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In the last lecture, we have seen how to obtain
the parameters of the approximate equivalent
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circuit of the single phase transformer; the
approximate equivalent circuit looks like
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this. This is
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the supply voltage V 1, magnetizing current
I 0, supply current I 1, R e q, j X e q is
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the referred load current I 2 dash, and this
is the referred load voltage V 2 dash. While
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this equivalent circuit is important for finding
out different performance parameters from
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a user point of view certain performance parameters
are very important. For example, one of the
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most important performance parameter is the
efficiency of transformer. Now efficiency
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as in the case of any other power processing
equipment for a transformer the efficiency
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is defined as the symbol is nu is defined
as output power
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divided by the input power.
Now what is the difference between output
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power and input power? It is the losses. So,
this can also be written as output power divided
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by output power plus losses. Now what are
the different losses that can occur in a single
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phase transformer? We have seen the first
element is the core loss resistance. This
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is the loss due to hysteresis and eddy current,
and in the equivalent circuit we represent
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it by a resistance. And the second major loss
that can occur is in the winding resistances
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R e q; therefore, the losses can be broken
down into two parts, plus core losses plus
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losses in the winding resistance, this is
called the copper loss.
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Now even from the equivalent circuit also
from the physical principle due to which these
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losses takes place we can argue that this
core loss depend on the applied voltage and
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frequency. But single phase transformers are
normally connected to voltage sources of constant
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magnitude and frequency; therefore, the core
loss occurring in a single phase transformer
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is more or less constant. It does not depend
on the load; however, the losses that are
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occurring in the winding resistance they do
depend on the load current magnitude and it
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given by I square Req.
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Therefore, in terms of equations we can write
output power is equal to V 2 dash I 2 dash
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cos phi where phi is the load power factor
angle. This is same as V 2 I 2 cos phi. Core
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loss is given by V 1 square by R 0; we indicate
it as P core or P iron, and this is usually
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constant. Copper loss
denoted as P Cu equal to I 2 dash square R
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e q or this I 2 square R e q dash same thing
I 2 square R e q dash. Therefore, efficiency
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nu can be written as which again can be written
as. In a general transformer the variation
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of V 2 dash with I 2 dash is usually very
small.
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So, for all practical purpose V 2 dash can
be assumed to be to remain constant at its
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rated value. Further more if we assume that
the load power factor also is constant, then
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we can say that the efficiency depends only
on the magnitude of the current I 2 dash.
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And it will be interesting to find out at
what value of I 2 dash this efficiency becomes
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maximum? Now if you look at the expression
of the efficiency we see that this V 2 dash
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cos phi divided by V 2 dash cos phi plus P
i by I 2 dash plus I 2 dash R e q of which
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the numerator is constant, the first term
of the denominator is constant; only these
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two terms are functions of I 2 dash.
For nu to be maximum the denominator should
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be minimum. Now up the denominator the first
term is constant. So, the sum of the second
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and third term should be minimum. So, differentiating
this quantity with respect to I 2 dash we
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get
the condition for maximum nu max is P i equal
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to I 2 dash square R e q; that is P i equal
to P cu. The load current at which the fixed
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iron loss becomes equal to the core loss is
the load current at which the efficiency of
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a single phase transformer is maximum. Now
while it is possible to if the load parameters
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are given, it is possible to find out I 2
dash and the efficiency under any given load
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condition. It is the practice usually is to
find out these values from the test data.
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The other day we have seen how to find this
loss component.
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We have seen that in the open circuit test;
we connect a wattmeter. The wattmeter reading
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is equal to the core loss, because there is
no current almost no current flowing in the
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transformer. The no load current is very small.
Similarly, on the short circuit test data
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we argued that the magnetizing current is
very small; therefore, the core loss is very
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small. So, the wattmeter reading actually
represented the copper loss. Now short circuit
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test is usually done at the rated current;
therefore, the value we get is the full load
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copper loss from short circuit test. Similarly
from the open circuit test we get the iron
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loss which is independent of the load current
and constant.
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So, let us say from the no load test data
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we get the wattmeter reading to be P i; short
circuit test data
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we get the reading to be P cu. Then from this
data we can find out the load at which the
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transformer efficiency will be maximum. The
load the KVA load on a transformer is equal
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to V 2 I 2 at any load current I 2; the corresponding
copper loss
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equal to I 2 square R eq dash. The full load
copper loss as obtained for circuit test P
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cu, let us say this is at x fraction of load;
P cu at full load is equal to I 2 full load
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square R eq dash. So, P cu x by P cu equal
to I 2 divided by I 2 full load square. This
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is same as V 2 I 2 divided by V 2 I 2 full
load square which is same as the fractional
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load x square.
Therefore, the efficiency at any fractional
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load x can be written as efficiency at a fractional
load x equal to x times the KVA rating of
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the transformer into cos phi divided by x
times rated KVA to cos phi plus P i plus x
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square P cu where P cu is the full load copper
loss of the transformer. And we have seen
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that the fractional load at which the efficiency
will be maximum then P i is equal to x square
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P cu or x equal to
square root P i by P cu. So, this is the fractional
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load at which the transformer efficiency will
be maximum. Now this realization has something
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to do with the design of a transformer. It
is natural that we would like the transformer
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to be as efficient as possible, because that
will reduce the energy loss which means that
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transformer should ideally be always operating
at the load at which its efficiency is maximum.
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However, it is not possible to guarantee that
for all types of transformers; for example,
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the transformers that are used in transmission
are almost always loaded up to their full
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capacity. Therefore, while designing a transmission
transformer we like to make sure that its
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maximum efficiency occurs near the rated load;
however, the situation is very different for
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distribution transformer. Distribution transformer
for a large part of the day remains lightly
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loaded or even remains on no load. Therefore
for a distribution transformer it is important
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to design the transformers such that the maximum
efficiency does not occur at full load but
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at a reduced load.
Usually for a distribution transformer it
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is not unusual to have the loading corresponding
to the maximum efficiency at around 60 to
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65 percent. From this discussion we also can
conclude one more thing that getting the transformer
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efficiency at a particular load does not give
the full picture, because the loading of a
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transformer can vary over a day. Normally
it is expected that the loading pattern of
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a transformer will repeat over every day in
a season at least.
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Therefore more useful parameter in order to
find out the performance of a transformer
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is called the all day efficiency; how do I
define all day efficiency? This is defined
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as energy output of the transformer over a
day divided by energy input to the transformer
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over the day. Now had the transformer been
uniformly loaded over the day then this all
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day efficiency would have been same as the
efficiency of the transformer at a given load;
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however, it is not so. Let us say the transformer
is loaded, say 25 percent of the time at let
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us say 50 percent of the load time load that
is x. Then 50 percent of the time let us say
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is loaded at 80 percent, and let us say 10
percent of the time it is loaded at 100 percent,
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and for the rest of the 15 percent of the
time it remains at no load.
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Now if this transformer has a copper loss
of P cu at full load, and P i is the core
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loss, then the losses will be copper losses
particularly
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will be 0.5 square P cu, here it will be 0.8
square P cu, and here it will be P cu, and
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here it will be 0. The core loses, however,
will always be P i, P i, P i, here also P
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i. Let us further assume that these loads
have power factor let us say at 50 percent,
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let us say this power factor is 0.8 lagging,
this power factor is 0.9 lagging, this power
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factor is 0.95 lagging, and here since there
is no current we do not mention the load power
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factor. So, let us calculate, what will be
the all day efficiency?
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The energy output over a day let us say the
transformer kVA rating is kVA rated then the
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energy output over a day is equal to 50 percent
0.5 into 25 percent of a day that is 6 hours
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one-forth is 6 hours which remains loaded
at 50 percent. So, 0.5 into KVA rated into
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power factor is 0.8 into 6 hours, so many
kilowatt hour at 25 percent of the loading
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plus 85 percent of the loading into KVA rated
into power factor it is 0.9 into 12 hours
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plus 100 percent of the loading that is 1
or KVA rated into 0.95 is the power factor
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into 10 percent of the time that is 2.4 hours
plus 0 percent loading that is no output for
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another 3.6 hours.
So, this is the total energy output of the
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transformer. What is the total energy input?
E in equal to E out plus E loss; how do I
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calculate the losses? Losses we can find since
we know that the core loss remain constant
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at any loading, and the copper loss is given
by these values. So, total energy loss is
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equal to E loss equal to 24 hours into P i
plus 6 hours at 50 percent of the load. So,
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0.25 into P cu into 6 hours plus 50 percent
of the time at 80 percent of the load; so
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0.64 into P cu into 12 hours plus rate in
full load for 10 percent of the time that
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is P cu into 2.4 hours. So, this is the total
energy loss. So, nu all day
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equal to E out divided by E out plus E loss.
This is how one should calculate the all day
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efficiency of a single phase transformer.
And that the only data we need to calculate
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this all day efficiency are the copper loss
at full load which can be obtained from short
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circuit test data and the core loss which
can be obtained from open circuit data and
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of course, the loading pattern of the transformer.
If this data are given then the all day efficiency
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of a transformer can be found. Now one of
the effect of these losses is to generate
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heat and increase the temperature of the transformer.
We have found out how from open circuit and
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short circuit test we can find out the core
loss and the copper loss of a transformer;
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however, during open circuit test there is
no load current going. So, the transformer
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is subjected to only core loss. Similarly
in the short circuit test the core loss is
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negligible; the transformer is subjected to
only copper loss.
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So, in none of these tests this transformer
usually sees both the loss occurring simultaneously;
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therefore, it is not possible to find out
experimentally the temperature rise that will
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occur in a transformer when it is actually
used. However, this test is very important
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from user point of view because during use
the transformer will be supplying load. Hence
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it will be subjected to core loss and copper
loss simultaneously, and as a result the final
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statistic temperature rise will be determined
by the total loss. Now it is possible to find
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out the temperature rise of a single phase
transformer by test but not from no load test
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or a short circuit test. The test by which
we can find out the temperature rise of a
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single phase transformer is called the back-to-back
test or Sumpnerâ€™s test.
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As we will see shortly this test cannot be
performed with a single transformer; we need
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at least two transformers, and these two transformers
should preferably be identical at least their
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voltages should be very close or identical.
The way it is done is the two identical transformers
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are taken. Their primary are connected in
parallel, and it is supplied from a rated
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voltage source, Of course, some meters are
to be connected means we have to connect an
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ammeter; we have to also connect a wattmeter.
The secondaries of the transformers are connected
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in series with this polarity. Here again another
ammeter is connected, another wattmeter is
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connected, and these secondaries are supplied
from a variable voltage source.
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So, how do you subject the transformers to
simultaneous core loss and copper loss just
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like this. Let us assume that let us apply
superposition theory; let us assume one source
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at a time. So, you replace this by a short
circuit. Since these two transformers are
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identical the voltage generated across the
secondary winding will be same provided since
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they are supplied from the same source; therefore,
there will be no circulating current flowing
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in the secondaries, because the induced voltage
in one will oppose the other. The current
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drawn by these transformers will be only the
no load current of these two transformers.
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Similarly, if we short the source V 1 and
consider the contribution of the source V
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2 we see that the current flowing in the secondary
windings we will see to short at primary couple
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to them.
Therefore, the current flowing through the
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transformer can be controlled by controlling
this voltage cells to be the rated current
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of these two transformers, and the power drawn
will be the copper loss of the two transformers.
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Therefore, when source V 1 is present, V 2
is shorted; the power drawn is simply the
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core loss of the two transformers. Now V 1
is shorted and V 2 is applied, then the power
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loss power drawn is the copper loss of the
two transformers. When both are applied let
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us say if current is circulated I 2 current
circulates this way; therefore, the reflection
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current here it will be I 2 dash flowing in
this direction, and similarly I 2 dash flowing
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in this direction. However, if these two transformers
are identical then this I 2 dashed current
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will circulate only in the transformer primary
winding the current drawn from the source
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we still be 2I 0.
Therefore during this test the total power
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drawn from the two sources V 1 and V 2 will
be the sum of the core loss and the copper
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loss and since the transformers will be subjected
simultaneously to core loss and copper loss
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their temperature rise will be same as that
when they will be in actual use. So, in this
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case we see that we can simulate actual loading
of the transformers without drawing the required
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power from the source; that is why it is called
a back-to-back test, and this is a general
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principle. In many electrical machines we
will see such equivalent circuits for back-to-back
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tests, because it is important for any electrical
equipment to be tested at its rated capacity.
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However, it is also inconvenient particularly
the equipment is of large rating to actually
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load the equipment and test it. Therefore
for almost all electrical machines some form
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of back to back test becomes important, and
this is the way the back to back test is performed
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on a single phase transformer. We will end
this lecture by looking at an exercise of
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how to find out the equivalent circuit parameters
of a single phase transformer from test data
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and how to find the efficiency.
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Let us say we have a 200 volts by 400 volt
transformer single phase transformer; KVA
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rating is 4 KVA. Let us say the no load test
data of this transformer is the voltage rating
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is no load test is of course done on the LV
side. The test data is voltage reading is
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200 volt since rated voltage is to be applied;
the current drawn is 0.7 ampere, and the wattmeter
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reading is 35 watts. The short circuit test
data this is done on the HV side; the applied
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voltage is 30 volts. The circulating current
is 10 ampere, and the wattmeter reading is
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90 watts. How do I find out the equivalent
circuit parameters of this transformer? Then
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at no load the approximate equivalent circuit
of the transformer looks like this.
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This voltage applied is 200 volts. This is
j X m, this is R 0, this is the no load current
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I 0, this is the magnetization current I m,
and this is the core loss current I c. The
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phasor relationship is. So, if this is
the applied voltage V LV then I m will lag
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00:39:45,640 --> 00:40:04,230
V LV by 90 degree, and I c will be in phase
with V LV; therefore, the total no load current
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00:40:04,230 --> 00:40:13,880
I 0 will be at a power factor angle phi 0
respect to the applied voltage. Now from the
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00:40:13,880 --> 00:40:24,470
no load test data these voltage is given to
be 200 volt, and the no load current I 0 equal
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00:40:24,470 --> 00:40:50,600
to 0.7 amperes. Therefore, the no load power
factor cos phi 0 equal to the V LV I 0 W 0
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00:40:50,600 --> 00:41:09,390
divided by V LV I 0; this is equal to 35 divided
by 200 into 0.7. Let us say 0.25, but what
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00:41:09,390 --> 00:41:38,820
is I m? I m equal to V LV by X m, but this
is also equal to I 0 sin phi 0, and I c equal
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00:41:38,820 --> 00:41:52,680
to V LV by R 0 equal to I 0 cos phi 0.
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00:41:52,680 --> 00:42:18,880
From the given data therefore, X m equal to
V LV divided by I 0 sin phi 0. This comes
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00:42:18,880 --> 00:42:50,290
as 200 and 95.08 ohm, and R 0 equal to V LV
by I 0 cos phi 0. This comes to 1.143 kilo
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00:42:50,290 --> 00:43:00,300
ohms. It should be remembered that both that
both these values are referred to LV side,
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since that is where the test was performed.
As we have seen that there are some advantages
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in expressing them in per unit; for that we
will have to find out what is Z base on the
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00:43:21,600 --> 00:43:48,180
LV side. So, Z base equal to V LV rated square
divided by rated KVA of the transformer. Hence
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00:43:48,180 --> 00:44:18,270
R 0 p. u equal to R 0 into rated KVA of the
transformer divided by V LV rated voltage
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00:44:18,270 --> 00:44:45,480
square. This comes to 114.3. Similarly by
the same formula X m p. u equal to X m into
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00:44:45,480 --> 00:45:08,280
KVA base rated KVA divided by; this comes
to 29.508. So, we have seen how to find out
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00:45:08,280 --> 00:45:14,970
the shunt parameter branches of the equivalent
circuit and express them in per unit. Let
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00:45:14,970 --> 00:45:19,300
us now shift our attention to the series parameters.
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00:45:19,300 --> 00:45:30,610
In the short circuit test we can neglect the
shunt branch
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00:45:30,610 --> 00:45:38,900
the equivalent circuit is somewhat like this.
The applied voltage is V sc, the current is
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00:45:38,900 --> 00:45:55,040
I sc, this resistance is R eq, this is j X
eq. Hence Z sc that is mod Z sc equal to V
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00:45:55,040 --> 00:46:08,750
sc by I sc; from the given data this is 30
volts by 10 amperes equal to 3 ohms. Therefore,
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00:46:08,750 --> 00:46:22,261
Z sc equal to square root R eq square plus
X eq square equal to 3 ohm. Similarly the
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00:46:22,261 --> 00:46:32,210
power P sc short circuit power equal to I
sc square R eq; this is given to be 90 watts.
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00:46:32,210 --> 00:46:53,810
I sc equal to 10 ampere; hence, R eq equal
to 0.9 ohm, X eq then will be Z sc square
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00:46:53,810 --> 00:47:08,900
minus R eq square; this comes to 2.862 ohms.
Now both the values of R eq and X eq are found
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00:47:08,900 --> 00:47:16,410
referred to the HV side, since short circuit
test is done on the HV side; again this can
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00:47:16,410 --> 00:47:29,430
also be converted to per unit. So, R eq per
unit will be equal to R eq into rated KVA
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00:47:29,430 --> 00:47:49,730
of the transformer divided by now HV side
rated voltage square. This comes to 0.0225
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00:47:49,730 --> 00:48:14,450
and X eq per unit will be equal to X eq into
KVA rated divided by V HV square rated. This
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00:48:14,450 --> 00:48:21,210
will give you 0.0716.
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00:48:21,210 --> 00:49:02,440
Therefore, the complete equivalent circuit
is per unit of this transformer will be. This
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00:49:02,440 --> 00:49:16,119
is X n p.u, this is R 0 p.u, this is R eq
p.u; this is j X eq p.u. Here you will get
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00:49:16,119 --> 00:49:32,900
V 2 in per unit, here you get I 2 in per unit,
here you will get I 1 in per unit, this is
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00:49:32,900 --> 00:49:43,270
I 0 in per unit. So, this is how you get the
equivalent circuit parameters of a single
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00:49:43,270 --> 00:49:52,560
phase transformer from short circuit and open
circuit test data. Now if I want to find out
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00:49:52,560 --> 00:49:58,810
for the same transformer what will be the
efficiency, let us say at 50 percent of the
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00:49:58,810 --> 00:50:04,740
load? First let us find out what will be the
efficiency at 100 percent of the load? At
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00:50:04,740 --> 00:50:15,420
100 percent of the transformer KVA rating
is 4 KVA. So, at 100 percent of the load,
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00:50:15,420 --> 00:50:29,360
and let us say at 0.8 power factor the output
power is 4 into 10 to the power 3 watts into
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00:50:29,360 --> 00:50:42,310
power factor is 0.8 divided by the output
power is 4 into 10 to the power 3 into 0.8
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00:50:42,310 --> 00:50:50,960
plus the copper loss; the core loss is 35
watts plus 100 percent; that is at full load
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00:50:50,960 --> 00:51:04,410
the copper loss is 90 watts.
So, at full load the efficiency is 3200 divided
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00:51:04,410 --> 00:51:28,930
by 3200 plus 125, 32 divided by 33.25. This
will be the full load efficiency nu full load.
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00:51:28,930 --> 00:51:39,840
If I reduce it to half load then nu half load
will be, at half load the power output will
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00:51:39,840 --> 00:51:48,260
be 0.5 and at the same power factor 4 into
10 to the power 3 into 0.8 divided by 0.5
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00:51:48,260 --> 00:52:07,500
into 4 into 10 to the power 3 into 0.8 plus
35 plus 0.5 square into 90. This will be the
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00:52:07,500 --> 00:52:14,360
efficiency at half load. It will be interesting
to find out at what load the efficiency of
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00:52:14,360 --> 00:52:17,840
the transformer will be maximum; let that
load be x.
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00:52:17,840 --> 00:52:23,850
Then we know from the condition of maximum
efficiency at the loading at which efficiency
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00:52:23,850 --> 00:52:33,720
is maximum; there P i equal to x square P
cu where x is the fractional load. From the
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00:52:33,720 --> 00:52:47,040
short circuit test data P cu at full load
equal to 90 watts, and P i equal to 35 watts.
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00:52:47,040 --> 00:52:56,710
So, at what load the efficiency will be maximum?
The maximum efficiency will occur at x equal
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00:52:56,710 --> 00:53:12,590
to P i by P cu square root; this is equal
to 35 by 90 square root. So, at this load
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00:53:12,590 --> 00:53:19,380
the efficiency will be maximum. It is to be
noted that although the actual value of this
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00:53:19,380 --> 00:53:26,080
efficiency is dependent on the power factor
the load at which efficiency over the maximum
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00:53:26,080 --> 00:53:31,940
efficiency occurs is independent of the power
factor. So, the plot of efficiency versus
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00:53:31,940 --> 00:53:45,020
percentage load
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00:53:45,020 --> 00:53:56,900
will look somewhat like this, at 0 it is 0,
but at a different power factor the efficiency
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00:53:56,900 --> 00:54:09,660
curves will pick at the same point. So, these
are for decreasing cos phi.