KECE 361 POWER SYSTEM ANALYSIS Ch. 4: Transmission Lines Professor Gilsoo Jang School of Electrical Engineering KOREA UNIVERSITY Power & Energy Laboratory Development of Line Models ▪ Goals of this chapter are 1) develop a simple model for transmission lines 2) gain an intuitive feel for how the geometry of the transmission line affects the model parameters KOREA UNIVERSITY 1 Power & Energy Laboratory Primary Methods for Power Transfer ▪ The most common methods for transfer of electric power are 1) Overhead ac 2) Underground ac 3) Overhead dc 4) Underground dc 5) other KOREA UNIVERSITY 2 Power & Energy Laboratory Magnetics Review ▪ Ampere’s circuital law: F = H dl = I e F = mmf = magnetomotive force (amp - turns) H = magnetic field intensity (amp - turns/meter) dl = Vector differential path length (meters) = Line integral about closed path (dl is tangent t o path) I e = Algebraic sum of current linked by KOREA UNIVERSITY 3 Power & Energy Laboratory Line Integrals ▪ Line integrals are a generalization of traditional integration b f ( x)dx dx Integration along the x-axis a dl Integration along a general path, which may be closed H dl Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line KOREA UNIVERSITY 4 Power & Energy Laboratory Magnetic Flux Density ▪ Magnetic fields are usually measured in terms of flux density KOREA UNIVERSITY 5 Power & Energy Laboratory Magnetic Flux Total flux passing through a surface A is 𝟇 = න 𝑩 𝑑𝐚 A (𝑑𝐚 = vector with direction normal to the surface) If flux density B is uniform and perpendicular to an area A. then 𝟇 = 𝑩𝑨 KOREA UNIVERSITY 6 Power & Energy Laboratory Magnetic Fields from Single Wire ▪ Assume 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? H Direction of H is given by the “Right-hand” Rule 4m 1000A 1m Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose a circle with a radius of x. KOREA UNIVERSITY 7 Power & Energy Laboratory Single Line Example , cont’d 2xH = I → H = B = 0 H I 2x 0 I = 0 H dA = dx A 4 2x 5 I 5 5 −7 = 0 ln = 2 10 I ln 2 4 4 = 4.46 10−5Wb For reference, the earth’s magnetic field is about 0.6 Gauss (Central US) −4 2 10 2 B= T = Gauss x x KOREA UNIVERSITY 8 Power & Energy Laboratory Flux linkages and Faraday’s law Where V = voltage, λ = flux linkages KOREA UNIVERSITY 9 Power & Energy Laboratory Inductance ▪ For a linear magnetic system, that is one where B = H ▪ we can define the inductance, L, to be the constant relating the current and the flux linkage = Li • where L has units of Henrys (H) KOREA UNIVERSITY 10 Power & Energy Laboratory Inductance Example ▪ Calculate 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. Assume 1) all flux is within the coil 2) all flux links each turn KOREA UNIVERSITY 11 Power & Energy Laboratory Inductance Example , cont’d I e = H dl NI = H 2R ( path length is 2R) NI H= B = H = r 0 H 2R = AB = N = LI NI = NAB = NA r 0 2R N 2 A r 0 L= H 2R KOREA UNIVERSITY 12 Power & Energy Laboratory Inductance 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 • flux linkages outside of the wire • flux linkages within the wire ▪ We’ll assume that the current density within the wire is uniform and that the wire has a radius of r. KOREA UNIVERSITY 13 Power & Energy Laboratory Flux Linkage outside of the wire R I r 2x = B dA = length 0 A KOREA UNIVERSITY 14 dx Power & Energy Laboratory Flux linkages outside , cont’d I = = AB dA = lengthr 0 dx 2x R 0 R = 0 dx = I ln r meter 2x 2 r KOREA UNIVERSITY R I 15 Power & Energy Laboratory Flux linkages inside of wire KOREA UNIVERSITY 16 Power & Energy Laboratory Flux linkages inside , cont’d Wire cross section x r KOREA UNIVERSITY 17 Power & Energy Laboratory Line Total Flux & Inductance KOREA UNIVERSITY 18 Power & Energy Laboratory Inductance Simplification − r Where r ' = re 4 0.78r for r = 1 KOREA UNIVERSITY 19 Power & Energy Laboratory Many-Conductor Case Now 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 flux linkage, k , depends upon its own current and the current in all the other conductors. To derive 1 we’ll be integrating from conductor 1 (at origin) to the right along the x-axis. KOREA UNIVERSITY 20 Power & Energy Laboratory Many-Conductor Case , cont’d 0 R1 R2 1 = + i1 ln ' + i2 ln 2 d12 r1 Rn + in ln d1n 0 1 1 1 = i1 ln ' + i2 ln + 2 d12 r1 1 + in ln + d1n 0 i1 ln R1 + i2 ln R2 + 2 + in ln Rn As R1 goes to infinity R1 = R2 = Rn so the second 0 n term from above can be written = i j ln R1 2 j =1 KOREA UNIVERSITY 21 Power & Energy Laboratory Many-Conductor Case , cont’d n Therefore if i j = 0, which is true in a balanced j =1 three phase system, then the second term is zero and 0 1 1 1 = i1 ln ' + i2 ln + 2 d12 r1 1 = L11i1 + L12i2 1 + in ln d1n + L1nin System has self and mutual inductance. However the mutual inductance can be canceled for balanced 3 systems with symmetry. KOREA UNIVERSITY 22 Power & Energy Laboratory Line Inductance Example Calculate the reactance for a balanced 3, 60Hz transmission line with a conductor geometry of an equilateral triangle with D = 5m, r = 1.24cm (Rook conductor) and a length of 5 miles. a D D c Since system is assumed balanced D b ia = −ib − ic 0 1 1 1 a = ia ln( ) + ib ln( ) + ic ln( ) 2 r' D D KOREA UNIVERSITY 23 Power & Energy Laboratory Line Inductance Example , cont’d Substituting ia = −ib − ic Hence 0 1 1 a = ia ln − ia ln 2 r ' D 0 D = ia ln 2 r' 0 D 4 10−7 5 La = ln = ln −3 2 r ' 2 9.67 10 = 1.25 10−6 H/m KOREA UNIVERSITY 24 Power & Energy Laboratory Line Inductance Example , cont’d La = 1.25 10−6 H/m Converting to reactance X a = 2 60 1.25 10−6 −4 = 4.71 10 /m = 0.768 /mile X Total for 5 mile line = 3.79 (this is the total per phase) The reason we did NOT have mutual inductance was because of the symmetric conductor spacing KOREA UNIVERSITY 25 Power & Energy Laboratory Conductor Bundling To increase the capacity of high voltage transmission lines it is very common to use a number of conductors per phase. This is known as conductor bundling. Typical values are two conductors for 345 kV lines, three for 500 kV and four for 765 kV. 2 1 A 4 3 9 10 5 6 B C 11 KOREA UNIVERSITY 7 12 26 8 Power & Energy Laboratory Bundled Conductor Flux Linkages 1 2 For the line shown on the left, define 𝑑𝑖𝑗 as the distance between conductors i and j. We can then determine for each A 3 9 10 4 5 6 B C 11 12 7 8 i 1 1 1 1 a + ln + ln + ln + ln d12 d13 d14 4 r' 0 ib 1 1 1 1 1 = + ln + ln + ln ln + 2 4 d15 d16 d17 d 18 ic 1 1 1 1 + ln + ln + ln ln 4 d d d d 19 1,10 1,11 1,12 KOREA UNIVERSITY 27 Power & Energy Laboratory Bundled Conductors , cont’d Simplifying 1 i ln + 1 a 4 (r ' d12 d13d14 ) 0 1 1 = i ln + b 2 ( d d d d ) 14 15 16 17 18 1 ic ln 1 4 ( d d d d ) 19 1,10 1,11 1,12 KOREA UNIVERSITY 28 Power & Energy Laboratory Bundled Conductors , cont’d = = KOREA UNIVERSITY 29 Power & Energy Laboratory Inductance of Bundle But remember each bundle has b conductors in parallel (4 in this example). So 𝐿1 𝜇0 𝐷 𝐿𝑎 = = ln 𝑏 2𝜋 𝑅𝑏 KOREA UNIVERSITY 30 Power & Energy Laboratory Bundle Inductance Example Consider the previous example of the three phases symmetrically spaced 5 meters apart using wire with a radius of r = 1.24 cm. Except now assume each phase has 4 conductors in a square bundle, spaced 0.25 meters apart. What is the new inductance per meter? r = 1.24 10−2 m 0.25 M 0.25 M ( UNIVERSITY −3 R b = 9.67 10 0.25 0.25 2 0.25 0.25 M )4 1 = 0.12 m (ten times bigger!) La = KOREA r ' = 9.67 10−3 m 31 0 5 ln = 7.46 10−7 H/m 2 0.12 Power & Energy Laboratory Transmission Tower Configurations The 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. Therefore in general Dab Dac Dbc Typical Transmission Tower Configuration KOREA UNIVERSITY 32 Unless something was done this would result in unbalanced phases Power & Energy Laboratory Transmission Tower Examples 230 kV wood pole H-frame 230 kV lattice steel tower double circuit Source: Tom Ernst, Minnesota Power KOREA UNIVERSITY 33 Power & Energy Laboratory Transposition ▪ To 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. position 1 2 3 a b c b c a c a b Aerial or side view of conductor positions over the length of the transmission line. KOREA UNIVERSITY 34 Power & Energy Laboratory Transposition Impact on Flux Linkages For a uniformly transposed line we can calculate the flux linkage for phase "a" 1 0 1 1 1 a = I a ln + I b ln + I c ln + 3 2 r' d12 d13 “a” phase in position “1” 1 0 1 1 1 I a ln + I b ln + I c ln + 3 2 r' d13 d 23 “a” phase in position “3” 1 0 1 1 1 I a ln + I b ln + I c ln 3 2 r' d 23 d12 “a” phase in position “2” KOREA UNIVERSITY 35 Power & Energy Laboratory Transposition Impact , cont’d Recognizing that 1 1 (ln a + ln b + ln c) = ln(abc) 3 3 We can simplify so 1 I ln 1 + I ln + 1 a r' b 3 ( d12 d13d 23 ) 0 a = 2 1 I c ln 1 ( d12 d13d 23 ) 3 KOREA UNIVERSITY 36 Power & Energy Laboratory Inductance of Transposed Line Define the geometric mean distance (GMD) 1 3 D m = ( d12 d13d 23 ) Then for a balanced 3 system ( I a = - I b - I c ) 0 Dm 1 1 0 a = I a ln − I a ln = I a ln 2 r' Dm 2 r' Hence 0 Dm Dm −7 La = ln = 2 10 ln H/m 2 r' r' KOREA UNIVERSITY 37 Power & Energy Laboratory Inductance with Bundling If the line is bundled with a geometric mean radius, R b , then 0 Dm a = I a ln 2 Rb 0 Dm Dm −7 La = ln = 2 10 ln H/m 2 Rb Rb KOREA UNIVERSITY 38 Power & Energy Laboratory Inductance Example ▪ Calculate the per phase inductance and reactance of a balanced 3, 60 Hz, line with horizontal phase spacing of 10 m using three conductor bundling with a spacing between conductors in the bundle of 0.3 m. Assume the line is uniformly transposed and the conductors have a 1 cm radius. 10m 10m 0.3m Answer: Dm = 12.6 m, Rb= 0.0889 m Inductance = 9.9 x 10–7 H/m; reactance = 0.4 /km KOREA UNIVERSITY 39 Power & Energy Laboratory Review of Electric Fields To develop a model for line capacitance we first need to review some electric field concepts. Gauss’s law : ( 𝑒𝑞 = 𝑎𝑑 𝐷 𝐴ׯintegrate over closed surface) Where 𝐷 = electric flux density, coulombs/𝑚2 𝑑𝑎 = differential area, with normal to surface 𝐴 = total closed surface area, 𝑚2 𝑞𝑒 = total charge in coulombs enclosed KOREA UNIVERSITY 40 Power & Energy Laboratory 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. Since D comes radially out integrate over the cylinder bounding the wire da == DD22Rh Rh==qqe e= =qhqh AADD da D = KOREA UNIVERSITY q 2 R ar where ar radially directed unit vector 41 Power & Energy Laboratory 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) = 0 r 0 = 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) KOREA UNIVERSITY 42 Power & Energy Laboratory Voltage Difference P V = − E dl P R V = − dR = ln R 2 R 20 R 0 R KOREA UNIVERSITY q q 43 Power & Energy Laboratory Voltage Difference , cont’d R V = − dR = ln R 2 R 20 R 0 R KOREA UNIVERSITY q q 44 Power & Energy Laboratory Multi-Conductor Case KOREA UNIVERSITY 45 Power & Energy Laboratory Multi-Conductor Case , cont’d KOREA UNIVERSITY 46 Power & Energy Laboratory Absolute Voltage Defined KOREA UNIVERSITY 47 Power & Energy Laboratory Three Conductor Case A C B Assume we have three infinitely long conductors, A, B, & C, each with radius r 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 KOREA UNIVERSITY 48 Power & Energy Laboratory Line Capacitance KOREA UNIVERSITY 49 Power & Energy Laboratory Line Capacitance, cont’d We will not be considering these case 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 𝑞𝑎 = 𝐶𝑉𝑎 → 𝐶 = KOREA UNIVERSITY 50 𝑞𝑎 𝑉𝑎 = 2𝜋𝜀 𝐷 ln 𝑟 Power & Energy Laboratory 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 1ൗ c R b = (𝑟𝑑12 ⋯ 𝑑1𝑛 ) 𝑛 Note for the capacitance equation we use r rather than r′ which was used for R b in the inductance equation KOREA UNIVERSITY 51 Power & Energy Laboratory Line Capacitance, cont’d For the case of uniformly transposed lines we use the same GMR, Dm , as before. 2𝜋𝜀 𝐶= 𝐷𝑚 ln 𝑐 𝑅𝑏 where 1ൗ Dm = (𝑑𝑎𝑏 𝑑𝑎𝑐 𝑑𝑏𝑐 ) 3 1Τ 𝑐 𝑅𝑏 = (𝑟𝑑12 ⋯ 𝑑1𝑛 ) 𝑛 (note r NOT r′) 𝜀 in air = 𝜀0 = 8.854 × 10−12 F/m KOREA UNIVERSITY 52 Power & Energy Laboratory Line Capacitance Example ▪ Calculate the per phase capacitance and reactance 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 1cm radius. 10m 10m 0.3m KOREA UNIVERSITY 53 Power & Energy Laboratory 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 10−12 −11 = = 1.141 10 F/m 12.6 ln 0.0963 1 1 = = C 2 60 1.141 10−11 F/m = 2.33 10 -m (not / m) 8 KOREA UNIVERSITY 54 Power & Energy Laboratory 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. Aluminum strands On outside KOREA UNIVERSITY Steel strand core 55 Power & Energy Laboratory Line Conductors , cont’d ▪ Total conductor area is given in square millimeters ▪ Example: what is the area of a solid, 2.54-cm diameter circular wire? Answer: 500𝐦𝐦𝟐 ▪ 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 (1 ft = 0.3 m). KOREA UNIVERSITY 56 Power & Energy Laboratory Line Resistance KOREA UNIVERSITY 57 Power & Energy Laboratory 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. KOREA UNIVERSITY 58 Power & Energy Laboratory ACSR Table Data (Similar to Table A.4) Inductance and Capacitance assume a Dm of 1 ft. GMR is equivalent to r′ 1 Ohm/mile = 0.625 Ohms/km; 1 ft = 0.3 m; 1 kcmil = 0.5 mm2 KOREA UNIVERSITY 59 Power & Energy Laboratory ACSR Data , cont’d Dm X LX 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 10−3 f ln Dm GMR −7 Term from table assuming a one foot (0.3 m) spacing KOREA UNIVERSITY 60 Term independent of conductor with 𝐷𝑚 in feet (1 ft = 0.3 m) Power & Energy Laboratory ACSR Data , cont’d To use the phase to neutral capacitance from table 2 0 1 X CX c = -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 (0.3m) spacing 𝐷𝑚 in feet (1 ft = 0.3m) KOREA UNIVERSITY 61 Power & Energy Laboratory Dove Example GMR = 0.0313 feet Outside Diameter = 0.07725 feet (radius = 0.03863) Assuming a one foot spacing at 60 Hz 1 −7 X a = 2 60 2 10 1609 ln Ω/mile 0.0313 X a = 0.420 Ω/mile, which matches the table For the capacitance 1 1 6 X C = 1.779 10 ln = 9.65 104 Ω-mile f r KOREA UNIVERSITY 62 Power & Energy Laboratory 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 4800 km of underground ac cables in U.S. Cables are primarily used in urban areas. In a cable the conductors are tightly spaced, (< 0.3m) with oil impregnated paper commonly used to provide insulation • inductance is lower • capacitance is higher, limiting cable length KOREA UNIVERSITY 63 Power & Energy Laboratory 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! KOREA UNIVERSITY 64 Power & Energy Laboratory 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 (> 640 km) • long cable power transfer such as underwater • providing an asynchronous means of joining different power systems (such as the Eastern and Western grids). KOREA UNIVERSITY 65 Power & Energy Laboratory Homework #4 ▪ Ch4 problems: ▪ 마감: KOREA UNIVERSITY 66 Power & Energy Laboratory Question? KOREA UNIVERSITY 67 Power & Energy Laboratory KECE 361 POWER SYSTEM ANALYSIS Ch. 5 Transmission Lines: Steady-State Operation Professor Gilsoo Jang School of Electrical Engineering KOREA UNIVERSITY Power & Energy Laboratory 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. KOREA UNIVERSITY 1 Power & Energy Laboratory Transmission Line Equivalent Circuit ▪ Our current model of a transmission line is shown below I+dI r jwL + I + V+dV y V dI - - Units on z and y are per unit length! Transmission line equivalent circuits For operation at frequency , let z = r + j L and y = g +jC (with g usually equal 0) KOREA UNIVERSITY 2 Power & Energy Laboratory Derivation of V , I Relationships I+dI jwL r + V+dV I + y V dI - Transmission line equivalent circuits We can then derive the following relationships: dV = I z dx dI dV ( x) dx KOREA UNIVERSITY = (V + dV ) y dx = zI V y dx dI ( x) = yV dx 3 Power & Energy Laboratory 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 KOREA UNIVERSITY 4 Power & Energy Laboratory 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 KOREA UNIVERSITY 5 Power & Energy Laboratory 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 e x + e − x e x − e − x V ( x) = K1 ( ) + K2 ( ) 2 2 = K1 cosh( x) + K 2 sinh( x) KOREA UNIVERSITY 6 Power & Energy Laboratory Real Hyperbolic Functions ▪ For real x the cosh and sinh functions have the following form: cosh 𝒙 1 sinh 𝒙 x x d cosh( x) = sinh( x) dx KOREA UNIVERSITY d sinh( x) = cosh( x) dx 7 Power & Energy Laboratory Complex Hyperbolic Functions ▪ For x = + j the cosh and sinh functions have the following form cosh x sinh x KOREA UNIVERSITY = cosh cos + j sinh sin = sinh cos + j cosh sin 8 Power & Energy Laboratory Determining Line Voltage The voltage along the line is determined based upon the current/voltage relationships at the terminals. Assuming we know V and I at one end (say the "receiving end" with VR and I R where x = 0) we can determine the constants K1 and K 2 , and hence the voltage at any point on the line. KOREA UNIVERSITY 9 Power & Energy Laboratory Determining Line Voltage , cont’d V ( x) = K1 cosh( x) + K 2 sinh( x) V (0) = VR = K1 cosh(0) + K 2 sinh(0) Since cosh(0) = 1 & sinh(0) = 0 K1 = VR dV ( x) dx = zI = K1 sinh( x) + K 2 cosh( x) K2 = zI R IR z z = = IR y yz V ( x) = VR cosh( x) + I R Z c sinh( x) where Zc = KOREA UNIVERSITY z y = characteristic impedance 10 Power & Energy Laboratory Determining Line Current By similar reasoning we can determine I(x) VR I ( x) = I R cosh( x) + sinh( x) Zc where x is the distance along the line from the receiving end. Pout Define transmission efficiency as = Pin KOREA UNIVERSITY 11 Power & Energy Laboratory Transmission Line Example Assume we have a 765 kV transmission line with a receiving end voltage of 765 kV(line to line), a receiving end power SR = 2000 + j1000 MVA and z = 0.0201 + j0.535 = 0.53587.8 km y = j 7.75 10−6 = 7.75 10−6 90.0 Siemensൗ km km Then KOREA UNIVERSITY = zy = 2.036 88.9/km c = z y = 262.7 -1.1 12 Power & Energy Laboratory Transmission Line Example, cont’d Do per phase analysis, using single phase power and line to neutral voltages. Then VR = 765 = 441.70 kV 3 6 * (2000 + j1000) 10 IR = = 1688 − 26.6 A 3 3 441.70 10 V ( x) = VR cosh( x) + I R Z c sinh( x) = 441, 7000 cosh( x 2.03688.9) + 443, 440 − 27.7 sinh( x 2.03688.9) KOREA UNIVERSITY 13 Power & Energy Laboratory Transmission Line Example , cont’d km KOREA UNIVERSITY 14 Power & Energy Laboratory Lossless Transmission Lines For a lossless line the characteristic impedance, Zc , is known as the surge impedance. Zc = jwl l = (a real value) jwc c If a lossless line is terminated in impedance VR Zc = IR Then I R Z c = VR so we get... KOREA UNIVERSITY 15 Power & Energy Laboratory Lossless Transmission Lines V ( x) = VR cosh x + VR sinh x I ( x) = I R cosh x + I R sinh x V ( x) = Zc I ( x) 2 V(x) Define as the surge impedance load (SIL). Zc Since the line is lossless this implies V ( x) = VR I ( x) = I R KOREA UNIVERSITY 16 Power & Energy Laboratory Transmission Matrix Model ▪ Oftentimes we’re only interested in the terminal characteristics of the transmission line. Therefore we can model it as a “blackbox”. + VS - IS IR Transmission Line + VR - VS A B VR With = I I C D R S KOREA UNIVERSITY 17 Power & Energy Laboratory Transmission Matrix Model , cont’d VS A B VR With = I I C D R S Use voltage/current relationships to solve for A,B,C,D VS = VR cosh l + Z c I R sinh l VR I S = I R cosh l + sinh l Zc cosh l A B 1 T = = sinh l C D Z c KOREA UNIVERSITY 18 Z c sinh l cosh l Power & Energy Laboratory Equivalent Circuit Model The common representation is the equivalent circuit Is IR Z′ + + VS Y' 2 Y' 2 VR - - Next we’ll use the T matrix values to derive the parameters Z' and Y'. KOREA UNIVERSITY 19 Power & Energy Laboratory Equivalent Circuit Parameters Is VS − VR Y' Z′ − VR = I R + Z' 2 Y V Z 'Y ' S 2 VS = 1 + V + Z ' I R R 2 Y' Y' I S = VS + VR + I R 2 2 Z 'Y ' Z 'Y ' I S = Y ' 1 + VR + 1 + IR 4 2 1 + Z 'Y ' Z' VR VS 2 I = Z 'Y ' Z 'Y ' I R S Y ' 1 + 1 + 4 2 ' KOREA UNIVERSITY 20 IR + ' Y 2 VR - Power & Energy Laboratory Equivalent Circuit Parameters We now need to solve for Z' and Y'. Using the B element solving for Z' is straightforward B = Z C sinh l = Z ' Then using A we can solve for Y' Z 'Y ' A = cosh l = 1 + 2 Y' cosh l − 1 1 l = = tanh 2 Z c sinh l Z c 2 KOREA UNIVERSITY 21 Power & Energy Laboratory Simplified Parameters These values can be simplified as follows: Z ' = Z C sinh l sinh l =Z l with Z = zl (recalling = Y' 1 l = tanh 2 Zc 2 l tanh Y 2 = 2 l 2 KOREA UNIVERSITY = zl z sinh l yl z = zy ) yl y l tanh zl y 2 with Y = yl 22 Power & Energy Laboratory Simplified Parameters For short lines make the following approximations: sinh l Z' = Z (assumes 1) l Y' Y tanh( l / 2) = (assumes 1) 2 2 l /2 sinhγl tanh(γl/2) Length γl γl/2 80 km 0.9980.02 1.001 − 0.01 160 km 0.9930.09 1.004 − 0.04 320 km 0.9720.35 1.014 − 0.18 KOREA UNIVERSITY 23 Power & Energy Laboratory Three Line Models Long Line Model (longer than 320 km) l tanh sinh l Y ' Y 2 use Z ' = Z , = l 2 2 l 2 Medium Line Model (between 80 and 320 km) Y use Z and 2 Short Line Model (less than 80 km) use Z (i.e., assume Y is zero) KOREA UNIVERSITY 24 Power & Energy Laboratory Power Transfer in Short Lines ▪ Often we'd like to know the maximum power that could be transferred through a short transmission line V1 + - I1 I1 Transmission Line with Impedance Z S12 S21 + - V2 * V1 − V2 * S12 = V1I1 = V1 Z with V1 = V1 1 , V2 = V2 2 Z = Z Z 2 S12 KOREA UNIVERSITY V1 V1 V2 = Z − Z + 12 Z Z 25 Power & Energy Laboratory Power Transfer in Lossless Lines If we assume a line is lossless with impedance jX and are just interested in real power transfer then: 2 P12 + jQ12 V1 V1 V2 = 90 − 90 + 12 Z Z Since - cos(90 + 12 ) = sin 12 , we get V1 V2 P12 = sin 12 X Hence the maximum power transfer is P12Max KOREA UNIVERSITY V1 V2 = X 26 Power & Energy Laboratory Limits Affecting Max. Power Transfer ▪ Thermal limits • limit is due to heating of conductor and hence depends heavily on ambient conditions. • For many lines, sagging is the limiting constraint. • Newer conductors limit can limit sag. • Trees grow, and they will eventually hit lines if they are planted under the line. KOREA UNIVERSITY 27 Power & Energy Laboratory Other Limits Affecting Power Transfer ▪ Angle limits • while the maximum power transfer occurs when line angle difference is 90°, actual limit is substantially less due to multiple lines in the system ▪ Voltage stability limits • as power transfers increases, reactive losses increase as I2X. As reactive power increases the voltage falls, resulting in a potentially cascading voltage collapse. KOREA UNIVERSITY 28 Power & Energy Laboratory Homework ▪ Ch5 problems: ▪ 마감: ▪ 제출장소: KOREA UNIVERSITY 29 Power & Energy Laboratory Question Question? KOREA UNIVERSITY 30 Power & Energy Laboratory KECE 361 POWER SYSTEM ANALYSIS Ch. 6 Power Flows Professor Gilsoo Jang School of Electrical Engineering KOREA UNIVERSITY Power & Energy Laboratory Two Bus System Example Z=0.03+j0.04 Yc/2=j0.1 Z=0.03+j0.04 I1 I2 + + V1 j0.1 j0.1 1. I1 I2 같이 안주어지고 전류는 다른 형태로 주어질 것 주로 복소전력 형태 2. 거기서 I값을 구해서 3. V값을 구하기 (Y버스 매트릭스 이용) I1 = V2 - Y (V1 − V2 ) + V1 c Z 2 1 = 12 − j16 0.03 + j 0.04 I1 12 − j15.9 −12 + j16 V1 I = −12 + j16 12 − j15.9 V 2 2 KOREA UNIVERSITY 1 Power & Energy Laboratory Using the Ybus If the voltages are known then we can solve for the current injections: Ybus V = I If the current injections are known then we can solve for the voltages: −1 Ybus I = V = Zbus I where Z bus is the bus impedance matrix KOREA UNIVERSITY 2 Power & Energy Laboratory Power Flow Analysis ▪ When analyzing power systems we know neither the complex bus voltages nor the complex current injections ▪ Rather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes ▪ Therefore we can not directly use the Ybus equations, but rather must use the power balance equations KOREA UNIVERSITY 3 Power & Energy Laboratory Power Balance Equations From KCL we know at each bus i in an n bus system the current injection, I i , must be equal to the current that flows into the network I i = I Gi − I Di = n Iik k =1 Since I = Ybus V we also know I i = I Gi − I Di = n YikVk RE k =1 The network power injection is then Si = Vi I i* KOREA UNIVERSITY 4 Power & Energy Laboratory Power Balance Equations, cont’d * n n Si = Vi I i* = Vi YikVk = Vi Yik*Vk* k =1 k =1 This is an equation with complex numbers. Sometimes we would like an equivalent set of real power equations. These can be derived by defining Yik = Gik + jBik Vi = Vi e ji = Vi i ik = i − k Recall e j = cos + j sin KOREA UNIVERSITY 5 Power & Energy Laboratory Real Power Balance Equations Si = = n n j ik * * Pi + jQi = Vi YikVk = Vi Vk e (Gik − jBik ) k =1 k =1 n Vi Vk (cos ik + j sinik )(Gik − jBik ) k =1 Resolving into the real and imaginary parts Pi = Qi = KOREA UNIVERSITY n Vi Vk (Gik cos ik + Bik sinik ) = PGi − PDi k =1 n Vi Vk (Gik sinik − Bik cosik ) = QGi − QDi k =1 6 Power & Energy Laboratory Power Flow Requires Iterative Solution In the power flow we assume we know Si and the Ybus . We would like to solve for the V's. The problem is the below equation has no closed form solution: * n n * * Si = = Vi YikVk = Vi YikVk k =1 k =1 Rather, we must pursue an iterative approach. * Vi I i KOREA UNIVERSITY 7 Power & Energy Laboratory Gauss Iteration There are a number of different iterative methods we can use. We'll consider two: Gauss and Newton. With the Gauss method we need to rewrite our equation in an implicit form: x = h(x) To iterate we first make an initial guess of x, x (0) , and then iteratively solve x = h( x ) until we X find a "fixed point", x, X 늿such that x = h(x). ? (v +1) KOREA UNIVERSITY 8 (v) Power & Energy Laboratory Gauss Iteration Example Example: Solve x - x − 1 = 0 점은 주로문제에서 제시 X(0) x ( v +1) = 1 + x ( v ) 출발 = Let v = 0 and arbitrarily guess x (0) = 1 and solve v 0 1 2 3 4 KOREA UNIVERSITY x( v ) 1 2 2.41421 2.55538 2.59805 v 5 6 7 8 9 9 x( v ) 2.61185 2.61612 2.61744 2.61785 2.61798 Power & Energy Laboratory Stopping Criteria A key problem to address is when to stop the iteration. With the Guass iteration we stop when with x ( v ) = x ( v +1) − x ( v ) x ( v ) If x is a scalar this is clear, but if x is a vector we need to generalize the absolute value by using a norm x ( v ) j Two common norms are the Euclidean & infinity x 2 = KOREA UNIVERSITY n 2 x i x = max i x i i =1 10 Power & Energy Laboratory infinity norm 요소 값중에서 최대값 Euclidean norm 원소들을 다 제곱해서 더하고 그 결과를 루트 씌워서 얻음. Gauss Power Flow We first need to put the equation in the correct form * n Si = Vi I i* n = Vi YikVk = Vi Yik*Vk* k =1 k =1 = Vi* I i = Vi YikVk = Vi YikVk S*i S*i = * Vi * n * k =1 n k =1 n n k =1 k =1,k i YikVk = YiiVi + YikVk n 1 S*i Vi = * − YikVk Yii V k =1,k i i KOREA UNIVERSITY 11 Power & Energy Laboratory Gauss Two Bus Power Flow Example ▪ A 100 MW, 50 MVAR load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 MVAR on each end (100 MVA base). Also, there is a 25 MVAR capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2? Z=0.02+j0.06 V2 V1 = 1.00 Ycap=j0.25 SLoad = 1.0 + j0.5 p.u. KOREA UNIVERSITY 12 Power & Energy Laboratory Gauss Two Bus Example, cont’d The unknown is the complex load voltage, V2 . To determine V2 we need to know the Ybus . 1 = 5 − j15 0.02 + j 0.06 5 − j14.95 −5 + j15 Hence Ybus = − 5 + j 15 5 − j 14.70 ( Note B22 = - j15 + j 0.05 + j 0.25) KOREA UNIVERSITY 13 Power & Energy Laboratory Gauss Two Bus Example, cont’d n 1 S*2 V2 = * − YikVk Y22 V2 k =1,k i -1 + j 0.5 1 V2 = − (−5 + j15)(1.00) * 5 − j14.70 V2 Guess V2(0) = 1.00 (this is known as a flat start) v KOREA UNIVERSITY V2( v ) 0 1 1.000 + j 0.000 0.9671 − j 0.0568 2 0.9624 − j 0.0553 v 3 4 14 V2( v ) 0.9622 − j 0.0556 0.9622 − j 0.0556 Power & Energy Laboratory Gauss Two Bus Example, cont’d V2 = 0.9622 − j 0.0556 = 0.9638 − 3.3 Once the voltages are known all other values can be determined, such as the generator powers and the line flows S1* = V1* (Y11V1 + Y12V2 ) = 1.023 − j 0.239 In actual units P1 = 102.3 MW, Q1 = 23.9 Mvar 2 The capacitor is supplying V2 25 = 23.2 Mvar KOREA UNIVERSITY 15 Power & Energy Laboratory Slack Bus ▪ In previous example we specified S2 and V1 and then solved for S1 and V2. ▪ We can not arbitrarily specify S at all buses because total generation must equal total load + total losses ▪ We also need an angle reference bus. ▪ To solve these problems we define one bus as the "slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection. KOREA UNIVERSITY 16 Power & Energy Laboratory Gauss with Many Bus Systems With multiple bus systems we could calculate new Vi ' s as follows: S* n 1 ( vi )* − YikVk( v ) Vi( v +1) = Yii V k = 1, k i i (v) (v) (v) = hi (V1 ,V2 ,...,Vn ) But after we've determined Vi( v +1) we have a better estimate of its voltage , so it makes sense to use this new value. This approach is known as the Gauss-Seidel iteration. KOREA UNIVERSITY 17 Power & Energy Laboratory Gauss – Seidel Iteration Immediately use the new voltage estimates: V2( v +1) = h2 (V1 ,V2( v ) ,V3( v ) ,,Vn( v ) ) V3( v +1) = h2 (V1 ,V2( v +1) ,V3( v ) ,,Vn( v ) ) V4( v +1) = h2 (V1 ,V2( v +1) ,V3( v +1) ,V4( v ) ,Vn( v ) ) Vn( v +1) = h2 (V1 ,V2( v +1) ,V3( v +1) ,V4( v +1) ,Vn( v ) ) The Gauss-Seidel works better than the Gauss, and is actually easier to implement. It is used instead of Gauss. KOREA UNIVERSITY 18 Power & Energy Laboratory Three Types of Power Flow Buses ▪ There are three main types of power flow buses • Load (PQ) at which P/Q are fixed; iteration solves for voltage magnitude and angle. • Slack at which the voltage magnitude and angle are fixed; iteration solves for P/Q injections • Generator (PV) at which P and |V| are fixed; iteration solves for voltage angle and Q injection ✓ special coding is needed to include PV buses in the Gauss-Seidel iteration 목적 1. 모든모선의전압을 찾는거 2. 모든선로의 Data를알고있음->Ybus Matrix를알고있음 3.각 모선별로 4개정보중에 2개정보는알고있음-> KOREA UNIVERSITY 19 4개정보는??(복소전력P,복소전력Q,|V|,theta(전압의 각도) Power & Energy Laboratory Inclusion of PV Buses in G-S To solve for Vi at a PV bus we must first make a guess of Qi : Si* = Vi* n YikVk = Pi + jQi k =1 Hence Qi( v ) = − Im Vi( v )* (v) Y V ik k k =1 n In the iteration we use Si( v ) = Pi + jQi( v ) Tentatively solve for Vi( v +1) S( v )* n 1 (v) ( v +1) i ( v )* − YikVk Vi = Yii V k =1,k i i But since Vi is specified, replace Vi( v +1) by Vi KOREA UNIVERSITY 20 Power & Energy Laboratory Two Bus PV Example Consider the same two bus system from the previous example, except the load is replaced by a generator z = 0.02 + j 0.06 Bus 1 (slack bus) Bus 2 V1 = 1.0 V2 = 1.05 P2 = 0 MW KOREA UNIVERSITY 21 Power & Energy Laboratory Two Bus PV Example, cont’d 1 S2* V2 = * − Y21V1 Y22 V2 Q2 = * * − Im[Y21V1V2 + Y22V2V2 ] Guess V2 = 1.050 v S2( v ) V2( v +1) V2( v +1) 0 0 + j 0.457 1.045 − 0.83 1.050 − 0.83 1 0 + j 0.535 1.049 − 0.93 1.050 − 0.93 2 0 + j 0.545 1.050 − 0.96 1.050 − 0.96 KOREA UNIVERSITY 22 Power & Energy Laboratory Generator Reactive Power Limits ▪ The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter ▪ To maintain higher voltages requires more reactive power ▪ Generators have reactive power limits, which are dependent upon the generator's MW output ▪ These limits must be considered during the power flow solution. KOREA UNIVERSITY 23 Power & Energy Laboratory Generator Reactive Limits, cont’d ▪ During power flow once a solution is obtained check to make generator reactive power output is within its limits ▪ If the reactive power is outside of the limits, fix Q at the max or min value, and resolve treating the generator as a PQ bus • this is known as "type-switching" • also need to check if a PQ generator can again regulate ▪ Rule of thumb: to raise system voltage we need to supply more reactive powers KOREA UNIVERSITY 24 Power & Energy Laboratory Accelerated G-S Convergence Previously in the Gauss-Seidel method we were calculating each value x as x ( v +1) = h( x ( v ) ) To accelerate convergence we can rewrite this as x ( v +1) = x ( v ) + h( x ( v ) ) − x ( v ) Now introduce acceleration parameter x ( v +1) = x ( v ) + (h( x ( v ) ) − x ( v ) ) With = 1 this is identical to standard gauss-seidel. Larger values of may result in faster convergence. KOREA UNIVERSITY 25 Power & Energy Laboratory Accelerated Convergence, cont’d Consider the previous example: x - x − 1 = 0 x ( v +1) = x ( v ) + (1 + x ( v ) − x ( v ) ) Comparison of results with different values of k =1 = 1.2 = 1.5 = 2 0 1 1 1 1 1 2 2.20 2.5 3 2 2.4142 2.5399 2.6217 2.464 3 2.5554 2.6045 2.6179 2.675 4 2.5981 2.6157 2.6180 2.596 5 2.6118 2.6176 2.6180 2.626 KOREA UNIVERSITY 26 Power & Energy Laboratory Gauss – Seidel Advantages ▪ Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system) ▪ Relatively easy to program KOREA UNIVERSITY 27 Power & Energy Laboratory Gauss – Seidel Disadvantages ▪ Tends to converge relatively slowly, although this can be improved with acceleration ▪ Has tendency to miss solutions, particularly on large systems ▪ Tends to diverge on cases with negative branch reactances (common with compensated lines) ▪ Need to program using complex numbers KOREA UNIVERSITY 28 Power & Energy Laboratory Homework 6-1 ▪ Ch6 problems: ▪ 마감: ▪ 제출장소: KOREA UNIVERSITY 29 Power & Energy Laboratory Question? KOREA UNIVERSITY 30 Power & Energy Laboratory KECE 361 POWER SYSTEM ANALYSIS Ch. 6 Power Flows Professor Gilsoo Jang School of Electrical Engineering KOREA UNIVERSITY Power & Energy Laboratory Newton – Raphson Algorithm ▪ The second major power flow solution method is the Newton-Raphson algorithm ▪ Key idea behind Newton-Raphson is to use sequential linearization General form of problem : Find an X such that ෝ =𝟎 𝒇 𝒙 KOREA UNIVERSITY 1 Power & Energy Laboratory Newton – Raphson Method (scalar) 1. For each guess of xˆ , x ( v ) , define x ( v ) = xˆ - x ( v ) 2. Represent f ( xˆ ) by a Taylor series about f ( x) (v) df ( x ) (v) (v) f ( xˆ ) = f ( x ) + x + dx ( ) d 2 f ( x(v) ) (v) 2 + x + higher order terms 2 dx KOREA UNIVERSITY 2 Power & Energy Laboratory Newton – Raphson Method, cont’d 3. Approximate f ( xˆ ) by neglecting all terms except the first two (v) df ( x ) (v) (v) f ( xˆ ) = 0 f ( x ) + x dx 4. Use this linear approximation to solve for x ( v ) −1 df ( x ) (v) x = − f ( x ) dx 5. Solve for a new estimate of x̂ (v) (v) x ( v +1) = x ( v ) + x ( v ) KOREA UNIVERSITY 3 Power & Energy Laboratory Newton – Raphson Example Use Newton-Raphson to solve f ( x) = x 2 - 2 = 0 The equation we must iteratively solve is −1 df ( x ) (v) x = − f ( x ) dx 1 (v) 2 (v) x = − ( v ) (( x ) - 2) 2x (v) (v) x ( v +1) = x ( v ) + x ( v ) x KOREA UNIVERSITY ( v +1) = x (v) 1 (v) 2 − ( v ) (( x ) - 2) 2x 4 Power & Energy Laboratory Newton – Raphson Example, cont’d x ( v +1) = x (v) 1 (v) 2 − ( v ) (( x ) - 2) 2x Guess x (0) = 1. Iteratively solving we get v 0 1 x(v) 1 1.5 f ( x(v) ) −1 0.25 x ( v ) 0.5 −0.08333 2 1.41667 6.953 10−3 −2.454 10−3 3 KOREA UNIVERSITY 1.41422 6.024 10 5 −6 Power & Energy Laboratory Sequential Linear Approximations At each iteration the N-R method uses a linear approximation to determine the next value for x Function is f(x) = x2 - 2 = 0. Solutions are points where f(x) intersects f(x) = 0 axis KOREA UNIVERSITY 6 Power & Energy Laboratory Newton-Raphson Comments ▪ When close to the solution the error decreases quite quickly - method has quadratic convergence ▪ f(x(v)) is known as the mismatch, which we would like to drive to zero ▪ Stopping criteria is when f(x(v)) < ▪ Results are dependent upon the initial guess. What if we had guessed x(0)= 0, or x(0) = -1? ▪ A solution’s region of attraction (ROA) is the set of initial guesses that converge to the particular solution. The ROA is often hard to determine KOREA UNIVERSITY 7 Power & Energy Laboratory Multi-Variable Newton-Raphson Next we generalize to the case where x is an ndimension vector, and f (x) is an n-dimension function x1 x x = 2 x n f1 (x) f ( x) f ( x) = 2 f ( x) n X so f ( x) = 0 and Again define the solution x늿 x = xˆ − x KOREA UNIVERSITY 8 Power & Energy Laboratory Multi-Variable Case, cont’d The Taylor series expansion is written for each fi (x) f1 (x) f1 ( x) f1 (xˆ ) = f1 ( x) + x1 + x2 + x1 x2 f1 (x) xn + higher order terms xn f n (x) f n ( x) f n (xˆ ) = f n ( x) + x1 + x2 + x1 x2 f n (x) xn + higher order terms xn KOREA UNIVERSITY 9 Power & Energy Laboratory Multi-Variable Case, cont’d This can be written more compactly in matrix form f1 (x) x 1 f ( x ) 1 f (x) f 2 (x) 2 + x1 f (xˆ ) = f ( x) n f (x) n x1 f1 (x) x2 f 2 (x) x2 f n (x) x2 f1 (x) xn x1 f 2 (x) x xn 2 xn f n (x) xn + higher order terms KOREA UNIVERSITY 10 Power & Energy Laboratory Jacobian Matrix The n by n matrix of partial derivatives is known as the Jacobian matrix, J (x) f1 (x) x 1 f 2 (x) J (x) = x1 f (x) n x1 KOREA UNIVERSITY f1 (x) x2 f 2 (x) x2 f n (x) x2 11 f1 ( x) xn f 2 ( x) xn f n ( x) xn Power & Energy Laboratory Multi-Variable N-R Procedure Derivation of N-R method is similar to the scalar case f (xˆ ) = f (x) + J (x)x + higher order terms f (xˆ ) = 0 f (x) + J (x)x x − J (x) −1 f (x) x( v +1) = x( v ) + x( v ) x( v +1) = x( v ) − J (x( v ) ) −1 f (x( v ) ) Iterate until f (x( v ) ) KOREA UNIVERSITY 12 Power & Energy Laboratory Multi-Variable Example x1 Solve for x = such that f (x) = 0 where x2 2 2 f1 (x) = 2 x1 + x2 − 8 = 0 f 2 (x) = x12 − x22 + x1 x2 − 4 = 0 First symbolically determine the Jacobian f1 (x) x 1 J (x) = f 2 (x) x1 KOREA UNIVERSITY f1 ( x) x2 f 2 ( x) x2 13 Power & Energy Laboratory Multi-Variable Example, cont’d 4 x1 J (x) = 2 x1 + x2 Then x1 4 x1 x = − 2 x + x 2 1 2 x1 − 2 x2 2 x2 −1 f1 (x) x1 − 2 x2 f 2 (x) 1 (0) Arbitrarily guess x = 1 2 x2 −1 x(1) KOREA UNIVERSITY 1 4 2 −5 2.1 = − = 1 3 −1 −3 1.3 14 Power & Energy Laboratory Multi-Variable Example, cont’d −1 2.1 8.40 2.60 2.51 1.8284 (2) x = − = 1.3 5.50 − 0.50 1.45 1.2122 Each iteration we check f (x) to see if it is below our specified tolerance 0.1556 f (x ) = 0.0900 If = 0.2 then we would be done. Otherwise we'd continue iterating. (2) KOREA UNIVERSITY 15 Power & Energy Laboratory N-R Application to Power Flow We first need to rewrite complex power equations as equations with real coefficients * n n Si = Vi I i* = Vi YikVk = Vi Yik*Vk* k =1 k =1 These can be derived by defining Yik = Gik + jBik Vi = Vi e ji = Vi i ik = i − k Recall e j = cos + j sin KOREA UNIVERSITY 16 Power & Energy Laboratory Real Power Balance Equations n n k =1 k =1 Si = Pi + jQi = Vi Yik*Vk* = Vi Vk e jik (Gik − jBik ) = n Vi Vk (cos ik + j sinik )(Gik − jBik ) k =1 Resolving into the real and imaginary parts Pi = Qi = KOREA UNIVERSITY n Vi Vk (Gik cos ik + Bik sinik ) = PGi − PDi k =1 n Vi Vk (Gik sinik − Bik cosik ) = QGi − QDi k =1 17 Power & Energy Laboratory Newton-Raphson Power Flow In the Newton-Raphson power flow we use Newton's method to determine the voltage magnitude and angle at each bus in the power system. We need to solve the power balance equations Pi = Qi = KOREA UNIVERSITY n Vi Vk (Gik cosik + Bik sin ik ) = PGi − PDi k =1 n Vi Vk (Gik sin ik − Bik cosik ) = QGi − QDi k =1 18 Power & Energy Laboratory Power Flow Variables Assume the slack bus is the first bus (with a fixed voltage angle/magnitude). We then need to determine the voltage angle/magnitude at the other buses. 2 n x = V2 V n KOREA UNIVERSITY P2 (x) − PG 2 + PD 2 Pn (x) − PGn + PDn f ( x) = Q (x) − QG 2 + QD 2 2 Qn (x) − QGn + QDn 19 Power & Energy Laboratory N-R Power Flow Solution The power flow is solved using the same procedure discussed last time: Set v = 0; make an initial guess of x, x( v ) While f (x( v ) ) Do x( v +1) = x( v ) − J (x( v ) ) −1 f (x( v ) ) v = v +1 End While KOREA UNIVERSITY 20 Power & Energy Laboratory Power Flow Jacobian Matrix The most difficult part of the algorithm is determining and inverting the n by n Jacobian matrix, J (x) f1 (x) x 1 f 2 (x) J (x) = x1 f (x) n x1 KOREA UNIVERSITY f1 (x) x2 f 2 (x) x2 f n (x) x2 21 f1 (x) xn f 2 (x) xn f n (x) xn Power & Energy Laboratory Power Flow Jacobian Matrix, cont’d Jacobian elements are calculated by differentiating each function, fi ( x), with respect to each variable. For example, if fi (x) is the bus i real power equation fi ( x) = fi ( x) = i n Vi Vk (Gik cosik + Bik sin ik ) − PGi + PDi k =1 n Vi Vk (−Gik sin ik + Bik cosik ) k =1 k i fi ( x) = Vi V j (Gik sin ik − Bik cos ik ) ( j i ) j KOREA UNIVERSITY 22 Power & Energy Laboratory Two Bus Newton-Raphson Example For the two-bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assume that bus one is the slack and SBase = 100 MVA. Line Z = 0.1j One 1.000 pu Two 0 MW 0 MVR 2 x = V2 KOREA UNIVERSITY 1.000 pu 200 MW 100 MVR − j10 j10 = j 10 − j 10 Ybus 23 Power & Energy Laboratory Two Bus Example, cont’d General power balance equations Pi = Qi = n Vi Vk (Gik cosik + Bik sin ik ) = PGi − PDi k =1 n Vi Vk (Gik sin ik − Bik cosik ) = QGi − QDi k =1 Bus two power balance equations P2 = V2 V1 (10sin 2 ) + 2.0 = 0 Q2 = V2 V1 (−10 cos 2 ) + V2 (10) + 1.0 = 0 2 KOREA UNIVERSITY 24 Power & Energy Laboratory Two Bus Example, cont’d P2 (x) = V2 (10sin 2 ) + 2.0 = 0 Q2 (x) = V2 (−10 cos 2 ) + V2 (10) + 1.0 = 0 2 Now calculate the power flow Jacobian P2 (x) 2 J ( x) = Q 2 (x) 2 P2 ( x) V 2 Q 2 ( x) V 2 10 V2 cos 2 = 10 V2 sin 2 KOREA UNIVERSITY 25 10sin 2 −10 cos 2 + 20 V2 Power & Energy Laboratory Two Bus Example, First Iteration Set v = 0, guess x (0) 0 = 1 Calculate V (10sin ) + 2.0 2 2 (0) f(x ) = = 2 V2 (−10 cos 2 ) + V2 (10) + 1.0 10sin 2 10 V2 cos 2 (0) J (x ) = = 10 V2 sin 2 −10 cos 2 + 20 V2 −1 Solve x KOREA UNIVERSITY (1) 0 10 0 2.0 = − 1.0 1 0 10 26 2.0 1.0 10 0 0 10 −0.2 = 0.9 Power & Energy Laboratory Two Bus Example, Next Iterations 0.9 (10sin( −0.2)) + 2.0 0.212 f(x ) = = 2 0.279 0.9( − 10 cos( − 0.2)) + 0.9 10 + 1.0 8.82 −1.986 (1) J (x ) = − 1.788 8.199 (1) −1 −0.2 8.82 −1.986 0.212 −0.233 (2) x = − = 0.9 − 1.788 8.199 0.279 0.8586 0.0145 −0.236 (2) (3) f(x ) = x = 0.0190 0.8554 0.0000906 f(x ) = 0.0001175 (3) KOREA UNIVERSITY Done! 27 V2 = 0.8554 − 13.52 Power & Energy Laboratory Two Bus Solved Values Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values, such as the line flows and the generator reactive power output 200.0 MW 168.3 MVR One -200.0 MW -100.0 MVR Line Z = 0.1j 1.000 pu Two 200.0 MW 168.3 MVR KOREA UNIVERSITY 0.855 pu -13.522 Deg 200 MW 100 MVR 28 Power & Energy Laboratory Two Bus Case Low Voltage Solution This case actually has two solutions! The second "low voltage" is found by using a low initial guess. Set v = 0, guess x (0) 0 = 0.25 Calculate V2 (10sin 2 ) + 2.0 f(x ) = = 2 V2 (−10 cos 2 ) + V2 (10) + 1.0 10sin 2 10 V2 cos 2 (0) J (x ) = = 10 V2 sin 2 −10 cos 2 + 20 V2 (0) KOREA UNIVERSITY 29 2 −0.875 2.5 0 0 −5 Power & Energy Laboratory Low Voltage Solution, cont’d −1 0 2.5 0 2 −0.8 (1) Solve x = − = 0.25 0 −5 −0.875 0.075 1.462 (2) −1.42 −0.921 (2) (3) f (x ) = x = x = 0.534 0.2336 0.220 Low voltage solution (not feasible) 200.0 MW 831.7 MVR One -200.0 MW -100.0 MVR Line Z = 0.1j 1.000 pu Two 200.0 MW 831.7 MVR KOREA UNIVERSITY 0.261 pu -49.914 Deg 200 MW 100 MVR 30 Power & Energy Laboratory PV Buses ▪ Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x or write the reactive power balance equations • the reactive power output of the generator varies to maintain the fixed terminal voltage (within limits) • optionally these variations/equations can be included by just writing the explicit voltage constraint for the generator bus |Vi | – Vi, set point = 0 KOREA UNIVERSITY 31 Power & Energy Laboratory Three Bus PV Case Example For this three bus case we have 2 x = 3 V2 P2 (x) − PG 2 + PD 2 f (x) = P3 (x) − PG 3 + PD 3 = 0 Q2 (x) + QD 2 Line Z = 0.1j 0.941 pu One 170.0 MW 68.2 MVR 1.000 pu Line Z = 0.1j Three Two Line Z = 0.1j -7.469 Deg 200 MW 100 MVR 1.000 pu 30 MW 63 MVR KOREA UNIVERSITY 32 Power & Energy Laboratory Modeling Voltage Dependent Load So far we've assumed that the load is independent of the bus voltage (i.e., constant power). However, the power flow can be easily extended to include voltage depedence with both the real and reactive load. This is done by making PDi and Q Di a function of Vi : n Vi Vk (Gik cosik + Bik sin ik ) − PGi + PDi ( Vi ) = 0 k =1 n Vi Vk (Gik sin ik − Bik cosik ) − QGi + QDi ( Vi ) = 0 k =1 KOREA UNIVERSITY 33 Power & Energy Laboratory Voltage Dependent Load Example In previous two bus example now assume the load is constant impedance, so P2 (x) = V2 (10sin 2 ) + 2.0 V2 2 = 0 Q2 (x) = V2 (−10 cos 2 ) + V2 (10) + 1.0 V2 = 0 2 2 Now calculate the power flow Jacobian 10 V2 cos 2 J ( x) = 10 V2 sin 2 KOREA UNIVERSITY 10sin 2 + 4.0 V2 −10 cos 2 + 20 V2 + 2.0 V2 34 Power & Energy Laboratory Voltage Dependent Load, cont’d Again set v = 0, guess x (0) 0 = 1 Calculate 2 V2 (10sin 2 ) + 2.0 V2 2.0 (0) f(x ) = = 2 2 1.0 V2 (−10 cos 2 ) + V2 (10) + 1.0 V2 10 4 (0) J (x ) = 0 12 −1 0 10 4 2.0 (1) Solve x = − 1 0 12 1.0 KOREA UNIVERSITY 35 −0.1667 = 0.9167 Power & Energy Laboratory Voltage Dependent Load, cont’d With constant impedance load the MW/Mvar load at bus 2 varies with the square of the bus 2 voltage magnitude. This if the voltage level is less than 1.0, the load is lower than 200/100 MW/Mvar 160.0 MW 120.0 MVR One Line Z = 0.1j 1.000 pu Two 160.0 MW 120.0 MVR KOREA UNIVERSITY -160.0 MW -80.0 MVR 0.894 pu -10.304 Deg 160 MW 80 MVR 36 Power & Energy Laboratory Solving Large Power Systems ▪ The most difficult computational task is inverting the Jacobian matrix • inverting a full matrix is an order n3 operation, meaning the amount of computation increases with the cube of the size • this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix • using sparse matrix methods results in a computational order of about n1.5. • this is a substantial savings when solving systems with tens of thousands of buses KOREA UNIVERSITY 37 Power & Energy Laboratory Newton-Raphson Power Flow ▪ Advantages • fast convergence as long as initial guess is close to solution • large region of convergence ▪ Disadvantages • each iteration takes much longer than a Gauss-Seidel iteration • more complicated to code, particularly when implementing sparse matrix algorithms ▪ Newton-Raphson algorithm is very common in power flow analysis KOREA UNIVERSITY 38 Power & Energy Laboratory ″DC″ Power Flow ▪ The ″DC″ power flow makes the most severe approximations: • completely ignore reactive power, assume all the voltages are always 1.0 per unit, ignore line conductance ▪ This makes the power flow a linear set of equations, which can be solved directly θ=B KOREA UNIVERSITY 39 −1 P Power & Energy Laboratory Power System Control ▪ A major problem with power system operation is the limited capacity of the transmission system • lines/transformers have limits (usually thermal) • no direct way of controlling flow down a transmission line (e.g., there are no valves to close to limit flow) • open transmission system access associated with industry restructuring is stressing the system in new ways ▪ We need to indirectly control transmission line flow by changing the generator outputs KOREA UNIVERSITY 40 Power & Energy Laboratory Homework 6-2 ▪ Ch6 problems: ▪ 마감: ▪ 제출장소: KOREA UNIVERSITY 41 Power & Energy Laboratory Question? KOREA UNIVERSITY 42 Power & Energy Laboratory
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