CN101615783A - Zero-sequence current longitudinal differential protection method based on star-connection delta line transformer - Google Patents

Abstract

The invention discloses a kind of zero-sequence current longitudinal differential protection method based on star-connection delta line transformer; the longitudinal difference protection scheme based on zero-sequence current that still can sensitive action when the present invention proposes in the transformer district weak fault utilizes the zero-sequence current in each side winding to constitute differential protection.External area error, normally operation and during magnetizing inrush current, the zero sequence operating current is approximately zero; When interior earth fault in district and turn-to-turn fault, operating current is much larger than zero.With this criterion as differentiation inside and outside fault.The present invention has higher sensitivity in reflection single-phase earthing and little turn-to-turn short circuit, and it is littler influenced by magnetizing inrush current.

Description

Zero-sequence current longitudinal differential protection method based on star-connection delta line transformer

Technical field

The present invention relates to the differential protection research field of transformer, more specifically relate to Zero sequence current differential protection based on star-connection delta line transformer.

Background technology

Transformer in system in occupation of consequence; because not only can the right area subregion inside and outside fault of Current Differential Protection; and do not need to cooperate with the protection of other element; can excise various faults in the district without delay; thereby be widely used as the main protection of transformer, but traditional longitudinal difference protection exists deficiency in the following aspects:

When (1) slight shorted-turn fault takes place transformer, be reflected to the fault current of each side of transformer even also littler than load current, owing to comprise the transformer load electric current in the stalling current, conventional differential protection does not have enough sensitivity to determine this type of fault;

(2) for star-triangle Yn/ Δ transformer, do not comprise zero-sequence current in traditional longitudinal difference protection, longitudinal difference protection sensitivity reduced when this caused internal ground fault.Meanwhile, during the single-line to ground fault of Yn winding generation winding inside, traditional longitudinal difference protection both sides electric current may present the phase property of external short circuit, and single-phase short circuit is more near neutral point, and this feature is just obvious more, and the sensitivity of protection greatly reduces;

(3) being extensive use of along with the large value capacitor of the increase of the increase of power system development, superhigh pressure long-distance transmission line direct-to-ground capacitance, transformer capacity and static reactive; when adopting traditional secondary harmonic brake principle identification magnetizing inrush current, protection responsiveness problem slow and poor reliability becomes increasingly conspicuous.

Summary of the invention

The objective of the invention is to overcome the weak point of existing transformer differential protection, a kind of zero-sequence current longitudinal differential protection method based on star-connection delta line transformer be provided, can sensitively differentiate in the transformer district a little less than fault and influenced by magnetizing inrush current very little.

Technical scheme of the present invention is achieved in that

1) measured value by star side side current transformer TA, voltage transformer TV calculates star side residual voltage U ₀And zero-sequence current

And according to external area error equivalent electric circuit calculating triangle side winding zero-sequence current

${\stackrel{\&CenterDot;}{U}}_{\left(0\right)}=\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_{\mathrm{mA}}+{\stackrel{\&CenterDot;}{U}}_{\mathrm{mB}}+{\stackrel{\&CenterDot;}{U}}_{\mathrm{mC}}),$ ${\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}=\frac{1}{3}({\stackrel{\&CenterDot;}{I}}_{\mathrm{mA}}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{mB}}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{mC}}),$ ${\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}=-\frac{\stackrel{\&CenterDot;}{U_{\left(0\right)}}-\stackrel{\&CenterDot;}{I_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}}*X_{1\mathrm{\&sigma;}}}{X_{2\mathrm{\&sigma;}}}$

Wherein

And The three-phase current that measures for the star side winding TA of place, wherein

And Be the three-phase voltage that the star side winding TV of place measures, X _{1 σ}, X _{2 σ}Be respectively the leakage reactance of transformer both sides winding;

2) utilize star side, triangle side zero-sequence current

Calculate operating current I _OpWith stalling current I _Res,

$I_{\mathrm{op}}=|{\stackrel{\&CenterDot;}{I}}_{m0}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|,$ $I_{\mathrm{res}}=|{\stackrel{\&CenterDot;}{I}}_{m0}-{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|;$

3) determine starting current $I_{\mathrm{op}0}=0.1{\stackrel{\&CenterDot;}{I}}_n,$ Restraint coefficient K _Res=0.75, keen current $I_{\mathrm{res}0}=\frac{I_{\mathrm{op}0}}{K_{\mathrm{res}}}=0.133I_n,$ Wherein

Be the transformer rated current, obtain two broken line characteristic curves of Zero sequence current differential protection then:

$\left\{\begin{array}{c}{I}_{\mathrm{op}}>I_{\mathrm{op}0}\\ I_{\mathrm{op}}\&GreaterEqual;I_{\mathrm{op}0}+K_{\mathrm{res}}(I_{\mathrm{res}}-I_{\mathrm{res}0})\end{array}\right.;$

4) determine fault type according to the size of differential current; if differential current is positioned at the characteristic top of two broken lines; then transformer generating region internal fault is judged in protection; the protection tripping operation; if differential current is positioned at two broken line characteristic curves below; then protection judges that transformer does not have troubles inside the sample space and takes place, and protection is failure to actuate.

Compare with existing tranformer protection, the present invention has the following advantages:

(1) since the zero-sequence current longitudinal difference protection with zero-sequence current as the braking amount, removed the influence of load current in the stalling current, so this programme has higher sensitivity for the little turn-to-turn short circuit in the transformer district;

(2) because the zero-sequence current longitudinal difference protection is based on the longitudinal difference protection of fault component, protection can the correct decision internal fault external fault, and when the inner single phase ground fault of winding, protection sensitivity can not reduce;

(3) owing to the circuit topological structure in transformer zero sequence equivalent loop when magnetizing inrush current is constant,, can be good at distinguishing internal fault and magnetizing inrush current so Zero sequence current differential protection is subjected to the influence of exciting current very little.

Description of drawings

Fig. 1 is the hardware block diagram of transformer microcomputer protection,

Fig. 2 protects decision flow chart;

Fig. 3 is a PSCAD analogue system illustraton of model.

Zero sequence equivalent circuit diagram when Fig. 4 is the transformer external area error;

Zero sequence equivalent circuit diagram when Fig. 5 is power transformer interior fault;

Fig. 6 is transformer zero-sequence current longitudinal difference protection two a broken lines protection performance chart;

Zero sequence equivalent circuit diagram when Fig. 7 is transformer excitation flow.

Embodiment

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

Referring to Fig. 1, for protection algorithm computing module, constitute in the frame of broken lines by data acquisition system, microcomputer main system, input-output system three parts; Es1, Es2 are system's both end voltage, Z _S1, Z _S2Be two ends system impedance, Z _LBe the substitutional connection impedance, transformer is star-triangle connected mode, and TA is a transformer star side current transformer, and TV is a transformer star side voltage transformer, the invention belongs to transformer microcomputer protection, and transformer Y side is equipped with the protective device that uses element of the present invention.Input variable is a transformer star side voltage U and electric current I and transformer both sides leakage reactance when normally moving, wherein electric current and voltage are obtained by system power instrument transformer TA and voltage transformer TV secondary side respectively, the both sides leakage reactance is calculated by the nominal parameter of transformer, has provided simple signal among Fig. 1.Be algorithm computing module of the present invention in the frame of broken lines among Fig. 1, key algorithm of the present invention is realized in the DSP microsystem by program.A; B, the analog quantity input of C three-phase voltage and electric current is delivered to the microcomputer main system after phasor filtering, sampling maintenance and A/D conversion; go out operating current and stalling current by corresponding algorithm computation, the protection criterion according to the dual slope characteristic judges whether to take place internal fault then.

Referring to Fig. 2, protection is judged as internal fault when having only the diagram of satisfying logical condition, sends trip command, on the contrary the protection locking.Promptly determine two broken line characteristics of transformer Zero sequence current differential protection, if differential current in active region, is then protected action, if differential current in the braking district, is then protected and is failure to actuate.

Referring to Fig. 3, Es1, Es2 are respectively the both sides electromotive force, and system impedance is respectively Z _S1, Z _S2, Z _LBe the substitutional connection impedance, transformer connects for the Yn/ Δ,

Be respectively the zero-sequence current of transformer star side and triangle side, note wherein

Zero sequence circulation in the Δ winding can't flow to the winding outside.

Referring to Fig. 4, X wherein _{1 σ}, X _{2 σ}Be respectively the leakage reactance of both sides winding, X _M0Be the zero sequence excitation reactance,

Residual voltage for installing place of star side winding voltage instrument transformer;

Be the zero-sequence current of Y winding,

Be the zero sequence circulation in the Δ winding.

Because X _M0Much larger than X _{2 σ}, thinking the field excitation branch line open circuit has at this moment ${\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}=0.$

Referring to Fig. 5, it is when be that transformer is inner turn-to-turn short circuit or earth fault take place, this moment system the zero sequence equivalent network, f is the fault point,

Be the fault branch residual voltage,

Be fault branch zero-sequence current, X _{1 σ}', X _{2 σ}' the equivalent leakage reactance of both sides winding when being respectively internal fault, X _M0Be the zero sequence excitation reactance, Zf is the fault branch impedance,

Be the residual voltage of star side winding voltage instrument transformer installing place,

Be the zero-sequence current of Y winding,

Be the zero sequence circulation in the Δ winding.

At this moment, according to Kirchhoff's current law (KCL), ${\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}={\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="k"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">k</figure-callout>}0}.$

According to above analysis, can select $I_{\mathrm{op}}=|{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|$ As the operating current of zero sequence longitudinal difference protection, choose $I_{\mathrm{res}}=|{\stackrel{\&CenterDot;}{I}}_{m0}-{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|$ Be stalling current, adopt the protection scheme of ratio-restrained characteristic, the protection criterion is:

$\left\{\begin{array}{c}{I}_{\mathrm{op}}>I_{\mathrm{op}0}\\ I_{\mathrm{op}}\&GreaterEqual;I_{\mathrm{op}0}+K_{\mathrm{res}}(I_{\mathrm{res}}-I_{\mathrm{res}0})\end{array}\right.$

Referring to Fig. 6, it is two broken line characteristic curves of transformer zero sequence differential protection, and this curve has been determined the active region and the braking district of protection.I _Op0Be starting current, I _Res0Be keen current, I _OpBe operating current, I _ResBe stalling current, K _ResBe restraint coefficient.The broken line top is an active region, and the broken line below is the braking district.

Determine each parameter in the two broken line characteristics below:

Protect not malfunction when wherein normally moving, set for the assurance transformer As starting current,

We are the transformer rated current, so can get $I_{\mathrm{op}0}=0.1{\stackrel{\&CenterDot;}{I}}_n;$ Choosing restraint coefficient according to experience is K _Res=0.75, because the differential protection curve of fault component is crossed initial point, so can get keen current $I_{\mathrm{<figure-callout\; id="0"\; label="res"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">res</figure-callout>}0}=\frac{I_{\mathrm{op}0}}{K_{\mathrm{res}}}=0.133I_n.$

When the zero sequence differential current is positioned at active region shown in Figure 6, judge transformer generation internal fault, the protection tripping operation when the zero sequence differential current is positioned at the braking district, judges that transformer is in normal operation, external fault or magnetizing inrush current state, protection is failure to actuate.

Referring to Fig. 7, X wherein _{1 σ}, X _{2 σ}Be respectively the leakage reactance of both sides winding, X _M0' be the zero sequence excitation reactance under the saturated conditions,

Be the residual voltage of star side winding voltage instrument transformer installing place,

Be the zero-sequence current of Y winding,

Be the zero sequence circulation in the Δ winding,

Zero-sequence current for field excitation branch line.

When magnetizing inrush current takes place the zero sequence equivalent circuit of transformer as shown in Figure 7, the X among this moment figure _M0' littler than the zero sequence excitation reactance value under the normal operation, but because and X _M0' reactance in parallel is Circuit Fault on Secondary Transformer leakage reactance X _{2 σ}, X is still arranged _M0The X of '＞＞ _{2 σ}So, the zero-sequence current of field excitation branch line

Very little, this moment, differential current still approached zero, and the influence of magnetizing inrush current can be ignored.

In order to reach above-mentioned technical purpose, technical scheme of the present invention is achieved in that

Step 1: the Y side three-phase phase voltage sampled value sequence U that gathers latter two cycle of fault _a, U _b, U _cWith three-phase phase current sampling value sequence I _a, I _b, I _c, be translated into the phasor sequence by the phasor filter.Calculate Y side residual voltage And zero-sequence current

Step 2: calculate the zero sequence circulation in the Δ side winding

Step 3: calculate operating current I _OpWith stalling current I _Res

Step 4: determine starting current, keen current and restraint coefficient are determined two broken line characteristic curves of Zero sequence current differential protection;

Step 5: when the zero sequence differential current is positioned at the characteristic active region of two broken lines, judge that internal fault takes place in protection, the protection action; When differential current was positioned at the braking district, this moment, transformer did not have internal fault, and protection is failure to actuate.

Provided below simulation results of the present invention:

Analogue system as shown in Figure 2.Transformer is formed by connecting by three single-phase transformers, and no-load voltage ratio is 525kV/230kV, S=900MVA, and leakage reactance 10%, winding connects for the Yn/ Δ, and sample frequency is 2kHz.The operating current of conventional differential protection when in table 1, providing the various internal fault of star-like side and homodyne protection and the ratio of stalling current.

As the above analysis, under the condition that the zero sequence longitudinal difference protection starts, when $\frac{I_{\mathrm{op}}}{I_{\mathrm{res}}}=\frac{|{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="m"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">m</figure-callout>}0}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|}{|{\stackrel{\&CenterDot;}{I}}_{m0}-{\stackrel{\&CenterDot;}{I}}_{\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2009100234830010H1.png,A2009100234830010H2.png"\; state="\{\{state\}\}">n</figure-callout>}0}|}>k_{\mathrm{res}}$ Time protection action.

Operating current and stalling current ratio during table 1 internal fault

We can see from table 1, and during weak fault, the K value of conventional differential protection is very little in the inner generating region of transformer, can't correctly reflect these faults, and the K value of zero sequence longitudinal difference protection has very high sensitivity all greater than 1, little turn-to-turn short circuit of reliably protecting and fault all over the ground; For the catastrophe failure in the district, the K value of zero sequence longitudinal difference protection is equally greater than 1, and the homodyne protection can be moved equally; And for line to line fault and three-phase shortcircuit, can not start because the zero sequence differential current less than starting current, is protected.

Conventional differential protection and the operating current of homodyne protection and the ratio of stalling current when in table 2, providing the various external area error of star-like side.As can be seen from Table 2, star-like side generating region is in addition during the ground connection branch trouble, and the ratio of the operating current of zero sequence longitudinal difference protection and stalling current is less than restraint coefficient 0.75, and protection can not moved; When fault, not containing the ground connection branch road,, the zero sequence operating current also can not start because less than starting current, protecting.By above analysis as can be known, the zero sequence longitudinal difference protection can malfunction when the transformer external area error.

Operating current and stalling current ratio during table 2 external fault

Claims (1)

1, a kind of zero-sequence current longitudinal differential protection method based on star-connection delta line transformer is characterized in that:

1) measured value by star side side current transformer TA, voltage transformer TV calculates star side residual voltage U ₀And zero-sequence current

And according to external area error equivalent electric circuit calculating triangle side winding zero-sequence current

${\stackrel{\&CenterDot;}{U}}_{\left(0\right)}=\frac{1}{3}({\stackrel{\&CenterDot;}{U}}_{\mathrm{mA}}+{\stackrel{\&CenterDot;}{U}}_{\mathrm{mB}}+{\stackrel{\&CenterDot;}{U}}_{\mathrm{mC}}),{\stackrel{\&CenterDot;}{I}}_{m0}=\frac{1}{3}({\stackrel{\&CenterDot;}{I}}_{\mathrm{mA}}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{mB}}+{\stackrel{\&CenterDot;}{I}}_{\mathrm{mC}}),{\stackrel{\&CenterDot;}{I}}_{n0}=-\frac{\stackrel{\&CenterDot;}{U_{\left(0\right)}}-\stackrel{\&CenterDot;}{I_{m0}}*X_{1\mathrm{\&sigma;}}}{X_{2\mathrm{\&sigma;}}}$

Wherein

And The three-phase current that measures for the star side winding TA of place, wherein

And Be the three-phase voltage that the star side winding TV of place measures, X _{1 σ}, X _{2 σ}Be respectively the leakage reactance of transformer both sides winding;

2) utilize star side, triangle side zero-sequence current

Calculate operating current I _OpWith stalling current I _Res,

$I_{\mathrm{op}}=|{\stackrel{\&CenterDot;}{I}}_{m0}+{\stackrel{\&CenterDot;}{I}}_{n0}|,I_{\mathrm{res}}=|{\stackrel{\&CenterDot;}{I}}_{m0}-{\stackrel{\&CenterDot;}{I}}_{n0}|;$

3) determine starting current $I_{\mathrm{op}0}=0.1{\stackrel{\&CenterDot;}{I}}_n,$ Restraint coefficient K _Res=0.75, keen current $I_{\mathrm{res}0}=\frac{I_{\mathrm{op}0}}{K_{\mathrm{res}}}=0.133I_n,$ Wherein

Be the transformer rated current, obtain two broken line characteristic curves of Zero sequence current differential protection then:

$\left\{\begin{array}{c}{I}_{\mathrm{op}}>I_{\mathrm{op}0}\\ I_{\mathrm{op}}\&GreaterEqual;I_{\mathrm{op}0}+K_{\mathrm{res}}(I_{\mathrm{res}}-I_{\mathrm{res}0})\end{array}\right.;$

4) determine fault type according to the size of differential current; if differential current is positioned at the characteristic top of two broken lines; then transformer generating region internal fault is judged in protection; the protection tripping operation; if differential current is positioned at two broken line characteristic curves below; then protection judges that transformer does not have troubles inside the sample space and takes place, and protection is failure to actuate.

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