EE 369 POWER SYSTEM ANALYSIS Lecture 5 Development of Transmission Line Models Tom Overbye and Ross Baldick 1 Reading • For lectures 5 through 7 read Chapter 4 – we will not be covering sections 4.7, 4.11, and 4.12 in detail. • HW 4 is 2.31, 2.41, 2.48, 4.8, 4.10, 4.12, 4.13, 4.15, 4.19, 4.20, 4.22, due Thursday 9/24. • Mid-term I is Thursday, October 1, covering up to and including material in HW 4. • HW 5 is Problems 4.24, 4.25 (assume Cardinal conductor and look up GMR in Table A.4), 4.26, 4.33, 4.36, 4.38, 4.49, 4.1, 4.3, 4.6; due Thursday 10/8. 2 Substation Bus 3 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 3) Radius of each turn is negligible compared to R Circular path Γ of radius R within the iron core encloses all N turns of the coil and hence links total enclosed current of Ie = NI. Since the radius of each turn is negligible compared to R, all circular paths within the iron core have radius approximately equal to R. 4 Inductance Example, cont’d Ie H dl NI H 2 R (path encloses I e NI ; path length is 2 R,) NI H (H varies somewhat with R, but ignore,) 2 R B H r 0 H (linear magnetic material,) A B (assuming H and therefore B constant,) N (each of the N turns link flux ,) NI NAB NAr 0 2 R N 2 Ar 0 L /I H 2 R 5 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: 1. flux linkages outside of the wire 2. flux linkages within the wire •We’ll assume that the current density within the wire is uniform and that the wire is solid with a radius of r. •In fact, current density is non-uniform, and conductor is stranded, so our calculations will be approximate. 6 Flux Linkages outside of the wire We'll think of the wire as a single loop "closed" by a return wire that is infinitely far away. Therefore = since there is N = 1 turn. The flux linking a length of wire outside it to a distance of R from the wire center is derived as follows: I A B da length R r 0 I 2 x H dl dx 7 Flux Linkages outside, cont’d A B da length R r 0 I dx 2 x Since length = we'll deal with per unit length values, assumed to be per meter. meter R r 0 R 0 dx I ln 2 x 2 r I Note, this quantity still goes to infinity as R 8 Flux linkages inside of wire Current inside conductor tends to travel on the outside of the conductor due to the skin effect. The penetration of the current into the conductor is approximated using the skin depth = 1 where f is the frequency in Hz f and is the conductivity in siemens/meter. 0.066 m For copper skin depth 0.33 inch at 60Hz. f For derivation we'll assume a uniform current density. 9 Flux linkages inside, cont’d Wire cross section Current enclosed within distance x2 x from center I e 2 I r x Ie Ix Hx r 2 x 2 r 2 However, situation is not as simple as outside wire case since flux only links part of wire (need Biot-Savart law to derive): inside Ix x 2 B da (length) dx 2 2 A 0 2 r r r 0 r r Ix 3 (length) 4 dx (length) I. 2 0 r 8 10 Line Total Flux & Inductance 0 R 0 r Total (per meter) I ln I 2 r 8 0 R r Total (per meter) I ln 2 r 4 0 R r L(per meter) ln 2 r 4 Note, this value still goes to infinity as we let R go to infinity. Note that inductance depends on logarithm of ratio of lengths. 11 Inductance Simplification Inductance expression can be simplified using two exponential identities: a ln(ab)=ln a + ln b ln ln a ln b a ln( e a ) b r 0 R r 0 4 L ln R ln r ln e ln 2 r 4 2 r 4 0 R 0 L ln R ln re ln 2 2 r ' Where r' re r 4 0.78r for r 1 We call r' the "effective radius" of the conductor. 12 Two Conductor Line Inductance •Key problem with the previous derivation is we assumed no return path for the current. Now consider the case of two wires, each carrying the same current I, but in opposite directions; assume the wires are separated by distance D. D Creates counterclockwise field To determine the inductance of each conductor we integrate as before. However now we get some field cancellation. Creates a clockwise field 13 Two Conductor Case, cont’d D D R Direction of integration Key Point: Flux linkage due to currents in each conductor tend to cancel out. Use superposition to get total flux linkage. Consider flux linked by left conductor from distance 0 to R. For a distance from left conductor that is greater than 2D, flux due to the right conductor from distance 0 to D cancels the flux due to the right conductor from D to 2 D : Summing the fluxes yields: left 0 R 0 R D I ln I ln 2 r ' 2 D Left Current 14 Right Current Two Conductor Inductance Simplifying (with equal and opposite currents) left Lleft 0 R R D I ln ln 2 r ' D 0 I ln R ln r ' ln( R D ) ln D 2 0 D R I ln ln 2 r ' RD 0 D I ln as R 2 r ' 0 D ln H/m 2 r ' 15 Many-Conductor Case Now assume we now have n conductors, with the k-th conductor having current ik, and 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. For example, to derive the flux linkage for conductor 1, 1, we’ll be integrating from conductor 1 (at origin) to the right along the x-axis. 16 Many-Conductor Case, cont’d Rk is the distance from conductor k to point c. At point b the net contribution to 1 from ik , 1k, is zero. We’d like to integrate the flux crossing between b to c. But the flux crossing between a and c is easier to calculate and provides a very good approximation of 1k. Point a is at distance d1k from conductor k. 17 Many-Conductor Case, cont’d 0 R1 R2 1 i1 ln ' i2 ln 2 d12 r1 0 1 1 i1 ln ' i2 ln 2 d12 r1 Rn in ln , d1n 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 18 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 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. 19 Symmetric Line Spacing – 69 kV 20 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. Since system is assumed balanced ia ib ic 0 1 1 1 a ia ln( ) ib ln( ) ic ln( ) 2 r' D D 21 Line Inductance Example, cont’d Substituting ia ib ic , obtain: 0 1 1 a i ln i ln a a 2 r ' D 0 D ia ln . 2 r' 0 D 4 107 5 La ln ln 3 2 r ' 2 9.67 10 1.25 106 H/m. Again note logarithm of ratio of distance between phases to the size of the conductor. 22 Line Inductance Example, cont’d La 1.25 10 6 H/m Converting to reactance Xa 2 60 1.25 106 4.71 104 /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 23 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. 24 Bundled Conductor Flux Linkages •For the line shown on the left, define dij as the distance between conductors i and j. •We can then determine k for conductor k. •Assuming ¼ of the phase current flows in each of the four conductors in a given phase bundle, then for conductor 1: i 1 1 1 1 ln ln a ln ln d12 d13 d14 4 r' 0 ib 1 1 1 1 1 ln ln ln ln 2 4 d15 d16 d17 d18 ic 1 1 1 1 ln ln ln ln d1,10 d1,11 d1,12 4 d19 25 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 26 Bundled Conductors, cont’d Rb geometric mean radius (GMR) of bundle (r ' d12 d13d14 ) (r ' d12 D1b d1b 1 4 1 ) b for our example in general geometric mean distance (GMD) of conductor 1 to phase b. (d15d16 d17 d18 ) D1c 1 4 D2b D3b D4b Dab (d19 d1,10 d1,11d1,12 ) 1 4 D2c D3c D4c Dac 27 Inductance of Bundle If Dab Dac Dbc D and ia ib ic Then 1 0 1 1 ia ln ia ln 2 D Rb D 0 I a ln 2 Rb D 0 4 I1 ln 2 Rb D 0 L1 4 ln , which is the 2 Rb self-inductance of wire 1. 28 Inductance of Bundle, cont’d But remember each bundle has b conductors in parallel (4 in this example). So, there are four inductances in parallel: 0 D La L1 / b ln . 2 Rb Again note that inductance depends on the logarithm of the ratio of distance between phases to the size of bundle of conductors. Inductance decreases with decreasing distance between phases and increasing bundle size. 29 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 1.24 102 m r ' 9.67 103 m 0.25 M 0.25 M 3 Rb 9.67 10 0.25 0.25 ( 2 0.25) 0.25 M 4 0.12 m (ten times bigger than r!) 0 5 La ln 7.46 107 H/m 2 0.12 Bundling reduces inductance. 30 1 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 configuration is seldom practical. • Therefore in general Dab Dac Dbc Typical Transmission Tower Configuration •Unless something was done this would result in unbalanced Phases. 31 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. Aerial or side view of conductor positions over the length of the transmission line. 32 Line Transposition Example 33 Line Transposition Example 34 Transposition Impact on Flux Linkages For a uniformly transposed line we can calculate the flux linkage for phase "a" 1 0 a 3 2 1 1 1 I a ln r ' I b ln d I c ln d 12 13 “a” phase in position “1” 1 0 3 2 1 1 1 I a ln r ' I b ln d I c ln d 13 23 “a” phase in position “3” 1 0 3 2 1 1 1 I a ln r ' I b ln d I c ln d 23 12 “a” phase in position “2” 35 Transposition Impact, cont’d Recognizing that 1 1 (ln a ln b ln c ) ln ( abc ) 3 3 We can simplify so 1 1 I a ln I b ln 1 d d d 3 r' 0 12 13 23 a 2 1 I c ln 1 d d d 3 12 13 23 36 Inductance of Transposed Line Define the geometric mean distance (GMD) Dm d12d13d 23 1 3 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' Again, logarithm of ratio of distance between phases to conductor size. 37 Inductance with Bundling If the line is bundled with a geometric mean radius, Rb , then 0 Dm a I a ln 2 Rb 0 Dm Dm 7 La ln 2 10 ln H/m 2 Rb Rb 38 Inductance Example Calculate the per phase inductance and reactance of a balanced 3, 60 Hz, line with: – horizontal phase spacing of 10m – using three conductor bundling with a spacing between conductors in the bundle of 0.3m. Assume the line is uniformly transposed and the conductors have a 1cm radius. 39 Inductance Example Dm d12 d13d 23 1 3, (10 (2 10) 10)1/ 3 12.6m, r'=r e r 4 0.0078m, 1 Rb ( r ' d bundle d bundle ) 3 , where d bundle is the distance between conductors in bundle ( r ' 0.3 0.3)1/ 3 0.0888m, 0 Dm La ln 2 Rb 9.9 107 H/m, X a 2 fLa (1600m/mile) = 0.6/mile. 40