500 MAGNETIC CIRCUITS AND TRANSFORMERS
kVA, 2400:480 V, following the notation of Fig-
ure 11.3.4, the equivalent circuit impedances in
ohms areR 1 = 0 .06,R 2 = 0 .003,RC=2000,
andX 1 = 0 .3,X 2 = 0 .012,Xm=500. The
load connected across the low-voltage terminals
draws rated current at 0.8 lagging power factor
with rated voltage at the terminals.
(a) Calculate the high-voltage winding terminal
voltage, current, and power factor.
(b) Determine the transformer series equivalent
impedance for the high-voltage and low-
voltage sides, neglecting the exciting current
of the transformer.
(c) Considering the T-equivalent circuit, find the
Thévenin equivalent impedance of the trans-
former under load as seen from the primary
high-voltage terminals.
11.3.6A 20-kVA, 2200:220-V, 60-Hz, single-phase
transformer has these parameters:
Resistance of the 2200-V windingR 1 = 2. 50
Resistance of the 220-V windingR 2 = 0. 03
Leakage reactance of the 2200-V windingX 1 =
0. 1
Leakage reactance of the 220-V windingX 2 =
0. 1
Magnetizing reactance on the 2200-V sideXm=
25,000
(a) Draw the equivalent circuits of the trans-
former referred to the high-voltage and low-
voltage sides. Label impedances numerically
in ohms.
(b) The transformer is supplying 15 kVA at 220
V and a lagging power factor of 0.85. Deter-
mine the required voltage at the high-voltage
terminals of the transformer.
11.4.1These data were obtained from tests carried out
on a 10-kVA, 2300:230-V, 60-Hz distribution
transformer:- Open-circuit test, with low-voltage winding ex-
cited: applied voltage 230 V, current 0.45 A,
input power 70 W - Short-circuit test, with high-voltage winding
excited: applied voltage 120 V, current 4.35 A,
input power 224 W
(a) Compute the efficiency of the transformer
when it is delivering full load at 230 V and
0.85 power factor lagging.(b) Find the efficiency of the transformer when
it is delivering 7.5 kVA at 230 V and 0.85
power factor lagging.
(c) Determine the fraction of rating at which the
transformer efficiency is a maximum, and
calculate the efficiency corresponding to that
load if the transformer is delivering the load
at 230 V and a power factor of 0.85.
(d) The transformer is operating at a constant
load power factor of 0.85 on this load cycle:
0.85 full load for 8 hours, 0.60 full load for 12
hours, and no load for 4 hours. Compute the
transformer’s all-day (or energy) efficiency.
(e) If the transformer is supplying full load at
230 V and 0.8 lagging power factor, de-
termine the voltage regulation of the trans-
former. Also, find the power factor at the
high-voltage terminals.
*11.4.2A 3-kVA, 220:110-V, 60-Hz, single-phase trans-
former yields these test data:- Open-circuit test: 200 V, 1.4 A, 50 W
- Short-circuit test: 4.5 V, 13.64 A, 30 W
Determine the efficiency when the transformer
delivers a load of 2 kVA at 0.85 power factor
lagging.
11.4.3A 75-kVA, 230/115-V, 60-Hz transformer was
tested with these results: - Open-circuit test: 115 V, 16.3 A, 750 W
- Short-circuit test: 9.5 V, 326 A, 1200 W
Determine:
(a) The equivalent impedance in high-voltage
terms.
(b) The voltage regulation at rated load, 0.8
power factor lagging.
(c) The efficiency at rated load, 0.8 power factor
lagging, and at half-load, unity power factor.
(d) The maximum efficiency and the current at
which it occurs.
11.4.4A 300-kVA transformer has a core loss of 1.5 kW
and a full-load copper loss of 4.5 kW.
(a) Calculate its efficiency corresponding to 25,
50, 75, 100, and 125% loads at unity power
factor.
(b) Repeat the efficiency calculations for the
25% load at power factors of 0.8 and 0.6.