1. **INTRODUCTION**
The invention of the power transformer towards the end of the nineteenth
century made possible the development of the modern constant voltage
AC supply system, with power stations often located many miles from centers
of electrical load. Before that, in the early days of public electricity
supplies, these were DC systems with the source of generation, of necessity,
close to the point of loading.
Pioneers of the electricity supply industry were quick to recognize
the benefits of a device which could take the high current relatively
low voltage out put of an electrical generator and transform this to
a voltage level which would enable it to be transmitted in a cable of
practical dimensions to consumers who, at that time, might be a mile
or more away and could do this with an efficiency which, by the standards
of the time, was nothing less than phenomenal.
Today’s transmission and distribution systems are, of course, vastly
more extensive and greatly dependent on transformers which themselves
are very much more efficient than those of a century ago; from the enormous
generator transformers, stepping
up the output of up to 19,000 A at 23.5 kV, of a large generating unit
in the UK, to 400 kV, thereby reducing the current to a more manageable
1200 A or so, to the thou sands of small distribution units which operate
almost continuously day in day out, with little or no attention, to provide
supplies to industrial and domestic consumers.
The main purpose of this guide is to examine the current state of transformer
technology, but in the rapidly shrinking
and ever more competitive world of technology it is not possible to retain
one's place in it without a knowledge of all that is going on the other
side of the globe, so the viewpoint will, hopefully, not be an entirely
parochial one.
For a reasonable understanding of the subject it is necessary to make
a brief review of transformer theory together with the basic formulae
and simple phasor diagrams.
2. **THE IDEAL TRANSFORMER: VOLTAGE RATIO**
A power transformer normally consists of a pair of windings, primary
and secondary, linked by a magnetic circuit or core. When an alternating
voltage is applied to one of these windings, generally by definition
the primary, a current will flow which sets up an alternating m.m.f.
and hence an alternating flux in the core. This alternating flux in linking
both windings induces an e.m.f. in each of them. In the primary winding
this is the 'back-e.m.f' and, if the transformer were perfect, it would
oppose the primary applied voltage to the extent that no current would
flow. In reality, the current which flows is the transformer magnetizing
current. In the secondary winding the induced e.m.f. is the secondary
open-circuit voltage. If a load is connected to the secondary winding
which permits the flow of secondary current, then this current creates
a demagnetizing m.m.f. thus destroying the balance between primary applied
voltage and back-e.m.f. To restore the balance an increased primary current
must be drawn from the supply to provide an exactly equivalent m.m.f.
so that equilibrium is once more established when this additional primary
current creates ampere-turns balance with those of the secondary.
Since there is no difference between the voltage induced in a single
turn whether it is part of either the primary or the secondary winding,
then the total voltage induced in each of the windings by the common
flux must be proportional to the number of turns. Thus the well-known
relationship is established that:
E1/E2 = N1/N2 (eqn. 1)
and, in view of the need for ampere-turns balance:
I1/N2 = I2/N2 (eqn. 2)
where E, I and N are the induced voltages, the currents and number of
turns respectively in the windings identified by the appropriate subscripts.
Hence, the voltage is transformed in proportion to the number of turns
in the respective windings and the currents are in inverse proportion
(and the relationship holds true for both instantaneous and r.m.s. quantities).
The relationship between the induced voltage and the flux is given by
reference to Faraday's law which states that its magnitude is proportional
to the rate of change of flux linkage and Lenz's law which states that
its polarity such as to oppose that flux linkage change if current were
allowed to flow. This is normally expressed in the form:
e =-N(dΦ/dt)
but, for the practical transformer, it can be shown that the voltage
induced per turn is
E/N = KΦm *f* (eqn. 3)
...where K is a constant, Fm is the maximum value of total flux in Webers
linking that turn and f is the supply frequency in Hertz.
The above expression holds good for the voltage induced in either primary
or secondary windings, and it is only a matter of inserting the correct
value of N for the winding under consideration. Figure 1 shows the
simple phasor diagram corresponding to a transformer on no-load (neglecting
for the moment the fact that the transformer has reactance) and the symbols
have the significance shown on the diagram. Usually in the practical
design of transformer, the small drop in voltage due to the flow of the
no-load current in the primary winding is neglected.
FIG. 1 Phasor diagram for a single-phase transformer on open circuit.
Assumed turns ratio 1:1
If the voltage is sinusoidal, which, of course, is always assumed, K
is 4.44 and Eq. (eqn. 3) becomes
E = 4.44f phi N
For design calculations the designer is more interested in volts per
turn and flux density in the core rather than total flux, so the expression
can be rewritten in terms of these quantities thus:
E/N = 4.44BmAf x 10^-6 (eqn. 4)
where E/N= volts per turn, which is the same in both windings
Bm = maximum value of flux density in the core, Tesla
A = net cross-sectional area of the core, mm^2
f = frequency of supply, Hz.
For practical designs Bm will be set by the core material which the
designer selects and the operating conditions for the transformer, A
will be selected from a range of cross-sections relating to the standard
range of core sizes produced by the manufacturer, whilst f is dictated
by the customer's system, so that the volts per turn are simply derived.
It is then an easy matter to determine the number of turns in each winding
from the specified voltage of the winding.
3. **LEAKAGE REACTANCE: TRANSFORMER IMPEDANCE**
Mention has already been made in the introduction of the fact that the
trans formation between primary and secondary is not perfect. Firstly,
not all of the flux produced by the primary winding links the secondary
so the transformer can be said to possess leakage reactance. Early transformer
designers saw leakage reactance as a shortcoming of their transformers
to be minimized to as great an extent as possible subject to the normal
economic constraints. With the growth in size and complexity of power
stations and transmission and distribution systems, leakage reactance
-- or in practical terms since transformer windings also have resistance
- impedance, gradually came to be recognized as a valuable aid in the
limitation of fault currents. The normal method of expressing transformer
impedance is as a percentage voltage drop in the transformer at full-load
current and this reflects the way in which it is seen by system designers.
For example, an impedance of 10 percent means that the voltage drop
at full-load current is 10 percent of the open-circuit voltage, or,
alternatively, neglecting any other impedance in the system, at 10 times
full load current, the voltage drop in the transformer is equal to the
total system voltage. Expressed in symbols this is:
VZ=%Z = I_FL Z / E = 100
where Z is √(R^{2} + X^{2} ), R and X being the transformer
resistance and leak age reactance respectively and I_{FL} and E are the
full-load current and open circuit voltage of either primary or secondary
windings. Of course, R and X may themselves be expressed as percentage
voltage drops, as explained below.
The 'natural' value for percentage impedance tends to increase as the
rating of the transformer increases with a typical value for a medium
sized power transformer being about 9 or 10 percent. Occasionally some
transformers are deliberately designed to have impedances as high as
22.5 percent. More will be said about transformer impedance in the following
section.
4. **LOSSES IN CORE AND WINDINGS**
The transformer also experiences losses. The magnetizing current is
required to take the core through the alternating cycles of flux at a
rate determined by system frequency. In doing so energy is dissipated.
This is known variously as the core loss, no-load loss or iron loss.
The core loss is present whenever the transformer is energized. On open
circuit the transformer acts as a single winding of high self-inductance,
and the open-circuit power factor averages about 0.15 lagging. The flow
of load current in the secondary of the trans former and the m.m.f. which
this produces is balanced by an equivalent primary load current and its
m.m.f., which explains why the iron loss is independent of the load.
The flow of a current in any electrical system, however, also generates
loss dependent upon the magnitude of that current and the resistance
of the system.
Transformer windings are no exception and these give rise to the load
loss or copper loss of the transformer. Load loss is present only when
the transformer is loaded, since the magnitude of the no-load current
is so small as to pro duce negligible resistive loss in the windings.
Load loss is proportional to the square of the load current.
**Reactive and resistive voltage drops and phasor diagrams **
The total current in the primary circuit is the phasor sum of the primary
load current and the no-load current. Ignoring for the moment the question
of resistance and leakage reactance voltage drops, the condition for
a transformer sup plying a non-inductive load is shown in phasor form
in Fig. 2. Considering now the voltage drops due to resistance and
leakage reactance of the trans former windings it should first be pointed
out that, however the individual voltage drops are allocated, the sum
total effect is apparent at the secondary terminals. The resistance drops
in the primary and secondary windings are easily separated and determinable
for the respective windings. The reactive volt age drop, which is due
to the total flux leakage between the two windings, is strictly not separable
into two components, as the line of demarcation between the primary and
secondary leakage fluxes cannot be defined. It has therefore become a
convention to allocate half the leakage flux to each winding, and similarly
to dispose of the reactive voltage drops. Figure 3 shows the phasor
relationship in a single-phase transformer supplying an inductive load
having a lagging power factor of 0.80, the resistance and leakage reactance
drops being allocated to their respective windings. In fact the sum total
effect is a reduction in the secondary terminal voltage. The resistance
and reactance voltage drops allocated to the primary winding appear on
the diagram as additions to the e.m.f. induced in the primary windings.
FIG. 2 Phasor diagram for a single-phase transformer supplying a
unity power factor load. Assumed turns ratio 1:1
FIG. 4 shows phasor conditions identical to those in Fig. 3, except
that the resistance and reactance drops are all shown as occurring on
the secondary side.
Of course, the drops due to primary resistance and leakage reactance
are converted to terms of the secondary voltage, that is, the primary
voltage drops are divided by the ratio of transformation n, in the case
of both step-up and step-down transformers. In other words the percentage
voltage drops considered as occurring in either winding remain the same.
To transfer primary resistance values R1 or leakage reactance values
X1 to the secondary side, R1 and X1 are divided by the square of the
ratio of trans formation n in the case of both step-up and step-down
transformers.
FIG. 3 Phasor diagram for a single-phase transformer supplying an
inductive load of lagging power factor cos Φ2. Assumed turns ratio 1:1.
Voltage drops divided between primary and secondary sides
The transference of impedance from one side to another is made as follows.
Let Zs =total impedance of the secondary circuit including leakage
and load characteristics
Z's = equivalent value of Zs when referred
to the primary winding.
Then
(eqn. 5)
FIG. 4 Phasor diagram for a single-phase transformer supplying an
inductive load of lagging power factor cos Φ2. Assumed turns ratio 1:1.
Voltage drops transferred to secondary side
Also,
(eqn. 6)
Comparing Eqs (eqn. 5) and (eqn. 6) it will be seen that Z's= Zs(N1/N2)^{2}.
The equivalent impedance is thus obtained by multiplying the actual
impedance of the secondary winding by the square of the ratio of transformation
n, that is, (N1/N2) ^{2}
This, of course, holds good for secondary winding leakage reactance
and secondary winding resistance in addition to the reactance and resistance
of the external load.
FIG. 5 Phasor diagram for a single-phase transformer supplying a
capacitive load of leading power factor cos Φ2. Assumed turns ratio 1:1.
Voltage drops transferred to secondary side
FIG. 5 is included as a matter of interest to show that when the
load has a sufficient leading power factor, the secondary terminal voltage
increases instead of decreasing. This happens when a leading power factor
current passes through an inductive reactance.
Preceding diagrams have been drawn for single-phase transformers, but
they are strictly applicable to polyphase transformers also so long as
the conditions for all the phases are shown. For instance Fig. 6 shows
the complete phasor diagram for a three-phase star/star-connected transformer,
and it will be seen that this diagram is only a threefold repetition
of Fig. 4, in which primary and secondary phasors correspond exactly
to those in Fig. 4, but the three sets representing the three different
phases are spaced 120º apart.
5. **RATED QUANTITIES**
The output of a power transformer is generally expressed in megavolt-amperes
(MVA), although for distribution transformers kilovolt-amperes (kVA)
is generally more appropriate, and the fundamental expressions for determining
these, assuming sine wave functions, are as follows.
**Single-phase transformers**
Output = 4.44 f Φm NI with the multiplier 10^{-3} for kVA and 10^{-6} for MVA.
FIG. 6 Phasor diagram for a three-phase transformer supplying an
inductive load of lagging power factor cos Φ2. Assumed turns ratio 1:1.
Voltage drops transferred to secondary side. Symbols have the same significance
as in Fig. 4 with the addition of A, B and C subscripts to indicate
primary phase phasors, and a, b and c subscripts to indicate secondary
phase phasors
**Three-phase transformers**
Output = 4.44 f Φ_{m} NI x _/3 with the multiplier 10^{-3} for
kVA and 10^{-6} for MVA.
In the expression for single-phase transformers, I is the full-load
current in the transformer windings and also in the line; for three-phase
transformers, I is the full-load current in each line connected to the
transformer. That part of the expression representing the voltage refers
to the voltage between line terminals of the transformer. The constant
_3 is a multiplier for the phase voltage in the case of star-connected
windings, and for the phase current in the case of delta-connected windings,
and takes account of the angular displacement of the phases.
Alternatively expressed, the rated output is the product of the rated
secondary (no-load) voltage E2 and the rated full-load output current
I2 although these do not, in fact, occur simultaneously and, in the case
of polyphase transformers, by multiplying by the appropriate phase factor
and the appropriate constant depending on the magnitude of the units
employed. It should be noted that rated primary and secondary voltages
do occur simultaneously at no load.
**Single-phase transformers**
Output = E_{2}I_{2} with the multiplier 10^-3 for kVA
and 10^-6 for MVA.
**Three-phase transformers**
Output = E_{2}I_{2} x _/3 with the multiplier 10^-3 for kVA and 10^-6
for MVA.
The relationships between phase and line currents and voltages for star
and for delta-connected three-phase windings are as follows:
**Three-phase star-connection**
phase current = line current I = VA/(E x _/3)
phase voltage = E/ _/3
**Three-phase delta-connection**
phase current = I/ _/3 _ VA/(E x 3)
phase voltage = line voltage = E
E and I = line voltage and current respectively.
6. **REGULATION**
The regulation that occurs at the secondary terminals of a transformer
when a load is supplied consists, as previously mentioned, of voltage
drops due to the resistance of the windings and voltage drops due to
the leakage reactance between the windings. These two voltage drops are
in quadrature with one another, the resistance drop being in phase with
the load current. The percent age regulation at unity power factor load
may be calculated by means of the following expression:
copper loss x 100 /output = (percentage reactance )^{2}/ 200
This value is always positive and indicates a voltage drop with load.
The approximate percentage regulation for a current loading of a times
rated full-load current and a power factor of cos phi 2 is given by the
following expression:
(eqn. 7)
where
V_{R} = percentage resistance voltage at full load = copper
loss x 100 / rated kVA
V=percentage reactance voltage = I_{2}X"_{e}/V_{2} x100
Equation (eqn. 7) is sufficiently accurate for most practical transformers,
however for transformers having reactance values up to about 4 percent
a further simplification may be made by using the expression:
percentage regulation = a(VR cos Phi 2 + Vx sin Phi 2) (eqn. 8)
and for transformers having high reactance values, say 20 percent or
over, it is sometimes necessary to include an additional term as in the
following expression:
(eqn. 9)
At loads of low power factor the regulation becomes of serious consequence
if the reactance is at all high on account of its quadrature phase relationship.
This question is dealt with more fully in another section of this guide.
FIG. 7 Simplified equivalent circuit of leakage impedance of two-winding
transformer
Copper loss in the above expressions is measured in kilowatts. The expression
for regulation is derived for a simplified equivalent circuit as shown
in Fig. 7, that is a single leakage reactance and a single resistance
in series between the input and the output terminals. The values have
been represented in the above expressions as secondary winding quantities
but they could equally have been expressed in primary winding terms.
Since the second term is small it is often sufficiently accurate to take
the regulation as equal to the value of the first term only, particularly
for values of impedance up to about 4 percent or power factors of about
0.9 or better.
VX may be obtained theoretically by calculation or actually from the
tested impedance and losses of the transformer. It should be noted that
the percent resistance used is that value obtained from the transformer
losses, since this takes into account eddy current losses and stray losses
within the transformer. This is sometimes termed the AC resistance, as
distinct from the value which would be measured by passing direct current
through the windings and measuring the voltage drop. |