Elektrik enerji sistemlerindeki dengesizliklerin optimal işletme bakımından etkinlikleri

Abstract

ÖZET Günümüzde, elektrik enerjisinin ilk üretildiği yerden tüketiciye ulaştığı noktaya kadar her aşamada optimal tasarım ve işletmenin büyük önem taşıdığı artık bilinen bir gerçektir. Enerji sistemindeki güçlerin dağılımıyla yakından ilgili olan optimizasyon analizleri genellikle, her üç fazı aynı güçle yüklenen dengeli enerji sistemleri için yapılmaktadır. Enerji sisteminin fazlarının farklı güçlerle yüklenmesiyle ortaya çıkan dengesiz yüklenme koşullarında, sistemdeki güç dağılımı değişir. Bunun sonucu olarak, dengeli yüklenme koşulları için öngörülmüş optimizasyon analizleri de geçerliliğini kaybetmiş olur. Bu çalışmada ilk olarak, üretim, maliyet, kâr vb. genel işletme ve ekonomi kavramları ışığında elektrik enerji sistemlerinin birer işletme özelliği taşıdıkları gösterilmiştir. Bundan sonra enerji sistemlerinde maliyet ve optimizasyon uygulaması ile dengesizlikler ele alınmıştır. Enerji sistemlerinde geçici ve sürekli dengesizlikler görülmektedir; simetrik olmayan arızalar geçici dengesizliğe, yükün karakterinden ortaya çıkan dengesizlikler ise sürekli dengesizliğe örnek gösterilebilir. Burada, dengesiz yüklenme halinde, üç fazlı yükün gücünden ayrı olarak şebekeden çekilen ek gücün ifadesi verilmiştir. Bu ek gücün niteliği Fortescue`nin Simetrili Bileşenleri yardımıyla incelenmiş, bu ek güç dolayısıyla hat kayıplarındaki artış ortaya konulmuştur. Ayrıca enerji iletim sistemlerinin çeşitli besleme türleri için, toprak yolunda ortaya çıkan kayıplar hesap edilmiştir. Son olarak, dengesiz yüklenmenin optimal işletme üzerindeki etkinliğini sayısal değerlerle ortaya koymak amacıyla, birisi Türkiye'deki dengesiz yüklenen enerji sistemi olmak üzere, iki ayrı enerji sistemi üzerinde hesaplamalar yapılmıştır. Bulunan sonuçlardan, dengesiz yüklenmenin enerji sistemleri üzerinde ihmal edilemeyecek ek kayıplara yol açtığı ve bu kayıpların, sistemin optimal çalışmasını olumsuz yönde etkilediği, bir başka deyişle sistemi optimal çalışmadan uzaklaştırdığı gözlenmiştir.

SUMMARY THE EFFECTS OF UNBALANCES IN ELECTRICAL POWER SYSTEMS WITH REFERENCE TO OPTIMAL OPERATION Authorities concerned with generation, transmission and distribution of electrical energy aim at serving the consumers with quality and in an economic way. Therefore it is very important what optimal values be used during planning and operation of generation, transmission and distribution systems. Conventional optimization analyses and techniques are concerned with balanced systems where the phases are equally loaded. Power distribution and power flow analyses of unbalanced power systems where the three phases are not equally loaded differ from balanced loading. It is clear that optimization analysis also differs in case of unbalanced loading since optimization of operating costs generally depends on the distribution of power flow in power systems. The each step from generation of electrical energy to its distribution has its own operational feature. Therefore the general operation and economy concepts such as production, cost, depreciation, interest, inflation, efficiency etc. will have their meanings also for power systems. First the cost of energy generation, transmission lines and substations have been given, secondly phase components method and symmetrical components methods have been summarized to search the effect of unbalancing on optimal operation, and current, voltage and impedance unbalances have been introduced. In Chapter 5 kinds of power in single phase systems and then in balanced and unbalanced three phase systems have been explained. If V.., V`, V are complex positive, negative and zero sequence values of phase voltages and I.., Iji I are complex positive, negative and zero sequence values of currents, the total apparent power received from the network is obtained, as follows with the aid of Fortescue's Symmetrical Components Transformation Matrix and fundamental relationships: S = 3V,I* + 3V~I* + 3V I* (1) - _1-1 -1 - JL -O-Oand also complex pulsating power is Ç = 3V`I. + 3V,I0 + 3V I (2) - -2-1 -1-2 -o-o v ' where the asterisk denotes the complex conjugate, and the terms of apparent power Ş are independent from time but the terms of pulsating power vary with time by angular frequency 2w (where ü)=27rf, f is supply frequency). An instantaneous power corresponds to the components (Yi. I].). (V2,I2), ^o'^' (Î2»îl)» (^1'Î2) in each phase with the aid of equations 1 and 2. The instantaneous powers corresponding to zero sequence current I with positive sequence phase voltage V.. and negative sequence phase voltage V2 must be also considered, although these powers are not seen in equations 1 and 2. By using the instantaneous power expression for a phase P = Re {VI* + yiejwt} = Re(Ş + ÇeJ2ü)t} (3) and the properties of symmetrical components, time variations of the instantaneous powers for the all pairs of symmetrical components are shown in figure 1. (Re means real part of any complex number). Assuming that phase, voltages of power system are sufficiently balanced (i.e. V`=V =0), we can write: ş - sy^* = su C - 3X^2 = £l2 y (4) Ş'= 3Wo - £io Hence the resultant apparent power received from the network becomes ;res =Vl3WÎI2 + Isy^l2 + Isy^ı*! (5) By defining a. = I9/I1 and g. = I /I- this expression is transformed to s - St a V1 + a? + P2 <6> res load > ı Ki which is in terms of S,, = 3VnIn. load 1 1 XXV3. S »? t AP21 >- t Figure 1 The time variations of instantaneous powers of o- o'' - z'-i'' xV1,X2), (V^Ij), (v2.ı2), (vo,ıo), (.Vj.ip. (V^,IQ) and (^'^o^ components, In the computation of line losses, the result is unchanged regardless of whether the physical currents I`,Ig,IT or the symmetrical component currents I.,]!»»! are used. Namely if R is the conductor resistance, the following equality is satisfied R(I^ + l2 + i2) = 3R(I2 + j-2 + j2} (7) If the neutral of the power system is insulated, the zero sequence current I will be zero. Then the negative sequence current In must be multiplied by a coefficient given in equation ö to satisfy the equality in equation 7. XXVIa. 6? «i Phasor summation of currents of phases R,S,T is termed as Residual Current. If the neutral of the power system was not insulated, the residual current would cause additional losses along the earth path. This path is different for overhead lines and for underground cables. The apparent power in the earth path due to the residual current is defined as S. = 6S (9) oö lo then equation 6 is repeated as S `S.,Jl + a? + 3? (1 + 5 res load/ ı x 2) (10) where 6 is a real number. In a power system which has n consumer busbars and r generation busbars the power flow analysis must be made to compute phase conductor losses in case of unbalanced loading. For unbalanced loaded systems various power flow methods may be used. Here with the aid of equation 6, we have a new power angle defined as cos<j> cos* = -====- (11) res V1 + ai + 32 where fa is power angle of the S.,. With above new power angle A, it is assumed that balanced power is * ö *res r received from the supply (Figure 2). Namely with the new power angle <j> mentioned above, total active power P will be constant but the total reactive power Q will increase, and in terms of line losses the following equality is valid. f(R, cos6R, cos&g, cos<j>T) = f(R, cos<J>res) (12) where AT, 6S, Am are power angles of the phases R,S,T respectively. Having run the power flow, the total losses for the system is computed as r n P, - 3 [<P + I PP4) - 1 P,] (13) loss L sw._. Gi.. iJ ' XXVIIwhere P sw PG and P mean phase powers of the swing bus, generation bus and consumer bus, respectively Figure 2 The phasor diagram for the positive, negative and zero sequence powers -in case of unbalanced loading (the voltages are assumed to be balanced). The earth path in overhead transmission lines is in the form of a `Lattice` which consists of ground wire between the towers and ground resistances of the towers, connected with one another along the line. The residual current distribution along the earth path for the i th consumer bus to which t transmission branches are connected is shown in figure 3. If the magnetic coupling between the phase conductors and ground wire may be neglected, the impedance of the lattice for the j th transmission branch is Z. =-v/R.Z. (in ohms) (14) Z. is the impedance of ground wire between the towers for j th transmission branch. Providing that span a in meters is an average value along the transmission line, Z. is written as -ej Z. = a zf. x 10 -ej (15) where z'. is unit length impedance of the ground wire in ohms per3 kilometer. Rt. is the ground resistance of the.. - __ ^.. -.. ti..Rt - ılJjP RT = ip2p (16) where iJ, and ij- are real numbers in 1/ohms. From figure 3 for the i th 1>us t 31. = I`. + II. (17) -oi -Ti.,-zi Therefore earth path loss for the i th bus is Po6i *.f^^e^ lî.j!2'+RTİ İÎTİİ2 (18) If m consumer buses are insulated in the system, the total earth path loss in the system is n-m IP ` = I P oö i=l o6i (19) 111)111)11)11)1 Figure 3 Distribution of the residual current 31 in i th busbar along the earth path. ~ The first sample system which is chosen as a numerical application is shown in figure 4. In this system which is originally balanced (i.e. SD=S =S,p=S, K S I S,Qaj=3S), a parameter such as K is used to cause unba lancing for the phases. Therefore the phase powers are Sr ?s - T K 100 load (66.67-K),<, v 100 -load; 100 l-load; (20) XXIXFigure 4 The sample power system chosen to analyse un balanced loading. Here K is assumed to vary in the range 23.33 to 42.33. The variation of line losses depending on the parameter K is given in figure 5 as a result of the power flow analysis carried out using equations 11,12,13 for the sample system which is converted to unbalanced loading via parameter K. As seen from figure 5, the losses are minimum at the point K=33.33 which defines balanced loading according to equation 20. Three different spans such as a=300, A00, 500 m and three different soil resistivities such as p=100,500, 1000 fim are assumed for the sample system, and the earth path losses were computed by equations 14, 15, 16, 17, 18. For ijj, and /m, 0.01 and 0.003 are chosen. The variation of earth path losses depending on the parameter K is given in figure 6. Annual energy losses are also computed assuming that the load is such that it is full load for 2000 hours and half load for 6000 hours per year, respectively. K.58 U.38 K 28 28 30 2 J 34 unbalanced balanced loading Figure 5 For the sample system the variation of the line losses depending on a parameter K. XXXFigure 6 The variation of the earth path losses depending on the parameter K (the span is 300 meters). The second sample system is Diliskelesi Substation in Turkey (figure 7). The daily unbalanced load values (for March 16, 1988) for six different consumers in this system had been recorded, and from them the system losses were computed. To show how these losses vary, daily losses variation of the load `Çolakoğlu 1` is shown in figure 8. Considering the numerical results in the sample systems, additional line losses occur. In addition earth path losses due to the residual current appear in case of unbalanced loading. Increase of losses due to unbalan ced loading in power systems will increase the variable operating costs, and then the optimal operation of power systems will be affected. XXXI-gdH 154/3^5 kV 100 MVA Dil iskelesi Substation. Çolakoçjlu-1 J-` £. Çolakoğlu-2 f Çolakoğlu-3 ÇoIakogiu-4 İ Yazıcı -1 Yazıcı - 2 Tuzla- 1 Tuzla- 2 Figure 7 The single-line diagram for Diliskelesi Substation. ÇOLAKOSUI- I * I 234 5 S 7 89SlianKSSI718BZ)2!Z2a UM Figure 8 The time variations of power losses due to the positive, negative and zero sequence currents in the load `Çolakoglu 1`. XXXII