Impedance of Overhead Lines By : Dr. Atul R. Phadke Associate Professor in Electrical Engineering College of Engineering Pune CONDUCTOR FOR OVERHEAD LINE: Copper Aluminium: Cheaper and lighter but less conductivity and less tensile strength than copper. ACSR (Aluminium Reinforced) Conductor Steel AAC (All Alumunium Conductor) AAAC (All Aluminium Alloy Conductor) ACAR (Alluminium Reinforced) Conductor Alloy 2 ACSR CONDUCTOR: ACSR: βͺ Internal steel strands increase the tensile strength βͺ Outer aluminium strands carry the current βͺ Stranded conductor with twisted wires for strength and flexibility 3 RESISTANCE OF CONDUCTOR: DC resistance of conductor at specified temperature is: ππ π π = ο π΄ ππ = the resistivity of the conductor at temperature, π β π = the length of the conductor in π π΄ = the cross-sectional area of the conductor in π2 The resistance of conductor at any temperature can be determined by – π π2 1 + πΌ0 π2 = π π1 1 + πΌ0 π1 πΌ0 = temperature coefficient of resistance at 0β 4 RESISTANCE OF STRANDED CONDUCTOR: ππ π π = ο π΄ ππ π π = ο ππ π΄ π ππ 1.05 × 103 π = ο/ππ π 2 π π ππ 4 ππ π = 1337 2 ο/ππ π π ππ 5 CONDUCTOR DATA SHEET: 6 INDUCTANCE OF TRANSPOSED THREE PHASE LINE: Inductance/phase/m = 2 × Or92.9 ππ 25 πππ΄ Inductance/phase/m = 2 × πΊππ· ππ π·ππ = πΊππ· −7 10 ln πΊππ 3 π»/π 37.5 πππ΄ π·ππ −7 10 ln πΊππ 50 πππ΄ π»/π πππ πππ πππ πΊππ = 0.7788 π π = πππππ’π ππ πππππ’ππ‘ππ Line impedance in ο/km for π = 50 Hz π§ = π + πππΏ π§ = π + π × 2 × π × 50 × 2 × 10−7 ln πΊππ· π§ = π + π0.0628 ln ο/ππ πΊππ πΊππ· × 1000 ο/ππ πΊππ 7 SELF IMPEDANCE OF UNTRANSPOSED DISTRIBUTION LINES: Flux linkages of conductor π due to it own current πΌπ 92.9 ππ 25 πππ΄ 37.5 πππ΄ 1 −7 ο¬ππ = 2 × 10 ln πΌπ πΊππ π 50 πππ΄ ο¬ππ 1 −7 πΏππ = = 2 × 10 ln π»/π πΌπ πΊππ π π§ππ = ππ + πππΏπ π§ππ = ππ + π2π50 × 2 × 10−7 ln 1 × 1000 ο/ππ πΊππ π 1 π§ππ = ππ + π0.0628 ln ο/ππ πΊππ π 8 MUTUAL IMPEDANCE UNTRANSPOSED DISTRIBUTION LINES: Flux linkages of conductor π due to current in 92.9 ππ π, πΌπ conductor 25 πππ΄ 37.5 πππ΄ 1 −7 ο¬ππ = 2 × 10 ln πΌπ π·ππ 50 πππ΄ ο¬ππ 1 −7 πΏππ = = 2 × 10 ln π»/π πΌπ π·ππ π§ππ = πππΏππ π§ππ = π2π50 × 2 × 10−7 ln 1 × 1000 ο/ππ π·ππ 1 π§ππ = π0.0628 ln ο/ππ π·ππ 9 IMPEDANCE OF DISTRIBUTION LINE: Transposed line: π§π = ππ + ππ × 2 × 10−7 ln πΊππ· ο/π πΊππ πΊππ· π§π = ππ + π0.0628 ln ο/ππ πΊππ Untransposed line: 1 π§ππ = ππ + π0.0628 ln ο/ππ πΊππ π 1 π§ππ = π0.0628 ln ο/ππ π·ππ a 50 ππ b ππππ πππ = πππ πππ c πππ πππ πππ πππ πππ πππ 10 IMPEDANCE OF DISTRIBUTION LINE: We can calculate the impedance of distribution line if they are transposed or untransposed. In case of transposed line, we consider one impedance only. 50 ππ In case of untransposed line, we get a matrix of size π × π. However, these formulas are applicable if there is no ground return current. In 1926, John Carson has developed a set of equations for computing the selfand mutual impedances of lines considering the return path of current through ground. 11 UNTRANSPOSED AND UNBALANCED SYSTEM WITH GROUND RETURN PATH: 50 ππ A line with two conductors π and π carrying currents πΌπ and πΌπ with the remote ends of the conductors tied to ground. A fictitious “dirt” conductor carrying current πΌπ is used to present the return path. The voltage between conductor π and ground – πππ = π§ππ β πΌπ + π§ππ β πΌπ + π§ππ β πΌπ − π§ππ β πΌπ − π§ππ β πΌπ − π§ππ β πΌπ 12 UNTRANSPOSED AND UNBALANCED SYSTEM WITH GROUND RETURN PATH: 50 ππ πππ = π§ππ β πΌπ + π§ππ β πΌπ + π§ππ β πΌπ − π§ππ β πΌπ − π§ππ β πΌπ − π§ππ β πΌπ πΌπ = −πΌπ − πΌπ πππ = π§ππ β πΌπ + π§ππ β πΌπ + π§ππ β −πΌπ − πΌπ − π§ππ β −πΌπ − πΌπ − π§ππ β πΌπ − π§ππ β πΌπ πππ = π§ππ + π§ππ − π§ππ − π§ππ πΌπ + π§ππ + π§ππ − π§ππ − π§ππ πΌπ 13 EQUIVALENT PRIMITIVE CIRCUIT: 50 ππ πππ = π§ππ + π§ππ − π§ππ − π§ππ πΌπ + π§ππ + π§ππ − π§ππ − π§ππ πΌπ πππ = παππ πΌπ + παππ β πΌπ παππ = π§ππ + π§ππ − π§ππ − π§ππ παππ = π§ππ + π§ππ − π§ππ − π§ππ 14 EQUIVALENT PRIMITIVE CIRCUIT: 50 ππ Primitive self impedance παππ = π§ππ + π§ππ − π§ππ − π§ππ παππ = ππ + ππ₯ππ + ππ + ππ₯ππ − ππ₯ππ − ππ₯ππ παππ = ππ + ππ + ππ × 2 × 10−7 1 1 1 1 ln + ln − ln − ln πΊππ π πΊππ π π·ππ π·ππ παππ = ππ + ππ + ππ × 2 × 10−7 ln 1 π·ππ β π·ππ + ln πΊππ π πΊππ π 15 EQUIVALENT PRIMITIVE CIRCUIT: 50 ππ Primitive mutual impedance παππ = π§ππ + π§ππ − π§ππ − π§ππ παππ = ππ₯ππ + ππ + ππ₯ππ − ππ₯ππ − ππ₯ππ παππ = ππ + ππ × 2 × 10−7 ln 1 1 1 1 + ln − ln − ln π·ππ πΊππ π π·ππ π·ππ −7 ln πΰ·’ = π + ππ × 2 × 10 ππ π π·ππ β π·ππ 1 + ln π·ππ πΊππ π 16 CARSON’S EQUATIONS: The earth is a conductor of enormous dimension, and it has non-uniform conductivity. Thus, earth current distribution is non-uniform. 50 ππ To calculate the impedance of conductor with earth return, it is necessary to know the distribution of current returning through earth. ππ , π·ππ , π·ππ , πΊππ π are not available. In 1926, Carson developed a technique to obtain the self and mutual impedance for an arbitrary number of overhead conductors. Carson’s equations consist of infinite series which is difficult to calculate. Therefore, many people have modified these equations so that we can easily use them for distribution system studies. 17 MODIFIED CARSON’S EQUATIONS: Cond 1 Cond 2 In the modified Carson’s formula, the image of conductor is assumed for ground current. However, the height of image conductor is not same as height of conductor. β1 β2 Earth Equivalent height of image conductor depends upon resistivity of earth and frequency of operation. Images of conductors are considered at distance π·π from the earth surface. π·π π·π π·π is called as equivalent depth of earth return. Equivalent Earth return conductors 18 MODIFIED CARSON’S EQUATIONS: Cond 1 π π·π = 2160 π Cond 2 ππ‘ Equivalent depth of return in meter – β1 β2 π π·π = 0.305 × 2160 π π Earth If frequency increases, the equivalent depth of earth return goes down. π·π π·π Soil π Damp Earth 100 Dry Earth 1000 Sea Water 1 Equivalent Earth return conductors 19 MODIFIED CARSON'S EQUATIONS: Primitive self impedance παππ = ππ + ππ + ππ × 2 × 10−7 ln 1 π·ππ β π·ππ + ln πΊππ π πΊππ π 50 ππ Primitive mutual impedance παππ = ππ + ππ × 2 × 10−7 ln π·ππ β π·ππ 1 + ln π·ππ πΊππ π π π·π = 0.305 × 2160 π π Modified equations: παππ = ππ + 9.86 × 10 π + ππ × 2 × 10 −7 −7 παππ = 9.86 × 10−7 π + ππ × 2 × 10−7 ln 1 ln + ln π·π πΊππ π 1 + ln π·π π·ππ 20 MODIFIED CARSON'S EQUATIONS (50 HZ AND KM): Modified equations: 1 −7 −7 α πππ = ππ + 9.86 × 10 π + ππ × 2 × 10 ln + ln π·π ο/π πΊππ π 50 ππ 1 −7 −7 α πππ = ππ + 9.86 × 10 × 50 + π2π × 50 × 2 × 10 ln + ln π·π × 1000 ο/ππ πΊππ π παππ = ππ + 0.0493 + π0.0628 ln π π·π = 0.305 × 2160 π 1 + ln π·π πΊππ π ο/ππ π 1 παππ = ππ + 0.0493 + π0.0628 ln + 6.843 πΊππ π ο/ππ 21 MODIFIED CARSON'S EQUATIONS (50 HZ AND KM): Modified equations: παππ = 9.86 × 10−7 π + ππ × 2 × 10−7 ln παππ = 9.86 × 10−7 1 + ln π·π π·ππ × 50 + π2π × 50 × 2 × παππ = 0.0493 + π0.0628 ln π π·π = 0.305 × 2160 π 1 + ln π·π π·ππ 10−7 ο/π 1 ln + ln π·π × 1000 ο/ππ π·ππ ο/ππ π 1 παππ = 0.0493 + π0.0628 ln + 6.843 π·ππ ο/ππ 22 MODIFIED CARSON'S EQUATIONS (50 HZ AND KM): παππ = ππ + 0.0493 + π0.0628 ln 1 + 6.843 πΊππ π 1 παππ = 0.0493 + π0.0628 ln + 6.843 π·ππ ο/ππ ο/ππ Size of primitive impedance matrix = π × π = 4 × 4 παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ 23 MODIFIED CARSON'S EQUATIONS (50 HZ AND KM): παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ 24 “KRON” REDUCTION: For most of the cases, it is required to reduce the primitive impedance matrix to a 3x3 matrix consisting of the self and mutual impedances for the three phases. One standard method of reduction is the “Kron” reduction. 50 ππ It is assumed that the line has a multi-grounded neutral. 25 “KRON” REDUCTION: 50 ππ 26 “KRON” REDUCTION: 50 ππ In partitioned form – ππππ πππ = π′πππ π′ππ παππ + παππ παππ παππ πΌπππ πΌπ 27 “KRON” REDUCTION: ππππ πππ = π′πππ π′ππ παππ + παππ Two separate equations – ππππ = π′πππ + παππ παππ παππ πΌπππ πΌπ 50 ππ πΌπππ + παππ πΌπ πππ = π′ππ + παππ πΌπππ + παππ πΌπ As the neutral is grounded, the voltages πππ and π′ππ are equal to zero. 0 = 0 + παππ πΌπππ + παππ πΌπ ∴ πΌπ = − παππ −1 παππ πΌπππ 28 “KRON” REDUCTION: πΌπ = − παππ −1 παππ πΌπππ Once the line currents have been computed it is possible to determine the current flowing in the neutral conductor. 50 ππ The “neutral transformation matrix” is defined as – π‘π = − παππ −1 παππ πΌπ = π‘π πΌπππ ππππ = π′πππ + παππ πΌπππ + παππ πΌπ Substituting πΌπ in the equation for ππππ ππππ = π′πππ + παππ − παππ παππ −1 παππ πΌπππ ππππ = π′πππ + ππππ πΌπππ ππππ = παππ − παππ παππ −1 παππ 29 “KRON” REDUCTION: 50 ππ ππππ = παππ − παππ παππ −1 παππ 30 EXAMPLE: Determine the phase impedance matrix. The phase conductors are 26/7 ACSR, and neutral conductor is 4/06/1 ACSR. Conductor GMR (m) 26/7 ACSR 0.0074 Resistance (ο/km) 0.1901 4/06/1 ACSR 0.0025 0.3679 50 ππ π·ππ = 0.76 π, π·ππ = 1.37 π, π·ππ = 2.13 π·ππ = 1.72 π, π·ππ = 1.30 π, π·ππ = 1.52 31 EXAMPLE: παππ = ππ + 0.0493 + π0.0628 ln παππ = 0.0493 + π0.0628 ln 1 + 6.843 πΊππ π ο/ππ 1 + 6.84350 ππ ο/ππ π·ππ παππ = 0.2394 + π0.7378 παππ = 0.0493 + π0.4467 π·ππ = 0.76 π, π·ππ = 1.37 π, π·ππ = 2.13 π·ππ = 1.72 π, π·ππ = 1.30 π, π·ππ = 1.52 Conductor GMR (m) 26/7 ACSR 0.0074 Resistance (ο/km) 0.1901 4/06/1 ACSR 0.0025 0.3679 32 EXAMPLE: Primitive impedance matrix: παππ παππ = παππ παππ = παππ παππ παππ παππ παππ παππ παππ παππ 50 ππ παππ παππ παππ παππ 0.2394 + π0.7378 0.0493 + π0.4467 0.0493 + π0.3819 0.0493 + π0.3953 0.0493 + π0.4467 0.2394 + π0.7378 0.0493 + π0.4097 0.0493 + π0.4129 0.0493 + π0.3819 0.0493 + π0.4097 0.2394 + π0.7378 0.0493 + π0.4031 0.0493 + π0.3953 0.0493 + π0.4129 0.0493 + π0.4031 0.4172 + π0.8060 33 EXAMPLE: = 0.2394 + π0.7378 0.0493 + π0.4467 0.0493 + π0.3819 0.0493 + π0.3953 0.0493 + π0.4467 0.2394 + π0.7378 0.0493 + π0.4097 0.0493 + π0.4129 0.0493 + π0.3819 0.0493 + π0.4097 0.2394 + π0.7378 0.0493 + π0.4031 0.0493 + π0.3953 0.0493 + π0.4129 0.0493 50 ππ + π0.4031 0.4172 + π0.8060 παππ 0.2394 + π0.7378 0.0493 + π0.4467 0.0493 + π0.3819 = 0.0493 + π0.4467 0.2394 + π0.7378 0.0493 + π0.4097 0.0493 + π0.3819 0.0493 + π0.4129 0.2394 + π0.7378 παππ 0.0493 + π0.3953 = 0.0493 + π0.4129 0.0493 + π0.4031 παππ = 0.0493 + π0.3953 0.0493 + π0.4129 0.0493 + π0.4031 παππ = 0.4172 + π0.8060 ππππ = παππ − παππ παππ −1 παππ 34 EXAMPLE: 50 ππ 35 SHUNT ADMITTANCE OF OVERHEAD LINES: Shunt admittance of distribution lines is very small and many times it is neglected. However, if the feeder length is long, we can consider shunt admittance for accurate calculations. Shunt conductance is generally neglected50 asππ it is very small. So, admittance will consist of only the capacitive part. For calculation of the capacitor, we need to take ground effect into account. The method of conductors and their images is employed in the calculation of the shunt capacitance of overhead lines. This is the same concept as Carson’s equations. 36 SHUNT ADMITTANCE OF OVERHEAD LINES: It is assumed that – ππ′ = −ππ 50 ππ ππ′ = −ππ 37 SHUNT ADMITTANCE OF OVERHEAD LINES: Equations 1 and 2 in matrix form – 50 ππ πππ , πππ and πππ are called potential coefficients. 38 SHUNT ADMITTANCE OF OVERHEAD LINES: 50 ππ 39 SHUNT ADMITTANCE OF OVERHEAD LINES: For overhead line with n conductors, the primitive coefficient matrix of nxn elements can be constructed. 50 ππ Partitioned primitive matrix - 40 SHUNT ADMITTANCE OF OVERHEAD LINES: As the neutral conductor is grounded, the matrix can be reduced using the “Kron reduction” method to an n-phase × n-phase phase potential coefficient matrix. 50 ππ The inverse of the potential coefficient matrix will give the n-phase × n-phase capacitance matrix - Neglecting the shunt conductance, the phase shunt admittance matrix is given by - 41 TAPE SHIELDED CABLE: 50 ππ The cable consists of a central “phase conductor” covered by a thin layer of nonmetallic semiconducting screen to which the insulating material is bonded. The insulation is covered by a semiconducting insulation screen. The shield is bare copper tape helically applied around the insulation screen. An insulating “jacket” encircles the tape shield. 42 TAPE SHIELDED CABLE: Parameters of the tape-shielded cable are 50 ππ ππ = the diameter of phase conductor ππ = the outside diameter of the tape shield πππ = the outside diameter over jacket π = the thickness of copper tape shield 43 TAPE SHIELDED CABLE: 50 ππ Modified Carson’s equations can be used to calculate the self impedance of the phase conductors and the tape-shield and also the mutual impedance. 1 παππ = ππ + 0.0493 + π0.0628 ln + 6.843 πΊππ π 1 παππ = 0.0493 + π0.0628 ln + 6.843 π·ππ ο/ππ ο/ππ 44 SERIES IMPEDANCE OF UNDERGROUND CABLES: 50 ππ This configuration will result in 7x7 primitive impedance matrix. For ungrounded circuits, neutral conductor is absent and the primitive impedance matrix will be 6x6. 45 SERIES IMPEDANCE OF UNDERGROUND CABLES: a b l c 50 ππ m Dab Dbc παππ = ππ + 0.0493 + π0.0628 ln παππ = 0.0493 + π0.0628 ln n 1 + 6.843 πΊππ π 1 + 6.843 π·ππ ο/ππ ο/ππ 46 SERIES IMPEDANCE OF UNDERGROUND CABLES: a l παπππ παππ παππ παππ = παππ παππ παππ παππ παππ παππ παππ παππ b παππ παππ παππ παππ παππ παππ παππ 50 ππ α α α πππ m πππ πππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ παππ c παππ παππ n παππ παππ Using “Kron” reduction: ππππ = παππ − παππ παππ ππππ πππ = πππ πππ πππ πππ πππ −1 παππ πππ πππ πππ 47
