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7/28/2019 Chapter 2 342 Transformer 4
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7/28/2019 Chapter 2 342 Transformer 4
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Fida Muhammad (Air University)
Induction heating Non-contact heating process uses high frequency electricity to heat materials that areelectrically conductive
1. What makes the conductive material to heat up?
Eddy current in conductive non-magnetic materials (Cu) & eddy& Hysterisis in conductive-magnetic materials (iron & steel)
2. What about plastics?
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Fida Muhammad (Air University)
100
load_fullV
load_fullVload_noVVR%
s
ss
Transformer (T/F) Characteristics (distinctiveness)
Efficiency () Voltage Regulation (VR)
InputPower
LossesInputPower
InputPower
OutputPower
Eddy Current Losses
Hysterisis Losses
Copper Losses(RP& RS)
Voltage Drop
Across XP & XS
Voltage Drop
(RP& RS)
In seriesvoltage
drop
Question: What about XM
Not used for Efficiency. Why?
XM - Why not for VR ?
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Page#90 Determining the Values of Components in the
Transformer Model
How to find Values of Components ?
Open Circuit Test Short Circuit Test
Do the Following Test
Reqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IP
find RC & XM find Reqp, Xeqp OR Reqs , Xeqs
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Fida Muhammad (Air University)
IP(t)
Wattmeter
VP(t)
V
A
V(t)
Transformer
Open
Circuit
Ampere
Meter
Volt
Meter
Figure 2-19 Connection fortransformer open-circuit test
Open Circuit Test
Apply rated primary (input) Voltage
IO No-Load current flows primary
Wattmeter reads ..?
Core losses
small
Scale
(IO)
Why?
Large
Scale
Why ?
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IP(t)
Wattmeter
VP(t)
V
A
V(t)
Transformer
Ampere
Meter
Volt
Meter
Figure 2-20 Connection for
transformer short-circuit test
A
Secondary
Short Circuit
Short Circuit Testlargescale
(IO)
Why?
small
Scale
Why ?
Apply rated current
Full load IP & IS current flows
Wattmeter reads ?.....I2R (Copper)
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Fida Muhammad (Air University)
Page#90 Determining the Values of Components in the
Transformer ModelTests : Find Values of Components
Open Circuit Test Short Circuit Test
IP(t)Wattmeter
VP(t)V
A
V(t)
Transformer
Open
Circuit
Ampere
Meter
Volt
Meter
Figure 2-19 Connection for
transformer open-circuit test
IP(t)Wattmeter
VP(t)V
A
V(t)
Transformer
Open
Circuit
Ampere
Meter
Volt
Meter
Figure 2-19 Connection for
transformer open-circuit test
IP(t)Wattmeter
VP(t)V
A
V(t)
Transformer
Ampere
Meter
Volt
Meter
Figure 2-20 Connection for
transformer short-circuit test
A
Secondary
Short
CircuitIP(t)Wattmeter
VP(t)V
A
V(t)
Transformer
Ampere
Meter
Volt
Meter
Figure 2-20 Connection for
transformer short-circuit test
A
Secondary
Short
Circuit
Apply rated primary (input) Voltage
IO No-Load current flows primary
Wattmeter reads ..?Core losses
Apply rated current
Full load IP & IS current flows
Wattmeter reads I2R (Copper)
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A 20-kVA, 8000/240-V, 60-Hz T/F. The open-circuit test (OCT) and
the short-circuit test (SCT) were performed on the primary side of
the T/F, and the following data were taken:
Find the impedances of the approximate equivalent circuit referred
to the primary side, and sketch the circuit.
OCT (on primary)
Voc= 8000 V
Ioc
= 0.214 A
Poc= 400 W
SCT (on primary)
Vsc= 489 V
Isc
= 2.5 A
Psc = 240 W
Example 2.2 (page 92)
IP(t)Wattmeter
VP(t)V
A
V(t)
IP(t)Wattmeter
VP(t)V
A
V(t)
IP(t)Wattmeter
VP(t)V
A
V(t)
A
IP(t)Wattmeter
VP(t)V
A
V(t)
A
W tt tW tt t
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Answer
OCT (on primary)
Voc = 8000 V
Ioc =Io= 0.214 A
Poc = 400 W ??
IP(t)Wattmeter
VP(t)V
A
V(t)
IP(t)Wattmeter
VP(t)V
A
V(t)
IOC=I0=IP=0.214
VOC8000
IC IM
RC XM
POC=400W
VOC
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OCT (on primary)
Voc= 8000 V
Ioc=Io= 0.214 A
Poc= 400 W ???
IOC=I0=0.214
VP=8000
IC IM
RC XM
POC=400W
ococococ cosIVP
YY
2336021408000
400.
.xcos oc
01
57623360 ..cosoc
mhox..
Y 510728000
2140 ococ
ococ
IV
Pcos
ocOC
.OCV
I
8000
2140 05
5761072 .x.
Y=1/Z
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OCT (on primary)
Voc= 8000 V
Ioc=Io= 0.214 A
Poc= 400 W
IOC=I0=0.214
VP=8000
IC IM
RC XM
YY ocOC
.OC
V
I
8000
2140 05 5761072 .x.
oc=76.5o
jBM
Gc=Yoccosoc
=YocsinocYoc=2.7x10-5mho
)circuitsparallel(cetanAdmit
Yoc=Yoccosoc -jYoc sinoc
C
CG
R1
M
MB
X1
MCoc jBGY
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R= Zcos
Z X= Zsin
G=Ycos
YB=Ysin
jBGY
Y=Ycos -jY sin
Y
I
R jX
V
jXRZ
Z=Zcos +jZ sin
Z
Z
Y=1/ZR
G1
jXjB
1
Why
Negative ?
Why
Negative ?
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OCT (on primary)
Voc= 8000 V
Ioc=Io= 0.214 AP
oc= 400 W
IOC=I0=0.214
VP=8000
IC IM
RC XM
MCoc jBGY
Yoc=Yoccosoc -jYoc sin oc
MCoc Xj
RZ
111
Yoc=2.7x10-5 cos76.5 -J2.7x10-5 sin76.5
Y=6.25x10-6 -j 2.53x10-6
Z=159000 + j38400
RC=159kXM=
38.4k
M
M
C
C
oc
oc
X
B;
R
G;
Z
Y111
oc=76.5o
jBM
Gc
Yoc=2.7x10-5mho
W tt tW tt t
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Answer IP(t)Wattmeter
VP(t)V
A
V(t)
A
IP(t)Wattmeter
VP(t)V
A
V(t)
A
SCT (on primary)
Vsc= 489 V
Isc
= 2.5 A
Psc
= 240 W ??
ISC=Irated=215
Reqp jXeqp
VSC=489
PSC=240W
VSC
R
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SCT (on primary)
Vsc= 489 V
Isc = 2.5 A
Psc
= 240 W
ISC=Irated=215
Reqp jXeqp
VSC=489
scscscsc cosIVP
scscsc ZZ
196052489
240.
.xcos sc
01
68761960 ..cossc
ohms.
.
Zsc 619552
489
scsc
sc
sc IV
P
cos
sc.sc
scI
V
52
489 0
68766195 ..
R
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Fida Muhammad (Air University)
SCT (on primary)
Vsc= 489 V
Isc = 2.5 A
Psc
= 240 W
ISC=Irated=215
Reqp jXeqp
VSC=489
scscsc ZZ 068766195 ..
=Zsc
cossc
)eqp(SC)eqp(SC)eqp(SC jXRZ oc=76.68
RSC
jXSC =jZscsinsc
ZSC
Zsc=Zsccossc+ jZsc sin sc
Zsc=195.6
cos76.68 +j195.6sin76.68
Zeqp=38.4
+j 192
=38.4 =j192
=195.6
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Question:- For the same transformer given in the previous
example, if the OCT & SCT are performed on the secondary
side, (a) what will be the Voltage, current and power readings
for the OCT and SCT.
(b) find the impedances of the approximate equivalent circuit
referred to the secondary side. Compare the data with part (a)
and sketch the circuit diagram.
Assignment ( Due Friday for section B & Saturday for section A
Reqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IP
38.4 j192
159k 38.4k
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100
loadfullVloadfullVloadnoV
/aVVload-noatsince
100loadfullV
loadfullVloadnoVRegulationVoltage%
s
sp/a
PS
s
ss
The voltage regulation of a transformer is the change in the
magnitude of the secondary terminal voltage from no-load to
full-load.
2.7 (page 100) Transformer Voltage Regulation(VR) & Efficiency
Transformer Voltage Reg lation & Vector Diagram
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100loadfullV
loadfullVloadnoVRegulationVoltage%
S
SP/a
SSeqSeqa
VVIjXIRP
VReq VjXeq
SjXeqqReaV VVVP
=ReqIS =XeqIS
IS
Reqp jXeqp
L
O
Ad
VS
a
VP
Transformer Voltage Regulation & Vector Diagram
Transformer Voltage Regulation & Vector Diagram
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100loadfullV
loadfullVloadnoVRV%
S
SP/a
Transformer Voltage Regulation & Vector Diagram
IS VReq VjXeq
=ReqIS =XeqIS
Reqp jXeqp
L
O
A
d
VSa
VP
SeqSeqSa
VIjXIRVP loadfullVloadnoV SP/a
100
loadfullVRV%
s
SeqSeq
IjXIR
The VR of the
T/F depends on
the voltage
magnitude of
these seriesparameters.
Fig 2 26 (page 101) Vector Dig
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IS
VReq
IS
VReq VjXeq
=ReqIS =XeqIS
Reqp jXeqp
L
oa
d
VSa
VP
VS
SjXeqqReaV VVVP
VjXeq
VP/a
Phase relation
between IS &
VReq ?Phase relation
between IS &
VXeq ?
Fig 2-26 (page 101) Vector Dig
Lagging Power Factor
Lagging
Power FactorLoad
?100V
VVVR%
sfl
sflp/a
Positive
VP/a > VS
-
VS is assumed to
be at angle 0o
(
VS-). All othervoltages & currents
are compared to
this reference
IR
Vector Diagram
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Fida Muhammad (Air University)
IS
VReq VjXeq
Reqp jXeqp
L
o
ad
VSa
VPVeq
:Angle between Vs & VP/a
Veq
: Angle between VReq & VXeq
: Power Factor angle (Between Vs & Is)
Vector Diagram
Fig 2 27(a) Vector Dig
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Fida Muhammad (Air University)
-
VS
IS
VReq VjXeq
=ReqIS =XeqIS
Reqp jXeqp
L
oa
d
VSa
VP
SjXeqqReaV VVVP
VP/a
Fig 2-27(a) Vector Dig
Unity Power Factor
UnityPower Factor
Load
?100V
VV
VR%sfl
sflp/a
Positive
VP/a > VS
Fig 2 27(a) Vector Dig
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VS
IS
VReq VjXeq
=ReqIS =XeqIS
Reqp jXeqp
L
oa
d
VSa
VP
SjXeqqReaV VVVP
VP/a
Fig 2-27(a) Vector Dig
Leading PF
VReq
LeadingPower Factor
Load
?100V
VV
VR%sfl
sflp/a
Negative
VP/a
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InputPower
OutputPower
The efficiency for a power transformer is between 0.9 to 0.99.
The higher the rating of a transformer, the greater is its efficiency.
Transformer Efficiency
InputPower
Losses
1
InputPower
LossesInputPower
cosIVPP
PP
sslosscorelosscopper
losscorelosscopper1
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A 15 kVA 2300/230 V 60 Hz T/F The open circuit test (OCT) andExample 2 5 (page 103)
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A 15-kVA, 2300/230-V, 60-Hz T/F. The open-circuit test (OCT) and
the short-circuit test (SCT) were performed on the primary side of
the T/F, and the following data were taken:
OCT (on primary)
Voc= 2300 V
Ioc
= 0.21 A
Poc= 50 W
SCT (on primary)
Vsc= 47 V
Isc
= 6.0 A
Psc = 160 W
Example 2.5 (page 103)
IP(t)Wattmeter
VP(t)V
A
V(t)
IP(t)Wattmeter
VP(t)V
A
V(t)
IP(t)Wattmeter
VP(t)V
A
V(t)
A
IP(t)Wattmeter
VP(t)V
A
V(t)
A
(a) Find the impedances of the approximate equivalent circuit
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(b) Find the impedances of the approximate equivalent circuit
referred to the LV-side. (You can do this part: refer to slide#9) .
(a) Find the impedances of the approximate equivalent circuit
referred to the HV-side (You can do this part: refer to slide#9) .
38.4 j192
159k
Reqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IPReqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IP
j11k105k
4.45 J6.45
Figure:2-29(a)
Reqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIPReqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIP
38.4
159k 38.4k
Figure:2-29(b)
j110k1050k
0.0445 J0.0645
( ) C l l t th f ll l d ( t d l d) lt l ti
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(c) Calculate the full-load (rated-load) voltage regulation:
(i) at 0.8 lagging power factor
(ii) at unity (1) power factor
(iii) at 0.8 leading power factor
Reqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIPReqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIP
38.4
159k 38.4k
Figure:2-29(b)
j110k1050k
0.0445 J0.0645
Reqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIPReqs jXeqs
VS
IS
VP/a
IO
IMIC
RC jXM
aIP
38.4
159k 38.4k
Figure:2-29(b)
j110k1050k
0.0445 J0.0645
We can use any of the
referred circuits to find
VR. Its easy to use
figure referred to thesecondary side.
SjXeqqRea
VVVVP
SjeqSeqSaV IjXIRVP
?100
V
VVVR%
sfl
sflp/a
VS=Vsf=230V - Given
Vp/a=unknown
To get Vp/a find ReqIS & XeqIS
Req=0.0445; Xeq=0.0645
from SCT & OCT.
IS=IS(rated) un-known
dthtl dllFHow to find I I ?
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Va
V oP0230
rateds,
ratedrated,S
V
SIisF/Tthisofside
ondarysectheoncurrentloadullF
How to find IS =IS(rated) ?
IS VReq VjXeq
ReqIS,rated XeqIS,rated
Reqp jXeqp
L
O
A
d
VS
a
VP
0.0445 0.0645
IS=IS,rated
Parallel circuit is omitted.
Why ??
26500015
.230
VA,I rated,S
lagging.PFat
..I rated,S
80
9362650
0936265 ..
)A..)(.( o93626504450 )A..)(.(j o93626506450 Va
V oP0230
SjeqSeqSa
VIjXIRVP
V.. o153214 V.. o93692
?100V
VVVR%
sfl
sflp/a
).()VA( 8015000
V
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%1.23085.34
2100
230V
2VR%
sfl
VV Angles are not used in
VR has no unit why?
InputPowerOutputPower
What is the
problem with
this equation ?
(e) What is the efficiency of the T/F at full load with a PF=0.8
lagging?
coreCuSS
SS
PPcosIVcosIV
InputPower
LossesInputPower
Use this
equation
VS IS =S (apparent power) and cos=0.8 lagging are known
Power input is unknown. We can use input equation, if the T/F
circuit is referred to the primary side. - Try & find efficiency
Va
V oP0230 V..
o153214 V..
o93692 V.. o4085234
IV
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PCu= (IS)2
Req = (65.2A)2
(0.0445) = 189 Watts
Watts.
V.
R
aV
PC
P
core 5521050
852342
2
000398100
5521898015000
8015000.x
W.W).()VA(
).()VA(
InputPower
OutputPower
coreCuSS
SS
PPcosIV
cosIV
Why the SCT cupper losses(160W) and OCT core losses(50W)are not used in efficiency calculation ?
Open circuit test at the primary is performed at 2300 V. Where
as with rate-load (PF=0.8 lagging) VP=234.85x10=2348.5V, i.e
48.5V more than the OCT.
SCT is performed
with ISC=6A, where
as IP(rated)= 6.52A
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What is the condition for the maximum efficiency in the T/F?
Core Losses = Copper LossesPcore = Pcopper
In the previous problem find the current at maximum efficiency?
I2Rseq= 52.5 W
Amps..
.
Imax3434
04450
552
T/F will operate at almost half the rated load (half of T/F capacity)
Example 2-5 Page # 103
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p g
Solution
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38.4 j192
159k
Reqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IPReqp jXeqp
aVS
IS/a
VP
IO
IMIC
RC jXM
IP
j11k105k
4.45 J6.45
Figure:2-29(a)
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Continued
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Practice Problem
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A single-phase, 100-kVA, 1000:100-V, 60-Hz transformer has thefollowing test results:
Open-circuit test (HV side open): 100 V, 6 A, 400 W
Short-circuit test (LV side shorted): 50 V, 100 A, 1800 W
Draw the equivalent circuit of the transformer referred to the high-voltage
side. Label impedances numerically in ohms and in per unit. Determine the voltage regulation at rated secondary current with 0.6
power factor lagging. Assume the primary is supplied with rated voltage
Determine the efficiency of the transformer when the secondary current is75% of its rated value and the power factor at the load is 0.8 lagging witha secondary voltage of 98 V across the load
Example 4
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Fida Muhammad (Air University)
A 20-kVA, 8000:277-V distribution transformer has the followingresistances and reactances:
The excitation branch impedances are referred to the high-voltage side.
a) Find the equivalent circuit of the transformer referred to the high-voltageside.
b)
b) Find the per unit equivalent circuit of this transformer.
c) Assume that the transformer is supplying rated load at 277 V and 0.8power factor lagging. What is this transformers input voltage? What isits voltage regulation?
d) What is this transformers efficiency under the conditions of part (c)?
RP= 32 ohm
XP= 45 ohm
RC= 250,000 ohm
RS
= 0.05 ohm
XS= 0.06 ohm
XM
= 30,000 ohm
Neglect
PU S t
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Fida Muhammad (Air University)
PU System
QuantityBase
QuantityActualValuePU
base
ohm
PU
base
basebase
base
base
base
base
base
base
basebasebase
basebasebasebasebasebase
base
basebase
Z
ZZ
V
IY
VA
V
S
V
I
V
ZXR
IVVASQP
V
VAI
22
Per unit system, a system of dimensionless parameters, is used for
computational convenience and for readily comparing the performance
of a set of transformers or a set of electrical machines.
Where actual quantity is a value in volts, amperes, ohms, etc.
[VA]base and [V]base are chosen first.
ratioturnsVV
VAVA
secbase
pr ibase
secbasepr ibase
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Power Electronics
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