Line model and performance

1 Line Model and Performance • Short Line Model • Medium-length Line Model • Long Line Model • Lossless Lines • Voltage Profiles • Complex Power Flow Through Transmission Lines • Power Transmission Capability • Line Compensation • Transmission Line Transients • Short Line • Medium-length Line • Long Line Types of Lines : Line Model and Performance (l < 80 km) (80 km < l < 240 km) (l > 240 km) 2-Port Network : A , D : dimensionless B : ohms C : siemens I S V S I R 2-PORT NETWORK V R R R S RRS DICVI BI AV V += + = Line Model and Performance z = series impedance per unit length per phase y = shunt admittance per unit length per phase to neutral Notation : Line Model and Performance l = length of line Z = zl = total series impedance per phase Y = yl = total shunt admittance per phase to neutral 2 Short Line Model : • The line length is less than about 80 km (50 miles) long. Short Line Model • The shunt admittance is neglected. • The series resistance and reactance are treated as lumped parameters. The ABCD parameter : A =1 , B = Z , C =0 , D =A =1 LjRZ ω + = R R S ZI V V + =             =       ⇒ R R S S I V Z I V 10 1 R S I I = Short Line Model Voltage Regulation : 100% × − = RFL RFLRNL V V V VR |V RFL | = the magnitude of receiving-end voltage at full load |V RNL | = the magnitude of receiving-end voltage at no load |V RNL | = |V S | / | A | Short Line Model R I S V R V RI R LR XI R I S V R V RI R LR XI R I S V R V RI R LR XI Short Line Model 3 Example : A 220-kV , 60 Hz , three phase transmission line is 40 km long. The resistance per phase is 0.15 Ω per km and the inductance per phase is 1.3263 mH per km. (a) Use the short line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying a three-phase load of 381 MVA at 0.8 power factor lagging at 220 kV. (b) Determine the voltage regulation when the line is supplying a three-phase load of 381 MVA at 0.8 power factor leading at 220 kV. Short Line Model The series impedance per phase is l L j r Z ) ( ω + = The receiving end voltage per phase is = R V Solution : km 40 , mH/km 3263 . 1 , /km 15 . 0 = = Ω = l L r 40)103263.160215.0( 3 × × × × + = − π j Ω + = 20 6 j kV 01270 3 220 oo ∠=∠ Short Line Model (a) The receiving end power is The current per phase is given by kA 87.360.1)8.0cos( 127 ) 3 / 381 ( 1 o −∠=−∠= − R I The sending end voltage is R R S ZI V V + = kV 250 )( = ⇒ −LLS V kV 93 . 4 33 . 144 )87.361)(206(0127 o oo ∠ = −∠++∠= j MVA 6.2288.30487.36381 j + = ∠ = o )8.0(cos381 1 )3( − ∠ = φ R S Short Line Model The sending end power is Voltage regulation is % 6.13100 220 220250 =× − = Transmission line efficiency is % 4.94100 8.322 8.304 =×== S R P P η ∗ = SSS IVS 3 )3( φ )87.360.1()93.433.144(3 oo ∠ × ∠ × = MVA 8.414336.2888.322 o ∠ = + = j Short Line Model 100% × − = RFL RFLRNL V VV VR 4 (b) The current per phase is given by kA 87.360.1)8.0(cos 127 ) 3 / 381 ( 1 o ∠=∠= − R I The sending end voltage is R R S ZI V V + = kV 3.210 )( = ⇒ −LLS V kV 29 . 9 39 . 121 o ∠ = Voltage regulation is 4.4%100 220 2203.210 % −=× − =VR ) 87 . 36 1 )( 20 6 ( 0 127 oo ∠ + + ∠ = j Short Line Model Medium-length Line Model Medium-length Line Model : • The line length is between 80 km (50 miles) and 240 km (150 miles) • The circuit is called a nominal π circuit • Half of the shunt capacitance is considered to be lumped at each end of the line • The series resistance and reactance are treated as lumped parameters RR I YZ V YZ Y       ++       += 2 1 4 1 2 RRRS VZI Y VV +       += RR ZIV YZ +       += 2 1 2 2 RRSS I Y V Y VI ++= LjRZ ω + = 2 Y 2 Y Medium-length Line Model The ABCD parameter : 4 1 , , 2 1       +==+== YZ YCZB YZ DA LjRZ ω + = 2 Y 2 Y Medium-length Line Model 5 Example : A three-phase completely transposed 345-kV, 200-km line has two 795,000-cmil 26/2 ACSR conductors per bundle and the following line constants : S/km 104.2 , /km 35.0032.0 6- × = Ω + = jyjz Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at 95% of rated voltage , determine the following : (a) ABCD parameters of the nominal π circuit (b) Sending-end voltage , current , and real power (c) Percent voltage regulation (d) Transmission-line efficiency at full load Medium-length Line Model Solution : The ABCD parameters are S/km 104.2, /km 35.0032.0 6- × = Ω + = jyjz S 104.8200)102.4( 46 − − × = × × = jj Y Ω ∠ = + = × + = 78.8429.70704.6200)35.0032.0( o jjZ o 159.09706.0 2 )704.6)(104.8( 1 4 ∠= + × +== − jj DA Ω ∠ = = 78 . 84 29 . 70 o Z B S 08 . 90 10 277 . 8 4 )104.8)(704.6( 1)104.8( 4 4 4 o ∠ × =       ×+ +×= − − − jj jC Medium-length Line Model The receiving end current is kV 0189.20 3 345 95 . 0 oo ∠=∠ × = R V kA 11.8246.1)99.0(cos 99 . 0 2 . 189 3 / 700 1 o ∠=∠ × = − R I The receiving end voltage per phase is Medium-length Line Model The sending end voltage and current are )11.8246.1)(78.8429.70()02.189)(159.09706.0( oooo ∠ ∠ + ∠ ∠ = S V kA 5 . 15 241 . 1 o ∠ = kV 14 . 26 6 . 199 o ∠ = )11.8246.1)(159.09706.0()02.189)(08.9010277.8( 4 oooo ∠∠+∠∠×= − S I kV 8.345 )( = ⇒ −LLS V The sending end real power is MW 5 . 730 )5.1514.26cos()241.1)(6.199(3 )3( = −= oo φ S P Medium-length Line Model 6 The no-load receiving end voltage is Voltage regulation is Transmission line efficiency is % 8.95100 5 . 730 700 =×= η A V V S RNL = kV 6.205 9706 . 0 6 . 199 == % 7.8100 2 . 189 2 . 189 6 . 205 % =× − =VR Medium-length Line Model Long Line Model Long Line Model : • The line length is greater than 240 km (150 miles) • Accuracy obtained by using distributed parameters V(x) I(x)I(x+∆x) V(x+∆x) ∆x z∆x y∆x x L L S V R V l S I R I y∆x Long Line Model From KVL : ) ( ) ( ) ( ) ( x I x z x V x x V ∆ + = ∆ + )( ) ( ) ( xzI x x V x x V = ∆ − ∆ + )( ) ( xzI dx x dV = V(x) I(x)I(x+∆x) V(x+∆x) ∆x z∆x y∆x x L L S V R V l S I R I y∆x Long Line Model From KCL : ) ( ) ( ) ( ) ( x x V x y x I x x I ∆ + ∆ + = ∆ + )( ) ( ) ( xxyV x x I x x I ∆+= ∆ − ∆ + )( ) ( xyV dx x dI = )()( xVxxV ≈ ∆ + V(x) I(x)I(x+∆x) V(x+∆x) ∆x z∆x y∆x x L L S V R V l S I R I y∆x 7 Long Line Model )( ) ( xyV dx x dI = )( ) ( xzI dx x dV = dx xdI z dx xVd )()( 2 2 = dx xdV y dx xId )()( 2 2 = ) ( x zyV = ) ( x zyI = The solution of the equation is x c x c xx e Z A e Z A xIeAeAxV γγγγ −− −=+= 21 21 )( and )( Long Line Model α is the attenuation constant β is the phase constant (rad/m) βαγ jyz +== Z c is the characteristic impedance , y z Z c = γ is the propagation constant (m -1 ) , x c x c xx e Z A e Z A xIeAeAx V γγγγ −− −=+= 21 21 )( and )( Long Line Model At receiving end , x = 0 ; 2 and 2 21 cRRcRR Z I V A Z I V A − = + =⇒ R R I I V V = = and x c x c xx e Z A e Z A xIeAeAx V γγγγ −− −=+= 21 21 )( and )( c c RR Z A Z A IAA V 21 21 and −=+= x RcR x RcR x cRR x cRR e IZV e IZV xI e Z I V e Z I V x V γγ γγ − −       − −       + =       − +       + = 2 / 2 / )( 22 )( Long Line Model R xx cR xx x cRR x cRR I ee ZV ee e Z I V e Z I V x V         − +         + =       − +       + = −− − 22 22 )( γγγγ γγ R xx R xx c x RcR x RcR I ee V ee Z e I Z V e I Z V xI         + +         − =       − −       + = −− − 22 1 2 / 2 / )( γγγγ γγ Rearranging RR c RcR IxVx Z xI I x Z V x x V )cosh()sinh( 1 )( ) sinh( ) cosh( ) ( γγ γ γ += + = 2 sinh and 2 cosh θθθθ θθ −− − = + = eeee 8 Long Line Model At sending end , x = l ; The ABCD parameters : )sinh( 1 ; )sinh( ; )cosh( l Z ClZBlDA c c γγγ ==== RR c S RcRS IlVl Z I I l Z V l V ) cosh() sinh( 1 ) sinh( ) cosh( γγ γ γ += + = S S I I V V = = and RR c RcR IxVx Z xI I x Z V x x V )cosh()sinh( 1 )( ) sinh( ) cosh( ) ( γγ γ γ += + = Long Line Model The computation of hyperbolic function of complex arguments : )sin()cosh()cos()sinh( )sinh()sinh( )sin()sinh()cos()cosh( ) cosh( ) cosh( lljll ljll lljll l j l l βαβα βαγ βαβα β α γ += += += + = Long Line Model Example : A 50 Hz transmission line 300 km long has a total series impedance of 40+j125 Ω and a total shunt admittance of j10 -3 S. The receiving-end load is 50 MW at 220 kV with 0.8 lagging power factor. Find the sending-end voltage, current, power and power factor using exact transmission line equations . Long Line Model The total series impedance and shunt admittance are S 901010 3.722.13112540 33 o o ∠== Ω∠=+= −− j Y jZ Solution : o o 5.81354.035.0052.0)sinh( 2.1938.002.0938.0)cosh( ∠=+= ∠=+= jl jl γ γ ZYl = γ o 2.81362.03577.00554.0 ∠=+= j Y Z Z c = o 85 . 8 21 . 362 − ∠ = 9 Long Line Model kV 01270 3 220 oo ∠=∠= R V o 9.36 8 . 0 220 3 50 −∠ × × = R I kA 9 . 36 164 . 0 o − ∠ = The receiving-end load is 50 MW at 220 kV with 0.8 lagging pf. o 2.1938.0)cosh( ∠=== lDA γ o 65.722.128)sinh( ∠== lZB c γ o 4 . 90 10 77 . 9 )sinh( 1 4 ∠ × = = − l Z C c γ Long Line Model MW 16 . 52 987 . 0 1286 . 0 97 . 136 3 = × × × = S P leading 987.0)3.152.6cos(pf =−= oo S R R S BI AV V + = kA 3 . 15 1286 . 0 )9.36164.0)(2.1938.0()0127)(4.901077.9( 4 o oooo ∠ = −∠∠+∠∠×= − S I kV 23.237 kV 2.697.136 )9.36164.0)(65.722.128()0127)(2.1938.0( )( =⇒∠= −∠∠+∠∠= −LLS V o oooo Long Line Model The ABCD parameter : Equivalent π ππ π Circuit : Z ′ V S V R I R I S 2 Y ′ 2 Y ′       ′ ′ + ′ = ′ = ′ ′ +== 4 1 , , 2 1 Z Y YCZB Z Y DA )sinh( 1 )sinh( )cosh( l Z C lZB lDA c c γ γ γ = = = = Long Line Model Equating the coefficients We get ) sinh( l Z Z c γ = ′ )2/tanh( 1 l Z c γ = )cosh( 2 1 and )sinh( l Z Y lZZ c γγ = ′ ′ += ′ )sinh(= l y z γ )sinh( l z γ γ = l l Z γ γ ) sinh( = )sinh( 1)cosh( =)2/tanh( l l l γ γ γ − Z l Y ′ − = 1 ) cosh( 2 ' γ )sinh( 1 ) cosh( lZ l c γ γ − = )2/( ) 2 / tanh( 2 = l l Y γ γ )2/tanh( l z y γ = 10 Long Line Model Equivalent π ππ π Circuit )2/( )2/tanh( 2 l lY γ γ l l Z γ γ )sinh( For a lossless line , R = G = 0 , and S/m , /m CjyLjz ω ω = Ω = y z Z c = LC ω β α = = ⇒ , 0 Ω== C L Cj Lj ω ω zy = γ ))(( CjLj ω ω = 1- m LCj ω = Lossless Lines (Surge impedance) • Lossless Lines The equations for voltage and current for a long transmission line RR c S RcRS IlVl Z I I l Z V l V ) cosh() sinh( 1 ) sinh( ) cosh( γγ γ γ += + = For a lossless line , RR c S RcRS IlV Z lj I I l jZ V l V )cos( )sin( ) sin( ) cos( β β β β += + = β γ α j = → = 0 Lossless Lines The velocity of propagation of voltage or current wave The wavelength is the distance required to change the phase of the voltage or current by 2π radians or 360 o β π λ 2 = β π λ f fv 2 == LC ωβ = For a lossless line , 1 and 1 LC v LCf == λ Lossless Lines [...]... transmission line is 300 km long The line inductance is 0.97 mH/km per phase and its capacitance is 0.0115 µF/km per phase Assume a lossless line Determine the line phase constant β , the surge impedance Zc , the velocity of propagation v , and the line wavelength λ 2 Vrated Zc 11 Lossless Lines Lossless Lines Voltage Profiles : V ( x) = cos(βx)VR + jZ c sin( βx) I R Solution : For a lossless line , β... lightning surges - reduce line loadability if they are not removed under full-load condition IS 100 300 500 700 Long line IR 1100 Line length (km) VS VR jX Lsh IR = VR jX Lsh 15 Line Compensation Line Compensation Example : A three-phase, 60 Hz, 765-kV transmission line is 400 km long The line inductance is 0.88853 mH/km per phase and its capacitance is 0.01268 µF/km per phase The line resistance is 0.011... full-load current Lossless Lines The velocity of propagation is v= v= 1 LC 1 = 2.994 × 105 km/s −6 0.97 × 10 × 0.0115 × 10 −3 The line wavelength is λ= λ= Lossless Lines v f VS Sh or tc irc uit 2.994 × 105 = 4990 km 60 12 Complex Power Flow Through Transmission Lines Summary Transmission -line ABCD parameters A=D Type Short line C B 1 Z Long line cosh(γ l ) Z c sinh(γ l ) Lossless line cos(βl ) jZ c sin(... 1023 Ω 571.8 − jX Cser VS VR jX Csh 17 Line Compensation Line Compensation Example : A three-phase, 60 Hz, 765-kV transmission line is 400 km long The line inductance is 0.88853 mH/km per phase and its capacitance is 0.01268 µF/km per phase The line resistance is 0.011 Ω/km The line supplies a load of 1600 MW , 0.8 power factor lagging at 765 kV Determine the MVAR and the capacitance of the shunt capacitors... bypass high current during faults and to reinsert the capacitor banks after fault clearing - may be excite low-frequency oscillations 18 Line Compensation Line Compensation Example : A three-phase, 60 Hz, 765-kV transmission line is 400 km long The line inductance is 0.88853 mH/km per phase and its capacitance is 0.01268 µF/km per phase The line resistance is 0.011 Ω/km The line supplies a load of 1600... Synchronous Compensator (STATCOM) • Unified Power Flow Controller (UPFC) 20 Line Compensation Line Compensation SVC TCSC Line Compensation Line Compensation • Thyristor-controlled Series Compensator (TCSC) • Static Synchronous Compensator (STATCOM) 21 Line Compensation Line Compensation STATCOM UPFC Line Compensation Transmission Line Transients • Unified Power Flow Controller (UPFC) Transient overvoltage... Diagram Single -line diagram (or one -line diagram) is a simplified notation for representing a three-phase power system Single -line Diagram Single -line diagram symbols : Electrical elements such as generators, transformers , transmission lines, loads, circuit breakers, capacitors, etc., are shown by standardized schematic symbols The single -line diagram of a very simple power system 1 Single -line Diagram... iR vR / Z c vR If the line is terminated in its characteristic impedance , ρ R = 0 At the sending end : ρS = Zs − Zc Zs + Zc Transmission Line Transients The lattice diagram is a tool/technique to simplify the accounting of reflections and waveforms ρR ρS • Diagram shows the boundaries (x =0 and x=l) and the reflection coefficients • Time axis shown vertically • Slope of the line should indicate flight... Transmission Line Transients Example : A dc source of 120 V with negligible resistance is connected through a switch S to a lossless transmission line having Zc = 30 Ω Plot vR versus time until t = 5T where T is the time for a voltage wave to travel the length of the line If the line is terminated in a resistance of (a) 90 Ω (b) 10 Ω 25 F F Transmission Line Transients (a) ZR = 90 Ω , Zs = 0 Ω and Zc =... 77.11∠87o The line supplies a load of 1600 MW , 0.8 power factor lagging at 765 kV The receiving end voltage per phase is VR = From equivalent π model Z ′ = Z c sinh(γl ) = 128.4∠88.197o = 4.04 + j128.34 = 4.04 + j 77 Line Compensation Line Compensation IS ′ B = Z new 765 o ∠0 = 441.7∠0o kV 3 The receiving end current is IR = 1600 ∠(− cos −1 0.8) = 1.509∠ − 36.87o kA 3 × 765 × 0.8 19 Line Compensation Line