ECE 476 POWER SYSTEM ANALYSIS

1. ECE 476POWER SYSTEM ANALYSIS Lecture 19 Balance Fault Analysis, Symmetrical Components Professor Tom Overbye Department of Electrical andComputer Engineering

2. 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.

3. 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 • On Tuesday Quinn called it a “smart greed” plan • Law authorizes a ten year, $3 billion project to “modernization” the electric grid in Illinois • 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

4. 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

5. 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

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

7. Network Fault Example, cont'd

8. 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

9. 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

10. 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!

11. 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.

12. Two Bus Superposition Solution This matches what we calculated earlier

13. Extension to Larger Systems However to use this approach we need to first determine If

14. Determination of Fault Current

15. Determination of Fault Current

16. Three Gen System Fault Example

17. Three Gen Example, cont’d

18. Three Gen Example, cont’d

19. Three Gen Example, cont’d

20. PowerWorld Example 7.5: Bus 2 Fault

21. Problem 7.28

22. 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.

23. 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 • safety • protect equipment

24. 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:

25. How good a conductor is dirt?

26. How good a conductor is dirt?

27. 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.

28. 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:

29. Calculation of grounding R, cont’d The actual values will be substantially less since we’ve assumed no current flowing downward into the ground