CN100508325C - Three-phase unbalance load compensation method - Google Patents

Abstract

The invention discloses a method about 3-phase asymmetric load compensation, which includes: Step one, it improves Steinmetz tenets of compensation claims, such as lambda = 1, Im [I1r I1 +] = 0, taking power or wattless into according, according to the request of network reactive regulation of -100 to +100%. Step two, it calculates the three-phase calculation of compensation to the satisfaction of Brab, Brbc, Brca on Ia, Ib, Ic relationship according to Steinmetz principle and the improved the compensation requirements. Step three, it calculates Im [ia], Im [alpha Ib], Im [alpha2Ic], Re [ia] Re [alpha Ib], Re [alpha2Ic] through the sampling method to get an instantaneous value of compensation required. Step four, it can calculate the relationship between iaq, ibq, icq and iap, ibp, icp with the compensation through the Brab, Brbc, Brca in the step two and step three about the relationship of symmetrical sequence I1 and negative sequence I2. The compensation can be shown with the instantaneous sampling values of the three-phase current.

Description

A kind of three-phase unbalance load compensation method

Technical field

The present invention relates to a kind of three-phase unbalance load compensation method.

Background technology

Grid voltage quality is weighed with indexs such as stability, symmetry and sines usually, along with nonlinear-loads such as modern power electronics devices insert electrical network in a large number, grid supply quality is had a strong impact on, the extensive application of wherein various electronic power switch devices and the frequent fluctuation of load are topmost interference sources, caused a series of such as line voltage fall, many harmful effects such as voltage fluctuation and flickering, power factor are low, electrical network three-phase imbalance.

Addressing this problem the method for generally using at present is to insert silent oscillation dynamic reactive reactive power compensator simultaneously at the disturbance load access point, be Static Var Compensator (SVC), in order to eliminate reactive power impact, the filtering high order harmonic component, balance three phase network, its typical case representative be fixed capacitor+thyristor-controlled reactor (Fixed Capacitor+Thyristor Controlled Reactor---FC+TCR).The key property of TCR type dynamic reactive compensation device (below be designated as TCR-SVC) is can be according to the real-time requirement of electric network reactive-load, regulate the idle of compensation arrangement continuously by the trigger delay angle of regulating the TCR thyristor, thereby the idle dynamic compensation of realization system makes the voltage of compensation point near remaining unchanged.The most important character of TCR-SVC is that it can be kept its terminal voltage and does not change, have fast-response frequent movement and phase splitting compensation ability, can be applicable to dynamic passive compensation field large-scale impact, fast cycle fluctuation variation, imbalance, nonlinear-load (as arc furnace, rolling mill, city secondary transformer station, remote distance power transmission, electric locomotive power supply etc.).It can effectively suppress these caused voltage fluctuation problems of loading, and solves voltage distortion, fluctuation and flickering problem significantly, plays a part to improve the quality of power supply.

Make the dynamic passive compensation controller give full play to its design function, adopting accurately and efficiently, analysis is vital with control method.At first to obtain in time, relevant " source " information accurately, as three-phase voltage, three-phase current etc., then these information are carried out in real time, analyzed fast, obtain required control information, controller is according to these control informations, adopt suitable control method to produce corresponding action, finally just can obtain perfect compensation effect.Research about the dynamic passive compensation controller mainly concentrates on following two aspects:

The first, the extraction of disturbing signal.For voltage fluctuation and flickering, harmonic wave, relative slower, the long-term power quality problem of these variations of three-phase imbalance, symmetrical component method, harmonic analysis method are the most frequently used time domain analysis methods.Their characteristics are that mathematic(al) representation is simple, and physical concept is clear and definite.

The second, control strategy.In case detect, analyze the information of the relevant power quality problem of existence, just must adopt effective control method elimination or suppress these information.Adopt which kind of control method and power quality problem type and control device closely related.

With the compensation of symmetrical component method analysis load, right title load is realized the phase-splitting adjusting, eliminates negative-sequence current, the balance three phase network.The equilibrating principle of the unbalanced threephase load that C.P.Steinmetz proposes has provided the method that single-phase burden with power equilibrium is transferred to three-phase circuit, and it has constituted the network theory basis that reactive power compensator is realized equal lotus control.But this principle only makes power factor with a full remuneration that realizes reactive power aspect compensating reactive power just be 1 serve as finally to compensate target, and the power factor after can not realizing compensating and idlely can reach any set point, thereby the idle adjustable range of compensation back system is narrow, and many bucking-out systems require to have the idle adjustable range of broad.

Summary of the invention

The invention provides a kind of three-phase unbalance load compensation method, it can make the power factor after the compensation and idlely can reach any set point, thereby can enlarge adjustable range.

The present invention has adopted following technical scheme: a kind of three-phase unbalance load compensation method, it may further comprise the steps: step 1, the requirement of scope-100%～+ 100% of regulating according to electric network reactive-load, to the compensation requirement of Steinmetz principle can from the angle of power factor (PF) or from idle angle to λ=1 $\mathrm{Im}[\stackrel{\·}{I_1}+\stackrel{\·}{I_{1r}}]=0$ Improve, wherein λ is a power factor (PF),

Be the positive phase-sequence symmetrical component of triple line electric current,

Positive phase-sequence symmetrical component for the triple line electric current of the reactive-load compensator of delta connection; Step 2, according to the Steinmetz principle calculate electrical network and improve after compensation require the compensation susceptance of the three-phase that calculates to be About

Relational expression; Step 3 is found the solution by sampling method then

Thereby extrapolate the instantaneous value of required compensation susceptance; Detailed process is as follows: establish ${\stackrel{\·}{I}}_a=I_{\mathrm{aR}}+{\mathrm{jI}}_{\mathrm{ax}},$ Expression formula according to transient current has,

A, as sin ω t=0, during cos ω t=1 $\mathrm{Im}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{ax}}=i_a/\sqrt{2},$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt},$ Work as u _aZero passage becomes timing, promptly as ω t=2k π, and the transient current i that collects when k=0,1,2... _aBe Doubly

Therefore, work as u _a, u _b, u _cZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will Be made as i successively _Aq, i _Bq, i _Cq

B, as sin ω t=1, during cos ω t=0 $\mathrm{Re}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{aR}}=i_a/\sqrt{2},$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt},$ As ω t=pi/2+2k π, when k=0,1,2..., the transient current i that collects _aBe

Doubly

Because u _BcHysteresis u _aThe radian pi/2, therefore, u _BcZero passage becomes the transient current i that timing collects _aAlso be

Doubly

Work as u _Bc, u _Ca, u _AbZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will

Be made as i successively _Ap, i _Bp, i _CpStep 4 is in a described in the step 3, b and step 2

About line current symmetrical component positive sequence

Negative phase-sequence

Relational expression i as can be known _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWith the relation of required compensation susceptance, the compensation susceptance can show by the instantaneous sampling value of three-phase current, to i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWhen sampling, sampling instant is u _a, u _b, u _c, u _Bc, u _Ca, u _AbZero passage becomes the positive moment, must prepare an artificial neutral point in the measuring circuit.

The compensation of Steinmetz principle can be required λ≤1 that be transformed to from the angle of power factor (PF) in step 1 of the present invention and the step 2, then $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ Become $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=+\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}$ Or $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=-\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ},$ Get positive sign, system is idle less than zero, shows that system is capacitive, idlely send to electrical network; When getting negative sign, system is idle greater than zero after the compensation, shows that system is perception, absorbs electric network reactive-load, and the compensation susceptance that draws three-phase in conjunction with the Steinmetz principle is

About

Relational expression, concrete solution procedure is: establish $\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}=k,$

For formula $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=k$ Can get after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]-k\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=-(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right])$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$ Substitution $\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]-k\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=-(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]),$ By

In $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0,$ Have: $B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[{\stackrel{\·}{I}}_1\right]\right)$

By condition ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0,$ Convolution Steinmetz principle

${\stackrel{\·}{I}}_0=0$

${\stackrel{\·}{I}}_1=(Y_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+Y_l^{\mathrm{ca}})U\sqrt{3}$

${\stackrel{\·}{I}}_2=-({\mathrm{\α}}^2{Y}_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+{\mathrm{\αY}}_l^{\mathrm{ca}})U\sqrt{3}$

And

$\stackrel{\·}{I_{0r}}=0$

$\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

$\stackrel{\·}{I_{2r}}=-j({{\mathrm{\α}}^{2}B}_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+{\mathrm{\αB}}_r^{\mathrm{ca}})U\sqrt{3}$

Can try to achieve following two formulas:

$R_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Combination again

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[{\stackrel{\·}{I}}_1\right])$

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Can ask

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

With in the Steinmetz principle

${\stackrel{\·}{I}}_0=({\stackrel{\·}{I}}_a+{\stackrel{\·}{I}}_b+{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_1=({\stackrel{\·}{I}}_a+{\mathrm{\α}\stackrel{\·}{I}}_b+{\mathrm{\α}}^2{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_2=({\stackrel{\·}{I}}_a+{{\mathrm{\α}}^2\stackrel{\·}{I}}_b+\mathrm{\α}{\stackrel{\·}{I}}_c)/\sqrt{3}$

Substitution

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]),$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

Can ask

About

Equation

$B_r^{\mathrm{ab}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[-(k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]+(-k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{kRe}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{bc}}=-\frac{1}{9U}\{[-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[-k\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-(k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]+(-k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{ca}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[(-k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-k\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-(k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

Check: with k=0, λ=1 substitution following formula can get the result of Steinmetz principle when power compensates entirely.

Can be with the compensation requirement of Steinmetz principle in step 1 and the step 2 among the present invention from idle angle $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ The imaginary part that is transformed to the bus current positive sequence component after the compensation is set at setting m, corresponding following formula: $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m$ When m be on the occasion of, show that system is capacitive, idle can sending to electrical network; When m is a negative value, show that system is perception, idlely do not send to electrical network; According in the Steinmetz principle ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0,$ And $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m,$ Can ask shape as

About

Relational expression, concrete computational process is as follows: will $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m$ Can get after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]=m-\mathrm{Im}\left[\stackrel{\·}{I_1}\right]$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$ The substitution formula $\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]=m-\mathrm{Im}\left[\stackrel{\·}{I_1}\right],$ Because $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0$ , have:

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-m\right)$

By condition ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0$ , convolution Steinmetz principle

${\stackrel{\·}{I}}_0=0$

${\stackrel{\·}{I}}_1=(Y_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+Y_l^{\mathrm{ca}})U\sqrt{3}$

${\stackrel{\·}{I}}_2=-({\mathrm{\α}}^2{Y}_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+{\mathrm{\αY}}_l^{\mathrm{ca}})U\sqrt{3}$

And

$\stackrel{\·}{I_{0r}}=0$

$\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

$\stackrel{\·}{I_{2r}}=-j({{\mathrm{\α}}^{2}B}_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+{\mathrm{\αB}}_r^{\mathrm{ca}})U\sqrt{3}$

Can try to achieve following two formulas:

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

By $B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-m)$

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Can ask

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

With in the Steinmetz principle

${\stackrel{\·}{I}}_0=({\stackrel{\·}{I}}_a+{\stackrel{\·}{I}}_b+{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_1=({\stackrel{\·}{I}}_a+{\mathrm{\α}\stackrel{\·}{I}}_b+{\mathrm{\α}}^2{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_2=({\stackrel{\·}{I}}_a+{{\mathrm{\α}}^2\stackrel{\·}{I}}_b+\mathrm{\α}{\stackrel{\·}{I}}_c)/\sqrt{3}$

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-m),$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

Can as

About

Relational expression

$B_r^{\mathrm{ab}}=-\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3U}(-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)$

$B_r^{\mathrm{ca}}=\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m).$

U in the step 4 of the present invention _a, u _b, u _cIt is to occur with 120 ° interval that zero passage becomes positive moment, and the phase upgrades three times weekly substantially in order to the signal of determining each needed compensation susceptance.

After having adopted above technical scheme, on the basis according to Steinmtz three-phase unbalance load compensation principle of the present invention, by improvement to its compensation condition, derive compensation susceptance formula, make behind the power network compensation power factor and idlely can reach any set point, thereby idle adjustable range can be set in-100%～+ 100% the scope, compensation susceptance of the present invention in addition is to have realized parameter

Sampling, can realize the hardware trigger circuit like this, and very big sampling error can not occur, thereby avoid effectively final control precision is exerted an influence.

Description of drawings

Fig. 1 for three-phase symmetric voltage in the Steinmtz principle of the present invention to three-phase asymmetric load power supply schematic diagram

Embodiment

The present invention is a kind of three-phase unbalance load compensation method based on the Steinmetz principle, it can have two kinds of embodiment: embodiment one, step 1, the requirement of scope-100%～+ 100% of regulating according to electric network reactive-load, the compensation of Steinmetz principle can be required to be transformed to λ≤1 from the angle of power factor (PF) $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ Become $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=+\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}$ Or $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=-\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ},$ , get positive sign, system is idle after the compensation shows that less than zero system is capacitive, idlely send to electrical network; When getting negative sign, system is idle greater than zero after the compensation, shows that system is perception, absorbs electric network reactive-load, and wherein λ is a power factor (PF),

Be the positive phase-sequence symmetrical component of triple line electric current,

Positive phase-sequence symmetrical component for the triple line electric current of the reactive-load compensator of delta connection.Steinmetz principle of the present invention is: according to Fig. 1, asymmetric triangle load is by three symmetrical power voltage supplies, and the Y-connection load for isolated neutral can be shown as the triangle connected load by Y-Δ map table and analyze.Compensating network also is the triangle method of attachment.If three-phase phase voltage is: ${\stackrel{\·}{U}}_a=\stackrel{\·}{U};$ ${\stackrel{\·}{U}}_b={\mathrm{\α}}^2\stackrel{\·}{U};$ ${\stackrel{\·}{U}}_c=\mathrm{\α}\stackrel{\·}{U},$ α=e wherein ^{J2 π/3}

Order $\stackrel{\·}{U}=U\∠0^O,$ Then can ask each phase line current to be:

${\stackrel{\·}{I}}_a=\stackrel{\·}{I_{\mathrm{ab}}}-\stackrel{\·}{I_{\mathrm{ca}}}=Y_l^{\mathrm{ab}}\stackrel{\·}{U_{\mathrm{ab}}}-Y_l^{\mathrm{ca}}\stackrel{\·}{U_{\mathrm{ca}}}=[Y_l^{\mathrm{ab}}(1-{\mathrm{\α}}^2)-Y_l^{\mathrm{ca}}(\mathrm{\α}-1)]U$

${\stackrel{\·}{I}}_b=\stackrel{\·}{I_{\mathrm{bc}}}-\stackrel{\·}{I_{\mathrm{ab}}}=Y_l^{\mathrm{bc}}\stackrel{\·}{U_{\mathrm{bc}}}-Y_l^{\mathrm{ab}}\stackrel{\·}{U_{\mathrm{ab}}}=[Y_l^{\mathrm{bc}}({\mathrm{\α}}^2-\mathrm{\α})-Y_l^{\mathrm{ab}}(1-{\mathrm{\α}}^2)]U---\left(1\right)$

${\stackrel{\·}{I}}_c=\stackrel{\·}{I_{\mathrm{ca}}}-\stackrel{\·}{I_{\mathrm{bc}}}=Y_l^{\mathrm{ca}}\stackrel{\·}{U_{\mathrm{ca}}}-Y_l^{\mathrm{bc}}\stackrel{\·}{U_{\mathrm{bc}}}=[Y_l^{\mathrm{ca}}(\mathrm{\α}-1)-Y_l^{\mathrm{bc}}({\mathrm{\α}}^2-\mathrm{\α})]U$

If the symmetrical component of triple line electric current are: zero sequence

Positive sequence

Negative phase-sequence

Then have:

${\stackrel{\·}{I}}_0=({\stackrel{\·}{I}}_a+{\stackrel{\·}{I}}_b+{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_1=({\stackrel{\·}{I}}_a+{\mathrm{\α}\stackrel{\·}{I}}_b+{\mathrm{\α}}^2{\stackrel{\·}{I}}_c)/\sqrt{3}---\left(2\right)$

${\stackrel{\·}{I}}_2=({\stackrel{\·}{I}}_a+{{\mathrm{\α}}^2\stackrel{\·}{I}}_b+\mathrm{\α}{\stackrel{\·}{I}}_c)/\sqrt{3}$

With formula (1) substitution (2), can get:

${\stackrel{\·}{I}}_0=0$

${\stackrel{\·}{I}}_1=(Y_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+Y_l^{\mathrm{ca}})U\sqrt{3}---\left(3\right)$

${\stackrel{\·}{I}}_2=-({\mathrm{\α}}^2{Y}_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+{\mathrm{\αY}}_l^{\mathrm{ca}})U\sqrt{3}$

The symmetrical component of triple line electric current that other establishes the reactive-load compensator of delta connection are: zero sequence Positive sequence

Negative phase-sequence

Have similar in appearance to following formula: $\stackrel{\·}{I_{0r}}=0$

$\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

$\stackrel{\·}{I_{2r}}=-j({{\mathrm{\α}}^{2}B}_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+{\mathrm{\αB}}_r^{\mathrm{ca}})U\sqrt{3}$

In the formula

Be three-phase compensation susceptance.

In the Steinmetz principle, be provided with two compensation requirements, respectively corresponding two conditionals:

A. the three-phrase burden balance after compensating, promptly compensation back bus current negative sequence component is zero, has

${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0---\left(5\right)$

B. the power factor (PF) after the compensation is λ=1, and promptly the imaginary part of compensation back bus current positive sequence component is for having

$\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0---\left(6\right)$

Convolution (4) (5) (6), can ask:

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right])---\left(7\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

With (2) substitution formula (7), can ask

About

Equation:

$B_r^{\mathrm{ab}}=-\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3U}(-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right])---\left(8\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right])$

Step 2, according to $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=+\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}---\left(9\right)$ Or $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=-\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}$

The compensation susceptance of the three-phase that (9*) calculates is

About

Relational expression;

Concrete computational process is as follows:

If $\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}=k,$ For formula $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=k$ Can get after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]-k\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=-(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right])---\left(10\right)$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

In the substitution (10), because $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0$ , have:

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[{\stackrel{\·}{I}}_1\right]\right)---\left(11\right)$

By condition ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0,$ (3) (4) of convolution Steinmetz principle try to achieve following two formulas:

$R_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]---\left(12\right)$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]---\left(13\right)$

In conjunction with above-mentioned (11) (12) (13), can ask again

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right])---\left(14\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

(2) substitution (14) in the Steinmetz principle can be got:

$B_r^{\mathrm{ab}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[-(k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]+(-k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{kRe}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{bc}}=-\frac{1}{9U}\{[-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+---\left(15\right)$

$[-k\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-(k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]+(-k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{ca}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[(-k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-k\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-(k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

Check: with k=0 (λ=1) substitution (15) formula, can get the result of Steinmetz principle when power compensates entirely, i.e. formula (8).

Step 3 is found the solution by sampling method then

Thereby extrapolate the instantaneous value of required compensation susceptance;

Detailed process is as follows: establish ${\stackrel{\·}{I}}_a=I_{\mathrm{aR}}+{\mathrm{jI}}_{\mathrm{ax}},$ Expression formula according to transient current has,

A, as sin ω t=0, during cos ω t=1 $\mathrm{Im}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{ax}}=i_a/\sqrt{2}.$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt}.$ As seen, work as u _aZero passage becomes timing, promptly as ω t=2k π, and the transient current i that collects when k=0,1,2... _aBe

Doubly

Therefore, work as u _a, u _b, u _cZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will

Be made as i successively _Aq, i _Bq, i _Cq

B, as sin ω t=1, during cos ω t=0 $\mathrm{Re}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{aR}}=i_a/\sqrt{2}.$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt}.$ As seen, as ω t=pi/2+2k π, when k=0,1,2..., the transient current i that collects _aBe

Doubly

Because u _BcHysteresis u _aThe radian pi/2, therefore, u _BcZero passage becomes the transient current i that timing collects _aAlso be

Doubly

Therefore, work as u _Bc, u _Ca, u _AbThe current instantaneous value i that collects during zero passage successively _a, i _b, i _cRespectively

And will

Be made as i successively _Ap, i _Bp, i _Cp

Step 4, by in conjunction with a, b and formula (15), kind of the required equivalent compensation susceptance of each branch road of scheme TCR of can winning:

$B_r^{\mathrm{ab}}=-\frac{1}{9\sqrt{2}U}\{[{2i}_{\mathrm{aq}}+{2i}_{\mathrm{bq}}-i_{\mathrm{cq}}]+[-(k+\sqrt{3})i_{\mathrm{ap}}+(-k+\sqrt{3})i_{\mathrm{bp}}-{\mathrm{ki}}_{\mathrm{cp}}]\}$

$B_r^{\mathrm{bc}}=-\frac{1}{9\sqrt{2}U}\{[{-i}_{\mathrm{aq}}+{2i}_{\mathrm{bq}}+{2i}_{\mathrm{cq}}]+[-{\mathrm{ki}}_{\mathrm{ap}}-(k+\sqrt{3})i_{\mathrm{bp}}+(-k+\sqrt{3})i_{\mathrm{cp}}]\}---\left(22\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{9\sqrt{2}U}\{[{2i}_{\mathrm{aq}}-i_{\mathrm{bq}}+{2i}_{\mathrm{cq}}]+[(-k+\sqrt{3})i_{\mathrm{ap}}-{\mathrm{ki}}_{\mathrm{bp}}-(k+\sqrt{3})i_{\mathrm{cp}}]\}$

By i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWith the relation of required compensation susceptance, the compensation susceptance can show by the instantaneous sampling value of three-phase current, to i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWhen sampling, sampling instant is u _a, u _b, u _c, u _Bc, u _Ca, u _AbZero passage becomes the positive moment, must prepare an artificial neutral point in the measuring circuit, because u _a, u _b, u _cIt is to occur with 120 ° interval that zero passage becomes positive moment, and the phase upgrades three times weekly substantially in order to the signal of determining each needed compensation susceptance.

Embodiment two, and it comprises following a few step:

Step 1, the requirement of scope-100%～+ 100% of regulating according to electric network reactive-load requires to improve from idle angle to the compensation of Steinmetz principle, and the compensation of Steinmetz principle is required to compensating requirement in the Steinmetz principle $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ The imaginary part of the bus current positive sequence component after the compensation is set at setting m, corresponding following formula:

$\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m---\left(16\right)$

When m be on the occasion of, show that system is capacitive, idle can sending to electrical network; When m is a negative value, show that system is perception, idlely do not send to electrical network.Wherein λ is a power factor (PF),

Be the positive phase-sequence symmetrical component of triple line electric current,

Positive phase-sequence symmetrical component for the triple line electric current of the reactive-load compensator of delta connection;

Step 2, according to (5) in the Steinmetz principle and (16), can ask shape as

About

Relational expression, concrete computational process is as follows:

Will $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m$ Can get after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]=m-\mathrm{Im}\left[\stackrel{\·}{I_1}\right]---\left(17\right)$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$ Substitution formula (17), because $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0,$ Have:

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-m\right)---\left(18\right)$

Can ask in conjunction with (12) (13) and (18) of implementing in one

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-m)---\left(19\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

With (2) substitution (19) in the Steinmetz principle

Can as

About Relational expression

$B_r^{\mathrm{ab}}=-\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3U}(-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)---\left(20\right)$

$B_r^{\mathrm{ca}}=-\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m).$

Step 3 is found the solution by sampling method then

Thereby extrapolate the instantaneous value of required compensation susceptance;

Detailed process is as follows: establish ${\stackrel{\·}{I}}_a=I_{\mathrm{aR}}={\mathrm{jI}}_{\mathrm{ax}},$ Expression formula according to transient current has,

A, as sin ω t=0, during cos ω t=1 $\mathrm{Im}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{ax}}=i_a/\sqrt{2}.$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt}.$ As seen, work as u _aZero passage becomes timing, promptly as ω t=2k π, and the transient current i that collects when k=0,1,2... _aBe

Doubly

Therefore, work as u _a, u _b, u _cZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will

Be made as i successively _Aq, i _Bq, i _Cp

B, as sin ω t=1, during cos ω t=0 $\mathrm{Re}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{aR}}=i_a/\sqrt{2}.$ Still get u _aBe the reference sinusoidal quantity, $u_a=\sqrt{2}U\sin \mathrm{\ωt}.$ As seen, as ω t=pi/2+2k π, when k=0,1,2..., the transient current i that collects _aBe

Doubly

Because u _BcHysteresis u _aThe radian pi/2, therefore, u _BcZero passage becomes the transient current i that timing collects _aAlso be

Doubly

Therefore, work as u _Bc, u _Ca, u _AbThe current instantaneous value i that collects during zero passage successively _a, i _b, i _cRespectively

And will

Be made as i successively _Ap, i _Bp, i _Cp

Step 4 in conjunction with a, b and formula (20), can get second kind of required equivalent compensation of each branch road of scheme TCR:

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{2}U}(i_{\mathrm{aq}}+i_{\mathrm{bq}}-i_{\mathrm{cq}}-\frac{\sqrt{6}}{3}m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{2}U}(-i_{\mathrm{aq}}+i_{\mathrm{bq}}+i_{\mathrm{cq}}-\frac{\sqrt{6}}{3}m)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{2}U}(i_{\mathrm{aq}}-i_{\mathrm{bq}}+i_{\mathrm{cq}}-\frac{\sqrt{6}}{3}m)$

By i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWith the relation of required compensation susceptance, the compensation susceptance can show by the instantaneous sampling value of three-phase current, to i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWhen sampling, sampling instant is u _a, u _b, u _c, u _Bc, u _Ca, u _AbZero passage becomes the positive moment, must prepare an artificial neutral point in the measuring circuit, because u _a, u _b, u _cIt is to occur with 120 ° interval that zero passage becomes positive moment, and the phase upgrades three times weekly substantially in order to the signal of determining each needed compensation susceptance.

Claims (4)

1, a kind of three-phase unbalance load compensation method, it may further comprise the steps: step 1, the requirement of scope-100%～+ 100% of regulating according to electric network reactive-load, to the compensation of Steinmetz principle require from the angle of power factor (PF) or from idle angle to λ=1 $\mathrm{Im}[\stackrel{\·}{I_1}+\stackrel{\·}{I_{1r}}]=0$ Improve, wherein λ is a power factor (PF),

Be the positive phase-sequence symmetrical component of triple line electric current,

Positive phase-sequence symmetrical component for the triple line electric current of the reactive-load compensator of delta connection;

Step 2, according to the Steinmetz principle calculate electrical network and improve after compensation require the compensation susceptance of the three-phase that calculates to be About

Relational expression;

Step 3 is found the solution by sampling method then

Thereby extrapolate the instantaneous value of required compensation susceptance;

Detailed process is as follows: establish ${\stackrel{\·}{I}}_a=I_{\mathrm{aR}}+{\mathrm{jI}}_{\mathrm{ax}},$ Expression formula according to transient current has,

$=\sqrt{2}(I_{\mathrm{aR}}\sin \mathrm{\ωt}+I_{\mathrm{ax}}\cos \mathrm{\ωt})$

(1), as sin ω t=0, during cos ω t=1 $\mathrm{Im}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{ax}}=i_a/\sqrt{2},$ Still get u _aBe the reference sine $u_a=\sqrt{2}U\sin \mathrm{\ωt},$ Work as u _aZero passage becomes timing, promptly as ω t=2k π, and the transient current i that collects when k=0,1,2... _aBe

Doubly

Therefore, work as u _a, u _b, u _cZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will

Be made as i successively _Aq, i _Bq, i _Cq

(2), as sin ω t=1, during cos ω t=0 $\mathrm{Re}\left[{\stackrel{\·}{I}}_a\right]=I_{\mathrm{aR}}=i_a/\sqrt{2},$ Still get u _aBe the reference sine $u_a=\sqrt{2}U\sin \mathrm{\ωt},$ As ω t=pi/2+2k π, when k=0,1,2..., the transient current i that collects _aBe

Doubly Because u _BcHysteresis u _aThe radian pi/2, therefore, u _BcZero passage becomes the transient current i that timing collects _aAlso be

Doubly

Work as u _Bc, u _Ca, u _AbZero passage becomes the current instantaneous value i that timing collects successively _a, i _b, i _cBe respectively

And will

Be made as i successively _Ap, i _Bp, i _Cp

Step 4 is in (1) described in the step 3, (2) and step 2

About line current symmetrical component positive sequence

Negative phase-sequence

Relational expression learn i _Ap, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWith the relation of required compensation susceptance, the compensation susceptance shows by the instantaneous sampling value of three-phase current, to i _Aq, i _Bq, i _CqAnd i _Ap, i _Bp, i _CpWhen sampling, sampling instant is u _a, u _b, u _c, u _Bc, u _Ca, u _AbZero passage becomes the positive moment, must prepare an artificial neutral point in the measuring circuit.

2, a kind of three-phase unbalance load compensation method according to claim 1 is characterized in that in step 1 and the step 2 from the angle of power factor (PF) the compensation of Steinmetz principle being required to be transformed to λ≤1, then $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ Become $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=+\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}$ Or

$\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=-\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ},$ Get positive sign, system is idle less than zero, shows that system is capacitive, idlely send to electrical network; When getting negative sign, system is idle greater than zero after the compensation, shows that system is perception, absorbs electric network reactive-load, and the compensation susceptance that draws three-phase in conjunction with the Steinmetz principle is

About

Relational expression, concrete solution procedure is:

If $\sqrt{1-{\mathrm{\λ}}^2}/\mathrm{\λ}=k,$

For formula $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]/\mathrm{Re}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=k$ Obtain after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]-k\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=-(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right])$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$ Substitution $\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]-k\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=-(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]),$ Because $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0,$ Have: $B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[{\stackrel{\·}{I}}_1\right]\right)$

By condition ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0,$ In conjunction with the Steinmetz principle

${\stackrel{\·}{I}}_0=0$

${\stackrel{\·}{I}}_1=(Y_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+Y_l^{\mathrm{ca}})U\sqrt{3}$

${\stackrel{\·}{I}}_2=-({\mathrm{\α}}^2{Y}_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+{\mathrm{\αY}}_l^{\mathrm{ca}})U\sqrt{3}$

And

$\stackrel{\·}{I_{0r}}=0$

$\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

$\stackrel{\·}{I_{2r}}=-j({{\mathrm{\α}}^{2}B}_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+{\mathrm{\αB}}_r^{\mathrm{ca}})U\sqrt{3}$

Try to achieve following two formulas:

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Combination again

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[{\stackrel{\·}{I}}_1\right])$

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Promptly try to achieve

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

With in the Steinmetz principle

${\stackrel{\·}{I}}_0=({\stackrel{\·}{I}}_a+{\stackrel{\·}{I}}_b+{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_1=({\stackrel{\·}{I}}_a+{\mathrm{\α}\stackrel{\·}{I}}_b+{\mathrm{\α}}^2{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_2=({\stackrel{\·}{I}}_a+{{\mathrm{\α}}^2\stackrel{\·}{I}}_b+\mathrm{\α}{\stackrel{\·}{I}}_c)/\sqrt{3}$

Substitution

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]),$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-\mathrm{kRe}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right])$

Try to achieve

About

Equation

$B_r^{\mathrm{ab}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[-(k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]+(-k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{kRe}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{bc}}=-\frac{1}{9U}\{[-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+2\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[-k\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-(k+\sqrt{3})\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]+(-k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

$B_r^{\mathrm{ca}}=-\frac{1}{9U}\{[2\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\mathrm{\α}{\stackrel{\·}{I}}_b\right]+2\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]+$

$[(-k+\sqrt{3})\mathrm{Re}\left[\stackrel{\·}{I_a}\right]-k\mathrm{Re}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-(k+\sqrt{3})\mathrm{Re}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]]\}$

Check: with k=0, λ=1 substitution following formula obtains the result of Steinmetz principle when power compensates entirely.

3, a kind of three-phase unbalance load compensation method according to claim 1 is characterized in that in step 1 and the step 2 from the compensation requirement of idle angle with the Steinmetz principle $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=0$ The imaginary part that is transformed to the bus current positive sequence component after the compensation is set at setting m, corresponding following formula:

$\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m$

When m be on the occasion of, show that system is capacitive, idlely give to electrical network; When m is a negative value, show that system is perception, idlely do not send to electrical network;

According in the Steinmetz principle ${\stackrel{\·}{I}}_2+{\stackrel{\·}{I}}_{2r}=0$ And $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m,$ Ask shape as

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Relational expression, concrete computational process is as follows:

Will $\mathrm{Im}[{\stackrel{\·}{I}}_1+\stackrel{\·}{I_{1r}}]=m$ Obtain after the expansion:

$\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]=m-\mathrm{Im}\left[\stackrel{\·}{I_1}\right]$

Will $\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$ The substitution formula $\mathrm{Im}\left[\stackrel{\·}{I_{1r}}\right]=m-\mathrm{Im}\left[\stackrel{\·}{I_1}\right],$ Because $\mathrm{Re}\left[\stackrel{\·}{I_{1r}}\right]=0,$ Have:

$B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-m]\right)$

By condition ${\stackrel{\·}{I}}_2+\stackrel{\·}{I_{2r}}=0,$ Convolution Steinmetz principle

${\stackrel{\·}{I}}_0=0$

${\stackrel{\·}{I}}_1=(Y_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+Y_l^{\mathrm{ca}})U\sqrt{3}$

${\stackrel{\·}{I}}_2=-({\mathrm{\α}}^2{Y}_l^{\mathrm{ab}}+Y_l^{\mathrm{bc}}+{\mathrm{\αY}}_l^{\mathrm{ca}})U\sqrt{3}$

And

$\stackrel{\·}{I_{0r}}=0$

$\stackrel{\·}{I_{1r}}=j(B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}})U\sqrt{3}$

$\stackrel{\·}{I_{2r}}=-j({{\mathrm{\α}}^{2}B}_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+{\mathrm{\αB}}_r^{\mathrm{ca}})U\sqrt{3}$

Try to achieve following two formulas:

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

By $B_r^{\mathrm{ab}}+B_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{1}{U\sqrt{3}}\left(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-m]\right)$

$B_r^{\mathrm{ab}}-{2B}_r^{\mathrm{bc}}+B_r^{\mathrm{ca}}=-\frac{2}{U\sqrt{3}}\mathrm{Im}\left[\stackrel{\·}{I_2}\right]$

$B_r^{\mathrm{ab}}-B_r^{\mathrm{ca}}=\frac{2}{\sqrt{3}}\frac{1}{U\sqrt{3}}\mathrm{Re}\left[{\stackrel{\·}{I}}_2\right]$

Promptly try to achieve

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

With in the Steinmetz principle

${\stackrel{\·}{I}}_0=({\stackrel{\·}{I}}_a+{\stackrel{\·}{I}}_b+{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_1=({\stackrel{\·}{I}}_a+{\mathrm{\α}\stackrel{\·}{I}}_b+{\mathrm{\α}}^2{\stackrel{\·}{I}}_c)/\sqrt{3}$

${\stackrel{\·}{I}}_2=({\stackrel{\·}{I}}_a+{{\mathrm{\α}}^2\stackrel{\·}{I}}_b+\mathrm{\α}{\stackrel{\·}{I}}_c)/\sqrt{3}$ Substitution

$B_r^{\mathrm{ab}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[{\stackrel{\·}{I}}_1\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]-2\mathrm{Im}\left[\stackrel{\·}{I_2}\right]-m),$

$B_r^{\mathrm{ca}}=-\frac{1}{3\sqrt{3}U}(\mathrm{Im}\left[\stackrel{\·}{I_1}\right]+\mathrm{Im}\left[\stackrel{\·}{I_2}\right]+\sqrt{3}\mathrm{Re}\left[\stackrel{\·}{I_2}\right]-m)$

Obtain as

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Relational expression

$B_r^{\mathrm{ab}}=\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\mathrm{\α}\stackrel{\·}{I_b}\right]-\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)$

$B_r^{\mathrm{bc}}=-\frac{1}{3U}(-\mathrm{Im}\left[\stackrel{\·}{I_a}\right]+\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m)$

$B_r^{\mathrm{ca}}=\frac{1}{3U}(\mathrm{Im}\left[\stackrel{\·}{I_a}\right]-\mathrm{Im}\left[\stackrel{\·}{\mathrm{\α}{I}_b}\right]+\mathrm{Im}\left[{\mathrm{\α}}^2\stackrel{\·}{I_c}\right]-\frac{\sqrt{3}}{3}m).$

4, a kind of three-phase unbalance load compensation method according to claim 1 is characterized in that u in the step 4 _a, u _b, u _cIt is to occur with 120 ° interval that zero passage becomes positive moment, and the phase upgrades three times weekly substantially in order to the signal of determining each needed compensation susceptance.