CN106056479A

CN106056479A - Three-phase modeling method for distribution network

Info

Publication number: CN106056479A
Application number: CN201610431553.1A
Authority: CN
Inventors: 韩冰; 张志昌; 赵家庆; 钱科军; 张玉林; 杜红卫; 徐春雷; 韩涛; 田江; 陈连杰; 吕洋
Original assignee: NARI Technology Co Ltd; Suzhou Power Supply Co of State Grid Jiangsu Electric Power Co Ltd; State Grid Corp of China SGCC
Current assignee: NARI Technology Co Ltd; Suzhou Power Supply Co of State Grid Jiangsu Electric Power Co Ltd; State Grid Corp of China SGCC
Priority date: 2016-06-15
Filing date: 2016-06-15
Publication date: 2016-10-26

Abstract

The invention discloses a three-phase modeling method for a distribution network, comprising the steps of: (1) establishing distribution network line models including a Bergeron model and a PI equivalent model of a transmission line; (2) establishing a three-phase balance model of a distribution transformer of the distribution network; and (3) expressing the static characteristics of active power and reactive power of a load by using a polynomial in combination with three-phase modeling for establishing the load of the distribution network. The three-phase modeling method for the distribution network provides a theoretical basis for three-phase model establishment and parameter setting of equipment in an intelligent distribution network management system, and provides a theoretical foundation for on-line three-phase asymmetrical load flow analysis.

Description

Three-phase Modeling Method for Distribution Network

技术领域technical field

本发明涉及建模技术领域，尤其涉及配电网三相建模方法。The invention relates to the technical field of modeling, in particular to a three-phase modeling method of a power distribution network.

背景技术Background technique

单相等值电路是基于三相元件参数完全对称，三相电流、电压完全对称的条件下得到的。它以无穷远处为零电位点，并且计及另外两相的影响之后得到以零电位点为公共端的单相等值电路。基于系统三相电气量、网络参数完全对称的，这与现代电力系统三相参数越来越不对称的状况以及运行状况导致的不对称等等情况不相符合。The single-phase equivalent circuit is obtained under the condition that the parameters of the three-phase components are completely symmetrical, and the three-phase current and voltage are completely symmetrical. It takes infinity as the zero potential point, and after taking into account the influence of the other two phases, a single-phase equivalent circuit with the zero potential point as the common terminal is obtained. Based on the complete symmetry of the three-phase electrical quantities and network parameters of the system, this is inconsistent with the increasingly asymmetrical three-phase parameters of the modern power system and the asymmetry caused by operating conditions.

电力系统中发电机、变压器、对称输电线路都可由三序解耦的等值电路表示，配电网的参数不对称性比较突出，线路不换位，各相负荷不对称，以及中性点接地方式的多样性，包括直接接地、经电阻接地、消弧线圈接地等等，配电网元件参数不对称导致配电网的电气量(包括电压、电流)不对称，从而中性点电压偏移比较严重，因而不宜采用传统的对称分量法进行计算分析，而宜采用相分量法。从电路的基本原理出发，从基本物理现象入手，建立符合实际物理意义的三相元件的相分量等值电路，方便用电路形式表示，并用相分量法对电网进行分析、计算。In the power system, generators, transformers, and symmetrical transmission lines can all be represented by three-sequence decoupled equivalent circuits. The parameter asymmetry of the distribution network is relatively prominent, the lines are not transposed, the loads of each phase are asymmetrical, and the neutral point is grounded. Diversity of ways, including direct grounding, resistance grounding, arc suppressing coil grounding, etc. The asymmetry of the distribution network component parameters leads to the asymmetry of the electrical quantity (including voltage and current) of the distribution network, so that the neutral point voltage shifts Therefore, it is not suitable to use the traditional symmetrical component method for calculation and analysis, but the phase component method should be used. Starting from the basic principle of the circuit and starting from the basic physical phenomena, the phase component equivalent circuit of the three-phase components in line with the actual physical meaning is established, which is conveniently expressed in the form of a circuit, and the power grid is analyzed and calculated by the phase component method.

发明内容Contents of the invention

针对上述问题，本发明提出一种配电网建模方法。In view of the above problems, the present invention proposes a distribution network modeling method.

实现上述技术目的，达到上述技术效果，本发明通过以下技术方案实现：Realize above-mentioned technical purpose, reach above-mentioned technical effect, the present invention realizes through the following technical solutions:

一种配电网三相建模方法，包括以下步骤：A three-phase modeling method for a distribution network, comprising the following steps:

(1)建立配电网线路模型，包括建立输电线路的贝杰龙模型和PI型等值模型；(1) Establish distribution network line models, including Bergeron model and PI equivalent model of transmission lines;

(2)利用对称向量法建立配电网配电变压器三相平衡模型；(2) Establish the three-phase balance model of the distribution transformer in the distribution network by using the symmetrical vector method;

(3)结合建立配电网负荷三相建模，利用多项式表示负荷的有功功率和无功功率静态特性。(3) Combining with the establishment of three-phase modeling of distribution network load, polynomials are used to represent the static characteristics of active power and reactive power of the load.

所述步骤(1)中建立输电线路贝杰龙模型具体包括：In the described step (1), setting up the transmission line Bergeron model specifically includes:

1a：建立传输线方程，1a: Establish the transmission line equation,

$- - \frac{\partial \partial u u}{\partial \partial x x} = = L L \frac{\partial \partial i i}{\partial \partial t t} + + R R i i$

$- - \frac{\partial \partial i i}{\partial \partial x x} = = L L \frac{\partial \partial u u}{\partial \partial t t} + + R R u u$

式中：u和i代表在距离为x处的线路的电压和电流，R是线路每单位长度的串联阻抗，L是单位长度串联电感；In the formula: u and i represent the voltage and current of the line at a distance of x, R is the series impedance per unit length of the line, and L is the series inductance per unit length;

1b：利用传输线方程推到出输电线路的贝杰龙模型，贝杰龙模型方程式为：1b: Use the transmission line equation to deduce the Bergeron model of the transmission line. The Bergeron model equation is:

${i i}_{k k} ((t t)) = = \frac{11}{Z Z} {u u}_{k k} ((t t)) - - \frac{11}{Z Z} {u u}_{m m} ((t t - - τ τ)) - - {i i}_{m m} ((t t - - τ τ))$

${i i}_{m m} ((t t)) = = \frac{11}{Z Z} {u u}_{m m} ((t t)) - - \frac{11}{Z Z} {u u}_{k k} ((t t - - τ τ)) - - {i i}_{k k} ((t t - - τ τ))$

式中：下标k和m分别代表发送端和接受端，i_k为时域沿传输线电流，u_k为时域沿传输线电压，t表示为时间，τ表示为电磁波由线路一段到另一端所需要的时间。In the formula, the subscripts k and m stand for the sending end and the receiving end respectively, _{ik is the current along the transmission line in the time domain, u k} _is the voltage along the transmission line in the time domain, t is time, τ is the electromagnetic wave from one end of the line to the other end time needed.

所述步骤(1)中，PI型等值模型包括：三相不平衡模型和三相平衡模型，In the step (1), the PI type equivalent model includes: a three-phase unbalanced model and a three-phase balanced model,

所述三相不平衡模型为：The three-phase unbalanced model is:

$L L \frac{{di di}_{i i j j} ((t t))}{d d t t} + + {Ri Ri}_{i i j j} ((t t)) = = {u u}_{i i} ((t t)) - - {u u}_{j j} ((t t))$

$C C \frac{{dU U}_{i i} ((t t))}{d d t t} = = {i i}_{i i} ((t t))$

i_ij为线路从端点i流入端点j的电流，u_i、u_j分别为线路端点i和j的电压，R、L为线路的电阻和电感矩阵，则线路阻抗矩阵为：i _ij is the current of the line flowing from terminal i to terminal j, u _i and u _j are the voltages of line terminal i and j respectively, R and L are the resistance and inductance matrix of the line, then the line impedance matrix is:

$R R + + j j w w L L = = [\begin{matrix} {R R}_{1111} + + {jwL wxya}_{1111} & {R R}_{1212} + + {jwL wxya}_{1212} & {R R}_{1313} + + {jwL wxya}_{1313} \\ {R R}_{21 twenty one} + + {jwL wxya}_{21 twenty one} & {R R}_{22 twenty two} + + {jwL wxya}_{22 twenty two} & {R R}_{23 twenty three} + + {jwL wxya}_{23 twenty three} \\ {R R}_{3131} + + {jwL wxya}_{3131} & {R R}_{3232} + + {jwL wxya}_{3232} & {R R}_{3333} + + {jwL wxya}_{3333} \end{matrix}]$

式中：对角元素分别为三相线路的自阻和自电感，非对角元素为三相线路的互电阻和互电感。In the formula: the diagonal elements are the self-resistance and self-inductance of the three-phase line, and the off-diagonal elements are the mutual resistance and mutual inductance of the three-phase line.

线路并联电容元件矩阵：Line shunt capacitive element matrix:

$[\begin{matrix} {C C}_{11 g g} & - - {C C}_{1212} & - - {C C}_{1313} \\ - - {C C}_{1212} & {C C}_{22 g g} & - - {C C}_{23 twenty three} \\ - - {C C}_{1313} & - - {C C}_{23 twenty three} & {C C}_{33 g g} \end{matrix}] = = [\begin{matrix} {C C}_{1010} + + {C C}_{1212} + + {C C}_{1313} & - - {C C}_{1212} & - - {C C}_{1313} \\ - - {C C}_{1212} & {C C}_{2020} + + {C C}_{1212} + + {C C}_{23 twenty three} & - - {C C}_{23 twenty three} \\ - - {C C}_{1313} & - - {C C}_{23 twenty three} & {C C}_{3030} + + {C C}_{1313} + + {C C}_{23 twenty three} \end{matrix}]$

式中，C₁₀、C₂₀、C₃₀分别为三相线路对地点容，C₁₂、C₁₃、C₂₃分别为三相线路相间电容。In the formula, C ₁₀ , C ₂₀ , and C ₃₀ are the three-phase line-to-ground capacitances, and C ₁₂ , C ₁₃ , and C ₂₃ are the phase-to-phase capacitances of the three-phase lines.

所述三相平衡模型为：The three-phase equilibrium model is:

R_s＝(R_z+2*R_p)/3R _s =(R _z +2*R _p )/3

X_s＝(X_z+2*X_p)/3X _s =(X _z +2*X _p )/3

B_s＝(B_z+2*B_p)/3B _s =(B _z +2*B _p )/3

式中：R_s为自电阻，X_s为自电抗值，B_s为电纳值，R_z、X_z、B_z分别是正序单位长度电阻、正序串联感性电抗、正序并联电纳。R_p、X_p、B_p分别是零序单位长度电阻、零序串联感性电抗、零序并联电纳；In the formula: R _s is the self-resistance, X _s is the self-reactance value, B _s is the susceptance value, R _z , X _z , and B _z are the positive sequence unit length resistance, positive sequence series inductive reactance, and positive sequence parallel susceptance respectively. R _p , X _p , and B _p are the zero-sequence unit length resistance, zero-sequence series inductive reactance, and zero-sequence parallel susceptance, respectively;

R_m＝(R_z-R_p)/3R _m =(R _z -R _p )/3

X_m＝(X_z-X_p)/3X _m = (X _z −X _p )/3

B_m＝(B_z-B_p)/3B _m =(B _z -B _p )/3

式中：R_m为互电阻，X_m为互抗值，B_m为互电纳值。In the formula: R _m is the mutual resistance, X _m is the mutual reactance value, and B _m is the mutual susceptance value.

所述步骤(2)中，利用对称相量法进行配电网配电变压器三相建模，具体为：In the step (2), the three-phase modeling of the distribution transformer of the distribution network is carried out using the symmetrical phasor method, specifically:

$[\begin{matrix} V V 11 \\ V V 22 \end{matrix}] = = [\begin{matrix} L L 11 + + L L 1212 & 11 / / a a * * L L 1212 \\ 11 / / a a * * L L 1212 & 11 / / {a a}^{22} * * ((L L 22 + + L L 1212)) \end{matrix}] d d / / d d t t [\begin{matrix} i i 11 \\ i i 22 \end{matrix}]$

式中：L1和L2表示漏电感，L12为励磁支路电感，a为变比，V1、V2分别为变压器一次和二次绕组电压，i1、i2为电流。In the formula: L1 and L2 represent the leakage inductance, L12 is the excitation branch inductance, a is the transformation ratio, V1 and V2 are the primary and secondary winding voltages of the transformer respectively, and i1 and i2 are the current.

所述步骤(3)中，配电网负荷三相建模的模型具体为：In the step (3), the model of three-phase modeling of distribution network load is specifically:

P_L＝(a_pV²+b_pV+C_p)(1+k_pΔf)P _L ＝(a _p V ² +b _p V+C _p )(1+k _p Δf)

Q_L＝(a_qV²+b_qV+C_q)(1+k_qΔf)Q _L ＝(a _q V ² +b _q V+C _q )(1+k _q Δf)

式中P_L、Q_L、V为负荷有功、无功和端电压的标么值，分别以给定的初始值P₀、Q₀、V₀为基准，且a_p+b_p+c_p＝1、a_q+b_q+c_q＝1，k_p＝dP_L/df、k_q＝dP_L/df，f和Δf分别为表示频率和频差。In the formula, P _L , Q _L , V are the unit values of load active power, reactive power and terminal voltage, which are based on the given initial values P ₀ , Q ₀ , V ₀ respectively, and a _p +b _p +c _p =1, a _q +b _q +c _q =1, k _p =dP _L /df, k _q =dP _L /df, f and Δf represent frequency and frequency difference respectively.

本发明的有益效果：Beneficial effects of the present invention:

本发明提供的配电网三相建模为智能配电网管理系统实现设备三相模型建立，参数设置提供理论依据，为在线三相不对称负荷潮流分析提供理论基础。The three-phase modeling of the distribution network provided by the present invention provides a theoretical basis for the intelligent distribution network management system to realize the establishment of the three-phase model of the equipment, parameter setting, and provides a theoretical basis for the analysis of the online three-phase asymmetrical load flow.

附图说明Description of drawings

图1为本发明一种实施例的流程示意图。Fig. 1 is a schematic flow chart of an embodiment of the present invention.

图2为本发明一种实施例的频变模型传输线路的时间域等值电路。FIG. 2 is a time-domain equivalent circuit of a frequency-varying model transmission line according to an embodiment of the present invention.

图3为本发明一种实施例的Zeq实现原理图。Fig. 3 is a schematic diagram of Zeq implementation in an embodiment of the present invention.

图4为本发明一种实施例的贝杰龙模型对应的时间域等值电路。FIG. 4 is a time-domain equivalent circuit corresponding to the Bergeron model according to an embodiment of the present invention.

图5为本发明一种实施例的三相耦合的PI结构模型图。Fig. 5 is a diagram of a PI structure model of three-phase coupling according to an embodiment of the present invention.

图6为本发明一种实施例的三相平衡集中参数等值电路。Fig. 6 is a three-phase balanced lumped parameter equivalent circuit of an embodiment of the present invention.

图7为本发明一种实施例的线性耦合变压器等值电路。FIG. 7 is an equivalent circuit of a linear coupling transformer according to an embodiment of the present invention.

图8为本发明一种实施例的带变比的线性耦合变压器等值电路。Fig. 8 is an equivalent circuit of a linear coupling transformer with a transformation ratio according to an embodiment of the present invention.

具体实施方式detailed description

为了使本发明的目的、技术方案及优点更加清楚明白，以下结合实施例，对本发明进行进一步详细说明。应当理解，此处所描述的具体实施例仅仅用以解释本发明，并不用于限定本发明。In order to make the object, technical solution and advantages of the present invention more clear, the present invention will be further described in detail below in conjunction with the examples. It should be understood that the specific embodiments described here are only used to explain the present invention, not to limit the present invention.

下面结合附图对本发明的应用原理作详细的描述。The application principle of the present invention will be described in detail below in conjunction with the accompanying drawings.

如图1所示，一种配电网三相建模方法，包括以下步骤：As shown in Figure 1, a three-phase modeling method for distribution network includes the following steps:

(2)建立配电网配电变压器三相平衡模型；(2) Establish the three-phase balance model of the distribution transformer in the distribution network;

1a：建立传输线路方程，1a: Establish transmission line equations,

$- - \frac{\partial \partial u u}{\partial \partial x x} = = L L \frac{\partial \partial i i}{\partial \partial t t} + + R R i i - - - - - - ((11))$

$- - \frac{\partial \partial i i}{\partial \partial x x} = = L L \frac{\partial \partial u u}{\partial \partial t t} + + R R u u - - - - - - ((22))$

由于很难直接写出在时域的传输线路方程式的求解，当考虑频率参数和损耗分布式特征，在频域很容易求解，在频域线路方程式的求解方程式为：Since it is difficult to directly write the solution of the transmission line equation in the time domain, when frequency parameters and loss distribution characteristics are considered, it is easy to solve in the frequency domain. The solution equation of the line equation in the frequency domain is:

U_k(ω)＝cosh[γ(ω)l]U_m(ω)-Z_c(ω)sinh[γ(ω)l]I_m(ω) (3)U _k (ω)＝cosh[γ(ω)l]U _m (ω)-Z _c (ω)sinh[γ(ω)l]I _m (ω) (3)

I_k(ω)＝sinh[γ(ω)l]U_m(ω)/Z_c(ω)-cosh[γ(ω)l]I_m(ω) (4)I _k (ω)＝sinh[γ(ω)l]U _m (ω)/Z _c (ω)-cosh[γ(ω)l]I _m (ω) (4)

下标k和m分别代表发送端和接受端，l表示线路的长度，Z_c(ω)anγ(ω)分别为阻抗特征和传播常数，定义如下：The subscripts k and m represent the transmitting end and the receiving end respectively, l represents the length of the line, and Z _c (ω)anγ(ω) are impedance characteristics and propagation constants respectively, which are defined as follows:

${Z Z}_{C C} ((ω ω)) = = \sqrt{\frac{R R + + j j ω ω L L}{G G + + j j ω ω C C}} - - - - - - ((55))$

$γ γ ((ω ω)) = = \sqrt{((R R + + j j ω ω L L)) ((G G + + j j ω ω C C))} - - - - - - ((66))$

在频域中的电流与电压之间的联系如下所述：The relationship between current and voltage in the frequency domain is as follows:

F_k(ω)＝U_k(ω)+Z_c(ω)I_k(ω) (7)F _k (ω)＝U _k (ω)+Z _c (ω)I _k (ω) (7)

F_m(ω)＝U_m(ω)+Z_c(ω)I_m(ω) (8)F _m (ω)＝U _m (ω)+Z _c (ω)I _m (ω) (8)

B_k(ω)＝U_k(ω)-Z_c(ω)I_k(ω) (9)B _k (ω)＝U _k (ω)-Z _c (ω)I _k (ω) (9)

B_m(ω)＝U_m(ω)-Z_c(ω)I_m(ω) (10)B _m (ω)＝U _m (ω)-Z _c (ω)I _m (ω) (10)

F表示向前，B表示向后，F_k(ω)、F_m(ω)、B_k(ω)和B_m(ω)为频域向前和向后电压行波函数。F means forward, B means backward, and F _k (ω), F _m (ω), B _k (ω) and B _m (ω) are forward and backward voltage traveling wave functions in frequency domain.

排除U_k(ω)和U_m(ω)，得出：Excluding U _k (ω) and U _m (ω), we get:

B_k(ω)＝A₁(ω)F_m(ω) (11)B _k (ω)＝A ₁ (ω)F _m (ω) (11)

B_m(ω)＝A₁(ω)F_k(ω) (12)B _m (ω)＝A ₁ (ω)F _k (ω) (12)

A₁(ω)＝e^-γ(ω)l＝1/{cosh[γ(ω)l]+sinh[γ(ω)l]} (13)A ₁ (ω)＝e ^-γ(ω)l ＝1/{cosh[γ(ω)l]+sinh[γ(ω)l]} (13)

如图2所示，为频变模型传输线路的时间域等值电路。As shown in Figure 2, it is the time-domain equivalent circuit of the frequency-varying model transmission line.

1b：利用传输线方程推到出输电线路的贝杰龙模型，及与贝杰龙模型对应的时间域等值电路，贝杰龙模型方程式为：1b: Use the transmission line equation to deduce the Bergeron model of the transmission line, and the time-domain equivalent circuit corresponding to the Bergeron model. The Bergeron model equation is:

具体为：Specifically:

a)Z_c(ω)的测定a) Determination of Z _c (ω)

特征阻抗Z_c(ω)是频率的函数，不能直接地用于图3，因为它通常是频率的列表函数，而且，在指定频率用单一阻抗也是不合理的。如果能找到个等值网络阻抗Zeq，该网络的频率响应与线路的特征阻抗值是相同的，那么Z_c(ω)能用Zeq替换。Zeq由J.Marti[1]发展起来的，由不依赖频率的无源电路元件R和C组成。采用了渐进式技术发展Zeq网络，网络由一系列电阻和电容并列模块组成。如图3所示：The characteristic impedance, Z _c (ω), is a function of frequency and cannot be used directly in Figure 3, since it is usually a tabular function of frequency, and it is not reasonable to use a single impedance at a given frequency. If an equivalent network impedance Zeq can be found, the frequency response of the network is the same as the characteristic impedance value of the line, then Z _c (ω) can be replaced by Zeq. Zeq was developed by J.Marti[1] and consists of frequency-independent passive circuit elements R and C. A progressive technique is used to develop the Zeq network, which consists of a series of parallel modules of resistors and capacitors. As shown in Figure 3:

图3的Zeq实现原理如下：The implementation principle of Zeq in Figure 3 is as follows:

Bode’s的渐进式技术能用于频域接近于Zc(ω)，采用合理的传递函数，如下所示：Bode's asymptotic technique can be used to approximate Zc(ω) in the frequency domain, using a reasonable transfer function as follows:

$H h ((s the s)) = = \frac{H h ((11 + + s the s / / {Z Z}_{11})) ((11 + + s the s / / {Z Z}_{22})) ... ... ((11 + + s the s / / {Z Z}_{n no}))}{((11 + + s the s / / {p p}_{11})) ((11 + + s the s / / {p p}_{22})) ... ... ((11 + + s the s / / {p p}_{n no}))}$

传递函数H(s)的所有极点和零点是实数，并位于复平面的左侧。近似函数能够通过幅值|H(ω)|以近似方式轨迹跟踪。|H(ω)|位于定义的渐近线的边界线内。在0或者正负20db/decade的直线段组成一个包迹。All poles and zeros of the transfer function H(s) are real numbers and lie on the left side of the complex plane. The approximation function can be tracked in an approximate fashion by the magnitude |H(ω)|. |H(ω)| lies within the boundary line of the defined asymptote. Line segments at 0 or plus or minus 20db/decade form an envelope.

有理函数Zeq包含在Z_c(ω)和渐进线中，渐进包迹的相交点定义了有理函数极点和零点。程序从水平参考面开始，每个步长加一个极值点，斜率会下降20db/decade。当加一个零点，斜率会增加20db/decade。极点和零点的个数取决于在加下一个点时，渐近线与数据之间被允许有多远的距离。偏差由下面的值指定。The rational function Zeq is contained in Z _c (ω) and the asymptote, and the intersection points of the asymptotic envelope define the rational function poles and zeros. The program starts from the horizontal reference plane, and an extreme point is added to each step, and the slope will drop by 20db/decade. When adding a zero, the slope will increase by 20db/decade. The number of poles and zeros depends on how far the asymptote is allowed to be from the data when adding the next point. The bias is specified by the value below.

有理函数Zeq发展为分式之和表达式。The rational function Zeq is developed as the expression of the sum of fractions.

$H h ((s the s)) = = {A A}_{00} + + \frac{{A A}_{11}}{s the s + + {p p}_{11}} + + \frac{{A A}_{22}}{s the s + + {p p}_{22}} + + .... .... + + \frac{{A A}_{n no}}{s the s + + {p p}_{n no}}$

R₀＝A₀ R ₀ =A ₀

R_i＝A_i/P_i i＝1,2,3…nR _i =A _i /P _i i=1,2,3...n

C_i＝1/A_i i＝1,2,3…nC _i =1/A _i i =1,2,3...n

利用计算得到的R₀、R_i、C_i通过图3中RC网络可以计算得到线路的R值和C值。The R value and C value of the line can be calculated by using the calculated R ₀ , R _i , and C _i through the RC network in Figure 3 .

b)b_k(t)和b_m(t)的测定b) Determination of b _k (t) and b _m (t)

贝杰龙模型线路模型既用于频变模型，也可以用于非频变模型，而不需要修改通用公式，The Bergeron model line model can be used for both frequency-dependent and non-frequency-dependent models without modifying the general formula,

I_k(ω)＝U_k(ω)/Z_c(ω)-B_k(ω)/Z_c(ω) (14)I _k (ω)＝U _k (ω)/Z _c (ω)-B _k (ω)/Z _c (ω) (14)

在时域中，In the time domain,

$\begin{matrix} {i i}_{k k} ((t t)) = = {u u}_{k k} ((t t)) / / {Z Z}_{c c} - - {b b}_{k k} ((ω ω)) / / {Z Z}_{c c} \\ = = {u u}_{k k} ((t t)) / / {Z Z}_{c c} - - {I I}_{k k h h i i s the s} \end{matrix} - - - - - - ((1515))$

其中I_khis＝b_k(ω)/Z_c，这里I_khis为贝杰龙模型电流源，如图4所示，Wherein I _khis =b _k (ω)/Z _c , where I _khis is a Bergeron model current source, as shown in Figure 4,

Bk(ω)＝A₁(ω)Fm(ω)Bk(ω)＝A ₁ (ω)Fm(ω)

Bm(ω)＝A₁(ω)Fk(ω)Bm(ω)＝A ₁ (ω)Fk(ω)

在时域等价方程式为：The equivalent equation in the time domain is:

${b b}_{k k} ((t t)) = = {&Integral; &Integral;}_{τ τ}^{t t} {f f}_{m m} ((t t - - u u)) {a a}_{11} ((u u)) d d u u - - - - - - ((1616))$

${b b}_{m m} ((t t)) = = {&Integral; &Integral;}_{τ τ}^{t t} {f f}_{k k} ((t t - - u u)) {a a}_{11} ((u u)) d d u u - - - - - - ((1717))$

衰减函数a₁(t)(或权重函数)可以描述为指数函数和的形式。递归卷积用于求解b_k(t)和b_m(t)，如果极点数为0，a₁(t)在频域为常数，在时域对应脉冲函数有τ的延时。The decay function a ₁ (t) (or weight function) can be described as a sum of exponential functions. Recursive convolution is used to solve b _k (t) and b _m (t). If the number of poles is 0, a ₁ (t) is a constant in the frequency domain, and there is a delay of τ in the time domain corresponding to the pulse function.

c)i_k(t)和i_m(t)的测定c) Determination of i _k (t) and i _m (t)

等式(11)和(12)在时域为：Equations (11) and (12) in the time domain are:

b_t(t)＝δ(t-τ)f_m(t)＝f_m(t-τ) (18)b _t (t)=δ(t-τ)f _m (t)=f _m (t-τ) (18)

b_m(t)＝δ(t-τ)f_k(t)＝f_k(t-τ) (19)b _m (t) = δ(t-τ) f _k (t) = f _k (t-τ) (19)

${I I}_{k k h h i i s the s} = = \frac{{b b}_{k k} ((t t)) - - {E E.}_{e e q q k k} ((t t))}{{R R}_{e e q q}} - - - - - - ((2020))$

${I I}_{m m h h i i s the s} = = \frac{{b b}_{m m} ((t t)) - - {E E.}_{e e q q m m} ((t t))}{{R R}_{e e q q}} - - - - - - ((21 twenty one))$

当E_eqk(t)＝0.0，E_eqm(t)＝0.0(没有Z_c(ω))When E _eqk (t) = 0.0, E _eqm (t) = 0.0 (without Z _c (ω))

${I I}_{k k h h i i s the s} = = \frac{{b b}_{k k} ((t t))}{z z} = = \frac{{f f}_{m m} ((t t - - τ τ))}{z z} = = \frac{11}{z z} (({u u}_{m m} ((t t - - τ τ)) + + {Zi Zi}_{m m} ((t t - - τ τ)))) = = \frac{11}{z z} (({u u}_{m m} ((t t - - τ τ)) + + {i i}_{m m} ((t t - - τ τ)) - - - - - - ((22 twenty two))$

那么(15)就改变为Then (15) is changed to

${i i}_{k k} ((t t)) = = \frac{11}{Z Z} {u u}_{k k} ((t t)) - - \frac{11}{Z Z} {u u}_{m m} ((t t - - τ τ)) - - {i i}_{m m} ((t t - - τ τ)) - - - - - - ((23 twenty three))$

在节点m，相似的方程为：At node m, the similar equation is:

${i i}_{m m} ((t t)) = = \frac{11}{Z Z} {u u}_{m m} ((t t)) - - \frac{11}{Z Z} {u u}_{k k} ((t t - - τ τ)) - - {i i}_{k k} ((t t - - τ τ)) - - - - - - ((24 twenty four))$

式(23)和(24)是典型的贝杰龙表达式。Equations (23) and (24) are typical Bergeron expressions.

衰减函数A₁(ω)＝e^-γl，曲线起始点幅度不是1，而是更小些为0.9963。Attenuation function A ₁ (ω)=e ^-γl , the amplitude of the starting point of the curve is not 1, but smaller at 0.9963.

${I I}_{k k h h i i s the s} = = \frac{{b b}_{k k} ((t t))}{Z Z} = = \frac{{f f}_{m m} ((t t))}{Z Z} = = \frac{0.9963 0.9963}{Z Z} {u u}_{m m} ((t t - - τ τ)) + + {i i}_{m m} ((t t - - τ τ)) - - - - - - ((2525))$

是贝杰龙模型电流源的近似值。is an approximation of the Bergeron model current source.

所述步骤(1)中，PI型等值模型包括：三相不平衡模型和三相平衡模型，具体为：In the step (1), the PI type equivalent model includes: a three-phase unbalanced model and a three-phase balanced model, specifically:

将线路作为集中参数处理并等值为一个π型电路，由于其仅能较近似地反映较短线路的工频特性，从而使其应用范围受到限制。当线路的长度受到限制，线路的行波时间小于计算的步长时间，这种模型技术仍然被应用，而不能使用行波模型。假定行波传播速度等于光速，最小计算步长为50微妙，那么线路的最长距离不超过15公里。也就是说当计算步长为50微妙，小于15公里的线路都可以利用PI模型分析计算。The line is treated as a concentrated parameter and equivalent to a π-type circuit, because it can only reflect the power frequency characteristics of a short line approximately, so its application range is limited. When the length of the line is limited and the travel time of the line is less than the calculated step time, this modeling technique is still applied instead of the traveling wave model. Assuming that the propagation speed of traveling waves is equal to the speed of light, and the minimum calculation step size is 50 microseconds, then the longest distance of the line does not exceed 15 kilometers. That is to say, when the calculation step is 50 microseconds, the lines less than 15 kilometers can be analyzed and calculated by using the PI model.

如图5所示，所述三相不平衡模型为：As shown in Figure 5, the three-phase unbalanced model is:

$C C \frac{{dU U}_{i i} ((t t))}{d d t t} = = {i i}_{i i} ((t t))$

线路并联电容元件矩阵：Line shunt capacitive element matrix:

所述三相平衡模型为：The three-phase equilibrium model is:

注：三相参数不对称，则矩阵的非对角元素不全为0，各序对称分量将不具有独立性。这时不能按序进行独立计算。Note: If the three-phase parameters are not symmetrical, the off-diagonal elements of the matrix are not all 0, and the symmetrical components of each sequence will not be independent. In this case, independent calculations cannot be performed sequentially.

如图6所示，所述三相平衡模型为：As shown in Figure 6, the three-phase equilibrium model is:

R_s＝(R_z+2*R_p)/3R _s =(R _z +2*R _p )/3

X_s＝(X_z+2*X_p)/3X _s =(X _z +2*X _p )/3

B_s＝(B_z+2*B_p)/3B _s =(B _z +2*B _p )/3

R_m＝(R_z-R_p)/3R _m =(R _z -R _p )/3

X_m＝(X_z-X_p)/3X _m = (X _z −X _p )/3

B_m＝(B_z-B_p)/3B _m =(B _z -B _p )/3

模型适用性分析：Model suitability analysis:

π模型适用于短距离架空线路或电缆。π模型往往用于电力系统的稳态计算，当考虑动态过程或者进行动模实验和TNA模拟时，则往往采用π模型，它将一条长线路分段为若干段，每段线路用一个π结构模型进行模拟，由多个π结构组成的π型链能够较好的反映出线路的暂态特性。在暂态过程中，由于π结构模型是由集中参数组成，其会产生虚假暂态振荡。The π model is suitable for short-distance overhead lines or cables. The π model is often used for steady-state calculations of power systems. When considering dynamic processes or performing dynamic model experiments and TNA simulations, the π model is often used. It divides a long line into several sections, and each section uses a π structure The π-type chain composed of multiple π structures can better reflect the transient characteristics of the line. In the transient process, since the π structure model is composed of lumped parameters, it will produce false transient oscillations.

在电磁暂态计算程序中通常不推荐使用π结构模型，主要是因为π结构模型中是采用某一固定频率(通常为工频)下的参数，它不能反映其他频率的线路特性，并且在暂态过程中，由于π结构的集中参数性质，计算结果容易出现虚假振荡。另外π结构模型由集中参数构成，在进行计算时需要增加相关节点，影响计算时间和效率。因此π结构模型主要应用于稳态计算以及用于模拟一些非常短的线路(由于线路太短，采用行波模型无法计算)。It is usually not recommended to use the π-structure model in the electromagnetic transient calculation program, mainly because the π-structure model uses parameters at a certain fixed frequency (usually power frequency), which cannot reflect the line characteristics of other frequencies, and in the transient In the state process, due to the lumped parameter nature of the π structure, the calculation results are prone to spurious oscillations. In addition, the π structure model is composed of concentrated parameters, and related nodes need to be added during calculation, which affects calculation time and efficiency. Therefore, the π structure model is mainly used in steady-state calculations and for simulating some very short lines (because the lines are too short, the traveling wave model cannot be used for calculation).

模型适用频带Model applicable frequency band

对于一定频率以上的高频信号，使用集中参数π模型产生的误差较大，会造成模型的精确度下降，影响区段定位的灵敏度和可靠性。元件模型的适用频带就是指元件数学模型的响应和实际物理模型响应差异较小(满足工程需要)的频带。For high-frequency signals above a certain frequency, the error generated by using the lumped parameter π model will be large, which will cause the accuracy of the model to decrease and affect the sensitivity and reliability of section positioning. The applicable frequency band of the component model refers to the frequency band in which the difference between the response of the component mathematical model and the response of the actual physical model is small (meeting engineering needs).

对配网线路π模型的适用频带进行分析可知，在一定截止频率之下的频带内，集中参数π模型和分布参数模型的相频、幅频特性非常接近，且均呈现容性；除了首段容性频带以外不再存在公共的容性频带，所以将输电线路的首段容性频带选定为适用频带，只不过随着输电线路的增长适用频带的上限截至频率将逐渐下降。The analysis of the applicable frequency band of the distribution network π model shows that, in the frequency band below a certain cut-off frequency, the phase-frequency and amplitude-frequency characteristics of the concentrated parameter π model and the distributed parameter model are very close, and both are capacitive; except for the first paragraph There is no public capacitive frequency band other than the capacitive frequency band, so the first section of the capacitive frequency band of the transmission line is selected as the applicable frequency band, but the upper limit cut-off frequency of the applicable frequency band will gradually decrease with the growth of the transmission line.

从以上分析可知，在一定频带内，线路可等效为集中参数π模型；事实上，在此基础上，线路可进一步简化为简单的对地电容模型。From the above analysis, it can be seen that within a certain frequency band, the line can be equivalent to a lumped parameter π model; in fact, on this basis, the line can be further simplified to a simple capacitance-to-ground model.

所述步骤(2)中，进行配电网配电变压器三相建模，具体为：In the step (2), the three-phase modeling of the distribution transformer of the distribution network is carried out, specifically:

$[\begin{matrix} V V 11 \\ V V 22 \end{matrix}] = = [\begin{matrix} L L 11 + + L L 1212 & 11 / / a a * * L L 1212 \\ 11 / / a a * * L L 1212 & 11 / / {a a}^{22} * * ((L L 22 + + L L 1212)) \end{matrix}] d d / / d d t t [\begin{matrix} i i 11 \\ t t 22 \end{matrix}]$

在电磁暂态分析中，变压器基本表达式使用T型模型，如图7所示，L1和L2表示漏电感，L12位励磁支路电感，In the electromagnetic transient analysis, the basic expression of the transformer uses a T-type model, as shown in Figure 7, L1 and L2 represent the leakage inductance, L12 is the excitation branch inductance,

$V V 11 - - L L 11 \frac{d d i i 11}{d d t t} = = V V m m$

$V V 22 - - L L 22 \frac{d d i i 22}{d d t t} = = V V m m$

i1+i2+im＝0i1+i2+im=0

$V V 11 - - L L 11 \frac{d d i i 11}{d d t t} = = - - L L 1212 \frac{d d ((- - i i 11 - - i i 22))}{d d t t}$

$V V 22 - - L L 22 \frac{d d i i 22}{d d t t} = = - - L L 1212 \frac{d d ((- - i i 11 - - i i 22))}{d d t t}$

$V V 11 = = L L 11 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d i i 22}{d d t t}$

$V V 22 = = L L 22 \frac{d d i i 22}{d d t t} + + L L 1212 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d i i 22}{d d t t}$

矩阵形式为：The matrix form is:

$[\begin{matrix} V V 11 \\ V V 22 \end{matrix}] = = [\begin{matrix} L L 11 + + L L 1212 & L L 1212 \\ L L 1212 & L L 22 + + L L 1212 \end{matrix}] d d / / d d t t [\begin{matrix} i i 11 \\ i i 22 \end{matrix}]$

其中：in:

$[[L L]] = = [\begin{matrix} L L 11 + + L L 1212 & L L 1212 \\ L L 1212 & L L 22 + + L L 1212 \end{matrix}]$

$d d / / d d t t [\begin{matrix} i i 11 \\ i i 22 \end{matrix}] = = {[[L L]]}^{- - 11} [\begin{matrix} V V 11 \\ V V 22 \end{matrix}]$

变压器变比可以在电路中通过添加理想的比率变换器，如图8所示：The transformer ratio can be added to the circuit by adding an ideal ratio converter, as shown in Figure 8:

引入变比值a后，则表达式变为：After introducing the variable ratio value a, the expression becomes:

$V V 11 - - L L 11 \frac{d d i i 11}{d d t t} = = V V m m$

$a a V V 22 - - L L 22 \frac{d d ((i i 22 / / a a))}{d d t t} = = V V m m$

i1+i2/a+im＝0i1+i2/a+im=0

$V V 11 - - L L 11 \frac{d d i i 11}{d d t t} = = - - L L 1212 \frac{d d ((- - i i 11 - - i i 22 / / a a))}{d d t t}$

$a a V V 22 - - L L 22 \frac{d d ((i i 22 / / a a))}{d d t t} = = - - L L 1212 \frac{d d ((- - i i 11 - - i i 22 / / a a))}{d d t t}$

$V V 11 = = L L 11 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d ((i i 22 / / a a))}{d d t t}$

$a a V V 22 = = L L 22 \frac{d d ((i i 22 / / a a))}{d d t t} + + L L 1212 \frac{d d i i 11}{d d t t} + + L L 1212 \frac{d d ((i i 22 / / a a))}{d d t t}$

$V V 22 = = 11 / / a a * * L L 22 * * 11 / / a a \frac{d d i i 22}{d d t t} + + 11 / / a a * * L L 1212 \frac{d d i i 11}{d d t t} + + 11 / / a a * * L L 1212 * * 11 / / a a \frac{d d i i 22}{d d t t}$

矩阵形式为：The matrix form is:

注意电感矩阵包括L1+L12，L2+L12，如果磁化电流非常的小，意味着L12的值非常的大，L12>>L1或者L2。Note that the inductance matrix includes L1+L12, L2+L12, if the magnetizing current is very small, it means that the value of L12 is very large, L12>>L1 or L2.

(1)变压器基本参数设置(1) Transformer basic parameter setting

变压器一次绕组及二次绕组类型：星型或者三角形，变压器模型类型：线性变压器，理想变压器，饱和变压器。Transformer primary winding and secondary winding type: star or delta, transformer model type: linear transformer, ideal transformer, saturated transformer.

(2)变压器电气参数设置(2) Transformer electrical parameter setting

变压器一次及二次绕组的线电压有效值L-L RMS，单位为KV，是否分接头调整：起始调整电压标幺值，调整幅度，调整最高标幺值，调整最低标幺值，变压器容量，单位MVA。The effective value of the line voltage L-L RMS of the primary and secondary windings of the transformer, the unit is KV, whether to adjust the tap: the initial adjustment voltage per unit value, the adjustment range, the highest adjustment per unit value, the lowest adjustment per unit value, transformer capacity, unit MVA.

P_L＝(a_pV²+b_pV+C_p)(1+k_pΔf)P _L ＝(a _p V ² +b _p V+C _p )(1+k _p Δf)

Q_L＝(a_qV²+b_qV+C_q)(1+k_qΔf)Q _L ＝(a _q V ² +b _q V+C _q )(1+k _q Δf)

式中P_L、Q_L、V为负荷功率和端电压的标么值，分别以给定的初始值P₀、Q₀、V₀为基准，且a_p+b_p+c_p＝1、a_q+b_q+c_q＝1，k_p＝dP_L/df、k_q＝dP_L/df，f和Δf分别为表示频率和频差。In the formula, P _L , Q _L , and V are the unit values of load power and terminal voltage, which are based on the given initial values P ₀ , Q ₀ , and V ₀ respectively, and a _p +b _p +c _p =1, a _q +b _q +c _q =1, k _p =dP _L /df, k _q =dP _L /df, f and Δf represent frequency and frequency difference respectively.

以上显示和描述了本发明的基本原理和主要特征和本发明的优点。本行业的技术人员应该了解，本发明不受上述实施例的限制，上述实施例和说明书中描述的只是说明本发明的原理，在不脱离本发明精神和范围的前提下，本发明还会有各种变化和改进，这些变化和改进都落入要求保护的本发明范围内。本发明要求保护范围由所附的权利要求书及其等效物界定。The basic principles and main features of the present invention and the advantages of the present invention have been shown and described above. Those skilled in the industry should understand that the present invention is not limited by the above-mentioned embodiments. What are described in the above-mentioned embodiments and the description only illustrate the principle of the present invention. Without departing from the spirit and scope of the present invention, the present invention will also have Variations and improvements are possible, which fall within the scope of the claimed invention. The protection scope of the present invention is defined by the appended claims and their equivalents.

Claims

1. a distribution network three-phase modeling method, is characterized in that, comprises the following steps:

(1) Establish distribution network line models, including Bergeron model and PI equivalent model of transmission lines;

(2) Establish the three-phase balance model of the distribution transformer in the distribution network;

(3) Combining with the establishment of three-phase modeling of distribution network load, polynomials are used to represent the static characteristics of active power and reactive power of the load.

2. a kind of distribution network three-phase modeling method according to claim 1, is characterized in that: in described step (1), setting up transmission line Bergeron model specifically comprises:

1a: Establish the transmission line equation,

- - \frac{\partial \partial u u}{\partial \partial x x} = = L L \frac{\partial \partial i i}{\partial \partial t t} + + R R i i

- - \frac{\partial \partial i i}{\partial \partial x x} = = L L \frac{\partial \partial u u}{\partial \partial t t} + + R R u u

In the formula: u and i represent the voltage and current of the line at a distance of x, R is the series impedance per unit length of the line, and L is the series inductance per unit length;

1b: Use the transmission line equation to deduce the Bergeron model of the transmission line. The Bergeron model equation is:

{i i}_{k k} ((t t)) = = \frac{11}{Z Z} {u u}_{k k} ((t t)) - - \frac{11}{Z Z} {u u}_{m m} ((t t - - τ τ)) - - {i i}_{m m} ((t t - - τ τ))

{i i}_{m m} ((t t)) = = \frac{11}{Z Z} {u u}_{m m} ((t t)) - - \frac{11}{Z Z} {u u}_{k k} ((t t - - τ τ)) - - {i i}_{k k} ((t t - - τ τ))

In the formula, the subscripts k and m represent the transmitting end and the receiving end respectively, _{ik is the current along the transmission line in the time domain, u k} _is the voltage along the transmission line in the time domain, t is time, τ is the electromagnetic wave from one end of the line to the other end time needed.

3. a kind of distribution network three-phase modeling method according to claim 1, is characterized in that: in described step (1), PI type equivalent model comprises: three-phase unbalanced model and three-phase balanced model,

The three-phase unbalanced model is:

L L \frac{{di di}_{i i j j} ((t t))}{d d t t} + + {Ri Ri}_{i i j j} ((t t)) = = {u u}_{i i} ((t t)) - - {u u}_{j j} ((t t))

C C \frac{{dU U}_{i i} ((t t))}{d d t t} = = {i i}_{i i} ((t t))

i _ij is the current of the line flowing from terminal i to terminal j, u _i and u _j are the voltages of line terminal i and j respectively, R and L are the resistance and inductance matrix of the line, then the line impedance matrix:

R R + + j j w w L L = = [\begin{matrix} {R R}_{1111} + + {jwL wxya}_{1111} & {R R}_{1212} + + {jwL wxya}_{1212} & {R R}_{1313} + + {jwL wxya}_{1313} \\ {R R}_{21 twenty one} + + {jwL wxya}_{21 twenty one} & {R R}_{22 twenty two} + + {jwL wxya}_{22 twenty two} & {R R}_{23 twenty three} + + {jwL wxya}_{23 twenty three} \\ {R R}_{3131} + + {jwL wxya}_{3131} & {R R}_{3232} + + {jwL wxya}_{3232} & {R R}_{3333} + + {jwL wxya}_{3333} \end{matrix}]

In the formula: the diagonal elements are the self-resistance and self-inductance of the three-phase line, and the off-diagonal elements are the mutual resistance and mutual inductance of the three-phase line. Line shunt capacitive element matrix:

[\begin{matrix} {C C}_{11 g g} & - - {C C}_{1212} & - - {C C}_{1313} \\ - - {C C}_{1212} & {C C}_{22 g g} & - - {C C}_{23 twenty three} \\ - - {C C}_{1313} & - - {C C}_{23 twenty three} & {C C}_{33 g g} \end{matrix}] = = [\begin{matrix} {C C}_{1010} + + {C C}_{1212} + + {C C}_{1313} & - - {C C}_{1212} & - - {C C}_{1313} \\ - - {C C}_{1212} & {C C}_{2020} + + {C C}_{1212} + + {C C}_{23 twenty three} & - - {C C}_{23 twenty three} \\ - - {C C}_{1313} & - - {C C}_{23 twenty three} & {C C}_{3030} + + {C C}_{1313} + + {C C}_{23 twenty three} \end{matrix}]

In the formula, C ₁₀ , C ₂₀ , and C ₃₀ are the three-phase line-to-ground capacitances, and C ₁₂ , C ₁₃ , and C ₂₃ are the phase-to-phase capacitances of the three-phase lines.

The three-phase equilibrium model is:

R _s =(R _z +2*R _p )/3

X _s =(X _z +2*X _p )/3

B _s =(B _z +2*B _p )/3

In the formula: R _s is the self-resistance, X _s is the self-reactance value, B _s is the susceptance value, R _z , X _z , and B _z are the positive sequence unit length resistance, positive sequence series inductive reactance, and positive sequence parallel susceptance respectively. R _p , X _p , and B _p are the zero-sequence unit length resistance, zero-sequence series inductive reactance, and zero-sequence parallel susceptance, respectively;

R _m =(R _z -R _p )/3

X _m = (X _z −X _p )/3

B _m =(B _z -B _p )/3

In the formula: R _m is the mutual resistance, X _m is the mutual reactance value, and B _m is the mutual susceptance value.

4. a kind of distribution network three-phase modeling method according to claim 1, is characterized in that: in described step (2), utilize symmetrical phasor method to carry out distribution network distribution transformer three-phase modeling, specifically for:

[\begin{matrix} V V 11 \\ V V 22 \end{matrix}] = = [\begin{matrix} L L 11 + + L L 1212 & 11 / / a a * * L L 1212 \\ 11 / / a a * * L L 1212 & 11 / / {a a}^{22} * * ((L L 22 + + L L 1212)) \end{matrix}] d d / / d d t t [\begin{matrix} i i 11 \\ t t 22 \end{matrix}]

In the formula: L1 and L2 represent the leakage inductance, L12 is the excitation branch inductance, a is the transformation ratio, V1 and V2 are the primary and secondary winding voltages of the transformer respectively, and i1 and i2 are the current.

5. a kind of distribution network three-phase modeling method according to claim 1, is characterized in that: in described step (3), the model of distribution network load three-phase modeling is specifically:

P _L ＝(a _p V ² +b _p V+C _p )(1+k _p Δf)

Q _L ＝(a _q V ² +b _q V+C _q )(1+k _q Δf)

In the formula, P _L , Q _L , V are the unit values of load active power, reactive power and terminal voltage, which are based on the given initial values P ₀ , Q ₀ , V ₀ respectively, and a _p +b _p +c _p =1, a _q +b _q +c _q =1, k _p =dP _L /df, k _q =dP _L /df, f and Δf represent frequency and frequency difference respectively.