Transcript SKEU2413 - Chapter 1 Part 2 transformer

Topic 1 : Magnetic Concept and
Transformer
Introduction
Two winding transformers
Construction and principles
Equivalent circuit
Determination of equivalent circuit parameters
Voltage regulation
Efficiency
Auto transformer
3 phase transformer
Introduction
Different variety of transformers
Introduction
Introduction
The word “Transformer” means an electromagnetic
device which transforms electrical power from one end to
another at different voltages and different currents
keeping frequency constant.
Unlike motor and generator it is static machine with
different turns ratio of primary and secondary windings
through which voltage/current is changed.
The transfer of energy takes place through the magnetic
field and all currents and voltages are AC.
The rating of transformer is either in kVA or MVA
because load to be connected is unknown
Introduction
 Transformers are adapted to numerous engineering
applications and may be classified in many ways:
Power level (from fraction of a volt-ampere (VA) to over a
thousand MVA),
Application (power supply, impedance matching, circuit
isolation),
Frequency range (power, audio, radio frequency (RF))
Voltage class (a few volts to about 750 kilovolts)
Cooling type (air cooled, oil filled, fan cooled, water cooled,
etc.)-ONAN, ONAF
Purpose (distribution, rectifier, arc furnace, amplifier output,
etc.).
Introduction
Examples of transformer classifications:
Power Three phase transformers (Step up) used for transmission
of power (3-phase) at a distance
Distribution transformer (Step down) used for utilization of power
3-pahse/1-phase
Instrument Transformer (VT & CT) used for
measurement/practical
Auto Transformer (Single limb, electrically connected) used for
measurement, practical, supply/utilization
Isolation Transformer (having winding ratio of 1:1) used for safety
of human and equipment for sensitive appliances or practical
purpose
Introduction
The invention of transformer caused transmission
of heavy AC electrical power possible thus plays
important role in electrical power technology
Functions of transformer:
Raise or lower voltage or current in AC circuit
Isolate circuit from each other
Enable to transmit electrical power energy over large
distances at about >1200kV
Provides electrical power according to the utilization
needs
Transformer- Introduction
Power transmission
Introduction
A typical power system consists of generation, transmission
and distribution.
Power from plant/station is generated around 11-13-20-30kV
(depending upon manufacturer and demand).
This voltage is carried out at a distance to reach for utilization
through transmission line system by step up transformer at
different voltage levels depending upon distance and losses.
Its distribution is made through step down transformer
according to the consumer demand.
Here again at this stage, a transformer play an important role
to reduce the voltage to suit the consumer need.
Introduction
Power Transmission
Introduction
Transformer is a device that makes use of the
magnetically coupled coils to transfer energy.
It is typically consists of one primary winding coil
and one or more secondary windings.
The primary winding and its circuit is called the
Primary Side of the transformer.
The secondary winding and its circuit is called the
Secondary Side of the transformer.
A magnetic circuit provides the link between
primary and secondary.
Introduction
 When an AC voltage Vp is applied to the primary winding of the
transformer, an AC current Ip will result.
 Ip sets up a time-varying magnetic flux Ф in the core.
 A voltage Vs is induced in the secondary circuit according to the
Faraday’s law.
Construction
The magnetic (iron) core is made
of thin laminated steel sheet.
to minimize the eddy current loss by
reducing thickness.
There are two common cross
section of core
square or (rectangular) for small
transformers
circular (stepped) for the large and 3
phase transformers.
Construction
Core (U/I) Type:
Constructed from a stack of U and I shaped laminations.
The primary and secondary windings are wound on two
different legs of the core.
Shell Type:
Constructed from a stack of E and I shaped laminations.
The primary and secondary windings are wound on the
same leg of the core, as concentric windings, one on top
of the other.
Construction
Construction
Construction
Construction
Ideal Transformer
Winding resistances are zero, no leakage inductance and iron
loss
Magnetization current generates a flux that induces voltage in
both windings
Current, voltages and flux in an unloaded ideal transformer
Ideal Transformer
Currents and fluxes in a loaded ideal
transformer
Ideal Transformer
Turn ratio
If the primary winding has N1 turns and secondary
winding has N2 turns, then:
The input and output complex powers are equal
Ideal Transformer
Functional description of a transformer:
a = 1
Isolation Transformer
| a | < 1
Step-Up Transformer
Voltage is increased from Primary side to secondary side
| a | > 1
Step-Down Transformer
Voltage is decreased from Primary side to secondary side
Ideal Transformer
Transformer Rating
Practical transformers are usually rated
based on:
Voltage Ratio (V1/V2) which gives us
the turns-ratio
Power Rating, small transformers are
given in Watts (real power) and Larger
ones (Power Transformers) are given in
kVA (apparent power)
Ideal Transformer
Example 1
Determine the turns-ratio of a 5 kVA
2400V/120V Power Transformer
Turns-Ratio = a = V1/V2 = 2400/120 = 20/1 = 20
This means it is a Step-Down transformer
Ideal Transformer
Example 2
A 480/2400 V (r.m.s) step-up ideal transformer
delivers 50 kW to a resistive load. Calculate:
the turns ratio,
(0.2)
the primary current,
(104.17A)
the secondary current.
(20.83A)
Ideal Transformer
Exercise 1
A 250kVA, 1100V/400v, 50Hz single-phase
transformer has 80 turns on the secondary.
Calculate:
the approximate values of the primary and secondary
currents
(227A, 625A)
the approximate number of primary turns
(220)
 the maximum value of the flux
(22.5mWb)
Ideal Transformer
Nameplate of a transformer
Ideal Transformer
Equivalent circuit of an ideal transformer
Ideal Transformer
Equivalent circuit of an ideal transformer
Transferring impedances through a transformer
Ideal Transformer
 Equivalent circuit when
secondary impedance is
transferred to primary side
and ideal transformer
eliminated.
 Equivalent circuit when
primary source is
transferred to secondary
side and ideal transformer
eliminated.
Practical Transformer
Equivalent Circuit
In a practical magnetic core having finite
permeability, a magnetizing current Im is
required to establish a flux in the core.
This effect can be represented by a magnetizing
inductance Lm.
The core loss can be represented by a
resistance Rc.
Equivalent Circuit
Rc :core loss component
Xm : magnetization component
Equivalent Circuit
Phasor diagram of an unloaded transformer
Equivalent Circuit
Winding resistance and leakage flux
The effects of winding resistance and leakage flux are
respectively accounted for by resistance R and leakage
reactance X (2πfL).
Equivalent Circuit
Rc :core loss component
Xm : magnetization component
R1 and R2 are resistance of the primary and secondary winding
X1 and X2 are reactance of the primary and secondary winding
Equivalent Circuit
Phasor diagram of a loaded transformer
(secondary)
Equivalent Circuit
Phasor diagram of a loaded transformer
(primary)
Approximate Equivalent Circuit
Since no load current is very small(3-5% of full load), the
parallel circuit of Rc and Xm can be moved close to the
supply without significant error in calculation.
Calculations becomes easier
Approximate Equivalent Circuit
Calculations will be much more easy if the primary and
secondary circuit are combined.
Transfer the secondary circuit to the primary circuit
Approximate Equivalent Circuit
Phasor diagram of a loaded transformer
(primary)
Approximate Equivalent Circuit
 For convenience, the turns is usually not shown
The resistance and reactance can be lumped together
 We can also transfer the primary circuit to the secondary circuit
Approximate Equivalent Circuit
Example 3
 A 100kVA transformer has 400 turns on the primary and 80
turns on the secondary. The primary and secondary
resistance are 0.3 ohm and 0.01 ohm respectively and the
corresponding leakage reactances are 1.1 ohm and 0.035
ohm respectively. The supply voltage is 2200V. Calculate:
the equivalent impedance referred to the primary circuit
(2.05 ohm)
 the equivalent impedance referred to the secondary
circuit
(0.082)
Transformer Test
The equivalent circuit model for the actual transformer
can be used to predict the behavior of the transformer.
The parameters Rc, Xm, R1, X1, R2, X2 and N1/N2 must
be known so that the equivalent circuit model can be
used.
These parameters can be directly and more easily
determined by performing tests:
No load test (or open circuit test)
Short circuit test
Transformer Test
No load/Open circuit test
Provides magnetizing reactance (Xm) and core loss
resistance (Rc)
Obtain components are connected in parallel
Short circuit test
Provides combined leakage reactance and winding
resistance
Obtain components are connected in series
Transformer Test – Open Circuit
 Equivalent circuit for open circuit test, measurement at the primary
side
 Simplified equivalent circuit
Transformer Test – Open Circuit
Open circuit test evaluation
.
.
Transformer Test – Short
Circuit
Short circuit test
Secondary (normally the LV winding) is shorted, that
means there is no voltage across secondary
terminals; but a large current flows in the secondary.
Test is done at reduced voltage (about 5% of rated
voltage, with full-load current in the secondary.
Hence the induced flux are also 5%) The core losses is
negligible since it is approximately proportional to the square
of the flux.
So, the ammeter reads the full-load current; the wattmeter
reads the winding losses, and the voltmeter reads the applied
primary voltage.
Transformer Test – Short
Circuit
 Equivalent circuit for short circuit test, measurement at the primary
side
 Simplified equivalent circuit
Transformer Test – Short
Circuit
 Simplified circuit for calculation of series impedance
 Primary and secondary impedances are combined
.
.
Transformer Test – Short
Circuit
Short circuit test evaluation
.
.
Transformer Test
Equivalent circuit for a real transformer resulting
from the open and short circuit tests.
Transformer Test
Example 4
Obtain the equivalent circuit of a 200/400V,
50Hz 1-phase transformer from the following test
data:O/C test : 200V, 0.7A, 70W - on L.V. side(LV data)
S/C test : 15V, 10A, 85W - on H.V. side(HV data)
(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm,
Xe=0.31 ohm)
Voltage Regulation
Most loads connected to the secondary of a transformer
are designed to operate at essentially constant voltage.
However, as the current is drawn through the
transformer, the load terminal voltage changes because
of voltage drop in the internal impedance.
To reduce the magnitude of the voltage change, the
transformer should be designed for a low value of the
internal impedance Zeq
The voltage regulation is defined as the change in
magnitude of the secondary voltage as the load current
changes from the no-load to the loaded condition.
Voltage Regulation
The voltage regulation is expressed as follows:
V2NL= secondary voltage (no-load condition)
 V2L = secondary voltage (full-load condition)
Voltage Regulation
For the equivalent circuit referred to the primary:
V1 = no-load voltage
V2’ = secondary voltage referred to the primary (full-load condition)
Voltage Regulation
Consider the equivalent circuit referred to the
secondary,
(-) : leading power factor
(+) : lagging power factor
Voltage Regulation
Consider the equivalent circuit referred to the
primary,
(-) : leading power factor
(+) : lagging power factor
Voltage Regulation
Example 5
Based on Example 3 calculate the voltage
regulation and the secondary terminal voltage for
full load having a power factor of
0.8 lagging
(0.0336pu,425V)
0.8 leading
(-0.0154pu,447V)
Approximate Equivalent Circuit
Example 3
 A 100kVA transformer has 400 turns on the primary and 80
turns on the secondary. The primary and secondary
resistance are 0.3 ohm and 0.01 ohm respectively and the
corresponding leakage reactances are 1.1 ohm and 0.035
ohm respectively. The supply voltage is 2200V. Calculate:
the equivalent impedance referred to the primary circuit
(2.05 ohm)
 the equivalent impedance referred to the secondary
circuit
(0.082)
Voltage Regulation
Efficiency
Losses in a transformer
Copper losses in primary and secondary
windings
Core losses due to hysteresis and eddy
current. It depends on maximum value of flux
density, supply frequency and core
dimension. It is assumed to be constant for all
loads
Efficiency
 Equipment is desired to operate at a high efficiency.
 Efficiency is defined as
 Since it is a static device, losses in transformers are
small
 The losses in the transformer are the core loss (Pc) and
copper loss (Pcu)
Efficiency
 The copper loss can be determined if the winding currents and
their resistances are known:
The copper loss is a function of the load current.
 The core loss depends on the peak flux density in the core,
which in turn depends on the voltage applied to the
transformer
Since a transformer remains connected to an essentially constant
voltage, the core loss is almost constant
Efficiency
If the parameters of the equivalent circuit of a
transformer are known, the efficiency of the
transformer under any operating condition may be
determined
.
Normally, load voltage remains fixed
Therefore efficiency depends on load current and load
power factor
Efficiency
Efficiency on full load
where S is the apparent power (in volt amperes)
Efficiency for any load equal to n x full load
where corresponding total loss =
Efficiency
Example 6
The following results were obtained on a 50 kVA
transformer:
open circuit test – primary voltage, 3300 V; secondary
voltage, 400 V; primary power, 430W.
Short circuit test – primary voltage, 124V;primary current,
15.3 A; primary power, 525W; secondary current, full load
value.
Calculate the efficiency at full load and half load for
0.7 power factor.
(97.3%, 96.9%)
Efficiency
Exercise 2
The primary and secondary windings of a 500kVA
transformer have resistance of 0.42 ohm and 0.0019
ohm respectively. The primary and secondary
voltages are 11000V and 400V respectively and the
core loss is 2.9kW, assuming the power factor of the
load to be 0.8. Calculate the efficiency on
Full load
half load
(98.3%, 98.1%)
Efficiency
For constant values of the terminal voltage V2
and load power factor angle θ2 , the maximum
efficiency occurs when
If this condition is applied, the condition for
maximum efficiency is
 that is, core loss = copper loss.
Efficiency
Exercise 3
Assuming the power factor of the load to be 0.8,
find the output power at which the efficiency of the
transformer of Exercise 2 is a maximum and
calculate its value
(346.4kW, 98.4%)