Lecture 1.pdf

Prof. Ahmed Bayoumy 1  J. Glover, M Sarma, and T Overbye, “Power system analysis and design” Thomson Publishing, 4th ed., 2008.  J. Grainger, and W. Stevenson, “Power system analysis” McGraw-Hill, int. ed., 1994. or newer edition.  Turan Gonen,...

Prof. Ahmed Bayoumy 1  J. Glover, M Sarma, and T Overbye, “Power system analysis and design” Thomson Publishing, 4th ed., 2008.  J. Grainger, and W. Stevenson, “Power system analysis” McGraw-Hill, int. ed., 1994. or newer edition.  Turan Gonen, Electric Power Distribution Engineering, CRC Press, 3rd ed., 2014. 2 3  Main components of power systems  Single-phase and Three-Phase Systems 4  An electric power system consists of three principle components: ◦ The generating stations ◦ The transmission lines  Are the connection links between all the generating stations ◦ The distribution systems  Connect all individual loads in a given area to the transmission lines 5 [https://www.webpages.uidaho.edu] 6  The electrical power is generated in generating stations using three phase synchronous generator.  The power is generated at medium voltage level (e.g. 11kV, 13.8kV, etc... )  Generating stations can generate power in the range of hundreds of MW.  They can be thermal stations, hydraulic stations, or nuclear stations.  Modern renewable energy stations are included in power systems now like wind stations, photovoltaic stations, concentrated solar, etc …  Generating stations are connected to transmission lines using step up transformers. 7  Transmission is made up by either: ◦ Overhead transmission lines ◦ High-voltage underground cables  Overhead transmission lines are more common than underground cables transmission.  Overhead transmission lines are more cheaper than underground cables transmission.  When the transmission system becomes near towns or customer centers, the voltage is stepped down to sub- transmission level. Sub- transmission lines belong to transmission system, but with lower voltages. 8  When the Sub-transmission ( secondary transmission) system is very close to load centers (cities), voltage is reduced using step down transformer stations for distribution of electrical power to loads.  One transmission line can feed many distribution systems. Distribution systems are located inside cities while transmission system are located outside cities.  Distribution system voltage level is lower than that of transmission.  The end of distribution system is your building at which the voltage can 380/220 V , 400/231 V,or 415/240 (Oman), 50 Hz. 9 11/220 kV 220/66 kV 66/11 kV 11 kV/415 V 10  In interconnected systems, the systems are connected by tie lines.  It is preferable to interconnect the systems to: ◦ Improve the overall stability. ◦ Improve the overall reliability. ◦ Be more economical 11 [Ref 1] 12 13  Transmission is made up by either: ◦ Overhead transmission lines ◦ High-voltage underground cables  Overhead lines are preferred because they are cheaper and easier to maintain.  An overhead transmission line consists of: ◦ Towers ◦ Phase conductors ◦ Insulators 14  [https://www.hydroquebec.com] [indiamart.com] [general source wikipidia.com] 15  The tower function is to carry the conductors and insulators.  The phase conductors function is to carry the current (electrical power).  The insulator function is to isolate energized phase conductors from the grounded towers.  The common transmission line voltages are 66, 69, 132, 220, 230, 345, 400, 500, and 765 kV. 16  Cap and bin insulators  They are made from porcelain or glass  A number of them can made a string to isolate larger voltage. [general source wikipidia.com] 17  Aluminum has replaced copper conductor due to its light Wight and law cost.  Conductors are made of strands in order to be able to have larger conductor sizes by simply adding more strands. Stranded conductors are more flexible than solid conductors.  Most common conductor type is ACSR (Aluminum conductor steel reinforced) which consists of a steel core surrounded by aluminum strands 18  Conductors of overhead transmission lines are bare (no insulation).  Air is used as an insulator 19  In order to study the performance of transmission lines, transmission line parameters should be studied.  Line parameters include: ◦ Resistance ◦ Inductance ◦ Capacitance 20 ⚫ The resistance (R) of a transmission line is the first parameter of a transmission line. ⚫ It determines the transmission losses. ⚫ The per phase dc line resistance is given by  Rdc = ohm per meter ⚫ Where A ⚫Rdc is the dc resistance per phase ⚫ρ is the resistivity of the conductor material in ohm. Meter ⚫A is the cross sectional area of the conductor  The ac conductor resistance, which differs from the dc resistance due to the non uniformity of current density in the conductor cross section is given by Rac  1.1Rdc ⚫ This is due to skin effect ⚫ In skin effect, the current tends to concentrate near the conductor service ⚫ In general, the conductor resistance is found from the manufacturers tables 22  Find the AC resistance per phase per km for a copper conductor. The copper resistivity is 1.68×10-8 Ωm. The cross sectional area is 3×10-3 m2. The AC resistance is 1.1 times the DC resistance.  Solution 𝜌 𝑅𝑑𝑐 = Ω/m 𝐴 A = 3 × 10−3 m2 1.68 × 10−8 𝑅𝐷𝐶 = = 5.6 × 10−6 Ω/m 3 × 10−3 𝑅𝐴𝐶 = 𝑅𝐷𝐶 × 1.1 = 6.16 × 10−6 Ω/m 𝑅𝐴𝐶 = 6.16 × 10−6 × 1000 = 6.16 × 10−3 Ω/km 23  Calculate the AC resistance in ohms for 10 km aluminum conductor with a 3 cm diameter. The aluminum resistivity is 2.83×10-8 Ωm. The AC resistance is 1.1 times the DC resistance 𝜌 𝑅𝑑𝑐 = Ω/m 𝐴 A = 𝜋 ⥂ r 2 = 𝜋 × (0.015 )2 = 7.069 × 10−4 m2 2.83 × 10−8 𝑅𝐷𝐶 = = 4.003 × 10−5 Ω/m 7.069 × 10−4 𝑅𝐴𝐶 = 𝑅𝐷𝐶 × 1.1 = 4.4033 × 10−5 Ω/m 𝑅𝐴𝐶−𝑡𝑜𝑡𝑎𝑙 = 4.4033 × 10−5 × 10000 = 0.44 Ω 24 ⚫ The inductance (L) of a transmission line is the most important line parameter. ⚫ In some applications, all other parameters are neglected. ⚫ The inductive reactance X = ωL is the dominating impedance element due to its effect on the voltage drop and transmission capacity. ⚫ The current –carrying conductor produces a magnetic field. ⚫ When the current changes, the flux changes. ⚫ By definition of the inductance, an inductance will exist for the conductor due to this situation. ⚫ In what follows, GMR stands for geometric mean radius. 26  a D D r c b D −7  D  L ph = 2 10 ln  H/m  GMR  GMR = 0.7788  r 27  The problem with the line analysis we’ve done so far is that we have assumed a symmetrical tower configuration. Such a tower figuration is rarely used.  Therefore in general Dab Dac  Dbc 28  a Dab Dca b r c Dbc Deq (GMD) = 3 Dab Dbc Dca  Deq  L ph = 2 10 ln −7  H/m  GMR  GMR = 0.7788r 29 ⚫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. Transposition is done by transposition tower. Cond. a Position 1 Cond. b Position 2 Cond. c Position 3 l/3 l/3 l/3 Aerial or side view of conductor positions over the length of the transmission line. Calculate the reactance for a balanced 3, 50Hz transmission line with a conductor geometry of an equilateral triangle with D = 5m, r = 1.24cm and a length of 5 km.  Solution  Deq   5  L ph = 2 10 ln −7  = 2 10 ln −7  H/m  GMR   1.24  0.01 0.7788  L ph = 1.25 H/m L ph total = 1.25 10−6  5000 = 6.25 mH 2  6.25  50 X= = 1.96 ohm 1000 31  A completely transposed 60 Hz line has flat horizontal configuration with10m spacing between adjacent conductors. The conductor radius is 2.04 cm. the line length is 200 km. Determine the line inductance and inductive reactance in ohm.  Solution  10m 10m GMR = 0.0204 0.7788  0.0159m 32 If the total inductance for a transmission line is 0.1875 H. Calculate the transmission line length if the transmission line has equilateral triangle formation with D = 10m, r = 2.5cm. 𝐷𝑒𝑞 𝐿𝑝ℎ = 2 × 10−7 ln × 𝑙𝑒𝑛𝑔𝑡ℎ 𝐺𝑀𝑅 −7 10 0.1875 = 2 × 10 ln × 𝑙𝑒𝑛𝑔𝑡ℎ 2.5 × 0.01 × 0.7788 0.1875 𝑙𝑒𝑛𝑔𝑡ℎ = =150000 m=150 km 1.25×10−6 33  It is common in extra high voltage (EHV) transmission lines to use more than conductor per phase to reduce the series impedance  This technique is called bundling.  Bundling increases the transmitted power. Moreover, it reduces the electric field, as result, it reduces corona losses, communication interference and audible noise 34  Bundling can be realized by two, three, or four bundled conductors. 35   Deq  L ph = 2 10 ln −7  H/m  Dr  Deq : Geometric Mean Distance (GMD) Deq = 3 Dab Dbc Dca 36 Dr : Geometric Mean Radius in case of bundled conductors Dr = r '  d (two bundeld conductors) Dr = r  d 3 ' 2 (three bundeld conductors) Dr = r  2d 4 ' 3 (four bundeld conductors) r = 0.7788  r ' 37  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? 0.25 M 0.25 M 0.25 M 38   Deq   5  L ph = 2 10 ln −7  = 2 10 ln −7  H/m  Dr   4 1.24  0.01 0.7788  2  0.253    L ph = 744.51 nH/m L ph total = 744.5110−9  5000 = 3.72 mH 2  3.72  50 X= = 1.17 ohm 1000 39 ⚫ The capacitance and conductance form the shunt admittance of the transmission line. ⚫ The capacitance represents a source of reactive power. ⚫ Its importance increases with the magnitude of the operating voltage and length of the line. ⚫ In high voltage cables, the shunt capacitance is very high.  a D D r c b D 2 o C ph = F/m ,  o = 8.85 10 −12 ln (D r ) 41  a Dab Dca b r c Dbc Deq (GMD) = 3 Dab Dbc Dca 2 o C ph = ln (Deq r ) F/m 42  2 o = ln (Deq Dr ) C ph F/m Deq : Geometric Mean Distance (GMD) Deq (GMD) = 3 Dab Dbc Dca Dr : Geometric Mean Raduis (GMR ) 43 Dr = r  d (two bundeldconductors) Dr = 3 r  d 2 (three bundeld conductors) Dr = r  2d 4 3 (four bundeld conductors) 44  Consider the previous example of the three phases symmetrically spaced 5 meters apart using wire with a radius of r = 1.24 cm, 4 bundled conductors in a square bundle, spaced 0.25 meters apart. What is the capacitance per meter? 45  2𝜋𝜀𝑜 𝐶𝑝ℎ = F/m ln 𝐷𝑒𝑞 Τ𝐷𝑟 4 𝐷𝑟 = 1.24 × 0.01 × 2 × 0.253 = 0.129 m 2𝜋𝜀𝑜 𝐶𝑝ℎ = = 15.2 pF/m ln 5Τ0.129 46