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TRANSFORMER RATINGS
Transformers carry ratings related to the primary and
secondary windings. The ratings refer to the power in kVA and
primary/secondary voltages.
A rating of 10 kVA, 1100/110 V means that the primary is
rated for 1100 V while the secondary is rated for 110 V (a
=10). The kVA rating gives the power information.
With a kVA rating of 10 kVA and a voltage rating of 1100 V,
the rated current for the primary is 10,000/1100 = 9.09 A while
the secondary rated current is 10,000/110 = 90.9 A.
NON-IDEAL TRANSFORMER EQUIVALENT
CIRCUITS
The non-ideal transformer equivalent circuit shown in Fig. 1
accounts for all of the loss terms that are neglected in the ideal
transformer model. The individual loss terms in the equivalent
circuit are:
Rw1, Rw2 - primary and secondary winding resistances.
(losses in the windings due to the resistance of the
wires)
Xl1, Xl2 - primary and secondary leakage reactances. (losses
due to flux leakage out of the transformer core)
Rc1 - core resistance. (core losses due to hysteresis loss and
eddy current loss)
Xm1 - magnetizing reactance. (magnetizing current
necessary to establish magnetic flux)
Fig. 1. Non-ideal transformer equivalent circuit.
Using the impedance reflection technique, all the quantities on
the secondary side of the transformer can be reflected back to
the primary side of the circuit. The resulting equivalent circuit
is shown in the figure below. The primed quantities represent
those values that equal the original secondary quantity
multiplied by a (voltages), divided by a (currents) or multiplied
by a2 (impedance components).
(1)
(2)
(3)
(4)
(5)
The primed quantities represent those values that equal the
original primary quantity divided by a (voltages), multiplied by
a (currents) or divided by a2 (impedance components).
(6)
(7)
(8)
(9)
(10)
APPROXIMATE TRANSFORMER EQUIVALENT
CIRCUITS
Given that the voltage drops across the primary winding
resistance and the primary leakage reactance are typically quite
small, the shunt branch of the core loss resistance and the
magnetizing reactance (excitation branch) can be shifted to the
primary input terminal. The primary voltage is then applied
directly across this shunt impedance and allows for the winding
resistances and leakage reactances to be combined as shown in
the figure below.
Simplifying the impedances in the series arms
We obtain the equivalent primary winding resistance and
primary leakage reactance given by
(11)
(12)
A further approximation to the transformer equivalent circuit
can be made by eliminating the excitation branch. This
approximation removes the core losses and the magnetizing
current from the transformer model. The resulting equivalent
circuit is shown below.
Note that this equivalent circuit is referred to the primary side
of the transformer (V1 and V’2). This circuit can easily be
modified so that it is referred to the secondary side of the
transformer (V’1 and V2) as shown in the figure below.
We obtain the equivalent secondary winding resistance and
secondary leakage reactance given by
(13)
(14)
TRANSFORMER VOLTAGE REGULATION
For a given input (primary) voltage, the output (secondary)
voltage of an ideal transformer is independent of the load
attached to the secondary. As seen in the transformer
equivalent circuit, the output voltage of a realistic transformer
depends on the load current. Assuming that the current through
the excitation branch of the transformer equivalent circuit is
small in comparison to the current that flows through the
winding loss and leakage reactance components, the
transformer approximate equivalent circuit referred to the
primary is shown below. Note that the load on the secondary
(Z2) and the resulting load current (I2) have been reflected to
the primary (Z’2, I’2).
The percentage voltage regulation (VR) is defined as the
percentage change in the magnitude of the secondary voltage
as the load current changes from the no-load to the loaded
condition.
(15)
The transformer equivalent circuit above gives only the
reflected secondary voltage. The actual loaded and no-load
secondary voltages are equal to the loaded and no-loaded
reflected secondary values divided by the turns ratio that are
given by
(16)
Thus, the percentage voltage regulation may be written in
terms of the reflected secondary voltages.
(17)
According to the approximate transformer equivalent circuit,
the reflected secondary voltage under no-load conditions is
equal to the primary voltage, so that
(18)
The secondary voltage for the loaded condition is taken as the
rated voltage.
(19)
Inserting the previous two equations into the percentage
voltage regulation equation gives
(20)
Note that this equation is defined in terms of the voltages given
in the transformer approximate equivalent circuit. Also note
that the rated secondary voltage reflected to the primary is the
rated primary voltage.
(21)
To determine the percentage voltage regulation, we may use
the reflected secondary voltage as the voltage reference,
(22)
and determine the corresponding value of V1 from the
approximate equivalent circuit.
The voltages V1 and V’2 in the approximate equivalent circuit
are related by
(23)
where
(24)
The reflected secondary current can be written as
(25)
The expression for V1 becomes
(26)
We can draw the phasor diagram relating the voltages V1 and
V’2 to determine how the phase angles of the load and the
transformer impedance affect the percentage voltage
regulation. Note that the percentage voltage regulation can be
positive or negative and the sign of VR is affected by the phase
angle in the expression above. Thus, the power factor of the
load will affect the voltage regulation of the transformer.
The percentage voltage regulation is positive if |V1|> V2,rated and
negative if |V1| < V2,rated. Note that with the limits on the angles
of
(27)
(28)
The worst case scenario for the percentage voltage regulation
occurs when
(29)
or when the load has a lagging power factor with the power
factor angle equal to the transformer impedance angle of Zeq1.
LOSSES IN THE TRANSFORMER
1. Copper Loss
The copper loss is due to the power wasted in the form of I2R
loss due to the resistances of the primary and secondary
windings. The copper loss depends on the magnitude of the
currents flowing through the windings.
The copper loss in a transformer may be written in terms of
both the primary and secondary currents as
(30)
or in terms of only one of these currents based on the
relationship I2 = aI1. In terms of I1, we have
(31)
and then in terms of I2
(32)
The copper losses are denoted as PCu. If the current through the
windings is full load current, we get copper losses at full load.
If the load on transformer is half then we get copper losses at
half load which are less than full load copper losses. Thus
copper losses are called variable losses. The transformer VA
rating is V1I1 or V2I2. We can say that the copper losses are
proportional to the square of the current
(33)
Copper loss or I2R loss in the windings is a variables loss as
this loss depends upon the value of load current, I. It may be
noted that sometimes the windings of transformers are made of
aluminum wires to reduce the cost of production. The copper
loss for a given fraction of the load can be defined as
Let x - the I2R loss full-load in Watts.
At half load, its load value will be x/4 Watts.
At one third of the full load, its load value will be x/9 Watts.
Therefore, copper loss is proportional to the square of the given fraction of the load.
2. Core Loss
Core loss or iron loss is called constant loss as this loss does
not depend upon the amount of electrical load connected to the
transformer. This means that the core loss remains the same as
on no-load and on full-load or on any other load.
Core loss is composed of hysteresis loss and eddy current loss.
2.1. Hysteresis Loss
The power loss is dissipated as heat from the core.
It is due to the alternate magnetization of the atoms forming
domains in the magnetic material of the core. Each domain
behaves as a very small magnet oriented in different directions
as shown in Fig. 2(a).
Due to the application of magnetizing force in Fig. 2(b), these
tiny magnets orient themselves in the direction of
magnetization. If the magnetizing force is alternating, the small
magnets will orient themselves alternately in opposite
directions.
Fig. 2. (a) Orientation of grains in different directions in an
unmagnetized magnetic material and (b) Orientation of grains an
application of magnetic force.
2.2.Eddy Current Loss
The two thick laminated sheets forming a part of the
transformer-core limb is shown in Fig. 3. However, the
laminated sheets are actually very thin and insulated from each
other.
In these sheets, EMF is induced due to the presence of an
alternating flux in the core. The EMF induced in laminated
sheet will produce circulating currents as shown. These are
called eddy current and they produces power loss in the
resistance of all iron path.
To reduce the eddy current loss in the core, the core is made up
of thin laminated sheets (instead of a solid mass) so that the
resistance to the eddy-current path is increased and hence the
value of the eddy current is reduced.
Fig. 3. Eddy currents flowing in laminated sheets of a transformer
core when current flows through the winding.
DETERMINATION OF EQUIVALENT CIRCUIT
PARAMETERS
The equivalent circuit model for the non-ideal transformer is
shown in Fig. 4.
An ideal transformer with resistors and inductors in parallel
and series replaces the non-ideal transformer. This model is
called the high side equivalent circuit model because all
parameters have been moved to the primary side of the ideal
transformer.
Series Parameters
The series resistance, Req, is the resistance of the copper
winding.
The series inductance, Xeq, accounts for the flux leakage.
That is, a small amount of flux travels through the air
outside the magnetic core path.
Shunt parameters
The shunt resistance, Rc, represents the core loss of the
magnetic core material due to hysteresis.
The shunt inductance, Xm, called the magnetizing
inductance, accounts for the finite permeability of the
magnetic core.
Fig. 4. The equivalent model for the non-ideal transformer.
The parallel parameter values are found with no load connected
to the secondary (open circuit) and the series parameter values
are found with the secondary terminals shorted (short circuit).
It is possible to make the tests on either the primary or the
secondary.
For the open (no-load) circuit test shown in Fig. 6, the series
parameters are neglected for convenience. This is reasonable
since the voltage drops are across Req and Xeq are normally
small.
Measurements of current, voltage and real power are made on
the input winding (most often the LV winding, for
convenience).
Fig. 6. Equivalent open (no-load) circuit test.
The purpose of short circuit test is to determine the series
branch parameters of the equivalent circuit as shown in Fig. 7.
The excitation current which is only 1% or less even at rated
voltage becomes negligibly small during this test and it is
neglected. The shunt branch is thus assumed to be absent.
Measurements of current, voltage and real power are made on
the input winding (most often the HV winding, for
convenience, since a relatively low voltage is necessary to
obtain rated current under short-circuit conditions).
Fig. 7. Equivalent short circuit test.
EFFICIENCY OF A TRANSFORMER
Due to the losses in a transformer, the output power of a
transformer is less than the input power supplied. Therefore
Pout = Pin – Total losses (34)
Pin = Pout + Total losses (35)
The efficiency of the device is defines as the ratio of the power
output to power input. So for a transformer the efficiency can
be expressed as
(36)
Now, we let where is the load power
factor.
The transformer supplies full load of current I2 and with
terminal voltage V2. Then the copper losses PCu on full load is
given by
(37)
Then the efficiency is given by
(38)
But V2I2 = VA rating of a transformer
(39)
This is full load percentage with I2 = full-load secondary
current
But if the transformer is subjected to fractional load then using
the appropriate values of various quantities, the efficiency can
be obtained.
Let n = fraction by which load is less than full load given by
(40)
For example, if transformer is subjected to half load then
When load changes, the load current changes by same
proportion
(41)
Similarly the output also reduces by the same fraction. The
fraction of VA rating is available at the output.
Similarly as copper losses are proportional to square of current
then,
(42)
So the copper losses get reduced by n2.
In general for fractional load, the efficiency is given by
(43)
where n is the fraction by which load is less than full load.
CONDITION FOR MAXIMUM EFFICIENCY
When a transformer works on a constant input voltage and
frequency then efficiency varies with the load. As load
increases, the efficiency increases. At a certain load current, it
achieves a maximum value. If the transformer is loaded further
efficiency starts decreasing. The graph of efficiency against
load current I2 as shown in the Fig. 8.
The load current at which the efficiency attains maximum
value is denoted as I2,ηmax and maximum efficieny is denoted as
ηmax.
Fig. 8. The graph of efficiency against load I2.
The efficiency is a function of load i.e. load current I2
assuming cos φ2 constant. The secondary terminal voltage V2 is
also assumed constant for maximum efficiency
Given the efficiency
Perform the differentiation of η with respect to I2 then equate to
0, we have
Canceling out from both the terms we get,
So to achieve maximum efficiency is that
Core Losses = Copper Losses
(44)
Solving for the load current at maximum efficiency.
For ηmax, but
but
(45)
Let (I2) FL be the full load current and divide Eq. (44) by (I2)
FL
(46)
For constant V2, the kVA supplied is function of load current
(47)
Substituting condition for ηmax in the expression of efficiency,
we can write expression for ηmax as
as (48)
ALL-DAY EFFICIENCY
For a transformer, the efficiency is defined as the ratio of
output power to input power. This is its power efficiency. But
power efficiency is not the true measure of the performance of
some special types of transformers such as distribution
transformers.
For instance, distribution transformers used for supplying
lighting loads have their primaries energized all the 24 hours in
a day but the secondaries supply little or no load during the
major portion of the day. It means that a constant loss (i.e., iron
loss) occurs during the whole day but copper loss occurs only
when the transformer is loaded and would depend upon the
magnitude of load.
Consequently, the copper loss varies considerably during the
day and the commercial efficiency of such transformers will
vary from a low value (or even zero) to a high value when the
load is high.
The performance of such transformers is judged on the basis of
energy consumption in kWh during the whole day (i.e., 24
hours). This is known as all-day or energy efficiency.
The all-day efficiency is defined as
(49)
SAMPLE PROBLEMS
1. The approximate equivalent circuit parameters for a single
phase 10 kVA, 2200/220 V, 60 Hz transformer are
required.
No-load and short circuit tests are performed on the
transformer with the following results:
For the open (no-load) circuit test:
HV winding open
The voltage is at full-load rated voltage on the LV side:
220 V
Current on the LV side: 2.5 A
Power on the LV side: 100 W (This is what you call
core loss)
For the short circuit test:
LV winding shorted
The current is at full-load rated current on the HV side:
10,000/2,200 = 4.55 A
Voltage on the HV side: 150 V
Power on the HV side: 215 W
(a) Determine the approximate equivalent circuit parameters
from the test data. Draw the equivalent circuit for this
transformer referred to the LV side.
(b) Draw the equivalent circuit for this transformer referred to
the HV side.
(c) From the no-load test results, express the excitation current
as a percentage of the rated current in the LV winding.
(d) Determine the power factor for the no-load and short-circuit
tests.
(e) Determine the percentage voltage regulation for a load
drawing 75% of rated current at a power factor of 0.6
lagging.
(f) Determine the percentage voltage regulation for a load
drawing 75% of rated current at a power factor of 0.6
leading.
(g) Determine the transformer efficiency at 75% of rated output
power with a power factor of 0.6 lagging.
(h) Determine the output power at maximum efficiency and the
value of maximum efficiency at unity power factor.
(i) Determine the percentage of full load power where
maximum efficiency occurs.
2. A 100 kVA distribution transformer supplying light and fan
loads has full-load copper loss iron loss of 1.5 and 2 kW
respectively. During 24 h in a day the transformer is loaded
as follows:
6 AM to 10 AM (4 h) Half load
10 AM to 6 PM (8 h) One-fourth load
6 PM to 10 PM (4 h) Full-load
10 PM to 6 AM (8 h) Negligible load
Calculate the all-day efficiency of the transformer.
3. A 400 kVA, distribution transformer has full load iron loss
of 2.5 kW and copper loss of 3.5 kW. During a day, its load
cycle for 24 hours is
6 hours 300 kW at 0.8 lagging pf
10 hours 200 kW at 0.7 lagging pf
4 hours 100 kW at 0.9 lagging pf
4 hours no load
Determine its all day efficiency.