The shifts in phase of positive- and negative-sequence currents and
voltages in passing through a Y-zigzag transformer bank are the same
as those through a Y-A bank. The Y-zigzag bank will therefore
operate in parallel with a Y-A bank.
— VA_ ~IA aflMQflo. h- VB — .-vc_ oj b, C| C2 CJ TOT) •Vos] ^ •IF ** •Vcj let *I« H - r \ \ » (a) (b)
FIG. 18. A-zigzag and Y-zigzag transformer banks• (a) Wiring diagram of zigzag
connections with windings of A or Y indicated, but not connected. (6), (c) Normal
voltage vector diagrams of A-zigzag and Y-zigzag banks, respectively.
A-zigzag and Y-zigzag transformers will be analyzed as three-
winding transformers with exciting currents neglected. The equiva-
lent leakage impedances of any three windings ai, a2, and a3 on the
same core are the three short-circuit impedances between windings
taken two at a time with the third winding open. Let
Zi2 = leakage impedance between windings ai and a2
Zis = leakage impedance between windings ai and a3
Z2s = leakage impedance between windings a2 and 03
where all impedances are in per unit based on rated kva per phase and
rated winding voltages which are in direct proportion to the number
of turns. Base voltage in winding ai of the A is a line-to-line voltage;
170
TRANSFORMERS AND AUTOTRANSFORMERS [Ch. IV]
in the Y it is a line-to-neutral voltage; in windings a2 and a3 of the
zigzag which are assumed to have the same number of turns, it is
1/V3 times base line-to-neutral voltage at the terminals of the
zigzag.
The equivalent leakage impedances Zx, Z„, and Zt of windings a\,
a2, and 03 can be obtained from [5] for the three-winding transformer
if Zx, Zy, and Zt replace Zp, Z„ and Zt and if Zi2, Z13, and Z23 replace
Zp„ Zpt, and Z„t, respectively. Then,
Zx = \(Z12 + Zl3 - Z23) [75]
Zy = iCZu + Z23 - Z13) [76]
Z. - i(Zi8 + ZM - Z12) [77]
The identical equivalent circuit for the three phases are shown in
Figs. 19(a), (6), and (c) with the voltages and currents at their
jic-ib-lo
(a)
(b)
(c)
Fig. 19. (a), (6), (c) Identical equivalent circuits to replace each of the three-
winding transformers, with currents and voltages indicated.
terminals indicated. The notation in Fig. 19 corresponds to that of
Fig. 18(a). In Fig. 18(a), the assumed direction of voltage rise and
of current flow in the windings are indicated by arrows, and currents
and voltages are represented by symbols.
From Fig. 18(a),
Va = Vb3 - Vc2 [78]
From Figs. 19(6) and 19(c),
Vb3 - VB - (/a - Ic)ZX - IaZt [79]
Vc2 = Vc - (/» - Ia)Zx + IaZy [80]
Substitution of [79] and [80] in [78] gives
Va = VB - Vc - Ia(2Zx + Zy + Zt) + (h + Ic)ZX [81]
[CH. IV] A-ZIGZAG AND Y-ZIGZAG TRANSFORMERS 171
The transformer is symmetrical in the three phases; if it supplies a
symmetrical circuit and if VA, VB, and Vc are taken as positive-
sequence applied voltages, VB — Vc = _j^/SVjLi, Va = 7ai, /0 = /Oi»
Ib + IK — —lai, and [81] becomes
V.i = -jV3 VAl - /.i (3Z, + Zv + Z,) [82]
If the applied voltages are of zero sequence, VB _ Vc = 0,
VA = VaQ, /0 = /0o, /& + Ic = 2/oo, and [81] becomes
700= -/a0(2v + Zz) [83]
In [82] and [83], 7ai and Vao are the positive- and zero-sequence
voltages, respectively, at the zigzag terminals in per unit of a base
voltage equal to 1/V3 times rated line-to-neutral voltage; /ai and /a0
are in per unit of a base current equal to rated kva per phase divided
by 1/V3 times line-to-neutral kv.
Let V'ai and 7^o be the positive- and zero-sequence voltages at the
zigzag terminals in per unit of line-to-neutral voltage, and l'ai and I'M
the corresponding currents in per unit of rated kva per phase divided
by line-to-neutral kv. Then
01 = ;o0
If these substitutions are made in [82] and [83] and both sides of the
equations are divided by V3, there results
''^Zc + Zv + Z,
*\ rl Z™
J- -Jio-y
[85]
From [84] at no load, V'ai = — jVAi. At no load the voltage-to-
neutral at the terminals of the zigzag lags VA of the A or Y [see Figs.
18(6) and 18(c)] by 90°.
The per unit positive-sequence impedance Zi which is the same
referred to either side of the transformer bank is
Z, - Z. + ?*±^ [86]
If Zx, Zv, and Zz are replaced by their values given in [75]-[77],
Zi = J(Zla + Zls -iZM) [87]
172 TRANSFORMERS AND AUTOTRANSFORMERS [CH. IV]
The per unit zero-sequence impedance of the bank viewed from the
terminals of the zigzag, based on rated kva per phase and rated line-
to-neutral voltage at the zigzag terminals is
Z0 = [88]
o
where Z23 is the per unit leakage impedance between any two windings
of the zigzag on the same core, based on rated kva per phase and rated
winding voltage.
Zero-sequence currents flowing in opposite direction in windings of
the zigzag on the same core cause no resultant flux in the core, and
therefore induce no voltage in the windings of the A or Y. The
zero-sequence impedance viewed from the terminals of the A or un-
grounded Y is infinite; if the Y is grounded, it is the self -impedance
of the Y.
Problem 6. In a A-zigzag transformer bank, rated 7500 kva, rated voltage of the
A is 13.8 kv, that of the zigzag windings is 23 kv. Line-to-line voltages at the zigzag
terminals is 69 kv; line-to-neutral voltage is 69/V 3 kv. Per cent leakage reactance
between a A winding and each of the zigzag windings on the same core is 12% based
on 2500 kva per phase and rated winding voltages; leakage impedance between two
zigzag windings on the same core is 9% based on 1250 kva and rated winding voltages,
or 18% based on 2500 kva. Determine the impedances Zi and Zo for use in the
positive-sequence and zero-sequence networks of the system, neglecting resistance
and exciting currents.
Solution. Let A windings be ai and zigzag windings a2 and a3 as in Fig. 18 (a).
Based on 2500 kva per phase and rated winding voltages,
x12 = x13 = 12%
x23 = 18%
xi, based on 2500 kva per phase and line-to-neutral voltage at either set of terminals
of the bank, from [87], is
xi = 5 (xn + xu - ix23) = 9%
From [88], Zo viewed from the zigzag terminals based on 2500 kva per phase and a
line- to-neutral voltage of 69/V 3 is
Z*> 18
BIBLIOGRAPHY
1. Transformer Engineering, by L. F. BLUME, G. CAMILLI, A. BOYAJIAN, and
V. M. MONTSINGER, edited by L. F. Blume, John Wiley and Sons, 1938.
2. "An Equivalent Circuit for the Four-Winding Transformer," by F. M. STARR,
Gen. Elec. Rev., Vol. 36, No. 3, March, 1933, pp. 150-152.
[Ch. IV] BIBLIOGRAPHY 173
3. "A Useful Equivalent Circuit for a Five-Winding Transformer," by L. C. Aicher,
Jr., A.I.E.E. Trans., Vol. 62, 1943, pp. 66-70.
4. " Progress in the Study of System Stability," by I. H. Summers and J. B.
McClure, A.I.E.E. Trans., Vol. 49, 1930, pp. 132-158.
5. "Zero-Phase-Sequence Characteristics of Transformers," Parts I and II, by
A. N. Garin, Gen. Elec. Rev., Vol. 43, Nos. 3 and 4, March and April, 1940, pp.
131-136, 174-179.
CHAPTER V
TRANSFORMERS IN SYSTEM STUDIES
In the usual system study, base voltage is arbitrarily chosen in a
specified circuit; base voltages in the other circuits of the system are
then determined by transformer turn ratios. There are systems,
however, in which transformers of unequal turn ratios are operated in
parallel or — a more usual condition — a transmission circuit is sup-
plied at two or more different locations by transformers of unequal turn
ratios, or of unequal equivalent turn ratios when several circuits are
involved. For such cases, an equivalent circuit has been developed
which is applicable to two-winding transformers when the ratio of
transformation is different from the ratio of system base voltages on
the two sides of the transformer.1 This equivalent circuit is especially
useful in the determination of load division when an a-c network
analyzer is not available. Although developed for two-winding trans-
formers, it has wider application and can be extended to transformers
of more than two windings.
The usual procedure for representing such transformers on the
a-c network analyzer is to use an autotransformer to take care of the
difference in ratios between the turn ratio of the transformer and the
ratio of the system base voltages. The autotransformer used has very
low core loss and exciting current, and approximate compensation for
its leakage reactance is secured by a series capacitor. For most
problems it can be considered an ideal unit. The transformer leakage
impedance may be placed on either side of the autotransformer;
this impedance is expressed in per unit of system base ohms in the
circuit in which it is placed.
EQUIVALENT CIRCUITS FOR TRANSFORMERS WITH TURN RATIO
DIFFERENT FROM THE RATIO OF SYSTEM BASE VOLTAGES
AT THEIR TERMINALS
Equivalent circuits will be developed for two- and three-winding
transformers with exciting currents neglected. It will be assumed
that rated winding voltages are in direct proportion to the number of
turns.
174
[CH.V]
175
TWO-WINDING TRANSFORMER
Two-Winding Transformer. With exciting current neglected, there
will be no current in either winding with the other winding open.
Let the two windings be designated 1 and
2. The proposed equivalent circuit is
shown in Fig. 1 with branch impedances
Zx, Zv, and Z, to be evaluated. In Fig.
1, the voltages at 1 and 2 referred to
neutral N represent the voltages across
windings 1 and 2, respectively; the cur-
rents at 1 and 2 are the currents in wind-
ings 1 and 2, respectively, positive
direction of current being from 1 to 2.
Let ei be rated voltage of transformer
winding 1 divided by base system voltage
in winding 1. Let e2 be a similar ratio in
winding 2. The no-load voltage ratio of
the transformer is then ei/e2 based on
system base voltages.
Let Z| = transformer leakage impedance in per unit based on its
rated winding voltages and system base kva per phase.
If the no-load voltage ratios in Fig. 1 are equated to those in the
transformer based on system voltages,
NEUTRAL N
FIG. 1. Positive-sequence
equivalent circuit (resonant A)
for two-winding transformer
bank, exciting current neg-
lected, for use where trans-
former turn ratio differs from
ratio of system base voltages
at bank terminals. See [4]_[6]
for evaluation of Zx> Zv,and Z,.
l
[1]
[2]
With winding 1 short-circuited, the impedance viewed from winding
2 based on system base voltage in winding 2 is e\Zt. If this value
of impedance is equated to that in Fig. 1 viewed from 2 with 1 shorted
to neutral,
*7 rr
[3]
Simultaneous solution of [l]-[3] gives
Zx
'[4]
[5]
Z, = (eie2Z,)
From [1] and [2], or from [4]-[6],
Zx + Zv + Z, = 0
[6]
[7]
176 TRANSFORMERS IN SYSTEM STUDIES [CH. V]
From [7], Fig. 1 is a resonant A since the sum of its branch im-
pedances is zero. A resonant A could have been assumed in the first
place and [7] used instead of [1] or [2] to evaluate Zx, Zv, and Z,.
With exciting current neglected, there will be no current flowing to
neutral when a voltage is applied to one terminal with the other open.
A resonant A satisfies this condition.
It will be noted from [4]-[6] that, if ei is greater than e2, Zv will be
negative; if e2 is greater than ei, Z, will be negative. If transformer
resistance is included, the negative branch Zv or Z, will consist of a
negative resistance and a capacitive reactance. Although the res-
onant A cannot be replaced by an equivalent Y of finite impedances,
its branches can be combined with other system impedances in the
usual manner to simplify calculations. In an analytic solution,
negative resistances are as easily handled as positive.
If ei and e2 are both unity, Zv = Z, = <*> and Fig. 1 reduces
to the series impedance Zx = Zt = leakage impedance between
windings.
If ei and e2 are equal but not unity, as is the case when base voltages
are in proportion to rated voltages but not equal to them, Zv and Z,
become infinite, ei = e2 = e, and the per unit leakage impedance
Zx = e2Zt is expressed in terms of base system quantities.
If base and rated voltage are the same in one of the windings, e
in that winding is unity. In many cases, base system voltage can be
so chosen that ei or e2 is unity.
Problem 1. Determine Zx, Zv, and Zz in per cent for insertion in the equivalent
circuit of Fig. 1 for a transformer bank of three single-phase units each rated 9000 kva
121 kv/31.5 kv, with a reactance of 9% and a resistance of 0.9% based on its rating.
System base three-phase kva is 30,000 kva; system base voltages in the circuits at
the transformer bank terminals are 1 10 kv and 35 kv.
Solution. By definition,
Zt = (0.9 +J9.0) - = 1 +J10%
In per cent, based on system base quantities,
1.10 X 0.90Z, = 0.99 +J9.90 Zy = Cl -Zx = 1'™Zz = -S.SZz = -5.445 - J54.45 e2 — ei —0.20 €•> 0.90 Z, = — Zx = - — Zx = 4.SZx = 4.455 +j44.55 ei — e<i 0.20
[CH. V]
177
THREE-WINDING TRANSFORMER
Three-Winding Transformer with Base System Voltage in One
Winding Different from Rated Voltage. Let the three windings be
designated 1, 2, and 3. With a three-winding transformer, it is
reasonable to suppose that the arbitrarily chosen system base voltage
will be such that in two of the windings (windings 1 and 2) system
base voltages will be rated winding voltages. As rated voltages and
base voltages are equal, employing the notation used for the two-
winding transformer, ei = 62 = 1. Base voltage in the third winding
will be different from rated winding voltage, and 63 will not be unity.
NEUTRAL N
FIG. 2. Positive-sequence equivalent circuit for three-winding transformer bank,
exciting current neglected, for use where transformer turn ratios differ from ratios
of system base voltages at bank terminals.
Let Zi2, Zi3, and Z23 be the per unit leakage impedances between
windings indicated by the subscripts, taken two at a time with the
other winding open, based on rated winding voltages and base system
kva per phase.
To satisfy no-load conditions, there must be a resonant A in the
equivalent circuit between terminals 1 and 3, and between 2 and 3,
but not between 1 and 2. Figure 2 shows the proposed equivalent
circuit, with branch impedances Zi, Z2, Zx, Zv, and Zz to be evaluated.
It follows from the equivalent circuit of a two-winding transformer that
[8]
z,
[9]
The impedance between windings 1 and 2 with winding 3 open in
the equivalent circuit of Fig. 2 must equal that in the transformer:
Zi
[10]
178 TRANSFORMERS IN SYSTEM STUDIES [CH. V]
With terminal 3 shorted to neutral and voltage applied first at 1
with 2 open, and then at 2 with 1 open,
Z, + Z. + ^ = Z13 [11]
£iv
'
= Z23 [12]
Simultaneous solution of [8]-[12] gives
+ Z23 _ 7 _ Z,= 2(e3 - 2(1 - Z2 = 2e3 -I- Z23 — Zi3 2ft,
If e3 = 1 in [13], Ztf and Z, become infinite, and the per unit equiva-
lent circuit is reduced to that of the three-winding transformer based
on rated winding voltages and system kva. See [5], Chapter IV.
The equivalent circuits for two- and three-winding transformers
given in Figs. 1 and 2 can be used to replace a transformer bank of
identical single-phase units or a three-phase transformer in the positive-
and negative-sequence networks when ratios of system base voltages
in the windings are not the same as the corresponding turn ratios.
Zero-sequence equivalent circuits depend upon the manner in which
the windings are connected and the method of grounding; however,
they can be derived from Figs. 1 and 2 by the modifications given and
illustrated in Chapter IV for banks of single-phase units. In a trans-
former bank of single-phase units, the impedance to positive-, nega-
tive-, and zero-sequence currents is the same, provided that there is a
path for zero-sequence current. In a three-phase transformer, the
zero-sequence leakage impedances between windings may differ
appreciably from the positive. (See "Three-Phase Transformers,"
Chapter IV.)
[CH. V] OPEN CONDUCTORS IN CIRCUITS 179
OPEN CONDUCTORS IN CIRCUITS SUPPLYING UNGROUNDED
TRANSFORMERS
In the equivalent circuits of Figs. 1 and 2, and in many of those in
Chapter IV, exciting currents are neglected. This can be done when
the exciting impedance is so large relative to the other impedances
under consideration that it may be regarded as infinite within the
degree of precision required in calculations. Such is usually the case
in short-circuit calculations where there are no open phases and short-
circuit currents are large relative to transformer exciting currents.
When one or two conductors are open in circuits which supply
ungrounded transformers, the conditions may be quite different. The
transmission circuit may be a low-voltage feeder of short length and the
transformer bank of low kva rating, so that the capacitive reactances
of the circuit and the magnetizing reactances of the transformer bank
are infinite relative to the resistances and other reactances of the
system. But the capacitive reactances are negative, and the mag-
netizing reactances are positive; therefore, for certain ratios of these
reactances a current path of relatively low impedance may be provided
with resultant high voltages on the open phase or phases. If such is
the case, the resistances and other reactances of the system may be
so small relative to the capacitive reactances of the circuit and the
magnetizing reactances of the transformer bank that they can be
regarded as zero within the degree of precision required in calculations.
Under certain conditions, high sustained voltages have resulted
from the opening of one or two conductors in circuits which supply
ungrounded transformer banks. Voltages of sufficient magnitude to
damage equipment have been reported, and induction motors have
been observed to reverse their direction of rotation. Cases of the
breakage of a line conductor with resulting high sustained voltages
have been noted, and abnormal voltage conditions have occurred on
potential transformers.
Open conductors may be due to the blowing of fuses, operation of
single-pole switches, or of any interrupting device in which the in-
terval of time between the opening or closing of the first and last
phases is of sufficient length to permit ultimate steady-state voltage
conditions to be attained. Included also is the accidental breaking
of a conductor where one severed end may fall to ground.
To determine the conditions under which abnormal voltages may
occur and motors reverse their direction of rotation, with one or two
conductors open in circuits supplying ungrounded transformers, an
investigation was made by means of calculations and laboratory
tests, and the results were reported.2
180 TRANSFORMERS IN SYSTEM STUDIES [CH. V]
The phenomena encountered may be summarized as follows:
When one conductor of a three-phase circuit supplying an ungrounded,
unloaded transformer bank is open, there is a path for currents from
the closed conductors through the exciting impedance of the bank and
the capacitance between conductors to the open conductor, and
thence to ground through the capacitance-to-ground of the open