Computer Application in Power system: Chapter three - fault studies
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this chapter deals with fault analysis of a power system. under this topic, only symmetrical fault analysis is given. it will describe the methods used to determine fault current and voltage values.
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Computer Application in Power system: Chapter three - fault studies
- 1. ASTU SCHOOL OF ELECTRICAL ENGINEERING AND COMPUTING DEPT. OF POWER AND CONTROL ENGINEERING COMPUTER APPLICATION IN POWER SYSTEM (PCE5307) CHAPTER THREE FAULT STUDIES BY: MESFIN M.
- 2. OUTLINE οΆ Three Phase Fault Analysis οΆAdmittance Matrix equations οΆImpedance Matrix equations οΆFault Calculation οΆAnalysis of Unbalanced Fault οΆShort Circuit Faults οΆOpen Circuit Faults
- 3. INTRODUCTION οΆ The main objective of fault analysis is to calculate fault currents and voltages for the determination of circuit breaker capacity and protective relay performance. οΆEarly methods used in the calculation of fault levels involved the following approximations. οΆAll voltage sources assumed a one pu magnitude and zero relative phase, which is equivalent to neglecting the pre-fault load current contribution. οΆTransmission plant components included only inductive parameters. οΆTransmission line shunt capacitance and transformer magnetizing impedance were ignored.
- 4. CONT.β¦ οΆBased on the above assumptions, simple equivalent sequence impedance networks were calculated and these were interconnected according to the fault specification. Conventional circuit analysis was then used to calculate the sequence voltage and currents οΆAlthough the basic procedure of the computer solution is still the same, the need for the various approximations has disappeared.
- 5. ANALYSIS OF THREE-PHASE FAULTS οΆ A preliminary stage to the analysis is the collection of appropriate data specifying the system to be analyzed in terms of: οΆPre-fault voltage, οΆloading and οΆGenerating conditions. οΆ Such data is then processed to form a nodal equivalent network constituted by admittances and injected currents.
- 6. CONT.β¦ οΆGenerator model and its converted nodal equivalent representation.
- 7. CONT.β¦. οΆThe injected nodal current for the above circuit is given by: π°π = π¬π π΄ ππ π΄ But using KVL π¬π π΄ = π½π + π°π π΄ ππ π΄ Thus, π°π = π½π ππ π΄ + π°π π΄ οΆπ°π π΄ is the current required at the voltage π½π to produce the machine power π·π π΄ + ππΈπ π΄ , so
- 8. CONT.β¦ (πΌπ π )β ππ = ππ π + πππ π Thus, from load flow data for P, Q and V we can calculate the injected nodal current πΌπ. πΌπ = π½π ππ π΄ + ππ π β πππ π π½π β
- 9. ADMITTANCE MATRIX EQUATION οΆ Let us take as a reference the small system given below,
- 10. CONT.β¦ οΆSubstituting the above network with equivalent model
- 11. CONT.β¦. οΆThe final equivalent circuit of the network with branch elements
- 12. CONT.β¦ οΆThe following equations may then be written for the final equivalent network. πΌ1 = π¦11 π£1 + π¦12 π£1 β π£2 πΌ2 = π¦22 π£2 + π¦21 π£2 β π£1 + π¦23 π£2 β π£3 + π¦24 π£2 β π£4 πΌ3 = π¦33 π£3 + π¦32 π£3 β π£2 + π¦34 π£3 β π£4 πΌ4 = π¦44 π£4 + π¦42 π£4 β π£2 + π¦43 π£4 β π£3 + π¦45 π£4 β π£5 πΌ3 = π¦55 π£5 + π¦54 π£5 β π£4
- 13. CONT.β¦ οΆ or in matrix form after grouping together the terms common to each voltage πΌ1 πΌ2 πΌ3 πΌ4 πΌ5 = π11 π12 π13 π14 π15 π21 π22 π23 π24 π25 π31 π41 π51 π32 π42 π52 π33 π43 π53 π34 π44 π54 π35 π45 π55 β π1 π2 π3 π4 π5 Where πππ = π π¦ππ and πππ = βπ¦ππ π β π
- 14. CONT.β¦ οΆ The matrix is usually written as; πΌ = π β π οΆwhere [I] and [V] are the current and voltage vectors and [Y] is the nodal admittance matrix of the system. οΆIn the admittance matrix nonzero elements only occur where branches exist between nodes. οΆSuch sparsity is exploited by only storing and processing the nonzero elements. οΆMoreover, the symmetry of the matrix ( Yij = Yji) permits using only the upper right-hand terms in the calculations.
- 15. IMPEDANCE MATRIX EQUATION οΆ The nodal admittance equation is inefficient as it requires a complete iterative solution for each fault type and location. Thus, π = π β1 β πΌ π = π β πΌ
- 16. FAULT CALCULATIONS οΆ From the initial machine data, the values of [I] are first calculated from equation for injected nodal current using one per unit voltages. οΆThese may now be used to obtain a better estimate of [V], the pre-fault voltage at every node from equation π = π β πΌ . οΆThe program now has sufficient information to calculate the voltages and currents during a fault. ππ π = π π πΌ π where k is the bus to be faulted, Zf is the fault impedance and πΌ π is the fault current.
- 17. CONT.β¦ οΆThe equation π = π β πΌ for a fault at K bus can be expanded as, π1 π2 β ππ β ππ = π11 π12 β π1π β π1π π21 π22 β π2π β π2π β π π1 β π π1 β π π2 β π π2 β β β β β π ππ β π ππ β β β π ππ β β β π ππ β πΌ1 πΌ2 β πΌ π β πΌ π Selecting row k and expanding gives ππ = π π1 πΌ1 + π π2 πΌ2 + β― + π ππ πΌ π + β― + π ππ πΌ π οΆThis equation describes the voltage at bus k prior to the fault.
- 18. CONT.β¦ οΆDuring a fault a large fault current If flows out of bus k thus, ππ π = π π πΌ π = π π1 πΌ1 + π π2 πΌ2 + β― + π ππ πΌ π + β― + π ππ πΌ π β π ππ πΌ π Or π π πΌ π = ππ β π ππ πΌ π Thus, the fault current can be found as, πΌ π = ππ π π + π ππ οΆAlso during fault voltage at any other bus j is. ππ π = ππ β πππ πΌ π
- 19. ANALYSIS OF UNBALANCED FAULT ο Reading Assignment οShort circuit οOpen circuit