\[\boxed{V_{V}=m_{v} g_{o} \delta
r\left(1+\frac{0 \cdot 3}{\sqrt{\delta r}}\right) \log _{e}
\frac{d}{r}}~ \mathrm{kV/phase}\]
\[\begin{aligned}
\text{Irregularity factor}~m_v &= 1.0~(\text{polished
conductor}) \\
&= 0.72 - 0.82~
(\text{rough conductors})
\end{aligned}\]
\[\boxed{P=242 \cdot
2\left(\frac{f+25}{\delta}\right)
\sqrt{\frac{r}{d}}\left(V-V_{c}\right)^{2} \times 10^{-5}}~
\mathrm{kW/km/phase}\]
\[\begin{aligned}
f &=\text { supply frequency in } \mathrm{Hz} \\
V &=\text { phase-neutral voltage }(r \cdot m . s .) \\
V_{c} &=\text { disruptive voltage }(r . m . s .) \text { per
phase }
\end{aligned}\]
Solution-2
Assume the line is 3 -phase.
Conductor radius, \(r=1 \cdot 956 / 2=0
\cdot 978~ \mathrm{cm}\)
Dielectric strength of air, \(g_{o}=30
/ \sqrt{2}=21 \cdot 2 \mathrm{kV}(\mathrm{r}\).m.s.\()\) per \(\mathrm{cm}\)
Disruptive voltage/phase, \(V_{c}=210 /
\sqrt{3}=121 \cdot 25 \mathrm{kV}\)
Assume smooth conductors (i.e., irregularity factor \(m_{o}=1\) ) and standard pressure and
temperature for which air density factor \(\delta=1\)
Let \(d ~\mathrm{cm}\) be the
spacing between the conductors.
\[\begin{aligned}
V_{c} &=m_{0} g_{o} \delta r \log _{e}(d / r) \mathrm{kV} \\
&=1 \times 21 \cdot 2 \times 1 \times 0.978 \times \log _{e}(d /
r) \\
121 \cdot 25 &=20 \cdot 733 \log _{e}(d / r) \\
\log _{e} \frac{d}{r} &=\frac{121 \cdot 25}{20 \cdot 733}=5.848
\\
2 \cdot 3 \log _{10} d / r &=5 \cdot 848 \\
\log _{10} d / r &=5 \cdot 848 / 2 \cdot 3=2 \cdot 5426 \\
d / r &=\text { Antilog } 2 \cdot 5426 \\
d / r &=348 \cdot 8
\end{aligned}\]
Disruptive voltage (r.m.s) per phase is Conductor spacing, \(d=348 \cdot 8
\times r=348 \cdot 8 \times 0.978=341 \mathrm{cm}\)
Problem-3
A 3-phase, \(220 \mathrm{kV}, 50
\mathrm{Hz}\) transmission line consists of \(1.5 \mathrm{cm}\) radius conductor spaced 2
metres apart in equilateral triangular formation. If the temperature is
\(40^{\circ} \mathrm{C}\) and
atmospheric pressure is \(76
\mathrm{cm},\) calculate the corona loss per km of the line. Take
\(\mathrm{m}_{0}=0.85 .\)
Solution-3
\[\begin{aligned}
P &=\frac{242 \cdot 2}{\delta}(f+25)
\sqrt{\frac{r}{d}}\left(V-V_{c}\right)^{2} \times 10^{-5} ~\mathrm{kW} /
\mathrm{km} / \mathrm{phase} \\
\delta &=\frac{3 \cdot 92 b}{273+t}=\frac{3 \cdot 92 \times
76}{273+40}=0 \cdot 952\\
g_{o}&=21 \cdot 2~ \mathrm{kV} / \mathrm{cm}(\mathrm{rm.s.})
\end{aligned}\]
corona loss is given by: \[\begin{aligned}
V_{c} &=m_{0} g_{0} \delta r \log _{e} d / r \mathrm{kV} \\
&=0.85 \times 21 \cdot 2 \times 0.952 \times 1 \cdot 5
\times \log _{e} 200 / 1 \cdot 5=125 \cdot 9 \mathrm{kV}
\end{aligned}\]
Critical disruptive voltage per phase is Supply voltage per phase, \(\quad V=220
/ \sqrt{3}=127 \mathrm{kV}\)
\[\begin{aligned}
P&=\frac{242 \cdot 2}{0 \cdot 952}(50+25) \times \sqrt{\frac{1
\cdot 5}{200}} \times(127-125 \cdot 9)^{2} \times 10^{-5} \mathrm{kW} /
\mathrm{phase} / \mathrm{km} \\
&=\frac{242 \cdot 2}{0 \cdot 952} \times 75 \times 0 \cdot 0866
\times 1 \cdot 21 \times 10^{-5} \mathrm{kW} / \mathrm{km} /
\mathrm{phase} \\
&=0 \cdot 01999 \mathrm{kW} / \mathrm{km} / \mathrm{phase}
\end{aligned}\]
\[=3 \times 0 \cdot 01999 \mathrm{kW}=0 \cdot 05998
\mathrm{kW}\]
\(\mathrm{km}\)\(\therefore
\quad\)Substituting the above values, we have corona loss as:
Problem-4
A certain 3 -phase equilateral transmission line has a total corona
loss of \(53~ \mathrm{kW}\) at \(106 ~\mathrm{kV}\) and a loss of \(98~ \mathrm{kW}\) at \(110.9~ \mathrm{kV}\).
Solution-4
\[\boxed{P=3 \times \frac{242 \cdot
2(f+25)}{\delta} \sqrt{\frac{r}{d}}\left(V-V_{c}\right)^{2} \times
10^{-5}}~\mathrm{kW} / \mathrm{km}\]
\[\therefore \quad P
\propto\left(V-V_{c}\right)^{2}\]
\(d\)\(f, \delta, r\)The power loss due to corona for 3 phases is given by :
For first case, \(P=53~
\mathrm{kW}\) and \(V=106 / \sqrt{3}=61
\cdot 2 ~\mathrm{kV}\)
For second case, \(P=98~
\mathrm{kW}\) and \(V=110 \cdot 9 /
\sqrt{3}=64~\mathrm{kV}\)
\(\therefore\)
\[\begin{aligned}
53 & \propto\left(61 \cdot 2-V_{c}\right)^{2} \\
98 & \propto\left(64-V_{c}\right)^{2}
\end{aligned}\]
\[\frac{98}{53}=\frac{\left(64-V_{c}\right)^{2}}{\left(61
\cdot 2-V_{c}\right)^{2}}\]
\(113 \mathrm{kV}\)\(W\)\[V_{c}=54 \mathrm{kV}\]
Dividing [(ii)/(i)] we get,
\[W
\propto\left(\frac{113}{\sqrt{3}}-V_{c}\right)^{2}\]
Dividing
[(iii)/(i)], we get,
\[\begin{aligned}
\frac{W}{53} &=\frac{(65 \cdot 2-54)^{2}}{(61 \cdot 2-54)^{2}} \\
W &=(11 \cdot 2 / 7 \cdot 2)^{2} \times 53=128~ \mathrm{kW}
\end{aligned}\]
Problem-5
Find the disruptive critical and visual corona voltage of a grid-line
operating at \(132~ \mathrm{kV}\)
\[\begin{aligned}
\text {conductor dia} &=1.9~ \mathrm{cm} \quad ; \quad \text {
conductor spacing}=3.81~ \mathrm{m}\\
\text { temperature } &=44^{\circ} \mathrm{C} \quad ; \quad
\text{barometric pressure} =73.7~ \mathrm{cm}
\end{aligned}\]
conductor surface factor:
\[\begin{array}{lll}
\text{fine weather}=0.8 & ; \quad \text { rough weather } &
=0.66 .
\end{array}\]
Solution-5
\[\begin{aligned}
& \boxed{V_{c} =48.8m_{0}\delta
r\log_{10}\left(D/r\right)}~\mathrm{kV/phase}\\
m_{0} & =0.8\\
\delta & =3.92\times73.7/\left(273+44\right)=0.91\\
\log_{10}\left(381/1.9\right) & =\log_{10}\left(200.4\right)=2.302\\
V_{c} & =48.8\times0.8\times0.91\times1.9\times2.302\\
& =155.3~\mathrm{kV/phase}\\
&\boxed{V_{v} =48.8m_{v}\delta r\left(1+\dfrac{0.3}{\sqrt{\delta
r}}\right)\log_{10}\left(D/r\right)}~\mathrm{kV/phase}\\
&
=48.8\times0.66\times0.91\times1.9\left(1+\dfrac{0.3}{1.314}\right)\times2.302\\
& =157.5~\mathrm{kV/phase}
\end{aligned}\]