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# Lecture 09 transmission lines

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• Pls am a college student and am having issues solving these questions on lossless lines, will be glad if you can help me out; QUE; A 1meter long lossless line has a characteristic impedance of 400 ohms. if the line is operating at a freq of 200Mhz and has an input impedance of (200-j200)ohms , use both analytical and smith chart method to determine i. VSWR II. The load impedance ZL III. the distance of the nearest voltage max from the load. 2. A current of 20mA is sent over a transmission line of length 10km, attenuation constant of 0.1 neper/km and phase constant of 0.05rad/km. Determine the received voltage and current if the receiving end is short circuited. thanks i will be very grateful if i can get d solution soon; my email : adegokeseun2032@gmail.com

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### Lecture 09 transmission lines

1. 1. 1 Lecture 9: Transmission linesInstructor:Dr. Gleb V.TcheslavskiContact:gleb@ee.lamar.eduOffice Hours:TBD; Room 2030Class web site:http://ee.lamar.edu/gleb/Index.htmELEN 3441 Fundamentals of Power Engineering Spring 2008
2. 2. 2 Preliminaries Generators and loads are connected together through transmission lines transporting electric power from one place to another. Transmission line must, therefore, take power from generators, transmit it to location where it will be used, and then distribute it to individual consumers. A ROT: the power capability of a transmission line is proportional to the square of the voltage on the line. Therefore, very high voltage levels are used to transmit power over long distances. Once the power reaches the area where it will be used, it is stepped down to a lower voltages in distribution substations, and then delivered to customers through distribution lines.ELEN 3441 Fundamentals of Power Engineering Spring 2008
3. 3. 3 Preliminaries Distribution line with no ground wire. Dual 345 kV transmission lineELEN 3441 Fundamentals of Power Engineering Spring 2008
4. 4. 4 PreliminariesThere two types of transmission lines:overhead lines and buried cables.ELEN 3441 Fundamentals of Power Engineering Spring 2008
5. 5. 5 PreliminariesAn overhead transmission line usually consists of three conductors or bundles ofconductors containing the three phases of the power system. The conductors areusually aluminum cable steel reinforced (ACSR), which are steel core (for strength)and aluminum wires (having low resistance) wrapped around the core.ELEN 3441 Fundamentals of Power Engineering Spring 2008
6. 6. 6 PreliminariesIn overhead transmission lines, the conductors are suspended from a poleor a tower via insulators.ELEN 3441 Fundamentals of Power Engineering Spring 2008
7. 7. 7 PreliminariesIn addition to phase conductors, a transmission line usually includes one or twosteel wires called ground (shield) wires. These wires are electrically connected tothe tower and to the ground, and, therefore, are at ground potential.In large transmission lines, thesewires are located above the phaseconductors, shielding them fromlightning.ELEN 3441 Fundamentals of Power Engineering Spring 2008
8. 8. 8 PreliminariesCable lines are designed to be placedunderground or under water. The conductors areinsulated from one another and surrounded byprotective sheath. Cable lines are usually moreexpensive and harder to maintain. They also havecapacitance problem – not suitable for longdistance.Transmission lines are characterized by a series resistance, inductance, andshunt capacitance per unit length. These values determine the power-carryingcapacity of the transmission line and the voltage drop across it at full load.ELEN 3441 Fundamentals of Power Engineering Spring 2008
9. 9. 9 ResistanceThe DC resistance of a conductor is given by ρl RDC = (9.9.1) AWhere l is the length of conductor; A – cross-sectional area, ρ is the resistivity ofthe conductor. Therefore, the DC resistance per meter of the conductor is ρ Ω rDC = (9.9.1) A m  The resistivity of a conductor is a fundamental property of the material that theconductor is made from. It varies with both type and temperature of the material.At the same temperature, the resistivity of aluminum is higher than the resistivity ofcopper.ELEN 3441 Fundamentals of Power Engineering Spring 2008
10. 10. 10 ResistanceThe resistivity increases linearly with temperature over normal range oftemperatures. If the resistivity at one temperature is known, the resistivity atanother temperature can be found from M + T2 ρT 2 = ρT 1 (9.10.1) M + T1Where T1 and ρT1 are temperature 1 in oC and the resistivity at that temperature, T2and ρT2 are temperature 2 in oC and the resistivity at that temperature, and M is thetemperature constant. Material Resistivity at 20oC [Ω⋅m] Temperature constant [oC] Annealed copper 1.72⋅10-8 234.5 Hard-drawn copper 1.77⋅10-8 241.5 Aluminum 2.83⋅10-8 228.1 Iron 10.00⋅10-8 180.0 Silver 1.59⋅10-8 243.0ELEN 3441 Fundamentals of Power Engineering Spring 2008
11. 11. 11 ResistanceWe notice that silver and copper would be among the best conductors. However,aluminum, being much cheaper and lighter, is used to make most of thetransmission line conductors. Conductors made out of aluminum should havebigger diameter than copper conductors to offset the higher resistivity of thematerial and, therefore, support the necessary currents.AC resistance of a conductor is always higher than its DC resistance due to theskin effect forcing more current flow near the outer surface of the conductor. Thehigher the frequency of current, the more noticeable skin effect would be.At frequencies of our interest (50-60 Hz), however, skin effect is not very strong.Wire manufacturers usually supply tables of resistance per unit length at commonfrequencies (50 and 60 Hz). Therefore, the resistance can be determined fromsuch tables.ELEN 3441 Fundamentals of Power Engineering Spring 2008
12. 12. 12 Inductance and inductive reactanceThe series inductance of a transmission line consists of two components: internaland external inductances, which are due the magnetic flux inside and outside theconductor respectively. The inductance of a transmission line is defined as thenumber of flux linkages [Wb-turns] produced per ampere of current flowing throughthe line: λ L= (9.12.1) I 1. Internal inductance:Consider a conductor of radius r carrying a current I. At adistance x from the center of this conductor, the magneticfield intensity Hx can be found from Ampere’s law: ÑH x ×dl = I x ∫ (9.12.2)ELEN 3441 Fundamentals of Power Engineering Spring 2008
13. 13. 13 Inductance and inductive reactanceWhere Hx is the magnetic field intensity at each point along a closed path, dl is aunit vector along that path and Ix is the net current enclosed in the path. For thehomogeneous materials and a circular path of radius x, the magnitude of Hx isconstant, and dl is always parallel to Hx. Therefore: Ix 2π xH x = I x ⇒ Hx = (9.13.1) 2π xAssuming next that the current is distributed uniformly in the conductor: π x2 Ix = 2 I (9.13.2) πrThus, the magnetic intensity at radius x inside the conductor is x Hx = I [ H m] (9.13.3) 2π r 2ELEN 3441 Fundamentals of Power Engineering Spring 2008
14. 14. 14 Inductance and inductive reactanceThe flux density at a distance x from the center of the conductor is µ xI Bx = µ H x = [T ] (9.14.1) 2π r 2The differential magnetic flux contained in a circular tube of thickness dx and at adistance x from the center of the conductor is µ xI dφ = dx [Wb m] (9.14.2) 2π r 2The flux linkages per meter of length due to flux in the tube is the product of thedifferential flux and the fraction of current linked: π x2 µ x3 I d λ = 2 dφ = dx [Wb − turns m] (9.14.3) πr 2π r 4ELEN 3441 Fundamentals of Power Engineering Spring 2008
15. 15. 15 Inductance and inductive reactanceThe total internal flux linkages per meter can be found via integration… µ x3 I r µI λint = ∫ d λ = ∫ dx = [Wb − turns m ] (9.15.1) 0 2π r 4 8πTherefore, the internal inductance per meter is λint µ lint = = [ H m] (9.15.2) I 8πIf the relative permeability of the conductor is 1 (non-ferromagnetic materials, suchas copper and aluminum), the inductance per meter reduces to µ0 4π × −7 10 lint = = = 0.5× −7 [ H m ] 10 (9.15.3) 8π 8πELEN 3441 Fundamentals of Power Engineering Spring 2008
16. 16. 16 External inductance between 2 points outside of the lineTo find the inductance external to a conductor, we needto calculate the flux linkages of the conductor due onlythe portion of flux between two points P1 and P2 that lieat distances D1 and D2 from the center of the conductor.In the external to the conductor region, the magneticintensity at a distance x from the center of conductor is Ix I Hx = = (9.16.1) 2π x 2π xsince all the current is within the tube.The flux density at a distance x from the center of conductor is µI Bx = µ H x = (9.16.2) 2π xELEN 3441 Fundamentals of Power Engineering Spring 2008
17. 17. 17 External inductance between 2 points outside of the lineThe differential magnetic flux contained in a circular tube of thickness dx and at adistance x from the center of the conductor is µI dφ = dx [Wb m] (9.17.1) 2π xThe flux links the full current carried by the conductor, therefore: µI d λ = dφ = dx [Wb − turns m] (9.17.2) 2π xThe total external flux linkages per meter can be found via integration… D2 D2 µI µ I D1 λext = ∫ d λ = ∫ dx = ln [Wb − turns m ] (9.17.3) D1 D1 2π x 2π D2The external inductance per meter is λex t µ D2 lex t = = ln [ H m] (9.17.4) I 2π D1ELEN 3441 Fundamentals of Power Engineering Spring 2008
18. 18. 18 Inductance of a single-phase 2-wire transmission lineWe determine next the series inductance of asingle-phase line consisting of two conductors ofradii r spaced by a distance D and both carryingcurrents of magnitude I flowing into the page in theleft-hand conductor and out of the page in the right-hand conductor.Considering two circular integration paths, wenotice that the line integral along x1 produces a netmagnetic intensity since a non-zero net current isenclosed by x1. Thus: ÑH ∫ x ×dl = I x (9.18.1)Since the path of radius x2 encloses both conductors and the currents are equal andopposite, the net current enclosed is 0 and, therefore, there are no contributions tothe total inductance from the magnetic fields at distances greater than D.ELEN 3441 Fundamentals of Power Engineering Spring 2008
19. 19. 19 Inductance of a single-phase 2-wire transmission lineThe total inductance of a wire per unit length in this transmission line is a sum ofthe internal inductance and the external inductance between the conductor surface(r) and the separation distance (D): µ 1 D l = lint + lext =  + ln ÷ [ H m ] (9.19.1) 2π 4 r By symmetry, the total inductance of the other wire is the same, therefore, the totalinductance of a two-wire transmission line is µ1 D l =  + ln ÷ [ H m ] (9.19.2) π 4 r Where r is the radius of each conductor and D is the distance between conductors.ELEN 3441 Fundamentals of Power Engineering Spring 2008
20. 20. 20 Inductance of a transmission lineEquations similar to (9.19.2) can be derived for three-phase lines and for lines withmore phases… In most of the practical situations, the inductance of thetransmission line can be found from tables supplied by line developers.Analysis of (9.19.2) shows that:1.The greater the spacing between the phases of a transmission line, the greaterthe inductance of the line. Since the phases of a high-voltage overheadtransmission line must be spaced further apart to ensure proper insulation, a high-voltage line will have a higher inductance than a low-voltage line. Since thespacing between lines in buried cables is very small, series inductance of cables ismuch smaller than the inductance of overhead lines.2.The greater the radius of the conductors in a transmission line, the lower theinductance of the line. In practical transmission lines, instead of using heavy andinflexible conductors of large radii, two and more conductors are bundled togetherto approximate a large diameter conductor. The more conductors included in thebundle, the better the approximation becomes. Bundles are often used in the high-voltage transmission lines.ELEN 3441 Fundamentals of Power Engineering Spring 2008
21. 21. 21 Inductance of a transmission lineA two-conductor A four-conductorbundle bundleELEN 3441 Fundamentals of Power Engineering Spring 2008
22. 22. 22 Inductive reactance of a lineThe series inductive reactance of a transmission line depends on both theinductance of the line and the frequency of the power system. Denoting theinductance per unit length as l, the inductive reactance per unit length will be xI = jωl = j 2π fl (9.22.1)where f is the power system frequency. Therefore, the total series inductivereactance of a transmission line can be found as X I = xI d (9.22.2)where d is the length of the line.ELEN 3441 Fundamentals of Power Engineering Spring 2008
23. 23. 23 Capacitance and capacitive reactanceSince a voltage V is applied to a pair of conductors separated by a dielectric (air),charges of equal magnitude but opposite sign will accumulate on the conductors: q = CV (9.23.1)Where C is the capacitance between the pair of conductors.In AC power systems, a transmission line carries a time-varying voltagedifferent in each phase. This time-varying voltage causes the changes incharges stored on conductors. Changing charges produce a changingcurrent, which will increase the current through the transmission line andaffect the power factor and voltage drop of the line. This changing currentwill flow in a transmission line even if it is open circuited.ELEN 3441 Fundamentals of Power Engineering Spring 2008
24. 24. 24 Capacitance and capacitive reactanceThe capacitance of the transmission line can be found using the Gauss’s law: ÒD ×dA = q ∫∫ A (9.24.1)where A specifies a closed surface; dA is the unit vector normal to the surface; q isthe charge inside the surface; D is the electric flux density at the surface: D =εE (9.24.2)where E is the electric field intensity at that point; ε is the permittivity of thematerial: ε = ε rε 0 (9.24.3) Relative permittivity of the materialThe permittivity of free space ε0 = 8.85⋅10-12 F/mELEN 3441 Fundamentals of Power Engineering Spring 2008
25. 25. 25 Capacitance and capacitive reactanceElectric flux lines radiate uniformly outwards from thesurface of the conductor with a positive charge on itssurface. In this case, the flux density vector D isalways parallel to the normal vector dA and isconstant at all points around a path of constantradius r. Therefore: DA = Q ⇒ D (2π xl ) = ql (9.25.1)were l is the length of conductor; q is the chargedensity; Q is the total charge on the conductor. qThen the flux density is D= (9.25.2) 2π x qThe electric field intensity is E= (9.25.3) 2πε xELEN 3441 Fundamentals of Power Engineering Spring 2008
26. 26. 26 Capacitance and capacitive reactanceThe potential difference between two points P1 and P2 can be found as P2 V12 = ∫ E ×dl (9.26.1) P1where dl is a differential element tangential to the integration path between P1 andP2. The path is irrelevant.Selection of path can simplify calculations.For P1 - Pint, vectors E and dl are parallel;therefore, E⋅dl = Edx. For Pint – P2 vectorsare orthogonal, therefore E⋅dl = 0. D2 D2 q q D2 V12 = ∫ Edx = ∫ D1 D1 2πε x dx = ln 2πε D1 (9.26.2)ELEN 3441 Fundamentals of Power Engineering Spring 2008
27. 27. 27 Capacitance of a single phase two-wire transmission lineThe potential difference due to thecharge on conductor a can be found as qa D Vab ,a = ln (9.27.1) 2πε rSimilarly, the potential difference due to the charge on conductor b is qb D Vba ,b = ln (9.27.2) 2πε r qb Dor Vab ,b = − ln (9.27.3) 2πε rELEN 3441 Fundamentals of Power Engineering Spring 2008
28. 28. 28 Capacitance of a single phase two-wire transmission lineThe total voltage between the lines is qa D qb D Vab = Vab ,a + Vab ,b = ln − ln (9.28.1) 2πε r 2πε rSince q1 = q2 = q, the equation reduces to q D Vab = ln (9.28.2) πε rThe capacitance per unit length between the two conductors of the line is q q cab = = (9.28.3) V q D ln πε rELEN 3441 Fundamentals of Power Engineering Spring 2008
29. 29. 29 Capacitance of a single phase two-wire transmission lineThus: πε cab = (9.29.1) D ln rWhich is the capacitance per unit length of a single-phase two-wire transmissionline.The potential difference between each conductor and the ground (or neutral) isone half of the potential difference between the two conductors. Therefore, thecapacitance to ground of this single-phase transmission line will be 2πε cn = can = cbn = (9.29.2) D ln rELEN 3441 Fundamentals of Power Engineering Spring 2008
30. 30. 30 Capacitance of a single phase two-wire transmission lineSimilarly, the expressions for capacitance of three-phase lines (and for lines withmore than 3 phases) can be derived. Similarly to the inductance, the capacitanceof the transmission line can be found from tables supplied by line developers.Analysis of (9.29.1) shows that:1.The greater the spacing between the phases of a transmission line, the lower thecapacitance of the line. Since the phases of a high-voltage overhead transmissionline must be spaced further apart to ensure proper insulation, a high-voltage linewill have a lower capacitance than a low-voltage line. Since the spacing betweenlines in buried cables is very small, shunt capacitance of cables is much largerthan the capacitance of overhead lines. Cable lines are normally used for shorttransmission lines (to min capacitance) in urban areas.2.The greater the radius of the conductors in a transmission line, the higher thecapacitance of the line. Therefore, bundling increases the capacitance. Goodtransmission line is a compromise among the requirements for low seriesinductance, low shunt capacitance, and a large enough separation to provideinsulation between the phases.ELEN 3441 Fundamentals of Power Engineering Spring 2008
31. 31. 31 Shunt capacitive admittanceThe shunt capacitive admittance of a transmission line depends on both thecapacitance of the line and the frequency of the power system. Denoting thecapacitance per unit length as c, the shunt admittance per unit length will be yC = jωc = j 2π fc (9.31.1)The total shunt capacitive admittance therefore is YC = yC d = j 2π fcd (9.31.2)where d is the length of the line. The corresponding capacitive reactance is thereciprocal to the admittance: 1 1 ZC = =−j (9.31.3) YC 2π fcdELEN 3441 Fundamentals of Power Engineering Spring 2008
32. 32. 32 ExampleExample 9.1: An 8000 V, 60 Hz, single-phase, transmission line consists of twohard-drawn aluminum conductors with a radius of 2 cm spaced 1.2 m apart. If thetransmission line is 30 km long and the temperature of the conductors is 200C,a.What is the series resistance per kilometer of this line?b.What is the series inductance per kilometer of this line?c.What is the shunt capacitance per kilometer of this line?d.What is the total series reactance of this line?e.What is the total shunt admittance of this line?a. The series resistance of the transmission line is ρl R= AIgnoring the skin effect, the resistivity of the line at 200 will be 2.83⋅10-8 Ω-m andthe resistance per kilometer of the line is ρ l 2.83 × −8 × 10 1000 r= = = 0.0225 Ω km A π ×0.02 2ELEN 3441 Fundamentals of Power Engineering Spring 2008
33. 33. 33 Example b. The series inductance per kilometer of the transmission line is µ1 D µ1 1.2  l=  + ln ÷× 1000 =  + ln ÷ 1000 = 1.738 × −3 H km × 10 π 4 r  π 4 0.02  c. The shunt capacitance per kilometer of the transmission line is πε π×8.854 × −12 10 cab = × 1000 = 1000 = 6.794 × −9 F km × 10 D 1.2 ln ln r 0.02 d. The series impedance per kilometer of the transmission line iszse = r + jx = r + j 2π fl = 0.0225 + j 2π ×60 × 1.738 × −3 = 0.0225 + j 0.655 Ω km 10 Then the total series impedance of the line is Z se = ( 0.0225 + j 0.655 ) × = 0.675 + j19.7 Ω 30 ELEN 3441 Fundamentals of Power Engineering Spring 2008
34. 34. 34 Examplee. The shunt admittance per kilometer of the transmission line is yC = j 2π fc = j 2π ×60 ×6.794 × −9 = j 2.561 × −6 S m 10 10The total shunt admittance will be Yse = ( j 2.561 × −6 ) × = j 7.684 × −5 S 10 30 10The corresponding shunt capacitive reactance is 1 1 Z sh = = −5 = − j13.0 k Ω Ysh j 7.684 × 10ELEN 3441 Fundamentals of Power Engineering Spring 2008
35. 35. 35 Transmission line modelsUnlike the electric machines studied so far, transmission lines are characterized bytheir distributed parameters: distributed resistance, inductance, and capacitance.The distributed series and shunt elements of the transmission line make it harderto model. Such parameters may be approximated by many small discreteresistors, capacitors, and inductors.However, this approach is not very practical, since it would require to solve forvoltages and currents at all nodes along the line. We could also solve the exactdifferential equations for a line but this is also not very practical for large powersystems with many lines.ELEN 3441 Fundamentals of Power Engineering Spring 2008
36. 36. 36 Transmission line modelsFortunately, certain simplifications can be used…Overhead transmission lines shorter than 80 km (50 miles) can be modeled as aseries resistance and inductance, since the shunt capacitance can be neglectedover short distances.The inductive reactance at 60 Hz for – overheadlines – is typically much larger than the resistanceof the line.For medium-length lines (80-240 km), shuntcapacitance should be taken into account.However, it can be modeled by two capacitors ofa half of the line capacitance each.Lines longer than 240 km (150 miles) are long transmission lines and are to bediscussed later.ELEN 3441 Fundamentals of Power Engineering Spring 2008
37. 37. 37 Transmission line modelsThe total series resistance, series reactance, and shunt admittance of atransmission line can be calculated as R = rd (9.37.1) X = xd (9.37.2) Y = yd (9.37.3)where r, x, and y are resistance, reactance, and shunt admittance per unit lengthand d is the length of the transmission line. The values of r, x, and y can becomputed from the line geometry or found in the reference tables for the specifictransmission line.ELEN 3441 Fundamentals of Power Engineering Spring 2008
38. 38. 38 Short transmission lineThe per-phase equivalent circuit of a short lineVS and VR are the sending and receiving endvoltages; IS and IR are the sending and receivingend currents. Assumption of no line admittanceleads to IS = IR (9.38.1)We can relate voltages through the Kirchhoff’s voltage law VS = VR + ZI = VR + RI + jX L I (9.38.2) VR = VS − RI − jX L I (9.38.3)which is very similar to the equation derived for a synchronous generator.ELEN 3441 Fundamentals of Power Engineering Spring 2008
39. 39. 39 Short transmission line: phasor diagramAC voltages are usually expressed as phasors.Load with lagging power factor.Load with unity power factor.Load with leading power factor.For a given source voltage VS and magnitude ofthe line current, the received voltage is lower forlagging loads and higher for leading loads.ELEN 3441 Fundamentals of Power Engineering Spring 2008
40. 40. 40 Transmission line characteristicsIn real overhead transmission lines, the line reactance XL is normally much largerthan the line resistance R; therefore, the line resistance is often neglected. Weconsider next some important transmission line characteristics…1. The effect of load changesAssuming that a single generatorsupplies a single load through atransmission line, we considerconsequences of increasing load.Assuming that the generator is ideal, an increase of load will increase a real and(or) reactive power drawn from the generator and, therefore, the line current, whilethe voltage and the current will be unchanged.1) If more load is added with the same lagging power factor, the magnitude of theline current increases but the current remains at the same angle θ with respect toVR as before.ELEN 3441 Fundamentals of Power Engineering Spring 2008
41. 41. 41 Transmission line characteristicsThe voltage drop across the reactance increases but stays at the same angle.Assuming zero line resistance and remembering thatthe source voltage has a constant magnitude: VS = VR + jX L I (9.41.1)voltage drop across reactance jXLI will stretchbetween VR and VS.Therefore, when a lagging load increases, the received voltage decreases sharply.2) An increase in a unity PF load, on the other hand,will slightly decrease the received voltage at the endof the transmission line.ELEN 3441 Fundamentals of Power Engineering Spring 2008