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1、1 Instructor:Kai Sun Fall 2014 ECE 421/599 Electric Energy Systems 6 Power Flow Analysis 2 Introduction Power flow(load flow)analysis Steady-state analysis of an interconnected power system To solve the power flow equations for Vi(i.e.|Vi|and i)and Si(i.e.Pi and Qi)Basis of power system analysis and
2、 design Assumptions for a power system to be studied Balanced conditions Represented by a single-phase network Modeled by nodes(buses)and branches All impedances/admittances are in per unit on a common MVA base(typically 100MVA)3 Bus Admittance Matrix Branch admittance Apply KCL to nodes(buses)1-4 I
3、i:current injected by an equivalent current source 11ijijijijyzrjx=+1101121213132202122123232332133134343443()()()()0()()()0()Iy VyVVyVVIy VyVVyVVyVVyVVyVVyVV=+=+=+=z12=z10=z13=z20=z23=z34=110121311221332121201223223313 12321323343344343344()()0()0 IyyyVy Vy VIy VyyyVy Vy Vy VyyyVy Vy Vy V=+=+=+=+11
4、10121322201223331323344434YyyyYyyyYyyyYy=+=+=+=122112133113233223344334YYyYYyYYyYYy=14410YY=42240YY=0111101 101101110()(0)(0)IEV yE yV yIV y=+=+E1 V1 E2 I01 0,1,=nniiijijjjj ijj iIVyy V4 Ibus:currents injected by external current sources Vbus:bus voltages relative to a reference(not included),usuall
5、y the ground Ybus:bus admittance matrix(symmetric and sparse)Diagonal elements:called self-admittances or driving point admittances Off-diagonal elements:called mutual admittances or transfer admittances Zbus=Y-1bus:bus impedance matrix 111 1122133144221 1222233244331 1322333344441 1422433444IY VY V
6、Y VY VIY VY VY VY VIY VY VY VY VIY VY VY VY V=+=+=+=+11111211222122221212.:.:.ininiiiiiniinninnnnnIVYYYYYYYYIVYYYYIVYYYYIV=busbusbusI=YV0,niiijjj iYy=ijjiijYYy=-1busbus busbus bus=V=Y IZIji0,1,1 =nnniiijijjijjjj ijj ijIVyy VY V5 If it is known that E1=1.10o pu and E2=1.00o pu 8.502.505.0002.508.755.
7、0005.005.0022.5012.500012.5012.50jjjjjjjjjjjj=busY10.500.400.450.450.400.480.440.440.450.440.5450.5450.450.440.5450.625jjjjjjjjjjjjjjjj=busbusZYHow many non-zero elements for a system with N buses(not including the ground)and M branches?111022201.11.1 pu1.01.01.25 pu0.8EIjzjEIjzj=12341.11.0501.251.0
8、4001.04501.045VjVjVV=bus=Zz12=z10=z13=z20=z23=z34=2M+N E1 E2 Why are they same?6 A more general power flow study How to solve all bus voltages and line real and reactive power flows?z12=z10=z13=z20=z23=j0.2 z34=|V1|=1|V2|=1.05pu P4+jQ4=1+j0.2 pu P2=0.5pu 7 Power Flow Equation Consider a typical bus
9、of an n-bus system All lines represented by equivalent models Admittances are in pu on a common MVA base Apply KCL Solve|Vi|,i,Pi,Qi and then calculate Pij,Qij 011220121 122()().().()(.).iiiiiiiijijininiiiiniiiijjinnIy Vy VVyVVy VVyVVyyyyVy Vy Vy Vy Vji=+=+01 nniiijijjjjIVyy Vji=*j=+=iiiiiSPQV I*j=i
10、iiiPQIV*01 j =nniiiijijjjjiPQVyy VjiV01|j=nniiiiijijjjjjiiPQVyyVjiVn complex nonlinear algebraic equations(2n real equations)with 4n real quantities can be solved by iterative techniques 8 Power Flow Solution Determining:|Vi|and i(magnitude and phase angle of each bus voltage)Pij and Qij(real and re
11、active power flows in each line)The system is assumed to be operating under balanced conditions and a single-phase model is used 4 quantities,i.e.|Vi|,i,Pi and Qi,are associated with each bus System buses are usually classified into three types Slack bus (swing bus or V-bus)Selected as the reference
12、 having|Vi|and i fixed Pi and Qi are usually unlimited and can take any values to make up the gap between system generation and load Load buses (P-Q buses)Pi and Qi are specified Regulated buses (generator buses or P-V buses)Pi and Vi are specified.Limits of Qi are also specified 9 More Thinking on
13、Types of Buses|Vi|i Pi Qi V-X X P-Q X X P-V X X Q-V X X P-X X Q-X X Need to know two of|Vi|,i,Pi and Qi(either constant or observed)Relax the other two within upper and lower limits May assume more types of buses for a variety of natures of buses 10 A more general power flow study z12=z10=z13=z20=z2
14、3=j0.2 z34=|V1|=1|V2|=1.05pu P4+jQ4=1+j0.2 pu P2=0.5pu 1=0 (slack bus)11 Solution of Nonlinear Algebraic Equations Gauss-Seidel Method Newton-Raphson Method 12 Gauss-Seidel Method:Example 6.2 32()6940f xxxx=+=32164()999xxxg x=+=3 roots:x1,2=1 and x3=4(0)2x=(0)32164()(2)(2)2.2222999g x=+=x(2)=g(x(1)=
15、2.5173|x(2)-x(1)|=0.2951 x(3)=g(x(2)=2.8966|x(3)-x(2)|=0.3793 x(4)=g(x(3)=3.3376|x(4)-x(3)|=0.4410 x(5)=g(x(4)=3.7398|x(5)-x(4)|=0.4022 x(6)=g(x(5)=3.9568|x(6)-x(5)|=0.2170 x(7)=g(x(6)=3.9988|x(7)-x(6)|=0.0420 x(8)=g(x(7)=4.0000|x(8)-x(7)|=0.0012 x(9)=g(x(8)=4.0000|x(9)-x(8)|=|V0|,i 0,Load buses:|Vi
16、|V0|,i 0 Initial guess could be Vi(0)=10o without a better estimation()*()1)1,(kjkschschniiijjjkiiiiiPjQYYVVV=+=x(k+1)=g(x(k)PQ Buses 21 PV Buses Pi=Pisch and|Vi|are specified Starting from an initial estimate of i(0)Vi(0)=|Vi|i(0)Since|Vi|is specified,only VI,i(k+1)=ImVci(k+1)is retained Continue t
17、he iterations until or,the power mismatch,i.e.the largest element in P and Q Using acceleration factor=1.31.7 *()()1)1(Im nijjkkkijiVY VQ=+=()*(1(1)1,=+=kjkkikcschiniijjj iiiiPjYVVYQV2(1)(1),2|()+=kkR iI iiVVV(1)(1),()(,),|+kkkR iIkiIiRiVVVV Update Vi(k+1)=VR,i(k+1)+j VI,i(k+1)()*()()(1)(1,()1)()+=+
18、kjkkkischniijjjiiiikkiciiPjYYVVVVVQ22*1 niiiijjjPjQVY V=Slack Bus Line Flows and Losses At bus i:At bus j:Power loss in line i j:00()ijliijijiiIIIy VVy V=+=+00()jiljijjijjIIIy VVy V=+=+ijiijSV I=jijjiSV I=LijijjiSSS=+yij 23 Tap Changing Transformers a is the per unit off-nominal tap position(usually
19、,|a|=0.91.1)Complex number for phase shifting transformers 1xjVVa=*ijIaI=()itixIy VV=ttijyyVVa=*1jiIIa=*2|ttijyyVVaa=+*2|ttiittjjyyIVayyIVaa=ST=VxIi*=-Vj Ij*Non-tap side Tap side Ybus is not symmetrical with a phase shifting transformer Equivalent circuit if a is real(ignoring phase shifting)24 Exam
20、ple 6.7 (V-and P-Q buses)y23=10-j20 y13=10-j30 y23=16-j32()*()1)1,(kjkschschniiijjjkiiiiiPjQYYVVV=+=Using the G-S method to find the power flow solution:(a)Determine the voltage phasors at P-Q buses 2 and 3 accurate to 4 decimal places(b)Find the slack bus real and reactive power(c)Determine the lin
21、e flows and losses.Show line flow directions in a power-flow diagram(Solve P1,Q1,|V2|,2,|V3|,3,Sij and Slij)P1,Q1|V2|,2|V3|,3 Step 1.Check what are given Step 2.Set initial estimates and iterate 25*1 niiiijjjPjQVY V=Step 3.Calculate P and Q of the slack bus 26 Step 4.Calculate line flows and losses
22、27 Example 6.8(V-,P-Q and P-V buses)Line charging susceptances are neglected.Obtain the power flow solution by the G-S method including line flows and line losses(Solve P1,Q1,|V2|,2,Q3,3,Sij and Slij)y23=10-j20 y13=10-j30 y23=16-j32 Note:|Vc3(1)|=1.0378 1.04=|V3|P1,Q1|V2|,2 Q3,3 Step 1.Check what ar
23、e given Step 2.Set initial estimates and iterate(1),3RV28()*()1)1,(kjkschschniiijjjkiiiiiPjQYYVVV=+=*()()1)1(Im nijjkkkijiVY VQ=+=2(1)3(21)31.0Im4Re+=kckVV(3)3(4)3(5)3(6)3(7)31.039540.008331.039780.008731.039890.008931.039930.009001.039950.00903cccccVjVjVjVjVj=()*(1(1)1,=+=kjkkcschniijjjiiiiikiQY VV
24、PYVjBus 2(P-Q):Solve|V2|,2 Bus 3(P-V):Solve Q3,3 29 Newton-Raphson Method Based on Taylors series expansion at an initial estimate of the solution Ignore all terms with orders 2 ()f xc=(0)(0)()f xxc+=2(0)(0)(0)(0)(0)221()()()()2!dfd ff xxxcdxdx+=(0)(0)(0)(0)()()dfxcf xcdx(0)(0)(0)()cxdfdx=Comparison
25、:G-S method ignores all differential terms(orders 1)(1)(0)(0)=+xxx30 Iteration 1:(0)(1)Iteration k+1:(k)(k+1)Until c=f(x)is actually approximated by the tangent line on the curve at x(k).(0)(0)(1)(0)(0)(0)(0)(0)(0)()()()ccf xxxxxxdfdfdxdx=+()()(1)()()()()()()()()()kkkkkkkkkccf xxxxxxdfdfdxdx+=+()()(
26、)()()kkkcf xxxdfdx=+()()()()kkkcxdfdx=(1)()|kkxx+It is straight line function c=kx+b 31 Example 6.4 2()3129df xxxdx=+(0)(0)32()0(6)6(6)9(6)450ccf x=+=Let x(0)=6(0)2()3(6)12(6)945dfdx=+=(0)(0)(0)501.111145()cxdfdx=x(0)x(1)(1)(0)(0)61.11114.8889xxx=+=(2)(1)(1)13.44314.88894.278922.037xxx=+=(5)(4)(4)0.
27、00954.00114.00009.0126xxx=+=(3)(2)(2)2.99814.27894.040512.5797xxx=+=(4)(3)(3)0.37484.04054.00119.4914xxx=+=x(2)(0)c(0)x()(1)()()()()kkkkcf xxxdfdx+32 N-dimensional System 1(1)()(1)()()()kkkkkkXXXXJC+=+=+()1()()2()kkkknxxXx=()11()()22()()()()kkkknncfcfCcf=()f xc=1121212212(,)(,)(,)nnnnnfx xxcfx xxcfx
28、 xxc=()()()11112()()()222()12()()()12()()()()()()()()()kkknkkkknkkknnnnfffxxxfffxxxJfffxxx=Jacobian Matrix:1(1)()()()()()()kkkkkkdfxxxxcdx+=+()()()kkccf x=33 Example 6.5 Use the N-R method to find the intersections of the curves 22124xx+=121xex+=112221xxxJe=If x1(0)=2,x2(0)=-2:k C J x x 1 -4.0000 4.
29、0000 -4.0000 -0.6424 1.3576 -4.3891 7.3891 1.0000 0.3576 -1.6424 2 -0.5406 2.7152 -3.2848 -0.2989 1.0587 -1.2445 3.8869 1.0000 -0.0825 -1.7249 3 -0.0962 2.1173 -3.4499 -0.0530 1.0056 -0.1576 2.8825 1.0000 -0.0047 -1.7296 4 -0.0028 2.0112 -3.4592 -0.0014 1.0042 -0.0040 2.7336 1.0000 -0.0000 -1.7296 5
30、 -0.0000 2.0083 -3.4593 -0.0000 1.0042 -0.0000 2.7296 1.0000 -0.0000 -1.7296 J tells the fastest direction(gradient)toward a solution 34 Compared to the Gauss-Seidel Method Since higher-order terms are ignored,the N-R method also needs the initial estimation to be sufficiently close to the actual so
31、lution The N-R method converges much faster N-R method:quadratic convergence(ignoring the 2nd and higher orders)G-S method:linear convergence(ignoring the 1st and higher orders)The N-R method has some computational issues:Needs J(k)-1 during each iteration,which is computationally intense 35 Dealing
32、 with J(k)-1 Try not to update J(k)so often(at least not in every iteration)Apply LU decomposition(triangular factorization)1(1)()()kkkXJC+=()1)kkkJXC+=()()1kkkkCL UX+=In MATLAB,the solution of JX=C can be obtained by X=J C 36*1,1 =niiiiiiijjjj iinijjjPjQIVYY VVY V=+()()()*(1)1),nijjjkkkiikkiiiiijjY
33、PQVYVV=+=*(1)()(1)(1)kkikkjjiinijVVYPQj+=G-S method X=G(X)N-R method F(X)=C JX=C =1234|V|PQJJJJPi+jQi Newton-Raphson Power Flow Solution 1*=ijnijiijVPVYjQ*1niiiijjjPjQVY V=1|niiijjijjjVYV=+Use polar forms:|iiiijijijVVYY=37 1(,)eq.(1)|cos()=+iinjiijijijjPfVVY|V|1(,)eq.(2)|sin()=+iinjiijijijjQgVVY|V|(
34、)()()()222222()()()()2()2(2()()()()2222)(22()2()2|=kkkknnkkkkkknnnnnnkkknkknknnknnPPPPVVPPPPVVQQQQPQQVQQPV()2()()2()()(2)|kknkknkkknnnnQQVVVV 1234JJP=JJQ|V|Algorithm:For the slack bus(say bus 1):No need to include bus 1 in J calculate P1 and Q1 by(1)and(2)at the end For m voltage-controlled(P-V)buse
35、s:solve i by(1)calculate Qi by(2)For the n-1-m load(P-Q)buses left:solve|Vi|and i by(1)and(2)There are 2n-2-m independent(1)s and(2)s J:(2n-2-m)(2n-2-m)P-V bus,(1)(1),(1)(1),(1)(1),(1)(1)=nnnnmnmnnmnm1234JJJJ|V|38 J1 (n-1)(n-1)i slack bus J2 (n-1)(n-1-m)i PV and slack buses J3 (n-1-m)(n-1)i PV and s
36、lack buses J4 (n-1-m)(n-1-m)i PV and slack buses Jacobian Matrix|sin()iijijijijj iiPVVY=+1234JJP=JJQ|V|sin()iijijijijjPV VY=+2|cos|cos()iiiiiiijijijijj iPVYVVY=+|cos()|iiijijijjPVYV=+jiji Diagonal and off-diagonal elements of J1J2 (i slack bus)ji2|sin|sin()iiiiiiijijijijj iQVYVVY=+|sin()|iiijijijjQV
37、YV=+|cos()iijijijijj iiQVVY=+|cos()iijijijijjQVVY=+ji39 Procedure for Power Flow Solution by N-R Method 1.Initial values Load buses:Pisch and Qisch specified,|Vi(0)|i(0)=10 or equal to the slack bus Voltage-regulated buses:|Vi|and Pisch specified,i(0)=0 or the slack bus angle 2.Calculate Pi(k),Pi(k)
38、,Qi(k)and Qi(k)Load buses:calculate Pi(k)and Qi(k)by eq.(1)and(2)and then Voltage-controlled buses:calculate Pi(k)by eq.(1)and 3.Calculate J1,J2,J3 and J4 and solve|Vi(k)|and i(k)from 4.Calculate 5.Iterate Steps 25 until ()()kschkiiiPPP=()()kschkiiiQQQ=(1)()()kkkiii+=+(1)()()|kkkiiiVVV+=+1234JJP=JJQ
39、|V|()()kschkiiiPPP=()()|kkiiPQ(applying triangular factorization and Gaussian elimination)40 How to accelerate the N-R method?Original N-R method J(k)is computed at each iteration Computing J(k)-1 is expensive Ideas:approximate J Constant matrix Block-diagonal matrix(ignoring small off-diagonal elem
40、ents)x(0)x(1)x(2)41 Fast Decoupled Power Flow Solution Some elements of J may be close to 0 Transmission lines usually have a high X/R ratio(close to lossless lines),P is less sensitive to|V|than it is to J2=P/|V|0 Q is less sensitive to than it is to|V|J3=Q/0 F-D method:1234JJP=JJQ|V|=1PPJ 4QQ=J|V|
41、=|V|V|cos()iijijijijVQVVXsin()ijijijijV VPX0 014JPJQ|V|42 J1 (n-1)(n-1)i slack bus J4 (n-1-m)(n-1-m)i PV and slack buses Jacobian Matrix|sin()=+ijijijijijiiVVYP0 014JP=JQ|V|ji2|sin|si|n()=+iiiiijijijiiiijjVYVYQV|sin()=+iiijjijijVVYQji21|sin()|sinnijijijijiiiiijV VYVY=+2|iiiiQVB=|iiiVB2,(|1)iiiiiQBVV
42、|iijVB|sin()=+ijijjijjiiVVYP|sinijijijVVY P=B|V|2|iiiiiiVBVB+|=iiiVB|iijVBQ=B|V|V|B(n-1)(n-1)about P-Q&P-V buses B”(n-1-m)(n-1-m)about P-Q buses -1P=B|V|-1Q|V|=B|V|Dot division(./)|=iiiijijijVVYY43 Proof:cos()cos()1 cos()1iijijijijiiijijijjijijjijiijVQVVXQVyVVyyB In Example 6.7,0.02iiiQBiiiQBPi+jQi
43、44 Compared to the N-R Method F-D method deals with constant Jacobian matrix:no need to update in every iteration It requires more iterations than the N-R method,but each iteration requires considerably less time Overall,the F-D method is much faster The F-D method is very useful in fast contingency
44、 screening 1100=P|V|B|V|QB|V|B(n-1)(n-1)about P-Q&P-V buses B”(n-1-m)(n-1-m)about P-Q buses 45 Examples 6.10&6.12 Obtain the powerflow solution(V2,2,3,P1,Q1 and Q3)by the N-R method and the F-D method y23=10-j20 y13=10-j30 y23=16-j32 205010201030102026521632103016322662busjjjYjjjjjj+=+53.851651.9029
45、22.36068 2.034431.62278 1.892522.36068 2.034458.137771.107135.77709 2.034431.62278 1.892535.77709 2.034467.230951.1737=52323262B=10.0281820.0145450.0145450.023636B=52B=Slack P-Q P-V For F-D Method:2(400250)4.02.5 pu100schjSj+=32002.0 pu100schP=P1,Q1|V2|,2 Q3,3 46 N-R Method 2n-2-m=3 independent equa
46、tions:J has 3x3 elements 222232223333223222222232|PPPVPPPPPVQVQQQV=222121212122222|cos()|cosPVVYVY=+23232323|cos()VVY+33131313132323232|cos()|cos()PVVYVVY=+233333|cos()VY+222121212122222|sin()|sinQVVYVY=+23332323|sin()VVY+221212121232323232|sin()|sin()PVVYVVY=+2232323233|sin()PVVY=+21212121222223232
47、3232|cos()2|cos|cos()|PVYVYVYV=+3322332322|sin()PVVY=+331313131323232323|sin()|sin()PVVYVVY=+333232322|cos()|PVYV=+221212121232323232|cos()|cos()QVVYVVY=+2232323233|cos()QVVY=+212121212222232323232|sin()2|sin|sin()|QVYVYVYV=+Eq(1)Eq(2)47 Procedure of the N-R Method 1.Initial values:2.Calculate P2(k)
48、,P3(k)and Q2(k)by eq.(1)and(2)and then P2(k),P3(k),and Q2(k),e.g.:3.Calculate J,solve|V2(k)|,2(k)and 3(k)and then|V2(k+1)|,2(k+1)and 3(k+1),e.g.:4.Stop after 3 iterations:5.Calculate:11.05 0 puV=31.04 puV=(0)2|1.0V=(0)20.0=(0)30.0=(0)(0)222(0)(0)333(0)(0)2224.0(1.14)2.86002.0(0.5616)1.43842.5(2.28)0
49、.2200schschschPPPPPPQQQ=(0)2(0)3(0)22.860054.2800033.2800024.860001.438433.2800066.0400016.640000.220027.140016.6400049.72000|V=(0)(1)22(0)(1)33(0)(1)220.045263(0)(0.045263)0.04526530.007718(0)(0.007718)0.007718|0.026548|1(0.026548)0.97345VV=+=+=+=(1)2(1)3(1)20.09921851.72467531.76561821.3025670.021
50、71532.98164265.65638315.3790860.05091428.53857717.40283848.103589|V=(1)(2)22(1)(2)33(1)(2)220.0017950.045263(0.001795)0.047060.0009850.007718(0.000985)0.00870|0.001767|0.973451(0.001767)0.971684VV=+=+=+=20.971682.696V=31.040.4988V=2331313131323232323333321111111212121213131313211111112121212|sin()|s