(a) In practice, one deals with the capacitive reactance of the line in ohms $\cdot$ mi to neutral. Show that Eq. (4.9.15) of the text can be rewritten as

$$\begin{array}{cc}{X}_C& =k^{\prime }\log \frac{D}{r}\text{ohms}\cdot \text{mi to netural}\\ =x_d^{\prime }+x_a^{\prime }\end{array}$$

where $x_d^{\prime }=k^{\prime }\log D$ is the capacitive reactance spacing factor

$x_a^{\prime }=k^{\prime }\log \frac{1}{r}$ is the capacitive reactance at $1-ft$ spacing

$k^{\prime }=(4.1\times {10}^6)/f=0.06833\times {10}^6$ at $f=60Hz$

(b) Determine the capacitive reactance in $\Omega \cdot mi$. for a single-phase line of Problem 4.14. If the spacing is doubled, how does the reactance change?

Eq. (4.9.15)

Problem 4.14

(a) In practice, one deals with the inductive reactance of the line per phase per mile and use the logarithm to the base 10. Show that Eq. (4.5.9) of the text can be rewritten as

$$\begin{array}{cc}x& =k\log \frac{D}{r^{\prime }}\text{ohms per mile per phase}\\ =x_d+x_a\end{array}$$

where $x_d=k\log D$ is the inductive reactance spacing factor in ohms per mile $x_a=k\log \frac{1}{r^{\prime }}$ is the inductive reactance at 1 -ft spacing in ohms per mile $k=4.657\times {10}^{-3}f=0.2794$ at $60Hz$

(b) Determine the inductive reactance per mile per phase at $60Hz$ for a single-phase line with phase separation of $10ft$ and conductor radius of $0.06677ft$. If the spacing is doubled, how does the reactance change?

Eq. (4.5.9)

Transcribed Image Text:

Can = qa In F/m line-to-neutral