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EET 103EET 103
Single Phase TransformerSingle Phase Transformer
Chapter 5Chapter 5
((Lecture 1)Lecture 1)
1
A transformer is a device that changes ac electric A transformer is a device that changes ac electric energy at one voltage level to ac electric energy energy at one voltage level to ac electric energy at another voltage level through the action of a at another voltage level through the action of a magnetic field.magnetic field.
Introduction to TransformerIntroduction to Transformer
2
• Transformers are constructed of two coils or more placed around the common freeomagnetic core so that the charging flux developed by one will link to the other.
• The coil to which the source is applied is called the primary coil.
• The coil to which the load is applied is called the secondary coil.
Introduction to TransformerIntroduction to Transformer
3
The most important tasks performed by The most important tasks performed by transformers are:transformers are:
•Changing voltage and current levels in electric power systems.
•Matching source and load impedances for maximum power transfer in electronic and control circuitry.
•Electrical isolation (isolating one circuit from another or isolating DC while maintaining AC continuity between two circuits).
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Mutual Inductance• Mutual inductance exits between coils of the
same or different dimensions.
• Mutual inductance is a phenomenon basic to the operation of the transformer.
6
Mutual Inductance• A transformer is constructed of 2 coils placed so that
the changing flux developed by one will link the other.
• The coil to which the source is applied is called primary• The coil which the load is applied is called secondary.
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Mutual Inductance (Cont…)
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An ideal transformer is a lossless device with an An ideal transformer is a lossless device with an input winding and output winding. input winding and output winding.
Ideal TransformerIdeal Transformer
aN
N
tv
tv
s
p
s
p )(
)(
a = turns ratio of the transformera = turns ratio of the transformer
)t(iN)t(iN sspp
ati
ti
s
p 1
)(
)(
10
Power in ideal transformerPower in ideal transformer
cosppinout IVPP
sinppinout IVQQ
Ppssinout IVIVSS
Where Where is the angle between voltage and current is the angle between voltage and current
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Impedance transformation through the Impedance transformation through the transformertransformer
The impedance of a device – the The impedance of a device – the ratioratio of the of the phasor voltage across it in the phasor current phasor voltage across it in the phasor current flowing through itflowing through it
L
LL I
VZ
LL ZaZ 2'
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Non-ideal or actual transformerNon-ideal or actual transformer
Mutual flux
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Losses in the transformerLosses in the transformer
• Copper (I2R) losses: Copper losses are the resistive heating in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
• Eddy current losses: Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
14
• Hysteresis losses: Hysteresis losses are associated with the arrangement of the magnetic domain in the core during each half cycle. They are complex, nonlinear function of the voltage applied to the transformer.
• Leakage flux: The fluxes ΦLP and ΦLS which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self inductance in the primary and secondary coils, and the effects of this inductance must be accounted for.
Losses in the transformerLosses in the transformer
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Transformer equivalent circuitTransformer equivalent circuit
EEpp = primary induced voltage = primary induced voltage EEss = secondary induced voltage = secondary induced voltage
VVpp = primary terminal voltage = primary terminal voltage VVss = secondary terminal voltage = secondary terminal voltage
IIpp = primary current = primary current IIss = secondary current = secondary current
IIee = excitation current = excitation current IIMM = magnetizing current = magnetizing current
XXMM = magnetizing reactance = magnetizing reactance IICC = core current = core current
RRCC = core resistance = core resistance RRpp = resistance of primary winding = resistance of primary winding
RRss = resistance of the secondary winding = resistance of the secondary winding XXpp = primary leakage reactance = primary leakage reactance
XXss = secondary leakage reactance = secondary leakage reactance
Ie
ImIc
16
Dot conventionDot convention
1.1. If the primary voltage is positive at the dotted end of If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to also. Voltage polarities are the same with respect to the dots on each side of the core.the dots on each side of the core.
2.2. If the primary current of the transformer flows into the If the primary current of the transformer flows into the dotted end of the primary winding, the secondary dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary current will flow out of the dotted end of the secondary winding.winding.
17
Exact equivalent circuit the actual transformerExact equivalent circuit the actual transformer
a.a. The transformer model referred to primary sideThe transformer model referred to primary side
b.b. The transformer model referred to secondary The transformer model referred to secondary sideside
18
Approximate equivalent circuit the actual Approximate equivalent circuit the actual transformertransformer
a. The transformer model referred to primary side
b. The transformer model referred to secondary side
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Exact equivalent circuit of a transformer Exact equivalent circuit of a transformer refer to primary siderefer to primary side
Ep = primary induced voltage Es = secondary induced voltageVp = primary terminal voltage Vs = secondary terminal voltageIp = primary current Is = secondary currentIe = excitation current IM = magnetizing currentXM = magnetizing reactance IC = core currentRC = core resistance Rp = resistance of primary winding
Rs = resistance of the secondary winding Xp = primary leakage reactance
Xs = secondary leakage reactance
20
a/III sep
MCe III
ppppp E)jXR(IV
CCp RIE
)jX(IE MMp
)jX//R(IE MCep
Primary sidePrimary side Secondary sideSecondary side
ssssS V)jXR(IE
Lss ZIV
s
p
p
s
s
p
s
p
N
N
I
I
E
E
V
Va
The Equation:The Equation:
21
R pI p X p
V p E p aV s
a 2 R sI s /aa 2 X s
I e
Exact equivalent circuit of a transformer referred to Exact equivalent circuit of a transformer referred to primary sideprimary side
V p /a
aI pR p /a 2 X p /a 2 I s
E p /a = E sV s
R s X s
aI c
R c/a 2X M /a 2
aI m
aI e
Exact equivalent circuit of a transformer referred to Exact equivalent circuit of a transformer referred to secondary sidesecondary side
22
Approximate equivalent circuit of a transformer referred Approximate equivalent circuit of a transformer referred to primary sideto primary side
Approximate equivalent circuit of a transformer referred to Approximate equivalent circuit of a transformer referred to secondary sidesecondary side
Vp/a
aIp
+
-
ReqsIsjXeqs
Rc/a2 jXM/a2 Vs
+
-
Reqs=Rp / a2+Rs
Xeqs=Xp / a2+Xs
Vp
Ip
+
-
ReqpIs/ajXeqp
RcjXM
aVs
+
-
Reqp=Rp+a2Rs
Xeqp=Xp+a2Xs
23
Example 1Example 1A single phase power system consists of a 480V 60Hz generator supplying a load Zload=4+j3 through a transmission line ZLine=0.18+j0.24 Answer the following question about the system.
a) If the power system is exactly as described below (figure 1(a)), what will be the voltage at the load be? What will the transmission line losses be?
b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line (figure 1(b)). What will the load voltage be now? What will the transmission line losses be now?
24
Figure 1 (a)
Figure 1 (b)
ZLoad=4+j3V=48000V
IG+
-
VLoad
ZLoad=0.18+j0.24
ILine
ILoad
ZLoad=4+j3
25
Solution 5.1Solution 5.1
(a) From figure 1 (a) shows the power system (a) From figure 1 (a) shows the power system without transformers. Hence Iwithout transformers. Hence IGG = I = ILINELINE = I = ILoadLoad. .
The line current in this system is given byThe line current in this system is given by
8.378.90
8.3729.5
0480
24.318.4
0480
)34()24.018.0(
0480
j
jj
V
ZZ
VI
loadlineline
26
Therefore the load voltage isTherefore the load voltage is
9.0454
)9.365)(8.378.90(
)34)(8.378.90(
A
jA
ZIV loadlineload
and the line losses areand the line losses are
W
A
RIP linelineloss
1484
)18.0()8.90( 2
2
27
(b) (b) From figure 1 (b) shows the power system with the From figure 1 (b) shows the power system with the transformers. To analyze the system, it is necessary to transformers. To analyze the system, it is necessary to convert it to a common voltage level. This is done in convert it to a common voltage level. This is done in two stepstwo steps
i) Eliminate transformer T2 by referring the load over to i) Eliminate transformer T2 by referring the load over to the transmission’s line voltage level.the transmission’s line voltage level.
ii) Eliminate transformer T1 by referring the transmission ii) Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the line’s elements and the equivalent load at the transmission line’s voltage over to the source side.transmission line’s voltage over to the source side.
The value of the load’s impedance when reflected to the The value of the load’s impedance when reflected to the transmission system’s voltage istransmission system’s voltage is
300400
)34()1
10(
'
2
2
j
j
ZaZ loadload
28
The total impedance at the transmission line level is nowThe total impedance at the transmission line level is now
88.363.500
24.30018.400
'
j
ZZZ loadlineeq
The total impedance at the transmission line level The total impedance at the transmission line level (Z(Zlineline+Z’l+Z’loadoad) is now reflected across T) is now reflected across T11 to the source’s to the source’s
voltage levelvoltage level
88.36003.5
)340024.00018.0(
)30040024.018.0()10
1(
)'(
'
2
2
2
jj
jj
ZZa
ZaZ
loadline
eqeq
29
Notice that Z’’Notice that Z’’loadload = 4+j3 = 4+j3 and Z’ and Z’line line =0.0018+j0.0024 =0.0018+j0.0024 . .
The resulting equivalent circuit is shown below. The The resulting equivalent circuit is shown below. The generator’s current isgenerator’s current is
AV
IG
88.3694.95
88.36003.5
0480
a) System with the load a) System with the load referred to the transmission referred to the transmission system voltage levelsystem voltage level
b) System with the load and b) System with the load and transmission line referred to transmission line referred to the generator’s voltage the generator’s voltage levellevel
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Knowing the current IKnowing the current IG, G, we can now work back we can now work back
and find Iand find Iline line and Iand ILoad.Load. Working back through T1, we Working back through T1, we
getget
A
A
IN
NI
ININ
GS
pline
linesGp
88.36594.9
)88.3694.95(10
11
1
11
31
Working back through T2 gives Working back through T2 gives
A
A
IN
NI
ININ
lines
pload
loadslinep
88.3694.95
)88.36594.9)(1
10(
2
2
22
It is now possible to answer the questions. The load voltage is given by
V
A
ZIV loadloadload
01.07.479
)87.365)(88.36594.9(
32
the line losses are given bythe line losses are given by
W
A
RIP linelineloss
7.16
)18.0()594.9(
)(2
2
Notice that raising the transmission voltage of the Notice that raising the transmission voltage of the power system reduced transmission losses by a power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers dropped much less in the system with transformers compared to the system without transformers.compared to the system without transformers.
33
Open circuit testOpen circuit test
Provides magnetizing reactance and core loss Provides magnetizing reactance and core loss resistanceresistance
Obtain components are connected in parallelObtain components are connected in parallel
Parameter determination of the transformerParameter determination of the transformer
34
Experiment SetupExperiment Setup
In the open circuit test, In the open circuit test, transformer rated voltagetransformer rated voltage is applied to the is applied to the primary voltage sideprimary voltage side of the of the transformer with the secondary side left open. transformer with the secondary side left open. Measurements of power, current, and voltageMeasurements of power, current, and voltage are are made on the made on the primaryprimary side. side.
Since the secondary side is open, the input Since the secondary side is open, the input current Icurrent IOCOC is equal to the excitation current is equal to the excitation current
through the shunt excitation branch. Because this through the shunt excitation branch. Because this current is very small, about 5% of rated value, current is very small, about 5% of rated value, the voltage drop across the secondary winding the voltage drop across the secondary winding and the winding copper losses are neglected.and the winding copper losses are neglected.
35
oc
ococ V
IY
ococ
oc
IV
PcosPF
ococ
oc
IV
Pcos 1
MCMC
oc
ococ X
jR
jBGV
IY
11
AdmittanceAdmittance
Open circuit Power FactorOpen circuit Power Factor
Open circuit Power Factor AngleOpen circuit Power Factor Angle
Angle of current always lags angle of voltage by Angle of current always lags angle of voltage by
36
Short circuit testShort circuit test
– Provides combined leakage reactance and Provides combined leakage reactance and winding resistancewinding resistance
– Obtain components are connected in seriesObtain components are connected in series
37
Experiment SetupExperiment Setup
In the short circuit test, In the short circuit test, the secondary side is the secondary side is short circuitedshort circuited and the and the primaryprimary sideside is is connected toconnected to a variable, a variable, low voltage sourcelow voltage source. .
Measurements of power, current, and voltage Measurements of power, current, and voltage are made on the primary side. The are made on the primary side. The applied applied voltagevoltage is adjusted until rated short circuit is adjusted until rated short circuit currents flows in the windings. currents flows in the windings.
This voltage is generally much smaller than the This voltage is generally much smaller than the rated voltage.rated voltage.
38
sc
scsc I
VZ
scsc
sc
IV
PcosPF
scsc
sc
IV
Pcos 1
spspeqeqsc XaXjRaRjXRZ 22
Impedances referred to the primary side Impedances referred to the primary side
Power Factor of the currentPower Factor of the current
Angle Power FactorAngle Power Factor
00
00
sc
sc
sc
scsc I
V
I
VZ
ThereforeTherefore
39
The equivalent circuit impedances of a 20-kVA, The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60-Hz transformer are to be 8000/240-V, 60-Hz transformer are to be determined. The open circuit test and the short determined. The open circuit test and the short circuit test were performed on the primary side of circuit test were performed on the primary side of the transformer and the following data were taken:the transformer and the following data were taken:
Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit
Open- circuit test (on Open- circuit test (on primary)primary)
Short- circuit test (on Short- circuit test (on primary)primary)
Voc = 8000 VVoc = 8000 V Vsc = 489 VVsc = 489 V
Ioc = 0.214 AIoc = 0.214 A Isc = 2.5 AIsc = 2.5 A
Poc = 400 WPoc = 400 W Psc = 240 WPsc = 240 W
40
Per unit SystemPer unit System
quantity of valueBase
Quantity Actual pu Per Unit,
The per unit value of any quantity is defined asThe per unit value of any quantity is defined as
Quantity – may be Quantity – may be power, voltage, current power, voltage, current or or impedanceimpedance
41
1.1. It eliminates the need for conversion of the It eliminates the need for conversion of the voltages, currents, and impedances across voltages, currents, and impedances across every transformer in the circuit; thus, there is every transformer in the circuit; thus, there is less chance of computational errors.less chance of computational errors.
2.2. The need to transform from three phase to The need to transform from three phase to single phase equivalents circuits, and vise single phase equivalents circuits, and vise versa, is avoided with the per unit quantities; versa, is avoided with the per unit quantities; hence, there is less confusion in handling and hence, there is less confusion in handling and manipulating the various parameters in three manipulating the various parameters in three phase system.phase system.
Two major advantages in using a per unit Two major advantages in using a per unit systemsystem
42
Per Unit (pu) in Single Phase SystemPer Unit (pu) in Single Phase System
basebasebasebasebase IVS,Q,P
base
basebase I
VZ
base
basebase V
IY
base
basebase S
)V(Z
2
43
Voltage Regulation (VR)Voltage Regulation (VR)
The voltage regulation of a transformer is defined The voltage regulation of a transformer is defined as the change in the magnitude of the secondary as the change in the magnitude of the secondary voltage as the current changes from full load to voltage as the current changes from full load to no load with the primary held fixed. no load with the primary held fixed.
%100,
,, XV
VVVR
flS
flSnlS
%100,
,
XV
Va
V
VRflS
flSp
aVV P
S
At no load,At no load,
I s
V p /a
R eq X eq
V s
+
-
+
-
44
V s
I sR eq
jI sX eq
V p /a
I s
Phasor DiagramPhasor Diagram
Lagging power factorLagging power factor
V s I sR eq
jI sX eq
V p /a
I s
Unity power factorUnity power factor 45
I s
V s
I sR eq
jI sX eq
V p /a
Leading power factorLeading power factor
46
EfficiencyEfficiencyThe efficiency of a transformer is defined as the ratio of the The efficiency of a transformer is defined as the ratio of the power output (Ppower output (Poutout) to the power input (P) to the power input (Pinin).).
%100XP
P
in
out
%100XPP
P
lossesout
out
%100cos
cosX
PPIV
IV
corecuss
ss
Pcore = Peddy current + Physteresis
And
Pcu= Pcopper losses 47
48
PPcu cu = Copper losses are resistive losses in the primary and = Copper losses are resistive losses in the primary and
secondary winding of the transformer core. They are modeled by secondary winding of the transformer core. They are modeled by placing a resistor placing a resistor RRpp in the primary circuit of the transformer and in the primary circuit of the transformer and
resistor resistor RRss in the secondary circuit. in the secondary circuit.
PPCORECORE = Core loss is resistive loss in the primary winding of the = Core loss is resistive loss in the primary winding of the
transformer core. It can be modeled by placing a resistor transformer core. It can be modeled by placing a resistor RRcc in the in the
primary circuit of the transformer.primary circuit of the transformer.
Example Example
A 15 kVA, 2400/240-V transformer is to be tested A 15 kVA, 2400/240-V transformer is to be tested to determine its excitation branch components, to determine its excitation branch components, its series impedances and its voltage regulation. its series impedances and its voltage regulation. The following test data have been taken from the The following test data have been taken from the primary side of the transformerprimary side of the transformer
Open- circuit testOpen- circuit test Short- circuit testShort- circuit test
Voc = 2400 VVoc = 2400 V Vsc = 48 VVsc = 48 V
Ioc = 0.25 AIoc = 0.25 A Isc = 6.0 AIsc = 6.0 A
Poc = 50 WPoc = 50 W Psc = 200 WPsc = 200 W
49
The data have been taken by using the The data have been taken by using the connections of connections of open circuit test open circuit test and and short circuit short circuit testtest
a.a.Find the equivalent circuit of this transformer Find the equivalent circuit of this transformer referred to the high voltage side.referred to the high voltage side.
b.b.Find the equivalent circuit of this transformer Find the equivalent circuit of this transformer referred to the low voltage side.referred to the low voltage side.
c.c.Calculate the full load voltage regulation at 0.8 Calculate the full load voltage regulation at 0.8 lagging power factor and 0.8 leading power lagging power factor and 0.8 leading power factor.factor.
d.d.What is the efficiency of the transformer at full What is the efficiency of the transformer at full load with a power factor of 0.8 lagging? load with a power factor of 0.8 lagging?
50
