Question

The relay in Problem $10.2$has a time-dial setting of $\mathbf{4}$. Determine the relay operating time if the primary fault current is$\mathbf{400}\mathbf{}\mathit{A}$.

Short answer

The relay operating time is$3.6sec$.
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Step-by-step solution

write the given data from the question.

The CT ratio$n=100:5$.

Current tap setting$TS=5A$.

Primary fault current$I=400A$.

Determine the formula to calculate the relay operating time.

The equation to calculate the secondary voltage of the CT is given as follows.

$\mathit{E}\mathbf{\text{'}}\mathbf{=}\mathit{I}\mathbf{\text{'}}\left(Z\text{'}+Z_B\right)$ …… (1)

Hererole="math" localid="1655403292867" $\mathit{I}\mathbf{\text{'}}$is the relay current, and ${\mathit{Z}}_{\mathbf{B}}$is the relay impedance.

The equation to calculate the minimum primary fault current is given as follows.

$\mathit{I}\mathbf{=}\mathit{n}\left(I\text{'}+I_c\right)$ …… (2)

Here,${\mathit{I}}_{\mathbf{c}}$is the exciting current.

Calculate the relay operating time.

Consider the CT equivalent circuit as,

Refer to the excitation curve for CO-8 relay in the text book.

The relay current, $I\text{'}=10A$

Relay impedance, $Z_B=1\Omega$

Calculate the secondary voltage of the CT,

Substitute $10A$for $I\text{'},0.082\Omega$, for $Z\text{'}$and $1\Omega$for $Z_B$into equation (1).

$$\begin{gathered} E\text{'}=10\left(0.082+1\right) \\ E\text{'}=10\times 1.082 \\ E\text{'}=10.82V \end{gathered}$$

The exciting current corresponding to the secondary voltage of the CT $10.82V$is $0.6A$.

$I_c=0.6A$

Calculate the minimum primary fault current.

Substitute $100:5$for $n$, $10A$for $I\text{'}$and $0.6A$for $I_c$into equation (2).

$$\begin{gathered} I=\frac{100}{5}\left(10+0.6\right) \\ I=20\times 10.6 \\ I=212A \end{gathered}$$

From the problem 10.2, consider the table as,

Write the minimum fault current for the different value of input current.

$I\text{'}$	$5$	$8$	$10$	$13$	$15$
I	$105$	$168$	$212$	$296$	$700$

Draw the graph between the input current and primary minimum fault current as,

The relay $I\text{'}$current corresponding to the primary fault current $I=400A$from the above graph is localid="1655404620175" $14A$.

Consider the characteristics curve of relay from the text book’s figure 10.12.

Calculate the relay input current multiple of tap setting.

$$\begin{gathered} \frac{I\text{'}}{I_p}=\frac{14}{5} \\ \frac{I\text{'}}{I_p}=2.8 \end{gathered}$$

The tap setting value is $5A$, so, the pick-up current, $I_p=5A$.

Select the curve corresponding to the TDS value and the curve comes out to be 4.

Calculate the operating time of the curve using the curve 4 and input relay multiplier$2.8$. therefore, the operating time of the relay is $3.7sec.$