CN102867085A - Method for calculating short-circuit current of power system with double-fed wind generator set - Google Patents

Abstract

The invention discloses a method for calculating short-circuit current of a power system with a double-fed wind generator set. An electromagnetic transient model of the double-fed wind generator set and a fault transient process of a stator magnetic chain and a rotor magnetic chain are analyzed; an expression of the short-circuit current of a stator of the double-fed wind generator set under different stator voltage drop degrees is deduced; and a stator voltage drop coefficient correction method by which the short-circuit current calculation error can be effectively reduced and the practical method for calculating the short-circuit current of the power system with the double-fed wind generator set are proposes. A fault point is used as a boundary point; the power system is simplified and equalized to be a two-branch network with a generator branch and a system branch; calculation curves of short-circuit current periodic components of the generator branch and the system branch are calculated; amplitude values of the short-circuit current periodic components are determined; and expressions of a short-circuit current non-periodic component and a rotor frequency component of a short-circuit point are deduced. The calculation process is simple, accurate and practical; and a good foundation is laid for protection and arrangement of the power system of the double-fed wind generator set.

Description

The Power System Shortcuts Current calculation method that contains the double-fed fan motor unit

Technical field

The present invention relates to a kind of Power System Shortcuts Current calculation method, relate in particular to a kind of Power System Shortcuts Current calculation method that contains the double-fed fan motor unit.

Background technology

The wind energy turbine set location generally is in the end of supply network; the ability that withstands shocks is very weak; the wind energy turbine set off-grid that causes because of the system failure; to bring electric system and seriously influence; might cause the collapse of whole distract electrical network; this adjusts to protecting electrical power system and has proposed Secretary, and short circuit curve is prerequisite and the foundation of protection configuration.About the research of the fault transient process of the widely used double-fed induction wind driven generator group of China's electrical network still not deeply, not yet form systemtheoretical short-circuit current calculation method, and the short-circuit current practical calculation method that contains the electric system of double-fed fan motor unit rarely has still and mentions.

The subject matter that the engineering calculation of short-circuit current of the electric system of at present wind turbine group access faces is:

1) time-domain-simulation and two kinds of methods of analytic derivation are mainly adopted in the analysis of double-fed fan motor unit fault transient process.The double-fed fan motor unit short-circuit current expression formula that has proposed has only considered that stator voltage amplitude drops to zero situation, and the transient state process of all different line voltages not being fallen degree double-fed fan motor unit is studied.

2) utilize the electro-magnetic transient model of network element to carry out calculation of short-circuit current, the result is accurate, but method complexity and calculated amount are large, can not satisfy engineering demand.

The short-circuit current practical calculation method should be taken into account accuracy and the property simplified, and does not form so far the systemtheoretical short-circuit current practical calculation method that contains the electric system of double-fed fan motor unit.

Summary of the invention

Purpose of the present invention is exactly in order to address the above problem; a kind of Power System Shortcuts Current calculation method that contains the double-fed fan motor unit is provided; it is simple, accurately practical that it has computation process, establishes the advantage of good basis for the protection work of adjusting of the electric system that contains the double-fed fan motor unit.

To achieve these goals, the present invention adopts following technical scheme:

A kind of Power System Shortcuts Current calculation method that contains the double-fed fan motor unit, concrete steps are:

Step 1: utilize the electro-magnetic transient model of the lower double-fed fan motor unit of static coordinate axle system, derive the relation of double-fed fan motor unit stator and rotor electric current and magnetic linkage;

The electro-magnetic transient model of double-fed fan motor unit:

${\stackrel{\→}{u}}_s=R_s{\stackrel{\→}{i}}_s+\frac{d{\stackrel{\→}{\mathrm{\ψ}}}_s}{\mathrm{dt}}---\left(1\right)$

${\stackrel{\→}{u}}_r=R_r{\stackrel{\→}{i}}_r+\frac{d{\stackrel{\→}{\mathrm{\ψ}}}_r}{\mathrm{dt}}+{\mathrm{j\ω}}_r{\stackrel{\→}{\mathrm{\ψ}}}_r---\left(2\right)$

${\stackrel{\→}{\mathrm{\ψ}}}_s=L_s{\stackrel{\→}{i}}_s+L_m{\stackrel{\→}{i}}_r---\left(3\right)$

${\stackrel{\→}{\mathrm{\ψ}}}_r=L_m{\stackrel{\→}{i}}_s+L_r{\stackrel{\→}{i}}_r---\left(4\right)$

Wherein, subscript s and r represent respectively stator winding and rotor winding;

Be stator voltage,

Be rotor voltage,

Be stator current,

Be rotor current,

Be stator magnetic linkage, Be rotor flux, R _sBe stator resistance, R _rBe rotor resistance, L _sBe stator inductance, L _rBe inductor rotor; L _s=L _{S σ}+ L _m, L _r=L _{R σ}+ L _m, L _{S σ}, L _{R σ}Be respectively the leakage inductance of stator, rotor winding, L _mBe magnetizing inductance; ω _rElectric angle speed for rotor.Be the stator and rotor electric current of description double-fed fan motor unit and the relation of magnetic linkage, (3) formula and (4) formula be rewritten as:

${\stackrel{\→}{i}}_s=\frac{{\stackrel{\→}{\mathrm{\ψ}}}_s}{{L_s}^{\′}}-k_r\frac{{\stackrel{\→}{\mathrm{\ψ}}}_r}{{L_s}^{\′}}---\left(5\right)$

${\stackrel{\→}{i}}_r={-k}_s\frac{{\stackrel{\→}{\mathrm{\ψ}}}_s}{{L_r}^{\′}}+\frac{{\stackrel{\→}{\mathrm{\ψ}}}_r}{{L_r}^{\′}}---\left(6\right)$

In the formula, L _s' and L _r' be respectively the transient state inductance of stator and rotor winding, k _sAnd k _rBe respectively stator, inductor rotor coupling coefficient.And ${L_s}^{\′}=L_s-\frac{{L_m}^2}{L_r}=L_{\mathrm{s\σ}}+\frac{L_{\mathrm{r\σ}}\·L_m}{L_{\mathrm{r\σ}}+L_m},$ $k_r=\frac{L_m}{L_r},$ ${L_r}^{\′}=L_r-\frac{{L_m}^2}{L_s}=L_{\mathrm{r\σ}}+\frac{L_{\mathrm{s\σ}}\·L_m}{L_{\mathrm{s\σ}}+L_m},$ $k_s=\frac{L_m}{L_s}.$

Step 2: the fault transient process of analyzing the double-fed wind turbine magnetic linkage;

When the double-fed fan motor unit normally moves, the stator voltage space vector

Amplitude be expressed as U _s, rotating speed is the synchronous electric angular velocity omega _sRotating vector:

${\stackrel{\→}{u}}_s=U_s{e}^{\mathrm{j\α}}{e}^{j{\mathrm{\ω}}_{s}t}={\stackrel{\·}{U}}_s{e}^{{\mathrm{j\ω}}_{s}t}---\left(7\right)$

In the formula,

Be the stator voltage vector, α is the initial phase angle of a phase voltage;

T represents the time, t ₀The expression fault occurs constantly, t＞0, t ₀0; According to formula (1) voltage equation, if ignore the effect of stator resistance, stator magnetic linkage steady-state component over time rate is approximately equal to stator voltage, and then before the electric network fault, stator magnetic linkage is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t<t_0)=\frac{{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(8\right)$

If t=t ₀Constantly, electrical network generation three phase short circuit fault, stator voltage amplitude drops to steady-state value U _Sf, the stator voltage Drop coefficient is defined as:

$K=\frac{U_s-U_{\mathrm{sf}}}{U_s}---\left(9\right)$

The stator voltage Drop coefficient has embodied the electrical distance between trouble spot and the wind-powered electricity generation unit, is to represent the electrical distance infinity of trouble spot and double-fed fan motor unit at 0 o'clock or do not break down, and is to represent machine end three phase short circuit fault at 1 o'clock; U _Sf=(1-K) U _s, then stator voltage is expressed as after the fault:

${\stackrel{\&RightArrow;}{u}}_s(t\&GreaterEqual;t_0)=(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}---\left(10\right)$

The stator magnetic linkage steady-state component is after getting fault by above stator voltage:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(11\right)$

Electrical network generation three phase short circuit fault moment, the double-fed wind turbine voltage magnitude falls, and the stator magnetic linkage steady-state component reduces; Because magnetic linkage can not suddenly change, for reducing of compensation stator magnetic linkage steady-state component, will produce the aperiodic component of transient state in the stator magnetic linkage; The initial value of this aperiodic component is the poor of stator magnetic linkage steady-state component amplitude before and after the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}\left({\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_{0+}\right)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s\left(t_{0-}\right)-{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}\left(t_0\right)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}---\left(12\right)$

Because the existence of stator resistance, the stator magnetic linkage aperiodic component will be decayed gradually, and damping time constant is:

$T_s^{\&prime;}=\frac{L_s^{\&prime;}}{R_s}---\left(13\right)$

Then behind the electric network fault, the stator magnetic linkage aperiodic component is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(14\right)$

To sum up, after the electrical network generation three phase short circuit fault, double-fed wind turbine magnetic linkage expression formula is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)+{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}+\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{j{\mathrm{\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(15\right)$

Step 3: the fault transient process of analyzing double-fed fan motor unit rotor flux;

During normal operation, by double-fed fan motor unit electro-magnetic transient model, know that rotor flux is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=L_m{\stackrel{\&RightArrow;}{i}}_s+L_r{\stackrel{\&RightArrow;}{i}}_r=L_r({\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r)-L_{\mathrm{r\&sigma;}}{\stackrel{\&RightArrow;}{i}}_s---\left(16\right)$

Got by (1), (2) formula:

${\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r=\frac{{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}){\stackrel{\&RightArrow;}{i}}_s}{{\mathrm{j\&omega;}}_s{L}_m}---\left(17\right)$

With (17) formula substitution (16) formula, the rotor flux expression formula that must be represented by stator voltage, electric current:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}){\stackrel{\&RightArrow;}{i}}_s]-L_{\mathrm{r\&sigma;}}{\stackrel{\&RightArrow;}{i}}_s=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}+{\mathrm{j\&omega;}}_s\frac{L_{\mathrm{r\&sigma;}}{L}_m}{L_r}){\stackrel{\&RightArrow;}{i}}_s]---\left(18\right)$

The stator current space vector

Amplitude be expressed as I _s,

Be the stator current vector, then the double-fed fan motor unit rotor flux instantaneous value of moment is before the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_0{t}_0}-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}+{\mathrm{j\&omega;}}_s\frac{L_{\mathrm{r\&sigma;}}{L}_m}{L_r}){\stackrel{\&CenterDot;}{I}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}]---\left(19\right)$

The control of frequency converter during ignoring thinks that the crow bar protection was moved at once when fault occured, and rotor voltage sports zero, and corresponding rotor flux steady-state component bust is zero; According to the magnetism chain conservation principle, induce the rotor flux aperiodic component of relative stationary rotor in the rotor winding, because the existence of rotor resistance, this aperiodic component will be decayed gradually, damping time constant is:

$T_r^{\&prime;}=\frac{L_r^{\&prime;}}{R_r}---\left(20\right)$

The initial value of rotor flux aperiodic component is the instantaneous value of fault moment rotor flux Then the rotor flux expression formula is behind the electric network fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(21\right)$

Step 4: set up double-fed wind turbine short-circuit current expression formula;

With stator magnetic linkage expression formula (15) behind the electric network fault and rotor flux expression formula (21) substitution formula (5), get the double-fed wind turbine short-circuit current space vector expression formula after the electrical network three phase short circuit fault:

${\stackrel{\&RightArrow;}{i}}_s={\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}+\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(22\right)$

In the formula,

It is the short-circuit current fundamental frequency cycles component of forcing magnetic linkage to produce by stator;

It is the DC component of short-circuit current that is produced by the stator magnetic linkage aperiodic component;

Be the short-circuit current transient state component with rotor frequency, decay that is produced by the rotor flux aperiodic component, be called short-circuit current rotor frequency component;

Step 5: the stator voltage Drop coefficient is revised;

In in the past double-fed wind turbine magnetic linkage fault transient process research, when calculating stator voltage Drop coefficient K, think that all the instant of failure stator voltage amplitude namely drops to steady-state value.In step 2 and step 4, the derivation of double-fed wind turbine short-circuit current expression formula supposes temporarily that also stator voltage drops to steady-state value at once at instant of failure, so that explain the transient state physical process of stator magnetic linkage.In fact, because the existence of nonlinear element in the electrical network, after fault occured, the generator unit stator voltage magnitude did not drop to steady-state value at once, and there is the oscillatory extinction of short time in the centre.A large amount of emulation experiments show, this oscillatory extinction generally continues the time about 1 cycle (0.02s), therefore, only utilize that stator voltage Drop coefficient K corresponding to stator voltage steady-state value calculates the double-fed wind turbine short-circuit current after the fault, will produce error.

According to the physical significance of double-fed wind turbine magnetic linkage, should utilize the stable state amplitude of stator voltage after the fault to calculate the stator magnetic linkage steady-state component, utilize the instant of failure stator voltage amplitude to calculate the initial value of stator magnetic linkage aperiodic component.Now the stator voltage Drop coefficient in the double-fed wind turbine short-circuit current expression formula is done following correction:

When 1) calculating the stator magnetic linkage aperiodic component, adopt instant of failure stator voltage amplitude U _s(t ₀) corresponding stator voltage Drop coefficient $K_{t_{}}$ ${{}_0}_{}=\frac{U_s-U_s\left(t_0\right)}{U_s};$

When 2) calculating stator magnetic linkage fundamental frequency steady-state component, stator voltage amplitude U when adopting after the fault 0.02s _s(t ₀+ 0.02) corresponding stator voltage Drop coefficient $K_{t_{}}$ ${{}_0+0.02}_{}=\frac{U_s-U_s({\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0+0.02)}{U_s};$

Based on above-mentioned correction, after the electrical network generation three phase short circuit fault, the double-fed wind turbine magnetic linkage is modified to:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)+{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}+\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(23\right)$

With following formula substitution formula (5), get revised double-fed wind turbine short-circuit current space vector expression formula:

${\stackrel{\&RightArrow;}{i}}_s={\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}+\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(24\right)$

Step 6: the electric system simplification equivalence that will contain the double-fed fan motor unit is two branch network;

Take the trouble spot as boundary, the electric system of complexity is reduced to the network that is comprised of some equivalent generators and infinitely great power generatrix, this network is carried out further abbreviation, with all cancellations of all intermediate nodes, only keep power supply node and short circuit node; Select capacity weighting multimachine equivalence method, the double-fed fan motor unit that will be incorporated into the power networks is simplified equivalence; By above network abbreviation, the simplification equivalence of the electric system that the double-fed fan motor unit is accessed is two branch network; Supplied with the branch road of short-circuit current by the double-fed fan motor unit, be called the generator branch road; Supplied with the branch road of short-circuit current by electric system, be called system branch; The short-circuit current at short dot F place Be generator branch, short-circuit electric current

With the system branch short-circuit current

Sum;

Step 7: to the basic assumption of calculation of short-circuit current; In order to simplify calculating, electric system, wind-powered electricity generation unit, short-circuit conditions and load etc. are made following hypothesis:

1) Three-phase Power Systems that accesses of wind-powered electricity generation unit is symmetrical;

2) phase angle of all generator electromotive forces is all identical in the electric system, and the simplification of being convenient to system branch is equivalent;

3) magnetic circuit of each element is unsaturated in the electric system, and namely the parameter of each element does not change with electric current, uses superposition principle in computation process;

4) resistance of each element of system branch all omits and disregards, and only when calculating the damping time constant of aperiodic component, just takes into account the effect of resistance;

5) the generator electromotive force of wind-powered electricity generation unit of access electric system is identical, and the wind-powered electricity generation unit of all accesses is carried out the multimachine equivalence;

6) three phase short circuit fault occurs suddenly in electric system under specified running status;

7) load is only done approximate estimation, and represents with constant impedance; The load of load distribution employing 50% is connected on the high voltage bus, and all the other are in the modes of connection in the short dot outside;

Step 8: ask for the short dot periodic component of short-circuit current;

Ask for the computing curve of the periodic component of short-circuit current of generator branch road according to the double-fed wind turbine short-circuit current expression formula of having derived, adopt the amplitude of traditional calculation curve method computing system branch, short-circuit current cycle component, and then get the amplitude of short dot periodic component of short-circuit current;

1) asks for generator branch, short-circuit current cycle component

Known by double-fed wind turbine short-circuit current expression formula (24), the periodic component of generator branch, short-circuit electric current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}---\left(25\right)$

Its amplitude is:

$I_{\mathrm{sf}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}})U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}=\frac{U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}/(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}})}=\frac{U_s}{X_K}---\left(26\right)$

Wherein ω _sL ' _sReflected the transient state reactance of equivalent double-fed fan motor unit,

The electrical distance that has reflected trouble spot and double-fed fan motor unit; Known amplitude and the X of generator branch, short-circuit current cycle component by (26) formula _KRelevant, form one with X _KFor the transverse axis variable, with periodic component of short-circuit current amplitude I _SfChange curve for longitudinal axis variable;

The parameter of double-fed fan motor unit may be different with fault condition owing to being incorporated into the power networks, same X _KBe worth corresponding I _SfMay be different, will form many change curves, ask for the mean value of these curves as the computing curve of generator branch, short-circuit current cycle component.

2) ask for the system branch periodic component of short-circuit current

As equivalent reactance, disregarding loads in the network makes equivalent network, is the simplified network that contains generator electromotive force node and short dot with its abbreviation, gets each power supply to the transfer impedance X between short dot with each generator subtranient reactance in the electric system _{J δ}The amplitude of system branch periodic component of short-circuit current is:

$I_{\mathrm{gf}}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(27\right)$

I _GfBe transfer impedance X _{I δ}Function, form with X _{I δ}Be the transverse axis variable, with the amplitude I of system branch periodic component of short-circuit current _GfComputing curve for longitudinal axis variable.

3) amplitude of short dot periodic component of short-circuit current

The short dot periodic component of short-circuit current is generator branch road and system branch periodic component of short-circuit current sum.Therefore the amplitude I of short dot periodic component of short-circuit current _FfFor:

$I_{\mathrm{ff}}=I_{\mathrm{sf}}+I_{\mathrm{gf}}=\frac{U_s}{X_K}+\frac{1}{X_{\mathrm{j\&delta;}}}---\left(28\right)$

According to ω _sL ' _sWith Ask for X _KAfter, determine I from generator branch, short-circuit current cycle components operation curve _SfTry to achieve transfer impedance X _{I δ}After, determine I from the computing curve of system branch periodic component of short-circuit current _Gf, and then the amplitude I of short dot periodic component of short-circuit current _Ff

Step 9: ask for the short dot DC component of short-circuit current;

Known by double-fed wind turbine short-circuit current expression formula (24), generator branch, short-circuit electric current aperiodic component be:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}=\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}---\left(29\right)$

Instant of failure, the initial value of this aperiodic component is:

$I_{{\mathrm{sdt}}_0}=\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}---\left(30\right)$

The initial value of the aperiodic component of system branch short-circuit current

Poor for the periodic component initial value of the system branch current instantaneous value of the normal operation of fault moment and system branch short-circuit current; The system branch calculation of short-circuit current is more paid close attention to the maximum aperiodic component that can occur; Therefore, get

Amplitude for the system branch periodic component of short-circuit current:

$I_{{\mathrm{gdt}}_0}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(31\right)$

The aperiodic component of system branch short-circuit current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}=\frac{1}{X_{\mathrm{j\&delta;}}}{e}^{-{\mathrm{\&omega;}}_{s}t/T_g}---\left(32\right)$

T wherein _g=X/R is the damping time constant of system branch, and X is the system branch equivalent reactance, and R is the system branch substitutional resistance.The X/R of system branch is different and different with system situation, short-circuit current practical calculation method in the past carries out statistical study to several large power systems of China and shows, the variation range of X/R is 8 ~ 22, and X/R≤15 in most cases, therefore the X/R with system branch regards a constant as, and is taken as 15.

The initial value of short dot DC component of short-circuit current is the initial value sum of generator branch road and system branch aperiodic component:

$I_{{\mathrm{fdt}}_0}=I_{{\mathrm{sdt}}_0}+I_{{\mathrm{gdt}}_0}=\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}}---\left(33\right)$

The aperiodic component of short dot short-circuit current is generator branch road and system branch aperiodic component sum:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fd}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}=\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}}{e}^{-{\mathrm{\&omega;}}_{s}t/T_g}=(\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}})e^{-{\mathrm{\&omega;}}_{s}t/T_a}---\left(34\right)$

Wherein, T _aFor the equivalent time constant of short dot aperiodic component, get

Then

$T_a=\frac{-{\mathrm{\&omega;}}_{s}t}{\ln [{\mathrm{Me}}^{-t/T_s^{\&prime;}}+(1-M)e^{-{\mathrm{\&omega;}}_{s}t/T_g}]}---\left(35\right)$

Step 10: ask for short dot short-circuit current rotor frequency component;

There is the rotor frequency component of decay in the double-fed wind turbine short-circuit current, and do not have this component in the system branch; Therefore, the rotor frequency component of short dot short-circuit current is the short-circuit current rotor frequency component of generator branch road:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fr}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(36\right)$

Its initial value is:

$I_{\mathrm{fr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)---\left(37\right)$

Beneficial effect of the present invention:

1. propose to effectively reduce the modification method of the stator voltage Drop coefficient of calculation of short-circuit current error, revised short-circuit current expression formula have a higher accuracy;

2. proposed to contain the Power System Shortcuts electric current practical calculation method of double-fed fan motor unit, the method computation process is simple, accurately practical, for good basis is established in the protection of the electric system that contains the double-fed fan motor unit work such as adjust.

Description of drawings

Fig. 1 is the result of calculation of double-fed wind turbine short-circuit current expression formula after revising and realistic model result's contrast;

Fig. 2 is the change curve of wind-powered electricity generation unit parameter generator branch, short-circuit current cycle component when identical;

Fig. 3 is the not change curve of generator branch, short-circuit current cycle component simultaneously of wind-powered electricity generation unit parameter.

Embodiment

The invention will be further described below in conjunction with accompanying drawing and embodiment.

A kind of Power System Shortcuts electric current practical calculation method that contains the double-fed fan motor unit, concrete steps are:

Step 1: utilize the electro-magnetic transient model of the lower double-fed fan motor unit of static coordinate axle system, derive the relation of double-fed fan motor unit stator and rotor electric current and magnetic linkage;

The electro-magnetic transient model of double-fed fan motor unit:

${\stackrel{\&RightArrow;}{u}}_s=R_s{\stackrel{\&RightArrow;}{i}}_s+\frac{d{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{\mathrm{dt}}---\left(1\right)$

${\stackrel{\&RightArrow;}{u}}_r=R_r{\stackrel{\&RightArrow;}{i}}_r+\frac{d{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{\mathrm{dt}}+{\mathrm{j\&omega;}}_r{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r---\left(2\right)$

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s=L_s{\stackrel{\&RightArrow;}{i}}_s+L_m{\stackrel{\&RightArrow;}{i}}_r---\left(3\right)$

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=L_m{\stackrel{\&RightArrow;}{i}}_s+L_r{\stackrel{\&RightArrow;}{i}}_r---\left(4\right)$

Wherein, subscript s and r represent respectively stator winding and rotor winding; Be stator voltage,

Be rotor voltage,

Be stator current,

Be rotor current,

Be stator magnetic linkage,

Be rotor flux, R _sBe stator resistance, R _rBe rotor resistance, L _sBe stator inductance, L _rBe inductor rotor; L _s=L _{S σ}+ L _m, L _r=L _{R σ}+ L _m, L _{S σ}, L _{R σ}Be respectively the leakage inductance of stator, rotor winding, L _mBe magnetizing inductance; ω _rBe the electric angle speed of rotor, be the stator and rotor electric current of description double-fed fan motor unit and the relation of magnetic linkage, (3) formula and (4) formula are rewritten as:

${\stackrel{\&RightArrow;}{i}}_s=\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{{L_s}^{\&prime;}}-k_r\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{{L_s}^{\&prime;}}---\left(5\right)$

${\stackrel{\&RightArrow;}{i}}_r={-k}_s\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{{L_r}^{\&prime;}}+\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{{L_r}^{\&prime;}}---\left(6\right)$

In the formula, L _s' and L _r' be respectively the transient state inductance of stator and rotor winding, k _sAnd k _rBe respectively the stator and rotor inductive coupling coefficient.And ${L_s}^{\&prime;}=L_s-\frac{{L_m}^2}{L_r}=L_{\mathrm{s\&sigma;}}+\frac{L_{\mathrm{r\&sigma;}}\&CenterDot;L_m}{L_{\mathrm{r\&sigma;}}+L_m},$ $k_r=\frac{L_m}{L_r},$ ${L_r}^{\&prime;}=L_r-\frac{{L_m}^2}{L_s}=L_{\mathrm{r\&sigma;}}+\frac{L_{\mathrm{s\&sigma;}}\&CenterDot;L_m}{L_{\mathrm{s\&sigma;}}+L_m},$ $k_s=\frac{L_m}{L_s}.$

Step 2: the fault transient process of analyzing the double-fed wind turbine magnetic linkage;

When the double-fed fan motor unit normally moves, the stator voltage space vector

Amplitude is expressed as U _s, rotating speed is the synchronous electric angular velocity omega _sRotating vector:

${\stackrel{\&RightArrow;}{u}}_s=U_s{e}^{\mathrm{j\&alpha;}}{e}^{j{\mathrm{\&omega;}}_{s}t}={\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}---\left(7\right)$

In the formula, Be the stator voltage vector, α is the initial phase angle of a phase voltage;

T represents the time, t ₀The expression fault occurs constantly, t＞0, t ₀0; According to formula (1) voltage equation, if ignore the effect of stator resistance, stator magnetic linkage steady-state component over time rate is approximately equal to stator voltage, and then before the electric network fault, stator magnetic linkage is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t<t_0)=\frac{{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(8\right)$

If t=t ₀Constantly, electrical network generation three phase short circuit fault, stator voltage amplitude drops to steady-state value U _Sf, the stator voltage Drop coefficient is defined as:

$K=\frac{U_s-U_{\mathrm{sf}}}{U_s}---\left(9\right)$

The stator voltage Drop coefficient has embodied the electrical distance between trouble spot and the wind-powered electricity generation unit, is to represent the electrical distance infinity of trouble spot and double-fed fan motor unit at 0 o'clock or do not break down, and is to represent machine end three phase short circuit fault at 1 o'clock.U _Sf=(1-K) U _s, then stator voltage is expressed as after the fault:

${\stackrel{\&RightArrow;}{u}}_s(t\&GreaterEqual;t_0)=(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}---\left(10\right)$

The stator magnetic linkage steady-state component is after getting fault by above stator voltage:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(11\right)$

Electrical network generation three phase short circuit fault moment, the double-fed wind turbine voltage magnitude falls, and the stator magnetic linkage steady-state component reduces; Because magnetic linkage can not suddenly change, for reducing of compensation stator magnetic linkage steady-state component, will produce the aperiodic component of transient state in the stator magnetic linkage; The initial value of this aperiodic component is the poor of stator magnetic linkage steady-state component amplitude before and after the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}\left({\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_{0+}\right)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s\left(t_{0-}\right)-{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}\left(t_0\right)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}---\left(12\right)$

Because the existence of stator resistance, the stator magnetic linkage aperiodic component will be decayed gradually, and damping time constant is:

$T_s^{\&prime;}=\frac{L_s^{\&prime;}}{R_s}---\left(13\right)$

Then behind the electric network fault, the stator magnetic linkage aperiodic component is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(14\right)$

To sum up, after the electrical network generation three phase short circuit fault, double-fed wind turbine magnetic linkage expression formula is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)+{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}+\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{j{\mathrm{\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(15\right)$

Step 3: the fault transient process of analyzing double-fed fan motor unit rotor flux:

During normal operation, by double-fed fan motor unit electro-magnetic transient model, know that rotor flux is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=L_m{\stackrel{\&RightArrow;}{i}}_s+L_r{\stackrel{\&RightArrow;}{i}}_r=L_r({\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r)-L_{\mathrm{r\&sigma;}}{\stackrel{\&RightArrow;}{i}}_s---\left(16\right)$

Got by (1), (2) formula:

${\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r=\frac{{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}){\stackrel{\&RightArrow;}{i}}_s}{{\mathrm{j\&omega;}}_s{L}_m}---\left(17\right)$

With (17) formula substitution (16) formula, the rotor flux expression formula that must be represented by stator voltage, electric current:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}){\stackrel{\&RightArrow;}{i}}_s]-L_{\mathrm{r\&sigma;}}{\stackrel{\&RightArrow;}{i}}_s=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}+{\mathrm{j\&omega;}}_s\frac{L_{\mathrm{r\&sigma;}}{L}_m}{L_r}){\stackrel{\&RightArrow;}{i}}_s]---\left(18\right)$

The stator current space vector

Amplitude be expressed as I _s,

Be the stator current vector, then the double-fed fan motor unit rotor flux instantaneous value of moment is before the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_0{t}_0}-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}+{\mathrm{j\&omega;}}_s\frac{L_{\mathrm{r\&sigma;}}{L}_m}{L_r}){\stackrel{\&CenterDot;}{I}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}]---\left(19\right)$

The control of frequency converter during ignoring thinks that the crow bar protection was moved at once when fault occured, and rotor voltage sports zero, and corresponding rotor flux steady-state component bust is zero; According to the magnetism chain conservation principle, induce the rotor flux aperiodic component of relative stationary rotor in the rotor winding, because the existence of rotor resistance, this aperiodic component will be decayed gradually, damping time constant is:

$T_r^{\&prime;}=\frac{L_r^{\&prime;}}{R_r}---\left(20\right)$

The initial value of rotor flux aperiodic component is the instantaneous value of fault moment rotor flux

Then the rotor flux expression formula is behind the electric network fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(21\right)$

Step 4: set up double-fed wind turbine short-circuit current expression formula;

With stator magnetic linkage expression formula (15) behind the electric network fault and rotor flux expression formula (21) substitution formula (5), get the double-fed wind turbine short-circuit current space vector expression formula after the electrical network three phase short circuit fault:

${\stackrel{\&RightArrow;}{i}}_s={\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}+\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(22\right)$

In the formula, It is the short-circuit current fundamental frequency cycles component of forcing magnetic linkage to produce by stator;

It is the DC component of short-circuit current that is produced by the stator magnetic linkage aperiodic component;

Be the short-circuit current transient state component with rotor frequency, decay that is produced by the rotor flux aperiodic component, be called short-circuit current rotor frequency component;

Step 5: the stator voltage Drop coefficient is revised:

As shown in Figure 1, be the result of calculation of revising rear double-fed wind turbine short-circuit current expression formula and realistic model result's contrast.

Step 6: the electric system simplification equivalence that will contain the double-fed fan motor unit is two branch network:

Take the trouble spot as boundary, the electric system of complexity is reduced to the network that is comprised of some equivalent generators and infinitely great power generatrix, this network is carried out further abbreviation, with all cancellations of all intermediate nodes, only keep power supply node and short circuit node; Select capacity weighting multimachine equivalence method, the double-fed fan motor unit that will be incorporated into the power networks is simplified equivalence; By above network abbreviation, the simplification equivalence of the electric system that the double-fed fan motor unit is accessed is two branch network; Supplied with the branch road of short-circuit current by the double-fed fan motor unit, be called the generator branch road; Supplied with the branch road of short-circuit current by electric system, be called system branch; The short-circuit current at short dot F place

Be generator branch, short-circuit electric current

With the system branch short-circuit current

Sum;

Step 7: to the basic assumption of calculation of short-circuit current: in order to simplify calculating, this paper makes following hypothesis to electric system, wind-powered electricity generation unit, short-circuit conditions and load:

1) Three-phase Power Systems that accesses of wind-powered electricity generation unit is symmetrical;

2) phase angle of all generator electromotive forces is all identical in the electric system, and the simplification of being convenient to system branch is equivalent;

3) magnetic circuit of each element is unsaturated in the electric system, and namely the parameter of each element does not change with electric current, in computation process to use superposition principle;

4) resistance of each element of system branch all omits and disregards, and only when calculating the damping time constant of aperiodic component, just takes into account the effect of resistance;

5) the generator electromotive force of wind-powered electricity generation unit of access electric system is identical, and the wind-powered electricity generation unit of all accesses is carried out the multimachine equivalence;

6) three phase short circuit fault occurs suddenly in electric system under specified running status;

7) load is only done approximate estimation, and represents with constant impedance; The load of load distribution employing 50% is connected on the high voltage bus, and all the other are in the modes of connection in the short dot outside;

Step 8: ask for the short dot periodic component of short-circuit current;

As shown in Figure 2, be the change curve of wind-powered electricity generation unit parameter generator branch, short-circuit current cycle component when identical;

As shown in Figure 3, be the wind-powered electricity generation unit parameter change curve of generator branch, short-circuit current cycle component simultaneously not.

Step 9: ask for the short dot DC component of short-circuit current;

Known by double-fed wind turbine short-circuit current expression formula (24), generator branch, short-circuit electric current aperiodic component be:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}=\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}---\left(29\right)$

Instant of failure, the initial value of this aperiodic component is:

$I_{{\mathrm{sdt}}_0}=\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}---\left(30\right)$

The initial value of the aperiodic component of system branch short-circuit current Poor for the periodic component initial value of the normal running current instantaneous value of fault moment and system branch short-circuit current; The system branch calculation of short-circuit current is more paid close attention to the maximum aperiodic component that may occur; Therefore, get

Amplitude for the system branch periodic component of short-circuit current:

$I_{{\mathrm{gdt}}_0}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(31\right)$

The aperiodic component of system branch short-circuit current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}=\frac{1}{X_{\mathrm{j\&delta;}}}{e}^{-{\mathrm{\&omega;}}_{s}t/T_g}---\left(32\right)$

T wherein _g=X/R is the damping time constant of system branch, and X is the system branch equivalent reactance, and R is the system branch substitutional resistance.The X/R of system branch is different and different with system situation.Short-circuit current practical calculation method in the past carries out statistical study to several large power systems of China and shows, the variation range of X/R is 8 ~ 22, and X/R≤15 in most cases, and therefore the X/R with system branch regards a constant as, and is taken as 15.

The initial value of short dot DC component of short-circuit current is the initial value sum of generator branch road and system branch aperiodic component:

$I_{{\mathrm{fdt}}_0}=I_{{\mathrm{sdt}}_0}+I_{{\mathrm{gdt}}_0}=\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}}---\left(33\right)$

The aperiodic component of short dot short-circuit current is generator branch road and system branch aperiodic component sum:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fd}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}=\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}}{e}^{-{\mathrm{\&omega;}}_{s}t/T_g}=(\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}+\frac{1}{X_{\mathrm{j\&delta;}}})e^{-{\mathrm{\&omega;}}_{s}t/T_a}---\left(34\right)$

Wherein, T _aFor the equivalent time constant of short dot aperiodic component, get

Then

$T_a=\frac{-{\mathrm{\&omega;}}_{s}t}{\ln [{\mathrm{Me}}^{-t/T_s^{\&prime;}}+(1-M)e^{-{\mathrm{\&omega;}}_{s}t/T_g}]}---\left(35\right)$

Step 10: ask for short dot short-circuit current rotor frequency component:

There is the rotor frequency component of decay in the double-fed wind turbine short-circuit current, and do not have this component in the system branch; Therefore, the rotor frequency component of short dot short-circuit current is the short-circuit current rotor frequency component of generator branch road:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fr}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(36\right)$

Its initial value is:

$I_{\mathrm{fr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)---\left(37\right)$

The concrete steps of described step 5 are:

In in the past double-fed wind turbine magnetic linkage fault transient process research, when calculating stator voltage Drop coefficient K, think that all the instant of failure stator voltage amplitude namely drops to steady-state value.In step 2 and step 4.The derivation of double-fed wind turbine short-circuit current expression formula supposes temporarily that also stator voltage drops to steady-state value at once at instant of failure, so that explain the transient state physical process of stator magnetic linkage.In fact, because the existence of nonlinear element in the electrical network, after fault occured, the generator unit stator voltage magnitude did not drop to steady-state value at once, and there is the oscillatory extinction of short time in the centre.A large amount of emulation experiments show, the time of this oscillatory extinction about generally lasting 1 cycle (0.02s).Therefore, only utilize that stator voltage Drop coefficient K corresponding to stator voltage steady-state value calculates the double-fed wind turbine short-circuit current after the fault, will produce error.

According to the physical significance of double-fed wind turbine magnetic linkage, should utilize the stable state amplitude of stator voltage after the fault to calculate the stator magnetic linkage steady-state component, utilize the instant of failure stator voltage amplitude to calculate the initial value of stator magnetic linkage aperiodic component.Now the stator voltage Drop coefficient in the double-fed wind turbine short-circuit current expression formula is done following correction:

When 1) calculating the stator magnetic linkage aperiodic component, adopt instant of failure stator voltage amplitude U _s(t ₀) corresponding stator voltage Drop coefficient $K_{t_{}}$ ${{}_0}_{}=\frac{U_s-U_s\left(t_0\right)}{U_s};$

When 2) calculating stator magnetic linkage fundamental frequency steady-state component, stator voltage amplitude U when adopting after the fault 0.02s _s(t ₀+ 0.02) corresponding stator voltage Drop coefficient $K_{t_{}}$ ${{}_0+0.02}_{}=\frac{U_s-U_s({\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0+0.02)}{U_s};$

Based on above-mentioned correction, after the electrical network generation three phase short circuit fault, the double-fed wind turbine magnetic linkage is modified to:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)+{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}+\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(23\right)$

With following formula substitution formula (5), get revised double-fed wind turbine short-circuit current space vector expression formula:

${\stackrel{\&RightArrow;}{i}}_s={\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}+\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{\mathrm{<figure-callout\; id="0"\; label="t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">t</figure-callout>}}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(24\right)$

The concrete steps of described step 8 are:

Ask for the computing curve of the periodic component of short-circuit current of generator branch road according to the double-fed wind turbine short-circuit current expression formula of having derived, adopt the amplitude of traditional calculation curve method computing system branch, short-circuit current cycle component, and then get the amplitude of short dot periodic component of short-circuit current;

1) asks for generator branch, short-circuit current cycle component

Known by double-fed wind turbine short-circuit current expression formula (24), the periodic component of generator branch, short-circuit electric current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}---\left(25\right)$

Its amplitude is:

$I_{\mathrm{sf}}=\frac{(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}})U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}=\frac{U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}/(1-{\mathrm{<figure-callout\; id="0"\; label="K\; t"\; filenames="HDA00002098720200011.png"\; state="\{\{state\}\}">K</figure-callout>}}_{t_{}})}=\frac{U_s}{X_K}---\left(26\right)$

Wherein

ω _sL ' _sReflected the transient state reactance of equivalent double-fed fan motor unit,

The electrical distance that has reflected trouble spot and double-fed fan motor unit; Known amplitude and the X of generator branch, short-circuit current cycle component by (26) formula _KRelevant, form one with X _KFor the transverse axis variable, with periodic component of short-circuit current amplitude I _SfChange curve for longitudinal axis variable;

The parameter of double-fed fan motor unit may be different with fault condition owing to being incorporated into the power networks, same X _KBe worth corresponding I _SfMay be different, will form many change curves, ask for the mean value of these curves as the computing curve of generator branch, short-circuit current cycle component.

2) ask for the system branch periodic component of short-circuit current

As equivalent reactance, disregarding loads in the network makes equivalent network, is the simplified network that contains generator electromotive force node and short dot with its abbreviation, gets each power supply to the transfer impedance X between short dot with each generator subtranient reactance in the electric system _{J δ}The amplitude of system branch periodic component of short-circuit current is:

$I_{\mathrm{gf}}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(27\right)$

I _GfBe transfer impedance X _{I δ}Function, form with X _{I δ}Be the transverse axis variable, with the amplitude I of system branch periodic component of short-circuit current _GfComputing curve for longitudinal axis variable.

3) amplitude of short dot periodic component of short-circuit current

The short dot periodic component of short-circuit current is generator branch road and system branch periodic component of short-circuit current sum.Therefore the amplitude I of short dot periodic component of short-circuit current _FfFor:

$I_{\mathrm{ff}}=I_{\mathrm{sf}}+I_{\mathrm{gf}}=\frac{U_s}{X_K}+\frac{1}{X_{\mathrm{j\&delta;}}}---\left(28\right)$

According to ω _sL ' _sWith Ask for X _KAfter, determine I from generator branch, short-circuit current cycle components operation curve _SfTry to achieve transfer impedance X _{J δ}After, determine I from the computing curve of system branch periodic component of short-circuit current _Gf, and then the amplitude I of short dot periodic component of short-circuit current _Ff

Although above-mentionedly by reference to the accompanying drawings the specific embodiment of the present invention is described; but be not limiting the scope of the invention; one of ordinary skill in the art should be understood that; on the basis of technical scheme of the present invention, those skilled in the art do not need to pay various modifications that creative work namely makes or distortion still in protection scope of the present invention.

Claims (8)

1. a Power System Shortcuts Current calculation method that contains the double-fed fan motor unit is characterized in that, concrete steps are:

Step 1: utilize the electro-magnetic transient model of the lower double-fed fan motor unit of static coordinate axle system, set up the relation of double-fed fan motor unit stator and rotor electric current and magnetic linkage;

Step 2: after the electrical network three phase short circuit fault occurs, analyze the fault transient process of double-fed wind turbine magnetic linkage and the fault transient process of rotor flux, determine stator voltage, stator magnetic linkage and rotor flux behind the electric network fault, thereby set up double-fed wind turbine short-circuit current space vector expression formula;

Step 3: the electric system simplification equivalence that will contain the double-fed fan motor unit is two branch network, is generator branch road and system branch;

Step 4: periodic component, aperiodic component and the rotor frequency component of determining generator branch road stator short-circuit current according to stator short-circuit current space vector expression formula;

Step 5: with each generator subtranient reactance in the electric system as equivalent reactance, disregarding loads in the network makes equivalent network, be the simplified network that contains generator electromotive force node and short dot with its abbreviation, get each power supply to the transfer impedance between short dot, determine the amplitude of system branch periodic component of short-circuit current and the aperiodic component of system branch short-circuit current according to transfer impedance;

Step 6: with periodic component of short-circuit current, the respectively addition of aperiodic component of periodic component, aperiodic component and the system branch of generator branch road stator short-circuit current, be expressed as again the short-circuit current at short dot F place with the rotor frequency component of generator branch road

2. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, the concrete steps of described step 1 are:

The electro-magnetic transient model of double-fed fan motor unit:

${\stackrel{\&RightArrow;}{u}}_s=R_s{\stackrel{\&RightArrow;}{i}}_s+\frac{d{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{\mathrm{dt}}---\left(1\right)$

${\stackrel{\&RightArrow;}{u}}_r=R_r{\stackrel{\&RightArrow;}{i}}_r+\frac{d{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{\mathrm{dt}}+{\mathrm{j\&omega;}}_r{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r---\left(2\right)$

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s=L_s{\stackrel{\&RightArrow;}{i}}_s+L_m{\stackrel{\&RightArrow;}{i}}_r---\left(3\right)$

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=L_m{\stackrel{\&RightArrow;}{i}}_s+L_r{\stackrel{\&RightArrow;}{i}}_r---\left(4\right)$

Wherein, subscript s and r represent respectively stator winding and rotor winding; Be stator voltage,

Be rotor voltage, Be stator current,

Be rotor current,

Be stator magnetic linkage, Be rotor flux, R _sBe stator resistance, R _rBe rotor resistance, L _sBe stator inductance, L _rBe inductor rotor; L _s=L _{S σ}+ L _m, L _r=L _{R σ}+ L _m, L _{S σ}, L _{R σ}Be respectively the leakage inductance of stator, rotor winding, L _mBe magnetizing inductance; ω _rElectric angle speed for rotor; For the stator and rotor electric current of describing the double-fed fan motor unit and the relation of magnetic linkage, (3) formula and (4) formula are rewritten as:

${\stackrel{\&RightArrow;}{i}}_s=\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{{L_s}^{\&prime;}}-k_r\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{{L_s}^{\&prime;}}---\left(5\right)$

${\stackrel{\&RightArrow;}{i}}_r={-k}_s\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s}{{L_r}^{\&prime;}}+\frac{{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r}{{L_r}^{\&prime;}}---\left(6\right)$

In the formula, L _s' and L _r' be respectively the transient state inductance of stator, rotor winding, k _sAnd k _rBe respectively the inductive coupling coefficient of stator, rotor; And ${L_s}^{\&prime;}=L_s-\frac{{L_m}^2}{L_r}=L_{\mathrm{s\&sigma;}}+\frac{L_{\mathrm{r\&sigma;}}\&CenterDot;L_m}{L_{\mathrm{r\&sigma;}}+L_m},$ $k_r=\frac{L_m}{L_r},$ ${L_r}^{\&prime;}=L_r-\frac{{L_m}^2}{L_s}=L_{\mathrm{r\&sigma;}}+\frac{L_{\mathrm{s\&sigma;}}\&CenterDot;L_m}{L_{\mathrm{s\&sigma;}}+L_m},$ $k_s=\frac{L_m}{L_s}.$

3. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, try to achieve stator current behind the electric network fault and the concrete steps of electronics magnetic linkage in the described step 2 and be:

When the double-fed fan motor unit normally moves, stator voltage

Amplitude is expressed as U _s, rotating speed is the synchronous electric angular velocity omega _sRotating vector:

${\stackrel{\&RightArrow;}{u}}_s=U_s{e}^{\mathrm{j\&alpha;}}{e}^{j{\mathrm{\&omega;}}_{s}t}={\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}---\left(7\right)$

In the formula,

Be the stator voltage vector, α is the initial phase angle of a phase voltage, and j is plural number, and t represents the time;

T represents the time, t ₀The expression fault occurs constantly, t＞0, t ₀0; According to formula (1) voltage equation, if ignore the effect of stator resistance, stator magnetic linkage steady-state component over time rate is approximately equal to stator voltage, and then before the electric network fault, stator magnetic linkage is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t<t_0)=\frac{{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(8\right)$

If t=t ₀Constantly, electrical network generation three phase short circuit fault, stator voltage amplitude drops to steady-state value U _Sf, the stator voltage Drop coefficient is defined as:

$K=\frac{U_s-U_{\mathrm{sf}}}{U_s}---\left(9\right)$

The stator voltage Drop coefficient has embodied the electrical distance between trouble spot and the wind-powered electricity generation unit, is to represent the electrical distance infinity of trouble spot and double-fed fan motor unit at 0 o'clock or do not break down, and is to represent machine end three phase short circuit fault at 1 o'clock; U _Sf=(1-K) U _s, then stator voltage is expressed as after the fault:

${\stackrel{\&RightArrow;}{u}}_s(t\&GreaterEqual;t_0)=(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}---\left(10\right)$

The stator magnetic linkage steady-state component is after getting fault by above stator voltage:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)=\frac{(1-K){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}---\left(11\right)$

Electrical network generation three phase short circuit fault moment, the double-fed wind turbine voltage magnitude falls, and the stator magnetic linkage steady-state component reduces; Because magnetic linkage can not suddenly change, for reducing of compensation stator magnetic linkage steady-state component, will produce the aperiodic component of transient state in the stator magnetic linkage; The initial value of this aperiodic component is the poor of stator magnetic linkage steady-state component amplitude before and after the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}\left(t_{0+}\right)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s\left(t_{0-}\right)-{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}\left(t_0\right)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}}{{\mathrm{j\&omega;}}_s}---\left(12\right)$

Because the existence of stator resistance, the stator magnetic linkage aperiodic component will be decayed gradually, and damping time constant is:

$T_s^{\&prime;}=\frac{L_s^{\&prime;}}{R_s}---\left(13\right)$

Then behind the electric network fault, the stator magnetic linkage aperiodic component is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{K{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(14\right)$

To sum up, after the electrical network generation three phase short circuit fault, double-fed wind turbine magnetic linkage expression formula is stator magnetic linkage steady-state component and stator magnetic linkage aperiodic component sum after the fault:

4. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, determine in the described step 2 that the concrete steps of rotor flux are behind the electric network fault:

During normal operation, by double-fed fan motor unit electro-magnetic transient model, know that rotor flux is:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r=L_m{\stackrel{\&RightArrow;}{i}}_s+L_r{\stackrel{\&RightArrow;}{i}}_r=L_r({\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r)-L_{\mathrm{r\&sigma;}}{\stackrel{\&RightArrow;}{i}}_s---\left(16\right)$

Got by (1), (2) formula:

${\stackrel{\&RightArrow;}{i}}_s+{\stackrel{\&RightArrow;}{i}}_r=\frac{{\stackrel{\&RightArrow;}{u}}_s-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}){\stackrel{\&RightArrow;}{i}}_s}{{\mathrm{j\&omega;}}_s{L}_m}---\left(17\right)$

With (17) formula substitution (16) formula, obtain the rotor flux expression formula that is represented by stator voltage, stator current;

The stator current space vector

Amplitude be expressed as I _s,

Be the stator current vector, then the double-fed fan motor unit rotor flux instantaneous value of moment is before the fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)=\frac{L_r}{{\mathrm{j\&omega;}}_s{L}_m}[{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_0{t}_0}-(R_s+{\mathrm{j\&omega;}}_s{L}_{\mathrm{s\&sigma;}}+{\mathrm{j\&omega;}}_s\frac{L_{\mathrm{r\&sigma;}}{L}_m}{L_r}){\stackrel{\&CenterDot;}{I}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}]---\left(19\right)$

According to the magnetism chain conservation principle, induce the rotor flux aperiodic component of relative stationary rotor in the rotor winding, because the existence of rotor resistance, this aperiodic component will be decayed gradually, damping time constant is:

$T_r^{\&prime;}=\frac{L_r^{\&prime;}}{R_r}---\left(20\right)$

The initial value of rotor flux aperiodic component is the instantaneous value of fault moment rotor flux

Then the rotor flux expression formula is behind the electric network fault:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(21\right).$

5. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, determine in the described step 2 that the concrete steps of stator short-circuit current space vector expression formula are behind the electric network fault:

When 1) calculating the stator magnetic linkage aperiodic component, adopt instant of failure stator voltage amplitude U _s(t ₀) corresponding stator voltage Drop coefficient $K_{t_0}=\frac{U_s-U_s\left(t_0\right)}{U_s};$

When 2) calculating stator magnetic linkage fundamental frequency steady-state component, stator voltage amplitude U when adopting after the fault 0.02s _s(t ₀+ 0.02) corresponding stator voltage Drop coefficient $K_{t_0+0.02}=\frac{U_s-U_s(t_0+0.02)}{U_s};$

Based on above-mentioned correction, after the electrical network generation three phase short circuit fault, the double-fed wind turbine magnetic linkage is modified to:

${\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_s(t\&GreaterEqual;t_0)={\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sf}}(t\&GreaterEqual;t_0)+{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_{\mathrm{sd}}(t\&GreaterEqual;t_0)=\frac{(1-K_{t_0+0.02}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s}+\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}}{{\mathrm{j\&omega;}}_s}{e}^{-t/T_s^{\&prime;}}---\left(23\right)$

After determining electrical network generation three phase short circuit fault according to the expression formula of revised stator magnetic linkage and rotor flux, revised double-fed wind turbine short-circuit current.

6. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, the concrete steps of described step 4 are:

1) asks for generator branch road stator periodic component of short-circuit current;

After electrical network generation three phase short circuit fault, according to revised double-fed wind turbine short-circuit current, determine that the periodic component of generator branch current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sf}}=\frac{(1-K_{t_0+0.02}){\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_{s}t}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}---\left(25\right)$

Its amplitude is:

$I_{\mathrm{sf}}=\frac{(1-K_{t_0+0.02})U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}=\frac{U_s}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}/(1-K_{t_0+0.02})}=\frac{U_s}{X_K}---\left(26\right)$

Wherein

ω _sL ' _sReflected the transient state reactance of equivalent double-fed fan motor unit,

The electrical distance that has reflected trouble spot and double-fed fan motor unit;

2) by after the electrical network generation three phase short circuit fault, the double-fed wind turbine current expression knows, the aperiodic component of generator branch current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}=\frac{K_{t_0}{\stackrel{\&CenterDot;}{U}}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}}{{\mathrm{j\&omega;}}_s{L}_s^{\&prime;}}{e}^{-t/T_s^{\&prime;}}---\left(29\right)$

Instant of failure, the initial value of this aperiodic component is:

$I_{{\mathrm{sdt}}_0}=\frac{K_{t_0}{U}_s{e}^{{\mathrm{j\&omega;}}_s{t}_0}}{{\mathrm{\&omega;}}_s{L}_s^{\&prime;}}---\left(30\right)$

3) have the rotor frequency component of decay in the double-fed wind turbine short-circuit current, after electrical network generation three phase short circuit fault, the double-fed wind turbine current expression knows, generator branch, short-circuit electric current rotor frequency component is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fr}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(36\right)$

Its initial value is:

$I_{\mathrm{fr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)---\left(37\right).$

7. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, the concrete steps of described step 5 are:

1) asks for the system branch periodic component of short-circuit current;

As equivalent reactance, disregarding loads in the network makes equivalent network, gets each power supply to the transfer impedance X between short dot with each generator subtranient reactance in the electric system _{J δ}The amplitude of system branch periodic component of short-circuit current is:

$I_{\mathrm{gf}}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(27\right)$

I _GfBe transfer impedance X _{J δ}Function, form with X _{J δ}Be the transverse axis variable, with the amplitude I of system branch periodic component of short-circuit current _GfComputing curve for longitudinal axis variable;

2) initial value of the aperiodic component of system branch electric current

Poor for the periodic component initial value of the system branch current instantaneous value of the normal operation of fault moment and system branch short-circuit current; Get

Amplitude for the system branch DC component of short-circuit current:

$I_{{\mathrm{gdt}}_0}=\frac{1}{X_{\mathrm{j\&delta;}}}---\left(31\right)$

After the electrical network generation three phase short circuit fault, the aperiodic component of system branch electric current is:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}=\frac{1}{X_{\mathrm{j\&delta;}}}{e}^{-{\mathrm{\&omega;}}_{s}t/T_g}---\left(32\right)$

T wherein _g=X/R is the damping time constant of system branch, and X is the system branch equivalent reactance, and R is the system branch substitutional resistance.

8. contain as claimed in claim 1 the Power System Shortcuts Current calculation method of double-fed fan motor unit, it is characterized in that, the concrete steps of described step 6 are:

1) amplitude of short dot periodic component of short-circuit current;

The short dot periodic component of short-circuit current is generator branch, short-circuit electric current and system branch periodic component of short-circuit current sum; Therefore the amplitude I of short dot periodic component of short-circuit current _FfFor:

$I_{\mathrm{ff}}=I_{\mathrm{sf}}+I_{\mathrm{gf}}=\frac{U_s}{X_K}+\frac{1}{X_{\mathrm{j\&delta;}}}---\left(28\right)$

2) initial value of short dot DC component of short-circuit current is the initial value sum of generator branch, short-circuit circuit and system branch short circuit current aperiodic component:

The aperiodic component of short dot short-circuit current is generator branch, short-circuit circuit and system branch short circuit current aperiodic component sum:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fd}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sd}}+{\stackrel{\&RightArrow;}{i}}_{\mathrm{gd}}---\left(34\right)$

3) have the rotor frequency component of decay in the double-fed wind turbine short-circuit current, and do not have this component in the system branch, therefore, the rotor frequency component of short dot short-circuit current is the short-circuit current rotor frequency component of generator branch road:

${\stackrel{\&RightArrow;}{i}}_{\mathrm{fr}}={\stackrel{\&RightArrow;}{i}}_{\mathrm{sr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)e^{{\mathrm{j\&omega;}}_{r}t}{e}^{-t/T_r^{\&prime;}}---\left(36\right)$

Its initial value is:

$I_{\mathrm{fr}}=-\frac{k_r}{L_s^{\&prime;}}{\stackrel{\&RightArrow;}{\mathrm{\&psi;}}}_r\left(t_0\right)---\left(37\right).$

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