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##### ECE 476 POWER SYSTEM ANALYSIS

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**ECE 476POWER SYSTEM ANALYSIS**Lecture 5 Development of Transmission Line Models Professor Tom Overbye Department of Electrical andComputer Engineering**Reading and Homework**• 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 • Go through Section 1.5, building the PowerWorld case • HW 2 is 2.32, 43, 47 • (You can download the latest educational version of PowerWorld (version 13) at http://www.powerworld.com/gloversarma.asp • The Problem 2.32 case will also be on the website**In the News**• 9/2/08: Kansas utilities agree to cooperate on major transmission system project**Special Guest Talk**• Linda Brown is the director of Transmission Planning with San Diego Gas and Electric (SDGE)**Inductance Example**• Calculate the inductance of an N turn coil wound tightly on a torodial iron core that has a radius of R and a cross-sectional area of A. Assume • 1) all flux is within the coil • 2) all flux links each turn**Inductance of a Single Wire**• To development models of transmission lines, we first need to determine the inductance of a single, infinitely long wire. To do this we need to determine the wire’s total flux linkage, including • 1. flux linkages outside of the wire • 2. flux linkages within the wire • We’ll assume that the current density within the wire is uniform and that the wire has a radius of r.**x**r Flux linkages inside, cont’d Wire cross section**R**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. To determine the inductance of each conductor we integrate as before. However now we get some field cancellation Creates a clockwise field Creates counter- clockwise field**Two Conductor Case, cont’d**R R Rp Direction of integration 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. Right Current Left Current**Many-Conductor Case**Now assume we now have k conductors, each with current ik, arranged in some specified geometry. We’d like to find flux linkages of each conductor. Each conductor’s flux linkage, lk, depends upon its own current and the current in all the other conductors. To derive l1 we’ll be integrating from conductor 1 (at origin) to the right along the x-axis.**Many-Conductor Case, cont’d**Rk is the distance from con- ductor k to point c. 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 l1k. Point a is at distance d1k from conductor k. At point b the net contribution to l1 from ik , l1k, is zero.**Line Inductance Example**Calculate the reactance for a balanced 3f, 60Hz transmission line with a conductor geometry of an equilateral triangle with D = 5m, r = 1.24cm (Rookconductor) and a length of 5 miles.**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. Book coverhas a transmissionline withtwo conductorbundling**Bundled Conductor Flux Linkages**For the line shown on the left, define dij as the distance bet- ween conductors i and j. We can then determine l for each**0.25 M**0.25 M 0.25 M Bundle Inductance Example 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?**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. Therefore in general Dab Dac Dbc Unless something was done this would result in unbalanced phases Typical Transmission Tower Configuration**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.**Transposition Impact on Flux Linkages**“a” phase in position “1” “a” phase in position “3” “a” phase in position “2”**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: Dm = 12.6 m, Rb= 0.0889 m Inductance = 9.9 x 10-7 H/m, Reactance = 0.6 /Mile