2.TRANSMISSION LINES The electric parameters of transmission lines (i.e. resistance, inductance, and capacitance) can be determined from the specifications for the conductors, and from the geometric arrangements of the conductors. 2.1 Transmission Line Resistance Resistance to d.c. current is given by A R dc ρ · where ρ is the resistivity at 20 o C is the length of the conductor A is the cross sectional area of the conductor Because of skin effect, the d.c. resistance is different from ac resistance. The ac resistance is referred to as effective resistance, and is found from power loss in the conductor R power loss I · 2 The variation of resistance with temperature is linear over the normal temperature range resistance (Ω ) R 2 R 1 T T 1 T 2 temperature ( o C) Figure 9 Graph of Resistance vs Temperature ( ) ( ) ( ) ( ) R T T R T T 1 1 2 2 0 0 − − · − − R T T T T R 2 2 1 1 · − − 1 2.2 Transmission Line Inductive Reactance Inductance of transmission lines is calculated per phase. It consists of self inductance of the phase conductor and mutual inductance between the conductors. It is given by: L GMD GMR · × − 2 10 7 ln [H/m] where GMR is the geometric mean radius (available from manufacturer’s tables) GMD is the geometric mean distance (must be calculated for each line configuration) Geometric Mean Radius: There are magnetic flux lines not only outside of the conductor, but also inside. GMR is a hypothetical radius that replaces the actual conductor with a hollow conductor of radius equal to GMR such that the self inductance of the inductor remains the same. If each phase consists of several conductors, the GMR is given by 1 2 3 n GMR d d d d d d d d d d n n n n nn n · ( .... ). ( . .... )......( . ..... ) 11 12 13 1 21 22 2 1 2 2 where d 11 =GMR 1 d 22 =GMR 2 . . . d nn =GMR n Note: for a solid conductor, GMR = r.e -1/4 , where r is the radius of the conductor. Geometric Mean Distance replaces the actual arrangement of conductors by a hypothetical mean distance such that the mutual inductance of the arrangement remains the same b c a’ b’ a 2 m c’ n’ GMD D D D D D D D D D aa ab an ba bb bn ma mb mn mn · ( ... ). ( ... ). ....( ..... ) ' ' ' ' ' ' ' ' ' ' where D aa’ is the distance between conductors “a” and “a’” etc. Inductance Between Two Single Phase Conductors r 1 r 2 D L D r 1 7 1 2 10 · × × − ln ' L D r 2 7 2 2 10 · × × − ln ' where r 1 ’ is GMR of conductor 1 r 2 ’ is GMR of conductor 2 D is the GMD between the conductors The total inductance of the line is then L L L D r D r D r r D r r L D r r D r r T T · + · × × + ] ] ] · × × · × × × × · × × ] ] ] · × × − − − − − 1 2 7 1 2 7 2 1 2 7 2 1 2 7 2 1 2 1 2 7 1 2 2 10 2 10 2 10 2 1 2 4 10 4 10 ln ' ln ' ln ' ' ln ' ' ln ' ' ln ' ' / If r 1 = r 2 , then L D r T · × × − 4 10 7 1 ln ' 3 Example: Find GMD, GMR for each circuit, inductance for each circuit, and total inductance per meter for two circuits that run parallel to each other. One circuit consists of three 0.25 cm radius conductors. The second circuit consists of two 0.5 cm radius conductor 9 m a a’ 6 m 6 m b b’ 6 m circuit B c circuit A Solution: m = 3, n’ = 2, ∴m⋅ n’ = 6 ( )( )( ) GMD D D D B D D aa ab ba bb ca cb · ' ' ' ' ' ' 6 where D D m D D D m D m aa bb ab ba cb ca ' ' ' ' ' ' · · · · · + · · + · 9 6 9 117 12 9 15 2 2 2 2 ∴ GMD = 10.743 m Geometric Mean Radius for Circuit A: m 481 . 0 12 6 e 10 25 . 0 D D D D D D D D D GMR 9 2 4 3 4 1 2 3 cc cb ca bc bb ba ac ab aa A 2 · × × , ` . | × × · · − − Geometric Mean Radius for Circuit B: m 153 . 0 6 e 10 5 . 0 D D D D GMR 4 2 2 4 1 2 2 ' a ' b ' b ' b ' b ' a ' a ' a B 2 · × , ` . | × × · · − − Inductance of circuit A L GMD GMR H m A A · × · × · × − − − 2 10 2 10 10 743 0 481 6 212 10 7 7 7 ln ln . . . / 4 Inductance of circuit B L GMD GMR H m B B · × · × · × − − − 2 10 2 10 10 743 0153 8503 10 7 7 7 ln ln . . . / The total inductance is then L L L H m T A B · + · × − 14 715 10 7 . / The Use of Tables Since the cables for power transmission lines are usually supplied by U.S. manufacturers, the tables of cable characteristics are in American Standard System of units and the inductive reactance is given in Ω /mile. X fL f GMD GMR m X f GMD GMR m X f GMD GMR mile X f GMD GMR mile X f GMR f GMD mile L L L L L · · × × · × · × × × · × × × · × × × + × × × − − − − − − 2 2 2 10 4 10 4 10 1609 2 022 10 2 022 10 1 2 022 10 7 7 7 3 3 3 π π π π ln / ln / ln / . ln / . ln . ln / Ω Ω Ω Ω Ω X a X d If both, GMR and GMD are in feet, then X a represents the inductive reactance at 1 ft spacing, and X d is called the inductive reactance spacing factor. 5 Example: Find the inductive reactance per mile of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacings between the conductor centers. D = 20 ft Solution: From the Tables, for Partridge conductor, GMR = 0.0217 ft and inductive reactance at 1 ft spacing X a = 0.465 Ω /mile. The spacing factor for 20 ft spacing is X d = 0.3635 Ω /mile. The inductance of the line is then X X X mile L a d · + · + · 0 465 0 3635 08285 . . . / Ω Inductance of Balanced Three Phase Line Average inductance per phase is given by: L D GMR eq · × − 2 10 7 ln where D eq is the geometric mean of the three spacings of the three phase line. D D D D eq ab ac bc · 3 Example: A three phase line operated at 60 Hz is arranged as shown. The conductors are ACSR Drake. Find the inductive reactance per mile. 20ft 20 ft 38 ft Solution: For ACSR Drake conductor, GMR = 0.0373 ft 6 D ft L H m X mile OR from the tables X mile eq L a · × × · · × · × · × × × × · · − − − 20 20 38 24 8 2 10 24 8 0 0373 13 10 2 60 1609 13 10 0 788 0 399 3 7 7 7 . ln . . / . / . / π Ω Ω The spacing factor is calculated for spacing equal the geometric mean distance between the conductors, that is, X mile d · × × · − 2 022 10 60 248 0 389 3 . ln . . / Ω Then the line inductance is X X X mile per phase line a d · + · 0 788 . / Ω Example: Each conductor of the bundled conductor line shown in the figure is 1272 MCM Pheasant. Find: a) the inductive reactance in Ω /km and Ω /mile per phase for d = 45 cm b) the p.u. series reactance if the length of the line is 160 km and the base is 100 MVA, 345 kV. d 8 m 8 m Solution: a) The distances in ft are d ft D ft · · · · 0 45 0 3048 1476 8 0 3048 26 25 . . . . . For Pheasant conductors, GMR = 0.0466 ft. GMR b for a bundle of conductors is GMR GMR d ft b · × · × · 0 0466 1476 0 2623 . . . The geometric mean of the phase conductor spacing is D ft eq · × × · 26 25 26 25 52 49 33 07 3 . . . . The inductance of the line is then 7 L D GMR H m eq b · × · × · × − − − 2 10 2 10 3307 0 2623 9 674 10 7 7 7 ln ln . . . / The inductive reactance is X fL m km mile L · · × × × · × · · − − 2 2 60 9 674 10 3647 10 0 3647 05868 7 4 π π . . / . / . / Ω Ω Ω b) Base impedance Z V S b b b · · · 2 2 345 100 1190 Ω Total impedance of the 160 km line is X X X Z p u L Lp u L b · × · · · · 160 0 3647 58 35 58 35 1190 0 049 . . . . . . . . Ω 2.3 Transmission Line Capacitive Reactance Conductors of transmission lines act like plates of a capacitor. The conductors are charged, and there is a potential difference between the conductors and between the conductors and the ground. Therefore there is capacitance between the conductors and between the conductors and the ground. The basic equation for calculation of the capacitance is the definition of the capacitance as the ratio of the charge and the potential difference between the charged plates: [ ] C Q V F · where Q is the total charge on the conductors (plates) V is the potential difference between the conductors or a conductor and ground (i.e. plates) For transmission lines, we usually want the capacitance per unit length [ ] C q V F m · / where q is the charge per unit length in C/m V is the potential difference between the conductors or a conductor and ground (i.e. plates) Capacitance of a Single Phase Line r 1 r 2 D 8 For a two conductor line, the capacitance between the conductors is given by [ ] C D r r F m · π ε 0 2 1 2 ln / where ε o is the permittivity of free space and is equal to 8.85× 10 -12 F/m D is the distance between the conductors, center to center r 1 and r 2 are the radii of the two conductors Formally, this equation corresponds to the equation for inductance of a two conductor line. The equation was derived for a solid round conductor and assuming a uniform distribution of charge along the conductors. The electric field, and therefore the capacitance of stranded conductors is not the same as for solid conductors, but if the radii of the conductors are much smaller than the distance between the conductors, the error is very small and an outside radii of the stranded conductors can be used in the equation. For most single phase lines, r 1 = r 2 . In this case, half way between the conductors there is a point where E = 0. This is the neutral point n a n b C an C bn The capacitance from conductor a to point n is C an and is the same as the capacitance from conductor b to n, C bn . C an and C bn are connected in series, therefore C C C an bn ab · · 2 . 1 It follows that C D r an o · 2πε ln [F/m] Since X fC c · 1 2π 1 C C C C C C C ab an bn an bn an bn · + · + 1 1 1 If C C an bn · , then C C C C ab an an an · · 2 2 2 Therefore, C C an ab · 2 9 ∴ [ ] X f D r f D r m c o · · × ⋅ 1 2 2 2 862 10 9 π π ε ln . ln Ω The capacitive reactance in Ω⋅ mile is [ ] X f D r f D r mile c · × × · × ⋅ 2862 10 1 1609 1779 10 9 6 . ln . ln Ω Similarly as for inductive reactance, this expression can be split into two terms that are called capacitive reactance at 1 ft spacing (X a ’) and the capacitive reactance spacing factor (X d ’). [ ] X f r f D mile c · × + × ⋅ 1779 10 1 1779 10 6 6 . ln . ln Ω X a ’ X d ’ X a ’ is given in the tables for the standard conductors, X d ’ is given in the tables for the capacitive reactance spacing factor. Example: Find the capacitive reactance in MΩ⋅ miles for a single phase line operating at 60 Hz. The conductor used for the line is Partridge, and the spacing is 20 ft. D = 20 ft 10 Solution: The outside radius of the Partridge conductor is r in ft · · 0 642 2 00268 . . The capacitive reactance is X f D r f M mile C · × · × · 1779 10 1779 10 20 0 0268 01961 6 6 . ln . ln . . . Ω OR From tables X M mile a ' . . · 01074 Ω X M mile d ' . . · 0 0889 Ω for 20’ spacing ∴ · + · X X X M m i l e C a d ' ' . . 01 9 6 3 Ω This is the capacitive reactance between the conductor and the neutral. Line-to-line capacitive reactance is X X M mile C L L C − · · 2 0 0981 . . Ω Capacitance of Balanced Three Phase Line between a phase conductor and neutral is given by [ ] C D D F m n o eq b · 2π ε ln / where D D D D eq ab bc ca · 3 and D ab , D bc , and D ca are the distances between the centers of the phase conductors, and D b is the geometric mean radius for the bundled conductors. (in the expression for D b the outside radius of the conductor is used, rather than the GMR from the tables.) The capacitive reactance to neutral than becomes [ ] X f D D mile cn eq b · × 1779 10 6 . ln . Ω 11 Example: a) A three phase 60 Hz line is arranged as shown. The conductors are ACSR Drake. Find the capacitive reactance for 1 mile of the line. b) If the length of the line is 175 miles and the normal operating voltage is 220 kV, find the capacitive reactance to neutral for the entire length of the line, the charging current for the line, and the charging reactive power. 20ft 20 ft 38 ft Solution: The outside radius for Drake conductors is r in ft · · 1108 2 0 0462 . . The geometric mean distance for this line is D ft eq · × × · 20 20 38 24 8 3 . From tables, X M mile a ' . . · 0 0912 Ω X f D M mile X X X M mile d eq cn a d ' ' ' . ln . ln . . . . . · × · × · ∴ · + · 1779 10 1779 10 60 248 0 0952 01864 6 6 Ω Ω This is the capacitive reactance to neutral. For the length of 175 miles, X X Ctotal cn · · 175 1065 Ω Charging current is I V X k A C LN Ctotal · · · 220 3 1065 119 Reactive power to charge the line is Q V I k MVAr C LL C · · × × · 3 3 220 119 4545 . 12 2.4 Transmission Line Equivalent Circuits a) Short lines (less than 80 km) b) Medium length lines (80 km to 200 km) c) Long lines (over 200 km) 2.5 Transmission Line Losses and Thermal Limits The power losses of a transmission line are proportional to the value of resistance of the line. The value of the resistance is determined by the type and length of the conductor. The current in the line is given by the power being delivered by the transmission line. P E I R R equiv R · cos Φ ∴ I P E equiv R R R · cosΦ From that, P I R P E R loss equiv R R R · · | . ` , 2 2 cosΦ Power utilities usually strive to maintain the receiving end voltage constant. The power delivered by the transmission line is determined by the load connected to the line and cannot be changed without changing the load. The only term in the above equation that can be regulated is the power factor. If the power factor can be adjusted to be equal to 1, the power losses will be minimum. 13 Efficiency of the transmission line is given by η % · ⋅ P P R S 100% Thermal Limits on equipment and conductors depend on the material of the insulation of conductors. The I 2 R losses are converted into heat. The heat increases the temperature of the conductors and the insulation surrounding it. Some equipment can be cooled by introducing circulation of cooling media, other must depend on natural cooling. If the temperature exceeds the rated value, the insulation will deteriorate faster and at higher temperatures more immediate damage will occur. The power losses increase with the load. It follows that the rated load is given by the temperature limits. The consequence of exceeding the rated load for short periods of time or by small amounts is a raised temperature that does not destroy the equipment but shortens its service life. Many utilities routinely allow short time overloads on their equipment - for example transformers are often overloaded by up to 15% during peak periods that may last only 15 or 30 minutes. 14