10. Write the expression for the self inductance of a coil wound on an iron core
3.14 PARALLEL OPERATION OF TRANSFORMERS
When the load outgrows the capacity of an existing transformer, it may be economical to install another one in parallel with it rather than replacing it with a single larger unit. Also, sometimes in a new installation, two units in parallel, though more expensive, may be preferred over a single unit for reasons of reliability—half the load can be supplied with one unit out. Further, the cost of maintaining a spare is less with two units in parallel. However, when spare units are maintained at a central location to serve transformer installations in a certain region, single-unit installations would be preferred. It is, therefore, seen that parallel operation of the transformer is quite important and desirable under certain circumstances.
The satisfactory and successful operation of transformers connected in parallel on both sides requires that they fulfil the following conditions:
(i) The transformers must be connected properly as far as their polarities are concerned so that the net voltage around the local loop is zero. A wrong polarity connection results in a dead short circuit.
(ii) Three-phase transformers must have zero relative phase displacement on the secondary sides and must be connected in a proper phase sequence. Only the transformers of the same phase group can be paralleled. For example, Y/Y and Y/D transformers cannot be paralleled as their secondary voltages will have a phase difference of 30°. Transformers with +30° and –30° phase shift can, however, be paralleled by reversing the phase-sequence of one of them.
(iii) The transformers must have the same voltage-ratio to avoid no-load circulating current when transformers are in parallel on both primary and secondary sides. Since the leakage impedance is low, even a small voltage difference can give rise to considerable no-load circulating current and extra I2R loss.
(iv) There should exist only a limited disparity in the per-unit impedances (on their own bases) of the transformers. The currents carried by two transformers (also their kVA loadings) are proportional to their ratings if their ohmic impedances (or their pu impedances on a common base) are inversely proportional to their ratings or their per unit impedances on their own ratings are equal. The ratio of equivalent leakage reactance to equivalent resistance should be the same for all the transformers.
A difference in this ratio results in a divergence of the phase angle of the two currents, so that one transformer will be operating with a higher, and the other with a lower power factor than that of the total output; as a result, the given active load is not proportionally shared by them.
Parallel Transformers on No-load
The parallel operation of transformers can be easily conceived on a per phase basis. Figure 3.59 shows two transformers paralleled on both sides with proper polarities but on no-load. The primary voltages V1 and V2 are obviously equal. If the voltage-ratio of the two transformers are not identical, the secondary induced emf’s, E1 and E2 though in phase will not be equal in magnitude and the difference (E1 – E2) will appear across the switch S. When secondaries are paralleled by closing the switch, a circulating current appears even though the secondaries are not supplying any load. The circulating current will
V1 Primary
Secondary
Load
1 2
S E1
ZL
V2
E2
Fig. 3.59
depend upon the total leakage impedance of the two transformers and the difference in their voltage ratios.
Only a small difference in the voltage-ratios can be tolerated.
Equal voltage-ratios
When the transformers have equal voltage ratio, E1 = E2 in Fig. 3.59, the equivalent circuit of the two transformers would then be as shown in Fig. 3.60 on the assumption that the exciting current can be neglected in comparison to the load current. It immediately follows from the sinusoidal steady-state circuit analysis that
I1 = Z Z 2Z IL
1+ 2 (3.83)
and I2 = Z
Z 1Z IL
1+ 2 (3.84)
Of course I1+ = II2 L (3.85)
Taking VL as the reference phasor and defining complex power as V I* , the multiplication of VL* on both sides of Eqs (3.83) and (3.84) gives
S1 = Z Z 2Z SL
1+ 2 (3.86)
S2 = Z Z 1Z SL
1+ 2 (3.87)
where S1 = VL*I1
S2 = VL*I2 SL = VL*IL
These are phasor relationships giving loadings in the magnitude and phase angle. Equations (3.86) and (3.87) also hold for pu loads and leakage impedances if all are expressed with reference to a common base.
It is seen from Eqs (3.83) and (3.84) that the individual currents are inversely proportional to the respective leakage impedances. Thus, if the transformers are to divide the total load in proportion to their kVA ratings, it is necessary that the leakage impedances be inversely proportional to the respective kVA ratings, i.e.
Z Z12 = S
S21
( )
( )
rated
rated (3.88)
This condition is independent of the power factor of the total load. The condition of Eq. (3.88) can be written as
Z
Z12 = V I V ILL 2 1
( )
( )
rated rated
V1 Z1 I1
Z2 I2
IL ZL
VL
Fig. 3.60
or Z I
It means that if individual transformer loadings are to be in the ratio of their respective kVA ratings, their pu impedances (on their own ratings) should be equal. If
Z1 < Z2S
the transformer 1 will be the first to reach its rated loading as the total kVA load is raised. The maximum permissible kVA loading of the two in parallel without overloading anyone is then given by
S1(rated) = Z
In either case (Eq. (3.90a) or (3.91a))
SL(max) < S1(rated) + S2(rated) (3.92) Unequal Voltage Ratios
It has already been mentioned that a small difference in voltage ratios can be tolerated in the parallel operation of transformers. Let E1 and E2 be the no-load secondary emfs of two transformers in parallel. With reference to Fig. 3.58, if a load current IL is drawn at voltage VL, two mesh voltage balance equations can be written as
E1 = I Z1 1+I ZL L=I Z1 1+(I1+I Z2) L (3.93) and E2 = I Z2 2+I ZL L =I Z2 2+(I1+I Z2) (3.94)L
\ E E1- 2 = I Z1 1-I Z2 2 (3.95)
On no-load IL = 0, so that the circulating current between the two transformers is given by I1 = -
Substituting for I1: in Eq. (3.94) we get
Normally E1 and E2 are in phase or their phase difference is insignificant. Severe results of paralleling transformers not belonging to the same phase-groups (say Y/Y and Y/D transformers) are immediately obvious from Eq. (3.96) for no-load circulating current. When many transformers are in parallel, their load sharing can be found out using the Millman theorem [2].
EXAMPLE 3.24 A 600-kVA, single-phase transformer with 0.012 pu resistance and 0.06 pu reactance is connected in parallel with a 300-kVA transformer with 0.014 pu resistance and 0.045 pu reactance to share a load of 800 kVA at 0.8 pf lagging. Find how they share the load (a) when both the secondary voltages are 440 V and (b) when the open-circuit secondary voltages are respectively 445 V and 455 V.
SOLUTION
(a) The pu impedances expressed on a common base of 600 kVA are Z1 = 0.012 + j 0.06 = 0.061 –79°
It may be noted that the transformers are not loaded in proportion to their ratings. At a total load of 800 kVA, the 300 kVA transformer operates with 5% overload because of its pu impedance (on common kVA base) being less than twice that of the 600 kVA transformer.
The maximum kVA load the two transformers can feed in parallel without any one of them getting overloaded can now be determined. From above it is observed that the 300 kVA transformer will be the first to reach its full-load as the total full-load is increased. In terms of magnitudes
0 061
while the sum of the ratings of the two transformers is 900 kVA. This is consequence of the fact that the transformer impedances (on common base) are not in the inverse ratio of their ratings.
(b) In this case it is more convenient to work with actual ohmic impedances. Calculating the impedances referred to secondary
Z1(actual) = (0.012 + j 0.06) ¥ 440 600 1000
440
¥ = 0.0039 + j 0.0194
= 0.0198 –79°
Z2(actual) = (0.028 + j 0.09) ¥ 440 600 1000
440
¥ = 0.009 + j 0.029
= 0.0304 –73°
Z1+Z2 = 0.0129 + j 0.0484 = 0.05 –75°
The load impedance ZL must also be estimated. Assuming an output voltage on load of 440 V, V IL L* ¥ 10–3 = (V ZL2/ L) ¥ 10–3 = 800 ––37°
\ ZL = (440)
800 10 37
2
¥ 3– - ∞ = 0.242 –37°
= 0.1936 + j 0.1452
From Eqs (3.98) and (3.99)
I1 = 445 0 0304 73 10 0 242 37 0 0198 79 0 0304 73 0 242 37
¥ – ∞ - ¥ – ∞
– ∞ ¥ – ∞ + – ∞
. .
. . . ¥¥0 05 75. – ∞
= 940 ––34° A
I2 = 445 0 0198 79 10 0 242 37 0 0198 79 0 0304 73 0 242 37
- – ∞ - ¥ – ∞
– ∞ ¥ – ∞ + – ∞
. .
. . . ¥¥0 05 75. – ∞
= 883 ––44° A
The corresponding kV As are
S1 = 440 ¥ 940 ¥ 10–3––34° = 413.6 ––34°
S2 = 440 ¥ 883 ¥ 10–3––4444° = 388 ––44°
The total output power will be
413.6 cos 34° + 388 cos 44° = 621.5 kW
This is about 3% less than 800 ¥ 0.8 = 640 kW required by the load because of the assumption of the value of the output voltage in order to calculate the load impedance.
The secondary circulating current on no-load is
( )
| |
E E
Z11 Z22
-+ = -10
0 05. = – 200 A which corresponds to about 88 kVA and a considerable waste as copper-loss.
Transformers may be built with a third winding, called the tertiary, in addition to the primary and secondary.
Various purposes which dictate the use of a tertiary winding are enumerated below:
(i) To supply the substation auxiliaries at a voltage different from those of the primary and secondary windings.
(ii) Static capacitors or synchronous condensers may be connected to the tertiary winding for reactive power injection into the system for voltage control.
(iii) A delta-connected tertiary reduces the impedance offered to the zero sequence currents thereby allowing a larger earth-fault current to flow for proper operation of protective equipment. Further, it limits voltage imbalance when the load is unbalanced. It also permits the third harmonic current to flow thereby reducing third-harmonic voltages.
(iv) Three windings may be used for interconnecting three transmission lines at different voltages.
(v) Tertiary can serve the purpose of measuring voltage of an HV testing transformer.
When used for purpose (iii) above the tertiary winding is called a stabilizing winding.
The star/star transformer comprising single-phase units or a single unit with a 5-limb core offers high reactance to the flow of unbalanced load between the line and neutral. Any unbalanced load can be divided into three 3-phase sets (positive, negative and zero sequence components). The zero-sequence component (cophasal currents on three lines, I0 = In/3) caused by a line-to-neutral load on the secondary side cannot be balanced by primary currents as the zero-sequence currents cannot flow in the isolated neutral star-connected primary. The zero-sequence currents on the secondary side therefore set up magnetic flux in the core. Iron path is available for the zero sequence flux* in a bank of single-phase units and in the 5-limb core and as a consequence the impedance offered to the zero-sequence currents is very high (0.5 to 5 pu) inhibiting the flow of these currents. The provision of a delta-connected tertiary permits the circulation of zero-sequence currents in it, thereby considerably reducing the zero-sequence impedance.
This is illustrated in Fig. 3.61.
0 I0
0
0 Stabilizing
tertiary winding I0
I0
I0 In=3l0
Fig. 3.61
Equivalent Circuit
The equivalent circuit of a 3-winding transformer can be represented by the single-phase equivalent circuit of Fig. 3.62 wherein each winding is represented by its equivalent resistance and reactance. All the values are reduced to a common rating base and respective voltage bases. The subscripts 1, 2 and 3 indicate the primary, secondary and tertiary respectively. For simplicity, the effect of the exciting current is ignored in the equivalent circuit. It may be noted that the load division between the secondary and tertiary is completely
* In a 3-limb core the zero-sequence flux (directed upwards or downwards in all the limbs) must return through the air path, so that only a small amount of this flux can be established; hence a low zero-sequence reactance.
arbitrary. Three external circuits are connected between terminals 1, 2 and 3 respectively and the common terminal labelled 0. Since the exciting current is neglected, I1+ + = 0.I2 I3
I1 1 +
–
Common 0 A
3 2
+
+
– –
V1 Z1
Z2 l2
Z3 l3
V3 V2
Fig. 3.62
The impedance of Fig. 3.62 can be readily obtained from three simple short-circuit tests. If Z12 indicates the SC impedance of windings 1 and 2 with winding 3 open, then from the equivalent circuit,
Z12 = Z1+Z2 (3.100)
Similarly Z23 = Z2+Z3 (3.101)
Z13 = Z1+Z3 (3.102)
where Z23 = SC impedance of windings 2 and 3 with winding 1 open.
Z13 = SC impedance of windings 1 and 3 with winding 2 open.
All the impedances are referred to a common base.
Solving Eq. (3.100) to Eq. (3.102) yields Z1 = 1
2(Z12+Z13-Z23) (3.103)
Z2 = 1
2(Z23+Z12-Z13) (3.104)
Z3 = 1
2(Z13+Z23-Z12) (3.105)
The open-circuit test can be performed on anyone of the three windings to determine the core-loss, magnetizing impedance and turn-ratio.
EXAMPLE 3.25 The primary, secondary and tertiary windings of a 50-Hz, single-phase, 3-winding transformer are rated as 6.35 kV, 5 MVA; 1.91 kV, 2.5 MVA; 400 V, 2.5 MVA respectively. Three SC tests on this transformer yielded the following results:
(i) Secondary shorted, primary excited: 500 V, 393.7 A (ii) Tertiary shorted, primary excited: 900 V, 393.7 A (iii) Tertiary shorted, secondary excited: 231 V, 21 312.1 A
Resistances are to be ignored.
(a) Find the pu values of the equivalent circuit impedances of the transformer on a 5 MVA, rated voltage base.
(b) Three of these transformers are used in a 15 MVA, Y-Y-D, 3-phase bank to supply 3.3 kV and 400 V auxiliary power circuits in a generating plant. Calculate the pu values of steady-state short-circuit currents and of the voltage at the terminals of the secondary windings for a 3-phase balanced short-circuit at the tertiary terminals. Use 15 MVA, 3-phase rated voltage base.
SOLUTION
(a) Let us first convert the SC data to pu on 5 MVA base/phase.
For primary, VB = 6.35 kV
IB = 5000
6 35. = 787.4 A For secondary, VB = 1.91 kV
IB = 5000
1 91. = 2617.8A Converting the given test data to pu yields:
Test No. Windings involved V I
1 2 3
P and S P and T S and T
0.0787 0.1417 0.1212
0.5 0.5 0.5 From tests 1,2 and 3, respectively. 0.0787
X12 = 0 0787
0 5 .
. = 0.1574 pu
X13 = 0 1417
0 5 .
. = 0.2834
X23 = 0 1212
0 5 .
. = 0.2424 pu From (3.103) – (3.105)
X1 = 0.5(0.1574 + 0.2834 – 0.2424) = 0.0992 pu X2 = 0.5(0.2424 + 0.1574 – 0.2834) = 0.05825 pu X3 = 0.5(0.2834 + 0.2424 – 0.1574) = 0.1842 pu
(b) The base line-to-line voltage for the Y-connected primaries is 3 ¥ 6.35 = 11 kV, i.e. the bus voltage is 1 pu.
From Fig. 3.62, for a short-circuit at the terminals of the tertiary, V3 = 0. Then
ISC = V
X1+1X3 = V
X131 = 1 00 0 2834
.
. = 3.53 pu SC current primary side = 3.53 ¥ 787.4 = 2779.5 A
SC current tertiary side = 3.53 ¥ 5000 1000¥
400 3/ = 76424 A (line current)
Neglecting the voltage drops due to the secondary load current, the secondary terminal voltage is the voltage at the junction point A (Fig. 3.63), i.e
VA = ISC X3 = 3.53 ¥ 0.1842 = 0.6502 pu VA(actual) = 0.6502 ¥ 1.91 3 = 2.15 kV (line-to-line)
V1
0.0992
0.05825
0.1842 A
lsc V3= 0
V2
Fig. 3.63 3.16 PHASE CONVERSION
Phase conversion from three to two phase is needed in special cases, such as in supplying 2-phase electric arc furnaces.
The concept of 3/2-phase conversion follows from the voltage phasor diagram of balanced 3-phase supply shown in Fig. 3.64(b). If the point M midway on VBC could be located, then VAM leads VBC by 90°. A 2-phase supply could thus be obtained by means of transformers; one connected across AM, called the teaser transformer and the other connected across the lines B and C. Since VAM = ( 3 /2) VBC, the transformer primaries must have 3 N1/2 (teaser) and N1 turns; this would mean equal voltage/turn in each transformer.
A balanced 2-phase supply could then be easily obtained by having both secondaries with equal number of turns, N2. The point M is located midway on the primary of the transformer connected across the lines B and C. The connection of two such transformers, known as the Scott connection, is shown in Fig. 3.64(a), while the phasor diagram of the 2-phase supply on the secondary side is shown in Fig. 3.64(c).
The neutral point on the 3-phase side, if required, could be located at the point N which divides the primary winding of the teaser in the ratio 1 : 2 (refer Fig. 3.64(b)).
A a2
lA
lA
la
Va
lA/2 lAl2
lC N l12 N l12 IBC
b2 b1
N2 Vb
Va
Vb
N N2
a1
lB
+
–
B
M C
– C +
(a) (b)
A
N
M B
(c) 3 N1/2
Ib
Fig. 3.64 Scott connection
Load Analysis
If the secondary load currents are Ia and Ib, the currents can be easily found on the 3-phase side from Fig. 3.64(a).
IA = 2 3
2 1
N
N Ia = 2
3Ia(for N1/N2 = 1) IBC = N
N2 Ib Ib
1 = (for N1/N2 = 1) IB = IBC- /2IA
IC = –IBC- /2IA
The corresponding phasor diagram for balanced secondary side load of unity power factor is drawn in Fig. 3.65 from which it is obvious that the currents drawn from the 3-phase system are balanced and cophasal with the star voltages. The phasor diagram for the case of an unbalanced 2-phase load is drawn in Fig. 3.66.
–lBC
–lAl2
lC lb Vb
Va
la lBC= 1
A
C B
1
1 IA= 2 3
3 –IA/2= 1/
3
= 2/
1+1/3 IB= Fig. 3.65
–lBC A
lA
lC
IB
IBC lA/2
Va la
fa
fb Vb
lb
C B
–lA/2
Fig. 3.66
Three/One-phase Conversion
A single-phase power pulsates at twice the frequency, while the total power drawn by a balanced 3-phase load is constant. Thus a 1-phase load can never be transferred to a 3-phase system as a balanced load without employing some energy-storing device (capacitor, inductor or rotating machine). Suitable transformer connections can be used in distributing a 1-phase load on all the three phases though not in a balanced fashion. For large 1-phase loads, this is better than allowing it to load one of the phases of a 3-phase system.
A variety of transformer connections are possible. Figure 3.67(a) shows how Scott-connected transformers could be used for this purpose and Fig. 3.67(b) shows the corresponding phasor diagram.
B A C
V l
(a)
VB lBC= 1
–lBC= 1
(b) Va A
B C
l = 1
V 3 – 1
3
2 I 3
I 3 + 1
3 I
3 – 1 3 I 1 =
IB = 1 – 3
3 + 1 3 1 = IC = 1 + 3
2 IA = 3
Fig. 3.67
Three Phase/Six-phase Conversion
Each secondary phase is divided into two equal halves with polarity labelling as in Fig. 3.50. Six-phase voltages (characteristic angle 360°/6 = 60°) are obtained by means of two stars in phase opposition, each star being formed from three respective half-windings as shown in Fig. 3.68. This connection is employed in rectifiers and thyristor circuits where a path for the dc current is needed.
C
B2 C2
A2 B
A
c4
a1
b1
a4
b4 c1
Fig. 3.68
EXAMPLE 3.26 Two single-phase furnaces A and B are supplied at 100 V by means of a Scott-connected transformer combination from a 3-phase 6600 V system. The voltage of furnace A is leading. Calculate the line currents on the 3-phase side, when the furnace A takes 400 kW at 0.707 pf lagging and B takes 800 kW at unity pf.
SOLUTION With reference to Fig. 3.64(a) N
N12 = 6600 100 = 66
\