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PowerPoint Slides
to accompany
Electric Machinery
Sixth Edition
A.E. Fitzgerald
Charles Kingsley, Jr.
Stephen D. Umans
Chapter 2
Transformers
2-0
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Transformer with open secondary.
Figure 2.4
2-1
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No-load phasor diagram.
Figure 2.5
2-2
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Ideal transformer and load.
Figure 2.6
2-3
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Three circuits which are identical at
terminals ab when the transformer
is ideal.
Figure 2.7
2-4
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Equivalent circuits for Example 2.2 (a) Impedance
in series with the secondary. (b) Impedance referred
to the primary.
Figure 2.8
2-5
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Schematic view of mutual
and leakage fluxes in a
transformer.
Figure 2.9
2-6
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Steps in the
development
of the
transformer
equivalent
circuit.
Figure 2.10
2-7
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Equivalent circuits for transformer of Example 2.3
referred to (a) the high-voltage side and (b) the
low-voltage side.
Figure 2.11
2-8
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Approximate transformer equivalent circuits.
Figure 2.12
2-9
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Cantilever equivalent circuit for Example 2.4.
Figure 2.13
2-10
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(a) Equivalent circuit and (b) phasor diagram
for Example 2.5.
Figure 2.14
2-11
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Equivalent circuit with short-circuited secondary.
(a) Complete equivalent circuit. (b) Cantilever equivalent
circuit with the exciting branch at the transformer
secondary.
Figure 2.15
2-12
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Equivalent circuit with open-circuited secondary.
(a) Complete equivalent circuit. (b) Cantilever equivalent
circuit with the exciting branch at the transformer primary.
Figure 2.16
2-13
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(a) Two-winding transformer. (b) Connection
as an autotransformer.
Figure 2.17
2-14
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(a) Autotransformer connection for Example 2.7.
(b) Currents under rated load.
Figure 2.18
2-15
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Common three-phase transformer connections; the
transformer windings are indicated by the heavy lines.
Figure 2.19
2-16
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2.8 VOLTAGE AND CURRENT TRANSFORMERS
Used in instrumentation applications
765 000 Volt, 10 000 Ampers can not be measured directly
Most instruments range: Voltage (Potantial) Transformers (PT)
0-120 V rms; Current Transformers (CT) 0-5 A rms
Equivalent circuit for an instrumentation transformer. Figure 2.21
Rc (core loss resistance) neglected in equivalent circuit.
Zb is referred to as the BURDEN on that transformer.
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FOR PT:
Z eq Z b
Vˆ2  N 2 



ˆ
V1  N1  ( R1  j X 1 )( Z eq  Z b  R2  j X 2 )
Z eq 
j X m ( R1  j X 1 )
R1  j ( X m  X 1 )
Zb  ( N1 / N2 )2 Zb
FOR CT:
j Xm
Iˆ2  N1 

 
Iˆ1  N 2  Zb  R2  j ( X 2  X m )
2-18
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Example 2.10: A 2400:120 V, 60 Hz potential transformer has the
following parameter values (referred to the 2400 V winding
side):
X1 143 
X 2  164 
R1 128 
X m  163 k
R2  141 
a) Assuming a 2400 V input, which ideally should produce a
voltage of 120 V at the low voltage winding, calculate the
magnitude and relative phase-angle errors of the secondary
voltage if the secondary winding is open-circuited.
b) Assuming the burden impedance to be purely resistive
(Zb=Rb), calculate the minimum resistance (maximum burden)
that can be applied to the secondary such that the magnitude
error is less than 0.5 percent.
c) Repeat part (b) but find the minimum resistance such that the
phase-angle error is less than 1 degree.
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2.9
THE PER-UNIT SYSTEM
Computations relating to machines, transformers, and systems
of machines are often carried out in per-unit form.
Quantities are expressed as ratios to chosen BASE values.
V, I, P, Q, S, R, X, Z, G (conductance), B (susceptance), Y
can be translated.
Quantity in per  unit 
Single Phase:
Sbase  Vbase I base
Base Change:
Z pu2
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Actual quantity
Base value of quantity
Zbase  Vbase / I base
2
Vbase
1 Sbase2
 Z pu1 2
Vbase2 Sbase1
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Example 2.12: The equivalent circuit for a 100 MVA, 7.97 kV:79.7
kV transformer is shown in Fig. 2.22a. Convert the equivalent
circuit parameters to per-unit using the transformer rating as
base.
7.97 kV:79.7 kV
X L  0.04 
X H  3.75 
X m  114 
RL  0.76 m
RH  0.085 
2-21
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Example 2.8: Three single-phase, 50 kV 2400:240 V transformers,
each identical with an impedance of 1.42+j1.82 Ohm referred to
high voltage side is connected Wye-Delta in a three-phase 150
kVA bank to step down the voltage at the load end of a feeder
whose impedance is 0.15+j1 Ohm/phase. The voltage at the
sending end of the feeder is 4160 V line-to-line. On their
secondary sides, the transformer supply a balanced threephase load through a feeder whos impedance is 0.0005+j
0.0020 Ohm/phase. Find the line-to-line voltage at the load
when the load draws rated current from the transformer at a
power factor of 0.8 lagging.
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