Question

An overcurrent relay set to operate at $10A$is connected to the CT in Figure 10.8 with a $500:5$ CT ratio. Determine the minimum primary fault current that the relay will detect if the burden $Z_B$is (a)$\Omega 0$, (b) $ῼ0$, and (c)$𝛀0$.

Short answer

(a) The minimum primary fault current is $1004A$.

(b) The minimum primary fault current is $1008A$.

(c) The minimum primary fault current is $1009.5A$.

Step-by-step solution

Write the given data from the question.

Overcurrent relay setting, $I\text{'}=10A$

CT ratio,$n=500:5$

The secondary resistance corresponding to CT ratio from the figure $10.8,Z\text{'}-0.242𝛺$.

Write the relay impedance.

(a)$Z_B=1ῼ$

(b)$Z_B=4𝝮$

(c)role="math" localid="1655199073435" $Z_B=5\Omega$

Determine the equations to calculate the minimum primary fault current.

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)

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 minimum primary fault current for .ØB=1

(a)

Calculate the secondary voltage of the CT,

Substitute $10A$for $I\text{'}$, $0.242\Omega$for $Z\text{'}$and $1\Omega$for$Z_B$ into equation (1).

$$\begin{gathered} E\text{'}=10\left(0.242+1\right) \\ E\text{'}=10\times 1.242 \\ E\text{'}=12.42V \end{gathered}$$

The exciting current corresponding to the secondary voltage of the CT$12.42V$is $0.04A$.

$l_c=0.04A$

Calculate the minimum primary fault current.

Substitute $500:5$for $n$, $10A$for $l\text{'}$and $0.04A$for $l_c$into equation (2).

$$\begin{gathered} 1=\frac{500}{5}\left(10+0.04\right) \\ 1=100\times 0.04 \\ 1=1004A \end{gathered}$$

Hence, the minimum primary fault current is $1004A$.

Calculate the minimum primary fault current for ØB=4.

(b)

Calculate the secondary voltage of the CT,

Substitute $10A$for $I\text{'}$, $0.242\Omega$for $Z\text{'}$and $4\Omega$for $Z_B$into equation (1).

$$\begin{gathered} E\text{'}=10\left(0.242+4\right) \\ E\text{'}=10\times 4.242 \\ E\text{'}=42.42V \end{gathered}$$

The exciting current corresponding to the secondary voltage of the CT $42.42V$is $0.08A$.

$l_c=0.08A$

Calculate the minimum primary fault current.

Substitute $500:5$for $n$, $10A$for $I\text{'}$and $0.08A$for $l_c$into equation (2).

$$\begin{gathered} l=\frac{500}{5}\left(10+0.08\right) \\ 1=100\times 10.08 \\ 1=1008A \end{gathered}$$

Hence, the minimum primary fault current is $1008A$.

Calculate the minimum primary fault current for ØB=5.

(c)

Calculate the secondary voltage of the CT,

Substitute $10A$for $l\text{'}$, $0.242\Omega$for $Z\text{'}$and $5\Omega$for $Z_B$into equation (1).

$$\begin{gathered} E\text{'}=10\left(0.242+5\right) \\ E\text{'}=10\times 5.242 \\ E\text{'}=52.42V \end{gathered}$$

The exciting current corresponding to the secondary voltage of the CT $52.42V$is $0.095A$.

$l_c=0.095A$

Calculate the minimum primary fault current.

Substitute $500:5$for $n$, $10A$for $l\text{'}$and $0.095A$for $l_c$into equation (2).

$$\begin{gathered} l=\frac{500}{5}\left(10+0.095\right) \\ l=100\times 10.095 \\ l=1009.5A \end{gathered}$$

Hence, the minimum primary fault current is $1009.5A$.