TRANFORMERS

1. Transformers

2. What is a Transformer?
• A transformer is a static electrical machine which transfers electrical
energy from one circuit to another without changing the frequency.
• Which raises or lowers voltage or current at the same frequency.
• It works on the principle of MUTUAL INDUCTION.

3. Transformer
• It consists of two windings insulated from each other and wound on a
common core made up of a magnetic material.
• AC voltage is connected across one of the windings called primary
winding.
• Load is connected to the other winding called the secondary winding.
• In both windings, EMF is induced by electromagnetic induction.

4. Transformer

5. Constructional details
Main Components of a Transformer are,
 Magnetic core
 Primary & Secondary windings
 Insulation of windings
 Conservator tank & Explosion vent
 Bushings
 Buchholz relay
 Breather
 Cooling arrangements

6. Magnetic Core
• Magnetic circuit consists of an iron
core.
• Core is made up of stacks of thin
laminations (0.35mm thickness) of Cold
Rolled Grain Oriented (CRGO) silicon
steel.
• These laminations are lightly insulated
with varnish.
• Two types of magnetic circuit are core
type and shell type.

7. Magnetic Core

8. Core type construction
• In the core type, the windings are wound around two legs of a
rectangular magnetic core.
• Windings surround the core & it has only one magnetic path.

9. Shell type construction
• In shell type, the windings are wound around the center leg of a
three-legged core
• Core surrounds the windings.

10. Windings
• A transformer has two windings namely primary and secondary.
• These windings consist of a series of turns called coils, wound around
the core.
• Transformer windings are made of solid or stranded copper or
aluminium strip conductors.

11. Conservator and Explosion Vent
• Conservator is used to provide adequate
space for the expansion of oil when
transformer is loaded or when ambient
temperature changes.
• Explosion Vent is used to discharge excess
pressure developed inside the transformer
during loading, to the atmosphere.

12. Breather
• It sucks the moisture from the air which is taken by transformer so that
dry air is taken by transformer.

13. Bushings
• Transformers are connected to high voltage
lines.
• Extreme care should be taken to prevent the
conductors touching the transformer tank.
• So the connections in and out of the
transformer are made by the use of bushings.
• Bushings are normally porcelain insulators.

14. Buchholz Relay
• It is a safety device connected between main tank and
conservator tank.
• In case of slow developing faults, it sounds an alarm to
alert the operator.
• If serious fault occur in the transformer, it disconnects the
transformer to protect it.

15. Methods of Cooling of Transformers
• Air natural
• Air Blast
• Oil natural
• Oil blast
• Forced circulation of oil
• Oil and water cooled
• Forced oil and water cooled

16. Losses in a Transformer
• The power losses in a transformer are of two types, namely;
 Core or Iron losses
 Copper losses

17. Core or Iron losses (Pi)
• This loss consists of hysteresis and eddy current loss and occur in
the transformer core due to the alternating flux.
• These losses can be determined by open-circuit test.
Hysteresis loss, Ph = Kh Bmax
1.6 f v watts
Eddy current loss, Pe = Ke Bmax
2 f2 t2v watts
• Both the above losses depend on Bm and frequency which are
constant.
• Hence, core or iron losses are practically the same at all loads.

18. Copper losses (PC)
• These losses occur in both the primary and secondary windings due to
their ohmic resistance.
• These losses can be determined by short-circuit test.
𝑃𝐶 = 𝐼1
2
𝑅1 + 𝐼2
2
𝑅2 = 𝐼1
2
𝑅01 = 𝐼2
2
𝑅02
• Copper losses vary as the square of load current.
• Copper losses account for about 90% of the total losses.

19. Summary
Core loss
It is the Constant loss
Does not change even as the
load current changes
Proportional to supply voltage
and frequency
Copper loss or I2R loss
It is a variable loss
Also called as I2R loss
Proportional to square of the load
current
Occurs in the winding resistances
It is dissipated as heat

20. Impedance Ratio {K}

21. • When transferring resistance or reactance from primary to secondary,
multiply it by K2.
• When transferring resistance or reactance from secondary to primary,
divide it by K2.
Shifting Impedances

22. Simplified Equivalent Circuit of a Transformer

23. Equivalent Circuit of a Transformer

24. Equivalent Circuit Referred to Primary Side

25. Equivalent Circuit Referred to Primary Side

26. Equivalent Circuit Referred to Secondary Side

27. Equivalent Circuit Referred to Secondary Side

28. Testing of Transformers
• The circuit constants, efficiency and voltage regulation of a
transformer can be determined by two simple tests.
(i) Open-circuit test
(ii) Short-circuit lest

29. Open Circuit Test
 This test is conducted to determine R0 & X0
 Rated voltage is applied on LV side & HV side is kept open.
 At no load, current taken by the transformer is 3-5% of full load
current. So I2R loss is negligible.
 Therefore power consumed by the transformer on no load is
considered as core loss.

30. Open Circuit Test
Data observed from the test
 Supply voltage = V0 volts
 No load current = I0 amps
 Iron losses = W0 watts
W0 = V0I0 CosФ0
CosФ0 = W0/(V0I0)
IW = I0 CosФ0
Im = I0 SinФ0
R0 = V1/IW
X0 = V1/Im

31. Short Circuit Test
 This test is conducted to determine R02 & X02
 LV side of the Tfr is short circuited & the test is conducted on HV side.
 A low voltage is applied on the HV side to circulate the rated current
on both the windings.
 Power drawn during this test is considered as copper loss.

32. Short Circuit Test
Data observed from the test
 Applied voltage = VSC volts
 Short circuit current = ISC amps
 Copper losses = WSC watts
WSC = ISC
2R02
R02= WSC/ISC
2
Z02=VSC/ISC
X02=[Z02
2-R02
2]1/2

33. Efficiency
• F.L. Iron loss = Pi ...from open-circuit test
• F.L. Cu loss = PC ...from short-circuit test
• Total losses = Pi + PC
• Full-load efficiency of the transformer at any p.f.
F. L. efficiency, ηfl =
Full load VA × P. F
Full load VA × P. F + Pi + PC

34. Efficiency
• At any load (X times full-load), the total losses will be
𝑃𝑇 = 𝑃𝑖 + 𝑋2
𝑃𝐶
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑡 𝑋 𝑙𝑜𝑎𝑑, 𝜂𝑋 =
(𝑋 × 𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑉𝐴 × 𝑃. 𝐹)
𝑋 × 𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑉𝐴 × 𝑃. 𝐹 + 𝑃𝑖 + 𝑋2𝑃𝐶
• Note that iron loss remains the same at all loads.

35. Condition for Maximum Efficiency
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑉2𝐼2 𝑐𝑜𝑠 𝛷2
If R02 is the total resistance of the transformer referred to secondary, then,
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠, 𝑃𝐶 = 𝐼2
2
. 𝑅02
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 𝑃𝑖 + 𝑃𝐶
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 =
𝑉2𝐼2 𝑐𝑜𝑠 𝛷2
𝑉2𝐼2 𝑐𝑜𝑠 𝛷2 + 𝑃𝑖 + 𝐼2
2
. 𝑅02
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 =
𝑉2 𝑐𝑜𝑠 𝛷2
𝑉2 𝑐𝑜𝑠 𝛷2 +
𝑃𝑖
𝐼2
+ 𝐼2 . 𝑅02

36. Condition for Maximum Efficiency
𝑑
𝑑𝐼2
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = 0
𝑑
𝑑𝐼2
V2 cos Φ2 +
Pi
I2
+ I2 . R02 = 0
0 −
𝑃𝑖
𝐼2
2 + R02 = 0
𝑃𝑖 = 𝐼2
2
R02
• i.e, Iron loss = Copper loss

37. Condition for Maximum Efficiency
• Hence efficiency of a transformer will be maximum when copper
losses are equal to iron losses.
• From above equation, the load current I2 corresponding to maximum
efficiency is given by,
𝐼2 =
𝑃𝑖
R02

38. Output kVA Corresponding to Maximum Efficiency
• PC = Copper losses at full-load kVA
• Pi = Iron losses
• X = Fraction of full-load kVA at which efficiency is maximum
• Total Cu losses = X2 PC
• For maximum efficiency, Pi = X2 PC
∴ X =
Pi
PC

39. Output kVA Corresponding to Maximum Efficiency
Output kVA corresponding to max. efficiency = 𝑋 × Full load kVA
Output kVA corresponding to max. efficiency = Full load kVA ×
Pi
PC
• It may be noted that the value of kVA, at which the efficiency is
maximum, is independent of p.f. of the load.

40. Voltage Regulation
• Change in secondary terminal voltage, when full load at a given power
factor and at rated voltage is thrown off, is expressed as a percentage
of rated terminal voltage.
• The change in secondary terminal voltage from no load to full load
expressed as a percentage of full load voltage.
% 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝑹𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 =
𝑽𝟐 𝑵.𝑳 − 𝑽𝟐 𝑭.𝑳
𝑽𝟐 𝑭.𝑳
× 𝟏𝟎𝟎%

41. Voltage Regulation at Different Power factors
• Voltage regulation for lagging p.f at load X,
% 𝒗𝒐𝒍𝒕𝒂𝒈𝒆 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 =
𝑿. 𝑰𝟐 𝑹𝟎𝟐 𝐜𝐨𝐬 𝝓𝟐 + 𝑿𝟎𝟐 𝐬𝐢𝐧 𝝓𝟐
𝑽𝟐
× 𝟏𝟎𝟎%
• Voltage regulation at leading p.f at load X,
% 𝒗𝒐𝒍𝒕𝒂𝒈𝒆 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 =
𝑿. 𝑰𝟐 𝑹𝟎𝟐 𝐜𝐨𝐬 𝝓𝟐 − 𝑿𝟎𝟐 𝐬𝐢𝐧 𝝓𝟐
𝑽𝟐
× 𝟏𝟎𝟎%
• Voltage regulation at Unity p.f at load X,
% 𝒗𝒐𝒍𝒕𝒂𝒈𝒆 𝒓𝒆𝒈𝒖𝒍𝒂𝒕𝒊𝒐𝒏 =
𝑿. 𝑰𝟐. 𝑹𝟎𝟐
𝑽𝟐
× 𝟏𝟎𝟎%

42. In a 25 kVA, 2000 V / 200 V transformer, the constant and
variable losses are 350 W and 400 W respectively. Calculate the
efficiency on unity power factor at full load and half the full
load.

43. Calculate the efficiency at half and full load of a 100 kVA
transformer for unity and 0.8 p.f. The copper loss is 1000 W at
full load and iron loss is 1000 W.

44. Obtain the equivalent circuit of a 200 / 400 V, 50 Hz, 1 phase
transformer from the following test data:
O.C. test: 200 V, 0.7 A, 70 W – on L.V side.
S.C. test: 15 V, 10 A, 85 W – on H.V side.
Calculate the secondary voltage when delivering 5 kW at 0.8 p.f
lagging, the primary voltage being 200 V.

45. From OC Test
P0 = V0I0 cos ϕ0
cos ϕ0 =
P0
V0I0
=
70
200 × 0.7
cos ϕ0 = 0.5
sin ϕ0 = 0.866
Iw = I0 cos ϕ0 = 0.7 × 0.5 = 0.35𝐴
Im = I0 sin ϕ0 = 0.7 × 0.866 = 0.606𝐴
R0 =
V0
Iw
=
70
0.35
= 200 Ω
X0 =
V0
Im
=
70
0.606
= 115.5 Ω
From SC Test
Psc = Isc
2 R02
R02 =
Psc
Isc
2 =
85
102
= 0.85 Ω
Z02 =
Vsc
Isc
=
15
10
= 1.5 Ω
X02 = Z02
2
− R02
2
X02 = 1.52 − 0.852
X02 = 1.235 Ω

46. Equivalent Circuit Referred to Primary Side
𝐾 =
400
200
= 2
R01 =
0.85
22
= 0.212 Ω
X01 =
1.235
22
= 0.308 Ω

47. Equivalent Circuit Referred to Secondary Side
𝐾 =
400
200
= 2
𝑅0
′
= 200 × 22 = 800 Ω
𝑋0
′
= 115.5 × 22 = 462 Ω

48. • Load kVA corresponding to 5 kW is,
=
5000
0.8
= 6250 𝑉𝐴
• Load current I2 while delivering 6250 VA is,
=
6250
400
= 15.625 𝐴
• Total voltage drop in secondary when it carries 15.625 A is,
= 𝐼2 𝑅02 𝑐𝑜𝑠 𝜙2 + 𝑋02 𝑠𝑖𝑛 𝜙2
= 15.625 0.85 × 0.8 + 1.235 × 0.6
= 22.20 𝑉
• Hence the secondary voltage is,
𝑉2 = 400 − 22.2 = 377.8 𝑉

49. All Day Efficiency
• The ordinary or commercial efficiency of a transformer is defined as
the ratio of output power to the input power i.e.,
Commercial efficiency =
Output power
Input power
Primaries of distribution transformers are energized all the 24 hours in a
day but the secondary windings supply little or no load during the major
portion of the day.

50. All Day Efficiency
• Constant loss occurs during the whole day but copper loss occurs only
when the transformer is loaded.
• The performance of such transformers is judged on the basis of energy
consumption during the whole day (i.e., 24 hours).
• This is known as all-day or energy efficiency.

51. All Day Efficiency
• The ratio of output in kWh to the input in kWh of a transformer over
a 24-hour period is known as all-day efficiency i.e.,
𝜂𝑎𝑙𝑙−𝑑𝑎𝑦 =
𝑘𝑊ℎ 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 24 ℎ𝑜𝑢𝑟𝑠
𝑘𝑊ℎ 𝑖𝑛𝑝𝑢𝑡 𝑖𝑛 24 ℎ𝑜𝑢𝑟𝑠
• In the design of such transformers, efforts should be made to reduce
the iron losses which continuously occur during the whole day.

52. All Day Efficiency
• A 40kVA distribution transformer has iron loss of 500 W and full load
copper loss of 500 W. the transformer is supplying a lighting load. The
load cycle is as under: Full load for 4 hours, half load for 8 hours and
no load for 12 hours. Calculate the all day efficiency.

53. All Day Efficiency
• A transformer has its maximum efficiency of 0.98 at 15 kVA at UPF.
During the day it is loaded as follows:
Duration Load Power Factor
12 hours 2 kW at 0.5 p.f
6 hours 12 kW at 0.8 p.f
4 hours 18 kW at 0.9 p.f
2 hours No load
• Find the “All Day Efficiency”.

54. Sumpner’s Test

55. Polarity Test
• Similar polarity ends of two windings are those ends that acquire
positive and negative polarity of emf induced in them simultaneously.

56. Auto Transformer
• An autotransformer has a single winding on an iron core and a part of
winding is common to both the primary and secondary circuits.

57. Auto Transformer
• Primary and secondary windings are connected electrically as well as
magnetically.
• Therefore, power from the primary is transferred to the secondary
conductively as well as inductively (transformer action).
• The voltage transformation ratio K of an ideal autotransformer is,
𝐸1
𝐸2
=
𝑁1
𝑁2
=
𝑉1
𝑉2
=
𝐼2
𝐼1
= 𝐾

58. Advantages of Autotransformers
• An autotransformer requires less Cu than a two -winding transformer of
similar rating.
• Autotransformer operates at a higher efficiency than a two-winding
transformer of similar rating.
• An autotransformer has better voltage regulation than a two-winding
transformer of the same rating.
• An autotransformer has smaller size than a two-winding transformer of the
same rating.

59. Advantages of Autotransformers
• An autotransformer requires smaller exciting current than a two-
winding transformer of the same rating.
• These advantages decrease as the ratio of transformation increases. So
an autotransformer has advantages only for low values of
transformation ratio.

60. Disadvantages of Autotransformers
• There is a direct connection between the primary and secondary.
Therefore, the output is no longer isolated from the input.
• It is not safe for stepping down a high voltage to a low voltage.
• The short - circuit current is much larger than for the two-winding
transformer of the same rating.
• This reduces the effective resistance and reactance.

61. Applications of Autotransformers
• Autotransformers are used to compensate for voltage drops in
transmission and distribution lines. When used for this purpose, they
are known as booster transformers.
• Autotransformers are used for reducing the voltage supplied to a.c.
motors during the starting period.
• Autotransformers are used for continuous variable supply.

62. Three Phase Transformers
• Large scale generation of electric power is usually 3 phase at 13.2 kV
or higher.
• But transmission voltage is 110 kV, 132 kV and 400 kV.
• Generated voltage needs to be increased.
• Hence 3 phase transformers are used.
• 3 single phase transformers can be used to construct a 3 phase
transformer.
• But it occupies more space and 15% more costlier than using a single
unit.

63. Three Phase Transformers

64. Three Phase Transformer Connections
• Star / Star (Y – Y)
• Delta / Delta (Δ – Δ)
• Star / Delta(Y – Δ)
• Delta / Star (V – Y)