CN101572405A - Method for ranging computer distance protection of transmission line - Google Patents

Abstract

The invention discloses a method for ranging computer distance protection of a transmission line. The method comprises the following steps: processing a sampling value of measuring current by a digital capacitance voltage transformator, if voltage is measured by a capacitance voltage transformator; supposing that a certain point on a protected circuit has a failure, calculating a voltage sampling value of the failure point; if the voltage is measured by the capacitance voltage transformator, processing the voltage sampling value of the failure point by the same digital capacitance voltage transformator; after processing the measuring voltage sampling value, the current sampling value and the failure voltage sampling value through low-pass filtering, introducing the measuring voltage sampling value, the current sampling value and the failure voltage sampling value into an algorithm for solving differential equation of the transmission line; improving velocity of convergence and range accuracy by adopting an iterative calculation method; and finally, determining that whether to trip protection or not according to a failure range result fulfilling an error range, if the failure range result reaching certain time cannot fulfill the error range, finish the ranging method of the round, wait and continue the iterative computation when a new failure occurs.

Description

The distance-finding method of a kind of transmission line microcomputer distance protection

Technical field

The invention belongs to the relay protection of power system technology, be specifically related to the distance-finding method of a kind of transmission line microcomputer distance protection.

Background technology

Extensively adopt the backup protection of distance protection as circuit on superhigh pressure and ultra high voltage long distance transmission line, capacitance type potential transformer transient process, initial high order harmonic component and the aperiodic component of fault are to cause main cause super, that extra high voltage line distance protection transient state surmounts.

The impedance relay of distance protection normally obtains phasor by Fourier transform; this algorithm has 2 deficiencies; the DC component that the first is decayed influences very big, secondly is that regular time window width (being generally a power frequency period) causes the transient response time of this algorithm longer.

Separate Differential Equation Algorithm and be the algorithm that extensively adopts in the microcomputer distance protection, the influence that it is not subjected to DC component and the low frequency component that caused by the decay aperiodic component causes, and be not subjected to the influence of mains frequency fluctuation.The shortcoming of this algorithm is, because the differential equation has been ignored the distributed capacitance of transmission line, makes its high frequency characteristics relatively poor, and when containing high fdrequency component in electric current and the voltage, computational accuracy will be affected.

On superhigh pressure and the ultra high voltage long distance transmission line, the voltage of microcomputer distance protection obtains usually from capacitance type potential transformer.But compare with electromagnetic potential transformer, its transient characterisitics are relatively poor.When system breaks down; serious transient process can take place in secondary side voltage; the measure error that causes can cause surmounting of conventional distance protection; obtained researcher's concern; to the transient process of CVT with surmount solution and done a large amount of research, but not solving transient state as yet fully surmounts problem.

Summary of the invention

The object of the present invention is to provide the distance-finding method of a kind of transmission line microcomputer distance protection, the transient state that this method can solve distance protection surmounts problem, realizes the range finding of distance protection ultrahigh speed.

The invention discloses the distance-finding method of a kind of transmission line microcomputer distance protection, its step comprises:

The 1st step was gathered measuring voltage sampled value, the measurement current sampling data of the transmission line that obtains each sampling instant in real time, if transmission line adopts capacitance type potential transformer, the measurement current sampling data of each sampling instant is handled through the digit capacitance voltage transformer; To the measuring voltage of each sampling instant, measure current sampling data by identical low-pass filtering treatment, obtain new measuring voltage, the measurement current sampling data of each sampling instant;

The 2nd step definition t ₀For fault takes place constantly, T ₃For the required minimum time width of protection tripping operation after the fault, get 5～20ms; Definition t _JsBe the current iterative computation moment, do not upgrade t when iterative computation finishes each time before the iterative computation end each time of this moment _JsBe updated to the sampling instant of up-to-date sampled data;

If t _Js-t ₀〉=T ₃, entered for the 3rd step, otherwise wait for;

The 3rd step was provided with fault distance iterative value D ', and iterative initial value is more than or equal to 0 and smaller or equal to the arbitrary value of transmission line length; The initial value of iterations Q is 0;

The 4th step was calculated t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period, 40ms≤T ₁≤ 500ms; t ₀-T ₁To t ₀The fault point voltage sampled value of each sampling instant is by calculating t with it with measuring voltage constantly, measurement current sampling data and fault distance iterative value D ' in time period ₀To t _JsThe fault point voltage sampled value value of each sampling instant in the time period is 0;

If the 5th step transmission line adopts capacitance type potential transformer, with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period obtains t by digit capacitance voltage transformer and the low-pass filtering treatment identical with step (1) ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in time period; Otherwise with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period obtains t by going on foot identical low-pass filtering treatment with the 1st ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in the time period;

The 6th step is with t _Js-T ₃To t _JsThe new measuring voltage of each sampling instant in the time period, measurement current sampling data and new fault point voltage sampled value substitution transmission line distance protecting are separated Differential Equation Algorithm, at t _Js-T ₃To t _JsChoosing length in time period is T ₂Time period, 5ms≤T ₂≤ T ₃Find the solution and obtain t _Js-T ₃To t _JsFault distance D in time period ₁And should interior range error E and T of time period ₂Fault distance D in time period ₂

If the 7th step t _Js-t ₀＜T _End, changed for the 8th step over to; T _EndBe the maximum time width that whole iterative computation finish, value is 60～100ms;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE＞E in time period _d, jumped to for the 10th step; E _dBe the range error setting value;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE≤E in time period _d, with D ₁Value compose to fault distance end value D _z, jumped to for the 9th step;

If the 8th step Q＞S is with D ₂Value is composed and is given D ' stand-by period T ₄, this iterative computation finishes, with t _JsBe updated to the sampling instant of up-to-date sampled data, Q is changed to zero, jumps to for the 4th step to carry out next iteration and calculate;

S is the iterations setting value, T ₄Stand-by period width for iterations during more than or equal to setting value;

If Q≤S, t _Js-T ₃To t _JsE＞E in time period _d, with D ₂Value is composed and is given D ', makes Q=Q+1, proceeds this iterative computation, jumps to for the 4th step;

If Q≤S, t _Js-T ₃To t _JsE≤E in time period _d, should D in the time period ₁Value compose to D ₂, changed for the 9th step over to;

The 9th step is according to fault distance end value D _zJudge whether it is troubles inside the sample space.

If troubles inside the sample space then sends trip signal, changed for the 10th step then over to, otherwise directly changed for the 10th step over to;

The 10th step waited for that fault took place next time, when transmission line breaks down once more, changed for the 2nd step over to.

The preferred version of separating Differential Equation Algorithm in above-mentioned the 6th step is as follows:

T represents t _Js-T ₃To t _JsArbitrary sampling instant of time period, set up suc as formula the differential equation shown in (I):

U (t)-u _f(t)=D * Δ u (t)+i _f(t) * r _gFormula (I)

Wherein, D is a t fault distance constantly, r _gBe transition resistance;

For phase fault, u (t) is the difference between two fault phase t new measuring voltage sampled value constantly, u _f(t) be difference between the constantly new fault point voltage sampled value of two fault phase t, Δ u (t) is the difference between the voltage drop of transmission line unit length of two fault phases, i _f(t) be t constantly fault mutually in arbitrary mutually new measurement current sampling data; For single-line to ground fault, u (t), u _f(t) be respectively fault constantly new measuring voltage sampled value of t, new fault point voltage sampled value and the voltage drop of transmission line unit length mutually, i with Δ u (t) _f(t) get 1/3rd of the t measurement current sampling data sum that three-phase is new constantly;

For t _Js-T ₃To t _JsAll sampling instants of time period are all set up suc as formula the differential equation shown in (I), obtain differential equation group;

Write above-mentioned differential equation group as matrix form AX=C;

Wherein $A=\left[\begin{array}{cc}\mathrm{\Δu}\left(t_1\right)& i_f\left(t_1\right)\\ .& .\\ .& .\\ \mathrm{\Δu}\left(t_n\right)& i_f\left(t_n\right)\end{array}\right]$ $X=\left[\begin{array}{c}{D}_1\\ r_g\end{array}\right]$ C＝[u(t ₁)-u _f(t ₁)…u(t _n)-u _f(t _n)] ^T

Wherein, subscript T representing matrix transposition;

Fault distance D is obtained in calculating ₁

Calculate fault distance D according to above-mentioned identical method ₂

It is very serious that the transient state of present superhigh pressure and extra high voltage line distance protection surmounts phenomenon; the present invention is directed to the deficiencies in the prior art; a kind of distance-finding method of super, extra high voltage line microcomputer distance protection is proposed; this method can solve distance protection transient state and surmount problem, realizes that metallicity fault distance error in the 10ms is less than 5% performance index.

Description of drawings

Fig. 1 is the electric power system line chart;

Fig. 2 is the capacitance type potential transformer principle assumption diagram;

Fig. 3 is the capacitance type potential transformer equivalent circuit diagram;

Fig. 4 is inductive element circuit figure, wherein, (a) is side circuit, (b) is the equivalent counting circuit of transient state;

Fig. 5 is the capacity cell circuit diagram, wherein, (a) is side circuit, (b) is the equivalent counting circuit of transient state;

The capacitance type potential transformer equivalent circuit diagram of Fig. 6 for representing with current source;

Fig. 7 is the capacitance type potential transformer equivalent circuit diagram behind the abbreviation;

Fig. 8 is the progress of disease link schematic diagram of the needed electric parameters of this paper distance-finding method.

Embodiment

The present invention is further detailed explanation below in conjunction with accompanying drawing and example.

As shown in Figure 1, the inventive method comprises the steps:

(1) following three steps (a) (b) (c) are all carried out in each sampling instant in proper order.

(a) collection obtains measuring voltage sampled value, the measurement current sampling data of each sampling instant of transmission line;

Both-end power source model as shown in Figure 1, transmission line MS is a protected circuit.500kV and above electric pressure transmission line generally all adopt capacitance type potential transformer (CVT), as shown in Figure 2.Fig. 3 is the capacitance type potential transformer equivalent circuit diagram.The primary voltage of transmission line enters protective device through after the actual CVT progress of disease, enters protective device after the progress of disease of primary current process current transformer.The electric parameters signal that enters protective device forms the measuring voltage sampled value of each sampling instant and measures current sampling data through the inner little converter of device and two progress of disease links of AD sampling again.The protective device sampling interval is Δ t.

T M end three-phase voltage sampled value constantly is expressed as u _MA(t), u _MB(t), u _MC(t)

T M end three-phase current sampled value constantly is expressed as i _MA(t), i _MB(t), i _MC(t)

(b) judge whether transmission line adopts capacitance type potential transformer, if, earlier the measurement current sampling data of each sampling instant is handled through digital CVT, enter step (c) again, otherwise directly enter step (c);

The measurement current sampling data of each sampling instant is handled through digital CVT, the differential equation and the integral equation discretization that are exactly the electric capacity that will characterize capacitance type potential transformer, inductance and resistive element voltage and current relation are handled, the differential equation represents that with difference formula integral equation is represented with trapezoidal sum formula.To measure current sampling data is input variable, and the later digital CVT linear circuit of discretization is found the solution in sampling instant one by one, obtains the measurement current sampling data through each sampling instant of digital CVT processing.

For make the measuring voltage sampled value voltage that protective device obtains and measure current sampling data the progress of disease link of process consistent, after the measurement current sampling data that protective device is obtained passes through the digital CVT progress of disease, the calculating of protecting again.The digital CVT circuit of measurement current sampling data process and the actual CVT equivalent circuit of voltage process are identical, as shown in Figure 3.CVT partly is made up of dividing potential drop electric capacity, compensation reactor, middle reactor, damper etc.

U ₁＝[C ₁/(C ₁+C ₂)]U Ce＝C ₁+C ₂ (1)

U is actual CVT input voltage, i.e. phase voltage one sub-value, U ₁Be CVT equivalent circuit input voltage; C ₁, C ₂Be actual CVT dividing potential drop electric capacity; Ce is equivalent dividing potential drop electric capacity; L ₁Be the leakage inductance sum of compensating inductance and intermediate transformer, R ₁Be corresponding resistance; R _f, C _f, L _fAnd r _fParameter for the mode of resonance damper; L _bAnd R _bThen be load inductance and resistance; Lm and Rm then are excitatory branch road inductance and resistance (CVT intermediate transformer iron core is unsaturated usually).

The concrete steps that to be carried out digital CVT processing by the measurement current sampling data that protective device AD sampling obtains are as follows:

At first four inductance elements and two electric capacity in the circuit are done following processing with methods of numerical:

The inductance differential equation is: Ldi _Jk/ dt=u _j-u _k(2)

Use the trapezoidal integration formula, turn to difference equation.

$i_{\mathrm{jk}}\left(t\right)=\frac{1}{R_L}[u_j\left(t\right)-u_k\left(t\right)]+I_L(t-\mathrm{\Δt})---\left(3\right)$

R wherein _L=2L/ Δ t

$I_L(t-\mathrm{\Δt})=i_{\mathrm{jk}}(t-\mathrm{\Δt})+\frac{1}{R_L}[u_j(t-\mathrm{\Δt})-u_k(t-\mathrm{\Δt})]---\left(4\right)$

The voltage and current relation of t moment inductive branch can replace with equivalent circuit shown in Figure 4 like this.

Make t capacitive branch equivalent circuit constantly with similar method, as shown in Figure 5.

The electric capacity differential equation is: Cd (u _j-u _k)/dt=i _Jk(5)

Use the trapezoidal integration formula, turn to difference equation.

$i_{\mathrm{jk}}\left(t\right)=\frac{1}{R_C}[u_j\left(t\right)-u_k\left(t\right)]+I_C(t-\mathrm{\Δt})---\left(6\right)$

R wherein _C=Δ t/2C

$I_C(t-\mathrm{\Δt})=-i_{\mathrm{jk}}(t-\mathrm{\Δt})-\frac{1}{R_C}[u_j(t-\mathrm{\Δt})-u_k(t-\mathrm{\Δt})]---\left(7\right)$

Just become the simple straight circuit that includes only three kinds of electric components at each sampling instant CVT transient state equivalent circuit like this: input voltage source, the current source of inductance capacitance equivalent circuit, resistance.Input voltage source U ₁=[C ₁/ (C ₁+ C ₂)] U, the numerical value of each sampling instant of U equals the measurement current sampling data of each phase of input protection device.Inductance, electric capacity equivalent circuit current source, its initial value can be made as zero, the numerical value of current source is pressed (4) formula and is calculated by voltage, the current value of previous moment inductance equivalent circuit in each inductance equivalent circuit constantly later on, and the numerical value of current source is calculated by (7) formula by voltage, the current value of previous moment electric capacity equivalent circuit in each electric capacity equivalent circuit constantly.

To measure the current sampling data calculation procedure constantly as follows for t after the digital CVT progress of disease, and protected circuit M end three-phase is measured current sampling data and all carried out following three steps in each sampling instant, and measuring current sampling data mutually with M end A is example:

(I) sampled value of measuring electric current mutually according to t A is constantly calculated t input voltage source U constantly ₁Numerical value;

U ₁＝[C ₁/(C ₁+C ₂)]×i _MA(t)

(II) according to the known t numerical value of current source in input voltage source and electric capacity, the inductance equivalent circuit constantly, find the solution the t equivalent DC circuit of CVT transient state constantly, obtain t each node voltage and branch current, wherein CVT load branch pressure drop u constantly _Fz(t) be that A measures current sampling data mutually through the measurement current sampling data constantly of t after the digital CVT progress of disease.

The CVT Equivalent Model of representing with current source according to (3), (4), (6), the described method of (7) formula is shown in figure (6).Concrete computing formula is as follows:

Current source equivalent transformation in each inductance, the electric capacity equivalent circuit is become voltage source, and formula is as follows:

u _ce(t)＝i _ce(t-Δt)×r _ce，u _lf(t)＝i _lf(t-Δt)×r _lf，u _l1(t)＝i _l1(t-Δt)×r _l1，u _cf(t)＝i _cf(t-Δt)×r _cf；

R wherein _Cf=Δ t/2C _f, r _Lf=2L _f/ Δ t, r _L1=2L ₁/ Δ t, r _Ce=Δ t/2C _e, u _Ce(t), u _Cf(t), u _L1(t), u _Lf(t) be respectively t capacitor C constantly _e, C _fAnd inductance L ₁, L _fVoltage source after the current source conversion in the equivalent circuit, i _Ce(t-Δ t), i _Cf(t-Δ t), i _L1(t-Δ t), i _Lf(t-Δ t) is respectively t capacitor C constantly _e, C _fAnd inductance L ₁, L _fEquivalent circuit in current source.

The CVT equivalent circuit that to scheme (6) carries out abbreviation, and the CVT equivalent circuit diagram behind the abbreviation is shown in figure (7), and concrete simplifying method is as follows:

u ₁₁(t)＝U ₁+u _ce(t)+u _l1(t)，i ₁₁(t)＝u ₁₁(t)/r ₁₁，r ₁₁＝r _ce+r _l1+R ₁；

i ₂₂(t)=u _Lf(t)/r ₂₂, r wherein ₂₂=r _Lf+ r _f

i ₃₃(t)=i ₂₂(t)+i _Cf(t-Δ t), u ₃₃(t)=r ₃₃* i ₃₃(t), r wherein ₃₃=1/ (1/r ₂₂+ 1/r _Cf);

i ₄₄(t)=u ₃₃(t)/r ₄₄, r wherein ₄₄=r ₃₃+ R _f

u ₅₅=i _Lb(t-Δ t) * r _Lb, i ₅₅(t)=u ₅₅(t)/r ₅₅, r wherein ₅₅=r _Lb+ R _b

i _w(t)＝i ₁₁(t)-i _lm(t)-i ₄₄(t)-i ₅₅(t)，r _w＝1/(1/r ₁₁+1/R _m+1/r _lm+1/r ₄₄+1/r ₅₅)，

Can measure current sampling data u mutually in the hope of the A after handling through digital CVT by above-mentioned formula _Fz(t).

u _fz(t)＝r _w×i _w(t)

(III) by obtaining the t voltage and the branch current of each inductance, electric capacity equivalent circuit constantly in (II), calculate the t+ Δ t value of each inductance, electric capacity equivalent circuit current source constantly respectively according to (4) formula and (7) formula.Concrete computing formula is as follows:

T flows through capacitor C constantly _eCurrent i ₆₆(t)=(U ₁+ u _Ce(t)+u _L1(t)-u _Fz(t))/r ₁₁, flow through inductance L _bElectric current: i ₇₇(t)=(u _Fz(t)+u ₅₅(t))/r ₅₅, flow through capacitor C _fElectric current: i ₈₈(t)=(u _Fz(t)+u ₃₃(t))/r ₃₃

T is capacitor C constantly _eBoth end voltage:

duce(t)＝(U ₁+u _l1(t)-i ₆₆(t)×(r _l1+R ₁)-u _fz(t))；

T is inductance L constantly ₁Both end voltage: dul1 (t)=(U ₁+ u _Ce(t)-i ₆₆(t) * (r _Ce+ R ₁)-u _Fz(t));

T is inductance L constantly _mThe voltage at two ends: dulm (t)=u _Fz(t);

T is inductance L constantly _bThe voltage at two ends: dulb (t)=u _Fz(t)-r _b* i ₇₇(t);

T is capacitor C constantly _fThe voltage at two ends: ducf (t)=u _Fz(t)-i ₈₈(t) * R _f

T flows through inductance L constantly _fElectric current be: i ₉₉(t)=(ducf (t)+u _Lf(t))/r ₂₂, inductance L _fThe voltage at two ends is: dulf (t)=ducf (t)-i ₉₉(t) * r _f

Can get t+ Δ t inductance L constantly by (3) and (4) formula ₁, L _b, L _mAnd L _fThe value of equivalent circuit current source be respectively:

i _l1(t)＝-i _l1(t-Δt)-2×dull(t)/r _l1；

i _lb(t)＝-i _lb(t-Δt)-2×dulb(t)/r _lb；

i _lm(t)＝-i _lm(t-Δt)-2×dulm(t)/r _lm；

i _lf(t)＝-i _lf(t-Δt)-2×dulf(t)/r _lf；

i _Lb(t-Δ t), i _Lm(t-Δ t) is respectively t inductance L constantly _b, L _mThe current source of equivalent circuit.

Can get t+ Δ t capacitor C constantly by (6) and (7) formula _e, C _fThe value of equivalent circuit current source, as follows respectively:

i _ce(t)＝-i _ce(t-Δt)-2×duce(t)/r _ce；

i _cf(t)＝-i _cf(t-Δt)-2×ducf(t)/r _cf；

(c) with the measuring voltage sampled value of each sampling instant with handle through digital CVT after the measurement current sampling data of each sampling instant by identical low-pass filtering treatment, obtain new measuring voltage, the measurement current sampling data of each sampling instant of protected circuit after the filtering.Step (c) can be carried out before step (b) order.

Low-pass filtering treatment is exactly with measuring voltage, measures current sampling data with conventional wave digital lowpass filter Filtering Processing, is example with the Butterworth filter, sample rate is every power frequency period 96 points, cut-off frequency is 100Hz, and initial value is made as zero, is example with M end t moment A phase current sampling value:

at＝0.00392；bt＝-1.81534；ct＝0.83101；

i _MA(t)＝at×i _{MA_X}(t)+2×at×i _{MA_X}(t-Δt)+att×i _{MA_X}(t-2Δt)

-bt×i _MA(t-Δt)-ct×i _MA(t-2Δt)；

I wherein _MA(t) be the output valve of Butterworth filter at t moment M end A phase current;

i _{MA_X}(t) be the input value of Butterworth filter at t moment M end A phase current;

(2) definition t ₀For fault takes place constantly, according to the relaying protection algorithm acquisition of routine, T ₃For the required minimum time width of protection tripping operation after the fault, get 5～20ms; Definition t _JsBe the current iterative computation moment, do not upgrade before the iterative computation end each time of this moment, be updated to the sampling instant of up-to-date sampled data when iterative computation finishes each time;

If t _Js-t ₀〉=T ₃, enter step (3), otherwise wait for;

(3) fault distance iterative value D ' is set, iterative initial value is more than or equal to 0 and smaller or equal to the arbitrary value of transmission line MS length; The initial value of iterations Q is 0;

(4) calculate t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period, 40ms≤T ₁≤ 500ms;

t ₀-T ₁To t _JsThe fault point voltage sampled value of interior each sampling instant before the time period internal fault takes place is by calculating t with it with measuring voltage constantly, measurement current sampling data and fault distance iterative value D ' ₀-T ₁To t _JsThe fault point voltage sampled value value of each sampling instant after the time period internal fault takes place is 0;

t ₀-T ₁To t _JsThe concrete computational methods of fault point voltage sampled value of each sampling instant before the time period internal fault takes place are as follows, to calculate t A phase fault point voltage sampled value u constantly _FA(t) be example:

u _fA(t)＝u _MA(t)-D′×Δu _MA(t)

Voltage drop Δ u on the transmission line MS section t moment A phase unit length _MA(t) computing formula is as follows:

Δu _MA(t)＝r _s×i _MA(t)+r _m×i _MB(t)+r _m×i _MC(t)

+l _s×di _MA(t)/dt+l _m×di _MB(t)/dt+l _m×di _MC(t)/dt

Rs and rm are respectively the self-resistance and the mutual resistance of transmission line unit length, and ls and lm are respectively the self-inductance and the mutual inductance of transmission line unit length, di _MA(t)/dt, di _MB(t)/dt, di _MC(t)/dt represents i respectively _MA(t), i _MB(t), i _MC(t) carry out differential calculation.

(5) if transmission line adopts capacitance type potential transformer, with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period obtains t by digit capacitance voltage transformer and the low-pass filtering treatment identical with step (1) ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in the time period; Otherwise with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period by with the identical low-pass filtering treatment of step (1), obtain t ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in the time period.

Step (1) to the progress of disease link of the electric parameters of step (5) shown in figure (8).

(6) with t _Js-T ₃To t _JsThe new measuring voltage of each sampling instant in the time period, measurement current sampling data and new fault point voltage sampled value substitution transmission line distance protecting are separated Differential Equation Algorithm, at t _Js-T ₃To t _JsChoosing length in time period is T ₂Time period, 5ms≤T ₂≤ T ₃Find the solution and obtain t _Js-T ₃To t _JsFault distance D in time period ₁And should interior range error E and T of time period ₂Fault distance D in time period ₂

That routine is separated the Differential Equation Algorithm employing is t _Js-T ₃To t _JsThe measuring voltage sampled value of each sampling instant in the time period is separated Differential Equation Algorithm and is replaced with t in this step _Js-T ₃To t _JsThe difference of measuring voltage sampled value that each sampling instant in the time period is new and new fault point voltage sampled value;

The present invention can be preferably as follows and separate Differential Equation Algorithm, is example to calculate t moment fault distance D:

u(t)-u _f(t)＝D×Δu(t)+i _f(t)×r _g

The concrete computational methods of Δ u (t) are seen step (4), but different with the measurement current sampling data of step (4) calculating employing, and it is new measurement current sampling data that this step is calculated the magnitude of current that adopts.

Find the solution this differential equation, at t _Js-T ₃To t _JsGet n different new measuring voltage, measurement current sampling data and new fault point voltage sampled value constantly in time period, list n independent equation, can obtain fault distance, i.e. t=t ₁, t ₂..., t _n, its equation group is:

u(t ₁)-u _f(t ₁)＝Δu(t ₁)×D+i _f(t ₁)×r _g(8)

u(t ₂)-u _f(t ₂)＝Δu(t ₂)×D+i _f(t ₂)×r _g(9)

......

u(t _n)-u _f(t _n)＝Δu(t _n)×D+i _f(t _n)×r _g(10)

Write as matrix form AX=C

Wherein $A=\left[\begin{array}{cc}\mathrm{\Δu}\left(t_1\right)& i_f\left(t_1\right)\\ .& .\\ .& .\\ \mathrm{\Δu}\left(t_n\right)& i_f\left(t_n\right)\end{array}\right],$ $X=\left[\begin{array}{c}{D}_1\\ r_g\end{array}\right]$ C＝[u(t ₁)-u _f(t ₁)…u(t _n)-u _f(t ⁿ)] ^T

With the least square method is example, and its least square solution is X=(A ^TA) ^-1A ^TC (11)

Error $E=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{\left[\right(u\left(t_i\right)-u_f\left(t_i\right))-(\mathrm{\Δu}\left(t_i\right)\×D_1+i_f\left(t_i\right)\×r_g\left)\right]}^2/\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{(u\left(t_i\right)-u_f\left(t_i\right))}^2---\left(12\right)$

Can obtain fault distance D by (11) formula ₁With transition resistance r _g, by the size of (12) formula error can failure judgement whether near actual value.Can try to achieve fault distance D with quadrat method ₂

(7) if t _Js-t ₀＜T _End, change step (8) over to; T _EndBe the maximum time width that whole iterative computation finish, value is 60～100ms usually;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE＞E in time period _d, jump to step (10); E _dFor the range error setting value, adjust by the user;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE≤E in time period _d, with D ₁Value compose to fault distance end value D _z, jump to step (9);

(8) if Q＞S, with D ₂Value compose to D ', T waits for a period of time ₄, with t _JsBe updated to the sampling instant of up-to-date sampled data, this iterative computation finishes, and Q is changed to zero, jumps to step (4) and carries out next iteration calculating;

S is the iterations setting value, and value is 2～10 usually; T ₄Stand-by period width for iterations during more than or equal to setting value, value is 1～5ms usually;

If Q≤S, t _Js-T ₃To t _JsE＞E in time period _d, with D ₂Value compose to D ', make Q=Q+1, proceed this iterative computation, jump to step (4);

If Q≤S, t _Js-T ₃To t _JsE≤E in time period _d, with D ₁Value compose to D _z, change step (9) over to;

(9) according to fault distance end value D _zJudge whether it is troubles inside the sample space.

If fault distance end value D _zLess than setting value (setting value get protected circuit MS length 90%～95%), then be judged as troubles inside the sample space, send trip signal, if be single phase ground fault, send the fault phase trip signal; Otherwise send the whole trip signals of three-phase, change step (10) then over to; If not troubles inside the sample space, then directly change step (10) over to;

(10) the epicycle range finding finishes.Wait for that fault takes place next time,, continue to wait for, otherwise change step (2) over to if transmission line does not break down.

Above-mentioned treatment step is the example calculating of finding range with transmission line one end (M end) all, equally also can adopt the data of the other end (S end) to find range.

The above is preferred embodiment of the present invention, but the present invention should not be confined to the disclosed content of this embodiment and accompanying drawing.So everyly do not break away from the equivalence of finishing under the spirit disclosed in this invention or revise, all fall into the scope of protection of the invention.

Claims (2)

1, the distance-finding method of a kind of transmission line microcomputer distance protection, its step comprises:

The 1st step was gathered measuring voltage sampled value, the measurement current sampling data of the transmission line that obtains each sampling instant in real time, if transmission line adopts capacitance type potential transformer, the measurement current sampling data of each sampling instant is handled through the digit capacitance voltage transformer; To the measuring voltage of each sampling instant, measure current sampling data by identical low-pass filtering treatment, obtain new measuring voltage, the measurement current sampling data of each sampling instant;

The 2nd step definition t ₀For fault takes place constantly, T ₃For the required minimum time width of protection tripping operation after the fault, get 5～20ms; Definition t _JsBe the current iterative computation moment, do not upgrade t when iterative computation finishes each time before the iterative computation end each time of this moment _JsBe updated to the sampling instant of up-to-date sampled data;

If t _Js-t ₀〉=T ₃, entered for the 3rd step, otherwise wait for;

The 3rd step was provided with fault distance iterative value D ', and iterative initial value is more than or equal to 0 and smaller or equal to the arbitrary value of transmission line length; The initial value of iterations Q is 0;

The 4th step was calculated t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period, 40ms≤T ₁≤ 500ms; t ₀-T ₁To t ₀The fault point voltage sampled value of each sampling instant is by calculating t with it with measuring voltage constantly, measurement current sampling data and fault distance iterative value D ' in time period ₀To t _JsThe fault point voltage sampled value value of each sampling instant in the time period is 0;

If the 5th step transmission line adopts capacitance type potential transformer, with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period obtains t by digit capacitance voltage transformer and the low-pass filtering treatment identical with step (1) ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in time period; Otherwise with t ₀-T ₁To t _JsThe fault point voltage sampled value of each sampling instant in the time period obtains t by going on foot identical low-pass filtering treatment with the 1st ₀-T ₁To t _JsThe new fault point voltage sampled value of each sampling instant in the time period;

The 6th step is with t _Js-T ₃To t _JsThe new measuring voltage of each sampling instant in the time period, measurement current sampling data and new fault point voltage sampled value substitution transmission line distance protecting are separated Differential Equation Algorithm, at t _Js-T ₃To t _JsChoosing length in time period is T ₂Time period, 5ms≤T ₂≤ T ₃Find the solution and obtain t _Js-T ₃To t _JsFault distance D in time period ₁And should interior range error E and T of time period ₂Fault distance D in time period ₂

If the 7th step t _Js-t ₀＜T _End, changed for the 8th step over to; T _EndBe the maximum time width that whole iterative computation finish, value is 60～100ms;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE＞E in time period _d, jumped to for the 10th step; E _dBe the range error setting value;

If t _Js-t ₀〉=T _End, t _Js-T ₃To t _JsE≤E in time period _d, with D ₁Value compose to fault distance end value D _z, jumped to for the 9th step;

If the 8th step Q＞S is with D ₂Value is composed and is given D ', stand-by period T ₄, this iterative computation finishes, with t _JsBe updated to the sampling instant of up-to-date sampled data, Q is changed to zero, jumps to for the 4th step to carry out next iteration and calculate;

S is the iterations setting value, T ₄Stand-by period width for iterations during more than or equal to setting value;

If Q≤S, t _Js-T ₃To t _JsE＞E in time period _d, with D ₂Value is composed and is given D ', makes Q=Q+1, proceeds this iterative computation, jumps to for the 4th step;

If Q≤S, t _Js-T ₃To t _JsE≤E in time period _d, should D in the time period ₁Value compose to D ₂, changed for the 9th step over to;

The 9th step is according to fault distance end value D _zJudge whether it is troubles inside the sample space.

If troubles inside the sample space then sends trip signal, changed for the 10th step then over to, otherwise directly changed for the 10th step over to;

The 10th step waited for that fault took place next time, when transmission line breaks down once more, changed for the 2nd step over to.

2, method according to claim 1 is characterized in that: it is as follows to separate Differential Equation Algorithm in the 6th step:

T represents t _Js-T ₃To t _JsArbitrary sampling instant of time period, set up suc as formula the differential equation shown in (I):

U (t)-u _f(t)=D * Δ u (t)+i _f(t) * r _gFormula (I)

Wherein, D is a t fault distance constantly, r _gBe transition resistance;

For phase fault, u (t) is the difference between two fault phase t new measuring voltage sampled value constantly, u _f(t) be difference between the constantly new fault point voltage sampled value of two fault phase t, Δ u (t) is the difference between the voltage drop of transmission line unit length of two fault phases, i _f(t) be t constantly fault mutually in arbitrary mutually new measurement current sampling data; For single-line to ground fault, u (t), u _f(t) be respectively fault constantly new measuring voltage sampled value of t, new fault point voltage sampled value and the voltage drop of transmission line unit length mutually, i with Δ u (t) _f(t) get 1/3rd of the t measurement current sampling data sum that three-phase is new constantly;

For t _Js-T ₃To t _JsAll sampling instants of time period are all set up suc as formula the differential equation shown in (I), obtain differential equation group;

Write above-mentioned differential equation group as matrix form AX=C;

Wherein $A=\left[\begin{array}{cc}\mathrm{\Δu}\left(t_1\right)& i_f\left(t_1\right)\\ \·& \·\\ \·& \·\\ \mathrm{\Δu}\left(t_n\right)& i_f\left(t_n\right)\end{array}\right]$ $X=\left[\begin{array}{c}{D}_1\\ r_g\end{array}\right]$ C＝[u(t ₁)-u _f(t ₁)...u(t _n)-u _f(t _n)] ^T

Wherein, subscript T representing matrix transposition;

Fault distance D is obtained in calculating ₁

Calculate fault distance D according to above-mentioned identical method ₂

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