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CN105977970B - A kind of power distribution network intelligent trouble restoration methods containing distributed generation resource - Google Patents

A kind of power distribution network intelligent trouble restoration methods containing distributed generation resource Download PDF

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CN105977970B
CN105977970B CN201610452688.6A CN201610452688A CN105977970B CN 105977970 B CN105977970 B CN 105977970B CN 201610452688 A CN201610452688 A CN 201610452688A CN 105977970 B CN105977970 B CN 105977970B
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CN105977970A (en
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杨强
蒋乐
赵海麟
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Zhejiang University ZJU
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    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for AC mains or AC distribution networks
    • HELECTRICITY
    • H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY
    • H02J3/00Circuit arrangements for AC mains or AC distribution networks
    • H02J3/001Methods to deal with contingencies, e.g. abnormalities, faults or failures

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Abstract

本发明公开了一种含分布式电源的配电网智能故障恢复方法,该方法针对目前配电网中发生的大部分是暂时性故障,考虑配电网中分布式电源的渗透率逐渐提高,故障恢复方法根据故障影响范围,以一级负荷以及负荷总恢复率为判据选择基于分布式电源或基于拓扑重构的恢复方法。在故障恢复过程中,为隔离故障点并恢复失电的非故障负荷,以故障恢复时间和故障恢复影响支路负荷率最小和可能恢复负荷最大为目标函数,逐步恢复配电网中的负荷。本发明以53节点网络为测试系统给出了详细的算法描述,并通过一系列的实验证明所提方法在提高故障恢复速度以及有效提高了一级负荷的恢复速度。

The invention discloses an intelligent fault recovery method for a distribution network containing distributed power sources. The method aims at most of the temporary faults occurring in the distribution network at present, and considering that the penetration rate of distributed power sources in the distribution network is gradually increased, The fault recovery method selects a recovery method based on distributed power or topology reconfiguration based on the scope of the fault, based on the primary load and the total recovery rate of the load. In the fault recovery process, in order to isolate the fault point and restore the non-fault loads that have lost power, the load in the distribution network is gradually restored with the minimum fault recovery time, the minimum load rate of the fault recovery affected branch and the maximum possible recovery load as the objective functions. The present invention provides a detailed algorithm description by using a 53-node network as a test system, and proves that the proposed method improves the fault recovery speed and effectively improves the recovery speed of the primary load through a series of experiments.

Description

一种含分布式电源的配电网智能故障恢复方法An Intelligent Fault Restoration Method for Distribution Network Containing Distributed Power

技术领域technical field

本发明涉及含分布式电源和故障恢复领域,尤其涉及一种含分布式电源的配电网智能故障恢复方法。The invention relates to the field of distributed power supply and fault recovery, in particular to an intelligent fault recovery method for distribution network with distributed power supply.

背景技术Background technique

近年来,为减少化石燃料燃烧造成的温室气体排放,提高配电网的稳定性,各种形式的分布式电源渗透率得到了显著的提高。但由于分布式电源接入配电网带来的双向潮流,可能造成系统故障率提高和保护失效等问题,这对配电网的运行控制与故障恢复速度提出了新的要求。同时,在分布式电源高渗透率的配网中,分布式电源也可以为故障恢复提供能源,加快故障恢复的速度。In recent years, in order to reduce greenhouse gas emissions caused by fossil fuel combustion and improve the stability of distribution networks, the penetration rate of various forms of distributed power has been significantly increased. However, due to the two-way power flow brought by the access of distributed power to the distribution network, it may cause problems such as increased system failure rate and protection failure, which puts forward new requirements for the operation control and fault recovery speed of the distribution network. At the same time, in a distribution network with a high penetration rate of distributed power, distributed power can also provide energy for fault recovery and speed up fault recovery.

一般来说,故障恢复是在尽可能短的时间内最大程度的恢复失电负荷的供应,特别是一级负荷。然而目前常用的故障恢复方法没有考虑故障的影响范围,在故障时均采用拓扑重构的方法进行故障恢复。但由于配网中暂时性故障较多,频繁的拓扑重构引起的潮流变化对于系统的安全性和稳定性也是不利的。此外,拓扑重构的方法需要操作关联开关,一般而言,关联开关的操作时间相比于支路上断路器的操作要更耗时。因此,从故障恢复的时间上考虑,在发生小范围故障的情况下,采用改变系统拓扑结构的方法进行故障恢复也是不利的。Generally speaking, fault recovery is to restore the supply of the lost load to the greatest extent in the shortest possible time, especially the primary load. However, the commonly used fault recovery methods do not consider the impact range of the fault, and the method of topology reconstruction is used for fault recovery when a fault occurs. However, because there are many temporary faults in the distribution network, the power flow changes caused by frequent topology reconfiguration are also unfavorable to the security and stability of the system. In addition, the topology reconfiguration method needs to operate the associated switch, and generally speaking, the operation time of the associated switch is more time-consuming than the operation of the circuit breaker on the branch. Therefore, considering the time of fault recovery, in the case of a small-scale fault, it is also unfavorable to adopt the method of changing the system topology for fault recovery.

在分布式电源高渗透率下的配网中,一些小范围故障可以在不改变系统拓扑结构的情况下,依靠附近的分布式电源对失电负荷的重新恢复供电。然而,由于配网中分布式电源的容量有限,在故障恢复的过程中需要对故障的范围和故障恢复过程中分布式供电能力合理的权衡找出合适的故障恢复方法,才能最大限度的利用分布式电源,提高故障恢复的速度。因此,针对上述问题,要想实现配网中智能故障恢复策略,就需要有一个合理的判据来区分基于分布式电源的故障恢复方法和基于拓扑重构的故障恢复方法。故障恢复率是评判故障恢复的一大重要指标,因此,一级负荷的故障恢复率以及所有失电负荷的恢复率被用作区分两种方法的使用范围。同时,确定故障恢复所采用的方法后,在选择恢复路径的时,需要同时考虑故障恢复的时间,故障恢复的速度以及故障恢复对于配网内其他正常线路的影响。基于这一认识,本发明建立了一个同时考虑这三个因素的故障恢复模型,实现在故障恢复过程中达到在尽可能短的时间内,恢复尽可能多的负荷,尤其是一级负荷,并且保证恢复过程中对于配网中其他线路的影响最小,即最终实现了故障恢复的快速性与安全性的折中最优。In the distribution network under the high penetration rate of distributed generation, some small-scale faults can rely on nearby distributed generation to restore power supply to the lost load without changing the system topology. However, due to the limited capacity of distributed power in the distribution network, it is necessary to find a suitable fault recovery method after a reasonable trade-off between the scope of the fault and the distributed power supply capacity in the process of fault recovery in order to maximize the use of distributed power. power supply to improve the speed of fault recovery. Therefore, in view of the above problems, in order to realize the intelligent fault recovery strategy in the distribution network, it is necessary to have a reasonable criterion to distinguish the fault recovery method based on distributed power supply and the fault recovery method based on topology reconstruction. The fault recovery rate is an important indicator for judging the fault recovery. Therefore, the fault recovery rate of the primary load and the recovery rate of all power-off loads are used to distinguish the scope of use of the two methods. At the same time, after determining the method of fault recovery, when selecting the recovery path, it is necessary to consider the time of fault recovery, the speed of fault recovery, and the impact of fault recovery on other normal lines in the distribution network. Based on this understanding, the present invention establishes a fault recovery model that simultaneously considers these three factors, so as to achieve as much load as possible in the shortest possible time during the fault recovery process, especially the primary load, and To ensure the minimum impact on other lines in the distribution network during the recovery process, that is, to finally realize the optimal compromise between the rapidity and safety of fault recovery.

发明内容Contents of the invention

针对现有故障恢复策略的不足,本发明的目的在于提出了一种含分布式电源的配电网智能故障恢复方法。Aiming at the deficiency of existing fault recovery strategies, the object of the present invention is to propose an intelligent fault recovery method for distribution network including distributed power sources.

本发明的目的是通过以下技术手段实现的,具体的实施步骤如下:一种含分布式电源的配电网智能故障恢复方法,该方法包括以下步骤:The purpose of the present invention is achieved through the following technical means, and the specific implementation steps are as follows: A method for recovering intelligent faults of distribution networks containing distributed power sources, the method comprising the following steps:

步骤(1):每隔时间Δt检测配电网系统中所有节点是否有故障;Step (1): Detect whether all nodes in the distribution network system are faulty at intervals Δt;

步骤(2):当故障发生后,根据配电网系统中的拓扑结构,断开受故障影响的支路隔离开关,并计算故障时刻的潮流分布;Step (2): When a fault occurs, according to the topological structure of the distribution network system, disconnect the branch circuit isolation switch affected by the fault, and calculate the power flow distribution at the time of the fault;

步骤(3):根据潮流分布,判断失电区域内分布式电源容量是否可以满足失电负荷的供应;具体为:若分布式电源容量可以满足所有处于失电状态的一级负荷以及处于失电状态的非一级负荷的80%,即满足:Step (3): According to the power flow distribution, judge whether the capacity of the distributed power source in the power-off area can meet the supply of the power-off load; specifically: if the capacity of the distributed power source can meet all the primary loads in the power-off state and the power-off 80% of the state's non-Level 1 load, that meets:

∑PDGs≥∑Pcritical loads+80%×∑Pnoncritical loads ∑P DGs ≥∑P critical loads +80%×∑P noncritical loads

其中,∑PDGs表示失电区域内所有分布式电源的可用容量;∑Pcritical loads表示失电状态的一级负荷,∑Pnoncritical loads失电区域内的非一级负荷;则在此次故障恢复过程中,采用基于分布式电源的故障恢复方法进行恢复;否则,改变系统的关联开关状态对故障进行恢复。Among them, ∑P DGs represents the available capacity of all distributed power sources in the de-energized area; ∑P critical loads represents the primary load in the de-energized state, and ∑P noncritical loads represents the non-primary load in the de-energized area; During the recovery process, the fault recovery method based on distributed power is used to recover; otherwise, the system's associated switch state is changed to recover from the fault.

进一步地,所述步骤3中,基于分布式电源的故障恢复方法,具体为:Further, in the step 3, the fault recovery method based on distributed power supply is specifically:

(A1)计算当前状态下所有可恢复支路的负荷恢复能力,其中包括直接恢复负荷大小以及每条可恢复支路对剩余待恢复负荷的位置关系;所述可恢复支路不包括系统关联开关,具体为:(A1) Calculate the load recovery capability of all recoverable branches in the current state, including the size of the direct recovery load and the positional relationship of each recoverable branch to the remaining load to be restored; the recoverable branches do not include system-associated switches ,Specifically:

(A1.1)假设当前状态下共有n条支路可恢复,若恢复其中一条支路,就可以使一部分负荷恢复供应,则该部分负荷为该条支路的直接恢复负荷;第i条可恢复支路的直接恢复负荷用表达式(1)计算:(A1.1) Assuming that there are a total of n branches that can be restored in the current state, if one of the branches is restored, a part of the load can be restored to supply, then this part of the load is the direct recovery load of the branch; the i can The direct restoration load of the restoration branch is calculated by expression (1):

其中,k为第i条可恢复支路的直接恢复负荷的个数;ρj为第j个直接恢复负荷的权重,若为一级负荷ρj=100;否则,ρj=1;Pj为第j个直接恢复负荷的大小;Among them, k is the number of direct recovery loads of the i-th recoverable branch; ρ j is the weight of the j-th direct recovery load, if it is a primary load ρ j =100; otherwise, ρ j =1; P j is the size of the jth direct recovery load;

(A1.2)第i条可恢复支路的间接恢复负荷Pindirect loads(i)用表达式(2)计算:(A1.2) The indirect restoration load P indirect loads (i) of the i-th restorable branch is calculated by expression (2):

其中,m为选择恢复第i条可恢复支路后,剩余待恢复负荷的个数;Pindirect loads,ij,恢复第i条可恢复支路后,第q个待恢复负荷的大小;distanceiq为在恢复第i条可恢复支路后,恢复第q个待恢复负荷需恢复的支路数量;Among them, m is the number of remaining loads to be restored after restoring the i-th recoverable branch; P indirect loads,ij is the size of the q-th load to be restored after restoring the i-th recoverable branch; distance iq After restoring the i-th recoverable branch, the number of branches to be restored to restore the q-th load to be restored;

(A1.3)综合考虑直接/间接恢复负荷,第i条可恢复支路的负荷恢复能力,用表达式(3)计算:(A1.3) Considering the direct/indirect restoration load comprehensively, the load restoration capacity of the i-th restorable branch is calculated by expression (3):

Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (3)P recover loads (i)=P direct loads (i)+P indirect loads (i) (3)

(A2)将每条可恢复支路的负荷恢复能力和时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路;其中,可恢复支路的恢复代价,用表达式(5)计算:(A2) Weight the normalized value of the load recovery capability and time cost of each recoverable branch according to the importance level, and then use it as the recovery cost matrix element of the recoverable branch, and each time from the matrix Select a branch with the least cost; among them, the recovery cost of the recoverable branch is calculated by expression (5):

(A3)将每条可恢复支路的负荷恢复能力和时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路;其中,可恢复支路的恢复代价,用表达式(5)计算:(A3) Weight the normalized value of the load recovery capability and time cost of each recoverable branch according to the importance level, and then use it as the recovery cost matrix element of the recoverable branch, and each time from the matrix Select a branch with the least cost; among them, the recovery cost of the recoverable branch is calculated by expression (5):

Cost(i)=β1φ(f1i)+β3φ(f3i) (5)Cost(i)=β 1 φ(f 1i )+β 3 φ(f 3i ) (5)

其中,f1i为时间成本的归一化值;f3i为支路负荷恢复能力的归一化值;β1为时间成本的权重,β1=1;β3为支路负荷恢复能力的权重,β3=5;而φ为惩罚函数,对于x∈[0,1]有Among them, f 1i is the normalized value of time cost; f 3i is the normalized value of branch load restoration ability; β 1 is the weight of time cost, β 1 = 1; β 3 is the weight of branch load restoration ability , β 3 =5; and φ is a penalty function, for x∈[0,1]

(A4)计算并检验系统的恢复此条支路后是否满足潮流运行约束和线路安全约束;对于所有的支路,其线路安全约束为:(A4) Calculate and check whether the system meets the power flow operation constraints and line safety constraints after restoring this branch; for all branches, the line safety constraints are:

其中,V为线路电压的下限;为线路电压的下限;V为线路电压;为线路电流的上限;I为线路电流;Among them, V is the lower limit of the line voltage; is the lower limit of the line voltage; V is the line voltage; Is the upper limit of the line current; I is the line current;

(A5)若不满足,则从可恢复支路中排除当前不满足约束的支路;重复步骤(A3)和(A4),直到找到满足约束条件的最小可恢复支路;恢复该满足约束条件的最小可恢复支路,更新系统的拓扑结构。(A5) If not satisfied, then exclude the current branch that does not satisfy the constraint from the recoverable branch; repeat steps (A3) and (A4), until finding the minimum recoverable branch that meets the constraint condition; restore the constraint condition The minimum recoverable branch of , update the topology of the system.

(A6)重复步骤(A1)~(A5),直到恢复所有可恢复负荷。(A6) Repeat steps (A1) to (A5) until all recoverable loads are recovered.

进一步地,所述步骤3中,改变系统拓扑结构对故障进行恢复,具体为:Further, in the step 3, the system topology is changed to restore the failure, specifically:

(B1)计算当前状态下所有可恢复支路的负荷恢复能力,其中包括直接恢复负荷大小以及每条可恢复支路对剩余待恢复负荷的位置关系;所述可恢复支路包括系统关联开关,具体为:(B1) Calculating the load recovery capabilities of all recoverable branches in the current state, including the size of the direct recovery load and the positional relationship of each recoverable branch to the remaining load to be restored; the recoverable branches include system-associated switches, Specifically:

(B1.1)假设当前状态下共有n条支路可恢复,若恢复其中一条支路,就可以使一部分负荷恢复供应,则该部分负荷为该条支路的直接恢复负荷;第i条可恢复支路的直接恢复负荷用表达式(1)计算:(B1.1) Assuming that there are a total of n branches that can be restored in the current state, if one of the branches is restored, a part of the load can be restored to supply, and this part of the load is the direct recovery load of the branch; The direct restoration load of the restoration branch is calculated by expression (1):

其中,k为第i条可恢复支路的直接恢复负荷的个数;ρj为第j个直接恢复负荷的权重,若为一级负荷ρj=100;否则,ρj=1;Pj为第j个直接恢复负荷的大小;Among them, k is the number of direct recovery loads of the i-th recoverable branch; ρ j is the weight of the j-th direct recovery load, if it is a primary load ρ j =100; otherwise, ρ j =1; P j is the size of the jth direct recovery load;

(B1.2)第i条可恢复支路的间接恢复负荷Pindirect loads(i)用表达式(2)计算:(B1.2) The indirect restoration load P indirect loads (i) of the i-th restorable branch is calculated by expression (2):

其中,m为选择恢复第i条可恢复支路后,剩余待恢复负荷的个数;Pindirect loads,ij,恢复第i条可恢复支路后,第q个待恢复负荷的大小;distanceiq为在恢复第i条可恢复支路后,恢复第q个待恢复负荷需恢复的支路数量;Among them, m is the number of remaining loads to be restored after restoring the i-th recoverable branch; P indirect loads,ij is the size of the q-th load to be restored after restoring the i-th recoverable branch; distance iq After restoring the i-th recoverable branch, the number of branches to be restored to restore the q-th load to be restored;

(B1.3)综合考虑直接/间接恢复负荷,第i条可恢复支路的负荷恢复能力,用表达式(3)计算:(B1.3) Considering the direct/indirect restoration load comprehensively, the load restoration capacity of the i-th restorable branch is calculated by expression (3):

Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (3)P recover loads (i)=P direct loads (i)+P indirect loads (i) (3)

(B2)计算当前状态下所有可恢复支路的风险系数,即为由于恢复某条支路而可能造成其他非故障支路负荷率增长的归一化加权值;具体为:(B2) Calculate the risk coefficient of all recoverable branches in the current state, which is the normalized weighted value that may cause the load rate increase of other non-faulty branches due to the restoration of a certain branch; specifically:

第i条可恢复支路的风险系数用表达式(4)计算:The risk coefficient of the i-th recoverable branch is calculated by expression (4):

其中,L为恢复第i条可恢复支路后,引起潮流变化的支路数量;Il为第l条支路的实际电流;Ilimit,l为第l条支路的电流容量;Among them, L is the number of branches that cause power flow changes after restoring the i-th recoverable branch; I l is the actual current of the l-th branch; I limit,l is the current capacity of the l-th branch;

(B3)计算当前状态下所有可恢复支路的操作时间成本;将系统中所有节点按照其连接的母线分成不同区域,若支路上的开关为关联开关,则时间成本为0.4;若支路上开关不是关联开关,根据支路两端连接的节点是否处于同一区域判断时间成本,若在同一区域内,则断路器时间成本为0.1;若不在同一区域内,则断路器时间成本为0.2;(B3) Calculate the operating time cost of all recoverable branches in the current state; divide all nodes in the system into different areas according to the bus they are connected to, if the switch on the branch is an associated switch, the time cost is 0.4; if the switch on the branch It is not an associated switch. The time cost is judged according to whether the nodes connected at both ends of the branch are in the same area. If they are in the same area, the time cost of the circuit breaker is 0.1; if they are not in the same area, the time cost of the circuit breaker is 0.2;

(B4)将每条可恢复支路的负荷恢复能力,风险系数以及时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路;其中,可恢复支路的恢复代价,用表达式(5)计算:(B4) Weight the normalized value of the load recovery ability, risk coefficient and time cost of each recoverable branch according to the importance level, and then use it as the restoration cost matrix element of the recoverable branch, each time Select a branch with the least cost from the matrix; among them, the recovery cost of the recoverable branch is calculated by expression (5):

Cost(i)=β1φ(f1i)+β2φ(f2i)+β3φ(f3i) (5)Cost(i)=β 1 φ(f 1i )+β 2 φ(f 2i )+β 3 φ(f 3i ) (5)

其中,f1i为时间成本的归一化值;f2i为支路风险系数的归一化值;f3i为支路负荷恢复能力的归一化值;β1为时间成本的权重,β1=1;β2为支路风险系数的权重,β2=3;β3为支路负荷恢复能力的权重,β3=5;而φ为惩罚函数,对于x∈[0,1]有Among them, f 1i is the normalized value of time cost; f 2i is the normalized value of branch risk coefficient; f 3i is the normalized value of branch load recovery ability; β 1 is the weight of time cost, β 1 = 1; β 2 is the weight of branch risk coefficient, β 2 = 3; β 3 is the weight of branch load recovery ability, β 3 = 5; and φ is the penalty function, for x∈[0,1]

(B5)计算并检验系统的恢复此条支路后是否满足潮流运行约束和线路安全约束;对于所有的支路,其线路安全约束为:(B5) Calculate and check whether the system meets the power flow operation constraints and line safety constraints after restoring this branch; for all branches, the line safety constraints are:

其中,V为线路电压的下限;为线路电压的下限;V为线路电压;为线路电流的上限;I为线路电流;Among them, V is the lower limit of the line voltage; is the lower limit of the line voltage; V is the line voltage; Is the upper limit of the line current; I is the line current;

(B6)若不满足,则从可恢复支路中排除当前不满足约束的支路;重复步骤(B4)~(B5),直到找到满足约束条件的最小可恢复支路;恢复该满足约束条件的最小可恢复支路,更新系统的拓扑结构。(B6) If not satisfied, then exclude the current branch that does not satisfy the constraint from the recoverable branch; repeat steps (B4) to (B5) until the minimum recoverable branch that meets the constraint is found; restore the satisfying constraint The minimum recoverable branch of , update the topology of the system.

(B7)重复步骤(B1)~(B6),直到恢复所有可恢复负荷。(B7) Steps (B1) to (B6) are repeated until all recoverable loads are recovered.

本发明的有益效果在于:本发明根据故障影响范围和故障内分布式电源对失电负荷的恢复能力,充分考虑一级负荷的重要性以及恢复过程中对于系统安全性和稳定性的影响,实现了智能故障恢复算法。该智能恢复算法,可以实现在满足一级负荷的快速恢复的前提下,充分利用故障区域内的分布式电源选择性的恢复一部分非一级负荷,可以有效减少故障恢复的时间代价以及频繁系统拓扑结构变化对于系统稳定性和安全性的影响。The beneficial effect of the present invention is that: the present invention fully considers the importance of the primary load and the impact on system security and stability during the recovery process according to the scope of the fault and the recovery ability of the distributed power supply within the fault to the power-off load, and realizes Intelligent fault recovery algorithm. This intelligent recovery algorithm can realize the selective recovery of some non-level one loads by making full use of the distributed power supply in the fault area under the premise of satisfying the fast recovery of the first level load, which can effectively reduce the time cost of fault recovery and the frequent system topology The impact of structural changes on system stability and security.

附图说明Description of drawings

图1是本发明的处理流程图;Fig. 1 is a process flowchart of the present invention;

图2是本方法的测试系统(53节点网络)拓扑示意图;Fig. 2 is the test system (53 node network) topological schematic diagram of this method;

图3是测试系统中各节点的有功功率和无功功率;Figure 3 is the active power and reactive power of each node in the test system;

图4是本发明实施例1中,两种故障恢复方法下,一级负荷/所有负荷恢复率变化图;Fig. 4 is a graph showing the variation of primary load/all load recovery rates under two fault recovery methods in Embodiment 1 of the present invention;

图5是本发明实施例1中,两种故障恢复方法下,节点过电压率曲线;Fig. 5 is the node overvoltage rate curve under two fault recovery methods in Embodiment 1 of the present invention;

图6是本发明实施例1中,两种故障恢复方法下,线路电流裕量曲线;Fig. 6 is in embodiment 1 of the present invention, under two kinds of fault recovery methods, line current margin curve;

图7是本发明实施例1中,两种故障恢复方法下,恢复成本曲线;Fig. 7 is the recovery cost curve under two fault recovery methods in Embodiment 1 of the present invention;

图8是本发明实施例2中,两种故障恢复方法下,一级负荷/所有负荷恢复率变化图;Fig. 8 is a graph showing the variation of primary load/all load recovery rates under two fault recovery methods in Embodiment 2 of the present invention;

图9是本发明实施例2中,两种故障恢复方法下,节点过电压率曲线;Fig. 9 is the node overvoltage rate curve under two fault recovery methods in Embodiment 2 of the present invention;

图10是本发明实施例2中,两种故障恢复方法下,线路电流裕量曲线;Fig. 10 is the line current margin curve under two fault recovery methods in Embodiment 2 of the present invention;

图11是本发明实施例2中,两种故障恢复方法下,恢复成本曲线。Fig. 11 is a recovery cost curve under two fault recovery methods in Embodiment 2 of the present invention.

具体实施方案specific implementation plan

下面结合附图对本发明做进一步详述:Below in conjunction with accompanying drawing, the present invention is described in further detail:

图1所示的是该发明的处理流程图。其具体的实施将结合具体实例描述如下。以53节点测试网络对其具体步骤进行描述,其拓扑结构如图2所示。What Fig. 1 shows is the processing flowchart of this invention. Its specific implementation will be described as follows in conjunction with specific examples. The specific steps are described with a 53-node test network, and its topology is shown in Figure 2.

系统中各节点的负荷数据曲线如图3所示。其中,节点8、9、12、18、22、30、33、39、40为一级负荷节点;而在节点19、24、35、49、28、37、39、27上有分布式电源。The load data curve of each node in the system is shown in Figure 3. Among them, nodes 8, 9, 12, 18, 22, 30, 33, 39, and 40 are primary load nodes; and nodes 19, 24, 35, 49, 28, 37, 39, and 27 have distributed power sources.

(1)每隔时间Δt进行检测,在当前时刻,配电网系统中所有节点是否有有故障。(1) Detect every time Δt, at the current moment, whether all nodes in the distribution network system are faulty.

(2)当故障发生后,根据配电网系统中的拓扑结构,断开受故障影响的支路隔离开关,并计算故障时刻的潮流分布。(2) When a fault occurs, according to the topological structure of the distribution network system, the disconnector of the branch affected by the fault is disconnected, and the power flow distribution at the time of the fault is calculated.

(3)为了减少由系统拓扑结构频繁变化引起系统潮流频繁变化,和系统中关联开关的频繁操作,在故障恢复前,根据潮流分布,判断失电区域内分布式电源容量是否可以满足失电负荷的供应。具体为:若分布式电源容量可以满足所有处于失电状态的一级负荷以及处于失电状态的非一级负荷的80%,即满足:(3) In order to reduce the frequent changes of system power flow caused by frequent changes of system topology and the frequent operation of associated switches in the system, before the fault is restored, according to the power flow distribution, it is judged whether the distributed power supply capacity in the power-off area can meet the power-off load supply. Specifically: if the capacity of the distributed power supply can meet 80% of all primary loads in a power-off state and non-primary loads in a power-off state, then:

∑PDGs≥∑Pcritical loads+80%×∑Pnoncritical loads ∑P DGs ≥∑P critical loads +80%×∑P noncritical loads

其中,∑PDGs表示失电区域内所有分布式电源的可用容量;∑Pcritical loads表示失电状态的一级负荷,∑Pnoncritical loads失电区域内的非一级负荷;则在此次故障恢复过程中,采用基于分布式电源的故障恢复方法进行恢复;否则,改变系统的关联开关状态对故障进行恢复。Among them, ∑P DGs represents the available capacity of all distributed power sources in the de-energized area; ∑P critical loads represents the primary load in the de-energized state, and ∑P noncritical loads represents the non-primary load in the de-energized area; During the recovery process, the fault recovery method based on distributed power is used to recover; otherwise, the system's associated switch state is changed to recover from the fault.

(4)根据步骤3中的条件判断选择相应的拓扑重构方法若为基于分布式电源的故障恢复方法,具体为:(4) According to the conditions in step 3, select the corresponding topology reconfiguration method. If it is a fault recovery method based on distributed power supply, specifically:

(4.1)计算当前状态下所有可恢复支路的负荷恢复能力,其中包括直接恢复负荷大小以及每条可恢复支路对剩余待恢复负荷的位置关系;所述可恢复支路包括系统关联开关,具体为:(4.1) Calculate the load recovery capability of all recoverable branches in the current state, including the size of the direct recovery load and the positional relationship of each recoverable branch to the remaining load to be restored; the recoverable branch includes a system associated switch, Specifically:

(4.1.1)假设当前状态下共有n条支路可恢复,若恢复其中一条支路,就可以使一部分负荷恢复供应,则该部分负荷为该条支路的直接恢复负荷;第i条可恢复支路的直接恢复负荷用表达式(1)计算:(4.1.1) Assuming that there are a total of n branches that can be restored in the current state, if one of the branches is restored, a part of the load can be restored to supply, then this part of the load is the direct recovery load of the branch; the i item can be The direct restoration load of the restoration branch is calculated by expression (1):

其中,k为第i条可恢复支路的直接恢复负荷的个数;ρj为第j个直接恢复负荷的权重,若为一级负荷ρj=100;否则,ρj=1;Pj为第j个直接恢复负荷的大小;Among them, k is the number of direct recovery loads of the i-th recoverable branch; ρ j is the weight of the j-th direct recovery load, if it is a primary load ρ j =100; otherwise, ρ j =1; P j is the size of the jth direct recovery load;

(4.1.2)第i条可恢复支路的间接恢复负荷Pindirect loads(i)用表达式(2)计算:(4.1.2) The indirect restoration load P indirect loads (i) of the i-th restorable branch is calculated by expression (2):

其中,m为选择恢复第i条可恢复支路后,剩余待恢复负荷的个数;Pindirect loads,ij,恢复第i条可恢复支路后,第q个待恢复负荷的大小;distanceiq为在恢复第i条可恢复支路后,恢复第q个待恢复负荷需恢复的支路数量;Among them, m is the number of remaining loads to be restored after restoring the i-th recoverable branch; P indirect loads,ij is the size of the q-th load to be restored after restoring the i-th recoverable branch; distance iq After restoring the i-th recoverable branch, the number of branches to be restored to restore the q-th load to be restored;

(4.1.3)综合考虑直接/间接恢复负荷,第i条可恢复支路的负荷恢复能力,用表达式(3)计算:(4.1.3) Considering the direct/indirect recovery load comprehensively, the load recovery capacity of the i-th recoverable branch is calculated by the expression (3):

Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (3)P recover loads (i)=P direct loads (i)+P indirect loads (i) (3)

(4.2)计算当前状态下所有可恢复支路的操作时间成本;将系统中所有节点按照其连接的母线分成不同区域,若支路上的开关为关联开关,则时间成本为0.4;若支路上开关不是关联开关,根据支路两端连接的节点是否处于同一区域判断时间成本,若在同一区域内,则断路器时间成本为0.1;若不在同一区域内,则断路器时间成本为0.2;(4.2) Calculate the operating time cost of all recoverable branches in the current state; divide all nodes in the system into different areas according to the bus they are connected to, if the switch on the branch is an associated switch, the time cost is 0.4; if the switch on the branch It is not an associated switch. The time cost is judged according to whether the nodes connected at both ends of the branch are in the same area. If they are in the same area, the time cost of the circuit breaker is 0.1; if they are not in the same area, the time cost of the circuit breaker is 0.2;

(4.3)将每条可恢复支路的负荷恢复能力,风险系数以及时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路;其中,可恢复支路的恢复代价,用表达式(5)计算:(4.3) Weight the normalized value of the load recovery ability, risk coefficient and time cost of each recoverable branch according to the importance level, and then use it as the recovery cost matrix element of the recoverable branch, each time Select a branch with the least cost from the matrix; among them, the recovery cost of the recoverable branch is calculated by expression (5):

Cost(i)=β1φ(f1i)+β2φ(f2i)+β3φ(f3i) (5)Cost(i)=β 1 φ(f 1i )+β 2 φ(f 2i )+β 3 φ(f 3i ) (5)

其中,f1i为时间成本的归一化值;f2i为支路风险系数的归一化值;f3i为支路负荷恢复能力的归一化值;β1为时间成本的权重,β1=1;β2为支路风险系数的权重,β2=3;β3为支路负荷恢复能力的权重,β3=5;而φ为惩罚函数,对于x∈[0,1]有Among them, f 1i is the normalized value of time cost; f 2i is the normalized value of branch risk coefficient; f 3i is the normalized value of branch load recovery ability; β 1 is the weight of time cost, β 1 = 1; β 2 is the weight of branch risk coefficient, β 2 = 3; β 3 is the weight of branch load recovery ability, β 3 = 5; and φ is the penalty function, for x∈[0,1]

(4.4)计算并检验系统的恢复此条支路后,是否满足潮流运行约束和线路安全约束;对于所有的支路,其线路安全约束为:(4.4) Calculate and check whether the system meets the power flow operation constraints and line safety constraints after restoring this branch; for all branches, the line safety constraints are:

其中,V为线路电压的下限;为线路电压的下限;V为线路电压;为线路电流的上限;I为线路电流;Among them, V is the lower limit of the line voltage; is the lower limit of the line voltage; V is the line voltage; Is the upper limit of the line current; I is the line current;

(4.5)若不满足,则从可恢复支路中排除当前不满足约束的支路,重复步骤(4.3),(4.4)直到找到满足约束条件的最小可恢复支路;恢复该满足约束条件的最小可恢复支路,更新系统的拓扑结构。(4.5) If not satisfied, then exclude the current branch that does not satisfy the constraint from the recoverable branch, repeat steps (4.3), (4.4) until the minimum recoverable branch that meets the constraint is found; restore the satisfying constraint The minimum recoverable branch updates the topology of the system.

(4.6)重复步骤(4.1)到(4.5),直到恢复所有可恢复负荷。(4.6) Repeat steps (4.1) to (4.5) until all recoverable loads are restored.

(5)根据步骤3中的条件判断选择相应的拓扑重构方法若为基于拓扑重构的故障恢复方法,具体为:(5) Select the corresponding topology reconstruction method according to the conditions in step 3. If it is a fault recovery method based on topology reconstruction, it is specifically:

(5.1)计算当前状态下所有可恢复支路的负荷恢复能力,其中包括直接恢复负荷大小以及每条可恢复支路对剩余待恢复负荷的位置关系;所述可恢复支路包括系统关联开关,具体为:(5.1) Calculate the load recovery capability of all recoverable branches in the current state, including the size of the direct recovery load and the positional relationship of each recoverable branch to the remaining load to be restored; the recoverable branch includes a system-associated switch, Specifically:

(5.1.1)假设当前状态下共有n条支路可恢复,若恢复其中一条支路,就可以使一部分负荷恢复供应,则该部分负荷为该条支路的直接恢复负荷;第i条可恢复支路的直接恢复负荷用表达式(1)计算:(5.1.1) Assuming that there are a total of n branches that can be restored in the current state, if one of the branches is restored, a part of the load can be restored to supply, and this part of the load is the direct recovery load of the branch; the i item can be The direct restoration load of the restoration branch is calculated by expression (1):

其中,k为第i条可恢复支路的直接恢复负荷的个数;ρj为第j个直接恢复负荷的权重,若为一级负荷ρj=100;否则,ρj=1;Pj为第j个直接恢复负荷的大小;Among them, k is the number of direct recovery loads of the i-th recoverable branch; ρ j is the weight of the j-th direct recovery load, if it is a primary load ρ j =100; otherwise, ρ j =1; P j is the size of the jth direct recovery load;

(5.1.2)第i条可恢复支路的间接恢复负荷Pindirect loads(i)用表达式(2)计算:(5.1.2) The indirect restoration load P indirect loads (i) of the i-th restorable branch is calculated by expression (2):

其中,m为选择恢复第i条可恢复支路后,剩余待恢复负荷的个数;Pindirect loads,ij,恢复第i条可恢复支路后,第q个待恢复负荷的大小;distanceiq为在恢复第i条可恢复支路后,恢复第q个待恢复负荷需恢复的支路数量;Among them, m is the number of remaining loads to be restored after restoring the i-th recoverable branch; P indirect loads,ij is the size of the q-th load to be restored after restoring the i-th recoverable branch; distance iq After restoring the i-th recoverable branch, the number of branches to be restored to restore the q-th load to be restored;

(5.1.3)综合考虑直接/间接恢复负荷,第i条可恢复支路的负荷恢复能力,用表达式(3)计算:(5.1.3) Considering the direct/indirect restoration load comprehensively, the load restoration capacity of the i-th restorable branch is calculated by expression (3):

Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (3)P recover loads (i)=P direct loads (i)+P indirect loads (i) (3)

(5.2)计算当前状态下所有可恢复支路的风险系数,即为由于恢复某条支路而可能造成其他非故障支路负荷率增长的归一化加权值;具体为:(5.2) Calculate the risk coefficient of all recoverable branches in the current state, that is, the normalized weighted value that may cause the load rate increase of other non-faulty branches due to the restoration of a certain branch; specifically:

第i条可恢复支路的风险系数用表达式(4)计算:The risk coefficient of the i-th recoverable branch is calculated by expression (4):

其中,L为恢复第i条可恢复支路后,引起潮流变化的支路数量;Il为第l条支路的实际电流;Ilimit,l为第l条支路的电流容量;Among them, L is the number of branches that cause power flow changes after restoring the i-th recoverable branch; I l is the actual current of the l-th branch; I limit,l is the current capacity of the l-th branch;

(5.3)计算当前状态下所有可恢复支路的操作时间成本;将系统中所有节点按照其连接的母线分成不同区域,若支路上的开关为关联开关,则时间成本为0.4;若支路上开关不是关联开关,根据支路两端连接的节点是否处于同一区域判断时间成本,若在同一区域内,则断路器时间成本为0.1;若不在同一区域内,则断路器时间成本为0.2;(5.3) Calculate the operating time cost of all recoverable branches in the current state; divide all nodes in the system into different areas according to the bus they are connected to, if the switch on the branch is an associated switch, the time cost is 0.4; if the switch on the branch It is not an associated switch. The time cost is judged according to whether the nodes connected at both ends of the branch are in the same area. If they are in the same area, the time cost of the circuit breaker is 0.1; if they are not in the same area, the time cost of the circuit breaker is 0.2;

(5.4)将每条可恢复支路的负荷恢复能力,风险系数以及时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路;其中,可恢复支路的恢复代价,用表达式(5)计算:(5.4) Weight the normalized value of the load recovery ability, risk coefficient and time cost of each recoverable branch according to the importance level, and then use it as the recovery cost matrix element of the recoverable branch, each time Select a branch with the least cost from the matrix; among them, the recovery cost of the recoverable branch is calculated by expression (5):

Cost(i)=β1φ(f1i)+β2φ(f2i)+β3φ(f3i) (5)Cost(i)=β 1 φ(f 1i )+β 2 φ(f 2i )+β 3 φ(f 3i ) (5)

其中,f1i为时间成本的归一化值;f2i为支路风险系数的归一化值;f3i为支路负荷恢复能力的归一化值;β1为时间成本的权重,β1=1;β2为支路风险系数的权重,β2=3;β3为支路负荷恢复能力的权重,β3=5;而φ为惩罚函数,对于x∈[0,1]有Among them, f 1i is the normalized value of time cost; f 2i is the normalized value of branch risk coefficient; f 3i is the normalized value of branch load recovery ability; β 1 is the weight of time cost, β 1 = 1; β 2 is the weight of branch risk coefficient, β 2 = 3; β 3 is the weight of branch load recovery ability, β 3 = 5; and φ is the penalty function, for x∈[0,1]

(5.5)计算并检验系统的恢复此条支路后,是否满足潮流运行约束和线路安全约束;对于所有的支路,其线路安全约束为:(5.5) Calculate and check whether the system meets the power flow operation constraints and line safety constraints after restoring this branch; for all branches, the line safety constraints are:

其中,V为线路电压的下限;为线路电压的下限;V为线路电压;为线路电流的上限;I为线路电流;Among them, V is the lower limit of the line voltage; is the lower limit of the line voltage; V is the line voltage; Is the upper limit of the line current; I is the line current;

(5.6)若不满足,则从可恢复支路中排除当前不满足约束的支路,重复步骤(5.1),(5.5)直到找到满足约束条件的最小可恢复支路;恢复该满足约束条件的最小可恢复支路,更新系统的拓扑结构。(5.6) If not satisfied, then exclude the current branch that does not satisfy the constraint from the recoverable branch, repeat steps (5.1), (5.5) until the minimum recoverable branch that meets the constraint is found; restore the satisfying constraint The minimum recoverable branch updates the topology of the system.

(5.7)重复步骤(5.1)到(5.6),直到恢复所有可恢复负荷。(5.7) Repeat steps (5.1) to (5.6) until all recoverable loads are restored.

下面结合附图和实施例对本发明的实施例进行详细的阐述,以下实施例中,在考虑基于智能恢复故障的方法下,强调一级负荷的有限性,实现了一级负荷的快速恢复(实施例1);在考虑一级负荷的前提下,考虑智能恢复方法,实现了分布式能源在故障恢复上的利用最大化(实施例2),意识本发明的有点和特征易于被本领域的技术人员理解,从而对本发明的保护范围做出更为清楚明确的界定。Below in conjunction with accompanying drawing and embodiment, the embodiment of the present invention is described in detail, in the following embodiment, under the consideration based on the method for recovering failure intelligently, emphasize the limitation of primary load, realize the fast recovery of primary load (implementation Example 1); under the premise of considering the primary load, consider the intelligent recovery method, and realize the utilization maximization (embodiment 2) of distributed energy resources on fault recovery, and realize that the points and characteristics of the present invention are easy to be understood by those skilled in the art Personnel understand, thereby make more clear and definite definition to the scope of protection of the present invention.

实施例1Example 1

本实施例分析了在考虑一级负荷的前提下,当发生单节点故障时候,一级负荷以及所有负荷的恢复程度,恢复代价以及恢复过程中对于系统稳定性和安全性的影响。In this embodiment, under the premise of considering the primary load, when a single node failure occurs, the recovery degree of the primary load and all loads, the recovery cost and the impact on system stability and security during the recovery process are analyzed.

系统分别运行在是否考虑一级负荷优先性和分区的条件下,分析系统中每个单节点故障的恢复情况,以节点1发生故障为例,具体的过程如下:The system runs under the condition of whether to consider the first-level load priority and partition, and analyzes the recovery situation of each single node failure in the system. Taking the failure of node 1 as an example, the specific process is as follows:

(1)每隔时间Δt进行检测,在当前时刻,配电网系统中所有节点是否有有故障。(1) Detect every time Δt, at the current moment, whether all nodes in the distribution network system are faulty.

(2)当故障发生在节点1时,根据配电网系统中的拓扑结构,断开受故障影响的支路隔离开关1-9,防止分布式电源运行于孤岛状态,并计算故障时刻的潮流分布。(2) When a fault occurs at node 1, according to the topological structure of the distribution network system, disconnect the branch isolation switches 1-9 affected by the fault to prevent the distributed power from running in an island state, and calculate the power flow at the time of the fault distributed.

(3)为了减少由系统拓扑结构频繁变化引起系统潮流频繁变化,和系统中关联开关的频繁操作,在故障恢复前,根据步骤(2)中故障时刻的拓扑结构,判断在失电区域内分布式电源容量是否可以满足失电负荷的供应;通过判断可知,分布式电源容量无法满足所有一级负荷以及80%的失电负荷的供应,系统的拓扑结构必须改变,即改变系统的关联开关状态。(3) In order to reduce the frequent changes of the system power flow caused by the frequent changes of the system topology and the frequent operation of the associated switches in the system, before the fault is restored, according to the topology structure at the time of the fault in step (2), the distribution in the power-off area is judged Whether the distributed power supply capacity can meet the supply of power-off loads; through judgment, it can be seen that the distributed power supply capacity cannot meet the supply of all primary loads and 80% of power-off loads, and the topology of the system must be changed, that is, the associated switch status of the system must be changed .

(4)首先计算当前状态下所有可恢复支路(支路6、7、51、52和57)的负荷恢复能力。(4) First calculate the load restoration capacity of all recoverable branches (branches 6, 7, 51, 52 and 57) in the current state.

(5)计算当前状态下各条可恢复支路的风险系数。(5) Calculate the risk coefficient of each recoverable branch in the current state.

(6)计算当前状态下各条可恢复支路恢复的时间成本。(6) Calculate the recovery time cost of each recoverable branch in the current state.

(7)将每条可恢复支路的负荷恢复能力,风险系数以及时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路。(7) Weight the normalized value of the load recovery ability, risk coefficient and time cost of each recoverable branch according to the importance level, and then use it as the restoration cost matrix element of the recoverable branch, each time Choose a branch with the least cost from the matrix.

(8)计算并检验系统的恢复此条支路后的潮流运行约束和线路安全约束;(8) Calculate and verify the system's power flow operation constraints and line safety constraints after restoring this branch;

(9)若不满足所有约束条件,则重复步骤(7),(8)直到找到满足约束条件的可恢复支路;若满足所有约束条件,则更新系统的拓扑结构(9) If all constraints are not satisfied, repeat steps (7), (8) until a recoverable branch that satisfies the constraints is found; if all constraints are satisfied, update the topology of the system

(10)重复步骤(1)~(9),直到恢复所有可恢复负荷。(10) Repeat steps (1) to (9) until all recoverable loads are recovered.

考虑系统中所有单节点故障情况,重复上述实验步骤(1)-(10)。比较两组对比实验结果,如图4~图7所示。由实施例1可知,当考虑一级负荷的优先性和由于分区导致之路操作时间代价不同这两个因素,明显提高了一级负荷的恢复速度,但对于总体故障恢复的速度并没有影响,对于系统的安全性和稳定性也稍有提高。由于一级负荷的优先性,导致少数单节点故障情况下恢复成本提高,但由于考虑分区,可以有效利用同一区域内的非故障节点恢复供电对于降低系统的故障恢复成本有显著作用。Considering all single-node failures in the system, repeat the above experimental steps (1)-(10). Compare the experimental results of the two groups, as shown in Figures 4 to 7. It can be seen from Example 1 that when considering the priority of the primary load and the difference in road operation time cost due to partitioning, the recovery speed of the primary load is significantly improved, but it has no effect on the overall fault recovery speed. The security and stability of the system are also slightly improved. Due to the priority of the first-level load, the recovery cost increases in the case of a small number of single-node failures. However, due to the consideration of partitions, the non-faulty nodes in the same area can be effectively used to restore power supply, which has a significant effect on reducing the system failure recovery cost.

实施例2Example 2

本实施例分析了在智能故障恢复方法的前提下,当发生单节点故障时候,一级负荷以及所有负荷的恢复程度,恢复代价以及恢复过程中对于系统稳定性和安全性的影响。This embodiment analyzes on the premise of the intelligent fault recovery method, when a single node fault occurs, the recovery degree of the primary load and all loads, the recovery cost and the impact on system stability and security during the recovery process.

系统分别运行在考虑是否采用智能故障恢复方法的条件下,分析系统中每个节点发生故障的恢复情况,以节点1发生故障为例,具体的过程如下:The system runs separately under the condition of considering whether to adopt the intelligent fault recovery method, and analyzes the recovery situation of each node in the system when a fault occurs. Taking the fault of node 1 as an example, the specific process is as follows:

(1)每隔时间Δt进行检测,在当前时刻,配电网系统中所有节点是否有有故障。(1) Detect every time Δt, at the current moment, whether all nodes in the distribution network system are faulty.

(2)当故障发生在节点1时,根据配电网系统中的拓扑结构,断开受故障影响的支路隔离开关1~9,防止分布式电源运行于孤岛状态,并计算故障时刻的潮流分布。(2) When a fault occurs at node 1, according to the topological structure of the distribution network system, disconnect the branch isolation switches 1 to 9 affected by the fault to prevent the distributed power from running in an island state, and calculate the power flow at the time of the fault distributed.

(3)若考虑智能恢复方法,在故障恢复前,根据潮流分布判断在失电区域内分布式电源容量是否可以满足失电负荷的供应;通过判断可知,分布式电源容量无法满足所有一级负荷以及80%的失电负荷的供应,系统的拓扑结构必须改变,即改变系统的关联开关状态。若不考虑智能恢复方法,直接采用基于拓扑重构的故障恢复方法。(3) If the intelligent restoration method is considered, before the fault is restored, it is judged according to the power flow distribution whether the distributed power supply capacity in the power-off area can meet the supply of power-off loads; through the judgment, it can be seen that the distributed power supply capacity cannot meet all primary loads As well as supplying 80% of the dead loads, the topology of the system has to be changed, ie change the associated switch states of the system. If the intelligent recovery method is not considered, the fault recovery method based on topology reconstruction is adopted directly.

(4)首先计算当前状态下所有可恢复支路(支路6、7、51、52和57)的负荷恢复能力,计算步骤如下:(4) Firstly calculate the load restoration capacity of all recoverable branches (branches 6, 7, 51, 52 and 57) in the current state, the calculation steps are as follows:

(5)计算当前状态下各条可恢复支路的风险系数。(5) Calculate the risk coefficient of each recoverable branch in the current state.

(6)计算当前状态下各条可恢复支路恢复的时间成本。(6) Calculate the recovery time cost of each recoverable branch in the current state.

(7)将每条可恢复支路的负荷恢复能力,风险系数以及时间成本,按照重要性等级加权三者的归一化值,进而以此作为可恢复支路的恢复代价矩阵元素,每次从矩阵中选择一条代价最小的支路。(7) Weight the normalized value of the load recovery ability, risk coefficient and time cost of each recoverable branch according to the importance level, and then use it as the restoration cost matrix element of the recoverable branch, each time Choose a branch with the least cost from the matrix.

(8)计算并检验系统的恢复此条支路后的潮流运行约束和线路安全约束;(8) Calculate and verify the system's power flow operation constraints and line safety constraints after restoring this branch;

(9)若不满足所有约束条件,则重复步骤(7),(8)直到找到满足约束条件的可恢复支路;若满足所有约束条件,则更新系统的拓扑结构.(9) If all constraints are not satisfied, repeat steps (7), (8) until a recoverable branch that satisfies the constraints is found; if all constraints are satisfied, update the topology of the system.

(10)重复步骤(1)~(9),直到恢复所有可恢复负荷。(10) Repeat steps (1) to (9) until all recoverable loads are recovered.

考虑系统中所有单节点发生故障情况,重复上述实验步骤(1)-(10),比较两组实验结果,如图8~图11所示。由实施例2可知,当采用智能故障恢复方法,明显提高了一级负荷的恢复率和恢复速度,但对于总体故障恢复的速度并没有影响,而对系统的安全性和稳定性也稍有提高。由于采用智能恢复算法可以有效利用故障范围内内的分布式电源恢复供电,对降低故障恢复成本有显著作用。Considering the failure of all single nodes in the system, repeat the above experimental steps (1)-(10), and compare the experimental results of the two groups, as shown in Figures 8 to 11. It can be seen from Example 2 that when the intelligent fault recovery method is adopted, the recovery rate and recovery speed of the primary load are obviously improved, but it has no effect on the overall fault recovery speed, and the security and stability of the system are also slightly improved . Since the intelligent recovery algorithm can effectively use the distributed power supply within the fault range to restore power supply, it has a significant effect on reducing the cost of fault recovery.

Claims (1)

1. A method for recovering intelligent faults of a power distribution network with distributed power supplies is characterized by comprising the following steps:
step (1): detecting whether all nodes in the power distribution network system have faults at intervals of delta t;
step (2): when a fault occurs, disconnecting the branch isolating switch affected by the fault according to a topological structure in the power distribution network system, and calculating the load flow distribution at the moment of the fault;
and (3): judging whether the capacity of the distributed power supply in the power loss region can meet the supply of the power loss load or not according to the power flow distribution; the method specifically comprises the following steps: if the capacity of the distributed power supply can satisfy 80% of all the primary loads in the power loss state and the non-primary loads in the power loss state, that is, the following conditions are satisfied:
∑PDGs≥∑Pcritical loads+80%×∑Pnoncritical loads
wherein, Sigma PDGsRepresenting the available capacity of all distributed power sources in the power loss area; sigma Pcritical loadsFirst-order load, SIG P, representing a power-loss conditionnoncritical loadsA non-primary load representing a power loss condition; in the fault recovery process, a fault recovery method based on a distributed power supply is adopted for recovery; otherwise, changing the topological structure of the system to recover the fault;
in the step (3), the fault recovery method based on the distributed power supply specifically includes:
(A1) calculating the load recovery capacity of all recoverable branches in the current state, wherein the load recovery capacity comprises the size of a direct recovery load and the position relation of each recoverable branch to the residual load to be recovered; the recoverable branch does not include a system association switch, and specifically comprises:
(A1.1) assuming that n branches can be recovered in the current state, if one branch is recovered, a part of load can be recovered and supplied, and the part of load is the direct recovery load of the branch; the direct recovery load of the i-th recoverable branch is calculated using the expression (A-1):
<mrow> <msub> <mi>P</mi> <mrow> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>k</mi> </munderover> <msub> <mi>&amp;rho;</mi> <mi>j</mi> </msub> <mo>&amp;times;</mo> <msub> <mi>P</mi> <mi>j</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>A</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow>
wherein k is the number of direct recovery loads of the ith recoverable branch; rhojFor the weight of the jth direct recovery load, if it is the first-order load ρj100; otherwise, ρj=1;PjThe magnitude of the direct recovery load for the jth;
(A1.2) Indirect restoration load P of i-th recoverable Branchindirect loads(i) Calculated using expression (A-2):
<mrow> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>n</mi> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>q</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mi>q</mi> </msub> <mo>&amp;times;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>n</mi> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> <mo>,</mo> <mi>i</mi> <mi>q</mi> </mrow> </msub> </mrow> <mrow> <msub> <mi>distance</mi> <mrow> <mi>i</mi> <mi>q</mi> </mrow> </msub> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>A</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow>
wherein m is the number of the residual loads to be recovered after the ith recoverable branch is selected to be recovered; pindirect loads,iqAfter the ith recoverable branch is recovered, the size of the qth load to be recovered; distanceiqRecovering the number of the branch circuits required to be recovered for the q-th load to be recovered after recovering the ith recoverable branch circuit;
(A1.3) comprehensively considering the direct/indirect recovery load, and calculating the load recovery capacity of the ith recoverable branch by using an expression (A-3):
Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (A‐3)
(A2) calculating the operation time cost of all recoverable branches in the current state; dividing all nodes in the system into different areas according to buses connected with the nodes, wherein if the switches on the branches are related switches, the time cost is 0.4; if the switch on the branch is not the correlated switch, judging the time cost according to whether the nodes connected with the two ends of the branch are in the same region, and if the nodes are in the same region, the time cost of the breaker is 0.1; if the circuit breakers are not in the same area, the time cost of the circuit breaker is 0.2;
(A3) weighting the normalized values of the load recovery capacity and the time cost of each recoverable branch according to the importance level, taking the normalized values as recovery cost matrix elements of the recoverable branches, and selecting a branch with the minimum cost from the matrix each time; wherein, the recovery cost of the recoverable branch is calculated by an expression (A-5):
Cost(i)=β1φ(f1i)+β3φ(f3i) (A‐5)
wherein f is1iNormalized value for time cost; f. of3iLoad recovery for a branchnormalized value of Complex ability,. beta1as a weight of the time cost, β1=1;β3weight of the load recovery capability of the branch, beta3(ii) 5; and φ is a penalty function for x ∈ [0,1 ]]Is provided with
<mrow> <mi>&amp;phi;</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mn>1</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>0</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>3</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>2</mn> <mo>/</mo> <mn>3</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>10</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>2</mn> <mo>/</mo> <mn>3</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>9</mn> <mo>/</mo> <mn>10</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>70</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>9</mn> <mo>/</mo> <mn>10</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>1</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>A</mi> <mo>-</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow>
(A4) Calculating and checking whether the system meets the load flow operation constraint and the line safety constraint after recovering the branch; for all branches, the line safety constraints are:
<mrow> <munder> <mi>V</mi> <mo>&amp;OverBar;</mo> </munder> <mo>&amp;le;</mo> <mi>V</mi> <mo>&amp;le;</mo> <mover> <mi>V</mi> <mo>&amp;OverBar;</mo> </mover> </mrow>
<mrow> <msup> <mi>I</mi> <mn>2</mn> </msup> <mo>&amp;le;</mo> <msup> <mover> <mi>I</mi> <mo>&amp;OverBar;</mo> </mover> <mn>2</mn> </msup> </mrow>
wherein,Vis the lower limit of the line voltage;is the upper limit of the line voltage; v is the line voltage;is the upper limit of the line current; i is line current;
(A5) if not, then excluding the branch which does not satisfy the constraint currently from the recoverable branch; repeating steps (A3) and (a4) until the smallest recoverable branch satisfying the constraint is found; recovering the minimum recoverable branch meeting the constraint condition, and updating the topological structure of the system;
(A6) repeating steps (a1) - (a5) until all recoverable loads are recovered;
in the step (3), the system topology is changed to recover the fault, specifically:
(B1) calculating the load recovery capacity of all recoverable branches in the current state, wherein the load recovery capacity comprises the size of a direct recovery load and the position relation of each recoverable branch to the residual load to be recovered; the recoverable branch comprises a system association switch, and specifically comprises:
(B1.1) assuming that n branches can be recovered in the current state, if one branch is recovered, a part of load can be recovered and supplied, and the part of load is the direct recovery load of the branch; the direct recovery load of the i-th recoverable branch is calculated using the expression (B-1):
<mrow> <msub> <mi>P</mi> <mrow> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>k</mi> </munderover> <msub> <mi>&amp;rho;</mi> <mi>j</mi> </msub> <mo>&amp;times;</mo> <msub> <mi>P</mi> <mi>j</mi> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>B</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow>
wherein k is the number of direct recovery loads of the ith recoverable branch; rhojFor the weight of the jth direct recovery load, if it is the first-order load ρj100; otherwise, ρj=1;PjThe magnitude of the direct recovery load for the jth;
(B1.2) Indirect restoration load P of i-th recoverable Branchindirect loads(i) Calculated using expression (B-2):
<mrow> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>n</mi> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>q</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>m</mi> </munderover> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mi>q</mi> </msub> <mo>&amp;times;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>n</mi> <mi>d</mi> <mi>i</mi> <mi>r</mi> <mi>e</mi> <mi>c</mi> <mi>t</mi> <mi>l</mi> <mi>o</mi> <mi>a</mi> <mi>d</mi> <mi>s</mi> <mo>,</mo> <mi>i</mi> <mi>q</mi> </mrow> </msub> </mrow> <mrow> <msub> <mi>distance</mi> <mrow> <mi>i</mi> <mi>q</mi> </mrow> </msub> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>B</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow>
wherein m is the number of the residual loads to be recovered after the ith recoverable branch is selected to be recovered; pindirect loads,iqAfter the ith recoverable branch is recovered, the size of the qth load to be recovered; distanceiqTo recover the ith recoverable branch after recovering the ith recoverable branchq branches of the load to be recovered need to be recovered;
(B1.3) comprehensively considering the direct/indirect recovery load, and calculating the load recovery capacity of the ith recoverable branch by using an expression B- (3):
Precover loads(i)=Pdirect loads(i)+Pindirect loads(i) (B‐3)
(B2) calculating risk coefficients of all recoverable branches in the current state, namely, calculating a normalized weighted value which is a normalized weighted value which can cause the increase of the load rate of other non-fault branches due to the recovery of a certain branch; the method specifically comprises the following steps:
the risk factor of the i-th recoverable branch is calculated by the expression (B-4):
<mrow> <mi>R</mi> <mi>i</mi> <mi>s</mi> <mi>k</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>l</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>L</mi> </munderover> <mfrac> <mrow> <mo>|</mo> <msub> <mi>I</mi> <mi>l</mi> </msub> <mo>-</mo> <msub> <mi>I</mi> <mrow> <mi>lim</mi> <mi>i</mi> <mi>t</mi> <mo>,</mo> <mi>l</mi> </mrow> </msub> <mo>|</mo> </mrow> <msub> <mi>I</mi> <mrow> <mi>lim</mi> <mi>i</mi> <mi>t</mi> <mo>,</mo> <mi>l</mi> </mrow> </msub> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>B</mi> <mo>-</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow>
wherein, L is the number of branch circuits causing the trend change after the ith recoverable branch circuit is recovered; i islThe actual current of the l branch; i islimit,lElectricity for the l branchA flow capacity;
(B3) calculating the operation time cost of all recoverable branches in the current state; dividing all nodes in the system into different areas according to buses connected with the nodes, wherein if the switches on the branches are related switches, the time cost is 0.4; if the switch on the branch is not the correlated switch, judging the time cost according to whether the nodes connected with the two ends of the branch are in the same region, and if the nodes are in the same region, the time cost of the breaker is 0.1; if the circuit breakers are not in the same area, the time cost of the circuit breaker is 0.2;
(B4) weighting the normalized values of the load recovery capacity, the risk coefficient and the time cost of each recoverable branch according to the importance level, further taking the normalized values as recovery cost matrix elements of the recoverable branches, and selecting a branch with the minimum cost from the matrix each time; wherein, the recovery cost of the recoverable branch is calculated by an expression (B-5):
Cost(i)=β1φ(f1i)+β2φ(f2i)+β3φ(f3i) (B‐5)
wherein f is1iNormalized value for time cost; f. of2iThe normalized value of the branch risk coefficient is obtained; f. of3iis a normalized value of the branch load recovery capacity, beta1as a weight of the time cost, β1=1;β2as a weight of the branch risk factor, β2=3;β3weight of the load recovery capability of the branch, beta3(ii) 5; and φ is a penalty function for x ∈ [0,1 ]]Is provided with
<mrow> <mi>&amp;phi;</mi> <mo>&amp;prime;</mo> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mn>1</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>0</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>3</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>2</mn> <mo>/</mo> <mn>3</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>10</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>2</mn> <mo>/</mo> <mn>3</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>9</mn> <mo>/</mo> <mn>10</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>70</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mn>9</mn> <mo>/</mo> <mn>10</mn> <mo>&amp;le;</mo> <mi>x</mi> <mo>&lt;</mo> <mn>1</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mi>B</mi> <mo>-</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow>
(B5) Calculating and checking whether the system meets the load flow operation constraint and the line safety constraint after recovering the branch; for all branches, the line safety constraints are:
<mrow> <munder> <mi>V</mi> <mo>&amp;OverBar;</mo> </munder> <mo>&amp;le;</mo> <mi>V</mi> <mo>&amp;le;</mo> <mover> <mi>V</mi> <mo>&amp;OverBar;</mo> </mover> </mrow>
<mrow> <msup> <mi>I</mi> <mn>2</mn> </msup> <mo>&amp;le;</mo> <msup> <mover> <mi>I</mi> <mo>&amp;OverBar;</mo> </mover> <mn>2</mn> </msup> </mrow>
wherein V is the lower limit of the line voltage;is the upper limit of the line voltage; v is the line voltage;is the upper limit of the line current; i is line current;
(B6) if not, then excluding the branch which does not satisfy the constraint currently from the recoverable branch; repeating the steps (B4) to (B5) until the smallest recoverable branch meeting the constraint condition is found; recovering the minimum recoverable branch meeting the constraint condition, and updating the topological structure of the system;
(B7) repeating steps (B1) - (B6) until all recoverable loads are recovered.
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