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Good morning. In the last class we have shown
that in alternating current circuits the voltage
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level can be easily stepped up or stepped
down using a transformer. The transformer
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we have shown is a set of coils that are magnetically
coupled; in order to couple a set of coils
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we make the following arrangement.
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We choose a core made up of ferromagnetic
material and put winding on them. So, if we
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connect a voltage source v1 here, there will
be a current flowing into the coil i1, and
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this will generate a voltage v 2 in the second
coil by the principle of induction that is
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Faraday’s law. The current in the first
coil produces a flux that circulates through
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the core and links the second coil. Since
in alternating circuits, the flux is also
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alternating, there is a rate of change of
flux linkage in the second coil as a result
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of which a voltage v 2 is generated, but this
is the basic operating principle of a transformer.
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We can get good insight into the operation
of this transformer, if we make certain simplifying
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assumption. For example, the first assumption
we make that the windings are lossless; windings
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are lossless that is they have zero resistance,
of course that is an ideal winding.
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The second assumption is regarding the core
we assume that the permeability of the core
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is infinity which means in order to establish
any finite amount of flux the current drawn
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will be 0. This implies the permeability of
the core infinity implies the B H curve of
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the material is the y axis; that is for any
finite B the corresponding H is 0; therefore,
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the required ampere turns is 0, and hence
the current is 0. As a corollary of this second
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assumption, since the permeability of the
core is infinity there will also be no leakage
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in this construction; that is all the flux
produced by coil one will be confined inside
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the core, and there will be no flux that will
be flowing through the air. Since all the
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flux will be flowing through the core all
the flux will also link coil two.
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So, the third assumption which is not really
a new one, but a consequence of the second
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assumption is that there is no leakage flux.
So, the transformer for which these assumptions
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are made is called an ideal transformer. Now
this ideal transformer is a conceptual device;
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you can understand that no such practical
transformer exist for which all these three
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assumptions hold. However, the introduction
of an ideal transformer as a conceptual device
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does help in understanding how a practical
transformer works, and we will later on relax
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this assumption and try to arrive at the model
of a practical single phase transformer. But
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for the time being let us spend some time
to find out what are the consequences of these
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three assumptions, and how we can analyze
a single phase ideal transformer.
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As we have mentioned when a voltage v 1 is
applied across coil one of an ideal transformer,
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a flux phi m is established in the core; however,
in order to just establish this flux no current
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is drawn from the source; that is under this
condition I 1 is identically equal to zero.
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Then we should also conclude that
the voltage induced across coil one e 1 must
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exactly balance the applied voltage v 1, so
that the current is 0. Now by Faraday’s
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law e 1 is given by N 1 d phi m dt where N
1 is the number of turns of coil one. Let
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us assume that phi m in steady state is given
by the expression phi max sin omega t then
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e1 is equal to N 1 omega phi max cosine omega
t.
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Similarly, the induced voltage e 2 in coil
two will be given by e 2 equal to N 2 d phi
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m dt. It is to be noted since in ideal transformer
there is no leakage the total flux linking
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the coil one and coil two are same. Both are
given by phi max sin omega t. If I write in
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phasor notation that is I want to find out
the rms value of e 1. So, v 1 equal to e 1
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equal to 2 pi by root 2 N 1 f phi max. This
is the well known formula 4.44 N 1 f phi max,
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and v 2 similarly e 2 equal to 4.44 into N
2 f phi max.
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Therefore what we can say about an ideal transformer?
If we have an ideal transformer
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where we apply a voltage v 1 across its first
coil with number of turns N1 then the induced
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voltage on the second coil with turns N 2;
it is related to the voltage induced in the
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first coil by the following relationship.
E 1 equal to v 1 and e 1 by e 2 equal to N
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1 by N2; not only that e 1 and v 1 are in
phase, e 2 is also in phase with e 1.
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The phasor diagram can be drawn as follows.
The previous calculation we have shown if
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the flux waveform is given by the time function
phi max sine omega t then the induced voltage
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is given by the function N 1 omega phi max
cos omega t. Now cos omega t leads sine omega
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t by an angle of 90 degree; therefore, you
can draw the phasor diagram as follows. Taking
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the flux to be the reference phasor the voltage
is v 1 e 1 and e 2 leads it by 90 degree.
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So, this is the phasor diagram of an ideal
transformer at no load; no load since no load
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has been connected on the secondary side.
Now suppose, I connect a load on the secondary
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side of impedance Z L; obviously, there is
an induced voltage here current I 2 will flow
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which will be determined by the magnitude
and phase angle of the impedance Z L. However,
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as soon as the current I 2 starts flowing
in the second coil, by Faraday’s law it
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will try to oppose the very cause which it
is due; that is it will try to send a flux
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which is opposite to the flux established
by coil one; as a result of which the total
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flux linking coil one will become phi m minus
phi m 2. So, the total flux linking coil one
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will reduce, hence e 1 will reduce; however,
since there is no impedance between the applied
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voltage and the coil there will be current
drawn from the source v 1 of such magnitude
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so that the induced voltage across coil one
always balances the applied voltage v 1.
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Otherwise this current will keep on increasing,
which means the flux produced by coil one
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will now change to phi m 1, and in steady
state phi m 1 minus phi m 2 will still be
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equal to phi m so that the induced voltage
e1 and e2 does not change. So, what will be
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the magnitude of the current I 1? Obviously,
if phi m 1 minus phi m 2 is same as phi m
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then phi m 2 is produced due to an ampere
turn of N 2 I 2 the ampere turns N 2 I 2 produces
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the flux phi m 2; therefore, this ampere turn
must be cancelled by the ampere turns of the
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primary coil so that the net flux still remains
at phi m. So, this should be equal to. So,
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the current drawn from the source must produce
a cancelling ampere turns N 1 I 1, so that
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the flux produced in the core remains constant
at phi m.
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Therefore, we can say the N 1 I 1 should be
equal to N 2 I 2 or I 1 equal to N 2 by N
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1 I 2. This is sometimes called the reflected
load current I 2 dash; therefore, when we
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put a load on the secondary of an ideal transformer
a reflection current I 2 dash flows in the
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primary side so which completely cancels the
ampere turns of the secondary side. The relationship
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is N 1 I 2 dash is equal to N 1 I 1 equal
to N 2 I 2. Therefore we have the following
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relation in the phasor; again taking the net
flux linkage phi m to be the reference phasor
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we have seen that under load condition E 1
and E 2 remains unchanged; that is E 1 still
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is equal to v 1, and E 2 is given by E 1 by
E 2 is equal to N 1 by N 2. The current I
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2 is determined by the loading Z L; the current
I 1 will be in phase with I 2, its magnitude
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will be I 1 equal to N 2 by N 1 I 2. This
is the phasor diagram of the ideal transformer
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under loaded condition.
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So, for an ideal transformer we have seen
E 1 equal to v 1, E 1 by E 2 equal to N 1
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by N 2, E 2 equal to v 2 equal to Z L I 2,
and N 1 I 1 equal to N 2 I 2. In other words
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from this relation we can show v 1 I 1 equal
to v 2 I 2; that is in ideal transformer the
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pure energy convertor whose input VA is always
equal to the output VA, and the voltage and
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current are transformed by the turns ratio
that is v 1 by v 2 equal to E 1 by E 2 equal
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to N 1 by N 2. Similarly, I 1 by I 2 equal
to N 2 by
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N 1; this relation holds for an ideal transformer.
As we have mentioned an ideal transformer
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is a pure energy convertor, and it is a conceptual
device; it cannot be made in practice. So,
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a practical single phase transformer is an
approximation of an ideal transformer; where
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are the differences? We had made three assumptions
while formulating the idea of an ideal transformer.
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That is we said that the windings are lossless;
no practical winding will be lossless. There
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will be some resistance in every winding.
How do we? Similarly the core of a practical
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transformer will not have infinite permeability.
It will have very high permeability but not
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infinite; therefore, even under no load condition
a practical single phase transformer will
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draw some current to establish the flux in
the core. This is a called the magnetization
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current. Then third, as a result of finite
permeability of the core material there will
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invariably be some leakage flux; in the single
phase practical transformer the leakage flux
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will circulate through the air, and it will
link only the coil that produces that flux
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and not the other coil.
We will shortly see how to incorporate this
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into the model of a practical single phase
transformer, and how to develop a model of
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a practical single phase transformer incorporating
the ideal transformer, but before that let
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us turn our attention to some of the constructional
details of a practical single phase transformer.
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The figure on your screen shows, the left
hand side shows the outward appearance of
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a practical single phase transformer. This
drum, this can like appearance is due to the
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casing. The actual transformer is inside the
casing; the actual transformer is inside this
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casing. These protruding parts are basically
insulators through which external connection
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is made to the transformer inside. Now if
this is cut open, inside we will find the
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actual transformer which is this object.
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Let us have a closer look of how this looks
like. If we take a closer look the practical
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transformer consists of an iron core, this
member, on which a set of windings are placed.
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So, the main parts are core and set of windings.
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Taking a closer look at the core we will find
that this core is not a solid block of iron;
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rather it is made of a large number of thin
plates of steel sheets which are cut in a
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particular fashion. Here it is shown that
these steel sheets are cut in E and I shapes;
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then many such E and I steel sheets are placed
together to form the core. Now this is done;
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these are called transformer laminations.
The transformers are laminated in order to
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reduce the eddy current loss.
Now, you are all familiar with the mechanism
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of eddy current loss. If there is a coil which
produces an alternating flux the flux flows
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through the core different parts of the core.
Since this is an alternating flux it produces
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circulating current around them; iron being
a conducting material a finite amount of current
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flows through the core, and it causes loss.
By laminating we tend to increase the resistance
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of the circulating current path so that the
loss is reduced. This is called laminating
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the transformer core almost all practical
or frequency transformers are laminated.
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Now this shows the different types of lamination
that can be used. Now here we should say that
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this is just one shape of the core. There
are basically two types of core used that
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are used with single phase transformer; one
is called the core type core, another is the
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shell type. Difference is here the windings
are placed on the core; that is the windings
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around the core in the shell type construction
it is rather the core that surrounds the winding.
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So, these are the two different construction
of the core; one is the core type and other
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is the shell type.
In both cases in order to have good coupling
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between the primary or the primary and the
secondary order the HV and LV winding; they
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are both made into different sections and
in terming out. Here you can see that both
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the winding are divided into two parts and
placed on both the limbs of the core. In the
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shell type construction each of the windings
the primary and the secondary are divided
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into different sections, and then they are
alternately placed, so that the coupling between
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the total winding is good, and the leakage
is kept to a minimum.
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If we look at the cross-section of the central
line then this is what we will find. The inside
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H portion shows the transformer laminations;
on the lamination usually a cylinder will
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be placed which is made up of insulating material
so that it provides insulation between the
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core and the winding that will be placed next.
The low voltage winding is placed on that
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insulating cylinder, and then, once all complete
low voltage winding is placed then there will
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be another cylinder which will enclose the
low voltage winding, and on top of which the
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high voltage winding will be placed by keeping
a gap between the low voltage and the high
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voltage winding which provides for the a duct
for the cooling medium.
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The transformers while operation carry current,
and hence there will be I square or R loss
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in both the windings that heat needs to be
taken out of the transformer so that the temperature
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rise of the transformers can be maintained.
Therefore, paths must be provided for the
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cooling medium usually oil to flow through
the core and the windings; it is the purpose
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of the duct in between. So, the core as we
can see has a strange set like a cross. This
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is called the cruciform core; the reason is
simple. The LV and HV winding both are circular,
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and they are placed on cylindrical surfaces.
The purpose of making this a cruciform rather
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than a square or a rectangle is to be able
to utilize the most of the area inside the
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cylinder for very small transformers such
as these even rectangular or square shape
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cores are used, but as the transformer rating
goes higher and higher, higher number of steps
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are used so that the periphery of the transformer
core approximates a circle, and the area inside
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the winding cross-section is fully utilized.
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Now coming to the windings, there are many
different types of windings that are used
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with transformers depending on their voltage
and current rating. For single phase transformer
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usually three types of windings are used.
For the low voltage winding for a given cable
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you can understand the low voltage winding;
voltage rating is less, but the current rating
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is large for which normally two types of winding
either a cylindrical multilayer or a helical
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winding is used, while for the high voltage
either a cylindrical or a crossover winding
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is used. This picture shows the diagram of
different winding arrangement. The low voltage
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winding are usually made up of rectangular
conductors may be a single conductor or a
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large number of strips taken together and
wound on the insulating cylinder of the low
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voltage winding.
Between turns another set of duct and insulation
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will be provided on which a second layer can
be wound. For the high voltage winding, high
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voltage winding will usually have a large
number of turns, but its current rating will
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be small. There can be still made up of rectangular
wires or even circular wires. In the crossover
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winding generally high voltage windings are
divided into several groups, and the groups
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are wound separately, and then they are connected
in series to form the high voltage winding.
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Each group can have many numbers of turns,
and they can be arranged in multiple layers.
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Therefore to summarize the constructional
details of a single phase transformer we can
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say that the main parts of a single phase
transformer are the core. This is usually
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made of silicon steel and can be either hot
rolled or cold rolled; for large rated transformers
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normally cold rolled grain oriented or CRGO
steel is used. For smaller transformer hot
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rolled steel is used; this is silicon steel.
The purpose of the core is to provide the
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path for flux, high permeability path for
the magnetic flux, and the cores are of two
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types. The transformer core can be of core
type or shell type. The next important part
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of a transformer is the windings; there are
two windings in a transformer.
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Normally a two winding transformer there will
be two windings; one is the LV winding, another
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is the HV winding. The windings are made up
of aluminum or copper conductors either round
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or square cross-section. Square cross-section
conductors can be made up of solid conductor
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or multiple strips connected in parallel.
The LV windings are usually of two types;
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one is the cylindrical or helix. The HV winding
can also be of two type either cylindrical
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or crossover. The next important constructional
feature is the insulator. Insulators are used
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in different places of a transformer, and
they are classified as major insulation
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which are between let us say core and the
LV winding, LV winding and the HV winding,
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and the minor insulation
which are usually between the turns of the
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same winding or the layers of the same winding.
Now there are different types of insulation
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materials that are used. For example, pressboard
cylinders, then paper, transformer oil at
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different purpose, different insulations are
used. The insulation material are characterized
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by the maximum temperature at which they can
maintain their insulation properties, and
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this is the weakest constructional component
of a transformer, and more often than not,
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the quality of the insulator will determine
the reliability or even the size of a transformer.
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Other than these three main constructional
elements, there are many other elements that
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are used; for example, the bushings. The bushings
are these protruding elements through which
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external connection is made to the transformer.
These are usually porcelain cylinders hollow
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porcelain cylinders which disc sets through
which the connecting conductors are placed.
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They can be for high voltage, they can be
oil field, or they can be of capacitor type.
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So, bushing is provided for external connection
and then cooling arrangement. For small transformer
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separate cooling arrangements are not provided.
The cooling is provided by the tank itself
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through in which the transformer is placed.
For large transformers, they are usually immersed
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in transformer oil so that heat can be extracted
from the inner parts of the transformer that
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is the core, and the LV winding which is inside
the HV winding. For even larger transformer
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the external cooling pipes or fins may be
provided. For even larger three phase transformer,
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forced cooling by oil and with fan, cooling
can be provided. So, a large number of cooling
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arrangements are possible, and for depending
on the power rating of the transformer, a
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suitable cooling method is chosen, and then
we will find protection arrangements which
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are found in the form of, for oil field transformer,
for conservators, and then explosion vents,
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Buchholz relay and then lightening arresters
and different circuit breakers and relays.
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So, these are the essential constructional
elements of a single phase transformer that
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we will normally find. Now coming to the comparison
with an ideal transformer there are as we
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said that for conceptualizing an ideal transformer
we made three assumptions; that is the windings
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are lossless, the permeability of the core
material is infinity, and there is no leakage.
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Now none of these three assumptions hold good
for a practical single phase transformer.
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So, we need to relax these assumptions one
by one and arrive at a model of a practical
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transformer. So, let us see.
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Let this be the ideal transformer. It is easy
to account for the resistance of the practical
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transformer coils. They will appear as a lump
resistance in series with both the coils;
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let us call it r 1 and the resistance on the
other side r 2. Here the resistances of coil
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one with turns N 1 and coil two with turn
N 2 respectively; however, it is more difficult
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to relax the condition of the infinite permeability.
We said due to infinite permeability in ideal
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transformer all the flux produced by the coil
remains inside the core. So, in a practical
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in an ideal transformer let this be the flux
phi m 1 produced by current I 1, and this
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is the flux phi m 2 produced by current I
2. We have seen that phi m 1 minus phi m 2
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remains constant if the applied voltage on
coil one is constant.
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However, in a practical single phase transformer
since the core is not infinitely permeable
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there will always be some flux which will
not flow entirely through the core but part
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of it will flow through the air. This is called
the leakage flux, call it phi l one. Similarly
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there will be the leakage flux produced by,
current flowing in the second coil phi l two
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which will link only coil two and not coil
one. So, what is the total flux linking coil
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one? Phi 1 equal to phi m 1 plus phi l 1 minus
phi m 2; similarly, the total flux linking
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00:49:15,550 --> 00:49:53,660
coil 2 phi 2 is equal to. Therefore, the voltage
induced in coil 1 is N 1 d phi 1 d t equal
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to N 1 d d t of phi m 1 minus phi N 2 plus
N1 d d t of phi l 1. Keeping the analogy with
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our discussion with the ideal transformer
let us call this term to be the induced voltage
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e 1 then N 1 d phi 1 d t equal to e 1 plus
N 1 d phi l 1 d t.
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Similar equation can be written for the other
coil N 2 d phi 2 d t equal to N 2 d d t of
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phi m 1 minus phi m 2 minus N 2 d phi l 2
d t. Now phi l 1 equal to N 1 I 1 divided
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by the reluctance of the leakage path R l
1. Similarly phi l 2 equal to N 2 I 2 by the
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reluctance of the second leakage path R l
2. This can be written as therefore N 1 d
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00:52:12,220 --> 00:52:38,420
phi l 1 d t equal to N 1 square by R l 1 d
i1 d t, and N 2 d phi l 2 d t equal to N 2
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00:52:38,420 --> 00:52:51,730
square by R l 2 d I 2 d t. The term N 1 square
by R l 1 is called the leakage inductance
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00:52:51,730 --> 00:53:05,720
of coil one l 1, and N2 square R l 2 is called
the leakage inductance in l 2.
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Therefore N 1 d phi 1 d t equal to e 1 plus
l 1 d i 1 dt and N 2 d phi 2 d t equal to
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e 2 minus l 2 d i 2 dt. Calling the applied
voltage v 1 and the terminal voltage v 2,
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then we can finally write v 1 equal to r 1
i 1 plus l 1 d i 1 d t plus e 1, whereas e
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2 equal to v 2 plus r 2 i 2 plus l 2 d i 2
dt. These set of equations give the model
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of a single phase transformer.