- Email: [email protected]
The Mathematical Model of Electrical Field Distribution in Optical Voltage Transformer
Available online at www.sciencedirect.com Available online at www.sciencedirect.com
Procedia Engineering
ProcediaProcedia Engineering 00 (2011) Engineering 29 000–000 (2012) 2661 – 2666 www.elsevier.com/locate/procedia
2012 International Workshop on Information and Electronics Engineering (IWIEE)
The Mathematical Model of Electrical Field Distribution in Optical Voltage Transformer Yi-Nan Zhao*,Guo-Qing Zhang, Zhi-Zhong Guo, Song Cheng School of Electrical Engineering and Automation,Harbin Institute of Technology, Harbin, China
Abstract The distributed optical voltage transformer measures voltage through the numerical integration, the choice of the integration points and weights rely on the distribution function of the electrical field. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through calculating electric potential. The simulation results show that the mathematical model reflects the main features of the axial electric field on the central axis of the distributed optical voltage transformer, which indicates that the method to solve the model is feasible. For a optical voltage transformer with a determined structure, the electric field distribution function can be directly obtained by the mathematical model, which could improve the accuracy of measuring the voltage through the numerical integration.
© 2011 Published by Elsevier Ltd. Selection and/or peer-review under responsibility of Harbin University of Science and Technology Keywords: Optical Voltage Transformer; Mathematical model; Electrical field distribution.
1. Introduction To Power system, it is important to accurately measure the voltage. The voltage measurement method could be divided into electrical measurement method and optical measurement method. Comparing with the conventional voltage transformer based on capacitive or inductive coupling, the optical voltage transformer relies on electro-optical principles such as the Pockels effect ,thus it has many advantages: the observer is completely electrically isolated from the point of measurement, and the measurement itself is nonintrusive to the power circuit; optical sensor generally offer immunity to electromagnetic interference; the using of optical element and optical fiber reduces the dependence of the isolation costs on the voltage
*
Corresponding author. E-mail address: [email protected]
1877-7058 © 2011 Published by Elsevier Ltd. doi:10.1016/j.proeng.2012.01.368
22662
Yi-Nan Zhao et al. / Procedia Engineering 29 000–000 (2012) 2661 – 2666 Author name / Procedia Engineering 00 (2011)
level, especially for the power network with voltage level above 110kV, optical voltage transformer has a greater economic advantage; the optical sensing technology makes the transformer have a very wide bandwidth and large dynamic range, as well as reduces the transformer size and weight; optical voltage transformer outputs both analog and digital signals to adapt the automation developing trend of power system. Distributed optical voltage transformer obtains the measured voltage by the numerical integration of the electric field, the choice of the integration points and weights rely on the electric field distribution on the central axis. How to obtain the electrical field distribution function on the center axis in the transformer becomes an important problem in designing the distributed optical voltage transformer, especially for the three-phase voltage transformer system, where is still never has the mathematical description of its electrical field distribution. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through the calculation of electric potential, which could be available to improve the precision of numerical integration algorithm to calculate the measured voltage, and to improve the accuracy of distributed optical voltage transformer. 2. Problem description To solve the mathematical model of electric field distribution, a simple structure of three-phase voltage transformer is analyzed. Three-phase transformer has the same electrode structure and separated from each other by 2.5m, the structure of each phase transformer consists of two identical, disc-shaped electrodes which have a diameter of 200mm and are separated from each other by 2m. The lower electrode is grounded, and the upper electrode is energized. The dimensions of this structure are similar to the basic dimensions of 110kV stand-offs found in high-voltage substations.
Figure1: The schematic of three-phase transformer system
To the three-phase transformer system shown in figure 1, electrical field of each point inside the three phase transformer is the superposition of the electric field which is generated by ABC three-phase voltage at this point, that is formula (1) as shown. ⎡ Ex ⎤ ⎡ k Ax k Bx kCx ⎤ ⎡u A ⎤ ⎢ E ⎥ = ⎢k ⎥⎢ ⎥ (1) ⎢ y ⎥ ⎢ Ay k By kCy ⎥ ⎢uB ⎥ ⎢⎣ Ez ⎥⎦ ⎢⎣ k Az k Bz kCz ⎥⎦ ⎢⎣uC ⎥⎦ In this formula, kij is the electric field distribution coefficient (i = A, B, C; j = x, y, z),which is generated by the i-phase transformer voltage about the j-direction at this point, its value depends on the coordination of this point. This equation is the general form of mathematical model of electric field distribution when considering the actual operating of the three –phase voltage transformer.
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Yi-Nan Zhao et al./ Procedia / ProcediaEngineering Engineering0029(2011) (2012)000–000 2661 – 2666 Author name
For the distributed optical voltage transformer, the effective electric field is the axial electric field component on the central axis of the transformer, that is the z-direction electric field distribution, therefore the electrical field distribution mathematical model could be simplified as follow: ⎡u A ⎤ (2) E i = [ k Ai k B i k C i ] ⎢⎢ u B ⎥⎥ ⎢⎣ u C ⎥⎦ In this formula, k ji should be a function of z (j=A,B,C； i=A,B,C). At the same time, it could obtain the following equation by the symmetry of transformer structure. (3) = k AB k= kCC = ；k AC k= k BC CB； k AA CA； k BA Obviously, it only need to solve the five coefficients k AA , k AB , k AC , k BA , k BB of the nine electric field distribution coefficients. 3. Mathematical model of electric field distribution 3.1. Model simplification Because of the edge effects and other factors, the actual electric field distribution model of the voltage transformer is very complicated. So when calculating the electrical field distribution coefficients, the three-phase voltage transformer physical model has been simplified as follows: • Due to the d L , considering the transformer its own electric field distribution, it seemed the plate of the transformer as the surface electrode, and deemed the charge to evenly distribute on the surface electrode; • Due to the R D , considering the impact of each phase is relative to the others when the transformer plate is as a point charge. After the above simplification, it can ensure that the established mathematical model which can reflect the main feature of electric field distribution. And the axial electric field calculation physical model of three-phase voltage transformer was shown in Figure2 after the simplified operation.
QB1
QA1
Qc1
L− z
QA1
QB1
L− z
z
QA2
Qc1
Qc2
QA2
QB2
QB1
Qc1
L− z
z
QB2
QA1
z
Qc2
QA2
QB2
Qc2
Figure2: The electrical field calculation physical model
At this point, transformer internal electrical field was solely determined by the amount of electric charge on each phase plate. Making QA1 、 QB1 、 QC1 to be the amount of electric charge on the up-plate of each phase, and QA 2 、 QB 2 、 QC 2 to be the amount of electric charge on the down-plate of each phase. It could obtain the relationship function between potential on the central axis of each phase transformer and the amount of electric charge of each plate through the integration and summation operations.
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Yi-Nanname Zhao/ et al. / Procedia Engineering 29 (2012) 2661 – 2666 Author Procedia Engineering 00 (2011) 000–000
)
(
QA1 Q ⎧ R2 + 4( L − z )2 − 2L + 2 z + A22 = ⎪ϕ A R2πε R ε0 π 0 ⎪ 2 2 2 2 ⎪ 4D + ( L − z) QB1 + D + ( L − z) QC1 ⎪ + + ⎪ 4πε 0 D2 + ( L − z )2 4D2 + ( L − z)2 ⎪ QB1 Q ⎪ R2 + 4( L − z)2 − 2L + 2 z + B22 = ⎪ϕB π R2ε R ε0 π 0 ⎪ ⎨ QA1 + QC1 QA2 + QC 2 ⎪ + + ⎪ 4πε 0 D2 + ( L − z)2 4πε 0 D2 + z 2 ⎪ QC1 Q ⎪ = R2 + 4( L − z)2 − 2L + 2 z + C22 ⎪ϕC π R2ε π R ε0 0 ⎪ ⎪ 4D2 + ( L − z)2 QB1 + D2 + ( L − z)2 QA1 ⎪ + + ⎪ 4πε 0 D2 + ( L − z )2 4D2 + ( L − z)2 ⎩
(
R2 + 4 z 2 − 2 z
)
4D2 + z 2 QB 2 + D2 + z 2 QC 2 4πε 0 D2 + z 2 4D2 + z 2
(
)
(
R2 + 4 z 2 − 2 z
)
(
)
(
R2 + 4 z 2 − 2 z
)
(4)
4D2 + z 2 QB 2 + D2 + z 2 QA2 4πε 0 D2 + z 2 4D2 + z 2
As can be seen from the function, the electrical potential distribution on the center axis would only depend on the amount of electric charge carried by each plate with determined structure. However the amount of electric charge could be obtained by potential boundary conditions at the location of each plate. Taking the obtained amount of electric charge into the equation (4) ,which could obtain potential distribution function, and further more could obtain the electrical field distribution coefficients. 3.2. The calculation for the amount of electric charge on the each plate To calculate the amount of electric charge on the each plate, it should ensure the potential boundary conditions firstly. When calculating the k AA , k AB , k AC , the potential boundary conditions were shown as follow: = ϕ A u A= , ϕ B 0,= ϕC 0 ⎧ z L，= ⎨ = z 0 ， = ϕ 0, = ϕ 0, = ϕ 0 A B C ⎩ When solving the k BA , k BB , the potential distribution function satisfied the boundary conditions shown as follow: = ϕ A 0,= ϕ B uB= , ϕC 0 ⎧ z L，= ⎨ = ϕ A 0,= ϕ B 0,= ϕC 0 ⎩ z 0，= Taking the above boundary conditions into the equation (4) , the amount of electric field carried by each plate on the two conditions could be obtained. It should be noted that the electric charge carried by the plate of the B-phase and C-phase transformer was inductive charge ,to express which we used QB1¢、 QB 2¢、 QC1¢、 QC 2¢ when solving the k AA , k AB , k AC . However, the electric charge carried by the plate of
the A-phase and C-phase transformer was inductive charge ,to express which we used QA1′ 、 QA 2′ 、 QC1¢、 QC 2¢ when solving the k BA , k BB . Finally the electric charge carried by each plate was obtained , as shown in the table 1. Table1.The amount of electric charge carried by each plate on the two conditions
u A ≠ 0, uB = uC = 0
plate
amount of electric charge （10-14）
plate
uB ≠ 0, u= u= 0 A C amount of electric charge （10-14）
2665 5
Yi-Nan Zhaoname et al./ /Procedia ProcediaEngineering Engineering00 29(2011) (2012)000–000 2661 – 2666 Author A1 A2 B1` B2` C1` C2`
556.73 u A
A1`
－5.0240 u A
B1
－5.0946 u A
C1`
－13.571 u A
A2`
－8.3072 u A
B2
－4.6615 u A
C2`
－10.542 u B －7.8896 u B 556.86 u B
－13.424 u B －10.542 u B －7.8896 u B
Taking the amount of electric charge on the two conditions into the potential distribution function respectively, which could obtain the electrical field distribution coefficient. ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
k
AA
⎛ ⎜ kCC == − 2 .0 6 4 − z ⎜ ⎜ ⎝ ⎛ ⎜ + (2 − z )⎜ ⎜ ⎝
k
AB
k
k
AC
BA
BB
− z
(4 z
2
⎛ ⎜
)⎜ ⎜ ⎝
2
+ 0 .0 1 )
− 1 6 z + 1 6 .0 4
)
(z
2
2
(z
− 1 6 z + 1 6 .0 4
)
+
+ 6 .2 5
2
2
−4
)
3 2
(z
+ 6 .2 5
2
−4
)
3 2
)
(z 3 2
2
− 4 z + 1 0 .2 5
+
+ 25
2
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
1 .1 4 5 2 × 1 0
(z
2
−4
− 4 z + 29
⎛ ⎜ k BC = = − 0 .0 4 7 8 + z ⎜ ⎜ ⎝
(2
− z
⎛ ⎜
)⎜
⎜ ⎝
(2
− z
(4 z
⎛ ⎜
)⎜ ⎜ ⎝
2
− 1 6 z + 1 6 .0 4 0 .0 6 0 3
(z
2
1
+ 0 . 0 1 )2
)
1 2
−
5 .0 0 7 2
(4 z
2
−
1
+ 0 . 0 1 )2
0 .3 7 9 2 2
⎛ ⎜ = − 2 .5 6 3 9 + z ⎜ ⎜ ⎝ +
0 .1 4 1 8
(z
− 1 6 z + 1 6 .0 4
)
1 2
+
0 .0 0 1 2
(z
2
+ 6 .2 5
)
3 2
+
3 2
−
0 .0 5 0 1
(z
2
− 4 z + 1 0 .2 5
0 .0 0 0 3
(z
)
2
+ 6 .2 5
)
3 2
)
3 2
(z
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎠
)
3 2
7 .0 9 4 2 × 1 0
(z −
+ 25
2
)
3 2
−4
⎞ ⎟ ⎟ ⎟ ⎠
9 .4 7 9 2 × 1 0
(z
2
⎞ ⎟ ⎟ ⎟ ⎠
−4
− 4 z + 29
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎠
0 .0 0 0 4 2
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
0 .0 4 9 6
(z
−4
1 .0 4 7 8 × 1 0
+
−4
− 4 z + 1 0 .2 5
8 .0 1 1 2 × 1 0
−
1
+ 0 . 0 1 )2
1 2
(z
1 .1 2 9 3 × 1 0
0 .1 4 9 4
0 .1 8 0 8
(4 z
+
1 2
1 .8 6 7 4 × 1 0
+
1 2
⎛ ⎞ ⎜ ⎟ 0 .0 8 3 8 7 .4 6 9 7 × 1 0 −4 0 .0 0 1 2 = − 0 .0 0 7 8 + z ⎜ − − kCA = 1 3 3 ⎟ ⎜ ( z 2 + 0 . 0 1 )2 ( z 2 + 6 . 2 5 )2 ( z 2 + 2 5 ) 2 ⎟⎠ ⎝ ⎛ ⎜ 0 .0 5 0 1 0 .1 8 3 2 4 .5 1 7 5 × 1 0 −4 − (2 − z )⎜ − + 3 1 2 ⎜ (4 z 2 − 1 6 z + 1 6 . 0 4 )2 ( z − 4 z + 1 0 . 2 5 )2 ( z 2 − 4 z + 2 9 ⎝
−
k
(2
(z
5 .0 0 6
⎛ ⎜ = − 0 .0 5 9 + z ⎜ kCB = ⎜ ⎝
−
0 .0 6 1
− 4 z + 1 0 .2 5
⎞ ⎟ ⎟ ⎟ ⎠
(5)
4. Simulation It used the electromagnetic simulation software Ansoft to simulate and analyze the axial electrical field distribution on the center axis of B-phase transformer. The simulation parameters were set as : u B = 1 V, u A = - 0.5 V, u C = - 0.5 V. The electric potential distribution is shown in Figure3(a).
62666
Yi-Nan Zhao et al. / Procedia Engineering 29 000–000 (2012) 2661 – 2666 Author name / Procedia Engineering 00 (2011) 0
Eb(V/m)
-0.5 -1 -1.5 -2 -2.5 -3 0
Mathematical model Ansoft simulation 0.5
1
1.5
z(m)
(a)
2
(b)
Figure3: (a)The potential distribution and (b) axial electrical field distribution on the central axis
Figure3(b) shows the axial electric field distribution curve on the central axis of B-phase transformer each by the Ansoft simulation and mathematical model. Obviously, the mathematical model more accurately reflect the axial electric field distribution. 5. Conclusion The distribution optical voltage transformer is one of the important types of OVT. It depends on the numerical integration of electrical field to obtain the measured voltage. And the choice of integration points and integration weights depends on the axial electrical field distribution on the center axis. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through calculating electric potential, calculated the electric field distribution coefficient according to structural parameter of 110kV distributed optical voltage transformer, and simulated electric field on the central axis of B-phase transformer using Ansoft. The simulation results show that the mathematical model more accurately reflect the feature of the axial electric field distribution. References [1] Guo Zhizhong. Review of electronic instrument transformer. Power System Protection and Contro. 2008, 36(15):1-5l [2] PING Shao-xun，YU Bo，HUANG Ren-Shan．Recent Progress and Development in Optical Transformer．High Voltage Engineering， 2003，(01) [3] Hongxing Wang, Guoqing Zhang, Zhizhong Guo, et al. Application of Electronic Transformers in Digital Substation. 2008 Joint International Conference on Power System Technology POWERCON and IEEE Power India Conference, New Delhi City, India, 2008 [4]
XIAO Yue-yu ． Influence of the Electric Field Distribution on the Optical Voltage Transformer ． High Voltage
Engineering，2007，33(05)：37-40． [5] ZHU Ping-ping，ZHANG Gui-xin，LUO Cheng-mu．Application and Error Analysis of the Quadrature in the Voltage Transformer．High Voltage Engineering，2008，34(5)：919-923．
Procedia Engineering
ProcediaProcedia Engineering 00 (2011) Engineering 29 000–000 (2012) 2661 – 2666 www.elsevier.com/locate/procedia
2012 International Workshop on Information and Electronics Engineering (IWIEE)
The Mathematical Model of Electrical Field Distribution in Optical Voltage Transformer Yi-Nan Zhao*,Guo-Qing Zhang, Zhi-Zhong Guo, Song Cheng School of Electrical Engineering and Automation,Harbin Institute of Technology, Harbin, China
Abstract The distributed optical voltage transformer measures voltage through the numerical integration, the choice of the integration points and weights rely on the distribution function of the electrical field. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through calculating electric potential. The simulation results show that the mathematical model reflects the main features of the axial electric field on the central axis of the distributed optical voltage transformer, which indicates that the method to solve the model is feasible. For a optical voltage transformer with a determined structure, the electric field distribution function can be directly obtained by the mathematical model, which could improve the accuracy of measuring the voltage through the numerical integration.
© 2011 Published by Elsevier Ltd. Selection and/or peer-review under responsibility of Harbin University of Science and Technology Keywords: Optical Voltage Transformer; Mathematical model; Electrical field distribution.
1. Introduction To Power system, it is important to accurately measure the voltage. The voltage measurement method could be divided into electrical measurement method and optical measurement method. Comparing with the conventional voltage transformer based on capacitive or inductive coupling, the optical voltage transformer relies on electro-optical principles such as the Pockels effect ,thus it has many advantages: the observer is completely electrically isolated from the point of measurement, and the measurement itself is nonintrusive to the power circuit; optical sensor generally offer immunity to electromagnetic interference; the using of optical element and optical fiber reduces the dependence of the isolation costs on the voltage
*
Corresponding author. E-mail address: [email protected]
1877-7058 © 2011 Published by Elsevier Ltd. doi:10.1016/j.proeng.2012.01.368
22662
Yi-Nan Zhao et al. / Procedia Engineering 29 000–000 (2012) 2661 – 2666 Author name / Procedia Engineering 00 (2011)
level, especially for the power network with voltage level above 110kV, optical voltage transformer has a greater economic advantage; the optical sensing technology makes the transformer have a very wide bandwidth and large dynamic range, as well as reduces the transformer size and weight; optical voltage transformer outputs both analog and digital signals to adapt the automation developing trend of power system. Distributed optical voltage transformer obtains the measured voltage by the numerical integration of the electric field, the choice of the integration points and weights rely on the electric field distribution on the central axis. How to obtain the electrical field distribution function on the center axis in the transformer becomes an important problem in designing the distributed optical voltage transformer, especially for the three-phase voltage transformer system, where is still never has the mathematical description of its electrical field distribution. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through the calculation of electric potential, which could be available to improve the precision of numerical integration algorithm to calculate the measured voltage, and to improve the accuracy of distributed optical voltage transformer. 2. Problem description To solve the mathematical model of electric field distribution, a simple structure of three-phase voltage transformer is analyzed. Three-phase transformer has the same electrode structure and separated from each other by 2.5m, the structure of each phase transformer consists of two identical, disc-shaped electrodes which have a diameter of 200mm and are separated from each other by 2m. The lower electrode is grounded, and the upper electrode is energized. The dimensions of this structure are similar to the basic dimensions of 110kV stand-offs found in high-voltage substations.
Figure1: The schematic of three-phase transformer system
To the three-phase transformer system shown in figure 1, electrical field of each point inside the three phase transformer is the superposition of the electric field which is generated by ABC three-phase voltage at this point, that is formula (1) as shown. ⎡ Ex ⎤ ⎡ k Ax k Bx kCx ⎤ ⎡u A ⎤ ⎢ E ⎥ = ⎢k ⎥⎢ ⎥ (1) ⎢ y ⎥ ⎢ Ay k By kCy ⎥ ⎢uB ⎥ ⎢⎣ Ez ⎥⎦ ⎢⎣ k Az k Bz kCz ⎥⎦ ⎢⎣uC ⎥⎦ In this formula, kij is the electric field distribution coefficient (i = A, B, C; j = x, y, z),which is generated by the i-phase transformer voltage about the j-direction at this point, its value depends on the coordination of this point. This equation is the general form of mathematical model of electric field distribution when considering the actual operating of the three –phase voltage transformer.
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Yi-Nan Zhao et al./ Procedia / ProcediaEngineering Engineering0029(2011) (2012)000–000 2661 – 2666 Author name
For the distributed optical voltage transformer, the effective electric field is the axial electric field component on the central axis of the transformer, that is the z-direction electric field distribution, therefore the electrical field distribution mathematical model could be simplified as follow: ⎡u A ⎤ (2) E i = [ k Ai k B i k C i ] ⎢⎢ u B ⎥⎥ ⎢⎣ u C ⎥⎦ In this formula, k ji should be a function of z (j=A,B,C； i=A,B,C). At the same time, it could obtain the following equation by the symmetry of transformer structure. (3) = k AB k= kCC = ；k AC k= k BC CB； k AA CA； k BA Obviously, it only need to solve the five coefficients k AA , k AB , k AC , k BA , k BB of the nine electric field distribution coefficients. 3. Mathematical model of electric field distribution 3.1. Model simplification Because of the edge effects and other factors, the actual electric field distribution model of the voltage transformer is very complicated. So when calculating the electrical field distribution coefficients, the three-phase voltage transformer physical model has been simplified as follows: • Due to the d L , considering the transformer its own electric field distribution, it seemed the plate of the transformer as the surface electrode, and deemed the charge to evenly distribute on the surface electrode; • Due to the R D , considering the impact of each phase is relative to the others when the transformer plate is as a point charge. After the above simplification, it can ensure that the established mathematical model which can reflect the main feature of electric field distribution. And the axial electric field calculation physical model of three-phase voltage transformer was shown in Figure2 after the simplified operation.
QB1
QA1
Qc1
L− z
QA1
QB1
L− z
z
QA2
Qc1
Qc2
QA2
QB2
QB1
Qc1
L− z
z
QB2
QA1
z
Qc2
QA2
QB2
Qc2
Figure2: The electrical field calculation physical model
At this point, transformer internal electrical field was solely determined by the amount of electric charge on each phase plate. Making QA1 、 QB1 、 QC1 to be the amount of electric charge on the up-plate of each phase, and QA 2 、 QB 2 、 QC 2 to be the amount of electric charge on the down-plate of each phase. It could obtain the relationship function between potential on the central axis of each phase transformer and the amount of electric charge of each plate through the integration and summation operations.
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Yi-Nanname Zhao/ et al. / Procedia Engineering 29 (2012) 2661 – 2666 Author Procedia Engineering 00 (2011) 000–000
)
(
QA1 Q ⎧ R2 + 4( L − z )2 − 2L + 2 z + A22 = ⎪ϕ A R2πε R ε0 π 0 ⎪ 2 2 2 2 ⎪ 4D + ( L − z) QB1 + D + ( L − z) QC1 ⎪ + + ⎪ 4πε 0 D2 + ( L − z )2 4D2 + ( L − z)2 ⎪ QB1 Q ⎪ R2 + 4( L − z)2 − 2L + 2 z + B22 = ⎪ϕB π R2ε R ε0 π 0 ⎪ ⎨ QA1 + QC1 QA2 + QC 2 ⎪ + + ⎪ 4πε 0 D2 + ( L − z)2 4πε 0 D2 + z 2 ⎪ QC1 Q ⎪ = R2 + 4( L − z)2 − 2L + 2 z + C22 ⎪ϕC π R2ε π R ε0 0 ⎪ ⎪ 4D2 + ( L − z)2 QB1 + D2 + ( L − z)2 QA1 ⎪ + + ⎪ 4πε 0 D2 + ( L − z )2 4D2 + ( L − z)2 ⎩
(
R2 + 4 z 2 − 2 z
)
4D2 + z 2 QB 2 + D2 + z 2 QC 2 4πε 0 D2 + z 2 4D2 + z 2
(
)
(
R2 + 4 z 2 − 2 z
)
(
)
(
R2 + 4 z 2 − 2 z
)
(4)
4D2 + z 2 QB 2 + D2 + z 2 QA2 4πε 0 D2 + z 2 4D2 + z 2
As can be seen from the function, the electrical potential distribution on the center axis would only depend on the amount of electric charge carried by each plate with determined structure. However the amount of electric charge could be obtained by potential boundary conditions at the location of each plate. Taking the obtained amount of electric charge into the equation (4) ,which could obtain potential distribution function, and further more could obtain the electrical field distribution coefficients. 3.2. The calculation for the amount of electric charge on the each plate To calculate the amount of electric charge on the each plate, it should ensure the potential boundary conditions firstly. When calculating the k AA , k AB , k AC , the potential boundary conditions were shown as follow: = ϕ A u A= , ϕ B 0,= ϕC 0 ⎧ z L，= ⎨ = z 0 ， = ϕ 0, = ϕ 0, = ϕ 0 A B C ⎩ When solving the k BA , k BB , the potential distribution function satisfied the boundary conditions shown as follow: = ϕ A 0,= ϕ B uB= , ϕC 0 ⎧ z L，= ⎨ = ϕ A 0,= ϕ B 0,= ϕC 0 ⎩ z 0，= Taking the above boundary conditions into the equation (4) , the amount of electric field carried by each plate on the two conditions could be obtained. It should be noted that the electric charge carried by the plate of the B-phase and C-phase transformer was inductive charge ,to express which we used QB1¢、 QB 2¢、 QC1¢、 QC 2¢ when solving the k AA , k AB , k AC . However, the electric charge carried by the plate of
the A-phase and C-phase transformer was inductive charge ,to express which we used QA1′ 、 QA 2′ 、 QC1¢、 QC 2¢ when solving the k BA , k BB . Finally the electric charge carried by each plate was obtained , as shown in the table 1. Table1.The amount of electric charge carried by each plate on the two conditions
u A ≠ 0, uB = uC = 0
plate
amount of electric charge （10-14）
plate
uB ≠ 0, u= u= 0 A C amount of electric charge （10-14）
2665 5
Yi-Nan Zhaoname et al./ /Procedia ProcediaEngineering Engineering00 29(2011) (2012)000–000 2661 – 2666 Author A1 A2 B1` B2` C1` C2`
556.73 u A
A1`
－5.0240 u A
B1
－5.0946 u A
C1`
－13.571 u A
A2`
－8.3072 u A
B2
－4.6615 u A
C2`
－10.542 u B －7.8896 u B 556.86 u B
－13.424 u B －10.542 u B －7.8896 u B
Taking the amount of electric charge on the two conditions into the potential distribution function respectively, which could obtain the electrical field distribution coefficient. ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
k
AA
⎛ ⎜ kCC == − 2 .0 6 4 − z ⎜ ⎜ ⎝ ⎛ ⎜ + (2 − z )⎜ ⎜ ⎝
k
AB
k
k
AC
BA
BB
− z
(4 z
2
⎛ ⎜
)⎜ ⎜ ⎝
2
+ 0 .0 1 )
− 1 6 z + 1 6 .0 4
)
(z
2
2
(z
− 1 6 z + 1 6 .0 4
)
+
+ 6 .2 5
2
2
−4
)
3 2
(z
+ 6 .2 5
2
−4
)
3 2
)
(z 3 2
2
− 4 z + 1 0 .2 5
+
+ 25
2
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
1 .1 4 5 2 × 1 0
(z
2
−4
− 4 z + 29
⎛ ⎜ k BC = = − 0 .0 4 7 8 + z ⎜ ⎜ ⎝
(2
− z
⎛ ⎜
)⎜
⎜ ⎝
(2
− z
(4 z
⎛ ⎜
)⎜ ⎜ ⎝
2
− 1 6 z + 1 6 .0 4 0 .0 6 0 3
(z
2
1
+ 0 . 0 1 )2
)
1 2
−
5 .0 0 7 2
(4 z
2
−
1
+ 0 . 0 1 )2
0 .3 7 9 2 2
⎛ ⎜ = − 2 .5 6 3 9 + z ⎜ ⎜ ⎝ +
0 .1 4 1 8
(z
− 1 6 z + 1 6 .0 4
)
1 2
+
0 .0 0 1 2
(z
2
+ 6 .2 5
)
3 2
+
3 2
−
0 .0 5 0 1
(z
2
− 4 z + 1 0 .2 5
0 .0 0 0 3
(z
)
2
+ 6 .2 5
)
3 2
)
3 2
(z
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎠
)
3 2
7 .0 9 4 2 × 1 0
(z −
+ 25
2
)
3 2
−4
⎞ ⎟ ⎟ ⎟ ⎠
9 .4 7 9 2 × 1 0
(z
2
⎞ ⎟ ⎟ ⎟ ⎠
−4
− 4 z + 29
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎠
0 .0 0 0 4 2
)
3 2
⎞ ⎟ ⎟ ⎟ ⎠
0 .0 4 9 6
(z
−4
1 .0 4 7 8 × 1 0
+
−4
− 4 z + 1 0 .2 5
8 .0 1 1 2 × 1 0
−
1
+ 0 . 0 1 )2
1 2
(z
1 .1 2 9 3 × 1 0
0 .1 4 9 4
0 .1 8 0 8
(4 z
+
1 2
1 .8 6 7 4 × 1 0
+
1 2
⎛ ⎞ ⎜ ⎟ 0 .0 8 3 8 7 .4 6 9 7 × 1 0 −4 0 .0 0 1 2 = − 0 .0 0 7 8 + z ⎜ − − kCA = 1 3 3 ⎟ ⎜ ( z 2 + 0 . 0 1 )2 ( z 2 + 6 . 2 5 )2 ( z 2 + 2 5 ) 2 ⎟⎠ ⎝ ⎛ ⎜ 0 .0 5 0 1 0 .1 8 3 2 4 .5 1 7 5 × 1 0 −4 − (2 − z )⎜ − + 3 1 2 ⎜ (4 z 2 − 1 6 z + 1 6 . 0 4 )2 ( z − 4 z + 1 0 . 2 5 )2 ( z 2 − 4 z + 2 9 ⎝
−
k
(2
(z
5 .0 0 6
⎛ ⎜ = − 0 .0 5 9 + z ⎜ kCB = ⎜ ⎝
−
0 .0 6 1
− 4 z + 1 0 .2 5
⎞ ⎟ ⎟ ⎟ ⎠
(5)
4. Simulation It used the electromagnetic simulation software Ansoft to simulate and analyze the axial electrical field distribution on the center axis of B-phase transformer. The simulation parameters were set as : u B = 1 V, u A = - 0.5 V, u C = - 0.5 V. The electric potential distribution is shown in Figure3(a).
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Yi-Nan Zhao et al. / Procedia Engineering 29 000–000 (2012) 2661 – 2666 Author name / Procedia Engineering 00 (2011) 0
Eb(V/m)
-0.5 -1 -1.5 -2 -2.5 -3 0
Mathematical model Ansoft simulation 0.5
1
1.5
z(m)
(a)
2
(b)
Figure3: (a)The potential distribution and (b) axial electrical field distribution on the central axis
Figure3(b) shows the axial electric field distribution curve on the central axis of B-phase transformer each by the Ansoft simulation and mathematical model. Obviously, the mathematical model more accurately reflect the axial electric field distribution. 5. Conclusion The distribution optical voltage transformer is one of the important types of OVT. It depends on the numerical integration of electrical field to obtain the measured voltage. And the choice of integration points and integration weights depends on the axial electrical field distribution on the center axis. This paper presents a mathematical model of electrical field distribution in distributed optical voltage transformer through calculating electric potential, calculated the electric field distribution coefficient according to structural parameter of 110kV distributed optical voltage transformer, and simulated electric field on the central axis of B-phase transformer using Ansoft. The simulation results show that the mathematical model more accurately reflect the feature of the axial electric field distribution. References [1] Guo Zhizhong. Review of electronic instrument transformer. Power System Protection and Contro. 2008, 36(15):1-5l [2] PING Shao-xun，YU Bo，HUANG Ren-Shan．Recent Progress and Development in Optical Transformer．High Voltage Engineering， 2003，(01) [3] Hongxing Wang, Guoqing Zhang, Zhizhong Guo, et al. Application of Electronic Transformers in Digital Substation. 2008 Joint International Conference on Power System Technology POWERCON and IEEE Power India Conference, New Delhi City, India, 2008 [4]
XIAO Yue-yu ． Influence of the Electric Field Distribution on the Optical Voltage Transformer ． High Voltage
Engineering，2007，33(05)：37-40． [5] ZHU Ping-ping，ZHANG Gui-xin，LUO Cheng-mu．Application and Error Analysis of the Quadrature in the Voltage Transformer．High Voltage Engineering，2008，34(5)：919-923．