Question

Select a station-class metal-oxide surge arrester from Table 13.2 for the high-voltage side of a three-phase$\mathbf{400}\mathbf{}\mathit{M}\mathit{V}\mathit{A}\mathbf{}\mathbf{,}\mathbf{}\mathbf{345}\mathbf{-}\mathit{k}\mathit{V}\mathbf{}\mathbf{}\mathbf{}\mathit{Y}\mathbf{/}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{-}\mathit{k}\mathit{V}$$\mathbf{∆}$transformer. The maximum $\mathbf{60}\mathbf{-}\mathit{H}\mathit{z}$operating voltage of the transformer under normal operating conditions is$\mathbf{1}\mathbf{.}\mathbf{03}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$. The high-voltage windings of the transformer have a BIL of $\mathbf{1300}\mathbf{}\mathit{k}\mathit{V}$and a solidly grounded neutral. A minimum protective margin of $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$based on a $\mathbf{10}\mathbf{-}\mathit{k}\mathit{V}$impulse current wave cresting in $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathit{\mu }\mathit{s}$is required. (Note: Additional considerations for the selection of metal-oxide surge arresters are given in reference $\left[22\right]$ (www.cooperpower.com).

Short answer

Answer

The value of protective margin ranges from $\left(3.86pu-6.25pu\right)$of MCOV.

Step-by-step solution

Write the given data from the question.

Repeat example 13.8.

Consider the following data:

The maximum 60 Hz operating voltage for a 345 kV system under specific system circumstances is 1.03 per unit, under normal operating conditions and with a 1300-kV BIL. A minimum protective margin of 1.4 per unit based on an 10 - kV impulse current wave cresting in$0.5\mu s$is required.

Determine the value of protective margin ranges.

Determine the maximum voltage available on the high voltage side of the transformer.

$$\begin{gathered} V_{HV}=\left(\frac{345k}{\sqrt{3}}\right)\left(1.03\right) \\ =\left(199.19k\right)\left(1.03\right) \\ 205.17kV \end{gathered}$$

Therefore, the value of the maximum voltage available on the high voltage side of the transformer is 205.17 kV.

Refer to Table 13.2 station class surge arrestor in the text book.

The maximum rms L - G voltage is given in kVin the second column of table 13.2 from the textbook.

Therefore, choose a station-class surge arrestor with a 209 kV maximum continuous over-voltage (MCOV) value and an AZES metal oxide surge arrestor with a 258 kVMCOV value.

From table 13.2 in the text book, the fifth column states the $0.5\mu s$FOW protective level.

The safe margin of discharge voltage for the chosen surge arrestor is (2.19 - 2.36) per unit of MCOV for an 10 kV impulse current cresting in 0.5 ms.

Determine the minimum value of the protective margin.

$$\begin{gathered} V_{min}=1300k-\left(2.36\right)\left(209k\right) \\ =1300k-493.24k \\ =806.76kV \end{gathered}$$

Therefore, the minimum value of the protective margin is, 806.76 kV.

Determine the maximum value of the protective margin.

$$\begin{gathered} V_{max}=2000k-\left(2.19\right)\left(318k\right) \\ =2000k-693.24k \\ =1306.76kV \end{gathered}$$

Therefore, the maximum value of the protective margin is 1306.76 kV.

Therefore, the range of the protective margin is (506.76 kV - 1306.76 kV).

Determine the protective margin ranges:

$\frac{1306.76}{209}=6.25pu$to $\frac{806.76}{209}=3.86pu$which is greater than 1.4, as required.

Therefore, the value of protective margin ranges from (3.86 pu - 6.25 pu)of MCOV.