CN102081132B - Two-end distance measuring method of transmission line fault under dynamic condition - Google Patents

Two-end distance measuring method of transmission line fault under dynamic condition Download PDF

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CN102081132B
CN102081132B CN 201010573147 CN201010573147A CN102081132B CN 102081132 B CN102081132 B CN 102081132B CN 201010573147 CN201010573147 CN 201010573147 CN 201010573147 A CN201010573147 A CN 201010573147A CN 102081132 B CN102081132 B CN 102081132B
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CN102081132A (en
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何正友
何文
张姝
麦瑞坤
林圣�
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Southwest Jiaotong University
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Abstract

The invention relates to a two-end distance measuring method of transmission line fault under a dynamic condition, comprising the following steps of: acquiring two-end voltage and current signals through two-end phasor measurement units of the transmission line; then acquiring dynamic positive sequence impedance, admittance, wave impedance and propagation coefficients according to line parameter estimation and a dynamic line parameter method; and finally solving a nonlinear equation related to a fault distance according to the fact that positive sequence voltages derived from two ends to a fault point are equal by applying a Newton iteration method so as to obtain a dynamic distance measuring result of the transmission line fault. The two-end distance measuring method can more effectively solve the problem of line parameter and fault distance estimation under the dynamic condition and has accurate and reliable fault distance measuring result.

Description

Transmission line malfunction both-end distance measuring method under a kind of dynamic condition
Technical field
The present invention relates to the both-end distance measuring method of transmission line malfunction in the electric system.
Background technology
Along with the development of WAMS, increasing phasor measurement unit has been installed in the electrical network, for the Two-terminal Fault Location method based on synchronous phasor measurement provides the foundation.Both-end distance measuring method based on phasor measurement unit can utilize failure message fully, and online computational scheme parameter is avoided the impact of the factors such as line parameter circuit value is worn out, environment temperature, humidity, can obtain accurately localization of fault.But in the application of reality, consider economy, electric system is in stable edge usually; Have inevitably disturbance on the power transmission line of long distance, the impact of the uncertain factors such as fault causes electrical network often to be in the current intelligence of vibration.Difficult point based on the Two-terminal Fault Location of synchronous phasor measurement is when the amplitude frequency of voltage and current signal occurs along with system oscillation to change fast, the electric signal that original distance-finding method is described can't have the ability of spatial variations characteristic and time behavior, larger error can appear in the fault localization result, be difficult to realize accurate localization of fault, affect the timely eliminating of fault.
Summary of the invention:
The purpose of this invention is to provide the transmission line malfunction both-end distance measuring method under a kind of dynamic condition, the method can be carried out more effectively line parameter circuit value and fault distance estimation under dynamic condition, and the fault localization result accurately, reliably; Can in various fault types, fault distance, fault angle situation, use.
The present invention is for solving its technical matters, and the technical scheme that adopts is:
A, pre-service
Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after filtering is processed, and obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides S1, I R1With positive sequence failure voltage signal U S1, U R1, wherein S represents left end, R represents right-hand member;
B, line parameter circuit value are estimated
Homogeneous line model according to distribution parameter calculates following positive order parameter:
B1, unit line length positive sequence wave impedance
Figure BDA0000036122510000011
Unit line length positive sequence propagation coefficient γ 1=l -1Ln[(U S1+ Z C1I S1)/(U R1-Z C1I R1)], wherein l is transmission line length;
B2, unit line length impedance Z 1=Z C1γ 1, unit line length admittance Y 11/ Z C1
C, Two-terminal Fault Location
Under C1, the dynamic condition, the forward-order current amplitude of transmission line of electricity left end
Figure BDA0000036122510000021
The positive sequence voltage amplitude
Figure BDA0000036122510000022
The forward-order current amplitude of transmission line of electricity right-hand member The positive sequence voltage amplitude
Figure BDA0000036122510000024
T is the time, ω 0Angular frequency for system.
The dynamic positive sequence impedance of definition left end is
Z ~ S 1 = Z 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition left end is
Figure BDA0000036122510000027
Y ~ S 1 = Y 1 + C 1 U S 1 ( t ) [ u S 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u S 1 ( t ) ∂ [ u S 1 ( t ) ] ∂ t
The dynamic positive sequence impedance of definition right-hand member is
Figure BDA0000036122510000029
Z ~ R 1 = Z 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition right-hand member is
Figure BDA00000361225100000211
Y ~ R 1 = Y 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t
Wherein, R 1, L 1, G 1And C 1The positive sequence resistance, inductance, the electricity that are respectively unit line length are led and electric capacity;
C2, the dynamic positive sequence wave impedance of circuit right-hand member
Figure BDA00000361225100000213
Dynamic positive sequence propagation coefficient
Figure BDA00000361225100000214
Trouble spot positive sequence voltage U by the right-hand member derivation RF1For:
U RF 1 = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Wherein x is that the trouble spot is apart from the distance of circuit right-hand member, i.e. fault distance;
The dynamic positive sequence wave impedance of circuit right-hand member Dynamic positive sequence propagation coefficient
Figure BDA00000361225100000217
Trouble spot positive sequence voltage U by the left end derivation SF1For:
U SF 1 = U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
C3, fault localization
Circuit is U under failure condition RF1=U SF1, the nonlinear equation of setting up about fault distance x is:
U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x ) = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Use the iteration function F (x) that Newton iteration method solves following formula:
F ( x ) = x - γ ~ S 1 U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + γ ~ S 1 U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
- ( γ ~ R 1 U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x - γ ~ R 1 U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x )
Following formula is carried out interative computation can obtain accurately fault distance x.
Compared with prior art, the invention has the beneficial effects as follows:
One, the two ends of electric transmission line system voltage phasor and the electric current phasor that obtain by synchronized sampling, can calculate surge impedance of a line and propagation coefficient, thereby can estimate accurate line parameter circuit value, comprise that resistance, inductance, electricity lead and electric capacity, avoid line parameter circuit value to wear out or the different problems that cause variation of environment, improved the reliability of fault localization.
Two, surge impedance of a line and propagation coefficient are subjected to frequency and signal amplitude variable effect, traditional time domain approach can not solve the problem of wave impedance and propagation coefficient variation, and the present invention is after sampling obtains the positive sequential signal in transmission line of electricity two ends and frequency, can accurately calculate dynamic positive sequence wave impedance and dynamic positive sequence propagation coefficient, can obtain optimum fault distance solution by introducing Newton iteration method at last, fast convergence rate and result are unique, greatly improved system's counting accuracy to various failure conditions below dynamic condition, its range finding result accurately, reliable.To in time searching and the processing circuitry fault, guarantee the safe operation of electrical network, improve stability of power system and reduce operating cost, have important society and economic worth.
Three, among the present invention, can think when electric system normally moves symmetrical, namely each element three-phase impedance phase with, three-phase voltage, size of current equate to have normal phase sequence separately.In the uneven situation of calculating electric system, quoted symmetrical component method, any asymmetric three-phase phasor A, B, C can be decomposed into three groups of symmetrical components that phase sequence is different.It is i.e. positive and negative, the zero phase-sequence of phasor composition that electric current, voltage or the impedance of any three-phase imbalance can be decomposed into three balances.Why hinder all existence under the type because the As positive-sequence component is in office, so method of the present invention utilizes symmetrical component method to only adopting positive sequence voltage component U after the circuit decoupling zero 1With forward-order current component I 1Carry out fault localization, computing is simple, quick, and can carry out accurate fault localization under various short trouble type cases.
Four, time domain distance-finding method under the existing static condition, sample frequency was 2400Hz when the present invention extracted the transmission line malfunction signal, to sample devices without specific (special) requirements, convenient enforcement.
In the above-mentioned A step pre-service, carry out decoupling change with symmetrical component method and get transmission line of electricity positive sequence fault current signal I in return 1With positive sequence failure voltage signal U 1The decoupling zero transformation for mula be:
U 1 U 2 U 0 = 1 3 1 a a 2 1 a 2 a 1 1 1 U a U b U c
I 1 I 2 I 0 = 1 3 1 a a 2 1 a 2 a 1 1 1 I a I b I c
In the formula, U a, U b, U cAnd I a, I b, I cBe phase voltage and the phase current that collects, U 1, U 2, U 0And I 1, I 2, I 0Be respectively sequence voltage and order electric current, and
The present invention is described in further detail below in conjunction with the drawings and specific embodiments.
Description of drawings
Fig. 1 is dynamic fault distance-finding method of the present invention from existing static failure distance-finding method to the relative error curve of a concrete circuit in different moment range finding results, and wherein horizontal ordinate is the time, and ordinate is range finding result's relative error.
Fig. 2 is that dynamic fault distance-finding method of the present invention and existing static failure distance-finding method are to the range finding result of concrete circuit under different faults position and fault type relative error curve, wherein horizontal ordinate is the fault distance, and ordinate is the average relative error of (t=50ms-1000ms) range finding result in a period of time.
Embodiment
Embodiment
Transmission line malfunction both-end distance measuring method under a kind of dynamic condition of the present invention the steps include:
A, pre-service
Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after filtering is processed, and obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides S1, I R1With positive sequence failure voltage signal U S1, U R1, wherein S represents left end, R represents right-hand member;
Carry out decoupling change with symmetrical component method and get transmission line of electricity positive sequence fault current signal I in return 1With positive sequence failure voltage signal U 1The decoupling zero transformation for mula be:
U 1 U 2 U 0 = 1 3 1 a a 2 1 a 2 a 1 1 1 U a U b U c
I 1 I 2 I 0 = 1 3 1 a a 2 1 a 2 a 1 1 1 I a I b I c
In the formula, U a, U b, U cAnd I a, I b, I cBe phase voltage and the phase current that collects, U 1, U 2, U 0And I 1, I 2, I 0Be respectively sequence voltage and order electric current, and
B, line parameter circuit value are estimated
Homogeneous line model according to distribution parameter calculates following positive order parameter:
B1, unit line length positive sequence wave impedance
Figure BDA0000036122510000044
Unit line length positive sequence propagation coefficient γ 1=l -1Ln[(U S1+ Z C1I S1)/(U R1-Z C1I R1)], wherein l is transmission line length;
B2, unit line length impedance Z 1=Z C1γ 1, unit line length admittance Y 11/ Z C1
C, Two-terminal Fault Location
Under C1, the dynamic condition, the forward-order current amplitude of transmission line of electricity left end The positive sequence voltage amplitude The forward-order current amplitude of transmission line of electricity right-hand member
Figure BDA0000036122510000047
The positive sequence voltage amplitude
Figure BDA0000036122510000051
T is the time, ω 0Angular frequency for system.
Can obtain on the circuit positive sequence voltage U of 1 m (to left end, the span of m corresponds to 0 to l from the right-hand member of circuit) under dynamic condition according to the homogeneous line model of distribution parameter 1Differential equation:
∂ U 1 ∂ m = I 1 R 1 + L 1 ∂ I 1 ∂ t
= I 1 R 1 + L 1 j ω 0 i 1 ( t ) e j ω 0 t ∂ t + e j ω 0 t ∂ [ i 1 ( t ) ] ∂ t
= I 1 R 1 + j ω 0 L 1 I 1 + L 1 ∂ [ i 1 ( t ) ] i 1 ( t ) ∂ t i 1 ( t ) e j ω 0 t
= I 1 ( R 1 + j ω 0 L 1 + L 1 i 1 ( t ) ∂ [ i 1 ( t ) ] ∂ t )
= I 1 Z ~ 1
∂ I 1 ∂ m = U 1 G 1 + C 1 ∂ U 1 ∂ t
= U 1 G 1 + C 1 j ω 0 u 1 ( t ) e j ω 0 t ∂ t + e j ω 0 t ∂ [ u 1 ( t ) ] ∂ t
= U 1 G 1 + j ω 0 C 1 U 1 + C 1 ∂ [ u 1 ( t ) ] u 1 ( t ) ∂ t u 1 ( t ) e j ω 0 t
= U 1 ( G 1 + j ω 0 C 1 + C 1 u 1 ∂ [ u 1 ( t ) ] ∂ t )
= U 1 Y ~ 1
According to above derivation, definable goes out following parameter:
The dynamic positive sequence impedance of definition left end is
Figure BDA00000361225100000512
Z ~ S 1 = Z 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition left end is
Figure BDA00000361225100000514
Y ~ S 1 = Y 1 + C 1 U S 1 ( t ) ∂ [ u S 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u S 1 ( t ) ∂ [ u S 1 ( t ) ] ∂ t
The dynamic positive sequence impedance of definition right-hand member is
Z ~ R 1 = Z 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition right-hand member is
Y ~ R 1 = Y 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t
Wherein, R 1, L 1, G 1And C 1The positive sequence resistance, inductance, the electricity that are respectively unit line length are led and electric capacity.
Such as the t=0 dynamic positive sequence impedance of left end constantly
Figure BDA00000361225100000520
The dynamic positive sequence admittance of left end
Figure BDA00000361225100000521
The dynamic positive sequence impedance of right-hand member
Figure BDA00000361225100000522
With the dynamic positive sequence admittance of right-hand member
Figure BDA00000361225100000523
Specifically be calculated as:
Report the constantly positive sequence voltage signal U of (τ reporting period) according to continuous 3 phasor measurement unit that circuit right-hand member (m=0) is located to gather R1(t) and forward-order current signal I R1(t), obtain t=0, positive sequence voltage amplitude change rate during m=0
Figure BDA0000036122510000061
With the forward-order current amplitude change rate For:
∂ [ u R 1 ( 0 ) ] ∂ t = U R 1 ( τ ) e - j ω 0 · τ - U R 1 ( - τ ) e j ω 0 · σ 2 τ
∂ [ i R 1 ( 0 ) ] ∂ t = I R 1 ( τ ) e - j ω 0 · τ - I R 1 ( - τ ) e j ω 0 · σ 2 τ
So, the dynamic positive sequence impedance of circuit right-hand member (m=0)
Figure BDA0000036122510000065
With dynamic positive sequence admittance
Figure BDA0000036122510000066
When t=0, be respectively:
Z ~ R 1 = R 1 + j ω 0 L 1 + L 1 I R 1 ( 0 ) I R 1 ( τ ) e - j ω 0 · τ - I R 1 ( - τ ) e j ω 0 · τ 2 τ
Y ~ R 1 = G 1 + j ω 0 C 1 + C 1 U R 1 ( 0 ) U R 1 ( τ ) e - j ω 0 · τ - U R 1 ( - τ ) e j ω 0 · τ 2 τ
In like manner, can obtain the dynamic positive sequence impedance of circuit left end (m=l)
Figure BDA0000036122510000069
With dynamic positive sequence admittance
Figure BDA00000361225100000610
When t=0, be respectively:
Z ~ S 1 = R 1 + j ω 0 L 1 + L 1 I S 1 ( 0 ) I S 1 ( τ ) e - j ω 0 · τ - I S 1 ( - τ ) e j ω 0 · τ 2 τ
Y ~ S 1 = G 1 + j ω 0 C 1 + C 1 U S 1 ( 0 ) U S 1 ( τ ) e - j ω 0 · τ - U S 1 ( - τ ) e j ω 0 · τ 2 τ
C2, the dynamic positive sequence wave impedance of circuit right-hand member
Figure BDA00000361225100000613
Dynamic positive sequence propagation coefficient
Figure BDA00000361225100000614
Trouble spot positive sequence voltage U by the right-hand member derivation RF1For:
U RF 1 = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Wherein x is that the trouble spot is apart from the distance of circuit right-hand member, i.e. fault distance;
The dynamic positive sequence wave impedance of circuit left end
Figure BDA00000361225100000616
Dynamic positive sequence propagation coefficient
Figure BDA00000361225100000617
Trouble spot positive sequence voltage U by the left end derivation SF1For:
U SF 1 = U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
C3, fault localization
Circuit is U under failure condition RF1=U SF1, the nonlinear equation of setting up about fault distance x is:
U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x ) = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Can obtain function f (x) by following formula is:
f ( x ) = U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x ) - U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x - U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
According to the iteration function F (x) in the Newton iteration method=x-f (x)/f ' (x), set up iteration function F (x):
F ( x ) = x - γ ~ S 1 U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + γ ~ S 1 U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
- ( γ ~ R 1 U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x - γ ~ R 1 U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x )
Following formula is carried out interative computation can obtain accurately fault distance x.
Adopt the method for the present embodiment, the result of the fault localization that carries out for a concrete transmission line of electricity system is as follows:
This transmission line of electricity system is the dual power supply double-circuit system, need to a ground short circuit fault F be set at circuit 1 in order to force system to be in current intelligence 1, power transmission line total length 200km, the circuit pattern is selected TOWER 3H5 shaft tower, power supply 1: voltage 230 ∠ 0kV, frequency 50Hz, power supply 2: voltage 230 ∠ 10kV, frequency 50Hz, the positive sequence resistance R of circuit unit 1=0.03468 Ω, unit positive sequence inductance L 1=1.3476mH, unit positive sequence electricity are led G 1=1.0 * 10 -7S, unit positive sequence capacitor C 1=8.6771 * 10 -9F, the zero sequence resistance R of unit 0=0.30000 Ω, unit zero sequence inductance L 0=3.6371mH, unit zero sequence electricity are led G 0=1.0 * 10 -7S and unit zero sequence capacitor C 0=6.1610 * 10 -9F, soil resistivity R are 100 Ω * m, the power supply 1 equiva lent impedance Z of place 1=0, distance measuring equipment is installed in respectively on the bus of power supply 1 and power supply 2 and carries out both-end distance measuring.
Fault localization result's relative error E EstBe calculated as follows:
E est = x - x 0 l × 100 %
Wherein, x 0Fault distance for reality.
For verifying the transmission line malfunction both-end distance measuring method accuracy under a kind of dynamic condition of the present invention, carried out the test in 4 kinds of situations, wherein adopted fault to occur after in the t=50ms-1000ms each result's that constantly finds range average relative error and maximum relative error estimate.
Test 1: the test of different faults position and transition resistance;
A phase earth fault occurs at t=0ms in circuit constantly, and the fault angle is 0 °, and fault resstance is respectively Rf=0 Ω, 5 Ω, 20 Ω, 50 Ω, physical fault apart from the circuit right-hand member apart from x 0Be 0km~200km, the result is as shown in table 1.
Table 1: 50-1000ms carries out the average relative error of fault localization after the fault of diverse location and transition resistance occurs
Figure BDA0000036122510000072
Figure BDA0000036122510000081
As can be seen from Table 1, its range error of dynamic fault distance-finding method of the present invention is between 0.0002% to 0.1765%, and its range error scope of existing static failure distance-finding method is 0.0001% to 0.5375%, and method of the present invention on the whole error is less, finds range more accurate.
Wherein, line fault occur in apart from the circuit right-hand member apart from x 0Be the 40km place, fault type is A phase earth fault, and the fault angle is 0 °, and transition resistance is 20 Ω, utilizes existing static failure distance-finding method and dynamic fault distance-finding method of the present invention at the different relative error curves of constantly testing the result that obtains finding range, sees Fig. 1.Among Fig. 1, dotted line is the relative error curve of existing static failure distance-finding method, and solid line is the relative error curve of dynamic fault distance-finding method of the present invention.
As can be seen from Figure 1, the range error of static failure distance-finding method is vibrated along with the dynamic oscillation of system, and variation range is large, is difficult to obtain accurate fault localization result; And method range error of the present invention is stable, and the dynamic oscillation with system does not change, and error is little, and distance accuracy is high.
Test 2: the test of different faults type and transition resistance;
The fault distance circuit right-hand member that circuit occurs constantly at t=0 apart from x 0Be 80km, the fault angle is 0 °, and fault resstance is respectively Rf=0 Ω, 5 Ω, 20 Ω, 50 Ω, fault type has A phase earth fault (Ag), AB phase-to phase fault (AB), AB double earthfault (ABg), ABC three-phase ground fault (ABCg), and the result is as shown in table 2.
Table 2: 50ms-1000ms carries out fault localization after the fault of different faults type and transition resistance occurs
Average relative error and maximum relative error
Figure BDA0000036122510000082
As can be seen from Table 2, the average relative error scope that existing static failure distance-finding method obtains is 0.0661% to 1.0812%, and the average relative error scope that dynamic fault distance-finding method of the present invention obtains only is 0.0058% to 0.0525%, and its relative error significantly reduces.
Test 3: the test under the different noises;
A phase earth fault occurs at t=0 in circuit constantly, and the fault angle is 0 °, and fault resstance is respectively Rf=50 Ω, physical fault apart from the circuit right-hand member apart from x 0Be 0km~200km, noise is 40dB, 50dB, and 60dB, the result is as shown in table 3.
Table 3: 50ms-1000ms carries out the average relative error of fault localization after the fault of different faults position and noise occurs
Figure BDA0000036122510000091
As can be seen from Table 3, the average relative error scope that existing static failure distance-finding method obtains is 0.15990% to 0.56697%, and the average relative error scope that dynamic fault distance-finding method of the present invention obtains only is 0.02366% to 0.21288%.
Test 4: the test of different faults position and fault type.
Circuit breaks down constantly at t=0ms, the fault angle is 0 °, and fault resstance is respectively Rf=50 Ω, and physical fault is 0km~200km apart from the bus of power supply R, fault type has A phase earth fault, AB phase-to phase fault, AB double earthfault, ABC three-phase ground fault, the results are shown in Figure 2.Horizontal ordinate is fault distance among Fig. 2, and ordinate is range finding result's average relative error; Upper part solid line is the average relative error curve of existing fault distance-finding method when A phase earth fault, dotted line is the average relative error curve of existing fault distance-finding method when the AB double earthfault, and dot-and-dash line is the average relative error curve of existing fault distance-finding method when ABC three-phase ground fault; Lower part solid line is the average relative error curve of fault distance-finding method of the present invention when A phase earth fault, dotted line is the average relative error curve of fault distance-finding method of the present invention when the AB double earthfault, and dot-and-dash line is the average relative error curve of fault distance-finding method of the present invention when ABC three-phase ground fault.
As can be seen from Figure 2, under all fault distances of 3 kinds of fault types, the average relative error of dynamic fault distance-finding method of the present invention all is starkly lower than existing static failure distance-finding method.
In a word, above four evidences, under the same line system condition, method of the present invention is than existing method, and it is subjected to the impact of fault distance, fault type, fault angle and noise little, and distance accuracy is high.

Claims (1)

1. the transmission line malfunction both-end distance measuring method under the dynamic condition the steps include:
A, pre-service
Phasor measurement unit obtains phase current and phase voltage is sent into main frame from the transmission line of electricity synchronous acquisition, uses symmetrical component method and carry out the decoupling zero conversion after filtering is processed, and obtains the positive sequence fault current signal I at two ends, the transmission line of electricity left and right sides S1, I R1With positive sequence failure voltage signal U S1, U R1, wherein S represents left end, R represents right-hand member;
B, line parameter circuit value are estimated
Homogeneous line model according to distribution parameter calculates following positive order parameter:
B1, unit line length positive sequence wave impedance Z C 1 = ( U S 1 2 - U R 1 2 ) / ( I S 1 2 - I R 1 2 ) ;
Unit line length positive sequence propagation coefficient γ 1=l -1Ln[(U S1+ Z C1I S1)/(U R1-Z C1I R1)], wherein l is transmission line length;
B2, unit line length impedance Z 1=Z C1γ 1, unit line length admittance Y 11/ Z C1
C, Two-terminal Fault Location
Under the C1 dynamic condition, the forward-order current amplitude of transmission line of electricity left end
Figure FDA00002056480500012
The positive sequence voltage amplitude The forward-order current amplitude of transmission line of electricity right-hand member
Figure FDA00002056480500014
The positive sequence voltage amplitude
Figure FDA00002056480500015
T is the time, ω 0Angular frequency for system:
The dynamic positive sequence impedance of definition left end is
Figure FDA00002056480500016
Z ~ S 1 = Z 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i S 1 ( t ) ∂ [ i S 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition left end is
Figure FDA00002056480500018
Y ~ S 1 = Y 1 + C 1 u S 1 ( t ) ∂ [ u S 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u S 1 ( t ) ∂ [ u S 1 ( t ) ] ∂ t
The dynamic positive sequence impedance of definition right-hand member is
Figure FDA000020564805000110
Z ~ R 1 = Z 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t = R 1 + j ω 0 L 1 + L 1 i R 1 ( t ) ∂ [ i R 1 ( t ) ] ∂ t
The dynamic positive sequence admittance of definition right-hand member is
Y ~ R 1 = Y 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t = G 1 + j ω 0 C 1 + C 1 u R 1 ( t ) ∂ [ u R 1 ( t ) ] ∂ t
Wherein, R 1, L 1, G 1And C 1The positive sequence resistance, positive sequence inductance, the positive sequence electricity that are respectively unit line length are led and positive sequence electric capacity;
C2, the dynamic positive sequence wave impedance of circuit right-hand member
Figure FDA000020564805000114
Dynamic positive sequence propagation coefficient
Figure FDA000020564805000115
Trouble spot positive sequence voltage U by the right-hand member derivation RF1For:
U RF 1 = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Wherein x is that the trouble spot is apart from the distance of circuit right-hand member, i.e. fault distance;
The dynamic positive sequence wave impedance of circuit right-hand member Dynamic positive sequence propagation coefficient
Figure FDA00002056480500023
Trouble spot positive sequence voltage U by the left end derivation SF1For:
U SF 1 = U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
C3, fault localization
Circuit is U under failure condition RF1=U SF1, the nonlinear equation of setting up about fault distance x is:
U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x ) = U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x + U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x
Use the iteration function F (x) that Newton iteration method solves following formula:
F ( x ) = x - γ ~ S 1 U S 1 - I S 1 Z ~ SC 1 2 e γ ~ S 1 ( l - x ) + γ ~ S 1 U S 1 + I S 1 Z ~ SC 1 2 e - γ ~ S 1 ( l - x )
- ( γ ~ R 1 U R 1 - I R 1 Z ~ RC 1 2 e γ ~ R 1 x - γ ~ R 1 U R 1 + I R 1 Z ~ RC 1 2 e - γ ~ R 1 x )
Following formula is carried out interative computation can obtain accurately fault distance x.
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