英文版电力系统分析上册第五章
电力系统自动化5 电力系统最小二乘法状态估计
；
X 1 ( H R H ) H R [ Z h( X 1 )]
T -1 v T -1 v
1
iii. 求
x1
iii. 求 X 1 ； iv. 第一次迭代结果
( x) ( x )
x x1
；
v. 重复ii ~ iv，直到获得较满意的 X 。
线性方程组的计算机解法之一 ——平方根因子分解法（略）
X 2 X1 X1
iv. 第一次迭代结果
x 2 x1 x1
v. 重复ii ~ iv，直到获得 较满意的 x。
第五章 电力系统运行的状态估计
第四节 电力系统最小二乘法状态估计
最小二乘法状态估计程序框图（图5-12） 例：图5-13，5-14，表5-4~5-7
Y13 Y31 y13 YT k 1 1 1.05 j7.5 j0.1269
T
T
电力系统中，Z 的元素包括状态变量的测量读值 Z X 和其他 系统变量的测量读值 Z Z 。
Z Z X
ZZ
Z Z 为 X 的非线性函数，故电力系统的量测方程式为
Z = h (X) + V
第五章 电力系统运行的状态估计
第四节 电力系统最小二乘法状态估计
注意：i. m n Z ii. 相角一般不能直接测量（*PMU）， Z 维数高于 Z X 。
故
H 12
P U
0,
H 21
Q θ
0
H 11 H (θ, U) 0
再经一些近似，可得
H 11 H (θ, U) 0
0 H 22
0 U 02 H 1 H 22 0
电力系统分析理论课后第五章答案
确定调相机容量p180(5-34)
Qc
V2c max X
(V2c max
V2'cmax kT
)kT2
10.4 (10.4 93.8823) 102 M var
80
10
13.156M var
取补偿电容量为QC 13M var ，验算最大负荷时受端低压侧的实际电 压：
（1）求最大、最小负荷时各绕组的电压损耗
最大负荷时：
V max
P R Q X V max
12 3 9 65 KV 112
5.5446KV
VⅡmax
PⅡRⅡ QⅡXⅡ V max V max
6 4 15 112 5.5446
KV
0.1785KV
KV
6KV
VⅢ min
112.2273 6.6 115.5
K压母线的调压要求，由高中压两侧选择中压绕组的分接 头。最大最小负荷时中压母线调压要求为35-38KV。
最大、最小负荷时中压绕组分接头电压为：
VtⅡmax
V' tⅡmax
Vt VtⅡmax
35 115.5 KV 106.2769
V' 2C max
117
12.798
104 .2(kV)
V2C max
V' 2C max K
104 .2 10
10.42kV
最小负荷时调相机按50％额定容量欠励磁运行，因而有：
QC 6.5M var
S m in
12 2
(9 6.5)2 110 2
英文版电力系统分析(上册)第一章
Chapter 1 Power Systems Basics
1.3 The Requirements for Power Systems Operation
Features: • Electric Energy can not be stored in bulk • Short transit • Has close relationship with every industry and everyday life
1.3 The Requirements for Power Systems Operation
Features: • Electric Energy can not be stored in bulk • Short transit • Has close relationship with every industry and everyday life
Power Systems Analysis
• Basic knowledge • Steady-state analysis • Transient analysis electromagnetic transient analysis electromechanical transient analysis (Stability analysis)
3.15 6.3 10.5 13.8 15.75 18 20
10.5 or
35 110 220 330 500
Chapter 1 Power Systems Basics
1.2 The Rated Voltage and Rated Frequency of Power System
Chapter 1 Power Systems Basics
电力系统分析第五章
•
9
§5.2电力系统中有功功率的最优分配
一、 各类发电厂的运行特点 1 火电厂
支付燃料费用 技术最小负荷
火 电 厂 锅炉 25%—70% 汽机 10%—15% 承担急剧变化负荷时与投、退相似，额外耗能、费时 效率与蒸汽 参数有关 高温高压：效率高、调节范围小 中温中压：较前者低、但调节范围大
电力系统分析基础 Power System Analysis Basis （五）
主讲人：朱晓荣
1
第五章电力系统的有功功率和频率调整 1、电力系统中有功功率的平衡 2、有功功率的最优分配
各类发电厂的运行特点和合理组合 不计网损时水、火电厂间的经济功率分配
计网及损时各发电厂间的经济功率分配
水煤换算系数 1m3水量相当于2吨煤的消耗 丰水期小，枯水期大
F d F 1 i 经过网损修正后 d iL i P P d 的等微增率准则 d P G i G i L 1 P G i
网损微增率
( i 1 ,2 ,
网损修正系数
,n )
网损微增率的物理意义：某发电厂所发功率的变化引起的 26 网路总损耗的变化——阻抗矩阵法
i1
• 等式约束条件：
P
i1
n
Gi
P LD 0
• 这是多元函数求条件极值问题，可以利用拉格朗日乘数法求 解。 n F P P • 构造拉格朗日函数 L Gi LD 1 i
23
• 求拉格朗日函数的无条件极值
n L F P P Gi LD 1 i
§5.1电力系统中有功功率的平衡 二 频率与有功平衡的关系
(完整版)电气工程及其自动化专业英语第五章课文翻译
Most people can formulate a mental picture of a computer, but computers do so many things and come in such a variety of shapes and sizes that it might seem difficult to distill their common characteristics into an all-purpose definition. At its core, a computer is a device that accepts input， processes data， stores data, and produces output, all according to a series of stored instructions.Computer input is whatever is put into a computer system. Input can be supplied by a person， by the environment, or by another computer。
Examples of the kinds of input that a computer can accept include the words and symbols in a document, numbers for a calculation, pictures， temperatures from a thermostat，audio signals from a microphone， and instructions from a computer program. An input device, such as a keyboard or mouse， gathers input and transforms it into a series of electronic signals for the computer.In the context of computing, data refers to the symbols that represent facts, objects， and ideas. Computers manipulate data in many ways， and we call this manipulation processing。
机电英语Unit5
机电英语Unit5在现代工业领域，机电一体化技术的应用日益广泛，而机电英语作为这一领域中重要的交流工具，对于相关专业人员来说至关重要。
Unit 5 主要涵盖了机电领域中的一些关键概念和技术，包括电气系统、机械传动以及自动化控制等方面。
首先，让我们来探讨一下电气系统。
在机电设备中，电气系统就如同设备的“神经中枢”，负责控制和驱动各种机械部件的运行。
从简单的电路原理到复杂的电力控制系统，都需要我们用准确的英语术语来描述和理解。
例如，“voltage”（电压）、“current”（电流）、“resistance”（电阻）等基础概念，是理解电气系统的基石。
当涉及到机械传动时，各种传动方式及其相关的英语表述也需要我们熟练掌握。
比如，“belt drive”（皮带传动）、“gear drive”（齿轮传动）和“chain drive”（链条传动）等。
这些传动方式在不同的机电设备中发挥着重要作用，而能够准确地用英语表达它们的特点、工作原理以及应用场景，对于国际间的技术交流和合作是必不可少的。
自动化控制是当今机电领域的核心发展方向之一。
诸如“programmable logic controller”（可编程逻辑控制器，简称 PLC）、“sensor”（传感器）和“actuator”（执行器）等词汇频繁出现在相关的技术文档和交流中。
了解这些术语以及它们所代表的技术，能够帮助我们更好地理解和描述自动化控制系统的工作流程和性能。
在学习机电英语 Unit 5 的过程中，我们还会遇到大量的专业词汇和短语，如“motor control”（电机控制）、“power supply”（电源）、“feedback loop”（反馈回路）等等。
这些词汇不仅要求我们能够准确地识别和理解，更要能够在实际的交流和写作中正确运用。
为了更好地掌握这部分知识，我们可以通过阅读相关的英文技术文献、观看英文的技术讲解视频以及参与国际技术交流会议等方式来提高自己的机电英语水平。
电力系统分析教学课件第五章
（元/h）
C2 =5000+250 PG 2 +110 10-3 PG22
（元/h）
式中，P1、P2的单位为MW。
当系统总负荷PL在分别为500、800、1400MW时，为使在不同总负荷下总发电成本
C最小，试求各发电厂的功率、发电成本微增率以及运行总成本。这里不计发电机
抽水蓄能
无调节水库
日、月、年、多年调节水库
各类发电厂的运行特点
3 核电厂
一次建设投资达、运行费用小
核
电
厂
技术最小负荷取决于汽轮机10%-15%
投、退、负荷急剧变化时耗能、费时，易于损坏设备厂
4 可再生能源电厂
风
电
、
光
伏
发电具有间歇性与波动性，光伏受昼夜、阴晴影响，风电受风资源影响
单机装机容量小、且分散
G2
dP
G2
依据等发电成本微增率原则，当分配两发电厂发电功率使它们具有相等的发电成本微增
率时，两电站运行总成本最低，于是有
300+210 10-3 PG1 =250+220 10-3 PG 2
300+210 10-3 PG1 =250+220 10-3 ( PL PG1 )
220 103 PL 50
✓ 备用容量配置的大小对电力系统产生哪些影响？
讨论
✓ 储能技术的飞速发展使得电能的大量存储成为可能，这会给有功功率平衡带
来哪些影响？
2
PART
有功功率的经济分配
问题
电力系统运行的基本要求：
保证可靠持续供电、良好的电能质量、运行经济性。
powersystemanalysis电力系统分析
Power System AnalysisThe one line diagramModern power systems are invariably three phase. In a balanced three phase systems is always solved as a single phase circuit composed of one of the three lines and a neutral return and this is sufficient to give a complete analysis. Such a simplified diagram of an electric system is called a one–line diagram. Figure 1 below shows the symbols for representing the components of a three phase power system.Rotating Machine( Generator or Motor).Circuit Breaker.3-phase delta connection.3-phase star connection (neutral ungrounded).3-phase star connection (neutral grounded).Two winding power transformer .Figure-1-symbols of components of a 3-phase power system.Figure-2shows the one line diagram of a simple 3-phase power system.Notes1. Generators neutrals are usually grounded through high resistances and somtimes through inductance coils in order to limit the flow of current to ground during a fault.2. Most transformer neutrals in transmission systems are solidly grounded .Per unit quantityWhen making calculations on a power system network having two or more voltages levels, it is very cumbersome to convert currents to different voltage levels at each point where they flow through a transformer, the change in current being inversely proportional to the transformer turns ratio. In order to simplify these calculations we can use per unit system.In this system a base quantities are assumed for each voltage level and theper unit quantities are calculated as follow:Actual quantityPer unit quantity =ــــــــــــــــــــــــــــــــــــــــ---------------------------(1)Base quantityThe four electrical quantity (voltages, current, power, and impedance) are so related that selection of base values for any two of them determine the base values of the remaining two.Usually base apperant power in megavoltamperes and base voltage in KV are quantities selected to specify the base values.For single phase system or 3-phase on per phase basis , the following relationships hold:Base voltamperesBase current =ــــــــــــــــــــــــــــــــــــــ-----------------------------------(2)Base voltageBase voltageBase impedance =ــــــــــــــــــــــــــ--------------------------------------(3)Base currentActual voltagePer unit voltage =ـــــــــــــــــــــــــــــ-------------------------------------(4)Base voltageActual currentPer unit current =ـــــــــــــــــــــــــــ--------------------------------------(5)Base currentActual impedancePer unit impedance =ـــــــــــــــــــــــــــــــــــــــ---------------------------(6)Base impedanceOrBase KVA 1-phaseBase current =ــــــــــــــــــــــــــــــــــــــــــ--------------------------------(7)Base voltage KV-phBase voltage VphBase impedance =ــــــــــــــــــــــــــــــــــــــ------------------------------(8)Base current Amp2( Base voltage KVph)×1000Base impedance =ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــ------------------(9)Base KVA-ph2(Base voltage KV-ph)Base impedance =ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ------------------(10)Base MVA-phBelow is given annual work summary, do not need friends can download after editor deleted Welcome to visit againXXXX annual work summaryDear every leader, colleagues:Look back end of XXXX, XXXX years of work, have the joy of success in your work, have a collaboration with colleagues, working hard, also have disappointed when encountered difficulties and setbacks. Imperceptible in tense and orderly to be over a year, a year, under the loving care and guidance of the leadership of the company, under the support and help of colleagues, through their own efforts, various aspects have made certain progress, better to complete the job. For better work, sum up experience and lessons, will now work a brief summary.To continuously strengthen learning, improve their comprehensive quality. With good comprehensive quality is the precondition of completes the labor of duty and conditions. A year always put learning in the important position, trying to improve their comprehensive quality. Continuous learning professional skills, learn from surrounding colleagues with rich work experience, equip themselves with knowledge, the expanded aspect of knowledge, efforts to improve their comprehensive quality.The second Do best, strictly perform their responsibilities. Set up the company, to maximize the customer to the satisfaction of the company's products, do a good job in technical services and product promotion to the company. And collected on the properties of the products of the company, in order to make improvement in time, make the products better meet the using demand of the scene.Three to learn to be good at communication, coordinating assistance. On‐site technical service personnel should not only have strong professional technology, should also have good communication ability, a lot of a product due to improper operation to appear problem, but often not customers reflect the quality of no, so this time we need to find out the crux, and customer communication, standardized operation, to avoid customer's mistrust of the products and even the damage of the company's image. Some experiences in the past work, mentality is very important in the work, work to have passion, keep the smile of sunshine, can close the distance between people, easy to communicate with the customer. Do better in the daily work to communicate with customers and achieve customer satisfaction, excellent technical service every time, on behalf of the customer on our products much a understanding and trust.Fourth, we need to continue to learn professional knowledge, do practical grasp skilled operation. Over the past year, through continuous learning and fumble, studied the gas generation, collection and methods, gradually familiar with and master the company introduced the working principle, operation method of gas machine. With the help of the department leaders and colleagues, familiar with and master the launch of the division principle, debugging method of the control system, and to wuhan Chen Guchong garbage power plant of gas machine control system transformation, learn to debug, accumulated some experience. All in all, over the past year, did some work, have also made some achievements, but the results can only represent the past, there are some problems to work, can't meet the higher requirements. In the future work, I must develop the oneself advantage, lack of correct, foster strengths and circumvent weaknesses, for greater achievements. Looking forward to XXXX years of work, I'll be more efforts, constant progress in their jobs, make greater achievements. Every year I have progress, the growth of believe will get greater returns, I will my biggest contribution to the development of the company, believe in yourself do better next year!I wish you all work study progress in the year to come.Actual impedance Per unit impedance=ــــــــــــــــــــــــــــــــــــــــ-----------------------(11)Base impedanceIf we choose base kilovoltamperes and base voltage in kv to mean kilovoltamperes for the total of the 3-phases base voltage for line to line, we find:Base KVA-3-ph Base current =ــــــــــــــــــــــــــــــــــــــــــــ---------------------------(12)3×Base KVA-lineFor 3-phase 2(Base voltage KV)Base impedance =ــــــــــــــــــــــــــــــــــــــــــــــ---------------------(13)Base MVA-3-ph Note:In a 3-phase system, the per unit 3-phase kVA and voltage on the 3-phase basis is equal to the per unit per phase kVA and voltage on the per phase basis.Example1-1: Consider a 3-phase wye –connected 50000 kVA, 120kV system.Express, 40000 kva three phase apperant power and 115 kv line t line voltage in per unit values on (i) 3-phase basis and (ii) per phase basis.Solution;(i) Three phase basis Base kva = 50000 kva Base kv = 120 kv (line to line)P.u kva =8.05000040000=p.u voltage =96.0120115=(ii) per phase basisBase kva =kva 166********1=⨯Base KV =3120=69.28 KV Per unit KVA=166673/40000=0.8Per unit voltage =28.693/115=0.96Change of Base Sometimes it is necessary to convert per-unit quantities from one base to another.The conversion formula for the impedance can be written as follow:()()2⎪⎪⎭⎫ ⎝⎛⨯⎪⎪⎭⎫ ⎝⎛⨯=----new b old b old b new b old pu new pu V V S S Z Z .----------------------------------(14)Example1-2The reactance of a generator Xg is given as 0.25pu based on generator nameplate rating of 18KV , 500MVA.The base for calculations is 20KV And 800MVA.Find Xg on the new base.Solution Xg-new=0.25×22018500100⎪⎭⎫ ⎝⎛⨯⎪⎭⎫ ⎝⎛=0.0405puPer unit impedance of transformer unitThe ohmic values of impedance of a transformer depend on whether they are measured on the high or low tension side of transformer In the per unit system, the per unit impedances of a transformer is the same regardless of whether it is determined from ohmic values referred to the high tension or low tension side of the transformer.Z1=2221N N ohm ----------------(14) . Z1=2221V V ohm ------------------(15 ,The impedance bases on the two sides of the transformer are from equation (13).Z1base =bMVA KV 21-------------------------------(16).Where KV1 is the 1st side base voltage.Z2base =bMVA KV 22-----------------------------(17).Where KV2 is the 2nd side base voltage.To prove that Z1pu=Z2pu22212221222121N N V V MVA KV MVA KV Z Z b b b b b bb b ====.pu b b pu Z Z Z N N Z N N Z Z Z 222222122221111=ΩΩ==ΩΩ=Example 1-3A single phase transformer is rated 110/440, 2.5 kva, leakage reactance measured from low tension side is 0.06Ω. Determine leakage reactance in per unit.Solution From the low tension side Base KV1=110×103-Base MVA= 2.5×103-Z1b =Ω=⨯⨯--84.4105.2)10110323(base impedance on low tension side).Z1pu=pu Z Z b 0124.084.406.011==ΩFrom the high tension side Base KV2=440 V Base MVA=2.5×103-Z2=Ω=⨯96.006.011044022Z2b =Ω=⨯⨯--44.77105.2)10440323Z2pu =pu Z Z b 0124.044.7796.022==ΩNote:1. In per unit calculations involving transformer in three phase system, we follow the same principles developed for single phase system and require the base voltage on the two sides of the transformer to have the same ratio as the rated line to line voltage on the two sides of the transformer. The base kva is the same on each side.2. To transfer the ohmic value of impedances from the voltage level on one side of 3 phase transformer to the voltage level on other, the multiplying factor is the square of the ratio of line to line voltages regardless of whether the transformer connection is Example 1-4The three single phase transformers each rated 25 Mva, 38.1/3.81 kv are connected as shown in figure with a balanced load of 0.6Ω-connected resistors. Choose a base of 75 Mva, 66 kv for the high tension side of transformer and specify the base for the low tension side. Determine the per unit resistance of the load on the base for the low tension side. Then determine the load resistance referred to high tension side and the per unit and the per unit value of this resistance on the chosen base.Solution:(1) on the low tension side The base for low tension side is 75 Mva,3.81 kvActual value = 0.6ΩBase value =()Ω=1935.07581.32Per unit value of R L 1.31935.06.0==(2) on the high tension side The base for high tension side is 75 Mva, 66kv Actual value =Ω=⨯18018.3666.022Base value =Ω=08.5875662Per unit value of R L =1.308.58180=Per unit impedance of 3-winding transformersGenerally, large power transformers have three windings. The third winding is known as a tertiary winding which may be used for the following purposes 1.To supply a load at a voltage different from the secondary voltage.2.To provide a low impedance for the flow of certain abnormal currents, such a third harmonic currents.3.To provide for the excitation of a regulating transformer.Note: When one winding is left open, the three winding transformer behaves as two winding transformer and standard short circuit tests can be used to evaluate per unit leakage impedances which are defined as followsZ ps = per unit leakage impedance measured from primary with secondary shorted and tertiary open.Z pt = per unit leakage impedance measured from primary with tertiary shorted and secondary open.Z st= per unit leakage impedance measured from secondary with tertiary shorted and primary pen.Z ps= Z p+ Z s……………(a)Z pt= Z p+ Z t………………(b)Z st= Z s+ Z t………………(c)Where Z p, Z s, and Z t: the impedances of primary, secondary and tertiary. Solving these equations we findZ p= ½ ( Z ps+ Z pt-Z st) ……..(d)Z s= ½ ( Z ps+Zst-Z pt) ……….. (e)Z t= ½ ( Z pt+ Z st–Z ps) ………..(f)These equations can be used to evaluate the per unit series impedances Z p, Z s, and Z t of three winding transformer equivalent circuit from the per unit impedances Z ps, Z pt and Z st which in turn are determined from short circuit tests.Note.The impedances Z p, Z s, and Z t of the three windings are connected in star .Example1-5The 3 phase rating of 3 winding transformer are:Primary:Star-connected, 66KV ,15MVA.Secondary: Star-connected ,13.2KV, 10MVA Tertiary: Delta-connected ,2.3KV,5MVA.Neglecting resistance, the leakage impedances areZ ps = 7% on 15 Mva, 66 kv base Z pt = 9% on 15 Mva, 66kv base Z st = 8% on 10 Mva, 13.2 kv baseFind the per unit impedances of the star connected equivalent circuit for a base of 15 Mva, 66 kv in the primary circuitSolution:With the base 15 Mva, 66 kv S base = 15 Mva for all three terminals is the same and V b1= 66 kv V b2= 13.2 kv V b3= 2.3 kvZ ps = 0.07 (no change)Z pt = 0.09 (no change)Z st = 0.08 ×(15/10)=0.12()pu j j j j Z p 02.012.009.007.021=-+=()pu j j j j Z s 05.009.012.007.021=-+=()puj j j j Z t 07.007.012.009.0=-+=Impedance and Reactance DiagramsLet use take a sample power system network as shown in figure belowThe impedance diagram of this sample network is shown in figure 4In many studies, like faults calculations study, in order to simplify the calculations we can neglect all static loads, all resistances, the magnetizing current of each transformer and the capacitance of transmission line and thus we obtain the reactances diagram as shown in figure.Notes: The impedance and reactance diagrams are sometimes called the positive sequence diagrams.。
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t
iap0 = sub − max imum
t
•
Em
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
i[0] ip0
iap 0
ϕ' α
•
ϕ Im
•
•
I m − I pm
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2
t
Short-circuit
of
ConstanE•t
Voltage
m
Source
Circuit
ϕ' α
i[0]
•
Im
t
•
Em
e = Em sin(ωt + α ) i = I pm sin(ωt + α −ϕ)
Im =
Em
(R)2 + ω 2L2
ϕ ' = arctg ωL
R
i[0] ip0
ϕ' α
iap0 = 0
t
t
•
Em
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
i[0] ip0
iap 0
ϕ' α
•
ϕ Im
•
•
I m − I pm
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2 Short-circuit of Constant Voltage Source Circuit
Chapter 5 3-phase Short-circuit
Transient of Power Systems
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-1 The Overall Concept of Short-circuit
ϕ ' = arctg ωL
R
i[0] ip0
iap 0
•
Em
ϕ' α
•
ϕ Im
•
•
I m − I pm
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2 Short-circuit of Constant Voltage Source Circuit
5-2 Short-circtuit of Constant Voltage Source Circuit
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
−t
= I pm sin(ωt + 0o − 90o ) + I pme Ta
5-2 Short-circtuit of Constant Voltage Source Circuit
iap0 = max imum
t
•
Em
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
i[0] ip0
iap 0
ϕ' α
•
ϕ Im
•
•
I m − I pm
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
•
ϕ Im
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2 Short-circuit of Constant Voltage Source Circuit
t
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
e = Em sin(ωt + α ) i = I pm sin(ωt + α −ϕ)
Im =
Em
(R)2 + ω 2L2
ϕ ' = arctg ωL
R
i[0] ip0
iap 0
•
Em
ϕ' α
•
ϕ Im
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2 Short-circuit of Constant Voltage Source Circuit
t
−t
i = I pm sin(ωt + α − ϕ) + iap0e Ta
e = Em sin(ωt + α ) i = I pm sin(ωt + α −ϕ)
Im =
Em
(R)2 + ω 2L2
Im =
Em
(R + R' )2 + ω 2 (L + L' )2
ϕ
'
=
arctg
ω(L + L' )
R + R'
t
•
Em
ϕ' α
i[0]
•
Im
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2
t
Short-circuit
of
ConstanE•t
Voltage
m
Source
Circuit
ϕ' α
i[0]
•
Im
t
•
Em
e = Em sin(ωt + α ) i = I pm sin(ωt + α −ϕ)
Im =
Em
(R)2 + ω 2L2
ϕ = arctg ωL
R
α
ip0
ϕ
•
I pm
Chapter 5 3-phase Short-circuit Transient of Power Systems
• The Cause of Short-circuit • The Type of Short-circuit • The Consequences of Short-circuit • The Purpose of Short-circuit Calculation
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-1 The Overall Concept of Short-circuit
Types of Short-circuit
Type
Schematic diagram
Symbol
3-phase short-circuit
f (3)
Double line-ground f (3) Line-line
Single line-ground
f (1,1)
f (2) f (1)
Chapter 5 3-phase Short-circuit Transient of Power Systems
5-2 Short-circuit of Constant Voltage Source Circuit
e = Em sin(ωt + α )
i = Im sin(ωt + α −ϕ ' )