Question

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

$$\begin{gathered} X_c=k\text{'}\log \frac{D}{r}ohms-mi \\ =x{\text{'}}_d+x{\text{'}}_a \end{gathered}$$

Where $x{\text{'}}_d=k\text{'}\log D$is the capacitive reactance spacing factor

$x{\text{'}}_d=k\text{'}log\frac{1}{r}$is the capacitive reactance at 1-ft spacing

role="math" localid="1655202080897" $k\text{'}=\left(4.1\times {10}^6\right)/f=0.06833\times {10}^6$at f=60Hz

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

Short answer

(a) The capacitive reactance $X_c$is $0.06833\times {10}^6\log \frac{D}{r}$.

(b) The value of capacitive reactance is $\Omega -14mi869\times {10}^4$and the capacitive reactance is increased by 13.78% when the spacing is doubled.

Step-by-step solution

Write the given data of the question

The capacitive reactance at 1-ft spacing is given by the expression $x{\text{'}}_d=k\text{'}\log \frac{1}{r}$.

The capacitive reactance spacing factor is given by $x{\text{'}}_d=k\text{'}\log D$.

Write the formulae for capacitance to neutral and admittance to neutral;

The capacitance to neutral is given by,

$C_n=\frac{\epsilon }{In\left({\displaystyle \frac{D}{r}}\right)}$ …… (1)

Here D is the conductor spacing, r is the radius of each solid cylinder and $C_n$is the capacitance to neutral.

The capacitive reactance is given by,

$X_c=\frac{1}{\omega C}$

Here $X_c$is the capacitive reactance, $\omega$is the angular frequency and C is the capacitance.

Calculate the capacitive reactance.

(a)

The capacitance to neutral is given by,

$C_n=\frac{2\mathrm{\pi \epsilon }}{In\left({\displaystyle \frac{D}{r}}\right)}F/m$

Here D is the conductor spacing, r is the radius of each solid cylinder and $C_n$is the capacitance to neutral.

Convert the above equation in F/mi units.

Substitute $8.85\times {10}^{-12}F/m$for $\epsilon$in the above equation.

$$\begin{gathered} C_n=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{D}{r}}\right)}\times \frac{1}{1609}F/mi \\ =\frac{8.946\times {10}^{-8}}{In\left({\displaystyle \frac{D}{r}}\right)}F/mi \end{gathered}$$

Now convert the above equation in terms of log to base 10.

$$\begin{gathered} C_n=\frac{8.946\times {10}^{-8}}{2.302\log \left({\displaystyle \frac{D}{r}}\right)}F/mi \\ =\frac{3.886\times {10}^{-8}}{\log \left({\displaystyle \frac{D}{r}}\right)} \end{gathered}$$

The capacitive reactance at 60Hz is given by,

$$\begin{gathered} X_c=\frac{1}{\omega C_n} \\ =\frac{1}{2{\mathrm{\pi fC}}_{\mathrm{n}}} \end{gathered}$$

Substitute $\frac{3.886\times {10}^{-8}}{\log \left(\begin{array}{c}D\\ r\end{array}\right)}$for $C_n$in the above equation.

$$\begin{gathered} X_c=\frac{1}{2\mathrm{\pi f}{\displaystyle \frac{3.886\times {10}^{-8}}{\log \left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}}} \\ =\frac{4.1\times {10}^6}{\mathrm{f}}\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \\ =\frac{4.1\times {10}^6}{60\mathrm{Hz}}\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \\ =0.06833\times {10}^6\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \end{gathered}$$

Therefore the capacitive reactance is $\frac{4.1\times {10}^6}{f}\log \left(\frac{D}{r}\right)$.

Calculate the increment in capacitive reactance.

(b)

The capacitive reactance in terms of log is given by,

$X_c=k\text{'}\Omega -\log mi\frac{D}{r}$to neutral.

$X_c=x{\text{'}}_d+x{\text{'}}_a$

Here

$x{\text{'}}_d=k\text{'}\log D$is the capacitive reactance spacing factor.

$x{\text{'}}_a=k\text{'}\log \frac{1}{r}$is the capacitive reactance at 1-ft spacing.

The value of k' is,

$k\text{'}=\frac{4.1\times {10}^6}{f}$

Substitute 60Hz in the above equation.

$$\begin{gathered} k\text{'}=\frac{4.1\times {10}^6}{60} \\ =0.06833\times {10}^6 \end{gathered}$$

The phase separation between the conductors is,

D=10ft

=10(0.3)m

=3m

The conductor radius is,

$$\begin{gathered} r=0.06677ft \\ =0.06677\times 0.3m \\ =0.02m \end{gathered}$$

$$\begin{gathered} x{\text{'}}_d=0.06833\times {10}^6\log 3 \\ =3\Omega -260mi\times {10}^4 \end{gathered}$$

$$\begin{gathered} x{\text{'}}_a=0.06833\times {10}^6\log \frac{1}{0.02} \\ =\Omega 11-mi609\times {10}^4 \end{gathered}$$

$$\begin{gathered} X_c=x{\text{'}}_d+x{\text{'}}_a \\ =3.260\times {10}^4+11.609\times {10}^4 \\ =\Omega 14-mi869\times {10}^4 \end{gathered}$$

Therefore the value of capacitive reactance is $\Omega 14-mi869\times {10}^4$.

When the spacing is getting doubled, that is D=6m.

$$\begin{gathered} x{\text{'}}_d=0.06833\times {10}^6\log 6 \\ =\Omega 5-31mi\times {10}^4 \end{gathered}$$

$$\begin{gathered} x{\text{'}}_a=0.06833\times {10}^6\log \frac{1}{0.02} \\ =\Omega 11-mi609\times {10}^4 \end{gathered}$$

$$\begin{gathered} X_c=x{\text{'}}_d+x{\text{'}}_a \\ =5.31\times {10}^4+11.609\times {10}^4 \\ =\Omega 16-mi919\times {10}^4 \end{gathered}$$

The change in reactance is calculated below,

$$\begin{gathered} Changeinreac\tan ce=\frac{\left(16.919\times {10}^4\right)-\left(14.869\times {10}^4\right)}{14.869\times {10}^4} \\ =13.78\% \end{gathered}$$

Therefore the capacitive reactance is increased by 13.78% when the spacing is doubled.