Tutorial questions and solutions for EEPS02 Overhead Lines

Tutorial questions and solutions for EEPS02 Overhead Lines

Q2. A 50Hz single-phase line consists of two parallel conductors 30cm apart. If each conductor has a diameter of 4mm, calculate the inductive reactance of a 500m length of the line.

Solution:

Using equation 8 in lecture note, noting that we assume the conductor is a round conductor (given as diameter of 4mm in the question). Equation 8 is about how to calculate the inductance of ONE conductor in a circuit. The inductance for the line is the sum of the inductance of the go and the return conductors.

Q3. A 220kV, 3-phase, 50Hz overhead line is 50km long and consists of 3 conductor groups a,b,c arranged in horizontal flat formation, with spacings ab=5m, bc=5m, and ac=10m. Each group consists of two round sub-conductors spaced 20cm apart, and the radius of each sub-conductor is 2cm. The line is fully transposed.

(a) Calculate the positive-sequence and zero-sequence inductive reactance of the line per phase

(b) Calculate the positive-sequence and zero-sequence capacitive reactance of the line per phase.

Assume that the lines are at a height of 15m above a perfectly-conducting earth.

Solution:

For sub-conductors or the bundled conductor or stranded conductor please see the lecture note page 8 to help in understanding the mutual GMD (geometric mean distance between the conductor group and the imaging group) and the self GMD (geometric mean distance of the conductor group).

(a). Refer to lecture note page 15, for the fully transposed line the positive sequence inductance is calculated as

For the phase conductor which consists of two sub-conductors (r = 0.02m) spaced 20cm apart, the Geometric Mean Radius of the phase conductor is

Since the distance between the phase conductors is much bigger than the spacing between the sub-conductors, the Geometric Mean Spacing of the three phase conductors is calculated as

For the fully transposed line the positive-sequence inductance is calculated as

For the fully transposed line the zero-sequence inductance is calculated as

The distance between conductor and imaging conductor is calculated as

GMD between the conductor group and the image conductor group is

GMD of the conductor group is

(b). In calculating capacitance, there is no such a thing as r, which is used to include self -inductance of the conductor. Therefore

For the fully transposed line the positive-sequence capacitance is calculated as

For the fully transposed line the zero-sequence capacitance is calculated as

Q4. A 3-phase fully-transposed overhead line is to be designed with a geometric mean spacing between phases of 8m. Calculate the inductance per km, if a single round conductor with a cross-sectional area of 200mm2 is used for each phase. What will be the percentage reduction in inductance if

(a) The phase conductors are split into 2 (twin) sub-conductors, with a spacing of 150mm?

(b) The phase conductors are split into 4 (quad) sub-conductors, arranged at the corner of a square of side 150mm?

Assume a constant design current density.

Solution:

Under the balanced three-phase situation, analysis is carried out for a single-phase system and positive-sequence inductance L1 is used as the series impedance of the line.

For round conductor wire,

(a). The same current density means the same cross-sectional area. For 2 sub-conductors with spacing of 0.15m,

Percentage reduction

(b). The same current density means the same cross-sectional area. For 4 sub-conductors (quad) arranged as a square of side 0.15m,

Percentage reduction

Q5. An 11kV unearthed single-phase overhead line is 12km long and consists of two conductors of 12mm diameter spaced 0.8m apart in flat formation 6m above the ground. Derive the elements of an equivalent capacitance network which will represent the line.

Solution:

Q6. A telephone conductor (t) runs parallel to a 3-phase 50Hz power line (a,b,c) as shown. If the power line carries a balanced 3-phase current of 1000A. Calculate the voltage (in magnitude and phase angle relative to phase a) induced in a 1km length of the telephone conductor. Assume a perfectly conducting earth.

Solution:

The mutual inductance between t and a,b,c will produce the inductive induced voltage will appear on the telephone wire:

The distances between conductor t and phase conductors are

The distances between conductor t and images of phase conductors are

|C21|*12000=0.0359uF

(|C11|-|C21|)*12000

=0.0647uF

(|C22|-|C12|)*12000

=0.0647uF

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