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ECE 476 POWER SYSTEM ANALYSIS Lecture7 Development of Transmission Line Models Professor Tom Overbye Department of Electrical and Computer Engineering Announcements For next two lectures read Chapter 5. 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 [email protected] for more information or to sign up. 1 SDGE Transmission Grid (From CALISO 2009 Transmission Plan) 2 Line Conductors Typical transmission lines use multi-strand conductors ACSR (aluminum conductor steel reinforced) conductors are most common. A typical Al. to St. ratio is about 4 to 1. 3 Line Conductors, cont’d Total conductor area is given in circular mils. One circular mil is the area of a circle with a diameter of 0.001 = 0.00052 square inches Example: what is the the area of a solid, 1” diameter circular wire? Answer: 1000 kcmil (kilo circular mils) Because conductors are stranded, the equivalent radius must be provided by the manufacturer. In tables this value is known as the GMR and is usually expressed in feet. 4 Line Resistance Line resistance per unit length is given by R = where is the resistivity A Resistivity of Copper = 1.68 10-8 Ω-m Resistivity of Aluminum = 2.65 10-8 Ω-m Example: What is the resistance in Ω / mile of a 1" diameter solid aluminum wire (at dc)? 2.65 10-8 Ω-m m R 1609 0.084 2 mile mile 0.0127m 5 Line Resistance, cont’d Because ac current tends to flow towards the surface of a conductor, the resistance of a line at 60 Hz is slightly higher than at dc. Resistivity and hence line resistance increase as conductor temperature increases (changes is about 8% between 25C and 50C) Because ACSR conductors are stranded, actual resistance, inductance and capacitance needs to be determined from tables. 6 Variation in Line Resistance Example 7 Review of Electric Fields To develop a model for line capacitance we first need to review some electric field concepts. Gauss's law: A D da = qe (integrate over closed surface) where D = electric flux density, coulombs/m 2 da = differential area da, with normal to surface A = total closed surface area, m 2 q e = total charge in coulombs enclosed 8 Gauss’s Law Example Similar 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. A D da D D 2 Rh q e qh q 2 R Since D comes radially out integrate over the cylinder bounding the wire ar where ar radially directed unit vector 9 Electric Fields The electric field, E, is related to the electric flux density, D, by D = E where E = electric field (volts/m) = permittivity in farads/m (F/m) = o r o = permittivity of free space (8.85410-12 F/m) r = relative permittivity or the dielectric constant (1 for dry air, 2 to 6 for most dielectrics) 10 Voltage Difference The voltage difference between any two points P and P is defined as an integral V P P E dl In previous example the voltage difference between points P and P , located radial distance R and R from the wire is (assuming = o ) V R R R dR ln 2 o R 2 o R q q 11 Voltage Difference, cont’d With V R R R dR ln 2 o R 2 o R q q if q is positive then those points closer in have a higher voltage. Voltage is defined as the energy (in Joules) required to move a 1 coulomb charge against an electric field (Joules/Coulomb). Voltage is infinite if we pick infinity as the reference point 12 Multi-Conductor Case Now assume we have n parallel conductors, each with a charge density of qi coulombs/m. The voltage difference between our two points, P and P , is now determined by superposition V n R i qi ln 2 i 1 R i 1 where R i is the radial distance from point P to conductor i, and R i the distance from P to i. 13 Multi-Conductor Case, cont’d n If we assume that qi 0 then rewriting i=1 V 1 1 n qi ln qi ln R i 2 i 1 R i 2 i 1 1 n n We then subtract qi ln R1 0 i 1 V R i 1 1 n qi ln qi ln 2 i 1 R i 2 i 1 R 1 1 n R i As we more P to infinity, ln 0 R 1 14 Absolute Voltage Defined Since the second term goes to zero as P goes to infinity, we can now define the voltage of a point w.r.t. a reference voltage at infinity: V 1 n 1 qi ln 2 i 1 R i This equation holds for any point as long as it is not inside one of the wires! 15 Three Conductor Case Assume we have three infinitely long conductors, A, B, & C, each with radius r C B and distance D from the other two conductors. Assume charge densities such that qa + qb + qc = 0 1 1 1 1 Va q ln q ln q ln a b c 2 r D D qa D Va ln 2 r A 16 Line Capacitance For a single line capacitance is defined as qi CiVi But for a multiple conductor case we need to use matrix relationships since the charge on conductor i may be a function of Vj q1 C11 qn Cn1 q CV C1n V1 Cnn Vn 17 Line Capacitance, cont’d In ECE 476 we will not be considering theses cases with mutual capacitance. To eliminate mutual capacitance we'll again assume we have a uniformly transposed line. For the previous three conductor example: Va V Since q a = C Va qa 2 C Va ln D r 18 Bundled Conductor Capacitance Similar to what we did for determining line inductance when there are n bundled conductors, we use the original capacitance equation just substituting an equivalent radius R cb (rd12 1 d1n ) n Note for the capacitance equation we use r rather than r' which was used for R b in the inductance equation 19 Line Capacitance, cont’d For the case of uniformly transposed lines we use the same GMR, D m , as before. ln 2 Dm d ab d ac dbc C c Rb where Dm c Rb (rd12 1 d1n ) n 1 3 (note r NOT r') ε in air o 8.854 10-12 F/m 20 Line Capacitance Example Calculate 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. 21 Line Capacitance Example, cont’d Rbc Dm C Xc 1 (0.01 0.3 0.3) 3 1 (10 10 20) 3 0.0963 m 12.6 m 2 8.854 1012 1.141 1011 F/m 12.6 ln 0.0963 1 1 11 C 2 60 1.141 10 F/m 2.33 108 -m (not / m) 22 ACSR Table Data (Similar to Table A.4) GMR is equivalent to r’ Inductance and Capacitance assume a Dm of 1 ft. 23 ACSR Data, cont’d Dm X L 2 f L 4 f 10 ln 1609 /mile GMR 1 3 2.02 10 f ln ln Dm GMR 1 3 2.02 10 f ln 2.02 103 f ln Dm GMR 7 Term from table assuming a one foot spacing Term independent of conductor with Dm in feet. 24 ACSR Data, Cont. To use the phase to neutral capacitance from table 2 0 1 XC -m where C Dm 2 f C ln r Dm 1 6 1.779 10 ln -mile (table is in M-mile) f r 1 1 1 1.779 ln 1.779 ln Dm M-mile f r f Term independent Term from table assuming of conductor with a one foot spacing Dm in feet. 25 Dove Example GMR 0.0313 feet Outside Diameter = 0.07725 feet (radius = 0.03863) Assuming a one foot spacing at 60 Hz 1 X a 2 60 2 10 1609 ln Ω/mile 0.0313 X a 0.420 Ω/mile, which matches the table 7 For the capacitance 1 1 6 X C 1.779 10 ln 9.65 104 Ω-mile f r 26 Additional Transmission Topics Multi-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, (< 1ft) with oil impregnated paper commonly used to provide insulation – – inductance is lower capacitance is higher, limiting cable length 27 Additional Transmission topics Ground wires: Transmission lines are usually protected from lightning strikes with a ground wire. This topmost wire (or wires) helps to attenuate the transient voltages/currents that arise during a lighting strike. The ground wire is typically grounded at each pole. Corona discharge: Due to high electric fields around lines, the air molecules become ionized. This causes a crackling sound and may cause the line to glow! 28 Additional Transmission topics Shunt conductance: Usually ignored. A small current may flow through contaminants on insulators. DC Transmission: Because of the large fixed cost necessary to convert ac to dc and then back to ac, dc transmission is only practical for several specialized applications – – – long distance overhead power transfer (> 400 miles) long cable power transfer such as underwater providing an asynchronous means of joining different power systems (such as the Eastern and Western grids). 29 Tree Trimming: Before 30 Tree Trimming: After 31 Transmission Line Models Previous lectures have covered how to calculate the distributed inductance, capacitance and resistance of transmission lines. In this section we will use these distributed parameters to develop the transmission line models used in power system analysis. 32 Transmission Line Equivalent Circuit Our current model of a transmission line is shown below Units on z and y are per unit length! For operation at frequency , let z = r + jL and y = g +jC (with g usually equal 0) 33 Derivation of V, I Relationships We can then derive the following relationships: dV I z dx dI (V dV ) y dx V y dx dV ( x) dI ( x) zI yV dx dx 34 Setting up a Second Order Equation dV ( x) dI ( x) zI yV dx dx We can rewrite these two, first order differential equations as a single second order equation d 2V ( x) dI ( x) z zyV 2 dx dx d 2V ( x) zyV 0 2 dx 35 V, I Relationships, cont’d Define the propagation constant as yz j where the attenuation constant the phase constant Use the Laplace Transform to solve. System has a characteristic equation ( s 2 2 ) ( s )( s ) 0 36 Equation for Voltage The general equation for V is V ( x) k1e x k2e x Which can be rewritten as e x e x e x e x V ( x) (k1 k2 )( ) (k1 k2 )( ) 2 2 Let K1 k1 k2 and K 2 k1 k2 . Then x x x e e e V ( x) K1 ( ) K2 ( 2 2 K1 cosh( x) K 2 sinh( x) e x ) 37 Real Hyperbolic Functions For real x the cosh and sinh functions have the following form: d cosh( x) sinh( x) dx d sinh( x) cosh( x) dx 38 Complex Hyperbolic Functions For x = + j the cosh and sinh functions have the following form cosh x cosh cos j sinh sin sinh x sinh cos j cosh sin 39