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 = ωL = 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 = ωC = 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 = jωC = 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 = ωC = 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.