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Good morning. Load on a practical transformer
such as a distribution transformer it grows
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over time. As the, as more and more houses
are electrified more on the load of each house
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increases, so does the load on the transformer.
So, it may very well happen that within the
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life time of a distribution transformer or
any other transformer, a transformer that
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was designed to the sufficient for a given
load over a period of time; the transformer
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may reach its rated load and the load may
even exceed it. So, there are 2 methods by
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which such a situation can be handled; one
is replace the transformer by a larger transformer.
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However, it has some practical limitations.
For example, the larger transformer should
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again, should be planned for future expansion;
which means, that the transformer rating should
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be relatively large compared to the present
pick load. This also means, particularly for
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distribution transformer that, the new transformer
will be operating at light load for most of
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the time which is not a very efficient utilization
of the transformer.
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The other approach is to connect a similarly
rated transformer in parallel to the existing
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transformer; this has certain advantages.
For example, even if the new transformer is
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of the same rating as the old transformer;
then at least it allows for a growth of 100
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percent in the load before a further replacement
becomes necessary. Second thing, this new
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transformer need not always be connected.
When the load is not sufficient only one transformer
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may be connected and it can supply at its
rated capacity. Maintenance also becomes easy;
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because you need to keep only one spare small
transformer. So, for these reasons parallel
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operation of transformers are important. And
in this lecture we will find out how to connect
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two single phase transformer in parallel;
what are the conditions for that and how to
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analyze parallel connected transformers.
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So, let us see. Suppose, we have 1 single
phase transformer connected to the load, to
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the source on the primary side and supplies
the load on the secondary side. This is let
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us say transformer 1. If I want to connect
another transformer parallel to this; the
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primary of the 2 transformer will be connected
in parallel to the source side, also the secondary.
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However, while connecting the secondary in
parallel one should be careful about the instantaneous
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polarity of the voltages.
For example, this is the instantaneous positive
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terminal of the parallel connected primaries
and this is the instantaneous positive terminal
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of the secondary of the first transformer.
Then the instantaneous positive terminal of
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the second transformer must be connected to
the instantaneous positive terminal of the
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first transformer and the other terminal should
be connected like this. If not, then one can
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easily understand if this polarity is reversed;
then even without this load, there will be
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a once this switch S is closed, there will
be a short circuit through the transformer
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winding and a very large current will flow.
So, this should be protected against.
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So, before connecting the transformers in
parallel, their instantaneous polarities should
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be checked and connected accordingly. This
is the first condition for connecting 2 transformers
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in parallel, but this is not all. Obviously,
one can appreciate that the no load voltages
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of the transformers should not be very different
for the same reason. Because even if we assume
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that there is no load and the transformer
secondary no load voltages are different.
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Then when you close this switch, there will
be a circulating current flowing through the
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secondaries, which will be limited only by
the series impedance of the transformers which
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we have shown to be, which we have seen they
are small; within typically, within 5 to 10
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percent. So, even at no load, a very large
current can flow; if the no load voltages
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of the 2 transformers are very different.
So, the transformers to be connected in parallel,
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should have instantaneous the correct polarity
terminals connected together and then no load
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voltages should be very close; if not exactly
equal. There are few other conditions which
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are not essential, but desirable and we will
derive them. And of course, the total KVA
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rating of the 2 transformers must be greater
than the total KVA demand of the load. It
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is for that purpose only that we are connecting
the transformers in parallel. Now, let us
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see what other desirable conditions we can
find out. For that we will need to analyze
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this parallel connected transformer and in
the beginning, we will assume that the transformers
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have identical; no load primary and secondary
voltages.
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The equivalent circuit of the first transformer
with respect to the load side can be shown
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to be like this; we have derived this before.
Let us say this series impedance is Z 1 equal
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to R 1 plus j X 1. Since, the secondary transformer,
the second transformer input are and output
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are connected to the same source and the same
load; effectively the series impedance of
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the second transformer will come in parallel
to the first transformer. Therefore, the total
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equivalent circuit of the parallel connected
transformers can be represented something
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like this. So, this is Z 2 equal to R 2 plus
j X 2. Let us say the total current drawn
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by the first transformer is I 1; the current
drawn by the second transformer is I 2 and
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total load current is I L. Then one can easily
derive that Z 1 I 1 equal to Z 2 I 2 and I
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L equal to I 1 plus I 2. Therefore, in terms
of I L, I 1 equal to Z 2 divided by Z 1 plus
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Z 2 into I L and I 2 equal to Z 1 divided
by Z 1 plus Z 2 I L.
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Now, let us try to see the phasor relationship
between these currents and voltages. Let us
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say this is the load voltage phasor V L and
this is the load current phasor I L. The vector
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sum of phasor sum of I 1 and I 2 is equal
to I L. So, let us say this is I 1 and this
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is I 2. Then from this equivalent circuit
we
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can show, this is V 1 dash. Similarly, this
is equal to I 1 Z 1 equal to I 2 Z 2. This
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is how the phasor diagram will look like.
Now, it is obvious from the phasor diagram
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that the current in the 2 transformers will
be minimum when I 1 and I 2 are in same phase;
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because then I L will be the algebraic sum
of I 1 and I 2. Therefore, one thing we can
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immediately say that one desirable condition
for parallel connected transformer is that,
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when they are connected in parallel the currents
drawn by them are in phase. That will happen
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only if; if I 1 and I 2 are to be in parallel,
then I 1 by I 2 is equal to Z 2 by Z 1.
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Therefore, if I 1 and I 2 have to be in phase,
then the ratio of I 1 by I 2 will be a scalar
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and that will happen only if angle of Z 2
is equal to angle of Z 1. That is the ratio
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of R 1 by X 1 is equal to R 2 by X 2. So,
this is a desirable condition; although not
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essential. This will ensure that for a given
transformer current rating, the maximum load
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current can be supplied.
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From the previous circuit diagram we have
seen I 1 equal to Z 2 by Z 1 plus Z 2 into
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I L and I 2 bar equal to Z 1 divided by Z
1 plus Z 2 I L. Now, multiplying both sides
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of this equation with V L star; we get, where
star denotes complex conjugate equal to Z
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2 divided by Z 1 plus Z 2 into I L V L star.
Now, V L star I 1 is the KVA complex, KVA
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supplied by transformer 1; while V L star
I L is the complex KVA demanded by the load.
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Therefore, we can write S 1 equal to Z 2 divided
by Z 1 plus Z 2 into S L. Similarly, S 2 equal
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to Z 1 divided by Z 1 plus Z 2 into S L or
S 1 bar by S 2 bar equal to Z 2 by Z 1. Hence,
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if you look at the magnitude alone, S 1 by
S 2 equal to mod Z 2 divided by mod Z 1.
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Now, it is desirable that under any given
load, under any given total load; the transformers
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share the load in proportion to their individual
KVA rating, which means, for any given load
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we would want S 1 divided by S 1 rated, should
be equal to S 2 divided by S 2 rated. Therefore,
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this translates to S 1 divided by S 2 equal
to S 1 rated divided by S 2 rated. That is,
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the transformers share the KVA, total load
KVA in proportion to their individual KVA
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rating; but this we have seen to be equal
to mod Z 2 divided by mod Z 1.
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Hence, for sharing the load in proportion
to their KVA, we find that the series impedance
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of the transformer; the magnitude of the series
impedance of the transformers should be inversely
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proportional to their KVA rating. That is
Z, mod Z should be inversely proportional
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to S rated; so that is another desirable condition
for connecting transformers in parallel. That
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is the series impedance of the transformer
should be inversely proportional to their
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KVA rating. This can be stated even in a simpler
form.
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If we note that, the original relation is
S 1 rated divided by S 2 rated equal to mod
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Z 2 rated mod Z 2 divided by mod Z 1 or S
1 rated can be written as V L I 1 rated divided
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by V L I 2 rated equal to mod Z 2 divided
by mod Z 1. Or we can write I 2 rated into
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mod Z 2 divided by V L equal to I 1 rated
mod Z 1 by V L. Now, if both the transformer
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has a rated secondary voltage of V L, then
this is Z 2 per unit; p u is equal to Z 1
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p u. Therefore, in simple terms the parallel
connected transformer should have equal per
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unit series impedance on their own base. That
will ensure that the transformers will share
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the total load KVA in proportion to their
individual KVA rating. Now, suppose this condition
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is not satisfied. That is let us say, S 1
rated divided by S 2 rated is greater than
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mod Z 2 by mod Z 1; then what will happen?
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This thing can be written as into mod Z 1
by mod Z 2 larger than 1 or S 1 rated by S
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2 rated into Z 1 is greater than Z 2. In which
case, it can be easily shown that it is a
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second transformer which will share more load
and reaches rated KVA earlier. Because we
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have seen, S 2 to be Z 1 divided by Z 1 plus
Z 2 into S L.
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Now, let us say under certain load condition
S 2 has reached its rated KVA, S 2 rated.
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Therefore, S 2 rated equal to mod Z 1 divided
by mod Z 1 plus Z 2 into mod S L; let us say
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S L max. S L max is the load KVA that is reached
when S 2 reaches its rated KVA. In other words
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S L max is equal to mod Z 1 plus mod plus
Z 2 divided by mod Z 1 into S 2 rated. So,
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what is the load on the first transformer
then? S 1. S 1 is given by S 1 equal to mod
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Z 2 divided by mod Z 1 plus Z 2 into S L.
And when S 2 reaches its rated KVA, then the
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load KVA is S L max; so this should be S L
max. So, this can be written as substituting
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S L max on the previous equation, we get these
to be mod Z 2 divided by mod Z 1 into S 2
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rated.
But we have seen mod Z 2 from this relation
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we have seen mod Z 2 by mod Z 1 is less than
S 1 rated divided by S 2 rated. Therefore,
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S 1 equal to mod Z 2 by mod Z 1 S 2 rated
will be less than S 1 rated divided by S 2
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rated into S 2 rated. Or the conclusion is
S 1 in that case will be less than S 1 rated.
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Therefore, we see that if this relation is
not satisfied, then one of the transformer
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with the lower series impedance reaches its
KVA limit first and beyond which it will not
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be possible to load the parallel combination.
Therefore, the, but the other transformer
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will still have some capacity left. Therefore,
in such a situation we will not be able to
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utilize the total KVA capacity of the 2 transformers.
So, it is desirable that the per unit impedance
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of the transformers be same on their own base
the impedance angle of the 2 transformers
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also be same; so that the total KVA can be
maximized. Now, this was the analysis of parallel
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connected transformer assuming the 2 transformers
have identical no load voltage. But in practical
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situation this is not always possible; there
will be a small difference between the no
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load voltages of the 2 transformer in which
case the question arises, how do we analyze
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parallel connected transformers?
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Again, this is the first transformer; the
second transformer again is connected parallel
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to it with appropriate polarity. Now, these
2 transformer primaries are connected to the
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same source, but their induced voltages are
slightly different. That is because their
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turns ratios are not exactly same. So, if
we want to draw an equivalent circuit of this
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situation; then we must recognize that now
this impedance will not exactly be connected
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in parallel. But for the first transformer
we can still draw the equivalent circuit with
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respect to the load side. So, this will be
V 1 1 dash; that is with respect to the secondary
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of the first transformer. This is the impedance
of the first transformer series and this is
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the load impedance.
Similarly, we can draw the equivalent circuit
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of the second transformer. Now, since their
turns ratio is different, the applied source
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referred to the secondary side of the second
transformer will not be V 1 1 dash, but it
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will some other voltage V 1 2 dash; but they
will be connected on the load side. Again,
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let us say this current is I 1, this current
is I 2 and this is the total load current
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I L. So, this will be the equivalent circuit
of the parallel connected transformer, when
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their no load voltages are not identical.
The problem is to find out the individual
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currents I 1 and I 2 contributed by the transformer
when the total load current is I L. In order
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to achieve that, we can write the node voltage
equation at this point. Assuming this voltage
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to be V L, we can write V L divided by Z L
equal to V 1 1 dash minus V L by Z 1 plus
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V 1 2 dash minus V L by Z 2. Or V L Y L equal
to V 1 1 dash Y 1 plus V 1 2 dash Y 2 minus
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V L Y 1 plus Y 2. Where Y L equal to 1 by
Z L, Y 1 equal to 1 by Z 1 and Y 2 equal to
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1 by Z 2.
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Hence, we can write V L into Y L plus Y 1
plus Y 2 equal to V 1 1 dash Y 1 plus V 1
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2 dash Y 2. Or V L equal to V 1 1 dash Y 1
divided by Y L plus Y 1 plus Y 2 plus Y 2
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divided by Y L plus Y 1 plus Y 2 V 1 2 dash.
Therefore, the current I 1 equal to V 1 1
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dash minus V L Y 1; this will be given by
Y 1 into. Or I 1 equal
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to Y 1 into Y 1 plus Y L plus Y 2 into V 1
1 dash divided by minus Y 2 V 1 2 dash divided
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by Y L plus Y 1 plus Y 2. Or I 1 equal to
Y L Y 1 V 1 1 dash plus Y 1 Y 2 V 1 1 dash
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minus Y 1 divided by Y L plus Y 1 plus Y 2.
The value of I 2 can be found in a similar
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manner and once Y 1, I 1 and I 2 are found
out; the KVA supplied by them, can also be
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determined. Now, let us look at a problem.
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Let us say there are 2 single phase transformers
with voltage rating open circuit voltage both
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of them. That is they have identical no load
voltage; 11000 volts by 3.3 kV. Now, on short
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circuit test with the HV winding shorted the
first transformer 1 get the following data.
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The applied voltage is 200 volts, the circulating
current is 400 ampere
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and the dissipated power is 15 kilo watt.
For the transformer 2, again with the HV winding
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shorted; the test data is applied voltage
100 volt, circulating current 400 ampere and
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power consumed is 20 kilo watt.
Now, these 2 transformers are connected in
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parallel and they supply a load current of
I L of 750 ampere at a 0.8 p.f. lagging power
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factor. We have to find out the KVA supplied
by each of these transformers and the load
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voltage. In order to do this, let us first
find out the series impedance of the transformers.
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So, for transformer 1 mod Z 1 equal to 200
volts by 400 ampere is equal to 0.5 ohm. And
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R 1 equal to 15 into 10 to the power 3 divided
by 16 into 10 to the power 4 ohm; that is
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equal to 0.094 ohm. For transformer 2 mod
Z 2 equal to 100 volts by 400 ampere equal
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to 0.25 ohm. And R 2 equal to 20 into 10 to
the power 3 divided by 16 into 10 to the power
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4 equal to 0.125 ohm. Now, X 1 equal to square
root Z 1 square minus R 1 square equal to
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0.491 ohm; X 2 equal to square root Z 2 square
minus R 2 square equal to 0.2165 ohm.
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Therefore, we can write the angle theta 1
equal to cos inverse R 1 by Z 1; this comes
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to 79.1 degree. And theta 2 equal to cos inverse
0 point R 2 divided by Z 2; this comes to
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60 degrees. So, Z 1 equal to 0.5 ohm angle
79.1 degrees and Z 2 equal to 0.25 ohm angle
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00:43:04,750 --> 00:43:19,190
60 degrees. Please note that here the impedances
do not have the same angle. Nevertheless,
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00:43:19,190 --> 00:43:43,030
we can still proceed in a similar manner.
So, what is Z 1 plus Z 2? This comes to 0.7407
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00:43:43,030 --> 00:44:00,490
ohm angle 72.8 degrees. So, what will be the
current I 1? I 1, we have seen to be equal
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00:44:00,490 --> 00:44:18,480
to Z 2 divided by Z 1 plus Z 2 into I L; that
is equal to 750 amperes 0.8 lagging power
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00:44:18,480 --> 00:44:30,270
factor; that corresponds to minus 36.8 degree
into 25 angle 60 degrees divided by 0.7407
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00:44:30,270 --> 00:44:45,750
angle 72.8 degrees. This gives you the current
I 1 2. So, I 1 equal to 253 amperes angle
193
00:44:45,750 --> 00:45:03,640
minus 49.6 degrees.
Similarly, I 2 equal to Z 1 divided by Z 1
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00:45:03,640 --> 00:45:26,190
plus Z 2 I L equal to 750 angle minus 36.8
degrees into Z 1 is 0.5 angle 79.1 degrees
195
00:45:26,190 --> 00:45:45,360
divided by 0.7407 angle 72.8 degrees. This
gives you I 2 equal to 506 amperes angle minus
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00:45:45,360 --> 00:45:58,170
30.5 degree. Please note that the algebraic
sum of these 2 currents are not in phase;
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00:45:58,170 --> 00:46:09,450
therefore, their phasor sum instead of being
759 ampere, it is just 750 ampere. So, that
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00:46:09,450 --> 00:46:17,280
is what happens that if the angle impedance
angles of the series impedance are not same;
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00:46:17,280 --> 00:46:27,420
the total current that you can supply reduces.
Next, let us try to find out what will be
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00:46:27,420 --> 00:46:38,260
the load voltage. For that let us try to draw
the phasor diagram.
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00:46:38,260 --> 00:47:03,250
So, let us say this is the load voltage V
L, I 1 is at a larger lagging angle of 49.6
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00:47:03,250 --> 00:47:14,790
degrees, but has smaller magnitude. I 2 have
a larger magnitude, but a smaller angle. This
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00:47:14,790 --> 00:47:52,320
is I 1, this is I 2 and this is the total
current I L. So, the impedance drops like
204
00:47:52,320 --> 00:48:08,920
this and this is the applied voltage V 1 dash;
let this angle be delta. Now, we know V 1
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00:48:08,920 --> 00:48:26,000
dash magnitude is 3.3 k V; because the applied
voltage is 11 k V. Now, V 1 dash here equal
206
00:48:26,000 --> 00:48:38,660
to 3300 volts at an angle of delta. V L equal
to magnitude of the load voltage at an angle
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00:48:38,660 --> 00:48:52,030
at the 0; that is we are assume this to be
the reference phasor I 1 equal to 253 amperes
208
00:48:52,030 --> 00:49:04,680
angle minus 49.6 degrees, I 2 equal to 506
amperes angle minus 30.5 degrees and I L,
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00:49:04,680 --> 00:49:22,120
their phasor sum equal to 750 amperes angle
minus 36.8 degrees. And this phasor is equal
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00:49:22,120 --> 00:49:37,870
to I 1 Z 1 equal to I 2 Z 2.
Now, if this angle is phi 1 and this angle
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00:49:37,870 --> 00:50:30,420
is phi 2; then we can write V 1 dash equal
to V L plus Z 1 I 1 equal to V L plus Z 2
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00:50:30,420 --> 00:50:54,550
I 2. Now, this angle is the impedance angle
of Z 1; that is this is theta 1 and this angle
213
00:50:54,550 --> 00:50:57,810
is the impendence angle of Z 2; that is theta
2. And this angle is phi 1 and this angle
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00:50:57,810 --> 00:51:26,410
is phi 2. So, theta 1 minus phi 1 equal to
79.1 degrees minus 49.6 degrees equal to 29.5
215
00:51:26,410 --> 00:52:10,240
degrees and theta 2 minus phi 2, that is also
equal to 29.5 degrees. So, we can write from
216
00:52:10,240 --> 00:52:31,920
this phasor diagram 3300 cos delta minus 126.5
which is 0.5 into this, 0.5 into I 1 Z 1 I
217
00:52:31,920 --> 00:53:05,100
1 cosine of 29.5 degrees. That will give you
V L; also 3300 sine delta equal to 62.3. From
218
00:53:05,100 --> 00:53:16,140
which we will find out V L to be equal to
3190 volt.
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00:53:16,140 --> 00:53:30,140
The KVA supplied by transformer 1, will be
V L I 1 this comes
220
00:53:30,140 --> 00:53:50,470
to 807.5 KVA at a lagging power factor of
0.648 lagging power factor. And S 2 equal
221
00:53:50,470 --> 00:54:04,420
to V L I 2 this comes to 1615 KVA at 0.862
lagging power factor.
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00:54:04,420 --> 00:54:08,010
Thank you.