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Equivalent circuit of transformers with control of voltage and phase angle
Electric Power Systems Research 81 (2011) 1349–1356
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Equivalent circuit of transformers with control of voltage and phase angle Juan A. Martinez-Velasco ∗ Dept. d’Eng. Elèctrica, Universitat Politècnica de Catalunya, Diagonal 647, 08028 Barcelona, Spain
a r t i c l e
i n f o
Article history: Received 8 October 2010 Received in revised form 19 January 2011 Accepted 31 January 2011 Available online 24 February 2011 Keywords: Power transformer Regulating transformer Equivalent circuit Steady-state analysis
a b s t r a c t This paper presents a methodology aimed at obtaining the equivalent circuit of regulating transformers. This aspect is insufficiently treated in most current textbooks in which the equivalent circuit of a tapped transformer is not well justified. The document shows how to obtain the equivalent circuit of a regulating transformer with control of voltage magnitude and phase angle, and how to apply it using either physical quantities or per unit (pu) quantities. The resulting equivalent circuit is adequate for steady-state calculations under balanced conditions and neglects transformer core parameters. The usage of per unit values can be made by selecting the base quantities in an arbitrary manner, which is an obvious advantage when the turns ratio must be estimated. © 2011 Elsevier B.V. All rights reserved.
1. Introduction The regulating transformer, with control of voltage magnitude and/or phase angle, is a common device in power systems. The control of the voltage magnitude is a useful capability to control voltage and reactive power flow, whereas the control of the phase angle can be used to control active power flow [1–7]. A problem that arises when dealing with transformers, with or without regulating capabilities, is the way in which the parameters of the equivalent circuit have to be calculated. This problem is complicated when the calculations must be performed with per unit quantities, and transformer ratings are not equal to base values. There are several reasons to select base values different from transformer ratings. For instance, when the system under study has several transformers and each transformer has different ratings. The representation of a regulating transformer in which either the turns ratio or the phase shift (or even both) is unknown is different from that of a conventional transformer since the parameters of the equivalent circuit are expressed as a function of the unknown values. Graduate and undergraduate students learn how to obtain the per unit parameters of the equivalent circuit of a transformer from the values measured in standardized tests following well established procedures. In addition, textbooks usually refine the procedure when system base values may be different from the nameplate ratings of any particular transformer. However, the
derivation of the equivalent circuit is not usually well justified when arbitrary system base values are used. The representation of an actual transformer when calculations are performed with physical quantities requires the use of the ideal transformer in the equivalent circuit. In general, this device is not needed when the calculations are made with per unit values. A tapped transformer with parameters in per unit is represented in textbooks with or without the ideal transformer, being the second approach (without ideal transformer) the preferred one when a load flow solution is used to determine the turns ratio. However, an approach without using the ideal transformer has not been proposed when the transformer can control the phase angle. The goal of this paper is to justify the derivation of the equivalent circuit of a two-winding regulating transformer with control of both voltage magnitude and phase angle, with parameters expressed in either physical quantities or per unit values, and using arbitrary system base values. The paper shows that it is possible to obtain more than one equivalent circuit using basic circuit components for the most general case, and that simplified representations are easily derived when only one regulating capability (e.g., load-tap-changing) is implemented in the transformer under study. The equivalent circuits considered in this paper are only valid for steady-state analysis, and when the system runs under balanced conditions, since only a single-phase representation is assumed. 2. Representation of two-winding transformers 2.1. Equivalent circuits
∗ Tel.: +34 934016725. E-mail address: [email protected] 0378-7796/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.epsr.2011.01.019
Fig. 1 shows the equivalent circuit of a single-phase twowinding transformer. This circuit can be also used to represent
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J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
For instance, using the following symbols: • • • •
Fig. 1. Equivalent circuit of a single-phase two-winding transformer.
Rated power Sr , in MVA Rated voltages Vr1 , Vr2 , in kV Short-circuit impedance z, in pu Short-circuit losses Wsh , in MW
and assuming that the short-circuit current is the rated current, the parameters of the circuit shown in Fig. 2b may be calculated as follows: Z2 = z
2 Vr2
Sr
X2 = ωL2 =
R2 =
2 Wsh Vr2 Sr Sr
Z22 − R22
(ω = 2f )
(1a) (1b)
where Z2 is the transformer impedance module reduced to the secondary side and f is the power frequency. To obtain the parameters of the model depicted in Fig. 2a (i.e., reduced to the primary side), just change subscript “2” by “1” in Eqs. (1). In the above expressions, rated voltages are the phase-to-phase voltages for a three-phase transformer and the phase-to-neutral voltage for a single-phase transformer. The rated power must include the power of all transformer phases. In the remaining sections of this paper the following symbols will be used: Zsh1 = R1 + jX1
Z1 = |Zsh1 |
(2a)
Zsh2 = R2 + jX2
Z2 = |Zsh2 |
(2b)
2.2. Discussion Taking into account the transformer models shown in Fig. 2, the calculation of the transformer impedance reduced to either the primary or to the secondary side may be obtained as follows: Fig. 2. Simplified equivalent circuit of a two-winding transformer. (a) Parameters reduced to the primary side. (b) Parameters reduced to the secondary side.
Z1 = z Z2 = z
a three-phase two-winding transformer under balanced steadystate conditions. In case of requiring a model for a more general analysis (e.g., transient response under unbalanced conditions), the representation of the three phases and both the primary and the secondary connections should be included [8]. The physical meaning of the parameters to be specified in this representation is as follows: • Rm and Lm are the core parameters; Rm represents core losses and Lm represents the core flux, common to both primary and secondary windings. • Rp , Lp , Rs and Ls are the winding parameters; Rp and Rs represent the losses of the primary and the secondary windings due to Joule effect, while Lp and Ls represent the leakage fluxes of primary and secondary windings, respectively. • NP /NS is the turns ratio, which is also the ratio of the rated voltages and currents In some cases, the influence of the core parameters can be neglected and then the winding parameters are merged into two basic components. Fig. 2 shows the two possible representations, which are the models used in this paper. The parameters to be specified in any of the models depicted in Fig. 2 can be obtained from nameplate ratings and values derived from the standard short-circuit test.
2 Vr1
Sr 2 Vr2
Sr
Zsh1 = ZP + rt2 ZS
(3a)
ZP
(3b)
Zsh2 =
rt2
+ ZS
where rt is the turns ratio rt =
NP Vr1 Ir2 = = NS Vr2 Ir1
(4)
For practical purposes, the variation of the per unit transformer impedance with the operating tap is assumed constant. In practice, this impedance value depends on the tap position and is not always known. Since transformer taps can be on either side of the transformer, it is important to account for the consequences of assuming that the per unit transformer impedance is constant. The following results may be derived then from Eqs. (3a) and (3b) when z is assumed constant: a. When the transformer taps are at the primary winding, Vr2 is known and constant; therefore, Z2 (=|Zsh2 |) can be determined by using (3b) and is also constant, while Vr1 is unknown and Z1 (=|Zsh1 |) cannot be determined. b. When the transformer taps are at the secondary winding, Vr1 is known and constant; therefore, Z1 (=|Zsh1 |) can be determined by using (3a) and is also constant, while Vr2 is unknown and Z2 (=|Zsh2 |) cannot be determined. A conclusion from these results is that the regulating transformer must be represented with the transformer impedance
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Fig. 3. Equivalent circuit of a regulating transformer with transformer impedance reduced to the primary side.
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Fig. 4. Equivalent circuit of a regulating transformer – first option.
reduced to the secondary side (i.e., using Eq. (3b)) when the taps are at the primary side, and with the transformer impedance reduced to the primary side (i.e., using Eq. (3a)) when the taps are at the secondary side. 3. Equivalent circuit of a transformer with voltage and phase angle control Fig. 5. Equivalent circuit of a regulating transformer – second option.
Assume that the transformer can control both voltage magnitude and phase angle. In such case, its turns ratio is a complex value that may be expressed as: rt = |rt |∠˛ = |rt | · (cos ˛ + j sin ˛)
(5)
The relationship between ratings in a two-winding transformer is ∗ ∗ = Vr2 Ir2 Vr1 Ir1
(6)
From where rt =
I∗ Vr1 = r2 ∗ Vr2 Ir1
(7)
which can be also expressed as: Vr1 = rt Vr2
Ir1 =
Ir2 rt∗
3.1. Transformer with control of voltage and phase 3.1.1. Transformer impedance reduced to the primary side Fig. 3 shows the equivalent circuit of the transformer with the transformer impedance reduced to the primary side. The equations of this circuit may be written as follows: V1 = rt V2 + Zsh1 I1 = rt V2 + I2 rt∗
Zsh1 I2 rt∗
(9)
Taking into account the convention for the admittance equations, the above equations can be written as follows:
I1 −I2
=
1 Zsh1
1 −rt∗
−rt |rt |2
V1 V2
I1 −I2
=
1 Zsh1
1 −rt
−rt |rt |2
V1 V2
+
0 (rt − rt∗ )V1
(11)
Zsh1 rt rt − rt∗ J= V1 Zsh1 Z2 =
(12)
b. Eq. (10) can be also rewritten as follows: I1 −I2
=
1 Zsh1
1 −rt∗
−rt∗ |rt |2
V1 V2
+
(rt∗ − rt )V2 0
(13)
The off-diagonal of the new admittance matrix are equal, so it can be also synthesized as a passive circuit. As for the second part of the right hand side, it is again a voltage-controlled current source, but this time the controlling voltage is V2 . The new equivalent circuit is that depicted in Fig. 5. The parameters of the new circuit are as follows: 1 − rt∗ Z Z2 = sh1 Zsh1 rt∗ |rt |2 − rt∗ r ∗ − rt Y3 = J= t V2 Zsh1 Zsh1 Y1 =
(14)
It is important to note that expressions (12) and (14) can lead to values of Y1 and Y3 whose real part is negative. The procedure to obtain the equations in per unit will be based on Eq. (10), which can be rewritten as:
(10)
Since the turns ratio is a complex value and the off-diagonal elements are not equal, a passive RLC circuit cannot be synthesized directly from this admittance matrix. However, the equations can be manipulated to obtain a matrix with equal off-diagonal elements. There are, at least, two options. a. Eq. (10) is rewritten as follows:
1 − rt Zsh1 |r|2 − rt Y3 = Zsh1
Y1 =
(8)
These equations are valid for any voltage and load level; that is, the relationship between voltages and currents at both sides of the ideal transformer is the same that for rated values. The rest of this section is dedicated to deduce the equivalent circuits of regulating transformers, beginning with the most general case, for which the turns ratio is a complex value.
I1 =
The matrix of the right hand side of this equation has equal off-diagonal elements, so it can be now synthesized as a passive circuit. On the other hand, the second part of the right hand side can be synthesized as a current source controlled by the voltage V1 . The equivalent circuit of the regulating transformer may be that depicted in Fig. 4. The parameters of this circuit can be obtained as follows:
Ib1 0
0
Ib2
I1 pu −I2 pu
=
1 Zsh1
1 −rt∗
−rt |rt |2
Vb1 0
0 Vb2
V1 pu V2 pu
(15) where Vb1 , Vb2 , Ib1 and Ib2 are the base values for voltages and currents, respectively. Upon solving for currents, the above equation becomes:
I1 pu −I2 pu
=
1 Zsh1
Ib1 0
0 Ib2
−1
1 −rt∗
−rt |rt |2
Vb1 0
0 Vb2
V1 pu V2 pu (16)
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Fig. 6. Equivalent circuits of a regulating transformer with transformer impedance reduced to the primary side and parameters in per unit. (a) First option. (b) Second option.
Fig. 8. Equivalent circuits of a regulating transformer with transformer impedance reduced to the secondary side and parameters in per unit. (a) First option. (b) Second option.
Table 1 Per unit parameters of a regulating transformer – a = 1∠␣ . First option (Figs. 6a and 8a) Y1 pu = Y3 pu =
Second option (Figs. 6b and 8b)
1−a Zsh pu
Y1 pu = Y3 pu =
Z2 pu = a∗ Zsh pu a − a∗ Jpu = V2 pu Zsh pu
1 − a∗ Zsh pu
Z2 pu = aZsh pu a∗ − a Jpu = V2 pu Zsh pu
Fig. 7. Equivalent circuit of a regulating transformer with transformer impedance reduced to the secondary side.
while the admittance equations may be expressed as:
Taking into account that Zb1 =
2 Vb1
Zsh1
Sb
pu
it results
I1 pu −I2 pu
=
1 Zsh1 pu
=
Zsh1 Zb1 1 −a∗
(17)
−a |a|2
V1 pu V2 pu
(18)
where a=
rt rb
rb =
Vb1 Vb2
(19)
The proof of this result is presented in Appendix A. After applying the same procedure that for the equations in physical quantities, the following two forms are derived:
I1 pu −I2 pu
=
1
Zsh1 pu
1 −a −a |a|2
V1 pu V2 pu
+
0 (a − a∗ )V1 pu
(20a)
I1 pu −I2 pu
=
1
Zsh1 pu
1 −a∗
−a∗ |a|2
V1 pu V2 pu
+
(a∗ − a)V2 pu 0
(20b)
The equivalent circuits that correspond to these equations are those depicted in Fig. 6. 3.1.2. Transformer impedance reduced to the secondary side Fig. 7 shows the equivalent circuit of the regulating transformer with the impedance reduced to the secondary side. The equations of this circuit may be written as follows: V1 = rt (V2 + Zsh2 I2 ) I2 I1 = ∗ rt
(21)
I1 −I2
=
1 Zsh2
1/|rt |2 −1/rt
−1/rt∗ 1
V1 V2
(22)
As for the previous case, the off-diagonal elements of the admittance matrix are not equal and a passive RLC circuit cannot be synthesized. The equations can be manipulated considering again the same two options that for the previous case. The procedure to obtain the parameters of these circuits in per unit will be the same that was applied to the previous case. Fig. 8 depicts the new equivalent circuits. All the equivalent circuits have been derived with system base values selected in an arbitrary manner. However, it is possible to obtain different and simplified equivalent circuits when the selection of base values is made following a certain criterion. For instance, the rating voltages of a transformer with control of phaseangle only are known and the base values can be selected such that rt = rb , then a = 1∠␣ and Zsh1
pu
= Zsh2 pu = Zsh pu = z
Vr2 Sb V 2 Sn
(23)
b
where Vr and Vb can be respectively the rating voltage and the base voltage of either the primary or the secondary side. Table 1 shows the results. Note that the parameters of one column can be deduced from those shown in the other column just by changing a per a* , and vice versa. 3.2. Transformer with control of voltage magnitude The regulating transformer with control of voltage magnitude only can be seen as a particular case of the transformer analyzed in the previous section, for which ˛ = 0, and rt = rt∗ . The equations obtained upon substitution of these conditions are presented below.
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Fig. 10. Example 1A: Diagram of the test system.
Fig. 9. Equivalent circuit of a transformer with control of voltage magnitude.
Z2 Table 2 Calculation of parameters of a regulating transformer with control of voltage magnitude. Parameters
Transformer impedance reduced to primary side 1 − rt Zsh1 Zsh1 Z2 = rt rt − 1 Y3 = rt Zsh1 1−a Y1 pu = Zsh1 pu Zsh1 pu Z2 pu = a a−1 Y3 pu = a Zsh1 pu Y1 =
Physical quantities
Per unit values
Transformer impedance reduced to secondary side Y1 =
1 − rt
rt2 Zsh2 Z2 = rt Zsh2 rt − 1 Y3 = rt Zsh2
I1 −I2
1
=
Zsh1
1 −rt
−rt |rt |2
V1 V2
1−a a2 Zsh2 pu Z2 pu = aZsh2 pu a−1 Y3 pu = aZsh2 pu
Y1 pu =
(24)
Note that this results from Eqs. (10), (11) or (13). b. Transformer impedance reduced to the secondary side:
I1 −I2
=
1
Zsh2
1/|rt |2 −1/rt
−1/rt 1
V1 V2
(25)
When the transformer does not control the phase angle, the voltage-controlled current source is cancelled out, and only one equivalent circuit can be synthesized. Note that the off-diagonal elements of the admittance matrix are now equal in both cases; this means that a passive RLC circuit can be now synthesized. Fig. 9 shows the new equivalent circuit, which is the circuit that results from removing the current source from the circuits shown in Figs. 4 and 5. The derivation of the parameters of the new transformer model in per unit may be made by following the procedure used with the previous transformer model. Table 2 shows the forms to be applied for calculating the parameters of the transformer in physical quantities and in per unit when the transformer impedance is reduced to either side. An important aspect is the equivalent circuit that must be used when analyzing a transformer with control of voltage magnitude. This has been already discussed in Section 2. It can be proved that the parameters of the two equivalent circuits are the same when the turns ratio is known. For instance, the parameter Z2 pu in both circuits can be expressed as shown below.
pu
=
Zsh1 pu a
=
Vr1 Vr2 Sb rb Zsh1 =z rt Zb1 Vb1 Vb2 Sn
• Transformer impedance reduced to secondary side
rt Zsh2 Vr1 Vr2 Sb =z rb Zb2 Vb1 Vb2 Sn
(26b)
That is, when system base values and nameplate ratings are known, it is irrelevant the side at which the transformer impedance has been reduced, irrespectively of the value of parameter a. However, this does not mean that either circuit can be used when the goal is to obtain the value of a, as shown in Example 1, see next section.
The examples analyzed in this section present the application of the equivalent circuit of a transformer with control of voltage magnitude and a transformer with control of phase angle, respectively. Either example can be solved by following more than one procedure; in this paper, the first example illustrates how to implement the equivalent circuit of a regulating transformer with control of voltage magnitude only in load flow equations, the second example shows how to use the equivalent circuit of a phase-shifting transformer. Both cases or similar configurations have been frequently studied in the literature, see [6,7,9,10]; the objectives pursued here are to emphasize the importance of selecting the adequate equivalent circuit for the first example, and the possibility of using the new equivalent circuit of a regulating transformer for the second example. The calculations are performed with per unit values in both examples. 4.1. Example 1 This example has two parts; both are based on the same configuration and the same operating conditions. The case analyzed in the first part assumes that the transformer taps are at the primary side, while the taps are located at the secondary side in the second part. As reasoned above, the calculations of parameters should be based on the tap location. The example is solved with an ideal lossless model of the transformer, using load flow equations and the Newton-Raphson method. 4.1.1. Voltage at the primary side is unknown Consider the system depicted in Fig. 10. The goal is to determine the rated primary voltage to achieve a secondary voltage of 25 kV, when the voltage at the primary side is 220 kV and the load at the secondary side is that shown in the figure. Note that the problem can be solved by considering that the transformer is either tapped or untapped. System base values and nameplate ratings are • Voltages: rated = Vr1 /25 kV, Base = 220/25 kV. • Power: rated = 20 MVA, Base = 100 MVA.
• Transformer impedance reduced to primary side Z2
= aZsh2 pu =
4. Examples
a. Transformer impedance reduced to the primary side:
pu
(26a)
Since the known rated voltage is the secondary one, the transformer impedance is obtained by means of Eq. (3b); it results Zsh2 = j0.40 pu.
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The operating conditions in per unit are as follows: • Node 1 voltage: V1 = 1 pu. • Node 2 voltage: V2 = 1 pu. • Node 2 power demand: SD2 = 0.16 + j0.12 pu. As for parameter a, it is a=
rt Vr1 /25 Vr1 = = rb 220 220/25
Fig. 11. Example 1B: Diagram of the test system.
from where Vr1 = 220a. The admittance matrix of the system is that of the transformer. Using the expressions for the per unit parameters of the equivalent circuit shown in Table 2 (transformer impedance reduced to the secondary side), the following admittance matrix is obtained [6,7]:
⎡
1
−
2 YBUS = ⎣ Zsh2 a 1 − Zsh2 a
⎤
1 Zsh2 a 1 Zsh2
⎦ pu
• Voltages: rated = 220/Vr2 kV, Base = 220/25 kV. • Power: rated = 20 MVA, Base = 100 MVA.
Since Zsh2 = j0.4, it results
⎡
YBUS = ⎣
2.5 a2 2.5 j a
−j
j
⎤
2.5 a
4.1.2. Voltage at the secondary side is unknown Consider again the same case, but assume now that the unknown rated voltage is that of the secondary side. The system to be analyzed is then that depicted in Fig. 11. The equivalent circuit is again that of Fig. 9, but per unit parameters are obtained now by using the transformer impedance reduced to its primary side. System base values and nameplate ratings are now:
Since the known rated voltage is now the primary one, the transformer impedance is obtained by means of Eq. (3a). The resulting value is, however, the same; i.e., Zsh1 = j0.40 pu. The operating conditions in per unit are the same that for the previous case. As for parameter a, it is now
⎦ pu
−j2.5
Power equations for node 2 are as follows: P2 = V2 [V1 (G21 cos ı21 + B21 sin ı21 ) + V2 G22 ] Q2 = V2 [V1 (G21 sin ı21 − B21 cos ı21 ) − V2 B22 ]
a=
pu
where V1 and V2 are node voltages, ı1 and ı2 are voltage phase angles (ı21 = ı2 − ı1 ), Gij and Bij are respectively the real and imaginary part of the admittance matrix elements, in per unit. Since the transformer is lossless (i.e., conductances are neglected), the resulting power equations are P2 = V2 V1 B21 sin ı21 Q2 = −V2 [V1 B21 cos ı21 + V2 B22 ]
220/Vr2 25 rt = = Vr2 rb 220/25
from where Vr2 = 25/a. Using again the expressions for the per unit parameters of the equivalent circuit shown in Table 2 (transformer impedance reduced to the primary side), the resulting admittance matrix is
⎡ YBUS = ⎣
pu
Letting ı1 = 0, then ı21 = ı2 − ı1 = ı2 = ı.Since V1 = 1
V2 = 1
B21 =
2.5 a
2.5 sin ı a
Q2 = −
pu YBUS =
2.5 cos ı + 2.5 a
pu
To apply the Newton-Raphson method, the equations to be solved are written as follows: fp = P2specified − P2cal = 0 fq = Q2specified − Q2cal = 0
pu
Upon substitution of the above expressions for P2 and Q2 , the final equations are 2.5 sin ı = 0 a 2.5 fq = −2.62 + cos ı = 0 a fp = −0.16 −
pu
(27)
It is a system of nonlinear algebraic equations whose unknown values are ı and a. The application of the Newton-Raphson method will be based on the following algorithm:
ı a
=
(m+1)
ı a
−
(m)
Zsh1 a Zsh1
−
−
a
Zsh1 a2 Zsh1
⎤ ⎦
Taking into account that Zsh1 = j0.4, it results B22 = −2.5
it finally results P2 =
1
∂fp /∂ı ∂fq /∂ı
∂fp /∂a ∂fq /∂a
−1 (m)
fp fq
(m)
The results of these equations are ı = −3.495◦ and a = 0.952, from where Vr1 = 220a ≈ 209.5 kV.
−j2.5 j2.5a
j2.5a −j2.5a2
The power equations for node 2 are obviously the same that were obtained for the previous case. Letting again ı1 = 0, the equations to be solved are now as follows: fp = −0.16 − 2.5a sin ı = 0 fq = −0.12 + 2.5a cos ı − 2.5a2 = 0
pu
(28)
Upon application of the Newton-Raphson method in the same manner as for the previous case the results are now ı = −3.875◦ and a = 0.947, from where Vr2 ≈ 26.4 kV. One can observe that the operating conditions in both cases are the same but the power flow equations to be solved, (27) and (28), are different, and so they are the results. The application of one or the other circuit is dictated by the rated voltage that is known in advance. Note, however, that Eqs. (27) and (28) would be the same if a = 1, which would occur if system base values were equal to nameplate ratings. 4.2. Example 2 Fig. 12 depicts the diagram of the system to be analyzed in this example. The goal is to estimate the phase shifting that must be provided by the regulating transformer to obtain the same active
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The voltage equation for the transformer can be written as follows: V1 = V2 + Z2 I12 = V2 + jXTR · 1∠−˛ · (I23 + J + I20 )
pu
where I20 = V2 Y3 = V2
1 − 1∠˛ jXTR
pu
Upon substitution of the expression of the current source J, it yields
Fig. 12. Example 2: Diagram of the test system.
V1 = V2 + jXTR I23 · 1∠−˛ + V1 (1 − 1∠−2˛ ) + V2 (1∠−˛ − 1)
pu
This equation is reordered as V1 · 1∠−2˛ = V2 · 1∠−˛ + jXTR I23 · 1∠−˛
pu
From which one obtains V1 · 1∠−˛ = V2 + jXTR I23
pu
Since V2 = V3 + jX23 I23
pu
the equation for line L2 may be finally written as follows: V1 · 1∠−˛ = V3 + j(XTR + X23 )I23 Fig. 13. Example 2: Equivalent circuit.
The active power equation from this result can be expressed as follows: P23 =
power flow in lines L1 and L2. The parameters of the system components, the voltages at nodes 1 and 3, and the active power demand at node 3 are shown in the figure. The study will be made by assuming an ideal lossless system, and neglecting the shunt admittances of the lines. The calculations will be performed by using Vb = 220 kV, Sb = 100 MVA as base values. The phase-angle regulating transformer is represented by means of the equivalent circuit of Fig. 8a. Fig. 13 depicts the equivalent circuit of the system. Since system base values are equal to nameplate ratings of the transformer, then a = 1∠␣ . The per unit values of the line parameters are X13 = j0.0413 X23 = 0.0723
pu
As for the transformer parameters, their per unit values are 1−a 1 − 1∠˛ = ZTR jXTR jXTR ZTR = Z2 = = jXTR · 1∠−˛ a 1∠˛ 1∠˛ − 1∠−˛ a − a∗ V1 = V1 J= ZTR jXTR Y1 = Y3 =
pu
where XTR = 0.06 pu. The operating conditions in per unit are as follows: V1 = 1.0∠ı1 V3 = 0.9545∠ı3 P3 = 1.6 P13 = P23 = 0.80
pu
From the active power equation of line L1 P13 =
V1 V3 sin ı13 X13
pu
the power angle ı13 can be deduced since all other values are known. This equation yields ı13 = 1.98◦ .
pu
V1 V3 sin(ı13 − ˛) XTR + X23
pu
The resulting value is ˛ = −4.39◦ , which means that the regulating transformer must advance the secondary voltage 4.39◦ with respect to the primary voltage. Other approaches can be used to solve this case; e.g., load flow equations following the procedure presented in [11]. In fact, the problem may be solved more easily by using a transformer model based on an ideal phase-shifting transformer, see for instance [9]. However, the new model may represent an advantage when using some computer programs to solve cases in which the regulating transformer is part of a larger system and it can control both the voltage magnitude and the phase angle, and both the turns ratio and the phase shifting must be determined. 5. Conclusions This paper has presented a method to obtain the equivalent circuit of a regulating transformer with control of both voltage magnitude and phase angle. The model is adequate for steadystate calculations under balanced conditions and does include a representation of the transformer core. It has been shown that it is possible to consider more than one equivalent circuit and that the selection of the most adequate representation depends on the case under study. The most important conclusions may be summarized as follows: 1. The equivalent circuit of the most general regulating transformer (i.e., that with control of voltage magnitude and phase angle) includes a voltage-controlled current source, which cancels out when the phase angle is not controlled. 2. The calculations with parameters in per unit can be made by selecting system base values in an arbitrary manner, although a more simple representation of the transformer is obtained when nameplate ratings and base values are equal, or when certain
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conditions are fulfilled (e.g., rb = rt ). This conclusion is not new, see for instance [7]. 3. The calculation of the transformer parameters is different if the transformer impedance must be reduced to the primary or to the secondary side. This fact is emphasized in very few textbooks. The selection of one or the other side depends on the side at which the transformer taps are installed. Acknowledgement The author wants to express his gratitude to Dr. Miquel Salichs for his help in the preparation of this paper. He is behind some of the main contributions presented in the paper.
Element (2,1) – after multiplying numerator and denominator by Vb1 Vb2 , it results −rt∗ ·
1 Zsh1
= −a∗ ·
V2 V r∗ Z Vb1 Vb1 Vb2 1 · · = −rt∗ · · b1 · b2 = − t · b1 Ib2 Vb1 Vb2 Zsh1 Sb Vb1 rb Zsh1
1 Zsh1 pu
|rt |2 ·
1 Zsh1
·
2 V2 V2 Vb2 Vb1 Vb2 1 |rt |2 Z · 2 · = |rt |2 · · b1 · b2 = 2 · b1 2 Ib2 V Vb2 Zsh1 Sb V Zsh1 rb b1 b1
1 Zsh1 pu
This appendix shows the derivation of Eq. (18) from Eq. (16), which may be expressed as follows:
where a = rt /rb and rb = Vb1 /Vb2 .
References
I1
pu
−I2 pu
⎡
Vb1 1 ⎢ Ib1 = ⎣ V Zsh1 −rt∗ b1 Ib2
Vb2 Ib1 V |rt |2 b2 Ib2 −rt
⎤
⎥ V1 pu ⎦ V
(A.1)
2 pu
The derivation of the matrix elements is presented below. Element (1,1) – after multiplying numerator and denominator by Vb1 , it results 1 Zsh1
·
V2 Vb1 Vb1 1 Z 1 · = · b1 = b1 = Ib1 Vb1 Zsh1 Sb Zsh1 Zsh1 pu
(A.2)
Element (1,2) – after multiplying numerator and denominator 2 , it results by Vb1 −rt ·
1 Zsh1
= −a ·
·
2 V2 V Vb2 Vb1 1 rt Z · 2 = −rt · · b1 · b2 = − · b1 Ib1 V Zsh1 Sb1 Vb1 rb Zsh1 b1
1 Zsh1 pu
(A.3)
(A.4)
Element (2,2) – after multiplying numerator and denominator 2 V , it results by Vb1 b2
= |a|2 ·
Appendix A.
·
(A.5)
[1] R.M. del Vecchio, B. Poulin, P.T. Feghali, D.M. Shah, R. Ahuja, Transformer Design Principles, second edition, CRC Press, 2010. [2] P.S. Georgilakis, Spotlight on Modern Transformer Design, Springer, 2009. [3] S.V. Kulkarni, S.A. Khaparde, Transformer Engineering. Design and Practice, Marcel Dekker, 2005. [4] J.H. Harlow, Electric Power Transformer Engineering, CRC Press, 2004. [5] The J & P Transformer Book, Newnes, 1998. [6] J.J. Grainger, W.D. Stevenson, Power System Analysis, McGraw-Hill, 1994. [7] J. Duncan Glover, M. Sarma, Power System Analysis and Design, PWS Publishers, 1987. [8] J.A. Martinez, R. Walling, B.A. Mork, J. Martin-Arnedo, D. Durbak, Parameter determination for modeling system transients – part III: transformers, IEEE Transactions on Power Delivery 20 (July (3)) (2005) 2051–2062. [9] O.I. Elgerd, Electric Energy Systems Theory. An Introduction, McGraw-Hill, 1982. [10] J.C. Das, Power System Analysis, Marcel Dekker, 2002. [11] N.M. Peterson, W. Scott Meyer, Automatic adjustment of transformer and phase-shifter taps in the Newton power flow, IEEE Transactions on Power Apparatus and Systems 90 (January/February (1)) (1971) 103–108.
Contents lists available at ScienceDirect
Electric Power Systems Research journal homepage: www.elsevier.com/locate/epsr
Equivalent circuit of transformers with control of voltage and phase angle Juan A. Martinez-Velasco ∗ Dept. d’Eng. Elèctrica, Universitat Politècnica de Catalunya, Diagonal 647, 08028 Barcelona, Spain
a r t i c l e
i n f o
Article history: Received 8 October 2010 Received in revised form 19 January 2011 Accepted 31 January 2011 Available online 24 February 2011 Keywords: Power transformer Regulating transformer Equivalent circuit Steady-state analysis
a b s t r a c t This paper presents a methodology aimed at obtaining the equivalent circuit of regulating transformers. This aspect is insufficiently treated in most current textbooks in which the equivalent circuit of a tapped transformer is not well justified. The document shows how to obtain the equivalent circuit of a regulating transformer with control of voltage magnitude and phase angle, and how to apply it using either physical quantities or per unit (pu) quantities. The resulting equivalent circuit is adequate for steady-state calculations under balanced conditions and neglects transformer core parameters. The usage of per unit values can be made by selecting the base quantities in an arbitrary manner, which is an obvious advantage when the turns ratio must be estimated. © 2011 Elsevier B.V. All rights reserved.
1. Introduction The regulating transformer, with control of voltage magnitude and/or phase angle, is a common device in power systems. The control of the voltage magnitude is a useful capability to control voltage and reactive power flow, whereas the control of the phase angle can be used to control active power flow [1–7]. A problem that arises when dealing with transformers, with or without regulating capabilities, is the way in which the parameters of the equivalent circuit have to be calculated. This problem is complicated when the calculations must be performed with per unit quantities, and transformer ratings are not equal to base values. There are several reasons to select base values different from transformer ratings. For instance, when the system under study has several transformers and each transformer has different ratings. The representation of a regulating transformer in which either the turns ratio or the phase shift (or even both) is unknown is different from that of a conventional transformer since the parameters of the equivalent circuit are expressed as a function of the unknown values. Graduate and undergraduate students learn how to obtain the per unit parameters of the equivalent circuit of a transformer from the values measured in standardized tests following well established procedures. In addition, textbooks usually refine the procedure when system base values may be different from the nameplate ratings of any particular transformer. However, the
derivation of the equivalent circuit is not usually well justified when arbitrary system base values are used. The representation of an actual transformer when calculations are performed with physical quantities requires the use of the ideal transformer in the equivalent circuit. In general, this device is not needed when the calculations are made with per unit values. A tapped transformer with parameters in per unit is represented in textbooks with or without the ideal transformer, being the second approach (without ideal transformer) the preferred one when a load flow solution is used to determine the turns ratio. However, an approach without using the ideal transformer has not been proposed when the transformer can control the phase angle. The goal of this paper is to justify the derivation of the equivalent circuit of a two-winding regulating transformer with control of both voltage magnitude and phase angle, with parameters expressed in either physical quantities or per unit values, and using arbitrary system base values. The paper shows that it is possible to obtain more than one equivalent circuit using basic circuit components for the most general case, and that simplified representations are easily derived when only one regulating capability (e.g., load-tap-changing) is implemented in the transformer under study. The equivalent circuits considered in this paper are only valid for steady-state analysis, and when the system runs under balanced conditions, since only a single-phase representation is assumed. 2. Representation of two-winding transformers 2.1. Equivalent circuits
∗ Tel.: +34 934016725. E-mail address: [email protected] 0378-7796/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.epsr.2011.01.019
Fig. 1 shows the equivalent circuit of a single-phase twowinding transformer. This circuit can be also used to represent
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For instance, using the following symbols: • • • •
Fig. 1. Equivalent circuit of a single-phase two-winding transformer.
Rated power Sr , in MVA Rated voltages Vr1 , Vr2 , in kV Short-circuit impedance z, in pu Short-circuit losses Wsh , in MW
and assuming that the short-circuit current is the rated current, the parameters of the circuit shown in Fig. 2b may be calculated as follows: Z2 = z
2 Vr2
Sr
X2 = ωL2 =
R2 =
2 Wsh Vr2 Sr Sr
Z22 − R22
(ω = 2f )
(1a) (1b)
where Z2 is the transformer impedance module reduced to the secondary side and f is the power frequency. To obtain the parameters of the model depicted in Fig. 2a (i.e., reduced to the primary side), just change subscript “2” by “1” in Eqs. (1). In the above expressions, rated voltages are the phase-to-phase voltages for a three-phase transformer and the phase-to-neutral voltage for a single-phase transformer. The rated power must include the power of all transformer phases. In the remaining sections of this paper the following symbols will be used: Zsh1 = R1 + jX1
Z1 = |Zsh1 |
(2a)
Zsh2 = R2 + jX2
Z2 = |Zsh2 |
(2b)
2.2. Discussion Taking into account the transformer models shown in Fig. 2, the calculation of the transformer impedance reduced to either the primary or to the secondary side may be obtained as follows: Fig. 2. Simplified equivalent circuit of a two-winding transformer. (a) Parameters reduced to the primary side. (b) Parameters reduced to the secondary side.
Z1 = z Z2 = z
a three-phase two-winding transformer under balanced steadystate conditions. In case of requiring a model for a more general analysis (e.g., transient response under unbalanced conditions), the representation of the three phases and both the primary and the secondary connections should be included [8]. The physical meaning of the parameters to be specified in this representation is as follows: • Rm and Lm are the core parameters; Rm represents core losses and Lm represents the core flux, common to both primary and secondary windings. • Rp , Lp , Rs and Ls are the winding parameters; Rp and Rs represent the losses of the primary and the secondary windings due to Joule effect, while Lp and Ls represent the leakage fluxes of primary and secondary windings, respectively. • NP /NS is the turns ratio, which is also the ratio of the rated voltages and currents In some cases, the influence of the core parameters can be neglected and then the winding parameters are merged into two basic components. Fig. 2 shows the two possible representations, which are the models used in this paper. The parameters to be specified in any of the models depicted in Fig. 2 can be obtained from nameplate ratings and values derived from the standard short-circuit test.
2 Vr1
Sr 2 Vr2
Sr
Zsh1 = ZP + rt2 ZS
(3a)
ZP
(3b)
Zsh2 =
rt2
+ ZS
where rt is the turns ratio rt =
NP Vr1 Ir2 = = NS Vr2 Ir1
(4)
For practical purposes, the variation of the per unit transformer impedance with the operating tap is assumed constant. In practice, this impedance value depends on the tap position and is not always known. Since transformer taps can be on either side of the transformer, it is important to account for the consequences of assuming that the per unit transformer impedance is constant. The following results may be derived then from Eqs. (3a) and (3b) when z is assumed constant: a. When the transformer taps are at the primary winding, Vr2 is known and constant; therefore, Z2 (=|Zsh2 |) can be determined by using (3b) and is also constant, while Vr1 is unknown and Z1 (=|Zsh1 |) cannot be determined. b. When the transformer taps are at the secondary winding, Vr1 is known and constant; therefore, Z1 (=|Zsh1 |) can be determined by using (3a) and is also constant, while Vr2 is unknown and Z2 (=|Zsh2 |) cannot be determined. A conclusion from these results is that the regulating transformer must be represented with the transformer impedance
J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
Fig. 3. Equivalent circuit of a regulating transformer with transformer impedance reduced to the primary side.
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Fig. 4. Equivalent circuit of a regulating transformer – first option.
reduced to the secondary side (i.e., using Eq. (3b)) when the taps are at the primary side, and with the transformer impedance reduced to the primary side (i.e., using Eq. (3a)) when the taps are at the secondary side. 3. Equivalent circuit of a transformer with voltage and phase angle control Fig. 5. Equivalent circuit of a regulating transformer – second option.
Assume that the transformer can control both voltage magnitude and phase angle. In such case, its turns ratio is a complex value that may be expressed as: rt = |rt |∠˛ = |rt | · (cos ˛ + j sin ˛)
(5)
The relationship between ratings in a two-winding transformer is ∗ ∗ = Vr2 Ir2 Vr1 Ir1
(6)
From where rt =
I∗ Vr1 = r2 ∗ Vr2 Ir1
(7)
which can be also expressed as: Vr1 = rt Vr2
Ir1 =
Ir2 rt∗
3.1. Transformer with control of voltage and phase 3.1.1. Transformer impedance reduced to the primary side Fig. 3 shows the equivalent circuit of the transformer with the transformer impedance reduced to the primary side. The equations of this circuit may be written as follows: V1 = rt V2 + Zsh1 I1 = rt V2 + I2 rt∗
Zsh1 I2 rt∗
(9)
Taking into account the convention for the admittance equations, the above equations can be written as follows:
I1 −I2
=
1 Zsh1
1 −rt∗
−rt |rt |2
V1 V2
I1 −I2
=
1 Zsh1
1 −rt
−rt |rt |2
V1 V2
+
0 (rt − rt∗ )V1
(11)
Zsh1 rt rt − rt∗ J= V1 Zsh1 Z2 =
(12)
b. Eq. (10) can be also rewritten as follows: I1 −I2
=
1 Zsh1
1 −rt∗
−rt∗ |rt |2
V1 V2
+
(rt∗ − rt )V2 0
(13)
The off-diagonal of the new admittance matrix are equal, so it can be also synthesized as a passive circuit. As for the second part of the right hand side, it is again a voltage-controlled current source, but this time the controlling voltage is V2 . The new equivalent circuit is that depicted in Fig. 5. The parameters of the new circuit are as follows: 1 − rt∗ Z Z2 = sh1 Zsh1 rt∗ |rt |2 − rt∗ r ∗ − rt Y3 = J= t V2 Zsh1 Zsh1 Y1 =
(14)
It is important to note that expressions (12) and (14) can lead to values of Y1 and Y3 whose real part is negative. The procedure to obtain the equations in per unit will be based on Eq. (10), which can be rewritten as:
(10)
Since the turns ratio is a complex value and the off-diagonal elements are not equal, a passive RLC circuit cannot be synthesized directly from this admittance matrix. However, the equations can be manipulated to obtain a matrix with equal off-diagonal elements. There are, at least, two options. a. Eq. (10) is rewritten as follows:
1 − rt Zsh1 |r|2 − rt Y3 = Zsh1
Y1 =
(8)
These equations are valid for any voltage and load level; that is, the relationship between voltages and currents at both sides of the ideal transformer is the same that for rated values. The rest of this section is dedicated to deduce the equivalent circuits of regulating transformers, beginning with the most general case, for which the turns ratio is a complex value.
I1 =
The matrix of the right hand side of this equation has equal off-diagonal elements, so it can be now synthesized as a passive circuit. On the other hand, the second part of the right hand side can be synthesized as a current source controlled by the voltage V1 . The equivalent circuit of the regulating transformer may be that depicted in Fig. 4. The parameters of this circuit can be obtained as follows:
Ib1 0
0
Ib2
I1 pu −I2 pu
=
1 Zsh1
1 −rt∗
−rt |rt |2
Vb1 0
0 Vb2
V1 pu V2 pu
(15) where Vb1 , Vb2 , Ib1 and Ib2 are the base values for voltages and currents, respectively. Upon solving for currents, the above equation becomes:
I1 pu −I2 pu
=
1 Zsh1
Ib1 0
0 Ib2
−1
1 −rt∗
−rt |rt |2
Vb1 0
0 Vb2
V1 pu V2 pu (16)
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Fig. 6. Equivalent circuits of a regulating transformer with transformer impedance reduced to the primary side and parameters in per unit. (a) First option. (b) Second option.
Fig. 8. Equivalent circuits of a regulating transformer with transformer impedance reduced to the secondary side and parameters in per unit. (a) First option. (b) Second option.
Table 1 Per unit parameters of a regulating transformer – a = 1∠␣ . First option (Figs. 6a and 8a) Y1 pu = Y3 pu =
Second option (Figs. 6b and 8b)
1−a Zsh pu
Y1 pu = Y3 pu =
Z2 pu = a∗ Zsh pu a − a∗ Jpu = V2 pu Zsh pu
1 − a∗ Zsh pu
Z2 pu = aZsh pu a∗ − a Jpu = V2 pu Zsh pu
Fig. 7. Equivalent circuit of a regulating transformer with transformer impedance reduced to the secondary side.
while the admittance equations may be expressed as:
Taking into account that Zb1 =
2 Vb1
Zsh1
Sb
pu
it results
I1 pu −I2 pu
=
1 Zsh1 pu
=
Zsh1 Zb1 1 −a∗
(17)
−a |a|2
V1 pu V2 pu
(18)
where a=
rt rb
rb =
Vb1 Vb2
(19)
The proof of this result is presented in Appendix A. After applying the same procedure that for the equations in physical quantities, the following two forms are derived:
I1 pu −I2 pu
=
1
Zsh1 pu
1 −a −a |a|2
V1 pu V2 pu
+
0 (a − a∗ )V1 pu
(20a)
I1 pu −I2 pu
=
1
Zsh1 pu
1 −a∗
−a∗ |a|2
V1 pu V2 pu
+
(a∗ − a)V2 pu 0
(20b)
The equivalent circuits that correspond to these equations are those depicted in Fig. 6. 3.1.2. Transformer impedance reduced to the secondary side Fig. 7 shows the equivalent circuit of the regulating transformer with the impedance reduced to the secondary side. The equations of this circuit may be written as follows: V1 = rt (V2 + Zsh2 I2 ) I2 I1 = ∗ rt
(21)
I1 −I2
=
1 Zsh2
1/|rt |2 −1/rt
−1/rt∗ 1
V1 V2
(22)
As for the previous case, the off-diagonal elements of the admittance matrix are not equal and a passive RLC circuit cannot be synthesized. The equations can be manipulated considering again the same two options that for the previous case. The procedure to obtain the parameters of these circuits in per unit will be the same that was applied to the previous case. Fig. 8 depicts the new equivalent circuits. All the equivalent circuits have been derived with system base values selected in an arbitrary manner. However, it is possible to obtain different and simplified equivalent circuits when the selection of base values is made following a certain criterion. For instance, the rating voltages of a transformer with control of phaseangle only are known and the base values can be selected such that rt = rb , then a = 1∠␣ and Zsh1
pu
= Zsh2 pu = Zsh pu = z
Vr2 Sb V 2 Sn
(23)
b
where Vr and Vb can be respectively the rating voltage and the base voltage of either the primary or the secondary side. Table 1 shows the results. Note that the parameters of one column can be deduced from those shown in the other column just by changing a per a* , and vice versa. 3.2. Transformer with control of voltage magnitude The regulating transformer with control of voltage magnitude only can be seen as a particular case of the transformer analyzed in the previous section, for which ˛ = 0, and rt = rt∗ . The equations obtained upon substitution of these conditions are presented below.
J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
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Fig. 10. Example 1A: Diagram of the test system.
Fig. 9. Equivalent circuit of a transformer with control of voltage magnitude.
Z2 Table 2 Calculation of parameters of a regulating transformer with control of voltage magnitude. Parameters
Transformer impedance reduced to primary side 1 − rt Zsh1 Zsh1 Z2 = rt rt − 1 Y3 = rt Zsh1 1−a Y1 pu = Zsh1 pu Zsh1 pu Z2 pu = a a−1 Y3 pu = a Zsh1 pu Y1 =
Physical quantities
Per unit values
Transformer impedance reduced to secondary side Y1 =
1 − rt
rt2 Zsh2 Z2 = rt Zsh2 rt − 1 Y3 = rt Zsh2
I1 −I2
1
=
Zsh1
1 −rt
−rt |rt |2
V1 V2
1−a a2 Zsh2 pu Z2 pu = aZsh2 pu a−1 Y3 pu = aZsh2 pu
Y1 pu =
(24)
Note that this results from Eqs. (10), (11) or (13). b. Transformer impedance reduced to the secondary side:
I1 −I2
=
1
Zsh2
1/|rt |2 −1/rt
−1/rt 1
V1 V2
(25)
When the transformer does not control the phase angle, the voltage-controlled current source is cancelled out, and only one equivalent circuit can be synthesized. Note that the off-diagonal elements of the admittance matrix are now equal in both cases; this means that a passive RLC circuit can be now synthesized. Fig. 9 shows the new equivalent circuit, which is the circuit that results from removing the current source from the circuits shown in Figs. 4 and 5. The derivation of the parameters of the new transformer model in per unit may be made by following the procedure used with the previous transformer model. Table 2 shows the forms to be applied for calculating the parameters of the transformer in physical quantities and in per unit when the transformer impedance is reduced to either side. An important aspect is the equivalent circuit that must be used when analyzing a transformer with control of voltage magnitude. This has been already discussed in Section 2. It can be proved that the parameters of the two equivalent circuits are the same when the turns ratio is known. For instance, the parameter Z2 pu in both circuits can be expressed as shown below.
pu
=
Zsh1 pu a
=
Vr1 Vr2 Sb rb Zsh1 =z rt Zb1 Vb1 Vb2 Sn
• Transformer impedance reduced to secondary side
rt Zsh2 Vr1 Vr2 Sb =z rb Zb2 Vb1 Vb2 Sn
(26b)
That is, when system base values and nameplate ratings are known, it is irrelevant the side at which the transformer impedance has been reduced, irrespectively of the value of parameter a. However, this does not mean that either circuit can be used when the goal is to obtain the value of a, as shown in Example 1, see next section.
The examples analyzed in this section present the application of the equivalent circuit of a transformer with control of voltage magnitude and a transformer with control of phase angle, respectively. Either example can be solved by following more than one procedure; in this paper, the first example illustrates how to implement the equivalent circuit of a regulating transformer with control of voltage magnitude only in load flow equations, the second example shows how to use the equivalent circuit of a phase-shifting transformer. Both cases or similar configurations have been frequently studied in the literature, see [6,7,9,10]; the objectives pursued here are to emphasize the importance of selecting the adequate equivalent circuit for the first example, and the possibility of using the new equivalent circuit of a regulating transformer for the second example. The calculations are performed with per unit values in both examples. 4.1. Example 1 This example has two parts; both are based on the same configuration and the same operating conditions. The case analyzed in the first part assumes that the transformer taps are at the primary side, while the taps are located at the secondary side in the second part. As reasoned above, the calculations of parameters should be based on the tap location. The example is solved with an ideal lossless model of the transformer, using load flow equations and the Newton-Raphson method. 4.1.1. Voltage at the primary side is unknown Consider the system depicted in Fig. 10. The goal is to determine the rated primary voltage to achieve a secondary voltage of 25 kV, when the voltage at the primary side is 220 kV and the load at the secondary side is that shown in the figure. Note that the problem can be solved by considering that the transformer is either tapped or untapped. System base values and nameplate ratings are • Voltages: rated = Vr1 /25 kV, Base = 220/25 kV. • Power: rated = 20 MVA, Base = 100 MVA.
• Transformer impedance reduced to primary side Z2
= aZsh2 pu =
4. Examples
a. Transformer impedance reduced to the primary side:
pu
(26a)
Since the known rated voltage is the secondary one, the transformer impedance is obtained by means of Eq. (3b); it results Zsh2 = j0.40 pu.
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J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
The operating conditions in per unit are as follows: • Node 1 voltage: V1 = 1 pu. • Node 2 voltage: V2 = 1 pu. • Node 2 power demand: SD2 = 0.16 + j0.12 pu. As for parameter a, it is a=
rt Vr1 /25 Vr1 = = rb 220 220/25
Fig. 11. Example 1B: Diagram of the test system.
from where Vr1 = 220a. The admittance matrix of the system is that of the transformer. Using the expressions for the per unit parameters of the equivalent circuit shown in Table 2 (transformer impedance reduced to the secondary side), the following admittance matrix is obtained [6,7]:
⎡
1
−
2 YBUS = ⎣ Zsh2 a 1 − Zsh2 a
⎤
1 Zsh2 a 1 Zsh2
⎦ pu
• Voltages: rated = 220/Vr2 kV, Base = 220/25 kV. • Power: rated = 20 MVA, Base = 100 MVA.
Since Zsh2 = j0.4, it results
⎡
YBUS = ⎣
2.5 a2 2.5 j a
−j
j
⎤
2.5 a
4.1.2. Voltage at the secondary side is unknown Consider again the same case, but assume now that the unknown rated voltage is that of the secondary side. The system to be analyzed is then that depicted in Fig. 11. The equivalent circuit is again that of Fig. 9, but per unit parameters are obtained now by using the transformer impedance reduced to its primary side. System base values and nameplate ratings are now:
Since the known rated voltage is now the primary one, the transformer impedance is obtained by means of Eq. (3a). The resulting value is, however, the same; i.e., Zsh1 = j0.40 pu. The operating conditions in per unit are the same that for the previous case. As for parameter a, it is now
⎦ pu
−j2.5
Power equations for node 2 are as follows: P2 = V2 [V1 (G21 cos ı21 + B21 sin ı21 ) + V2 G22 ] Q2 = V2 [V1 (G21 sin ı21 − B21 cos ı21 ) − V2 B22 ]
a=
pu
where V1 and V2 are node voltages, ı1 and ı2 are voltage phase angles (ı21 = ı2 − ı1 ), Gij and Bij are respectively the real and imaginary part of the admittance matrix elements, in per unit. Since the transformer is lossless (i.e., conductances are neglected), the resulting power equations are P2 = V2 V1 B21 sin ı21 Q2 = −V2 [V1 B21 cos ı21 + V2 B22 ]
220/Vr2 25 rt = = Vr2 rb 220/25
from where Vr2 = 25/a. Using again the expressions for the per unit parameters of the equivalent circuit shown in Table 2 (transformer impedance reduced to the primary side), the resulting admittance matrix is
⎡ YBUS = ⎣
pu
Letting ı1 = 0, then ı21 = ı2 − ı1 = ı2 = ı.Since V1 = 1
V2 = 1
B21 =
2.5 a
2.5 sin ı a
Q2 = −
pu YBUS =
2.5 cos ı + 2.5 a
pu
To apply the Newton-Raphson method, the equations to be solved are written as follows: fp = P2specified − P2cal = 0 fq = Q2specified − Q2cal = 0
pu
Upon substitution of the above expressions for P2 and Q2 , the final equations are 2.5 sin ı = 0 a 2.5 fq = −2.62 + cos ı = 0 a fp = −0.16 −
pu
(27)
It is a system of nonlinear algebraic equations whose unknown values are ı and a. The application of the Newton-Raphson method will be based on the following algorithm:
ı a
=
(m+1)
ı a
−
(m)
Zsh1 a Zsh1
−
−
a
Zsh1 a2 Zsh1
⎤ ⎦
Taking into account that Zsh1 = j0.4, it results B22 = −2.5
it finally results P2 =
1
∂fp /∂ı ∂fq /∂ı
∂fp /∂a ∂fq /∂a
−1 (m)
fp fq
(m)
The results of these equations are ı = −3.495◦ and a = 0.952, from where Vr1 = 220a ≈ 209.5 kV.
−j2.5 j2.5a
j2.5a −j2.5a2
The power equations for node 2 are obviously the same that were obtained for the previous case. Letting again ı1 = 0, the equations to be solved are now as follows: fp = −0.16 − 2.5a sin ı = 0 fq = −0.12 + 2.5a cos ı − 2.5a2 = 0
pu
(28)
Upon application of the Newton-Raphson method in the same manner as for the previous case the results are now ı = −3.875◦ and a = 0.947, from where Vr2 ≈ 26.4 kV. One can observe that the operating conditions in both cases are the same but the power flow equations to be solved, (27) and (28), are different, and so they are the results. The application of one or the other circuit is dictated by the rated voltage that is known in advance. Note, however, that Eqs. (27) and (28) would be the same if a = 1, which would occur if system base values were equal to nameplate ratings. 4.2. Example 2 Fig. 12 depicts the diagram of the system to be analyzed in this example. The goal is to estimate the phase shifting that must be provided by the regulating transformer to obtain the same active
J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
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The voltage equation for the transformer can be written as follows: V1 = V2 + Z2 I12 = V2 + jXTR · 1∠−˛ · (I23 + J + I20 )
pu
where I20 = V2 Y3 = V2
1 − 1∠˛ jXTR
pu
Upon substitution of the expression of the current source J, it yields
Fig. 12. Example 2: Diagram of the test system.
V1 = V2 + jXTR I23 · 1∠−˛ + V1 (1 − 1∠−2˛ ) + V2 (1∠−˛ − 1)
pu
This equation is reordered as V1 · 1∠−2˛ = V2 · 1∠−˛ + jXTR I23 · 1∠−˛
pu
From which one obtains V1 · 1∠−˛ = V2 + jXTR I23
pu
Since V2 = V3 + jX23 I23
pu
the equation for line L2 may be finally written as follows: V1 · 1∠−˛ = V3 + j(XTR + X23 )I23 Fig. 13. Example 2: Equivalent circuit.
The active power equation from this result can be expressed as follows: P23 =
power flow in lines L1 and L2. The parameters of the system components, the voltages at nodes 1 and 3, and the active power demand at node 3 are shown in the figure. The study will be made by assuming an ideal lossless system, and neglecting the shunt admittances of the lines. The calculations will be performed by using Vb = 220 kV, Sb = 100 MVA as base values. The phase-angle regulating transformer is represented by means of the equivalent circuit of Fig. 8a. Fig. 13 depicts the equivalent circuit of the system. Since system base values are equal to nameplate ratings of the transformer, then a = 1∠␣ . The per unit values of the line parameters are X13 = j0.0413 X23 = 0.0723
pu
As for the transformer parameters, their per unit values are 1−a 1 − 1∠˛ = ZTR jXTR jXTR ZTR = Z2 = = jXTR · 1∠−˛ a 1∠˛ 1∠˛ − 1∠−˛ a − a∗ V1 = V1 J= ZTR jXTR Y1 = Y3 =
pu
where XTR = 0.06 pu. The operating conditions in per unit are as follows: V1 = 1.0∠ı1 V3 = 0.9545∠ı3 P3 = 1.6 P13 = P23 = 0.80
pu
From the active power equation of line L1 P13 =
V1 V3 sin ı13 X13
pu
the power angle ı13 can be deduced since all other values are known. This equation yields ı13 = 1.98◦ .
pu
V1 V3 sin(ı13 − ˛) XTR + X23
pu
The resulting value is ˛ = −4.39◦ , which means that the regulating transformer must advance the secondary voltage 4.39◦ with respect to the primary voltage. Other approaches can be used to solve this case; e.g., load flow equations following the procedure presented in [11]. In fact, the problem may be solved more easily by using a transformer model based on an ideal phase-shifting transformer, see for instance [9]. However, the new model may represent an advantage when using some computer programs to solve cases in which the regulating transformer is part of a larger system and it can control both the voltage magnitude and the phase angle, and both the turns ratio and the phase shifting must be determined. 5. Conclusions This paper has presented a method to obtain the equivalent circuit of a regulating transformer with control of both voltage magnitude and phase angle. The model is adequate for steadystate calculations under balanced conditions and does include a representation of the transformer core. It has been shown that it is possible to consider more than one equivalent circuit and that the selection of the most adequate representation depends on the case under study. The most important conclusions may be summarized as follows: 1. The equivalent circuit of the most general regulating transformer (i.e., that with control of voltage magnitude and phase angle) includes a voltage-controlled current source, which cancels out when the phase angle is not controlled. 2. The calculations with parameters in per unit can be made by selecting system base values in an arbitrary manner, although a more simple representation of the transformer is obtained when nameplate ratings and base values are equal, or when certain
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J.A. Martinez-Velasco / Electric Power Systems Research 81 (2011) 1349–1356
conditions are fulfilled (e.g., rb = rt ). This conclusion is not new, see for instance [7]. 3. The calculation of the transformer parameters is different if the transformer impedance must be reduced to the primary or to the secondary side. This fact is emphasized in very few textbooks. The selection of one or the other side depends on the side at which the transformer taps are installed. Acknowledgement The author wants to express his gratitude to Dr. Miquel Salichs for his help in the preparation of this paper. He is behind some of the main contributions presented in the paper.
Element (2,1) – after multiplying numerator and denominator by Vb1 Vb2 , it results −rt∗ ·
1 Zsh1
= −a∗ ·
V2 V r∗ Z Vb1 Vb1 Vb2 1 · · = −rt∗ · · b1 · b2 = − t · b1 Ib2 Vb1 Vb2 Zsh1 Sb Vb1 rb Zsh1
1 Zsh1 pu
|rt |2 ·
1 Zsh1
·
2 V2 V2 Vb2 Vb1 Vb2 1 |rt |2 Z · 2 · = |rt |2 · · b1 · b2 = 2 · b1 2 Ib2 V Vb2 Zsh1 Sb V Zsh1 rb b1 b1
1 Zsh1 pu
This appendix shows the derivation of Eq. (18) from Eq. (16), which may be expressed as follows:
where a = rt /rb and rb = Vb1 /Vb2 .
References
I1
pu
−I2 pu
⎡
Vb1 1 ⎢ Ib1 = ⎣ V Zsh1 −rt∗ b1 Ib2
Vb2 Ib1 V |rt |2 b2 Ib2 −rt
⎤
⎥ V1 pu ⎦ V
(A.1)
2 pu
The derivation of the matrix elements is presented below. Element (1,1) – after multiplying numerator and denominator by Vb1 , it results 1 Zsh1
·
V2 Vb1 Vb1 1 Z 1 · = · b1 = b1 = Ib1 Vb1 Zsh1 Sb Zsh1 Zsh1 pu
(A.2)
Element (1,2) – after multiplying numerator and denominator 2 , it results by Vb1 −rt ·
1 Zsh1
= −a ·
·
2 V2 V Vb2 Vb1 1 rt Z · 2 = −rt · · b1 · b2 = − · b1 Ib1 V Zsh1 Sb1 Vb1 rb Zsh1 b1
1 Zsh1 pu
(A.3)
(A.4)
Element (2,2) – after multiplying numerator and denominator 2 V , it results by Vb1 b2
= |a|2 ·
Appendix A.
·
(A.5)
[1] R.M. del Vecchio, B. Poulin, P.T. Feghali, D.M. Shah, R. Ahuja, Transformer Design Principles, second edition, CRC Press, 2010. [2] P.S. Georgilakis, Spotlight on Modern Transformer Design, Springer, 2009. [3] S.V. Kulkarni, S.A. Khaparde, Transformer Engineering. Design and Practice, Marcel Dekker, 2005. [4] J.H. Harlow, Electric Power Transformer Engineering, CRC Press, 2004. [5] The J & P Transformer Book, Newnes, 1998. [6] J.J. Grainger, W.D. Stevenson, Power System Analysis, McGraw-Hill, 1994. [7] J. Duncan Glover, M. Sarma, Power System Analysis and Design, PWS Publishers, 1987. [8] J.A. Martinez, R. Walling, B.A. Mork, J. Martin-Arnedo, D. Durbak, Parameter determination for modeling system transients – part III: transformers, IEEE Transactions on Power Delivery 20 (July (3)) (2005) 2051–2062. [9] O.I. Elgerd, Electric Energy Systems Theory. An Introduction, McGraw-Hill, 1982. [10] J.C. Das, Power System Analysis, Marcel Dekker, 2002. [11] N.M. Peterson, W. Scott Meyer, Automatic adjustment of transformer and phase-shifter taps in the Newton power flow, IEEE Transactions on Power Apparatus and Systems 90 (January/February (1)) (1971) 103–108.