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ECE 476 POWER SYSTEM ANALYSIS Lecture 19 Balance Fault Analysis, Symmetrical Components Professor Tom Overbye Department of Electrical and Computer Engineering Announcements Be reading Chapter 8 HW 8 is 7.6, 7.13, 7.19, 7.28; due Nov 3 in class. Start working on Design Project. Tentatively due Nov 17 in class. 1 In the News • Earlier this week the Illinois House and Senate overrode the Governor’s veto of the Smart Grid Bill; it is now law • • Law authorizes a ten year, $3 billion project to “modernization” the electric grid in Illinois • • On Tuesday Quinn called it a “smart greed” plan All ComEd customers and most of Ameren will get “smart” meters Effort will be paid for through rate increases over the period; ComEd estimated the average increase of $36 per year would be offset by electricity savings 2 Network Fault Analysis Simplifications To simplify analysis of fault currents in networks we'll make several simplifications: 1. 2. 3. 4. 5. Transmission lines are represented by their series reactance Transformers are represented by their leakage reactances Synchronous machines are modeled as a constant voltage behind direct-axis subtransient reactance Induction motors are ignored or treated as synchronous machines Other (nonspinning) loads are ignored 3 Network Fault Example For the following network assume a fault on the terminal of the generator; all data is per unit except for the transmission line reactance generator has 1.05 terminal voltage & supplies 100 MVA with 0.95 lag pf Convert to per unit: X line 19.5 0.1 per unit 2 138 100 4 Network Fault Example, cont'd Faulted network per unit diagram To determine the fault current we need to first estimate the internal voltages for the generator and motor For the generator VT 1.05, SG 1.018.2 * I Gen 1.018.2 0.952 18.2 1.05 ' Ea 1.1037.1 5 Network Fault Example, cont'd The motor's terminal voltage is then 1.050 - (0.9044 - j 0.2973) j 0.3 1.00 15.8 The motor's internal voltage is 1.00 15.8 (0.9044 - j 0.2973) j 0.2 1.008 26.6 We can then solve as a linear circuit: 1.1037.1 1.008 26.6 If j 0.15 j 0.5 7.353 82.9 2.016 116.6 j 9.09 6 Fault Analysis Solution Techniques Circuit models used during the fault allow the network to be represented as a linear circuit There are two main methods for solving for fault currents: 1. 2. Direct method: Use prefault conditions to solve for the internal machine voltages; then apply fault and solve directly Superposition: Fault is represented by two opposing voltage sources; solve system by superposition – first voltage just represents the prefault operating point – second system only has a single voltage source 7 Superposition Approach Faulted Condition Exact Equivalent to Faulted Condition Fault is represented by two equal and opposite voltage sources, each with a magnitude equal to the pre-fault voltage 8 Superposition Approach, cont’d Since this is now a linear network, the faulted voltages and currents are just the sum of the pre-fault conditions [the (1) component] and the conditions with just a single voltage source at the fault location [the (2) component] Pre-fault (1) component equal to the pre-fault power flow solution Obvious the pre-fault “fault current” is zero! 9 Superposition Approach, cont’d Fault (1) component due to a single voltage source at the fault location, with a magnitude equal to the negative of the pre-fault voltage at the fault location. I g I (1) I g(2) g I m I m(1) I m(2) (2) (2) I f I (1) I 0 I f f f 10 Two Bus Superposition Solution Before the fault we had E f 1.050, I (1) 0.952 18.2 and I m(1) 0.952 18.2 g Solving for the (2) network we get Ef 1.050 (2) Ig j7 j0.15 j0.15 E f 1.050 (2) Im j 2.1 j0.5 j0.5 I (2) f j 7 j 2.1 j 9.1 I g 0.952 18.2 j 7 7.35 82.9 This matches what we calculated earlier 11 Extension to Larger Systems The superposition approach can be easily extended to larger systems. Using the Ybus we have Ybus V I For the second (2) system there is only one voltage source so I is all zeros except at the fault location 0 I I f 0 However to use this approach we need to first determine If 12 Determination of Fault Current Define the bus impedance matrix Z bus as Z bus Z11 Then Z n1 1 Ybus V Z busI (2) V 1 (2) Z1n 0 V2 I f Z nn 0 V (2) n 1 (2) Vn For a fault a bus i we get -If Zii V f Vi (1) 13 Determination of Fault Current Hence Vi(1) If Zii Where Zii driving point impedance Zij (i j ) transfer point imepdance Voltages during the fault are also found by superposition Vi Vi(1) Vi(2) Vi(1) are prefault values 14 Three Gen System Fault Example For simplicity assume the system is unloaded before the fault with E g1 Eg 2 Eg 3 1.050 Hence all the prefault currents are zero. 15 Three Gen Example, cont’d Ybus 0 15 10 j 10 20 5 5 9 0 1 Zbus 0 15 10 j 10 20 5 5 9 0 0.1088 0.0632 0.0351 j 0.0632 0.0947 0.0526 0.0351 0.0526 0.1409 16 Three Gen Example, cont’d 1.05 For a fault at bus 1 we get I1 j 9.6 I f j 0.1088 V (2) 0.1088 0.0632 0.0351 j 9.6 j 0.0632 0.0947 0.0526 0 0.0351 0.0526 0.1409 0 1.050 0.600 0.3370 17 Three Gen Example, cont’d 1.050 1.050 00 V 1.050 0.6060 0.4440 1.050 0.3370 0.7130 18 PowerWorld Example 7.5: Bus 2 Fault One Five Four Three 7 pu 11 pu slack 0.724 pu 0.000 deg 0.579 pu 0.000 deg 0.687 pu 0.000 deg 0.798 pu 0.000 deg 0.000 pu 0.000 deg Two 19 Problem 7.28 SLACK345 0.79 pu 5 pu RAY345 sla ck 0.78 pu SLACK138 TIM345 0.70 pu RAY138 0.83 pu TIM138 0.61 pu 0.79 pu 0.64 pu 0.52 pu TIM69 RAY69 PAI69 0.56 pu 0.58 pu GROSS69 FERNA69 MORO138 0.50 pu WOLEN69 HISKY69 0.52 pu 0.59 pu PETE69 0.64 pu BOB138 DEMAR69 HANNAH69 0.50 pu UIUC69 BOB69 0.43 pu 0.00 pu 9 pu LYNN138 0 pu 0.564 pu 0.56 pu BLT138 0.61 pu AMANDA69 SHIMKO69 HOMER69 0.24 pu 0.82 pu BLT69 0.32 pu HALE69 9 pu 0.62 pu 0.61 pu 0.35 pu 0.62 pu PATTEN69 ROGER69 0.64 pu LAUF69 0.68 pu WEBER69 0 pu 3 pu 0.70 pu LAUF138 0.77 pu 0.75 pu SAVOY69 0.77 pu 2 pu JO138 JO345 BUCKY138 0.77 pu 3 pu SAVOY138 3 pu 0.84 pu 0.91 pu 20 Grounding When studying unbalanced system operation how a system is grounded can have a major impact on the fault flows Ground current does not come into play during balanced system analysis (since net current to ground would be zero). Becomes important in the study of unbalanced systems, such as during most faults. 21 Grounding, cont’d Voltages are always defined as a voltage difference. The ground is used to establish the zero voltage reference point – ground need not be the actual ground (e.g., an airplane) During balanced system operation we can ignore the ground since there is no neutral current There are two primary reasons for grounding electrical systems 1. 2. safety protect equipment 22 How good a conductor is dirt? There is nothing magical about an earth ground. All the electrical laws, such as Ohm’s law, still apply. Therefore to determine the resistance of the ground we can treat it like any other resistive material: conductor length Resistance R cross sectional area 2.65 10 8 -m for aluminum 1.68 108 -m for copper where is the resistivity 23 How good a conductor is dirt? 2.65 108 -m for aluminum 5 1016 -m for quartz (insulator!) 160 -m for top soil 900 -m for sand/gravel 20 -m for salt marsh What is resistance of a mile long, one inch diameter, circular wire made out of aluminum ? 2.65 108 1609 R= 0.083 mile 0.01282 24 How good a conductor is dirt? What is resistance of a mile long, one inch diameter, circular wire made out of topsoil? 160 1609 6 R= 500 10 mile 0.01282 In order to achieve 0.08 with our dirt wire mile we would need a cross sectional area of 160 1609 3.2 106 m 2 (i.e., a radius of about 1000 m) 0.08 But what the ground lacks in , it makes up for in A! 25 Calculation of grounding resistance Because of its large cross sectional area the earth is actually a pretty good conductor. Devices are physically grounded by having a conductor in physical contact with the ground; having a fairly large area of contact is important. Most of the resistance associated with establishing an earth ground comes within a short distance of the grounding point. Typical substation grounding resistance is between 0.1 and 1 ohm; fence is also grounded, usually by connecting it to the substation ground grid. 26 Calculation of grounding R, cont’d Example: Calculate the resistance from a grounding rod out to a radial distance x from the rod, assuming the rod has a radius of r: In general we have R dR dx 2 length x x dxˆ x cross sectional area but now area changes with length. x R ln 2 length xˆ 2 length r r 27 Calculation of grounding R, cont’d For example, if r 1.5 inches, length = 10 feet, and 160 m we get the following values as a function of x (in meters) 160 x R ln 2 3.05 0.038 x R The actual values will be 1m 27.2 substantially less since 10 m 46.4 we’ve assumed no current 100 m 65.6 flowing downward into the ground 100 km 83.4 28