): 1.31
System: Mc
Peak amplitude: 1.06
Overshoot (%): 5.74
At time (seconds): 0.293
System: Mc
Rise time (seconds): 0.142
System: Mc
Settling time (seconds): 0.401 System: Mc
Final value: 1
Peak amplitude: >= 0.0322
Overshoot (%): 0
At time (seconds): > 3.5
Figure 4.22: Kp = 200, Ki = 300 the step response of the system
Observe that steady state error is close to 0 now. But increasing Ki has resulted in
overshoot.
Further increasing Ki will only increase overshoot.
68

Conclusion
Kieliminates the steady state error.
After certain limit, increasing Kiwill only increase overshoot. Ki reduces rise time.
MATLAB CODE
%G=1/(s^2+20s+30) // t r a n s f e r function of plant
num = [ 1 ]
denom = [1 20 30]
Gp=t f (num, denom)
H=[ 1 ]
%step ( feedback (Gp,H))
%hold on
M_noController = feedback ( Gp, H)
step ( M_noController )
hold on
Kp = 200;
Ki = 300;
Kd = 0;
Gc = pid (Kp, Ki , Kd)
Mc = feedback ( Gc ∗ Gp, H)
step (Mc)
grid on
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
num =
1
denom =
1 20 30
Gp =
1
−−−−−−−−−−−−−−−
s^2 + 20 s + 30
Continuous−time t r a n s f e r function .
H =
69

1
M_noController =
1
−−−−−−−−−−−−−−−
s^2 + 20 s + 31
Gc =
1
Kp + Ki ∗ −−−
s
with Kp = 200 , Ki = 300
Continuous−time PI c o n t r o l l e r in p a r a l l e l form .
Mc =
200 s + 300
−−−−−−−−−−−−−−−−−−−−−−−−−−
s^3 + 20 s^2 + 230 s + 300
70

Acknowledgments
I would like to express my thanks to my supervisor, Assoc. Prof. Dr. Izudin Džaﬁć,
for his motivation, encouragement, sincerity,valuable guidance and assistance in the
preparation, and completion of this project. I would also like to express my gratitude
to Senior Assist. Tarık Namas for his supports,valuable guidance and suggestions
throughout the study. I would like to thank my family for their continuous support.
Lastly, I want to thank all who gave a hand during my project.
71

Chapter Work and Result
.1 Generator
Figure .23: Synchronous Generator
72

.1 Generator
Figure .24
Figure .25: Generator
73

Figure .26
Figure .27
74

.1 Generator
Figure .28
Figure .29: diameters
75

Figure .30: 3D Stator and Rotor by AutoCAD
76

.1 Generator
Figure .31: 4 poles 1 slot rotor winding
77

78

.1 Generator
79

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82

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