附录1：外文资料翻译A1.1外文资料题目26.22 接地故障电路开关我们目前为止报道的接地方法通常是充分的, 但更加进一步的安全措施在某些情况下是必要的。

假设例如, 有人将他的手指伸进灯口（如Fig.26.45示）。

虽然金属封入物安全地接地, 但那人仍将受到痛苦的震动。

或假设1个120V 的电炉掉入游泳池。

发热设备和联络装置将导致电流流入在水池中的危害，即使电路的外壳被安全地接地，现在已经发展为当这样的事件发生时，设备的电源将被切断。

如果接地电流超过5mA ，接地开关将在5 ms 内跳掉，这些装置怎么运行的？如Fig.26.46所示，一台小变流器缠绕上导线 ，第二步是要连接到可能触发开合120 V 线的一台敏感电子探测器。

在正常情况下流过导体的电流W I 与中性点上的电流N I 准切的相等，因此流经核心的净潮流(N W I I -)是零。

结果，在核心没有产生电流，导致的电压F E 为零，并且开关CB 没有动作。

假设如果某人接触了一个终端(图Fig.26.45示)，故障电流F I 将直接地从载电线漏到地面，这是可能发生的。

如果绝缘材料在马达和它的地面封入物之间断开，故障电流也会被产生。

在以下任何情况下，流经CT 的孔的净潮流等于F I 或L I ，不再是零。

电流被产生，并且产生了可以控制CB 开关的电压F E 。

由于5 mA 不平衡状态只必须被检测出，变压器的核心一定是非常有渗透性的在低通量密度。

Supermalloy 是最为常用的，因为它有相对渗透性典型地70000在通量密度仅4mT 。

26.23 t I 2是导体迅速发热的因素它有时发生于导体短期内电流远大于正常值的情况下，R I 2损失非常大并且导体的温度可以在数秒内上升几百度。

例如，当发生严重短路时，在保险丝或开关作用之前，会有很大的电流流过导体和电缆。

此外，热量没有时间被消散到周围，因此导体的温度非常迅速地增加。

在这些情况下什么是温度上升？ 假设导体有大量m ，电阻R 和热量热容量c 。

而且，假设电流是I ，并且那它流动在t 少于15秒期间。

在导体上引起的热Rt I Q 2=从Eq.3.17，在功率一定的情况下我们可以计算导体上升的温度差：t mc Q ∆=因此由t mc Rt I ∆=2得出 )(2t I mcR t =∆ 因而断定一个导体的温度上升取决于t I 2因素。

众所周知：高温会破坏包裹在导体表面的绝缘。

因此确定温度上升因素t I 2是非常重要的，因为它在短路情况下，决定导体的温度上升。

例如，一个最初在温度C ︒90的No.2AWG 铜导体，如果它的温度在短路时必须控制在C ︒250，不可能承受t I 2因素超出s A 261022⨯。

一般来说，t I 2因素可以被用于计算知道(a)导体的横断面， (b)它的构成(铜或铝)和(c)它能承受的最高温度。

铜和铝的t I 2因素给出以下等式：对于铜导体， )234234(log 105.11010242θθ++⨯=m A t I 对于铝导体， )234234(log 102.5010242θθ++⨯=m A t I 其中：I=短路电流（A ）t=短路的时间（s ）A =导体没有计算空的空间的网横断面（mm 2） =0θ导体的最初温度(c ︒)=m θ导体的最后温度(c ︒)例子26-1___________________________________________一条由铝导体NO3.AWG 制成且横断面为26.6mm 的架空线。

在正常情况下这个导体可能连续运载电流160 A 。

a.在短路期间，计算最大可允许的t I 2因素，知道最初的温度是c ︒80，并且最大温度不应该超出c ︒250。

b.在此架空线上有2000A 的最大短路电流，问它在无需超出c ︒250温度极限的情况下运行多久？ 解答a.由Eq.26.5我们发现 )234234(log 102.5010242θθ++⨯=m A t I =80234250234(log 102.51024++⨯A =7s A 2610⨯b.2000A 电流可运行的时间tt I 2=7610⨯ 2000t 2 =7610⨯t = 1.75s例子26-2___________________________________________它提议使用NO.30AWG 铜丝作为一根临时保险丝。

如果它最初的温度是50c ︒，计算以下： a.需要熔化导线的t I 2 (铜熔化在1083c ︒)b.如果短路电流是30A ，需要熔化导线的时间解答a.由我们有的Eq.26.4 )502341083234(log 0507.0105.1110242++⨯⨯=t I =197 A s 2b.从电流为30A 可得到t I 2=197301972=tt=0.22s因此，保险丝可能在220ms 后被熔断。

26.24 保险丝的角色为了保护导体在短路期间免受过热温度的毁坏，一系列保险丝必须安置在导体上。

保险丝是经过选择的，目的是它的规定值t I 2比将保护导体上升的过热温度要少。

实际上，我们要保险丝在导体获得一个危险温度之前熔断，通常采取是250c ︒。

实践上，保险丝的规定值t I 2远在产生导体温度的t I 2最大限度之下。

尽管如此，导体的规定值t I 2在保险丝的选择时是一个重要的因素。

26.25 在建筑内的电子设备安装电子分布式系统在建筑内是消费者和电能初始源之间最后的链接。

所有这一类建筑内部的分布式系统，不管他们大或小，必须符合一些基本要求：1. 安全性a. 电击的保护b. 对导体物理损伤的保护c. 对超载保护d. 对恶劣环境的保护2. 导体的电压降它不应该超出1或2%3. 估计寿命分布式系统应该持续最小限度于50年4. 经济性电力设施的费用应该是在考虑到当前一般标准的最小值。

标准由全国电子代码规定，并且电子设施在它可以被放入服务之前必须由审查员批准。

26.26 电力设施的主要成分许多成分组成了电子设施。

结构图Figs.26.47和26.48，与以下定义一起，将帮助读者了解某些更加重要的项目的意义。

1. 服务导体这些是在消费者之前从街道主馈线延长或从变压器对服务设备的导体。

2. 服务设备这些是必要的设备，通常包括开关或开关和保险丝以及他们的辅助部件，位于大厦或者其它结构服务导体的入口处，或是一个其他方面被定义的区域和意欲构成主控制和切断供应的方法。

3. 会议设备各种各样的记录在这个前提下预测电能消耗。

4 镶板一个盘区或小组盘区单位为汇编设计了以一个唯一盘区的形式; 包括公共汽车，自动过载电流设备和有或没有为光、热或者功率电路控制的开关；27.1 核电的安全尽管核电厂的安全与设计阶段的决策.施工前的谨慎的研究以及施工本身有着直接关系，但只有在电厂真正进入运行阶段，才有可能在各方面看到其核安全状况，如：由于其设计缺陷出现的事故危险质量不佳或运行故障。

保持安全水平为了保持安全水平，首先必须遵守设计所规定的运行限制。

这些限制在《安全分析报告》的“技术操作说明”中作了明确的陈述。

按照安全重要系统（IPS）进行一项详尽无疑的分析，同样也是为了确定哪一种设备需要进行定期的检测，哪一项无需检测（正常运行就是运行状况良好的充分证明）。

对每个系统和每次检测来说，《检测程序》文件对操作程序，操作标准和要求检测的周期等都作了明确的规定，这些规定成了运行人员制定检测计划表的基础。

提高安全水平，操作经验反馈由EDF采用的标准政策（EDF是一种主要PWR系统类型和有限的单个电厂序列数量）可保证通过特别重视预兆的分析极大的提高操作经验的价值。

这有双重目的：1．减少能够引起发电设备不能用的事故发生频率，即使这些事故可能对安全没有直接关系；2．减少可能对安全造成后果的严重事故的发生频率，实际上，以“多层保护”和故障存在为基础的核机组设计原理，认为严重的事故只能是那些单独事故和那些所谓不可能发生的事故同时发生的产物。

从核安全的角度考虑，操作反馈的主要目标如下：1确认可导致更为严重事故的预兆事故，以明确和贯彻在这类事故发生前所需采取的改正措施；2利用标准的机组（事故的一般方面，从修改研究等方面获取最大的效益）3如有必须修改之处，那么在将原理推广到整个电厂序列之前，要保证不会产生相反的副作用；4要利用在机组实际运行中所获得的数据，以统一安全标准，特别是对新厂更需要如此。

EDF是按如下方针制定其操作经验反馈系统的：1系统地收集尽可能多的数据资料，尽可能广泛地交流收集到的数据，特别是在EDF范围内；2在数据分析过程中，要尽可能地让设计人员，运行人员，制造厂商和安全机构一起参与；3吸取经验，这不仅有益于那些正在计划和建设中的机组，而且有益于那些目前正在运行的机组。

A1.2外文资料题目26.22 Ground-fault circuit breakerThe grounding methods we have covered so far are usually adequate,but further safety measures are needed in some cases.Suppose foe example , that a person sticks his finger into a lamp socket (Fig.26.45).Although the metal enclosure is securely grounded, the person will still receive a painful shock. Or suppose a 120 V eletric toster tumbles into a swimming pool. The heating elements and contacts will produce a hazardos leakage current throughout the pool ,ven if the frame of the toaster is securely grounded. Devices have been developed that will cut the source of power as soon as such accidents occur. These ground-fault circuit breaker will typically trip in 25 ms if the leakage current exceeds 5 mA. How do these protective devices operate ?A small current transformer surrounds the live and neutral wires as shown in Fig.26.46. The secondary is connected to a sensitive electronic detetor that can trigged a circuit breaker CB that is in series with the 120 V line.Under normal conditions the current I w in the line conductor is exactly equal to thecurrent I n in the neutral ,and so the net current (I W-I N) flowing through the hole in the toroidal core is zero. Consequently, no flux is produced in the core, the induced voltage E F is zero, and breaker CB does not trip.Suppose now that a fault current I F leaks directly from the live wire to ground . This could happen if someone touched a live terminal (Fig.26.45). A fault current I L would also be produced if the insulation broke down between a motor and its grounded enclosure . Under any of these conditions, the net current flowing through the hole of the CT is no longer zero but equal to I F or I L. A flux is set up and a voltage E F is induced, which trips CB. Because an imbalance of only 5 mA has to be detected, the core of the transformer must be verypermeable at low flux densities. Supermalloy TM is often used for this purpose because it has a relative permeability of typically 70000 at a flux density of only 4mT .26.23 rapid conductor heating: the I2t factorIt sometimes happens that a current far greater than normal flows for a brief period in a conductor . The I 2losses are than very large and the temperature of the conductor can rise several hundred degrees in a fraction Rof a second . For example, during a severe short-circuit, intense currents can flow in conductors and cables before the circuit is opened by the fuse or circuit breaker.Furthermore, the heat does not have time to be dissipated to the surroundings and so the temperature of the conductor increases very rapidly. What is the temperature rise under these condition?Suppose the conductor has a mass m, a resistance R, and a thermal heat capacity c. Moreover, suppose the current is I and that it flows for a period t that is typically less than 15 seconds. The heat generated in the conductor is given byRt I Q 2=Form Eq.3.17,we can calculate the temperature rise t ∆for a given value of Q :t mc Q ∆=hencet mc Rt I ∆=2from which)(2t I mcR t =∆ If follows that for a given conductor the temperature rise depends upon the I t 2 factor .It is well known that high temperature damage the insulation that covers a conductor. The I t 2 factor is ,therefore ,very important because it determines the temperature rise under short-circuit conditions. For example , a No.2AWG copper conductor, initially at a temperature of 90C ︒,cannot endure an It 2 factor in eccess of 222610A ⨯s if its temperature is to be limited to 250C ︒during a short-circuit.In general, the I t 2 factor can be calculated knowing (a) the cross section of the conductor, (b) itscomposition(copper or aluminum), and (c) the maximum temperature it can tolerate. The It 2 factor for copperand aluminum conductors are given by the following equations:for copper conductors, )234234(log 105.11010242θθ++⨯=m A t Ifor aluminum conductors,)234234(log 102.5010242θθ++⨯=m A t I whereI =short-circuit current []A=t duration of the short-circuit []sA = net cross-section of conductor without counting the empty spaces []2mm =0θ initial temperature of conductor []C ︒=m θfinal temperature of conductor []C ︒Example 26-1___________________________________________An overhead line made of aluminum conductor No.3 AWG has a cross-section of 26.6mm 2. Under normal conditions this conductor can continuously carry a current of 160 A. ,a Calculate the manximum permissible I t 2 factor during a short-circuit, knowing that the initial temperature is 80C ︒and that the manximum temperature should not exceed 250C ︒.,b A manximum short-circuit of 2000 A is foreseen on this overhead line. For how long can it circulate without exceeding the 250C ︒ temperature limit?Solution,a Using Eq.26.5 we find)234234(log 102.5010242θθ++⨯=m A t I =80234250234(log 102.51024++⨯A )=7s A 2610⨯,b The 2000 A current can flow for a time t given byI t 2=7610⨯2000t 2 =7610⨯t = 1.75sExample 26-1___________________________________________It is proposed to use a No.30 AWG copper wire as a temporary fuse. If its initial temperature is 50C ︒, calculate the following :,a The I t 2 needed to melt the wire ( copper melts at 1083C ︒),b The time needed to melt the wire if the short circuit current is 30 ASolutionForm Eq.26.4 we have)502341083234(log 0507.0105.1110242++⨯⨯=t I =197 A s 2,b For a current of 30 A we obtainI t 2=197301972=tt=0.22sThus, the fuse will blow in approximately 220ms.26.24 The role of fusesIn order to protect a conductor from excessive temperature during a short-circuit, a fuse must be placed in series with the conductor. The fuse must be selected so that its I t2rating is less than that which will protect an excessive temperature rise of the conductor. In effect, we want the fuse to blow before the conductor attains a dangerous temperature, usually taken to be 250C . In practice, the I t2rating of the fuse is such as to product conductor temperatures far below this maximum limit. Nevertheless, the I t2 rating of the conductor is an important element in the choice of the fuse.26.25 Electrical installation in buildingsThe electrical distribution system in a building is the final link between the consumer and the original source of electrical energy. All such in-house distribution systems, be they large or small, must meet certain basic requirements:1.Safetya.Protection against electric shockb.Protection of conductors against physical damagec.Protection against overloadsd.Protection against hostile environments2.Conductor voltage dropIt should not exceed 1 or 2 percent3.Life expectancyThe distribution system should last a minimum of 50 years4.EconomyThe cost of the installation should be minimized while observing the pertinent standards.Standards are set by the National Electrical Code and every electrical installation must be approved by an inspector before it can be put into service.26.26 Principal components of an electrical installationMany components are used in the markup of an electrical installation. The block diagrams of Figs.26.47 and 26.48, together with the following definitions, will help the reader understand the purpose of some of the more important items.1.Service Conductors. These are the conductors that extend form the street main or form atransformer to the service equipment on the consumer premises.2.Service Equipment. The necessary equipment, usually consisting of a circuit breaker or switch andfuses, and their accessories, located near the point of entrance of service conductors to a building orother structure, or an otherwise defined area, and intended to constitute the main control and means ofcutoff of the supply.3.Meeting Equipment. Various meters and recorders to indicate the electrical energy consumed on thepremises.4.Panel Board.A single panel or group of panel units designed for assembly in the form of a singlepanel; including buses, automatic over current devices, and with or without switches for the control oflight, heat, or power circuits;5.SafetyAlthough the safety of a nuclear power plant is directly linked to the decision taken at the design stage and to the care taken in pre-construction studies and construction itself, it is only really in operation that nuclear safety can be seen in all its various facets, i.e. the risk of accident due to weaknesses in the design, insufficient quality or operational error.Maintaining the Safety levelIn order to maintain the level of safety, it is first of all necessary to observe the operational limits of the design. These limitatio ns are set out in the “Technical Operating Specifications” volume of the Safety Analysis Report.An exhaustive analysis is also carried out on systems Important For Safety (IPS) to identify which equipment is required to undergo periodic testing and to justify which items do not require testing (normal operation is sufficient proof of good working order). For each system and test the “Test Procedure” document specifies the operating procedure, criteria and the required testing intervals. These instructions are used by the operators as the basis for preparing “Test Schedules”.27.1 Safety of Nuclear electricityThe standardization policy adopted by EDF (one major PWR system type and a limited number of individual plant series) justifies a major commitment to maximizing the value of operational experience by placing particular emphasis on the analysis of precursor incidents. This has a dual purpose:●To reduce the frequency of serious incidents liable to result in the generating equipment becomingunavailable, even if these is no direct bearing on safety.●To reduce the frequency of incidents likely to have consequences to safety , in practice, the design principlesfor nuclear units based on Defence In Depth and the presence of barriers are such that serious accidents could only result from a simultaneous combination of independent and improbable incidents.In terms of nuclear safety, the principal objectives of operational feedback are as follows:# to identify precursor incidents leading to more serious accidents in order to define and implement anycorrective measures required before any such accidents occur.# to take adventage of standardized units (the generic aspects of incidents, deriving maximum benefit from modification studies, ect )# to ensure, in cases where modification is necessary, that these are no adverse secondary effects before principle is extended to the entire plant series.# to utilize data gathered during actual operation of the installation in order to unify the level of safety, especially in the case of a new plant series.The guidelines on which EDF has based its operational experience feedback system are as follows:i.To systematically gather the maximum amount of data and circulate it as widely as possible, particularlywithin EDF.ii.To bring together as many as possible Designers. Operators, Manufacturers and Safety Authorities in the data analysis process.iii.To lean from experience on order to benefit not only those units planned or under construction but also units currently in service.。