* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download PSA NOTES
Scattering parameters wikipedia , lookup
Buck converter wikipedia , lookup
Mechanical-electrical analogies wikipedia , lookup
Stray voltage wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Immunity-aware programming wikipedia , lookup
History of electric power transmission wikipedia , lookup
Power engineering wikipedia , lookup
Mains electricity wikipedia , lookup
Ground (electricity) wikipedia , lookup
Electrical substation wikipedia , lookup
Alternating current wikipedia , lookup
Distribution management system wikipedia , lookup
Two-port network wikipedia , lookup
Three-phase electric power wikipedia , lookup
Nominal impedance wikipedia , lookup
Fault tolerance wikipedia , lookup
Network analysis (electrical circuits) wikipedia , lookup
Zobel network wikipedia , lookup
EE6501 POWER SYSTEM ANALYSIS 1 UNIT I INTRODUCTION 2 Power system network 3 SINGLE LINE DIAGRAM It is a diagrammatic representation of a power system in which the components are represented by their symbols. 4 COMPONENTS OF A POWER SYSTEM 1.Alternator 2.Power transformer 3.Transmission lines 4.Substation transformer 5.Distribution transformer 6.Loads 5 MODELLING OF GENERATOR AND SYNCHRONOUS MOTOR 1Φ equivalent circuit of generator 1Φ equivalent circuit of synchronous motor 6 MODELLING OF TRANSFORMER K E2 N 2 I 1 E1 N1 I 2 R2 K2 X X 1 X 2 ' X 1 22 K R01 R1 R2 ' R1 X 01 =Equivalent resistance referred to 1o =Equivalent reactance referred to 1o 7 MODELLING OF TRANSMISSION LINE Π type T type 8 MODELLING OF INDUCTION MOTOR 1 1) =Resistance representing load s R RS Rr ' =Equivalent resistance referred to stator Rr ' ( X X S X r' =Equivalent reactance referred to stator 9 per unit=actual value/base value Let KVAb=Base KVA kVb=Base voltage Zb=Base impedance in Ω Zb kVb 2 MVAb kVb 2 KVAb 1000 10 Changing the base of per unit quantities Let z = actual impedance(Ω) Z b = base impedance (Ω) Z p.u Let Z * MVAb Z Z 2 2 Zb kVb kVb MVAb kVb,old & MVBb,old kVb,new & MVBb,new represent old base values represent new base values 11 Z p.u ,old Z * MVAb ,old kV b , old Z (1) 2 Z p.u ,old * MVAb ,old kV b , old Z p.u ,new (2) 2 Z * MVAb ,new kV kV * kV 2 (3) b , new Z p.u ,new Z p.u ,old b , old b , new 2 2 * MVAb ,new MVAb ,old 12 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 YY,Δ-Y p.u value always less than unity. 13 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. 2. Shunt branches of the transformer equivalent circuit neglected. 14 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 REACTANCE DIAGRAM FOR THE GIVEN POWER SYSTEM NETWORK 16 PROCEDURE TO FORM REACTANCE DIAGRAM FROM SINGLE DIAGRAM 1.Select a base power kVAb or MVAb 2.Select a base voltage kVb 3. The voltage conversion is achieved by means of transformer kVb on LT section= kVb 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 kV * kV 2 X p.u ,new X p.u ,old b , old b , new 2 * MVAb ,new MVAb ,old 17 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 1 Z p Z ps Z pt Z st ' 2 1 ' Z s Z ps Z st ' Z pt 2 1 ' Zt Z ps Z pt Z st ' 2 Z ps = Leakage impedance measured in 1o with 2o short circuited and tertiary open. Z pt = Leakage impedance measured in 1o with tertiary short circuited and 2o open. Z st ' = Leakage impedance measured in 2o with tertiary short circuited and 1o open and referred to primary 19 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 20 BUS ADMITTANCE(Y BUS) MATRIX Y BUS can be formed by 2 methods 1.Inspection method 2.Singular transformation Y BUS = Y11 Y12 Y1n Y Y Y 21 22 2n Y Y Y nn n1 n 2 21 INSPECTION METHOD For n bus system Diagonal element of Y BUS n Yii yij j 1 Off Diagonal element of Y BUS Yij yij 22 SINGULAR TRANSFORMATION METHOD A y A T Y BUS = Where [y]=primitive admittance A=bus incidence matrix 23 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 zb from a new bus p to the reference bus Addition of new bus will increase the order the Zbus matrix by 1 zoriginal 0 Zbus,new zb 0 (n+1)th column and row elements are zero except the diagonal diagonal element is zb 24 Case 2: adding a branch impedance zb from a new bus p to the existing bus q Addition of new bus will increase the order the Zbus 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 Zqq+Zb Case 3:adding a branch impedance zb 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 Zqq+Zb and (n+1)th row and column should be eliminated using the following formula Z jk ,act Z jk Z j ( n1) Z ( n1) k Z ( n1)( n1) j 1,2...n; k 1,2..n 25 Case 4:adding a branch impedance zb 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 b Z hh Z qq 2 Z hq and (n+1)th row and column should be eliminated using the following formula Z jk ,act Z jk Z j ( n1) Z( n1) k Z ( n1)( n1) j 1,2...n; k 1,2..n 26 UNIT II POWER FLOW ANALYSIS 27 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. 2.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 28 ITERATIVE METHOD n I p YpqVq q 1 S p Pp jQp VP I p * Pp jQp VP * n YpqVq q 1 The above Load flow equations are non linear and can be solved by following iterative methods. 1.Gauss seidal method 2.Newton Raphson method 3.Fast Decoupled method 29 GAUSS SEIDAL METHOD For load bus calculate |V| and δ from Vpk+1 equation p 1 n P jQ 1 p p k 1 k 1 k Vp YpqVq YpqVq k * Ypp (VP ) q 1 q p 1 For generator bus calculate Q from QPK+1 equation p 1 n k 1 k * k 1 k Qp 1*Im (VP ) YpqVq YpqVq q p q 1 30 • Check Qp,calk+1 with the limits of Qp • If Qp,calk+1 lies within the limits bus p remains as PV bus otherwise it will change to load bus • Calculate δ for PV bus from Vpk+1 equation • Acceleration factor α can be used for faster convergence • Calculate change in bus-p voltage k 1 k 1 Vp Vp Vp k • If |ΔVmax |< ε, find slack bus power otherwise increase the iteration count (k) • Slack bus power= S S G L 31 NEWTON RAPHSON METHOD n Pi Qi Vi V j Yij ij i j j 1 n Pi Vi V j Yij cos( ij i j ) j 1 n Qi Vi V j Yij sin( ij i j ) j 1 J 2 J 4 V Pi k P J 1 Q J 3 Pi k Pi sch Qi k Qi sch Qi k 32 • Calculate |V| and δ from the following equation i k 1 i k k Vi k 1 Vi k Vi k • If Pi k Qi k • stop the iteration otherwise increase the iteration count (k) 33 FAST DECOUPLED METHOD J2 & J3 of Jacobian matrix are zero 0 J 4 V P P J1 Q Q J 4 V V V P B ' Vi P J1 Q 0 Q B '' V Vi B ' 1 V B '' P V 1 Q V 34 i k 1 i k k Vi k 1 Vi k Vi k This method requires more iterations than NR method but less time per iteration It is useful for in contingency analysis 35 COMPARISION BETWEEN ITERATIVE METHODS Gauss – Seidal Method 1. Computer memory requirement is less. 2. 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 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 UNIT III FAULT ANALYSIS-BALANCED FAULT 38 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. 39 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. 40 Short Circuit Capacity (SCC) SCC V 0 I F IF VT ZT 2 SCC 1 Sb ,1 VT MVA / ZT ZT p.u SCC 3 If Sb ,3 ZT MVA p .u SCC 3 *106 3 *VL ,b *106 41 Procedure for calculating short circuit capacity and fault current Draw a single line diagram and select common base Sb MVA and kV Draw the reactance diagram and calculate the total p.u impedance from the fault point to source (Thevenin impedance ZT) Determine SCC and If 42 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 zf • Prefault voltages at all the buses are V1 (0) V (0) 2 . Vbus (0) Vk (0) . V (0) n • Draw the Thevenin equivalent circuit i.e Zeroing all voltage sources and add voltage source V (0) at faulted bus k and draw the reactance k diagram 43 • The change in bus voltage due to fault is Vbus V1 . . V k . Vn • The bus voltages during the fault is Vbus ( F ) Vbus (0) Vbus • 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 44 Vbus Z bus I bus Z11 . Z1k . . . . . Vbus Z k1 . Z kk . . . . . Z . Z . nk n1 Vk ( F ) Vk (0) Z kk I k ( F ) Z1n 0 . . Z kn I k ( F ) . . Z nn 0 Vk ( F ) Z f I k ( F ) Vk (0) Ik (F ) Z kk Z f Vi ( F ) Vi (0) Z ik I k ( F ) 45 UNIT IV FAULT ANALYSIS – UNBALANCED FAULTS 46 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) 2. 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) 47 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 2. Negative sequence components 3. Zero sequence components 48 Positive sequence components Three phasors with equal magnitudes, equally displaced from one another by 120o and phase sequence is same as that of original phasors. Va1 ,Vb1 ,Vc1 Negative sequence components Three phasors with equal magnitudes, equally displaced from one another by 120o and phase sequence is opposite to that of original phasors. V , V , V a2 b2 c2 Zero sequence components Three phasors with equal magnitudes and displaced from one another by 0o Va 0 ,Vb 0 ,Vc 0 49 RELATIONSHIP BETWEEN UNBALANCED VECTORS AND SYMMETRICAL COMPONENTS Va Va 0 Va1 Va 2 Vb Vb 0 Vb1 Vb 2 Vc Vc 0 Vc1 Vc 2 Va 1 1 V 1 a 2 b Vc 1 a 1 1 A 1 a 2 1 a 1 Va 0 a Va1 a 2 Va 2 1 a a 2 Similarly we can obtain for currents also 50 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. 51 SEQUENCE NETWORK SEQUENCE NETWORK FOR GENERATOR positive sequence network negative sequence network Zero sequence network 52 SEQUENCE NETWORK FOR TRANSMISSION LINE positive sequence network negative sequence network Zero sequence network 53 SEQUENCE NETWORK FOR TRANSFORMER positive sequence network negative sequence network Zero sequence network 54 SEQUENCE NETWORK FOR LOAD positive sequence network negative sequence network Zero sequence network 55 SINGLE LINE TO GROUND FAULT Ib 0 Ic 0 Va Zf Ia Ia1 Ia2 Ia0 Ia / 3 Consider a fault between phase a and ground through an impedance zf Ea Ia1 Z1 Z 2 Z 0 3Z f 56 LINE TO LINE (LL) FAULT Ia 0 I c Ib Vb Vc Ib Z f Ia2 Ia1 Ia 0 0 Consider a fault between phase b and c through an impedance zf Va1 Va2 Zf Ia1 Ia1 Ea Z1 Z 2 3Z f Ib I c jEa Z1 Z 2 3Z f 57 DOUBLE LINE TO GROUND (LLG) FAULT Ia0 0 Ia1 Ia2 Ia0 0 Vb =Vc Z f (Ib Ic ) 3Zf Ia0 Va0 Va1 Vb 3Zf Ia0 Ea Ia1 Z1 Z 2 (Z 0 3Z f ) / ( Z 2 Z 0 3Z f ) Consider a fault between phase b and c through an impedance zf to ground 58 UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX SINGLE LINE TO GROUND FAULT USING Zbus Consider a fault between phase a and ground through an impedance zf at bus k For a fault at bus k the symmetrical components of fault current Ik 0 Ik1 Ik 2 Vk (0) Z kk1 Z kk 2 Z kk 0 3Z f 59 LINE TO LINE (LL) FAULT Consider a fault between phase b and c through an impedance zf Ik 0 0 I k1 I k 2 Vk (0) Z kk 1 Z kk 2 Z f 60 DOUBLE LINE TO GROUND (LLG) FAULT Consider a fault between phase b and c through an impedance zf to ground Ik1 Ik 2 Ik 0 Vk (0) 2 0 f Z ( Z 3 Z ) Z kk 1 kk2 kk 0 Z kk Z kk 3Z f Vk (0) Z kk 1I k 1 Z kk 2 Vk (0) Z kk 1I k 1 Z kk 0 3Z f Ik ( F ) Ik b Ik c 61 BUS VOLTAGES AND LINE CURRENTS DURING FAULT Vi 0 ( F ) 0 Z ik 0 I k 0 Vi1 ( F ) Vi 0 (0) Z ik 1I k 1 Vi 2 ( F ) 0 Z ik 2 I k 2 Iij 0 Iij1 Iij 2 Vi 0 ( F ) V j 0 ( F ) Z ij 0 Vi1 ( F ) V j1 ( F ) Z ij1 Vi 2 ( F ) V j 2 ( F ) Z ij 2 62 UNIT V STABILITY ANALYSIS 63 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 64 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. 65 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. 66 Swing Equation for Single Machine Infinite Bus System • The equation governing the motion of the rotor of a synchronous machine d 2m J 2 Ta Tm Te dt where J=The total moment of inertia of the rotor(kg-m2) m =Singular displacement of the rotor Tm=Mechanical torque (N-m) Te=Net electrical torque (N-m) Ta=Net accelerating torque (N-m) 67 m smt m d m d m sm dt dt d 2 m d 2 m 2 2 dt dt d 2 m J m 2 pa pm pe dt • Where pm is the shaft power input to the machine pe is the electrical power pa is the accelerating power 68 J m M d 2 m M pa pm pe dt 2 2H M S machine sm pa pm pe 2 H d 2 m sm dt 2 S machine S machine 2 H d 2 pa pm pe s dt 2 H=machine inertia constant s 2 f H d 2 pa pm pe f 0 dt 2 f0 d 2 f 0 p p sin pa m 2 max dt 2 H H d dt d f 0 d 2 pa p.u dt H dt 2 p.u δ and ωs are in electrical radian 69 Swing Equation for Multimachine System Smachine =machine rating(base) S system =system base H system d 2 pa pm pe p.u 2 f dt Smachine H system H machine S system 70 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 2. Transient stability 71 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. 72 • Mid Term Stability It represents transition between short term and long term responses. Typical ranges of time periods. 1. Short term : 0 to 10s 2. 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. 2. Poor co-ordination of control and protection equipment. 3. Insufficient active/reactive power reserves. 73 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. 74 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 δ0 and δc 75 Once a fault occurs, the machine starts accelerating. Once the fault is cleared, the machine keeps on accelerating before it reaches its peak at δc , The area of accelerating A1 The area of deceleration is given by A2 If the two areas are equal, i.e., A1 = A2, then the power system will be stable 76 Critical Clearing Angle (δcr) maximum allowable value of the clearing time and angle for the system to remain stable are known as critical clearing time and angle. δcr expression can be obtained by substituting δc = δcr in the equation A1 = A2 Substituting Pe = 0 in swing equation Integrating the above equation 77 Replacing δ by δcr and t by tcr in the above equation, we get the critical clearing time as 78 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. 79 Numerical Integration methods Modified Euler’s method Runge-Kutta method 80 MODIFIED EULER’S METHOD • Using first derivative of the initial point next point is obtained dX • the step t1 t0 t X1 p X 0 t dt • Using this x1p dx/dt at x1p=f(t1, x1p) • Corrected value is X 1P X ic1 dx dx p dt dt X1 X0 X0 2 dx dx p dt dt X i1 Xi Xi 2 t t 81 Numerical Solution of the swing equation • Input power pm=constant • At steady state pe=pm, pm p1max 0 sin 1 p1max E' V X1 • At synchronous speed 0 0 p2 max E' V X2 82 The swing equation H d 2 pa pm pe 2 f 0 dt f0 d 2 f 0 pa pm p2max sin 2 dt H H d dt d f 0 d 2 pa 2 dt H dt Applying Modified Eulers method to above equation t1 t0 t d t dt i i p1 i d ip1 i t dt i 83 • The derivatives at the end of interval d p p i 1 dt i1 f0 d pa p dt i1 H p i 1 The corrected value d d p dt dt i1 c i i 1 i 2 t d d p dt dt i1 i ic1 i 2 t 84 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 K1k f1 ( k , k ) t l1k f 2 ( k , k ) t K1k l1k k K f1 ( , )t 2 2 Kk lk l2k f 2 ( k 1 , k 1 )t 2 2 k K2 l2k k k k K 3 f1 ( , )t 2 2 k k K l l3k f 2 ( k 2 , k 2 )t 2 2 k K3 l3k k k k K 4 f1 ( , )t 2 2 Kk lk l4k f 2 ( k 3 , k 3 )t 2 2 K1k 2 K 2k 2 K 3k K 4k k 6 k k l1 2l2 2l3k l4k k 6 k 2 k 85 • Compute the change in state vector k K k • k 1 2 K 2k 2 K3k K 4k 6 l1k 2l2k 2l3k l4k 6 Evaluate the new state vector k 1 k k k 1 k k • Evaluate the internal voltage behind transient reactance using the relation E pk 1 E pk cos pk 1 j E pk sin pk 1 • • • • Check if t<tc 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