Power System Analysis - Chapter 2: Transmission Line Parameters

nce, inductance and capacitance.Line ResistanceLine Resistance, cont’dConductor resistance depends on the following factors:1. Spiraling2. Temperature3. Frequency (‘‘skin effect’’)4. Current magnitude—magnetic conductors Spiraling makes the strands 1 or 2% longer than the actual conductor length. As a result, the dc resistance of a stranded conductor is 1 or 2% larger than that calculated for a specified conductor length.Resistivity of conductor metals varies linearly over normal operating temperatures according toLine Resistance, cont’dWhere ρT2 and ρT1 are resistivities at temperatures T2 and T1, respectively, (oC). T is a temperature constant (oC) that depends on the conductor material. Line Resistance, cont’dFor ac, the current distribution is nonuniform. As frequency increases, the current in a solid cylindrical conductor tends to crowd toward the conductor surface, with smaller current density at the conductor center. This phenomenon is called “skin effect”. This effect causes ac resistance higher dc resistance.For magnetic conductors, resistance depends on current magnitudeLine ConductanceLine conductance accounts for real power loss between conductors or between conductors and ground.For overhead lines, this power loss is due to leakage currents at insulators and to corona.Conductance is usually neglected in power studies because it is very small.Line InductanceMagnetics review and Line integralsMagnetic flux densityMagnetic fluxMagnetic field from single wireFlux linkages and Faraday’s lawInductanceMagnetics ReviewAmpere’s circuital law:Line IntegralsLine integrals are a generalization of traditional integrationIntegration along thex-axisIntegration along ageneral path, whichmay be closedAmpere’s law is most useful in cases of symmetry, such as with an infinitely long lineMagnetic Flux DensityMagnetic fields are usually measured in terms of flux densityMagnetic FluxMagnetic Fields from Single WireAssume we have an infinitely long wire with current of 1000A. How much magnetic flux passes through a 1 meter square, located between 4 and 5 meters from the wire?Direction of H is givenby the “Right-hand” RuleEasiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose acircle with a radius of x. Single Line Example, cont’dFor reference, the earth’smagnetic field is about0.6 Gauss (Central US)Flux linkages and Faraday’s lawInductanceFor a linear magnetic system, that is one where 	B	= m Hwe can define the inductance, L, to bethe constant relating the current and the fluxlinkage	l	= L iwhere L has units of Henrys (H)	Inductance ExampleCalculate 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. Assume1) all flux is within the coil 2) all flux links each turnInductance Example, cont’dInductance of a Single WireTo 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, including1.	flux linkages outside of the wireflux linkages within the wireWe’ll assume that the current density within the wire is uniform and that the wire has a radius of r. Flux Linkages outside of the wireFlux Linkages outside, cont’dFlux linkages inside of wireFlux linkages inside, cont’dWire cross sectionxrLine Total Flux & InductanceInductance SimplificationTwo Conductor Line InductanceKey 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. RCreates counter-clockwise fieldCreates aclockwise fieldTo determine theinductance of eachconductor we integrateas before. Howevernow we get somefield cancellationTwo Conductor Case, cont’dRRDirection of integrationRpKey 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 CurrentRight CurrentTwo Conductor InductanceMany-Conductor CaseNow 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 fluxlinkage, 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’dAt point b the net contribution to l1from ik , l1k, 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 l1k. Point a is at distance d1k from conductor k. Rk is the distancefrom con-ductor kto pointc.Many-Conductor Case, cont’dMany-Conductor Case, cont’dSymmetric Line Spacing – 69 kVLine Inductance ExampleCalculate the reactance for a balanced 3f, 60Hztransmission line with a conductor geometry of anequilateral triangle with D = 5m, r = 1.24cm (Rookconductor) and a length of 5 miles. Line Inductance Example, cont’dLine Inductance Example, cont’dConductor BundlingTo increase the capacity of high voltage transmissionlines it is very common to use a number of conductors per phase. This is known as conductorbundling. Typical values are two conductors for 345 kV lines, three for 500 kV and four for 765 kV.Fourth editionbook coverhad a transmissionline withtwo conductorbundlingBundled Conductor Pictures	The AEP Wyoming-JacksonFerry 765 kV line uses6-bundle conductors.Conductors in a bundle areat the same voltage!Photo Source: BPA and American Electric PowerBundled Conductor Flux LinkagesFor the line shown on the left,define dij as the distance bet-ween conductors i and j. Wecan then determine l for eachBundled Conductors, cont’dBundled Conductors, cont’dInductance of BundleInductance of Bundle, cont’dBundle Inductance Example0.25 M0.25 M0.25 MConsider the previous example of the three phasessymmetrically spaced 5 meters apart using wire with 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? Transmission Tower ConfigurationsThe 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 ConfigurationTherefore ingeneral Dab Dac  DbcUnless something was done this wouldresult in unbalancedphasesTranspositionTo 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. Line Transposition ExampleLine Transposition ExampleTransposition Impact on Flux Linkages“a” phase inposition “1”“a” phase inposition “3”“a” phase inposition “2”Transposition Impact, cont’dInductance of Transposed LineInductance with BundlingInductance ExampleCalculate 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 /MileLine ConductorsTypical transmission lines use multi-strand conductorsACSR (aluminum conductor steel reinforced) conductors are most common. A typical Al. to St. ratio is about 4 to 1.Line CapacitanceReview of electric fiedlsVoltage differenceLine capacitanceReview of Electric FieldsGauss’s Law ExampleSimilar to Ampere’s Circuital law, Gauss’s Law is most useful for cases with symmetry.Example: Calculate D about an infinitely long wire that has a charge density of q coulombs/meter. Since D comesradially out inte-grate over the cylinder bounding the wireElectric FieldsThe electric field, E, is related to the electric flux density, D, by D =  Ewhere E = electric field (volts/m)  = permittivity in farads/m (F/m)  = o r o = permittivity of free space (8.85410-12 F/m) r = relative permittivity or the dielectric constant	(1 for dry air, 2 to 6 for most dielectrics)Voltage Difference	Voltage Difference, cont’dMulti-Conductor CaseMulti-Conductor Case, cont’dAbsolute Voltage DefinedThree Conductor CaseABCAssume we have three infinitely long conductors, A, B, & C, each with radius r and distance D from the other two conductors. Assume charge densities suchthat qa + qb + qc = 0Line CapacitanceLine Capacitance, cont’dBundled Conductor CapacitanceLine Capacitance, cont’dLine Capacitance ExampleCalculate the per phase capacitance and susceptance of a balanced 3, 60 Hz, transmission 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 a 1cm radius.Line Capacitance Example, cont’dACSR Table Data (Similar to Table A.4)Inductance and Capacitance assume a Dm of 1 ft. GMR is equivalent to r’ACSR Data, cont’dTerm from table assuminga one foot spacingTerm independentof conductor withDm in feet. ACSR Data, Cont.Term from table assuminga one foot spacingTerm independentof conductor withDm in feet. Dove ExampleAdditional Transmission TopicsMulti-circuit lines: Multiple lines often share a common transmission right-of-way. This DOES cause mutual inductance and capacitance, but is often ignored in system analysis. Cables: There are about 3000 miles of underground ac cables in U.S. Cables are primarily used in urban areas. In a cable the conductors are tightly spaced, ( 400 miles)long cable power transfer such as underwaterproviding an asynchronous means of joining different power systems (such as the Eastern and Western grids).