電氣專(zhuān)業(yè)畢業(yè)設(shè)計(jì)外文翻譯--變壓器

1、<p><b>　　變壓器</b></p><p>　　摘要：變壓器是變電所的主要設(shè)備，功能是實(shí)現(xiàn)電網(wǎng)電壓的等級(jí)變換，基本工作原理是電磁感應(yīng)。</p><p>　　變配電所是實(shí)現(xiàn)電壓等級(jí)變換和電能分配的場(chǎng)所。對(duì)供電電源進(jìn)行電壓等級(jí)變換，應(yīng)對(duì)電能進(jìn)行重新分配的場(chǎng)所稱為變電所。建筑變電所是供配電系統(tǒng)的樞紐，供電電源由電網(wǎng)引到變電所，在變電所完成降壓，電能分配等功

2、能。</p><p>　　關(guān)鍵詞: 變電所；變壓器；繼電保護(hù)；</p><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>&l

3、t;b>　　2. 雙繞組變壓器</b></p><p>　　變壓器的最簡(jiǎn)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](méi)有旋轉(zhuǎn)損失，因此在電壓等級(jí)轉(zhuǎn)換的過(guò)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。&

4、lt;/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&g

5、t;<p>　　當(dāng)二次側(cè)電路開(kāi)路是，即使原邊被施以正弦電壓Vp，也是沒(méi)有能量轉(zhuǎn)移的。外加電壓在一次側(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(shuō)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

6、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(jiàn)電流分量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I0 = Im+ Ie&l

7、t;/p><p>　　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。</p><p>　　如果假定變壓器中沒(méi)有其他的電能損耗一次側(cè)的感應(yīng)電動(dòng)勢(shì)Ep和二次側(cè)的感應(yīng)電壓Es可以表示出來(lái)。因?yàn)橐淮蝹?cè)繞組中的磁通會(huì)通過(guò)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(shì)E，即E=NΔφ/Δt。相同的磁通會(huì)通過(guò)原邊自身，產(chǎn)生一個(gè)電動(dòng)勢(shì)Ep。正如前文中討論到的，所產(chǎn)生的電

8、壓必定滯后于磁通90º，因此，它于施加的電壓有180º的相位差。因?yàn)闆](méi)有電流流過(guò)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(shì)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>

9、　　即是法拉第定律在瞬時(shí)時(shí)間里的應(yīng)用。它遵循</p><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</p><p>　　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

10、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 = 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(zhǎng)的比率稱做變比。用字母a來(lái)表示這個(gè)比率

11、，如下式</p><p><b>　　a = = </b></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(méi)有任何損耗。因此</p><p><b>　　Pm = Pout</b></p><p><b

12、>　　或者</b></p><p>　　VpIp × primary 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

13、可以得知</b></p><p><b>　　= ≌ ≌ a</b></p><p>　　它表明端電壓比等于匝數(shù)比，換句話說(shuō)，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(duì)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,

14、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)副邊電壓Vs相對(duì)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(lèi)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

15、維持相同的電壓等級(jí)的狀況下。因此，我們把這種類(lèi)型的變壓器稱為絕緣型變壓器。</p><p>　　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(lái)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

16、據(jù)Vs和Is，可得Vp和Ip的比例是</p><p><b>　　= = </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(lái)的a2倍。我們把這種折算方式

17、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. 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(shuō)，極性的標(biāo)注可以表明當(dāng)電流流過(guò)兩側(cè)的線圈時(shí)，線圈中的磁動(dòng)勢(shì)會(huì)增加。</

18、p><p>　　總的來(lái)說(shuō)，變壓器為了保持磁通是常數(shù)，對(duì)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(lái)使電流Ip增加。</p><p>　　在一次側(cè)，電流Ip’在一次側(cè)流過(guò)以平衡Is產(chǎn)生的影響。它的磁動(dòng)勢(shì)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

19、流Ip’與I0’的和。</p><p>　　因?yàn)榭蛰d電流相對(duì)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(duì)于滿載電流是極其小的。</p><p>　　當(dāng)一個(gè)電流流過(guò)二次側(cè)繞組，它的磁動(dòng)勢(shì)（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(guò)一次側(cè)繞組，所以

20、它不是一個(gè)公共磁通。</p><p>　　另外，流過(guò)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)，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

21、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(duì)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>　　TRANSFORMER</p><p>　　Abstract：The transformer is the transformer substation major installa

22、tion, the function realizes the network voltage rank transformation, the key job principle is the electromagnetic induction.Changes the substation is realizes the voltage class transformation and the electrical energy as

23、signment place. Carries on the voltage class transformation to the electric power supply, deals with the place which the electrical energy carries on redistributes to be called the transformer substation. Constr</p>

24、;<p>　　Key-words: substation；transformer；Relay protection；</p><p>　　1. INTRODUCTION</p><p>　　The high-voltage transmission was need for the case electrical power is to be provided at con

25、siderable 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

26、 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 simplest form consists of tw

27、o 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 core transformers (to which th

28、is 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

29、 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 connected to the ac source is called

30、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 winding or simply secondary. T

31、he 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 secondary supplies </p>

32、;<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 impressed voltage causes a sma

33、ll 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 is the maximum value of the co

34、re 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 of the rated full-load curren

35、t 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. It is readily seen that the

36、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 phase with Im.</p><p

37、>　　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 no-load current, or</p>

38、<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. If it is assumed that there ar

39、e 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 seconda

40、ry winding in accordance with Faraday’s law, namely, E=NΔφ/Δt. This same flux also links the primary itself</p><p>　　Eavg = turns×</p><p>　　which is Faraday’s law applied to a finite time i

41、nterval. It follows 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

42、 is 1.11 times the 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

43、</p><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,

44、respectively. The 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

45、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

46、 no losses. Thus</p><p><b>　　Pm = Pout</b></p><p><b>　　or</b></p><p>　　VpIp × primary PF = VsIs × secondary PF</p><p>　　where PF is th

47、e power factor. 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>

48、<p><b>　　= ≌ ≌ 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 rela

49、ted 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.</p>

50、<p>　　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 vo

51、ltage under full-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 ra

52、ised, it is called 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

53、, the transformer 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 connectin

54、g 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.</p><p>　　Looking into

55、 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</p><p><

56、;b>　　= = </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 con

57、nected 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 o

58、f transformers that 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

59、 “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 instanta

60、neous 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 a</p><p>　　In

61、 general, it will be found that the transformer reacts almost instantaneously to keep the resultant core flux essentially constant. Moreover, the core flux φ0 drops very slightly between n o load and full load (about 1 t

62、o 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 flows in the primary to balance the demagnetizing effect of Is. It

63、s 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 at no load. The primary current Ip is therefore the sum of the curr

64、ent 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 secondary ampere-turns, since it is under this condition that the cor

65、e 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>　　When a current flows in the secondary winding, the resulting MMF

66、(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 winding and is therefore not a mutual flux.</p><p>　　In

67、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 leakage flux. The secondary- leakage flux gives rise to an induced volta

68、ge that is not counter balanced by an equivalent induced voltage in the primary. Similarly, the voltage induced in the primary is not counterbalanced in the secondary winding. Consequently, these two induced voltages beh