3. Working principle
•Physical basis of a transformer is mutual
induction between two circuits linked by a
common magnetic flux.
•If one coil is connected to a source of
alternating voltage, an alternating flux is set up
in the laminated core, most of which is linked
with the other coils in which it produces mutually
induced emf
4. • A transformer is a static device which is used to step up or step down
voltages at constant frequency
• It consists of two coils, that are electrically isolated but magnetically
linked
• The primary coil is connected to the power source and the secondary
coil is connected to the load
• Voltage is stepped up or stepped down proportional to turns ratio
• The turn’s ratio is the ratio between the number of turns on the
secondary (Ns)to the number of turns on the primary (Np).
5.
6. Transformer classification
Based on construction
Core type
Shell type
Berry type
Based on application
Power transformer
Distribution transformer
Based on cooling
Oil filled self cooled
Oil fill water cooled
Air blast type
7.
8. Losses in a transformer
• No load losses or core losses
• Load losses or copper losses
9. No load losses
No load losses remain the same irrespective of the load connected
to the transformer
It is the power consumed to sustain the magnetic field in the
transformer’s core
It is of two types – hysteresis loss and eddy current loss
Hysteresis loss is the energy lost by reversing the magnetizing field
in the core as the AC changes direction in every cycle.
Eddy current loss is a result of induced currents circulating in the
core
Hysteresis loss is minimized by using steel of high silicon content for
the core
Eddy current loss is minimized by using very thin laminations
polished with varnish
No load loss = IL( Va / Vr ) ²
10. Load losses
It is associated with load current flow in the transformer windings
Copper loss is power lost in the primary and secondary windings of
a transformer due to the ohmic resistance of the windings
load loss = I ² R
11. Problem
• Find the total losses taking place in a 250 KVA transformer operating
at 60% of its rated capacity whose No load loss = 500 W and
Full load loss = 4500 W
12. Problem
Transformer Rating 5 0 0 kVA, PF is 0 . 8, No
Load Loss =3.5 kW, Full Load Loss = 4 . 5 kW
No. of
hrs
Load kW PF
6 4 0 0 0 . 8
1 0 3 0 0 0 . 7 5
4 1 0 0 0 . 8
4 0 0
13. How to improve the efficiency of
transformer
By operating the transformer at optimum load
By operating the transformers in parallel
Voltage regulation of transformer
At optimum loading no load loss = Full load loss
Thus during max. efficiency no load loss = Full load loss
No Load Loss = 1600 W, Full Load Loss = 2 845 W
X = 100 √(No Load Loss/ Full Load Loss)
Load at max Eff = ( 1 6 0 0 / 2 8 4 5 ) 0 . 5
= 7 5 . 0 %
14. Parallel operation of transformer
•This is done for fluctuating loads, so that the load
can be optimized by sharing the load between the
transformers
•This way of operation provides high efficiency
•For parallel operation, both the transformers should
be technically identical and should have the same
impedance level.
15. Problem
•Power Required : 8 0 0 kVA
( 4 0 0 kVA x 2 )
•No of Transformers : 2
Rated Capacity : 1 2 5 0 kVA each
No Load Loss : 2 k W
Load Loss : 1 5 k W
16. Transformer selection
Calculate the connected load and diversity factor
Multiply Diversity Factor with connected load applicable
to particular industry and arrive at kVA rating of
transformers
Diversity factor is the ratio of sum of individual maximum
demand of various equipment to the overall maximum
demand of the plant
It will be always greater than one
17. Voltage regulation
• When the supply voltage changes, it causes tripping of voltage
sensitive load devices
• The voltage regulation in transformers is done by altering the voltage
transformation ratio with the help of tapping
• There are two methods of tap changing facility available
• Off-circuit tap changer
• On-load tap changer
18. System Distribution losses and
Optimization
• Relocating transformers and substations near to the load centres
• Re-routing the feeders and cables, where the line losses and voltage
drops are higher
• Power factor improvement by incorporating capacitors at the load
end.
• Optimum loading of the transformers in the system
• Opting for low resistance All Aluminium Alloy Conductors (AAAC)
instead of conventional Aluminium Cored Steel Reinforced (ACSR)
lines
19. Capacity
( kVA )
C o n v e n t i o n a l A m o r p h o u s
No Load
Loss
Copper
Loss at FL
No Load
Loss
Copper
Loss at FL
W
1 0 0 2 6 0 1 7 6 0 6 0 1 6 3 5
2 5 0 6 0 0 3 6 0 0 1 6 0 3 2 8 0
5 0 0 8 4 0 5 7 0 0 2 4 0 5 6 0 0
7 5 0 1 1 0 0 7 5 0 0 3 6 0 7 2 0 0
1 0 0 0 1 3 0 0 9 8 0 0 4 3 0 9 0 0 0
E N E R G Y L O S S E S C O N V E N T I O N A L
T R A N S F O R M E R V I S - A - V I S
A M O R P H O U S T R A N S F O R M E R
20. Energy conservation in transformer
% Loading of Transformer
Transformer Efficiency
No Load Losses
Operating Power Factor
Power Factor improvement in capacitor installation
On Load Tap Changer ( OLTC )
Parallel operation of Transformers
Idle transformer
Separate transformer for lighting
