Determine the voltage across each disc of suspension insulators as a percentage of the line voltage to earth. The self and capacitance to ground of each disc is C and 0.2C respectively. The capacitance between the link pin and the guard ring is 0.1C. (b) If the capacitance to the line of the lower

Question

Determine the voltage across each disc of suspension insulators as a percentage of the line voltage to earth. The self and capacitance to ground of each disc is C and 0.2C respectively. The capacitance between the link pin and the guard ring is 0.1C. (b) If the capacitance to the line of the lower link pin were increased to 0.3C by means of a guard ring, determine the redistribution of voltage. Also determine the string efficiency in each case.

Accepted Answer

(a) Let [latex]E_{1}, E_{2}[/latex] and [latex]E_{3}[/latex] be the voltage drops across the discs as shown in Fig. E. 8.2. Applying Kirchhoff's current law at node A, we have

[latex]\begin{aligned}I_{2}+I_{x}=I_{1}+I_{a} \E_{2} \omega C+\left(E_{2}+E_{3} ight) \omega 0.1 C-E_{1} \omega C-0.2 E_{1} \omega C &=0 \E_{2}+\left(E_{2}+E_{3} ight) 0.1-E_{1}-0.2 E_{1}=0\end{aligned}[/latex]

[latex]1.1 E_{2}+0.1 E_{3}-1.2 E_{2}=0[/latex]

or [latex]1.2 E_{1}-1.1 E_{2}-0.1 E_{3}=0[/latex] (1)

Again writing equation at node B

[latex]\begin{gathered}I_{3}+I_{y}-I_{2}-I_{b}=0 \E_{3} C \omega+E_{3} 0.1 \omega C-E_{2} \omega C-\left(E_{1}+E_{2} ight) 0.2 C=0 \E_{3}+0.1 E_{3}-1.2 E_{2}-0.2 E_{1}=0 (2) \0.2 E_{1}+1.2 E_{2}-1.1 E_{3}=0\end{gathered}[/latex]

There are three unknowns with two equations. We divide both of them by [latex]E_{3}[/latex] and rewrite them as

[latex] 12 x-11 y =1[/latex] (1a)

[latex] 2 x+12 y =11[/latex] (2a)

where [latex]x=\frac{E_{1}}{E_{3}}[/latex] and [latex]y=\frac{E_{2}}{E_{3}}[/latex].

[latex]\begin{aligned} 12 x-11 y =1\ \underline{12 x+72 y =66}\83 y =65 \end{aligned}[/latex]

[latex]y=\frac{65}{83}=0.783=\frac{E_{2}}{E_{3}}[/latex]

[latex] herefore \quad 2 x=11-12 imes \frac{65}{83}=11-9.39 ext { or } x=0.801=\frac{E_{1}}{E_{3}}[/latex]

[latex] herefore \quad E_{2}=0.783 E_{3} ext { and } E_{1}=0.801 E_{3}[/latex]

Now total voltage is

[latex]\begin{aligned} E &=E_{1}+E_{2}+E_{3}=0.801 E_{3}+0.783 E_{3}+E_{3}=2.584 E_{3} \ herefore \quad E_{3} &=\frac{E}{2.584} imes 100=38.70 \% \ E_{2} &=0.783 imes 38.70=30.3 \% \ E_{1} &=0.801 imes 38.7=31 \% \end{aligned}[/latex]

[latex] ext { The } \% ext { string efficiency }= \frac{1}{3 imes 0.387} imes 100=86.1 \% [/latex]

(b) If the capacitance to the line of the lower pin is increased to 0.3C the equation at node [latex]A[/latex] is unchanged whereas that at B now becomes

[latex]0.2 E_{1}+1.2 E_{2}-1.3 E_{3}=0[/latex] (3)

or [latex]2x + 12y = 13[/latex] (3a)

or [latex]12x – 11y = 1[/latex] (1a)

[latex]\begin{aligned}12 x+72 y=78 \ \underline{12 x-11 y=1} \ 83 y=77\end{aligned}[/latex]

[latex]\begin{aligned}y &=0.9277 \2 x &=13-12 imes 0.9277 \ ext {or} \quad x &=0.9337\end{aligned}[/latex]

[latex]\begin{aligned} herefore \quad \frac{E_{1}}{E_{3}} &=0.9337 ext { and } \frac{E_{2}}{E_{3}}=0.9277 \E_{1} &=0.9337 E_{3} ext { and } E_{2}=0.9277 E_{3} \E &=E_{1}+E_{2}+E_{3}=0.9337 E_{3}+0.9277 E_{3}+E_{3} \E_{3} &=0.3494 E ext { or } 34.94 \% \E_{2} &=0.9277 imes 34.94 \%=32.42 \% \E_{1} &=0.9337 imes 34.94=32.62 \%\end{aligned}[/latex]

The [latex]\%[/latex] string efficiency [latex]=\frac{1}{3 imes 0.3494} imes 100=95.4 \%[/latex].

Question 8.2: Determine the voltage across each disc of suspension insulat...

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