首页资源分类IC设计及制造 > Electrical Powersystem

Electrical Powersystem

已有 445466个资源

下载专区

上传者其他资源

文档信息举报收藏

标    签:powersystemenglish

分    享:

文档简介

english and german vision from Duisburg-Essen university

文档预览

0 Basics of Alternating Current Circuits Contents 0 Basics of Alternating Current Circuits ............................................................................... 1 0.1 Concept of Phasors......................................................................................................... 2 0.2 Three-Phase System ................................................................................................... 5 0.3 Electric Power ............................................................................................................ 9 0.3.1 Definition of Instantaneous Power ..................................................................... 9 0.3.2 Definition of Complex Power .......................................................................... 11 1 0.1 Concept of Phasors Today’s electric power systems bases on alternating currents and voltages. This is due to the fact, that alternating electricity can be generated very efficiently in rotating induction machines. Figure 0.1 shows the time behavior of alternating current and voltage. u(t) i(t) 0 5 10 15 20 25 ωt 30 Figure 0.1: Alternating voltage and current Thus, voltage and currents are described by the functions u(t) = uˆ sin(ωt + ϕu ) i(t) = iˆ sin(ωt + ϕi ) However, this time characteristic can be mapped by complex phasors rotating with the speed ω, where the real part of the phasors corresponds with the time functions. { } i(t ) = 2Re Ue jωt U = Ue jϕu = U∠ϕu = U (cos ϕu + j sin ϕu ) { } i(t ) = 2Re Ie jωt I = Ie jϕi = I∠ϕi = I (cos ϕi + j sin ϕi ) U and I are complex phasors fully characterizing the alternating voltages and currents, than it is always possible to derive U = Ue jϕu ⇒ u(t) = uˆ sin(ωt + ϕu ) uˆ = 2 U I = Ie jϕu ⇒ i(t) = ˆi sin(ωt + ϕu ) ˆi = 2 I Because of all voltage and current phasors Ue jωt and Ie jωt turn with the same speed ω, it is favorable to skip e jωt and use the fixed parts U and I only. 2 In power system analysis U and I are called voltage and current phasors without the extension e jωt . The main advantage of using phasors results from the fact that phasors are, in steady state, constant. The magnitudes of the phasors U and I are characterizing the Root Mean Square (rms) values of the corresponding waves according to the equations ∫ U = 1 T (uˆ sinωt)2 = uˆ T0 2 ( ) ∫ I = 1 T ˆi sinωt 2 = ˆi T0 2 Fig. 0.2 shows a representation of voltage and current phasors in the complex plain. Figure 0.2: Phasors in the complex plain The following useful definitions are introduced: ϕ = ϕu − ϕi Impedance Angel of the corresponding load Ia = I cos ϕ Active current producing/absorbing active power Ir = −I sinϕ reactive current producing/absorbing reactive power When the voltage and current are measured over a particular load, the corresponding complex load impedance can be calculated as Z = U = R + jX I It is important to realize that impedanzes never alternate with time like voltages and currents. 3 For a resistive load the current phasor is in phase with the voltage. For a purely inductive load, the current lags the voltage by π/2 and for a purely capacitive load, the current leads the voltage by π/2 as shown in Fig. 0.3. I I U U R ϕ=0 I U L U ϕ ϕ=+π/2 I I I U C ϕ=−π/2 ϕ U Figure 0.3: Voltage and current over different loads Loads are usually composed of different elements. Therefore, load impedances are described by serial or parallel connections of resistances and reactances. The latest can be inductive or capacitive depending on which load characteristic is dominating (see Fig. 0.4). 4 R jXL R -jXC Figure 0.4: Impedance characteristics Z jX ϕ R R ϕ Z -jXC 0.2 Three-Phase System We consider a three-phase system according to Fig. 0.5. Uqa Iqa Zaa ULa ILa Uqb Iqb Zac, Zca Zbb Uqc Iqc Zcc Zab, Zba Zbc, Zcb ULb ILb ULc ILc Source Transmission Load Figure 0.5: Three-phase system consisting of source, transmission and load The voltage sources of the three phases a,b,c are symmetrical, i.e. the voltage magnitudes are equal but the corresponding angel are shifted each by π/3 according to the Fig. 0.6. 5 Uqc = Uqe2π / 3 = a Uq e2π/3 e2π /3 e2π/3 Uqa = Uq Figure 0.6: Voltage phasors of a threephase system Uqb = Uqe−2π / 3 = a 2 Uq For simplicity the unity phasors a = e jπ / 3 a2 = e j2π / 3 are introduced. The time characteristic of a symmetrical three-phase system is shown in Fig. 0.7 abc 0 5 10 15 20 25 ωt 30 Figure 0.7: Three-phase characteristic 6 For the line to line voltages follows from Fig. 0.6: U ab = U a −U b = U a(1− a2 ) U ab = U ab = 3 U Υ U ac = U a −U c = U a(1− a ) U ac = U ac = 3 U Υ U cb = U c −U b = U a( a − a2 ) U cb = Ucb = 3 U Υ In general: UΔ = 3UΥ with UΥ phase voltage UΔ line to line voltage Further, we assume that the three-phase transmission line is characterized by self and mutual inductances. Taking into account that real lines always include resistive elements too so the line can be described by self and mutual impedances as shown in the Fig. 0.5. The following equation describes our three-phase system: ⎡U La ⎤ ⎡Z aa ⎢⎢U Lb ⎥ ⎥ = ⎢⎢Z ba Z ab Z bb Z Z ac bc ⎤ ⎥ ⎥ ⎡I ⎢⎢I aq bq ⎤ ⎥ ⎥ + ⎡U ⎢⎢U qa qb ⎤ ⎥ ⎥ ⎢⎣U Lc ⎥⎦ ⎢⎣Z ca Z cb Z cc ⎥⎦⎢⎣I cq ⎥⎦ ⎢⎣U qc ⎥⎦ Solving equation like above may be difficult without computer due to the matrix operations have to be carried out. However, we can assume that the phase wires are arranged symmetrically to each other and to the ground. In this case the impedance matrix becomes also symmetrical, i.e. Z aa = Z bb = Z cc = Z s self impedances Z ab = Z ba = Z bc = Z cb = Z ca = Z ac = Z m mutual impedances Then we can write for phase a ULa = Z s I aq + Z m I bq + Z m I cq + U qa With the assumption of symmetrical load the currents become also symmetrical 7 I aq = a I bq = a 2 I cq So we get U La = Z I aq + U qa where Zb = Zs −Zg . Z b is called “operation impedance” of the line. Similar relationships can be derived to the phases b and c: U Lb = Z b I qb + U qb U Lc = Z b I qc + U qc From the equations above it is obvious that symmetrical three-phase systems can be treated as three independent one-phase systems. However, solving all three equations is not necessary because a,b,c, voltages and currents are equal in magnitude. Besides, we know that quantities of phase b leg by π/3 and those of c lead by π/3. Therefore, from voltages and currents calculated for phase a, one always can conclude the corresponding quantities in b and c. It should be emphasize that the impedance Z b which is the only impedance in all three phase equations is not identical with the impedance of one circuit built by a simple phase. In fact, Z b represents the whole three-phase system. As we will see later, Z b = Z1 where Z1 is the positive sequence impedance of the transmission line. In general, power systems are usually described by positive, negative and zero sequence components of currents and voltages rather than by abc components. Positive, negative and zero sequence components are called Symmetrical Components. The mean advantage of using Symmetrical Components results from two facts: - Assuming symmetrical transmission lines and loadings only the positive sequence voltages, currents and the corresponding impedances exist - In the impedance matrix of symmetrical lines, positive, negative and zero sequence impedances are decoupled. Figure 0.8 shows the positive sequence equivalent circuit of the three-phase system. 8 Iqa Zb Positive Sequence Transmission Uqa Source Load ULa, ILa Figure 0.8: Positive Sequence equivalent circuit of the three-phase system 0.3 Electric Power 0.3.1 Definition of Instantaneous Power The Instantaneous Power s(t) is defined as the product of instantaneous voltage u(t) and current i(t). s(t) =u(t )i(t )=uˆ ˆi cos(ωt +ϕu )cos(ωt +ϕi ) after algebraic manipulation we get s(t ) = uˆ ˆi 2 [cos(ϕu − ϕi ) + cos(2ωt + ϕu + ϕi )] with ϕ = ϕu − ϕi and eliminating of ϕu s(t ) = uˆ ˆi 2 [cosϕ + cosϕ cos 2(ωt + ϕi ) − sinϕ sin 2(ϕt + ϕi )] s(t ) = uˆ ˆi 2 [cosϕ(1 + cos 2(ωt + ϕi )) − sinϕ sin 2(ωt + ϕi )] This equation consist of two swings: The magnitude of the first swing is uˆˆi cosϕ = UI cosϕ = P 2 and those of the second uˆˆi sinϕ = UI sinϕ = Q 2 9 The integral of the instantaneous power ∫ J = 1 t2 s(t )dt t2 − t1 t1 becomes for t2 − t1 >> T to P . The first part of the instantaneous power s1(t) = UI cosϕ[1+ cos 2(ωt +ϕi )] describes a non-zero energy transfer. Therefore it is called Active Power. It is important to realize, that instantaneous active power represents a quantity which alternate with 2ω. The mean value of the second component s2 (t) = −UI sinϕ sin 2(ωt + ϕi ) is equal to zero, i.e. in this case no energy transfer takes place in the long term. Therefore it is called Reactive Power. Also the reactive power is an alternating power with frequency 2ω. In the next, we will show that the reactive power component s2 corresponds with the derivative of electric and magnetic field energy ∂Wel resp. ∂Wmag ∂t ∂t For this, we consider a serial circuit consisting of R und X. The corresponding power 10 s2 = UI sinϕ sin 2(ωt + ϕi ) = − uˆ ˆi X R2 + X 2 sin(ωt + ϕi )cos(ωt +ϕi ) s2 = uˆ ˆi X R2 + X 2 cos⎜⎛ωt ⎝ + ϕi + π 2 ⎟⎞ cos(ωt ⎠ +ϕi ) for ϕ > 0 we get s2 = uˆ L cos⎜⎛ωt ⎝ + ϕi + π 2 ⎟⎞ˆi ⎠ cos(ωt + ϕi ) = uLi and for ϕ < 0 s2 = uˆ C cos⎜⎛ωt ⎝ + ϕi − π 2 ⎟⎞ˆi ⎠ cos(ωt +ϕi ) = uCi From the last two equations follows uLi = L ∂i ∂t i = ∂Wmag ∂i ∂i ∂t = ∂Wmag ∂t uCi = C ∂uC ∂t uC = ∂Wel ∂uC ∂uC ∂t = ∂Wel ∂t The following conclusions can be drawn: Q > 0 : Q corresponds with the magnitude of the magnetic field derivative Q < 0 : |Q | corresponds with the magnitude of the electric field derivative We can use P und Q as characteristic quantities for loads and generators, due to the fact that for a given voltage u( t ) = uˆ sin(ωt + ϕu ) the instantaneous power can be always determined from P and Q. P,Q; u(t) ⇒ s(t) Thus, the time characteristic of the current is specified too s(t); u(t) ⇒ i(t) 0.3.2 Definition of Complex Power Because rms-values are always used for the analysis of large AC power systems in the steady, respectively quasi steady states, it is suitable to define a Complex Power phasor S = U I* = U Ie j(ϕu−ϕi ) = S e jϕ = S(cosϕ + j sinϕ ) The real and imaginary parts can be interpreted as: 11 Re{S}= UI cosϕ = P Im{S}= UI sinϕ = Q Active Power Reactive Power S = P + jQ S termed as complex Apparent Power. For practical reason, it is suitable to define the active current Ia, which corresponds with the active power P Ia = I cosϕ and P = UIw Similarly we define the reactive current Ir Ir = I sinϕ Q = UIr Consider the sign of Ir in the following Fig. 0.9 Im Ir (positiv) Ia U Re ϕ I Im Ir (positiv) Ia ϕ U Re I Figure 0.9: Definition of active and reactive currents In a three-phase AC system the complex apparent power is calculated as the sum of the particular phase powers. S = U a I * a + U b I * b + U c I * c For symmetrical systems one can write for the magnitude S = S S = 3UΔI where UΔ is the line to line voltage. Further follows for active and reactive power P = S cosϕ Q = S sinϕ In three-phase systems the nominal voltage is always defined as line to line voltage. 12

Top_arrow
回到顶部
EEWORLD下载中心所有资源均来自网友分享,如有侵权,请发送举报邮件到客服邮箱bbs_service@eeworld.com.cn 或通过站内短信息或QQ:273568022联系管理员 高进,我们会尽快处理。