DET30043
ELECTRICAL MACHINE
TOPIC 1 DC GENERATOR
MEET THE LECTURER
ACADEMIC QUALIFICATION
Bachelor of Electrical Engineering Technology (Electrical Power) with
honours
Universiti Tun Hussein Onn Malaysia (UTHM)
M UA H M A A D A L I A S
BIN OMAR ABDUL AZIZ
2022 MUAHMAAD ALIAS BIN OMAR ABDUL AZIZ
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COURSE LEARNING OUTCOMES (CLO)
Upon completion of this course, students should be able to:
CLO1
Apply the concept, principle operation and motor control of
electrical machine to solve the related problems using standard
formula.
CLO2
Measure and record electrical and mechanical parameters related
to ac and dc electrical machine using appropriate electrical
equipment.
CLO3
Demonstrate ability to work in team to complete assigned tasks.
3
PROGRAMME LEARNING OUTCOMES (PLO)
PLO1:
Apply knowledge of applied mathematics, applied science,
engineering fundamentals and an engineering specialisation as
specified in DK1 to DK4 respectively to wide practical procedures
and practices
PLO5:
Apply appropriate techniques, resources, and modern engineering
and IT tools to well-defined engineering problems, with an
awareness of the limitations (DK6)
PLO9:
Function effectively as an individual, and as a member in diverse
technical teams
4
PROGRAMME LEARNING OUTCOMES (PLO)
5
MAPPING OF THE CLO TO THE PLO,
TEACHING METHODS AND ASSESSMENT
6
ASSESMENT
PLO1:
Apply knowledge of applied mathematics, applied science, engineering
fundamentals and an engineering specialisation as specified in DK1 to
DK4 respectively to wide practical procedures and practices
PLO5:
Apply appropriate techniques, resources, and modern engineering and
IT tools to well-defined engineering problems, with an awareness of the
limitations (DK6)
PLO9:
Function effectively as an individual, and as a member in diverse
technical teams
7
TOPIC 1 DC GENERATOR
• 1.1 Remember the construction and principle of DC generators.
• 1.2 Apply the principle operation of DC generators.
2022 MUAHMAAD ALIAS BIN OMAR ABDUL AZIZ
PRESENTATION TITLE
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TOPIC 1 DC GENERATOR
1.1 Remember the construction and principle of DC generators.
1.1.1 List the main parts of DC generators.
1.1.2 Identify the functions of components in section 1.1.1.
1.1.3 Define Faraday’s Law for the electromagnetic induction.
1.1.4 Define the Right Hand Rule as basic principle of operation for DC generator.
1.1.5 Identify the armature winding layout and the number of parallel path
1.1.6 Identify the field excitation methods for DC generators
1.1.7 Describe the self-excitation circuit
TOPIC 1 DC GENERATOR
1.2 Apply the principle operation of DC generators.
1.2.1 Apply the formula self-excitation DC generator
1.2.2 Solve the problems using formula in section 1.2.1.
1.2.3 Draw the various power stages in DC generator.
INTRODUCTION
Mechanical
energy
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Energy
conversion
device
Electrical
energy
An electrical generator is a
machine which converts
mechanical energy (or power)
into electrical energy (or power).
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INTRODUCTION
PRIME MOVERS
❑ All generators, large and small, ac and dc, require a source of
mechanical power to turn their rotors.
❑ This source of mechanical energy is called a prime mover
❑ divided into two classes for generators-high-speed and lowspeed.
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INTRODUCTION
PRIME MOVERS
❑ high-speed prime mover: Steam and gas turbines
❑ low-speed prime mover: internal-combustion engines, water,
and electric motors
Terminal Voltage
❑ Terminal voltage, in DC generators, is defined as the voltage
that can be measured at the output of the generator
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INTRODUCTION
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1.1.1 LIST THE MAIN PARTS OF DC GENERATORS.
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1.1.1 LIST THE MAIN PARTS OF DC GENERATORS.
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1.1.1 LIST THE MAIN PARTS OF DC GENERATORS.
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1.1.1 LIST THE MAIN PARTS OF DC GENERATORS.
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1.1.1 LIST THE MAIN PARTS OF DC GENERATORS.
1. Frame
2. Field pole
3. Armature
4. Commutator
5. Brush
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
1. Frame
❑ outer frame of a dc machine. It is made up of cast iron
(cheaper) or cast steel or rolled steel (large machine)
❑ provides mechanical strength to the whole assembly
❑ carries the magnetic flux produced by the field winding
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
❑ joined to the frame with the help of bolts or welding
❑ carry field winding and pole shoes are fastened to them
❑ function of field pole is to generate a magnetic field and
there are 3 parts of the field pole which is field winding, pole
core and pole shoes
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
a) field winding / pole coil
❑ consist of copper wire, former-wound for the correct
dimension. Then, the former is removed and wound coil is put
into place over the core on each pole and are connected in
series.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
a) field winding / pole coil
❑ When current is passed through these coils, they
electromagnetise the poles which produce the necessary flux
that is cut by revolving armature conductors. They are wound
in such a way that, when energized, they form alternate
North and South poles
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
a) field winding / pole coil
❑ It is wound around pole core and called as field coil.
❑ Field winding consist of 2 main field winding which are series
field winding and shunt field winding.
Series Field
Shunt Field
Coarse copper wire size, but less number
of turns of wire - very low resistance
wire size is fine but more number of turns
of wire - higher resistance
series with armature winding
parallel to the armature windings
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
b) Pole Core
❑ part that holds the field winding and the coil core serves as a
pole for producing a strong magnetic field.
❑ It is made of layers of annealed steel.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
2. Field pole
c) Pole Shoes
❑ support field coils.
❑ spread out the flux in air gap uniformly.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
3. Armature
❑ function of armature is to allow armature conductor cutting
magnetic flux and induces emf in it.
❑ It is composed of two parts, namely an armature core and
the armature windings.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
3. Armature
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
3. Armature
a) Armature Core
❑ cylindrical in shape with slots to carry armature winding.
❑ The armature is built up of thin laminated circular steel disks
for reducing eddy current losses.
❑ It houses the armature conductors or coils and causes them
to rotate and hence cut the magnetic flux of the field
magnets
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
3. Armature
a) Armature Core
❑ Usually, these laminations are perforated for air ducts which
permits axial flow of air through the armature for cooling
purposes
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
3. Armature
b) Armature winding
❑ It is usually a former wound copper coil which rests in
armature slots. The armature conductors are insulated from
each other and also from the armature core.
❑ Armature windings serves to cut the magnetic flux will
produce e.m.f
❑ wound by 2 methods; lap winding or wave winding.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
4. Commutator
❑ function of the commutator is to collect current from the
armature conductors.
❑ Physical connection to the armature winding is made through
a commutator-brush arrangement
❑ A commutator consists of a set of copper segments which are
insulated from each other by thin layers of mica.
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
4. Commutator
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
5. Brushes
❑ function is to collect current from commutator, are usually
made of carbon or graphite
❑ in the shape of a rectangular block.
❑ rest on commutator segments and slide on the segments
when the commutator rotates keeping the physical contact
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
5. Brushes
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
5. Brushes
Brush Contact Drop
❑ voltage drop over the brush contact resistance when current
passes from commutator segments to brushes and finally to
the external load.
❑ This drop is usually small and includes brushes of both
polarities
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1.1.2 IDENTIFY THE FUNCTIONS OF COMPONENTS IN
SECTION 1.1.1.
5. Brushes
Brush Contact Drop
❑ in practice, the brush contact drop is assumed to have
following constant values for all loads.
❑ 0.5 V for metal-graphite brushes.
❑ 2.0 V for carbon brushes.
❑ Either given in question or needed to be calculate.
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ Faraday’s law of electromagnetic induction, also known as
Faraday’s law, is the basic law of electromagnetism which
helps us predict how a magnetic field would interact with an
electric circuit to produce an electromotive force (EMF).
❑ Developed on the basis of experimental observations made in
1831 by the English scientist Michael Faraday
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ Faraday’s Laws of Electromagnetic Induction consists of two
laws. The first law describes the induction of emf in a
conductor.
❑ Whenever a conductor is placed in a varying magnetic field,
an electromotive force is induced. If the conductor circuit is
closed, a current is induced, which is called induced current.
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ Faraday’s Laws of Electromagnetic Induction consists of two
laws. The second law quantifies the emf produced in the
conductor.
❑ The induced emf in a coil is equal to the rate of change of
flux linkage.
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ The flux linkage is the product of the number of turns in the
coil and the flux associated with the coil. The formula of
Faraday’s law is given below:
❑ Where ε is the electromotive force, Φ is the magnetic flux,
and N is the number of turns. - negative sign for Lenz’s Law
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ Where ε is the electromotive force (e.m.f.) measured in volts
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1.1.3 DEFINE FARADAY’S LAW FOR THE
ELECTROMAGNETIC INDUCTION.
❑ - negative sign for Lenz’s Law. Lenz’s law states that “The
polarity of induced emf is such that it tends to produce a
current which opposes the change in magnetic flux that
produced it.”
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1.1.4 DEFINE THE RIGHT HAND RULE AS BASIC
PRINCIPLE OF OPERATION FOR DC GENERATOR.
❑ Fleming’s Right Hand Rule states that if we arrange our
thumb, forefinger and middle finger of the right-hand
perpendicular to each other, then the thumb points towards
the direction of the motion of the conductor relative to the
magnetic field, the forefinger points towards the direction of
the magnetic field and the middle finger points towards the
direction of the induced current.
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1.1.4 DEFINE THE RIGHT HAND RULE AS BASIC
PRINCIPLE OF OPERATION FOR DC GENERATOR.
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1.1.4 DEFINE THE RIGHT HAND RULE AS BASIC
PRINCIPLE OF OPERATION FOR DC GENERATOR.
❑ Fleming Right Hand rule is mainly applicable for electric
generator.
❑ used for finding the direction of the induced current in an
electric generator.
❑ The middle finger represents the direction of the induced
current
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1.1.4 DEFINE THE RIGHT HAND RULE AS BASIC
PRINCIPLE OF OPERATION FOR DC GENERATOR.
Fleming Right Hand rule @ Generator Rule
❑ Field : direction of magnetic field (N to S)
❑ F = BiL
❑ F = Force
❑ B = magnetic field
❑ i = current flowing through this conductor
❑ L = length of conductor
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
❑ There are two types of basic arrangement of the armature
winding in DC machines are:
a) Lap windings
b) Wave winding
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
a) Lap windings
❑ commonly used in machines that are intended to operate on
low voltage and high current, such as starter motors in
automobiles
❑ Total number of brushes needed is equal to the total number
of poles
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
a) Lap windings
❑ a = mP for lap winding with m = number of plex and P is the
number of poles
❑ m = 1, simplex
❑ m = 2, duplex
❑ m = 3, triplex
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
b) Wave windings
❑ intended for used in high voltage, low current machines, such
as high voltage generators
❑ total number of brushes required is 2 even though the actual
number equaling the total number of poles
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
b) Wave windings
❑ a = 2m for wave winding with m = number of plex
❑ If there is only one set of closed winding, it is called simplex
wave winding.
❑ If there are two such windings on the same armature, it is
called duplex winding and so on.
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1.1.5 IDENTIFY THE ARMATURE WINDING LAYOUT
AND THE NUMBER OF PARALLEL PATH.
b) Wave windings
❑ a = 2m for wave winding with m = number of plex
❑ m = 1, simplex
❑ m = 2, duplex
❑ m = 3, triplex
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
❑ Two type of excitation for DC generator:
a) self excitation
b) Separate excitation
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
a) self excitation
❑ energized the magnetic field generated by direct current in
the generator itself
❑ When the armature is rotated, balance magnetic flux will cut
conductors and induces emf and the current supplied to the
field winding to reinforce the balance magnetic flux until it
reaches saturation point.
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
a) self excitation
❑ There are three (3) types of self excitation generator:
i)
series generator
ii) shunt generator
iii) compound generator (long shunt and short shunt)
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
i) series generator
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
ii) shunt generator
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
iii) compound generator (long shunt and short shunt)
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
b) Separate excitation
❑ also known as a excited split or excited isolated. The
magnetic field is energized by a dc current of external supply
sources.
❑ There are two (2) types of separate excitation generator
i)
shunt connection
ii) compound connection
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
i) shunt connection
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1.1.6 IDENTIFY THE FIELD EXCITATION METHODS FOR
DC GENERATORS.
ii) compound connection
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1.1.7 DESCRIBE THE SELF-EXCITATION CIRCUIT.
❑ flux in a self-excited DC generator, builds up from a small
amount of residual magnetism present in the field poles.
From the theory of magnetism, this means that some of
magnetic are aligned in the same direction giving rise to a
residual magnetic field.
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1.1.7 DESCRIBE THE SELF-EXCITATION CIRCUIT.
❑ There must be residual magnetism in the machine pole.
❑ The field coil must provide flux in the same direction as the
residual flux. Otherwise, the machine will build zero volts.
❑ The field circuit resistance must not be excessive.
❑ The machine must have sufficient speed because voltage is a
function of speed.
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1.1.7 DESCRIBE THE SELF-EXCITATION CIRCUIT.
❑ equivalent circuit of self-excited DC generator
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1.2.1 APPLY THE FORMULA SELF-EXCITATION DC
GENERATOR.
emf generated, Eg
NZ P
Eg =
x
60
a
Eg
= emf generated in armature (V)
N
= rotational speed of the armature (rpm)
Z
= number of armature conductors
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1.2.1 APPLY THE FORMULA SELF-EXCITATION DC
GENERATOR.
Z
= number of slots x number of conductor per slot
Z
= 2CNc

= flux per pole (Weber / Wb)
P
= number of poles
a
= number of parallel parts by armature type
a
= mP
… for (armature) lap winding
a
= 2m
… for (armature) wave winding
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1.2.1 APPLY THE FORMULA SELF-EXCITATION DC
GENERATOR.
Shunt DC generator
Vt = Eg – IaRa – Vb
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1.2.1 APPLY THE FORMULA SELF-EXCITATION DC
GENERATOR.
Ia = Ifl + If
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.1
A duplex lap wound armature is used in a six pole DC machine.
There are 1728 conductors in this machine. The flux per pole in
the machine is 0.039Wb, the speed of the machine is 400rpm,
and the current path in this machine is 12. What is the
generated e.m.f. in this DC machine?.
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.1 Answer:
N = 400 rpm, Z = 1728 conductors, Փ = 0.039 Weber, P = 6 poles
𝑁𝑍
Eg =
Eg =
60
𝑥
𝑃
𝑎
, where a=mP for lap winding
400 𝑟𝑝𝑚 𝑥 1728 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑥 0.039 𝑊𝑏
60
𝑥
6 𝑝𝑜𝑙𝑒𝑠
6 𝑝𝑜𝑙𝑒𝑠 𝑥 2 (𝑑𝑢𝑝𝑙𝑒𝑥)
Generated voltage, Eg = 224.64 V #
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2
Vt = Eg – IaRa – Vb
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2
A shunt generator 150kW, 250V has a field circuit resistance of
50 ohms and an armature resistance of 0.05 ohm. Calculate:
a) Full load current, Ifl
b) Field current, If
c) Armature current, Ia
d) generated emf, Eg
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2 Answer:
a) Full load current, Ifl
P = VI
𝐼𝑓𝑙 =
𝑃𝑜𝑢𝑡
𝑉𝑖𝑛
=
150 𝑘𝑊
250 𝑉
Full load current, Ifl = 600 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2 Answer:
b) Field current, If
V = IR
𝐼𝑓 =
𝑉𝑖𝑛
𝑅𝑓
=
250 𝑉
50 Ω
Field current, If = 5 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2 Answer:
c) Armature current, Ia
Ia = Ifl + If
Ia = (600 + 5) A
Armature current, Ia = 605 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.2 Answer:
d) generated emf, Eg
Vt = Eg – IaRa – Vb, Vb = 0.5 V, graphite brush
Eg = Vt + IaRa + Vb = 250 V + (605 A x 0.05 Ω) + 0.5 V
Eg = 280.75 V
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3
4 pole DC generator shunt lap wound 0.25 kW with terminal
voltage of 220 V. The armature resistance is 0.05 Ohm and field
resistance is 20 Ohm. If the speed is 2000 rpm, flux per pole is
0.02 weber and the brush drop is 0.5 V.
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3
Calculate:
a) full load current
b) field current
c) armature current
d) emf generated
e) number of armature conductors
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3 Answer:
a) Full load current, Ifl
P = VI
𝐼𝑓𝑙 =
𝑃𝑜𝑢𝑡
𝑉𝑖𝑛
=
0.25 𝑘𝑊
220 𝑉
Full load current, Ifl = 1.14 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3 Answer:
b) Field current, If
V = IR
𝐼𝑓 =
𝑉𝑖𝑛
𝑅𝑓
=
220 𝑉
20 Ω
Field current, If = 11 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3 Answer:
c) Armature current, Ia
Ia = Ifl + If
Ia = (1.14 + 11) A
Armature current, Ia = 12.14 A
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3 Answer:
d) generated emf, Eg
Vt = Eg – IaRa – Vb,
Vb = 0.5 V x 2 = 1 V, carbon brush
Eg = Vt + IaRa + Vb = 220 V + (12.14 A x 0.05 Ω) + 1 V
Eg = 221.607 V
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Example 1.3 Answer:
e) number of armature conductors
Eg =
𝑁𝑍
60
𝑥
𝑃
𝑎
222.607 V =
Z=
, 𝑎 = 𝑚𝑃 𝑙𝑎𝑝 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 , m = 1 simplex
2000 𝑟𝑝𝑚 𝑥 𝑍 𝑥 0.02 𝑊𝑏
60
𝑥
4
1𝑥4
221.607 𝑉 𝑥 60
2000 𝑟𝑝𝑚 𝑥 0.02 𝑊𝑏
Z = 332.41
Z  332 number of conductor
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.1
A shunt generator delivers 450 A at 230 V and the resistance of
the shunt field and armature are 50 Ω and 0.03 Ω respectively.
Calculate the generated e.m.f.
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.1
Answer: Ish = 4.6 A, Ifl = 450 A, Ia = 454.6 A, Eg = 243.6 V
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.2
A long-shunt compound generator delivers a load current of 50
A at 500 V and has armature, series field and shunt field
resistances of 0.05 Ω, 0.03 Ω and 250 Ω respectively. Calculate
the armature current and the generated voltage. Allow 0.5 V per
brush for contact drop.
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.2
Vt = Eg – Ia(Ra+Rs) – Vb
Answer: Ia = 52 A, Eg = 505.16 V
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.3
4 pole, long shunt compound lap wound generator supplies 25
kW at a terminal voltage of 500 V. The armature resistance is
0.03 Ohm, series field resistance is 0.04 Ohm and shunt field
resistance is 200 Ohm. If the speed is 1200 rpm, flux per pole is
0.02 weber and the brush drop may be taken as 0.5 V.
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.3
Calculate:
a) full load current
b) field current
c) armature current
d) voltage generated
e) number of armature conductors
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.3
Answer: Ifl = 50 A, If = 2.5 A, Ia = 52.5 A, Eg = 504.675 V,
Z  1262
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1.2.2 SOLVE THE PROBLEMS USING FORMULA IN
SECTION 1.2.1.
Practice 1.4
A separately excited DC generator is rated at 172 kW, generated
voltage is 382 V, current armature of this machine is 360A, and
resistor armature is 0.05 Ω. Calculate the terminal voltage.
Answer: Vt = 364 V
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR.
❑ Not all power supplied will be derived. A bit of energy that is
converted into heat energy that is not useful called losses.
❑ 3 types of power loss in the DC generator are:
a) Copper Losses
b) Iron / Core / Magnet Losses
c) Mechanical losses (friction and wind)
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR.
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR.
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4
A long shunt generator running at 1000 rpm. supplies 20 kW at
a terminal voltage of 220 V. The resistance of armature, shunt
field, and series field are 0.04, 110 and 0.05 ohm respectively.
Overall efficiency at the above load is 85%.
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4
Calculate :
a) Copper loss,
b) Iron and friction loss,
c) Torque developed by the prime mover
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
N = 1000 rpm, Pout = 20kW, Vt = 220 V, Ra = 0.04 Ohm,
Rs = 0.05 Ohm, Rf = 110 Ohm,  = 85% = 0.85
a) Copper loss
Ifl = P / V = 20kW / 220 V = 90.91 A
If = Vin / Rf = 220 V / 110 Ohm = 2 A
(If = Ishunt)
Ia = Ifl + If = 90.91 A + 2 A = 92.91 A
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
a) Copper loss
Is = Ia = = 92.91 A kerana sambungan siri
Copper Loss = Pcu armature (Ia²Ra) + Pcu shunt field (If²Rf) +
Pcu series field (Is²Rs)
Copper Loss = (92.912 A x 0.04 Ohm) + (2 2 A x 110 Ohm) +
(92.912 A x 0.05 Ohm)
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
a) Copper loss
Copper Loss = (345.29 W) + (440 W) + (431.61 W)
Copper loss = 1216.9 W
Copper loss  1217 W #
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
b) Iron and friction loss
Efficiency,  =
0.85 =
Pin =
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
𝑥 100
20𝑘𝑊
𝑃𝑖𝑛
20𝑘𝑊
0.85
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
b) Iron and friction loss
Pin = 23529 kW
Total losses = P in - Pout = (23529 – 20000) W = 3529 W
Loss of iron and friction = Total losses - Loss of copper
Loss of iron and friction = (3529 – 1217) W
Loss of iron and friction = 2312 W
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.4 Answer:
c) Torque developed by the prime mover
Pin = T =
2𝜋𝑁
60
23529 W =
T=
2𝜋𝑁
60
23529 𝑊 𝑥 60
2𝜋𝑁
𝑥𝑇
𝑥𝑇
=
23529 𝑊 𝑥 60
2𝜋1000 𝑟𝑝𝑚
T = 224.69 Nm
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5
A shunt generator provides 24kW of power at a terminal voltage
of 200V. The armature resistance is 0.05 ohm with field
resistance of 40 ohms. If the iron and friction losses on the load
conditions equivalent to copper losses, calculate
a) Engine output to drives the generator
b) Efficiency
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
a) Engine output to drives the generator
Ifl = 24 kW / 200 V = 120 A
If = 200 V / 40 Ω = 5 A
Ia = 120 A + 5 A = 125 A
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
a) Engine output to drives the generator
Copper Loss = Pcu armature (Ia²Ra) + Pcu shunt field (If 2
Rf) + Pcu series field (Is 2Rs)
Pcu armature = Ia²Ra =125 2 A x 0.05 Ω = 781.25 W
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
a) Engine output to drives the generator
Pcu shunt field = If 2Rf = 52 A x 40 Ω = 1000 W = 1 kW
Pcu series field = Is 2Rs = 0 W
Copper Loss = 781.25 W + 1000 W = 1781.25 W
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
a) Engine output to drives the generator
Iron and friction losses = copper losses = 1781.25 W
Total losses = 1781.25 W + 1781.25 W = 3562.5 W
Pin = Pout + Total losses = 24 kW + 3562.5 W = 27562.5 W =
27.562 kW
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
a) Engine output to drives the generator
Engine output to drives the generator = Pin
Engine output to drives the generator = 27.562 kW #
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Example 1.5 Answer:
Pout = 24 kW, Vt = 200 V, Ra = 0.05 Ω, Rf = 40 Ω
b) Efficiency
Efficiency,  =
Pout
Pin
𝑥 100
Efficiency = (24 kW / 27.562 kW) x 100
Efficiency = 87.08% = 0.87 #
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.5
A 10 kW, 250 V, DC 6-pole shunt generator runs at 1000 rpm
when delivering full-load. The armature has 534 lap-connected
conductors. Full load Cu loss is 0.64 kW. The total brush drop is
1 V. Calculate the flux per pole. Neglect shunt current.
Answer:  = 0.03 Wb
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.6
A shunt generator delivers 195 A at terminal p.d. of 250 V. The
armature resistance and shunt field resistance are 0.02 Ω and
50 Ω respectively. The iron and friction losses equal 950 W.
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.6
Calculate
a) e.m.f. generated
b) Cu losses
Answer: Eg = 254 V, Total Cu loss = 2050 W
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.7
A long-shunt generator running at 1000 r.p.m. supplies 22 kW at
a terminal voltage of 220 V. The resistances of armature, shunt
field and the series field are 0.05, 110 and 0.06 Ω respectively.
The overall efficiency at the above load is 88%.
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.7
Find
a) Cu losses
b) iron and friction losses
c) the torque exerted by the prime mover
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1.2.3 DRAW THE VARIOUS POWER STAGES IN DC
GENERATOR
Practice 1.7
Answer: Total Cu losses = 1584.5 W,
Iron and friction losses = 1,415.5 W, T = 238.74 Nm
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THANK YOU
Muahmaad Alias Bin Omar Abdul Aziz
Bangunan Staf, JKE
maliaso@poliku.edu.my
018-2644034
121