ELE 2303 Power Generation and Transmission PDF

8/15/2021 ELE 2303: Power Generation and Transmission CLO4 : Develop expressions for resistance, inductance and capacitance of high voltage power transmission lines and determine the equivalent circuit of a three phase transmission line 1 Power Generation and Transmission ELE 2303 CLO4 Develop expressions for resistance, inductance and capacitance of high voltage power transmission lines and determine the equivalent circuit of a three phase transmission line 4.1 List the types of conductors used in power transmission line. 4.2 Develop the expression for the inductance and capacitance of a simple, single-phase, two wire transmission line composed of solid round conductors. 4.3 Calculate the inductance and capacitance of a simple, single-phase composite (stranded) conductor line. 4.4 Calculate the inductance and capacitance of three-phase lines having symmetrically and asymmetrically spacing and for bundled conductors. 4.1 List the types of conductors used in power transmission line. Conductor Selection Considerations i- General Properties of Transmission Conductors: Line conductors are one of the main parts of overhead lines. The important characteristics, which the line conductors must have are: 1. High tensile strength: The material for the conductor of an overhead line should have a high tensile strength (high breaking load) so that the spans between transmission line towers can be as long as possible and the sag as small as possible thus reducing the number and height of towers, and number of insulators. Conductor Selection Considerations 2. Low resistivity: The conductor should have low resistivity to reduce the power losses and voltage drop. 3. Low cost: The cost of its installation and maintenance should be low and it should have a long life. 4. Low Corrosion: Conductors must be stable against corrosion. 5. Low Skin Effect and Corona Losses: The structure of the conductor should be such that to minimize the additional losses due to skin effect or corona effect in case of high and extra high voltages. Conductor Materials The final choice of material is often a compromise. Copper, aluminum, steel and steel-cored aluminum conductors are generally employed in an overhead lines to transmit electrical energy. The following is a list of properties of each of these materials. Conductor Materials a) Copper. The most common conductor material used for transmission is hard drawn copper, because it is twice as strong as soft drawn copper. The merits of this metal as a line conductor are: 1. It has a best conductivity in comparison to other metals. 2. It has higher current density, so for the given current density (rating), less cross sectional area of conductor is required and hence it provides lesser cross sectional area to wind loads. 3. The metal is quite homogenous. 4. It has low specific resistance. 5. It is durable and has a higher scrap value. Conductor Materials b) Aluminum: Next to copper, aluminum is the conductor used in order of preference as far as the conductivity is concerned. Its merits and demerits are: 1. It is cheaper than copper. 2. It is lighter in weight. 3. It is second in conductivity among the metals used for transmission. 4. For same Ohmic resistance, its diameter is about 1.27 times that of copper. Aluminium 5. Since the diameter of the conductor is more, so it is subjected to greater wind pressure due to which greater is the swing of the conductor and greater is the sag. 6. Since the conductors are liable to swing, so it requires larger cross arms. 7. As the melting point of the conductor is low, so the short circuits and similar effects will damage it. 8. Joining of aluminum is much more difficult than that of any other material. Conductor Materials c) Steel: No doubt it has got the greater tensile strength, but it is least used for transmission of electrical energy as it has got high resistance. Bare steel conductors are not used since they deteriorate rapidly owing to rusting. Generally galvanized steel wires are used where high strength is desired. It has the following properties: 1. It has high internal reactance. 2. It is lowest in conductivity, 3. It is much subjected to eddy current and hysteresis losses. 4. In a damp atmosphere it rusts quickly. Hence its use is limited and the main application is for ground wires. Types of Conductors AAC : All Aluminium Conductor AAAC : All Aluminium Alloy Conductor ACSR : Aluminium Conductor, Steel Reinforced ACAR : Aluminium Conductor, Alloy Reinforced Types of Conductors AAC : All Aluminium Conductor This type is sometimes also referred as ASC (Aluminium Stranded Conductor). It is made up of strands of Electrical Conductor grade aluminium. AAC conductor has conductivity about 61% IACS (International Annealed Copper Standard). Despite having a good conductivity, because of its relatively poor strength, AAC has limited use in transmission and rural distribution lines. However, AAC can be seen in urban areas for distribution where spans are usually short but higher conductivity is required. AAAC : All Aluminium Alloy Conductor These conductors are made from aluminium alloy 6201 which is a high strength Aluminium-Magnesium-Silicon alloy. This alloy conductor offers good electrical conductivity (about 52.5% IACS) with better mechanical strength. Because of AAAC's lighter weight as compared to ACSR of equal strength and current capacity, AAAC may be used for distribution purposes. However, it is not usually preferred for transmission because of its low tensile strength. Also, AAAC conductors can be employed in coastal areas because of their excellent corrosion resistance. Conductor Materials d) Aluminum conductor with steel reinforced (ACSR). An aluminum conductor having a central core of galvanized steel wires is used in stranded conductors for high voltage transmission purposes. This is done to increase the tensile strength of aluminum conductors. The galvanized steel core is covered by one or more strands of aluminum wires. The steel conductors used are galvanized in order to prevent rusting and electrolytic corrosion. Examples of such conductors are shown in Figure 3. ACSR Lines Figure 3: Stranded conductors (a)Multi-core homogeneous conductor (b)Multi-core non-homogeneous conductor (c)Multi-core bimetallic conductor (d) Multi-core bundled conductor Single core or bundled? Conductors on transmission lines—especially high-voltage, high-capacity lines—are sometimes bundled, meaning that what is electrically a single conductor is actually composed of two, three, or four wires a few inches apart, held together every so often with connectors known as conducting frames. Bundling of Conductors There are several reasons for bundling conductors: Increasing heat dissipation as a result of increasing surface area, Reducing corona losses, Reducing inductance, Increasing the amount of current in the conductor due to reduced skin effect Bundling of Conductors For 220 kV lines, two-conductor bundles are usually used, for 380 kV lines usually three or even four. A bundle conductor results in lower reactance, compared to a single conductor. As a disadvantage, the bundle conductors have higher wind loading. 4.2 Develop the expression for the inductance and capacitance of a simple, single-phase, two wire transmission line composed of solid round conductors. Inductance of a single-phase two-wire transmission line  The total inductance of a wire per unit length in this transmission line is a sum of the internal inductance and the external inductance between the conductor surface (r) and the separation distance (D) and μ is the permeability : µ 1 D l lint + lext= =  + ln  [ H m] (1) 2π 4 r   By symmetry, the total inductance of the other wire is the same, therefore, the total inductance of a two-wire transmission line is µ1 D =l  + ln  [ H m] (2) π 4 r   Where r is the radius of each conductor and D is the distance between conductors. 19 Inductance of a transmission line  Equations similar to (2) can be derived for three-phase lines and for lines with more phases… In most of the practical situations, the inductance of the transmission line can be found from tables supplied by line developers. Analysis of (2) shows that: 1. The greater the spacing between the phases of a transmission line, the greater the inductance of the line. 2. The greater the radius of the conductors in a transmission line, the lower the inductance of the line. 20 A two-conductor A four-conductor bundle bundle 21 Inductive reactance of a line  The series inductive reactance of a transmission line depends on both the inductance of the line and the frequency of the power system. Denoting the inductance per unit length as l, the inductive reactance per unit length will be = ωl j 2π fl xI j= (24)  where f is the power system frequency. Therefore, the total series inductive reactance of a transmission line can be found as X I = xI d (25)  where d is the length of the line. 22 Inductance Simplification   D µ  D µ  1 µ 1 D  l =  + ln  =  ln e + ln  =  ln −1  4 π 4 r π r π 4   r e  µ = µ r µ0 ∴µ o = 4 π x10 −7 23  For air the µr=1   µ D  −7 D l =  ln −1  = 4 x 10 ln −1 π 4   r e  re4 −1  r′ = r e 4 D −7 ∴ l = 4 x 10 ln [ H / m] r′  The value of the inductance is called inductance per loop meter or per loop mile to distinguish. 24 Capacitance of a Single Phase Two- wire Transmission Line (16) The capacitance per unit length of a single-phase two-wire transmission line πε cab = (1) D ln r 25. The potential difference between each conductor and the ground (or neutral) is one half of the potential difference between the two conductors. Therefore, the capacitance to ground of this single-phase transmission line will be 2πε c= c= c= n an bn D (2) ln r where ε is the permittivity of the material 26 Capacitance in a Transmission Line  Similarly, the expressions for capacitance of three-phase lines (and for lines with more than 3 phases) can be derived. Similarly to the inductance, the capacitance of the transmission line can be found from tables supplied by line developers. 1. The greater the spacing between the phases of a transmission line, the lower the capacitance of the line. 2. The greater the radius of the conductors in a transmission line, the higher the capacitance of the line. 27 Shunt capacitive admittance The shunt capacitive admittance of a transmission line depends on both the capacitance of the line and the frequency of the power system. Denoting the capacitance per unit length as c, the shunt admittance per unit length will be = ω c j 2π fc yC j= (18) The total shunt capacitive admittance therefore is YC y= = Cd j 2π fcd (20) where d is the length of the line. The corresponding capacitive reactance is the reciprocal to the admittance: 1 1 ZC = = −j (21) YC 2π fcd 28 Electrical and Electronics Department Example An 8000 V, 60 Hz, single-phase, transmission line consists of two hard-drawn aluminum conductors with a radius of 2 cm spaced 1.2 m apart. If the transmission 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 impedance of this line? e. What is the total shunt admittance of this line? a. The series resistance of the transmission line is ρl R= A Ignoring the skin effect, the resistivity of the line at 200 will be 2.83⋅10-8 Ω-m and the resistance per kilometer of the line is ρl2.83 ⋅10−8 ⋅1000 r = = = 0.0225 Ω km A π ⋅ 0.02 2 29 Electrical and Electronics Department b. The series inductance per kilometer of the transmission line is µ1 D µ1 1.2  −3 =l  + ln  ⋅ 1000 =  + ln  ⋅ 1000 = 1.738 ⋅ 10 H km π 4 r  π 4 0.02  c. The shunt capacitance per kilometer of the transmission line is πε π ⋅ 8.854 ⋅10−12 cab = ⋅1000 = ⋅1000 = 6.794 ⋅10−9 F km D 1.2 ln ln r 0.02 d. The series impedance per kilometer of the transmission line is zse = r + j 2π fl = r + jx = 0.0225 + j 2π ⋅ 60 ⋅1.738 ⋅10−3 = 0.0225 + j 0.655 Ω km Then the total series impedance of the line is Z se = ( 0.0225 + j 0.655= ) ⋅ 30 0.675 + j19.7 Ω 30 e. The shunt admittance per kilometer of the transmission line is yC j 2π= = fc j 2π ⋅ 60 ⋅ 6.794 ⋅10 = −9 j 2.561 ⋅10−6 S m The total shunt admittance will be Yse = ( j 2.561 ⋅ 10 −6 = ) ⋅ 30 j 7.684 ⋅ 10 −5 S The corresponding shunt capacitive reactance is 1 1 Z sh = = −5 = − j13.0 k Ω Ysh j 7.684 ⋅10 31 4.3 Calculate the inductance and capacitance of a simple, single-phase composite (stranded) conductor line. Two Conductor Line Inductance  Key problem with the previous derivation is we assumed no return path for the current. Now consider the case of two wires, each carrying the same current I, but in opposite directions; assume the wires are separated by distance D. D To determine the inductance of each conductor we integrate as before. However now we get some field cancellation. Creates counter- Creates a clockwise field clockwise field 33 Line Inductance Example Calculate the reactance for a balanced 3φ, 60Hz transmission line with a conductor geometry of an equilateral triangle with D = 5m, r = 1.24cm (Rook conductor) and a length of 5 miles. Since system is assumed balanced ia = −ib − ic µ0  1 1 1  λa = i ln( ) + i ln( ) + i ln( ) 2π  D  a b c r' D 34 Substituting ia = −ib − ic , obtain: µ0   1  1  =λa  ia ln   − ia ln   2π   r'  D  µ0  D = ia ln  . 2π  r'  µ0  D  4π × 10−7  5  =La = ln   ln  −3  2π  r '  2π  9.67 × 10  = 1.25 × 10−6 H/m. Again note logarithm of ratio of distance between phases to the size of the conductor. 35 La = 1.25 × 10−6 H/m Converting to reactance X a= 2π × 60 × 1.25 × 10−6 −4 = 4.71 × 10 Ω/m = 0.768 Ω/mile = 3.79 Ω X Total for 5 mile line (this is the total per phase) The reason we did NOT have mutual inductance was because of the symmetric conductor spacing 36 4.4 Calculate the inductance and capacitance of three-phase lines having symmetrically and asymmetrically spacing and for bundled conductors Conductor Bundling To increase the capacity of high voltage transmission lines it is very common to use a number of conductors per phase. This is known as conductor bundling. Typical values are two conductors for 345 kV lines, three for 500 kV and four for 765 kV (in 60 Hz systems) 38 Bundle Inductance Example 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? r=1.24 × 10−2 m r ' = 9.67 × 10−3 m 1 0.25 M Rb = ( −3 9.67 × 10 × 0.25 × 0.25 × ( 2 × 0.25) ) 4 = 0.12 m (ten times bigger than r!) 0.25 M 0.25 M µ0 5 La = ln = 7.46 × 10−7 H/m 2π 0.12 Bundling reduces inductance. 39 Inductance Example  Calculate the per phase inductance and reactance of a balanced 3φ, 60 Hz, line with: – horizontal phase spacing of 10m – using three conductor bundling with a spacing between conductors in the bundle of 0.3m.  Assume the line is uniformly transposed and the conductors have a 1cm radius. 40 1 Dm = ( d12 d13d 23 ) 3 , = (10 × (2 × 10) × 10)1/ 3 = 12.6m, − µr r'=r e 4 = 0.0078m, 1 Rb = ( r ' d12 d13 ) 3, = ( r '× 0.3 × 0.3)1/ 3 = 0.0888m, µ0 Dm La = ln 2π Rb = 9.9 × 10−7 H/m, Xa 2π fLa (1600m/mile) = 0.6Ω/mile. 41 42 43 44 45 46 Example An 8000 V, 60 Hz, single-phase, transmission line consists of two hard-drawn aluminum conductors with a radius of 2 cm spaced 1.2 m apart. If the transmission 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 total series impedance of this line? 47 Example An 8000 V, 60 Hz, single-phase, transmission line consists of two hard-drawn aluminum conductors with a radius of 2 cm spaced 1.2 m apart. If the transmission 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 total series impedance of this line? a. The series resistance of the transmission line is ρl R= A Ignoring the skin effect, the resistivity of the line at 200 will be 2.83⋅10-8 Ω-m and the resistance per kilometer of the line is ρl2.83 ⋅10−8 ⋅1000 r = = = 0.0225 Ω km A π ⋅ 0.02 2 48 b. The series inductance per kilometer of the transmission line is µ1 D µ1 1.2  −3 =l  + ln  ⋅ 1000 =  + ln  ⋅ 1000 = 1.738 ⋅ 10 H km π 4 r  π 4 0.02  d. The series impedance per kilometer of the transmission line is zse = r + j 2π fl = r + jx = 0.0225 + j 2π ⋅ 60 ⋅1.738 ⋅10−3 = 0.0225 + j 0.655 Ω km Then the total series impedance of the line is Z se = ( 0.0225 + j 0.655= ) ⋅ 30 0.675 + j19.7 Ω 49 ACSR Table Data GMR is equivalent to r’ Inductance and Capacitance assuming a Dm of 1 ft. 50 Electrical and Electronics Department ACSR Table Data 51 ACSR Data Term from table, depending on Term independent of conductor type, but assuming a conductor, but with one foot spacing spacing Dm in feet. 52 ACSR Data If both GMR and GMD are in feet, then Xa represents the inductive reactance at 1 ft spacing, and Xd is called the inductive reactance spacing factor. 53 To use the phase to neutral capacitance from table 1 2πε 0 XC = Ω-m where C = 2π f C Dm ln r 1 6 Dm = × 1.779 × 10 ln Ω-mile (table is in MΩ-mile) f r 1 1 1 = × 1.779 × ln + × 1.779 × ln Dm MΩ-mile f r f Term from table, Term independent depending on conductor type, of conductor, but but assuming a one foot spacing with spacing Dm in feet. 54 Dove Example GMR = 0.0313 feet Outside Diameter = 0.07725 feet (radius = 0.03863) Assuming a one foot spacing at 60 Hz −7 1 X a 2π 60 × 2 × 10 × 1609 × ln = Ω/mile 0.0313 X a = 0.420 Ω/mile, which matches the table For the capacitance 1 6 1 X C = × 1.779 × 10 ln =9.65 × 104 Ω-mile f r 55 Example 1 Find the inductive reactance per mile and the capacitive reactance in MΩ.miles of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacing between the conductor centers. 56 Example 1 Find the inductive reactance per mile and the capacitive reactance in MΩ.miles of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacing between the conductor centers. From the Tables, for Partridge conductor, GMR = 0.0217 ft and inductive reactance at 1 ft spacing Xa = 0.465 Ω/mile, which matches the table. The spacing factor for 20 ft spacing is Xd = 0.3635 Ω /mile. The inductance of the line is then XL = Xa + Xd = 0.465 + 0.3635 = 0.8285 Ω/ mile 57 Example 1 Find the inductive reactance per mile and the capacitive reactance in MΩ.miles of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacing between the conductor centers. 58 Example 1 Find the inductive reactance per mile and the capacitive reactance in MΩ.miles of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacing between the conductor centers. This is the capacitive reactance between the conductor and the neutral. 59 Example 1 Find the inductive reactance per mile and the capacitive reactance in MΩ.miles of a single phase line operating at 60 Hz. The conductor used is Partridge, with 20 ft spacing between the conductor centers. This is the capacitive reactance between the conductor and the neutral. Line-to-line capacitive reactance is: 60 Example 2 A three phase line operated at 60 Hz is arranged as shown. The conductors are ACSR Drake. If the length of the line is 175 miles and the normal operating voltage is 220 kV, Find: 1. the inductive reactance per mile 2. the inductive reactance for the entire length of the line 3. the capacitive reactance for one mile 4. the capacitive reactance to neutral for the entire length of the line 5. the charging current for the line 6. the charging reactive power 61 Example 2 1. The inductive reactance per mile: 62 Example 2 1. The inductive reactance per mile: or, 63 Example 2 2. The inductive reactance the entire length of the line: XL = 0.788 x 175 = 137.9 Ω 64 Example 2 3. The capacitive reactance for one mile: 1 foot = 12 inch 65 Example 2 4. The capacitive reactance to neutral for the entire length of the line: 66 Example 2 5. the charging current for the line: 6. the charging reactive power: 67

ELE 2303 Power Generation and Transmission PDF

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