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Unformatted text preview: Overhead Conductor Overhead Spacer Cable Underground Cable Three-Conductor Cable Service Cables Power Systems I ￿ ￿ ￿ ￿ ￿ Transmission Lines ACSR Aluminum Conductor with inner Steel Reinforced strands ACAR Aluminum Conductor with inner Al allow Reinforced strands ACSR/AW Aluminum Conductor with inner Alumoweld Steel Reiforced strands Aluminum - current carrying member Steel - structural support Power Systems I ￿ ￿ ￿ ￿ ￿ Overhead Conductors Where conductor close proximity is required In s u la t in g ja c k e t surrounds each conductor Plastic spacers keep conductors from coming in c o n t a c t w i th o n e a n o t h e r Power Systems I ￿ ￿ ￿ Overhead Cable Power Systems I Cables Underground transmission and distribution cables Semiconducting material surrounds the conductor to grade the electric field Plastic jacket provides insulation and protection Neutral strands for an outer shell for protection and return currents Power Systems I ￿ ￿ ￿ ￿ Cables ￿ ￿ ￿ ρ = conductor resistivity l = conductor length A = conductor crosssectional area Rnew = Rold Rac = 1.02 ⋅ Rdc ρl Rdc = A TAl = 228°C T + told increased resistance at conductor temperature rises wiring is rated for 65°C, 75°C, or 90°C T + t new ambient temperature is 20°C Temperature effects ￿ skin effect at 60 Hz: ac resistance ￿ ￿ dc resistance ￿ Line resistance Power Systems I ￿ ￿ Transmission Line Parameters Power Systems I Integral of the flux density that is normal to a defined area Magnetic Flux A φ = ∫ B ⋅ da B = µH Integral of the scalar product of a closed path ie and the magnetic field equals the encircled current B A H φ A= B= H= F = ∫ H ⋅ d l = ie Γ Γ= Ampere’s circuital law Review of Magnetics and Inductance I I I Power Systems I ∑ φ = ∑ ∫ B ⋅ da = ∑ ∫ µH ⋅ da = λ L= I Inductance i =1 λ = ∑ φi N Flux Linkage Review of Magnetics and Inductance infinite straight wire is an approximation of a reasonably long wire ￿ ￿ ￿ ￿ ￿ Image the wire to close at +/- infinity, establishing a kind of “one-turn coil” with the return path at infinity Straight infinitely long wire of radius r Uniform current density in the wire. Total current is Ix Flux lines form concentric circles (i.e. H is tangential) Angular symmetry - it suffices to consider Hx Assumptions: ￿ Conditions: Power Systems I ￿ ￿ Inductance of a Single Conductor ⋅ dl = I x → Hx = µI I x → Bx = 0 2 x 2π r 2 2π r Power Systems I µ0 I r 3 µ0 I λint = ∫ dλx = x dx = 4∫ 2π r 0 8π 0 r µ0 → Lint = = 0.5 ×10 −7 8π x2 µ0 I µ0 I 3 x dx dφ x = Bx dx = xdx → dλx = 2 dφ x = 2 4 2π r r 2π r I I = x2 π r2 π x Case 1: Points inside of the conductor (x < r) 0 ∫H ￿ x Ix ⇒ H= 2π x General: ￿ 2πx Inductance of a Single Conductor D1 λext = ∫ dλx = D2 µ0 I 2π µI D 1 d x = 0 ln 2 ∫x 2π D1 D1 D2 → Lext = 2 × 10 −7 ln µ0 I I x = I → Bx = µ 0 H x = 2π x µI µI dφ x = Bx dx = 0 dx → dλx = dφ x = 0 dx 2π x 2π x Case 2: Points outside of the conductor (x > r) Power Systems I ￿ Inductance of a Single Conductor D2 D1 r + = re −1 4 = DS D r1 r1 Power Systems I −7 D ȹ D ȹ D ȹ − 7 ȹ − 7 ȹ ȹ ln L = 2 ×10 ȹ ln −1 4 ȹ = 2 × 10 ȹ ln ȹ = 2 × 10 ȹ ȹ re Ⱥ ȹ r + Ⱥ ȹ DS −7 ȹ ȹ 1 ȹ ln −1 4 + ln D ȹ L1 = 2 × 10 ȹ ȹ r1e Ⱥ ȹ r1 = r2 L1 = L2 = L L1 = L1(int ) + L1( ext ) = 0.5 × 10 −7 + 2 × 10 −7 ln D r1 ȹ ȹ ȹ Ⱥ conductors of radii r1 and r2, separated by a distance D L1( ext ) = 2 × 10 −7 ln ￿ Inductance of a Single-Phase Line D r2 ȹ 1 ȹ ȹ ln ȹ L22 = 2 ×10 −7 ȹ r + ȹ ȹ 1 Ⱥ Power Systems I ȹ 1 ȹ ȹ ln ȹ ȹ r + ȹ ȹ 2 Ⱥ ȹ 1 ȹ ȹ ln ȹ ȹ D Ⱥ λ1 = L11 I1 − L12 I1 I1 = − I 2 → λ2 = − L21 I 2 + L22 I 2 L12 = L21 = −2 ×10 −7 (ln D ) = 2 ×10 −7 L11 = 2 ×10 −7 1 = L11 I1 + L12 I 2 λ2 = L21 I1 + L22 I 2 From the 2 conductor case: Flux Linkage - Self and Mutual Inductances Power Systems I λi = 2 ×10 −7 j≠i ȹ 1n 1 ȹ ȹ I i ln + ∑ I j ȹ ȹ ri+ j =1 Dij ȹ ȹ Ⱥ j =1 λi = Lii I i + ∑ Lij I j n I1 + I 2 + ￿ + I i + ￿ + I n = 0 General Case: Total Inductance j≠i −7 1 1 1 ȹ ȹ λa = 2 × 10 ȹ I a ln + I b ln + I c ln ȹ r + D D Ⱥ ȹ 1 1 ȹ ȹ λa = 2 × 10 −7 ȹ I a ln − I a ln ȹ r + D Ⱥ ȹ D −7 λa = 2 × 10 I a ln r + D L = 0.2 ln DS I a + Ib + Ic = 0 Symmetrical spacing Power Systems I ￿ Ic Inductance of Three-Phase Lines D D D Ia Ib −7 1 D12 1 ln r + 1 D32 ln ln 1 Ⱥ Ⱥ D13 Ⱥ 1 Ⱥ ln D23 Ⱥ 1 Ⱥ ln Ⱥ r + Ⱥ Ⱥ ȹ 1 1 1 ȹ ȹ I a ln + I b ln ȹ + I c ln λa = 2 ×10 ȹ r + D12 D13 ȹ ȹ Ⱥ ȹ 1 1 1 ȹ ȹ + I c ln λb = 2 ×10 −7 ȹ I a ln + I b ln ȹ + r D21 D23 ȹ ȹ Ⱥ ȹ 1 1 1 ȹ ȹ + I c ln λc = 2 ×10 −7 ȹ I a ln + I b ln ȹ r + D31 D32 ȹ ȹ Ⱥ = LI Asymmetrical spacing Ⱥ 1 Ⱥ ln + Ⱥ r 1 L = 2 × 10 −7 Ⱥln Ⱥ D21 Ⱥ 1 Ⱥ Power Systems I Ⱥ D31 Ⱥ ￿ Inductance of Three-Phase Lines Ic D31 D23 D12 Ia Ib horizontal or vertical configurations are most popular Symmetry is lost - unbalanced conditions ￿ c a a b b c a b a c Each phase occupies each position for the same fraction of the total length of the line Power Systems I ￿ b c Average inductance of each phase will be the same restore balanced conditions by the method of transposition of lines ￿ ￿ The practice of equilateral arrangement of phases is not c o n v e n ie n t position a 1 b 2 c 3 ￿ ￿ Transposition c a b D2 Power Systems I q =Cv Capacitance D1 v12 = vD1 − vD2 = − ∫ E ⋅ dl Electric field D=εE Electric field A q e = ∫ D ⋅ da Gauss’s law R Review of Electric Fields D Gaussian Surface h q1 D ln 2πε 0 r q D2 q ln dx = 2πε 0 x 2πε 0 D1 R h Power Systems I q D ln v12 = πε 0 r q1 D q2 r v12 = v12(q1) + v21(q 2 ) = ln + ln 2πε 0 r 2πε 0 D q2 r v21(q 2 ) = ln 2πε 0 D v12 (q1) = D1 v12 = ∫ D2 Infinite Straight Wire D q C= v 2π ε C= D ln r Infinite wire of radius r c C C a n ȹ ȹ ȹ Ⱥ ￿ per mile per phase GMDφ = geometeric mean distance between conductors rφ= conductor radius 0.0389 ȹ GMDφ log10 ȹ ȹ r ȹ φ b C C= Equilateral spacing Power Systems I ￿ Three-Phase Capacitance 12 i n conductor R: 0.3263 Ω/mile GMR: 0.0244 ft Dia.: 0 .7 2 0 i n 44 i n Power Systems I 44 i n ￿ ) = 149.9 Ω mi ￿￿￿￿￿￿￿￿ 1 (0.720 in ) ⋅ 12 = 0.03 ft 0.0389 = 0.177 log10 (4.73 0.03) 1 2 X C = 1 (2π 60 ⋅ 0.177 C= rφ = 1 dia = 2 ȹ 4.73 ȹ ȹ 10 ȹ ȹ 0.0244 Ⱥ (60) log 60 = 0.326 + j 0.639 Ω / mi Z a = (0.3263) + j 0.2794 = 56.8 in = 4.73 ft GMDφ = 3 d12 d 23 d13 = 3 (45.6 ) (88) (45.6 ) Calculate the resistance, inductive reactance, and capacitive reactance per phase and rated current carrying capacity for the overhead line shown. Assume the line operates at 60 Hz Example D12 3 conductors D12 2 conductors D13 D12 4 conductors D13 D14 Commonly used to reduce the electric field strength at the conductor surface Used on overhead lines above 230 kV Conductors are connected in parallel Typical bundled conductor configurations Power Systems I ￿ ￿ ￿ ￿ Conductor Bundling n i=2 rφ+ = n rφ ⋅ ∏ D1i Equivalent radius i=2 + GMRφ = n GMRφ ⋅ ∏ D1i n The use of bundled conductors effects the impedance of the line, the GMRφ , the GMDφ , and the equivalent radius GMDφ : the distance between the center of each bundle is u sed GMRφ : Power Systems I ￿ ￿ ￿ ￿ Conductor Bundling Power Systems I 10 ft 30 ft 20 in 20 in conductor R: 0.1204 Ω/mile GMR: 0.0403 ft Dia.: 1 .1 9 6 i n Calculate the resistance, inductive reactance, and capacitive reactance of the overhead line shown. Assume the line operates at 60 Hz Example 1 2 Power Systems I X C = 1 (2π 60 ⋅ 0.177 ￿ ) = 116.85 0.0389 C= = 0.0227 log10 (39.15 0.7568) rφ = 1 dia = 2 Ω mi ￿￿￿￿￿￿￿￿ 1 (1.196 in ) ⋅ 12 = 0.0498 ft rφ+ = 4 (0.0498) (1.67 ) (1.414 ) (1.67 ) = 0.7568 ft ȹ 39.15 ȹ ȹ 10 ȹ ȹ 0.7178 Ⱥ (60) log 60 = 0.0301 + j 0.485 Ω / mi Z a = (0.0301) + j 0.2794 GMRφ = 4 (0.0403) (1.67 ) (1.414 ) (1.67 ) = 0.7178 ft GMDφ = 3 D12 D23 D13 = 3 (31.6 ) (60 ) (31.6 ) = 39.15 ft R+ = 1 ⋅ 0.1204 = 0.0301 Ω / mi 4 Example Voltages are expressed as phase-to-neutral Currents are expressed for one phase The three phase system is reduced to an equivalent single-phase Line parameters: R, L, C, & G ￿ ￿ depend on the length and the voltage level short, medium, and long length line models Three types of models ￿ All lines are made up of distributed series inductance and resistance, and shunt capacitance and conductance ￿ ￿ ￿ Transmission lines are represented by an equivalent circuit with parameters on a per-phase basis Power Systems I ￿ ￿ ￿ Transmission Line Modeling circuit equations matrix form ￿ ￿ ȺVS Ⱥ Ⱥ A B Ⱥ ȺVR Ⱥ Ⱥ I Ⱥ = ȺC D Ⱥ Ⱥ I Ⱥ Ⱥ Ⱥ R Ⱥ Ⱥ S Ⱥ Ⱥ I S = C VR + D I R VS = A VR + B I R All transmission line models may be described as a twoport network The ABCD two-port network is the most common representation The network is described by the four constants: A, B, C, & D Network equations: Power Systems I ￿ ￿ ￿ ￿ ABCD Two-Port Network The line length is less than 50 miles (80 km), or The line voltage is not over 69 kV VS RL XL VR The shunt capacitance and conductance are ignored The line resistance and reactance are treated as lumped parameters Circuit of the short model ￿ ￿ Modeling of the transmission line parameters ￿ ￿ The short transmission line model may be used when Power Systems I ￿ ￿ ￿ Short Transmission Line Model Gen. IS VS = VR + I R Z VS = VR + I R ( R + j ω s L) IS = IR Z=R+jωL Circuit analysis of the short line model Power Systems I ￿ Short Transmission Line Model VR Load IR A=1 B = Z line ABCD values: ￿ D = A=1 C =0 ȺVS Ⱥ Ⱥ1 Z line Ⱥ ȺVR Ⱥ Ⱥ I Ⱥ = Ⱥ0 1 Ⱥ Ⱥ I Ⱥ Ⱥ Ⱥ R Ⱥ Ⱥ S Ⱥ Ⱥ Matrix representation: IS = IR VS = VR + Z line I R ￿ Circuit Equations: Power Systems I ￿ Two-Port Representation R = 0.15 Ω/km L = 1.3263 mH/km ( 381 MVA load at 0.8 lagging pf at 220 kV ) ( ) Z = 6 + j 20 Ω 220,000∠0° VR = = 127,000∠0° 3 S R ( 3φ ) = 381∠ cos −1 0.8 = 381∠36.9° = 304.8 + j 228.6 MVA Z = (r + j ω L ) ￿ = 0.15 Ω + j 2 π × 60 ×1.3263 ×10 −3 ⋅ 40 ￿ Find V, S, V.R., and η at the sending end of the line for ￿ 40 km, 220 kV transmission line has per phase Power Systems I ￿ ￿ Short Transmission Line Example 3 VR* 381× 106 ∠ − 36.9° = = 1000∠ − 36.9° A 3 × 127,000∠0° Power Systems I 250 - 220 304.8 VR% = × 100% = 13.6% η = ×100% = 94.4% 220 322.8 = 322.8 + j 288.6 = 433∠41.8° MVA * S S (3φ ) = 3 ⋅ VS ⋅ I S = 3 ⋅ (144,330∠4.93°)(1000 ∠ − 36.9°) VS − LL = 3 ⋅ VS = 250 kV = 144,330∠4.93° VS = VR + Z I R = 127,000∠0° + (6 + j 20)(1000∠ − 36.9°) IR = * S R ( 3φ ) Short Transmission Line Example The line length is greater than 50 miles (80 km) The line length is less than 150 miles (250 km) VS YC/2 RL XL YC/2 VR Half of the shunt capacitance is considered to be lumped at each end of the line The line resistance and reactance are treated as lumped parameters Circuit model: ￿ ￿ Modeling of the transmission line parameters ￿ ￿ The medium transmission line model may be used when Power Systems I ￿ ￿ ￿ Medium Transmission Line Model Gen. IS VS ½ YC IS C R R R Z line YC 4 R YC 2 YC 2 Z line YC 2 S line R C VS = VR + Z line I R + Y2 VR ( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 + ½ YC Z=R+jωL Z line YC 2 Circuit analysis of the short line model Power Systems I ￿ )I Medium Transmission Line Model R VR Load IR Matrix representation: ABCD values: ￿ ￿ Circuit Equations: Power Systems I ￿ C R YC 2 Z line YC 4 ( A = 1+ C = YC 1 + Z line YC 2 Z line YC 4 ( ) Z line YC 2 )I R Z line Ⱥ ȺVR Ⱥ Ⱥ Z line YC Ⱥ Ⱥ 1 + 2 Ⱥ Ⱥ I R Ⱥ S line R B = Z line D = 1 + Zline YC 2 ) R R YC 2 R ȺVS Ⱥ Ⱥ 1 + Z line YC 2 = Ⱥ Ⱥ I Ⱥ YC 1 + Zline YC Ⱥ S Ⱥ Ⱥ 4 IS Z line YC 2 VS = VR + Z line I R + Y2C VR ( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 + Two-Port Representation R = 0.036 Ω/km L = 0.80 mH/km ( 270 MVA load at 0.8 lagging pf at 325 kV ) ( ) 325,000∠0° VR = = 187,600∠0° 3 S R (3φ ) = 270∠ cos −1 0.8 = 270∠36.9° = 216 + j162 MVA Y = ( j ω C ) ￿ = j 2 π × 60 × 0.0112 × 10 −6 ⋅130 = j 0.549 siemens ( = 4.68 + j 39.2 Ω Z = (r + j ω L ) ￿ = 0.036 Ω + j 2 π × 60 × 0.8 ×10 −3 ⋅130 ￿ ) C = 0.0112 uF/km F i n d V a n d S a t t h e s e n d i n g e n d o f th e l i n e f o r ￿ 130 km, 345 kV transmission line has per phase Power Systems I ￿ ￿ Medium Transmission Line Example 3 VR* 270 ×106 ∠ − 36.9° = = 480∠ − 36.9° A 3 ×187,600∠0° ( ) Power Systems I = 421.5∠ − 25.58° (480∠ − 36.9°)(0.989 + j 0.001284) I S = C VR + D I R = (187,600∠0°) − 3.53 ×10 −7 + j 5.46 ×10 − 4 + = 199,160∠4.02° (480∠ − 36.9°)(4.68 + j39.2) 0.989 + j 0.001284 4.68 + j 39.2 Ⱥ Ⱥ ABCD = Ⱥ − 3.53 ×10 −7 + j 5.46 ×10 − 4 0.989 + j 0.001284Ⱥ Ⱥ Ⱥ VS = A VR + B I R = (187,600∠0°)(0.989 + j 0.001284 ) + IR = * S R (3φ ) Medium Transmission Line Example Power Systems I = 325 345 0.989 + j 0.001284 - 325 ×100% = 7.3% = 218.9 + j124.2 MVA pf = 0.87 VR ( NL ) − VR ( FL ) VS ( FL ) / A − VR ( FL ) VR% = ×100% = × 100% VR ( FL ) VR ( FL ) * S S (3φ ) = 3 ⋅ VS ⋅ I S = 3 ⋅ (199,160∠4.02°)(421 ∠ − 25.58°) VS − LL = 3 ⋅ VS = 345 kV Medium Transmission Line Example The line length is greater than 150 miles (250 km) ￿ ￿ ￿ Accuracy obtained by using distributed parameters The series impedance per unit length is z The shunt admittance per unit length is y Modeling of the transmission line parameters ￿ The long transmission line model are used when Power Systems I ￿ ￿ Long Transmission Line Model V(x + Δx) I(x + Δx) y Δx Δx l y Δx I(x) V(x) x IR VR Power Systems I V ( x + Δx) = V ( x) + z Δx I ( x) I ( x + Δx) = I ( x) + y Δx V ( x + Δx) V ( x + Δx) − V ( x) I ( x + Δx) − I ( x) = z I ( x) = y V ( x + Δx) Δx Δx dV ( x) dI ( x) limit as Δx → 0 = z I ( x) limit as Δx → 0 = y V ( x) dx dx VS IS z Δx Long Transmission Line Model VS I + ΔI V+ΔV Δx V I x VR Load IR d 2V ( x) dI ( x) d 2 I ( x) dV ( x) =z =y 2 2 dx dx dx dx d 2V ( x) d 2 I ( x) = z ( y V ( x) ) = y ( z I ( x) ) 2 2 dx dx γ 2 = z y propagation constant Power Systems I Gen. IS Long Transmission Line Model (r + jωL )(g + jωC ) Power Systems I @x=0⇒ ( VR + I R Z c A1 = 2 ( ( ) VR − I R Z c A2 = 2 ) ) 1 dV ( x) γ I ( x) = = A1 eγ x − A2 e −γ x = y A1 eγ x − A2 e −γ x z z dx z 1 Zc = z y I ( x) = A1 eγ x − A2 e −γ x characteristic impedance Zc γ = α + jβ = z y = d 2V ( x) = γ 2V ( x) dx 2 V = A1 eγ x + A2 e −γ x Long Transmission Line Model 1e Zc ( −e 2 + e−x 2 x yz yz ) VR + VR + Z c −x y z yz e yz x yz ex ) y z)I +e 2 yz −x y z − e−x 2 Power Systems I c R R V ( x) = cosh x y z VR + Z c sinh x y z I R ( 1 I ( x) = sinh (x y z )V + cosh (x Z I ( x) = V ( x) = ex VR + Z c I R x y z VR − Z c I R − x y z V ( x) = e e + 2 2 V Z +I V Z −I I ( x) = R c R e x y z − R c R e − x y z 2 2 IR IR Long Transmission Line Model eθ + e −θ cosh θ = 2 eθ − e −θ sinh θ = 2 Hyperbolic Functions Power Systems I γ= zy Zc = z y Ⱥ cosh (γ ￿ ) Z c sinh (γ ￿ )Ⱥ Ⱥ ABCD = Ⱥ 1 Ⱥ sinh (γ ￿ ) cosh (γ ￿ ) Ⱥ Ⱥ Z c Ⱥ 1 IS = sinh (γ ￿ ) VR + cosh (γ ￿ ) I R Zc let x → ￿ VS = cosh (γ ￿ ) VR + Z c sinh (γ ￿ ) I R Two-Port Representation Y’/2 VR ) ( ) Y + 1 cosh (γ ￿ ) − 1 1 ȹ γ ￿ ȹ tanh ȹ = (cosh (γ ￿ ) − 1) = = ȹ Z c sinh (γ ￿ ) Z c 2 Z + ȹ 2 Ⱥ ( → ) + 1 + Z + 4Y + VR + 1 + Z + 2Y + I R IS = Y + + VS = 1 + Z 2Y VR + Z + I R ( → Z + = Z c sinh (γ ￿ ) Y’/2 F i n d th e v a l u e s fo r Z ’ a n d Y ’ VS Represent a long transmission line as a pi-model for circuit analysis Z’ The circuit: Power Systems I ￿ ￿ ￿ Pi-Model of a Long Transmission Line z = 0.045 + j 0.4 Ω/km Y = j 4. 0 uS/km γ = zy = (0.045 + j 0.4 )(4 × 10 −6 ) = 7.104 × 10 −5 + j 0.001267 Z + = Z c sinh (γ ￿ ) = 10.88 + j 98.36 Y + 1 ȹ γ ￿ ȹ = tanhȹ ȹ = j 0.001008 2 Zc ȹ 2 Ⱥ Zc = z 0.045 + j 0.4 = = 316.7 - j17.76 −6 4 × 10 y Find ABCD for a pi model of the long transmission line ￿ 250 km, 500 kV transmission line has per phase Power Systems I ￿ ￿ Long Transmission Line Example ) + + ) Power Systems I C = Y + 1 + Z 4Y = j 0.00100 ( B = Z + = 10.88 + j 98.36 ( Z + = 10.88 + j 98.36 Y + = j 0.001008 2 + + A = D = 1 + Z 2Y = 0.9504 + j 0.0055 Long Transmission Line Example ...
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This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.

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