CN103632031B

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

Info

Publication number: CN103632031B
Application number: CN201310500676.2A
Authority: CN
Inventors: 王�琦; 汤涌; 龙飞; 王建明; 侯俊贤; 赵兵; 易俊; 张健; 刘丽平
Original assignee: State Grid Corp of China SGCC; China Electric Power Research Institute Co Ltd CEPRI
Current assignee: State Grid Corp of China SGCC; China Electric Power Research Institute Co Ltd CEPRI
Priority date: 2013-10-22
Filing date: 2013-10-22
Publication date: 2016-08-31
Anticipated expiration: 2033-10-22
Also published as: CN103632031A

Abstract

The invention provides a load modeling method for rural load types based on load curve decomposition. The method includes the following steps: decomposing the rural load curve, and calculating the proportion of each type of load in the rural load; calculating the static load equivalent Parameters, dynamic load equivalent parameters and distribution network system impedance, combined with the topology data of the power supply area network of rural load nodes, the rural load type load model is obtained. The present invention provides a load modeling method for rural load types based on load curve decomposition. This method overcomes the shortcomings of the traditional statistical synthesis method and can quickly, conveniently and accurately model residential load sites. The load model generated by the present invention It can improve the accuracy of power grid simulation calculation and ensure the safe, reliable and economical operation of the power grid.

Description

A Load Modeling Method for Rural Load Types Based on Load Curve Decomposition

技术领域technical field

本发明涉及一种建模方法，具体讲涉及一种基于负荷曲线分解的农村负荷类型负荷建模方法。The invention relates to a modeling method, in particular to a load modeling method based on load curve decomposition in rural areas.

背景技术Background technique

电力系统数字仿真己成为电力系统规划设计、调度运行和分析研究的主要工具，电力系统各元件的数学模型以及由其构成的全系统数学模型是电力系统数字仿真的基础，模型的准确与否直接影响仿真结果和以此为基础的决策方案。仿真所采用的模型和参数是仿真准确性的重要决定因素，目前发电机、励磁系统、调速系统、变压器、输电线路的详细数学模型和建模技术已经得到了很好的发展，相对而言电力负荷模型仍较简单，往往是从基本物理概念出发而采用的实用化模型和参数。多年来，我国各大电网在电力系统分析计算时，通常按照经验选定某种常见的负荷模型（如电动机+恒阻抗模型或恒定功率+恒定阻抗模型）并定性地确定模型参数。Power system digital simulation has become the main tool for power system planning and design, dispatching operation and analysis research. The mathematical model of each component of the power system and the mathematical model of the whole system constituted by it are the basis of power system digital simulation. The accuracy of the model is directly related to Influence simulation results and decision-making schemes based on them. The model and parameters used in the simulation are important determinants of the accuracy of the simulation. At present, the detailed mathematical models and modeling techniques of generators, excitation systems, speed control systems, transformers, and transmission lines have been well developed. The electric load model is still relatively simple, and it is often a practical model and parameters based on basic physical concepts. For many years, when analyzing and calculating the power system of major power grids in my country, a common load model (such as motor + constant impedance model or constant power + constant impedance model) is usually selected according to experience, and the model parameters are qualitatively determined.

电力系统负荷建模由于其复杂性、分布性、时变性以及随机性等因素，决定了其数学模型建立的困难。目前的负荷建模方法基本上可分为3类，即统计综合法、总体辨测法和故障拟合法。这三种方法各有其优缺点。用统计综合法得到的负荷模型具有物理概念清晰、易于被工程人员理解的优点，但其核心是建立在“统计资料齐全，负荷特性精确”的基础之上，这一点往往难以做到，而且不可能经常进行统计，从而无法考虑负荷随时间变化的特性。总体测辨法避免了大量的统计工作，有可能得到随时间变化的在线实时负荷特性，其最大的问题在于过份依赖扰动事故，且辨识的模型参数物理意义不明确，另一个问题是难以在系统中所有变电站都安装有关装置。故障拟合法的优点是参数确定过程与现在程序计算时选择参数的过程一致，而且在某些故障下能获得重现。但实际上它是一种试凑的方法，在某些故障下的负荷参数是否适用于其他故障难以保证，而且认为全系统负荷参数相同、不变显然不符合负荷的实质。以上三种方法中统计综合法由于其物理模型清晰、概念明确，便于定性了解负荷特性，被广泛应用。Due to the complexity, distribution, time-varying and randomness of power system load modeling, it is difficult to establish its mathematical model. The current load modeling methods can basically be divided into three categories, namely statistical synthesis method, overall discrimination method and fault fitting method. All three methods have their pros and cons. The load model obtained by the statistical synthesis method has the advantages of clear physical concept and easy understanding by engineers, but its core is based on "complete statistical data and accurate load characteristics", which is often difficult to achieve, and it is not Statistics may be performed so often that the time-varying nature of the load cannot be considered. The overall measurement and identification method avoids a lot of statistical work, and it is possible to obtain the online real-time load characteristics that change over time. The biggest problem is that it relies too much on disturbance accidents, and the physical meaning of the identified model parameters is not clear. All substations in the system are equipped with relevant devices. The advantage of the fault fitting method is that the parameter determination process is consistent with the process of selecting parameters in the current program calculation, and can be reproduced under certain faults. But in fact it is a trial and error method. It is difficult to guarantee whether the load parameters under certain faults are applicable to other faults, and it is obviously not in line with the essence of the load to think that the load parameters of the whole system are the same and unchanged. Among the above three methods, the statistical synthesis method is widely used because of its clear physical model and clear concept, which is convenient for qualitative understanding of load characteristics.

但传统的统计综合法存在一定的缺点，包括：However, the traditional statistical synthesis method has certain shortcomings, including:

（1）调查所得负荷容量与实际负荷功率并不一致，因为存在同时率的问题，并非所有设备都是24小时投入使用，因此，需要进行分时段调查统计；(1) The load capacity obtained from the survey is not consistent with the actual load power, because there is a problem of simultaneous rate, not all equipment is put into use 24 hours a day, therefore, it is necessary to conduct surveys and statistics by time;

（2）随着时间的推移，实际负荷功率、负荷构成以及网络结构都可能发生变化，如果对负荷进行一次调查统计建模工作后就想一劳永逸，难以达到准确度要求；(2) As time goes by, the actual load power, load composition, and network structure may change. If you want to do it once and for all after conducting a survey and statistical modeling work on the load, it is difficult to meet the accuracy requirements;

（3）调查工作需统计成千上万个用户的负荷组成及参数，工作量巨大，而且难于获得准确的统计结果。(3) The survey work needs to count the load composition and parameters of thousands of users, the workload is huge, and it is difficult to obtain accurate statistical results.

综合负荷构成成分的随机时变是其负荷特性具有随机时变性的本质原因。这种时变性必然导致电力用户和变电站的日负荷曲线具有时变性。因此，用户和变电站的日负荷曲线必然含有反映负荷构成特性的丰富信息。The random time-varying components of the comprehensive load are the essential reasons for the random time-varying nature of the load characteristics. This time-varying will inevitably lead to time-varying daily load curves of power users and substations. Therefore, the daily load curves of users and substations must contain rich information reflecting the characteristics of load composition.

根据行业负荷分类的惯例，可将负荷划分为工业负荷、商业负荷和城市居民负荷、农村负荷及其他负荷这4类。其中农村负荷类型是比较复杂的负荷类型之一，也是比较重要的负荷类型，在各负荷类型站点中占有较大比重。农村负荷曲线具有如下特点：According to the practice of industry load classification, the load can be divided into four categories: industrial load, commercial load and urban residential load, rural load and other load. Among them, the rural load type is one of the more complex load types, and it is also a relatively important load type, which occupies a large proportion in all load type sites. The rural load curve has the following characteristics:

含有灌溉负荷的农村负荷的基本特点如下：The basic characteristics of rural loads including irrigation loads are as follows:

(1)农村负荷也随着季节和时间的变化而变化，春季和秋季白天含有大量的灌溉负荷，冬季含有一定的制热负荷，而夏季含有一定的制冷负荷，但是制热负荷和制冷负荷比例都不大。(1) Rural loads also change with seasons and time. Spring and autumn contain a large amount of irrigation load during the day, winter contains a certain amount of heating load, and summer contains a certain amount of cooling load, but the ratio of heating load to cooling load Not big.

(2)春季和秋季含有较大的灌溉负荷，是农村负荷的主要特点。灌溉负荷一般出现在白天的上午和下午，3、4、5月份和9、10、11月份灌溉负荷占有的比例较大；(2) Spring and autumn contain large irrigation loads, which are the main characteristics of rural loads. Irrigation load generally appears in the morning and afternoon of the day, and the proportion of irrigation load is larger in March, April, May and September, October, and November;

(3)农村负荷一般都比较小，负荷曲线的规律性不强，特殊的负荷曲线占有的比重较高。(3) Rural loads are generally small, the regularity of the load curve is not strong, and the proportion of special load curves is relatively high.

(4)不同的农村负荷线路灌溉负荷占有的比例有一定的差异。(4) There are certain differences in the proportion of irrigation load occupied by different rural load lines.

(5)对于每一天的负荷曲线，高峰负荷一般出现在晚上，部分线路白天的灌溉负荷较多，可能出现在白天。对于晚上负荷，一般春季和秋季的负荷较低，而冬季和夏季的负荷较高。对于白天的负荷，2月份的负荷比较低，其它月份负荷都比较高。(5) For the load curve of each day, the peak load generally occurs at night, and some lines have more irrigation load during the day, which may appear during the day. For evening loads, generally spring and autumn loads are lower, while winter and summer loads are higher. For the daytime load, the load in February is relatively low, and the load in other months is relatively high.

通过对用户和变电站日负荷曲线的特征分析，就能确定与之对应的各个典型用电行业用电设备构成比例及变电站的用电行业构成比例。由于数据采集与监控（SCADA）系统及负荷控制管理系统（简称负控系统）能够分别提供实时的变电站综合负荷的日负荷曲线和电力用户的日负荷曲线，因此，由此确定的综合负荷构成比例将具有在线、实时建模的性质，从而能够从根本上克服传统统计综合法负荷建模的固有缺陷。By analyzing the characteristics of the daily load curves of users and substations, it is possible to determine the proportion of electrical equipment in each typical power-consuming industry and the proportion of power-consuming industries in substations corresponding to it. Since the data acquisition and monitoring (SCADA) system and the load control management system (referred to as the load control system) can respectively provide real-time daily load curves of substation comprehensive loads and daily load curves of power users, the composition ratio of the comprehensive load thus determined It will have the nature of online and real-time modeling, so that it can fundamentally overcome the inherent defects of traditional statistical synthesis method load modeling.

发明内容Contents of the invention

为了克服上述现有技术的不足，本发明提供一种基于负荷曲线分解的农村负荷类型负荷建模方法，该方法克服了传统统计综合法的缺点，可以快速、方便、准确地为居民负荷站点建模，用本发明生成的负荷模型可提高电网仿真计算的准确度，保障电网安全、可靠、经济地运行。In order to overcome the deficiencies of the prior art above, the present invention provides a load modeling method for rural load types based on load curve decomposition. The load model generated by the invention can improve the accuracy of power grid simulation calculation and ensure safe, reliable and economical operation of the power grid.

为了实现上述发明目的，本发明采取如下技术方案：In order to realize the above-mentioned purpose of the invention, the present invention takes the following technical solutions:

本发明提供一种基于负荷曲线分解的农村负荷类型负荷建模方法，所述方法包括以下步骤：The invention provides a load modeling method for rural load types based on load curve decomposition. The method includes the following steps:

步骤1：分解农村负荷曲线，并计算农村负荷中每种负荷所占总量的比例；Step 1: Decompose the rural load curve and calculate the proportion of each load in the total rural load;

步骤2：计算静态负荷等值参数、动态负荷等值参数和配电网系统阻抗，并结合农村负荷节点的供电区域网络拓扑数据，得到农村负荷类型负荷模型。Step 2: Calculate static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and combine the topology data of the power supply area network of rural load nodes to obtain the rural load type load model.

所述步骤1包括以下步骤：Described step 1 comprises the following steps:

步骤1-1：分解农村负荷曲线；Step 1-1: Decompose the rural load curve;

步骤1-1-1：对所供负荷类型为农村负荷的变电站，合并该变电站10kV或6kV负荷出线的日负荷曲线；Step 1-1-1: For substations whose load type is rural load, merge the daily load curves of 10kV or 6kV load outgoing lines of the substation;

步骤1-1-2：计算该变电站负荷每月的有效负荷天数，如果某天负荷有零值，认为该天无效，如果某天负荷有反号现象，也认为该天无效；Step 1-1-2: Calculate the number of effective load days of the substation load per month. If the load on a certain day has a zero value, the day is considered invalid, and if the load on a certain day has a negative sign, the day is also considered invalid;

步骤1-1-3：计算该变电站负荷每月的平均日负荷曲线和平均月负荷曲线；Step 1-1-3: Calculate the monthly average daily load curve and average monthly load curve of the substation load;

步骤1-1-4：计算每月的每条日负荷曲线在白天时段，6～17点的最大值，并从2月份的平均日负荷曲线中选择白天时段的最大值，并确定满足基准负荷曲线要求的负荷曲线范围，其范围为计算出的白天时段的最大值0.8倍到1.15倍；Step 1-1-4: Calculate the maximum value of each daily load curve in the daytime period from 6 to 17 o'clock every month, and select the maximum value in the daytime period from the average daily load curve in February, and determine that the benchmark load is met The load curve range required by the curve, which ranges from 0.8 times to 1.15 times the maximum value of the calculated daytime period;

步骤1-1-5：从2月份开始对所有月份的负荷曲线进行分解，如果某个月的某条日负荷曲线在白天时段的最大值满足基准负荷曲线要求的日负荷曲线平均值范围，则认为该日负荷曲线为满足基准负荷曲线要求的曲线，选择分解月份中满足基准负荷曲线要求的曲线，计算曲线的平均曲线，作为基准负荷曲线；Step 1-1-5: Decompose the load curves of all months starting from February, if the maximum value of a certain daily load curve in a certain month during the daytime meets the average range of the daily load curve required by the benchmark load curve, then It is considered that the daily load curve is a curve that meets the requirements of the base load curve, and the curve that meets the requirements of the base load curve in the decomposition month is selected, and the average curve of the calculated curve is used as the base load curve;

步骤1-1-6：确定满足大负荷曲线要求的负荷曲线平均值下限，该下限值为白天时段的最大值的1.1倍，如果某条日负荷曲线在白天时段的最大值大于该下限值，则认为该条负荷曲线为满足大负荷曲线要求的曲线；Step 1-1-6: Determine the lower limit of the average value of the load curve that meets the requirements of the large load curve. The lower limit is 1.1 times the maximum value during the daytime period. If the maximum value of a certain daily load curve during the daytime period is greater than the lower limit value, it is considered that the load curve is a curve that meets the requirements of the large load curve;

步骤1-1-7：确定该月的大负荷曲线，并计算基准曲线负荷的平均值；Step 1-1-7: Determine the large load curve of the month, and calculate the average value of the benchmark curve load;

步骤1-2：计算农村负荷中每种负荷所占总量的比例；Step 1-2: Calculate the proportion of each load in the total rural load;

步骤1-2-1：计算各变电站每月的月平均负荷值和指定月份指定日的平均负荷值；Step 1-2-1: Calculate the monthly average load value of each substation and the average load value of the specified day in the specified month;

步骤1-2-2：计算指定月份、指定日、指定时刻的负荷值；Step 1-2-2: Calculate the load value of the specified month, specified day, and specified time;

步骤1-2-3：计算包括灌溉负荷、制冷负荷和制热负荷的分离部分负荷所占比例k_p0和基准负荷所占比例k_f。Step 1-2-3: Calculate the proportion k _p0 of the split part load including the irrigation load, the cooling load and the heating load and the proportion k _f of the base load.

所述步骤1-1-3中，计算该变电站负荷每月的平均日负荷曲线，将每月的所有有效日负荷曲线相加，再除以该月份的有效天数；In the step 1-1-3, calculate the monthly average daily load curve of the substation load, add all the effective daily load curves of each month, and then divide by the effective days of the month;

计算该变电站负荷每月的平均月负荷曲线，将每月每天日负荷曲线中24个时刻的负荷相加再除以24所得到的曲线；平均月负荷曲线中各有效天对应的值又称作日负荷曲线平均值，日负荷曲线平均值为每天24个时刻的负荷相加，再除以24所得到的平均值；Calculate the monthly average monthly load curve of the substation load, add the load at 24 moments in the daily load curve every month and divide it by 24 to obtain the curve; the value corresponding to each effective day in the average monthly load curve is also called The average value of the daily load curve, the average value of the daily load curve is the average value obtained by adding the load at 24 moments per day and dividing it by 24;

所述步骤1-1-5中，如果该月份满足基准曲线的负荷曲线的数目小于3条，并且前面没有计算过任何一个月份的基准曲线，则先不计算该月的曲线，进行下一个月曲线的计算；如果该月满足基准曲线要求的曲线数大于等于3条，计算基准曲线，将该月份的所有满足基准曲线要求的日负荷曲线相加，再除以该月满足基准曲线要求的曲线数，计算得到的基准曲线作为该月的小负荷曲线；否则，采用上一个月的基准曲线；In the step 1-1-5, if the number of load curves that meet the benchmark curves in this month is less than 3, and the benchmark curves of any month have not been calculated before, the curves of this month will not be calculated first, and the next month will be carried out Calculation of curves; if the number of curves that meet the requirements of the benchmark curve in the month is greater than or equal to 3, calculate the benchmark curve, add all the daily load curves that meet the requirements of the benchmark curve in the month, and then divide by the curve that meets the requirements of the benchmark curve in the month number, the calculated benchmark curve is used as the small load curve of the month; otherwise, the benchmark curve of the previous month is used;

所述步骤1-1-7中，确定该月的大负荷曲线，如果该月份满足大负荷曲线要求的负荷曲线条数小于3条，则忽略这些负荷曲线，该月份不计算大负荷曲线，不进行负荷曲线分离计算；如果该月份满足大负荷曲线要求的负荷曲线条数大于等于3条，计算大负荷曲线；In the step 1-1-7, determine the large load curve of the month, if the number of load curves that meet the requirements of the large load curve in the month is less than 3, then ignore these load curves, do not calculate the large load curve in this month, and do not Carry out load curve separation calculation; if the number of load curves that meet the requirements of large load curves in this month is greater than or equal to 3, calculate the large load curve;

计算基准曲线负荷的平均值，对大负荷曲线、基准负荷曲线进行标幺化，将三者的平均值变为1。Calculate the average value of the load of the reference curve, perform standard unitization on the large load curve and the reference load curve, and change the average value of the three to 1.

所述步骤1-2-1中，计算各变电站每月的月平均负荷值，将每月的平均月负荷曲线中各有效天对应的值相加，再除以该月份的有效天数；In the step 1-2-1, the monthly average load value of each substation is calculated, the value corresponding to each effective day in the average monthly load curve of each month is added, and then divided by the effective number of days in the month;

计算指定月份指定日的平均负荷值，将计算出的该月份的月平均负荷值乘以指定日的日负荷曲线平均值的标幺值；To calculate the average load value of the specified day in the specified month, multiply the calculated monthly average load value of the month by the per unit value of the average daily load curve of the specified day;

所述步骤1-2-2中，计算指定月份、指定日、指定时刻的负荷值；In described step 1-2-2, calculate the load value of specified month, specified day, specified moment;

1）如果只有小负荷曲线，则采用小负荷曲线进行计算；1) If there is only a small load curve, use the small load curve for calculation;

2）如果同时存在大负荷曲线和小负荷曲线，根据该月份指定日的日负荷曲线平均值大小确定是采用大负荷曲线或小负荷曲线进行计算；2) If there are large load curves and small load curves at the same time, it is determined whether to use the large load curve or the small load curve for calculation according to the average value of the daily load curve on the specified day of the month;

设定值为某月份的基准负荷平均值的1.1倍，则有：The set value is 1.1 times the average value of the base load in a certain month, then:

2-1）如果该月的日负荷曲线平均值小于该定值，采用小负荷曲线进行计算；2-1) If the average daily load curve of the month is less than the fixed value, use the small load curve for calculation;

2-2）如果日负荷曲线平均值大于该定值，采用大负荷曲线进行计算，大负荷计算结果为计算出的该月份的基准负荷平均值乘以该月份指定日的小/大负荷曲线中指定时刻对应的值；2-2) If the average daily load curve is greater than the fixed value, the large load curve is used for calculation, and the large load calculation result is the calculated average load of the month multiplied by the small/large load curve on the specified day of the month The value corresponding to the specified moment;

还需计算指定月份、指定日、指定时刻的分离负荷值，计算大负荷曲线和基准负荷曲线的差值，作为分离后的负荷曲线，该值等于计算出的指定月份、指定日、指定时刻的负荷值减去该月份的基准负荷平均值乘以该月份指定日的小负荷曲线中指定时刻对应的值，再除以计算出的指定月份、指定日、指定时刻的负荷值；It is also necessary to calculate the separated load value of the specified month, specified day, and specified time, and calculate the difference between the large load curve and the reference load curve, as the separated load curve, which is equal to the calculated specified month, specified day, and specified time. The load value minus the base load average value of the month multiplied by the value corresponding to the specified time in the small load curve of the specified day of the month, and then divided by the calculated load value of the specified month, specified day, and specified time;

所述步骤1-2-3中，计算分离部分负荷所占比例k_p0，将分离负荷值除以计算出的指定月份、指定日、指定时刻的负荷值；则基准负荷所占比例k_f为1-k_p0。In the step 1-2-3, the proportion k _p0 of the separated partial load is calculated, and the separated load value is divided by the calculated load value of the specified month, specified day, and specified time; then the proportion k _f of the reference load is 1-k _p0 .

所述步骤2中，静态负荷等值参数计算过程如下：In the step 2, the static load equivalent parameter calculation process is as follows:

将负荷功率与电压之间的关系描述为多项式方程形式的多项式静态负荷模型，有Describe the relationship between load power and voltage as a polynomial static load model in the form of a polynomial equation, with

P＝P₀[a×(V/V₀)²+b×(V/V₀)+c] （1）P＝P ₀ [a×(V/V ₀ ) ² +b×(V/V ₀ )+c] (1)

Q＝Q₀[α×(V/V₀)²+β×(V/V₀)+γ] （2）Q＝Q ₀ [α×(V/V ₀ ) ² +β×(V/V ₀ )+γ] (2)

其中，P和Q分别为静态负荷的有功功率和无功功率，V为静态负荷的实时电压，V₀为静态负荷的额定电压，P₀和Q₀分别为表示在V₀下静态负荷的额定有功功率和无功功率，a、b和c均为多项式静态负荷模型的有功功率系数，α、β和γ均为多项式静态负荷模型的无功功率系数；Among them, P and Q are the active power and reactive power of the static load respectively, V is the real-time voltage of the static load, V ₀ is the rated voltage of the static load, P ₀ and Q ₀ represent the rated voltage of the static load under V ₀ Active power and reactive power, a, b and c are the active power coefficients of the polynomial static load model, α, β and γ are the reactive power coefficients of the polynomial static load model;

对静态负荷的等值主要是对a、b、c、P₀和α、β、γ、Q₀的等值，对多项式静态负荷模型的等值是基于静态负荷功率对静态负荷端电压的灵敏度，有The equivalent value for static load is mainly the equivalent value for a, b, c, P ₀ and α, β, γ, Q ₀ , and the equivalent value for polynomial static load model is based on the sensitivity of static load power to static load terminal voltage ,Have

$\frac{&PartialD; &PartialD; P P}{&PartialD; &PartialD; V V} {| |}_{V V = = {V V}_{00}} = = {Σ Σ}_{i i = = 11}^{n no} \frac{{&PartialD; &PartialD; P P}_{i i}}{&PartialD; &PartialD; V V} {| |}_{V V = = {V V}_{00}} - - - - - - ((33))$

$\frac{&PartialD; &PartialD; Q Q}{&PartialD; &PartialD; V V} {| |}_{V V = = {V V}_{00}} = = {Σ Σ}_{i i = = 11}^{n no} \frac{{&PartialD; &PartialD; Q Q}_{i i}}{&PartialD; &PartialD; V V} {| |}_{V V = = {V V}_{00}} - - - - - - ((44))$

其中，n为静态负荷个数，P_i和Q_i分别为第i个静态负荷的有功和无功功率，和分别为第i个静态负荷有功功率相对于电压的偏微分和第i个静态负荷无功功率相对于电压的偏微分；Among them, n is the number of static loads, P _i and Q _i are the active and reactive power of the i-th static load, respectively, and Respectively, the partial differential of the i-th static load active power relative to the voltage and the partial differential of the i-th static load reactive power relative to the voltage;

当V＝V₀时，有：When V=V ₀ , there are:

${P P}_{00} = = {Σ Σ}_{i i = = 11}^{n no} {P P}_{00 i i} - - - - - - ((55))$

${Q Q}_{00} = = {Σ Σ}_{i i = = 11}^{n no} {Q Q}_{00 i i} - - - - - - ((66))$

其中，P_0i和Q_0i分别为第i个静态负荷的初始有功功率和无功功率；Among them, P _0i and Q _0i are the initial active power and reactive power of the i-th static load, respectively;

$a a = = \frac{{Σ Σ}_{i i = = 11}^{n no} {P P}_{00 i i} {a a}_{i i}}{{P P}_{00}} - - - - - - ((77))$

$b b = = \frac{{Σ Σ}_{i i = = 11}^{n no} {P P}_{00 i i} {b b}_{i i}}{{P P}_{00}} - - - - - - ((88))$

$c c = = \frac{{Σ Σ}_{i i = = 11}^{n no} {P P}_{00 i i} {c c}_{i i}}{{P P}_{00}} - - - - - - ((99))$

其中，a_i、b_i和c_i均为第i个静态负荷的多项式静态负荷模型有功功率系数；Among them, a _i , b _i and c _i are the active power coefficients of the polynomial static load model of the ith static load;

$α α = = \frac{{Σ Σ}_{i i = = 11}^{n no} {Q Q}_{00 i i} {α α}_{i i}}{{Q Q}_{00}} - - - - - - ((1010))$

$β β = = \frac{{Σ Σ}_{i i = = 11}^{n no} {Q Q}_{00 i i} {β β}_{i i}}{{Q Q}_{00}} - - - - - - ((1111))$

$γ γ = = \frac{{Σ Σ}_{i i = = 11}^{n no} {Q Q}_{00 i i} {γ γ}_{i i}}{{Q Q}_{00}} - - - - - - ((1212))$

其中，α_i、β_i和γ_i均为第i个静态负荷的多项式静态负荷模型无功功率系数。Among them, α _i , β _i and γ _i are the reactive power coefficients of the i-th static load polynomial static load model.

所述步骤2中具体包括以下步骤：The step 2 specifically includes the following steps:

步骤A：计算所有电动机的总吸收有功功率P_Σ与无功功率Q_Σ、总电磁功率P_Σem、总转子绕组铜耗P_Σcu2和总最大电磁功率P_{Σem_max}，具体有：Step A: Calculate the total absorbed active power P _Σ and reactive power Q _Σ , total electromagnetic power P _Σem , total rotor winding copper loss P _Σcu2 and total maximum electromagnetic power P _{Σem_max} of all motors, specifically:

${P P}_{Σ Σ} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{j j}$

${Q Q}_{Σ Σ} = = {Σ Σ}_{j j = = 11}^{m m} {Q Q}_{i i}$

${P P}_{Σem Σem} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{emj emj} - - - - - - ((1313))$

${P P}_{Σcu Σcu 22} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{cu cu 22 j j}$

${P P}_{Σem Σem__max max} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{em em__max max j j}$

其中P_j、Q_j、P_emj、P_cu2j和P_{em_maxj}分别表示第j台等值电动机的有功功率、无功功率、电磁功率、转子绕组铜耗和最大电磁功率；m为等值电动机台数；Among them, P _j , Q _j , P _emj , P _cu2j and P _{em_maxj} represent the active power, reactive power, electromagnetic power, rotor winding copper loss and maximum electromagnetic power of the jth equivalent motor respectively; m is the number of equivalent motors;

步骤B：计算等值电动机的定子绕组铜耗P_Σcu1及等值电动机的滑差S，分别表示为：Step B: Calculate the stator winding copper loss P _Σcu1 of the equivalent motor and the slip S of the equivalent motor, which are expressed as:

P_Σcu1=P_Σ-P_Σem （14）P _Σcu1 = P _Σ - P _Σem (14)

S=P_Σcu2/P_Σem （15）S=P _Σcu2 /P _Σem (15)

初始化等值电动机的最大电磁功率P_{emt_max}，使P_{emt_max}＝P_{Σem_max}；Initialize the maximum electromagnetic power P _{emt_max} of the equivalent electric motor, so that P _{emt_max} = P _{Σem_max} ;

步骤C：计算等值电动机的定子绕组电阻R_s；Step C: Calculate the stator winding resistance R _s of the equivalent motor;

先计算等值电动机的定子绕组相电流有First calculate the stator winding phase current of the equivalent motor Have

${\overset{\cdot \cdot}{I I}}_{Σ Σ} = = - - {((\frac{{P P}_{Σ Σ} + + {jQ jQ}_{Σ Σ}}{\sqrt{33} {\overset{\cdot &Center Dot;}{U u}}_{11}}))}^{* *} - - - - - - ((1616))$

其中，为电动机的机端电压；in, is the terminal voltage of the motor;

则等值电动机的定子绕组电阻R_s表示为：Then the stator winding resistance R _s of the equivalent motor is expressed as:

${R R}_{s the s} = = \frac{{P P}_{Σcu Σcu 11}}{33 {I I}_{Σ Σ}^{22}} - - - - - - ((1717))$

步骤D：计算等值机模型的等值阻抗Z_deq：Step D: Calculate the equivalent impedance Z _deq of the equivalent machine model:

${Z Z}_{deq deq} = = \frac{{U u}_{11}^{22}}{{P P}_{Σ Σ} - - j j {Q Q}_{Σ Σ}} - - - - - - ((1818))$

且有and have

R_deq=real(Z_deq) （19）R _deq = real(Z _deq ) (19)

X_deq=imag(Z_deq) （20）X _deq =imag(Z _deq ) (20)

其中，R_deq和X_deq为相应的等值电阻和等值电抗；Among them, _Rdeq and _Xdeq are the corresponding equivalent resistance and equivalent reactance;

步骤E：计算等值电动机的定子绕组漏抗X_s和转子绕组漏抗X_r，有Step E: Calculate the stator winding leakage reactance X _s and the rotor winding leakage reactance X _r of the equivalent motor, with

${X x}_{s the s} = = {X x}_{r r} = = \frac{X x}{22} = = - - \frac{11}{22} \sqrt{{((\frac{{U u}_{11}^{22}}{22 {P P}_{emt emt__max max}} - - {R R}_{s the s}))}^{22} - - {R R}_{s the s}^{22}} - - - - - - ((21 twenty one))$

步骤F：经过迭代对等值电动机的定子绕组漏抗X_s和转子绕组漏抗X_r进行修正；Step F: Iteratively correct the stator winding leakage reactance X _s and the rotor winding leakage reactance X _r of the equivalent motor;

步骤G：计算等值电动机的转子绕组电阻和等值激磁电抗；Step G: Calculating the rotor winding resistance and the equivalent exciting reactance of the equivalent motor;

根据计算得到的R_s、X_s、X_r和Z_deq，设K_r=R_deq-R_s，K_x=X_deq-X_s，则等值电动机的转子绕组电阻R_r和等值激磁电抗X_m分别表示为：According to the calculated R _s , X _s , X _r and Z _deq , let K _r =R _deq -R _s , K _x =X _deq -X _s , then the rotor winding resistance R _r and the equivalent excitation reactance of the equivalent motor _Xm are expressed as:

$\begin{matrix} {R R}_{r r} = = \frac{(({K K}_{r r} + + {K K}_{x x}^{22} / / {K K}_{r r} - - \sqrt{{(({K K}_{r r} + + {K K}_{x x}^{22} / / {K K}_{r r}))}^{22} - - {44 X x}_{s the s}^{22}})) S S}{22} \\ {X x}_{m m} = = \frac{{K K}_{r r} {X x}_{s the s} + + {K K}_{x x} \frac{{R R}_{r r}}{S S}}{\frac{{R R}_{r r}}{S S} - - {K K}_{r r}} \end{matrix} - - - - - - ((23 twenty three))$

步骤H：采用迭代法计算等值电动机的最大电磁功率，并对其进行修正；Step H: Calculating the maximum electromagnetic power of the equivalent electric motor by an iterative method, and correcting it;

1）计算第k次迭代中的等值电动机的最大电磁功率P_{emt_maxk}，表示为1) Calculate the maximum electromagnetic power P _{emt_maxk} of the equivalent electric motor in the kth iteration, expressed as

${P P}_{emt emt__max max k k} = = \frac{{U u}_{11}^{22}}{22 (({R R}_{s the s} + + \sqrt{{R R}_{s the s}^{22} + + {X x}^{22}}))} - - - - - - ((24 twenty four))$

基于戴维南定理，等值电动机的戴维南等值阻抗Z_dp表示为Based on Thevenin's theorem, the Thevenin equivalent impedance Z _dp of the equivalent motor is expressed as

${Z Z}_{dp dp} = = {jX wxya}_{r r} + + \frac{{jX wxya}_{m m} (({R R}_{s the s} + + {jX wxya}_{s the s}))}{{R R}_{s the s} + + j j (({X x}_{s the s} + + {X x}_{m m}))} - - - - - - ((2525))$

且有，R_dp=real(Z_dp)和X_dp=imag(Z_dp)，R_dp和X_dp分别为戴维南等值电阻和等值电抗；And, R _dp =real(Z _dp ) and X _dp =imag(Z _dp ), R _dp and X _dp are Thevenin equivalent resistance and equivalent reactance respectively;

等值电动机产生最大电磁功率的条件为：The conditions for the equivalent electric motor to generate the maximum electromagnetic power are:

${R R}_{pm pm} = = \frac{{R R}_{r r}}{{S S}_{m m}} = = \sqrt{{R R}_{dp dp}^{22} + + {X x}_{dp dp}^{22}} - - - - - - ((2626))$

其中，R_pm为对应产生最大电磁功率的戴维南等值阻抗幅值，S_m为临界滑差；Among them, R _pm is the Thevenin equivalent impedance amplitude corresponding to the maximum electromagnetic power, and S _m is the critical slip;

戴维南等值开路电压为Thevenin equivalent open circuit voltage for

${\overset{\cdot &Center Dot;}{V V}}_{00} = = \frac{{\overset{\cdot &Center Dot;}{U u}}_{11}}{\sqrt{33}} \times \times \frac{{jX wxya}_{m m}}{{R R}_{s the s} + + j j (({X x}_{s the s} + + {X x}_{m m}))} - - - - - - ((2727))$

则，第k次迭代中的戴维南等值电动机的最大电磁功率可表示为Then, the maximum electromagnetic power of the Thevenin equivalent motor in the kth iteration can be expressed as

${P P}_{em em__max max k k} = = \frac{33 {U u}_{00}^{22} {R R}_{pm pm}}{{(({R R}_{dp dp} + + {R R}_{pm pm}))}^{22} + + {X x}_{dp dp}^{22}} - - - - - - ((2828))$

2）修正等值电动机的最大电磁功率；2) Correct the maximum electromagnetic power of the equivalent motor;

计算第k次迭代中等值电动机最大电磁功率的修正系数τ_maxk，其表示为：Calculate the correction factor τ _maxk for the maximum electromagnetic power of the equivalent motor in the k-th iteration, which is expressed as:

${τ τ}_{max max k k} = = \frac{{P P}_{emt emt__max max k k}}{{P P}_{em em__max max k k}} - - - - - - ((2929))$

则修正后的等值电动机最大电磁功率P_{emt_max}表示为：Then the modified equivalent motor maximum electromagnetic power P _{emt_max} is expressed as:

P_{emt_max}＝τ_maxk×P_{Σem_max} （30）P _{emt_max} ＝τ _maxk ×P _{Σem_max} (30)

以P_{emt_max}与P_{emt_maxk}的绝对误差为等值电动机最大电磁功率的迭代误差有Take the absolute error of P _{emt_max} and P _{emt_maxk} as the iterative error of the maximum electromagnetic power of the equivalent motor Have

${Err Err}_{{P P}_{em em__max max}} = = | | {P P}_{emt emt__max max} - - {P P}_{emt emt__max max k k} | | - - - - - - ((3131))$

以为迭代收敛标准，若则重新计算X_s、R_r、X_r和X_m；by is the iterative convergence criterion, if Then recalculate X _s , R _r , X _r and X _m ;

步骤I：计算等值惯性时间常数；Step I: Calculate the equivalent inertial time constant;

等值惯性时间常数H表示为：The equivalent inertial time constant H is expressed as:

$H h = = \frac{{Σ Σ}_{j j = = 11}^{m m} {P P}_{nj nj} {H h}_{j j}}{{Σ Σ}_{j j = = 11}^{m m} {P P}_{nj nj}} - - - - - - ((3232))$

其中P_nj和H_j为第j台等值电动机的额定功率和惯性时间常数。Among them, P _nj and H _j are the rated power and inertia time constant of the jth equivalent electric motor.

所述步骤2中，配电网系统阻抗表示为：In the step 2, the impedance of the distribution network system is expressed as:

${Z Z}_{eq eq} = = [[{Σ Σ}_{f f = = 11}^{M m} {u u}_{f f}^{22} {((11 / / {Z Z}_{f f}))}^{* *}]] / / {(({Σ Σ}_{l l = = 11}^{N N} {I I}_{l l}))}^{22} - - - - - - ((3333))$

其中，Z_eq为配电网系统阻抗；u_f表示母线电压，Z_f表示变压器和配电线路阻抗；I_l表示负荷电流，其中，M和N分别为配电网系统中的节点总数和支路总条数。Among them, Z _eq is the impedance of the distribution network system; u _f is the bus voltage, Z _f is the transformer and distribution line impedance; I _l is the load current, where M and N are the total number of nodes and branch points The total number of roads.

所述步骤2中的供电区域网络拓扑数据包括配电线路数据、变压器数据和无功补偿数据。The power supply area network topology data in the step 2 includes distribution line data, transformer data and reactive power compensation data.

与现有技术相比，本发明的有益效果在于：Compared with prior art, the beneficial effect of the present invention is:

1.本方法基于电力系统营销部门负荷控制管理系统采集的电力用户的日负荷曲线及数据采集与监控（SCADA）系统提供的实时变电站综合负荷的日负荷曲线，通过负荷曲线分解获得任意时刻农村负荷变电站所供的各用电设备类型及比例，并利用调度部门提供的负荷节点的供电区域网络拓扑数据，采用统计综合法，实现农村负荷站点的在线负荷建模，以达到准确地为农村负荷站点建模的目的，提高电网仿真计算的准确度，保障电网安全、可靠、经济地运行。1. This method is based on the daily load curve of power users collected by the load control management system of the marketing department of the power system and the daily load curve of the real-time substation comprehensive load provided by the data acquisition and monitoring (SCADA) system, and the rural load at any time can be obtained by decomposing the load curve The type and proportion of each electrical equipment supplied by the substation, and using the topology data of the power supply area network of the load node provided by the dispatching department, the statistical synthesis method is used to realize the online load modeling of the rural load site, so as to achieve an accurate model for the rural load site. The purpose of modeling is to improve the accuracy of power grid simulation calculations and ensure safe, reliable and economical operation of the power grid.

2.本方法基于实际电网的日负荷曲线数据，开展了在线负荷建模，改变了传统统计综合法依靠离线人工调查的模式，可快捷、方便、准确地进行负荷建模，大大节省了人力、物力，且避免了由于调查结果不准确导致的负荷建模不准确问题，本方法是对传统建模方式的进一步提升，为负荷建模工作提供了重要的指导的作用。2. This method is based on the daily load curve data of the actual power grid, and carries out online load modeling, which changes the mode of traditional statistical synthesis method relying on offline manual investigation, and can quickly, conveniently and accurately carry out load modeling, greatly saving manpower, Material resources, and avoid the inaccurate load modeling problem caused by inaccurate survey results, this method is a further improvement to the traditional modeling method, and provides an important guiding role for the load modeling work.

3.本方法通过多样化的在线数据资源（如营销负荷控制管理系统采集的电力用户的日负荷曲线及数据采集与监控（SCADA）系统提供的实时变电站综合负荷的日负荷曲线及调度网络拓扑信息）实现统计综合法的在线化，对负荷节点可以进行实时SLM建模，模型物理意义明确，适应性强，克服了传统所有负荷建模方法难以适应负荷时变性的问题。3. This method uses diversified online data resources (such as the daily load curve of power users collected by the marketing load control management system and the daily load curve of real-time substation comprehensive load and dispatching network topology information provided by the data acquisition and monitoring (SCADA) system. ) realizes the online statistical synthesis method, and can carry out real-time SLM modeling for load nodes. The model has clear physical meaning and strong adaptability, and overcomes the problem that all traditional load modeling methods are difficult to adapt to the time-varying load.

附图说明Description of drawings

图1是漯河变35kV农村负荷1月份的负荷曲线图；Figure 1 is the load curve of Luohe Substation 35kV rural load in January;

图2是漯河变35kV农村负荷5月份的负荷曲线图；Figure 2 is the load curve of Luohe Substation 35kV rural load in May;

图3是漯河变35kV农村负荷7月份的负荷曲线图；Figure 3 is the load curve of Luohe Substation 35kV rural load in July;

图4是漯河变35kV农村负荷11月份的负荷曲线图；Figure 4 is the load curve of Luohe Substation 35kV rural load in November;

图5是漯河变35kV农村负荷1月份分解后的负荷曲线图；Figure 5 is the load curve diagram of Luohe Substation's 35kV rural load after decomposition in January;

图6是漯河变35kV农村负荷5月份分解后的负荷曲线图；Figure 6 is the load curve of Luohe Substation 35kV rural load decomposed in May;

图7是漯河变35kV农村负荷7月份分解后的负荷曲线图；Figure 7 is the load curve of Luohe Substation 35kV rural load after decomposition in July;

图8是为漯河变35kV农村负荷11月份分解后的负荷曲线图；Figure 8 is the load curve diagram after decomposition of the 35kV rural load of Luohe Substation in November;

图9是基于负荷曲线分解的农村负荷类型负荷建模方法流程图；Fig. 9 is a flow chart of a load modeling method for rural load types based on load curve decomposition;

图10是220kV负荷母线电压曲线的对比图；Figure 10 is a comparison diagram of the 220kV load bus voltage curve;

图11是220kV负荷母线负荷有功功率曲线的对比图；Figure 11 is a comparison diagram of the active power curve of the 220kV load bus load;

图12是220kV负荷母线负荷无功功率曲线的对比图。Figure 12 is a comparison chart of reactive power curves of 220kV load busbar load.

具体实施方式detailed description

下面结合附图对本发明作进一步详细说明。The present invention will be described in further detail below in conjunction with the accompanying drawings.

针对农村负荷的负荷曲线基本特点，采用的基本处理方法如下：According to the basic characteristics of the load curve of the rural load, the basic processing methods adopted are as follows:

(1)由于农村负荷一般都比较小，并且规律性较差，因此对于同一变电站的农村负荷，进行合并。(1) Since the rural loads are generally small and have poor regularity, the rural loads of the same substation are combined.

如果不进行合并，由于农村负荷的规律性较差，很难得到理想的分解结果。图1-图4中的负荷曲线就是漯河变所有35kV农村负荷线路合并后的负荷曲线，合并后总的负荷较大，负荷的变化规律也较强。If it is not merged, it is difficult to obtain ideal decomposition results due to the poor regularity of rural loads. The load curves in Figure 1-Figure 4 are the combined load curves of all 35kV rural load lines in Luohe Substation. After the combined load, the total load is relatively large, and the change law of the load is also strong.

(2)含有灌溉负荷的农村负荷曲线分解为基本负荷、制冷负荷、制热负荷和灌溉负荷。(2) The rural load curve including irrigation load is decomposed into base load, cooling load, heating load and irrigation load.

制冷负荷主要出现在夏季的6、7、8月份，制热负荷主要出现在冬季的12、1、2月份，灌溉负荷主要出现在春季的3、4、5月份和秋季的9、10、11月份。因此在分解时，分离出来的负荷，6、7、8月份的认为是制冷负荷，12、1、2月份的认为是制热负荷，3、4、5、9、10、11月份的认为是灌溉负荷。但是季节交替的部分月份分离出来的负荷可能同时存在两种负荷，由于无法再继续进行分解，因此近似认为只存在一种负荷，例如8月份分离出来的负荷可能同时存在制冷负荷和灌溉负荷，近似认为8月份分离的负荷都是制冷负荷。The cooling load mainly occurs in June, July and August in summer, the heating load mainly occurs in December, January and February in winter, and the irrigation load mainly occurs in March, April and May in spring and September, October and November in autumn. month. Therefore, when decomposing, the separated loads are considered to be cooling loads in June, July, and August, heating loads in December, January, and February, and considered to be heating loads in March, April, May, September, October, and November. irrigation load. However, the loads separated in some months during the alternation of seasons may have two loads at the same time. Since the decomposition can no longer be continued, it is approximately considered that there is only one load. For example, the loads separated in August may have cooling loads and irrigation loads at the same time, approximately The load separated in August is considered to be the cooling load.

(3)分离曲线的基本方法：根据白天时段最大值对曲线进行分解，将每个月的负荷曲线分为两条负荷曲线，一条为基准负荷曲线，另一条为大负荷曲线，两曲线的差值作为分离的负荷曲线。(3) The basic method of separating the curves: Decompose the curve according to the maximum value during the daytime, and divide the monthly load curve into two load curves, one is the base load curve, the other is the large load curve, and the difference between the two curves Values are used as separate load curves.

对于每个月的负荷曲线，根据负荷的大小将其分为两组曲线，计算这两组曲线的平均曲线，作为典型的负荷曲线，用于近似模拟每个月中不同负荷水平时的负荷曲线。其中的一条曲线为基本负荷曲线，该曲线代表了基本负荷的变化过程，应根据所有月份中平均负荷较小的曲线计算得出。农村居民负荷含有灌溉负荷，分离基本负荷曲线的基本方法是采用白天时段最大值进行分解，选择白天时段最大值较小的曲线作为基准负荷曲线。For the load curve of each month, it is divided into two groups of curves according to the size of the load, and the average curve of these two groups of curves is calculated as a typical load curve, which is used to approximate the load curve at different load levels in each month . One of the curves is the base load curve, which represents the change process of the base load and should be calculated from the curve with the lower average load in all months. The load of rural residents includes irrigation load. The basic method of separating the basic load curve is to use the maximum value of the daytime period for decomposition, and select the curve with a smaller maximum value during the daytime period as the base load curve.

分解主要考虑两个时段，一个是白天时段（6-17点），另一个为晚上时段，基准负荷曲线应该满足白天负荷较小，晚上负荷也较小。但是从农村负荷曲线的基本特点可知，①通常春季和秋季晚上的负荷较小，而夏季和冬季晚上的负荷较大，但是两者相差的并不很大；②白天负荷在不同季节变化较大，特别是春秋季节；③白天负荷较小的负荷曲线，晚上负荷并不一定比较小。如果选择白天负荷和晚上负荷都比较小的曲线作为基准负荷曲线，则满足基准曲线的曲线数较少，仅采用白天时段最大值能够比较准确分解基础负荷曲线，因此仅采用白天负荷最大值进行分解。The decomposition mainly considers two time periods, one is the daytime period (6-17 o'clock) and the other is the evening period. The base load curve should satisfy that the daytime load is small and the evening load is also small. However, from the basic characteristics of the rural load curve, it can be seen that ① usually the load in spring and autumn evenings is small, while the load in summer and winter is relatively large, but the difference between the two is not very large; ② the daytime load varies greatly in different seasons , especially in spring and autumn; ③ The load curve with a smaller load during the day does not necessarily have a smaller load at night. If a curve with relatively small daytime load and nighttime load is selected as the base load curve, the number of curves that meet the base curve is small, and only the maximum value during the daytime period can be used to decompose the base load curve more accurately. Therefore, only the maximum value of the daytime load is used for decomposition. .

确定基准负荷曲线白天时段最大值满足的范围采用两种方法：Two methods are used to determine the range within which the maximum value of the baseline load curve during the day is met:

计算所有日负荷曲线白天时段的最大值，从中选择一定范围内的值，计算它们的平均值； Calculate the maximum value of all daily load curves during the daytime, select values within a certain range from them, and calculate their average value;

由于2月份白天时段负荷最低，并且分散性较小，因此计算2月份所有日负荷曲线白天时段的最大值，然后计算其平均值。 Since the daytime load is the lowest in February and the dispersion is small, the maximum value of all daily load curves in February during the daytime is calculated, and then its average value is calculated.

在上述平均值的基础上，选择基准负荷曲线白天时段最大负荷满足的范围，对于每个月的负荷，白天时段最大负荷只要在此范围之内，认为是基础负荷曲线；大于此范围的最大值，认为是大负荷曲线。计算基础负荷曲线的平均负荷曲线作为小负荷典型负荷曲线，计算大负荷曲线的平均曲线作为大负荷曲线的典型曲线，两者的差值作为制冷、制热或灌溉负荷。On the basis of the above average value, select the range that the maximum load of the benchmark load curve meets during the daytime. For the load of each month, as long as the maximum load during the daytime is within this range, it is considered to be the base load curve; the maximum value greater than this range , considered to be a large load curve. Calculate the average load curve of the base load curve as the typical load curve of small load, calculate the average curve of the large load curve as the typical curve of the large load curve, and take the difference between the two as the cooling, heating or irrigation load.

上述的两种方法中，第一种方法需要确定计算基准曲线平均值的范围，该范围比较难于确定；第二种方法相对比较容易，并且具有一定的适应性。Among the above two methods, the first method needs to determine the range for calculating the average value of the benchmark curve, which is difficult to determine; the second method is relatively easy and has certain adaptability.

分解农村负荷曲线过程具体如下：The process of decomposing the rural load curve is as follows:

计算该变电站负荷每月的平均日负荷曲线，即将每月的所有有效日负荷曲线相加，再除以该月份的有效天数；Calculate the monthly average daily load curve of the substation load, that is, add all the effective daily load curves of each month, and then divide by the number of effective days in the month;

计算该变电站负荷每月的平均月负荷曲线，即将每月每天日负荷曲线中24个时刻的负荷相加再除以24所得到的曲线；平均月负荷曲线中各有效天对应的值又称作日负荷曲线平均值，日负荷曲线平均值即为每天24个时刻的负荷相加，再除以24所得到的平均值；Calculate the monthly average monthly load curve of the substation load, which is the curve obtained by adding the load at 24 moments in the daily load curve every month and dividing it by 24; the value corresponding to each valid day in the average monthly load curve is also called The average value of the daily load curve, the average value of the daily load curve is the average value obtained by adding the load at 24 moments per day and dividing it by 24;

步骤1-1-4：计算每月的每条日负荷曲线在白天时段，即6～17点的最大值，并从2月份的平均日负荷曲线中选择白天时段的最大值，并确定满足基准负荷曲线要求的负荷曲线范围，其范围为计算出的白天时段的最大值0.8倍到1.15倍；Step 1-1-4: Calculate the maximum value of each daily load curve in the daytime period of each month, that is, from 6 to 17 o'clock, and select the maximum value in the daytime period from the average daily load curve in February, and determine that the benchmark is met The load curve range required by the load curve, which ranges from 0.8 times to 1.15 times the maximum value of the calculated daytime period;

如果该月份满足基准曲线的负荷曲线的数目小于3条，并且前面没有计算过任何一个月份的基准曲线，则先不计算该月的曲线，进行下一个月曲线的计算；如果该月满足基准曲线要求的曲线数大于等于3条，计算基准曲线，即将该月份的所有满足基准曲线要求的日负荷曲线相加，再除以该月满足基准曲线要求的曲线数，计算得到的基准曲线作为该月的小负荷曲线；否则，采用上一个月的基准曲线；If the number of load curves that meet the benchmark curve in this month is less than 3, and the benchmark curve of any month has not been calculated before, the curve of this month will not be calculated first, and the calculation of the next month's curve will be performed; if the month meets the benchmark curve The number of required curves is greater than or equal to 3, and the calculation of the benchmark curve is to add all the daily load curves that meet the requirements of the benchmark curve in the month, and then divide by the number of curves that meet the requirements of the benchmark curve in the month, and the calculated benchmark curve is used as the month the small load curve; otherwise, use the baseline curve of the previous month;

确定该月的大负荷曲线，如果该月份满足大负荷曲线要求的负荷曲线条数小于3条，则忽略这些负荷曲线，该月份不计算大负荷曲线，不进行负荷曲线分离计算；如果该月份满足大负荷曲线要求的负荷曲线条数大于等于3条，计算大负荷曲线；Determine the large load curve of the month, if the number of load curves that meet the requirements of the large load curve in this month is less than 3, ignore these load curves, do not calculate the large load curve in this month, and do not perform separate calculation of the load curve; if the month meets The number of load curves required by the large load curve is greater than or equal to 3, and the large load curve is calculated;

计算基准曲线负荷的平均值，对大负荷曲线、基准负荷曲线进行标幺化，即将三者的平均值变为1。Calculate the average value of the load of the reference curve, and perform standard unitization on the large load curve and the reference load curve, that is, the average value of the three is changed to 1.

附图5-8为漯河变35kV农村负荷1、5、7、11月份分解后的负荷曲线。Attached Figures 5-8 are the decomposed load curves of Luohe Substation's 35kV rural load in January, May, July and November.

如附图1-4和附图5-8所示，1月份负荷较大，并且分散性较小，没有满足基准负荷曲线要求的负荷曲线，采用12月份的基础负荷曲线，形成了一条大负荷曲线，分离出来的负荷为制冷负荷。5月份和11月份灌溉负荷占有的比例较多，灌溉负荷主要出现在上午和下午。7月份分离出来的负荷为制冷负荷。As shown in attached drawings 1-4 and attached drawings 5-8, the load in January is relatively large, and the dispersion is small, and there is no load curve that meets the requirements of the base load curve. Using the base load curve in December, a large load is formed. curve, the separated load is the cooling load. In May and November, the proportion of irrigation load is relatively large, and the irrigation load mainly occurs in the morning and afternoon. The load separated in July is the cooling load.

计算农村负荷中每种负荷所占总量的比例过程如下：The process of calculating the proportion of each load in the total rural load is as follows:

计算各变电站每月的月平均负荷值，即将每月的平均月负荷曲线中各有效天对应的值相加，再除以该月份的有效天数；Calculate the monthly average load value of each substation, that is, add the values corresponding to each effective day in the monthly average monthly load curve, and then divide by the effective number of days in the month;

计算指定月份指定日的平均负荷值，即将计算出的该月份的月平均负荷值乘以指定日的日负荷曲线平均值的标幺值；Calculate the average load value of the specified day in the specified month, that is, multiply the calculated monthly average load value of the month by the per unit value of the average daily load curve of the specified day;

计算分离部分负荷所占比例k_p0，即将分离负荷值除以计算出的指定月份、指定日、指定时刻的负荷值；则基准负荷所占比例k_f为1-k_p0。Calculating the proportion k _p0 of the separated partial load is to divide the separated load value by the calculated load value of the specified month, specified day, and specified time; then the proportion k _f of the reference load is 1-k _p0 .

对静态负荷的等值主要是对a、b、c、P₀和α、β、γ、Q₀的等值，对多项式静态负荷模型的等值是基于静态负荷功率对静态负荷端电压的灵敏度，即The equivalent value for static load is mainly the equivalent value for a, b, c, P ₀ and α, β, γ, Q ₀ , and the equivalent value for polynomial static load model is based on the sensitivity of static load power to static load terminal voltage ,Right now

当V＝V₀时，有：When V=V ₀ , there are:

${P P}_{00} = = {Σ Σ}_{i i = = 11}^{n no} {P P}_{00 i i} - - - - - - ((55))$

${Q Q}_{00} = = {Σ Σ}_{i i = = 11}^{n no} {Q Q}_{00 i i} - - - - - - ((66))$

${P P}_{Σ Σ} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{j j}$

${Q Q}_{Σ Σ} = = {Σ Σ}_{j j = = 11}^{m m} {Q Q}_{i i}$

${P P}_{Σcu Σcu 22} = = {Σ Σ}_{j j = = 11}^{m m} {P P}_{cu cu 22 j j}$

P_Σcu1=P_Σ-P_Σem （14）P _Σcu1 = P _Σ - P Σ _em (14)

S=P_Σcu2/P_Σem （15）S=P _Σcu2 /P _Σem (15)

${\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} = = - - {((\frac{{P P}_{Σ Σ} + + {jQ jQ}_{Σ Σ}}{\sqrt{33} {\overset{\cdot \cdot}{U u}}_{11}}))}^{* *} - - - - - - ((1616))$

${Z Z}_{deq deq} = = \frac{{U u}_{11}^{22}}{{P P}_{Σ Σ} - - {jQ jQ}_{Σ Σ}} - - - - - - ((1818))$

且有and have

R_deq=real(Z_deq) （19）R _deq = real(Z _deq ) (19)

X_deq=imag(Z_deq) （20）X _deq =imag(Z _deq ) (20)

步骤H：采用迭代法计算等值电动机的最大电磁功率，并对其进行修正；Step H: Calculate the maximum electromagnetic power of the equivalent electric motor by an iterative method, and correct it;

戴维南等值开路电压为Thevenin equivalent open circuit voltage for

附图10-12为采用本发明得出的SLM等值模型参数的仿真曲线与实测曲线的对比图Accompanying drawing 10-12 adopts the comparison figure of the emulation curve of the SLM equivalent model parameter that the present invention draws and measured curve

附图10为220kV负荷母线电压曲线的对比图；其中实线为实测曲线，虚线为采用本发明得出的等值SLM模型参数时的仿真结果；Accompanying drawing 10 is the comparative figure of 220kV load busbar voltage curve; Wherein solid line is measured curve, and dotted line is the simulation result when adopting the equivalent SLM model parameter that the present invention draws;

附图11为220kV负荷母线负荷有功功率曲线的对比图；其中实线为实测曲线，虚线为采用本发明得出的等值SLM模型参数时的仿真结果；Accompanying drawing 11 is the comparative figure of 220kV load busbar load active power curve; Wherein solid line is measured curve, and dotted line is the simulation result when adopting the equivalent SLM model parameter that the present invention draws;

附图12为220kV负荷母线负荷无功功率曲线的对比图；其中实线为实测曲线，虚线为采用本发明得出的等值SLM模型参数时的仿真结果。Accompanying drawing 12 is the comparison chart of 220kV load bus bar load reactive power curve; Wherein the solid line is the measured curve, and the dotted line is the simulation result when adopting the equivalent SLM model parameter that the present invention obtains.

本发明的技术方案在河南省电力公司的河南电网稳定模型分析研究中得到应用。The technical scheme of the invention is applied in the analysis and research of the stability model of the Henan power grid of the Henan Electric Power Company.

最后应当说明的是：以上实施例仅用以说明本发明的技术方案而非对其限制，尽管参照上述实施例对本发明进行了详细的说明，所属领域的普通技术人员应当理解：依然可以对本发明的具体实施方式进行修改或者等同替换，而未脱离本发明精神和范围的任何修改或者等同替换，其均应涵盖在本发明的权利要求范围当中。Finally, it should be noted that the above embodiments are only used to illustrate the technical solutions of the present invention and not to limit them. Although the present invention has been described in detail with reference to the above embodiments, those of ordinary skill in the art should understand that: the present invention can still be Any modification or equivalent replacement that does not depart from the spirit and scope of the present invention shall be covered by the scope of the claims of the present invention.

Claims

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{\partial P}{\partial V} |_{V = V_{0}} = Σ_{i = 1}^{n} \frac{\partial P_{i}}{\partial V} |_{V = V_{0}} - - - (3)

\frac{\partial Q}{\partial V} |_{V = V_{0}} = Σ_{i = 1}^{n} \frac{\partial Q_{i}}{\partial V} |_{V = V_{0}} - - - (4)

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} = Σ_{i = 1}^{n} P_{0 i} - - - (5)

Q_{0} = Σ_{i = 1}^{n} Q_{0 i} - - - (6)

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

a = \frac{Σ_{i = 1}^{n} P_{0 i} a_{i}}{P_{0}} - - - (7)

b = \frac{Σ_{i = 1}^{n} P_{0 i} b_{i}}{P_{0}} - - - (8)

c = \frac{Σ_{i = 1}^{n} P_{0 i} c_{i}}{P_{0}} - - - (9)

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

α = \frac{Σ_{i = 1}^{n} Q_{0 i} α_{i}}{Q_{0}} - - - (10)

β = \frac{Σ_{i = 1}^{n} Q_{0 i} β_{i}}{Q_{0}} - - - (11)

γ = \frac{Σ_{i = 1}^{n} Q_{0 i} γ_{i}}{Q_{0}} - - - (12)

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{matrix} P_{Σ} = Σ_{j = 1}^{m} P_{j} \\ Q_{Σ} = Σ_{j = 1}^{m} Q_{i} \\ P_{Σ e m} = Σ_{j = 1}^{m} P_{e m j} \\ P_{Σ c u 2} = Σ_{j = 1}^{m} P_{c u 2 j} \\ P_{Σ e m_\max} = Σ_{j = 1}^{m} P_{e m_\max j} \end{matrix} - - - (13)

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

{\overset{\cdot}{I}}_{Σ} = - {(\frac{P_{Σ} + {jQ}_{Σ}}{\sqrt{3} {\overset{\cdot}{U}}_{1}})}^{*} - - - (16)

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_{Σ c u 1}}{3 {I_{Σ}}^{2}} - - - (17)

step D: equivalent impedance Z of computer equivalent model_deq：

Z_{d e q} = \frac{{U_{1}}^{2}}{P_{Σ} - {jQ}_{Σ}} - - - (18)

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}}{2 P_{e m t_m a x}} - R_{s})}^{2} - {R_{s}}^{2}} - - - (21)

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{matrix} 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{matrix} - - - (23)

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_{e m t_\max k} = \frac{{U_{1}}^{2}}{2 (R_{s} + \sqrt{{R_{s}}^{2} + X^{2}})} - - - (24)

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

Z_{d p} = {jX}_{r} + \frac{{jX}_{m} (R_{s} + {jX}_{s})}{R_{s} + j (X_{s} + X_{m})} - - - (25)

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_{p m} = \frac{R_{r}}{S_{m}} = \sqrt{{R_{d p}}^{2} + {X_{d p}}^{2}} - - - (26)

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

{\overset{\cdot}{V}}_{0} = \frac{{\overset{\cdot}{U}}_{1}}{\sqrt{3}} \times \frac{{jX}_{m}}{R_{s} + j (X_{s} + X_{m})} - - - (27)

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

P_{e m_\max k} = \frac{3 {V_{0}}^{2} R_{p m}}{{(R_{d p} + R_{p m})}^{2} + {X_{d p}}^{2}} - - - (28)

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:

τ_{\max k} = \frac{P_{e m t_\max k}}{P_{e m_\max k}} - - - (29)

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

{Err}_{P_{e m_m a x}} = | P_{e m t_m a x} - P_{e m t_\max k} | - - - (31)

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{Σ_{j = 1}^{m} P_{n j} H_{j}}{Σ_{j = 1}^{m} P_{n j}} - - - (32)

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_{e q} = [Σ_{f = 1}^{M} u_{f}^{2} {(1 / Z_{f})}^{*}] / {(Σ_{l = 1}^{N} I_{l})}^{2} - - - (33)

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.