Question

(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

$$\begin{gathered} \mathit{x}\mathbf{=}\mathit{k}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{D}}{\mathbf{r}\mathbf{\text{'}}}\mathit{o}\mathit{h}\mathit{m}\mathit{s}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{m}\mathit{i}\mathit{l}\mathit{e}\mathbf{}\mathit{p}\mathit{h}\mathit{a}\mathit{s}\mathit{e} \\ \mathbf{}\mathbf{}\mathbf{=}{\mathit{x}}_{\mathbf{d}}\mathbf{+}{\mathit{x}}_{\mathbf{a}} \end{gathered}$$

Where,${\mathit{X}}_{\mathbf{d}}\mathbf{=}\mathit{k}\mathit{l}\mathit{o}\mathit{g}\mathit{D}$ is the inductive reactance spacing factor in ohms per mile

${\mathit{X}}_{\mathbf{a}}\mathbf{=}\mathit{k}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{1}}{\mathbf{r}\mathbf{\text{'}}}$ is the inductive reactance at 1 ft spacing in Ohm’s mile per mile

$\mathit{k}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{657}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{f}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2794}$at $\mathbf{60}\mathit{H}\mathit{z}$

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

Short answer

Answer

(a) The inductive reactance spacing factor in ohms per mile is$x_d=0.2794\log \left(D\right)$, the inductive reactance at 1ft spacing in Ohm’s mile per mile $x_a=0.2794\log \left(\frac{1}{r\text{'}}\right)$and value of k at 60Hz frequency is 0.2794.

(b) The value of the per mile per phase reactance when the spacing is doubled is $0.722\frac{\Omega }{mi}perphase$.

Step-by-step solution

Write the given data by the question.

The inductive reactance of the line per mile per phase,$x=k\log \frac{D}{r\text{'}}$

The radius of the conductor, $r=0.06677ft$

The spacing between the conductor, $D=10ft$

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

Determine the formula to prove the algorithm and calculate the inductive reactance per mile per phase.

The expression to calculate the total inductance is given by,

$\mathit{L}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathit{I}\mathit{n}\left(\frac{D}{r\text{'}}\right)\frac{\mathbf{H}}{\mathbf{m}}$ …… (1)

The expression to calculate inductive reactance is given by,

$\mathit{X}\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{f}\mathit{L}$ …… (2)

The expression to calculate the change in the reactance is given by,

$\mathbf{∆}\mathit{X}\mathbf{=}\left(\frac{X_2-X_1}{X_1}\right)\mathbf{\times }\mathbf{100}$ …… (3)

Here, ${\mathit{X}}_{\mathbf{1}}$is the reactance when the spacing is $\mathbf{10}\mathbf{}\mathit{f}\mathit{t}$and ${\mathit{X}}_{\mathbf{2}}$is the reactance when the spacing is $\mathbf{20}\mathbf{}\mathit{f}\mathit{t}$.

The conversion frommeterto miles.

$\mathbf{1}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00621}\mathit{m}\mathit{i}$

The conversion from $\mathit{I}\mathit{n}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to $\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

$\mathit{I}\mathit{n}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3026}\mathit{l}\mathit{o}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Determine the inductive reactance spacing factor in ohms per mile and inductive reactance at 1ft spacing in Ohm’s mile per.

(a)

Calculate the inductive reactance.

Substitute $2k{10}^{-7}In\left(\frac{D}{r\text{'}}\right)\frac{H}{m}$for L in the equation (2).

$$\begin{gathered} X=2\pi f\times 2\times {10}^{-7}In\left(\frac{D}{r\text{'}}\right) \\ X=4\mathrm{\pi \pi }\times {10}^{-7}\mathrm{In}\left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right) \\ \mathrm{X}=4\mathrm{\pi }\times {10}^{-7}\mathrm{f}\mathrm{In}\left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)\frac{\mathrm{\Omega }}{\mathrm{m}} \end{gathered}$$

Convert the inductive reactance from per meter per phase to per mile per phase.

$X=4\mathrm{\pi }\times {10}^{-7}\times \mathrm{f}\hspace{0.17em}\mathrm{In}\left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)\times \frac{1}{0.000621}$

Substitute $2.3036\log \left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)$for $\mathrm{In}\left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)$into above equation.

$$\begin{gathered} X=4\mathrm{\pi }\times {10}^{-7}\times \mathrm{f}\times 2.3026\times \log \left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)\times \frac{1}{0.000621} \\ \mathrm{X}=2.023\times \mathrm{f}\times 2.3026\times \log \left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right) \\ \mathrm{X}=4.658\times {10}^{-3}\times \mathrm{f}\log \left(\frac{\mathrm{D}}{\mathrm{r}\text{'}}\right)\frac{\mathrm{\Omega }}{\mathrm{mi}}\mathrm{per}\mathrm{phase} \end{gathered}$$

Substitute 60Hz for f in the above expression.

$$\begin{gathered} X=4.658\times {10}^{-3}\times 60\times \log \left(\frac{D}{r\text{'}}\right) \\ X=0.2794\log \left(\frac{D}{r\text{'}}\right)\frac{\Omega }{mi}perphase \end{gathered}$$

Now comparing the above equation with $x=k\log \frac{D}{r\text{'}}$,

The value of role="math" localid="1656423609234" $k=0.2794at60Hz$.

Solve further as,

$$\begin{gathered} k=0.2794\log \left(\frac{D}{r\text{'}}\right) \\ X=0.2794\log \left(D\right)+0.2794\log \left(\frac{1}{r\text{'}}\right) \end{gathered}$$

Hence, the inductive reactance spacing factor in ohms per mile is$x_d=0.2794\log \left(D\right)$, the inductive reactance at 1ft spacing in Ohm’s mile per mile $x_a=0.2794\log \left(\frac{1}{r\text{'}}\right)$and value of k at 60Hz frequency is 0.2794.

Calculate the inductive reactance with the spacing between the conductors is  and change in inductive reactance when spacing is double.

(b)

The derived expression for the inductive reactance in the step (3),

$X=0.2794\log \left(\frac{D}{r\text{'}}\right)\frac{\Omega }{mi}perphase$

Substitute 0.7788r for r'in the above expression.

$X=0.2794\log \left(\frac{D}{0.7788r}\right)\frac{\Omega }{mi}perphase$

Substitute 10ft for D and 0.06677 ft for r into above expression.

$$\begin{gathered} X_1=0.2794\left(\frac{10}{0.7788\times 0.06677}\right) \\ {X}_1=0.2795\times \log \left(192.307\right) \\ {X}_1=0.2795\times 2.283 \\ {X}_1=0.637\frac{\Omega }{mi}perphase \end{gathered}$$

The spacing between the conductor is doubled.

Calculate the inductive reactance.

Substitute $D=20ft$for D and 0.06677 ft for r into role="math" localid="1656424528922" $$\begin{gathered} X=0.2794\log \left(\frac{D}{0.7788r}\right)\frac{\Omega }{mi}perphase \\ {X}_2=0.2794\log \left(\frac{20}{0.7788\times 0.0667}\right) \\ {X}_2=0.2795\times \log \left(384.614\right) \\ {X}_2=0.2795\times 2.585 \\ {X}_2=0.722\frac{\Omega }{mi}perphase \end{gathered}$$

Therefore, the value of the per mile per phase reactance when the spacing is doubled is $0.722\frac{\Omega }{mi}perphase$.