1. EE6352 - Electrical Engg & Instrumentation
Unit 2- Transformer
S. Arulmozhi, AP/EE
11/27/2018

2. Syllabus
TRANSFORMER 9
Introduction – Single phase transformer construction and principle of operation – EMF equation of transformer-Transformer no–load phasor diagram –– Transformer on–load phasor diagram – Equivalent circuit of transformer – Regulation of transformer –Transformer losses and efficiency-All day efficiency –auto transformers.
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3. Transformer
Transformer is a static device which transfers electric power from one circuit to another without changing its frequency
The principle of operation of a transformer depends upon Faraday's Law of Electomagnetic Induction.
Actually mutual inductance between 2 or more windings is responsible for transformation action in an electrical transformer.
According to Faraday's Law, “Rate of change of flux linkages with respect to time is directly proportional to the induced EMF in a conductor or a coil”.
e = MdI/dt or N.dϕ / dt
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4. Transformer
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5. Primary coil and Secondary coil
Consists of two inductive coils which are electrically seperate but magnetically linked through a path of low reluctance.
Primary coil and Secondary coil
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6. transfers electric power from one circuit to another
Transformer -
transfers electric power from one circuit to another
does so without change of frequency
accomplishes this by electromagnetic induction
where the two electric circuits are linked by mutual induction.
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7. Types of transformer Based on terminal voltage, Step-up transformer
Step down transformer
Based on supply voltage,
Single phase transformer
Three phase transformer
Based on construction,
Core type transformer
Shell type transformer
Spiral type or wound core type transformer (spirakore)
Based on type of cooling,
Oil-filled self cooled
Oil-filled water cooled
Air blast type
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8. Core Type Transformer
The windings surround a considerable part of the core
It is a single magnetic circuit
Rectangular in shape having 2 limbs
The coils used are of cylindrical type.
The coils are wound in helical layers with different layers insulated from each other by paper or micaor cloth.
The core is made up of large number of thin laminations.
LV winding is placed nearest to the core with interleaved structure.
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9. Lecture - 2 Review of Lecture 1 Construction & Types of Transformer
EMF equation of transformer
Summary
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10. Review - Transformers operate only on AC
Frequency of DC supply = 0Hz
Inductive reactance XL = 2πfL
The effective impedance of the winding -very low
= resistance of the copper used.
Winding draws very high current from the DC supply (I = V/R)
Overheat & Burnout of windings
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11. Construction - Core Core is either square or rectangular in size
The vertical portion in which the coils are wound are called the limb and carries the windings
The horizontal portion is called the yoke which carries the flux produced by one winding to another.

12. Core Core is made up of laminations to prevent eddy current losses.
Laminations are made up of silicon steel of 0.3 to 0.5mm thick
Laminations are insulated from each other by insulating materials like varnish.
Laminations are overlapped to avoid airgap at the joints.
'L', 'E' , 'I' or 'T' shaped laminations are used.

13. Laminations
In core type transformers, the steel core laminations are in the shape of 'L' , and these laminations are placed one above the other alternatively.

14. Laminations
In shell type transformers, the steel core laminations are in the shape of 'E and I' and placed similar to that of core type.

16. Types of laminations

18. Windings
Transformer windings are made of solid copper or aluminium strip conductors.
The coils used are wound on the limbs and are insulated from each other.
It carries the current and produces the flux necessary for the functioning of the transformer.

19. Core type
Windings are wrapped around two sides of a laminated square core.
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20. Shell type transformer
The core surrounds a considerable portion of the windings.
It is a double magnetic circuit
The core has 3 limbs.
The coils used are multilayered disc type or sandwich coils.
The core is laminated.
All the joints at alternate layers are staggered to avoid narrow airgaps at the joints. Such joints are called as overlapped or imbricated joints.
Windings are wrapped around the center leg of a laminated core.
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21. Shell type - sandwich coil structure
HV winding lies between two LV windings
Controls leakage
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22. Sectional view of transformer
Windings are wrapped around the center leg of a laminated core.
Windings are wrapped around two sides of a laminated square core.
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23. Comparison between Core & Shell Type
Description
Core Type
Shell Type
Construction
Easy to assemble & Dismantle
Complex
Mechanical Strength
Low
High
Leakage reactance
Higher
Smaller
Cooling
Better cooling of Winding
Better cooling of Core
Repair
Easy
Hard
Applications
High Voltage & Low output
Low Voltages & Large Output 23
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24. Berry type transformer
It has distributed magnetic circuit
The core type is like the spikes of a wheel.
The transformers are generally placed in tightly fitted sheet metal tanks.
Tanks are made up of high quality steel plate, formed and welded into a rigid structure.
The tanks are filled with special insulating oil.
All the joints are painted with light blue chalk solution disclosing even a minute leak.
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25. Symbol for ideal transformer
Ideal Transformers
Zero leakage flux:
-Fluxes produced by the primary and secondary currents are confined within the core
The windings have no resistance:
- Induced voltages equal applied voltages
The core has infinite permeability
- Reluctance of the core is zero
- Negligible current is required to establish
magnetic flux
Loss-less magnetic core
- No hysteresis or eddy currents
NP : NS VP VS
Primary
Secondary
Symbol for ideal transformer
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26. Ideal transformer V1 – supply voltage ; I1- noload input current ;
V2- output voltgae; I2- output current
Im- magnetising current;
E1-self induced emf ; E2- mutually induced emf
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27. Transformer Equations
Using Faraday’s law, expressions for the primary and secondary voltages is as follows.
Dividing the above equations we get,
Assuming that there is no power loss,
K - transformation ratio
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28. EMF Equation of a transformer
Let N1= No. of primary turns
N2= No. of secondary turns
φm= Maximum flux density in transformer core in Weber
=BmA where Bm-> flux density in the transformer core
A -> cross sectional Area of the transformer In an EMF equation, flux increases from its zero value to maximum value Фm in one quarter of cycle
Average rate of change of flux = Фm/(1/4f)=4f Фm wb/sec
The average value of emf induced / turn = 4f Фm
If flux Фm varies sinusoid ally, then R.M.S value of induced EMF is obtained by multiplying the average value with form factor.
……(1)
……(2)

29. EMF Equation of a transformer -contd
Form factor=R.M.S value/Average value =
= (for sine wave)
R.M.S value of EMF/turn=1.11*4fФm volts
Now,
R.M.S value of the induced EMF in the whole primary winding = (induced EMF/turn)*No of primary turns. E1 = 4.44fN1Фm
llly,
E2 = 4.44fN2Фm
……(3)
……(4)
……(5)
……(6)

30. EMF equation of transformer - contd
If ip is sinusoidal, the flux produced also sinusoidal, i.e
 = m sin 2ft
……(7)
therefore
v1 = N12fmcos 2ft = N12fmsin (2ft + /2)
……(8)
The peak value = Vpm = N12fm
and v1 is leading the flux by p/2.
……(9)
The rms value
……(10)
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31. Summary Construction - Core & Shell type EMF equation of a transformer
Queries????????
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32. Power Transmission (CONTENT BEYOND SYLLABUS)
Transformers play a key role in the transmission of electric power.
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33. Transformer with conservator and breather(CONTENT BEYOND SYLLABUS)
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34. Parts of a transformer (CONTENT BEYOND SYLLABUS)
Conservator
Oil is stored in the conservator
It prevents the oil from moisture contact in air during the expansion and contraction.
Breather
It is a device which contains silica gel crystals.
The gel absorbs the moisture in the atmosphere when the oil expands and contracts.
Explosive Vent
It bursts when pressure inside the transformer becomes excessive and protects the transformer from damage.
Transformer Tank filled with transformer Oil

35. Transformer-No load condition

36. Phasor diagram: Transformer on No-load
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37. Transformer – On load condition

38. Transformer on load assuming no voltage drop in the winding
Fig shows the Phasor diagram of a transformer on load by assuming
No voltage drop in the winding
Equal no. of primary and secondary turns
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39. Transformer on load Fig. a: Ideal transformer on load
Fig. b: Main flux and leakage flux in a transformer
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40. Phasor diagram of transformer with UPF load
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41. Phasor diagram of transformer with lagging p.f load
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42. Phasor diagram of transformer with leading p.f load
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43. Equivalent circuit of a transformer
No load equivalent circuit:
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44. Equivalent circuit parameters referred to primary and secondary sides respectively
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45. Contd.,
The effect of circuit parameters shouldn’t be changed while transferring the parameters from one side to another side
It can be proved that a resistance of R2 in sec. is equivalent to R2/k2 will be denoted as R2’(ie. Equivalent sec. resistance w.r.t primary) which would have caused the same loss as R2 in secondary,
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46. Transferring secondary parameters to primary side
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47. Equivalent circuit referred to secondary side
Transferring primary side parameters to secondary side
Similarly exciting circuit parameters are also transferred to secondary as Ro’ and Xo’
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48. equivalent circuit w.r.t primary
where
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49. Approximate equivalent circuit
Since the noload current is 1% of the full load current, the nolad circuit can be neglected
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50. Transformer Tests
The performance of a transformer can be calculated on the basis of equivalent circuit
The four main parameters of equivalent circuit are:
- R01 as referred to primary (or secondary R02)
- the equivalent leakage reactance X01 as referred to primary (or secondary X02)
- Magnetising susceptance B0 ( or reactance X0)
- core loss conductance G0 (or resistance R0)
The above constants can be easily determined by two tests
- Oper circuit test (O.C test / No load test)
- Short circuit test (S.C test/Impedance test)
These tests are economical and convenient
- these tests furnish the result without actually loading the transformer
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51. Open-circuit Test
In Open Circuit Test the transformer’s secondary winding is open-circuited, and its primary winding is connected to a full-rated line voltage.
Usually conducted on H.V side
To find
(i) No load loss or core loss
(ii) No load current Io which is helpful in finding Go(or Ro ) and Bo (or Xo )
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52. Short-circuit Test
In Short Circuit Test the secondary terminals are short circuited, and the primary terminals are connected to a fairly low-voltage source
The input voltage is adjusted until the current in the short circuited windings is equal to its rated value. The input voltage, current and power is measured.
Usually conducted on L.V side
To find
(i) Full load copper loss – to pre determine the efficiency
(ii) Z01 or Z02; X01 or X02; R01 or R02 - to predetermine the voltage regulation
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53. Contd…
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54. Transformer Voltage Regulation and Efficiency
The output voltage of a transformer varies with the load even if the input voltage remains constant. This is because a real transformer has series impedance within it. Full load Voltage Regulation is a quantity that compares the output voltage at no load with the output voltage at full load, defined by this equation:
Ideal transformer, VR = 0%.
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55. Voltage regulation of Transformer
recall
Secondary voltage on no-load
V2 is a secondary terminal voltage on full load
Substitute we have
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56. Transformer Phasor Diagram
To determine the voltage regulation of a transformer, it is necessary understand the voltage drops within it.
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57. Transformer Phasor Diagram
Ignoring the excitation of the branch (since the current flow through the branch is considered to be small), more consideration is given to the series impedances (Req +jXeq).
Voltage Regulation depends on magnitude of the series impedance and the phase angle of the current flowing through the transformer.
Phasor diagrams will determine the effects of these factors on the voltage regulation. A phasor diagram consist of current and voltage vectors.
Assume that the reference phasor is the secondary voltage, VS. Therefore the reference phasor will have 0 degrees in terms of angle.
Based upon the equivalent circuit, apply Kirchoff Voltage Law,
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58. Transformer Phasor Diagram
For lagging loads, VP / a > VS so the voltage regulation with lagging loads is > 0.
When the power factor is unity, VS is lower than VP so VR > 0.
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59. Transformer Phasor Diagram
For lagging loads, the vertical components of Req and Xeq will partially cancel each other. Due to that, the angle of VP/a will be very small, hence we can assume that VP/k is horizontal. Therefore the approximation will be as follows:
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60. Transformer Phasor Diagram
With a leading power factor, VS is higher than the referred VP so VR < 0
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61. Voltage regulation Lagging P.F. VP/ k > VS V.R. > 0 Unity P.F.
V.R. >0 (smaller)
Leading P.F.
VS > VP/ k
V.R. < 0
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62. Voltage regulation for Lagging Power Factor

63. Voltage Regulation for Leading Power Factor

64. Formula: voltage regulation
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65. Transformer Efficiency
Transformer efficiency is defined as (applies to motors, generators and transformers):
Types of losses incurred in a transformer:
Copper I2R losses
Hysteresis losses
Eddy current losses
Therefore, for a transformer, efficiency may be calculated using the following:
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66. Losses in a transformer
Core or Iron loss:
Copper loss:
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67. Condition for maximum efficiency
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68. Contd.,
The load at which the two losses are equal =
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69. AUTO TRANSFORMER
At some occasions it is desirable to change voltage level only by a small amount
i.e. may need to increase voltage from 110 to 120 V or from 13.2 to 13.8 kV
This may be due to small increase in voltage drop that occur in a power system with long lines
In such cases it is very expensive to hire a two full winding transformer, however a special transformer called: ”auto-transformer” can be used

70. AUTO TRANSFORMER
Diagram of a step-up auto-transformer shown in figure below:
C: common, SE: series

71. AUTO TRANSFORMER
A step-down auto-transformer :
IH=ISE
IL=ISE+IC

72. AUTO TRANSFORMER In step-up autotransformer: VC / VSE = NC / NSE (1)
NC IC = NSE ISE (2)
voltages in coils are related to terminal voltages as follows:
VL=VC (3)
VH=VC+VSE (4)
current in coils are related to terminal currents:
IL=IC+ISE (5)
IH=ISE (6)

73. AUTO TRANSFORMER Voltage & Current Relations in Autotransformer
VH=VC+VSE
since VC/VSE=NC/NSE  VH=VC+ NSE/NC . VC
Noting that: VL=VC 
VH=VL+ NSE/NC . VL= (NSE+NC)/NC . VL
VL / VH = NC / (NSE+NC) (7)
Current relations:
IL=IC+ISE employing Eq.(2)  IC=(NSE / NC)ISE
IL= (NSE / NC)ISE + ISE, since ISE=IH 
IL= (NSE / NC)IH +IH = (NSE + NC)/NC . IH 
IL / IH = (NSE + NC)/NC (8)

74. AUTO TRANSFORMER
Apparent Power Rating Advantage of Autotransformer
Note : not all power transferring from primary to secondary in autotransformer pass through windings
Therefore if a conventional transformer be reconnected as an autotransformer, it can handle much more power than its original rating
The input apparent power to the step-up autotransformer is : Sin=VLIL
And the output apparent power is:
Sout=VH IH

75. AUTO TRANSFORMER And : Sin=Sout=SIO
Apparent power of transformer windings:
SW= VCIC=VSE ISE
This apparent power can be reformulated:
SW= VCIC=VL(IL-IH) =VLIL-VLIH
employing Eq.(8)  SW= VLIL-VLIL NC/(NSE+NC)
=VLIL [(NSE+NC)-NC] /(NSE+NC)=SIO NSE /(NSE+NC)
SIO / SW = (NSE+NC) / NSE (9)

76. AUTO TRANSFORMER Internal Impedance of an Autotransformer
Another disadvantage: effective per unit impedance of an autotransformer w.r.t. the related conventional transformer is the reciprocal of power advantage
This is a disadvantage where the series impedance is required to limit current flows during power system faults (S.C.)

77. Example of Variable Auto-Transformer

78. All day efficiency
All day efficiency is defined as the ratio of total energy output of transformer to thetotal energy input in 24 hours.
All day efficiency is always less than the commercial efficiency
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79. Transformer Voltage Regulation and Efficiency - Tutorial
(a) Find the equivalent circuit referred to H.V. side
(b) Find the equivalent circuit referred to L. V. side
(c) Calculate the full-load voltage regulation at 0.8 lagging PF, 1.0 PF, and at 0.8 leading PF
(d) Find the efficiency at full load with PF 0.8 lagging
SOLUTION:
Open circuit impedance angle is:
Excitation admittance is:

80. Transformer Voltage Regulation and Efficiency
Impedance of excitation branch referred to primary:
Short Circuit Impedance angle:
Equivalent series Impedance:
Req=4.45 Ω, Xeq=6.45 Ω

81. Transformer Voltage Regulation and Efficiency
The equivalent circuits shown below:

82. Transformer Voltage Regulation and Efficiency
(b) To find eq. cct. Referred to L.V. side,
impedances divided by a²=NP/NS=10
RC=1050 Ω , XM=110 Ω
Req= Ω , Xeq= Ω
(c) full load current on secondary side:
IS,rated=Srated/ VS,rated=15000/230 =65.2 A
To determine V.R., VP/ a is needed
VP/a = VS + Req IS + j Xeq IS , and:
IS=65.2/_-36.9◦ A , at PF=0.8 lagging

83. Transformer Voltage Regulation and Efficiency
Therefore:
VP / a =
V.R.=( )/230 x 100 %=2.1 % for 0.8 lagging
At PF=0.8 leading  IS=65.2/_36.9◦ A

84. Transformer Voltage Regulation and Efficiency
V.R. = ( )/230 x 100%= %
At PF=1.0 , IS= 65.2 /_0◦ A
VP/a=
V.R. = ( )/230 x 100% = 1.28 % for PF=1

85. Transformer Voltage Regulation and Efficiency
Example: Phasor Diagrams …

86. Transformer Voltage Regulation and Efficiency
(d) to plot V.R. as a function of load is by repeating the calculations of part “c” for many different loads using MATLAB

87. Transformer Voltage Regulation and Efficiency
(e) Efficiency of Transformer:
- Copper losses:
PCu=(IS)²Req =(65.2)² (0.0445)=189 W
- Core losses:
PCore= (VP/a)² / RC= (234.85)² / 1050=52.5 W
output power:
Pout=VSIS cosθ=230x65.2xcos36.9◦=12000 W
η= VSIS cosθ / [PCu+PCore+VSIS cosθ] x 100%=
12000/ [ ] = %