* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download module 1
Variable-frequency drive wikipedia , lookup
War of the currents wikipedia , lookup
Power inverter wikipedia , lookup
Mercury-arc valve wikipedia , lookup
Stepper motor wikipedia , lookup
Induction motor wikipedia , lookup
Electric machine wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Current source wikipedia , lookup
Ground loop (electricity) wikipedia , lookup
Electrification wikipedia , lookup
Electric power system wikipedia , lookup
Opto-isolator wikipedia , lookup
Voltage optimisation wikipedia , lookup
Buck converter wikipedia , lookup
Power MOSFET wikipedia , lookup
Amtrak's 25 Hz traction power system wikipedia , lookup
Single-wire earth return wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Transformer wikipedia , lookup
Electric power transmission wikipedia , lookup
Electrical substation wikipedia , lookup
Ground (electricity) wikipedia , lookup
Rectiverter wikipedia , lookup
Power engineering wikipedia , lookup
Transformer types wikipedia , lookup
Stray voltage wikipedia , lookup
Transmission tower wikipedia , lookup
Mains electricity wikipedia , lookup
Aluminium-conductor steel-reinforced cable wikipedia , lookup
Earthing system wikipedia , lookup
History of electric power transmission wikipedia , lookup
Skin effect wikipedia , lookup
MODULE 1 MODELING POWER SYSTEM COMPONENTS The development of the modern day electrical energy system took a few centuries. Prior to 1800, scientists like William Gilbert, C. A. de Coulomb, Luigi Galvani, Benjamin Franklin, Alessandro Volta etc. worked on electric and magnetic field principles. However, none of them had any application in mind. They also probably did not realize that their work will lead to such an exciting engineering innovation. They were just motivated by the intellectual curiosity. Between 1800 and 1810 commercial gas companies were formed . first in Europe and then in North America. Around the same time with the research efforts of scientists like Sir Humphrey Davy, Andre Ampere, George Ohm and Karl Gauss the exciting possibilities of the use of electrical energy started to dawn upon the scientific community. In England Michael Faraday worked on his induction principle between 1821 and 1831. The modern world owe a lot to this genius. Faraday subsequently used his induction principle to build a machine to generate voltage. Around the same time American engineer Joseph Henry also worked independently on the induction principle and applied his work on electromagnets and telegraphs. For about three decades between 1840 and 1870 engineers like Charles Wheatstone, Alfred varley, Siemens brothers Werner and Carl etc. built primitive generators using the induction principle. It was also observed around the same time that when current carrying carbon electrodes were drawn apart, brilliant electric arcs were formed. The commercialization of arc lighting took place in the decade of 1870s. The arc lamps were used in lighthouses and streets and rarely indoor due to high intensity of these lights. Gas was still used for domestic lighting. It was also used for street lighting in many cities. From early 1800 it was noted that if a current carrying conductor could be heated to the point of incandescent. Therefore the idea of using this principle was very tempting and attracted attention. However the incandescent materials burnt very quickly to be of any use. To prevent them from burning they were fitted inside either vacuum globes or globes filled with inert gas. In October 1879 Thomas Alva Edison lighted a glass bulb with a carbonized cotton thread filament in a vacuum enclosed space. This was the first electric bulb that glowed for 44 hours before burning out. Edison himself improved the design of the lamp later and also proposed a new generator design. The Pearl Street power station in New York City was established in 1882 to sell electric energy for incandescent lighting. The system was direct current three-wire, 220/110 V and supplied Edison lamps for a total power requirement of 30 kW. The only objective of the early power companies was illumination. However we can easily visualize that this would have resulted in the under utilization of resources. The lighting load peaks in the evening and by midnight it reduces drastically. It was then obvious to the power companies that an elaborate and expensive set up would lay idle for a major amount of time. This provided incentive enough to improve upon the design of electric motors to make them commercially viable. The motors became popular very quickly and were used in many applications. With this the electric energy era really and truly started. However with the increase in load large voltage and unacceptable drops were experienced, especially at points that were located far away from the generating stations due to poor voltage regulation capabilities of the existing dc networks. One approach was to transmit power at higher voltages while consuming it at lower voltages. This led to the development of the alternating current. In 1890s the newly formed Westinghouse Company experimented with the new form electricity, the alternating current. This was called alternating current since the current changed direction in synchronism with the generator rotation. Westinghouse Company was lucky have Serbian engineer Nicla Tesla with them. He not only invented polyphase induction motor but also conceived the entire polyphase electrical power system. He however had to face severe objection from Edison and his General Electric Company who were the proponents of dc. The ensuing battle between ac and dc was won by ac due to the following factors: • Transformers could boost ac voltage for transmission and could step it down for distribution. • The construction of ac generators was simpler. • The construction of ac motors was simpler. Moreover they were more robust and cheaper than the dc motors even though not very sophisticated. With the advent of ac technology the electric power could reach more and more people. Also size of the generators started increasing and transmission level voltages started increasing. The modern day system contains hundreds of generators and thousands of buses and is a large interconnected network. Modern electric power systems have three separate components . generation, transmission and distribution. Electric power is generated at the power generating stations by synchronous alternators that are usually driven either by steam or hydro turbines. Most of the power generation takes place at generating stations that may contain more than one such alternator-turbine combinations. Depending upon the type of fuel used, the generating stations are categorized as thermal, hydro, nuclear etc. Many of these generating stations are remotely located. Hence the electric power generated at any such station has to be transmitted over a long distance to load centers that are usually cities or towns. This is called the power transmission. In fact power transmission towers and transmission lines are very common sights in rural areas. Modern day power systems are complicated networks with hundreds of generating stations and load centers being interconnected through power transmission lines. Electric power is generated at a frequency of either 50 Hz or 60 Hz. In an interconnected ac power system, the rated generation frequency of all units must be the same. In India the frequency is 50 Hz. The basic structure of a power system is shown in Fig. 1.1. It contains a generating plant, a transmission system, a subtransmission system and a distribution system. These subsystems are interconnected through transformers T1, T2 and T3. Let us consider some typical voltage levels to understand the functioning of the power system. The electric power is generated at a thermal plant with a typical voltage of 22 kV (voltage levels are usually specified line-to-line). This is boosted up to levels like 400 kV through transformer T1 for power transmission. Transformer T2 steps this voltage down to 66 kV to supply power through the subtransmission line to industrial loads that require bulk power at a higher voltage. Most of the major industrial customers have their own transformers to step down the 66 kV supply to their desired levels. The motivation for these voltage changes is to minimize transmission line cost for a given power level. Distribution systems are designed to operate for much lower power levels and are supplied with medium level voltages. Fig. 1.1 A typical power system. The power distribution network starts with transformer T3, which steps down the voltage from 66 kV to 11 kV. The distribution system contains loads that are either commercial type (like office buildings, huge apartment complexes, hotels etc) or residential (domestic) type. Usually the commercial customers are supplied power at a voltage level of 11 kV whereas the domestic consumers get power supply at 400-440 V. Note that the above figures are given for line-to-line voltages. Since domestic customers get single-phase supplies, they usually receive 230-250 V at their inlet points. While a domestic customer with a low power consumption gets a single-phase supply, both industrial and commercial consumers get three-phase supplies not only because their consumption is high but also because many of them use three-phase motors. For example, the use of induction motor is very common amongst industrial customers who run pumps, compressors, rolling mills etc. The main components of a power system are generators, transformers and transmission lines. In this module we shall we discuss of the models of these components that will be used subsequently in power system studies. 1.1 SERIES PARAMETERS OF TRANSMISSION LINES Overhead transmission lines and transmission towers are a common sight in rural India. The transmission towers are usually made of steel and are solidly erected with a concrete base. The three-phase conductors are supported by the towers through insulators. The conductors are usually made of aluminum or its alloys. Aluminum is preferred over copper as an aluminum conductor is lighter in weighted and cheaper in cost than copper conductor of the same resistance. The conductors are not straight wires but strands of wire twisted together to form a single conductor to give it higher tensile strength. One of the most common conductor is aluminum conductor, steel reinforced (ACSR). The cross sectional view of such a conductor is shown in Fig. 1.2. The central core is formed with strands of steel while two layers of aluminum strands are put in the outer layer. The other type of conductors that are in use are all aluminum conductor (AAC), all aluminum alloy conductor (AAAC), aluminum conductor, alloy reinforced (ACAR). Fig. 1.2 Cross sectional view of an ACSR conductor. 1.1.1 Line Resistance It is very well known that the dc resistance of a wire is given by l Rdc (1.1) A where is the resistivity of the wire in m , l is the length in m and A is the cross sectional area in m2. Unfortunately however the resistance of an overhead conductor is not the same as that given by the above expression. When alternating current flows through a conductor, the current density is not uniform over the entire cross section but is somewhat higher at the surface. This is called the skin effect and this makes the ac resistance a little more than the dc resistance. Moreover in a stranded conductor, the length of each strand is more that the length of the composite conductor. This also increases the value of the resistance from that calculated in (1.1). Finally the temperature also affects the resistivity of conductors. However the temperature rise in metallic conductors is almost linear in the practical range of operation and is given by R 2 T t1 (1.2) R1 T t 2 where R1 and R2 are resistances at temperatures t1 and t2 respectively and T is a constant that depends of the conductor material and its conductivity. Since the resistance of a conductor cannot be determined accurately, it is best to determine it from the data supplied by the manufacturer. 1.1.2 Inductance of a Straight Conductor From the knowledge of high school physics we know that a current carrying conductor produces a magnetic field around it. The magnetic flux lines are concentric circles with their direction specified by Maxwell’s right hand thumb rule (i.e., if the thumb of the right hand points towards the flow of current then the fingers of the fisted hand point towards the flux lines). The sinusoidal variation in the current produces a sinusoidal variation in the flux. The flux linkage is then expressed as L I where L is the inductance in Henry, λ is the flux linkage in Weber-turns and I is the phasor current in Ampere. A. Internal Inductance Consider a straight round (cylindrical) conductor as shown in Fig. 1.3. The conductor has a radius of r and carries a current I. Ampere’s law states that the magnetomotive force (mmf) in ampere-turns around a closed path is equal to the net current in amperes enclosed by the path. We then get the following expression mmf H ds I (1.4) where H is the magnetic field intensity in At/m, s is the distance along the path in m and I is the current in A. Let us denote the field intensity at a distance x from the center of the conductor by Hx. It is to be noted that Hx is constant at all points that is at a distance x from the center of the conductor. Therefore Hx is constant over the concentric circular path with a radius of r and is tangent to it. Denoting the current enclosed by Ix we can then write H dx I H x x x Ix 2 x (1.5) Fig. 1.3 Cross section of a round conductor. If we now assume that the current density is uniform over the entire conductor, we can write I Ix x2 I x I r2 x2 r2 (1.6) Substituting (1.6) in (1.5) we get I x (1.7) 2 r 2 Assuming a relative permeability of 1, the flux density at a distance of x from the center of the conductor is given by Hx 0I x 2 r 2 where µ0 is the permeability of the free space and is given by 4π x 10-7. Bx 0 Hx (1.8) The flux inside (or outside) the conductor is in the circumferential direction. The two directions that are perpendicular to the flux are radial and axial. Let us consider an elementary area that is of dimension dx m along the radial direction and 1 m along the axial direction. Therefore the area perpendicular to the flux at all angular positions is dx x 1 m2. Let the flux along the circular strip be denoted by dφx and this is given by 0I dx Bx dx 1 x dx (1.9) 2 r 2 Note that the entire conductor cross section does not enclose the above flux. The cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux dx . Therefore the flux linkage is x2 0I 3 (1.10) dx x dx 2 r 2 r 4 Integrating (1.10) over the range of x, i.e., from 0 to r, we get the internal flux linkage as dx r int 0 0I 3 0I I x dx 107 Wbt / m 4 2 r 8 2 (1.11) Then from (1.3) we get the internal inductance per unit length as 1 L int 107 H / m 2 (1.12) It is interesting to note that the internal inductance is independent of the conductor radius. B. External Inductance Let us consider an isolated straight conductor as shown in Fig. 1.4. The conductor carries a current I. Assume that the tubular element at a distance x from the center of the conductor has a field intensity Hx. Since the circle with a radius of x encloses the entire current, the mmf around the element is given by 2π xHx = I and hence the flux density at a radius x becomes 0I Bx 2 x (1.13) (1.14) Fig. 1.4 A conductor with two external points. The entire current I is linked by the flux at any point outside the conductor. Since the distance x is greater than the radius of the conductor, the flux linkage dλx is equal to the flux dx . Therefore for 1 m length of the conductor we get 0I dx (1.15) 2 x The external flux linkage between any two points D1 and D2, external to the conductor is D2 0I 1 (1.16) ext dx 2 107 2 x D1 From (1.3) we can then write the inductance between any two points outside the conductor as D2 Lext 2 107 ln H/m (1.17) D1 1.1.2 Inductance of a Single-phase Line dx dx Bx dx 1 Consider two solid round conductors with radii of r1 and r2 as shown in Fig. 1.5. One conductor is the return circuit for the other. This implies that if the current in conductor 1 is I then the current in conductor 2 is -I. First let us consider conductor 1. The current flowing in the conductor will set up flux lines. However, the flux beyond a distance D + r2 from the center of the conductor links a net current of zero and therefore does not contributed to the flux linkage of the circuit. Also at a distance less than D -r2 from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since D >> r2 we can make the following approximations D r1 D and D r1 D Therefore from (1.12) and (1.17) we can specify the inductance of conductor 1 due to internal and external flux as D 1 L1 2 ln 107 H / m r1 2 (1.18) Fig. 1.5 A single-phase line with two conductors. We can rearrange L1 given in (1.18) as follows D D D 1 L1 2 107 ln 2 107 ln e1/ 4 ln 2 107 ln 1/ 4 r1 r1 4 r1e Substituting r1’ = r1 e1/4 in the above expression we get D L1 2 107 ln ' H / m r1 (1.19) The radius r1’. can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r1. In a similar way the inductance due current in the conductor 2 is given by D L 2 2 107 ln ' H / m r2 Therefore the inductance of the complete circuit is D L L1 L 2 107 ln ' 2 107 r1 D ln ' r2 D2 D 2 10 ln ' ' 4 107 ln H/ m r1' r 2 ' r1 r 2 If we assume r1’= r2’= r’, then the total inductance becomes D L 4 107 ln H / m r' (1.20) (1.21) 7 (1.22) 1.1.3 Inductance of Three-Phase Lines with Symmetrical Spacing Consider the three-phase line shown in Fig. 1.6. Each of the conductors has a radius of r and their centers form an equilateral triangle with a distance D between them. Assuming that the currents are balanced, we have Ia + Ib + Ic = 0 (1.22) Also consider a point P external to the conductors. The distance of the point from the phases a, b and c are denoted by Dpa, Dpb and Dpc respectively. Let us consider the flux linked by the conductor of phase-a due to a current Ia including internal flux linkages but excluding flux linkages beyond the point P. From (1.18) we get Dpa Dpa 1 7 apa 2 ln Ia 2 10 Ia ln r r' 2 (1.23) The flux linkage with the conductor of phase-a due to the current Ib, excluding all flux beyond the point P, is given by (1.17) as Fig. 1.6 Three-phase symmetrically spaced conductors and an external point P. Dpb D Similarly the flux due to the current Ic is apb 2 107 Ib ln apc 2 107 Ic ln Dpc D (1.24) (1.25) Therefore the total flux in the phase-a conductor is Dpa Dpb Dpc a apa apb apc 2 107 Ia ln Ib ln Ic ln r' D D The above expression can be expanded as 1 1 1 a 2 107 Ia ln Ib ln Ic ln Ia ln Dpa Ib ln Dpb Ic ln Dpc r' D D From (1.22) we get (1.26) (1.27) Ib + Ic = -Ia Substituting the above expression in (1.27) we get 1 1 Dpb Dpc a 2 107 Ia ln Ia ln Ib ln Ic ln r' D Dpa Dpa (1.28) Now if we move the point P far away, then we can approximate D pa D pb D pc Therefore their logarithmic ratios will vanish and we can write (1.28) as 1 1 D a 2 107 Ia ln Ia ln 2 107 Ia ln r' D r' (1.29) Hence the inductance of phase-a is given as D (1.30) r' Note that due to symmetry, the inductances of phases b and c will be the same as that of phase-a given above, i.e., Lb = Lc = La. La 2 107 ln 1.1.4 Inductance of Three-Phase Lines with Asymmetrical Spacing It is rather difficult to maintain symmetrical spacing as shown in Fig. 1.6 while constructing a transmission line. With asymmetrical spacing between the phases, the voltage drop due to line inductance will be unbalanced even when the line currents are balanced. Consider the three-phase asymmetrical spaced line shown in Fig. 1.7 in which the radius of each conductor is assumed to be r. The distances between the phases are denoted by Dab, Dbc and Dca. We then get the following flux linkages for the three phases 1 1 1 (1.31) a 2 107 Ia ln Ib ln Ic ln r' Dab Dca 1 1 1 (1.32) b 2 107 Ib ln Ia ln Ic ln r' Dab Dbc 1 1 1 (1.33) c 2 107 Ic ln Ia ln Ib ln r' Dca Dbc Fig. 1.7 Three-phase asymmetrically spaced line. Let us define the following operator 0 1 3 a e j120 j 2 2 Note that for the above operator the following relations hold (1.34) 0 1 3 a 2 e j240 j and1 a a 2 0 2 2 Let as assume that the current are balanced. We can then write Ib = a2Ia and Ic = aIa (1.35) Substituting the above two expressions in (1.31) to (1.33) we get the inductance of the three phases as 1 1 1 (1.36) La 2 107 ln a 2 ln a ln r ' D D ab ca 1 1 1 (1.37) Lb 2 107 ln a ln a 2 ln Dab Dbc r' 1 1 1 (1.38) Lc 2 107 ln a 2 ln a ln r ' D D ca bc It can be seen that the inductances contain imaginary terms. The imaginary terms will vanish only when Dab = Dbc = Dca. In that case the inductance will be same as given by (1.30). 1.1.5 Transposed Line The inductances that are given in (1.36) to (1.38) are undesirable as they result in an unbalanced circuit configuration. One way of restoring the balanced nature of the circuit is to exchange the positions of the conductors at regular intervals. This is called transposition of line and is shown in Fig. 1.8. In this each segment of the line is divided into three equal sub-segments. The conductors of each of the phases a, b and c are exchanged after every sub-segment such that each of them is placed in each of the three positions once in the entire segment. For example, the conductor of the phase-a occupies positions in the sequence 1, 2 and 3 in the three sub-segments while that of the phase-b occupies 2, 3 and 1. The transmission line consists several such segments. Fig. 1.8 A segment of a transposed line. In a transposed line, each phase takes all the three positions. Therefore the per phase inductance is the average value of the three inductances calculated in (1.36) to (1.38). We therefore have L La Lb Lc 3 (1.39) This implies L 2 107 3 3 1 1 1 2 ln ln ln a a ln D bc D bc r ' Dab From (1.35) we have a + a2 = -1. Substituting in the above equation we get L 2 107 3 1 1 1 ln ln ln 3 r' Dab Dbc Dca (1.40) The above equation can be simplified as 2 107 1 1 ln ln L 1/ 3 3 r' Dab Dbc Dca 1/ 3 Dab Dbc Dca 7 2 10 ln r' (1.41) Defining the geometric mean distance (GMD) as GMD 3 Dab Dbc Dca equation (1.41) can be rewritten as (1.42) GMD H/m (1.43) r' Notice that (1.43) is of the same form as (1.30) for symmetrical spaced conductors. Comparing these two equations we can conclude that GMD can be construed as the equivalent conductor spacing. The GMD is the cube root of the product of conductor spacings. L 2 107 ln 1.1.5 Composite Conductors So far we have considered only solid round conductors. However as mentioned at the beginning of Section 1.1 stranded conductors are used in practical transmission line. We must therefore modify the equations derived above to accommodate stranded conductors. Consider the two groups of conductors shown in Fig. 1.9. Of these two groups conductor x contains n identical strands of radius rx while conductor y contains m identical strands of radius ry. Conductor x carries a current I the return path of which is through conductor y. Therefore the current through conductor y is -I. Fig. 1.9 Single-phase line with two composite conductors. Since the strands in a conductor are identical, the current will be divided equally among the strands. Therefore the current through the strands of conductor x is I/n and through the strands of conductor y is -I/m. The total flux linkage of strand a is given by a 2 10 7 I 1 1 1 1 ln .... ln ' ln n rx D ab D ac D an I 1 1 1 1 2 10 ln ln .... ln m D aa ' D ab' D ac' D am' 7 (1.44) We can write (1.44) as a 2 107 I ln m Daa ' Dab' Dac' ...Dam' (1.45) rx' Dab Dac ...Dan The inductance of the strand a is then given by m D D D ...D aa ' ab' ac' am' La a 2n 107 ln (1.46) ' n r D D ...D In x ab ac an In a similar way the inductances of the other conductors are also obtained. For example, m D D D ...D ba ' bb' bc' bm' L b 2n 107 ln n r ' D D ...D x ab bc bn (1.47) m D D D ...D ' ' ' ' ca cb cc cm Lc 2n 107 ln n r ' D D ...D x ac bc cn n The average inductance of any one of the strands in the group of conductor x is then Lav,x La Lb Lc ... Ln n (1.48) Conductor x is composed of n strands that are electrically parallel. Even though the inductance of the different is different, the average inductance of all of them is the same as Lav,x. Assuming that the average inductance given above is the inductances of n parallel strands, the total inductance of the conductor x is Lx Lav,x n La Lb Lc ... Ln (1.49) n2 Substituting the values of La, Lb etc. in the above equation we get L x 2 107 ln GMD GMR x (1.50) where the geometric mean distance (GMD) and the geometric mean radius (GMR) are given respectively by GMD mn Daa' Dab' Dac' ...Dam' ... Dna' Dnb' Dnc' ...Dnm' GMR x n2 rx' Dab Dac ...Dan' ... rx' Dna Dnb ...Dnn 1 (1.51) (1.52) The inductance of the conductor y can also be similarly obtained. The geometric mean radius GMRy will be different for this conductor. However the geometric mean distance will remain the same. 1.1.7 Bundled Conductors So far we have discussed three-phase systems that have only one conductor per phase. However for extra high voltage lines corona cause a large problem if the conductor has only one conductor per phase. Corona occurs when the surface potential gradient of a conductor exceeds the dielectric strength of the surrounding air. This causes ionization of the area near the conductor. Corona produces power loss. It also causes interference of the communication channels. Corona manifests itself with a hissing sound and ozone discharge. Since most long distance power lines in India are either 220 kV or 400 kV, avoidance of the occurrence of corona is desirable. The high voltage surface gradient is reduced considerably by having two or more conductors per phase in close proximity. This is called conductor bundling. The conductors are bundled in groups of two, three or four as shown in Fig. 1.10. The conductors of a bundle are separated at regular intervals with spacer dampers that prevent clashing of the conductors and prevent them from swaying in the wind. They also connect the conductors in parallel. The geometric mean radius (GMR) of two-conductor bundle is given by Ds,2b 4 Ds d Ds d 2 (1.53) where Ds is the GMR of conductor. The GMR for three-conductor and four-conductor bundles are given respectively by Ds,3b 9 Ds d d 3 Ds d 2 3 (1.54) Ds,b 9 Ds d d 2d 4 1.09 4 Ds d 3 (1.55) The inductance of the bundled conductor is then given by GMD (1.56) L 2 107 ln Ds,b where the geometric mean distance is calculated assuming that the center of a round conductor is the same as that of the center of the bundle. Fig. 1.10 Bundled conductors: (a) 2-conductor, (b) 3-conductor and (c) 4-conductor bundles. 1.2 SHUNT PARAMETERS OF TRANSMISSION LINES Capacitance in a transmission line results due to the potential difference between the conductors. The conductors get charged in the same way as the parallel plates of a capacitor. Capacitance between two parallel conductors depends on the size and the spacing between the conductors. Usually the capacitance is neglected for the transmission lines that are less than 50 miles (80 km) long. However the capacitance becomes significant for longer lines with higher voltage. In this section we shall derive the line capacitance of different line configuration. 1.2.1 Capacitance of a Straight Conductor Consider the round conductor is shown in Fig. 1.11. The conductor has a radius of r and carries a charge of q coulombs. The capacitance C is the ratio of charge q of the conductor to the impressed voltage, i.e., q (1.57) V The charge on the conductor gives rise to an electric field with radial flux lines where the total electric flux is equal to the charge on the conductor. By Gauss’s law, the electric flux density at a cylinder of radius x when the conductor has a length of 1 m is C Fig. 1.11 Cylindrical conductor with radial flux lines. q q C / m2 (1.58) A 2 x The electric filed intensity is defined as the ratio of electric flux density to the permittivity of the medium. Therefore D E q V/m 2 x 0 (1.59) Now consider the straight long conductor of Fig. 1.12 that is carrying a positive charge q C/m. Let two points P1 and P2 be located at distances D1 and D2 respectively from the center of the conductor. The conductor is an equipotential surface in which we can assume that the uniformly distributed charge is concentrated at the center of the conductor. The potential difference V12 between the points P1 and P2 is the work done in moving a unit of charge from P2 to P1. Therefore the voltage drop between the two points can be computed by integrating the field intensity over a radial path between the equipotential surfaces, i.e., D2 D2 D1 D1 V12 E dx D q ln 2 V 2 x0 D1 (1.60) Fig. 1.12 Path of integration between two point external to a round straight conductor. 1.2.2 Capacitance of a Single-Phase Line Consider the single-phase line consisting of two round conductors as shown in Fig. 1.5. The separation between the conductors is D. Let us assume that conductor 1 carries a charge of q1 C/m while conductor 2 carries a charge q2 C/m. The presence of the second conductor and the ground will disturb filed of the first conductor. However we assume that the distance of separation between the conductor is much larger compared to the radius of the conductor and the height of the conductor is much larger than D for the ground to disturb the flux. Therefore the distortion is small and the charge is uniformly distributed on the surface of the conductor. Assuming that the conductor 1 alone has the charge q1, the voltage between the conductors is q1 D ln V (1.61) 2 0 r1 Similarly if the conductor 2 alone has the charge q2, the voltage between the conductors is q2 D V21 q2 ln V (1.62) 2 0 r2 V12 q1 This implies that V12 q2 r q2 ln 2 V 2 0 D (1.62) From the principle of superposition we can write V12 V12 q1 V12 q2 r q1 D q2 ln ln 2 V 2 0 r1 2 0 D (1.63) For a single-phase line q1 = -q2 = q. We therefore have V12 r q D q q D2 ln ln 2 ln V 20 r1 20 D 20 r1r2 (1.64) Assuming r1 = r2 = r, we can rewrite (1.64) as q D ln V 0 r Therefore from (1.57) the capacitance between the conductors is given by V12 C12 0 Fm ln D r (1.65) (1.66) The above equation gives the capacitance between two conductors. For the purpose of transmission line modeling, the capacitance is defined between the conductor and neutral. This is shown in Fig. 1.13. Therefore the value of the capacitance is given from Fig. 1.13 as C 2C12 20 Fm ln D r (1.67) Fig. 1.13 (a) Capacitance between two conductors and (b) equivalent capacitance to ground. 1.2.3 Capacitance of a Three-Phase Transposed Line Consider the three-phase transposed line shown in Fig. 1.14. In this the charges on conductors of phases a, b and c are qa, qb and qc respectively. Since the system is assumed to be balanced we have qa qb qc 0 (1.68) Fig. 1.14 Charge on a three-phase transposed line. Using superposition the voltage Vab for the first, second and third sections of the transposition are given respectively as ()........++=cabccabbabaabDDqDrqrDqVlnlnln2110επ (1.69) ()........++=abcacbcbbcaabDDqDrqrDqVlnlnln2120επ (1.70) ()........++=bcabccabcaaabDDqDrqrDqVlnlnln2130επ (1.71) Then the average value of the voltage is ........++=cabcabcabcabccabcabbcabcabaabDDDDDDqDDDrqrDDDqVlnlnln21330επ (1.72) This implies Cond 1 Cond 2 C12 Cond 2 Cond 1 C C Neutral (a) (b) Dca Dbc qc qa Dab qb b c c b c b a a a ........+=330lnln21cabcabbcabcabaabDDDrqrDDDqVεπ (1.73) The GMD of the conductors are given in (1.42). We can therefore write ......+=GMDrqrGMDqVbaablnln210επ (1.74) Similarly the voltage is given as ......+=GMDrqrGMDqVcaaclnln210επ (1.75) Adding (1.74) and (1.75) and using (1.68) we get ()rGMDqGMDrqrGMDqGMDrqqrGMDqVVaaacbaacabln23lnln221lnln221000επεπεπ= .76) For a set of balanced three-phase voltages °...°.=°...°.=24001200ananacananabVVVVVV Therefore we can write °.=°...°...°.=+0324012002ananananacabVVVVVV (1.77) Combining (1.76) and (1.77) we get rGMDqVaanln210επ= (1.78) Therefore the capacitance to neutral is given by ()rGMDVF/m qCanaln20επ== (1.79) 1.2.4 Effect of Earth on the Calculation of Capacitance Earth affects the calculation of capacitance of three-phase lines as its presence alters the electric field lines. Usually the height of the conductors placed on transmission towers is much larger than the spacing between the conductors. Therefore the effect of earth can be neglected for capacitance calculations, especially when balanced steady state operation of the Vac power system is considered. However for unbalanced operation when the sum of the three line currents is not zero, the effect of earth needs to be considered. 1.3 SYNCHRONOUS MACHINE MODEL The schematic diagram of a synchronous generator is shown in Fig. 1.15. This contains three stator windings that are spatially distributed. It is assumed that the windings are wye-connected. The winding currents are denoted by ia, ib and ic. The rotor contains the filed winding the current through which is denoted by if. The field winding is aligned with the so-called direct (d) axis. We also define a quadrature (q) axis that leads the d-axis by 90°. The angle between the d-axis and the a-phase of the stator winding is denoted by θd. Fig. 1.15 Schematic diagram of a synchronous generator. Let the self-inductance of the stator windings be denoted by , and such that ccbbaasLLLL=== (1.80) and the mutual inductance between the windings be denoted as cabcabsLLLM===. (1.81) The mutual inductance between the field coil and the stator windings vary as a function of θd and are given by dfafMLθcos= (1.82) ()°.=120cosdfbfMLθ (1.83) ()°+=120cosdfcfMLθ (1.84) The self-inductance of the field coil is denoted by Lff. a c b d-axis q-axis if ib ia ic θd Laa Lbb Lcc The flux linkage equations are then given by ()fafcbsasfafccababaaaaiLiiMiLiLiLiLiL++.=+++=λ (1.85) ()fbfcasbsbiLiiMiL++.=λ (1.86) ()fafbbscsciLiiMiL++.=λ (1.87) fffccfbbfaaffiLiLiLiL+++=λ (1.88) For balanced operation we have 0=++cbaiii Hence the flux linkage equations for the stator windings (1.85) to (1.87) can be modified as ()fafassaiLiML++=λ (1.89) ()fbfbssbiLiML++=λ (1.90) ()fcfcssciLiML++=λ (1.91) For steady state operation we can assume constant==ffIi (1.92) Also assuming that the rotor rotates at synchronous speed ωs we obtain the following two equations dtddsθω= (1.93) 0dsdtθω+= θ(1.94) where θd0 is the initial position of the field winding with respect to the phase-a of the stator winding at time t = 0. The mutual inductance of the field winding with all the three stator windings will vary as a function of θd, i.e., ()0cosdsfaftMLθω+= (1.95) (°.+=120cos0dsfbftML)θω (1.96) (°++=120cos0dsfcftML)θω (1.97) Substituting (1.92), (1.94), (1.95), (1.96) and (1.97) in (1.89) to (1.91) we get ()(θω+++= λ(1.98) ()(°).+++=120cos0dsffbssbtIMiMLθωλ (1.99) ()(°)++++=120cos0dsffcssctIMiMLθωλ (1.100) Since we assume balanced operation, we need to treat only one phase. Let the armature resistance of the generator be R. The generator terminal voltage is given by 0cosdsffassatIMiML) dtdRivaaaλ..= (1.101) where the negative sign is used for generating mode of operation in which the current leaves the terminal. Substituting (1.98) in (1.101) we get ()()0sindsffassaatIMdtdiMLRivθω+++..= (1.102) The last term of (1.102) is the internal emf ea that is given by ()0sin2dsiatEeθω+= (1.103) where the rms magnitude .Ei. is proportional to the field current 2ffsiIMEω= (1.104) Since θd0 is the position of the d-axis at time t = 0, we define the position of the q-axis at that instant as °.=900dθδ (1.105) Therefore (1.94) can be rewritten as °++=90δωtsd θ(1.106) Substituting (1.105) in (1.103) we get ()δω+=tEesiacos2 (1.107) Hence (1.102) can be written as ()aassaaedtdiMLRiv++..= (1.108) The equivalent circuit is shown in Fig. 1.16. Let the current ia lag the internal emf ea by θa. The stator currents are then ()asaatIiθδω.+=cos2 (1.109) ()°..+=120cos2asabtIiθδω (1.110) ()°+.+=120cos2asactIiθδω (1.111) The single-phase equivalent circuit is shown in Fig. 1.17. The phase angle θa between ea and ia is rather difficult to measure under load as ea is the no load voltage. To avoid this, we define the phase angle between va and ia to be θ. We assume that ea leads va by δ. Therefore we can write δθθ.=a (1.112) Then the voltages and currents shown in Fig. 1.17 are given as tVvsaaωcos2= (1.113) ()δω.=tVesaacos2 (1.114) ()θω.=tVisaacos2 (1.115) Equations (1.113) to (1.115) implies that θδ..=.=°.=aaaaaaIIEEVVand,0 (1.116) The synchronous impedance is then defined as ()sssddMLjRjXRZ++=+=ω (1.117) The terminal voltage equation is then (adaaIjXREV++=) (1.118) Fig. 1.16 Three-phase equivalent circuit of a synchronous generator. Fig. 1.16 Single-phase equivalent circuit of a synchronous generator. + ea R Ls+ Ms va vb eb + ec + Ls+ Ms Ls + Ms Ls + Ms R R vc ia ic ib Ia Ls+ Ms R Ea + Va + 1.4 TRANSFORMER MODEL The equivalent circuit of a single-phase transformer is shown in Fig. 1.17. In this the primary voltage and currents are denoted by V1 and V2 respectively. The current entering the primary terminals is I1. The core loss component is represented by Rc while magnetizing reactance is denoted by Xm. The leakage inductance of the transformer is denoted by Xeq and Req is transformer winding resistance. It is to be noted that all the quantities are referred to the primary side. The turns ratio of the transformer is given by N1:N2. Fig. 1.17 Equivalent circuit of a single-phase transformer. The impedance of the shunt branch is much larger compared to that of the series branch. Therefore we neglect Rc and Xm. Again of the series parameters, Req is much smaller than Xeq. We can therefore neglect the series impedance. Therefore the transformer can be represented by the leakage reactance Xeq. The single-phase transformer equivalent circuit is then as shown is Fig. 1.18 when referred to the primary and secondary side where a = N1/N2. Fig. 1.18 Simplified equivalent circuit of a single-phase transformer: (a) when referred to the primary side and (b) when referred to the secondary side. 1.5 BALANCED OPERATION OF A THREE-PHASE CIRCUIT In the language of Power Systems, a three-phase circuit is said to be balanced if the following conditions are true. (1) If all the sources and loads are y-connected. (2) There is no mutual inductance between the phases. (3) All neutrals are at the same potential. (4) As a consequence of the points (2) and (3) above, all phases are decoupled. (5) All network variables are balanced sets in the same sequence as the sources. Rc Xm Req Xeq I1 Ic Im Iφ V1 V2 N1 N2 N1 N2 V2 V1 I1 V1 (a) I1 Xeq Xeq/a2 V2 N1 N2 (b) Consider the three-phase circuit shown in Fig. 1.19 that contains three balanced sources Ea, Eb and Ec along with three balanced source impedance ZS. The sources two balanced loads . one wye-connected with impedance of ZY and the other .-connected with impedance of Z.. Since this is a balanced network, the sum of the currents at the neutrals N (or n) is zero. Therefore the neutral are at the same potential. Transforming the .-connected load to an equivalent wye, we get the per phase equivalent circuit as shown in Fig. 1.20. In this fashion an entire power system can be converted into its per phase equivalent. The line diagram showing a per phase equivalent circuit is called a single-line diagram. Fig. 1.19 Three balanced sources supplying two balanced load through balanced source impedances. Fig. 1.20 Per phase equivalent circuit of the network of Fig. 1.19. 1.6 PER UNIT REPRESENTATION In a power system different power equipment with different voltage and power levels are connected together through various step up or step down transformers. However the presence of various voltage and power levels causes problem in finding out the currents (or voltages) at different points in the network. To alleviate this problem, all the system quantities are converted into a uniform normalized platform. This is called the per unit system. In a per unit system each system variable or quantity are normalized with respect to its own base value. The units of these normalized values are per unit (abbreviated as pu) and not Volt, Ampere or Ohm. The base quantities chosen are: • VA base (Pbase): This is the three-phase apparent power (Volt-Ampere) base that is common to the entire circuit. • Voltage Base (Vbase): This is the line-to-line base voltage. This quantity is not uniform for the entire circuit but gets changed by the turns ratio of the transformer. Based on the above two quantities the current and impedance bases can be defined as basebasebaseVPI.=3 (1.119) Ec Eb Ea + ZY ZY ZY ZS Z./3 Z. Z. Z. Ea + ZY n N ZS ZS ZS ()basebasebasePVZ2= (1.120) Assume that an impedance Z is defined as Z1 per unit in a base impedance of Zbase_old. Then we have ()oldbaseoldbaseZZZZZZ_1_1pu.=.= (1.121) The impedance now has to be represented in a new base value denoted as Zbase_new. Then we have ()newbaseoldbasenewbaseZZZZZZZ__12_2pu.=.= pu (1.122) From (1.120) Z2 can be defined in terms of old and new values of VA base and voltage base as oldbasenewbasenewbaseoldbasePPVVZZ__2__12.........= pu (1.123) Example 1.1: Let us consider the circuit shown in Fig. 1.18 (a) which contains the equivalent circuit of a transformer. Let the transformer rating be 500 MVA, 220/22 kV with a leakage reactance of 10%. Then the VA base of the transformer is 500 MVA and the voltage bases in the primary and secondary side are 200 kV and 22 kV respectively. Therefore the impedance bases of these two sides are ()8.9610500102206231=..=baseZ . and ()968.01050010226232=..=baseZ . where the subscripts 1 and 2 refer to the primary and secondary sides respectively. Assume that the leakage reactance is referred to the primary side. Then for 10%, i.e., 0.1 pu leakage reactance we have 68.98.961.01=.=eqX . The above reactance when referred to the secondary side is 0968.01068.968.9222===aXeq . Hence the per unit impedance in the secondary side is 0.0968/0.968 = 0.1. Therefore we see that the per unit leakage reactance is the same for both side of the transformer and, as a consequence, the transformer can be represented by only its leakage reactance. The equivalent circuit of the transformer is then as shown in Fig. 1.21. Since this diagram only shows the reactance (or impedance) of the circuit, this is called the reactance (or impedance) diagram. ... Fig. 1.21 The per unit equivalent circuit of a transformer. Example 1.2: Consider the 50 Hz power system the single-line diagram of which is shown in Fig. 1.22. The system contains three generators, three transformers and three transmission lines. The system ratings are Generator G1: 200 MVA, 20 kV, Xd = 15% Generator G2: 300 MVA, 18 kV, Xd = 20% Generator G3: 300 MVA, 20 kV, Xd = 20% Transformer T1: 300 MVA, 220Y/22. kV, 10% Transformer T2: Three single-phase units each rated 100 MVA, 130Y/25 kV, 10% Transformer T3: 300 MVA, 220/22 kV, 10% The transmission line reactances are as indicated in the figure. We have to draw the reactance diagram choosing the Generator 3 circuit as the base. Fig. 1.22 Single-line diagram of the power system of Example 1.2. As we have chosen the circuit of Generator 3 as the base, the base MVA for the circuit is 300. The high voltage side of transformer T2 is connected wye. Therefore its line to line voltage is .3 . 130 = 225 kV. Since the low voltage side is connected in ., its line to line voltage is 25 kV. The base voltages are chosen as discussed below. 200 kV 20 kV 200 kV 22.22 kV 200 kV 20 kV 1 T1 G3 j50 . j75 . V2 (pu) Xeq (pu) T3 j75 . V1 (pu) T2 G2 G1 Since the base voltage of G3 is 20 kV, the base voltage between T3 and bus 1 will be 20.10 = 200 kV. Also as there is no transformer connected in bus 1, the base voltage of 200 kV must be chosen for both the lines that are connected to either side of bus 1. Then the base voltage for the circuit of G1 will also be 20 kV. Finally since the turns ratio of T2 is 9 (= 225 . 25), the base voltage in the G2 side is 200 . 9 = 22.22 kV. The base voltages are also indicated in Fig. 1.22. Once the base voltages for the various parts of the circuit are known, the per unit values for the various reactances of the circuit are calculated according to (1.123) for a base MVA of 300. These are listed below. Generator G1: 225.020030015.01=.=GX pu Generator G2: 1312.022.22182.021=.......=GX pu Generator G3: pu 2.03=GX Transformer T1: 121.02002201.021=.......=TX pu Transformer T2: 1266.022.22251.022=.......=TX pu Transformer T3: 121.020221.023=.......=TX pu The base impedance of the transmission line is ()33.1331030010200623=..=baseZ . Therefore the per unit values of the line impedances are 5625.033.1337575==jX pu and 375.033.1335050==jX pu The impedance diagram is shown in Fig. 1.23. ... This completes our discussion on the modeling of power system components. In the subsequent portion of this course we shall use these models to construct a power system and use the per unit notation and the impedance diagram to represent the system. Therefore this module forms the basis for discussions for the subsequent modules. Fig. 1.23 The impedance diagram of the system of Fig. 1.22. j0.5625 XG3 j0.5625 XT2 j0.121 XT1 j0.375 j0.225 Xj50 Xj75 + XG2 XG1 XT3 + + EG1 EG3 EG2 Xj75 j0.121 j0.1266 j0.1312 j0.2