BEKP 4773 : POWER SYSTEM ANALYSIS BY DR AZIAH KHAMIS Introduction To determine the steady state analysis of an interconnected power system during normal operation This system is assumed to be operating under balanced condition and is represented by a single-phase network The network contains hundreds of nodes and branches with impedances specified in per-unit on a common MVA base LOAD/POWER FLOW STUDIES Load-flow studies are performed to determine the steady-state operation of an electric power system. It calculates the voltage drop on each feeder, the voltage at each bus, and the power flow in all branch and feeder circuits. It also determine if system voltages remain within specified limits under various contingency (unforeseen event) conditions, and whether the equipment such as transformers and conductors are overloaded. Load-flow studies are often used to identify the need for additional generation, capacitive, or inductive VAR support, or the placement of capacitors and/or reactors to maintain system voltages within specified limits. Losses in each branch and total system power losses are also calculated. Necessary for planning, economic scheduling, and control of an existing system as well as planning its future expansion THE IMPORTANCE OF POWER FLOW ANALYSIS ➢ Back bone of power system ➢ Planning ➢ Operation ➢ Economic scheduling ➢ Exchange of power between utilities ➢ Transient stability ➢ Contingency studies REQUIREMENT FOR SUCCESSFUL OPERATION OF POWER SYSTEM ▪ Generator supplies the demand & cater for the losses ▪ Bus voltages magnitudes → close to rated values ▪ Generators operates within specified real & reactive power limit ▪ Transmission lines & transformers are not overloaded Therefore, the utilities need some kind of computation tool or program in order to successfully obtained these information for operation purposes- it’s called power flow or load flow program SCOPE ▪ Model the power system using the nodal voltage method. ▪ Present a careful formulation of the basic power flow problem ▪ Investigate its solution by the: ▪ Gauss Seidel Method and Newton Raphson Method and analyze the solution characteristics THE IMPORTANT STEPS IN SOLVING THE POWER FLOW STUDIES: ▪ Previously (recall your earlier power engineering subjects), you have learned how to model the components of power system (Gen, Tx, T-line) from single line diagram ▪ Now, you’re going to learn the steady-state analysis of an interconnected power system during normal operation ▪ which have hundreds/thousands of nodes and branches ▪ impedances is converted into per units on a common MVA base. ▪ The way to analyze these are using “network equations” such as: ▪ If analyze Voltages & Current only and formulate the admittance matrix →”NodeVoltage Method” (if currents known and want to solve for voltages only , or vice versa) ▪ If analyze Power Flow → “Iteration Method” (if power are known, and will become nonlinear) thus, is called power flow/load flow equations. E.g. Gauss Seidel, Newton Raphson ▪ TO APPLY THE PER UNIT SYSTEM IN ORDER TO GENERATE THE IMPEDANCE, REACTANCE AND ADMITTANCE DIAGRAM FROM ONE LINE DIAGRAM. ▪ TO FURTHER DETERMINE THE ADMITTANCE MATRIX IN ORDER TO USE FOR FURTHER PERFORMANCE IN THE POWER SYSTEM ANALYSIS. G1 T1 G2 T2 Line 1 Bus 1 Bus 2 Line 2 Line 3 Using per unit system taught earlier Bus 3 Bus 4 1) single line diagram Analysis using Gauss Seidel/NR/FD to solve for P, Q & V. Using conversion Y=1/Z OR Analysis using Nodal Voltage again to solve V &I 2) reactance/impedance diagram 4) bus admittance matrices Using Nodalvoltage method (I=YV) NOTE : Figure 2,3 & 4 are in PER UNIT! 3) Admittance diagram ▪ The most common way to represent a power system network is to use the node-voltage method: “Given the voltages of generators at all generator nodes, and knowing all impedances of machines and loads, one can solve for all the currents in the typical node voltage analysis methods using Kirchoff's current law (KCL)”. ▪ First the generators are replaced by equivalent current sources and the node equations are written in the form: [I]=[Y][V] I = injected current vector/matrix, Y = is the admittance vector/matrix V = is the node voltage vector/matrices. Bus admittance matrix is just a matrix of all interconnected admittances between busses. ▪ Step 1: Number all the nodes of the system from 0 to n. Node 0 is the reference node (or ground node). ▪ Step 2: Replace all generators by equivalent current sources in parallel with an admittance. ▪ Step 3: Replace all lines, transformers, loads to equivalent admittances whenever possible. The rule for this is simple: where and y, z are generally complex numbers. ▪ Step 4: The bus admittance matrix Y is then formed by inspection as follows : ▪ The current vector/matrix is next found from the sources connected to nodes 0 to n. If no source is connected, the injected current would be 0. ▪ The equations which result are called the node-voltage equations and are given the "bus" subscript in power studies thus: or I = YV DERIVATION OF NODE VOLTAGE EQUATION In order to obtain the node voltage equations : • Impedances are expressed in per-unit on a common MVA base • Impedances are converted to admittance • Nodal solutions is based upon kirchoff’s current law DERIVATION OF NODE VOLTAGE EQUATION DERIVATION OF NODE VOLTAGE EQUATION (USING KCL) : I1 = y10V1 + y12 (V1 − V2 ) + y13 (V1 − V3 ) I 2 = y20V2 + y12 (V2 − V1 ) + y23 (V2 − V3 ) 0 = y23 (V3 − V2 ) + y13 (V3 − V1 ) + y34 (V3 − V4 ) 0 = y34 (V4 − V3 ) Rearrange, I1 = ( y10 + y12 + y13 )V1 − y12V2 − y13V3 I 2 = − y12V1 + ( y20 + y y12 + y23 )V2 − y23V3 0 = − y13V1 − y23V2 + ( y13 + y23 + y34 )V3 − y34V4 0 = − y34V3 + y34V4 Y11 = y10 + y12 + y13 Node equation reduces to: Y22 = y 20 + y y12 + y23 Y33 = y13 + y23 + y34 Y44 = y34 Y12 = Y21 = − y12 Y13 = Y31 = − y y13 Y23 = Y32 = − y23 Y34 = Y43 = − y34 Y14 = Y41 = 0 , Y24 = Y42 = 0 BUS ADMITTANCE MATRIX Ibus = Ybus Vbus Vector of the injected bus currents Vector of bus voltages measured from the reference node Bus Admittance Matrix Diagonal element ➔ self-admittance / driving point admittance Off-diagonal element ➔ mutual admittance / transfer admittance ➔ equal to the negative of the admittance between the node EXAMPLE (TO FIND THE Y BUS) Ibus = Ybus Vbus Bus admittance matrix for the network in previous figure is: EXAMPLE 1: 1 2 3 4 ▪ Figure above shows the one line diagram of a simple four-bus system. Table below gives the line impedances identified by the buses on which these terminate. The shunt admittance at all the buses is assumed negligible. (a) Find the Ybus assuming that the line shown dotted is not connected. (b) What modifications need to be carried out in Ybus if the line shown dotted is connected. Line, Bus to bus 1-2 1-3 2-3 2-4 3-4 R, pu X, pu 0.05 0.10 0.15 0.10 0.05 0.15 0.30 0.45 0.30 0.15 SOLUTION: ▪ (a) Ybus without dotted line. Line, Bus to bus 1-2 1-3 2-3 2-4 3-4 Ybus = 1 – j3 G, pu B, pu 2.0 1.0 0.666 1.0 2.0 - 6.0 - 3.0 - 2.0 - 3.0 - 6.0 0 -1 + j3 0 0 1.666 – j5 -0.666 + j2 -1 + j3 -1 + j3 -0.666 + j2 3.666 – j11 -2 + j6 0 -1 + j3 -2 + j6 3 – j9 SOLUTION ▪ (b) Ybus with dotted line. Line, Bus to bus 1-2 1-3 2-3 2-4 3-4 Ybus = 3 – j9 G, pu B, pu 2.0 1.0 0.666 1.0 2.0 - 6.0 - 3.0 - 2.0 - 3.0 - 6.0 -2 + j6 -1 + j3 0 -2 + j6 3.666 – j11 -0.666 + j2 -1 + j3 -1 + j3 -0.666 + j2 3.666 – j11 -2 + j6 0 -1 + j3 -2 + j6 3 – j9 EXAMPLE 2 SOLUTION: Solution: