Presentation on theme: "EE 369 POWER SYSTEM ANALYSIS"— Presentation transcript:
1EE 369 POWER SYSTEM ANALYSIS Lecture 5Development of Transmission Line ModelsTom Overbye and Ross Baldick
2Reading For lectures 5 through 7 read Chapter 4 Read Section 1.5, we will not be covering sections 4.7, 4.11, and 4.12 in detailRead Section 1.5,HW 4 is chapter 4 case study questions A through D, and Problems 2.31, 2.41, 2.43, 2.48, 4.1, 4.3, 4.6, due Thursday 9/26.HW 5 is Problems 4.9, 4.11, 4.13, 4.18, 4.21, 4.22, 4.24, 4.25 (assume Cardinal conductor and look up GMR in Table A.4); due Thursday 10/2.
4Inductance ExampleCalculate the inductance of an N turn coil wound tightly on a toroidal iron core that has a radius of R and a cross-sectional area of A. Assume1) all flux is within the coil2) all flux links each turn3) Radius of each turn is negligible compared to RCircular path Γ of radius R within the iron core encloses all N turns of the coil and hence links total enclosed current of Ie = NI.Since the radius of each turn is negligible compared to R, all circular paths within the iron core have radius approximately equal to R.
6Inductance of a Single Wire To develop 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 wire2. flux linkages within the wireWe’ll assume that the current density within the wire is uniform and that the wire is solid with a radius of r.In fact, current density is non-uniform, and conductor is stranded, so our calculations will be approximate.
13Two 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 D.DTo determine theinductance of eachconductor we integrateas before. Howevernow we get somefield cancellation.Creates counter-clockwise fieldCreates aclockwise field
14Two Conductor Case, cont’d Direction of integrationDRKey Point: Flux linkage due to currents in each conductor tendto cancel out. Use superposition to get total flux linkage.Left CurrentRight Current
16Many-Conductor Case Now assume we now have n conductors, with the k-th conductor having current ik, andarranged in some specified geometry.We’d like to find flux linkages of each conductor.Each conductor’s fluxlinkage, lk, depends uponits own current and thecurrent in all the otherconductors.For example, to derive the flux linkage for conductor 1, l1, we’ll be integrating fromconductor 1 (at origin) to the right along the x-axis.
17Many-Conductor Case, cont’d Rk is thedistancefrom con-ductor kto pointc.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 netcontribution to l1from ik , l1k, is zero.
21Line Inductance Example Calculate the reactance for a balanced 3f, 60Hztransmission line with a conductor geometry of anequilateral triangle with D = 5m, r = 1.24cm (Rook conductor) and a length of 5 miles.
24Conductor BundlingTo increase the capacity of high voltage transmissionlines it is very common to use a number ofconductors per phase. This is known as conductorbundling. Typical values are two conductors for345 kV lines, three for 500 kV and four for 765 kV.
25Bundled Conductor Flux Linkages For the line shown on the left,define dij as the distance betweenconductors i and j.We can then determine lk for conductor k.Assuming ¼ of the phase current flowsin each of the four conductors ina given phase bundle, then for conductor 1:
30Bundle Inductance Example Consider the previous example of the three phasessymmetrically spaced 5 meters apart using wirewith a radius of r = 1.24 cm. Except now assumeeach phase has 4 conductors in a square bundle,spaced 0.25 meters apart. What is the new inductanceper meter?0.25 M
31Transmission Tower Configurations The problem with the line analysis we’ve done so far is we have assumed a symmetrical tower configuration.Such a tower configuration is seldom practical.Therefore ingeneral Dab Dac DbcUnless somethingwas done this wouldresult in unbalancedPhases.Typical Transmission TowerConfiguration
32TranspositionTo 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 lengthof the transmission line.
39Inductance ExampleCalculate the per phase inductance and reactance of a balanced 3, 60 Hz, line with:horizontal phase spacing of 10musing 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.