CN102944809B - Method for rapidly identifying power grid fault - Google Patents

Abstract

The invention discloses a method for rapidly identifying a power grid fault. The method comprises the following steps of: acquiring instantaneous values of electric signals of user side busbars when a power system fault occurs; calculating each phase voltage parameter; judging the type of the power grid fault; and judging the position to which the power system fault occurs. By utilizing the method, whether the power system fault occurs to the power grid side or the user side can be rapidly identified. By utilizing the method, the time from the occurrence of the power system fault to the input of a standby power supply can be obviously shortened and is just within 10ms generally; and in a controlled area, even a high-capacity motor in use exists, power can be supplied in time by the standby power supply before the terminal voltage of the motor is greatly lowered, so that the continuous operation of the motor is ensured. By utilizing the method, a good foundation is laid for a standby power supply automatically-inputting device to judge whether to input the standby power supply when the power system fault occurs.

Description

A kind of method of quick identification electric network fault

Technical field

The present invention relates to Power System Faults Detection technology, particularly relate to a kind of method that position occurs electric power system fault that judges.

Background technology

In electric system, have many important regions to use automatic throw-in equipment of emergency power supply, at present, automatic throw-in equipment of emergency power supply usually with protective relaying device with the use of, power distribution network protective relaying device great majority employing current protection.When electric system is broken down, first, protective relaying device is cut off the electricity supply, and subsequently, automatic throw-in equipment of emergency power supply, after the feeder ear decompression in certain region being detected, automatically drops into standby power supply, is this block supply.This mode also exists following deficiency: current protection actuation time of protective relaying device is at more than 0.5s, automatic throw-in equipment of emergency power supply is detecting feeder ear decompression and is dropping into the use procedure of standby power supply, need more than 20ms, therefore, the overlong time required for whole process of powering is carried out from standby power supply of cutting off the electricity supply, during this period of time, when the certain user in this region is using large capacity motor, the set end voltage of motor can sharply decline, motor low-voltage variation can cut off the isolating switch of motor automatically, motor is caused to run continuously.Can run continuously in order to ensure motor in the whole process of carrying out powering at standby power supply of cutting off the electricity supply, just must shorten the time that this process spends as much as possible, therefore, relevant technologies personnel attempt only to adopt automatic throw-in equipment of emergency power supply to carry out independent operation, so, must improve automatic throw-in equipment of emergency power supply, to the problem that first the new automatic throw-in equipment of emergency power supply improved must solve be exactly: when electric system is broken down, must judge that the fault of electric system is positioned at grid side or user side fast, thus for judging whether that dropping into standby power supply lays a good foundation.

Summary of the invention

The object of this invention is to provide a kind of method of quick identification electric network fault, it can identify that the fault of electric system is positioned at grid side or user side rapidly, thus for automatic throw-in equipment of emergency power supply judge whether drop into standby power supply lay a good foundation.

The object of the invention is by such technical scheme realize, it includes following step:

(1), when electric power system fault occurs, user side bus electric signal instantaneous value is gathered:

(1), when electric power system fault occurs, user side bus electric signal instantaneous value is gathered:

On these 80 time periods of T (0)-T (79), the three-phase current momentary signal sequential value of Real-time Collection user side bus and three-phase voltage momentary signal sequential value, that is:

A phase current momentary signal sequential value is: i _a(t _-16), i _a(t _-15) ... i _a(t ₆₄);

B phase current momentary signal sequential value is: i _b(t _-16), i _b(t _-15) ... i _b(t ₆₄);

C phase current momentary signal sequential value is: i _c(t _-16), i _c(t _-15) ... i _c(t ₆₄);

A phase voltage momentary signal sequential value is: u _a(t ₁), u _a(t ₂) ... u _a(t ₆₄);

B phase voltage momentary signal sequential value is: u _b(t ₁), u _b(t ₂) ... u _b(t ₆₄);

C phase voltage momentary signal sequential value is: u _c(t ₁), u _c(t ₂) ... u _c(t ₆₄);

Wherein, t _-16, t _-15... t ₆₄represent that each gathers the moment of electric signal respectively, in T (0)-T (79), the end points of each time period is sampling instant successively, and the length of time period is T'(T'=3.125ms);

(2), each phase voltage parameter is calculated

A phase voltage first voltage parameter A is calculated respectively with the voltage data that step (1) gathers _uA, the second voltage parameter B _uA, tertiary voltage parameter C _uA, the 4th voltage parameter D _uA; B phase voltage first voltage parameter A _uB, the second voltage parameter B _uB, tertiary voltage parameter C _uB, the 4th voltage parameter D _uB; C phase voltage first voltage parameter A _uC, the second voltage parameter B _uC, tertiary voltage parameter C _uC, the 4th voltage parameter D _uC, concrete steps are as follows:

1., first setting X phase voltage voltage parameter initial value is A _uX, B _uX, C _uX, D _uX;

2. the residual vector e in each moment, is gone out with following formulae discovery _mX=[e _1X, e _2X... e _64X] ^t;

$e_{\mathrm{nX}}=A_{\mathrm{uX}}\cos \left({\mathrm{wt}}_n\right)+B_{\mathrm{uX}}\sin \left({\mathrm{wt}}_n\right)+C_{\mathrm{uX}}{e}^{-t_n{/D}_{\mathrm{uX}}}-u_X\left(t_n\right)$

In formula:

X is one of A phase, B phase, C phase;

N is positive integer, n=1 ~ 64;

U _x(t _n) be t _nmoment user side bus X phase voltage instantaneous value;

3., total residual epsilon is calculated _x

${\mathrm{\ϵ}}_X=e_{1X}^2+e_{2X}^2+\·\·\·+e_{64X}^2$

By ε _xwith set threshold residual value ε ₀compare:

Work as ε _x> ε ₀time, the increment calculating X phase voltage parameter is as follows:

$\left[\begin{array}{c}{\mathrm{\ΔA}}_{\mathrm{uX}}\\ {\mathrm{\ΔB}}_{\mathrm{uX}}\\ {\mathrm{\ΔC}}_{\mathrm{uX}}\\ {\mathrm{\ΔD}}_{\mathrm{uX}}\end{array}\right]=-{\left[{\▿e}_{\mathrm{MX}}^T{\▿e}_{\mathrm{MX}}\right]}^{-1}{\▿e}_{\mathrm{MX}}^T{e}_{\mathrm{MX}}$

Wherein:

${\▿e}_{\mathrm{MX}}=\left[\begin{array}{cccc}\frac{{\∂e}_{1X}}{{\∂A}_{\mathrm{uX}}}& \frac{{\∂e}_{1X}}{{\∂B}_{\mathrm{uX}}}& \frac{{\∂e}_{1X}}{{\∂C}_{\mathrm{uX}}}& \frac{{\∂e}_{1X}}{{\∂D}_{\mathrm{uX}}}\\ \frac{{\∂e}_{2X}}{{\∂A}_{\mathrm{uX}}}& \frac{{\∂e}_{2X}}{{\∂B}_{\mathrm{uX}}}& \frac{{\∂e}_{2X}}{{\∂C}_{\mathrm{uX}}}& \frac{{\∂e}_{2X}}{{\∂D}_{\mathrm{uX}}}\\ .& .& .& .\\ .& .& .& .\\ .& .& .& .\\ \frac{{\∂e}_{64X}}{{\∂A}_{\mathrm{uX}}}& \frac{{\∂e}_{64X}}{{\∂B}_{\mathrm{uX}}}& \frac{{\∂e}_{64X}}{{\∂C}_{\mathrm{uX}}}& \frac{{\∂e}_{64X}}{{\∂D}_{\mathrm{uX}}}\end{array}\right]$

By A _uX+ Δ A _uX, B _uX+ Δ B _uX, C _uX+ Δ C _uX, D _uX+ Δ D _uXrespectively as new X phase voltage voltage parameter A _uX, B _uX, C _uX, D _uX, 2. perform step (2) the walk;

Work as ε _x< ε ₀time, the X phase voltage voltage parameter A now participating in calculating is described _uX, B _uX, C _uX, D _uXit is exactly X phase voltage voltage parameter when there is electric network fault;

(3) electric network fault type, is judged

1., the voltage battery parameter value obtained in step (2) is gone out A phase voltage amplitude u with following formulae discovery _a, B phase voltage amplitude u _b, C phase voltage amplitude u _c:

$u_A=\sqrt{A_{\mathrm{uA}}^2+B_{\mathrm{uA}}^2}$

$u_B=\sqrt{A_{\mathrm{uB}}^2+B_{\mathrm{uB}}^2}$

$u_C=\sqrt{A_{\mathrm{uC}}^2+B_{\mathrm{uC}}^2}$

2., electric network fault type judges

By the u obtained in previous step _a, u _b, u _cwith set voltage magnitude threshold values 0.85u _pcompare, wherein u _pphase voltage amplitude:

Work as u _a<0.85u _p, u _b<0.85u _p, u _c>0.85u _ptime, electric network fault type belongs to generation AB two-phase short-circuit fault;

Work as u _a>0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation BC two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b>0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation CA two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation ABC three phase short circuit fault;

Wherein, AB two-phase short-circuit fault, BC two-phase short-circuit fault, CA two-phase short-circuit fault are commonly referred to as two-phase short-circuit fault;

(4) position that electric power system fault occurs, is judged

According to the fault type that step (3) is known, select the determination methods of following abort situation with selecting a property:

1., when there is two-phase short-circuit fault:

The user side comprehensive impedance Z of fault phase is gone out to occur with following formulae discovery _x1X2:

$Z_{X1X2}={\left[M_{\mathrm{IX}1}^T{M}_{\mathrm{IX}1}\right]}^{-1}{M}_{\mathrm{IX}1}^T(U_{X1}-U_{X2})$

Wherein:

$M_{\mathrm{IX}1}=\left[\begin{array}{cc}{i}_{X1}\left(t_{48}\right)& i_{X1}\left(t_{48}\right){-i}_{X1}\left(t_{-16}\right)\\ i_{X1}\left(t_{49}\right)& i_{X1}\left(t_{49}\right)-i_{X1}\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_{X1}\left(t_{64}\right)& i_{X1}\left(t_{64}\right)-i_{X1}\left(t_0\right)\end{array}\right]$

$U_{X1}=\left[\begin{array}{c}{u}_{X1}\left(t_{49}\right)\\ u_{X1}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X1}\left(t_{64}\right)\end{array}\right]$

$U_{X2}=\left[\begin{array}{c}{u}_{X2}\left(t_{49}\right)\\ u_{X2}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X2}\left(t_{64}\right)\end{array}\right]$

In formula: X1 refers to A phase, B phase or C phase; X2 refers to: when X1 is A phase time, X2 can only be B phase; When X1 is B phase time, X2 can only be C phase; When X1 is C phase time, X2 can only be A phase;

The user side comprehensive impedance Z calculated _x1X2the matrix of 2 × 1, the 1st element Z in matrix _x1X21resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _x1X21during >0, illustrate that the abort situation of this electric system occurs in user side;

Work as Z _x1X21during <0, illustrate that the abort situation of this electric system occurs in grid side;

2., when there is ABC three phase short circuit fault:

The user side comprehensive impedance Z of A phase is gone out with following formulae discovery _aN

$Z_{\mathrm{AN}}={\left[M_{\mathrm{IA}}^T{M}_{\mathrm{IA}}\right]}^{-1}{M}_{\mathrm{IA}}^T{U}_A$

Wherein:

$M_{\mathrm{IA}}=\left[\begin{array}{cc}{i}_A\left(t_{48}\right)& i_A\left(t_{48}\right)-i_A\left(t_{-16}\right)\\ i_A\left(t_{49}\right)& i_A\left(t_{49}\right)-i_A\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_A\left(t_{64}\right)& i_A\left(t_{64}\right)-i_A\left(t_0\right)\end{array}\right]$

$U_A=\left[\begin{array}{c}{u}_A\left(t_{49}\right)\\ u_A\left(t_{50}\right)\\ .\\ .\\ .\\ u_A\left(t_{64}\right)\end{array}\right]$

The user side comprehensive impedance Z calculated _aNthe matrix of 2 × 1, the 1st element Z in matrix _aN1resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _aN1during >0, illustrate that the abort situation of this electric system occurs in user side;

Work as Z _aN1during <0, illustrate that the abort situation of this electric system occurs in grid side.

The present invention and automatic throw-in equipment of emergency power supply usually with protective relaying device with the use of scheme compared with; the present invention can not only identify electric power system fault fast; and can identify that abort situation is positioned at grid side or user side fast, judge whether to start automatic throw-in equipment of emergency power supply.The short trouble that any one position in electric system is occurred, by the data gathered in 5ms, instantaneous voltage amplitude can be extrapolated, and user side comprehensive impedance.When the sampling time is constant, along with the increase of sampled data quantity, the instantaneous voltage amplitude calculated and actual value more close, and its close value significantly reduces along with the increase of sampled data quantity.User side comprehensive impedance only characterizing consumer side practical impedance positive and negative, not representative of consumer side practical impedance.According to instantaneous voltage amplitude and user side comprehensive impedance, can electric power system fault be identified and judge that electric power system fault occurs in grid side or user side, and along with the increase of sampled data quantity, instantaneous voltage amplitude is more accurate.Owing to identifying that abort situation is positioned at grid side or user side only needs to judge that user side comprehensive impedance value is positive number or negative, the data gathered within 5ms are enough to meet final purpose of the present invention.Thus, the present invention can identify abort situation to be positioned at grid side or user side completely in 10ms.

Method of the present invention is applied in automatic throw-in equipment of emergency power supply, cause the new automatic throw-in equipment of emergency power supply improved when electric system is broken down, identify the position that fault occurs quickly and accurately, when abort situation is when being positioned at user side, automatic throw-in equipment of emergency power supply need not drop into standby power supply and run; When abort situation is when being positioned at grid side, automatic throw-in equipment of emergency power supply automatic cut-off power, and in-put of spare power supply is run.This process need only be extremely short time, for the large capacity motor run, set end voltage is not before significantly declining, and standby power supply supplies stable electric energy in time, thus guarantees that motor keeps the state run continuously in the process of this power conversion.

Owing to have employed technique scheme, the present invention can identify that the fault of electric system is positioned at grid side or user side rapidly.Use it, can significantly reduce and occur to from electric power system fault the time that in-put of spare power supply spends, usually within only needing 10ms, in control area, even if there is the large capacity motor used, also before motor set end voltage declines to a great extent, can be powered by standby power supply in time, thus guarantee the continuous operation of motor.The present invention is that automatic throw-in equipment of emergency power supply judges whether that when electric system is broken down dropping into standby power supply lays a good foundation.

Accompanying drawing explanation

Fig. 1 is the equivalent circuit diagram of electric system;

In figure: 1. electrical network; 2. the equivalent resistance of grid side; 3. the equivalent inductance of grid side; 4. transformer; 5. bus; 6. the equivalent resistance of user side; 7. the equivalent inductance of user side.

Embodiment

Below in conjunction with accompanying drawing and embodiment, the present invention is described in further detail.

The present invention includes following step:

(1), when electric power system fault occurs, user side bus electric signal instantaneous value is gathered:

As shown in Fig. 1 in accompanying drawing, on these 80 time periods of T (0)-T (79), the three-phase current momentary signal sequential value of Real-time Collection user side bus and three-phase voltage momentary signal sequential value, that is:

A phase current momentary signal sequential value is: i _a(t _-16), i _a(t _-15) ... i _a(t ₆₄);

B phase current momentary signal sequential value is: i _b(t _-16), i _b(t _-15) ... i _b(t ₆₄);

C phase current momentary signal sequential value is: i _c(t _-16), i _c(t _-15) ... i _c(t ₆₄);

A phase voltage momentary signal sequential value is: u _a(t ₁), u _a(t ₂) ... u _a(t ₆₄);

B phase voltage momentary signal sequential value is: u _b(t ₁), u _b(t ₂) ... u _b(t ₆₄);

C phase voltage momentary signal sequential value is: u _c(t ₁), u _c(t ₂) ... u _c(t ₆₄);

Wherein, t _-16, t _-15... t ₆₄represent that each gathers the moment of electric signal respectively, in T (0)-T (79), the end points of each time period is sampling instant successively, and the length of time period is T'(T'=3.125ms);

(2), each phase voltage parameter is calculated

A phase voltage first voltage parameter A is calculated respectively with the voltage data that step (1) gathers _uA, the second voltage parameter B _uA, tertiary voltage parameter C _uA, the 4th voltage parameter D _uA; B phase voltage first voltage parameter A _uB, the second voltage parameter B _uB, tertiary voltage parameter C _uB, the 4th voltage parameter D _uB; C phase voltage first voltage parameter A _uC, the second voltage parameter B _uC, tertiary voltage parameter C _uC, the 4th voltage parameter D _uC, concrete steps are as follows:

1., first setting X phase voltage voltage parameter initial value is A _uX, B _uX, C _uX, D _uX;

2. the residual vector e in each moment, is gone out with following formulae discovery _mX=[e _1X, e _2X... e _64X] ^t;

$e_{\mathrm{nX}}=A_{\mathrm{uX}}\cos \left({\mathrm{wt}}_n\right)+B_{\mathrm{uX}}\sin \left({\mathrm{wt}}_n\right)+C_{\mathrm{uX}}{e}^{-t_n{/D}_{\mathrm{uX}}}-u_X\left(t_n\right)$

In formula:

X is one of A phase, B phase, C phase;

N is positive integer, n=1 ~ 64;

U _x(t _n) be t _nmoment user side bus X phase voltage instantaneous value;

3., total residual epsilon is calculated _x

${\mathrm{\&epsiv;}}_X=e_{1X}^2+e_{2X}^2+\&CenterDot;\&CenterDot;\&CenterDot;+e_{64X}^2$

By ε _xwith set threshold residual value ε ₀compare:

Work as ε _x> ε ₀time, the increment calculating X phase voltage parameter is as follows:

$\left[\begin{array}{c}{\mathrm{\&Delta;A}}_{\mathrm{uX}}\\ {\mathrm{\&Delta;B}}_{\mathrm{uX}}\\ {\mathrm{\&Delta;C}}_{\mathrm{uX}}\\ {\mathrm{\&Delta;D}}_{\mathrm{uX}}\end{array}\right]=-{\left[{\&dtri;e}_{\mathrm{MX}}^T{\&dtri;e}_{\mathrm{MX}}\right]}^{-1}{\&dtri;e}_{\mathrm{MX}}^T{e}_{\mathrm{MX}}$

Wherein:

${\&dtri;e}_{\mathrm{MX}}=\left[\begin{array}{cccc}\frac{{\&PartialD;e}_{1X}}{{\&PartialD;A}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{1X}}{{\&PartialD;B}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{1X}}{{\&PartialD;C}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{1X}}{{\&PartialD;D}_{\mathrm{uX}}}\\ \frac{{\&PartialD;e}_{2X}}{{\&PartialD;A}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{2X}}{{\&PartialD;B}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{2X}}{{\&PartialD;C}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{2X}}{{\&PartialD;D}_{\mathrm{uX}}}\\ .& .& .& .\\ .& .& .& .\\ .& .& .& .\\ \frac{{\&PartialD;e}_{64X}}{{\&PartialD;A}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{64X}}{{\&PartialD;B}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{64X}}{{\&PartialD;C}_{\mathrm{uX}}}& \frac{{\&PartialD;e}_{64X}}{{\&PartialD;D}_{\mathrm{uX}}}\end{array}\right]$

By A _uX+ Δ A _uX, B _uX+ Δ B _uX, C _uX+ Δ C _uX, D _uX+ Δ D _uXrespectively as new X phase voltage voltage parameter A _uX, B _uX, C _uX, D _uX, 2. perform step (2) the walk;

Work as ε _x< ε ₀time, the X phase voltage voltage parameter A now participating in calculating is described _uX, B _uX, C _uX, D _uXit is exactly X phase voltage voltage parameter when there is electric network fault;

(3) electric network fault type, is judged

1., the voltage battery parameter value obtained in step (2) is gone out A phase voltage amplitude u with following formulae discovery _a, B phase voltage amplitude u _b, C phase voltage amplitude u _c:

$u_A=\sqrt{A_{\mathrm{uA}}^2+B_{\mathrm{uA}}^2}$

$u_B=\sqrt{A_{\mathrm{uB}}^2+B_{\mathrm{uB}}^2}$

$u_C=\sqrt{A_{\mathrm{uC}}^2+B_{\mathrm{uC}}^2}$

2., electric network fault type judges

By the u obtained in previous step _a, u _b, u _cwith set voltage magnitude threshold values 0.85u _pcompare, wherein u _pphase voltage amplitude:

Work as u _a<0.85u _p, u _b<0.85u _p, u _c>0.85u _ptime, electric network fault type belongs to generation AB two-phase short-circuit fault;

Work as u _a>0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation BC two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b>0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation CA two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation ABC three phase short circuit fault;

Wherein, AB two-phase short-circuit fault, BC two-phase short-circuit fault, CA two-phase short-circuit fault are commonly referred to as two-phase short-circuit fault;

(4) position that electric power system fault occurs, is judged

According to the fault type that step (3) is known, select the determination methods of following abort situation with selecting a property:

1., when there is two-phase short-circuit fault:

The user side comprehensive impedance Z of fault phase is gone out to occur with following formulae discovery _x1X2:

$Z_{X1X2}={\left[M_{\mathrm{IX}1}^T{M}_{\mathrm{IX}1}\right]}^{-1}{M}_{\mathrm{IX}1}^T(U_{X1}-U_{X2})$

Wherein:

$M_{\mathrm{IX}1}=\left[\begin{array}{cc}{i}_{X1}\left(t_{48}\right)& i_{X1}\left(t_{48}\right){-i}_{X1}\left(t_{-16}\right)\\ i_{X1}\left(t_{49}\right)& i_{X1}\left(t_{49}\right)-i_{X1}\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_{X1}\left(t_{64}\right)& i_{X1}\left(t_{64}\right)-i_{X1}\left(t_0\right)\end{array}\right]$

$U_{X1}=\left[\begin{array}{c}{u}_{X1}\left(t_{49}\right)\\ u_{X1}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X1}\left(t_{64}\right)\end{array}\right]$

$U_{X2}=\left[\begin{array}{c}{u}_{X2}\left(t_{49}\right)\\ u_{X2}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X2}\left(t_{64}\right)\end{array}\right]$

In formula: X1 refers to A phase, B phase or C phase; X2 refers to: when X1 is A phase time, X2 can only be B phase; When X1 is B phase time, X2 can only be C phase; When X1 is C phase time, X2 can only be A phase;

The user side comprehensive impedance Z calculated _x1X2the matrix of 2 × 1, the 1st element Z in matrix _x1X21resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _x1X21during >0, illustrate that the abort situation of this electric system occurs in user side;

Work as Z _x1X21during <0, illustrate that the abort situation of this electric system occurs in grid side;

2., when there is ABC three phase short circuit fault:

The user side comprehensive impedance Z of A phase is gone out with following formulae discovery _aN

$Z_{\mathrm{AN}}={\left[M_{\mathrm{IA}}^T{M}_{\mathrm{IA}}\right]}^{-1}{M}_{\mathrm{IA}}^T{U}_A$

Wherein:

$M_{\mathrm{IA}}=\left[\begin{array}{cc}{i}_A\left(t_{48}\right)& i_A\left(t_{48}\right)-i_A\left(t_{-16}\right)\\ i_A\left(t_{49}\right)& i_A\left(t_{49}\right)-i_A\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_A\left(t_{64}\right)& i_A\left(t_{64}\right)-i_A\left(t_0\right)\end{array}\right]$

$U_A=\left[\begin{array}{c}{u}_A\left(t_{49}\right)\\ u_A\left(t_{50}\right)\\ .\\ .\\ .\\ u_A\left(t_{64}\right)\end{array}\right]$

The user side comprehensive impedance Z calculated _aNthe matrix of 2 × 1, the 1st element Z in matrix _aN1resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _aN1during >0, illustrate that the abort situation of this electric system occurs in user side;

Work as Z _aN1during <0, illustrate that the abort situation of this electric system occurs in grid side.

The invention will be further described for existing Binding experiment example:

This experimental example for be feeder line total length l=1km and voltage is 10kV, user side voltage is the electric system of 380V, as shown in Figure 1, this electric system actual parameter respectively:

Table 1 electric system actual parameter is arranged

In upper table:

R _lineit is the equivalent resistance of grid side;

L _lineit is the equivalent inductance of grid side;

R _loadit is the equivalent resistance of user side;

L _loadit is the equivalent inductance of user side.

Test example: under the dissimilar failure condition of generation, identifies that fault is positioned at grid side or user side

Feed line length gets l=1km, AB two-phase short-circuit fault, BC two-phase short-circuit fault, CA two-phase short-circuit fault and ABC three phase short circuit fault is there is respectively for grid side and user side, sampled data is 5ms data after fault occurs, use method of the present invention, every type Fault Identification time is no more than 10ms, and recognition result is as shown in the table:

Table 2 abort situation recognition result

As can be seen from Table 2, the present invention has for two-phase short-circuit fault, ABC three phase short circuit fault and well identifies accuracy.When fault is positioned at grid side, comprehensive impedance result of calculation is negative value; When fault is positioned at user side, comprehensive impedance result of calculation be on the occasion of.Identify that the order of accuarcy of abort situation is up to 100%.

From test example, according to the method for comprehensive impedance identification electric network fault position, user side, what the present invention proposed accurately can identify that electric network fault position is positioned at grid side or user side, to two-phase short-circuit fault, ABC three phase short circuit fault, there is good accuracy, and because the method only needs to judge user side comprehensive impedance symbol, can guarantee in 10ms, identify that electric network fault position is positioned at grid side or user side, thus reach the object quick and precisely identified.

Claims (1)

1. identify a method for electric network fault fast, it includes following step:

(1), when electric power system fault occurs, user side bus electric signal instantaneous value is gathered:

On 80 time periods of T (0)-T (79), the three-phase current momentary signal sequential value of Real-time Collection user side bus and three-phase voltage momentary signal sequential value, that is:

A phase current momentary signal sequential value is: i _a(t _-16), i _a(t _-15) ... i _a(t ₆₄);

B phase current momentary signal sequential value is: i _b(t _-16), i _b(t _-15) ... i _b(t ₆₄);

C phase current momentary signal sequential value is: i _c(t _-16), i _c(t _-15) ... i _c(t ₆₄);

A phase voltage momentary signal sequential value is: u _a(t ₁), u _a(t ₂) ... u _a(t ₆₄);

B phase voltage momentary signal sequential value is: u _b(t ₁), u _b(t ₂) ... u _b(t ₆₄);

C phase voltage momentary signal sequential value is: u _c(t ₁), u _c(t ₂) ... u _c(t ₆₄);

Wherein, t _-16, t _-15... t ₆₄represent that each gathers the moment of electric signal respectively, in T (0)-T (79), the end points of each time period is sampling instant successively, and the length of time period is T', T'=3.125ms;

(2), each phase voltage parameter is calculated

A phase voltage first voltage parameter A is calculated respectively with the voltage data that step (1) gathers _uA, the second voltage parameter B _uA, tertiary voltage parameter C _uA, the 4th voltage parameter D _uA; B phase voltage first voltage parameter A _uB, the second voltage parameter B _uB, tertiary voltage parameter C _uB, the 4th voltage parameter D _uB; C phase voltage first voltage parameter A _uC, the second voltage parameter B _uC, tertiary voltage parameter C _uC, the 4th voltage parameter D _uC, concrete steps are as follows:

1., first setting X phase voltage voltage parameter initial value is A _uX, B _uX, C _uX, D _uX;

2. the residual vector e in each moment, is gone out with following formulae discovery _mX=[e _1X, e _2X... e _64X] ^t;

$e_{\mathrm{nX}}=A_{\mathrm{uX}}\cos \left({\mathrm{wt}}_n\right)+B_{\mathrm{uX}}\sin \left({\mathrm{wt}}_n\right)+C_{\mathrm{uX}}{e}^{-t_n/D_{\mathrm{uX}}}-u_X\left(t_n\right)$

In formula:

X is one of A phase, B phase, C phase;

N is positive integer, n=1 ~ 64;

U _x(t _n) be t _nmoment user side bus X phase voltage instantaneous value;

3., total residual epsilon is calculated _x

${\mathrm{\&epsiv;}}_X=e_{1X}^2+e_{2X}^2+...+e_{64X}^2$

By ε _xwith set threshold residual value ε ₀compare:

Work as ε _x> ε ₀time, the increment calculating X phase voltage parameter is as follows:

$\left[\begin{array}{c}\mathrm{\&Delta;}{A}_{\mathrm{uX}}\\ \mathrm{\&Delta;}{B}_{\mathrm{uX}}\\ \mathrm{\&Delta;}{C}_{\mathrm{uX}}\\ {\mathrm{\&Delta;D}}_{\mathrm{uX}}\end{array}\right]=-{\left[{\&dtri;e}_{\mathrm{MX}}^T{\&dtri;e}_{\mathrm{MX}}\right]}^{-1}{\&dtri;e}_{\mathrm{MX}}^T{e}_{\mathrm{MX}}$

Wherein:

${\&dtri;e}_{\mathrm{MX}}=\left[\begin{array}{cccc}\frac{\&PartialD;e_{1X}}{\&PartialD;A_{\mathrm{uX}}}& \frac{\&PartialD;e_{1X}}{\&PartialD;B_{\mathrm{uX}}}& \frac{\&PartialD;e_{1X}}{\&PartialD;C_{\mathrm{uX}}}& \frac{\&PartialD;e_{1X}}{\&PartialD;D_{\mathrm{uX}}}\\ \frac{\&PartialD;e_{2X}}{\&PartialD;A_{\mathrm{uX}}}& \frac{\&PartialD;e_{2X}}{\&PartialD;B_{\mathrm{uX}}}& \frac{\&PartialD;e_{2X}}{\&PartialD;C_{\mathrm{uX}}}& \frac{\&PartialD;e_{2X}}{\&PartialD;D_{\mathrm{uX}}}\\ .& .& .& .\\ .& .& .& .\\ .& .& .& .\\ \frac{\&PartialD;e_{64X}}{\&PartialD;A_{\mathrm{uX}}}& \frac{\&PartialD;e_{64X}}{\&PartialD;B_{\mathrm{uX}}}& \frac{\&PartialD;e_{64X}}{\&PartialD;C_{\mathrm{uX}}}& \frac{\&PartialD;e_{64X}}{\&PartialD;D_{\mathrm{uX}}}\end{array}\right]$

By A _uX+ Δ A _uX, B _uX+ Δ B _uX, C _uX+ Δ C _uX, D _uX+ Δ D _uXrespectively as new X phase voltage voltage parameter A _uX, B _uX, C _uX, D _uX, 2. perform step (2) the walk;

Work as ε _x< ε ₀time, the X phase voltage voltage parameter A now participating in calculating is described _uX, B _uX, C _uX, D _uXit is exactly X phase voltage voltage parameter when there is electric network fault;

(3) electric network fault type, is judged

1., the voltage battery parameter value obtained in step (2) is gone out A phase voltage amplitude u with following formulae discovery _a, B phase voltage amplitude u _b, C phase voltage amplitude u _c:

$u_A=\sqrt{A_{\mathrm{uA}}^2+B_{\mathrm{uA}}^2}$

$u_B=\sqrt{A_{\mathrm{uB}}^2+B_{\mathrm{uB}}^2}$

$u_C=\sqrt{A_{\mathrm{uC}}^2+B_{\mathrm{uC}}^2}$

2., electric network fault type judges

By the u obtained in previous step _a, u _b, u _cwith set voltage magnitude threshold values 0.85u _pcompare, wherein u _pphase voltage amplitude:

Work as u _a<0.85u _p, u _b<0.85u _p, u _c>0.85u _ptime, electric network fault type belongs to generation AB two-phase short-circuit fault;

Work as u _a>0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation BC two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b>0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation CA two-phase short-circuit fault;

Work as u _a<0.85u _p, u _b<0.85u _p, u _c<0.85u _ptime, electric network fault type belongs to generation ABC three phase short circuit fault;

Wherein, AB two-phase short-circuit fault, BC two-phase short-circuit fault, CA two-phase short-circuit fault are commonly referred to as two-phase short-circuit fault;

(4) position that electric power system fault occurs, is judged

According to the fault type that step (3) is known, select the determination methods of following abort situation with selecting a property:

1., when there is two-phase short-circuit fault:

The user side comprehensive impedance Z of fault phase is gone out to occur with following formulae discovery _x1X2:

$Z_{X1X2}={\left[M_{\mathrm{IX}1}^T{M}_{\mathrm{IX}1}\right]}^{-1}{M}_{\mathrm{IX}1}^T(U_{X1}-U_{X2})$

Wherein:

$M_{\mathrm{IX}1}=\left[\begin{array}{cc}{i}_{X1}\left(t_{48}\right)& i_{X1}\left(t_{48}\right)-i_{X1}\left(t_{-16}\right)\\ i_{X1}\left(t_{49}\right)& i_{X1}\left(t_{49}\right)-i_{X1}\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_{X1}\left(t_{64}\right)& i_{X1}\left(t_{64}\right)-i_{X1}\left(t_0\right)\end{array}\right]$

$U_{X1}=\left[\begin{array}{c}{u}_{X1}\left(t_{49}\right)\\ u_{X1}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X1}\left(t_{64}\right)\end{array}\right]$

$U_{X2}=\left[\begin{array}{c}{u}_{X2}\left(t_{49}\right)\\ u_{X2}\left(t_{50}\right)\\ .\\ .\\ .\\ u_{X2}\left(t_{64}\right)\end{array}\right]$

In formula: X1 refers to A phase, B phase or C phase; X2 refers to: when X1 is A phase time, X2 can only be B phase; When X1 is B phase time, X2 can only be C phase; When X1 is C phase time, X2 can only be A phase;

The user side comprehensive impedance Z calculated _x1X2the matrix of 2 × 1, the 1st element Z in matrix _x1X21resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _x1X21during >0, illustrate that the abort situation of this electric system occurs in user side;

Work as Z _x1X21during <0, illustrate that the abort situation of this electric system occurs in grid side;

2., when there is ABC three phase short circuit fault:

The user side comprehensive impedance Z of A phase is gone out with following formulae discovery _aN

$Z_{\mathrm{AN}}={\left[M_{\mathrm{IA}}^T{M}_{\mathrm{IA}}\right]}^{-1}{M}_{\mathrm{IA}}^T{U}_A$

Wherein:

$M_{\mathrm{IA}}=\left[\begin{array}{cc}{i}_A\left(t_{48}\right)& i_A\left(t_{48}\right)-i_A\left(t_{-16}\right)\\ i_A\left(t_{49}\right)& i_A\left(t_{49}\right)-i_A\left(t_{-15}\right)\\ .& .\\ .& .\\ .& .\\ i_A\left(t_{64}\right)& i_A\left(t_{64}\right)-i_A\left(t_0\right)\end{array}\right]$

$U_A=\left[\begin{array}{c}{u}_A\left(t_{49}\right)\\ u_A\left(t_{50}\right)\\ .\\ .\\ .\\ u_A\left(t_{64}\right)\end{array}\right]$

The user side comprehensive impedance Z calculated _aNthe matrix of 2 × 1, the 1st element Z in matrix _aN1resistance value in the comprehensive impedance of characterizing consumer side;

Work as Z _aN1during >0, illustrate that the abort situation of this electric system occurs in user side; Work as Z _aN1during <0, illustrate that the abort situation of this electric system occurs in grid side.