CN105158592A - Real-time three-phase electrical network impedance detection method on the asymmetric electrical network condition - Google Patents

Abstract

The invention relates to a real-time three-phase electrical network impedance detection method on the asymmetric electrical network condition and is applicable to rapid electrical network impedance detection in solar energy grid-connection generation. The method comprises steps that, three-phase symmetric harmonic wave currents in 175Hz and 325Hz non-characteristic frequencies are injected into an electrical network; the two non-characteristic frequency harmonic wave current components in the currents are separated; the two non-characteristic frequency harmonic wave voltage components in electrical network voltages are separated; phases of the two non-characteristic frequency harmonic wave currents are respectively acquired; effective values of the two non-characteristic frequency harmonic wave voltages are respectively resolved; included angles of the two non-characteristic frequency harmonic wave voltages and the two non-characteristic frequency harmonic wave currents are respectively calculated; each-phase electrical network impedance is respectively calculated according to the acquired voltage, the acquired current and angle information. The real-time three-phase electrical network impedance detection method has advantages of rapid calculation speed and small computational complexity and is applicable to electrical network impedance detection on the condition of asymmetric electrical network voltages and asymmetric electrical network impedances.

Description

Three phase network impedance real-time detection method under electrical network asymmetrical

Technical field

The present invention relates to a kind of three phase network impedance real-time detection method, be particularly useful for solar grid-connected in quick three phase network impedance real-time detection method under the electrical network asymmetrical that uses of detection of grid impedance.

Background technology

Grid-connection converter is the key position in solar grid-connected generation technology, and network process in play very important status.Generally, grid-connection converter needs the function of islanding detect, and the voltage that gathers at work of grid-connection converter needs the landing of removing voltage on power network line, so carry out impedance detection to electrical network just seem particularly important.For single electrical network, because material is certain, and the line length of three-phase is the same, so line impedance value is merely certain.But along with various electrical equipment in the development low-voltage end parallel connection of electrical network, this makes three phase network impedance very likely be in asymmetrical state.Further, may to there is the rapid change from several ohm to tens ohm within a short period of time in electric network impedance, how effectively to detect electric network impedance fast and accurately, become the essential condition of grid-connection converter stable operation.

At present, the method detected for symmetrical electric network impedance is both at home and abroad little, has the symmetrical electric network impedance detection method that document proposes based on complex filter, but the method has no idea to detect three-phase electric network impedance value separately under unbalance grid impedance conditions.Have document to propose asymmetrical three-phase electric network impedance detection method based on FFT in addition, but the method calculated amount is comparatively large, detection time is longer, and not have the detection method of proposition impedance under unbalanced power supply condition.

Summary of the invention

The object of the invention is for above-mentioned technical matters, provide a kind of method simple, detection speed is fast, three phase network impedance real-time detection method under the electrical network asymmetrical that accuracy is high

For above-mentioned technical matters, under electrical network asymmetrical of the present invention, three phase network impedance real-time detection method utilizes three phase harmonic current feedback circuit unit to be connected by the three phase network of wire with grid-connection converter to be detected, and wire between harmonic current generator unit and three phase network installs current sensor and voltage sensor, its step is as follows:

A. by three phase harmonic current feedback circuit unit, in the three phase network of grid-connection converter, inject non-feature secondary frequencies is respectively f simultaneously ₁=175Hz, f ₂=325Hz, amplitude is the positive sequence harmonic electric current of 2A;

B. the component of voltage being separated two kinds of non-feature secondary frequencies harmonic waves in line voltage is obtained:

The voltage sensor be arranged on three phase network is utilized to obtain sampling grid side three-phase line voltage u _ab, u _bc, u _ca; Utilize formula: $\left[\begin{array}{c}{u}_{\mathrm{\α}}\\ u_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right],$ To three-phase line voltage u _ab, u _bc, u _cacarry out Clark conversion, obtain voltage u under two-phase rest frame _αand u _β;

By voltage u _αand u _βcarry out machine voltage decouples computation, obtain frequency respectively and be respectively f ₁=175Hz, f ₂voltage positive-sequence component u under two-phase rest frame of the three-phase symmetrical positive sequence harmonic of=325Hz _{p α 175}, u _{p β 175}, u _{p α 325}, u _{p β 325}and negative sequence component u _{n α 175}, u _{n β 175}, u _{n α 325}, u _{n β 325};

C. the current component being separated two kinds of non-feature secondary frequencies harmonic waves in electric current is obtained:

The current sensor be arranged on three phase network is utilized to obtain the three-phase current i of sampling grid side _a, i _b, i _c; Utilize formula: $\left[\begin{array}{c}{i}_{\mathrm{\α}}\\ i_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{i}_a\\ i_b\\ i_c\end{array}\right],$ To three-phase current i _a, i _b, i _ccarry out Clark conversion, obtain the current i under two-phase rest frame _αand i _β;

By current i _αand i _βcarry out current separation calculating, obtain frequency respectively and be respectively f ₁=175Hz, f ₂=325Hz amplitude is the positive-sequence component i of three-phase symmetrical positive sequence harmonic electric current under two-phase rest frame of 2A _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}, by positive-sequence component i _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}substitute into formula respectively: obtain the phase angle theta of 175Hz harmonic current ₁₇₅with the phase angle theta of 325Hz harmonic current ₃₂₅;

D. formula is utilized: $\left[\begin{array}{c}{u}_{\mathrm{\α}}\\ u_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right],$ To three-phase line voltage u _ab, u _bc, u _cacarry out Clark conversion, obtain voltage u under two-phase rest frame _αand u _β;

By voltage u _α, u _βcarry out voltage decouples computation, obtain frequency respectively and be respectively f ₁=175Hz, f ₂voltage positive-sequence component u under two-phase rest frame of the three-phase symmetrical positive sequence harmonic of=325Hz _{p α 175}, u _{p β 175}, u _{p α 325}, u _{p β 325}and negative sequence component u _{n α 175}, u _{n β 175}, u _{n α 325}, u _{n β 325};

E. the effective value of two kinds of non-feature secondary frequencies harmonic voltages is solved respectively:

Utilize formula: $\left[\begin{array}{c}{u}_{Pab}\\ u_{Pbc}\\ u_{Pca}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\end{array}\right],\left[\begin{array}{c}{u}_{Nab}\\ u_{Nbc}\\ u_{Nca}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right],$ By u _{p α 175}, u _{p β 175}, u _{n α 175}, u _{n β 175}and u _{p α 325}, u _{p β 325}, u _{n α 325}, u _{n β 325}carry out anti-Clark conversion respectively, obtain two groups of positive sequence line voltage u _pab175, u _pbc175, u _pca175and u _pab325, u _pbc325, u _pca325and two groups of negative phase-sequence line voltage u _nab175, u _nbc175, u _nca175and u _nab325, u _nbc325, u _nca325;

Utilize formula: $\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right]=\left[\begin{array}{c}{u}_{Pab}\\ u_{Pbc}\\ u_{Pca}\end{array}\right]+\left[\begin{array}{c}{u}_{Nab}\\ u_{Nbc}\\ u_{Nca}\end{array}\right],$ Just two groups of positive sequence line voltages and two groups of negative phase-sequence line voltage difference correspondences will be added, obtain line voltage u _ab175, u _bc175, u _ca175with line voltage u _ab325, u _bc325, u _ca325;

Utilize formula: calculate frequency respectively and be respectively f ₁=175Hz, f ₂the effective value of each line voltage: U in the three-phase symmetrical positive sequence harmonic of=325Hz _ab175, U _bc175, U _ca175and U _ab325, U _bc325, U _ca325, in formula, T is the cycle length of harmonic electric current;

F. harmonic voltage and harmonic current angle under 175Hz and 325Hz frequency is calculated respectively:

By the current angle { θ of 175Hz ₁₇₅,-θ ₁₇₅bring formula into:

$\left[\begin{array}{c}{u}_{Pd}\\ u_{Pq}\\ u_{Nd}\\ u_{Nq}\end{array}\right]=\left[\begin{array}{cccc}cos\mathrm{\θ}& sin\mathrm{\θ}& 0& 0\\ -\sin \mathrm{\θ}& cos\mathrm{\θ}& 0& 0\\ 0& 0& cos(-\mathrm{\θ})& sin(-\mathrm{\θ})\\ 0& 0& -sin(-\mathrm{\θ})& cos(-\mathrm{\θ})\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\\ u_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right],$ By voltage u _{p α}, u _{p β}, u _{n α}, u _{n β}carry out two-phase static coordinate and be tied to the conversion of two-phase rotating coordinate system, obtain according to current phase angle { θ ₁₇₅,-θ ₁₇₅rotate lower new component of voltage u under synchronous coordinate system _pd, u _pq, u _nd, u _nq, by the component of voltage u newly obtained _pd, u _pq, u _nd, u _nqsubstitute into formula:

obtain corresponding line voltage u _ab175vector with phase current i _a175vector between angle theta=θ _ab-a-175, in like manner, will by current phase angle { θ ₃₂₅,-θ ₃₂₅, solve corresponding component of voltage u respectively _pd, u _pq, u _nd, u _nq, and according to the component of voltage u newly obtained _pd, u _pq, u _nd, u _nqobtain line voltage u respectively _bc175vector with phase current i _b175vector between angle theta _bc-b-175; Line voltage u _ca175vector with phase current i _c175vector between angle theta _ca-c-175; Line voltage u _ab325vector with phase current i _a325vector between angle theta _ab-a-325; Line voltage u _bc325vector with phase current i _b325vector between angle theta _bc-b-325; Line voltage u _ca325vector with phase current i _c325vector between angle theta _ca-c-325; Wherein solve line voltage vector with phase current between angle time, δ=0; Solve line voltage vector with phase current between angle time, π/3, δ=2; Solve line voltage vector with phase current between angle time, π/3, δ=4;

G. each phase electric network impedance is calculated:

Utilize formula: Q _ab-a-175=U _ab175sin (θ _ab-a-175), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-175, utilize formula: Q _ab-a-325=U _ab325sin (θ _ab-a-325), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-325; Again according to formula: obtain B phase resistance value R _b, ω ₁₇₅, ω ₃₂₅be respectively the angular velocity of 175Hz and 325Hz electric current, I in formula _bit is the non-harmonics current effective value that b phase is injected;

H. formula is utilized: P _ab-a-175=U _ab175cos (θ _ab-a-175), obtain line voltage vector at phase current vector projection P on direction _ab-a-175, utilize formula: P _ab-a-325=U _ab325cos (θ _ab-a-325), obtain line voltage vector at phase current vector projection P on direction _ab-a-325, finally utilize formula: obtain B phase inductance value L _b;

In like manner, { the Q that projects is obtained _ca-c-172, Q _ca-c-325, { Q _bc-b-172, Q _bc-b-325after, utilize formula:

$\begin{array}{c}{R}_a=\frac{{\mathrm{\ω}}_{175}{Q}_{ca-c-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{ca-c-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_a\]/2}\\ L_a=\frac{P_{ca-c-{\mathrm{\ω}}_{175}}-P_{ca-c-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_a\]/2}\end{array},$ Solve and obtain A phase resistance value R _ainductance value L _a; Utilize formula:

$\begin{array}{c}{R}_c=\frac{{\mathrm{\ω}}_{175}{Q}_{bc-b-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{bc-b-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_c\]/2}\\ L_c=\frac{P_{bc-b-{\mathrm{\ω}}_{175}}-P_{bc-b-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_c\]/2}\end{array},$ Solve and obtain C phase resistance value R _cinductance value L _c.

Beneficial effect: this method injects two kinds of non-feature sub symmetry positive sequence harmonic electric currents by three phase harmonic current feedback circuit unit in electrical network, and with complex filter, non-harmonics voltage and harmonic current are extracted, resistance value and the inductance value of each phase of three phase network is obtained by computing.This kind of method calculated amount is little, and detection speed is fast, and accuracy is high, and this is applicable to detect for the electric network impedance under line voltage and the asymmetric situation of electric network impedance.

Accompanying drawing explanation

Fig. 1 is the overall schematic diagram of impedance detection of the present invention

Fig. 2 is the vector plot of three phase network impedance under the non-characteristic frequency of the present invention

Fig. 3 is complex filter schematic diagram of the present invention

Fig. 4 be the present invention for positive sequence m subfrequency, complex filter realize structural drawing

Fig. 5 is C phase resistance of the present invention, inductance is undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, inductance L detected _csituation of change

Fig. 6 is C phase resistance of the present invention, inductance is undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, resistance R detected _csituation of change

Fig. 7 is the present invention's vector when injecting 175Hz harmonic current with with with between angle

Fig. 8 is the present invention's vector when injecting 325Hz harmonic current with with with between angle

Fig. 9 is harmonic frequency of the present invention is under 175Hz, electric network impedance three-phase line voltage waveform

Figure 10 is harmonic frequency of the present invention is under 325Hz, electric network impedance three-phase line voltage waveform

Figure 11 is the harmonic current waveforms that the present invention injects to electrical network

When Figure 12 is three-phase voltage generation unbalanced fault of the present invention, three-phase line voltage waveform

When Figure 13 is three-phase voltage generation unbalanced fault of the present invention, C phase resistance, inductance are undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, resistance R detected _csituation of change

When Figure 14 is three-phase voltage generation unbalanced fault of the present invention, C phase resistance, inductance are undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, inductance L detected _csituation of change

Embodiment

Implementation step according to the present invention is described further below:

As shown in Figure 1, three phase network impedance real-time detection method under electrical network asymmetrical of the present invention, utilize three phase harmonic current feedback circuit unit to be connected by the three phase network of wire with grid-connection converter to be detected, and wire between phase harmonic current generator unit and three phase network installs current sensor and voltage sensor.

Method utilizes three phase harmonic current feedback circuit unit, Clark coordinate transformation unit; Current separation computing unit, current phase solves unit, voltage separate calculation unit, anti-Clark coordinate transformation unit; Effective value solves unit, the voltage of non-feature secondary frequencies and electric current angle calcu-lation unit, impedance computation unit;

By three phase harmonic current feedback circuit unit, in the three phase network of grid-connection converter, inject non-feature secondary frequencies is respectively f simultaneously ₁=175Hz, f ₂=325Hz, amplitude is the positive sequence harmonic electric current of 2A, wherein: the voltage of 175Hz and 325Hz harmonic wave, the vector of electric current are:

$\begin{array}{c}\frac{\sqrt{3}}{2}{\mathrm{\ω}}_x{L}_b{I}_b+\frac{1}{2}{R}_b{I}_b+R_a{I}_a=P_{ab-a}\\ {\mathrm{\ω}}_x{L}_a{I}_a-\frac{\sqrt{3}}{2}{R}_b{I}_b+\frac{1}{2}{\mathrm{\ω}}_x{L}_b{I}_b=Q_{ab-a}\end{array}---\left(1\right)$

P in formula (1) _ab-afor vector ? on projection, Q _ab-afor ? projection in vertical direction; ω _xrepresent 175Hz harmonic current angular velocity 350 π and 325Hz harmonic current angular velocity 650 π of injection respectively, four corresponding equations can be obtained when different harmonic current, and then solve the resistance of B phase harmonic wave and the formula of inductance:

$\begin{array}{c}{R}_b=\frac{{\mathrm{\ω}}_{175}{Q}_{ab-a-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{ab-a-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_b\]/2}\\ L_b=\frac{P_{ab-a-{\mathrm{\ω}}_{175}}-P_{ab-a-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_b\]/2}\end{array}---\left(2\right)$

In like manner, the expression formula of A phase of impedance and C phase of impedance is

$\begin{array}{c}{R}_a=\frac{{\mathrm{\ω}}_{175}{Q}_{ca-c-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{ca-c-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_a\]/2}\\ L_a=\frac{P_{ca-c-{\mathrm{\ω}}_{175}}-P_{ca-c-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_a\]/2}\end{array}---\left(3\right)$

$\begin{array}{c}{R}_c=\frac{{\mathrm{\ω}}_{175}{Q}_{bc-b-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{bc-b-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_c\]/2}\\ L_c=\frac{P_{bc-b-{\mathrm{\ω}}_{175}}-P_{bc-b-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_c\]/2}\end{array}---\left(4\right)$

Be injected with the symmetric harmonic electric current that valid value is identical for twice, i.e. I ₁₇₅=I ₃₂₅, from the expression formula of electric network impedance, can find out, only need to obtain under same frequency, grid line voltage effective value and the angle between line voltage and phase current.

The electric current obtained through sensor, voltage are all containing fundametal compoment, 3,5,7 inferior harmonic voltage and the electric currents that the harmonic component injected and electrical network exist, calculated by voltage decouples computation and current separation and be separated, for three-phase line voltage, three-phase line voltage is:

$\left\{\begin{array}{c}{u}_{ab}=\underset{y=-n}{\overset{y=n}{\Σ}}\left(\sqrt{2}{U}_{py}sin\right({\mathrm{\ω}}_{y}t+{\mathrm{\θ}}_p)+\sqrt{2}{U}_{ny}\sin \left(-{\mathrm{\ω}}_{y}t+{\mathrm{\θ}}_n\right))\\ u_{bc}=\underset{y=-n}{\overset{y=n}{\Σ}}\left(\sqrt{2}{U}_{py}sin\right({\mathrm{\ω}}_{y}t-\frac{2\mathrm{\π}}{3}+{\mathrm{\θ}}_p)+\sqrt{2}{U}_{ny}\sin \left(-{\mathrm{\ω}}_{y}t-\frac{2\mathrm{\π}}{3}+{\mathrm{\θ}}_n\right))\\ u_{ca}=\underset{y=-n}{\overset{y=n}{\Σ}}\left(\sqrt{2}{U}_{py}sin\right({\mathrm{\ω}}_{y}t-\frac{4\mathrm{\π}}{3}+{\mathrm{\θ}}_p)+\sqrt{2}{U}_{ny}\sin \left(-{\mathrm{\ω}}_{y}t-\frac{4\mathrm{\π}}{3}+{\mathrm{\θ}}_n\right))\end{array}\right.---\left(5\right);$

Under electrical network asymmetrical, three phase network impedance real-time detection method step is as follows:

A. by three phase harmonic current feedback circuit unit, in the three phase network of grid-connection converter, inject non-feature secondary frequencies is respectively f simultaneously ₁=175Hz, f ₂=325Hz, amplitude is the positive sequence harmonic electric current of 2A,

B. the component of voltage being separated two kinds of non-feature secondary frequencies harmonic waves in line voltage is obtained:

Three-phase line voltage u _ab, u _bc, u _cacarry out Clark conversion by formula (6), obtain voltage u under two-phase rest frame _αand u _β;

$\left[\begin{array}{c}{u}_{\mathrm{\α}}\\ u_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right]---\left(6\right)$

U _α, u _βcarry out each harmonic extraction by voltage decouples computation, voltage, current separation computation structure are identical, as shown in Figure 3, are separated different harmonic waves, thus effectively find the harmonic wave injecting electrical network;

Get wherein m subharmonic voltage positive sequence and export u _{m α} ⁺carry out mathematical analysis,

$F_m=\frac{{\mathrm{\ω}}_c}{s-{\mathrm{jm\ω}}_0}---\left(7\right)$

Wherein, ω _cfor cutoff frequency, ω ₀for fundamental frequency.Expression formula realizes structural drawing as shown in Figure 4.Set up u _{m α} ⁺with overall input voltage amount u _αmathematic(al) representation (8),

${u_{m\mathrm{\α}}}^{+}=\frac{F_m}{1+\underset{x=-n,x\≠0}{\overset{x=n}{\Σ}}{F}_x}{u}_{\mathrm{\α}}---\left(8\right)$

F _msubstitute into expression formula (8) to obtain,

$G\left(jC\right)=\frac{{u_{m\mathrm{\α}}}^{+}}{u_{\mathrm{\α}}}=\frac{{\mathrm{\ω}}_c}{(\mathrm{\ω}-{\mathrm{m\ω}}_0)\·(j+{\mathrm{\ω}}_c.\underset{x=-n,x\≠0}{\overset{x=n}{\Σ}}\frac{1}{\mathrm{\ω}+{\mathrm{x\ω}}_0})}---\left(9\right)$

Wherein ω=x ω ₀, C=ω/ω _m.By analyzing G (jC) expression formula, known as ω=m ω ₀=ω _mtime, | G (jC) | during=1, C ≠ m, | G (jC) | namely=0 be m ω for frequency ₀signal, without any amplitude attenuation by wave filter, but for other characteristic harmonics, can decay to 0 at this passage.So the 175Hz that will be able to be injected by this kind of method, the non-harmonics electric current of 325Hz, voltage leach;

C. the current component being separated two kinds of non-feature secondary frequencies harmonic waves in electric current is obtained:

The three-phase phase current utilizing the current sensor be arranged on three phase network to obtain sampling grid side carries out Clark conversion by formula (10), obtains the current i under two-phase rest frame _αand i _β;

$\left[\begin{array}{c}{i}_{\mathrm{\α}}\\ i_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{i}_a\\ i_b\\ i_c\end{array}\right]---\left(10\right)$

I _α, i _βcalculated by current separation and isolate i _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}, by i _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}bring current phase angle computing formula (11) into, obtain current phase angle θ ₁₇₅, θ ₃₂₅;

$\mathrm{\θ}=\arcsin \frac{i_{\mathrm{\β}}}{\sqrt{{i_{\mathrm{\α}}}^2+{i_{\mathrm{\β}}}^2}}---\left(11\right);$

D. formula is utilized: $\left[\begin{array}{c}{u}_{\mathrm{\α}}\\ u_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right],$ To three-phase line voltage u _ab, u _bc, u _cacarry out Clark conversion, obtain voltage u under two-phase rest frame _αand u _β;

By voltage u _α, u _βcarry out voltage decouples computation, obtain frequency respectively and be respectively f ₁=175Hz, f ₂voltage positive-sequence component u under two-phase rest frame of the three-phase symmetrical positive sequence harmonic of=325Hz _{p α 175}, u _{p β 175}, u _{p α 325}, u _{p β 325}and negative sequence component u _{n α 175}, u _{n β 175}, u _{n α 325}, u _{n β 325};

E. bring 175Hz and the 325Hz voltage leached into formula (12) and carry out anti-Clark conversion,

$\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\end{array}\right]+\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right]---\left(12\right);$

By u _{p α 175}, u _{p β 175}, u _{n α 175}, u _{n β 175}and u _{p α 325}, u _{p β 325}, u _{n α 325}, u _{n β 325}carry out anti-Clark conversion respectively, obtain two groups of positive sequence line voltage u _pab175, u _pbc175, u _pca175and u _pab325, u _pbc325, u _pca325and two groups of negative phase-sequence line voltages

U _nab175, u _nbc175, u _nca175and u _nab325, u _nbc325, u _nca325;

Utilize formula: $\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right]=\left[\begin{array}{c}{u}_{Pab}\\ u_{Pbc}\\ u_{Pca}\end{array}\right]+\left[\begin{array}{c}{u}_{Nab}\\ u_{Nbc}\\ u_{Nca}\end{array}\right],$ Just two groups of positive sequence line voltages and two groups of negative phase-sequence line voltage difference correspondences will be added, obtain line voltage u _ab175, u _bc175, u _ca175with line voltage u _ab325, u _bc325, u _ca325;

Three-phase line voltage effective value is solved again according to formula (13),

$U=\sqrt{\frac{4}{T}{\∫}_0^{\frac{1}{4T}}{u}^{2}dt}---\left(13\right)$

Obtain u _ab, u _bc, u _caeffective value be respectively U _ab, U _bc, U _ca, i.e. U _ab175, U _bc175, U _ca175and U _ab325, U _bc325, U _ca325, in formula, T is the cycle length of harmonic electric current;

F. in the hope of solution vector with phase current vector angle theta _ab-a-175for example is described: u _{p α}, u _{p β}, u _{n α}, u _{n β}with three-phase synthesis current vector angle degree, orientation is carried out to line voltage, concrete formula as shown in (14), by the current angle { θ of 175Hz ₁₇₅,-θ ₁₇₅bring formula into:

$\left[\begin{array}{c}{u}_{Pd}\\ u_{Pq}\\ u_{Nd}\\ u_{Nq}\end{array}\right]=\left[\begin{array}{cccc}cos\mathrm{\θ}& sin\mathrm{\θ}& 0& 0\\ -\sin \mathrm{\θ}& \cos \mathrm{\θ}& 0& 0\\ 0& 0& cos(-\mathrm{\θ})& sin(-\mathrm{\θ})\\ 0& 0& -sin(-\mathrm{\θ})& cos(-\mathrm{\θ})\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\\ u_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right]---\left(14\right),$

By u _pd, u _pq, u _nd, u _nqbring formula (15) into and obtain line voltage vector with phase current vector between angle theta _ab-a-175;

${\mathrm{\θ}}_{ab-a-175}=\arcsin \frac{u_{Pq}-u_{Nq}}{\sqrt{{(u_{Pd}+u_{Nd})}^2+{(u_{Pq}-u_{Nq})}^2}}+\mathrm{\δ}---\left(15\right)$

In like manner, will by current phase angle { θ ₃₂₅,-θ ₃₂₅, solve corresponding component of voltage u respectively _pd, u _pq, u _nd, u _nq, and according to the component of voltage u newly obtained _pd, u _pq, u _nd, u _nqobtain line voltage u respectively _bc175vector with phase current i _b175vector between angle theta _bc-b-175; Line voltage u _ca175vector with phase current i _c175vector between angle theta _ca-c-175; Line voltage u _ab325vector with phase current i _a325vector between angle theta _ab-a-325; Line voltage u _bc325vector with phase current i _b325vector between angle theta _bc-b-325; Line voltage u _ca325vector with phase current i _c325vector between angle theta _ca-c-325; Wherein solve line voltage vector with phase current between angle time, δ=0; Solve line voltage vector with phase current between angle time, π/3, δ=2; Solve line voltage vector with phase current between angle time, π/3, δ=4;

G. each phase electric network impedance is calculated:

Utilize formula: Q _ab-a-175=U _ab175sin (θ _ab-a-175), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-175, utilize formula: Q _ab-a-325=U _ab325sin (θ _ab-a-325), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-325; Again according to formula: obtain B phase resistance value R _b, ω ₁₇₅, ω ₃₂₅be respectively the angular velocity of 175Hz and 325Hz electric current, I in formula _bit is the non-harmonics current effective value that b phase is injected;

H. formula is utilized: P _ab-a-175=U _ab175cos (θ _ab-a-175), obtain line voltage vector at phase current vector projection P on direction _ab-a-175, utilize formula: P _ab-a-325=U _ab325cos (θ _ab-a-325), obtain line voltage vector at phase current vector projection P on direction _ab-a-325, finally utilize formula: obtain B phase inductance value L _b;

In like manner, { the Q that projects is obtained _ca-c-172, Q _ca-c-325, { Q _bc-b-172, Q _bc-b-325after, utilize formula:

$\begin{array}{c}{R}_a=\frac{{\mathrm{\ω}}_{175}{Q}_{ca-c-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{ca-c-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_a\]/2}\\ L_a=\frac{P_{ca-c-{\mathrm{\ω}}_{175}}-P_{ca-c-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_a\]/2}\end{array},$ Solve and obtain A phase resistance value R _ainductance value L _a; Utilize formula:

$\begin{array}{c}{R}_c=\frac{{\mathrm{\ω}}_{175}{Q}_{bc-b-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{bc-b-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_c\]/2}\\ L_c=\frac{P_{bc-b-{\mathrm{\ω}}_{175}}-P_{bc-b-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_c\]/2}\end{array},$ Solve and obtain C phase resistance value R _cinductance value L _c.

Fig. 5 is C phase resistance, inductance is undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, inductance L detected _csituation of change.

Fig. 6 is C phase resistance, inductance is undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, resistance R detected _csituation of change.

Fig. 7 is vector when injecting 175Hz harmonic current with with with between angle.

Fig. 8 is vector when injecting 325Hz harmonic current with with with between angle.

Fig. 9 is harmonic frequency is under 175Hz, electric network impedance three-phase line voltage waveform.

Figure 10 is harmonic frequency is under 325Hz, electric network impedance three-phase line voltage waveform.

Figure 11 is the harmonic current waveforms injected to electrical network.

When Figure 12 is three-phase voltage generation unbalanced fault, three-phase line voltage waveform.

When Figure 13 is three-phase voltage generation unbalanced fault, C phase resistance, inductance are undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, resistance R detected _csituation of change.

When Figure 14 is three-phase voltage generation unbalanced fault, C phase resistance, inductance are undergone mutation at 0.6s simultaneously, R _c1.33 Ω are sported, L by 1 Ω _cwhen sporting 2.4mH by 1.2mH, inductance L detected _csituation of change.

Claims (1)

1. three phase network impedance real-time detection method under an electrical network asymmetrical, three phase harmonic current feedback circuit unit is utilized to be connected by the three phase network of wire with grid-connection converter to be detected, and wire between phase harmonic current generator unit and three phase network installs current sensor and voltage sensor, it is characterized in that step is as follows:

A. by three phase harmonic current feedback circuit unit, in the three phase network of grid-connection converter, inject non-feature secondary frequencies is respectively f simultaneously ₁=175Hz, f ₂=325Hz, amplitude is the positive sequence harmonic electric current of 2A;

B. the component of voltage being separated two kinds of non-feature secondary frequencies harmonic waves in line voltage is obtained:

The voltage sensor be arranged on three phase network is utilized to obtain sampling grid side three-phase line voltage u _ab, u _bc, u _ca; Utilize formula: $\left[\begin{array}{c}{u}_{\mathrm{\α}}\\ u_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right],$ To three-phase line voltage u _ab, u _bc, u _cacarry out Clark conversion, obtain voltage u under two-phase rest frame _αand u _β;

By voltage u _αand u _βcarry out machine voltage decouples computation, obtain frequency respectively and be respectively f ₁=175Hz, f ₂voltage positive-sequence component u under two-phase rest frame of the three-phase symmetrical positive sequence harmonic of=325Hz _{p α 175}, u _{p β 175}, u _{p α 325}, u _{p β 325}and negative sequence component u _{n α 175}, u _{n β 175}, u _{n α 325}, u _{n β 325};

C. the current component being separated two kinds of non-feature secondary frequencies harmonic waves in electric current is obtained:

The current sensor be arranged on three phase network is utilized to obtain the three-phase current i of sampling grid side _a, i _b, i _c; Utilize formula: $\left[\begin{array}{c}{i}_{\mathrm{\α}}\\ i_{\mathrm{\β}}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{2}\\ 0& \frac{\sqrt{3}}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{i}_a\\ i_b\\ i_c\end{array}\right],$ To three-phase current i _a, i _b, i _ccarry out Clark conversion, obtain the current i under two-phase rest frame _αand i _β;

By current i _αand i _βcarry out current separation calculating, obtain frequency respectively and be respectively f ₁=175Hz, f ₂=325Hz amplitude is the positive-sequence component i of three-phase symmetrical positive sequence harmonic electric current under two-phase rest frame of 2A _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}, by positive-sequence component i _{α 175}, i _{β 175}, i _{α 325}, i _{β 325}substitute into formula respectively: obtain the phase angle theta of 175Hz harmonic current ₁₇₅with the phase angle theta of 325Hz harmonic current ₃₂₅;

D. the effective value of two kinds of non-feature secondary frequencies harmonic voltages is solved respectively:

Utilize formula: $\left[\begin{array}{c}{u}_{Pab}\\ u_{Pbc}\\ u_{Pca}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\end{array}\right],\left[\begin{array}{c}{u}_{Nab}\\ u_{Nbc}\\ u_{Nca}\end{array}\right]=\sqrt{\frac{2}{3}}\left[\begin{array}{cc}1& 0\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{c}{u}_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right],$ By u _{p α 175}, u _{p β 175}, u _{n α 175}, u _{n β 175}and u _{p α 325}, u _{p β 325}, u _{n α 325}, u _{n β 325}carry out anti-Clark conversion respectively, obtain two groups of positive sequence line voltage u _pab175, u _pbc175, u _pca175and u _pab325, u _pbc325, u _pca325and two groups of negative phase-sequence line voltage u _nab175, u _nbc175, u _nca175and u _nab325, u _nbc325, u _nca325;

Utilize formula: $\left[\begin{array}{c}{u}_{ab}\\ u_{bc}\\ u_{ca}\end{array}\right]=\left[\begin{array}{c}{u}_{Pab}\\ u_{Pbc}\\ u_{Pca}\end{array}\right]+\left[\begin{array}{c}{u}_{Nab}\\ u_{Nbc}\\ u_{Nca}\end{array}\right],$ By two groups of positive sequence line voltages and two groups of negative phase-sequence line voltage corresponding additions respectively, obtain line voltage u _ab175, u _bc175, u _ca175with line voltage u _ab325, u _bc325, u _ca325;

Utilize formula: calculate frequency respectively and be respectively f ₁=175Hz, f ₂the effective value of each line voltage: U in the three-phase symmetrical positive sequence harmonic of=325Hz _ab175, U _bc175, U _ca175and U _ab325, U _bc325, U _ca325, in formula, T is the cycle length of harmonic electric current;

E. harmonic voltage and harmonic current angle under 175Hz and 325Hz frequency is calculated respectively:

In the hope of solution vector with phase current vector angle theta _ab-a-175for example is described, utilize voltage u _{p α}, u _{p β}, u _{n α}, u _{n β}with three-phase synthesis current vector angle degree, orientation is carried out to line voltage,

By the current angle { θ of 175Hz ₁₇₅,-θ ₁₇₅bring formula into:

$\left[\begin{array}{c}{u}_{Pd}\\ u_{Pq}\\ u_{Nd}\\ u_{Nq}\end{array}\right]=\left[\begin{array}{cccc}cos\mathrm{\θ}& sin\mathrm{\θ}& 0& 0\\ -sin\mathrm{\θ}& cos\mathrm{\θ}& 0& 0\\ 0& 0& cos(-\mathrm{\θ})& sin(-\mathrm{\θ})\\ 0& 0& -sin(-\mathrm{\θ})& cos(-\mathrm{\θ})\end{array}\right]\left[\begin{array}{c}{u}_{P\mathrm{\α}}\\ u_{P\mathrm{\β}}\\ u_{N\mathrm{\α}}\\ u_{N\mathrm{\β}}\end{array}\right],$ By voltage u _{p α}, u _{p β}, u _{n α}, u _{n β}carry out two-phase static coordinate and be tied to the conversion of two-phase rotating coordinate system, obtain according to current phase angle { θ ₁₇₅,-θ ₁₇₅rotate lower new component of voltage u under synchronous coordinate system _pd, u _pq, u _nd, u _nq, by the component of voltage u newly obtained _pd, u _pq, u _nd, u _nqsubstitute into formula:

$\mathrm{\θ}=arcsin\frac{u_{Pq}-u_{Nq}}{\sqrt{{(u_{Pd}+u_{Nd})}^2+{(u_{Pq}-u_{Nq})}^2}}+\mathrm{\δ},$ Obtain corresponding line voltage u _ab175vector with phase current i _a175vector between angle theta=θ _ab-a-175, in like manner, will by current phase angle

$\{{\mathrm{\θ}}_{175}+\frac{4\mathrm{\π}}{3},-({\mathrm{\θ}}_{175}-\frac{4\mathrm{\π}}{3})\},\{{\mathrm{\θ}}_{325},-{\mathrm{\θ}}_{325}\},\{{\mathrm{\θ}}_{325}+\frac{2\mathrm{\π}}{3},-({\mathrm{\θ}}_{325}-\frac{2\mathrm{\π}}{3})\},$

solve corresponding component of voltage u respectively _pd, u _pq, u _nd, u _nq, and according to the component of voltage u newly obtained _pd, u _pq, u _nd, u _nqobtain line voltage u respectively _bc175vector with phase current i _b175vector between angle theta _bc-b-175; Line voltage u _ca175vector with phase current i _c175vector between angle theta _ca-c-175; Line voltage u _ab325vector with phase current i _a325vector between angle theta _ab-a-325; Line voltage u _bc325vector with phase current i _b325vector between angle theta _bc-b-325; Line voltage u _ca325vector with phase current i _c325vector between angle theta _ca-c-325; Wherein solve line voltage vector with phase current between angle time, δ=0; Solve line voltage vector with phase current between angle time, π/3, δ=2; Solve line voltage vector with phase current between angle time, π/3, δ=4;

F. each phase electric network impedance is calculated:

Utilize formula: Q _ab-a-175=U _ab175sin (θ _ab-a-175), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-175, utilize formula: Q _ab-a-325=U _ab325sin (θ _ab-a-325), obtain line voltage vector at phase current vector projection Q in vertical direction _ab-a-325; Again according to formula: obtain B phase resistance value R _b, ω ₁₇₅, ω ₃₂₅be respectively the angular velocity of 175Hz and 325Hz electric current, I in formula _bit is the non-harmonics current effective value that b phase is injected;

G. formula is utilized: P _ab-a-175=U _ab175cos (θ _ab-a-175), obtain line voltage vector at phase current vector projection P on direction _ab-a-175, utilize formula: P _ab-a-325=U _ab325cos (θ _ab-a-325), obtain line voltage vector at phase current vector projection P on direction _ab-a-325, finally utilize formula: obtain B phase inductance value L _b;

In like manner, { the Q that projects is obtained _ca-c-175, Q _ca-c-325, { Q _bc-b-175, Q _bc-b-325after, utilize formula:

$\begin{array}{c}{R}_a=\frac{{\mathrm{\ω}}_{175}{Q}_{ca-c-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{ca-c-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_a\]/2}\\ L_a=\frac{P_{ca-c-{\mathrm{\ω}}_{175}}-P_{ca-c-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_a\]/2}\end{array},$ Solve and obtain A phase resistance value R _awith inductance value L _a; Utilize formula:

$\begin{array}{c}{R}_c=\frac{{\mathrm{\ω}}_{175}{Q}_{bc-b-{\mathrm{\ω}}_{325}}-{\mathrm{\ω}}_{325}{Q}_{bc-b-{\mathrm{\ω}}_{175}}}{\[\sqrt{3}({\mathrm{\ω}}_{325}-{\mathrm{\ω}}_{175})I_c\]/2}\\ L_c=\frac{P_{bc-b-{\mathrm{\ω}}_{175}}-P_{bc-b-{\mathrm{\ω}}_{325}}}{\[\sqrt{3}({\mathrm{\ω}}_{175}-{\mathrm{\ω}}_{325})I_c\]/2}\end{array},$ Solve and obtain C phase resistance value R _cwith inductance value L _c.