TRANSMISSION LINES. Transmission Structures An overhead transmission line consists of conductor, insulators, support structures and in most cases shield wires. Overhead power transmission lines are classified in the electrical power industry by the range of voltages: Low voltage less than 000 volts, used for connection between a residential or small commercial customer and the utility. Medium Voltage (istribution) between 000 volts ( kv) and to about 33 kv, used for distribution in urban and rural areas. High Voltage (Subtransmission if 33-5kV and transmission if 5kV+) between 33 kv and about 30 kv, used for sub-transmission and transmission of bulk quantities of electric power and connection to very large consumers. Extra High Voltage (Transmission) over 30 kv, up to about 800 kv, used for long distance, very high power transmission. Ultra High Voltage higher than 800 kv... Conductor Aluminum has replaced copper as the most common conductor metal for overhead transmission. An aluminum conductor has the advantage of better resistivity/weight than copper, as well as being cheaper. Some copper cable is still used, especially at lower voltages and for grounding. Aluminum conductor steel-reinforced (ACSR) are primarily used for medium and high voltage lines and may also be used for overhead services to individual customers. All aluminum conductors (AAC) All aluminum alloy conductor (AAAC) Aluminum conductor alloy reinforced (ACAR)
.. Insulators Insulators for transmission lines above 33 kv. At higher voltages only suspension-type insulators are common for overhead conductors. Insulators are usually made of wet-process porcelain or toughened glass, with increasing use of glass-reinforced polymer insulators. However, with rising voltage levels and changing climatic conditions, polymer insulators (silicone rubber based) are seeing increasing usage. China has already developed polymer insulators having a highest system voltage of 00kV and India is currently developing a 00kV (highest system voltage) line which will initially be charged with 400kV to be upgraded to a 00kV line...3 Support Structures transmission lines employ a variety of support structures. In the fig. A self-supporting, lattice steel tower typically used for 500 and 765 kv lines. ouble circuit 345 kv lines usually have selfsupporting steel tower with the phases arranged either in triangular configuration to reduce tower height or in a vertical configuration to reduce tower width.
3 Transmission line 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.. Transmission Line Resistance Resistance to d.c. current is given by R dc A where is the resistivity at T 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 The variation of resistance with temperature is linear over the normal temperature range resistance () R R T T T temperature ( o C) Figure 9 Graph of Resistance vs Temperature
4 ( R 0 ) ( R 0) ( T T ) ( T T ) R T T T T R. 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: GM L 0 7 ln [H/m] where is the geometric mean radius (available from manufacturer s tables) GM 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. is a hypothetical radius that replaces the actual conductor with a hollow conductor of radius equal to such that the self inductance of the inductor remains the same. If each phase consists of several conductors, the is given by 3 n n ( d d d... d ).( d. d... d )...( d. d... d ) 3 n n n n nn where d= d=... dnn=n
5 Note: for a solid conductor, = r.e -/4, where r is the radius of the conductor. Geometric Mean istance 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 m c n GM mn ' (... ).(... )...(... ) aa' ab' an' ba' bb ' bn' ma' mb ' mn' where aa is the distance between conductors a and a etc. Inductance Between Two Single Phase Conductors r r L 7 0 ln L r ' 0 7 ln r ' where r is of conductor r is of conductor is the GM between the conductors The total inductance of the line is then 7 LT L L r r 7 r r 7 0 ln ln 0 ln 0 ln ' ' ' ' r ' r ' L T 7 r r 7 4 0 ln 4 0 ' ' ln / r ' r ' If r = r, then L T 4 0 7 ln r '
6 Example: Find GM, 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 0.5 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 =, mn = 6 GM 6 B aa' ab' ba' bb ' ca' cb' where 9 m 6 9 7 m aa' bb' ab' ba' cb' ca' 9 5 m GM = 0.743 m Geometric Mean Radius for Circuit A: 3 9 4 4 A aa ab ac ba bb bc ca cb cc 0.50 e 6 3 0.48m Geometric Mean Radius for Circuit B: 4 B a 'a ' a ' b' b' b' b' a ' 0.50 e 4 6 0.53m
7 Inductance of circuit A L A GM 0743. 0 7 ln 0 7 ln 6. 0 7 H / m 0. 48 A Inductance of circuit B L B GM 0. 743 0 7 ln 0 7 ln 8. 5030 7 H / m 053. B The total inductance is then L L L 4. 75 0 7 H / m T A B 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 /. 7 GM L fl f 0 ln / m 7 GM L 4 f 0 ln / m 7 GM L 4 f 0 609 ln / 3 GM L. 0 0 f ln / 3 3 L. 0 0 f ln. 0 0 f ln GM / a If both, and GM are in feet, then a represents the inductive reactance at ft spacing, and d is called the inductive reactance spacing factor. d
8 Example: Find the inductive reactance per of a single phase line operating at 60 Hz. The conductor used is Partridge, with 0 ft spacings between the conductor centers. = 0 ft Solution: From the Tables, for Partridge conductor, = 0.07 ft and inductive reactance at ft spacing a = 0.465 /. The spacing factor for 0 ft spacing is d = 0.3635 /. The inductance of the line is then 0. 465 03635. 0885. / L a d Inductance of Balanced Three Phase Line Average inductance per phase is given by: eq L 0 7 ln where eq is the geometric mean of the three spacings of the three phase line. eq 3 ab ac bc Example: A three phase line operated at 60 Hz is arranged as shown. The conductors are ACSR rake. Find the inductive reactance per. 0ft 0 ft 38 ft Solution: For ACSR rake conductor, = 0.0373 ft
9 eq 7 4. 8 7 L 0 ln 3 0 H / m 0. 0373 7 60 609 3 0 0. 788 / L 3 0 0 38 4. 8 ft OR from the tables a 0. 399 / The spacing factor is calculated for spacing equal the geometric mean distance between the conductors, that is, 3 d. 0 0 60 ln 4. 8 0389. / Then the line inductance is 0. 788 / per phase line a d Example: Each conductor of the bundled conductor line shown in the figure is 7 MCM Pheasant. Find: a) the inductive reactance in /km and / per phase for d = 45 cm b) the p.u. series reactance if the length of the line is 60 km and the base is 00 MVA, 345 kv. d 8 m 8 m Solution: a) The distances in ft are 0. 45 d. 476 ft 0. 3048 8 6. 5 ft 0. 3048 For Pheasant conductors, = 0.0466 ft. b for a bundle of conductors is b d 0. 0466 476. 0. 63 ft The geometric mean of the phase conductor spacing is
0 3 eq 6. 5 6. 5 5. 49 3307. ft The inductance of the line is then eq 3307. L 0 7 ln 0 7 ln 9. 674 0 7 H / m 0. 63 b The inductive reactance is fl 7 4 L 60 9. 674 0 3. 647 0 / m 0. 3647 / km 0. 5868 / b) Base impedance Z b Vb 345 90 S 00 b Total impedance of the 60 km line is L 60 0. 3647 58. 35 Lp. u. L 58. 35 0. 049 p. u. Z 90 b.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 F V 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 F / m V 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 r For a two conductor line, the capacitance between the conductors is given by C 0 ln r r F / m where o is the permittivity of free space and is equal to 8.850 - F/m is the distance between the conductors, center to center r and r 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 = r. In this case, half way between the conductors there is a point where E = 0. This is the neutral point n a n b Can Cbn 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 an C bn C ab. C ab C C an bn C anc bn C C an bn If C an C bn, then C ab Can C C an an Therefore, C an C ab
It follows that C an o ln r [F/m] Since c fc o f ln r. 86 0 9 f r ln m c The capacitive reactance in is c. 86 0 779. 0 ln f r 609 f 9 6 ln r Similarly as for inductive reactance, this expression can be split into two terms that are called capacitive reactance at ft spacing ( a ) and the capacitive reactance spacing factor ( d ). c 779. 0 779. 0 ln f r f 6 6 ln a d a is given in the tables for the standard conductors, d is given in the tables for the capacitive reactance spacing factor. Example: Find the capacitive reactance in Ms for a single phase line operating at 60 Hz. The conductor used for the line is Partridge, and the spacing is 0 ft. = 0 ft
3 Solution: 0. 64 The outside radius of the Partridge conductor is r in 0. 068 ft The capacitive reactance is 6 6 779. 0. 779 0 0 C ln ln 096. M. f r f 0. 068 OR ' From tables 0074. M. a ' d 00889. M. for 0 spacing ' ' 0963. M. C a d This is the capacitive reactance between the conductor and the neutral. Line-to-line capacitive reactance is C 0. 098 M. L L C Capacitance of Balanced Three Phase Line between a phase conductor and neutral is given by C n o eq ln b F / m where eq 3 abbc ca and ab, bc, and ca are the distances between the centers of the phase conductors, and b is the geometric mean radius for the bundled conductors. (in the expression for b the outside radius of the conductor is used, rather than the from the tables.) The capacitive reactance to neutral than becomes cn 779. 0 6 eq ln. f b
4 Example: a) A three phase 60 Hz line is arranged as shown. The conductors are ACSR rake. Find the capacitive reactance for of the line. b) If the length of the line is 75 s and the normal operating voltage is 0 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. 0ft 0 ft 38 ft Solution: 08. The outside radius for rake conductors is r in 0. 046 ft The geometric mean distance for this line is 3 eq 0 0 38 4. 8 ft ' From tables, 0. 09 M. a 6 6 ' 779. 0 779. 0 d ln eq ln 48. 0. 095 M. f 60 ' ' 0. 864 M. cn a d This is the capacitive reactance to neutral. For the length of 75 s, Ctotal cn 065 75 Charging current is I C 0k VLN 3 9 065 Ctotal A
5 Reactive power to charge the line is Q 3V I 3 0k 9 4545. MVAr C LL C.4 Transmission Line Equivalent Circuits a) Short lines (less than 80 km) b) Medium length lines (80 km to 00 km) c) Long lines (over 00 km).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 cos I R R equiv R equiv E R PR cos R From that, PR Ploss I equiv R R E cos R R
6 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, the power losses will be minimum. Efficiency of the transmission line is given by P % P R S 00% Thermal Limits on equipment and conductors depend on the material of the insulation of conductors. The I 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 5% during peak periods that may last only 5 or 30 minutes.