电力系统导论双语实验报告

合集下载

电力系统分析实验报告

电力系统分析实验报告

电力系统分析实验报告电力系统分析实验报告引言:电力系统是现代社会不可或缺的基础设施,它为我们的生活提供了稳定的电力供应。

为了确保电力系统的可靠性和安全性,对电力系统进行分析是非常重要的。

本实验旨在通过对电力系统的分析,探讨电力系统的性能和效能,以及可能存在的问题和改进措施。

一、电力系统的基本原理电力系统由发电厂、输电网和配电网组成。

发电厂负责将化学能、机械能等转化为电能,输电网将发电厂产生的电能输送到各个地区,配电网将电能供应给终端用户。

电力系统的基本原理是通过电压和电流的传输,实现电能的转换和分配。

二、电力系统的分析方法1. 潮流计算潮流计算是电力系统分析中最基本的方法之一。

通过潮流计算,可以确定电力系统中各节点的电压和电流分布情况,从而评估系统的稳定性和负载能力。

潮流计算需要考虑各个节点的功率平衡和电压平衡,以及各个元件的参数和状态。

2. 短路分析短路分析是评估电力系统安全性的重要手段。

通过短路分析,可以确定电力系统中各个节点和支路的短路电流,从而评估设备的额定容量和保护措施的有效性。

短路分析需要考虑系统的拓扑结构、设备参数和保护装置的动作特性。

3. 阻抗分析阻抗分析是评估电力系统稳定性和负载能力的重要方法。

通过阻抗分析,可以确定电力系统中各个节点和支路的阻抗,从而评估系统的电压稳定性和电力传输能力。

阻抗分析需要考虑系统的拓扑结构、设备参数和负载特性。

三、实验结果与讨论在本实验中,我们选取了一个具体的电力系统进行分析。

通过潮流计算,我们确定了系统中各个节点的电压和电流分布情况。

通过短路分析,我们评估了系统的安全性,并确定了保护装置的动作特性。

通过阻抗分析,我们评估了系统的稳定性和负载能力。

实验结果显示,系统中存在一些节点电压偏低的问题,可能会影响设备的正常运行。

为了解决这个问题,我们建议采取增加变压器容量、调整负载分配和优化配电网结构等措施。

此外,我们还发现系统中某些支路的短路电流超过了设备的额定容量,可能导致设备的损坏和安全事故。

电力系统分析实验报告

电力系统分析实验报告

电力系统分析实验报告本文主要介绍电力系统分析实验报告的相关内容,包括实验目的、实验原理、实验结果及分析等。

实验目的:本次实验旨在掌握电力系统的基本理论和分析方法,通过对电力系统的模拟和实验,深入理解电力系统的构成和工作原理,并提高对电力系统的分析和调试能力。

实验原理:电力系统是由发电机、变电站、电网和负载等组成的,其中发电机将燃料等能源转换为电能,经变电站进行升压变换后,输往各个地方的电网上,供相应的用户使用。

而电量的传输和分配过程中,会受到各种因素的影响,如短路故障、过流保护、功率因数等。

因此,在电力系统的设计、建设和维护过程中,需要对其进行详尽分析和性能评估。

主要实验器材:1. 变压器模型2. 电感器、电容器、电阻器等模型3. 处理器、仿真软件等实验过程:1. 构建电力系统模型,包括发电机、变电站、输电线路、配电站和负载等。

2. 对不同模型参数进行设置和调整,如线路长度、阻抗等。

3. 进行各种测试和实验,如短路故障测试、过流保护测试、功率因数测试等,并记录实验数据。

4. 使用仿真软件,对电力系统进行分析和模拟,得出相关结论。

5. 对实验数据和仿真结果进行分析和比较,并提出改进建议。

实验结果及分析:通过实验和仿真,我们得出了以下结论:1. 线路长度和阻抗大小会对电力系统的稳定性和传输效率产生影响。

2. 不同短路故障类型的处理方式不同,需要根据实际情况进行应对。

3. 过流保护的设置和参数调整需要根据负载情况和线路容量进行优化。

4. 功率因数的影响因素包括谐波、电路阻抗等,需要进行综合考虑。

总结:本次实验通过对电力系统的模拟和实验,深入了解了电力系统的构成和工作原理,并掌握了电力系统的分析和调试方法。

同时,也发现了在实际工作中需要注意的问题和改进方向。

在今后的工作中,我们将进一步加强对电力系统的学习和研究,提高自己专业能力和技能水平。

电力系统自动装置实验报告

电力系统自动装置实验报告

电力系统自动装置实验报告英文回答:Conducting experiments with power system automation equipment is a valuable experience that provides insights into the practical applications of automation in power systems. During the experiment, I had the opportunity to explore various aspects of power system automation, including:1. System Protection and Control:Analyzed the operation of protective relays to mitigate faults and maintain system stability.Examined the role of supervisory control and data acquisition (SCADA) systems in monitoring and controlling power system operations.2. Voltage and Frequency Regulation:Investigated the principles of voltage and frequency regulation using automated control systems.Implemented techniques for adjusting generator excitation and reactive power compensation to maintain voltage and frequency within acceptable limits.3. Load Shedding and Restoration:Studied the strategies for load shedding andrestoration during system emergencies.Simulated the operation of automatic load shedding and restoration schemes to mitigate the impact of power outages.4. System Simulation and Modeling:Developed models of power systems using simulation software.Conducted dynamic simulations to analyze systembehavior under various operating conditions and disturbances.These experiments provided me with a comprehensive understanding of the principles and practices of power system automation. Through hands-on experience, I gained valuable insights into the design, implementation, and operation of automated systems that ensure the reliable and efficient operation of power systems.中文回答:电力系统自动化装置实验是一次宝贵的经历,让我对自动化在电力系统中的实际应用有了深刻的理解。

电力系统实验报告

电力系统实验报告

电力系统综合实验实验报告1实验目的1.通过实验一,观察发电机的四种运行状态。

2.通过实验二,观察系统在不同电压和不同拓扑结构中的静稳极限,观察失稳之后各相电压和电流波形。

3.通过实验三,观察不同短路情况下,短路切除时间对于电力系统稳定性的影响。

2实验内容2.1实验一:发电机不同象限运行实验2.1.1实验内容通过改变发电机的转速和励磁分别改变发电机的有功功率P与无功功率Q,实现发电机在不同象限的运行。

2.1.2理论分析发电机的四种运行状态:1.迟相运行(常态运行):发电机向电网同时送出有功功率和无功功率(容性)。

2.进相运行(超前运行):发电机向电网送出有功功率,吸收电网无功功率。

3.调相运行:发电机吸收电网的有功功率维持同步运转,向电网送出无功功率(容性)。

4.电动机运行(非正常运行):发电机同时吸收电网的有功功率和无功功率维持同步运行。

2.1.3实验步骤1.按照双回线方式,依次接入断路器,双回线,电动机,无穷大电网,组成简易电力系统。

2.测试各个接线端子的是否能够正常使用,闭合断路器。

3.启动发电机,并网运行。

4.改变发电机设定转速改变其有用功率,改变发电机励磁改变其无功功率,使其运行在四个象限,四个象限各取三组数据。

在正常状态下,设定三组不同转速使其保持正常运行状态,记录机端电压,有功功率,无功功率;然后降低转速,使其运行于第二象限,再次记录三组调相数据;接着降低励磁电压,使发电机运行于第三象限,记录三组电动机数据;最后提高转速使点击运行与第四象限,获得3组进相数据。

2.1.4实验结果具体现象如图所示,图. 1转速设定值0.90图. 2转速设定值0.91图. 3转速设定值0.89图. 4转速设定值0.875图. 5转速设定值0.865图. 6转速设定值0.855图. 7转速设定值0.860 4.P > 0, Q < 0 第四象限图. 8转速设定值0.882图. 9转速设定值0.892图. 10转速设定值0.9022.2实验二:线路静态稳定极限测试实验2.2.1实验内容测试线路的静态稳定运行极限,测试不同电压等级和不同电抗条件下,电压静态稳定极限的变化情况。

电力系统分析实验报告

电力系统分析实验报告

一、实验目的1. 了解电力系统的基本组成和运行原理;2. 掌握电力系统潮流计算的方法和步骤;3. 熟悉电力系统故障计算的方法和步骤;4. 培养分析电力系统问题的能力。

二、实验原理1. 电力系统潮流计算:通过求解电力系统中的潮流方程,得到系统中各节点的电压、电流、功率等参数,从而分析电力系统的运行状态。

2. 电力系统故障计算:通过求解电力系统中的故障方程,得到故障点附近的电压、电流、功率等参数,从而分析电力系统故障的影响。

三、实验仪器与设备1. 电力系统分析软件:如PSCAD/EMTDC、MATLAB等;2. 电力系统仿真设备:如电力系统仿真机、计算机等;3. 电力系统相关教材和资料。

四、实验步骤1. 建立电力系统模型:根据实验要求,利用电力系统分析软件建立电力系统模型,包括发电机、变压器、线路、负荷等元件。

2. 潮流计算:(1)设置初始条件:根据实验要求,设置电力系统运行状态,如电压、功率等;(2)求解潮流方程:利用电力系统分析软件求解潮流方程,得到系统中各节点的电压、电流、功率等参数;(3)分析潮流计算结果:根据计算结果,分析电力系统的运行状态,如电压分布、潮流分布等。

3. 故障计算:(1)设置故障条件:根据实验要求,设置电力系统故障,如短路、断路等;(2)求解故障方程:利用电力系统分析软件求解故障方程,得到故障点附近的电压、电流、功率等参数;(3)分析故障计算结果:根据计算结果,分析电力系统故障的影响,如电压波动、潮流变化等。

五、实验结果与分析1. 潮流计算结果分析:(1)电压分布:根据潮流计算结果,分析系统中各节点的电压分布情况,判断电压是否满足运行要求;(2)潮流分布:根据潮流计算结果,分析系统中各线路的潮流分布情况,判断潮流是否合理。

2. 故障计算结果分析:(1)故障点电压:根据故障计算结果,分析故障点附近的电压变化情况,判断电压是否满足运行要求;(2)故障点电流:根据故障计算结果,分析故障点附近的电流变化情况,判断电流是否过大;(3)故障点功率:根据故障计算结果,分析故障点附近的功率变化情况,判断功率是否过大。

电力系统实训实验报告

电力系统实训实验报告

1. 熟悉电力系统的基本组成及工作原理。

2. 掌握电力系统中的基本设备及其操作方法。

3. 培养实际操作能力,提高对电力系统的认识。

4. 深入了解电力系统运行过程中的安全注意事项。

二、实验内容1. 电力系统基本组成及工作原理(1)电力系统组成:电力系统主要由发电厂、输电线路、变电站、配电线路和用户组成。

(2)电力系统工作原理:发电厂将机械能转化为电能,通过输电线路传输到变电站,再通过配电线路分配到用户,用户使用电能进行各种生产和生活活动。

2. 电力系统中基本设备及其操作方法(1)发电机:发电机是电力系统的动力源,通过旋转产生电能。

操作方法:启动发电机,调节励磁电流,使发电机稳定运行。

(2)变压器:变压器用于将高压电能降压至低压电能,以满足用户需求。

操作方法:检查变压器油位、温度,调整分接头,使变压器稳定运行。

(3)输电线路:输电线路用于将电能从发电厂传输到变电站。

操作方法:检查输电线路绝缘状况,确保线路安全运行。

(4)变电站:变电站是电力系统中的重要环节,负责将高压电能降压至低压电能,并通过配电线路分配给用户。

操作方法:检查设备运行状况,调整电压、电流,确保变电站稳定运行。

3. 电力系统运行过程中的安全注意事项(1)遵守安全操作规程,确保人身安全。

(2)熟悉设备操作方法,避免误操作。

(3)定期检查设备,确保设备正常运行。

(4)掌握触电急救知识,提高应急处理能力。

1. 熟悉电力系统基本组成及工作原理,了解电力系统中基本设备及其操作方法。

2. 按照实验要求,依次进行发电机、变压器、输电线路和变电站的操作。

3. 在操作过程中,密切观察设备运行状况,记录实验数据。

4. 分析实验数据,总结实验结果。

四、实验结果与分析1. 实验过程中,发电机、变压器、输电线路和变电站均能正常运行,实验数据符合预期。

2. 通过实验,掌握了电力系统中基本设备及其操作方法,提高了实际操作能力。

3. 了解了电力系统运行过程中的安全注意事项,增强了安全意识。

《电力系统分析综合实验》PSASP部分的实验报告要求(短路部分修改0430)

《电力系统分析综合实验》PSASP部分的实验报告要求(短路部分修改0430)

《电力系统分析综合实验》PSASP仿真实验报告要求1.实验报告请统一增加面处理,封面上注明实验课程名称、班级、学号、姓名。

封面格式请参见后文。

2.实验报告统一以A4页面布局,提交电子word文档,电子文档实验报告页数不超过10页,5月8号之前由所在班级的学委进行打包压缩,发送至whpxt@邮箱。

3.实验报告内容要求:(1)实验目的(不超过400字介绍)结合本次实验课程的试验内容要求,谈谈你对本次实验课程内容模块的设置目的的认识。

(2)潮流计算部分a)简单介绍本次实验的潮流计算的试验内容,叙述利用PSASP软件进行潮流计算需要输入哪些数据(不超过800字)。

b)绘制仿真实验任务指导书中要求的两个潮流计算拓扑结构图,并根据两种潮流方案的潮流计算结果在拓扑结构图上标注母线节点电压和相位,利用复数功率形式标注各支路功率(参考方向选择数据表格中各支路的i侧母线至j侧),统计发电机和负荷的功率大小并统计系统网损c)在规划潮流方式下,增加STNC-230母线负荷的有功至1.5p.u.,无功保持不变,试调试潮流,并在绘制的规划方式拓扑结构上标注母线节点电压和相位,利用复数功率形式标注各支路功率(参考方向选择数据表格中各支路的i侧母线至j侧),统计发电机和负荷的功率大小并统计系统网损。

d)利用psasp软件潮流结果输出的excel表格形式,进行上述b)中两种运行方式的母线节点电压和相位,利用复数功率形式标注各支路功率(参考方向选择数据表格中各支路的i侧母线至j侧),统计发电机和负荷的功率大小并统计系统网损等结果信息输出,,表中出现多余信息,此表格视为无效。

e)利用d)的输出形式,在同一个excel中对比给出b)中方式1/方式2采用牛顿拉夫逊和PQ解耦控制进行潮流求解的结果,比较每个运行方式基于两种潮流求解方法的求解迭代步骤和计算结果,说明两种求解方法的特点。

f)利用b)中两种运行方式的潮流求解结果,基于线路传输功率公式,估算GEN2-230和STNC-230、GEN2-230和STNA-230两条支路的等效阻抗,比较两次估算结果是否相同?(3) 短路计算部分a)简单介绍本次实验的短路计算的试验内容,利用PSASP软件进行其短路计算所需要输入的计算数据有哪些,结合实验过程简单分析一下潮流计算和短路计算之间的联系。

电力系统分析实验报告

电力系统分析实验报告

电力系统分析理论试验汇报一.单机—无穷大系统稳态运行试验(一)、试验目旳1.理解和掌握对称稳定状况下,输电系统旳多种运行状态与运行参数旳数值变化范围;2.理解和掌握输电系统稳态不对称运行旳条件;不对称度运行参数旳影响;不对称运行对发电机旳影响等。

(二)、原理与阐明电力系统稳态对称和不对称运行分析,除了包括许多理论概念之外,尚有某些重要旳“数值概念”。

为一条不一样电压等级旳输电线路,在经典运行方式下,用相对值表达旳电压损耗,电压降落等旳数值范围,是用于判断运行报表或监视控制系统测量值与否对旳旳参数根据。

因此,除了通过结合实际旳问题,让学生掌握此类“数值概念”外,试验也是一条很好旳、更为直观、易于形成深刻记忆旳手段之一。

试验用一次系统接线图如图2所示。

图2 一次系统接线图本试验系统是一种物理模型。

原动机采用直流电动机来模拟,当然,它们旳特性与大型原动机是不相似旳。

原动机输出功率旳大小,可通过给定直流电动机旳电枢电压来调整。

试验系统用原则小型三相似步发电机来模拟电力系统旳同步发电机,虽然其参数不能与大型发电机相似,但也可以当作是一种具有特殊参数旳电力系统旳发电机。

发电机旳励磁系统可以用外加直流电源通过手动来调整,也可以切换到台上旳微机励磁调整器来实现自动调整。

试验台旳输电线路是用多种接成链型旳电抗线圈来模拟,其电抗值满足相似条件。

“无穷大”母线就直接用试验室旳交流电源,由于它是由实际电力系统供电旳,因此,它基本上符合“无穷大”母线旳条件。

为了进行测量,试验台设置了测量系统,以测量多种电量(电流、电压、功率、频率)。

为了测量发电机转子与系统旳相对位置角(功率角),在发电机轴上装设了闪光测角装置。

此外,台上还设置了模拟短路故障等控制设备。

(三)、试验环节:1、开机环节:⑴进行冷检查,确定无误后启动发电机电源进行热检查,确定之后再进行下列环节;⑵启动励磁开关,励磁开机;⑶开机(手动调整励磁旋钮);⑷使发电机工作,并调整调速旋钮,使发电机旳功角指示器由一种角变成几种角(试验中旳功角指示器有四个角,表达电机为四极电机,p=2,额定转速为1500r/min ;8个角对应旳转速为1500r/min,当功角指示器旳几种角不稳定期,表达额定转速也许不小于或不不小于额定转速,此时应尽量调整调速器使转速为额定转速);⑸加励磁,调整机端电压与系统相似(本试验为380V);⑹进行投切操作,在操作时,由于有延误,因此应保留一种小余量,保证准时精确地投入系统;此时应调整原动机,当转动不太快,角度在0到5度时投入;2、关机环节:⑴调整调速器使输出功率(有功)P降为0;⑵调整励磁使励磁电流If降为0,虽然无功降为0;⑶此时会发既有功又增大了,因此应继续调整调速器使有功降为0;⑷解联(断开电机并网断路器);⑸调整励磁使电压U降为0;⑺调整调速器使转速降为0;⑻退出开机再关闭励磁。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

Experiment Introduction to powersystems学生姓名:学号:专业班级:实验名称:电力系统导论(双语)CONTENTS1、EXPERIMENT 1 (1)B US A DMITTANCE M A TRIX ...................................................................................................................... 1-82、EXPERIMENT 2 (8)B US I MPEDANCE M A TRIX....................................................................................................................... 8-213、EXPERIMENT 3 (21)G AUSS-S EIDEL AND NEWTON METHOD................................................................................................. 21-244、PERSENAL SUMMARY (24)1 Experiment 1Bus Admittance Matrix1. Objective• To write a simple program in MATLAB® for the algorithm of bus admittance matrix. 2. DiscussionBus Admittance matrixThe node-voltage equation of an n-bus power system can be written in matrix form as(1)Or(2)Where Ibus is the column vector of the injected bus currents. Vbusis the column vector of busvoltages measured from the reference node. Y busis known as the bus admittance matrix .Calculation method:1. The diagonal element of each node is the sum of admittances connected to it. It is knownas the self-admittance or driving point admittance , i.e.,(3)2. The off-diagonal element is equal to the negative of the admittance between the nodes. Itis known as the mutual admittance or transfer admittance , i.e.(4)Result:Base on t he equations (3) and (4), the algorithm of bus admittance matrix can be used to build the bus admi ttance matrix (Y bus). 3. Mathmatics modelThe mathematics models of power system elements are two kinds, one is transmission line or cable and another is transformer, and they are equivalent into pi type equivalent circuit, and2described as following: 1) Transmission line or cableAccording to the Pi sectional circuit of transmission line or cable, we can get the elements of nodal Admittance matrix value, as the followingFigure: The pi section equivalent circuit of transmission line or cable()ij ijio ii ii jX Rg Y Y +++=1(5) ()ij ijjo jj jj jX Rg Y Y +++=1(6)()ij ijij ij jX RY Y +-=1(7)ijji Y Y =(8)2) transformerAs show as the transformer circuit , we have k I Z V V jT j i -=(9)j i KI I -=(10)According to equations (10) and (9) , we have()()()i j Tj T j T i j T j T i j T j V V Z K V Z K V Z K V V Z K V Z KV V Z I -+-=--+=-=111(11)()()i Tj i T i T j T i T j i T j i V Z K V V Z K V Z K V Z K V Z K V KV Z K KI I --+=-=-=-=22()j i Ti T V V Z KV Z K K -+-=2(12)j3Figure: the Pi section equivalent circuit of TransformerAssume ijij T T jX R Z Y +==11 ()ij ijii ii jX R KY Y ++=2(13) ()ij ijjj jj jX RY Y ++=1(14) ()ij ijij ij jX RKY Y +-=(15)ijji Y Y =(16)4. System RequirementComputer with MATLAB® 6 or above installed. 5. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.3.0 Write a function Y = The_Node_Admittance_Matrix(TopoStructureAndBranchPara)for the formation of the bus admittance matrix.4.0 TopoStructureAndBranchPara is the transmission line, cable and transformer inputdata and contains five columns parameters. The first two columns are the line bus numbers and the remaining columns contain the line resistance and reactance in per-unit and transformer tap ratio or capacitor of transmission line.5.0 The function should return the bus admittance matrix.6. ExercisesUse the written function, Y = The_Node_Admittance_Matrix (TopoStructureAndBranchPara) to obtain the Y bus of the following power system network:Q1. You are required to write the Ybus topological structure and parameter into a text file.(Hint: use the matlab text compiler to write down the table 1 data, using the comma to separate the parameters, and save it use the name of 4_Power_System_Data.dbf)Q2. You are required to write out the program flow figure of forming a nodal admittance matrix.Hint. You are required to compile a program to form the Y bus Matrix, the following program isa reference program to you.Figure : One-line diagram of power system47.The flow chartFigure : The flow chart of Forming Nodal Admittance Matrix5程序是:%function OutPut=The_Node_Admittance_Matrix(handles)%is a subroutine of PowerSystemCalculationfunction OutPut=The_Node_Admittance_Matrix(handles)%the following program is open a data file and get the Number of% Node and Branch data to form a nodal addmittance matrix%the following code is open a file and read the data of power system network[fname,pname] = uigetfile('*.dbf','Select the network parametre data-file'); TopoStructureAndBranchPara= csvread(fname);[NumberOfBranch,NumberOfPara]=size(TopoStructureAndBranchPara);Temporary1=max(TopoStructureAndBranchPara(:,1));Temporary2=max(TopoStructureAndBranchPara(:,2));if Temporary1 > Temporary2NumberOfNode=Temporary1;elseNumberOfNode=Temporary2;end%The following program is to form the Nodal Admittance Matrix% and the Topologic structure and Branch Parametres are arranged% I,J,R,X,C/K, and pay attention to the inpedence of transformer is in the% side of Node J and the ratio of transformer 1:K is in the side of Node Ifor CircleNumber1=1:NumberOfBranchfor CircleNumber2=1:NumberOfBranchNodalAdmittanceMatrix(CircleNumber1,CircleNumber2)=0;endendfor CircleNumber=1:NumberOfBranchif TopoStructureAndBranchPara(CircleNumber,5) > 0.85NodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNum ber,1),TopoStructureAndBranchPara(CircleNumber,1)))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNum ber,1),TopoStructureAndBranchPara(CircleNumber,1)))+...TopoStructureAndBranchPara(CircleNumber,5)^2/...(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4)) ;NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranc hPara(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranc hPara(CircleNumber,2))+...61/((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumb er,4)));NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara(CircleNumber,2))...-TopoStructureAndBranchPara(CircleNumber,5)/...((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumber, 4)));NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranc hPara(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara(CircleNumber,2));elseNodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara(CircleNumber,1))+...+1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumbe r,5);NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranc hPara(CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranc hPara(CircleNumber,2))+...+1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumbe r,5)NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara( CircleNumber,2))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc7hPara( CircleNumber,2))...-1/(TopoStructureAndBranchPara(CircleNumber,3)+...j*TopoStructureAndBranchPara(CircleNumber,4));NodalAdmittanceMatrix(TopoStructureAndBranchPara( CircleNumber,2),TopoStructureAndBran chPara(CircleNumber,1))=...NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranc hPara( CircleNumber,2));endend运行结果:NodalAdmittanceMatrix =1.0421 - 8.2429i -0.5882 +2.3529i 0 +3.6667i -0.4539 + 1.8911i-0.5882 + 2.3529i 1.0690 - 4.7274i 0 00 + 3.6667i 0 0 - 3.3333i 0-0.4539 + 1.8911i 0 0 0.9346 - 4.2616i Experiment 2Bus Impedance Matrix1. Objective• To write a simple program in MATLAB® for the algorithm of bus impedance matrix.2. DiscussionBus Impedance matrixThe node-current equation of an n-bus power system can be written in matrix form as8(1) Or(2)Where Ibus is the vector of the injected bus currents. Vbusis the vector of bus voltagesmeasured from the reference node. Zbusis known as the bus impedance matrix.Z bus CALCULATIONThere are two major methods:1. Inversion of Y bus2. Z bus building algorithm3. LU factorization.Matrix inversion for very large-scale networks is very slow and avoided. On the other hand, the Z bus building algorithm is the acceptable method for large networks. It is much faster and easier to update.Z bus Building AlgorithmThis is a very systematic way of forming Z bus and allow for easy implementation on the digital computer.Advantages1. Easy to implement2. Avoids matrix inversion3. It is easy to handle Z bus modifications due to line switchings and changes in network topologySolution:NotationThe addition of a branch of impedance z b falls into 4 categories as follows. CASE 1:Adding z b from a new bus p to the reference bus 0Here, ●new Bus Z will be ()()11++n n matrix● Injected current p I will not change the bus voltages of the original network ●p b p I z V =Then,CASE 2:Adding z b from a new bus p to an existing bus kHere, ●new Bus Z will be ()()11++n n matrix● Injected current p I will change the bus voltages of the original network ● Injected current at bus k becomes p k I I + ●p b k p I z V V +=Then,Summary of Case 2CASE 3: Adding z b from an existing bus k to the reference bus 0Here, ●new Bus Z will be n n ⨯ matrix● This is a special case of CASE 2 except that 0=p VThen,Since Vp = 0, the last row can be eliminated by Kron reduction (one kind of mathematics method) to yield an n×n matrix asSummary of Case 3 • Proceed as in CASE 2 • Eliminate the last rowCASE 4: Adding z b from two existing buses j and kHere,●new Bus Z will be n n ⨯ matrix● Branch current p I will change the bus voltages of the original network ● Injected current at bus j becomes p j I I + ● Injected current at bus k becomes p k I I - ●p b k j p b j k I z V V I z V V +-=→=-0● will change the bus voltages of the original network ● Injected current at bus k becomes p k I I +p b k p I z V V +=Then,The last row can be eliminated by Kron reduction to yield an n×n matrix asSummary of Case 4 1. Proceed as follows2. Eliminate the last rowPROCEDURE FOR BUILDING ALGORITHMSteps1. Prepare a transmission line list●First line considered must have a tie to the reference●All subsequent lines in the list must have at least one bus already seen2.Algorithm●Determine which of the cases (1-4) apply and proceed accordingly.●Read a line. If the last line has been processed, then stop else consideranother lineComments on the algorithm1. Loops must be closed as early as possible2. The axes of the final matrix may not correspond with the buses in numerical order3. The line list order affects the efficiency of calculation4. If there is no tie to the reference, place a fictitious ground tie at one of the buses.5. Line outage is modeled by using –z b instead of z b3. System RequirementComputer with MATLAB® 6 or above installed.4. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.3.0 Write a function Z = znbus (z) for the formation of the bus impedance matrix.4.0 z is the line input and contains three columns. The first two columns are the line busnumbers and the remaining columns contain the line resistance in per-unit.5.0 The function should return the bus impedance matrix.5. Mathmatics modelThe mathmatics models of power system elements are two kinds, one is transmission line and another is transformation, and they are equivalented into pi type equivalent ciucuit,and dricribled as following:6. ExercisesUse the written function, Z = znbus(z) to obtain the Ybus of the following power system network:Example 1Figure 3: One-line diagram of power systemQ2. You are required to write the Z bus into a text file. (Hint: use the matlab text compiler) Example 2For the system shown, form Zbus matrix using the building algorithmjij1:KSolutionA line listApply Kron reduction to eliminate the last rowHint. You are required to compile a program to form the Z bus Matrix.the following programis a reference program to you.The program is:%function OutPut=The_Node_impedance_Matrix(handles)%is a subroutine of PowerSystemCalculationfunction OutPut=The_Node_impedance_Matrix(handles)%the following program is open a data file and get the Number of% Node and Branch data to form a nodal impedance matrix%the following code is open a file and read the data of power system network[fname,pname] = uigetfile('*.dbf','Select the network parametre data-file');Topo_Structure_And_Branch_Para= csvread(fname);%get the electric power system the number of branch and the parametre of% elements[NumberOfBranch,NumberOfPara]=size(Topo_Structure_And_Branch_Para);%Temporary1---temporary variable 1%Temporary2---temporary variable 2Temporary1=max(Topo_Structure_And_Branch_Para(:,1));Temporary2=max(Topo_Structure_And_Branch_Para(:,2));if Temporary1 > Temporary2NumberOfNode=Temporary1;elseNumberOfNode=Temporary2;end% The following program is to form the Nodal impedance Matrix% and the Topologic structure and Branch Parametres are arranged% I,J,R,X,C/K, and pay attention to the inpedence of transformer is in the% side of Node J and the ratio of transformer 1:K is in the side of Node %% set the initial value of Nodal Admittance Matrix to zerofor CircleNumber1=1:NumberOfNodefor CircleNumber2=1:NumberOfNodeNodal_impedance_Matrix(CircleNumber1,CircleNumber2)=0;endendfor CircleNumber=1:NumberOfBranchif Topo_Structure_And_Branch_Para(CircleNumber,5) > 0.85Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para (CircleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para (CircleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))+Topo_Structure_And_B ranch_Para(CircleNumber,5)^2/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4)) ;Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,2))+...1/((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(Ci rcleNumber,4)));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para(CircleNumber,2))...-Topo_Structure_And_Branch_Para(CircleNumber,5)/...((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(Circ leNumber,4)));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para(CircleNumber,2));elseNodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para(CircleNumber,1))+...+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Cir cleNumber,5);Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,2))+...+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Cir cleNumber,5)Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para( CircleNumber,2))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para( CircleNumber,2))...-1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+...j*Topo_Structure_And_Branch_Para(CircleNumber,4));Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para( CircleNumber,2),Topo_Structure_ And_Branch_Para(CircleNumber,1))=...Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_ And_Branch_Para( CircleNumber,2));endendformat shortNodal_impedance_Matrix*inv(Nodal_impedance_Matrix)运行结果:Nodal_impedance_Matrix =1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i 0 +3.6667e+000i 0-5.8824e-001 +2.3529e+000i 5.8824e-001 -2.3377e+000i 0 00 +3.6667e+000i 0 0 -3.3333e+000i 00 0 0 4.5386e-001 -1.8719e+000iNodal_impedance_Matrix =1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i 0 +3.6667e+000i -4.5386e-001 +1.8911e+000i-5.8824e-001 +2.3529e+000i 1.0690e+000 -4.7274e+000i 0 00 +3.6667e+000i 0 0 -3.3333e+000i 0-4.5386e-001 +1.8911e+000i 0 0 9.3463e-001 -4.2616e+000ians =1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 - 0.0000i-0.0000 - 0.0000i 1.0000 - 0.0000i -0.0000 + 0.0000i -0.0000 - 0.0000i-0.0000 - 0.0000i -0.0000 - 0.0000i 1.0000 - 0.0000i -0.00000 - 0.0000i 0 + 0.0000i 0.0000 - 0.0000i 1.0000 + 0.0000i以上就是对阻抗矩阵的验证,其和其逆相乘为单位对角矩阵3、Experiment 3Gauss-Seidel Method1. Objective• To write a simple program in MATLAB® for the algorithm to solution of nonlinear algebraic equations;• Known as the method of successive displacements.2. DiscussionThe most common methods for solving nonlinear algebraic equations are Gauss-Seidel,Newtow-Rahpson, and quasi-Newton-Raphson methods. We start with one dimensional equations and then generalize to n-dimensional equations.3. Mathmatics modelConsider the nonlinear equation 0)(=x f .The equation is broken into two parts thus:)(x g x =. We assume )0(x is an initial "guess" of the solution, then "refine" the solution using:)()0()1(x g x =This process is repeated thus)()1()2(x g x =and on the th n iteration we have: )()1()(-=n n x g x. If this process is convergent, then the successive solutions approach a value which is declared as the solution. Thus if at some step 1+k we have:ε≤-+)()1(k k x xwhere e ε is the desired "accuracy", then we claim the solution has been found to the accuracy specified.4. System RequirementComputer with MATLAB® 6 or above installed.5. Procedure1.0 Launch the MATLAB program.2.0 Go to FILE NEW M-file.3.0 Write a function program of Gauss Seidel Method.6. Exercises Example: Using the Gauss-Seidel method to obtain the roots of the equation:0496)(23=-+-=x x x x f First the equation is expressed in a different form thus())(469123x g x x x =---= And the iteration can proceed. Take a good look at the shape of the iterations! Below is the program showing the process graphically (later showing how to do it iteratively).7.The flow chart of Gauss Seidel method (Omitted)8.Reference Program and result.程序是:clear allclcx0=0.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)x0=-(x0^3-6*x0^2-4)/9;y=x0;n=n+1;end结果是:n=1627 y=x0=0.99818clear allclcx0=2.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)x0=-(x0^3-6*x0^2-4)/9;y=x0;n=n+1;end结果是:n=7 y=x0=4仿照高斯--赛德尔法,我们可以写出简单的牛顿法的程序,如下:牛顿法解方程x0=0.5;n=0;while (abs((x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9))>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是:dx0=1.7684e-005 n=15 x0=0.99998 y= -0.875x0=0.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是: dx0=0.0011305 n=9 x0=0.99887 y= -0.875x0=3.5;n=0;while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);x0=x0+dx0;n=n+1;end结果是:dx0= -2.5283e-006 n=5 x0=4 y= -0.875实验总结:在做实验前,我以为不会难做,就像以前做物理实验一样,做完实验,然后两下子就将实验报告做完.直到做完测试实验时,我才知道其实并不容易做,但学到的知识与难度成正比,使我受益匪浅.在做实验前,一定要将实验指导书上的知识吃透,因为这是做实验的基础,否则,在老师讲解时就会听不懂,这将使你在做实验时的难度加大,浪费做实验的宝贵时间。

相关文档
最新文档