Demonstrative Video

Corona in Transmission Lines

SECTION 01

Revision of Important Formulas

\[\boxed{\text { Critical disruptive voltage, } \quad V_{c}=m_{o} g_{o} \delta r \log _{e} \frac{d}{r}}~ \mathrm{kV/phase}\]

Under standard conditions, the value of

\[\boxed{V_{V}=m_{v} g_{o} \delta r\left(1+\frac{0 \cdot 3}{\sqrt{\delta r}}\right) \log _{e} \frac{d}{r}}~ \mathrm{kV/phase}\]

\[\begin{aligned} \text{Irregularity factor}~m_v &= 1.0~(\text{polished conductor}) \\ &= 0.72 - 0.82~ (\text{rough conductors}) \end{aligned}\]

\[\boxed{P=242 \cdot 2\left(\frac{f+25}{\delta}\right) \sqrt{\frac{r}{d}}\left(V-V_{c}\right)^{2} \times 10^{-5}}~ \mathrm{kW/km/phase}\]

\[\begin{aligned} f &=\text { supply frequency in } \mathrm{Hz} \\ V &=\text { phase-neutral voltage }(r \cdot m . s .) \\ V_{c} &=\text { disruptive voltage }(r . m . s .) \text { per phase } \end{aligned}\]

SECTION 02

Problem-1

A 3 -phase line has conductors \(2 \mathrm{cm}\) in diameter spaced equilaterally \(1 \mathrm{m}\) apart. If the dielectric strength of air is \(30 \mathrm{kV/cm}\) (max), find the disruptive critical voltage for the line. Take air density factor \(\delta=0.952\) and irregularity factor \(m_{o}=0.9 .\)

SECTION 03

Solution-1

Conductor radius, \(r=2 / 2=1 \mathrm{cm}\)
Conductor spacing \(d=1 \mathrm{m}=100 \mathrm{cm}\)
Dielectric strength of air, \(g_{o}=30~ \mathrm{kV} / \mathrm{cm}(\max .)=21 \cdot 2 ~\mathrm{kV}(\mathrm{rm.s.})\) per \(\mathrm{cm}\)
Disruptive critical voltage,

\[\begin{aligned} V_{c} &=m_{o} g_{o} \delta r \log _{e}(d / r)~ \mathrm{kV}^{*} / \mathrm{phase}(r . m . s . \text { value }) \\ &=0.9 \times 21 \cdot 2 \times 0.952 \times 1 \times \log _{e} 100 / 1=83 \cdot 64 \mathrm{kV} / \mathrm{phase} \end{aligned}\]
Line voltage (r.m.s) =\(\sqrt{3} \times 83 \cdot 64=144 \cdot 8 \mathrm{kV}\)

SECTION 04

Problem-2

A \(132 \mathrm{kV}\) line with \(1.956 \mathrm{cm}\) dia. conductors is built so that corona takes place if the line voltage exceeds \(210 \mathrm{kV}\) (r.m.s). If the value of potential gradient at which ionisation occurs can be taken as \(30 \mathrm{kV/cm}\), find the spacing between the conductors.

SECTION 05

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, \(d=348 \cdot 8 \times r=348 \cdot 8 \times 0.978=341 \mathrm{cm}\)

SECTION 06

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 \(\mathrm{m}_{0}=0.85 .\)

SECTION 07

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:

SECTION 08

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 \(110.9~ \mathrm{kV}\).

What is the disruptive critical voltage?
What is the corona loss at \(113~ \mathrm{kV}\) ?

SECTION 09

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,

\[\begin{aligned} \frac{W}{53} &=\frac{(65 \cdot 2-54)^{2}}{(61 \cdot 2-54)^{2}} \\ W &=(11 \cdot 2 / 7 \cdot 2)^{2} \times 53=128~ \mathrm{kW} \end{aligned}\]

SECTION 10

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:

\[\begin{array}{lll} \text{fine weather}=0.8 & ; \quad \text { rough weather } & =0.66 . \end{array}\]

SECTION 11

Solution-5

\[\begin{aligned} & \boxed{V_{c} =48.8m_{0}\delta r\log_{10}\left(D/r\right)}~\mathrm{kV/phase}\\ m_{0} & =0.8\\ \delta & =3.92\times73.7/\left(273+44\right)=0.91\\ \log_{10}\left(381/1.9\right) & =\log_{10}\left(200.4\right)=2.302\\ V_{c} & =48.8\times0.8\times0.91\times1.9\times2.302\\ & =155.3~\mathrm{kV/phase}\\ &\boxed{V_{v} =48.8m_{v}\delta r\left(1+\dfrac{0.3}{\sqrt{\delta r}}\right)\log_{10}\left(D/r\right)}~\mathrm{kV/phase}\\ & =48.8\times0.66\times0.91\times1.9\left(1+\dfrac{0.3}{1.314}\right)\times2.302\\ & =157.5~\mathrm{kV/phase} \end{aligned}\]