變電所畢業(yè)設(shè)計(jì)外文翻譯

1、<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><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

2、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)

3、。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

4、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

5、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

6、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(shì)Ep和二次側(cè)的感應(yīng)電壓Es可以表示出來。因

7、為一次側(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(shì)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(shì)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180º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；

8、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>　　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，其值為平均電壓的1.11

9、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 = 4.44fNpφm</p><p><b>　　并且</b></p><p>　　Es = 4.44fNsφm</

10、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來表示這個(gè)比率，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；這意味著它沒有任何損耗。因此</p

11、><p><b>　　Pm = Pout</b></p><p><b>　　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

12、</p><p>　　VpIp = VsIs</p><p><b>　　從上式我們可以得知</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，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

13、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

14、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>　　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&

15、gt;　　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>　　= = </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

16、) = 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. 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)

17、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(shì)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)載電勢(shì)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(shì)相當(dāng)于一個(gè)電壓源。二次側(cè)電流產(chǎn)生的磁動(dòng)勢(shì)NsIs會(huì)產(chǎn)生一個(gè)勵(lì)磁。這個(gè)磁通的

18、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(shì)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(shì)之間的差值更大，它將使初級(jí)線圈中流過更大的電流。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(shì)將增加，以此來抵消二次側(cè)中的磁動(dòng)勢(shì)減小磁通的趨勢(shì)。</p

19、><p>　　總的來說，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降落很多來使電流Ip增加。</p><p>　　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(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是電流

20、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)磁通不流過一次側(cè)繞組，所以它

21、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)，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

22、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>　　譯自<<科技英語>></p><p><b>　　A1.2原文</b></p><p>　　TRANSFORMER

23、</p><p>　　1. INTRODUCTION</p><p>　　The 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 h

24、igh 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 transfor

25、mer principles and applications.</p><p>　　2. TOW-WINDING TRANSFORMERS</p><p>　　A transformer in its simplest form consists of two stationary coils coupled by a mutual magnetic flux. The coils ar

26、e said to be mutually coupled because they link a common flux.</p><p>　　In power applications, laminated steel core transformers (to which this paper is restricted) are used. Transformers are efficient becau

27、se 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%

28、, the higher values applying to the larger power transformers.</p><p>　　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 φ i

29、n 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 second

30、ary by electromagnetic induction in accordance with Lenz’s law. Thus the primary receives its power from the source while the secondary supplies </p><p>　　3. TRANSFORMER PRINCIPLES</p><p>　　When

31、 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 Iθ to flow in the primary winding. This no-load current

32、 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 core flux; and (2) it provides a component to account for the hyster

33、esis 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 current of the transformer (about 2 to 5%). Since at no-load the primary

34、 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 current component Im= I0sinθ0, called the magnetizing current, is

35、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>　　The second component, Ie=I0sinθ0, is in phase with the prima

36、ry 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><p>　　I0 = Im+ Ie</p><p>　　It should be noted

37、that the no-load current is distortes and nonsinusoidal. This is the result of the nonlinear behavior of the core material.</p><p>　　If it is assumed that there are no other losses in the transformer, the in

38、duced 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 Faraday’s law,

39、 namely, E=NΔφ/Δt. 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 90º, therefore, they are 180º out of phase with the<

40、/p><p>　　Eavg = turns×</p><p>　　which is Faraday’s law applied to a finite time interval. It follows that</p><p>　　Eavg = N = 4fNφm</p><p>　　which N is the number of

41、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; thus</p><p>　　E = 4.44fNφm</p><p>　　Since the sam

42、e flux links with the primary and secondary windings, the voltage per turn in each winding is the same. Hence</p><p>　　Ep = 4.44fNpφm</p><p><b>　　and</b></p><p>　　Es = 4

43、.44fNsφm</p><p>　　where 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 th

44、is ratio by a, it is seen that</p><p><b>　　a = = </b></p><p>　　Assume that the output power of a transformer equals its input power, not a bad sumption in practice considering the h

45、igh efficiencies. What we really are saying is that we are dealing with an ideal transformer; that is, it has no losses. Thus</p><p><b>　　Pm = Pout</b></p><p><b>　　or</b>

46、</p><p>　　VpIp × primary PF = VsIs × secondary PF</p><p>　　where PF is the power factor. For the above-stated assumption it means that the power factor on primary and secondary sides a

47、re equal; therefore</p><p>　　VpIp = VsIs</p><p>　　from which is obtained</p><p><b>　　= ≌ ≌ a</b></p><p>　　It shows that as an approximation the terminal v

48、oltage 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

49、in relation to the primary voltage. To calculate the voltage regulation, we need more information.</p><p>　　The ratio of the terminal voltage varies somewhat depending on the load and its power factor. In pr

50、actice, the transformation ratio is obtained from the nameplate data, which list the primary and secondary voltage under full-load condition.</p><p>　　When the secondary voltage Vs is reduced compared to the

51、 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

52、 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 transformer secondary voltage equals the primary voltage. This is a special type of transformer used in i

53、nstances w</p><p>　　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

54、itself to the secondary load current when the transformer supplies a load.</p><p>　　Looking into the transformer terminals from the source, an impedance is seen which by definition equals Vp / Ip. From = ≌

55、 ≌ 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>　　= = </b></p><p>　　But Vs / Is is the load impedance ZL thus we can say that</p&g

56、t;<p>　　Zm (primary) = a2ZL</p><p>　　This 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 it

57、s 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.</p><p>　　4. TRANSFORMERS UNDER LOA

58、D</p><p>　　The 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 theo

59、ry; 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

60、current enters both windings at the marked terminals, the MMFs of the two windings a</p><p>　　Since the secondary voltage depends on the core flux φ0, it must be clear that the flux should not change appreci

61、ably 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 seconda

62、ry 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 p</p><p>　　In general, it will be found t

63、hat 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 to 3%), a necessary condition

64、 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. Its MMF NpIp’ sets up a flux l

65、inking 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 current Ip’ and I0.</p>&

66、lt;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 core flux is essentially consta

67、nt. 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 (NsIs) creates a separate fl

68、ux, 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 addition, the load current t

69、hat 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 voltage that is not counter balan

70、ced 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 behave like voltage drops, gene