Transcript Transformers and Power Flow





A transformer is an electrical device that is
used to raise or lower the voltage (or current)
level.
A single-phase transformer has two electrically
isolated windings.
The winding that is connected to the electrical
power source is called the “primary” winding.
The winding that is used to draw electrical
power is called the “secondary” winding.
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Figure below shows a schematic diagram of an
ideal single-phase transformer.
EP = induced voltage on the primary
side (RMS)
ES = induced voltage on the secondary
Φm
side (RMS)
NP = Number of turns on the primary
windings
NS = Number of turns on the secondary
windings
Φm = magnetic flux created by the
source voltage E
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

When the primary winding is connected to an AC
power source it produces an instantaneous
magnetic flux “Φm(t)” linking the primary as well
as secondary windings of the transformer.
The flux linkage induces an instantaneous voltage
“Ep(t)” in the primary winding in accordance with
the Faraday’s law of electromagnetic induction.
That is,
𝑑∅𝑚 (𝑡)
𝐸𝑝 (𝑡) = 𝑁𝑝
𝑑𝑡
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Since the voltage applied at the primary is
sinusoidal, therefore, the magnetic flux Φm(t) is
also sinusoidal and mathematically it can be
expressed as
∅𝑚 𝑡 = ∅𝑚𝑎𝑥 sin 𝜔𝑡.
Therefore,
𝑑∅𝑚 (𝑡)
𝑑(sin 𝜔𝑡)
𝐸𝑝 (𝑡) = 𝑁𝑝
= 𝑁𝑝 × ∅𝑚𝑎𝑥
𝑑𝑡
𝑑𝑡
or
𝐸𝑝 (𝑡) = 𝑁𝑝 × ∅𝑚𝑎𝑥 × 𝜔 cos 𝜔𝑡.
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Substituting cos 𝜔𝑡 = sin(𝜔𝑡 + 90𝑜 ) and 𝜔 = 2𝜋𝑓 we get
𝐸𝑝 𝑡 = 2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥 sin(𝜔𝑡 + 90𝑜 ) .
Above Eq. shows that the induced voltage Ep(t) leads the
magnetic flux Φm(t) by 90o. The RMS value of Ep(t),
denoted by Ep is given by
𝐸𝑝 = 𝑅𝑀𝑆 𝐸𝑝 𝑡
=
𝐸𝑝 𝑡
𝑚𝑎𝑥
2
=
2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥
2
or
𝐸𝑝 = 4.44 𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥 .
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(1)
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Since the magnetic flux Φm(t) also links the secondary
winding that has Ns turns, the RMS value of the induced
voltage at the secondary side of the transformer, denoted
by Es is given by
𝐸𝑠 = 4.44 𝑓 × 𝑁𝑠 × ∅𝑚𝑎𝑥 .
(2)
Comparing Eqs. (1) and (2) yields
𝐸𝑝
𝐸𝑠
=
4.44 𝑓 × 𝑁𝑝
4.44 𝑓 × 𝑁𝑠
or
𝐸𝑝
𝐸𝑠
=
𝑁𝑝 𝑁𝑠
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or
or
𝐸𝑝
𝑁𝑝
=
=𝑎
𝐸𝑠
𝑁𝑠
𝐸𝑝 = 𝑎 × 𝐸𝑠
where, “a” is called the turn ratio of the
transformer.
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Figure below shows an ideal transformer with a
load on the secondary side.
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Under ideal conditions no current flows through the
primary winding and the primary current “IP” is zero when
secondary side is open.
When a load is connected to the secondary side, a current,
Is, flows out of the secondary windings.
The secondary current, Is, creates a magnetic flux “Φs” in the
transformer’s core.
Since the source voltage, E, is unchanged, which mean the
flux, Φm, must remain the same as it was before the load was
connected.
Therefore, the new flux (Φs) due to the secondary the
current, Is, must be countered by an equal and opposite flux.
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
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The counter flux “Φp” is produced by causing a
current, Ip, to flow into the primary windings.
Under ideal conditions, the fluxes Φs and Φp cancel
each other and the total flux through the core
remains Φm.
Result: Primary and secondary induced voltages
(Ep and Es) remain unchanged.
The flow of the primary current, Ip, represents the
power transfer from primary side to the secondary
side of the transformer to support the load Zs.
Under ideal conditions there is no loss of power
during the power transfer from the primary side to
the secondary side of the transformer.
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Therefore, the total power, Sp, on the primary side of the
transformer must be equal to the total power, Ss, on the
secondary side of the transformer, where
𝑆𝑝 = 𝐸𝑝 × 𝐼𝑝 and 𝑆𝑠 = 𝐸𝑠 × 𝐼𝑠
but Sp = Ss, therefore,
𝐸𝑝 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠 .
But 𝐸𝑝 = 𝑎 × 𝐸𝑠 , therefore,
𝑎 × 𝐸𝑠 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠
or
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𝐼𝑝 = × 𝐼𝑠 .
𝑎
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The load impedance as seen by the primary side, denoted by
Zp, can be given by
𝑍𝑝 =
𝐸𝑝
𝐼𝑝
=
𝑎×𝐸𝑠
1
×𝐼
𝑎 𝑠
= 𝑎2 ×
𝐸𝑠
𝐼𝑠
but
𝑍𝑠 =
Therefore,
𝐸𝑠
.
𝐼𝑠
𝑍𝑝 = 𝑎2 × 𝑍𝑠 .
Above Eq. shows an important property of the transformer;
that is, it can be used to change the effective impedance of any
load and thus can also act as an impedance transformer.
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Using impedance transformation, the impedances on the
secondary side can be expressed as the equivalent
impedances on the primary side to greatly simplify the
circuit analysis involving transformers.
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
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The ideal transformer is lossless.
In a real transformer, however, the primary
current is not zero under no-load condition.
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Reason: Iron losses and magnetizing reactance

Causes of iron loss:

Eddy currents

Hysteresis losses
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Eddy Current Losses:
 When time-varying magnetic flux passes through a
solid metallic core, the solid metallic core acts like
a bundle of concentric conductors and a voltage is
induced in each of these sections of the metallic
core.
 In a solid metal plate each of these sections would
behave like a short-circuited conductor and,
therefore, the induce voltages would cause a
strong currents which swirl back and forth in the
core.
 These currents are called eddy currents.
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Eddy currents are a major source of generating heat in
transformers. In a solid metal core these currents can be so
strong that they render the core red hot.
 Solution: Design the core with
thin laminated strips to insulates
the strips from each other.
 Result: The length of the
conducting path is greatly
reduced.
 Magnitudes of the induced
voltages in the core thereby
diminishing eddy current losses.
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Hysteresis Losses
The magnetic flux is caused by magnetic field strength, H. The
resulting flux density, B, in any material is related to the
magnetic field strength (H) by the following equation:
𝐵 = 𝜇𝑜 × 𝜇𝑟 × 𝐻
where,
B = magnetic flux density (weber/m2 or tesla)
H = magnetic field strength (A-turns/m)
µo = permeability of free space = 4𝜋 × 10−7
µr = relative permeability of the material
The value of µr changes with flux and, therefore, the
relationship between B and H is expressed in terms of a B-H
curve.
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Figure below shows the closed B-H curve called
hysteresis loop.
 The area under this curve
represents the energy that
changes into heat during
each cycle (joules per cubic
meters).
 Iron losses are represented
as resistance (Rm) in
parallel to the primary
terminals of the
transformer.
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
Magnetizing Inductance: In order to generate the
magnetic flux Φm, a current must flow through the primary
windings.
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
This current is called magnetizing current and depends on
the permeability of the material used to build the core.
Higher permeability means lower magnetic current and vice
versa.
Permeability can be represented by a parallel inductance
(Xm) to the primary terminal because magnetizing current
does not affect the amount of induced voltage.
This inductance is called magnetizing inductance.
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Winding Resistance:
 In a real transformer, the winding of the
transformer have finite resistance that must be
taken into account while analyzing or
modeling a practical transformer.

Since these winding resistances result in a
potential drop when current passes through
the windings, they are represented as the series
resistors (Rp and RS), both at the primary as
well as at the secondary side of the
transformer.
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Leakage Flux:
 In a real transformer some of Φs and Φp does leak
in the air.
 This flux leakage induces voltages both in the
primary winding and secondary windings, which
is counter to the voltages induced by the flux Φm.
 It results in a voltage drop both at the primary as
well as the secondary side.
 Accordingly, the effect of the leakage flux is
represented as the series impedances (Xlp and Xls)
at the primary as well as the secondary side of the
transformer in the model of a practical
transformer.
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Figure below shows the complete model of practical
transformer that incorporate the iron losses, magnetizing
current, primary and secondary winding resistances, and
the effect of the leakage flux under load conditions.
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The model of the practical transformer is easier to
analyze if the elements on the secondary side are
shifted to the primary side as shown in Fig. below.
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No-load Condition:
Under no-load condition the secondary is open and,
therefore, no current flows through the secondary
terminals.
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Full-load Condition:
Under the full-load condition 𝐼𝑝 ≫ 𝐼𝑜 and, therefore,
we can ignore Rm and Xm and the model for the
practical transformer reduces to the one shown below.
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The model of the practical transformer can further
be simplified by adding the resistances and
inductances as follows:
𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠
and
𝑋𝑒𝑝 = 𝑋𝑙𝑝 + 𝑎2 × 𝑋𝑙𝑠 .
Furthermore, Rlp and Xlp can be represented by
the complex impedance Zp as follow:
𝑍𝑝 = 𝑅𝑒𝑝 + 𝑗𝑋𝑒𝑝
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Figure below shows the simplified circuit for the
practical transformer with full load.
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Example: Specifications of a power transformer are as
follows:
Compute the full-load Vs, Is, as well as heat dissipation in
the transformer using the simplified model. Also,
compute Iron losses using the model under no-load
condition .
Sn (KVA)
Enp (V) Ens (V) Inp (A) Ins (A)
Rp (Ω)
1000
69000
6900
27.2
Rs (Ω)
Xlp (Ω)
Xls (Ω) Xm (Ω) Rm (Ω)
Io (A)
0.25
151
1.5
0.210
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14.5
145
505000 432000
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Solution: Let’s assume it’s a pure resistive load and full-load
secondary voltage is the same as Vns and full-load secondary
current is the same as Ins. Now we can calculate the load
resistance, RL, as follows:
𝑅𝐿 =
𝐸𝑛𝑠
𝐼𝑛𝑠
=
6900
14.5
= 47.6 Ω
The turn ratio is given by
𝐸𝑛𝑝 69000
𝑎=
=
= 10
𝐸𝑛𝑠
6900
Now we can compute Rep and Xep
𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠 = 27.2 + 102 × 0.25 = 52.2 Ω
𝑋𝑒𝑝 = 𝑋𝑝 + 𝑎2 × 𝑋𝑠 = 151 + 102 × 1.5 = 301 Ω
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The resistance shifted to primary, RLp, is given by
𝑅𝐿𝑝 = 𝑎2 × 𝑅𝐿 = 102 × 47.6 = 4760 Ω.
Now we can compute the total impedance of the model,
ZTp, as follows
𝑍𝑇𝑝 = 𝑅𝑒𝑝 + 𝑅𝐿𝑝 + 𝑗𝑋𝑒𝑝 = 52.2 + 4760 + 𝑗301
= 4812.2 + 𝑗301 = 4821.6 0.0186𝑜 Ω
Now we can compute primary current, Ip, primary
voltage, Ep, secondary current, Is, and secondary voltage,
Es as follows
𝐸𝑛𝑝 69000
𝐼𝑝 =
=
= 14.31 𝐴
𝑍𝑇𝑝 4821.6
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𝐸𝑝 = 𝐼𝑝 × 𝑍𝑇𝑝 = 14.31 × 4821.6 = 68118.4 𝑉
𝐸𝑝 68118.4
𝐸𝑠 =
=
= 6811.84 𝑉
𝑎
10
𝐸𝑠 6811.84
𝐼𝑠 =
=
= 143.1 𝐴
𝑅𝐿
47.6
Now we can compute the heat loss due to the load as follows
𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 = 𝐼𝑝2 × 𝑅𝑒𝑝 = 14.312 × 52.2 = 10689.3 𝑊 = 10.69 𝐾𝑊
Finally, we’ll compute iron losses at no load and at full load as
follows
2
𝐸𝑛𝑝
690002
𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑 =
=
= 11020.83 𝑊 = 11.02 𝐾𝑊
𝑅𝑚 432000
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𝐸𝑝2 68118.42
𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 =
=
𝑅𝑚
432000
= 10741 𝑊 = 10.74 𝐾𝑊
Total power loss at full load = 10.69 + 10.74 =
21.43 KW.
This power loss produces significant amount of
heat that must be removed from the transformer
to avoid any damage to transformer.
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
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
There are a number of techniques that are used
in the transformers to remove heat.
Through natural circulation of air (type AA),
Through forced air circulation (type AFA),
Oil-immersed self-cooled (OA),
Oil immersed self-cooled/forced air cooled
(type OA/FA) to name a few.
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Figure below show the front view of an OA/FA
type power transformer.
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 Figure next shows the rear view of
the same transformer where
cooling fans can be seen which
remove heat from the transformer
through forced air circulation.
 A transformer is called a “stepdown” transformer if the
secondary voltage, ES, is lower
than the primary voltage, EP. On
the hand, if the secondary voltage
is higher than the primary voltage,
it is called a “step-up”
transformer.
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Terminal Marking on Power Transformers:
 In power transformers, the high voltage (HV)
terminals are marked by symbols H1 and H2.


The low voltage (LV) terminal are marked by
symbols X1 and X2.
These terminals are mounted on the
transformer tank in a way that they either have
additive polarity or they have subtractive
polarity.
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Figure below shows the terminal marking and
polarity of the power transformers.
 It is imperative to know the correct polarity of
the transformers if you are connecting them in
parallel or
configuring a 3phase
transformer
bank using
single phase
transformers.
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


Voltage Regulation:
Voltage regulation is measure of variation in
the secondary voltage from no-load to full-load
condition when the primary voltage is kept
constant.
Voltage regulation is expressed in percent (%)
and is defined by the following equation
𝐸𝑠 (𝑛𝑜−𝑙𝑜𝑎𝑑) − 𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =
× 100
𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
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In AC systems the power is divided into three
different categories:
Apparent Power:


It is the product of the voltage and complex
conjugate of the current and is expressed in “voltamperes” or VA.
If the current I = IR + j IX then the conjugate of I,
denoted by I*, is given by I*= IR – j IX. The apparent
power, PA, is given by
𝑃𝐴 = 𝑉 × 𝐼 ∗
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Real Power:
 The real power (PR) is the power utilized by a
user and is expressed in watts (W).
 When the apparent power is expressed in
rectangular form, the real part of the apparent
power is the real power.
 The real power always flows from the source to
the user.
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Reactive Power:
 The reactive power (PX) is needed to establish
magnetic field in inductors.
 Electric field in a capacitor acts as a source of
reactive power.
 Reactive power can be supplied as well as
absorbed by an electric power source. It means
that the reactive power can flow from the
source to the circuit or from the circuit to the
source.
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


When the reactive power flows from the source
to the circuit it is considered “positive” reactive
power.
However, it is considered “negative” reactive
power when it flows from the circuit to the
source.
Mathematically, the apparent power (PA), the
real power (PR), and the reactive power (PX) are
related as follow:
𝑃𝐴 = 𝑃𝑅 ± 𝑗𝑃𝑋
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