1. ECE 476 Power System Analysis Lecture 17: OPF, Symmetrical Faults Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu Special Guest Lecturer: TA Won Jang

2. Announcements Read Chapter 7 HW 8 is 12.19, 12.20, 12.26, 12.28; due October 29 in class (no quiz) – P. 12.26: PW case 12.8 – P. 12.28: PW case 12.28 – Both posted on the webpage Second exam is Thursday Nov. 12 during class. Closed book, closed notes, one note sheet and calculators allowed Abbott Power Plant tour is Tuesday Nov. 10 during class 1

3. “DC” Power Flow The “DC” power flow makes the most severe approximations: – completely ignore reactive power, assume all the voltages are always 1.0 per unit, ignore line conductance This makes the power flow a linear set of equations, which can be solved directly 2

4. Optimal Power Flow (OPF) OPF functionally combines the power flow with economic dispatch Minimize cost function, such as operating cost, taking into account realistic equality and inequality constraints Equality constraints – bus real and reactive power balance – generator voltage setpoints – area MW interchange 3

5. OPF, cont’d Inequality constraints – transmission line/transformer/interface flow limits – generator MW limits – generator reactive power capability curves – bus voltage magnitudes (not yet implemented in Simulator OPF) Available Controls – generator MW outputs – transformer taps and phase angles 4

6. Two Example OPF Solution Methods Non-linear approach using Newton’s method – handles marginal losses well, but is relatively slow and has problems determining binding constraints Linear Programming – fast and efficient in determining binding constraints, but can have difficulty with marginal losses. – used in PowerWorld Simulator 5

7. LP OPF Solution Method Solution iterates between – solving a full ac power flow solution enforces real/reactive power balance at each bus enforces generator reactive limits system controls are assumed fixed takes into account non-linearities – solving a primal LP changes system controls to enforce linearized constraints while minimizing cost 6

8. Two Bus with Unconstrained Line Transmission line is not overloaded With no overloads the OPF matches the economic dispatch Marginal cost of supplying power to each bus (locational marginal costs) 7

9. Two Bus with Constrained Line With the line loaded to its limit, additional load at Bus A must be supplied locally, causing the marginal costs to diverge. 8

10. Three Bus (B3) Example Consider a three bus case (bus 1 is system slack), with all buses connected through 0.1 pu reactance lines, each with a 100 MVA limit Let the generator marginal costs be – Bus 1: 10 $ / MWhr; Range = 0 to 400 MW – Bus 2: 12 $ / MWhr; Range = 0 to 400 MW – Bus 3: 20 $ / MWhr; Range = 0 to 400 MW Assume a single 180 MW load at bus 2 9

11. B3 with Line Limits NOT Enforced Line from Bus 1 to Bus 3 is over- loaded; all buses have same marginal cost

12. B3 with Line Limits Enforced LP OPF redispatches to remove violation. Bus marginal costs are now different. 11

13. Verify Bus 3 Marginal Cost One additional MW of load at bus 3 raised total cost by 14 $/hr, as G2 went up by 2 MW and G1 went down by 1MW 12

14. Why is bus 3 LMP = $14 /MWh All lines have equal impedance. Power flow in a simple network distributes inversely to impedance of path. – For bus 1 to supply 1 MW to bus 3, 2/3 MW would take direct path from 1 to 3, while 1/3 MW would “loop around” from 1 to 2 to 3. – Likewise, for bus 2 to supply 1 MW to bus 3, 2/3MW would go from 2 to 3, while 1/3 MW would go from 2 to 1to 3. 13

15. Why is bus 3 LMP $ 14 / MWh, cont’d With the line from 1 to 3 limited, no additional power flows are allowed on it. To supply 1 more MW to bus 3 we need – Pg1 + Pg2 = 1 MW – 2/3 Pg1 + 1/3 Pg2 = 0; (no more flow on 1-3) Solving requires we up Pg2 by 2 MW and drop Pg1 by 1 MW -- a net increase of $14. 14

16. Both lines into Bus 3 Congested For bus 3 loads above 200 MW, the load must be supplied locally. Then what if the bus 3 generator opens? 15

17. Fault Analysis The cause of electric power system faults is insulation breakdown This breakdown can be due to a variety of different factors – lightning – wires blowing together in the wind – animals or plants coming in contact with the wires – salt spray or pollution on insulators 16

18. Fault Types There are two main types of faults – symmetric faults: system remains balanced; these faults are relatively rare, but are the easiest to analyze so we’ll consider them first. – unsymmetric faults: system is no longer balanced; very common, but more difficult to analyze The most common type of fault on a three phase system by far is the single line-to-ground (SLG), followed by the line-to-line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults 17

19. Lightning Strike Event Sequence 1.Lighting hits line, setting up an ionized path to ground 30 million lightning strikes per year in US! a single typical stroke might have 25,000 amps, with a rise time of 10  s, dissipated in 200  s. multiple strokes can occur in a single flash, causing the lightning to appear to flicker, with the total event lasting up to a second. 2.Conduction path is maintained by ionized air after lightning stroke energy has dissipated, resulting in high fault currents (often > 25,000 amps!) 18

20. Lightning Strike Sequence, cont’d 3.Within one to two cycles (16 ms) relays at both ends of line detect high currents, signaling circuit breakers to open the line nearby locations see decreased voltages 4.Circuit breakers open to de-energize line in an additional one to two cycles breaking tens of thousands of amps of fault current is no small feat! with line removed voltages usually return to near normal 5.Circuit breakers may reclose after several seconds, trying to restore faulted line to service 19

21. Worldwide Lightning Strike Density Source: http://science.nasa.gov/science-news/science-at-nasa/2001/ast05dec_1/ Units are Lightning Flashes per square km per year; Florida is top location in the US; very few on the West Coast, or HI, AK. This is an important consideration when talking about electric reliability! 20

22. Fault Analysis Fault currents cause equipment damage due to both thermal and mechanical processes Goal of fault analysis is to determine the magnitudes of the currents present during the fault – need to determine the maximum current to insure devices can survive the fault – need to determine the maximum current the circuit breakers (CBs) need to interrupt to correctly size the CBs 21

23. RL Circuit Analysis To understand fault analysis we need to review the behavior of an RL circuit Before the switch is closed obviously i(t) = 0. When the switch is closed at t=0 the current will have two components: 1) a steady-state value 2) a transient value 22

24. RL Circuit Analysis, cont’d 23

25. RL Circuit Analysis, cont’d 24

26. Time varying current 25

27. RL Circuit Analysis, cont’d 26

28. RMS for Fault Current 27

29. Generator Modeling During Faults During a fault the only devices that can contribute fault current are those with energy storage Thus the models of generators (and other rotating machines) are very important since they contribute the bulk of the fault current. Generators can be approximated as a constant voltage behind a time-varying reactance 28

30. 3  bal. windings (a,b,c) – stator Field winding (fd) on rotor Damper in “d” axis (1d) on rotor 2 dampers in “q” axis (1q, 2q) on rotor 29 Synchronous Machine Modeling

31. Generator Modeling, cont’d 30

32. Generator Modeling, cont’d 31

33. Generator Modeling, cont'd 32

34. Generator Short Circuit Currents 33

35. Generator Short Circuit Currents 34

36. Generator Short Circuit Example A 500 MVA, 20 kV, 3  is operated with an internal voltage of 1.05 pu. Assume a solid 3  fault occurs on the generator's terminal and that the circuit breaker operates after three cycles. Determine the fault current. Assume 35

37. Generator S.C. Example, cont'd 36

38. Generator S.C. Example, cont'd 37

39. Network Fault Analysis Simplifications To simplify analysis of fault currents in networks we'll make several simplifications: – 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 38

40. 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 39

41. Network Fault Example, cont'd Faulted network per unit diagram 40

42. Network Fault Example, cont'd 41