CN103632031B - A kind of rural area based on load curve decomposition load type load modeling method - Google Patents

Abstract

The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, said method comprising the steps of: decomposes rural area load curve, and calculates the ratio of total amount shared by every kind of load in the load of rural area；Calculating static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and combine the power supply area network topology data of rural area load bus, obtain rural area load type load model.The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, the shortcoming the method overcoming conventional statistics synthesis, can be resident load Site Modeling quickly, easily and accurately, the load model generated by the present invention can improve the degree of accuracy that grid simulation calculates, and ensures power grid security, reliably, run economically.

Description

Rural load type load modeling method based on load curve decomposition

Technical Field

The invention relates to a modeling method, in particular to a rural load type load modeling method based on load curve decomposition.

Background

The digital simulation of the power system becomes a main tool for planning, designing, dispatching, operating and analyzing and researching the power system, mathematical models of all elements of the power system and a whole-system mathematical model formed by the mathematical models are the basis of the digital simulation of the power system, and whether the accuracy of the models directly influences simulation results and decision schemes based on the simulation results. The model and parameters adopted by simulation are important determining factors of simulation accuracy, detailed mathematical models and modeling technologies of the generator, the excitation system, the speed regulation system, the transformer and the power transmission line are well developed at present, and the power load model is still simple relatively and is often a practical model and parameters adopted from a basic physical concept. For many years, when various large power grids in China are used for analyzing and calculating power systems, a certain common load model (such as a motor + constant impedance model or a constant power + constant impedance model) is generally selected according to experience, and model parameters are determined qualitatively.

The power system load modeling determines the difficulty of establishing a mathematical model due to factors such as complexity, distribution, time variability and randomness. The current load modeling methods can be basically classified into 3 types, i.e., statistical synthesis methods, ensemble discrimination methods, and fault fitting methods. Each of these three methods has its advantages and disadvantages. The load model obtained by the statistical synthesis method has the advantages of clear physical concept and easy understanding by engineering personnel, but the core of the load model is established on the basis of complete statistical data and accurate load characteristics, which is difficult to realize and cannot be counted frequently, so that the characteristic of the load changing along with time cannot be considered. The overall measurement and identification method avoids a large amount of statistical work, the online real-time load characteristics which change along with time can be obtained possibly, the biggest problem is that disturbance accidents are excessively depended on, the physical significance of identified model parameters is not clear, and the other problem is that related devices are difficult to install in all substations in a system. The fault-fitting method has the advantages that the parameter determination process is consistent with the process of selecting parameters in the calculation of the program at present, and can obtain the recurrence under certain faults. But in fact, the method is a trial and error method, whether the load parameters under some faults are applicable to other faults is difficult to guarantee, and the fact that the load parameters of the whole system are the same and unchanged obviously does not accord with the essence of the load. The statistical synthesis method in the three methods is widely applied because the physical model is clear and the concept is clear, thereby being convenient for qualitatively understanding the load characteristics.

However, the conventional statistical synthesis method has certain disadvantages including:

(1) the load capacity obtained by investigation is not consistent with the actual load power, because the problem of concurrency exists, not all equipment is put into use in 24 hours, and therefore, time-interval investigation statistics needs to be carried out;

(2) with the lapse of time, the actual load power, the load composition and the network structure are likely to change, and if the load is subjected to investigation, statistics and modeling once and for all, the requirement on accuracy is difficult to achieve;

(3) the survey needs to count the load composition and parameters of thousands of users, the workload is huge, and accurate statistical results are difficult to obtain.

The random time variation of the combined load components is a substantial cause of the load characteristics having random time variation. This time-varying nature necessarily results in time-varying daily load curves for the power consumers and substations. Therefore, the daily load curves of the users and the substations inevitably contain rich information reflecting the load configuration characteristics.

According to the convention of industry load classification, loads can be classified into 4 categories of industrial, commercial and urban residential, rural and other loads. The rural load type is one of more complex load types and also is an important load type, and has a large proportion in each load type site. The rural load curve has the following characteristics:

the basic characteristics of the rural load containing irrigation load are as follows:

(1) rural loads also change along with the change of seasons and time, a large amount of irrigation loads are contained in spring and autumn in the daytime, a certain heating load is contained in winter, a certain refrigerating load is contained in summer, and the proportion of the heating load to the refrigerating load is not large.

(2) Spring and autumn have larger irrigation load, which is the main characteristic of rural load. The irrigation load generally occurs in the morning and afternoon of the day, and the proportion of the irrigation load in 3, 4 and 5 months and 9, 10 and 11 months is large;

(3) rural loads are generally small, regularity of load curves is not strong, and the proportion occupied by special load curves is high.

(4) The proportion of irrigation load of different rural load lines is different.

(5) For the load curve of each day, peak loads typically occur at night, and part of the line is heavily loaded during the day, possibly during the day. For evening loading, generally the loading is lower in spring and autumn, and higher in winter and summer. For daytime loads, the load is lower in month 2 and higher in other months.

Through characteristic analysis of daily load curves of the user and the transformer substation, the corresponding component proportion of the electric equipment in each typical electric industry and the corresponding component proportion of the electric industry in the transformer substation can be determined. Because the data acquisition and monitoring (SCADA) system and the load control management system (a negative control system for short) can respectively provide a real-time daily load curve of the comprehensive load of the transformer substation and a daily load curve of a power user, the determined comprehensive load composition proportion has the property of on-line and real-time modeling, and the inherent defect of the traditional statistical comprehensive load modeling can be fundamentally overcome.

Disclosure of Invention

In order to overcome the defects of the prior art, the invention provides a rural load type load modeling method based on load curve decomposition, which overcomes the defects of the traditional statistical synthesis method, can rapidly, conveniently and accurately model residential load stations, and can improve the accuracy of power grid simulation calculation and ensure the safe, reliable and economic operation of a power grid by using a load model generated by the method.

In order to achieve the purpose of the invention, the invention adopts the following technical scheme:

the invention provides a rural load type load modeling method based on load curve decomposition, which comprises the following steps:

step 1: decomposing a rural load curve, and calculating the proportion of each load in the rural loads to the total amount;

step 2: and calculating the static load equivalent parameter, the dynamic load equivalent parameter and the power distribution network system impedance, and combining power supply area network topology data of the rural load node to obtain a rural load type load model.

The step 1 comprises the following steps:

step 1-1: decomposing a rural load curve;

step 1-1-1: for a transformer substation with the supplied load type being rural load, combining daily load curves of 10kV or 6kV load outgoing lines of the transformer substation;

step 1-1-2: calculating the number of effective load days of the load of the transformer substation every month, considering the day as invalid if the load of a certain day has a zero value, and considering the day as invalid if the load of the certain day has a reverse sign phenomenon;

step 1-1-3: calculating an average daily load curve and an average monthly load curve of the load of the transformer substation every month;

step 1-1-4: calculating the maximum value of each daily load curve of each month at 6-17 points in the daytime period, selecting the maximum value of the daytime period from the average daily load curves of 2 months, and determining a load curve range meeting the requirement of a reference load curve, wherein the range is 0.8-1.15 times of the calculated maximum value of the daytime period;

step 1-1-5: decomposing the load curves of all months from month 2, if the maximum value of a certain daily load curve of a certain month in the daytime period meets the daily load curve average value range required by the reference load curve, considering the daily load curve as the curve meeting the reference load curve requirement, selecting the curve meeting the reference load curve requirement in the decomposed months, and calculating the average curve of the curves to serve as the reference load curve;

step 1-1-6: determining the lower limit of the average value of the load curves meeting the requirements of the large-load curves, wherein the lower limit is 1.1 times of the maximum value of the daily load curve in the daytime, and if the maximum value of a certain daily load curve in the daytime is greater than the lower limit, the certain daily load curve is considered to be the curve meeting the requirements of the large-load curves;

step 1-1-7: determining a large load curve of the month, and calculating an average value of loads of the reference curve;

step 1-2: calculating the proportion of each load in the rural load in the total amount;

step 1-2-1: calculating the monthly average load value of each transformer substation and the average load value of the appointed day of the appointed month;

step 1-2-2: calculating the load values of the designated month, the designated day and the designated moment;

step 1-2-3: the calculation includes irrigation load, refrigeration load and heatingFraction k of the load in the separated part of the load_p0K is a ratio to the reference load_f。

In the step 1-1-3, calculating an average daily load curve of the load of the transformer substation every month, adding all effective daily load curves every month, and dividing the sum by the effective days of the month;

calculating the monthly average monthly load curve of the load of the transformer substation, adding the loads at 24 moments in the daily load curve of each month, and dividing the added loads by 24 to obtain a curve; the value corresponding to each effective day in the average monthly load curve is also called as the average value of the daily load curve, and the average value of the daily load curve is the average value obtained by adding the loads at 24 moments each day and dividing by 24;

in the steps 1-1-5, if the number of the load curves of the month satisfying the reference curve is less than 3 and the reference curve of any month is not calculated in the front, the curve of the month is not calculated first, and the curve of the next month is calculated; if the number of the curves meeting the requirement of the reference curve in the month is more than or equal to 3, calculating the reference curve, adding all daily load curves meeting the requirement of the reference curve in the month, dividing the daily load curves by the number of the curves meeting the requirement of the reference curve in the month, and taking the calculated reference curve as a small load curve in the month; otherwise, adopting the reference curve of the last month;

in the step 1-1-7, determining the large load curves of the month, if the number of the load curves meeting the requirement of the large load curves of the month is less than 3, neglecting the load curves, not calculating the large load curves of the month, and not performing load curve separation calculation; if the number of the load curves meeting the requirement of the large load curve in the month is more than or equal to 3, calculating the large load curve;

and calculating the average value of the load of the reference curve, and converting the large load curve and the reference load curve into a unit, wherein the average value of the large load curve and the reference load curve is 1.

In the step 1-2-1, monthly average load values of each transformer substation are calculated, the values corresponding to each effective day in a monthly average monthly load curve are added, and then the added values are divided by the effective days of the month;

calculating the average load value of a specified day of a specified month, and multiplying the calculated average load value of the month of the specified month by the per unit value of the average value of the daily load curve of the specified day;

in the step 1-2-2, calculating the load values of the designated month, the designated day and the designated time;

1) if only a small load curve exists, calculating by adopting the small load curve;

2) if a large load curve and a small load curve exist at the same time, determining whether to adopt the large load curve or the small load curve for calculation according to the average value of the daily load curves of the appointed days of the month;

if the set value is 1.1 times of the average value of the reference load of a month, then:

2-1) if the average value of the daily load curve of the month is less than the fixed value, calculating by adopting a small load curve;

2-2) if the average value of the daily load curve is larger than the fixed value, calculating by adopting a large load curve, wherein the large load calculation result is the calculated average value of the reference load of the month multiplied by a value corresponding to the designated moment in the small/large load curve of the designated day of the month;

calculating the separated load values of the designated month, the designated day and the designated time, calculating the difference value between the large load curve and the reference load curve as the separated load curve, wherein the value is equal to the calculated load value of the designated month, the designated day and the designated time, subtracting the reference load average value of the month, multiplying the value corresponding to the designated time in the small load curve of the designated day of the month, and dividing the value by the calculated load values of the designated month, the designated day and the designated time;

in the step 1-2-3, the proportion k of the separated part load is calculated_p0Dividing the separated load value by the calculated load values of the designated month, the designated day and the designated time(ii) a The proportion k of the reference load_fIs 1-k_p0。

In the step 2, the static load equivalent parameter calculation process is as follows:

a polynomial static load model describing the relationship between load power and voltage as a polynomial equation has

P＝P₀[a×(V/V₀)²+b×(V/V₀)+c]（1）

Q＝Q₀[α×(V/V₀)²+β×(V/V₀)+γ]（2）

Wherein P and Q are respectively the active power and the reactive power of the static load, V is the real-time voltage of the static load₀Rated voltage, P, for static load₀And Q₀Are respectively shown at V₀Rated active power and reactive power of the lower static load, a, b and c are active power coefficients of a polynomial static load model, and α, β and gamma are reactive power coefficients of the polynomial static load model;

the equivalence to the static load is mainly to a, b, c and P₀And α, β, gamma, Q₀The equivalence to the polynomial static load model is based on the sensitivity of the static load power to the static load terminal voltage, having

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂P}_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂Q}_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein n is the number of static loads, P_iAnd Q_iRespectively the active and reactive power of the ith static load,andthe partial differential of the active power of the ith static load relative to the voltage and the partial differential of the reactive power of the ith static load relative to the voltage are respectively;

when V is equal to V₀In time, there are:

$P_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}---\left(6\right)$

wherein, P_0iAnd Q_0iRespectively the initial active power and the reactive power of the ith static load;

$a=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

wherein, a_i、b_iAnd c_iActive power coefficients of polynomial static load models of the ith static load;

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

wherein, α_i、β_iAnd gamma_iAnd the polynomial static load model reactive power coefficients are all the ith static load.

The step 2 specifically comprises the following steps:

step A: calculating the total absorbed active power P of all motors_ΣAnd reactive power Q_ΣTotal electromagnetic power P_ΣemTotal rotor winding copper loss P_Σcu2And total maximum electromagnetic power P_{Σem_max}The method specifically comprises the following steps:

$P_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j$

$Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i$

$P_{\mathrm{\Σem}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{emj}}---\left(13\right)$

$P_{\mathrm{\Σcu}2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{cu}2j}$

$P_{\mathrm{\Σem}\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{em}\_\max j}$

wherein P is_j、Q_j、P_emj、P_cu2jAnd P_{em_maxj}Respectively representing active power, reactive power, electromagnetic power, rotor winding copper loss and maximum electromagnetic power of the jth equivalent motorRate; m is the number of equivalent motors;

and B: calculating the copper loss P of the stator winding of an equivalent motor_Σcu1And the slip S of the equivalent motor, respectively:

P_Σcu1=P_Σ-P_Σem（14）

S=P_Σcu2/P_Σem（15）

initializing the maximum electromagnetic power P of an equivalent motor_{emt_max}Let P stand_{emt_max}＝P_{Σem_max}；

And C: calculating the stator winding resistance R of an equivalent motor_s；

Calculating the stator winding phase current of equivalent motorIs provided with

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,is the terminal voltage of the motor;

then the value of the stator winding resistance R of the motor is equal_sExpressed as:

$R_s=\frac{P_{\mathrm{\Σcu}1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

step D: equivalent impedance Z of computer equivalent model_deq：

$Z_{\mathrm{deq}}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-j{Q}_{\mathrm{\Σ}}}---\left(18\right)$

And is provided with

R_deq=real(Z_deq) （19）

X_deq=imag(Z_deq) （20）

Wherein R is_deqAnd X_deqCorresponding equivalent resistance and equivalent reactance;

step E: calculating leakage reactance X of stator winding of equivalent motor_sAnd rotor winding leakage reactance X_rIs provided with

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{\mathrm{emt}\_\max }}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: stator winding leakage reactance X of equivalent motor through iteration_sAnd rotor winding leakage reactance X_rCorrecting;

step G: calculating the rotor winding resistance and the equivalent excitation reactance of the equivalent motor;

according to the calculated R_s、X_s、X_rAnd Z_deqIs provided with K_r=R_deq-R_s，K_x=X_deq-X_sThen equal value of the rotor winding resistance R of the motor_rSum-equivalent excitation reactance X_mRespectively expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-{{4X}_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

step H: calculating the maximum electromagnetic power of the equivalent motor by adopting an iterative method, and correcting the maximum electromagnetic power;

1) calculating the maximum electromagnetic power P of the equivalent motor in the k iteration_{emt_maxk}Is shown as

$P_{\mathrm{emt}\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Thevenin equivalent impedance Z of equivalent motor based on Thevenin theorem_dpIs shown as

$Z_{\mathrm{dp}}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And has R_dp=real(Z_dp) And X_dp=imag(Z_dp)，R_dpAnd X_dpRespectively a Thevenin equivalent resistance and an equivalent reactance;

the conditions for generating the maximum electromagnetic power of the equivalent motor are as follows:

$R_{\mathrm{pm}}=\frac{R_r}{S_m}=\sqrt{{R_{\mathrm{dp}}}^2+{X_{\mathrm{dp}}}^2}---\left(26\right)$

wherein R is_pmFor the corresponding Thevenin equivalent impedance amplitude, S, which generates the maximum electromagnetic power_mCritical slip;

thevenin equivalent open circuit voltageIs composed of

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j(X_s+X_m)}---\left(27\right)$

The maximum electromagnetic power of the thevenin equivalent motor in the kth iteration can then be expressed as

$P_{\mathrm{em}\_\max k}=\frac{3{U_0}^2{R}_{\mathrm{pm}}}{{(R_{\mathrm{dp}}+R_{\mathrm{pm}})}^2+{X_{\mathrm{dp}}}^2}---\left(28\right)$

2) Correcting the maximum electromagnetic power of the equivalent motor;

calculating a correction coefficient tau of the maximum electromagnetic power of the equivalent motor in the kth iteration_maxkIt is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{\mathrm{emt}\_\max k}}{P_{\mathrm{em}\_\max k}}---\left(29\right)$

the corrected maximum electromagnetic power P of the equivalent motor_{emt_max}Expressed as:

P_{emt_max}＝τ_maxk×P_{Σem_max}（30）

with P_{emt_max}And P_{emt_maxk}The absolute error of (2) is the iterative error of the maximum electromagnetic power of the equivalent motorIs provided with

${\mathrm{Err}}_{P_{\mathrm{em}\_\max }}=|P_{\mathrm{emt}\_\max }-P_{\mathrm{emt}\_\max k}|---\left(31\right)$

To be provided withFor the criterion of iterative convergence, ifThen X is recalculated_s、R_r、X_rAnd X_m；

Step I: calculating an equivalent inertia time constant;

the equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}{H}_j}{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}}---\left(32\right)$

wherein P is_njAnd H_jIs the rated power and inertia time constant of the j-th equivalent motor.

In step 2, the system impedance of the power distribution network is represented as:

$Z_{\mathrm{eq}}=\left[\underset{f=1}{\overset{M}{\mathrm{\Σ}}}{u}_f^2{(1/Z_f)}^{*}\right]/{\left(\underset{l=1}{\overset{N}{\mathrm{\Σ}}}{I}_l\right)}^2---\left(33\right)$

wherein Z is_eqIs the system impedance of the power distribution network; u. of_fRepresenting bus voltage, Z_fRepresenting transformer and distribution line impedance; i is_lRepresents negativeAnd (4) carrying current, wherein M and N are the total number of nodes and the total number of branches in the power distribution network system respectively.

And the power supply area network topology data in the step 2 comprise distribution line data, transformer data and reactive compensation data.

Compared with the prior art, the invention has the beneficial effects that:

1. the method is based on a daily load curve of a power user acquired by a load control management system of a power system marketing department and a daily load curve of a real-time comprehensive load of a transformer substation provided by a data acquisition and monitoring (SCADA) system, obtains types and proportions of various electric equipment supplied by a rural load transformer substation at any moment through load curve decomposition, and realizes online load modeling of rural load stations by using power supply area network topology data of load nodes provided by a scheduling department and adopting a statistical synthesis method, so that the aim of accurately modeling the rural load stations is fulfilled, the accuracy of power grid simulation calculation is improved, and the safe, reliable and economic operation of a power grid is guaranteed.

2. The method develops online load modeling based on daily load curve data of an actual power grid, changes the mode that the traditional statistical synthesis method depends on offline manual investigation, can quickly, conveniently and accurately model the load, greatly saves manpower and material resources, avoids the problem of inaccurate load modeling caused by inaccurate investigation results, further improves the traditional modeling mode, and provides important guidance for load modeling work.

3. The method realizes the online of a statistical synthesis method through diversified online data resources (such as a daily load curve of a power user acquired by a marketing load control management system, a daily load curve of a real-time transformer substation comprehensive load provided by a data acquisition and monitoring (SCADA) system and scheduling network topology information), can perform real-time SLM modeling on load nodes, has clear model physical significance and strong adaptability, and solves the problem that all traditional load modeling methods are difficult to adapt to load time-varying property.

Drawings

FIG. 1 is a load curve diagram of a complex 35kV rural load for 1 month;

FIG. 2 is a load graph of the complex 35kV rural load for 5 months;

FIG. 3 is a load graph of the complex 35kV rural load for 7 months;

FIG. 4 is a load graph of the complex 35kV rural load for 11 months;

FIG. 5 is a load curve diagram after the complex 35kV rural load is decomposed for 1 month;

FIG. 6 is a load curve graph after the complex 35kV rural load is decomposed in 5 months;

FIG. 7 is a load curve graph after the complex 35kV rural load is decomposed for 7 months;

FIG. 8 is a load graph after decomposition for the rural load of 35kV at the time of the condensation for 11 months;

FIG. 9 is a flow chart of a rural load type load modeling method based on load curve decomposition;

FIG. 10 is a comparison of 220kV load bus voltage curves;

FIG. 11 is a comparison graph of a 220kV load bus load active power curve;

fig. 12 is a comparison graph of the load reactive power curve of the 220kV load bus.

Detailed Description

The present invention will be described in further detail with reference to the accompanying drawings.

The invention provides a rural load type load modeling method based on load curve decomposition, which comprises the following steps:

step 1: decomposing a rural load curve, and calculating the proportion of each load in the rural loads to the total amount;

step 2: and calculating the static load equivalent parameter, the dynamic load equivalent parameter and the power distribution network system impedance, and combining power supply area network topology data of the rural load node to obtain a rural load type load model.

Aiming at the basic characteristics of the load curve of the rural load, the adopted basic processing method is as follows:

(1) because the rural loads are generally small and have poor regularity, the rural loads of the same transformer substation are combined.

If not, because the regularity of the rural load is poor, an ideal decomposition result is difficult to obtain. The load curves in fig. 1-4 are combined load curves of all 35kV rural load lines, the combined total load is large, and the change rule of the load is strong.

(2) The rural load curve containing irrigation load is decomposed into basic load, refrigeration load, heating load and irrigation load.

The cooling load occurs mainly in the summer months of 6, 7 and 8, the heating load occurs mainly in the winter months of 12, 1 and 2, and the irrigation load occurs mainly in the spring months of 3, 4 and 5 and the autumn months of 9, 10 and 11. Thus, in the split, the split loads, in months 6, 7, and 8, are considered cooling loads, in months 12, 1, and 2, are considered heating loads, and in months 3, 4, 5, 9, 10, and 11, are considered irrigation loads. However, the load separated from part of months of alternate seasons may have two loads at the same time, and since the decomposition can not be continued, only one load is considered to exist approximately, for example, the load separated from 8 months may have both cooling load and irrigation load, and the load separated from 8 months is considered to be both cooling load approximately.

(3) Basic method of separation curves: and decomposing the curve according to the maximum value in the daytime, dividing the load curve of each month into two load curves, wherein one load curve is a reference load curve, the other load curve is a large load curve, and the difference value of the two curves is used as a separated load curve.

For the load curve of each month, the load curve is divided into two groups of curves according to the load size, and the average curve of the two groups of curves is calculated to be used as a typical load curve for approximately simulating the load curve at different load levels in each month. One of the curves is a base load curve, which represents the variation process of the base load and is calculated according to the curve with smaller average load in all months. The rural resident load contains irrigation load, the basic method for separating the basic load curve is to decompose by adopting the maximum value in the daytime, and the curve with smaller maximum value in the daytime is selected as the reference load curve.

The decomposition mainly considers two periods, one is a day period (6-17 points) and the other is a night period, and the reference load curve should satisfy that the load is smaller in the day and smaller in the night. However, as can be seen from the basic characteristics of the rural load curve, the load of the rural load curve is smaller in spring and autumn, and larger in summer and winter, but the difference between the load and the load is not very large; secondly, the load changes greatly in different seasons in the daytime, particularly in spring and autumn; and thirdly, the load curve with smaller load in the daytime is not necessarily smaller at night. If the curve with smaller daytime load and night load is selected as the reference load curve, the number of the curves meeting the reference curve is small, and the base load curve can be accurately decomposed only by adopting the maximum value of the daytime load, so that the decomposition is carried out only by adopting the maximum value of the daytime load.

Two methods are adopted for determining the range met by the maximum value of the reference load curve in the daytime:

calculating the maximum value of all daily load curves in the daytime, and selecting the maximum value within a certain rangeCalculating their average value;

since the day period of 2 months is the lowest in load and the dispersion is small, the maximum value of the day period of the load curve for all days of 2 months is calculated, and then the average value thereof is calculated.

On the basis of the average value, selecting a range satisfied by the maximum load of the reference load curve in the daytime, regarding the load of each month, regarding the maximum load in the daytime as long as the maximum load is within the range, considering the maximum load as the base load curve; above the maximum value of this range, a large load curve is considered. And calculating an average load curve of the base load curve as a small load typical load curve, calculating an average load curve of the large load curve as a large load typical load curve, and using the difference value of the two curves as a refrigerating, heating or irrigation load.

Of the two methods described above, the first method requires determining a range for calculating the mean value of the reference curve, which is difficult to determine; the second method is relatively easy and has some flexibility.

The process of decomposing the rural load curve is as follows:

step 1-1-1: for a transformer substation with the supplied load type being rural load, combining daily load curves of 10kV or 6kV load outgoing lines of the transformer substation;

step 1-1-2: calculating the number of effective load days of the load of the transformer substation every month, considering the day as invalid if the load of a certain day has a zero value, and considering the day as invalid if the load of the certain day has a reverse sign phenomenon;

step 1-1-3: calculating an average daily load curve and an average monthly load curve of the load of the transformer substation every month;

calculating the monthly average daily load curve of the load of the transformer substation, namely adding all the effective daily load curves of each month, and dividing the sum by the effective days of the month;

calculating the monthly average monthly load curve of the load of the transformer substation, namely adding the loads at 24 moments in the daily load curve of each month and dividing the added loads by 24 to obtain a curve; the value corresponding to each effective day in the average monthly load curve is also called as the average value of the daily load curve, and the average value of the daily load curve is the average value obtained by adding the loads at 24 moments each day and dividing the sum by 24;

step 1-1-4: calculating the maximum value of each daily load curve of each month at the daytime period, namely 6-17 points, selecting the maximum value of the daytime period from the average daily load curves of 2 months, and determining a load curve range meeting the requirement of a reference load curve, wherein the range is 0.8-1.15 times of the calculated maximum value of the daytime period;

step 1-1-5: decomposing the load curves of all months from month 2, if the maximum value of a certain daily load curve of a certain month in the daytime period meets the daily load curve average value range required by the reference load curve, considering the daily load curve as the curve meeting the reference load curve requirement, selecting the curve meeting the reference load curve requirement in the decomposed months, and calculating the average curve of the curves to serve as the reference load curve;

if the number of the load curves of the month satisfying the reference curves is less than 3 and the reference curve of any month is not calculated in the front, the curve of the month is not calculated firstly, and the curve of the next month is calculated; if the number of the curves meeting the requirement of the reference curve in the month is more than or equal to 3, calculating the reference curve, namely adding all daily load curves meeting the requirement of the reference curve in the month, dividing the daily load curves meeting the requirement of the reference curve in the month by the number of the curves meeting the requirement of the reference curve in the month, and taking the calculated reference curve as a small load curve in the month; otherwise, adopting the reference curve of the last month;

step 1-1-6: determining the lower limit of the average value of the load curves meeting the requirements of the large-load curves, wherein the lower limit is 1.1 times of the maximum value of the daily load curve in the daytime, and if the maximum value of a certain daily load curve in the daytime is greater than the lower limit, the certain daily load curve is considered to be the curve meeting the requirements of the large-load curves;

step 1-1-7: determining a large load curve of the month, and calculating an average value of loads of the reference curve;

determining the large load curves of the month, if the number of the load curves meeting the requirement of the large load curves of the month is less than 3, neglecting the load curves, not calculating the large load curves of the month, and not performing load curve separation calculation; if the number of the load curves meeting the requirement of the large load curve in the month is more than or equal to 3, calculating the large load curve;

and calculating the average value of the load of the reference curve, and performing per unit on the large load curve and the reference load curve, namely changing the average value of the large load curve and the reference load curve into 1.

Figures 5-8 are load curves after the load in rural areas of 1, 5, 7, 11 months of the complex of 35 kV.

As shown in fig. 1-4 and fig. 5-8, the load is large in 1 month, the dispersity is small, no load curve meeting the requirement of the reference load curve is adopted, a large load curve is formed by adopting the basic load curve of 12 months, and the separated load is the refrigeration load. The proportion of the irrigation load in 5 months and 11 months is large, and the irrigation load mainly occurs in the morning and afternoon. The load split in month 7 is the refrigeration load.

The process of calculating the proportion of each load in the rural load to the total amount is as follows:

step 1-2-1: calculating the monthly average load value of each transformer substation and the average load value of the appointed day of the appointed month;

calculating monthly average load values of each transformer substation, namely adding the values corresponding to each effective day in the monthly average monthly load curve, and dividing the sum by the effective days of the month;

calculating the average load value of a specified day of a specified month, namely multiplying the calculated average load value of the month of the specified month by the per unit value of the average value of the daily load curve of the specified day;

step 1-2-2: calculating the load values of the designated month, the designated day and the designated moment;

1) if only a small load curve exists, calculating by adopting the small load curve;

2) if a large load curve and a small load curve exist at the same time, determining whether to adopt the large load curve or the small load curve for calculation according to the average value of the daily load curves of the appointed days of the month;

if the set value is 1.1 times of the average value of the reference load of a month, then:

2-1) if the average value of the daily load curve of the month is less than the fixed value, calculating by adopting a small load curve;

2-2) if the average value of the daily load curve is larger than the fixed value, calculating by adopting a large load curve, wherein the large load calculation result is the calculated average value of the reference load of the month multiplied by a value corresponding to the designated moment in the small/large load curve of the designated day of the month;

calculating the separated load values of the designated month, the designated day and the designated time, calculating the difference value between the large load curve and the reference load curve as the separated load curve, wherein the value is equal to the calculated load value of the designated month, the designated day and the designated time, subtracting the reference load average value of the month, multiplying the value corresponding to the designated time in the small load curve of the designated day of the month, and dividing the value by the calculated load values of the designated month, the designated day and the designated time;

step 1-2-3: calculating the proportion k of the separated part load including irrigation load, refrigeration load and heating load_p0K is a ratio to the reference load_f。

Calculating the proportion k of the load of the separated part_p0Dividing the separated load value by the calculated load values of the designated month, the designated day and the designated time; the proportion k of the reference load_fIs 1-k_p0。

In the step 2, the static load equivalent parameter calculation process is as follows:

a polynomial static load model describing the relationship between load power and voltage as a polynomial equation has

P＝P₀[a×(V/V₀)²+b×(V/V₀)+c]（1）

Q＝Q₀[α×(V/V₀)²+β×(V/V₀)+γ]（2）

Wherein P and Q are respectively the active power and the reactive power of the static load, V is the real-time voltage of the static load₀Rated voltage, P, for static load₀And Q₀Are respectively shown at V₀Rated active power and reactive power of the lower static load, a, b and c are active power coefficients of a polynomial static load model, and α, β and gamma are reactive power coefficients of the polynomial static load model;

the equivalence to the static load is mainly to a, b, c and P₀And α, β, gamma, Q₀Is based on the sensitivity of the static load power to the static load terminal voltage, i.e. the equivalence of the polynomial static load model

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂P}_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂Q}_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein n is the number of static loads, P_iAnd Q_iRespectively the active and reactive power of the ith static load,andthe partial differential of the active power of the ith static load relative to the voltage and the partial differential of the reactive power of the ith static load relative to the voltage are respectively;

when V is equal to V₀In time, there are:

$P_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}---\left(6\right)$

wherein, P_0iAnd Q_0iRespectively the initial active power and the reactive power of the ith static load;

$a=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

wherein, a_i、b_iAnd c_iActive power coefficients of polynomial static load models of the ith static load;

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

wherein, α_i、β_iAnd gamma_iAnd the polynomial static load model reactive power coefficients are all the ith static load.

The step 2 specifically comprises the following steps:

step A: calculating the total absorbed active power P of all motors_ΣAnd reactive power Q_ΣTotal electromagnetic power P_ΣemTotal rotor winding copper loss P_Σcu2And total maximum electromagnetic power P_{Σem_max}The method specifically comprises the following steps:

$P_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j$

$Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i$

$P_{\mathrm{\Σem}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{emj}}---\left(13\right)$

$P_{\mathrm{\Σcu}2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{cu}2j}$

$P_{\mathrm{\Σem}\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{em}\_\max j}$

wherein P is_j、Q_j、P_emj、P_cu2jAnd P_{em_maxj}Respectively representing active power, reactive power, electromagnetic power, rotor winding copper loss and maximum electromagnetic power of a jth equivalent motor; m is the number of equivalent motors;

and B: calculating the copper loss P of the stator winding of an equivalent motor_Σcu1And the slip S of the equivalent motor, respectively:

P_Σcu1=P_Σ-P_Σem（14）

S=P_Σcu2/P_Σem（15）

initializing the maximum electromagnetic power P of an equivalent motor_{emt_max}Let P stand_{emt_max}＝P_{Σem_max}；

And C: calculating the stator winding resistance R of an equivalent motor_s；

Calculating the stator winding phase current of equivalent motorIs provided with

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,is the terminal voltage of the motor;

then the value of the stator winding resistance R of the motor is equal_sExpressed as:

$R_s=\frac{P_{\mathrm{\Σcu}1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

step D: equivalent impedance Z of computer equivalent model_deq：

$Z_{\mathrm{deq}}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-{\mathrm{jQ}}_{\mathrm{\Σ}}}---\left(18\right)$

And is provided with

R_deq=real(Z_deq) （19）

X_deq=imag(Z_deq) （20）

Wherein R is_deqAnd X_deqCorresponding equivalent resistance and equivalent reactance;

step E: calculating leakage reactance X of stator winding of equivalent motor_sAnd rotor winding leakage reactance X_rIs provided with

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{\mathrm{emt}\_\max }}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: stator winding leakage reactance X of equivalent motor through iteration_sAnd rotor winding leakage reactance X_rCorrecting;

step G: calculating the rotor winding resistance and the equivalent excitation reactance of the equivalent motor;

according to the calculated R_s、X_s、X_rAnd Z_deqIs provided with K_r=R_deq-R_s，K_x=X_deq-X_sThen equal value of the rotor winding resistance R of the motor_rSum-equivalent excitation reactance X_mRespectively expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-{{4X}_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

step H: calculating the maximum electromagnetic power of the equivalent motor by adopting an iterative method, and correcting the maximum electromagnetic power;

1) calculating the maximum electromagnetic power P of the equivalent motor in the k iteration_{emt_maxk}Is shown as

$P_{\mathrm{emt}\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Thevenin equivalent impedance Z of equivalent motor based on Thevenin theorem_dpIs shown as

$Z_{\mathrm{dp}}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And has R_dp=real(Z_dp) And X_dp=imag(Z_dp)，R_dpAnd X_dpRespectively a Thevenin equivalent resistance and an equivalent reactance;

the conditions for generating the maximum electromagnetic power of the equivalent motor are as follows:

$R_{\mathrm{pm}}=\frac{R_r}{S_m}=\sqrt{{R_{\mathrm{dp}}}^2+{X_{\mathrm{dp}}}^2}---\left(26\right)$

wherein R is_pmFor the corresponding Thevenin equivalent impedance amplitude, S, which generates the maximum electromagnetic power_mCritical slip;

thevenin equivalent open circuit voltageIs composed of

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j(X_s+X_m)}---\left(27\right)$

The maximum electromagnetic power of the thevenin equivalent motor in the kth iteration can then be expressed as

$P_{\mathrm{em}\_\max k}=\frac{3{U_0}^2{R}_{\mathrm{pm}}}{{(R_{\mathrm{dp}}+R_{\mathrm{pm}})}^2+{X_{\mathrm{dp}}}^2}---\left(28\right)$

2) Correcting the maximum electromagnetic power of the equivalent motor;

calculating a correction coefficient tau of the maximum electromagnetic power of the equivalent motor in the kth iteration_maxkIt is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{\mathrm{emt}\_\max k}}{P_{\mathrm{em}\_\max k}}---\left(29\right)$

the corrected maximum electromagnetic power P of the equivalent motor_{emt_max}Expressed as:

P_{emt_max}＝τ_maxk×P_{Σem_max}（30）

with P_{emt_max}And P_{emt_maxk}The absolute error of (2) is the iterative error of the maximum electromagnetic power of the equivalent motorIs provided with

${\mathrm{Err}}_{P_{\mathrm{em}\_\max }}=|P_{\mathrm{emt}\_\max }-P_{\mathrm{emt}\_\max k}|---\left(31\right)$

To be provided withFor the criterion of iterative convergence, ifThen X is recalculated_s、R_r、X_rAnd X_m；

Step I: calculating an equivalent inertia time constant;

the equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}{H}_j}{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}}---\left(32\right)$

wherein P is_njAnd H_jIs the rated power and inertia time constant of the j-th equivalent motor.

In step 2, the system impedance of the power distribution network is represented as:

$Z_{\mathrm{eq}}=\left[\underset{f=1}{\overset{M}{\mathrm{\Σ}}}{u}_f^2{(1/Z_f)}^{*}\right]/{\left(\underset{l=1}{\overset{N}{\mathrm{\Σ}}}{I}_l\right)}^2---\left(33\right)$

wherein Z is_eqIs the system impedance of the power distribution network; u. of_fRepresenting bus voltage, Z_fRepresenting transformer and distribution line impedance; i is_lRepresenting load current, where M and N are the total number of nodes and branches in the distribution grid system, respectivelyTotal number of strips.

And the power supply area network topology data in the step 2 comprise distribution line data, transformer data and reactive compensation data.

FIGS. 10-12 are comparison graphs of simulation curves and actual measurement curves of SLM equivalent model parameters obtained by the present invention

FIG. 10 is a graph comparing the voltage curves of a 220kV load bus; wherein the solid line is an actual measurement curve, and the dotted line is a simulation result when the equivalent SLM model parameters obtained by the method are adopted;

FIG. 11 is a comparison graph of the active power curve of a 220kV load bus load; wherein the solid line is an actual measurement curve, and the dotted line is a simulation result when the equivalent SLM model parameters obtained by the method are adopted;

FIG. 12 is a comparison graph of the load reactive power curve of a 220kV load bus; wherein the solid line is an actual measurement curve, and the dotted line is a simulation result when the equivalent SLM model parameters obtained by the method are adopted.

The technical scheme of the invention is applied to analysis and research of the Henan power grid stability model of the power company in Henan province.

Finally, it should be noted that: the above embodiments are only for illustrating the technical solutions of the present invention and not for limiting the same, and although the present invention is described in detail with reference to the above embodiments, those of ordinary skill in the art should understand that: modifications and equivalents may be made to the embodiments of the invention without departing from the spirit and scope of the invention, which is to be covered by the claims.

Claims (7)

1. A rural load type load modeling method based on load curve decomposition is characterized in that: the method comprises the following steps:

step 1: decomposing a rural load curve, and calculating the proportion of each load in the rural loads to the total amount;

step 2: calculating a static load equivalent parameter, a dynamic load equivalent parameter and the system impedance of the power distribution network, and combining power supply area network topology data of rural load nodes to obtain a rural load type load model;

the step 1 comprises the following steps:

step 1-1: decomposing a rural load curve;

step 1-1-1: for a transformer substation with the supplied load type being rural load, combining daily load curves of 10kV or 6kV load outgoing lines of the transformer substation;

step 1-1-2: calculating the number of effective load days of the load of the transformer substation every month, considering the day as invalid if the load of a certain day has a zero value, and considering the day as invalid if the load of the certain day has a reverse sign phenomenon;

step 1-1-3: calculating an average daily load curve and an average monthly load curve of the load of the transformer substation every month;

step 1-1-4: calculating the maximum value of each daily load curve of each month at 6-17 points in the daytime period, selecting the maximum value of the daytime period from the average daily load curves of 2 months, and determining a load curve range meeting the requirement of a reference load curve, wherein the range is 0.8-1.15 times of the calculated maximum value of the daytime period;

step 1-1-5: decomposing the load curves of all months from month 2, if the maximum value of a certain daily load curve of a certain month in the daytime period meets the daily load curve average value range required by the reference load curve, considering the daily load curve as the curve meeting the reference load curve requirement, selecting the curve meeting the reference load curve requirement in the decomposed months, and calculating the average curve of the curves to serve as the reference load curve;

step 1-1-6: determining the lower limit of the average value of the load curves meeting the requirements of the large-load curves, wherein the lower limit is 1.1 times of the maximum value of the daily load curve in the daytime, and if the maximum value of a certain daily load curve in the daytime is greater than the lower limit, the certain daily load curve is considered to be the curve meeting the requirements of the large-load curves;

step 1-1-7: determining a large load curve of the month, and calculating an average value of loads of the reference curve;

step 1-2: calculating the proportion of each load in the rural load in the total amount;

step 1-2-1: calculating the monthly average load value of each transformer substation and the average load value of the appointed day of the appointed month;

step 1-2-2: calculating the load values of the designated month, the designated day and the designated moment;

step 1-2-3: calculating the proportion k of the separated part load including irrigation load, refrigeration load and heating load_p0K is a ratio to the reference load_f。

2. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: in the step 1-1-3, calculating an average daily load curve of the load of the transformer substation every month, adding all effective daily load curves every month, and dividing the sum by the effective days of the month;

calculating the monthly average monthly load curve of the load of the transformer substation, adding the loads at 24 moments in the daily load curve of each month, and dividing the added loads by 24 to obtain a curve; the value corresponding to each effective day in the average monthly load curve is also called as the average value of the daily load curve, and the average value of the daily load curve is the average value obtained by adding the loads at 24 moments each day and dividing by 24;

in the steps 1-1-5, if the number of the load curves of the month satisfying the reference curve is less than 3 and the reference curve of any month is not calculated in the front, the curve of the month is not calculated first, and the curve of the next month is calculated; if the number of the curves meeting the requirement of the reference curve in the month is more than or equal to 3, calculating the reference curve, adding all daily load curves meeting the requirement of the reference curve in the month, dividing the daily load curves by the number of the curves meeting the requirement of the reference curve in the month, and taking the calculated reference curve as a small load curve in the month; otherwise, adopting the reference curve of the last month;

in the step 1-1-7, determining the large load curves of the month, if the number of the load curves meeting the requirement of the large load curves of the month is less than 3, neglecting the load curves, not calculating the large load curves of the month, and not performing load curve separation calculation; if the number of the load curves meeting the requirement of the large load curve in the month is more than or equal to 3, calculating the large load curve;

and calculating the average value of the load of the reference curve, and converting the large load curve and the reference load curve into a unit, wherein the average value of the large load curve and the reference load curve is 1.

3. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: in the step 1-2-1, monthly average load values of each transformer substation are calculated, the values corresponding to each effective day in a monthly average monthly load curve are added, and then the added values are divided by the effective days of the month;

calculating the average load value of a specified day of a specified month, and multiplying the calculated average load value of the month of the specified month by the per unit value of the average value of the daily load curve of the specified day;

in the step 1-2-2, calculating the load values of the designated month, the designated day and the designated time;

1) if only a small load curve exists, calculating by adopting the small load curve;

2) if a large load curve and a small load curve exist at the same time, determining whether to adopt the large load curve or the small load curve for calculation according to the average value of the daily load curves of the appointed days of the month;

if the set value is 1.1 times of the average value of the reference load of a month, then:

2-1) if the average value of the daily load curve of the month is less than the fixed value, calculating by adopting a small load curve;

2-2) if the average value of the daily load curve is larger than the fixed value, calculating by adopting a large load curve, wherein the large load calculation result is the calculated average value of the reference load of the month multiplied by a value corresponding to the designated moment in the small/large load curve of the designated day of the month;

calculating the separated load values of the designated month, the designated day and the designated time, calculating the difference value between the large load curve and the reference load curve as the separated load curve, wherein the value is equal to the calculated load value of the designated month, the designated day and the designated time, subtracting the reference load average value of the month, multiplying the value corresponding to the designated time in the small load curve of the designated day of the month, and dividing the value by the calculated load values of the designated month, the designated day and the designated time;

in the step 1-2-3, the meter is measuredCalculating the proportion k of the load of the separation part_p0Dividing the separated load value by the calculated load values of the designated month, the designated day and the designated time; the proportion k of the reference load_fIs 1-k_p0。

4. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: in the step 2, the static load equivalent parameter calculation process is as follows:

a polynomial static load model describing the relationship between load power and voltage as a polynomial equation has

P＝P₀[a×(V/V₀)²+b×(V/V₀)+c](1)

Q＝Q₀[α×(V/V₀)²+β×(V/V₀)+γ](2)

Wherein P and Q are respectively the active power and the reactive power of the static load, V is the real-time voltage of the static load₀Rated voltage, P, for static load₀And Q₀Are respectively shown at V₀Rated active power and reactive power of the lower static load, a, b and c are active power coefficients of a polynomial static load model, and α, β and gamma are reactive power coefficients of the polynomial static load model;

the equivalence to the static load is mainly to a, b, c and P₀And α, β, gamma, Q₀The equivalence to the polynomial static load model is based on the sensitivity of the static load power to the static load terminal voltage, having

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\Σ}}\frac{\∂P_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\Σ}}\frac{\∂Q_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein n is the number of static loads, P_iAnd Q_iRespectively the active and reactive power of the ith static load,andthe partial differential of the active power of the ith static load relative to the voltage and the partial differential of the reactive power of the ith static load relative to the voltage are respectively;

when V is equal to V₀In time, there are:

$P_0=\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}---\left(6\right)$

wherein, P_0iAnd Q_0iRespectively the initial active power and the reactive power of the ith static load;

$a=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

wherein, a_i、b_iAnd c_iActive power coefficients of polynomial static load models of the ith static load;

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

wherein, α_i、β_iAnd gamma_iAnd the polynomial static load model reactive power coefficients are all the ith static load.

5. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: the step 2 specifically comprises the following steps:

step A: calculating the total absorbed active power P of all motors_∑And reactive power Q_∑Total electromagnetic power P_∑emTotal rotor winding copper loss P_∑cu2And total maximum electromagnetic power P_{∑em_max}The method specifically comprises the following steps:

$\begin{array}{c}{P}_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j\\ Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i\\ P_{\mathrm{\Σ}em}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{emj}\\ P_{\mathrm{\Σ}cu2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{cu2j}\\ P_{\mathrm{\Σ}em\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{em\_\max j}\end{array}---\left(13\right)$

wherein P is_j、Q_j、P_emj、P_cu2jAnd P_em__maxjRespectively representing active power, reactive power, electromagnetic power, rotor winding copper loss and maximum electromagnetic power of a jth equivalent motor; m is the number of equivalent motors;

and B: calculating the copper loss P of the stator winding of an equivalent motor_∑cu1And the slip S of the equivalent motor, respectively:

P_∑cu1＝P_∑-P_∑em(14)

S＝P_∑cu2/P_∑em(15)

initializing the maximum electromagnetic power P of an equivalent motor_{emt_max}Let P stand_{emt_max}＝P_{∑em_max}；

And C: calculating the stator winding resistance R of an equivalent motor_s；

Calculating the stator winding phase current of equivalent motorIs provided with

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,is the terminal voltage of the motor;

then the value of the stator winding resistance R of the motor is equal_sExpressed as:

$R_s=\frac{P_{\mathrm{\Σ}cu1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

step D: equivalent impedance Z of computer equivalent model_deq：

$Z_{deq}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-{\mathrm{jQ}}_{\mathrm{\Σ}}}---\left(18\right)$

And is provided with

R_deq＝real(Z_deq) (19)

X_deq＝imag(Z_deq) (20)

Wherein R is_deqAnd X_deqCorresponding equivalent resistance and equivalent reactance;

step E: calculating leakage reactance X of stator winding of equivalent motor_sAnd rotor winding leakage reactance X_rIs provided with

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{emt\_max}}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: stator winding leakage reactance X of equivalent motor through iteration_sAnd rotor winding leakage reactance X_rCorrecting;

step G: calculating the rotor winding resistance and the equivalent excitation reactance of the equivalent motor;

according to the calculated R_s、X_s、X_rAnd Z_deqIs provided with K_r＝R_deq-R_s，K_x＝X_deq-X_sThen equal value of the rotor winding resistance R of the motor_rSum-equivalent excitation reactance X_mRespectively expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-4{X_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

step H: calculating the maximum electromagnetic power of the equivalent motor by adopting an iterative method, and correcting the maximum electromagnetic power;

1) calculating the maximum electromagnetic power P of the equivalent motor in the k iteration_{emt_maxk}Is shown as

$P_{emt\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Thevenin equivalent impedance Z of equivalent motor based on Thevenin theorem_dpIs shown as

$Z_{dp}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And has R_dp＝real(Z_dp) And X_dp＝imag(Z_dp)，R_dpAnd X_dpRespectively a Thevenin equivalent resistance and an equivalent reactance;

the conditions for generating the maximum electromagnetic power of the equivalent motor are as follows:

$R_{pm}=\frac{R_r}{S_m}=\sqrt{{R_{dp}}^2+{X_{dp}}^2}---\left(26\right)$

wherein R is_pmFor the corresponding Thevenin equivalent impedance amplitude, S, which generates the maximum electromagnetic power_mCritical slip;

thevenin equivalent open circuit voltageIs composed of

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j\left(X_s+X_m\right)}---\left(27\right)$

The maximum electromagnetic power of the thevenin equivalent motor in the kth iteration can then be expressed as

$P_{em\_\max k}=\frac{3{V_0}^2{R}_{pm}}{{(R_{dp}+R_{pm})}^2+{X_{dp}}^2}---\left(28\right)$

2) Correcting the maximum electromagnetic power of the equivalent motor;

calculating a correction coefficient tau of the maximum electromagnetic power of the equivalent motor in the kth iteration_maxkIt is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{emt\_\max k}}{P_{em\_\max k}}---\left(29\right)$

the corrected maximum electromagnetic power P of the equivalent motor_{emt_max}Expressed as:

P_{emt_max}＝τ_maxk×P_{∑em_max}(30)

with P_{emt_max}And P_{emt_maxk}The absolute error of (2) is the iterative error of the maximum electromagnetic power of the equivalent motorIs provided with

${\mathrm{Err}}_{P_{em\_max}}=|P_{emt\_max}-P_{emt\_\max k}|---\left(31\right)$

To be provided withFor the criterion of iterative convergence, ifThen X is recalculated_s、R_r、X_rAnd X_m；

Step I: calculating an equivalent inertia time constant;

the equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\Σ}}{P}_{nj}{H}_j}{\underset{j=1}{\overset{m}{\Σ}}{P}_{nj}}---\left(32\right)$

wherein P is_njAnd H_jIs the rated power and inertia time constant of the j-th equivalent motor.

6. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: in step 2, the system impedance of the power distribution network is represented as:

$Z_{eq}=\[\underset{f=1}{\overset{M}{\Σ}}{u}_f^2{(1/Z_f)}^{*}\]/{\left(\underset{l=1}{\overset{N}{\Σ}}{I}_l\right)}^2---\left(33\right)$

wherein Z is_eqIs the system impedance of the power distribution network; u. of_fRepresenting bus voltage, Z_fRepresenting transformer and distribution line impedance; i is_lRepresenting the load current, where M and N are the total number of nodes and the total number of branches in the distribution grid system, respectively.

7. The rural load type load modeling method based on load curve decomposition according to claim 1, characterized in that: and the power supply area network topology data in the step 2 comprise distribution line data, transformer data and reactive compensation data.