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So far we have discussed the analysis and
performance of single phase transformers.
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However, majority of the transmission network
and power distribution network is 3 phase.
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Therefore, for to be useful, one should ideally
consider 3 phase transformer, which we will
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introduce today.
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Now, that should not be a problem if one has
a 3 phase 4 wire incoming line. One can use
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three single phase transformers
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of the usual construction that we have been
familiar with and connect them to three input
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lines in this manner – A, B, C and neutral.
Similarly, the secondaries can also be connected
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in a similar manner and get a 3 phase, four
wire output, which is stepped up or step down.
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So, these three single phase transformer together
will constitute a 3 phase transformer. This
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is very much possible. And in many cases,
this is also done. But, something more can
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be done. Normally, the input voltages will
be balanced; that is, V A N, V B N and V C
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N form a balanced 3 phase system.
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This is V A N; this is V B N; this is V C
N. We know that, the flux produced – the
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magnetizing flux produced in each of this
transformer lags the corresponding applied
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voltage by 90 degree. So, let us call them
phi A, phi B and phi C. So, phi A, phi B,
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phi C are also balanced 3 phase phasors. And
we can draw them maintaining same phase relationship.
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So, phi A will lag V A N by 90 degree; this
will be phi A. Phi B will lag phase by 90
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degree. And phi C will lag V C N by 90 degree.
Just as V A N phasors V A N plus V B N plus
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V C N equal to 0; similarly, from geometry,
phi A plus phi B plus phi C is also equal
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to 0.
Now, this has some implication; which means
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the sum of these phasor sum of these three
fluxes are identically 0. Hence, instead of
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putting these secondaries on this arm, if
we put the secondaries on the same arm as
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the primary and merge the three of the remaining
arms, the total flux flowing through that
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arm – that common arm will be 0. And that
total… That common arm can be completely
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eliminated. Hence, we will arrive at a structure,
which will look like this.
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This is the arm for A phase – both primary
and secondary. And this merged arm… Still
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the flux produced… This will be phi A; this
will be phi B and this will be phi C. And
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the sum of this flux, which is expected to
flow through this merged return path – this
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will be phi A plus phi B plus phi C. And but
this is identically equal to 0. Therefore,
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this arm is not necessary.
Hence, the simple structure; it becomes something
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like this with three windings – both primary
and secondary. Here three arms carrying respectively
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3 phase windings. This is for a phase; this
is for b phase; this is for c phase. So, this
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keeps a 3 phase transformer as a single unit.
Therefore, we saw that, we can connect three
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single phase transformers separately to form
a 3 phase transformer. Or, we can build a
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3 phase transformer as a single unit. There
are advantages and disadvantages of both the
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approaches, which we will find it; we will
see shortly.
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Before that, let us look at the core structure
a little more carefully. The core structure
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that we had considered so far
looks somewhat like this. Ideally if we assume
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the core to have infinite permeability; then
this yoke section will have zero reluctance;
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and any flux let us say produced phi A will
divide equally here as phi A by 2 and here
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through this arm as another phi A by 2; similarly,
for the flux produced by other phases. Therefore,
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what will be the total flux flowing through
each of these limbs.
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So, let us say for the first limb, it will
be phi A in the upward direction minus phi
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B by 2 in the downward direction minus phi
C by 2 in the downward direction. This is
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the net flux flowing through limb 1 or limb-A.
But, we have seen phi A plus phi B plus phi
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C equal to 0 or phi B plus phi C equal to
minus phi A. Hence, phi limb-A will be
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phi A minus half of minus phi A equal to 3
by 2 phi A. Similar exercise can be done for
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the other limbs. And we will find that phi
limb-B will be equal to 3 by 2 phi B and phi
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limb-C will be equal to 3 by 2 phi C.
Now, this is an important conclusion – important
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observation. We find that, the flux flowing
through the limbs are in fact, in phase with
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the flux produced by the winding on that limb
even when there is mutual coupling between
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the fluxes. Therefore, the operation of the
coils on individual limbs will be independent
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of each other as long as this relationship,
that is, the balanced voltage relationship
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is maintained. Hence, as far as analysis of
a 3 phase transformer goes, each phase can
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be treated separately at least with balanced
applied voltage and in steady state. This
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makes the analysis of 3 phase transformer
very simple. They can be balanced; they can
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be performed just like single phase transformer.
Assume if we can find out what is the applied
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voltage across each winding.
Now, this comment should be taken with a caveat,
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because we have assumed the iron to have infinite
permeability; however, in actual practice,
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the iron will not have infinite permeability.
In fact, since the transformers are operated
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very close to their knee point, there will
be some amount of reluctance to flux path
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in the yoke. Therefore, the division of the
flux in an actual transformer may not be exactly
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phi by 2 and phi by 2; there will be more
flux flowing through the central limb in this
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case than the farther limb, because this reluctance
is higher than this reluctance path.
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So, the division will not be exactly phi A
by 2 and phi A by 2, except for the middle
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arm, where the division will possibly be exactly
phi B by 2 and phi B by 2. But, for the phase
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A and phase C, the division will not be exact.
Hence, this in a practical transformer, it
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will not be completely… – each phase will
not be completely decoupled, but there will
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be some amount of coupling; but usually that
is negligible. The coupling can be made a
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little better if we use a different type of
core. Incidentally, this is the core type
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transformer. We call it a 3-limbed transformer
core. More regarding construction of 3 phase
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transformer we will see in a later lecture.
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But, to continue with this, this is a 3-limb
core, which is most commonly used. However,
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for very large transformer, sometimes a 5-limb
core is also used. The windings are placed
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on the inner limbs; while these two limbs
do not carry any winding. In this case, we
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can easily see that, the magnetic circuit
for each of the phases are more symmetric
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than the previous case. And hence, the assumption
of independent operation is more valid in
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this case. But, as we can easily find out
that, this core is much larger than the core
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that we have discussed before; hence, it is
justified only for very large rated transformer.
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This is called 3-limb core type. This is 5-limb
core. There is another construction of the
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core, which is called the shell type construction;
but this central section holds the winding.
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This also gives for each phase an independent
non-interacting flux path.
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Here also, the three phases can be treated
separately. This is the shell type 3 phase
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transformer. So, although, 3 phase transformer
are most commonly found with 3-limb core;
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some very large transformers are also built
with 5-limb core or a shell type core, where
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the magnetic circuit of each phases are either
symmetric or they are independent. But, as
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you can see, at least for the shell type core,
this is little more than three single phase
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transformers connected together. So, this
does not offer the advantages of 3-limb core;
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it is seldom used.
Now, let us look at what are the advantages
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of using a 3 phase transformer built as a
single unit and three single phase transformers.
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It is the common experience is that, if you
build a 3 phase transformer as a single unit,
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then the volume, cost everything are lower
compared to if you build the same transformer
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with three single phase unit. That is a big
advantage. However, there are certain advantages
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of building a 3 phase transformer with three
single units.
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For example, for maintenance and reliability
purpose, it will be sufficient to keep one
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spare single phase transformer, so that if
there is a fault in one of the transformers,
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the spare single phase transformer can replace
the faulty transformer; and the transformer
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can keep supplying power. However, for a transformer,
which is built as a single unit in order to
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provide redundancy, a full 3 phase transformer
is to be kept as standby, which is a more
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expensive proposition.
In certain applications, such as in mines,
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this becomes very important, because the space
is restricted; therefore, it is convenient
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to keep a single phase transformer as a spare.
And that is why, in mining application, a
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single phase transformer is preferred. There
are several other advantages as we will shortly
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see; that a single phase transformer even
when one of the transformers developed a fault
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and is consequently removed, the remaining
transformer can keep supplying at least a
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part of the power, which is not possible with
a 3 phase transformer. Also, the maintenance
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and repair of 3 phase transformer are somewhat
more difficult. We have discussed how the
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core of a 3 phase transformer looks like.
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Now, let us look at how we can connect the
windings of a 3 phase transformer. We have
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already mentioned that, as far as the magnetic
circuit is concerned, the windings operate
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almost independent of each other. So, let
us schematically represent the
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transformer windings. As you can see, even
in the simplest of the 3 phase transformer,
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whether built with a 3 single phase transformer
or built as a single unit, there are six windings
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and twelve terminals. This can be potentially
confusing if we do not follow a specific terminal
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marking convention, that is, leveling convention.
Once you are familiar with the drawings and
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the connections, we can dispense with that;
but at least initially, it will be useful
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to follow a certain terminal leveling convention.
The convention is very simple. We denote the
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terminals of the high voltage side with capital
letter; for example, A 2 A 1, B 2 B 1, and
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C 2 C 1. And the low voltage side with small
letters; for example, a 2 a 1; this is capital
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A 2 and this is small a 2. The implication
is the instantaneous polarities of these two
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terminals are same; that is, the phasors V
A 2 A 1 indicating the voltage of the terminal
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A 2 with respect to A 1 is in phase with the
voltage V a 2 a 1; similarly, for the other
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windings. So, following the same logic, V
B 2 B 1 and V b 2 b 1 are co-phasor; V C 2
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C 1 and V c 2 c 1 are co-phasor.
Now, what is the simplest connection we can
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think of? We can connect all one terminals
together on the high voltage side to form
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a star point and supply to the two terminals
A, B, C; in which case, the voltages – the
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voltage phasors of A 2 A 1, B 1 B 2 and C
1 C 2 form a balanced 3 phase voltage system.
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This point is N. Now, since these transformers
operate independent of each other, the voltages
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V A 2 A 1, B 2 B 1 and C 2 C 1 will be co-phasor
to their respective high voltage phases – phase
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voltages. So, the voltage V a 2 a 1 will be
this; b 2 b 1 will be this; and c 2 c 1 will
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be this.
Hence, we can connect the one terminal on
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the low voltage side also in a star and obtain
the output voltage on the two terminals. That
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way the phase voltage and the line voltage
on the high voltage side and low voltage side
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will be co-phasor. What will be the relationship
between the voltages? Let us see if the transformer
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turns ratio is 1 is to n then and the line
voltage on the high voltage side is V L; then
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V phase equal to V L by root 3 on the high
voltage side. This is equal to mod V A 2 A
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1. But, we know mod V a 2 a 1 equal to V a
2 a 1 by mod V capital A 2 capital A 1 equal
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to 1 by n. Therefore, V a 2 a 1 equal to mod
V A 2 A 1 by n equal to V L by root 3 n. So,
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the line voltage on the low voltage side V
L will be root 3 mod V a 2 a 1 equal to V
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L by n.
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So, the phase voltages and line voltages in
this case are related by high voltage phase
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voltage is to low voltage side phase voltage
are related as 1 is to n; same as the turns
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ratio. Also, the same relation holds; V line
of the high voltage side and V line of the
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low voltage side are also related by 1 is
to n. How about the currents? Let us say the
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line current on the high voltage side are
I capital L and line currents on the low voltage
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side are I small l. Then if we assume each
of this transformer to be ideal; that is,
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the net mmf is required to establish the flux
is 0; then we can write 1 into I L since for
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this star connection, winding current and
line current are same; equal to n into I small
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l. Therefore I L by I small l is also equal
to n is to 1.
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Hence, we have for this connection V phase
into I phase; which is same as I line
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divided by V phase of the low voltage side
into line current of the low voltage side
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equal to 1 by n into n by 1 equal to 1. Or,
single phase KVA of the both sides are same.
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Hence, the 3 phase KVA of both sides are also
same. Therefore, as we saw that, if the transformers
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are ideal; as in the case of a single phase
transformer, the KVA of the two sides of a
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3 phase transformer are also same. This connection
is called the star-star connection or Y, Y
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connection. And since the phase angle between
the line to neutral voltage of both sides
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are 0; it is also called star-star, 0 degree
connection.
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Obviously, on the secondary side, had we connected
a 2, b 2 and c 2 together to form the neutral
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point and obtain the supply from the one terminal,
then the phase and line voltages should have
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been in phase opposition to the primary side.
These would have been V a 1 a 2; these would
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have been V b 1 b 2; and this would have been
V c 1 c 2; in which case, the connection would
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have still been YY. But, now the phase difference
between the primary side and the secondary
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side or the high voltage side and the low
voltage side would have been 180 degree. And
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hence, this would have been called YY 180
degree connection. Just as the transformer
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on both sides can be connected in star, they
can also be connected in delta.
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Let us see how the delta connection can be
done. As usual, the supply is given to the
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two terminals. And then we connect A 1 with
B 2, B 1 with C 2, and C 1 with A 2. Supply
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is given to A, B, C. So, taking the voltage
V B C as the reference phase, now B terminal
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is connected to B 2 and C terminals is connected
to C 2; which in turn is connected to B 1.
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So, V B C is nothing but the voltage V B 2
B 1, which is shown here – B 2 B 1. And
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then terminal A 1 is connected to B 2 and
A 2 is connected to C 1. This is voltage A
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1 A 2 and this is voltage C 1 C 2. On the
secondary side, these are the terminals a
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2, b 2, c 2; and these are the terminals a
1, b 1, c 1. Similar connection is made; results
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in a similar phasor diagram. Please note here
also the phase angle between line to neutral
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is 0. Therefore, this connection can be called
delta-delta 0 degree.
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Had we reversed the connection on the secondary
side; that is, instead of connecting a 1 to
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b 2, we had connected b 1 to a 2, c 1 to b
2 and taken the supplies from a 1 b 1 c 1
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instead of a 2 b 2 c 2; then this phasor diagram
would have been reversed and we would have
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got D d 180 degree connection. In either case,
let us see what will be the voltage and line
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conversion ratio. Again, we will assume the
turns ratio to be 1 is to n and the line voltage
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on the high voltage side is V L, line current
is I L. On the low voltage side, the line
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voltage we will take to be V l and the line
current to be I l.
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Here again we find that here for delta connection,
the winding voltage is same as the line voltage.
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Therefore, we can write V L by V l equal to
1 by n. But, winding current is now root 1
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by root 3 times the line current, which is
at a phase shift of 30 degree compared to
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the line current. This is phase current. But,
these two phase currents are also co-phasor.
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And we have I phase into – on the high voltage
side, into 1 equal to I phase on the low voltage
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side into n. But, I phase is I L by root 3
equal to – on the low voltage side, this
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is I small l into n by root 3. Here again
we have I L by I small l equal to n is to
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00:41:42,900 --> 00:41:52,270
1. Hence, the relations for the star-star
connection and the delta-delta connection
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00:41:52,270 --> 00:42:06,270
between the phase voltages, line voltages
and phase currents are same.
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However, this star-star, delta-delta connection
has one advantage. Let us assume that, we
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have a delta-delta connected transformer.
Sometimes in order to indicate delta connection,
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the windings are placed like this although
physically, in a transformer, it will not
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form a triangle; it will be rather what we
have shown in this diagram. But, this is just
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00:43:01,400 --> 00:43:27,350
a schematic representation. Suppose this was
being supplied from a balanced 3 phase 3 wire
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00:43:27,350 --> 00:43:44,570
supply and this was feeding a balanced 3 phase
load; now, assume that, these delta-delta
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00:43:44,570 --> 00:43:57,869
connected transformer is built with three
single phase transformers. While operating,
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let us assume one of the transformers develops
some fault and is consequently removed. Then
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the remaining transformer can still continue
to supply some amount of load. When you remove
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these windings, you get the structure what
is known as open delta or V-connected transformers.
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A V-connected transformer can continue to
supply the load. That is one advantage of
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using three single phase transformers separately
to form a 3 phase transformer rather than
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using a 3 phase transformer as a single unit.
Sometimes this is important.
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Let us see exactly how much kVA or power can
be supplied with this open delta connected
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transformer or V-connected transformer through
an example. Let us say the line voltage for
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this transformer was 11 kV; and on this side,
it was 400 volts. And let us say each of these
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00:45:26,490 --> 00:45:36,080
transformers are rated at 100 kVA. This is
made of three single phase transformers rated
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00:45:36,080 --> 00:45:49,289
at 100 kVA. And it was supplying a balanced
star connected resistive load of 240 kilowatt.
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Assume the transformers to be ideal.
Therefore, if this load was drawing 240 kilowatt,
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00:45:59,299 --> 00:46:08,130
then the input load was also 240 kilowatt.
So, what was the line current? LV side line
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00:46:08,130 --> 00:46:42,180
current was… This was 346.41 amperes. So,
what was the LV side transformer winding current,
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00:46:42,180 --> 00:47:09,579
which was LV side winding current? What was
the transformer turns ratio from the given
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00:47:09,579 --> 00:47:24,330
data? The transformer turns ratio was… On
this side, the winding voltage is 11 kV; on
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00:47:24,330 --> 00:47:30,020
this side, the winding voltage is 400 volts;
the turns ratio was 27.5.
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00:47:30,020 --> 00:47:46,800
So, what was the HV side winding current?
HV side winding current was 200 ampere by
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00:47:46,800 --> 00:48:35,730
turns ratio – 27.5; so 7.273 ampere. This
was when it was operating as a 3 phase transformer.
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00:48:35,730 --> 00:48:44,950
Now, let us say one of the transformers develops
fault and is removed from the circuit. So,
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the transformer becomes a V-connected transformer
or open delta connected transformer. The question
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is how much power I can still continue to
supply to the load? The constant of course
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is that, the transformer winding current should
not exceed its rating while being used as
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a V-connected transformer. So, what is the
transformer winding rated current? We have
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mentioned that, each of the transformers is
rated at 100 kVA. Therefore, the LV side rated
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00:49:33,690 --> 00:49:45,780
current is transformer winding rated current
is 100 kVA divided by 400 volt equal to 250
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00:49:45,780 --> 00:49:59,390
amperes.
So, when you are supplying this transformer
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with an open delta connection, the LV side
winding current should not exceed 250 amperes.
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However, it is easy to see with open V connection,
the line current on the LV side is same as
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00:50:22,710 --> 00:50:28,339
the phase current, because this winding is
not there; this line current flows like this
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00:50:28,339 --> 00:50:33,569
and comes back through this; similarly, for
the HV winding.
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00:50:33,569 --> 00:50:47,930
Therefore, the line current during open delta
or V connection should not exceed 250 amperes.
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00:50:47,930 --> 00:50:58,830
What about the phase? It is easy to see that,
even with open delta or V connection, the
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00:50:58,830 --> 00:51:07,520
line voltages will be balanced. This is so
because the supply voltage is balanced; hence,
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00:51:07,520 --> 00:51:16,000
the voltage across the remaining two transformers
do not change. So, let us say if this was
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00:51:16,000 --> 00:51:21,460
A, this was C, this was B; then this voltage
still remains BC, this voltage still remains
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00:51:21,460 --> 00:51:32,900
CA; therefore, this is still BC and CA. And
hence, this voltage is ab. So, this low voltage
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00:51:32,900 --> 00:51:42,029
side or LV side voltages still forms a balanced
3 phase system even during open delta connection.
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00:51:42,029 --> 00:51:50,960
So, if this supply is balanced with open delta
transformer, the LV side line voltages are
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00:51:50,960 --> 00:51:55,380
balanced.
Now, if this balanced line voltage supplies
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00:51:55,380 --> 00:52:02,900
a balanced 3 phase load; so the line currents
will also be balanced. Hence, with open delta
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00:52:02,900 --> 00:52:09,559
connection, we have seen that, that is, transformer
will be able to supply a balanced 250 amperes
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00:52:09,559 --> 00:52:18,109
of current. Hence, the total power that it
will be able to supply is maximum load in
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00:52:18,109 --> 00:52:33,849
kilowatt; or, load kVA should be equal to
root 3 into 400 into 250 amperes; which is
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00:52:33,849 --> 00:52:52,089
about 173.2 kilowatt. Please note that, the
total rating of these transformers are 300
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00:52:52,089 --> 00:53:05,560
kilowatt; total transformer kVA is 300 kVA.
Hence, with open delta, we can supply approximately
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00:53:05,560 --> 00:53:14,630
58 percent of the total transformer kVA. Please
note that, we have removed one transformer.
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So, it was expected that, we should be able
to supply about two-third or 67 percent of
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00:53:20,829 --> 00:53:28,299
the total kVA; but that is not the case. We
can supply only 58 percent of the remaining
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00:53:28,299 --> 00:53:33,769
kVA of the total installed transformer kVA.
Thank you.