某變電所畢業(yè)設(shè)計(jì)的中英文對照（中英文翻譯）

1、<p><b>　　摘 要</b></p><p>　　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(jí)的重要部分。其設(shè)計(jì)分為電氣一次部分和電氣二次部分設(shè)計(jì)。</p><p>　　一次部分由說明書，計(jì)算書與電氣工程圖組成，說明書和計(jì)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(jì);短路電流計(jì)算;電氣設(shè)備選擇;配電裝置選擇;變

2、電所總平設(shè)計(jì)及防雷保護(hù)設(shè)計(jì)。</p><p>　　二次部分由說明書，計(jì)算書與電氣工程圖組成。說明書和計(jì)算書包括整體概述;線路保護(hù)的整定計(jì)算;主變壓器的保護(hù)整定計(jì)算;電容器的保護(hù)整定計(jì)算;母線保護(hù)和所用變保護(hù)設(shè)計(jì)。</p><p>　　計(jì)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。</p><p>　　本變電所設(shè)計(jì)為畢業(yè)

3、設(shè)計(jì)課題，以鞏固大學(xué)所學(xué)知識(shí)。通過本次設(shè)計(jì)，使我對電氣工程及其自動(dòng)化專業(yè)的主干課程有一個(gè)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(shí)，鍛煉了我獨(dú)立分析和解決電力工程設(shè)計(jì)問題的能力，為未來的實(shí)際工作奠定了必要的基礎(chǔ)。</p><p>　　關(guān)鍵詞: Ⅰ、變電所 Ⅱ、變壓器 Ⅲ、繼電保護(hù)</p><p><b>　　Abstract</b>

4、</p><p>　　XF 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.</p><p>　　The design of

5、 the substation can be separated in two parts: primary part and secondary part of the electric design.</p><p>　　The first part consists of specifications, computation book and Electrical engineering drawings

6、 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

7、 electric devices, Power distribution devices, General design of substation plane and the design of thunderbolt protection.</p><p>　　The second part also consists of specifications, computation book and elec

8、trical 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

9、Self-using transformer protection.</p><p>　　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 h

10、and, 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.</p><p>　　This is a design of substation for

11、graduation design test. It can strengthen our specified knowledge.</p><p>　　Key-words: Ⅰsubstation Ⅱtransformer Ⅲ Relay protection</p><p><b>　　謝 辭</b></p><p>　　

12、首先，在設(shè)計(jì)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(shí)和實(shí)踐分析能力。在本次設(shè)計(jì)的近一個(gè)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(gè)個(gè)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！</p><p>　　在本次設(shè)計(jì)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(jì)變得輕松了許多。</p&

13、gt;<p>　　設(shè)計(jì)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(jì)，加之自身水平有限，設(shè)計(jì)及論述過程中難免有錯(cuò)誤，請各位老師批評指正。</p><p>　　附錄1：外文資料翻譯</p><p><b>　　A1.1譯文</b></p><p><b>　　變壓器</b></p>

14、<p><b>　　1. 介紹</b></p><p>　　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(gè)部分運(yùn)行在電壓不同的等級(jí)。本文我們討論的原則和電力變壓器的應(yīng)用。</p><p><b>　　2. 雙繞組變壓器</b></p><p>　　變壓器

15、的最簡單形式包括兩個(gè)磁通相互耦合的固定線圈。兩個(gè)線圈之所以相互耦合，是因?yàn)樗鼈冞B接著共同的磁通。</p><p>　　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(jí)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。</p><p>　　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

16、鐵圈中建立了磁通φ，它的幅值和方向都會(huì)發(fā)生周期性的變化。磁通連接的第二個(gè)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(shí)也在向副邊所帶的負(fù)荷輸送電能。這就是變壓器的作用。</p><p>　　3. 變壓器的工作原理</p><p>　　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

17、一次側(cè)繞組中產(chǎn)生一個(gè)小電流Iθ。這個(gè)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φm之間做正弦變化，φm是鐵芯磁通的最大值；（2）它的一個(gè)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。</p><p>　　變壓器空載電流Iθ一般大約只有滿載電流的2%—5%。因?yàn)樵诳蛰d時(shí)，原邊繞組中的鐵芯相當(dāng)于一個(gè)很大的電抗，空載電流的相位大約將滯后于原邊電壓相位90º。顯然可見電流分量

18、Im= I0sinθ0，被稱做勵(lì)磁電流，它在相位上滯后于原邊電壓VP 90º。就是這個(gè)分量在鐵芯中建立了磁通；因此磁通φ與Im同相。</p><p>　　第二個(gè)分量Ie=I0sinθ0，與原邊電壓同相。這個(gè)電流分量向鐵芯提供用于損耗的電流。兩個(gè)相量的分量和代表空載電流，即</p><p>　　I0 = Im+ Ie</p><p>　　應(yīng)注意的是空載電流是

19、畸變和非正弦形的。這種情況是非線性鐵芯材料造成的。</p><p>　　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(dòng)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(huì)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(gè)電動(dòng)勢E，即E=NΔφ/Δt。相同的磁通會(huì)通過原邊自身，產(chǎn)生一個(gè)電動(dòng)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90º，因此，它于施加的電壓有180

20、º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(dòng)勢Ep和Es的值是做正弦變化的。產(chǎn)生電壓的平均值如下</p><p>　　Eavg = turns×</p><p>　　即是法拉第定律在瞬時(shí)時(shí)間里的應(yīng)用。它遵循</p>

21、;<p>　　Eavg = N = 4fNφm</p><p>　　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(gè)正弦波，其值為平均電壓的1.11倍；因此</p><p>　　E = 4.44fNφm</p><p>　　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</p><p>　　Ep =

22、4.44fNpφm</p><p><b>　　并且</b></p><p>　　Es = 4.44fNsφm</p><p>　　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(gè)比率，如下式</p><p><b>　　a = = </b>&

23、lt;/p><p>　　假設(shè)變壓器輸出電能等于其輸入電能——這個(gè)假設(shè)適用于高效率的變壓器。實(shí)際上我們是考慮一臺(tái)理想狀態(tài)下的變壓器；這意味著它沒有任何損耗。因此</p><p><b>　　Pm = Pout</b></p><p><b>　　或者</b></p><p>　　VpIp × p

24、rimary PF = VsIs × secondary PF</p><p>　　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</p><p>　　VpIp = VsIs</p><p><b>　　從上式我們可以得知</b></p><p><b>　　= ≌ ≌

25、a</b></p><p>　　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(jì)算電壓，我們需要更多數(shù)據(jù)。</p><p>　　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(shí)牌數(shù)據(jù)獲得, 列出在滿載情況下原邊和副邊電壓。</p><p>　　當(dāng)副邊

26、電壓Vs相對于原邊電壓減小時(shí)，這個(gè)變壓器就叫做降壓變壓器。如果這個(gè)電壓是升高的，它就是一個(gè)升壓變壓器。在一個(gè)降壓變壓器中傳輸變比a遠(yuǎn)大于1(a>1.0)，同樣的，一個(gè)升壓變壓器的變比小于1(a<1.0)。當(dāng)a=1時(shí)，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(jí)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。</p><p

27、>　　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(huì)了解到當(dāng)變壓器帶負(fù)荷運(yùn)行時(shí)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。</p><p>　　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</p><p><b>　　= =

28、 </b></p><p>　　但是Vs / Is 負(fù)荷阻抗ZL，因此我們可以這樣表示</p><p>　　Zm (primary) = a2ZL</p><p>　　這個(gè)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(gè)公式應(yīng)用于變壓器的阻抗匹配。</p><p>　　4

29、. 有載情況下的變壓器</p><p>　　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(hào)表示。如果點(diǎn)號(hào)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(gè)負(fù)載時(shí)，在瞬間就有一個(gè)負(fù)荷電流沿著這個(gè)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(shí)，線圈中的磁動(dòng)勢會(huì)增加。</p><p>　　因?yàn)槎蝹?cè)電壓的大小取決于鐵芯磁通大小φ0，所以很顯然當(dāng)正常情況下負(fù)載電

30、勢Es沒有變化時(shí)，二次電壓也不會(huì)有明顯的變化。當(dāng)變壓器帶負(fù)荷運(yùn)行時(shí)，將有電流Is流過二次側(cè)，因?yàn)镋s產(chǎn)生的感應(yīng)電動(dòng)勢相當(dāng)于一個(gè)電壓源。二次側(cè)電流產(chǎn)生的磁動(dòng)勢NsIs會(huì)產(chǎn)生一個(gè)勵(lì)磁。這個(gè)磁通的方向在任何一個(gè)時(shí)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(dòng)勢會(huì)使主磁通φ0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(huì)增大。感應(yīng)電壓的減小將使外施電壓和感應(yīng)電動(dòng)勢之間的差值更大，它將使初級(jí)線圈中流過更大的電流

31、。初級(jí)線圈中的電流Ip的增大，意味著前面所說明的兩個(gè)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(jí)線圈中的磁動(dòng)勢將增加，以此來抵消二次側(cè)中的磁動(dòng)勢減小磁通的趨勢。</p><p>　　總的來說，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(shí)的。更重要的是，在空載和滿載時(shí)，主磁通φ0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。</p><p>

32、;　　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(dòng)勢NpIp’只停留在一次側(cè)。因?yàn)殍F芯的磁通φ0保持不變，變壓器空載時(shí)空載電流I0必定會(huì)為其提供能量。故一次側(cè)電流Ip是電流Ip’與I0’的和。</p><p>　　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。</p&

33、gt;<p>　　當(dāng)一個(gè)電流流過二次側(cè)繞組，它的磁動(dòng)勢（NsIs）將產(chǎn)生一個(gè)磁通，于空載電流I0產(chǎn)生的磁通φ0不同，它只停留在二次側(cè)繞組中。因?yàn)檫@個(gè)磁通不流過一次側(cè)繞組，所以它不是一個(gè)公共磁通。</p><p>　　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(gè)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(gè)增大的電壓具有電壓降的性質(zhì)，

34、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(gè)電阻壓降。把這些附加的電壓降也考慮在內(nèi)，這樣一個(gè)實(shí)際的變壓器的等值電路圖就完成了。由于分支勵(lì)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(jì)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(shí)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。</p><p>　　譯自<<

35、;科技英語>></p><p><b>　　A1.2原文</b></p><p>　　TRANSFORMER</p><p>　　1. INTRODUCTION</p><p>　　The high-voltage transmission was need for the case electrical p

36、ower 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

37、 a power system to operate at different voltage levels. In this paper we discuss power transformer principles and applications.</p><p>　　2. TOW-WINDING TRANSFORMERS</p><p>　　A transformer in its

38、 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.</p><p>　　In power applications, laminated steel c

39、ore 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 transf

40、orming 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.</p><p>　　The current flowing in the coil connecte

41、d to the ac source is called the primary winding or simply the primary. It sets up the flux φ in the core, which varies periodically both in magnitude and direction. The flux links the second coil, called the secondary w

42、inding or simply secondary. The flux is changing; therefore, it induces a voltage in the secondary by electromagnetic induction in accordance with Lenz’s law. Thus the primary receives its power from the source while the

43、 secondary supplies </p><p>　　3. TRANSFORMER PRINCIPLES</p><p>　　When a sinusoidal voltage Vp is applied to the primary with the secondary open-circuited, there will be no energy transfer. The i

44、mpressed voltage causes a small current Iθ 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 φm, where φm i

45、s 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</p><p>　　The no-load current Iθ is usually few percent

46、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 90º.

47、 It is readily seen that the current component Im= I0sinθ0, called the magnetizing current, is 90º in phase behind the primary voltage VP. It is this component that sets up the flux in the core; φ is therefore in ph

48、ase with Im.</p><p>　　The second component, Ie=I0sinθ0, 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

49、no-load current, or</p><p>　　I0 = Im+ Ie</p><p>　　It should be noted that the no-load current is distortes and nonsinusoidal. This is the result of the nonlinear behavior of the core material.&l

50、t;/p><p>　　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 wi

51、nding，there will be an induced EMF E in the secondary winding in accordance with Faraday’s law, namely, E=NΔφ/Δt. This same flux also links the primary itself, inducing in it an EMF, Ep. As discussed earlier, the induced

52、 voltage must lag the flux by 90º, therefore, they are 180º out of phase with the</p><p>　　Eavg = turns×</p><p>　　which is Faraday’s law applied to a finite time interval. It foll

53、ows that</p><p>　　Eavg = N = 4fNφm</p><p>　　which 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 t

54、he average voltage; thus</p><p>　　E = 4.44fNφm</p><p>　　Since the same flux links with the primary and secondary windings, the voltage per turn in each winding is the same. Hence</p><

55、p>　　Ep = 4.44fNpφm</p><p><b>　　and</b></p><p>　　Es = 4.44fNsφm</p><p>　　where Ep and Es are the number of turn on the primary and secondary windings, respectively. Th

56、e ratio of primary to secondary induced voltage is called the transformation ratio. Denoting this ratio by a, it is seen that</p><p><b>　　a = = </b></p><p>　　Assume that the output

57、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. Thus

58、</p><p><b>　　Pm = Pout</b></p><p><b>　　or</b></p><p>　　VpIp × primary PF = VsIs × secondary PF</p><p>　　where PF is the power factor.

59、For the above-stated assumption it means that the power factor on primary and secondary sides are equal; therefore</p><p>　　VpIp = VsIs</p><p>　　from which is obtained</p><p><b&

60、gt;　　= ≌ ≌ a</b></p><p>　　It 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

61、 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.</p><p>　　The

62、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

63、-load condition.</p><p>　　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 call

64、ed a step-up transformer. In a step-down transformer the transformation ratio a is greater than unity (a>1.0), while for a step-up transformer it is smaller than unity (a<1.0). In the event that a=1, the transforme

65、r secondary voltage equals the primary voltage. This is a special type of transformer used in instances w</p><p>　　As is apparent, it is the magnetic flux in the core that forms the connecting link between p

66、rimary 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.</p><p>　　Looking into the transformer

67、 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</p><p><b>　　= = <

68、;/b></p><p>　　But Vs / Is is the load impedance ZL thus we can say that</p><p>　　Zm (primary) = a2ZL</p><p>　　This equation tells us that when an impedance is connected to the se

69、condary 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 t

70、hat is used in impedance-matching applications.</p><p>　　4. TRANSFORMERS UNDER LOAD</p><p>　　The primary and secondary voltages shown have similar polarities, as indicated by the “dot-making” co

71、nvention. 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 curre

72、nt 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 a</p><p>　　Since the secondar

73、y voltage depends on the core flux φ0, 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

74、 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 th

75、e main flux that created it in the first p</p><p>　　In general, it will be found that the transformer reacts almost instantaneously to keep the resultant core flux essentially constant. Moreover, the core fl

76、ux φ0 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.</p><p>　　On the primary side, Ip’ is the current that

77、 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 φ0 remains constant. I0 must be the same current that energizes the transformer

78、at no load. The primary current Ip is therefore the sum of the current Ip’ and I0.</p><p>　　Because the no-load current is relatively small, it is correct to assume that the primary ampere-turns equal the se

79、condary ampere-turns, since it is under this condition that the core flux is essentially constant. Thus we will assume that I0 is negligible, as it is only a small component of the full-load current.</p><p>

80、　　When a current flows in the secondary winding, the resulting MMF (NsIs) creates a separate flux, apart from the flux φ0 produced by I0, which links the secondary winding only. This flux does no link with the primary wi

81、nding and is therefore not a mutual flux.</p><p>　　In addition, the load current that flows through the primary winding creates a flux that links with the primary winding only; it is called the primary leaka

82、ge flux. The secondary- leakage flux gives rise to an induced voltage that is not counter balanced by an equivalent induced voltage in the primary. Similarly, the voltage induced in the primary is not counterbalanced in

83、the secondary winding. Consequently, these two induced voltages behave like voltage drops, generally called leakage reactanc</p><p>　　這漫長的兩個(gè)小時(shí)怎么過？老班在考場外盯著他們這些差生呢，怕他們早早出去，影響了成績?！白魑囊欢ㄒ獙懲臧??！碑?dāng)她進(jìn)來的時(shí)候，老班站在門外，向她發(fā)出討

84、好的笑。想到這里，劉冰兒看了看作文題目：“陽光燦爛的日子”。她抬起頭，朝窗外望了望：陽光很純凈，很嫵媚，果然是個(gè)陽光燦爛的日子。</p><p>　　劉冰兒不討厭這個(gè)題目，雖然她很討厭作文課。它仿佛觸動(dòng)了她心底的一角，讓她聽到了冰層融化的聲音。</p><p>　　那一瞬間，劉冰兒做出了一個(gè)重大的決定，她要寫作文了。雖然進(jìn)考場時(shí)，她曾賭氣地想，要做一個(gè)白先生，把老師氣死。但現(xiàn)在，她改變注意

85、了。</p><p><b>　　一、相遇是個(gè)偶然</b></p><p>　　“曾經(jīng)，我的日子里充滿了陽光的味道。那兒流動(dòng)著鮮花的芳香，彌漫著歡樂的氣息……劉冰兒就這樣開始了自己的開頭，從小，她的文筆就不錯(cuò)，語文老師經(jīng)?？渌@讓她很自豪。</p><p>　　初三第一學(xué)期期末，我的成績有所進(jìn)步，開家長會(huì)的時(shí)候，老師表揚(yáng)了我，媽媽特高興?；氐?/p>

86、家，爸爸媽媽帶我去飯店吃飯,說是對我的獎(jiǎng)勵(lì)。再說，村里剛發(fā)了錢，他們高興，他們手里癢癢。自從我們村的土地被征用，自從我爸媽手里有了錢，我就發(fā)現(xiàn)我爸不是我爸我媽不是我媽。</p><p>　　我爸再也不去找工作了，他開了個(gè)棋牌室，名譽(yù)上是棋牌室，其實(shí)就是召集村里的閑人賭博；我媽整天在鏡子前描眉畫眼，試圖把自己捯飭成一個(gè)美女。但再怎么捯飭，也掩蓋不了臉上的那些褶子。</p><p>　　吃飯的

87、時(shí)候，我給爸爸說我想要電腦,他們很爽快的答應(yīng)了。第二天就帶我去選電腦,第三天就買回來了,第四天就有人來安裝寬帶,就這樣我的生活加入了電腦!</p><p>　　我陽光燦爛的日子來臨了。</p><p>　　可以上網(wǎng)了,我就天天呆在電腦前。聊天,看電影,玩游戲,成了我生活中不可缺少的。起初我有節(jié)制,會(huì)控制自己,玩到9點(diǎn)10點(diǎn)的不用爸媽催,那時(shí)候真的很乖啊。</p><p&

88、gt;　　那天，在去逛商場的路上，突然遇到了同年級(jí)的楊一一，我們乘坐同一輛公交車，下車時(shí)，他竟然要了我的**號(hào)，說要加我為好友，這讓我受寵若驚。</p><p>　　要知道，楊一一是我們學(xué)校有名的帥哥，不僅人長得帥，而且歌也唱得特別好。他還會(huì)彈吉它，去年在我們學(xué)校舉辦的藝術(shù)節(jié)上，他自彈自唱的歌曲讓全校師生沸騰了。為此，女孩子們都愿意和他交往。但在老師的心目中，他名聲卻不怎么好，因?yàn)樗粣蹖W(xué)習(xí)，經(jīng)常招惹事非，老師們