1. ECE 476 Power System Analysis Lecture 18: LMP Markets, Symmetrical Faults and Components Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu

2. Announcements Read Chapter 8 HW 8 is due now HW 9 is 7.6, 7.13, 7.19, 7.28, 8.4; will be covered by an in-class quiz on Nov 5 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 (we’ll be meeting at Abbott though) 1

3. Generator Costs As discussed in the economic dispatch section, generators have incremental costs that vary according to their output If an OPF is cost-based, then the costs curves can be expressed either as continuous functions or in piecewise linear blocks With a piecewise linear approach there can be a no load cost plus an incremental energy cost for blocks (first 50 MW at $20/MWh, second 50 MWs at $25/MWh) 2

4. Generator Offers Traditionally utility generator costs were based on the actual cost of operating the generator As power industry has restructured and become more competitive, there has been a move to price-based curves – Competitive business do not need to sell their products at their actual production cost (if they did they might not stay in business long) In many power markets the generators submit “offers” to sell power – Offers can either be cost-based or price-based 3

5. Electricity Markets Worldwide there are a large number of electricity markets, each with their own rules Electricity markets provide for competition – Market might have many competitive companies owning a portfolio of individual generators A third party (e.g., an independent system operator [ISO]) runs the market to get enough electricity to keep the lights on 4 Image: www.ferc.gov/market-oversight/mkt-electric/overview.asp

6. Electricity Markets, cont. Electricity markets trade a number of different commodities, with MWh being the most important Commonly a market has two settlement periods: day ahead and real-time – Day Ahead: Generators (and possibly loads) submit offers for the next day; OPF is used to determine who gets dispatched based upon forecasted conditions. Results are financially binding – Real-time: Modifies the day ahead market based upon real-time conditions. 5

7. Payment Generators are only paid if they are selected to run, otherwise they stay off-line In most markets generators are not paid their offer, rather they are paid the LMP at their bus, the loads pay the LMP At the residential/commercial level the LMP costs are usually not passed on directly to the end consumer. Rather, they these consumers typically pay a fixed rate. LMPs may differ across a system due to transmission system “congestion.” 6

8. MISO LMP Contour: Oct 23, 2015, 8:05 am CDT Image: https://www.misoenergy.org/LMPContourMap/MISO_All.html 7

9. Some Electricity Market Challenges Running a real-time electricity market can be quite challenging – the total generation must always match the total load plus losses; the load is continuously varying; generators can fail – the transmission limited capacity; transmission line and transformer flows must be within their limits; bus voltages must be within their limits Goal of the market is to allow for competition, so the generator owners are seeking to maximize their profits There is a need to prevent generators from exercising market power 8

10. Profit Maximization: 30 Bus Example For the below 30 bus case assume the generators at buses 13 and 23 have true costs of $30/MWh If they each offer at $30/MWh LMPs are shown; their net profit is 35*0.38 = $13.3/hr and 50*0.38 = $19/hr This is a rather low profit! 9

11. Profit Maximization: 30 Bus Example Now assume the generator at 13 offers $50/MWh The result is shown, which is good for the generator at 23 but bad for the one at 13 The bus 13 gen is off, but the one at 23 has a profit of 42.47*20 =$849.94/hr 10

12. Profit Maximization: 30 Bus Example If they both offer $200/MWh they end up making a lot of profit! The bus 13 gen makes $4472/hr and the one at 23 $687/hr They have market power! Market rules are set to minimize such situations 11

13. Generator Short Circuit Example Returning to balanced three-phase faults 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 12

14. Generator S.C. Example, cont'd 13

15. Generator S.C. Example, cont'd 14

16. 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 15

17. 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 16

18. Network Fault Example, cont'd Faulted network per unit diagram 17

19. Network Fault Example, cont'd 18

20. 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: – 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 19

21. 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 20

22. 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 Obviously the pre-fault “fault current” is zero! 21

23. 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. 22

24. Two Bus Superposition Solution This matches what we calculated earlier 23

25. Extension to Larger Systems However to use this approach we need to first determine I f 24

26. Determination of Fault Current 25

27. Determination of Fault Current 26