Question

The CT of Problem $10.5$is utilized in conjunction with a current sensitive device that will operate at current levels of role="math" localid="1655405578534" $\mathbf{8}\mathbf{}\mathit{A}$or above. Check whether the device will detect the $\mathbf{1300}\mathbf{}\mathit{A}$fault current for cases (b) and (d) in Problem $10.5$.

Short answer

(a) The secondary output current$\left(11.78A\right)$is greater than $8A$.Therefore, device will conduct.

(b) The secondary output current$\left(6.574A\right)$is less than $8A$.Therefore, device will not conduct.

Step-by-step solution

Write the given data from the question:

The CT current ratio,$n=500:5$

The secondary leakage impedance,$Z_{21}=4.9+j0.5\Omega$

The device operates equal or above the$8A$

The equation to calculate the CT secondary output current and CT error.

The Frohlich equation is given as,

$\mathit{E}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{A}{\mathbf{I}}_{\mathbf{e}}}{\mathbf{B}\mathbf{+}{\mathbf{I}}_{\mathbf{e}}}$ …… (1)

Here,$\mathit{A}$and $\mathit{B}$are the constant.

Hererole="math" localid="1655405982877" $\mathit{E}\mathbf{\text{'}}$is the secondary excitation voltage and ${\mathit{I}}_{\mathbf{e}}$is the secondary excitation current.

The equation to calculate the total termination impedance is given as follows,

${\mathit{Z}}_{\mathbf{T}}\mathbf{=}\mathit{Z}\mathbf{+}{\mathit{Z}}_{\mathbf{21}}$ …… (2)

Here role="math" localid="1655406069394" $\mathit{Z}$is the device termination impedance.

The equation to calculate the secondary input current in CT is given as follows.

$\mathit{I}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{I}}{\mathbf{n}}$ …… (3)

Here, role="math" localid="1655406239423" $\mathit{I}$ is the primary current of the CT.

The equation to calculate the voltage across the total termination voltage is given as follows.

${\mathit{E}}_{\mathbf{T}}\mathbf{=}\left|\begin{array}{c}I{\text{'}}_2\end{array}\right|\left|\begin{array}{c}{Z}_T\end{array}\right|$ …… (4)

The equation to calculate the excitation current is given as follows.

${\mathit{I}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{T}}}{{\sqrt{{\mathbf{5}}^{\mathbf{2}}\mathbf{+}\left(1+{\displaystyle \frac{E_2}{I_e}}\right)}}^{\mathbf{2}}}$ …… (5)

The equation to calculate the secondary output currentis given as follows.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{E}\mathbf{\text{'}}}{{\mathbf{Z}}_{\mathbf{T}}}$ …… (6)

Calculate the secondary output current of CT and CT error for the impedance of the terminating device is 𝞨9+j0.5 and the primary CT load current is 1200 A.

(b)

The device termination impedance,$Z=4.9+j0.5\Omega$

Primary CT load current,$I=1200A$

Take two point$\left(1,63\right)$and $\left(10,100\right)$from the given graph.

Substitute$63V$for $E\text{'}$into equation (1).

$63=\frac{A}{B+1}$ …… (7)

Substitute$100V$for $E\text{'}$into equation (1).

$100-\frac{10A}{B+10}$ …… (8)

By solving the equation (7) and (8).

$$\begin{gathered} A=107 \\ B=0.698 \end{gathered}$$

Calculate the excitation voltage.

Substitute $107$for $A$and $0.698$for $B$into equation (1).

$E\text{'}=\frac{107I_e}{0.698+I_e}$ …… (9)

Calculate the total termination impedance.

Substitute$4.9+j0.5\Omega$for $Z$and $0.1+j0.5\Omega$for $Z_{21}$into equation (2).

$$\begin{gathered} Z_T=4.9+j0.5+0.1+j0.5 \\ {Z}_T=5+j1\Omega \\ {Z}_T=5.099\angle 11.3°\Omega \end{gathered}$$

Calculate the secondary input current.

Substitute$1200A$for $I$and $500:5$for $n$into equation (3).

$$\begin{gathered} I{\text{'}}_2=\frac{1200}{500} \\ I{\text{'}}_2=\frac{6000}{500} \\ I{\text{'}}_2=12A \end{gathered}$$

Calculate the voltage across the termination impedance.

Substitute$12A$for $I{\text{'}}_2$and $5.099\Omega$for $Z_r$into equation (4).

$$\begin{gathered} E_T=\left|\begin{array}{c}12\end{array}\right|\times \left|\begin{array}{c}5.099\end{array}\right| \\ {E}_T=61.2V \end{gathered}$$

Calculate the excitation current.

Substituterole="math" localid="1655408409441" $\frac{107I_e}{0.698+I_e}$for $E_2$ and $61.2V$for $E_T$into equation(5).

$$\begin{gathered} I_e=\frac{61.2}{\sqrt{5^2+{\left(1+{\displaystyle \frac{107I_e}{0.698+I_e}}\right)}^2}} \\ {I}_e=\frac{61.2}{{\sqrt{25+\left(1+{\displaystyle \frac{107I_e}{0.698+I_e}}\right)}}^2} \end{gathered}$$

By solving the above equation by iteration process, the value of the excitation current is $0.894A$.

Calculate the excitation voltage.

Substitute$0.894A$into equation (9).

$$\begin{gathered} E\text{'}=\frac{107\times 0.894}{0.698+0.894} \\ E\text{'}=\frac{95.66}{1.592} \\ E\text{'}=60.16V \end{gathered}$$

Calculate the secondary output current.

Substitute$60.16V$for $E\text{'}$and $5.099\Omega$for $Z_T$in equation (6).

$$\begin{gathered} I_2=\frac{60.16}{5.099} \\ {I}_2=11.78A \end{gathered}$$

Since the secondary output current is greater than$8A$. Therefore, device will conduct.

Calculate the secondary output current of CT and CT error for the impedance of the terminating device is Ω4.9+j1.5 and the primary CT load current is 1200 A.

(d)

The device termination impedance,$Z=14.9+j1.5\Omega$

Primary CT load current,$I=1200A$

Calculate the total termination impedance.

Substitute$14.9+j1.5\Omega$for $Z$and $0.1+j0.5\Omega$for $Z_{21}$into equation (2).

$$\begin{gathered} Z_T=14.9+j1.5+0.1+j0.5 \\ {Z}_T=15+j2\Omega \\ {Z}_T=15.13\angle 7.59°\Omega \end{gathered}$$

Calculate the secondary input current.

Substitute$1200A$for $I$and $500:5$for $n$into equation (3).

$$\begin{gathered} I{\text{'}}_2=\frac{1200}{{\displaystyle \frac{500}{5}}} \\ I{\text{'}}_2=\frac{6000}{500} \\ I{\text{'}}_2=12A \end{gathered}$$

Calculate the voltage across the termination impedance.

Substitute$12A$for $I\text{'}$and $15.13\Omega$for $Z_r$into equation (4).

$$\begin{gathered} E_T=\left|\begin{array}{c}12\end{array}\right|\times \left|\begin{array}{c}15.13\end{array}\right| \\ {E}_T=181.56V \end{gathered}$$

Calculate the excitation current.

Substitute$\frac{107I_e}{0.698+I_e}$for$E_2$ and $181.56V$for $E_T$into equation(5).

$$\begin{gathered} I_e=\frac{181.56}{{\sqrt{5^2+\left(1+{\displaystyle \frac{107I_e}{0.698+I_e}}\right)}}^2} \\ {I}_e=\frac{181.56}{{\sqrt{25+\left(1+{\displaystyle \frac{107I_e}{0.698+I_e}}\right)}}^2} \end{gathered}$$

By solving the above equation by iteration process, the value of the excitation current is$9.21A$.

Calculate the excitation voltage.

Substitute$9.21A$into equation (9).

$$\begin{gathered} E\text{'}=\frac{107\times 9.21}{0.698+9.21} \\ E\text{'}=\frac{985.47}{9.908} \\ E\text{'}=99.46V \end{gathered}$$

Calculate the secondary output current.

Substitute$99.46V$for $E\text{'}$and $15.13\Omega$for $Z_T$in equation (6).

$$\begin{gathered} I_2=\frac{99.46}{15.13} \\ {I}_2=6.574A \end{gathered}$$

Since, the secondary output current is less than$8A$. Therefore, device will not conduct.