电力系统潮流计算软件设计外文原文及中文翻译

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1、外文原文及中文翻译Modelling and Analysis of Electric Power SystemsPower Flow Analysis Fault AnalysisPower Systems Dynamics and StabilityPreface In the lectures three main topics are covered,i.e. Power flow an analysis Fault current calculations Power systems dynamics and stability In Part I of these notes th

2、e two first items are covered,while Part II givesAn introduction to dynamics and stability in power systems. In appendices brief overviews of phase-shifting transformers and power system protections are given. The notes start with a derivation and discussion of the models of the most common power sy

3、stem components to be used in the power flow analysis.A derivation of the power ow equations based on physical considerations is then given.The resulting non-linear equations are for realistic power systems of very large dimension and they have to be solved numerically.The most commonly used techniq

4、ues for solving these equations are reviewed.The role of power flow analysis in power system planning,operation,and analysis is discussed. The next topic covered in these lecture notes is fault current calculations in power systems.A systematic approach to calculate fault currents in meshed,large po

5、wer systems will be derived.The needed models will be given and the assumptions made when formulating these models discussed.It will be demonstrated that algebraic models can be used to calculate the dimensioning fault currents in a power system,and the mathematical analysis has similarities with th

6、e power ow analysis,soitis natural to put these two items in Part I of the notes. In Part II the dynamic behaviour of the power system during and after disturbances(faults) will be studied.The concept of power system stability is dened,and different types of power system in stabilities are discussed

7、.While the phenomena in Part I could be studied by algebraic equations,the description of the power system dynamics requires models based on differential equations. These lecture notes provide only a basic introduction to the topics above.To facilitate for readers who want to get a deeper knowledge

8、of and insight into these problems,bibliographies are given in the text.Part IStatic Analysis1 IntroductionThis chapter gives a motivation why an algebraic model can be used to de scribe the power system in steady state.It is also motivated why an algebraic approach can be used to calculate fault cu

9、rrents in a power system.A power system is predominantly in steady state operation or in a state that could with sufficient accuracy be regarded as steady state.In a power system there are always small load changes,switching actions,and other transients occurring so that in a strict mathematical sen

10、se most of the variables are varying with the time.However,these variations are most of the time so small that an algebraic,i.e.not time varying model of the power system is justified.A short circuit in a power system is clearly not a steady state condition.Such an event can start a variety of diffe

11、rent dynamic phenomena in the system,and to study these dynamic models are needed.However,when it comes to calculate the fault current sin the system,steady state(static) model swith appropriate parameter values can be used.A fault current consists of two components,a transient part,and a steady sta

12、te part,but since the transient part can be estimated from the steady state one,fault current analysis is commonly restricted to the calculation of the steady state fault currents.1.1 Power Flow AnalysisIt is of utmost importance to be able to calculate the voltages and currents that different parts

13、 of the power system are exposed to.This is essential not only in order to design the different power system components such as generators,lines,transformers,shunt elements,etc.so that these can withstand the stresses they are exposed to during steady state operation without any risk of damages.Furt

14、hermore,for an economical operation of the system the losses should be kept at a low value taking various constraint into account,and the risk that the system enters into unstable modes of operation must be supervised.In order to do this in a satisfactory way the state of the system,i.e.all(complex)

15、 voltages of all nodes in the system,must be known.With these known,all currents,and hence all active and reactive power flows can be calculated,and other relevant quantities can be calculated in the system. Generally the power ow,or load ow,problem is formulated as a nonlinear set of equations f (x

16、, u, p)=0 (1.1)wheref is an n-dimensional(non-linear)functionx is an n-dimensional vector containing the state variables,or states,as components.These are the unknown voltage magnitudes and voltage angles of nodes in the systemu is a vector with(known) control outputs,e.g.voltages at generators with

17、 voltage controlp is a vector with the parameters of the network components,e.g.line reactances and resistancesThe power flow problem consists in formulating the equations f in eq.(1.1) and then solving these with respect to x.This will be the subject dealt with in the first part of these lectures.A

18、 necessary condition for eq.(1.1) to have a physically meaningful solution is that f and x have the same dimension,i.e.that we have the same number of unknowns as equations.But in the general case there is no unique solution,and there are also cases when no solution exists.If the states x are known,

19、all other system quantities of interest can be calculated from these and the known quantities,i.e. u and p.System quantities of interest are active and reactive power flows through lines and transformers,reactive power generation from synchronous machines,active and reactive power consumption by vol

20、tage dependent loads, etc.As mentioned above,the functions f are non-linear,which makes the equations harder to solve.For the solution of the equations,the linearization (1.2)is quite often used and solved.These equations give also very useful information about the system.The Jacobian matrix whose e

21、lements are given by (1.3) can be used form any useful computations,and it is an important indicator of the system conditions.This will also be elaborate on.1.2 Fault Current AnalysisIn the lectures Elektrische Energiesysteme it was studied how to calculate fault currents,e.g.short circuit currents,

22、for simple systems.This analysis will now be extended to deal with realistic systems including several generators,lines,loads,and other system components.Generators(synchronous machines) are important system components when calculating fault currents and their model will be elaborated on and discuss

23、ed.1.3 LiteratureThe material presented in these lectures constitutes only an introduction to the subject.Further studies can be recommended in the following text books:1. Power Systems Analysis,second edition,by Artur R.Bergen and Vijay Vittal.(Prentice Hall Inc.,2000,ISBN0-13-691990-1,619pages)2.

24、Computational Methods for Large Sparse Power Systems,An object oriented approach,by S.A.Soma,S.A.Khaparde,Shubba Pandit(Kluwer Academic Publishers, 2002, ISBN0-7923-7591-2, 333pages) 2 Net work ModelsIn this chapter models of the most common net work elements suitable for power flow analysis are der

25、ived.These models will be used in the subsequent chapters when formulating the power flow problem.All analysis in the engineering sciences starts with the formulation of appropriate models.A model,and in power system analysis we almost invariably then mean a mathematical model,is a set of equations

26、or relations,which appropriately describes the interactions between different quantities in the time frame studied and with the desired accuracy of a physical or engineered component or system.Hence,depending on the purpose of the analysis different models of the same physical system or components m

27、ight be valid.It is recalled that the general model of a transmission line was given by the telegraph equation,which is a partial differential equation, and by assuming stationary sinusoidal conditions the long line equations, ordinary differential equations,were obtained.By solving these equations

28、and restricting the interest to the conditions at the ends of the lines,the lumped-circuit line models (-models) were obtained,which is an algebraic model.This gives us three different models each valid for different purposes.In principle,the complete telegraph equations could be used when studying

29、the steady state conditions at the network nodes.The solution would then include the initial switching transients along the lines,and the steady state solution would then be the solution after the transients have decayed. However, such a solution would contain a lot more information than wanted and,

30、furthermore,it would require a lot of computational effort.An algebraic formulation with the lumped-circuit line model would give the same result with a much simpler model at a lower computational cost.In the above example it is quite obvious which model is the appropriate one,but in many engineerin

31、g studies these lection of the“correct”model is often the most difficult part of the study.It is good engineering practice to use as simple models as possible, but of course not too simple.If too complicated models are used, the analysis and computations would be unnecessarily cumbersome.Furthermore

32、,generally more complicated models need more parameters for their definition,and to get reliable values of these requires often extensive work.Figure2.1. Equivalent circuit of a line element of length dxIn the subsequent sections algebraic models of the most common power system components suitable f

33、or power flow calculations will be derived.If not explicitly stated,symmetrical three-phase conditions are assumed in the following.2.1 Lines and CablesThe equivalent -model of a transmission line section was derived in the lectures Elektrische Energie System, 35-505.The general distributed model is

34、 characterized by the series parametersR=series resistance/km per phase(/km)X=series reactance/km per phase(/km)and the shunt parametersB=shunt susceptance/km per phase(siemens/km)G=shunt conductance/km per phase(siemens/km)As depicted in Figure2.1.The parameters above are specific for the line or c

35、able configuration and are dependent on conductors and geometrical arrangements.From the circuit in Figure2.1the telegraph equation is derived,and from this the lumped-circuit line model for symmetrical steady state conditions,Figure2.2.This model is frequently referred to as the -model,and it is ch

36、aracterized by the parametersFigure2.2. Lumped-circuit model(-model)of a transmission line between nodes k and m.Note. In the following most analysis will be made in the p.u.system.For impedances and admittances,capital letters indicate that the quantity is expressed in ohms or siemens,and lower cas

37、e letters that they are expressed in p.u.Note.In these lecture notes complex quantities are not explicitly marked asunder lined.This means that instead of writing we will write when this quantity is complex. However,it should be clear from the context if a quantity is real or complex.Furthermore,we

38、will not always use specific type settings for vectors.Quite often vectors will be denoted by bold face type setting,but not always.It should also be clear from the context if a quantity is a vector or a scalar.When formulating the net work equations the node admittance matrix will be used and the s

39、eries admittance of the line model is needed (2.1)With (2.2)and (2.3)For actual transmission lines the series reactance and the series resistance are both positive,and consequently is positive and is negative.The shunt susceptancekm and the shunt conductance km are both positive for real line sectio

40、ns.In many cases the value of km is so small that it could be neglected.The complex currents and in Figure2.2 can be expressed as functions of the complex voltages at the branch terminal nodes k and m: (2.4) (2.5)Where the complex voltages are (2.6) (2.7)This can also be written in matrix form as (2

41、.8)As seen the matrix on the right hand side of eq.(2.8)is symmetric and the diagonal elements are equal.This reflects that the lines and cables are symmetrical elements.2.2 Transformers We will start with a simplified model of a transformer where we neglect the magnetizing current and the no-load l

42、osses .In this case the transformer can be modelled by an ideal transformer with turns ratioin series with a series impedance which represents resistive(load-dependent)losses and the leakage reactance,see Figure2.3.Depending on if is real ornon-real(complex)the transformer is in-phase or phase-shift

43、ing.Figure2.3. Transformer model with complex ratio （）Figure2.4. In-phase transformer model2.2.1In-Phase Transformers Figure2.4shows an in-phase transformer model indicating the voltage at the internalnon-physicalnode p.In this model the ideal voltage magnitude ratio(turns ratio)is (2.9)Since k = p,

44、this is also the ratio between the complex voltages at nodes k and p, (2.10) There are no power losses(neither active nor reactive)in the ideal transformer(the k-p part of the model),which yields (2.11)Then applying eqs.(2.9)and(2.10)gives (2.12) Figure2.5. Equivalent -model for in-phase transformer

45、which means that the complex currents and are out of phase by 180 since R. Figure2.5 represents the equivalent -model for thein-phase transformer in Figure2.4.Parameters A, B,and C of this model can be obtained by identifying the coefficients of the expressions for the complex currents and associate

46、d with the models of Figures2.4 and 2.5.Figure2.4 gives (2.13) (2.14)or in matrix form (2.15) As seen the matrix on the right hand side of eq.(2.15) is symmetric,but the diagonal elements are not equal when .Figure2.5 provides now the following: (2.16) (2.17)or in matrix form (2.18) Identifying the

47、matrix elements from the matrices in eqs. (2.15) and (2.18) yields (2.19) (2.20) (2.21)2.2.2 Phase-Shifting Transformers Phase-shifting transformers,such as the one represented in Figure2.6,are used to control active power flows;the control variable is the phase angle and the controlled quantity can

48、 be,among other possibilities,the active power flow in the branch where the shifter is placed.In Appendix A the physical design of phase-shifting transformer is described. A phase-shifting transformer affects both the phase and magnitude of the complex voltages and ,without changing their ratio,i.e.

49、, (2.22)Thus, and ,using eqs. (2.11) and (2.22) (2.23)Figure2.6. Phase-shifting transformer with As with in-phase transformers,the complex currents andcan be expressed in terms of complex voltages at the phase-shifting transformer terminals: (2.24) (2.25)Or in matrix form (2.26) As seen this matrix

50、is not symmetric if is non-real,and the diagonal matrix elements are not equal if .There is no way to determine parameters A, B,and C of the equivalent -model from these equations,since the coefficient of Em in eq.(2.24)differs from in eq.(2.25),as long as there is non zero phase shift,i.e. R.A phas

51、e-shifting transformer can thus not be represented by a -model.2.2.3Unified Branch ModelThe expressions for the complex currents and for both transformers and shifters derived above depend on the side where the tap is located;i.e., they are not symmetrical.It is how ever possible to develop unified

52、complex expressions which can be used for lines,transformers,and phase-shifters, regardless of the side on which the tap is located(or even in the case when there are taps on both sides of the device).Consider initially the model in Figure2.8 in which shunt elements have been temporarily ignored and

53、 and 。In this case (2.27)and (2.28)which together yield (2.29)and (2.30) (These expressions are symmetrical in the sense that if k and m are interchanged,as in the expression for ,the result is the expression for ,and vice-versa.) Figure2.9 shows the unified branch model.All the devices studied abov

54、e can be derived from this general model by establishing the appropriate definitions of the parameters that appear in the unified model.Thus,for instance,if = =1 is assumed,the result is an equivalent -model of a transmission line;or,if the shunt elements are ignored,and =1 andis assumed,then the re

55、sult is a phase shifting transformer with the tap located on the bus m side.The general expressions for and can be obtained from the model in Figure2.9: (2.31)and (2.32)or (2.33) The transformers model led above were all two-winding transformers.In power systems there are also three-winding and N-wi

56、nding(N 3)transformers,and they can be modelled in a similar way.In stead of linear relationships between two complex current sand two complex voltages,we will have linear relationships between these quantities model led by N N matrices.It should be notes that for N-winding transformers there are le

57、akage reactances between each pair of windings,i.e.in total N(N 1)/2reactances.Since the power ratings of the different windings are not necessarilyequal,as in the two-winding case,it should be noted that the power bases for expressing the reactance sin p.u.are not always equal.Figure2.8. Transforme

58、r symmetrical model.Figure2.9. Unified branch model extended(-model).2.3 Shunt Elements The modelling of shunt elements in the network equations is straightforward and the main purpose here is to introduce the notation and the sign convention to be used when formulating the network equations in the

59、coming chapters.As seen from Figure2.10 the current from a shunt is defined as positive when injected into the bus.This means that (2.34)with being the complex voltage at node k.Shunts are in all practical cases either shunt capacitors or reactors.From eq.(2.34) the injected complex power is (2.35)2

60、.4 Loads Load model is an important topic in power system analysis. When formulating the load flow equations for high voltage systems,a load is most often the in feed of power to a network at a lower voltage level,e.g.a distribution network.Often the voltage in the distribution systems is kept const

61、ant by controlling the tap-positions of the distribution transformers which means that power,active and reactive,in most cases can be regarded as independent of the voltage on the high voltage side.This means that the complex power is constant,i.e.independent of the voltage magnitude . Also in this case the current is defined as positive when injected into the bus,see Figure2.11.In the general case the complex load current can be writte