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BEE701 POWER SYSTEM ANALYSIS

BEE701 POWER SYSTEM ANALYSIS
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BEE701 POWER SYSTEM ANALYSISUNIT IPOWER SYSTEM COMPONENTSPower system analysisThe evaluation of power system is called as power system analysisFunctions of power system analysis To monitor the voltage at various buses, real and reactive power flow between buses.To design the circuit breakers.To plan future expansion of the existing systemTo analyze the system under different fault conditionsTo study the ability of the system for small and large disturbances (Stability studies)COMPONENTS OF A POWER SYSTEM1.Alternator2.Power transformer3.Transmission lines4.Substation transformer5.Distribution transformer6.LoadsSINGLE LINE DIAGRAMA single line diagram is diagrammatic representation of power system in which the components arerepresented by their symbols and interconnection between them are shown by a straightline9eventhough the system is three phase system0.The ratings and the impedances of thecomponents are also marked on the single line diagram.

Purpose of using single line diagramThe purpose of the single line diagram is to supply in concise form of the significant informationabout the system.Per unit value.The per unit value of any quantity is defined as the ratio of the actual value of the any quantity tothe base value of the same quantity as a decimal.per unit actual value/base valueNeed for base valuesThe components or various sections of power system may operate at different voltage and powerlevels. It will be convenient for analysis of power system if the voltage, power, current andimpedance rating of components of power system are expressed with reference to a common valuecalled base value.Advantages of per unit systemi. Per unit data representation yields valuable relative magnitude information.ii. Circuit analysis of systems containing transformers of various transformation ratios is greatlysimplified.iii. The p.u systems are ideal for the computerized analysis and simulation of complex power systemproblems.iv. Manufacturers usually specify the impedance values of equivalent in per unit of the equipmentsrating. If the any data is not available, it is easier to assume its per unit value than its numerical value.

v. The ohmic values of impedances are refereed to secondary is different from the value as referee toprimary. However, if base values are selected properly, the p.u impedance is the same on the twosides of the transformer.vi. The circuit laws are valid in p.u systems, and the power and voltages equations are simplifiedsince the factors of 3 and 3 are eliminated.Change the base impedance from one set of base values to another setLet Z Actual impedance ,ΩZb Base impedance ,ΩPer unit impedance of a circuit element 𝑍𝑍𝑏 𝑍2π‘˜π‘‰π‘ 𝑍 𝑀𝑉𝐴 π‘π‘˜π‘‰π‘ 2(1)𝑀𝑉𝐴 𝑏The eqn 1 show that the per unit impedance is directly proportional to basemegavoltampere and inversely proportional to the square of the base voltage.Using Eqn 1 we can derive an expression to convert the p.u impedance expressedin one base value ( old base) to another base (new base)Let kVb,oldand MVAb,old represents old base values and kVb,newand MVArepresent new base valueb ,newLet Zp.u,old p.u. impedance of a circuit element calculated on old baseZp.u,new p.u. impedance of a circuit element calculated on new baseIf old base values are used to compute the p.u.impedance of a circuit element ,withimpedance Z then eqn 1 can be written as𝑍𝑝.𝑒 ,π‘œπ‘™π‘‘ 𝑍 𝑑𝑍 𝑍𝑝.𝑒 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘22𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘(2)If the new base values are used to compute thep.u. impedance of a circuit elementwith impedance Z, then eqn 1 can be written as

𝑍𝑝.𝑒 ,𝑛𝑒𝑀 𝑍 𝑀𝑉𝐴 𝑏 ,π‘›π‘’π‘€π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀(3)2On substituting for Z from eqn 2 in eqn 3 we get2𝑍𝑝.𝑒 ,𝑛𝑒𝑀 οΏ½οΏ½ 𝑍𝑝𝑒 ,π‘œπ‘™π‘‘ π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘2π‘˜π‘‰ 𝑏,𝑛𝑒𝑀 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝐴𝑏,𝑛𝑒𝑀 �𝑀(4)The eqn 4 is used to convert the p.u.impedance expressed on one base value to another baseMODELLING OF GENERATOR AND SYNCHRONOUS MOTOR1Ξ¦ equivalent circuit of generator1Ξ¦ equivalent circuit of synchronous motor

MODELLING OF TRANSFORMERMODELLING OF TRANSMISSION LINET typeΞ  type

MODELLING OF INDUCTION MOTORImpedance diagram & approximations made in impedance diagramThe impedance diagram is the equivalent circuit of power system in which the variouscomponents of power system are represented by their approximate or simplified equivalentcircuits. The impedance diagram is used for load flow studies.Approximation:(i) The neutral reactances are neglected.(ii) The shunt branches in equivalent circuit of transformers are neglected.Reactance diagram & approximations made in reactance diagramThe reactance diagram is the simplified equivalent circuit of power system in which the variouscomponents of power system are represented by their reactances. The reactance diagram can beobtained from impedance diagram if all the resistive components are neglected. The reactancediagram is used for fault calculations.Approximation:(i) The neutral reactances are neglected.(ii) The shunt branches in equivalent circuit of transformers are neglected.(iii) The resistances are neglected.(iv) All static loads are neglected.(v) The capacitance of transmission lines are neglected.

PROCEDURE TO FORM REACTANCE DIAGRAM FROM SINGLEDIAGRAMLINE1.Select a base power kVAb or MVAb2.Select a base voltage kVb3. The voltage conversion is achieved by means of transformer kVb on LT section kVb on HT sectionx LT voltage rating/HT voltage rating4. When specified reactance of a component is in ohmsp.u reactance actual reactance/base reactancespecified reactance of a component is in p.uEXAMPLE1. The single line diagram of an unloaded power system is shown in Fig 1.The generator transformerratings are as follows.G1 20 MVA, 11 kV, X’’ 25%G2 30 MVA, 18 kV, X’’ 25%G3 30 MVA, 20 kV, X’’ 21%T1 25 MVA, 220/13.8 kV ( /Y), X 15%T2 3 single phase units each rated 10 MVA, 127/18 kV(Y/ ), X 15%T3 15 MVA, 220/20 kV(Y/ ), X 15%Draw the reactance diagram using a base of 50 MVA and 11 kV on the generator1.Fig 1SOLUTION

Base megavoltampere,MVAb,new 50 MVABase kilovolt kVb,new 11 kV ( generator side)FORMULAThe new p.u. reactance 𝑋𝑝𝑒 ,𝑛𝑒𝑀 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀2 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘Reactance of Generator GkVb,old 11 kVkVb,new 11 kVMVAb,old 20 MVAMVAb,new 50 MVAXp.u,old 0.25p.uThe new p.u. reactance of Generator G 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.25 π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘2π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀11 211 5020 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘ j0.625p.uReactance of Transformer T1kVb,old 11 kVkVb,new 11 kVMVAb,old 25 MVAXp.u,old 0.15p.uMVAb,new 50 MVAThe new p.u. reactance of Transformer T1 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.15 11 211π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀 50252 It is connected to the HT side of the Transformer T1𝐻𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 11 22011 220 π‘˜π‘‰Actual Impedance X actual 100ohmπ‘˜π‘‰Base impedance X base 𝑀𝑉𝐴𝑏 ,𝑛𝑒𝑀𝑏 ,𝑛𝑒𝑀2 220 250 968 π‘œ π‘šπ‘€π‘‰π΄ 𝑏,π‘œπ‘™π‘‘ j0.3 p.uReactance of Transmission LineBase kV on HT side of transformer T 1 π΅π‘Žπ‘ π‘’ π‘˜π‘‰ π‘œπ‘› 𝐿𝑇 𝑠𝑖𝑑𝑒 𝐿𝑇 π‘£π‘œπ‘™ π‘‘π‘Žπ‘”π‘’π‘€π‘‰π΄ 𝑏 ,π‘›π‘’π‘€π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”

p.u reactance of 100 Ω transmission line p.u reactance of 150 Ω transmission line π΄π‘π‘‘π‘’π‘Žπ‘™ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘šπ΅π‘Žπ‘ π‘’ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘šπ΄π‘π‘‘π‘’π‘Žπ‘™ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘šπ΅π‘Žπ‘ π‘’ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘š 100 150968968 𝑗0.103 𝑝. 𝑒 𝑗0.154 𝑝. 𝑒Reactance of Transformer T2kVb,old 127 * 3 kV 220 kVkVb,new 220 kVMVAb,old 10 * 3 30 MVAMVAb,new 50 MVAXp.u,old 0.15p.uThe new p.u. reactance of Transformer T2 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.15 220 2220 50302π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘ j0.25 p.uReactance of Generator G2It is connected to the LT side of the Transformer T2𝐿𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”Base kV on LT side of transformer T 2 π΅π‘Žπ‘ π‘’ π‘˜π‘‰ π‘œπ‘› 𝐻𝑇 𝑠𝑖𝑑𝑒 𝐻𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ 220 18220π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 18 π‘˜π‘‰kVb,old 18 kVkVb,new 18 kVMVAb,old 30 MVAMVAb,new 50 MVAXp.u,old 0.25 p.uThe new p.u. reactance of Generator G 2 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.25 18 218π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀 5030Reactance of Transformer T3kVb,old 20 kVkVb,new 20 kVMVAb,old 20 MVAMVAb,new 50 MVA2 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘ j0.4167 p.u

Xp.u,old 0.15p.uThe new p.u. reactance of Transformer T3 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.15 20 220 50302π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘ j0.25 p.uReactance of Generator G3It is connected to the LT side of the Transformer T3Base kV on LT side of transformer T 3 π΅π‘Žπ‘ π‘’ π‘˜π‘‰ π‘œπ‘› 𝐻𝑇 𝑠𝑖𝑑𝑒 220 20220𝐿𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”π»π‘‡ π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 20 π‘˜π‘‰kVb,old 20 kVkVb,new 20 kVMVAb,old 30 MVAMVAb,new 50 MVAXp.u,old 0.21 p.uThe new p.u. reactance of Generator G 3 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.21 20 220π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀 50302 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘ j0.35 p.u2) Draw the reactance diagram for the power system shown in fig 4 .Use a base of 50MVA 230 kV in 30Ξ© line. The ratings of the generator, motor and transformers areGenerator 20 MVA, 20 kV, X 20%Motor 35 MVA, 13.2 kV, X 25%T1 25 MVA, 18/230 kV (Y/Y), X 10%T2 45 MVA, 230/13.8 kV (Y/ ), X 15%

Fig 4SolutionBase megavoltampere,MVAb,new 50 MVABase kilovolt kVb,new 230 kV ( Transmission line side)FORMULAThe new p.u. reactance 𝑋𝑝𝑒 ,𝑛𝑒𝑀 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀2 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘Reactance of Generator GIt is connected to the LT side of the T1 transformer𝐿𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’Base kV on LT side of transformer T 1 π΅π‘Žπ‘ π‘’ π‘˜π‘‰ π‘œπ‘› 𝐻𝑇 𝑠𝑖𝑑𝑒 π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”π»π‘‡ π‘£π‘œπ‘™π‘‘π‘Ž 𝑔𝑒 π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”18 230 230 18 π‘˜π‘‰kVb,old 20 kVkVb,new 18 kVMVAb,old 20 MVAMVAb,new 50 MVAXp.u,old 0.2p.uThe new p.u. reactance of Generator G 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.2 20 218π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘2π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀 5020 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘ j0.617 p.uReactance of Transformer T1kVb,old 18 kVkVb,new 18 kVMVAb,old 25 MVAXp.u,old 0.1p.uMVAb,new 50 MVAThe new p.u. reactance of Transformer T1 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.1 Reactance of Transmission LineIt is connected to the HT side of the Transformer T118 218 π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏,𝑛𝑒𝑀50252 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘ j0.2 p.u

Actual Impedance X actual j30 ohmπ‘˜π‘‰Base impedance X base 𝑀𝑉𝐴𝑏 ,𝑛𝑒𝑀2𝑏 ,𝑛𝑒𝑀 230 250 1058 π‘œ π‘šp.u reactance of j30 Ω transmission line π΄π‘π‘‘π‘’π‘Žπ‘™ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘šπ΅π‘Žπ‘ π‘’ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘π‘’ ,π‘œ π‘šπ‘— 30 1058 𝑗0.028 𝑝. 𝑒Reactance of Transformer T2kVb,old 230 kVkVb,new 230 kVMVAb,old 45 MVAMVAb,new 50 MVAXp.u,old 0.15p.u 0.15 230 2230 2π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘The new p.u. reactance of Transformer T2 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ π‘˜π‘‰ 𝑏,𝑛𝑒𝑀50 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏,π‘œπ‘™π‘‘ j0.166 p.u45Reactance of Motor M2It is connected to the LT side of the Transformer T2𝐿𝑇 π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘”Base kV on LT side of transformer T 2 π΅π‘Žπ‘ π‘’ π‘˜π‘‰ π‘œπ‘› 𝐻𝑇 𝑠𝑖𝑑𝑒 𝐻𝑇 ��𝑔 230 230 13.8 π‘˜π‘‰kVb,old 13.2 kVMVAb,old 35 MVAkVb,new 13.8 kVMVAb,new 50 MVAXp.u,old 0.25 p.uThe new p.u. reactance of Generator G 2 𝑋𝑝𝑒 ,π‘œπ‘™π‘‘ 0.25 π‘˜π‘‰ 𝑏 ,π‘œπ‘™π‘‘π‘˜π‘‰ 𝑏 ,𝑛𝑒𝑀13.2 213.8 50352 𝑀𝑉𝐴 𝑏 ,𝑛𝑒𝑀𝑀𝑉𝐴 𝑏 ,π‘œπ‘™π‘‘ j0.326 p.u

BUSThe meeting point of various components in a power system is called a bus. The bus is aconductor made of copper or aluminum having negligible resistance. The buses are considered as pointsof constant voltage in a power system.BUS IMPEDANCE MATRIXThe matrix consisting of driving point impedances and impedances of the network of a powersystem is called bus impedance matrix. It is given by the inverse of bus admittance matrix and it isdenoted as Zbus . The bus impedance matrix is symmetrical.BUS ADMITTANCE MATRIXThe matrix consisting of the self and mutual admittances of the network of a power system iscalled bus admittance matrix. It is given by the admittance matrix Y in the node basis matrix equation of apower system and it is denoted as Ybus . The bus admittance matrix is symmetrical.EXAMPLE1. Find the bus admittance matrix for the given network in Fig 2. Determine the reduced admittancematrix by eliminating node 4. The values are marked in p.u.Fig 2π‘Œπ΅π‘ˆπ‘†π‘Œ11π‘Œ21 οΏ½οΏ½33π‘Œ43π‘Œ14π‘Œ24π‘Œ34π‘Œ44π‘Œ11 𝑦12 𝑦13 𝑦14 𝑗0.5 𝑗0.4 𝑗0.4 𝑗1.3π‘Œ22 𝑦12 𝑦23 𝑗0.5 𝑗0.6 𝑗1.1π‘Œ33 𝑦32 𝑦31 𝑦34 𝑗0.6 𝑗0.4 𝑗0.5 𝑗1.5

π‘Œ44 𝑦41 𝑦43 𝑗0.4 𝑗0.5 𝑗0.9π‘Œ12 𝑦12 𝑗0.5π‘Œ13 𝑦13 𝑗0.4π‘Œ14 𝑦14 𝑗0.4π‘Œ21 π‘Œ12 𝑗0.5π‘Œ23 𝑦23 𝑗0.6π‘Œ24 𝑦24 0π‘Œ31 π‘Œ13 𝑗0.4π‘Œ32 π‘Œ23 𝑗0.6π‘Œ34 𝑦34 𝑗0.5π‘Œ41 π‘Œ14 𝑗0.4π‘Œ42 π‘Œ24 0π‘Œ43 π‘Œ34 𝑗0.5π‘Œπ΅π‘ˆπ‘† 𝑗1.3 𝑗0.5𝑗0.4𝑗0.4𝑗0.5 𝑗1.1 𝑗0.60 𝑗0.4𝑗0.6 𝑗1.5 𝑗0.5𝑗0.40𝑗0.5 𝑗0.9Elements of new bus admittance matrix after eliminating 4th row and 4th columnπ‘Œπ‘—π‘˜ ,𝑛𝑒𝑀 π‘Œπ‘—π‘˜ N 4, j 1,2,3π‘Œπ‘—π‘› π‘Œπ‘›π‘˜π‘Œπ‘›π‘›k 1,2,3π‘Œ11,𝑛𝑒𝑀 π‘Œ11 π‘Œ12,𝑛𝑒𝑀 π‘Œ12 π‘Œ14 π‘Œ41π‘Œ44 𝑗1.3 𝑗 0.4 (𝑗 0.4) 𝑗 0.9π‘Œ14 π‘Œ42𝑗0.4 (𝑗0) 𝑗0.5 𝑗0.5π‘Œ44 𝑗0.9 𝑗1.12

π‘Œ13,𝑛𝑒𝑀 π‘Œ13 π‘Œ14 π‘Œ43𝑗0.4 (𝑗0.5) 𝑗0.4 𝑗0.622π‘Œ44 𝑗0.9π‘Œ21,𝑛𝑒𝑀 π‘Œ12,𝑛𝑒𝑀 𝑗0.5π‘Œ22,𝑛𝑒𝑀 π‘Œ22 π‘Œ24 π‘Œ42𝑗0 (𝑗0) 𝑗1.1 𝑗1.1π‘Œ44 𝑗0.9π‘Œ23,𝑛𝑒𝑀 π‘Œ23 π‘Œ24 π‘Œ43𝑗0 (𝑗0.5) 𝑗0.6 𝑗0.6π‘Œ44 𝑗0.9π‘Œ31,𝑛𝑒𝑀 π‘Œ13,𝑛𝑒𝑀 𝑗0.622π‘Œ32,𝑛𝑒𝑀 π‘Œ23,𝑛𝑒𝑀 𝑗0.6π‘Œ33,𝑛𝑒𝑀 π‘Œ33 π‘Œ34 π‘Œ43𝑗0.5 (𝑗0.5) 𝑗1.5 𝑗1.22π‘Œ44 𝑗0.9Reduced admittance matrix after eliminating 4 th row and 4th column\π‘Œπ΅π‘ˆπ‘† 𝑗1.12 𝑗0.5𝑗0.622𝑗0.5 𝑗1.1𝑗0.6 𝑗0.622 𝑗0.6 𝑗1.2222) Find the bus impedance matrix for the system whose reactance diagram is shown in fig 3. All theimpedances are in p,u.Step 1: connect bus 1 to ref bus through impedance j1.0

𝑍𝑏𝑒𝑠 𝑗1.0Step 2 connect bus 2 to the bus 1 through impedance j0.25𝑍𝑏𝑒𝑠 𝑗1.0𝑗1.0𝑗1.0 𝑗1.0 𝑗0.25𝑍𝑏𝑒𝑠 𝑗1.0 𝑗1.0𝑗1.0 𝑗1.25Step 3 connect bus 2 to ref bus through impedance j1.25

𝑗1.0 𝑗1.0𝑗1.0𝑗1.25𝑍𝑏𝑒𝑠 𝑗1.0 𝑗1.25𝑗1.0 𝑗1.25 𝑗1.25 𝑗1.25𝑍𝑏𝑒𝑠𝑗1.0 𝑗1.0𝑗1.0 𝑗1.0 𝑗1.25 𝑗1.25𝑗1.0 𝑗1.25 𝑗2.5Number of buses is only 2.But matrix size is 3*3.The matrix size ir reduced by eliminating 3 rd row and 3rdcolumnπ‘π‘—π‘˜ ,π‘Žπ‘π‘˜ π‘π‘—π‘˜ 𝑍𝑗 (𝑛 1) 𝑍 𝑛 1 π‘˜π‘(𝑛 1)(𝑛 1)Where n 2 j 1,2 k 1,2n 2 j 1 k 1𝑍11,π‘Žπ‘π‘˜ 𝑍11 𝑍11,π‘Žπ‘π‘˜ 𝑗1.0 𝑍13 𝑍31𝑍33𝑗1.0 𝑗1.0 𝑗0.6𝑗2.5n 2 j 1 k 2𝑍12,π‘Žπ‘π‘˜ 𝑍12 𝑍12,π‘Žπ‘π‘˜ 𝑗1.0 𝑍13 𝑍32𝑍33𝑗1.0 𝑗1.25 𝑗0.5𝑗2.5n 2 j 2 k 1𝑍21,π‘Žπ‘π‘˜ 𝑍12,π‘Žπ‘π‘˜ 𝑗0.5n 2 j 2 k 2𝑍22,π‘Žπ‘π‘˜ 𝑍22 𝑍22,π‘Žπ‘π‘˜ 𝑗1.25 The reduced matrix𝑍23 𝑍32𝑍33𝑗1.25 𝑗1.25 𝑗0.625𝑗2.5

𝑍𝑏𝑒𝑠 𝑗0.6 𝑗0.5𝑗0.5 𝑗. 625Step 4:connect bus 3 to bus 2 through impedance j0.0𝑗0.6𝑗0.5𝑗0.5𝑍𝑏𝑒𝑠 𝑗0.5 𝑗0.625 𝑗0.625𝑗0.5 𝑗. 625 𝑗. 675Symmetrical ComponentsAn unbalanced system of N related vectors can be resolved into N systems of balanced vectors.The N – sets of balanced vectors are called symmetrical components. Each set consists of N – vectorswhich are equal in length and having equal phase angles between adjacent vectors.Sequence Impedance and Sequence NetworkThe sequence impedances are impedances offered by the devices or components for the likesequence component of the current .The single phase equivalent circuit of a power system consisting ofimpedances to the current of any one sequence only is called sequence network.Positive Sequence ComponentsThe positive sequence components are equal in magnitude and displayed from each otherby 120o with the same sequence as the original phases. The positive sequence currents and voltagesfollow the same cycle order of the original source. In the case of typical counter clockwise rotationelectrical system, the positive sequence phasor are shown in Fig . The same case applies for thepositive current phasors. This sequence is also called the β€œabc” sequence and usually denoted bythe symbol β€œ ” or β€œ1”

120012001200Negative Sequence ComponentsThis sequence has components that are also equal in magnitude and displayed from eachother by 120o similar to the positive sequence components. However, it has an opposite phasesequence from the original system. The negative sequence is identified as the β€œacb” sequenceand usually denoted by the symbol β€œ-” or β€œ2” [9].The phasors of this sequence are shown in Figwhere the phasors rotate anti- clockwise. This sequence occurs only in case of an unsymmetricalfault in addition to the positive sequence components,120012001200Zero Sequence ComponentsIn this sequence, its components consist of three phasors which are equal in magnitude as before butwith a zero displacement. The phasor components are in phase with each other. This is illustrated inFig . Under an asymmetrical fault condition, this sequence symbolizes the residual electricity in thesystem in terms of voltages and currents where a ground or a fourth wire exists. It happens whenground currents return to the power system through any grounding point in the electrical system. Inthis type of faults, the positive and the negative components are also present. This sequence is knownby the symbol β€œ0” .

EXAMPLE1. The symmetrical components of a phase –a voltage in a 3-phase unbalanced system are00Va 0 10 180 0 V, Va1 50 0 V and Va 2 20 90 V. Determine the phase voltages Va ,Vb and VcThe phase voltages of π‘‰π‘Ž , 𝑉𝑏 π‘Žπ‘›π‘‘ π‘‰π‘π‘‰π‘Ž1𝑉𝑏 οΏ½οΏ½π‘Ž2π‘‰π‘Ž π‘‰π‘Ž0 π‘‰π‘Ž1 π‘‰π‘Ž2𝑉𝑏 π‘‰π‘Ž0 π‘Ž2 π‘‰π‘Ž1 π‘Žπ‘‰π‘Ž2𝑉𝑐 π‘‰π‘Ž0 π‘Žπ‘‰π‘Ž1 π‘Ž2 π‘‰π‘Ž2Va 0 10 1800 10 j 0 VVa1 50 00 50 j 0VVa 2 20 900 0 j 20Va 1 1200 π‘Ž2 1 2400π‘Ž2 π‘‰π‘Ž1 1 2400 50 00 50 2400 25 j43.30π‘Žπ‘‰π‘Ž1 1 1200 50 00 50 1200 25 j43.30π‘Ž2 π‘‰π‘Ž2 1 2400 20 900 20 233 17.32 j10π‘Žπ‘‰π‘Ž2 1 1200 20 900 20 2100 17.32 j10π‘‰π‘Ž π‘‰π‘Ž0 π‘‰π‘Ž1 π‘‰π‘Ž2 10 𝑗0 50 𝑗0 0 𝑗20 40 𝑗20 44.72 270 𝑉

𝑉𝑏 π‘‰π‘Ž0 π‘Ž2 π‘‰π‘Ž1 π‘Žπ‘‰π‘Ž2 10 𝑗0 25 j43.30 17.32 j10 52.32 𝑗53.90 74.69 1340 V𝑉𝑐 π‘‰π‘Ž0 π‘Žπ‘‰π‘Ž1 π‘Ž2 π‘‰π‘Ž2 25 j43.30 ( 25 j43.30) 17.32 j10 -17.68 j33.3 37.70 1180 𝑉THREE-SEQUENCE IMPEDANCES AND SEQUENCE NETWORKSPositive sequence currents give rise to only positive sequence voltages, the negative sequence currentsgive rise to only negative sequence voltages and zero sequence currents give rise to only zero sequencevoltages, hence each network can be regarded as flowing within in its own network through impedancesof its own sequence only.In any part of the circuit, the voltage drop caused by current of a certain sequence depends on theimpedance of that part of the circuit to current of that sequence.The impedance of any section of a balanced network to current of one sequence may be different fromimpedance to current of another sequence.The impedance of a circuit when positive sequence currents are flowing is called impedance, When onlynegative sequence currents are flowing the impedance is termed as negative sequence impedance. Withonly zero sequence currents flowing the impedance is termed as zero sequence impedance.The analysis of unsymmetrical faults in power systems is carried out by finding the symmetricalcomponents of the unbalanced currents.Since each sequence current causes a voltage drop of that sequence only, each sequence current can beconsidered to flow in an independent network composed of impedances to current of that sequenceonly.The single phase equivalent circuit composed of the impedances to current of any one sequence only iscalled the sequence network of that particular sequence. The sequence networks contain the generatedemfs and impedances of like sequence. Therefore for every power system we can form three- sequencenetwork s. These sequence networks, carrying current Ia1, Ia2 and Ia0 are then inter-connected torepresent the different fault conditions.SEQUENCE NETWORKS OF SYNCHRONOUS MACHINESAn unloaded synchronous machine having its neutral earthed through impedance, Zn, is shownin fig. below. A fault at its terminals causes currents Ia, Ib and Ic to flow in the lines. If faultinvolves earth, a current In flows into the neutral from the earth. This current flows through the

neutral impedance Zn. Thus depending on the type of fault, one or more of the line currentsmay be zero. Thus depending on the type of fault, one or more of the line currents may be zero.POSITIVE SEQUENCE NETWORKThe generated voltages of a synchronous machine are of positive sequence only since thewindings of a synchronous machine are symmetrical.The positive sequence network consists of an emf equal to no load terminal voltages and is inseries with the positive sequence impedance Z1 of the machine. Fig.2 (b) and fig.2(c) shows thepaths for positive sequence currents and positive sequence network respectively on a singlephase basis in the synchronous machine.The neutral impedance Zn does not appear in the circuit because the phasor sum of I a1, Ib1 andIc1 is zero and no positive sequence current can flow through Zn. Since its a balanced circuit, thepositive sequence N The reference bus for the positive sequence network is the neutral of thegenerator. The positive sequence impedance Z1 consists of winding resistance and direct axisreactance. The reactance is the sub-transient reactance X”d or transient reactance X’d orsynchronous reactance Xd depending on whether sub-transient, transient or stead

(1) The eqn 1 show that the per unit impedance is directly proportional to base megavoltampere and inversely proportional to the square of the base voltage. Using Eqn 1 we can derive an expression to convert the p.u impedance expressed in one base value ( old base) to another base (new base) Let kV b,old and MVA

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