ECE 476 { Power System Analysis Fall 2017 Homework 3 Reading: Chapter 4 from textbook. In-class quiz: Thursday, September 21, 2017 Problem 1. A 60-Hz single-phase, two-wire overhead line has solid cylindrical copper conductors with 1.5 cm di- ameter. The conductors are arranged in a horizontal configuration with 0.5 m spacing. Calculate in mH/km: a) The inductance of each conductor due to internal flux linkages only. The internal inductance of each conductor is given by (4.4.10): 1 1000 mH 1000 m L = × 10−7 H/m = 0:05 mH/km per conductor: int 2 1 H 1 km b) The inductance of each conductor due to both internal and external flux linkages. The inductance for each conductor due to both internal and external flux linkages is given by (4.4.21): D L = L = 2 × 10−7 ln x y r0 ! 0:5 H 1000 mH 1000 m = 2 × 10−7 ln : −1=4 0:015 e 2 m 1 H 1 km = 0:8899 mH/km per conductor c) The total inductance of the line. The total inductance of the single phase circuit is L = Lx + Ly = 1:780 mH/km per circuit: Problem 2. Rework Problem 1 if the diameter of each conductor is: a) Increased by 20% to 1.8 cm. 1 1000 mH 1000 m L = × 10−7 H/m = 0:05 mH/km per conductor: int 2 1 H 1 km D L = L = 2 × 10−7 ln x y r0 ! 0:5 H 1000 mH 1000 m = 2 × 10−7 ln −1=4 0:018 e 2 m 1 H 1 km = 0:8535 mH/km per conductor: L = Lx + Ly = 1:707 mH/km per circuit: 1 b) Decreased by 20% to 1.2 cm, without changing the phase spacing. 1 1000 mH 1000 m L = × 10−7 H/m = 0:05 mH/km per conductor: int 2 1 H 1 km D L = L = 2 × 10−7 ln x y r0 ! 0:5 H 1000 mH 1000 m = 2 × 10−7 ln −1=4 0:012 e 2 m 1 H 1 km = 0:9346 mH/km per conductor: L = Lx + Ly = 1:869 mH/km per circuit: Compare the results with those of Problem 1. • Lint is independent of conductor diameter. • A 20% increase in conductor diameter results in a 4.1% decrease in inductance. • A 20% decrease in conductor diameter results in a 5% increase in inductance. Problem 3. A 60-Hz three-phase, three-wire overhead line has solid cylindrical conductors arranged in the form of an equilateral triangle with 4 ft conductor spacing. Conductor diameter is 0.5 in. Calculate the positive-sequence inductance in H/m and the positive-sequence inductive reactance in Ω/km. The line inductance is given by D L = 2 × 10−7 ln r0 ! 4 = 2 × 10−7 ln = 1:101 × 10−6 H/m per phase: −1=4 0:5 1 ft e 2 12 in Then the inductive reactance is Ω 1000 m X = !L = 2π60 1:101 × 10−6 = 0:4153 Ω/km: L m km Problem 4. Calculate the capacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the three-phase line in Problem 3 (stated above). Neglect the effect of the earth plane. The capacitance-to-neutral per line length is −12 2π 2π(8:854 × 10 ) −11 Can = = = 1:058 × 10 F/m line-to-neutral: ln(D=r) 4 ln 0:25=12 Then, the admittance-to-neutral per line length is S 1000 m B = !C = 2π60(1:058 × 10−11) = 3:989 × 10−6 S/km line-to-neutral: an an m 1 km Problem 5. Rework Problem 4 if the phase spacing is: a) Increased by 20% to 4.8 ft. The capacitance-to-neutral per line length is −12 2π 2π(8:854 × 10 ) −11 Can = = = 1:023 × 10 F/m line-to-neutral: ln(D=r) 4:8 ln 0:25=12 Then, the admittance-to-neutral per line length is S 1000 m B = j!C = j2π60(1:023 × 10−11) = 3:855 × 10−6 S/km line-to-neutral: an an m 1 km b) Decreased by 20% to 3.2 ft. The capacitance-to-neutral per line length is −12 2π 2π(8:854 × 10 ) −11 Can = = = 1:105 × 10 F/m line-to-neutral: ln(D=r) 3:2 ln 0:25=12 Then, the admittance-to-neutral per line length is S 1000 m B = !C = j2π60(1:105 × 10−11) = 4:166 × 10−6 S/km line-to-neutral: an an m 1 km Compare the results with those of Problem 4. • A 20% increase in phase spacing results in a 3.4% decrease in shunt capacitance and admittance. • A 20% decrease in phase spacing results in a 4.4% increase in shunt capacitance and admittance..