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In the last lecture, we have seen how to represent
a single phase transformer by approximate
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equivalent circuit either refer to the primary
or refer to the secondary. Now it would have
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been convenient if we can derive a single
equivalent circuit without being too concerned
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about whether it is referred to the primary
or referred to the secondary. There are some
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practical advantages of having equivalent
circuit like that. The way we can do it is
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by converting this equivalent circuit to per
unit equivalent circuit where each parameter
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of the equivalent circuit and each variable
are expressed in power units. How do we do
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that?
In order to get a per unit as equivalent circuit
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we have to define certain base quantities;
although the bases can be selected arbitrarily
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the customer practice is to select the rated
voltage of a given side of a transformer to
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be the base voltage. Therefore, the rated
voltage of the primary will be the base voltage
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for the primary side variables and parameters.
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So, we call it V base 1 equal to rated voltage
of the primary. Similarly V base 2 would be
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the rated voltage of the secondary. Since
in a transformer the KVA of the primary and
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the secondary are the same; therefore, there
is only a single KVA base is the rated KVA
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of the transformer. From these bases we can
define the base currents; I base 1 equal to
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KVA base divided by V base 1. Similarly, I
base 2 equal to KVA base by V base 2. Next
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step we can define a base impedance Z base
1 equal to V base 1 by I base 1, and this
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is the V base 1 square by KVA base, and Z
base 2 equal to V base 2 divided by I base
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2 equal to V base 2 square by KVA base.
Once we define these base quantities voltage
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in per unit can be written as V 1 in per unit
equal to the actual value of V 1 divided by
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V 1 base or rather V base 1. Similarly V 2
in per unit will be equal to actual value
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of V 2 divided by V base 2. I 1 in per unit
will be equal to actual value of I 1 divided
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by I base 1. Similarly I 2 in per unit would
be equal to I 2 divided by I base 2. Any impedance
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let us say r 1 in per unit would be equal
to actual value of r 1 in ohms divided by
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Z base 1, and let us say X 2 in per unit will
be actual value of X 2 in ohms divided by
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Z base 2. Now how does defining this base
quantities and per unit values head to achieve
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a single unique equivalent circuit of a transformer
let us see.
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For that let us consider the exact equivalent
circuit of the transformer refers to the primary
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side. The set of equation that describes this
equivalent circuit are V 1 equal to r 1 plus
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j X 1 into I 1 plus E 1. E 1 equal to r 2
dash plus j X 2 dash I 2 dash plus V 2 dash,
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and V 2 dash equal to I 2 dash Z L dash. Similarly
I 0 equal to E 1, these are all phasors, by
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R 0 plus E 1 by j X m. I 1 equal to I 0 plus
I 2 dash. Now let us convert them to equations
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in per unit variables and parameters. In order
to do that let us first divide the left hand
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side by the first equation by V base 1. So,
it becomes V 1 by V base 1 equal to r 1 plus
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j X 1. This also divide I 1 by I base 1, and
since also multiply by I base 1 divided by
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V base 1 plus E 1 by V base 1, but this is
V 1 per unit equal to V base 1 by I base 1
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is Z base 1 and r 1 plus j X 1 by Z base 1
is r 1 per unit plus j X 1 per unit into I
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1 per unit plus E 1 per unit.
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Similarly, by dividing the second equation
by V base 1 we get, the right hand side can
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be rewritten as, but from our definition of
the base quantities, equal to N 1 by N 2,
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and also applying this relations we have E
1 per unit equal to r 2 plus j X 2 divided
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by Z base 2; that is r 2 per unit plus j X
2 per unit multiplied by I 2 by I 2 base is
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I 2 per unit. N 2 by N 1 cancels with 1 of
N 1 by N 2 square. So, this becomes N 1 by
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N 2 into V base 2 by V base 1 which is 1.
N 1 by N 2 into V base 2 by V base 1 which
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is equal to 1. Therefore this is what we get
from this, and on this side we have N 1 by
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N 2 into V base 2 by V base 1 which is again
1, and we are left with only V 2 per unit.
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So, we see
that we arrive at the same equation in per
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unit.
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Similarly, we can apply similar thing to the
last equation V 2 dash bar by V base 2 equal
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to I 2 dash bar by I base 2 into I base 2
by V base 2 into Z L dash, but V 2 dash equal
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to
N 1 by N 2 into V 2 divided by V base 2 equal
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to I 2 dash is N 2 by N 1 into I 2 divided
by I base 2 into 1 by Z base 2 into Z L into
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N 1 by N 2 square, or N 1 by N 2 V 2 per unit
equal to Z L per unit into I 2 per unit into
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N 1 by N 2.
Cancelling N 1 by N 2 from both sides we have
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the per unit relation V 2 per unit equal to
Z L per unit into I 2 per unit. Coming to
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the fourth equation I 0 equal to E 1 by R
0 plus E 1 by j X m, one can write I 0 by
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I base 1 equal to E 1 bar by V base 1 into
V base 1 by I base 1 into 1 by R 0 plus E
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1 bar by V base 1 into V base 1 by I base
1 into 1 by j X m or I 0 per unit equal to
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E 1 per unit divided by R 0 per unit plus
E 1 per unit divided by j X m per unit.
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For the last equation I 1 equal to I 0 plus
I 2 dash. One can write I 1 bar by I base
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1 equal to I 0 bar by I base 1 plus N 2 by
N 1 I 2 divided by I base 2 into I base 2
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by I base 1, or I 1 per unit equal to I 0
per unit plus I 2 per unit, since N 2 by N
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1 into I base 2 by I base 1 is unity. Therefore,
the complete set of equations in per unit
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can be written as E 1 per unit equal to r
2 per unit plus j X 2 per unit into I 2 per
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unit plus V 2 per unit. V 2 per unit equal
to I 2 per unit into Z L per unit. I 0 per
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unit equal to E 1 per unit by R 0 per unit
plus E 1 per unit by j X m per unit. And I
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1 per unit equal to I 0 per unit plus I 2
per unit. This is a unit set of equation;
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the r 1 and r 2 expressed in per units with
respect to their base are same on both sides.
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Therefore the equivalents per unit equivalent
circuit exact equivalent circuit will be similar
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to our familiar exact equivalent circuit except
that now all parameters are to be expressed
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in per units, and all variables are also expressed
in per units. And once they are expressed
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in per units there is no need to refer these
parameters to one side or the other by any
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turns ratio. Now this is advantageous in many
situations. It may not be so apparent when
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we consider a single transformer; however,
it will be easy to understand if we have several
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such transformer. For example, in a power
system, suppose we have a force V 1 which
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supplies some transformer supplies another
transformer which may in turn supply few more
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transformers
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which may in turn supply the load.
In such a situation, if we have to refer every
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impedance to a single side in order to solve
the problem, it will be too complicated, and
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for very large power system with literally
tens of thousands of transformer it will become
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completely impossible. Under such situation
representing the parameters of the transformer
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or the load or the transmission line in per
unit system is advantageous, because it does
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not involve referring the impedances to a
single side. Therefore, for almost all practical
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calculation, the transformer parameters are
normally represented in per unit. In fact,
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just as with exact parameters the approximate
equivalent circuit of the transformer can
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also be drawn in per unit
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where R eq per unit is simply r 1 per unit
plus r 2 per unit; similarly, X eq per unit
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equal to X 1 per unit plus X 2 per unit.
Now, a transformer equivalent circuit whether
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expressed in actual parameters referred to
the primary or secondary are expressed in
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per unit parameter will be useful only if
one knows the values of the parameter. Now
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therefore, it is important to be able to determine
the transformer equivalent circuit parameters
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either from the design data or by carrying
out certain tests. It is possible to estimate
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the parameters of the transformer from the
dimensions and the design data; however, in
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most cases the user will not have access to
such design data. And also the design data
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gives a first approximation of these parameters;
in order to get more accurate values the user
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must conduct certain tests on the transformer.
There are two tests that are normally used
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to find out the equivalent circuit parameters
of the transformer. One is no load test which
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finds outs the shunt branch parameters R 0
and X m are connected across the source. So,
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these are called the shunt branch parameters.
Similarly, a short circuit test to find out
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the approximate equivalent circuit parameter
R e q and X e q which are called the series
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branch parameter; we will see how these tests
are done.
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Let us first concentrate on no load test.
In order to perform a no load test you take
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the transformer under test; keep one side
open connected to a voltmeter, a. c. voltmeter.
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On the other side you connect an ammeter,
a wattmeter, and normally supply it from what
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we call a single phase variac which can supply
at least the rated voltage of this side of
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the transformer. In order to do the no load
test the variac is started from zero voltage
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position, then the power is applied; the voltage
supplied by the variac is slowly raised till
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the voltmeter connected on the other side
reads the rated voltage of this side.
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Under that condition you take the readings
of voltage V; there is also another voltmeter
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connected here. Let us call it V1; this is
V2. You take the readings of V1. This current
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I 0 which is the no load current and the wattmeter
reading W 0. From this we will see how to
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find out the shunt branch parameters very
shortly, but before that let us be a little
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more specific about how to choose this equipment
and which side should we choose? Should we
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keep the high voltage side open and apply
voltage to the low voltage side or the other
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way round? Now theoretically it does not make
a difference. It can be used at, the supply
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can be given to either side of the transformer,
and per unit parameters determined will be
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the same.
However there are some practical limitations.
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For example, this test is to be performed
by applying rated voltage of the side; that
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is the V 1 should read the rated voltage of
the transformer on this side. So, if this
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is the H V side then one, I will need a variac
which can apply the rated high voltage across
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this transformer, and for high voltages this
may be really large in kilovolts; that is
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one problem. The meters should also be rated
at the high voltage winding rating of the
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transformer. The second problem is since the
transformer is in open circuit condition it
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draws only no load current. Now the no load
current of a well designed transformer is
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less than 5 percent of the rated full load
current. As such the rated full load current
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of the high voltage side is much lower than
that of the low voltage side.
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No load current being even only 5 percent
of the high voltage rated current its value
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will be really small, and it will be difficult
to get an accurate n of ammeter to read that,
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since as we will see the same ammeter will
also be required to read the rated current
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in a later test. So, it is usually preferred
to supply the voltage to the low voltage side
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of the transformer and keep the high voltage
side open. Therefore, no load tests are usually
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performed on the low voltage side. One should
be also careful about choosing the wattmeter.
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The wattmeter potential coil should be rated
at the rated voltage of the low voltage side
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of the transformer; however, the current flowing
through the wattmeter current coil will only
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be the no load coil, no load current. Therefore,
the current rating of the current coil of
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the wattmeter should be low; not only that
at no load, the transformer per factor is
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very poor, it is almost 0. So, one must use
a low per factor wattmeter LPF wattmeter to
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get reliable readings. Now let us see how
we use the no load test data to find out the
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shunt branch parameters.
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At no load, the transformer equivalent circuit
the approximate equivalent circuit of the
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transformer looks like this. Since the load
is open, these terminals are open. We are
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applying voltage here voltage of V 1 here;
the current drawn is I 0, and the power reading
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is W 0. So, if I draw the phasor diagram then
this is V 1. This is the no load current I
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0. Hence, the component in phase with V 1
is I c flowing through R m, and the component
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perpendicular to V 1 and lagging it by 90
degree is the current being drawn by the magnetizing
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branch I m. Hence we can write this is the
no load power factor angle phi 0.
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We can write R m equal to V 1 by I c equal
to V 1 by I 0 cos phi 0, and X m equal to
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V 1 by I m equal to V 1 by I 0 sin phi 0.
So, from the meters V 1 I read the voltage;
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from the meter ammeter I read I 0, and from
the wattmeter we read the power reading W
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0. So, what is cos phi 0? Now W 0 equal to
V 1 I 0 cos phi 0; therefore, cos phi 0 equal
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to W 0 by V 1 I 0, and sin phi 0 equal to
square root 1 minus cos square phi 0. Therefore,
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R m equal to V 1 by I 0 cos phi 0 equal to
V 1 square by W 0, and X m equal to V 1 divided
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by I 0 sin where sin phi 0 equal to square
root 1 minus W 0 square by V 1 square I 0
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square. Now, since the test is performed on
the LV side R m and X m values that are obtained
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these will be referred to LV side.
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Now, if we want to convert them into per unit
then R m per unit will be R m divided by Z
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base of the LV side, now Z base of the LV
side equal to V of the transformer LV side
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rated divided by I base of the LV side. Now
I base of the LV side equal to KVA base divided
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by LV side voltage base. Therefore Z base
LV side equal to V LV rated square divided
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by KVA rated. Hence R m p u will be equal
to R m into KVA rated KVA of the transformer
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divided by rated LV voltage square.
Similarly X m per unit would be X m into rated
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KVA of the transformer divided to a V LV square
rated. So, thus we can find out the actual
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shunt branch parameter of the transformer
referred to the LV side and convert them to
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per unit parameter with LV side base quantities.
So, of the parameters of the equivalent circuit
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of a transformer approximate equivalent circuit
parameters we have found R 0 and X m. Next
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we have to find out the parameters R e q and
X e q, and this is done by short circuit test.
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In short circuit test what we do? We take
the single phase transformer and short one
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of the sides. On the other side again we connect
an ammeter, we connect a wattmeter and supply
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this again from a variac. It can be the same
variac or some other variac. Again as before
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in the case of no load test question arises,
which side should be shorted and which side
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the supply should be applied? Now the requirement
for short circuit test is that it should be
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performed at rated current which means the
ammeter should read the rated current of the
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side it is connected. Obviously, if this is
the LV side then we will have to supply a
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very large current, because LV side rated
current is much larger than the HV side rated
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current. Hence normally the short circuit
test is performed on the HV side.
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So, this is the HV side and the LV side is
shorted. Again coming to the question of wattmeter,
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the wattmeter current coil now should be capable
of carrying the rated HV side current; however,
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the voltage required to circulate rated HV
side current when the LV side is short circuited
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is very small. It is usually between 5 to
10 percent of the rated HV side voltage, and
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that is another reason why the short circuit
test is performed on the HV side, because
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one does not need to apply the full HV voltage;
only very small fraction of the HV voltage
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is required. But the wattmeter potential coil
should be sensitive enough to give accurate
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reading at that small voltage.
At no load we have said that the power factor
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is very poor; it is close to 0. At short circuit
power factor is still poor but may not be
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as bad as that in the case of an open circuit
case. Hence a normal wattmeter can possibly
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be used, but in some very special cases this
test may also require a LPF wattmeter, but
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00:50:38,589 --> 00:50:46,760
we will assume a normal wattmeter will do.
Now how do we find out the series branch parameter
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00:50:46,760 --> 00:50:52,239
from the short circuit test? For that let
us draw the approximate equivalent circuit
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00:50:52,239 --> 00:51:09,162
under short circuit condition. Now this is
the equivalent winding resistance, and this
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00:51:09,162 --> 00:51:16,579
is the equivalent leakage resistance. Now
refer to the HV side because the supply is
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00:51:16,579 --> 00:51:27,660
applied to the HV side, and this is the magnetization
branch again referred to the HV side.
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00:51:27,660 --> 00:51:40,979
At this point let us say this is the voltage
applied V, short circuit I short circuit,
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00:51:40,979 --> 00:51:55,709
and the wattmeter reading is W short circuit.
Now even at when V s is at it is rated value
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00:51:55,709 --> 00:52:04,829
we have seen the no load current I 0 is 5
to 7 percent of the rated current; that is
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00:52:04,829 --> 00:52:13,619
I 0 if it is connected to the HV side then
even if we apply rated voltage then I 0 is
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00:52:13,619 --> 00:52:24,239
only 5 to 7 percent of I s. Therefore, when
we reduce V s in order to and short the low
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00:52:24,239 --> 00:52:32,150
voltage side we restrict I s to its rated
value. The voltage that will be required for
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00:52:32,150 --> 00:52:37,539
that we will send very negligible current
through the magnetization branch, and hence
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00:52:37,539 --> 00:52:44,569
for all practical purpose the magnetization
branch can be neglected during short circuit;
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00:52:44,569 --> 00:52:53,140
then we have very simple relation.
The short circuit impedance Z s equal to square
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00:52:53,140 --> 00:53:05,619
root of R e q square plus X e q square. This
is given by V short circuit divided by I short
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00:53:05,619 --> 00:53:20,049
circuit. The power loss during short circuit
is assumed to be entirely due to the series
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00:53:20,049 --> 00:53:33,469
resistance; therefore, W s equal to I s square
R e q. Hence R e q equal to W s divided by
205
00:53:33,469 --> 00:53:49,190
I s square. Hence X e q equal to square root
Z s square minus R e q square. Just as in
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00:53:49,190 --> 00:53:56,859
this R e q and X e q are now found referred
to the HV side. Please remember the shunt
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00:53:56,859 --> 00:54:05,609
branch parameters R m and j X m were found
referred to the LV side. It is possible to
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00:54:05,609 --> 00:54:13,449
either refer R m X m to the HV side or R e
q and X e q to the LV side by the appropriate
209
00:54:13,449 --> 00:54:21,269
turns ratio and arrive at a complete approximate
equivalent circuit of the transformer referred
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00:54:21,269 --> 00:54:27,910
to either HV side or LV side.
However as I have mentioned earlier it is
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00:54:27,910 --> 00:54:35,630
customary to represent this parameter in per
unit and R e q, for example, can be represented
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00:54:35,630 --> 00:54:46,670
in per unit as R e q that we have got from
the test into KVA rated of the transformer
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00:54:46,670 --> 00:55:00,089
divided by V HV square now rated HV voltage
square. Similarly, X e q in per unit will
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00:55:00,089 --> 00:55:10,039
be the value obtained from the test into KVA
rated of the transformer divided by rated
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00:55:10,039 --> 00:55:16,160
H V voltage square of the transformer.
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00:55:16,160 --> 00:55:22,949
Hence the approximate equivalent circuit of
the transformer in per unit will look like.