Chapter 5 - Transformer
Chapter 5 - Transformer
Chapter 5 - Transformer
ELECTRICAL AND
INSTRUMENTATION TECHNOLOGY
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Week 9-10
Chapter 5: Transformers
Transform
!!!
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Lesson Outcome
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Transformers
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Overview
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CONSTRUCTION OF TRANSFORMER
Core Type
Shell Type
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PRINCIPLE OF
OPERATION
The
primary coil of a transformer is supplied by the
voltage source or AC.
When the current flows in a primary coil, the alternating
flux will be produced in a core.
If it can reach secondary coil, it will somehow induce the
emf at the same magnitude with the primary coil.
Therefore, for N1 and N2 numbers of windings/turns in a
primary and secondary coil, several relationship can be
established:
9 Turn
ratio a
STEP UP TRANSFORMERS
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STEP DOWN TRANSFORMERS
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EXAMPLE 1
aV2
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SIMPLIFICATION OF EQUIVALENT
CIRCUIT-SECONDARY REF.
V1=V1/a
I1=aI1
Rc=(1/a)2R
c
Xm=(1/a)2X
m
R1=(1/a)2R
1
17 2X
X1=(1/a)
APPROXIMATE EQUIVALENT CIRCUIT-
PRIMARY REF.
I2=I2/a
V2=aV2
RS1=R1+R2
XS1=X1+X2
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APPROXIMATE EQUIVALENT CIRCUIT-
SECONDARY REF.
Ze
I1=aI1
V1=V1/a
RS2=R1+R2
XS2=X1+X2
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EXAMPLE 2
200
150
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EXAMPLE 2
0.15 0.20
I2/a
(720/240)2(0.08)=
0.72
(720/240)2(0.0
3)= 0.27 aV2
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EXAMPLE 2
V1=V1/a
I1=aI1
Rc=(1/a)2Rc=(240/720)2(200)=
22.2
Xm=(1/a)2Xm=(240/720)2(150)
= 16.7
R1=(1/a)2R1=(240/720)2(0.15)
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= 0.017
EXAMPLE 2
RS1=R1+R2=0.15+0.27=
0.42
XS1=X1+X2=0.20+0.72=
0.92
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EXAMPLE 2
I1=aI1
V1=V1/a
RS2=R1+R2=0.017+0.03=0.
047
XS2=X1+X2=0.022+0.08=0.
102
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END OF TOPIC 1
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TOPIC 2:
EFFICIENCY
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EFFICIENCY
Transformers
Efficiency can be defined as
output power divided by input power
whereby the unit of those should be the
same (watt)
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EFFICIENCY
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EFFICIENCY
TRANSFORM
ERS LOSS
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EXAMPLE 3
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EXAMPLE 3
Input power=output
power+losses=500+10=510W
Efficiencies=(output/input power)x100%
=(500/510)x100%=98%
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END OF TOPIC 2
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TOPIC 3: VOLTAGE
REGULATIONS
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VOLTAGE REGULATIONS
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VOLTAGE REGULATIONS
(perunit)
(%)
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VOLTAGE REGULATIONS
Power factor
lagging (vice-
versa for
leading)
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EXAMPLE 4
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EXAMPLE 4
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EXAMPLE 4
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EXAMPLE 4
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EXAMPLE 4
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END OF TOPIC 3
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TOPIC 4: 3-PHASE
TRANSFORMERS
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3-PHASE
TRANSFORMERS
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3-PHASE
TRANSFORMERS
Transformer Bank
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3-PHASE
TRANSFORMERS
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PRIMARY AND SECONDARY
CONNECTION
Wye-Wye
Delta-Delta
Wye-Delta
Delta-Wye
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WYE-WYE CONNECTION
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DELTA-DELTA CONNECTION
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WYE-DELTA CONNECTION
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DELTA-WYE CONNECTION
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RELATIONSHIP BETWEEN LINE-
PHASE VOLTAGE-CURRENT
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EXAMPLE 5
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EXAMPLE 5
d) The current in LV lines:
IL = (24.4 x 106) /(3)(4160)= 3386 A
f) Since they are sharing the load equally so the load = 24.4 MVA/3 =
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EXAMPLE 6
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EXAMPLE 6
a) The primary line voltage, VL1 = 13.2 kV
Primary Vph = VL1 = 13.2 kV
Vph1/Vph2=13.2/80
So, Vph2=80/13.2 x 13.2 = 80 V
VL2=Vph2 x 3 = 80 x 3 = 138.56 kV
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THE WAY TO PREVENT HARMONICS
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END OF TOPIC 5
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TOPIC 6:
AUTOTRANSFORM
ER
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65 SOURCES: http://electrical-info.com/auto-
transformer/
AUTO TRANSFORMER
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AUTO TRANSFORMER
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AUTO TRANSFORMER
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AUTO TRANSFORMER
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AUTO TRANSFORMER
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Conclusion
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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PAST YEAR QUESTIONS
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