1. Lecture 6 Development of Transmission Line Models Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

2. 1 Announcements For lectures 5 through 7 please be reading Chapter 4 – we will not be covering sections 4.7, 4.11, and 4.12 in detail HW 2 is 4.10 (positive sequence is the same here as per phase), 4.18, 4.19, 4.23. Use Table A.4 values to determine the Geometric Mean Radius of the wires (i.e., the ninth column). Due September 15 in class. “Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit a coal power plant, a coal mine, a wind farm and a bio-diesel processing plant. Sponsored by Students for Environmental Concerns. Cost isn’t finalized, but should be between $10 and $20. Contact Rebecca Marcotte at marcott1@illinois.edu for more information or to sign up. marcott1@illinois.edu

3. 2 Two Conductor Line Inductance Key problem with the previous derivation is we assumed no return path for the current. Now consider the case of two wires, each carrying the same current I, but in opposite directions; assume the wires are separated by distance R. R Creates counter- clockwise field Creates a clockwise field To determine the inductance of each conductor we integrate as before. However now we get some field cancellation

4. 3 Two Conductor Case, cont’d R R Direction of integration Rp Key Point: As we integrate for the left line, at distance 2R from the left line the net flux linked due to the Right line is zero! Use superposition to get total flux linkage. Left Current Right Current

5. 4 Two Conductor Inductance

6. 5 Many-Conductor Case Now assume we now have k conductors, each with current i k, arranged in some specified geometry. We’d like to find flux linkages of each conductor. Each conductor’s flux linkage, k, depends upon its own current and the current in all the other conductors. To derive 1 we’ll be integrating from conductor 1 (at origin) to the right along the x-axis.

7. 6 Many-Conductor Case, cont’d At point b the net contribution to 1 from i k, 1k, is zero. We’d like to integrate the flux crossing between b to c. But the flux crossing between a and c is easier to calculate and provides a very good approximation of 1k. Point a is at distance d 1k from conductor k. R k is the distance from con- ductor k to point c.

8. 7 Many-Conductor Case, cont’d

9. 8

10. 9 Symmetric Line Spacing – 69 kV

11. 10 Birds Do Not Sit on the Conductors

12. 11 Line Inductance Example Calculate the reactance for a balanced 3 , 60Hz transmission line with a conductor geometry of an equilateral triangle with D = 5m, r = 1.24cm (Rook conductor) and a length of 5 miles.

13. 12 Line Inductance Example, cont’d

14. 13 Line Inductance Example, cont’d

15. 14 Conductor Bundling To increase the capacity of high voltage transmission lines it is very common to use a number of conductors per phase. This is known as conductor bundling. Typical values are two conductors for 345 kV lines, three for 500 kV and four for 765 kV. Fourth edition book cover had a transmission line with two conductor bundling

16. 15 Bundled Conductor Pictures The AEP Wyoming-Jackson Ferry 765 kV line uses 6-bundle conductors. Conductors in a bundle are at the same voltage! Photo Source: BPA and American Electric Power

17. 16 Bundled Conductor Flux Linkages For the line shown on the left, define d ij as the distance bet- ween conductors i and j. We can then determine  for each

18. 17 Bundled Conductors, cont’d

19. 18 Bundled Conductors, cont’d

20. 19 Inductance of Bundle

21. 20 Inductance of Bundle, cont’d

22. 21 Bundle Inductance Example 0.25 M Consider the previous example of the three phases symmetrically spaced 5 meters apart using wire with a radius of r = 1.24 cm. Except now assume each phase has 4 conductors in a square bundle, spaced 0.25 meters apart. What is the new inductance per meter?

23. 22 Transmission Tower Configurations The problem with the line analysis we’ve done so far is we have assumed a symmetrical tower configuration. Such a tower figuration is seldom practical. Typical Transmission Tower Configuration Therefore in general D ab  D ac  D bc Unless something was done this would result in unbalanced phases

24. 23 Transposition To keep system balanced, over the length of a transmission line the conductors are rotated so each phase occupies each position on tower for an equal distance. This is known as transposition. Aerial or side view of conductor positions over the length of the transmission line.

25. 24 Line Transposition Example

26. 25 Line Transposition Example

27. 26 Transposition Impact on Flux Linkages “a” phase in position “1” “a” phase in position “3” “a” phase in position “2”

28. 27 Transposition Impact, cont’d

29. 28 Inductance of Transposed Line

30. 29 Inductance with Bundling

31. 30 Inductance Example Calculate the per phase inductance and reactance of a balanced 3 , 60 Hz, line with horizontal phase spacing of 10m using three conductor bundling with a spacing between conductors in the bundle of 0.3m. Assume the line is uniformly transposed and the conductors have a 1cm radius. Answer: D m = 12.6 m, R b = 0.0889 m Inductance = 9.9 x 10 -7 H/m, Reactance = 0.6  /Mile

32. 31 Grid Weakness

33. 32 New Southwest Campus 138 kV Line In 2010 Ameren proposed building a new 138 kV transmission between the Southwest Campus and Bondville Substations Project cost is estimated to be about $14 million using the “preferred” nine mile route. An alternative seventeen mile route would cost about $22 million. Project should begin in 2012 and finish by 2014. Campus also wants to install at 138 kV line between Southwest Campus and North Champaign substations.

34. 33 The Local Grid

35. 34 Proposed and Alternative Route