某變電所畢業(yè)設(shè)計的中英文對照（中英文翻譯）.doc

xxx：xf110kv變電所設(shè)計摘 要xf 110kv變電所是地區(qū)重要變電所，是電力系統(tǒng)110kv電壓等級的重要部分。其設(shè)計分為電氣一次部分和電氣二次部分設(shè)計。一次部分由說明書，計算書與電氣工程圖組成，說明書和計算書包括變電所總體分析;負(fù)荷分析與主變選擇;電氣主接線設(shè)計;短路電流計算;電氣設(shè)備選擇;配電裝置選擇;變電所總平設(shè)計及防雷保護(hù)設(shè)計。二次部分由說明書，計算書與電氣工程圖組成。說明書和計算書包括整體概述;線路保護(hù)的整定計算;主變壓器的保護(hù)整定計算;電容器的保護(hù)整定計算;母線保護(hù)和所用變保護(hù)設(shè)計。計算書和電氣工程圖為附錄部分。其中一次部分電氣autocad制圖六張；二次部分為四張手工制圖。本變電所設(shè)計為畢業(yè)設(shè)計課題，以鞏固大學(xué)所學(xué)知識。通過本次設(shè)計，使我對電氣工程及其自動化專業(yè)的主干課程有一個較為全面，系統(tǒng)的掌握，增強(qiáng)了理論聯(lián)系實(shí)際的能力，提高了工程意識，鍛煉了我獨(dú)立分析和解決電力工程設(shè)計問題的能力，為未來的實(shí)際工作奠定了必要的基礎(chǔ)。關(guān)鍵詞: 、變電所 、變壓器 、繼電保護(hù)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 separated 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 the main transformer, layout of configuration, computation of short circuit; select of electric 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 electrical 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 and 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 substation for graduation design test. it can strengthen our specified knowledge.key-words: substation transformer relay protection謝 辭首先，在設(shè)計前的理論學(xué)習(xí)和實(shí)驗(yàn)環(huán)節(jié)中，劉憲林、王克文、陳根永、孔斌、包毅等專業(yè)課和實(shí)驗(yàn)指導(dǎo)老師的教導(dǎo)為我提供了豐富的專業(yè)理論知識和實(shí)踐分析能力。在本次設(shè)計的近一個學(xué)期中，和極其認(rèn)真負(fù)責(zé)的輔導(dǎo)和耐心的解答幫助我解決了一個個的難題。在此要對老師們不辭勞苦的工作和無私奉獻(xiàn)的精神表示衷心的感謝！在本次設(shè)計過程中，特別要感謝同寢室的同學(xué)、及同組的同學(xué)，他們的幫助讓這次設(shè)計變得輕松了許多。設(shè)計中雖然充分采納了老師和同學(xué)們的意見，幾經(jīng)修改，但由于是初次設(shè)計，加之自身水平有限，設(shè)計及論述過程中難免有錯誤，請各位老師批評指正。附錄1：外文資料翻譯a1.1譯文變壓器1. 介紹要從遠(yuǎn)端發(fā)電廠送出電能，必須應(yīng)用高壓輸電。因?yàn)樽罱K的負(fù)荷，在一些點(diǎn)高電壓必須降低。變壓器能使電力系統(tǒng)各個部分運(yùn)行在電壓不同的等級。本文我們討論的原則和電力變壓器的應(yīng)用。2. 雙繞組變壓器變壓器的最簡單形式包括兩個磁通相互耦合的固定線圈。兩個線圈之所以相互耦合，是因?yàn)樗鼈冞B接著共同的磁通。在電力應(yīng)用中，使用層式鐵芯變壓器(本文中提到的)。變壓器是高效率的，因?yàn)樗鼪]有旋轉(zhuǎn)損失，因此在電壓等級轉(zhuǎn)換的過程中，能量損失比較少。典型的效率范圍在92到99%，上限值適用于大功率變壓器。從交流電源流入電流的一側(cè)被稱為變壓器的一次側(cè)繞組或者是原邊。它在鐵圈中建立了磁通，它的幅值和方向都會發(fā)生周期性的變化。磁通連接的第二個繞組被稱為變壓器的二次側(cè)繞組或者是副邊。磁通是變化的；因此依據(jù)楞次定律，電磁感應(yīng)在二次側(cè)產(chǎn)生了電壓。變壓器在原邊接收電能的同時也在向副邊所帶的負(fù)荷輸送電能。這就是變壓器的作用。3. 變壓器的工作原理當(dāng)二次側(cè)電路開路是，即使原邊被施以正弦電壓vp，也是沒有能量轉(zhuǎn)移的。外加電壓在一次側(cè)繞組中產(chǎn)生一個小電流i。這個空載電流有兩項(xiàng)功能：（1）在鐵芯中產(chǎn)生電磁通，該磁通在零和m之間做正弦變化，m是鐵芯磁通的最大值；（2）它的一個分量說明了鐵芯中的渦流和磁滯損耗。這兩種相關(guān)的損耗被稱為鐵芯損耗。變壓器空載電流i一般大約只有滿載電流的2%5%。因?yàn)樵诳蛰d時，原邊繞組中的鐵芯相當(dāng)于一個很大的電抗，空載電流的相位大約將滯后于原邊電壓相位90。顯然可見電流分量im= i0sin0，被稱做勵磁電流，它在相位上滯后于原邊電壓vp 90。就是這個分量在鐵芯中建立了磁通；因此磁通與im同相。第二個分量ie=i0sin0，與原邊電壓同相。這個電流分量向鐵芯提供用于損耗的電流。兩個相量的分量和代表空載電流，即i0 = im+ ie應(yīng)注意的是空載電流是畸變和非正弦形的。這種情況是非線性鐵芯材料造成的。如果假定變壓器中沒有其他的電能損耗一次側(cè)的感應(yīng)電動勢ep和二次側(cè)的感應(yīng)電壓es可以表示出來。因?yàn)橐淮蝹?cè)繞組中的磁通會通過二次繞組，依據(jù)法拉第電磁感應(yīng)定律，二次側(cè)繞組中將產(chǎn)生一個電動勢e，即e=n/t。相同的磁通會通過原邊自身，產(chǎn)生一個電動勢ep。正如前文中討論到的，所產(chǎn)生的電壓必定滯后于磁通90，因此，它于施加的電壓有180的相位差。因?yàn)闆]有電流流過二次側(cè)繞組，es=vs。一次側(cè)空載電流很小，僅為滿載電流的百分之幾。因此原邊電壓很小，并且vp的值近乎等于ep。原邊的電壓和它產(chǎn)生的磁通波形是正弦形的；因此產(chǎn)生電動勢ep和es的值是做正弦變化的。產(chǎn)生電壓的平均值如下eavg = turns即是法拉第定律在瞬時時間里的應(yīng)用。它遵循eavg = n = 4fnm其中n是指線圈的匝數(shù)。從交流電原理可知，有效值是一個正弦波，其值為平均電壓的1.11倍；因此e = 4.44fnm因?yàn)橐淮蝹?cè)繞組和二次側(cè)繞組的磁通相等，所以繞組中每匝的電壓也相同。因此ep = 4.44fnpm并且es = 4.44fnsm其中np和es是一次側(cè)繞組和二次側(cè)繞組的匝數(shù)。一次側(cè)和二次側(cè)電壓增長的比率稱做變比。用字母a來表示這個比率，如下式a = = 假設(shè)變壓器輸出電能等于其輸入電能這個假設(shè)適用于高效率的變壓器。實(shí)際上我們是考慮一臺理想狀態(tài)下的變壓器；這意味著它沒有任何損耗。因此pm = pout或者vpip primary pf = vsis secondary pf這里pf代表功率因素。在上面公式中一次側(cè)和二次側(cè)的功率因素是相等的；因此vpip = vsis從上式我們可以得知 = a它表明端電壓比等于匝數(shù)比，換句話說，一次側(cè)和二次側(cè)電流比與匝數(shù)比成反比。匝數(shù)比可以衡量二次側(cè)電壓相對于一次惻電壓是升高或者是降低。為了計算電壓，我們需要更多數(shù)據(jù)。 終端電壓的比率變化有些根據(jù)負(fù)載和它的功率因素。實(shí)際上, 變比從標(biāo)識牌數(shù)據(jù)獲得, 列出在滿載情況下原邊和副邊電壓。 當(dāng)副邊電壓vs相對于原邊電壓減小時，這個變壓器就叫做降壓變壓器。如果這個電壓是升高的，它就是一個升壓變壓器。在一個降壓變壓器中傳輸變比a遠(yuǎn)大于1(a1.0)，同樣的，一個升壓變壓器的變比小于1(a1.0)。當(dāng)a=1時，變壓器的二次側(cè)電壓就等于起一次側(cè)電壓。這是一種特殊類型的變壓器，可被應(yīng)用于當(dāng)一次側(cè)和二次側(cè)需要相互絕緣以維持相同的電壓等級的狀況下。因此，我們把這種類型的變壓器稱為絕緣型變壓器。顯然，鐵芯中的電磁通形成了連接原邊和副邊的回路。在第四部分我們會了解到當(dāng)變壓器帶負(fù)荷運(yùn)行時一次側(cè)繞組電流是如何隨著二次側(cè)負(fù)荷電流變化而變化的。從電源側(cè)來看變壓器，其阻抗可認(rèn)為等于vp / ip。從等式 = a中我們可知vp = avs并且ip = is/a。根據(jù)vs和is，可得vp和ip的比例是 = = 但是vs / is 負(fù)荷阻抗zl，因此我們可以這樣表示zm (primary) = a2zl這個等式表明二次側(cè)連接的阻抗折算到電源側(cè)，其值為原來的a2倍。我們把這種折算方式稱為負(fù)載阻抗向一次側(cè)的折算。這個公式應(yīng)用于變壓器的阻抗匹配。4. 有載情況下的變壓器一次側(cè)電壓和二次側(cè)電壓有著相同的極性，一般習(xí)慣上用點(diǎn)記號表示。如果點(diǎn)號同在線圈的上端，就意味著它們的極性相同。因此當(dāng)二次側(cè)連接著一個負(fù)載時，在瞬間就有一個負(fù)荷電流沿著這個方向產(chǎn)生。換句話說，極性的標(biāo)注可以表明當(dāng)電流流過兩側(cè)的線圈時，線圈中的磁動勢會增加。因?yàn)槎蝹?cè)電壓的大小取決于鐵芯磁通大小0，所以很顯然當(dāng)正常情況下負(fù)載電勢es沒有變化時，二次電壓也不會有明顯的變化。當(dāng)變壓器帶負(fù)荷運(yùn)行時，將有電流is流過二次側(cè)，因?yàn)閑s產(chǎn)生的感應(yīng)電動勢相當(dāng)于一個電壓源。二次側(cè)電流產(chǎn)生的磁動勢nsis會產(chǎn)生一個勵磁。這個磁通的方向在任何一個時刻都和主磁通反向。當(dāng)然，這是楞次定律的體現(xiàn)。因此，nsis所產(chǎn)生的磁動勢會使主磁通0減小。這意味著一次側(cè)線圈中的磁通減少，因而它的電壓ep將會增大。感應(yīng)電壓的減小將使外施電壓和感應(yīng)電動勢之間的差值更大，它將使初級線圈中流過更大的電流。初級線圈中的電流ip的增大，意味著前面所說明的兩個條件都滿足：（1）輸出功率將隨著輸出功率的增加而增加（2）初級線圈中的磁動勢將增加，以此來抵消二次側(cè)中的磁動勢減小磁通的趨勢?？偟膩碚f，變壓器為了保持磁通是常數(shù)，對磁通變化的響應(yīng)是瞬時的。更重要的是，在空載和滿載時，主磁通0的降落是很少的（一般在）1至3%。其需要的條件是e降落很多來使電流ip增加。在一次側(cè)，電流ip在一次側(cè)流過以平衡is產(chǎn)生的影響。它的磁動勢npip只停留在一次側(cè)。因?yàn)殍F芯的磁通0保持不變，變壓器空載時空載電流i0必定會為其提供能量。故一次側(cè)電流ip是電流ip與i0的和。因?yàn)榭蛰d電流相對較小，那么一次側(cè)的安匝數(shù)與二次側(cè)的安匝數(shù)相等的假設(shè)是成立的。因?yàn)樵谶@種狀況下鐵芯的磁通是恒定的。因此我們?nèi)耘f可以認(rèn)定空載電流i0相對于滿載電流是極其小的。當(dāng)一個電流流過二次側(cè)繞組，它的磁動勢（nsis）將產(chǎn)生一個磁通，于空載電流i0產(chǎn)生的磁通0不同，它只停留在二次側(cè)繞組中。因?yàn)檫@個磁通不流過一次側(cè)繞組，所以它不是一個公共磁通。另外，流過一次側(cè)繞組的負(fù)載電流只在一次側(cè)繞組中產(chǎn)生磁通，這個磁通被稱為一次側(cè)的漏磁。二次側(cè)漏磁將使電壓增大以保持兩側(cè)電壓的平衡。一次側(cè)漏磁也一樣。因此，這兩個增大的電壓具有電壓降的性質(zhì)，總稱為漏電抗電壓降。另外，兩側(cè)繞組同樣具有阻抗，這也將產(chǎn)生一個電阻壓降。把這些附加的電壓降也考慮在內(nèi)，這樣一個實(shí)際的變壓器的等值電路圖就完成了。由于分支勵磁體現(xiàn)在電流里，為了分析我們可以將它忽略。這就符我們前面計算中可以忽略空載電流的假設(shè)。這證明了它對我們分析變壓器時所產(chǎn)生的影響微乎其微。因?yàn)殡妷航蹬c負(fù)載電流成比例關(guān)系，這就意味著空載情況下一次側(cè)和二次側(cè)繞組的電壓降都為零。譯自a1.2原文transformer1. introductionthe 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 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 common flux.in power applications, laminated steel core transformers (to which this paper is restricted) are used. transformers are efficient because the rotational losses normally associated with rotating machine are absent, so relatively little power 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.the 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 changing; 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 sinusoidal 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 sinusoidally 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 normally referred to as the core losses.the no-load current i is usually few percent of the 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, 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 components 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 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 induced 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 ep. the primary voltage and the resulting flux are sinusoidal; thus the induced quantities ep and es vary as a sine function. the average value of the induced voltage given byeavg = turnswhich is faradays law applied to a finite time interval. it follows thateavg = n = 4fnmwhich n is the number of turns on the winding. form ac circuit theory, the effective or root-mean-square (rms) voltage for a sine wave 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 = 4.44fnpmandes = 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 is called the transformation ratio. denoting this ratio by a, it is seen thata = = assume that the output power of a transformer equals its input power, not a bad sumption in practice considering the high efficiencies. what we really are saying is that we are dealing with an ideal transformer; that is, it has no losses. thuspm = poutorvpip primary pf = vsis secondary pfwhere pf is the power factor. for the above-stated assumption it means that the power factor on primary and secondary sides are equal; thereforevpip = vsisfrom which is obtained = ait shows that as an approximation the terminal voltage ratio equals the turns ratio. the primary and secondary current, on the other hand, are inversely related to the turns ratio. the turns ratio gives a measure of how much the secondary voltage is raised or lowered in relation to the primary voltage. to calculate the voltage regulation, we need more information.the ratio of the terminal voltage varies somewhat depending on the load and its power factor. in practice, the transformation ratio is obtained from the nameplate data, which list the primary and secondary voltage under full-load condition.when the secondary voltage vs is reduced compared to the primary voltage, the transformation is said to be a step-down transformer: conversely, if this voltage is raised, it is called a step-up transformer. in a step-down transformer the transformation ratio a is greater than unity (a1.0), while for a step-up transformer it is smaller than unity (a1.0). in the event that a=1, the transformer secondary voltage equals the primary voltage. this is a special type of transformer used in instances where electrical isolation is required between the primary and secondary circuit while maintaining the same voltage level. therefore, this transformer is generally knows as an isolation transformer.as is apparent, it is the magnetic flux in the core that forms the connecting link between primary and secondary circuit. in section 4 it is shown how the primary winding current adjusts itself to the secondary load current when the transformer supplies a load.looking into the transformer terminals from the source, an impedance is seen which by definition equals vp / ip. from = a , we have vp = avs and ip = is/a.in terms of vs and is the ratio of vp to ip is = = but vs / is is the load impedance zl thus we can say thatzm (primary) = a2zlthis equation tells us that when an impedance is connected to the secondary side, it appears from the source as an impedance having a magnitude that is a2 times its actual value. we say that the load impedance is reflected or referred to the primary. it is this property of transformers that is used in impedance-matching applications.4. transformers under loadthe primary and secondary voltages shown have similar polarities, as indicated by the “dot-making” convention. the dots near the upper ends of the windings have the same meaning as in circuit theory; the marked terminals have the same polarity. thus when a load is connected to the secondary, the instantaneous load current is in the direction shown. in other words, the polarity markings signify that when positive current enters both windings at the marked terminals, the mmfs of the two windings add.since the secondary voltage depends on the core flux 0, it must be clear that the flux should not change appreciably if es is to remain essentially constant under normal loading conditions. with the load connected, a current is will flow in the secondary circuit, because the induced emf es will act as a voltage source. the secondary current produces an mmf nsis that creates a flux. this flux has such a direction that at any instant in time it opposes the main flux that created it in the first place. of course, this is lenzs law in action. thus the mmf represented by nsis tends to reduce the core flux 0. this means that the flux linking the primary winding reduces and consequently the primary induced voltage ep, this reduction in induced voltage causes a greater difference between the impressed voltage and the counter induced emf, thereby allowing more current to flow in the primary. the fact that primary current ip increases means that the two conditions stated earlier are fulfilled: (1) the power input increases to match the power output, and (2) the primary mmf increases to offset the tendency of the secondary mmf to reduce the flux.in general, it will be found that the transformer reacts almost instantaneously to keep the resultant core flux essentially constant. moreover, the core flux 0 drops very slightly between n o load and full load (about 1 to 3%), a necessary condition if ep is to fall sufficiently to allow an increase in ip.on the primary side, ip is the current that flows in the primary to balance the demagnetizing effect of is. its mmf npip sets up a flux linking the primary only. since the core flux 0 remains constant. i0 must be the same current that energizes the transformer at no load. the primary current ip is therefore the sum of the current ip and i0.because the no-load current is relatively small, it is correct to assume that the primary ampere-turns equa