1、完整版某变电所毕业设计的中英文对照中英文翻译要点XXX：XF110KV变电所设计摘 要XF 110KV变电所是地区重要变电所，是电力系统110KV电压等级的重要部分。其设计分为电气一次部分和电气二次部分设计。总体分析;配电装置选择;二次部分由说明书，计算书与电气工程图组成。说明书和计算书包括整体概述;线路保护的整定计算;主变压器的保护整定计算;电容器的保护整定计算;母线保护和所用变保护设计。计算书和电气工程图为附录部分。其中一次部分电气AutoCAD制图六张；二次部分为四张手工制图。本变电所设计为毕业设计课题，以巩固大学所学知识。通过本次设计，使我对电气工程及其自动化专业的主干课程有一个较为全
2、面，系统的掌握，增强了理论联系实际的能力，提高了工程意识，锻炼了我独立分析和解决电力工程设计问题的能力，为未来的实际工作奠定了必要的基础。关键词: 、变电所 、变压器 、继电保护AbstractXF county 110KV substation is an important station in this distract, which is one of the extremely necessary parts of the 110KV network in electric power system.The design of the substation can be separa
3、ted in two parts: primary part and secondary part of the electric design.The first part consists of specifications, computation book and Electrical engineering drawings about the design. The specifications has several parts which are General analysis of the station, Load analysis, The selection of t
4、he main transformer, Layout of configuration, Computation of short circuit; Select of electric郑州大学电气工程学院毕业论文devices, Power distribution devices, General design of substation plane and the design of thunderbolt protection.The second part also consists of specifications, computation book and electrica
5、l drawings about the design。 Specifications and computation book include following section: General, The evaluation and calculate of line protection, Transformer protection, capacitor protection, Bus protection and Self-using transformer protection.Computation book, Electrical engineering drawings a
6、nd catalogue of drawings are attached in the end。 There are nine drawings total, in which four are prepared by hand, others are prepared by computer in which installed the software electrical AutoCAD. From other view, it also can be classified as first part and second part.This is a design of substa
7、tion for graduation design test. It can strengthen our specified knowledge.Key-words: substation transformer Relay protection谢 辞首先，在设计前的理论学习和实验环节中，刘宪林、王克文、陈根永、孔斌、包毅等专业课和实验指导老师的教导为我提供了丰富的专业理论知识和实践分析能力。在本次设计的近一个学期中，和极其认真负责的辅导和耐心的解答帮助我解决了一个个的难题。在此要对老师们不辞劳苦的工作和无私奉献的精神表示衷心的感谢！在本次设计过程中，特别要感谢同寝室的同学、及同组的同学，
8、他们的帮助让这次设计变得轻松了许多。设计中虽然充分采纳了老师和同学们的意见，几经修改，但由于是初次设计，加之自身水平有限，设计及论述过程中难免有错误，请各位老师批评指正。附录1：外文资料翻译A1.1译文变压器1. 介绍 2XXX：XF110KV变电所设计们讨论的原则和电力变压器的应用。2. 双绕组变压器变压器的最简单形式包括两个磁通相互耦合的固定线圈。两个线圈之所以相互耦合，是因为它们连接着共同的磁通。在电力应用中，使用层式铁芯变压器(本文中提到的)。变压器是高效率的，因为它没有旋转损失，因此在电压等级转换的过程中，能量损失比较少。典型的效率范围在92到99%，上限值适用于大功率变压器。从交流
9、电源流入电流的一侧被称为变压器的一次侧绕组或者是原边。它在铁圈中建立了磁通，它的幅值和方向都会发生周期性的变化。磁通连接的第二个绕组被称为变压器的二次侧绕组或者是副边。磁通是变化的；因此依据楞次定律，电磁感应在二次侧产生了电压。变压器在原边接收电能的同时也在向副边所带的负荷输送电能。这就是变压器的作用。3. 变压器的工作原理当二次侧电路开路是，即使原边被施以正弦电压Vp，也是没有能量转移的。外加电压在一次侧绕组中产生一个小电流I。这个空载电流有两项功能：（1）在铁芯中产生电磁通，该磁通在零和 m之间做正弦变化，m是铁芯磁通的最大值；（2）它的一个分量说明了铁芯中的涡流和磁滞损耗。这两种相关的损
10、耗被称为铁芯损耗。变压器空载电流I一般大约只有满载电流的2%5%。因为在空载时，原边绕组中的铁芯相当于一个很大的电抗，空载电流的相位大约将滞后于原边电压相位90。显然可见电流分量Im= I0sin0，被称做励磁电流，它在相位上滞后于原边电压VP 90。就是这个分量在铁芯中建立了磁通；因此磁通与Im同相。第二个分量Ie=I0sin0，与原边电压同相。这个电流分量向铁芯提供用于损耗0e应注意的是空载电流是畸变和非正弦形的。这种情况是非线性铁芯材料造成的。如果假定变压器中没有其他的电能损耗一次侧的感应电动势Ep和二次侧的感应电压Es可以表示出来。因为一次侧绕组中的磁通会通过二次绕组，依据法拉第电磁感
11、应定律，二次侧绕组中将产生一个电动势E，即E=N/t。相同的磁通会通过原边自身，产生一个电动势Ep。正如前文中讨论到的，所产生的电压必定滞后于磁通90，因此，它于施加的电压有180的相位差。因为没有电流流过二次侧绕组，Es=Vs。一次侧空载电流很小，仅为满载电流的百分之几。因此原边电压很小，并且Vp的值近乎等于Ep。原边的电压和它产生的磁通波形是正弦形的；因此郑州大学电气工程学院毕业论文产生电动势Ep和Es的值是做正弦变化的。产生电压的平均值如下给定时间内磁通变化量Eavg 给定时间即是法拉第定律在瞬时时间里的应用。它遵循Eavg = N2 m = 4fNm 1/(2f)其中N是指线圈的匝数。
12、从交流电原理可知，有效值是一个正弦波，其值为平均电压的1.11倍；因此E = 4.44fNm因为一次侧绕组和二次侧绕组的磁通相等，所以绕组中每匝的电压也相同。因此Ep = 4.44fNpm并且Es = 4.44fNsm其中Np和Es是一次侧绕组和二次侧绕组的匝数。一次侧和二次侧电压增长的比率称做变比。用字母a来表示这个比率，如下式 a = EpNp = NsEs假设变压器输出电能等于其输入电能这个假设适用于高效率的变压器。实际上我们是考虑一台理想状态下的变压器；这意味着它没有任何损耗。因此Pm = Pout或者VpIp primary PF = VsIs secondary PF这里PF代表功
13、率因素。在上面公式中一次侧和二次侧的功率因素是相等的；因此VpIp = VsIs从上式我们可以得知VpIpEp = a VsIsEs它表明端电压比等于匝数比，换句话说，一次侧和二次侧电流比与匝数比成当副边电压Vs相对于原边电压减小时，这个变压器就叫做降压变压器。如果这个电压是升高的，它就是一个升压变压器。在一个降压变压器中传输变比a远大于1(a1.0)，同样的，一个升压变压器的变比小于1(a1.0)。当a=1时，变压器 4XXX：XF110KV变电所设计的二次侧电压就等于起一次侧电压。这是一种特殊类型的变压器，可被应用于当一次侧和二次侧需要相互绝缘以维持相同的电压等级的状况下。因此，我们把这种
14、类型的变压器称为绝缘型变压器。显然，铁芯中的电磁通形成了连接原边和副边的回路。在第四部分我们会了解到当变压器带负荷运行时一次侧绕组电流是如何随着二次侧负荷电流变化而变化的。从电源侧来看变压器，其阻抗可认为等于Vp / Ip。从等式 VpIpEp = VsIsEsa中我们可知Vp = aVs并且Ip = Is/a。根据Vs和Is，可得Vp和Ip的比例是aVsVpa2Vs = = Is/aIpIs但是Vs / Is 负荷阻抗ZL，因此我们可以这样表示Zm (primary) = a2ZL这个等式表明二次侧连接的阻抗折算到电源侧，其值为原来的a2倍。我们把这种折算方式称为负载阻抗向一次侧的折算。这个
15、公式应用于变压器的阻抗匹配。4. 有载情况下的变压器一次侧电压和二次侧电压有着相同的极性，一般习惯上用点记号表示。如果点号同在线圈的上端，就意味着它们的极性相同。因此当二次侧连接着一个负载时，在瞬间就有一个负荷电流沿着这个方向产生。换句话说，极性的标注可以表明当电流流过两侧的线圈时，线圈中的磁动势会增加。因为二次侧电压的大小取决于铁芯磁通大小0，所以很显然当正常情况下负载电势Es没有变化时，二次电压也不会有明显的变化。当变压器带负荷运行时，将有电流Is流过二次侧，因为Es产生的感应电动势相当于一个电压源。二次侧电流产生的磁动势NsIs会产生一个励磁。这个磁通的方向在任何一个时刻都和主磁通反向。
16、当然，这是楞次定律的体现。因此，NsIs所产生的磁动势会使主磁通0减小。这意味着一次侧线圈中的磁通减少，因而它的电压Ep将会增大。感应电压的减小将使外施电压和感应电动势之间的差值更大，它将使初级线圈中流过更大的电流。初级线圈中的电流Ip的增大，意味着前面所说明的两个条件都满足：（1）输出功率将随着输出功率的增加而增加（2）初级线圈中的磁动势将增加，以此来抵消二次侧中的磁动势减小磁通的趋势。总的来说，变压器为了保持磁通是常数，对磁通变化的响应是瞬时的。更重要的是，在空载和满载时，主磁通0的降落是很少的（一般在）1至3%。其需要的条件是E降落很多来使电流Ip增加。在一次侧，电流Ip在一次侧流过以平
17、衡Is产生的影响。它的磁动势NpIp只停郑州大学电气工程学院毕业论文留在一次侧。因为铁芯的磁通0保持不变，变压器空载时空载电流I0必定会为其提供能量。故一次侧电流Ip是电流Ip与I0的和。因为空载电流相对较小，那么一次侧的安匝数与二次侧的安匝数相等的假设是成立的。因为在这种状况下铁芯的磁通是恒定的。因此我们仍旧可以认定空载电流I0相对于满载电流是极其小的。当一个电流流过二次侧绕组，它的磁动势（NsIs）将产生一个磁通，于空载电流I0产生的磁通0不同，它只停留在二次侧绕组中。因为这个磁通不流过一次侧绕组，所以它不是一个公共磁通。另外，流过一次侧绕组的负载电流只在一次侧绕组中产生磁通，这个磁通被称
18、为一次侧的漏磁。二次侧漏磁将使电压增大以保持两侧电压的平衡。一次侧漏磁也一样。因此，这两个增大的电压具有电压降的性质，总称为漏电抗电压降。另外，两侧绕组同样具有阻抗，这也将产生一个电阻压降。把这些附加的电压降也考虑在内，这样一个实际的变压器的等值电路图就完成了。由于分支励磁体现在电流里，为了分析我们可以将它忽略。这就符我们前面计算中可以忽略空载电流的假设。这证明了它对我们分析变压器时所产生的影响微乎其微。因为电压降与负载电流成比例关系，这就意味着空载情况下一次侧和二次侧绕组的电压降都为零。译自XXX：XF110KV变电所设计A1.2原文TRANSFORMER1. INTRODUCTIONThe
19、 high-voltage transmission was need for the case electrical power is to be provided at considerable distance from a generating station. At some point this high voltage must be reduced, because ultimately is must supply a load. The transformer makes it possible for various parts of a power system to
20、operate at different voltage levels. In this paper we discuss power transformer principles and applications.2. TOW-WINDING TRANSFORMERSA transformer in its simplest form consists of two stationary coils coupled by a mutual magnetic flux. The coils are said to be mutually coupled because they link a
21、common flux.In power applications, restricted) are used. associated with rotating machine are absent, so relatively little is lost when transforming power from one voltage level to another. Typical efficiencies are in the range 92 to 99%, the higher values applying to the larger power transformers.T
22、he current flowing in the coil connected to the ac source is called the primary winding or simply the primary. It sets up the flux in the core, which varies periodically both in magnitude and direction. The flux links the second coil, called the secondary winding or simply secondary. The flux is cha
23、nging; therefore, it induces a voltage in the secondary by electromagnetic induction in accordance with Lenzs law. Thus the primary receives its power from the source while the secondary supplies this power to the load. This action is known as transformer action.3. TRANSFORMER PRINCIPLESWhen a sinus
24、oidal voltage Vp is applied to the primary with the secondary open-circuited, there will be no energy transfer. The impressed voltage causes a small current I to flow in the primary winding. This no-load current has two functions: (1) it produces the magnetic flux in the core, which varies sinusoida
25、lly between zero and m, where m is the maximum value of the core flux; and (2) it provides a component to account for the hysteresis and eddy current losses in the core. There combined losses are 7郑州大学电气工程学院毕业论文normally referred to as the core losses.The no-load current I is usually few percent of t
26、he rated full-load current of the transformer (about 2 to 5%). Since at no-load the primary winding acts as a large reactance due to the iron core, the no-load current will lag the primary voltage by nearly 90. It is readily seen that the current component Im= I0sin0, called the magnetizing current,
27、 is 90 in phase behind the primary voltage VP. It is this component that sets up the flux in the core; is therefore in phase with Im.The second component, Ie=I0sin0, is in phase with the primary voltage. It is the current component that supplies the core losses. The phasor sum of these two component
28、s represents the no-load current, orI0 = Im+ IeIt should be noted that the no-load current is distortes and nonsinusoidal. This is the result of the nonlinear behavior of the core material.If it is assumed that there are no other losses in the transformer, the induced voltage In the primary, Ep and
29、that in the secondary, Es can be shown. Since the magnetic flux set up by the primary winding，there will be an induced EMF E in the secondary winding in accordance with Faradays law, namely, E=N/t. This same flux also links the primary itself, inducing in it an EMF, Ep. As discussed earlier, the ind
30、uced voltage must lag the flux by 90, therefore, they are 180 out of phase with the applied voltage. Since no current flows in the secondary winding, Es=Vs. The no-load primary current I0 is small, a few percent of full-load current. Thus the voltage in the primary is small and Vp is nearly equal to
31、 Ep. The primary voltage and the resulting flux Eavg = turns given time2 m = 4fNm 1/(2f)which is Faradays law applied to a finite time interval. It follows that Eavg = Nwhich N is the number of turns on the winding. Form ac circuit theory, the effective or root-mean-square (rms) voltage for a sine w
32、ave is 1.11 times the average voltage; thusE = 4.44fNmSince the same flux links with the primary and secondary windings, the voltage per turn in each winding is the same. HenceEp = 4.44fNpmXXX：XF110KV变电所设计andEs = 4.44fNsmwhere Ep and Es are the number of turn on the primary and secondary windings, respectively. The ratio of primary to secondary induced voltage i
