CN101969199B - Fault loss estimation method for risk assessment of transient power angle stability - Google Patents

Abstract

The invention relates to fault loss estimation of a power network and provides a practical and effective fault loss estimation method for the risk assessment of transient rotor angle stability. In the technical scheme adopted by the invention, the fault loss estimation method for the risk assessment of the transient rotor angle stability comprises the following steps of: performing the ordinary power flow calculation to obtain the initial condition; performing transient time domain simulation and calculating trace sensitivity; if instability is determined, resolving a transient stability minimum control cost model; and if the convergence conditions that the absolute value of delta PL is less than epsilon and that the absolute value of delta Pg is less than epsilon are not met, repeating the previous steps. The method is mainly applied to the fault loss estimation and management of the power network.

Description

The breakdown loss method of estimation that is used for the transient rotor angle stability risk assessment

Technical field

The present invention relates to the system fault estimated amount of damage, specifically relate to the breakdown loss method of estimation for the transient rotor angle stability risk assessment.

Background technology

Along with the development of China's electric power system, the continuous expansion of Interconnection Scale, the uncertain factor in the system constantly increases, and by risk analysis, the reasonable assessment uncertain factor is significant on the impact of system stable operation ^[1]The difficult point of risk analysis is how accurately to estimate the loss that causes after the fault initiating system unstability, utilizes optimum stable control cost to come the approximate evaluation system loss, for finding the solution of this problem provides a kind of feasible approach ^[1]Document [2] is taked to preset low pressure off-load strategy system is carried out emergency control, to obtain the control cost; Document [3] loss that adjustment expense and load emergency control expense replace system's unstability to cause of taking generator maintenance, start and exert oneself; Document [4] then utilizes Dynamic Security Region that system is carried out prevention and control and emergency control, thus the dynamic unsafe value of risk of the system that obtains.Said method all needs can have certain limitation for the default control strategy of fault when practical application.

Summary of the invention

For overcoming the deficiencies in the prior art, a kind of breakdown loss method of estimation that is used for the transient rotor angle stability risk assessment of practicability and effectiveness is provided, the technical scheme that the present invention takes is that the breakdown loss method of estimation for the transient rotor angle stability risk assessment comprises the steps:

Carry out the ordinary tides flowmeter and calculate, obtain initial condition;

Carry out the transient state time-domain-simulation and calculate trace sensitivity;

If unstable, find the solution the transient stability minimum model of controlling cost;

If do not satisfy the condition of convergence: ‖ Δ P _L‖＜ε and ‖ Δ P _g‖＜ε repeats abovementioned steps.

Described calculating trace sensitivity, computational methods are as follows:

Electrical Power System Dynamic can represent with following differential-algebraic equation:

In the formula, x is system state variables, and y is system's algebraically variable, and λ is system's control variables, x ₀, y ₀The initial value that represents respectively state variable and algebraically variable, ask local derviation to get to λ formula (1):

In the formula,

Represent system trajectory to the sensitivity matrix of control variables λ,

Be respectively differential equation group to the local derviation of state variable, algebraically variable and control variables,

Be respectively Algebraic Equation set to the partial derivative of state variable, algebraically variable and control variables, can try to achieve the preliminary examination condition x of trace sensitivity according to system's initial launch point _λ(t ₀), y _λ(t ₀), formula (1) and formula (2) simultaneous solution can be obtained each trace sensitivity constantly of system, and generator's power and angle is as follows with its meritorious sensitivity relation circular of exerting oneself and loading between meritorious: generator's power and angle is with the sensitivity between its meritorious exerting oneself:

When generator was taked classical model, the Electrical Power System Dynamic equation was as follows:

In the formula, δ _i, ω _i, M _i, P _Gi, E ' _iBe respectively generator amature angle, rotating speed, moment of inertia, mechanical output and built-in potential; G _Ii, G _Ij, B _IjAfter expression only keeps the generator interior nodes respectively, real part and the imaginary part of system's admittance matrix element, m is the generator nodes;

The meritorious P that exerts oneself of generator k is asked respectively at formula (3) two ends _GkPartial derivative can get:

In the formula, Be the sensitivity of generator i merit angle between exerting oneself with generator k is meritorious since in transient process E ' _i, P _GiRemain unchanged, only relevant with the state of system before the fault, therefore A constant in transient process, further, as can be known:

$\frac{{\∂P}_{\mathrm{gi}}}{{\∂P}_{\mathrm{gk}}}=\left\{\begin{array}{cc}1& k=i\\ 0& k\≠i\end{array}\right.---\left(5\right)$

Can be got by Generator end power equation:

Wherein, Be the electric current of generator injection network, S _i, α _iBe respectively apparent power and the power-factor angle of generator output, V _i, θ _iBe respectively generator terminal voltage amplitude and phase angle,

$\left\{\begin{array}{c}{S}_i=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="P\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_{\mathrm{gi}}^{}+{\mathrm{<figure-callout\; id="2"\; label="Q\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">Q</figure-callout>}}_{\mathrm{gi}}^{}}\\ {\mathrm{\&alpha;}}_i=\arctan \left(\frac{Q_{\mathrm{gi}}}{P_{\mathrm{gi}}}\right)\end{array}\right.---\left(7\right)$

Generator built-in potential E ' _iCan be expressed as:

Therefore, have:

$E_i^{\&prime;}=\sqrt{V_i^2+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}---\left(9\right)$

Formula (9) two ends are respectively to P _GkAsk partial derivative:

$\frac{{\&PartialD;E}_i^{\&prime;}}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{1}{2}\&CenterDot;\frac{1}{\sqrt{V_i^2+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}}---\left(10\right)$

$\&CenterDot;(2S_i{\left(\frac{x_d^{\&prime;}}{V_i}\right)}^2\frac{{\&PartialD;S}_i}{\&PartialD;P_{\mathrm{gk}}}+2x_d^{\&prime;}\frac{{\&PartialD;Q}_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})$

Formula (7) two ends are respectively to P _GkAsk partial derivative:

$\frac{{\&PartialD;S}_i}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{1}{\sqrt{{\mathrm{<figure-callout\; id="2"\; label="P\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_{\mathrm{gi}}^{}+Q_{\mathrm{gi}}^2}}(P_{\mathrm{gi}}\frac{{\&PartialD;P}_{\mathrm{gi}}}{{\&PartialD;P}_{\mathrm{gk}}}+Q_i\frac{{\&PartialD;Q}_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})---\left(11\right)$

The generator node is idle, and the injection equation is as follows:

$Q_{\mathrm{gi}}=Q_{\mathrm{Li}}+V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})---\left(12\right)$

N is the system node number, and the following formula two ends are respectively to P _GkAsk partial derivative:

$\begin{array}{c}\frac{\&PartialD;Q_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}=V_i\underset{j\&NotElement;G}{\overset{n}{\underset{j=1}{\mathrm{\&Sigma;}}}}\frac{{\&PartialD;V}_j}{\&PartialD;P_{\mathrm{gk}}}(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})\\ +V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}}+B_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}})(\frac{{\&PartialD;\mathrm{\&theta;}}_i}{\&PartialD;P_{\mathrm{gk}}}-\frac{{\&PartialD;\mathrm{\&theta;}}_j}{\&PartialD;P_{\mathrm{gk}}})\end{array}---\left(13\right)$

Wherein, G represents the generator node set,

Be power flow equation Jacobi inverse matrix element, formula (13), formula (11) substitution formula (10) can be tried to achieve

And then simultaneous solution formula (3)-(4) just can be tried to achieve

For the calculating of sensitivity coefficient initial value, because of ω _i(t ₀)=1, so have:

$\frac{{\&PartialD;\mathrm{\&omega;}}_i}{{\&PartialD;P}_{\mathrm{gk}}}\left(t_0\right)=0---\left(14\right)$

Generator machinery power when knowing stable state again is as follows:

$P_{\mathrm{gi}}=\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)---\left(15\right)$

Formula (15) both sides are respectively to P _GkAsk partial derivative, can get:

$\begin{array}{c}\frac{{\&PartialD;P}_{\mathrm{gi}}}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{V_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)\frac{{\&PartialD;E}_i^{\&prime;}}{{\&PartialD;P}_{\mathrm{gk}}}\\ +\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\cos ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{{\&PartialD;P}_{\mathrm{gk}}}-\frac{\&PartialD;{\mathrm{\&theta;}}_i}{{\&PartialD;P}_{\mathrm{gk}}})\end{array}---\left(16\right)$

Therefrom can solve

With what tried to achieve

Substitution can obtain

Sensitivity between generator's power and angle is meritorious with load:

Suppose that load takes the constant-impedance model, formula (3) two ends are respectively to the burden with power P of load bus k _LkAsk partial derivative:

Wherein,

Find the solution and find the solution

Similar, do not giving unnecessary details at this; G _Ii, G _Ij, B _IjAll can be expressed as each load bus Equivalent admittance

The function of (q=1L l, l represent the load bus number):

$\left\{\begin{array}{c}{B}_{\mathrm{ij}}=h_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ij}}=p_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ii}}=q_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\end{array}\right.---\left(18\right)$

Formula (18) two ends are respectively to P _LkAsk partial derivative:

$\left\{\begin{array}{c}\frac{\&PartialD;B_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{{\&PartialD;h}_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{\&PartialD;h_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}})\\ \frac{{\&PartialD;G}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;p_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{{\&PartialD;p}_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}})\\ \frac{{\&PartialD;G}_{\mathrm{ii}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;q_{\mathrm{ij}}}{{\&PartialD;B}_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{\&PartialD;q_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;G}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}})\end{array}\right.---\left(19\right)$

For the constant-impedance load, Equivalent admittance can be expressed as:

$\left\{\begin{array}{c}{G}_q^{\mathrm{eq}}=\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\\ B_q^{\mathrm{eq}}=\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\end{array}\right.---\left(20\right)$

Formula (20) two ends are respectively to P _LkAsk partial derivative, can get:

$\frac{{\&PartialD;G}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=\left\{\begin{array}{cc}\frac{1}{{\left(V_q\right)}^2}-2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{\&PartialD;V_q}{\&PartialD;P_{\mathrm{Lq}}}& q=k\\ -2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{{\&PartialD;V}_q}{\&PartialD;P_{\mathrm{Lk}}}& q\&NotEqual;k\end{array}\right.---\left(21\right)$

$\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=-2\&CenterDot;\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{{\&PartialD;V}_q}{\&PartialD;P_{\mathrm{Lk}}}---\left(22\right)$

Formula (21)-(22) substitution (19) can be tried to achieve

With

With what try to achieve

With

Substitution formula (17), vertical (3) in parallel can try to achieve generator's power and angle with the meritorious sensitivity of load, the computational methods of initial value and generated power are exerted oneself sensitivity calculations roughly the same, can obtain equally generator's power and angle with the sensitivity between reactive load according to this step

When system control parameters changes, utilize and to have tried to achieve trace sensitivity, can be by the variable quantity at following formula approximate evaluation generator i merit angle:

$\mathrm{\&Delta;}{\mathrm{\&delta;}}_i=\underset{k=1}{\overset{m}{\mathrm{\&Sigma;}}}\left(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{gk}}}\mathrm{\&Delta;}{P}_{\mathrm{gk}}\right)+\underset{k=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;Q_{\mathrm{Lk}}}\mathrm{\&Delta;}{P}_{\mathrm{Lk}}+\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;Q_{\mathrm{Lk}}}\mathrm{\&Delta;}{Q}_{\mathrm{Lk}})---\left(23\right).$

The described transient stability minimum model of controlling cost of finding the solution is, trace sensitivity and optimal load flow are combined, and finds the solution following nonlinear optimal problem, gets final product to such an extent that the transient stability minimum is controlled cost:

$\min f({\mathrm{\&Delta;P}}_g^{+},{\mathrm{\&Delta;P}}_g^{-},{\mathrm{\&Delta;P}}_L)=C_g^{+}\&CenterDot;\mathrm{\&Delta;}{P}_g^{+}+C_g^{-}\&CenterDot;\mathrm{\&Delta;}{P}_g^{-}+C_L\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(24a\right)$

$s.t.{\mathrm{<figure-callout\; id="0"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^{}+\mathrm{\&Delta;}{P}_g^{+}-\mathrm{\&Delta;}{P}_g^{-}-{\mathrm{<figure-callout\; id="0"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L^{}+\mathrm{\&Delta;}{P}_L-P(V,\mathrm{\&theta;})=0---\left(24b\right)$

$Q_g-{\mathrm{<figure-callout\; id="0"\; label="Q\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">Q</figure-callout>}}_L^{}+\mathrm{\&Delta;}{Q}_L-Q(V,\mathrm{\&theta;})=0---\left(24c\right)$

S(V，θ)≤S _max (24d)

V _min≤V≤V _max (24e)

$0\&le;\mathrm{\&Delta;}{P}_g^{+}\&le;P_{g_{\max }}^{+}---\left(24f\right)$

$0\&le;\mathrm{\&Delta;}{P}_g^{-}\&le;P_{g_{\max }}^{-}---\left(24g\right)$

$Q_{g_{\min }}\&le;Q_g\&le;Q_{g_{\max }}---\left(24h\right)$

$0\&le;\mathrm{\&Delta;}{P}_L\&le;P_{L_{\max }}---\left(24i\right)$

-δ _lim≤δ ⁰(t _sen)+S(t _sen)·Δu≤δ _lim (24j)

Wherein,

With

Represent respectively the generated power upper mediation downward modulation amount of exerting oneself, Δ P _LWith Δ Q _LThe expression load is gained merit and idle reduction amount respectively;

C _LRepresent respectively generated power exert oneself rise expense, downward modulation expense and load is meritorious cuts the cost; The front generated power of expression control is exerted oneself, is loaded and gains merit and idle initial value respectively;

Represent respectively generated power exert oneself rise amount, downward modulation amount and load reduction amount maximum; δ _LimThe limiting value at the relative merit of expression generator angle, t _SenThe relative merit angle that the constraint of expression transient state is selected and trace sensitivity are constantly; δ ⁰Expression t _SenRelative merit angle between the generator is a dimensional vector constantly; Δ u represents control variables, comprises that generated power exerts oneself and load meritorious, is the b dimensional vector; S represents t _SenConstantly the sensitivity matrix between generator relative merit angle and control variables is a*b dimension matrix, and a represents the number of the Transient Stability Constraints that comprises in the optimal models, and b represents the control variables number;

Above-mentioned model Chinese style (24a) expression is controlled cost for certain fault, comprises cost, the downward modulation generator output cost that raises generator output and reduces the load cost.For the sake of simplicity, this model all is thought of as three kinds of control costs the linear function of controlled quentity controlled variable, further also can adopt secondary, linear segmented cost function ^[17], its solution procedure and above-mentioned model class are seemingly; (24b), (24c) be respectively meritorious and the reactive power flow equation; (24d) be the Line Flow constraint; (24e) be the node voltage constraint; (24f), (24g), (24h) be respectively generated power exert oneself rise amount constraint, the constraint of downward modulation amount and idle constraint; (24i) expression load reduction amount constraint; (24j) expression transient state merit angle constraint.

For load bus, when loading meritorious the reduction, reactive load reduction amount is determined by following formula:

$\mathrm{\&Delta;}{Q}_L=\frac{Q_L^0}{{\mathrm{<figure-callout\; id="0"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L^{}}\&CenterDot;\mathrm{\&Delta;}{P}_L=\tan \mathrm{\&theta;}\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(25\right)$

Namely in control procedure, load power factor is thought of as constant, and θ represents the power-factor angle of loading in the following formula;

Formula (24) belongs to and typically contains the nonlinear restriction optimal model, utilizes ‖ Δ P _L‖＜ε and ‖ Δ P _g‖＜ε when two conditions satisfy simultaneously, shows system without adjustable generator and load as convergence criterion, then finishes.

The described transient stability minimum of finding the solution is controlled cost in the model, in formula (24j), has:

$\mathrm{\&Delta;u}={\left[\begin{array}{ccc}{\mathrm{\&Delta;P}}_g^{+}& \mathrm{\&Delta;}{P}_g^{-}& \mathrm{\&Delta;}{P}_L\end{array}\right]}^T---\left(26\right)$

Wherein:

$\left\{\begin{array}{c}{P}_g^{+}=\left[\begin{array}{cccc}\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^1& \mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^{2}L& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{+}L& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{+}\end{array}\right]\\ \mathrm{\&Delta;}{P}_g^{-}=\left[\begin{array}{cccc}\mathrm{\&Delta;}{P}_{g1}^{-}& \mathrm{\&Delta;}{P}_{g2}^{-}L& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{-}L& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{-}\end{array}\right]\\ \mathrm{\&Delta;}{P}_L=\left[\begin{array}{cccc}\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L& \mathrm{\&Delta;}{P}_{L2}L& \mathrm{\&Delta;}{P}_{\mathrm{Lq}}L& \mathrm{\&Delta;}{P}_{\mathrm{Ll}}\end{array}\right]\end{array}\right.---\left(27\right)$

The formation of sensitivity matrix S is suc as formula shown in (28):

$s\left(t_{\mathrm{sen}}\right)=\left[\begin{array}{cccccccccccc}{S}_{g1}^{12}& S_{g2}^{12}L& S_{\mathrm{gp}}^{12}L& S_{\mathrm{gm}}^{12}& -S_{g1}^{12}& -S_{g2}^{12}L& -S_{\mathrm{gp}}^{12}L& -S_{\mathrm{gm}}^{12}& S_{L1}^{12}& S_{L2}^{12}L& S_{\mathrm{Lq}}^{12}L& S_{\mathrm{Ll}}^{12}\\ S_{g1}^{13}& S_{g2}^{13}L& S_{\mathrm{gp}}^{13}L& S_{\mathrm{gm}}^{13}& -S_{g1}^{13}& -S_{g2}^{13}L& -S_{\mathrm{gp}}^{13}L& -S_{\mathrm{gm}}^{13}& S_{L1}^{13}& S_{L2}^{13}L& S_{\mathrm{Lq}}^{13}L& S_{\mathrm{Ll}}^{13}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{23}& S_{g2}^{23}L& S_{\mathrm{gp}}^{23}L& S_{\mathrm{gm}}^{23}& -S_{g1}^{23}& -S_{g2}^{23}L& -S_{\mathrm{gp}}^{23}L& -S_{\mathrm{gm}}^{23}& S_{L1}^{23}& S_{L2}^{23}L& S_{\mathrm{Lq}}^{23}L& S_{L1}^{23}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{\mathrm{ij}}& S_{g2}^{\mathrm{ij}}L& S_{\mathrm{gp}}^{\mathrm{ij}}L& S_{\mathrm{gm}}^{\mathrm{ij}}& -S_{g1}^{\mathrm{ij}}& -S_{g2}^{\mathrm{ij}}L& -S_{\mathrm{gp}}^{\mathrm{ij}}L& -S_{\mathrm{gm}}^{\mathrm{ij}}& S_{L1}^{\mathrm{ij}}& S_{L2}^{\mathrm{ij}}L& S_{\mathrm{Lq}}^{\mathrm{ij}}L& S_{\mathrm{Ll}}^{\mathrm{ij}}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{m-1,g}& S_{g2}^{m-1,g}L& S_{\mathrm{gp}}^{m-\mathrm{1,2}}L& S_{\mathrm{gm}}^{m-\mathrm{1,2}}& -S_{g1}^{m-\mathrm{1,2}}& -S_{g2}^{m-\mathrm{1,2}}L& -S_{\mathrm{gp}}^{m-\mathrm{1,2}}L& -S_{\mathrm{gm}}^{m-\mathrm{1,2}}& S_{L1}^{m-\mathrm{1,2}}& S_{L2}^{m-\mathrm{1,2}}L& S_{\mathrm{Lq}}^{m-\mathrm{1,2}}L& S_{\mathrm{Ll}}^{m-\mathrm{1,2}}\end{array}\right]---\left(28\right)$

Wherein,

Sensitivity between relative merit angle between expression generator i and the j is meritorious with generator p has

$S_{\mathrm{gp}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{gp}}}\left(t_{\mathrm{sen}}\right)---\left(29\right)$

Because load is processed according to constant power factor, therefore the sensitivity of the relative merit of generator angle with load or burden without work is folded in the burden with power sensitivity according to corresponding power factor,

Relative merit angle between expression generator i and the j has with the sensitivity between load q

$S_{\mathrm{Lq}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)+\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;Q_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)\&CenterDot;\tan \left({\mathrm{\&theta;}}_q\right)---\left(30\right)$

In formula (30), θ _qThe power-factor angle of expression load q.

${\mathrm{\&delta;}}^0\left(t_{\mathrm{sen}}\right)={\left[\begin{array}{ccccc}{\mathrm{\&delta;}}_{12}^0& {\mathrm{\&delta;}}_{13}^{0}L& {\mathrm{\&delta;}}_{23}^{0}L& {\mathrm{\&delta;}}_{\mathrm{ij}}^{0}L& {\mathrm{\&delta;}}_{m-1,m}^0\end{array}\right]}^T---\left(31\right)$

Wherein,

Initial phase between expression generator i and the j has the merit angle:

${\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="0"\; label="ij"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">ij</figure-callout>}}^0={\mathrm{\&delta;}}_{\mathrm{ij}}^0\left(t_{\mathrm{sen}}\right)---\left(32\right).$

Its characteristics of the present invention are:

1, utilizes the variable quantity at the relative merit of generator angle after the trace sensitivity technology approximate evaluation system parameter variations, and then Transient Stability Constraints introduced the optimal load flow model, reduce as control device with generator adjustment and load, realize finding the solution of minimum control cost.

2, by finding the solution the transient stability minimum model of controlling cost, when the generator reserve capacity is sufficient, preferentially adjusts generator output (controlling cost lower) and keep system stability; And when the generator reserve capacity is not enough, then guarantee system stability by the load reduction measure of controlling cost higher, its optimizing process reality of more fitting.

3, amount of calculation is reduced, and is practical reliable.

Description of drawings

Fig. 1 angle stability control flow chart.

The relative power-angle curve schematic diagram of Fig. 2 generator.

Figure 31 4 node system generator's power and angle curves.

Generator's power and angle curve (sight 1) after Figure 41 4 node system adjustment.

Generator's power and angle curve (sight 2) after Figure 51 4 node system adjustment.

Figure 63 9 node system generator's power and angle curves.

Generator's power and angle curve (sight 1) after Figure 73 9 node systems are adjusted.

Generator's power and angle curve (sight 2) after Figure 83 9 node system adjustment.

Embodiment

When carrying out the electric power system transient stability risk assessment, the system loss that science estimation fault causes is its most crucial problem.The loss that the present invention causes fault is regarded as system is converted into the needed minimum control cost of stable state from labile state, and by the trace sensitivity technology, the optimal power flow problems that the problems referred to above is converted into a considering transient scleronomic constraint realizes finding the solution.At last, correctness and the validity of this paper institute extracting method of having utilized the system verifications such as IEEE-14, New England-39 node.

Further describe the present invention below in conjunction with drawings and Examples.

Trace sensitivity can be in transient process each constantly provide system abundant operation information, and not limited by component models ^[5], in recent years at the optimal load flow (OTS) of considering transient scleronomic constraint ^[6-7]Optimization (comprises prevention and control with Transient Stability Control ^[8-12]And emergency control ^[13-14]) etc. the aspect application of succeeding.This paper causes fault and is considered as loss after system's unstability system is converted into the required minimum control cost of stable state from labile state, it is converted into an OTS problem is found the solution.Transient Stability Constraints in the fault takes in by the relative merit of generator angle, and latter utilizes that trace sensitivity is approximate to be estimated ^[15]Generator output is adjusted and load is reduced as control device, and different control devices has different control costs (cost).When the generator reserve capacity is sufficient, preferentially adjusts generator output (controlling cost lower) and keep system stability; And when the generator reserve capacity is not enough, then guarantee system stability by the load reduction measure of controlling cost higher, its optimizing process reality of more fitting.Wherein, the ordinary tides flowmeter is calculated with the transient state time-domain-simulation and all can be utilized existing computational tool to be calculated ^[18]The calculating that the calculating of trace sensitivity and transient stability minimum are controlled cost is its core.

1 trace sensitivity

Electric power system trace sensitivity method is the earliest by propositions such as M.A.Pai ^[5], it think in the transient process system state variables homologous ray parameter and the state variable initial value is approximate satisfies linear relationship, and linear coefficient is the trace sensitivity of system.

Electrical Power System Dynamic can represent with following differential-algebraic equation:

In the formula, x is system state variables, and y is system's algebraically variable, and λ is system's control variables, x ₀, y ₀The initial value that represents respectively state variable and algebraically variable.Ask local derviation to get to λ formula (1):

In the formula,

Represent system trajectory to the sensitivity matrix of control variables λ,

Be respectively differential equation group to the local derviation of state variable, algebraically variable and control variables,

Be respectively Algebraic Equation set to the partial derivative of state variable, algebraically variable and control variables.Can try to achieve the preliminary examination condition x of trace sensitivity according to system's initial launch point _λ(t ₀), y _λ(t ₀), formula (1) and formula (2) simultaneous solution can be obtained each trace sensitivity constantly of system.Generator's power and angle is as follows with its meritorious sensitivity relation circular of exerting oneself and loading between gaining merit.

1.1 generator's power and angle is with the sensitivity between its meritorious exerting oneself

When generator was taked classical model, the Electrical Power System Dynamic equation was as follows:

In the formula, δ _i, ω _i, M _i, P _Gi, E ' _iBe respectively generator amature angle, rotating speed, moment of inertia, mechanical output and built-in potential; G _Ii, G _Ij, B _IjAfter expression only keeps the generator interior nodes respectively, real part and the imaginary part of system's admittance matrix element, m is the generator nodes.

The meritorious P that exerts oneself of generator k is asked respectively at formula (3) two ends _GkPartial derivative can get:

In the formula,

Be the sensitivity of generator i merit angle between exerting oneself with generator k is meritorious.Because E ' in transient process _i, P _GiRemain unchanged, only relevant with the state of system before the fault, therefore

It is a constant in transient process.Further, as can be known:

$\frac{{\&PartialD;P}_{\mathrm{gi}}}{{\&PartialD;P}_{\mathrm{gk}}}=\left\{\begin{array}{cc}1& k=i\\ 0& k\&NotEqual;i\end{array}\right.---\left(5\right)$

Can be got by Generator end power equation:

Wherein,

Be the electric current of generator injection network, S _i, α _iBe respectively apparent power and the power-factor angle of generator output, V _i, θ _iBe respectively generator terminal voltage amplitude and phase angle.

$\left\{\begin{array}{c}{S}_i=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="P\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_{\mathrm{gi}}^{}+{\mathrm{<figure-callout\; id="2"\; label="Q\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">Q</figure-callout>}}_{\mathrm{gi}}^{}}\\ {\mathrm{\&alpha;}}_i=\arctan \left(\frac{Q_{\mathrm{gi}}}{P_{\mathrm{gi}}}\right)\end{array}\right.---\left(7\right)$

Generator built-in potential E ' _iCan be expressed as:

Therefore, have:

$E_i^{\&prime;}=\sqrt{{\mathrm{<figure-callout\; id="2"\; label="V\; i"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">V</figure-callout>}}_i^{}+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}---\left(9\right)$

Formula (9) two ends are respectively to P _GkAsk partial derivative:

$\frac{{\&PartialD;E}_i^{\&prime;}}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{1}{2}\&CenterDot;\frac{1}{\sqrt{{\mathrm{<figure-callout\; id="2"\; label="V\; i"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">V</figure-callout>}}_i^{}+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}}---\left(10\right)$

$\&CenterDot;(2S_i{\left(\frac{x_d^{\&prime;}}{V_i}\right)}^2\frac{{\&PartialD;S}_i}{\&PartialD;P_{\mathrm{gk}}}+2x_d^{\&prime;}\frac{{\&PartialD;Q}_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})$

Formula (7) two ends are respectively to P _GkAsk partial derivative:

$\frac{{\&PartialD;S}_i}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{1}{\sqrt{{\mathrm{<figure-callout\; id="2"\; label="P\; gi"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_{\mathrm{gi}}^{}+Q_{\mathrm{gi}}^2}}(P_{\mathrm{gi}}\frac{{\&PartialD;P}_{\mathrm{gi}}}{{\&PartialD;P}_{\mathrm{gk}}}+Q_i\frac{{\&PartialD;Q}_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})---\left(11\right)$

The generator node is idle, and the injection equation is as follows:

$Q_{\mathrm{gi}}=Q_{\mathrm{Li}}+V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})---\left(12\right)$

N is the system node number, and the following formula two ends are respectively to P _GkAsk partial derivative:

$\begin{array}{c}\frac{\&PartialD;Q_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}=V_i\underset{j\&NotElement;G}{\overset{n}{\underset{j=1}{\mathrm{\&Sigma;}}}}\frac{{\&PartialD;V}_j}{\&PartialD;P_{\mathrm{gk}}}(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})\\ +V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}}+B_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}})(\frac{{\&PartialD;\mathrm{\&theta;}}_i}{\&PartialD;P_{\mathrm{gk}}}-\frac{{\&PartialD;\mathrm{\&theta;}}_j}{\&PartialD;P_{\mathrm{gk}}})\end{array}---\left(13\right)$

Wherein, G represents the generator node set,

It is power flow equation Jacobi inverse matrix element.Formula (13), formula (11) substitution formula (10) can be tried to achieve And then simultaneous solution formula (3)-(4) just can be tried to achieve

For the calculating of sensitivity coefficient initial value, because of ω _i(t ₀)=1, so have:

$\frac{{\&PartialD;\mathrm{\&omega;}}_i}{{\&PartialD;P}_{\mathrm{gk}}}\left(t_0\right)=0---\left(14\right)$

Generator machinery power when knowing stable state again is as follows:

$P_{\mathrm{gi}}=\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)---\left(15\right)$

Formula (15) both sides are respectively to P _GkAsk partial derivative, can get:

$\begin{array}{c}\frac{{\&PartialD;P}_{\mathrm{gi}}}{{\&PartialD;P}_{\mathrm{gk}}}=\frac{V_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)\frac{{\&PartialD;E}_i^{\&prime;}}{{\&PartialD;P}_{\mathrm{gk}}}\\ +\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\cos ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{{\&PartialD;P}_{\mathrm{gk}}}-\frac{\&PartialD;{\mathrm{\&theta;}}_i}{{\&PartialD;P}_{\mathrm{gk}}})\end{array}---\left(16\right)$

Therefrom can solve With what tried to achieve

Substitution can obtain

The sensitivity between 1.2 generator's power and angle is meritorious with load

Suppose that load takes the constant-impedance model, formula (3) two ends are respectively to the burden with power P of load bus k _LkAsk partial derivative:

Wherein,

Find the solution and find the solution

Similar, do not giving unnecessary details at this; G _Ii, G _Ij, B _IjAll can be expressed as each load bus Equivalent admittance

The function of (q=1L l, l represent the load bus number):

$\left\{\begin{array}{c}{B}_{\mathrm{ij}}=h_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ij}}=p_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ii}}=q_{\mathrm{ij}}({\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">B</figure-callout>}}_1^{\mathrm{eq}},{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_2^{\mathrm{eq}},L,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},L,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\end{array}\right.---\left(18\right)$

Formula (18) two ends are respectively to P _LkAsk partial derivative:

$\left\{\begin{array}{c}\frac{\&PartialD;B_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{{\&PartialD;h}_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{\&PartialD;h_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}})\\ \frac{{\&PartialD;G}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;p_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{{\&PartialD;p}_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}})\\ \frac{{\&PartialD;G}_{\mathrm{ii}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;q_{\mathrm{ij}}}{{\&PartialD;B}_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{{\&PartialD;P}_{\mathrm{Lk}}}+\frac{\&PartialD;q_{\mathrm{ij}}}{{\&PartialD;G}_q^{\mathrm{eq}}}\&CenterDot;\frac{{\&PartialD;G}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}})\end{array}\right.---\left(19\right)$

For the constant-impedance load, Equivalent admittance can be expressed as:

$\left\{\begin{array}{c}{G}_q^{\mathrm{eq}}=\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\\ B_q^{\mathrm{eq}}=\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\end{array}\right.---\left(20\right)$

Formula (20) two ends are respectively to P _LkAsk partial derivative, can get:

$\frac{{\&PartialD;G}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=\left\{\begin{array}{cc}\frac{1}{{\left(V_q\right)}^2}-2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{\&PartialD;V_q}{\&PartialD;P_{\mathrm{Lq}}}& q=k\\ -2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{{\&PartialD;V}_q}{\&PartialD;P_{\mathrm{Lk}}}& q\&NotEqual;k\end{array}\right.---\left(21\right)$

$\frac{{\&PartialD;B}_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=-2\&CenterDot;\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{{\&PartialD;V}_q}{\&PartialD;P_{\mathrm{Lk}}}---\left(22\right)$

Formula (21)-(22) substitution (19) can be tried to achieve

With

With what try to achieve

With Substitution formula (17), vertical (3) in parallel can be tried to achieve generator's power and angle with the meritorious sensitivity of load.The computational methods of initial value and the generated power sensitivity calculations of exerting oneself is similar, repeats no more.Can obtain equally generator's power and angle with the sensitivity between reactive load according to this step.

When generator adopts than complex model, still can derive generator's power and angle with the generated power sensitivity relation meritorious and idle of exerting oneself, load ^[16]But for complex model, for ease of calculating, can utilize numerical method to carry out finding the solution of trace sensitivity ^[15]

When system control parameters changes, utilize and to have tried to achieve trace sensitivity, can be by the variable quantity at following formula approximate evaluation generator i merit angle:

$\mathrm{\&Delta;}{\mathrm{\&delta;}}_i=\underset{k=1}{\overset{m}{\mathrm{\&Sigma;}}}\left(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{gk}}}\mathrm{\&Delta;}{P}_{\mathrm{gk}}\right)+\underset{k=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;Q_{\mathrm{Lk}}}\mathrm{\&Delta;}{P}_{\mathrm{Lk}}+\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;Q_{\mathrm{Lk}}}\mathrm{\&Delta;}{Q}_{\mathrm{Lk}})---\left(23\right)$

2 based on the electric power system transient stability minimum of the sensitivity model of controlling cost

The model 2.1 the transient stability minimum is controlled cost

Trace sensitivity and optimal load flow are combined, find the solution following nonlinear optimal problem, get final product to such an extent that the transient stability minimum is controlled cost:

The described transient stability minimum model of controlling cost of finding the solution is, trace sensitivity and optimal load flow are combined, and finds the solution following nonlinear optimal problem, gets final product to such an extent that the transient stability minimum is controlled cost:

$\min f({\mathrm{\&Delta;P}}_g^{+},{\mathrm{\&Delta;P}}_g^{-},{\mathrm{\&Delta;P}}_L)=C_g^{+}\&CenterDot;\mathrm{\&Delta;}{P}_g^{+}+C_g^{-}\&CenterDot;\mathrm{\&Delta;}{P}_g^{-}+C_L\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(24a\right)$

$s.t.{\mathrm{<figure-callout\; id="0"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^{}+\mathrm{\&Delta;}{P}_g^{+}-\mathrm{\&Delta;}{P}_g^{-}-{\mathrm{<figure-callout\; id="0"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L^{}+\mathrm{\&Delta;}{P}_L-P(V,\mathrm{\&theta;})=0---\left(24b\right)$

$Q_g-{\mathrm{<figure-callout\; id="0"\; label="Q\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">Q</figure-callout>}}_L^{}+\mathrm{\&Delta;}{Q}_L-Q(V,\mathrm{\&theta;})=0---\left(24c\right)$

S(V，θ)≤S _max (24d)

V _min≤V≤V _max (24e)

$0\&le;\mathrm{\&Delta;}{P}_g^{+}\&le;P_{g_{\max }}^{+}---\left(24f\right)$

$0\&le;\mathrm{\&Delta;}{P}_g^{-}\&le;P_{g_{\max }}^{-}---\left(24g\right)$

$Q_{g_{\min }}\&le;Q_g\&le;Q_{g_{\max }}---\left(24h\right)$

$0\&le;\mathrm{\&Delta;}{P}_L\&le;P_{L_{\max }}---\left(24i\right)$

-δ _lim≤δ ⁰(t _sen)+S(t _sen)·Δu≤δ _lim (24j)

Wherein,

With

Represent respectively the generated power upper mediation downward modulation amount of exerting oneself, Δ P _LWith Δ Q _LThe expression load is gained merit and idle reduction amount respectively;

C _LRepresent respectively generated power exert oneself rise expense, downward modulation expense and load is meritorious cuts the cost;

The front generated power of expression control is exerted oneself, is loaded and gains merit and idle initial value respectively;

Represent respectively generated power exert oneself rise amount, downward modulation amount and load reduction amount maximum; δ _LimThe limiting value at the relative merit of expression generator angle, t _SenThe relative merit angle that the constraint of expression transient state is selected and trace sensitivity are constantly; δ ⁰Expression t _SenRelative merit angle between the generator is a dimensional vector constantly; Δ u represents control variables, comprises that generated power exerts oneself and load meritorious, is the b dimensional vector; S represents t _SenConstantly the sensitivity matrix between generator relative merit angle and control variables is a*b dimension matrix, and a represents the number of the Transient Stability Constraints that comprises in the optimal models, and b represents the control variables number;

Above-mentioned model Chinese style (24a) expression is controlled cost for certain fault, comprises cost, the downward modulation generator output cost that raises generator output and reduces the load cost.For the sake of simplicity, this model all is thought of as three kinds of control costs the linear function of controlled quentity controlled variable, further also can adopt secondary, linear segmented cost function ^[17], its solution procedure and above-mentioned model class are seemingly; (24b), (24c) be respectively meritorious and the reactive power flow equation; (24d) be the Line Flow constraint; (24e) be the node voltage constraint; (24f), (24g), (24h) be respectively generated power exert oneself rise amount constraint, the constraint of downward modulation amount and idle constraint; (24i) expression load reduction amount constraint; (24j) expression transient state merit angle constraint.

For load bus, when loading meritorious the reduction, reactive load reduction amount is determined by following formula:

$\mathrm{\&Delta;}{Q}_L=\frac{Q_L^0}{{\mathrm{<figure-callout\; id="0"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L^{}}\&CenterDot;\mathrm{\&Delta;}{P}_L=\tan \mathrm{\&theta;}\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(25\right)$

Namely in control procedure, load power factor is thought of as constant, and θ represents the power-factor angle of loading in the following formula;

Formula (24) belongs to and typically contains the nonlinear restriction optimal model, utilizes ‖ Δ P _L‖＜ε and ‖ Δ P _g‖＜ε when two conditions satisfy simultaneously, shows that system is without adjustable generator and load, then EP (end of program) as convergence criterion.

Fig. 1 angle stability control flow chart

2.2 some explanation

1, for the balancing machine in the system, only when other generators and load adjustment can not satisfy the system load flow balance, just participate in the trend adjustment, for reaching this purpose, this paper is taken as common generator with the meritorious adjustment expense of exerting oneself of balancing machine and adjusts 2 times of expense.

2, the formula in the model (24j) is the Transient Stability Constraints of system, when the relative merit of generator angle surpasses δ _LimThe time think system's generation Transient Instability.Might as well in the supposing the system m generator and l load be arranged, C is then arranged _m ²The relative merit of individual generator angle, and then following relation is arranged:

$\left\{\begin{array}{c}a=C_m^2\\ b=2m+l\end{array}\right.---\left(26\right)$

3, for formula (24j), because δ ₀, S all and t _SenRelevant, so t _SenChoose extremely important.With Fig. 2 simplicity of explanation t _SenSelection principle constantly, this figure has provided the relative merit angle of 3 machine single systems: arrange mutually at the merit angle of No. 2 machines and 1, No. 3 machine,

After, δ ₁₂, δ ₂₃Absolute value will be greater than δ _Lim, and relative merit angle δ ₁₃Absolute value always less than δ _LimSuppose at δ ₁₂, δ ₂₃In the curve, δ ₁₂Reach at first δ _Sen, this moment engrave into

In order to guarantee

All relative merit angles all are positioned at [δ constantly _Lim, δ _Lim] in, get

δ _SenThe threshold values at a predefined relative merit angle, different system δ _SenChoose may be different.

The relative power-angle curve schematic diagram of Fig. 2 generator

4, in formula (4j), have

$\mathrm{\&Delta;u}={\left[\begin{array}{ccc}{\mathrm{\&Delta;P}}_g^{+}& \mathrm{\&Delta;}{P}_g^{-}& \mathrm{\&Delta;}{P}_L\end{array}\right]}^T---\left(27\right)$

Wherein:

$\left\{\begin{array}{c}{P}_g^{+}=\left[\begin{array}{cccc}\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^1& \mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="P\; g"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_g^{2}L& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{+}L& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{+}\end{array}\right]\\ \mathrm{\&Delta;}{P}_g^{-}=\left[\begin{array}{cccc}\mathrm{\&Delta;}{P}_{g1}^{-}& \mathrm{\&Delta;}{P}_{g2}^{-}L& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{-}L& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{-}\end{array}\right]\\ \mathrm{\&Delta;}{P}_L=\left[\begin{array}{cccc}\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="P\; L"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">P</figure-callout>}}_L& \mathrm{\&Delta;}{P}_{L2}L& \mathrm{\&Delta;}{P}_{\mathrm{Lq}}L& \mathrm{\&Delta;}{P}_{\mathrm{Ll}}\end{array}\right]\end{array}\right.---\left(28\right)$

The formation of sensitivity matrix S is suc as formula shown in (29):

$s\left(t_{\mathrm{sen}}\right)=\left[\begin{array}{cccccccccccc}{S}_{g1}^{12}& S_{g2}^{12}L& S_{\mathrm{gp}}^{12}L& S_{\mathrm{gm}}^{12}& -S_{g1}^{12}& -S_{g2}^{12}L& -S_{\mathrm{gp}}^{12}L& -S_{\mathrm{gm}}^{12}& S_{L1}^{12}& S_{L2}^{12}L& S_{\mathrm{Lq}}^{12}L& S_{\mathrm{Ll}}^{12}\\ S_{g1}^{13}& S_{g2}^{13}L& S_{\mathrm{gp}}^{13}L& S_{\mathrm{gm}}^{13}& -S_{g1}^{13}& -S_{g2}^{13}L& -S_{\mathrm{gp}}^{13}L& -S_{\mathrm{gm}}^{13}& S_{L1}^{13}& S_{L2}^{13}L& S_{\mathrm{Lq}}^{13}L& S_{\mathrm{Ll}}^{13}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{23}& S_{g2}^{23}L& S_{\mathrm{gp}}^{23}L& S_{\mathrm{gm}}^{23}& -S_{g1}^{23}& -S_{g2}^{23}L& -S_{\mathrm{gp}}^{23}L& -S_{\mathrm{gm}}^{23}& S_{L1}^{23}& S_{L2}^{23}L& S_{\mathrm{Lq}}^{23}L& S_{L1}^{23}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{\mathrm{ij}}& S_{g2}^{\mathrm{ij}}L& S_{\mathrm{gp}}^{\mathrm{ij}}L& S_{\mathrm{gm}}^{\mathrm{ij}}& -S_{g1}^{\mathrm{ij}}& -S_{g2}^{\mathrm{ij}}L& -S_{\mathrm{gp}}^{\mathrm{ij}}L& -S_{\mathrm{gm}}^{\mathrm{ij}}& S_{L1}^{\mathrm{ij}}& S_{L2}^{\mathrm{ij}}L& S_{\mathrm{Lq}}^{\mathrm{ij}}L& S_{\mathrm{Ll}}^{\mathrm{ij}}\\ M& M& M& M& M& M& M& M& M& M& M& M\\ S_{g1}^{m-1,g}& S_{g2}^{m-1,g}L& S_{\mathrm{gp}}^{m-\mathrm{1,2}}L& S_{\mathrm{gm}}^{m-\mathrm{1,2}}& -S_{g1}^{m-\mathrm{1,2}}& -S_{g2}^{m-\mathrm{1,2}}L& -S_{\mathrm{gp}}^{m-\mathrm{1,2}}L& -S_{\mathrm{gm}}^{m-\mathrm{1,2}}& S_{L1}^{m-\mathrm{1,2}}& S_{L2}^{m-\mathrm{1,2}}L& S_{\mathrm{Lq}}^{m-\mathrm{1,2}}L& S_{\mathrm{Ll}}^{m-\mathrm{1,2}}\end{array}\right]---\left(29\right)$

Wherein,

Sensitivity between relative merit angle between expression generator i and the j is meritorious with generator p has

$S_{\mathrm{gp}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{gp}}}\left(t_{\mathrm{sen}}\right)---\left(30\right)$

Because load is processed according to constant power factor, therefore the sensitivity of the relative merit of generator angle with load or burden without work is folded in the burden with power sensitivity according to corresponding power factor,

Relative merit angle between expression generator i and the j has with the sensitivity between load q

$S_{\mathrm{Lq}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)+\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;Q_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)\&CenterDot;\tan \left({\mathrm{\&theta;}}_q\right)---\left(31\right)$

In formula (31), θ _qThe power-factor angle of expression load q.

${\mathrm{\&delta;}}^0\left(t_{\mathrm{sen}}\right)={\left[\begin{array}{ccccc}{\mathrm{\&delta;}}_{12}^0& {\mathrm{\&delta;}}_{13}^{0}L& {\mathrm{\&delta;}}_{23}^{0}L& {\mathrm{\&delta;}}_{\mathrm{ij}}^{0}L& {\mathrm{\&delta;}}_{m-1,m}^0\end{array}\right]}^T---\left(32\right)$

Wherein,

Initial phase between expression generator i and the j has the merit angle:

${\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="0"\; label="ij"\; filenames="HDA0000028978130000022.png,HDA0000028978130000023.png"\; state="\{\{state\}\}">ij</figure-callout>}}^0={\mathrm{\&delta;}}_{\mathrm{ij}}^0\left(t_{\mathrm{sen}}\right)---\left(33\right)$

3 sample calculation analysis

This paper verifies this paper method take IEEE-14 and New England-39 node system as example, gets δ in the calculating _Lim=π ^[10], ε=0.05MW.For 14 node systems, generator adopts time transient Model, and load is taked constant-impedance load, δ _Sen=5 π; For 39 node systems, generator adopts classical model, and load adopts constant-impedance model, δ _Sen=10 π.Other generator control expense: C except balance node _g ⁺=C _g ^-=10K/MW, the regulate expenditure of balance node: C _s ⁺=C _s ^-=20K/MW, load regulate expenditure: C _l=30K/MW, K are certain monetary unit.

3.1IEEE-14 node system

Suppose that circuit 2-4 near node 2 places three-phase shortcircuit, critical clearing time t occurs _Cr=0.29s establishes its actual mute time and is: t _Cl=1.5t _Cr=0.435s, the node voltage maximum is got 115p.u, and the node voltage minimum value is got 0.85p.u, and this moment, unstability situation as shown in Figure 3 appearred in system, can see clearly that from figure generator has been divided into two groups of { 1,2}, { 3,4,5}.

Figure 31 4 node system generator's power and angle curves

Sight 1: when generator has sufficient reserve (as shown in table 1), do not need to adjust load, only can realize system stability by adjusting generator output, the result is as shown in table 1 in control, and minimum control cost is 1184.50K.Each generator's power and angle curve can be found out on scheming as shown in Figure 4 after adjusting, and can keep transient stability through the generator Adjustment System.

Table 1 generator control result (sight 1)

Generator's power and angle curve (sight 1) after Figure 41 4 node system adjustment

Sight 2: when generator deficiency for subsequent use, only depend on the adjustment generator, can not reach the purpose of system stability, also should follow this moment load to reduce control.As example, might as well suppose that the maximum that G3, G4, G5 generated power are exerted oneself becomes 20MW by original 100MW, then calculate by optimizing, the adjustment result of generator and load is shown in table 2 and table 3.At this moment, generator and the total regulate expenditure of load are 3294.67K, and the power-angle curve of generator as shown in Figure 5 after adjusting, can find out that system also can keep stable, but because the generated power deposit is less, the reduction of need to loading, the control cost of sight 2 will be higher than sight 1.

Table 1 generator control result (sight 2)

Table 2 load control result's (sight 2)

Generator's power and angle curve (sight 2) after Figure 91 4 node system adjustment

3.2New England-39 node system example

Suppose that circuit 22-21 near node 22 places three-phase shortcircuit, critical clearing time t occurs _Cr=0.13s, because of certain reason, system's physical fault mute time becomes: t _Cl=1.5t _Cr=0.2s, the node voltage maximum is got 1.15p.u, and the node voltage minimum value is got 0.85p.u, and this moment, unstability situation as shown in Figure 6 appearred in system, and as can be seen from the figure 10 generators are divided into 4 groups of { 1,3,4,5,8,9}, { 2}, { 6,7}, { 10}.

Figure 103 9 node system generator's power and angle curves

Sight 1: when generator abundance for subsequent use (as shown in table 4), it is as shown in table 4 that generator is adjusted the result.Because it is for subsequent use that generator all has sufficient meritorious conciliation, so after the fault, can the assurance system recover stable operation by adjusting generator output, system's minimum is controlled cost and is 9564.39K, and system's power-angle curve as shown in Figure 7 after the control.

Table 3 generator control result (sight 1)

Generator's power and angle curve (sight 1) after Figure 11 39 node systems are adjusted.

Sight 2: suppose that No. 10 generated powers maximum of exerting oneself becomes 1400MW by 1600MW, system will not have enough generator reserve capacitys adjustable, therefore after same fault occurs, the adjustment of simple dependence generator can't guarantee system stability, must be aided with load and reduce, the control program of system and minimum are controlled cost and are shown in table 5 and table 6 at this moment.Be not difficult to find out, because generator lacks enough reserves, the Fault Control cost of system raises and becomes 11128.62K, and the system's power-angle curve after the control is shown in Fig. 8.

Table 4 generator control result (sight 2).

Table 5 load control result's (sight 2)

Generator's power and angle curve (sight 2) after Figure 12 39 node system adjustment

4 conclusion

This paper has proposed a kind of breakdown loss method of estimation for the transient stability risk assessment, breakdown loss is used system is converted into the required minimum control cost of stable state from labile state takes in.Utilize the variable quantity at the relative merit of generator angle after the trace sensitivity technology approximate evaluation system parameter variations, and then Transient Stability Constraints is introduced the optimal load flow model, reduce as control device with generator adjustment and load, realize finding the solution of minimum control cost.This paper not yet considers the impact of fault randomness, will be in following work Improvement and perfection in addition, thereby realize better the electric power system transient stability risk assessment.

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Claims (4)

1. a breakdown loss method of estimation that is used for the transient rotor angle stability risk assessment is characterized in that, comprises the steps:

Carry out the ordinary tides flowmeter and calculate, obtain initial condition;

Carry out the transient state time-domain-simulation and calculate trace sensitivity;

If unstable, find the solution the transient stability minimum model of controlling cost;

If do not satisfy the condition of convergence: || Δ P _L||＜ε and || Δ P _g||＜ε, repeat abovementioned steps.

2. a kind of breakdown loss method of estimation for the transient rotor angle stability risk assessment according to claim 1 is characterized in that, described calculating trace sensitivity, and computational methods are as follows:

Electrical Power System Dynamic can represent with following differential-algebraic equation:

$\left\{\begin{array}{c}\stackrel{\&CenterDot;}{x}=f(x,y,\mathrm{\&lambda;})\\ 0=g(x,y,\mathrm{\&lambda;})\\ x\left(t_0\right)=x_0\\ y\left(t_0\right)=y_0\end{array}\right.---\left(1\right)$

In the formula, x is system state variables, and y is system's algebraically variable, and λ is system's control variables, x ₀, y ₀The initial value that represents respectively state variable and algebraically variable, ask local derviation to get to λ formula (1):

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{x}}_{\mathrm{\&lambda;}}=\frac{\&PartialD;f}{\&PartialD;x}\&CenterDot;x_{\mathrm{\&lambda;}}+\frac{\&PartialD;f}{\&PartialD;y}\&CenterDot;y_{\mathrm{\&lambda;}}+\frac{\&PartialD;f}{\&PartialD;\mathrm{\&lambda;}}\\ 0=\frac{\&PartialD;g}{\&PartialD;x}\&CenterDot;x_{\mathrm{\&lambda;}}+\frac{\&PartialD;g}{\&PartialD;y}\&CenterDot;y_{\mathrm{\&lambda;}}+\frac{\&PartialD;g}{\&PartialD;\mathrm{\&lambda;}}\end{array}\right.---\left(2\right)$

In the formula,

Represent system trajectory to the sensitivity matrix of control variables λ,

Be respectively differential equation group to the local derviation of state variable, algebraically variable and control variables,

Be respectively Algebraic Equation set to the partial derivative of state variable, algebraically variable and control variables, can try to achieve the initial condition x of trace sensitivity according to system's initial launch point _λ(t ₀), y _λ(t ₀), formula (1) and formula (2) simultaneous solution can be obtained each trace sensitivity constantly of system, generator's power and angle is as follows with its meritorious sensitivity relation circular of exerting oneself and loading between gaining merit:

Generator's power and angle is with the sensitivity between its meritorious exerting oneself:

When generator was taked classical model, the Electrical Power System Dynamic equation was as follows:

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{\mathrm{\&delta;}}}_i={\mathrm{\&omega;}}_i\\ M_i{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_i=P_{\mathrm{gi}}-{\left(E_i^{\&prime;}\right)}^2{G}_{\mathrm{ii}}\\ -\underset{j=1,j\&NotEqual;i}{\overset{m}{\mathrm{\&Sigma;}}}(E_i^{\&prime;}{E}_j^{\&prime;}{B}_{\mathrm{ij}}\sin {\mathrm{\&delta;}}_{\mathrm{ij}}+E_i^{\&prime;}{E}_j^{\&prime;}{G}_{\mathrm{ij}}\cos {\mathrm{\&delta;}}_{\mathrm{ij}})\end{array}\right.---\left(3\right)$

In the formula, δ _i, ω _i, M _i, P _Gi, E ' _iBe respectively generator amature angle, rotating speed, moment of inertia, mechanical output and built-in potential; G _Ii, G _Ij, after expression only keeps the generator interior nodes, the real part of system's admittance matrix element, B _IjThe expression imaginary part, m is the generator nodes;

The meritorious P that exerts oneself of generator k is asked respectively at formula (3) two ends _GkPartial derivative can get:

$\left\{\begin{array}{c}\frac{\&PartialD;{\stackrel{\&CenterDot;}{\mathrm{\&delta;}}}_i}{\&PartialD;P_{\mathrm{gk}}}=\frac{\&PartialD;{\mathrm{\&omega;}}_i}{\&PartialD;P_{\mathrm{gk}}}\\ M\&CenterDot;\frac{\&PartialD;{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_i}{\&PartialD;P_{\mathrm{gk}}}=-\underset{\underset{j\&NotEqual;i}{j=1}}{\overset{m}{\mathrm{\&Sigma;}}}(E_i^{\&prime;}{E}_j^{\&prime;}{B}_{\mathrm{ij}}\cos {\mathrm{\&delta;}}_{\mathrm{ij}}-E_i^{\&prime;}{E}_j^{\&prime;}{G}_{\mathrm{ij}}\sin {\mathrm{\&delta;}}_{\mathrm{ij}})\&CenterDot;(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{gk}}}-\frac{\&PartialD;{\mathrm{\&delta;}}_j}{\&PartialD;P_{\mathrm{gk}}})\\ -\underset{\underset{j\&NotEqual;i}{j=1}}{\overset{m}{\mathrm{\&Sigma;}}}(B_{\mathrm{ij}}\sin {\mathrm{\&delta;}}_{\mathrm{ij}}+G_{\mathrm{ij}}\cos {\mathrm{\&delta;}}_{\mathrm{ij}})\&CenterDot;(\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{gk}}}{E}_j^{\&prime;}-\frac{\&PartialD;E_j^{\&prime;}}{\&PartialD;P_{\mathrm{gk}}}{E}_i^{\&prime;})\\ -2E_i^{\&prime;}{G}_{\mathrm{ii}}\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{gk}}}+\frac{\&PartialD;P_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}\end{array}\right.---\left(4\right)$

In the formula,

Be the sensitivity of generator i merit angle between exerting oneself with generator k is meritorious since in transient process E ' _i, P _GiRemain unchanged, only relevant with the state of system before the fault, therefore

A constant in transient process, further, as can be known:

$\frac{\&PartialD;P_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}=\left\{\begin{array}{cc}1& k=i\\ 0& k\&NotEqual;i\end{array}\right.---\left(5\right)$

Can be got by Generator end power equation:

${\stackrel{\&CenterDot;}{I}}_i=\frac{{\widehat{\stackrel{\&CenterDot;}{S}}}_i}{{\widehat{\stackrel{\&CenterDot;}{V}}}_i}=\frac{S_i\&angle;-{\mathrm{\&alpha;}}_i}{V_i\&angle;-{\mathrm{\&theta;}}_i}=\frac{S_i}{V_i}\&angle;{\mathrm{\&theta;}}_i-{\mathrm{\&alpha;}}_i---\left(6\right)$

Wherein,

Be the electric current of generator injection network, S _i, α _iBe respectively apparent power and the power-factor angle of generator output, V _i, θ _iBe respectively generator terminal voltage amplitude and phase angle,

$\left\{\begin{array}{c}{S}_i=\sqrt{P_{\mathrm{gi}}^2+Q_{\mathrm{gi}}^2}\\ {\mathrm{\&alpha;}}_i=\arctan \left(\frac{Q_{\mathrm{gi}}}{P_{\mathrm{gi}}}\right)\end{array}\right.---\left(7\right)$

Generator built-in potential E ' _iCan be expressed as:

${\stackrel{\&CenterDot;}{E}}_i^{\&prime;}={\stackrel{\&CenterDot;}{V}}_i+j{x}_d^{\&prime;}\&CenterDot;{\stackrel{\&CenterDot;}{I}}_i$

$=(V_i\cos {\mathrm{\&theta;}}_i-\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\sin ({\mathrm{\&theta;}}_i-{\mathrm{\&alpha;}}_i))---\left(8\right)$

$+j(V_i\sin {\mathrm{\&theta;}}_i+\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\cos ({\mathrm{\&theta;}}_i-{\mathrm{\&alpha;}}_i))$

Therefore, have:

$E_i^{\&prime;}=\sqrt{V_i^2+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}---\left(9\right)$

Formula (9) two ends are respectively to P _GkAsk partial derivative:

$\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{gk}}}=\frac{1}{2}\&CenterDot;\frac{1}{\sqrt{V_i^1+{\left(\frac{x_d^{\&prime;}\&CenterDot;S_i}{V_i}\right)}^2+2x_d^{\&prime;}{Q}_{\mathrm{gi}}}}---\left(10\right)$

$\&CenterDot;(2S_i{\left(\frac{x_d^{\&prime;}}{V_i}\right)}^2\frac{\&PartialD;S_i}{\&PartialD;P_{\mathrm{gk}}}+2x_d^{\&prime;}\frac{\&PartialD;Q_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})$

Formula (7) two ends are respectively to P _GkAsk partial derivative:

$\frac{\&PartialD;S_i}{\&PartialD;P_{\mathrm{gk}}}=\frac{1}{\sqrt{P_{\mathrm{gi}}^2+Q_{\mathrm{gi}}^2}}(P_{\mathrm{gi}}\frac{\&PartialD;P_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}+Q_i\frac{\&PartialD;Q_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}})---\left(11\right)$

The generator node is idle, and the injection equation is as follows:

$Q_{\mathrm{gi}}=Q_{\mathrm{Li}}+V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})---\left(12\right)$

N is the system node number, and the following formula two ends are respectively to P _GkAsk partial derivative:

$\frac{\&PartialD;Q_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}=V_i\underset{\underset{j\&NotElement;G}{j=1}}{\overset{n}{\mathrm{\&Sigma;}}}\frac{\&PartialD;V_j}{\&PartialD;P_{\mathrm{gk}}}(G_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}}-B_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}})$ (13)

$+V_i\underset{j=1}{\overset{n}{\mathrm{\&Sigma;}}}{V}_j(G_{\mathrm{ij}}\cos {\mathrm{\&theta;}}_{\mathrm{ij}}+B_{\mathrm{ij}}\sin {\mathrm{\&theta;}}_{\mathrm{ij}})(\frac{\&PartialD;{\mathrm{\&theta;}}_i}{\&PartialD;P_{\mathrm{gk}}}-\frac{\&PartialD;{\mathrm{\&theta;}}_j}{\&PartialD;P_{\mathrm{gk}}})$

Wherein, G represents the generator node set,

Be power flow equation Jacobi inverse matrix element, formula (13), formula (11) substitution formula (10) can be tried to achieve

And then simultaneous solution formula (3)-(4) just can be tried to achieve

For the calculating of sensitivity coefficient initial value, because of ω _i(t ₀)=1, so have:

$\frac{\&PartialD;{\mathrm{\&omega;}}_i}{\&PartialD;P_{\mathrm{gk}}}\left(t_0\right)=0---\left(14\right)$

Generator machinery power when knowing stable state again is as follows:

$P_{\mathrm{gi}}=\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)---\left(15\right)$

Formula (15) both sides are respectively to P _GkAsk partial derivative, can get:

$\frac{\&PartialD;P_{\mathrm{gi}}}{\&PartialD;P_{\mathrm{gk}}}=\frac{V_i}{x_d^{\&prime;}}\sin ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{gk}}}$ (16)

$+\frac{E_i^{\&prime;}{V}_i}{x_d^{\&prime;}}\cos ({\mathrm{\&delta;}}_i-{\mathrm{\&theta;}}_i)(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{gk}}}-\frac{\&PartialD;{\mathrm{\&theta;}}_i}{\&PartialD;P_{\mathrm{gk}}})$

Therefrom can solve

With what tried to achieve

Substitution can obtain

Sensitivity between generator's power and angle is meritorious with load:

Suppose that load takes the constant-impedance model, formula (3) two ends are respectively to the burden with power P of load bus k _LkAsk partial derivative:

$\left\{\begin{array}{c}\frac{\&PartialD;{\stackrel{\&CenterDot;}{\mathrm{\&delta;}}}_i}{\&PartialD;P_{\mathrm{Lk}}}=\frac{\&PartialD;{\mathrm{\&omega;}}_i}{\&PartialD;P_{\mathrm{Lk}}}\\ M\&CenterDot;\frac{\&PartialD;{\stackrel{\&CenterDot;}{\mathrm{\&omega;}}}_i}{\&PartialD;P_{\mathrm{Lk}}}=-\underset{\underset{j\&NotEqual;i}{j=1}}{\overset{m}{\mathrm{\&Sigma;}}}(E_i^{\&prime;}{E}_j^{\&prime;}{B}_{\mathrm{ij}}\cos {\mathrm{\&delta;}}_{\mathrm{ij}}-E_i^{\&prime;}{E}_j^{\&prime;}{G}_{\mathrm{ij}}\sin {\mathrm{\&delta;}}_{\mathrm{ij}})\&CenterDot;(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{Lk}}}-\frac{\&PartialD;{\mathrm{\&delta;}}_j}{\&PartialD;P_{\mathrm{Lk}}})\\ -\underset{\underset{j\&NotEqual;i}{j=1}}{\overset{m}{\mathrm{\&Sigma;}}}(E_i^{\&prime;}{E}_j^{\&prime;}\sin {\mathrm{\&delta;}}_{\mathrm{ij}}\frac{\&PartialD;B_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}+E_i^{\&prime;}{E}_j^{\&prime;}\cos {\mathrm{\&delta;}}_{\mathrm{ij}}\frac{\&PartialD;G_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}})\\ -\underset{\underset{j\&NotEqual;i}{j=1}}{\overset{m}{\mathrm{\&Sigma;}}}(B_{\mathrm{ij}}\sin {\mathrm{\&delta;}}_{\mathrm{ij}}+G_{\mathrm{ij}}\cos {\mathrm{\&delta;}}_{\mathrm{ij}})\&CenterDot;(\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{Lk}}}{E}_j^{\&prime;}-\frac{\&PartialD;E_j^{\&prime;}}{\&PartialD;P_{\mathrm{Lk}}}{E}_i^{\&prime;})\\ -2E_i^{\&prime;}{G}_{\mathrm{ii}}\frac{\&PartialD;E_i^{\&prime;}}{\&PartialD;P_{\mathrm{Lk}}}-{\left(E_i^{\&prime;}\right)}^2\frac{\&PartialD;G_{\mathrm{ii}}}{\&PartialD;P_{\mathrm{Lk}}}\end{array}\right.---\left(17\right)$

Wherein,

Find the solution and find the solution

Similar, do not giving unnecessary details at this; G _Ii, G _Ij, B _IjAll can be expressed as each load bus Equivalent admittance

(q=1 ... l, l represent the load bus number) function:

$\left\{\begin{array}{c}{B}_{\mathrm{ij}}=h_{\mathrm{ij}}(B_1^{\mathrm{eq}},G_2^{\mathrm{eq}},...,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},...,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ij}}=p_{\mathrm{ij}}(B_1^{\mathrm{eq}},G_2^{\mathrm{eq}},...,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},...,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\\ G_{\mathrm{ii}}=q_{\mathrm{ij}}(B_1^{\mathrm{eq}},G_2^{\mathrm{eq}},...,B_q^{\mathrm{eq}},G_q^{\mathrm{eq}},...,B_l^{\mathrm{eq}},G_l^{\mathrm{eq}})\end{array}\right.---\left(18\right)$

Formula (18) two ends are respectively to P _LkAsk partial derivative:

$\left\{\begin{array}{c}\frac{\&PartialD;B_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;h_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;B_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}+\frac{\&PartialD;h_{\mathrm{ij}}}{\&PartialD;G_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}})\\ \frac{\&PartialD;G_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;p_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;B_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}+\frac{\&PartialD;p_{\mathrm{ij}}}{\&PartialD;G_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}})\\ \frac{\&PartialD;G_{\mathrm{ii}}}{\&PartialD;P_{\mathrm{Lk}}}=\underset{m=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;q_{\mathrm{ij}}}{\&PartialD;B_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;B_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}+\frac{\&PartialD;q_{\mathrm{ij}}}{\&PartialD;G_q^{\mathrm{eq}}}\&CenterDot;\frac{\&PartialD;G_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}})\end{array}\right.---\left(19\right)$

For the constant-impedance load, Equivalent admittance can be expressed as:

$\left\{\begin{array}{c}{G}_q^{\mathrm{eq}}=\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\\ B_q^{\mathrm{eq}}=\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^2}\end{array}\right.---\left(20\right)$

Formula (20) two ends are respectively to P _LkAsk partial derivative, can get:

$\frac{\&PartialD;G_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=\left\{\begin{array}{cc}\frac{1}{{\left(V_q\right)}^2}-2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{\&PartialD;V_q}{\&PartialD;P_{\mathrm{Lq}}}& q=k\\ -2\&CenterDot;\frac{P_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{\&PartialD;V_q}{\&PartialD;P_{\mathrm{Lk}}}& q\&NotEqual;k\end{array}\right.---\left(21\right)$

$\frac{\&PartialD;B_q^{\mathrm{eq}}}{\&PartialD;P_{\mathrm{Lk}}}=-2\&CenterDot;\frac{Q_{\mathrm{Lq}}}{{\left(V_q\right)}^3}\&CenterDot;\frac{\&PartialD;V_q}{\&PartialD;P_{\mathrm{Lk}}}---\left(22\right)$

Formula (21)-(22) substitution (19) can be tried to achieve With

With what try to achieve

With Substitution formula (17), vertical (3) in parallel can try to achieve generator's power and angle with the meritorious sensitivity of load, the computational methods of initial value and generated power are exerted oneself sensitivity calculations roughly the same, can obtain equally generator's power and angle with the sensitivity between reactive load according to this step

When system control parameters changes, utilize and to have tried to achieve trace sensitivity, can be by the variable quantity at following formula approximate evaluation generator i merit angle:

$\mathrm{\&Delta;}{\mathrm{\&delta;}}_i=\underset{k=1}{\overset{m}{\mathrm{\&Sigma;}}}\left(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{gk}}}\mathrm{\&Delta;}{P}_{\mathrm{gk}}\right)+\underset{k=1}{\overset{l}{\mathrm{\&Sigma;}}}(\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;P_{\mathrm{Lk}}}\mathrm{\&Delta;}{P}_{\mathrm{Lk}}+\frac{\&PartialD;{\mathrm{\&delta;}}_i}{\&PartialD;Q_{\mathrm{Lk}}}\mathrm{\&Delta;}{Q}_{\mathrm{Lk}})---\left(23\right).$

3. a kind of breakdown loss method of estimation for the transient rotor angle stability risk assessment according to claim 1, it is characterized in that, the described transient stability minimum model of controlling cost of finding the solution is, trace sensitivity and optimal load flow are combined, find the solution following nonlinear optimal problem, get final product to such an extent that the transient stability minimum is controlled cost:

$\min f(\mathrm{\&Delta;}{P}_g^{+},\mathrm{\&Delta;}{P}_g^{-},\mathrm{\&Delta;}{P}_L)=C_g^{+},\mathrm{\&Delta;}{P}_g^{+}+C_g^{-}\&CenterDot;\mathrm{\&Delta;}{P}_g^{-}+C_L\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(24a\right)$

$s.t.P_g^0+\mathrm{\&Delta;}{P}_g^{+}-\mathrm{\&Delta;}{P}_g^{-}-P_L^0+\mathrm{\&Delta;}{P}_L-P(V,\mathrm{\&theta;})=0---\left(24b\right)$

$Q_g-Q_L^0+\mathrm{\&Delta;}{Q}_L-Q(V,\mathrm{\&theta;})=0---\left(24c\right)$

S(V，θ)≤S _max (24d)

V _min≤V≤V _max (24e)

$0\&le;\mathrm{\&Delta;}{P}_g^{+}\&le;P_{g_{\max }}^{+}---\left(24f\right)$

$0\&le;\mathrm{\&Delta;}{P}_g^{-}\&le;P_{g_{\max }}^{-}---\left(24g\right)$

$Q_{g_{\min }}\&le;Q_g\&le;Q_{g_{\max }}---\left(24h\right)$

$0\&le;\mathrm{\&Delta;}{P}_L\&le;P_{L_{\max }}---\left(24i\right)$

-δ _lim≤δ ⁰(t _sen)+S(t _sen)·Δu≤δ _lim (24j)

Wherein,

With

Represent respectively the generated power upper mediation downward modulation amount of exerting oneself, Δ P _LWith Δ Q _LThe expression load is gained merit and idle reduction amount respectively;

C _LRepresent respectively generated power exert oneself rise expense, downward modulation expense and load is meritorious cuts the cost;

The front generated power of expression control is exerted oneself, is loaded and gains merit and idle initial value respectively;

Represent respectively generated power exert oneself rise amount, downward modulation amount and load reduction amount maximum; δ _LimThe limiting value at the relative merit of expression generator angle, t _SenThe relative merit angle that the constraint of expression transient state is selected and trace sensitivity are constantly; δ ⁰Expression t _SenRelative merit angle between the generator is a dimensional vector constantly; Δ u represents control variables, comprises that generated power exerts oneself and load meritorious, is the b dimensional vector; S represents t _SenConstantly the sensitivity matrix between generator relative merit angle and control variables is a*b dimension matrix, and a represents the number of the Transient Stability Constraints that comprises in the optimal models, and b represents the control variables number;

Above-mentioned model Chinese style (24a) expression is controlled cost for certain fault, comprises cost, the downward modulation generator output cost that raises generator output and reduces the load cost; For the sake of simplicity, this model all is thought of as three kinds of control costs the linear function of controlled quentity controlled variable, further also can adopt secondary, linear segmented cost function ^[17], its solution procedure and above-mentioned model class are seemingly; (24b), (24c) be respectively meritorious and the reactive power flow equation; (24d) be the Line Flow constraint; (24e) be the node voltage constraint; (24f), (24g), (24h) be respectively generated power exert oneself rise amount constraint, the constraint of downward modulation amount and idle constraint; (24i) expression load reduction amount constraint; (24j) expression transient state merit angle constraint;

For load bus, when loading meritorious the reduction, reactive load reduction amount is determined by following formula:

$\mathrm{\&Delta;}{Q}_L=\frac{Q_L^0}{P_L^0}\&CenterDot;\mathrm{\&Delta;}{P}_L=\tan \mathrm{\&theta;}\&CenterDot;\mathrm{\&Delta;}{P}_L---\left(25\right)$

Namely in control procedure, load power factor is thought of as constant, and θ represents the power-factor angle of loading in the following formula;

Formula (24) belongs to and typically contains the nonlinear restriction optimal model, utilizes || Δ P _L||＜ε and || Δ P _g||＜ε when two conditions satisfy simultaneously, shows system without adjustable generator and load as convergence criterion, then finishes.

4. a kind of breakdown loss method of estimation for the transient rotor angle stability risk assessment according to claim 3 is characterized in that, the described transient stability minimum of finding the solution is controlled cost in the model, in formula (24j), has:

$\mathrm{\&Delta;u}={\left[\begin{array}{ccc}\mathrm{\&Delta;}{P}_g^{+}& \mathrm{\&Delta;}{P}_g^{-}& \mathrm{\&Delta;}{P}_L\end{array}\right]}^T---\left(26\right)$

Wherein:

$\left\{\begin{array}{c}\mathrm{\&Delta;}{P}_g^{+}=\left[\begin{array}{cccccc}\mathrm{\&Delta;}{P}_{g1}^{+}& \mathrm{\&Delta;}{P}_{g2}^{+}& ...& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{+}& ...& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{+}\end{array}\right]\\ \mathrm{\&Delta;}{P}_g^{-}=\left[\begin{array}{cccccc}\mathrm{\&Delta;}{P}_{g1}^{-}& \mathrm{\&Delta;}{P}_{g2}^{-}& ...& \mathrm{\&Delta;}{P}_{\mathrm{gp}}^{-}& ...& \mathrm{\&Delta;}{P}_{\mathrm{gm}}^{-}\end{array}\right]\\ \mathrm{\&Delta;}{P}_L=\left[\begin{array}{cccccc}\mathrm{\&Delta;}{P}_{L1}& \mathrm{\&Delta;}{P}_{L2}& ...& \mathrm{\&Delta;}{P}_{\mathrm{Lq}}& ...& \mathrm{\&Delta;}{P}_{\mathrm{Ll}}\end{array}\right]\end{array}\right.---\left(27\right)$

The formation of sensitivity matrix S is suc as formula shown in (28):

$S\left(t_{\mathrm{sen}}\right)=\left[\begin{array}{cccccccccccccccccc}{S}_{g1}^{12}& S_{g2}^{12}& ...& S_{\mathrm{gp}}^{12}& ...& S_{\mathrm{gm}}^{12}& -S_{g1}^{12}& -S_{g2}^{12}& ...& -S_{\mathrm{gp}}^{12}& ...& -S_{\mathrm{gm}}^{12}& S_{L1}^{12}& S_{L2}^{12}& ...& S_{\mathrm{Lq}}^{12}& ...& S_{\mathrm{Ll}}^{12}\\ S_{g1}^{13}& S_{g2}^{13}& ...& S_{\mathrm{gp}}^{13}& ...& S_{\mathrm{gm}}^{13}& -S_{g1}^{13}& -S_{g2}^{13}& ...& -S_{\mathrm{gp}}^{13}& ...& -S_{\mathrm{gm}}^{13}& S_{L1}^{13}& S_{L2}^{13}& ...& S_{\mathrm{Lq}}^{13}& ...& S_{\mathrm{Ll}}^{13}\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ S_{g1}^{23}& S_{g2}^{23}& ...& S_{\mathrm{gp}}^{23}& ...& S_{\mathrm{gm}}^{23}& -S_{g1}^{23}& -S_{g2}^{23}& ...& -S_{\mathrm{gp}}^{23}& ...& -S_{\mathrm{gm}}^{23}& S_{L1}^{23}& S_{L2}^3& ...& S_{\mathrm{Lq}}^{23}& ...& S_{\mathrm{Ll}}^{23}\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ S_{g1}^{\mathrm{ij}}& S_{g2}^{\mathrm{ij}}& ...& S_{\mathrm{gp}}^{\mathrm{ij}}& ...& S_{\mathrm{gm}}^{\mathrm{ij}}& -S_{g1}^{\mathrm{ij}}& -S_{g2}^{\mathrm{ij}}& ...& -S_{\mathrm{gp}}^{\mathrm{ij}}& ...& -S_{\mathrm{gm}}^{\mathrm{ij}}& S_{L1}^{\mathrm{ij}}& S_{L2}^{\mathrm{ij}}& ...& S_{\mathrm{Lq}}^{\mathrm{ij}}& ...& S_{\mathrm{Ll}}^{\mathrm{ij}}\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ .& .& & .& & .& .& .& & .& & .& .& .& & .& & .\\ S_{g1}^{m-1,g}& S_{g2}^{m-1,g}& ...& S_{\mathrm{gp}}^{m-1,m}& ...& S_{\mathrm{gm}}^{m-1,m}& -S_{g1}^{m-1,m}& -S_{g2}^{m-1,m}& ...& -S_{\mathrm{gp}}^{m-1,m}& ...& -S_{\mathrm{gm}}^{m-1,m}& S_{L1}^{m-1,m}& S_{L2}^{m-1,m}& ...& S_{\mathrm{Lq}}^{m-1,m}& ...& S_{\mathrm{Ll}}^{m-1,m}\end{array}\right]$

$\left(28\right)$

Wherein,

Sensitivity between relative merit angle between expression generator i and the j is meritorious with generator p has

$S_{\mathrm{gp}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{gp}}}\left(t_{\mathrm{sen}}\right)---\left(29\right)$

Because load is processed according to constant power factor, therefore the sensitivity of the relative merit of generator angle with load or burden without work is folded in the burden with power sensitivity according to corresponding power factor,

Relative merit angle between expression generator i and the j has with the sensitivity between load q

$S_{\mathrm{Lq}}^{\mathrm{ij}}=\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;P_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)+\frac{\&PartialD;{\mathrm{\&delta;}}_{\mathrm{ij}}}{\&PartialD;Q_{\mathrm{Lq}}}\left(t_{\mathrm{sen}}\right)\&CenterDot;\tan \left({\mathrm{\&theta;}}_q\right)---\left(30\right)$

In formula (30), θ _qThe power-factor angle of expression load q;

${\mathrm{\&delta;}}^0\left(t_{\mathrm{sen}}\right)={\left[\begin{array}{cccccccc}{\mathrm{\&delta;}}_{12}^0& {\mathrm{\&delta;}}_{13}^0& ...& {\mathrm{\&delta;}}_{23}^0& ...& {\mathrm{\&delta;}}_{\mathrm{ij}}^0& ...& {\mathrm{\&delta;}}_{m-1,m}^0\end{array}\right]}^T---\left(31\right)$

Wherein,

Initial phase between expression generator i and the j has the merit angle:

${\mathrm{\&delta;}}_{\mathrm{ij}}^0={\mathrm{\&delta;}}_{\mathrm{ij}}^0\left(t_{\mathrm{sen}}\right)---\left(32\right).$

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