Question

A 60Hz three-phase three-wire overhead line has solid cylindrical conductors arranged in the form of an equilateral triangle with 4ftconductor spacing. The conductor diameter is 0.5in. Calculate the positive-sequence inductance in $\frac{H}{m}$and the positive-sequence inductive reactance in $\frac{\Omega }{km}$.

Short answer

The positive sequence inductance and positive sequence inductive reactance are $1.10\times {10}^{-6}\frac{H}{m}$and $0.414\frac{\Omega }{km}$respectively.

Step-by-step solution

 Write the given data by the question.

The frequency of the system,$f=60Hz$

The diameter of the conductor,$d=0.5in$

The spacing between the conductors,$D=4ft$

 Determine the formulas to calculate the positive sequence inductance inHm and positive sequence inductive reactance inΩkm.

The expression to calculate the radius of the conductor is given by,

$r=\frac{d}{2}$ ……. (1)

The expression to calculate the positive sequence inductance is given by,

$L=2\times {10}^{-7}\ln \left(\frac{D}{0.7788r}\right)$ …… (2)

The expression to calculate the positive sequence inductive is given by,

…… (3)

The conversion from foot to meter.

$1ft=0.3048m$

The conversion from inches to meter.

$1ft=0.025m$

Calculate the positive sequence inductance and inductive reactance. 

Calculate the radius of the conductor.

Substitute $d=0.5in$for $d=0.5in$into equation (1).

$$\begin{gathered} r=\frac{0.5}{2} \\ r=0.25in \end{gathered}$$

Calculate the positive sequence inductance.

Substitute 4ft for D and 0.25in for r into equation (2).

$$\begin{gathered} L=2\times {10}^{-7}\ln \left(\frac{4\times 0.3048}{0.7788\times 0.25\times 0.0254}\right) \\ L=2\times {10}^{-7}\ln \left(246.533\right) \\ L=2\times {10}^{-7}\left(5.507\right) \\ L=1.10\times {10}^{-6}\frac{H}{m} \end{gathered}$$

Calculate the positive sequence inductive reactance.

Substitute $1.10\times {10}^{-6}\frac{H}{m}$ for L and 60Hz for finto equation (3).

$$\begin{gathered} X=2\pi \times 60\times 1.10\times {10}^{-6} \\ X=414.69\times {10}^{-6} \\ X=0.414\frac{\Omega }{km} \end{gathered}$$

Hence the positive sequence inductance and positive sequence inductive reactance are $1.10\times {10}^{-6}\frac{H}{m}$and $0.414\frac{\Omega }{km}$respectively.