EE2351 POWER SYSTEM ANALYSIS

Transcription

1 EE351 POWER SYSTEM ANALYSIS A.Ahamed Riazudeen EEE DEPARTMENT 1

2 UNIT I INTRODUCTION

3 Power system network 3

4 SINGLE LINE DIAGRAM It is a diagrammatic representation of a power system in which the components are represented by their symbols. 4

5 COMPONENTS OF A POWER SYSTEM 1.Alternator.Power transformer 3.Transmission lines 4.Substation transformer 5.Distribution transformer 6.Loads 5

6 MODELLING OF GENERATOR AND SYNCHRONOUS MOTOR 1Φ equivalent circuit of generator 1Φ equivalent circuit of synchronous motor 6

7 MODELLING OF TRANSFORMER K E N I E N I ' R R01 R1 R R1 K X X X X X K ' =Equivalent resistance referred to 1 o =Equivalent reactance referred to 1 o 7

8 MODELLING OF TRANSMISSION LINE Π type T type 8

9 MODELLING OF INDUCTION MOTOR ' 1 R r ( 1) s R R R S S ' r X X X =Resistance representing load ' r =Equivalent resistance referred to stator =Equivalent reactance referred to stator 9

10 per unit=actual value/base value Let KVA b =Base KVA kv b =Base voltage Z b =Base impedance in Ω Z b kv kvb KVAb 1000 b MVA b 10

11 Changing the base of per unit quantities Let z = actual impedance(ω) Z b = base impedance (Ω) Z Z Z * MVAb p. u Zb kvb kvb Z MVA b Let kvb, old & MVB, kv & MVB b old b, new b, new represent old base values represent new base values 11

12 Z Z Z b, old p. u, old Z Z * * b, old p. u, old b, old kv Z b, old * MVA kv MVA MVA b, new p. u, new kv b, new (1) () (3) Z kv b, old b, new p. u, new Z p. u, old * * kv MVA b, new b, old MVA 1

13 ADVANTAGES OF PER UNIT CALCULATIONS The p.u impedance referred to either side of a 1Φ transformer is same The manufacturers provide the impedance value in p.u The p.u impedance referred to either side of a 3Φ transformer is same regardless of the 3Φ connections Y- Y,Δ-Y p.u value always less than unity. 13

14 IMPEDANCE DIAGRAM This diagram obtained by replacing each component by their 1Φ equivalent circuit. Following approximations are made to draw impedance diagram 1. The impedance b/w neutral and ground omitted.. Shunt branches of the transformer equivalent circuit neglected. 14

15 REACTANCE DIAGRAM It is the equivalent circuit of the power system in which the various components are represented by their respective equivalent circuit. Reactance diagram can be obtained after omitting all resistances & capacitances of the transmission line from impedance diagram 15

16 REACTANCE DIAGRAM FOR THE GIVEN POWER SYSTEM NETWORK 16

17 PROCEDURE TO FORM REACTANCE DIAGRAM FROM SINGLE DIAGRAM 1.Select a base power kva b or MVA b.select a base voltage kv b 3. The voltage conversion is achieved by means of transformer kv b on LT section= kv b on HT section x LT voltage rating/ht voltage rating 4. When specified reactance of a component is in ohms p.u reactance=actual reactance/base reactance specified reactance of a component is in p.u X kv b, old b, new p. u, new X p. u, old * * kv MVA b, new b, old MVA 17

18 p.u. calculation of 3 winding transformer Zp=Impedance of primary winding Zs =Impedance of secondary winding Zt =Impedance of tertiary winding Short circuit test conducted to find out the above 3 impedances 18

19 1 ' Zp Zps Zpt Z st ' 1 ' Zs Zps Zst Zpt ' 1 ' Zt Z ps Zpt Z st Z ps Z pt ' Z st = Leakage impedance measured in 1 o with o short circuited and tertiary open. = Leakage impedance measured in 1 o with tertiary short circuited and o open. = Leakage impedance measured in o with tertiary short circuited and 1 o open and referred to primary 19

20 PRIMITIVE NETWORK It is a set of unconnected elements which provides information regarding the characteristics of individual elements. it can be represented both in impedance & admittance form 0

21 BUS ADMITTANCE(Y BUS) MATRIX Y BUS can be formed by methods 1.Inspection method.singular transformation Y BUS = Y Y Y Y Y Y Y 1 Y Y n 1 n n n nn 1

22 INSPECTION METHOD For n bus system Diagonal element of Y BUS Y ii n j1 1 Off Diagonal element of Y BUS y ij Y ij y ij

23 SINGULAR TRANSFORMATION METHOD Y BUS = T A y A Where [y]=primitive admittance A=bus incidence matrix 3

24 ALGORITHM FOR FORMATION OF THE BUS IMPEDANCE MATRIX Modification of Zbus matrix involves any one of the following 4 cases Case 1:adding a branch impedance z b from a new bus p to the reference bus Addition of new bus will increase the order the Z bus matrix by 1 Z bus, new z original 0 0 zb (n+1)th column and row elements are zero except the diagonal diagonal element is z b 4

25 Case : adding a branch impedance z b from a new bus p to the existing bus q Addition of new bus will increase the order the Z bus matrix by 1 The elements of (n+1)th column and row are the elements of qth column and row and the diagonal element is Z qq +Z b Case 3:adding a branch impedance z b from an existing bus p to the reference bus The elements of (n+1)th column and row are the elements of qth column and row and the diagonal element is Z qq +Z b and (n+1)th row and column should be eliminated using the following formula Z Z j( n1)( n1) k Zjk, act Zjk j 1,... n; k1,.. n Z( n1)( n 1) 5

26 Case 4:adding a branch impedance z b between existing buses h and q elements of (n+1)th column are elements of bus h column bus q column and elements of (n+1)th row are elements of bus h row bus q row the diagonal element= Z Z Z Z and (n+1)th row and column should be eliminated using the following formula b hh qq hq Z Z j( n1)( n1) k Zjk, act Zjk j1,... n; k1,.. n Z( n1)( n 1) 6

27 UNIT II POWER FLOW ANALYSIS 7

28 BUS CLASSIFICATION 1.Slack bus or Reference bus or Swing bus: V and δ are specified. P and Q are un specified, and to be calculated..generator bus or PV bus or Voltage controlled bus: P and V are specified. Q and δ are un specified, and to be calculated 3.Load bus or PQ bus: P and Q are specified. V and δ are un specified, and to be calculated 8

29 I p pq q q1 ITERATIVE METHOD * p p p P p p * P n Y V S P jq V I P V jq p n Y pq V q1 The above Load flow equations are non linear and can be solved by following iterative methods. q 1.Gauss seidal method.newton Raphson method 3.Fast Decoupled method 9

30 GAUSS SEIDAL METHOD For load bus calculate V and δ from V p k+1 equation V 1 P jq Y V Y V p1 n k1 p p k1 k p k * pq q pq q Y pp () V P q 1 q p 1 For generator bus calculate Q from Q P K+1 equation Q V Y V Y V p1 n k1 k * k1 k p 1*Im() P pq q pq q q1 qp 30

31 Check Q p,cal k+1 with the limits of Q p If Q p,cal k+1 lies within the limits bus p remains as PV bus otherwise it will change to load bus Calculate δ for PV bus from V p k+1 equation Acceleration factor α can be used for faster convergence Calculate change in bus-p voltage V V V k1 k1 k p p p If ΔV max < ε, find slack bus power otherwise increase the iteration count (k) Slack bus power= S S G L 31

32 NEWTON RAPHSON METHOD n P Q V V Y i i i j ij ij i j j 1 n P V V Y co s() i i j ij ij i j j 1 n Q V V Y sin() i i j ij ij i j j 1 P J J 1 Q J 3 J 4 V P P P k sch k i i i Q Q Q k sch k i i i 3

33 Calculate V and δ from the following equation k1 k k i i V V V k1 k k i i i If k Pi Q k i stop the iteration otherwise increase the iteration count (k) 33

34 FAST DECOUPLED METHOD J & J 3 of Jacobian matrix are zero J P V J Q P P J Q Q J V V V 34 ' '' 1 ' 1 '' i i P B V Q B V V P B V Q V B V

35 k1 k k i i V V V k1 k k i i i This method requires more iterations than NR method but less time per iteration It is useful for in contingency analysis 35

36 COMPARISION BETWEEN ITERATIVE METHODS Gauss Seidal Method 1. Computer memory requirement is less.. Computation time per iteration is less. 3. It requires less number of arithmetic operations to complete an iteration and ease in programming. 4. No. of iterations are more for convergence and rate of convergence is slow (linear convergence characteristic. 5. No. of iterations increases with the increase of no. of buses. 36

37 NEWTON RAPHSON METHOD Superior convergence because of quadratic convergence. It has an 1:8 iteration ratio compared to GS method. More accurate. Smaller no. of iterations and used for large size systems. It is faster and no. of iterations is independent of the no. of buses. Technique is difficult and calculations involved in each iteration are more and thus computation time per iteration is large. Computer memory requirement is large, as the elements of jacobian matrix are to be computed in each iteration. Programming logic is more complex. 37

38 FAST DECOUPLED METHOD It is simple and computationally efficient. Storage of jacobian matrix elements are60% of NR method computation time per iteration is less. Convergence is geometric, to 5 iterations required for accurate solutions Speed for iterations is 5 times that of NR method and -3 times of GS method 38

39 UNIT III FAULT ANALYSIS-BALANCED FAULT 39

40 Need for fault analysis To determine the magnitude of fault current throughout the power system after fault occurs. To select the ratings for fuses, breakers and switchgear. To check the MVA ratings of the existing circuit breakers when new generators are added into a system. 40

41 BALANCED THREE PHASE FAULT All the three phases are short circuited to each other and to earth. Voltages and currents of the system balanced after the symmetrical fault occurred. It is enough to consider any one phase for analysis. SHORT CIRCUIT CAPACITY It is the product of magnitudes of the prefault voltage and the post fault current. It is used to determine the dimension of a bus bar and the interrupting capacity of a circuit breaker. 41

42 Short Circuit Capacity (SCC) 0 SCC V I F I F SCC SCC I f V Z 1 3 T T SCC V T Z S Z 3 b,1 T T p. u L, b b,3 T p. u *10 3 * V * S Z MVA MVA / 4

43 Procedure for calculating short circuit capacity and fault current Draw a single line diagram and select common base S b MVA and kv Draw the reactance diagram and calculate the total p.u impedance from the fault point to source (Thevenin impedance Z T ) Determine SCC and I f 43

44 ALGORITHM FOR SHORT CIRCUIT ANALYSIS USING BUS IMPEDANCE MATRIX Consider a n bus network. Assume that three phase fault is applied at bus k through a fault impedance z f Prefault voltages at all the buses are V1 (0) V (0). Vbus (0) Vk (0). Vn (0) Draw the Thevenin equivalent circuit i.e Zeroing all voltage sources and add voltage source V at faulted bus k and draw the reactance k (0) diagram 44

45 The change in bus voltage due to fault is V1.. Vbus Vk. Vn The bus voltages during the fault is V ()(0) F V V bus bus bus The current entering into all the buses is zero.the current entering into faulted bus k is ve of the current leaving the bus k 45

46 V Z I bus bus bus Z11. Z1 k. Z1 n Vbus Zk1. Zkk.() Z kn Ik F Z. Z. Z 0 n1 nk nn V ()(0)() F V Z I F k k kk k V ()() F Z I F I k f k k () F Z V kk k (0) Z V ()(0)() F V Z I F i i ik k f 46

47 UNIT IV FAULT ANALYSIS UNBALANCED FAULTS 47

48 INTRODUCTION UNSYMMETRICAL FAULTS o One or two phases are involved o Voltages and currents become unbalanced and each phase is to be treated individually o The various types of faults are Shunt type faults 1.Line to Ground fault (LG). Line to Line fault (LL) 3. Line to Line to Ground fault (LLG) Series type faults Open conductor fault (one or two conductor open fault) 48

49 FUNDAMENTALS OF SYMMETRICAL COMPONENTS Symmetrical components can be used to transform three phase unbalanced voltages and currents to balanced voltages and currents Three phase unbalanced phasors can be resolved into following three sequences 1.Positive sequence components. Negative sequence components 3. Zero sequence components 49

50 Positive sequence components Three phasors with equal magnitudes, equally displaced from one another by 10 o and phase sequence is same as that of original phasors. V, V, V a1 b1 c1 Negative sequence components Three phasors with equal magnitudes, equally displaced from one another by 10 o and phase sequence is opposite to that of original phasors. V, V, V a b c Zero sequence components Three phasors with equal magnitudes and displaced from one another by 0 o V, V, V a0 b0 c0 50

51 RELATIONSHIP BETWEEN UNBALANCED VECTORS AND SYMMETRICAL COMPONENTS V V V V a a0 a1 a V V V V b b0 b1 b V V V V c c0 c1 c Va Va 0 V b 1 a a V a1 V 1 a a c V a A 1 a a 1 a a Similarly we can obtain for currents also 51

52 SEQUENCE IMPEDANCE Impedances offered by power system components to positive, negative and zero sequence currents. Positive sequence impedance The impedance of a component when positive sequence currents alone are flowing. Negative sequence impedance The impedance of a component when negative sequence currents alone are flowing. Zero sequence impedance The impedance of a component when zero sequence currents alone are flowing. 5

53 SEQUENCE NETWORK SEQUENCE NETWORK FOR GENERATOR positive sequence network negative sequence network Zero sequence network 53

54 SEQUENCE NETWORK FOR TRANSMISSION LINE positive sequence network negative sequence network Zero sequence network 54

55 SEQUENCE NETWORK FOR TRANSFORMER positive sequence network negative sequence network Zero sequence network 55

56 SEQUENCE NETWORK FOR LOAD positive sequence network negative sequence network Zero sequence network 56

57 SINGLE LINE TO GROUND FAULT Consider a fault between phase a and ground through an impedance z f I 0 b I 0 c V a f Z Ia I I I I /3 I a1 a a0 a a1 Ea Z Z Z 3Z 1 0 f 57

58 LINE TO LINE (LL) FAULT Consider a fault between phase b and c through an impedance z f I 0 I a a0 I V V I I a c I I 0 b Z f b c b a1 V V Z I I I f a1 a a1 a1 Ea Z Z 3Z I 1 a b c f Z1 Z 3Z f je 58

59 DOUBLE LINE TO GROUND (LLG) FAULT Consider a fault between phase b and c through an impedance z f to ground I 0 a0 I I I 0 a1 a a0 f V=V(I Z I) 3Z I V V V 3Z I I b c b c f a0 a1 b a0 a1 Ea f Z Z ( Z 3)/( Z Z Z 3) Z f a f 59

60 UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX SINGLE LINE TO GROUND FAULT USING Z bus Consider a fault between phase a and ground through an impedance z f at bus k For a fault at bus k the symmetrical components of fault current I I I V(0) 0 1 k k k k 1 0 f Zkk Zkk Zkk 3Z 60

61 LINE TO LINE (LL) FAULT Consider a fault between phase b and c through an impedance z f 0 Ik 0 I I V(0) 1 k k k 1 f Z kk Z kk Z 61

62 DOUBLE LINE TO GROUND (LLG) FAULT Consider a fault between phase b and c through an impedance z f to ground I I I 1 k k 0 k Z 1 kk V(0) V(0) Z ( Z f 3) Z Z Z 3Z k 0 kk kk 0 kk kk I 1 1 k kk k Zkk V(0) b I() F I I I 1 1 k kk k 0 f Zkk 3Z c k k k Z Z f 6

63 BUS VOLTAGES AND LINE CURRENTS DURING FAULT V () F 0 Z I i ik k V ()(0) F V I Z i i ik k V () F 0 Z I I I I i ik k 0 ij 1 ij ij 0 0 Vi ()() F V j F 0 Z ij 1 1 Vi ()() F V j F 1 Z Vi ()() F V j F Z ij ij 63

64 UNIT V STABILITY ANALYSIS 64

65 STABILITY The tendency of a power system to develop restoring forces equal to or greater than the disturbing forces to maintain the state of equilibrium. Ability to keep the machines in synchronism with another machine 65

66 CLASSIFICATION OF STABILITY Steady state stability Ability of the power system to regain synchronism after small and slow disturbances (like gradual power changes) Dynamic stability Ability of the power system to regain synchronism after small disturbances occurring for a long time (like changes in turbine speed, change in load) Transient stability This concern with sudden and large changes in the network conditions i.e.. sudden changes in application or removal of loads, line switching operating operations, line faults, or loss of excitation. 66

67 Steady state limit is the maximum power that can be transferred without the system become unstable when the load in increased gradually under steady state conditions. Transient limit is the maximum power that can be transferred without the system becoming unstable when a sudden or large disturbance occurs. 67

68 Swing Equation for Single Machine Infinite Bus System The equation governing the motion of the rotor of a synchronous machine where m d m J T T T dt a m e J=The total moment of inertia of the rotor(kg-m ) =Singular displacement of the rotor T m =Mechanical torque (N-m) T e =Net electrical torque (N-m) T a =Net accelerating torque (N-m) 68

69 t m sm m d m dt d sm d m m d dt dt dt J d dt p p p m m a m e Where p m is the shaft power input to the machine p e is the electrical power p a is the accelerating power m 69

70 J d m M p p p dt H M S machine m sm H d m pa pm pe sm dt S machine S machine H d dt s s H f d M f 0 d dt a m e p p p a m e p p p f a m e f p p p 0 0 m max sin dt H H d dt d f 0 d pa dt H dt p.u a H=machine inertia constant p.u δ and ω s are in electrical radian 70

71 Swing Equation for Multimachine System S S m a ch in e system =machine rating(base) =system base Hsystem d pa pm pe p.u H f system dt H machine S S machine system 71

72 Rotor Angle Stability It is the ability of interconnected synchronous machines of a power system to maintain in synchronism. The stability problem involves the study of the electro mechanical oscillations inherent in power system. Types of Rotor Angle Stability 1. Small Signal Stability (or) Steady State Stability. Transient stability 7

73 Voltage Stability It is the ability of a power system to maintain steady acceptable voltages at all buses in the system under normal operating conditions and after being subjected to a disturbance. The major factor for instability is the inability of the power system to meet the demand for reactive power. 73

74 Mid Term Stability It represents transition between short term and long term responses. Typical ranges of time periods. 1. Short term : 0 to 10s. Mid Term : 10 to few minutes 3. Long Term : a few minutes to 10 s of minutes Long Term Stability Usually these problem be associated with 1. Inadequacies in equipment responses.. Poor co-ordination of control and protection equipment. 3. Insufficient active/reactive power reserves. 74

75 Equal Area Criterion This is a simple graphical method to predict the transient stability of two machine system or a single machine against infinite bus. This criterion does not require swing equation or solution of swing equation to determine the stability condition. The stability conditions are determined by equating the areas of segments on power angle diagram. 75

76 Power-angle curve for equal area criterion multiplying swing equation by on both sides Multiplying both sides of the above equation by dt and then integrating between two arbitrary angles δ 0 and δ c 76

77 Once a fault occurs, the machine starts accelerating. Once the fault is cleared, the machine keeps on accelerating before it reaches its peak at δ c, The area of accelerating A1 The area of deceleration is given by A If the two areas are equal, i.e., A 1 = A, then the power system will be stable 77

78 Critical Clearing Angle (δ cr ) maximum allowable value of the clearing time and angle for the system to remain stable are known as critical clearing time and angle. δ cr expression can be obtained by substituting δ c = δ cr in the equation A1 = A Substituting P e = 0 in swing equation Integrating the above equation 78

79 Replacing δ by δ cr and t by t cr in the above equation, we get the critical clearing time as 79

80 Factors Affecting Transient Stability Strength of the transmission network within the system and of the tie lines to adjacent systems. The characteristics of generating units including inertia of rotating parts and electrical properties such as transient reactance and magnetic saturation characteristics of the stator and rotor. Speed with which the faulted lines or equipments can be disconnected. 80

81 Numerical Integration methods Modified Euler s method Runge-Kutta method 81

82 MODIFIED EULER S METHOD Using first derivative of the initial point next point is obtained p dx X the step 1 X0 t t1 t0 t dt Using this x 1p dx/dt at x 1p =f(t 1, x 1p ) Corrected value is d x d x d t p P X d t 0 X 1 X 1 X 0 t d x d x d t p c X d t i X i 1 X i 1 X i t 8

83 Numerical Solution of the swing equation Input power p m =constant At steady state p e =p m, p 0 1 m ax 1 p m sin p1 m ax E ' V X 1 At synchronous speed p 0 max 0 E ' X V 83

84 The swing equation H d f dt d m max sin dt H H p p p f a m e d dt d f 0 d pa dt H dt f p p p Applying Modified Eulers method to above equation p i t1 t0 t d 1 i t dt p d i 1 i t dt i i a 84

85 The derivatives at the end of interval p i p p i i p i a d dt f d p dt H The corrected value p i i p i i c i i c i i d d dt dt t d d dt dt t

86 Runge-Kutta Method Obtain a load flow solution for pretransient conditions Calculate the generator internal voltages behind transient reactance. Assume the occurrence of a fault and calculate the reduced admittance matrix Initialize time count K=0,J=0 Determine the eight constants K f (,) t k k k 1 1 l f (,) t k k k 1 k k k k K 1 k l1 K f1 (,) k k k k K 1 k l1 l f (,) t t k k k k K k l K 3 f1 (,) t k k k k K k l l3 f (,) t k k k k K 3 k l3 K 4 f1 (,) t k k k k K 3 k l3 l4 f (,) t k k k k K 1 K K k 3 K 4 l l l l k 6 6 k k k k

87 Compute the change in state vector Evaluate the new state vector k1 k k k1 k k Evaluate the internal voltage behind transient reactance using the relation E E cos j E sin k 1 k k 1 k k 1 p p p p p Check if t<t c yes K=K+1 Check if j=0,yes modify the network data and obtain the new reduced admittance matrix and set j=j+1 set K=K+1 Check if K<Kmax, yes start from finding 8 constants k k K K K K k k k k l l l l k k k k