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ELECTRICAL MACHINE
BMT 3013CHAPTER 1: SINGLE AND THREE PHASE
TRANSFORMER
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LEARNINGOUTCOMES
At the end of the chapter, students should be able to:
Understand the fundamental laws in the dynamic
magnetic systemsand their relation to the electrical
machines.
ELECTRIC
ALMACHINE(DEI202
3)
2
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INTRODUCTIONTOELECTRICALMACHINE
An electrical machine is a device which converts
electrical power (voltages and current) into mechanical
power (torque and rotational speed) and/or vice versa.
Electromechanical energy conversion the process
which an electrical machine deals with the energytransfer either from mechanical to electrical form or from
electrical to mechanical form.
Electrical motor an electrical machine which converts
an electrical energy into the mechanical energy.
Electric generator an electrical machine which
converts mechanical energy into an electrical energy.
3
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INTRODUCTIONTOELECTRICALMACHINE
(CONT..)
Many electric machines are capable of performing
both as motors and generators;
The capability of a machine performing as one or
the other is often through the action of a magnetic
field, to perform such conversions.
To understand how an electrical machines works,
the key is to understand how the electromagnet
works.
The principles of magnetismplay an important role
in the operation of an electrical machines.
4
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REVISIONOFMAGNETISM
Magnetismproperty by virtue of which a piece of
solid body (natural magnet) attracts iron piece and
piece of some other metals.
Polestwo ends of a magnet
The end of magnets which adjusting in the direction
of North is called N pole while other is S pole
Behavior of two magnets which are brought near to
each other is governed by Laws of Magnetism
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Thus,
=12
2
Where
1and 2are the pole strength
d is the distance between the two poles
K is constant which depends on the nature of thesurrounding
7
ELECTRIC
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3)
REVISIONOFMAGNETISM(CONT..)
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MAGNETICFIELD
Magnetic Fieldthe region around magnet within
which influence of the magnet can be experienced.
The presence of magnetic field is represented by
imaginary lines around magnet which called
magnetic lines of force.
The total number of lines of force existing in a
particular magnetic fieldmagnetic flux
Line of flux have a fixed direction
1.ExternalN-pole to S-pole
2.InternalS-pole to N-pole
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ELECTRIC
ALMACHINE(DEI202
3)
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MAGNETICFIELD(CONT..)
Unlike electric fields (which start on +q and end onq),magnetic field encircle their current source.
The field weakens as you move away from the wire Amperescircuital law - the integration path length
is longeridH
.
A circular magnetic field
develops around the wirefollows r ight-hand rules
field is perpendicular to
the wire and that the
field's direction dependson which direction the
current is flowing in the
wire
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MAGNETICFIELD(CONT..)
o Magnetic field due to circularconductor
Solenoid an arrangement inwhich a long conductor is woundwith number of turns on a core,close together to for a coil.
The right hand thumb rule : holdthe solenoid in the right handsuch that curled fingers point inthe direction of the currentthrough the curled conductor,then the outstretched thumbalong the axis of the solenoidpoints to the North pole of
solenoid or points in the directionof flux lines inside the core.
ELECTRIC
ALMACHINE(DEI202
3)
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12
REVIEWOFELECTROMAGNETISM
The basic idea behind an electromagnet is
extremely simple: a magnetic field around the
conductor can be produced when current flows
through a conductor.
In other word, the magnetic field only exists whenelectric current is flowing
By using this simple principle, you can create all
sorts of things, including motors, solenoids,
read/write heads for hard disks and tape drives,speakers, and so on
ELECTRIC
ALMACHINE(DEI202
3)
http://electronics.howstuffworks.com/motor.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/speaker.htmhttp://electronics.howstuffworks.com/speaker.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/motor.htm8/12/2019 CH 1 Transformer
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ELECTRICAL MACHINE (DEI 2023) 13
EXAMPLEOFELECTROMAGNETIC
An electromagnet can be made by winding theconductor into a coil and applying a DC voltage.
The lines of flux, formed by current flow through the
conductor, combine to produce a larger and stronger
magnetic field. The center of the coil is known as the core. In this
simple electromagnet the core is air.
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ELECTRICAL MACHINE (DEI 2023) 14
ADDINGANIRONCORE
Iron is a better conductor of fluxthan air. The air coreof an electromagnet can be replaced by a piece of soft
iron.
When a piece of iron is placed in the center of the coil
more lines of flux can flow and the magnetic field isstrengthened.
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15
STRENGTHOFMAGNETICFIELD(CONT)
Because the magnetic field around a wire iscircular and perpendicular to the wire, an easyway to amplify the wire's magnetic field is to coilthe wire
The strength of the magnetic field in the DC
electromagnet can be increased by increasing the
number of turns in the coil. The greater the
number of turns the stronger the magnetic
field will be.
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16
FARADAYSLAWANDLENZSLAW
FaradaysLaw :If a magnetic flux, , in a coil is changing intime (n turns),hence a voltage, V
ab
is induced
Lenzs Law : if the loop is closed, a connected to b, thecurrent would flow in the direction to produce the flux insidethe coil opposing the original flux change. (in other words,LenzsLaw will determine the polarity of the induced voltage)
ab
tNV
V = induced voltage
N = no of turns in coil
= change of flux in coil
t = time interval
Vt
If no turns :
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17
FARADAYSLAW
The effect of magnetic field:
Induced Voltage from a Time Changing
Magnetic Field
Production of Induced Force on a Wire Induced Voltage on a Conductor Moving
in a Magnetic Field
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ELECTRICAL MACHINE (DEI 2023) 18
VOLTAGEINDUCEDFROMATIME
CHANGINGMAGNETICFIELD
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19
VOLTAGEINDUCEDINACONDUCTOR
MOVINGINAMAGNETICFIELD
Faradays Law for moving conductors : For coils in which wire(conductor) is moving thru the magnetic flux, an alternateapproach is to separate the voltage induced by time-varying fluxfrom the voltage induced in a moving conductor.
This situation is indicates the presence of an electromagnetic fieldin a wire (conductor). This voltage described by Faradays Law is
called as the flux cutting or Electromotive force, or emf.
The value of the induced voltage is given by
E =Blvwhere
E= induced voltage (V)
B= flux density (T)
l= active length of the conductor in the magnetic field (m)
v= relative speedof the conductor (m/s)
The polarity of induced
voltage is given by theright-hand rule.
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20
INDUCEDFORCE
The electrical circuit consists of battery, resistor, two stationary rails, and movablebar that can roll or slide along the rails with electrical contact.
When switch is closed:
Current will not start immediately as inductance of the circuit.
(However time constant L/R is very small). Hence, current quickly
reach V/R.
Force is exerted on the bardue to interaction between current and
magnetic fluxto the right and made the bar move with certain velocity.
The mechanical power out of the bar.
F= ilB
Force induced on
the conductor:
Unit: (N)
The direction of
force is given by
the right-hand
rule.
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21
INDUCEDFORCE(CONT)
The motion of the bar produces an
electromagnetic force. The polarity of
the emfis+ve where the current enters
the moving bars. The moving bargenerates a back emfthat opposes the
current.
The instantaneous electrical power into the bar =
mechanical output power
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22
PRODUCTIONOFAMAGNETICFIELD
The production of a magnetic field by a current is
determine byAmpereslaw:
netIdlH
H = magnetic field intensity
dl = differential element of
length along the path of
integration
cl
NiH lc = mean path length
Magnetic field intensity:
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23
PRODUCTIONOFAMAGNETICFIELD
HB u = magnetic permeability ofmaterial
r0 u0= permeability of free spaceur= relative permeability of
material
m/H104 70
The strength of the magnetic field flux produced in the
core also depends on the material of the core.
Magnetic flux density:
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24
PRODUCTIONOFAMAGNETICFIELD
cl
NiHB
BA
cl
NiA
Total flux:
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25
MAGNETICCIRCUIT
iRV
Ni
Electric circuit equation:
Magnetic circuit equation:
Analogy: Electric circuit & Magnetic circuit
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26
MAGNETICSATURATION& HYSTERESIS
INACMAGNETICFIELD
unmagnetized Material
Iron becomes
magnetically
saturated
Magnetism increase as
magnetic field magnetized
unmagnetized iron
a
b
c
d
Applied field is reduced; the magnetism
reduced thru diff. curve since iron tends to
retains magnetized state - hence produced
permanent magnet, Residual Flux, res
AC increased in negative direction,
magnetic field reversed , the iron
reversely magnetized until saturated
againIf continue apply ac current, curve
continue to follow S-shaped curve(hysteresis curve)
The area enclosed by hysteresis curve is energy loss per unit volume per cycleheats the iron and is one
reason why electric machines become hot
Therefore, it is required to select magnetic materials that have a narrow hysteresis loop
Hm
Magnetic field density
Bm
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27
HYSTERESISLOSS
During a cycle of variation of i (hence H),
there is a net energy flow from the source to
the coil-core assembly and return to the
source.Energy flowing is greater than energy
returned.
This energy loss goes to heat the core.
The loss of power in the core due to thehysteresis effect is called hysteresis loss.
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28
EDDYCURRENTLOSS
Voltage will be induced in the path ofmagnetic core because of time variation of flux
enclosed by the path.
A current, known as an eddy current will flowaround the path.
Because core has resistance, a power loss
will be cause by the eddy current and appearas heat in the core.
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29
EDDYCURRENTLOSS
Eddy current can be reduced in 2 ways:
1. Adding a few percent of silicon to iron to
increase the resistivity.
2. Laminate core with thin laminations and
insulated from each other.
Hysteresis loss + eddy current loss = Core loss
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LEARNINGOUTCOMES
At the end of the lecture, student should to:
Understand the principle and the nature of
static machines of transformer.
Perform an analysis on transformers which
their principles are basic to the
understanding of electrical machines.
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INTRODUCTION
A transformer is a static machines. The word transformercomes form the word transform.
Transformer is not an energy conversion device, but is adevice that changes AC electrical power at one voltagelevel into AC electrical power at another voltage levelthrough the action of magnetic field, without a change infrequency.
It can be either to step-upor step down.
GenerationStation
TX1 TX1
Distributions
TX1
TX1
Transmission
System
33/13.5kV 13.5/6.6kV
6.6kV/415V
Consumer
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TRANSFORMERCONSTRUCTION
Two types of iron-core construction:a) Core - type construction
b) Shell - type construction
Core - type construction
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TRANSFORMERCONSTRUCTION
Shell - type construction
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IDEALTRANSFORMER
An ideal transformer is a transformer which has no loses,
i.e. its winding has no ohmic resistance, no magnetic
leakage, and therefore no I2 R and core loses.
However, it is impossible to realize such a transformer in
practice.
Yet, the approximate characteristic of ideal transformer will
be used in characterized the practical transformer.
V1 V2
N1 : N2
E1 E2
I1 I2
V1Primary Voltage
V2Secondary Voltage
E1Primary induced Voltage
E2secondary induced Voltage
N1:N2Transformer ratio
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ELECTRIC
ALMACHINE(DEI2023
)
TRANSFORMEREQUATION
Faradays Law states that,
If the flux passes through a coil of wire, a voltage will be
inducedin the turns of wire. This voltage is directly
proportional to the rate of change in the flux with respect of
time.
If we have Nturns of wire,
dt
tdEmfV indind
)(
dt
tdNEmfV indind
)(
Lenzs Law
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ELECTRIC
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TRANSFORMEREQUATION
For an ac sources,
Let V(t) = Vmsint
i(t) = imsint
Since the flux is a sinusoidal function;
Then:
Therefore:
Thus:
tt m sin)(
tN
dt
tdNEmfV
m
mindind
cos
sin
mmindind fNNEmfV 2(max)
mmm
rmsind fNfNN
Emf
44.42
2
2)(
36
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TRANSFORMEREQUATION
For an ideal transformer
In the equilibrium condition, both the input power will be equaled to the outputpower, and this condition is said to ideal condition of a transformer.
From the ideal transformer circuit, note that,
Hence, substitute in (i)
m
m
fNE
fNE
22
11
44.4
44.4
1
2
2
1
2211 coscos
I
I
V
V
IVIV
poweroutputpowerInput
(i)
2211 VEandVE 37
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ELECTRIC
ALMACHINE(DEI2023
)
TRANSFORMEREQUATION
aI
I
N
N
E
ETherefore
1
2
2
1
2
1,
Where, ais the Voltage Transformation Ratio; which will
determine whether the transformer is going to be step-upor step-down
E1> E2For a >1
For a
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ELECTRICALMACHINE(DEI2023
)
TRANSFORMERRATING
Transformer rating is normally writtenin terms ofApparent
Power.
Apparent power is actually the product of its rated current
and rated voltage.
2211 IVIVVA
Where,
I1and I2= rated current on primary and secondary winding.
V1and V2= rated voltage on primary and secondary winding.
Rated currents are actually the full load currentsin
transformer39
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ELECTRICALMACHINE(DEI2023
)
EXAMPLE
1. 1.5kVA single phase transformer has rated voltage of144/240 V. Finds its full load current.
Solution
AI
AI
FL
FL
250.6240
1500
417.10144
1500
2
1
40
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE
2. A single phase transformer has 400 primary and
1000 secondary turns. The net cross-sectional
area of the core is 60m2
. If the primary winding isconnected to a 50Hz supply at 520V, calculate:
a) The induced voltage in the secondary winding
b) The peak value of flux density in the core
Solution
N1=400 V1=520V A=60m2
N2=1000 V2=?
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE
3. A 25kVA transformer has 500 turns on the primary and 50
turns on the secondary winding. The primary is connected
to 3000V, 50Hz supply. Find:
a) Full load primary current
b) The induced voltage in the secondary winding
c) The maximum flux in the core
Solution
VA = 25kVA
N1=500 V1=3000VN2=50 V2=?
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ELECTRICALMACHINE(DEI2023
)
EXAMPLE3 (CONT)
a) Know that,
b) Induced voltage,
c) Max flux
VIIEE
AI
I
I
N
N
a
3003.8333.83000
3.8350
33.8500
2
112
2
1
2
2
1
AV
VAI
IVVA
FL 33.83000
1025 3
1
1
mWb
fNE
27
)50)(50(44.4300
44.4
44
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ELECTRICAL MACHINE (DEI 2023)
PRACTICALTRANSFORMER
(EQUIVALENTCIRCUIT)
V1 = primary supply voltage
V2 = 2ndterminal (load) voltage
E1 = primary winding voltageE2 = 2
ndwinding voltage
I1 = primary supply current
I2 = 2ndwinding current
I1 = primary winding current
Io = no load current
Ic = core current
Im = magnetism current
R1= primary winding resistanceR2= 2
ndwinding resistance
X1= primary winding leakage reactance
X2= 2ndwinding leakage reactance
Rc= core resistance
Xm= magnetism reactance
V1
I1 R1X1
RC
Ic
Xm
Im
Io
E1 E2
V2
I1
N1: N2R2
X2
Load
I2
45
SINGLE PHASE TRANSFORMER (REFERRED
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ELECTRICALMACHINE(DEI2023
)
SINGLEPHASETRANSFORMER(REFERRED
TOPRIMARY)
1) Transferred secondary parameter to primary winding
2
2
22
2
2
12 '' RaRORR
NNR
2
2
22
2
2
12 '' XaXORX
N
NX
a
II
aVVORVN
NVE
22
222
2
1'
21
'
'
V1
I1 R1 X1
RC
Ic
Xm
Im
Io
E1 E2 V2
I2 N1: N2
R2
X2
Load
I2
46
SINGLE PHASE TRANSFORMER (REFERRED
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ELECTRICALMACHINE(DEI2023
)
SINGLEPHASETRANSFORMER(REFERRED
TOPRIMARY)
2) Transferred core resistance and magnetism reatance
2
2
22
2
2
12 '' RaRORR
N
NR
V1
I1 R1X1
RC
Ic
Xm
Im
Io
E1 E2 V2
I2 N1: N2R2
X2
Load
I2
2
2
22
2
2
12 '' XaXORX
N
NX
a
II
aVVORVN
NVE
22
222
2
1'
21
'
'
47
SINGLE PHASE TRANSFORMER (REFERRED
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ELECTRICALMACHINE(DEI2023
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SINGLEPHASETRANSFORMER(REFERRED
TOPRIMARY)
3. Equivalent resistance and reactance
2
2
22
2
2
12 '' RaRORR
N
NR
V1
I1
R01 X01
aV2
2
2
22
2
2
12 '' XaXORX
N
NX
'
'
2101
2101
XXX
RRR
222
2
1'
2 ' aVVORVN
NV
In some application, the excitation
branch has a small current compared
to load current, thus it may be
neglected without causing seriouserror.
48
SINGLE PHASE TRANSFORMER (REFERRED
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ELECTRICALMACHINE(DEI2023
)
SINGLEPHASETRANSFORMER(REFERRED
TOSECONDARY)
2
111
2
1
21 ''
a
RRORR
N
NR
a
VVORV
N
NV 111
1
21 ''
I1 R1X1
RC
Ic
Xm
Im
IoI2 R2
X2
V2
2
111
2
1
21 ''
a
XXORX
N
NX
aV1
49
1) Transferred primary parameter to secondary winding
SINGLE PHASE TRANSFORMER (REFERRED
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ELECTRICALMACHINE(DEI2023
)
SINGLEPHASETRANSFORMER(REFERRED
TOSECONDARY)
2
111
2
1
21 ''
a
XXORX
N
NX
2102
2102
'
'
XXX
RRR
I1R02 X02
V2
a
V1
2
111
2
1
21 ''
a
RRORR
N
NR
a
V
VORVN
N
V
1
111
2
1 ''
11 ' aII
Neglect the excitation branch
50
2. Equivalent resistance and reactance
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE
4. For the parameters obtained from the test of 20kVA
2600/245 V single phase transformer, refer all the
parameters to the high voltage side if all the
parameters are obtained at lower voltage side.
Rc = 3.3, Xm=j1.5, R2= 7.5, X2= j12.4
Solution
Given
Rc
= 3.3, Xm
=j1.5,
R2= 7.5, X2= j12.4
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ELECTRICALMACHINE(DEI2023
)
EXAMPLE4 (CONT)
i) Refer to H.V side (primary)
R2=(10.61)2(7.5) = 844.65,
X2=j(10.61)2(12.4) = 1.396k
Rc=(10.61)2(3.3) = 371.6,
Xm=j(10.61)2(1.5) = j168.9
61.10245
2600
2
1
2
1 V
V
E
Ea
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ELECTRICAL MACHINE (DEI 2023)
POWERFACTOR
Power factor = angle between Current and voltage,cos
V
I
= -ve
V
I
= +ve
VI
= 1
Lagging Leading unity
Power factor always lagging for real
transformer.
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE
5. A 10 kVA single phase transformer 2000/440V has
primary resistance and reactance of 5.5and 12
respectively, while the resistance and reactance of
secondary winding is 0.2and 0.45 respectively.
Calculate:i. The parameter referred to high voltage side and draw the
equivalent circuit
ii. The approximate value of secondary voltage at full load of 0.8
lagging power factor, when primary supply is 2000V.
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ELECTRICA
LMACHINE(DEI2023
)
EXAMPLE5 (CONT)
SolutionR1=5.5 , X1=j12
R2=0.2 , X2=j0.45
i) Refer to H.V side (primary)
R2=(4.55)2(0.2) = 4.141,
X2=j(4.55)2
0.45 = j9.316
Therefore,
R01=R1+R2=5.5 + 4.141 = 9.641
X01=X1+X2=j12 + j9. 32 = j21.316
55.4440
2000
2
1
2
1 V
V
E
Ea
V1 aV2
R01 X01
21.329.64
I1
55
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ELECTRICA
LMACHINE(DEI2023
)
EXAMPLE5 (CONT)
Solutionii) Secondary voltage
p.f = 0.8
Cos = 0.8
=36.87o
Full load,
From eqn.
AV
VAIFL 5
2000
1010 3
1
o
oo
oo
V
Vj
aVIjXRV
7.1214.417
)55.4()87.365)(316.21641.9(02000
))((0
2
2
2101011
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ELECTRICA
LMACHINE(DEI2023
)
TRANSFORMERLOSSES
Generally, there are two types of losses;i. Iron losses:- occur in core parameters
ii. Copper losses:- occur in winding resistance
i. Iron Losses
ii. Copper Losses
circuitopenccciron PRIPP 2)(
02
2
201
2
1
2
2
21
2
1
)()(,
)()(
RIRIPreferredifor
PRIRIPP
cu
circuitshortcucopper
Pocand Pscwill be discusses later in transformer test
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ELECTRICA
LMACHINE(DEI2023
)
TRANSFORMEREFFICIENCY
To check the performance of the device, by comparing the
output with respect to the input.
The higher the efficiency, the better the system.
%100cos
cos
%100
%100,
22
22
cuc
lossesout
out
PPIVIV
PP
P
PowerInput
PowerOutputEff iciency
%100cos
cos
%100cos
cos
2)(
)(
cuc
nload
cuc
loadfull
PnPnVA
nVA
PPVA
VA
Where, if load, hence n = ,
load, n= ,
90% of full load, n =0.9Where Pcu = PscPc = Poc
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ELECTRICAL MACHINE (DEI 2023)
VOLTAGEREGULATION
The purpose of voltage regulation is basically to
determine the percentage of voltage drop
between no load and full load.
Voltage Regulation can be determine based on 3
methods:a) Basic Definition
b) Shortcircuit Test
c) Equivalent Circuit
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ELECTRICA
LMACHINE(DEI2023
)
VOLTAGEREGULATION(BASICDEFINITION)
In this method, all parameter are being referred to primary
or secondary side.
Can be represented in either
Downvoltage Regulation
UpVoltage Regulation
%100.
NL
FLNL
V
VVRV
%100.
FL
FLNL
V
VVRV
60
VOLTAGEREGULATION(SHORTCIRCUIT
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ELECTRICA
LMACHINE(DEI2023)
(
TEST)
In this method, direct formula can be used.
%100cos
.1
. V
VRV
fpscsc If referred to primary side
%100cos
.2
. V
VRV
fpscsc If referred to secondary side
Note that:
is for Lagging power factor
+ is for Leading power factor
Iscmust equal to IFL
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VOLTAGEREGULATION(EQUIVALENT
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ELECTRICA
LMACHINE(DEI2023)
(
CIRCUIT)
In this method, the parameters must be referred to primary orsecondary
%100sincos
.1
.01.011
V
XRIRV
fpfp If referred toprimary side
If referred to
secondary side
Note that:
+ is for Lagging power factor
is for Leading power factor
j terms ~0
%100sincos
.2
.02.022
V
XRIRV
fpfp
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE
6. In example 5, determine the Voltage regulation by usingdownvoltage regulation and equivalent circuit.
Solution
Downvoltage Regulation
Know that, V2FL=422.6V
V2NL=440V
Therefore,
%95.3
%100440
6.422440
%100.
NL
FLNL
V
VVRV
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE6 (CONT)
Equivalent Circuit
I1=5A R01=9.641 X01= 21.316 V1=2000V, 0.8 lagging p.f
%13.5
%1002000
)6.0(32.21)8.0(64.95
%100
sincos
. 1
.01.011
V
XRI
RV
fpfp
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE
7. A short circuit test was performed at the secondary side of 10kVA,
240/100V transformer. Determine the voltage regulation at 0.8lagging power factor if
Vsc=18V
Isc=100
Psc=240W
Solution
Check:
Hence, we can use short-circuit method
,
100100
10000
2
2
scFL
FL
II
AV
VAI
%100cos
.
2
.
V
VRV
fpscsc 65
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE7 (CONT)
o
scsc
scsc
scscscsc
o
fp
fpscsc
IV
P
IVPthatKnow
Hence
fpGiven
VVRV
34.82)100)(18(
240cos
cos
cos,
87.368.0cos,
8.0.
%100cos.
1
1
1
.
2
.
%62.12
%100100
87.3634.82cos18.
oo
RV 66
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE8 (CONT)
.732.7861.346.63643.8
643.833.8
72
46.63)33.8)(72(
268cos
cos
cos,
87.368.0cos,
8.0.
%100cos
.
0101
1
1
1
.
2
.
sideprimarytoconnectedbecausejXRjZ
I
VZ
IV
P
IVPthatKnow
Hence
fpGiven
V
VRV
o
sc
sc
scsc
o
scsc
scsc
scscscsc
o
fp
fpscsc
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE8 (CONT)
%68.2%100
2400
)6.0(72.7)8.0(86.32400
20000
%100
sincos
.,.2
%68.2%100
2400
87.364.63cos72.
%100cos
.,.1
1
.01.011
1
.
V
XRI
RVcircuitEquivalent
RV
V
VRVmethodCircuitShort
fpfp
oo
fpscsc
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE8 (CONT)
%68.2
%100240
58.233240
%100.
79.058.233
240
24004.6364.887.362400
2000002400
,.3
2
2
20111
NL
FLNL
o
ooo
VVVRV
VV
V
aVZIV
Def inationBasic
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE8 (CONT)
%12.97%100)268()5.0(170)8.0)(20000)(5.0(
)8.0)(20000)(5.0(
%34.97%100)268()1(170)8.0)(20000)(1(
)8.0)(20000)(1(
2)(
2)(
loadhalf
loadfull
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ELECTRICAL MACHINE (DEI 2023)
MEASUREMENTONTRANSFORMER
There are two test conducted on transformer.
i. Open Circuit Test
ii. Short Circuit test
The test is conducted to determine the parameter of
the transformer.
Open circuit test is conducted to determine
magnetism parameter, Rcand Xm.
Short circuit test is conducted to determine the
copper parameter depending where the test is
performed. If performed at primary, hence the
parameters are R01and X01and vice-versa.
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ELECTRICAL MACHINE (DEI 2023)
OPEN-CIRCUITTEST
Measurement are at high voltage side
From a given test parameters,
Rc XmVoc
Ic Im
Voc
Ioccosoc
Ioc
Voc
Ic
Im
Ioc
sinoc
oc
Note:
If the question asked parameters referred to
Low voltage side, the parameters (Rcand Xm) obtained
need to be referred to low voltage sidem
ocm
c
occ
mc
ococm
ococc
ococ
ococ
ococococ
I
VX
I
VR
XandRThen
II
II
HenceIV
P
IVP
,
,,
sin
cos
,
cos
cos
1
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ELECTRICAL MACHINE (DEI 2023)
SHORT-CIRCUITTEST
Measurement are at low voltage side
If the given test parameters are taken on primary
side, R01 and X01 will be obtained. Or else, vice-
versa.
X01R01
For a case referred to
Primary side
010101
01
1
,
cos
cos
jXRZ
I
VZ
Hence
IV
P
IVP
sc
sc
sc
scsc
scsc
scscscsc
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE9 (CONT)
A
II
A
II
IVP
o
ococm
o
ococc
o
oc
ococococ
2.49
6.69sin5.52
sin
26.18
6.69cos5.52
cos
6.69
)208)(5.52(
3800cos
cos
1
Ioccosoc
Ioc
Voc
Ic
Im
Iocsinoc
oc
From Open Circuit Test,
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE9 (CONT)
AV
VAIFL 4.217
2300
10500 3
11
o
sc
scscscsc IVP
53.72)4.217)(95(
6200cos
cos
1
From Short Circuit Test,First, check the Isc
Since IFL1=Isc , all reading are actually taken on the primary side
42.013.0
53.7244.053.724.217
95
01
j
I
VZ
oo
sc
sc
sc
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ELECTRICAL MACHINE (DEI 2023)
EXAMPLE9 (CONT)
Equivalent circuit referred to high voltage side,
V2=aV2V1Rc
1392
Xm
517.21
R010.13
X010.42
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ELECTRICA
LMACHINE(DEI2023)
EXAMPLE9 (CONT)
Efficiency,
%59.97
%1003800)5.0)(6200()866.0)(10500)(5.0(
)866.0)(10500)(5.0(
%100cos
cos
%74.97
%10038006200)866.0)(10500(
)866.0)(10500(
%100cos
cos
23
3
22
1
3
3
ocscL
ocsc
FL
PPnnVA
nVA
PPVA
VA
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