CN103983897A - Method for distinguishing fault types of non-transposition same-tower double-circuit line - Google Patents

Abstract

The invention discloses a method for distinguishing fault types of a non-transposition same-tower double-circuit line. The method comprises the steps that firstly, when the non-transposition same-tower double-circuit line fails, voltage information and current information at the protection installation position of the non-transposition same-tower double-circuit line are collected in real time through a current transformer and a voltage transformer; secondarily, positive-phase-sequence identical-phase-sequence current components, positive-phase-sequence differential-phase-sequence current components, negative-phase-sequence identical-phase-sequence current components, negative-phase-sequence differential-phase-sequence current components, the zero-phase-sequence identical-phase-sequence current components and zero-phase-sequence differential-phase-sequence current components of the double-circuit line are extracted by using the six-sequence component conversion method, and the relative phase angle between the differential-phase-sequence components and the identical-phase-sequence components of positive-phase-sequence currents and the relative phase angle between the differential-phase-sequence components of negative-phase-sequence currents and the differential-phase-sequence components of the positive-phase-sequence currents are calculated; then whether the non-transposition same-tower double-circuit line has a same-name-phase crossing-line fault or a single-circuit-line fault or a non-same-name-phase crossing-line fault is judged. By means of the method, the single-circuit-line fault, the same-name-phase crossing-line fault and the non-same-name-phase crossing-line fault of the non-transposition same-tower double-circuit line can be distinguished correctly. Excellent fault distinguishing capacity can be still kept when non-transposition parameters and line parameters of the same-tower double-circuit line are seriously unsymmetrical.

Description

A kind of for distinguishing the method for the common-tower double-return line fault type that do not replace

Technical field

The invention belongs to technical field of power systems, relate to the Fault Phase Selection technical field of the common-tower double-return circuit that do not replace, be specifically related to a kind of for distinguishing the method for the common-tower double-return line fault type that do not replace.

Background technology

Because parallel erected on same tower circuit through-put power is large, in the time there is cross line fault, how to avoid double-circuit line to excise as far as possible simultaneously, be the important topic of widely paying close attention in engineering application, and the solution of the problems referred to above depend on well behaved Novel Faulty Phase Selector.Meanwhile, the performance of Novel Faulty Phase Selector has direct impact to the correct work of auto recloser, and therefore, the Novel Faulty Phase Selector that research is applicable to parallel erected on same tower circuit has important theory and realistic meaning.

The current research for common-tower double-return line fault phase selection, except the phase selection principle of six order components, remaining phase selection scheme is all to improve on the basis of common phase selection principle, and these schemes are because principle defect is all difficult to meet the Fault Phase Selection requirement of common-tower double-return line complexity.The phase selection principle of six order components can be identified all kinds of single line down and the cross line fault that common-tower double-return line occurs well, have phase selection effect well, but it must utilize the electric parameters of two loop lines, therefore also do not obtain rig-site utilization.Can predict, along with the development of digital transformer substation and the widespread use of electronic mutual inductor, six order component method Fault Phase Selections can further be applied.

The achievement in research of existing six order component method Fault Phase Selections is all desirable transposition based on common-tower double-return line, main flow thinking be first according to the large wisp fault type of the value of two loop line electric current six order components be divided into single line down, with famous prime minister's cross line fault and non-same famous prime minister's cross line fault three classes, and then each type fault is carried out to deep differentiation.But in production reality, short distance common-tower double-return circuit often adopts not transposition form to set up, so just make mutual inductance parameter unbalance between alternate, line, the fault signature of six order components while also having changed desirable transposition simultaneously.Now distinguish fault type by the value size of six order components merely and become very unreliable, be difficult to meet the requirement of common-tower double-return line to Novel Faulty Phase Selector that do not replace of the large electrical network of high pressure.

Summary of the invention

The problem existing for solving above-mentioned prior art, the object of this invention is to provide a kind of for distinguishing the method for the common-tower double-return line fault type that do not replace, can correctly distinguish the common-tower double-return circuit list line down that do not replace, with famous prime minister's cross line fault and non-same famous prime minister's cross line fault, avoided the asymmetric incorrect operation of phase selection element when comparatively serious of line parameter circuit value; Can not replace at common-tower double-return line, line parameter circuit value is asymmetric still keeps good fault to screen ability when comparatively serious.

For achieving the above object, the technical solution used in the present invention is:

For distinguishing a method for the common-tower double-return line fault type that do not replace, comprise the steps:

Step 1: when not replacing after common-tower double-return line failure, by the do not replace voltage of common-tower double-return line protection installation place of current-voltage transformer Real-time Collection and the electric current of double loop and calculate line voltage

Step 2: calculate respectively double loop zero-sequence current separately forward-order current negative-sequence current and extract the same inverted sequence component of the positive and negative zero current of double loop by six order converter techniques wherein computing formula is:

$\stackrel{\·}{{3I}_{10}}=\stackrel{\·}{I_{1A}}+\stackrel{\·}{I_{1B}}+\stackrel{\·}{I_{1C}}---\left(1\right)$

$\stackrel{\·}{{3I}_{20}}=\stackrel{\·}{I_{2A}}+\stackrel{\·}{I_{2B}}+\stackrel{\·}{I_{2C}}---\left(2\right)$

$\stackrel{\·}{I_{11}}=1/3*(\stackrel{\·}{I_{1A}}+\mathrm{\α}*\stackrel{\·}{I_{1B}}+{\mathrm{\α}}^2*\stackrel{\·}{I_{1C}})---\left(3\right)$

$\stackrel{\·}{I_{21}}=1/3*(\stackrel{\·}{I_{2A}}+\mathrm{\α}*\stackrel{\·}{I_{2B}}+{\mathrm{\α}}^2*\stackrel{\·}{I_{2C}})---\left(4\right)$

$\stackrel{\·}{I_{12}}=1/3*(\stackrel{\·}{I_{1A}}+{\mathrm{\α}}^2*\stackrel{\·}{I_{1B}}+\mathrm{\α}*\stackrel{\·}{I_{1C}})---\left(5\right)$

$\stackrel{\·}{I_{22}}=1/3*(\stackrel{\·}{I_{2A}}+{\mathrm{\α}}^2*\stackrel{\·}{I_{2B}}+\mathrm{\α}*\stackrel{\·}{I_{2C}})---\left(6\right)$

$\left[\begin{array}{c}\stackrel{\·}{I_{T0}}\\ \stackrel{\·}{I_{F0}}\\ \stackrel{\·}{I_{T1}}\\ \stackrel{\·}{I_{F1}}\\ \stackrel{\·}{I_{T2}}\\ \stackrel{\·}{I_{F2}}\end{array}\right]={\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 1& {\mathrm{\α}}^2& {\mathrm{\α}}^2& \mathrm{\α}& \mathrm{\α}\\ 1& 1& \mathrm{\α}& \mathrm{\α}& {\mathrm{\α}}^2& {\mathrm{\α}}^2\\ 1& -1& 1& -1& 1& -1\\ 1& -1& {\mathrm{\α}}^2& {-\mathrm{\α}}^2& \mathrm{\α}& -\mathrm{\α}\\ 1& -1& \mathrm{\α}& -\mathrm{\α}& {\mathrm{\α}}^2& -{\mathrm{\α}}^2\end{array}\right]}^{-1}\left[\begin{array}{c}\stackrel{\·}{I_{1A}}\\ \stackrel{\·}{I_{1B}}\\ \stackrel{\·}{I_{1C}}\\ \stackrel{\·}{I_{2A}}\\ \stackrel{\·}{I_{2B}}\\ \stackrel{\·}{I_{2C}}\end{array}\right]---\left(7\right)$

In formula, α is label symbol,

Step 3: first judge that whether the positive and negative zero-sequence current inverted sequence of circuit component is zero and double loop forward-order current whether approximately equal; If above-mentioned two conditions meet simultaneously, enter step 4 and continue judgement; If have arbitrary condition not meet, enter step 5;

In protection, the criterion that the positive and negative zero-sequence current inverted sequence of common-tower double-return circuit that do not replace component is zero is:

$\left\{\begin{array}{c}{I}_{F0}<I_{T1}/8\\ I_{F1}<I_{T1}/5\\ I_{F2}<I_{T1}/5\end{array}\right.---\left(8\right)$

The approximately equalised criterion of two loop line forward-order currents is:

{0.7<I ₁₁/I ₂₁<1.3 (9)

Step 4: judge line voltage whether approximately equal, if unequal, is directly judged to be same famous prime minister's cross line fault; If equate, further judge whether double loop zero, negative-sequence current are zero, if now zero be judged to be occur with famous prime minister's cross line fault, otherwise think that non-same famous prime minister's cross line fault has occurred circuit;

Judge line voltage herein, approximately equalised algorithm is:

Max{U _AB-U _BC,U _BC-U _CA,U _CA-U _AB}<(U _N-Min{U _AB,U _BC,U _CA})/5 (10)

In formula (10), U _nfor circuit rated voltage;

The criterion that double loop zero, negative-sequence current are zero is:

$\left\{\begin{array}{c}{3I}_{10}<I_{T1}/3\\ {3I}_{20}<I_{T1}/3\\ I_{12}<I_{T1}/6\\ I_{22}<I_{T1}/6\end{array}\right.---\left(11\right)$

Step 5: calculate forward-order current instead with order component relative phase angle θ _fT1with negative-sequence current instead with order component relative phase angle θ _fT2; Judge θ _fT1whether be 0 ° or 180 ° and θ _fT2whether be 0 ° or 180 °, if both meet simultaneously, enter step 6; If have arbitrary condition not meet, be directly judged to be non-same famous prime minister's cross line fault; The criterion that completes this step in protection is:

Step 6: judge line voltage whether approximately equal, if unequal, enters step 7; If equate, secondly judge double loop zero-sequence current and negative-sequence current whether be approximately zero, if be approximately zero, fault type is regarded as single line down, otherwise thinks to have occurred non-same famous prime minister's cross line fault; Described judgement line voltage approximately equal and double loop zero, negative current are approximately zero method and see step 4;

Step 7: calculate the relative phase angle θ F21 of negative-sequence current inverted sequence component and forward-order current inverted sequence component, judge whether it is ± 60 ° or 180 °; If ± 60 ° or 180 °, enter 8. further judgement of step; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described judgement θ _f21the criterion that is whether ± 60 ° or 180 ° is:

{|θ _F21+60°|<30°U|θ _F21-60°|<30°U|θ _F21-180°|<30°U|θ _F21+180°|<30° (13)

Step 8: the inverted sequence component I that calculates zero-sequence current _f0whether be zero, if be zero, be judged to single line down; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described calculating zero-sequence current inverted sequence component I _f0be zero according to being:

{I _F0<I _T1/10 (14)

The present invention compared to the prior art, possesses following advantage:

The inventive method, at distinguished while not replacing common-tower double-return line fault type Integrated using six order current component relative phase angles, three-phase line voltage, the positive and negative zero-sequence current of two loop lines, has effectively promoted the ability of phase selection element tolerance circuit degree of asymmetry.Wherein order component relative phase angle adopts a larger threshold range, the angular deflection that can dissolve while transposition with respect to ideal than phase angle.The correctness distinguished of fault type that adopted three-phase line voltage, the positive and negative zero-sequence current multi-ensuring of two loop lines.

Brief description of the drawings

Fig. 1 is a kind of conspectus that is applicable to the inventive method.

Fig. 2 is the process flow diagram of realizing the inventive method.

Embodiment:

Below in conjunction with drawings and Examples, the present invention is described in further details.

As shown in Figure 1, provide the common-tower double-return line model that do not replace of both-end common bus, in the time that various fault type occurs circuit, used the method for the differentiation fault type of the present invention's proposition.

As shown in Figure 2, the present invention is a kind of for distinguishing the method for the common-tower double-return line fault type that do not replace, and comprises the steps:

Step 1: when not replacing after common-tower double-return line failure, by the do not replace voltage of common-tower double-return line protection installation place of current-voltage transformer Real-time Collection and the electric current of double loop and calculate line voltage

Step 2: calculate respectively double loop zero-sequence current separately forward-order current negative-sequence current and extract the same inverted sequence component of the positive and negative zero current of double loop by six order converter techniques wherein computing formula is:

$\stackrel{\&CenterDot;}{{3I}_{10}}=\stackrel{\&CenterDot;}{I_{1A}}+\stackrel{\&CenterDot;}{I_{1B}}+\stackrel{\&CenterDot;}{I_{1C}}---\left(1\right)$

$\stackrel{\&CenterDot;}{{3I}_{20}}=\stackrel{\&CenterDot;}{I_{2A}}+\stackrel{\&CenterDot;}{I_{2B}}+\stackrel{\&CenterDot;}{I_{2C}}---\left(2\right)$

$\stackrel{\&CenterDot;}{I_{11}}=1/3*(\stackrel{\&CenterDot;}{I_{1A}}+\mathrm{\&alpha;}*\stackrel{\&CenterDot;}{I_{1B}}+{\mathrm{\&alpha;}}^2*\stackrel{\&CenterDot;}{I_{1C}})---\left(3\right)$

$\stackrel{\&CenterDot;}{I_{21}}=1/3*(\stackrel{\&CenterDot;}{I_{2A}}+\mathrm{\&alpha;}*\stackrel{\&CenterDot;}{I_{2B}}+{\mathrm{\&alpha;}}^2*\stackrel{\&CenterDot;}{I_{2C}})---\left(4\right)$

$\stackrel{\&CenterDot;}{I_{12}}=1/3*(\stackrel{\&CenterDot;}{I_{1A}}+{\mathrm{\&alpha;}}^2*\stackrel{\&CenterDot;}{I_{1B}}+\mathrm{\&alpha;}*\stackrel{\&CenterDot;}{I_{1C}})---\left(5\right)$

$\stackrel{\&CenterDot;}{I_{22}}=1/3*(\stackrel{\&CenterDot;}{I_{2A}}+{\mathrm{\&alpha;}}^2*\stackrel{\&CenterDot;}{I_{2B}}+\mathrm{\&alpha;}*\stackrel{\&CenterDot;}{I_{2C}})---\left(6\right)$

$\left[\begin{array}{c}\stackrel{\&CenterDot;}{I_{T0}}\\ \stackrel{\&CenterDot;}{I_{F0}}\\ \stackrel{\&CenterDot;}{I_{T1}}\\ \stackrel{\&CenterDot;}{I_{F1}}\\ \stackrel{\&CenterDot;}{I_{T2}}\\ \stackrel{\&CenterDot;}{I_{F2}}\end{array}\right]={\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 1& {\mathrm{\&alpha;}}^2& {\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}& \mathrm{\&alpha;}\\ 1& 1& \mathrm{\&alpha;}& \mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2& {\mathrm{\&alpha;}}^2\\ 1& -1& 1& -1& 1& -1\\ 1& -1& {\mathrm{\&alpha;}}^2& {-\mathrm{\&alpha;}}^2& \mathrm{\&alpha;}& -\mathrm{\&alpha;}\\ 1& -1& \mathrm{\&alpha;}& -\mathrm{\&alpha;}& {\mathrm{\&alpha;}}^2& -{\mathrm{\&alpha;}}^2\end{array}\right]}^{-1}\left[\begin{array}{c}\stackrel{\&CenterDot;}{I_{1A}}\\ \stackrel{\&CenterDot;}{I_{1B}}\\ \stackrel{\&CenterDot;}{I_{1C}}\\ \stackrel{\&CenterDot;}{I_{2A}}\\ \stackrel{\&CenterDot;}{I_{2B}}\\ \stackrel{\&CenterDot;}{I_{2C}}\end{array}\right]---\left(7\right)$

In formula, α is label symbol,

Step 3: first judge that whether the positive and negative zero-sequence current inverted sequence of circuit component is zero and double loop forward-order current whether approximately equal; If above-mentioned two conditions meet simultaneously, enter step 4 and continue judgement; If have arbitrary condition not meet, enter step 5;

In protection, the criterion that the positive and negative zero-sequence current inverted sequence of common-tower double-return circuit that do not replace component is zero is:

$\left\{\begin{array}{c}{I}_{F0}<I_{T1}/8\\ I_{F1}<I_{T1}/5\\ I_{F2}<I_{T1}/5\end{array}\right.---\left(8\right)$

The approximately equalised criterion of two loop line forward-order currents is:

{0.7<I ₁₁/I ₂₁<1.3 (9)

Step 4: judge line voltage whether approximately equal, if unequal, is directly judged to be same famous prime minister's cross line fault; If equate, further judge whether double loop zero, negative-sequence current are zero, if now zero be judged to be occur with famous prime minister's cross line fault, otherwise think that non-same famous prime minister's cross line fault has occurred circuit;

Judge line voltage herein, approximately equalised algorithm is:

Max{U _AB-U _BC,U _BC-U _CA,U _CA-U _AB}<(U _N-Min{U _AB,U _BC,U _CA})/5 (10)

In formula (10), U _nfor circuit rated voltage;

The criterion that double loop zero, negative-sequence current are zero is:

$\left\{\begin{array}{c}{3I}_{10}<I_{T1}/3\\ {3I}_{20}<I_{T1}/3\\ I_{12}<I_{T1}/6\\ I_{22}<I_{T1}/6\end{array}\right.---\left(11\right)$

Step 5: calculate forward-order current instead with order component relative phase angle θ _fT1with negative-sequence current instead with order component relative phase angle θ _fT2; Judge θ _fT1whether be 0 ° or 180 ° and θ _fT2whether be 0 ° or 180 °, if both meet simultaneously, enter step 6; If have arbitrary condition not meet, be directly judged to be non-same famous prime minister's cross line fault; The criterion that completes this step in protection is:

Step 6: judge line voltage whether approximately equal, if unequal, enters step 7; If equate, secondly judge double loop zero-sequence current and negative-sequence current whether be approximately zero, if be approximately zero, fault type is regarded as single line down, otherwise thinks to have occurred non-same famous prime minister's cross line fault; Described judgement line voltage approximately equal and double loop zero, negative current are approximately zero method and see step 4;

Step 7: the relative phase angle θ that calculates negative-sequence current inverted sequence component and forward-order current inverted sequence component _f21, judge whether it is ± 60 ° or 180 °; If ± 60 ° or 180 °, enter 8. further judgement of step; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described judgement θ _f21the criterion that is whether ± 60 ° or 180 ° is:

{|θ _F21+60°|<30°U|θ _F21-60°|<30°U|θ _F21-180°|<30°U|θ _F21+180°|<30° (13)

Step 8: the inverted sequence component I that calculates zero-sequence current _f0whether be zero, if be zero, be judged to single line down; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described calculating zero-sequence current inverted sequence component I _f0be zero according to being:

{I _F0<I _T1/10 (14)

Adopting the fault type differentiating method of above-mentioned proposition can correctly judge the common-tower double-return line that do not replace is to occur with also right and wrong phase cross line fault of the same name of famous prime minister's cross line fault, single line down, there is the ability of stronger tolerance circuit degree of asymmetry, effectively ensured the correctness of the common-tower double-return line Fault Phase Selection that do not replace.

Claims (1)

1. for distinguishing a method for the common-tower double-return line fault type that do not replace, it is characterized in that: comprise the steps:

Step 1: when not replacing after common-tower double-return line failure, by the do not replace voltage of common-tower double-return line protection installation place of current-voltage transformer Real-time Collection and the electric current of double loop and calculate line voltage

Step 2: calculate respectively double loop zero-sequence current separately forward-order current negative-sequence current and extract the same inverted sequence component of the positive and negative zero current of double loop by six order converter techniques wherein computing formula is:

In formula, α is label symbol,

Step 3: first judge that whether the positive and negative zero-sequence current inverted sequence of circuit component is zero and double loop forward-order current whether approximately equal; If above-mentioned two conditions meet simultaneously, enter step 4 and continue judgement; If have arbitrary condition not meet, enter step 5;

In protection, the criterion that the positive and negative zero-sequence current inverted sequence of common-tower double-return circuit that do not replace component is zero is:

The approximately equalised criterion of two loop line forward-order currents is:

{0.7<I ₁₁/I ₂₁<1.3 (9)

Step 4: judge line voltage whether approximately equal, if unequal, is directly judged to be same famous prime minister's cross line fault; If equate, further judge whether double loop zero, negative-sequence current are zero, if now zero be judged to be occur with famous prime minister's cross line fault, otherwise think that non-same famous prime minister's cross line fault has occurred circuit;

Judge line voltage herein, approximately equalised algorithm is:

Max{U _AB-U _BC,U _BC-U _CA,U _CA-U _AB}<(U _N-Min{U _AB,U _BC,U _CA})/5 (10)

In formula (10), U _nfor circuit rated voltage;

The criterion that double loop zero, negative-sequence current are zero is:

Step 5: calculate forward-order current instead with order component relative phase angle θ _fT1with negative-sequence current instead with order component relative phase angle θ _fT2; Judge θ _fT1whether be 0 ° or 180 ° and θ _fT2whether be 0 ° or 180 °, if both meet simultaneously, enter step 6; If have arbitrary condition not meet, be directly judged to be non-same famous prime minister's cross line fault; The criterion that completes this step in protection is:

Step 6: judge line voltage whether approximately equal, if unequal, enters step 7; If equate, secondly judge double loop zero-sequence current and negative-sequence current whether be approximately zero, if be approximately zero, fault type is regarded as single line down, otherwise thinks to have occurred non-same famous prime minister's cross line fault; Described judgement line voltage approximately equal and double loop zero, negative current are approximately zero method and see step 4;

Step 7: the relative phase angle θ that calculates negative-sequence current inverted sequence component and forward-order current inverted sequence component _f21, judge whether it is ± 60 ° or 180 °; If ± 60 ° or 180 °, enter 8. further judgement of step; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described judgement θ _f21the criterion that is whether ± 60 ° or 180 ° is:

{|θ _F21+60°|<30°U|θ _F21-60°|<30°U|θ _F21-180°|<30°U|θ _F21+180°|<30° (13)

Step 8: the inverted sequence component I that calculates zero-sequence current _f0whether be zero, if be zero, be judged to single line down; Otherwise, think that non-same famous prime minister's cross line fault has occurred circuit; Described calculating zero-sequence current inverted sequence component I _f0be zero according to being:

{I _F0<I _T1/10 (14)。