ELE B7 Power Systems Engineering. Power Flow- Introduction

Transcription

1 ELE B7 Power Systems Engneerng Power Flow- Introducton

2 Introducton to Load Flow Analyss The power flow s the backbone of the power system operaton, analyss and desgn. It s necessary for plannng, operaton, economc schedulng and exchange power between utltes. G G G The power flow s also requred for many other applcatons such as short-crcut calculatons, transent stablty and contngency analyss. are Also Specfed at Dr. the A.M. Gaouda generator buses. For the network shown, there are some buses connected wth the generators and other buses are connected to the loads. The Real and Reactve power s known at each Load bus. The Generator oltages The Transmsson Lnes nterconnectng the buses have resstance and nductance. Therefore, the Electrc Current flowng through the lnes results n Electrcal Losses. The Generators n the System Must supply the Total Electrcal Loads pulse the Electrcal Losses. G Slde #

3 There are some constrans should be consdered whle runnng the system:. The Generators Must Operate wthn ther Generaton Capabltes. In Case of An Equpment Over-Loaded Or oltage-lmt olaton.. The Generators Must Delver the requred power at the Desred oltage at the Loads. 3. There should be no bus voltage ether above or below the specfed oltage operatng lmts. 4. There Should be no Over-Loadng of equpment, ncludng Transmsson Lnes and Transformers The Generaton Schedule have to be adjusted and Power Flow n the transmsson lnes have to be Re-routed or Capactor ELEC57, Banks 4/5 have to be swtched n order to brng the system nto ts Normal Operatng Condtons. To Satsfy all the prevous requrement for a Relable Power System Operaton, Power Flow Study s a MUST. The Power flow study s an essental part n power system Operaton, Plannng and Desgn. Slde # 3

4 Power Flow Concept Consder the three-bus power system. Generators (G and G ) are connected to the frst two buses and an electrc load s connected to the thrd bus. The real and reactve power demands are known for the load bus (3). The generator voltages are also specfed at bus and bus. G Bus Bus Bus 3 G The three transmsson lnes nterconnectng the buses contan both resstance and reactance, thus currents flow through these lnes results n electrcal-losses. The two generators (G and G ) must jontly supply the total load requrements and the power losses n the transmsson lnes. The generators are constraned to operate wthn ther power generaton capabltes. The generators are also constraned to delver the requred power at the desred voltage at the customer loads. In addton, there should be no over-loadng of the power system equpments ncludng transmsson lnes and transformers. Furthermore, there should be no bus voltage ether above or below specfed values of the bus voltage operatng lmts. ELEC57, Slde 4/5 # 4

5 Power Flow Analyss Consder the above crcut, f all the components and loads are expressed n terms of constant power loads (.e., n power systems, powers are known rather than currents), then the equaton to be solved are gven by I S * I * It s clear that the set of equatons are nonlnear and the soluton (bus voltages) can only obtaned by teratve technques Slde # 5

6 Power Flow Analyss It s the soluton for the statc operatng condton of a power system. The node voltage method s commonly used for the power system analyss. The formulaton of the network equatons results n complex lnear equatons n terms of node currents. In power systems, powers are known rather than currents. Thus, resultng equatons n terms of power become non-lnear and must be solved by teratve technques. These non-lnear equatons are known as power flow equatons or load flow equatons. The power flow programs compute the voltage magntude and phase angle at each bus bar n the system under steady-state operaton condton. These programs use the bus-voltage data to compute the power flow n the network and the power losses for all equpment and transmsson lnes. Slde # 6

7 What are the power flow equatons? What do you expect to get by solvng the power flow equatons? How do we beneft from the soluton of the power flow equatons? Slde # 7

8 Bus Admttance Matrx or bus Frst step n solvng the power flow s to formulate the bus admttance matrx, often call the bus. The bus gves the relatonshps between all the bus current njectons, I, and all the bus voltages,, I = bus The bus s developed by applyng KCL at each bus n the system to relate the bus current njectons, the bus voltages, and the branch mpedances and admttances Slde # 8

9 Formulate the bus admttance matrx for the network shown n the Fgure. The Impedance dagram of the system s as ndcated. Shunt elements are gnored. Soluton: The node voltage method s commonly used for the power system analyss. Where, Or I I I I 3 4 I = bus [ bus Example bus ] bus G j. j. 8 Bus Bus j. j.8 j.4 Bus 3 Bus 4 G j. Impedance dagram Slde # 9

10 The system can be represented n terms of ts admttance elements as shown, where: y y yj Z j j. y j. 5 j. j. 5 y y j j. 8 I y y Bus Bus y y 3 y 3 I Applyng KCL at each node (bus), then I ( y y y3 ) y y3 3 y 34 Bus 3 Bus 4 Admttance dagram I ( y y y3 ) y y3 3 ( y y y ) y y y y y Slde #

11 defned: y y y3 y y y3 33 y3 y3 y34 44 y Then, the Node oltage Equaton s: I I I I 3 4 = I [ ] Or bus bus bus and bus 3 4 [ bus ] y 3 3 y3 3 3 y y Substtutng the values, then the bus admttance matrx of the network s: I bus [ Z bus ] I bus bus j8.5 j.5 j5. j.5 j8.75 j5. j5. j5. j.5 j.5 j.5 j.5 Slde #

12 bus General Form The dagonal terms,, are the self admttance terms, equal to the sum of the admttances of all devces ncdent to bus. The off-dagonal terms, j, are equal to the negatve of the sum of the admttances jonng the two buses. Wth large systems bus s a sparse matrx (that s, most entres are zero) Shunt terms, such as wth the π-lne model, only affect the dagonal terms. I y y Bus Bus y y y 3 3 y 34 Bus 3 Bus 4 I ( y y ( y ( y 3 ( ) ) ) y 3 ) ( y ( y y ( y 3 ( ) ) ) y 3 ) ( y 3 ( y ( y y 3 ( y ) ) ) y 43 ) ( ) ( ) ( y ) 34 ( y43 ) Slde #

13 Two Bus System Example c ( ) I j6 Z.3 j.4 I j5.9 j6 I j6 j5.9 Slde # 3

14 Usng the bus If the voltages are known then we can solve for the current njectons: I bus If the current njectons are known then we can solve for the voltages: bus I Z I where Z s the bus mpedance matrx bus bus Slde # 4

15 Solvng for Bus Currents For example, n prevous case assume..8 j. Then j5.9 j j.7 j6 j5.9.8 j j.88 Therefore the power njected at bus s * I S. (5.6 * j.7) 5.6 j.7 S I (.8 j.) ( 5.58 j.88) 4.64 j.4 Slde # 5

16 Solvng for Bus oltages For example, n prevous case assume 5. I 4.8 Then * I j5.9 j j.9 j6 j j.98 Therefore the power njected s S (.738 j.9) 5.37 j4.5 * S I (.738 j.98) ( 4.8).35 j5.7 Slde # 6

17 Power Flow Analyss When analyzng power systems we know nether the complex bus voltages nor the complex current njectons Rather, we know the complex power beng consumed by the load, and the power beng njected by the generators plus ther voltage magntudes Therefore we can not drectly use the bus equatons, but rather must use the power balance equatons I S * I * Slde # 7

18 Defntons: Power Flow Analyss, cont d The startng pont s the sngle lne dagram from whch the nput data can be obtaned. The nput data: Bus data, transmsson lne data and transformer data As shown, 4 varables are assocated wth each bus k: the voltage magntude k, phase angle δ k, net power P k and reactve power k. At each bus, two of these varables are specfed as nput data and the other two are unknowns to computed by the power flow analyss Slde # 8

19 Power Balance Equatons From KCL we know at each bus n an n bus system the current njecton, I, must be equal to the current that flows nto the network I I I I G D k k Snce I = we also know bus I I I G D k k k n n The network power njecton s then S * I Slde # 9

20 Power Balance Equatons, cont d * * n n * * k k k k k k S I Ths s an equaton wth complex numbers. Sometmes we would lke an equvalent set of real power equatons. These can be derved by defnng = G jb k k k = k k j j e = Recall e cos jsn Slde #

21 Real Power Balance Equatons n n * * jk k k k k k k k S P j e ( G jb ) n (cos jsn )( G jb ) k Resolvng nto the real and magnary parts n P ( G cos B sn ) P P k n k ( Gk snk Bk cos ) k k k k k k k k k k k G D k G D Slde #

22 Power Flow Requres Iteratve Soluton In the power flow we assume we know S and the bus. We would lke to solve for the 's. The problem s the below equaton has no closed form soluton: * * n n * * k k k k k k S I Rather, we must pursue an teratve approach. Slde #

23 Gauss Iteraton There are a number of dfferent teratve methods we can use. We'll consder two: Gauss and Newton. Wth the Gauss method we need to rewrte our equaton n an mplct form: x = h(x) To terate we frst make an ntal guess of x, x, ( v+) ( v) and then teratvely solve x hx ( ) untl we fnd a "fxed pont", x, ˆ such that x ˆ h(x). ˆ () Slde # 3

24 Gauss Iteraton Example Example: Solve x- x x ( v) ( v) Let k = and arbtrarly guess x x () and solve ( v) ( v) k x k x Slde # 4

25 Stoppng Crtera A key problem to address s when to stop the teraton. Wth the Guass teraton we stop when ( v) ( v) ( v) ( v) x wth x = x x If x s a scalar ths s clear, but f x s a vector we need to generalze the absolute value by usng a norm x ( v) j Two common norms are the Eucldean & nfnty n x x x max x Slde # 5

26 Gauss Power Flow We frst need to put the equaton n the correct form S S S * * n n * * k k k k k k n n * * * * I kk kk k k * * I n k k k k k k, k * n S * kk k, k n Slde # 6

27 Gauss Two Bus Power Flow Example A MW, 5 Mvar load s connected to a generator through a lne wth z =. + j.6 p.u. and lne chargng of 5 Mvar on each end ( MA base). Also, there s a 5 Mvar capactor at bus. If the generator voltage s. p.u., what s? Slde # 7

28 Gauss Two Bus Example, cont d The unknown s the complex load voltage,. To determne we need to know the.. j.6 bus 5 j5 5 j j5 Hence bus 5 j5 5 j4.7 ( Note B - j5 j.5 j.5) Slde # 8

29 Gauss Two Bus Example, cont d * n S * k, k - +j.5 j ( 5 j 5)(. ) * 5 j4.7 () k Guess. (ths s known as a flat start) k ( v) ( v) v v. j j j j j.553 Slde # 9

30 Gauss Two Bus Example, cont d.96 j Once the voltages are known all other values can be determned, such as the generator powers and the lne flows * * j.3 MW S ( ).3.39 In actual unts P, 3.9 Mvar The capactor s supplyng 5 3. Mvar Slde # 3

31 Slack Bus In prevous example we specfed S and and then solved for S and. We can not arbtrarly specfy S at all buses because total generaton must equal total load + total losses We also need an angle reference bus. To solve these problems we defne one bus as the "slack" bus. Ths bus has a fxed voltage magntude and angle, and a varyng real/reactve power njecton. Slde # 3

32 Gauss wth Many Bus Systems Wth multple bus systems we could calculate new ' s as follows: * n ( v) S ( v) ( v)* kk k, k ( v) ( v) ( v) n h (,,..., ) ( v) But after we've determned we have a better estmate of ts voltage, so t makes sense to use ths new value. Ths approach s known as the Gauss-Sedel teraton. Slde # 3

33 Gauss-Sedel Iteraton Immedately use the new voltage estmates: ( v) ( v) ( v) ( v) 3 n h (,,,, ) ( v) ( v) ( v) ( v) 3 3 n h (,,,, ) ( v) ( v) ( v) ( v) ( v) n h (,,,, ) ( v) ( v) ( v) ( v) ( v) n (,, 3, 4, n h The Gauss-Sedel works better than the Gauss, and s actually easer to mplement. It s used nstead of Gauss. ) Slde # 33

34 Three Types of Power Flow Buses There are three man types of power flow buses Load (P) at whch P/ are fxed; teraton solves for voltage magntude and angle. Slack at whch the voltage magntude and angle are fxed; teraton solves for P/ njectons Generator (P) at whch P and are fxed; teraton solves for voltage angle and njecton specal codng s needed to nclude P buses n the Gauss-Sedel teraton Slde # 34

35 Three Types of Power Flow Buses Each bus can be categorzed nto one of the followng:. Load Bus: Input data: P k and k Output (soluton): k and δ k Load Bus wth no generaton: P k = P Lk k = Lk (nductve) ; k = + Lk (capactve). Swng (Reference) Bus: Only one swng bus (bus #) nput data: δ =. o Output (soluton): P and 3. oltage Controlled Bus: Generators, Swtched shunt capactors, statc var systems Input data: P k, k, kgmax and kgmn Output (soluton): k and δ k Slde # 35

36 Incluson of P Buses n G-S Assumng a power system has n buses, then; one bus wll be consdered as a slack bus and the other buses are load buses (P-buses) and voltage controlled buses (P-buses). Let the system buses be numbered as: Salck bus,3,..., m P buses m, m,..., n P buses For the voltage controlled buses, P and are known & and areunknown Specfed,mn,max The second requrement for the voltage controlled bus may be volated f the bus voltage becomes too hgh or too small. It s to be noted that we can control the bus voltage by controllng the bus reactve power. Slde # 36

37 Therefore, durng any teraton, f the P-bus reactve power volates ts lmts then set t accordng to the followng rule.,mn,max,max set,max,mn set,mn And treat ths bus as P-bus. P & NOTE For P bus and and θ are known are unknown Slde # 37

38 Load flow soluton when P buses are present a. Fnd To solve for at P buses, we must frst make a guess of S * * n k k k * ( ( v) ( v Im n the teraton, we use... n )* k k ( v) S ( v) k P... j ( v) n n ) P j b. Check v+ to see f t s wthn the lmts, mn,max Case : If the reactve power lmts are not volated, calculate v ( v) S ( v)* ( v)* n k k, k ( v) k Use the most updated value of to calculate S. New oltage magntude and angle are obtaned Slde # 38

39 Use Reset the magntude v and For the P-bus voltage. spec v Spec oltage magntude s known for P bus, therefore the new calculated magntude wll not be used. v Spec Case : If the reactve power lmts are volated, v Only the calculated angle wll be updated and used. Or v,max set v,max v,mn set v,mn Consder ths bus as a P-Bus, calculate bus voltage v ( v) v S ( v)* ( v)* v n k k, k v ( v) k The P-bus becomes P-bus and both oltage magntude and angle are calculated and used Slde # 39

40 EXAMPLE: 3 4 Lne Data for the 5 buses Network From Bus To Bus R X G Each lne has an mpedance of.5+j.5 G The shunt admttance s neglected Bus No. Bus code Bus Data for the the 5 buses Network Before load flow soluton olt Mag. olt Angle Load MW Load MAR Gen. MW Gen. MAR Mn. Max. Inject MAR Slack. 5?? P.?? 6 3 P?? P P???? 5 5 Slde # 4

41 For the 5 bus system Construct the bus admttance matrx Fnd,,3,4 and 5 max mn.6 pu. pu bus 3 4 G 5 G Slde # 4

42 SOLUTION: 3 4 bus Constructon y z.5 j.5 5 y y 4 j6 j G G 5 5 y y3 y y3 y y43 y 4 54 j j 55 y5 y5 y 6 j8 j8 y 5 5 y 4 3 j6 j6 bus 4. - J J J J J J J J J J J J J J J J J8. Slde # 4

43 The net scheduled power njected at each bus s: 3 4 Bus No. Bus code olt Mag. olt Angle Load MW Load MAR Gen. MW Gen. MAR Mn. Max. Inject MAR Slack. 5?? G G P P P P.??????? 5 5 5? 6 S,sch ( P,g P,d ) j(,g S,sch ( P,g P,d ) j(,g, d S,sch ( P,g. ) j(, g.5 ), d ) ) S,sch (. ) j(, g ) S, sch 3 (.5 ) j(. ) S, sch 3.5 S, sch 4.5 S, sch 5.5 j. j. j. Slde # 43

44 The known values are: o. spec. The bus admttance matrx s 4. - J J J J J J J J J J J J J J J J J8.. and,max. 6,mn Usng GS method, select the ntal values for the unknowns as: o o o o and o Bus s P Bus Check s wthn the lmts Start the frst teraton,mn,max P j * ( n n ) * o o o Im{ ( o 5 )}.448 Slde # 44

45 ,mn,max.e.; The reactve power lmts are not volated, Calculate: K P j L K o o o L o * L3 3 L4 4 L5 5 ( ) The values for K and L p are computed once n the begnnng and used n every teraton. S, sch. j.448 L 3 K = j.959 L = L 3 = L 4 4 L 5 5 L 4 =. L 5 = Reset the magntude Therefore, o Spec 5. 3 o. o.5.3 oltage magntude s known and fxed for a P bus, therefore the new calculated magntude wll not be used. Slde # 45

46 Bus 3 s P Bus K 3 o o 3 L3 o * L3 L34 4 L35 5 (3 ) K 3 P 3 33 j 3 L L L L K 3 = j.35 L 3 =. L 3 = -.5 L 34 = -.5 L 35 =. Bus 4 s P Bus K 4 P 4 44 j 4 L K 4 3 o o 4 L4 o * L4 L43 3 L45 5 (4 ) L L L K 4 = j.35 L 4 =. L 4 =. L 43 = -.5 L 45 = o Slde # 46

47 Bus 5 s P Bus K L5 o * L5 L53 3 L54 4 (5 ) K 5 = L 5 = L 5 = L 53 =. L 54 = o Start the second teraton Bus s P Bus Check s wthn the lmts.. 6 * Im{ ( The reactve power lmts are volated,mn set,mn. And treat ths bus as P-bus. j. S, sch 5 5 )} Use the most updated value of to calculate the constant K All Buses, 3, 4 and 5 are P Buses. Fnd the bus voltages usng GS method Slde # 47

48 Accelerated G-S Convergence Prevously n the Gauss-Sedel method we were calculatng each value x as x ( v) ( v) h( x ) To accelerate convergence we can rewrte ths as ( v) ( v) ( v) ( v) x x h( x ) x Now ntroduce acceleraton parameter x ( v) ( v) ( v) ( v) x ( h( x ) x ) Wth = ths s dentcal to standard gauss-sedel. Larger values of may result n faster convergence. Slde # 48

49 Accelerated Convergence, cont d Consder the prevous example: x- x ( v) ( v) ( v) ( v) x x ( x x ) Comparson of results wth dfferent values of k Slde # 49

50 Accelerated Convergence, cont d The Effect of Acceleraton Factor Adequate alues of the Acceleraton Factor:.5.7 Slde # 5

51 Gauss-Sedel Advantages Each teraton s relatvely fast (computatonal order s proportonal to number of branches + number of buses n the system Relatvely easy to program Slde # 5

52 Gauss-Sedel Dsadvantages Tends to converge relatvely slowly, although ths can be mproved wth acceleraton Has tendency to mss solutons, partcularly on large systems Tends to dverge on cases wth negatve branch reactances (common wth compensated lnes) Need to program usng complex numbers Slde # 5