The surge voltage of 1000 kV caused by lightning propagates along the transmission line, having the distributed parameters L = 1.34 mH/km and C =8.6 nF/km.
Determine: (a) The surge power in the line, (b) The surge current in the line.

Question

Accepted Answer

(a) The total energy stored in an electromagnetic field per unit length of the line is

[latex]W=C\upsilon ^2=8.6·1^{−9}(1000·103)²=8.6·10^3 J/km[/latex]

The surge velocity is

[latex] u =\frac{1}{\sqrt{LC} }=\frac{1}{\sqrt{1.34·10^{−3}·8.6·10^{−9}} } =295·10^3 km/s [/latex]

Therefore, the surge power is

[latex]P=8.6·10^3·295·103\cong 2500 MW. [/latex]

(b) The characteristic impedance of the line is

[latex]Z_c=\sqrt{\frac{L}{C} }=\sqrt{\frac{1.34·10^{−3}}{8.6·10^{−9 }} } =395 \Omega. [/latex]

Therefore the surge current is

[latex]i=\frac{\upsilon }{Z_c}=\frac{1000\cdot 10^3}{395} =2.53 kA. [/latex]

It should be noted that for the transient analysis of some problems, when the voltage and current change versus time is needed, it is more convenient to express the voltage and current-traveling waves in the form

[latex] \upsilon =\upsilon _f+\upsilon _b=\phi _1\left(t-\frac{x}{ u } ight)+\phi _2\left(t+\frac{x}{ u } ight) [/latex] (7.23a)

[latex] i=i_f-i_b=\frac{1}{Z_c}\left[\phi _1\left(t-\frac{x}{ u } ight) -\phi _2\left(t-\frac{x}{ u } ight) ight]. [/latex] (7.23b)

Question 7.1: The surge voltage of 1000 kV caused by lightning propagates ...

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