[James_H._Harlow]_Electric_Power_Transformer_Engin(BookSee.org)

FIGURE 2.2.36 Complete transformer equivalent circuit. (By permission of ABB Inc., Jefferson City, MO.)
2.2.10.2 Complete Equivalent Circuit
Figure 2.2.36 shows a complete equivalent circuit of the transformer. An ideal transformer is inserted to
represent the current- and voltage-transformation ratios. A parallel resistance and inductance representing
the magnetizing impedance are placed across the primary of the ideal transformer. Resistance and
inductance of the two windings are placed in the H 1 and X 1 legs, respectively.
2.2.10.3 Simplified Model
To create a simplified model, the magnetizing impedance has been removed, acknowledging that noload
loss is still generated and magnetizing current still flows, but it is so small that it can be ignored
when compared with the rated currents. The R and X values in either winding can be translated to the
other side by using percent values or by converting ohmic values with a factor equal to the turns ratio
squared (N P /N S ) 2 . To convert losses or ohmic values of R and X to percent, use Equation 2.2.1 or Equation
2.2.2:
Load Loss ( R)
kVA
% R
2
10 kVA kV
(2.2.1)
AW ( L)
kVA
% X
(2.2.2)
2
10 kVA kV
where AW is apparent watts, or the scalar product of applied voltage and exciting current in units of
amperes. Once the resistances and inductances are translated to the same side of the transformer, the
ideal transformer can be eliminated and the percent values of R and X combined. The result is the simple
model shown in Figure 2.2.37. A load, having power factor cos may be present at the secondary.
2.2.10.4 Impedance
The values of %R and %X form the legs of what is known as the “impedance triangle.” The hypotenuse
of the triangle is called the transformer’s impedance and can be calculated using Equation 2.2.3.
% Z % R %
X
2 2
(2.2.3)
A transformer’s impedance is sometimes called “impedance volts” because it can be measured by
shorting the secondary terminals and applying sufficient voltage to the primary so that rated current
flows in each winding. The ratio of applied voltage to rated voltage, times 100, is equal to the percent
impedance.
FIGURE 2.2.37 Simplified transformer model. (By permission of ABB Inc., Jefferson City, MO.)
2.2.10.5 Short-Circuit Current
If the load (right) side of the model of Figure 2.2.37 is shorted and rated voltage from an infinite source
is applied to the left side, the current I SC will be limited only by the transformer impedance:
I 100 I % Z
SC
(2.2.4)
For example, if the rated current, I R , is 100 A and the impedance is 2.0%, the short-circuit current
will be 100 100/2 = 5000 A.
2.2.10.6 Percent Regulation
When a transformer is energized with no load, the secondary voltage will be exactly the primary voltage
divided by the turns ratio (N P /N S ). When the transformer is loaded, the secondary voltage will be
diminished by an amount determined by the transformer impedance and the power factor of the load.
This change in voltage is called regulation and is actually defined as the rise in voltage when the load is
removed. One result of the definition of regulation is that it is always a positive number. The primary
voltage is assumed to be held constant at the rated value during this process. The exact calculation of
percent regulation is given in Equation 2.2.5:
05 .
2 2 2
% reg L % R % X 200 L % X sin% R cos10000 – 100
(2.2.5)
where cos is the power factor of the load and L is per unit load on the transformer. The most significant
portion of this equation is the cross products, and since %X predominates over %R in the transformer
impedance and cos predominates over sin for most loads, the percent regulation is usually less than
the impedance (at L = 1). When the power factor of the load is unity, then sin is zero and regulation
is much less than the transformer impedance. A much simpler form of the regulation calculation is given
in Equation 2.2.6. For typical values, the result is the same as the exact calculation out to the fourth
significant digit or so.
% X* cos – % R*sin
% reg L % R * cos % X *sin
2
200
(2.2.6)
2.2.10.7 Percent Efficiency
As with any other energy conversion device, the efficiency of a transformer is the ratio of energy delivered
to the load divided by the total energy drawn from the source. Percent efficiency is expressed as:
5
L kVAcos
10
% Efficiency
3 2
L kVAcos
10
NL L LL
R
(2.2.7)
© 2004 by CRC Press LLC
© 2004 by CRC Press LLC