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1、四川大学电气信息学院电能质量报告第二章王飞鹏1143031228108班2014/6/1在此处键入文档摘要。摘要通常为文档内容的简短概括。在此处键入文档摘要。摘要通常为文档内容的简短概括。ORIGIN OF POWER QUALITY VARIATIONS电能质量变化的来源This chapter describes the origin and some of the basic analysis tools of power quality variationsORIGIN电能质量变化这章介绍了电能质量变化的基本分析工具的起源The consecutive sections of the
2、chapter discuss (voltage) frequency variations, voltage (magnitude) variations, voltage unbalance, voltage fluctuations (and the resulting light flicker), and waveform distortion. 本章的连续的章节讨论（电压）频率变化，电压（幅度）的变化，电压不平衡度，电压波动（和由此产生的光线闪烁），和波形失真。A summary and conclusions for each of the sections will be gi
3、ven at the end of this chapter.一种用于每个部分的摘要和结论将在本章的最后给出。2.1 VOLTAGE FREQUENCY VARIATIONS2.1电压频率变化Variations in the frequency of the voltage are the first power quality disturbance to be discussed here. After a discussion on the origin of frequency variations (the power balance) the method for limitin
4、g the frequency variations (power frequency control) is discussed. The section closes with an overview of consequences of frequency variations and measurements of frequency variations in a number of interconnected systems.在该电压的频率变化是第一个电源质量扰动在这里讨论。后讨论在频率变化（功率平衡）的来源用于限制频率的变化（功率 - 频率控制）。本节用的频率变化，和许多相互连
5、接的系统的频率变化的测量结果相近。2.1.1 Power Balance2.1.1功率平衡Storage of electrical energy in large amounts for long periods of time is not possible, therefore the generation and consumption of electrical energy should be in balance. Any unbalance in generation and production results in a change in the amount of ene
6、rgy present in the system. The energy in the system is dominated by the rotating电能的大量用于长时间储存是不可能的，因此，电能的产生和消耗应该是平衡的。任何不平衡的产生和消耗在能量的系统中存在的量的变化。在系统中的能量以转动为主energy Erot of all generators and motors:所有的发电机和电动机的能量EROT： With J the total moment of inertia of all rotating machines and v the angular velocity
7、 at which these machines are rotating. An unbalance between generated power Pg and the total consumption and losses Pc causes a change in the amount of rotational energy and thus in angular velocity:令J为所有旋转电机的转动惯量和v为角速度在这些机器都旋转的总力矩。发电电力Pg和总消耗量和损失的PC之间的不平衡导致的旋转能量的量，从而在角速度的变化 ： The amount of inertia i
8、s normally quantified through the inertia constant H, which is defined as the ratio of the rotational energy at nominal speed v0 and a base power Sb:转动惯量的值会通过惯性常数H，其定义为旋转能量的额定速度v 0和基电源的Sb的比值通常量化 The base power is normally taken as the sum of the (apparent) rated powers of all generators connected to
9、 the system, but the mathematics that will follow is independent of the choice of base power. Typical values for the inertia constant of large systems are between 4 and 6 s.该基本功率通常取为连接到系统的所有发电机的（明显的）额定功率的总和，但这个数字是独立于基本功率的。对于大型系统的惯性时间常数典型值是4和6秒之间。Inserting (2.3) in (2.2), assuming that the frequency
10、remains close to the nominal frequency, and replacing angular velocity by frequency give the following expression:代入（2.3）在（2.2）中，假设频率保持接近标称频率，并置换角速度由频率给下面的表达式： Where Pg and Pc are per-unit (up) values on the same base as the inertia constant H.其中PG和PC为每单位（最多）在同一基地为惯性常数H。值Consider a 0.01-pu unbalance
11、 between generation and production in a system with an inertia constant of 5 s. This leads to a change in frequency equal to 0.05 Hz/s. If there would be a 0.01-pu surplus of generation, the frequency would rise to 51 Hz in 20 s; for a 0.01-pu deficit in generation the frequency would drop to 49 Hz
12、in 20 s. It is very difficult to predict the load with a 1% accuracy. To keep the frequency constant some kind of control is needed.考虑发电和生产之间的0.01pu不平衡在一个系统中的5秒的惯性时间常数。这导致了在频率0.05赫兹/秒的变化。如果将有0.01普盈余，频率将上升到51赫兹在20秒中;对于0.01-PU缺额产生的频率将降低到49赫兹在20秒中。这是用1的精度非常难于预测的负载。为了保持频率恒定的某种控制是必要的。The sudden loss of a
13、 large power station of 0.15 pu will result in a frequency drop of 1 Hz/s. In 1 s the frequency has dropped to 49 Hz. As the sudden unexpected loss of a large generator unit cannot be ruled out, there is obviously the need for an auto-matic control of the frequency and of the balance between generat
14、ion and consumption.大电站突然丧失0.15 PU将导致1赫兹/秒的频率下降。在1 s中频率已经下降到49赫兹。作为一家大型发电机组的突然意外损失无法排除，显然需要对生产和消费之间的平衡的频率和自动控制。For comparison we calculate the amount of electrical and magnetic energy present in 500 km of a 400-kV three-phase overhead line when transporting 1000 MW of active power at unity power fac
15、tor. Assuming 1 mH/km and12 nF/km as inductance and capacitance, respectively, gives for theelectrical energy 12 CU2 =320 kJ and for the magnetic energy 12Li2=1040kJ.For unity power factor the peaks in magnetic and electrical energy(current and voltage) occur at the same time, so that the maximum total electromagnetic energy equals 1360 kJ. As before we can express this as a time constant by dividing with the rated power. For a 1000-MVA base, we find a time constant of 1.4 ms. This is significantly
