Question

Repeat Example 13.8 for a 500 - kV system with a 1.08 per-unit maximum 60 - Hz voltage under normal operating conditions and with a 2000 - kVBIL.

Short answer

Answer

(a)

The value of operating voltage of the power system is 311.77 kV.

The value of station-class surge arrester with maximum continuous operating voltage rating is 318 kV.

(b)

The value of minimum protective margin is 1214.54 kV.

The value of maximum protective margin is 1360.82 kV.

Step-by-step solution

Write the given data from the question.

Repeat example 13.8.

Consider the following data:

The maximum 60 Hz voltage for a 500 kV system under specific system circumstances is 1.08 per unit, under normal operating conditions and with a 2000 - kVBIL.

(a) Determine the value of operating voltage and station-class surge arrester rating.

Determine the line to neutral operating voltage of the given system.

$$\begin{gathered} V=\left(\frac{500k}{\sqrt{3}}\right)\left(1.08\right) \\ =\left(288.68k\right)\left(1.08\right) \\ =311.77kV \end{gathered}$$

Therefore, the value of the operating voltage of the power system is 311.77 kV.

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

The line to neutral working voltage maximum is 311.77 kV; thus, choose a surge arrestor with the lowest MCOV that surpasses this value.

From the table 13.2 in the text book, the third column states the minimum MCOV rating in kV.

The value is, 318 kV.

Therefore, the value of station-class surge arrester with maximum continuous operating voltage rating is 318 kV.

(b) Determine the value of minimum and maximum protective margin.

The textbook lists the $0.5\mu s$FOW protective threshold in the fifth column of Table 13.2. The safe margin of discharge voltage for the chosen surge arrester is (2.01 - 2.47 per unit of MCOV for an 10 kV impulse current generated in$0.5\mu s$.

Determine the minimum value of the protective margin.

$$\begin{gathered} V_{min}=2000k-\left(2.47\right)\left(318k\right) \\ =2000k-785.46k \\ =1214.54kV \end{gathered}$$

Therefore, the minimum value of the protective margin is, 1214.54 kV.

Determine the maximum value of the protective margin.

$$\begin{gathered} V_{min}=2000k-\left(2.01\right)\left(318k\right) \\ =2000k-639.18k \\ =1360.82kV \end{gathered}$$

Therefore, the value of maximum protective margin is 1360.82 kV.

Therefore, the range of the protective margin is (1214.54 kV - 1360.82 kV.