Question

A CT with an excitation curve given in Figure 10.39 has a rated current ratio of$\mathbf{500}\mathbf{}\mathbf{:}\mathbf{5}\mathbf{}\mathit{A}$and a secondary leakage impedance of $\mathbf{\varnothing }\mathbf{1}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}$. Calculate the CT secondary output current and the CT error for the following cases: (a) The impedance of the terminating device is $\mathbf{\varnothing }\mathbf{9}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}$and the primary CT load current is $\mathbf{400}\mathbf{}\mathit{A}$. (b) The impedance of the terminating device is $\mathbf{\varnothing }\mathbf{9}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}$and the primary CT fault current is $\mathbf{1200}\mathbf{}\mathit{A}$. (c) The impedance of the terminating device is $\mathbf{\varnothing }\mathbf{4}\mathbf{.}\mathbf{9}\mathbf{+}\mathit{j}\mathbf{1}\mathbf{.}\mathbf{5}$and the primary CT load current is $\mathbf{400}\mathbf{}\mathit{A}$. (d) The impedance of the terminating device is role="math" localid="1655725625321" $\mathbf{\varnothing }\mathbf{4}\mathbf{.}\mathbf{9}\mathbf{+}\mathit{j}\mathbf{1}\mathbf{.}\mathbf{5}$and the primary CT fault current is$\mathbf{1200}\mathbf{}\mathit{A}$

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Short answer

(a) The secondary output current is$3.97A$and CT error is $0.75\%$

(b) The secondary output current is$11.78A$and CT error is $1.83\%$.

(c) The secondary output current is$3.807A$and CT error is data-custom-editor="chemistry" $4.81\%$.

(d) The secondary output current is$6.574A$and CT error is $45.21\%$.

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 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.

Here $\mathit{E}\mathbf{\text{'}}$is the secondary excitation voltage and role="math" localid="1655726028613" ${\mathit{I}}_{\mathbf{e}}$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$\mathit{Z}$is the device termination impedance.

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

$\mathit{I}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{I}}{\mathbf{n}}$ …… (3)

Here,$\mathit{I}$is the primary current of the CT.

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

$\mathit{E}\mathbf{=}\left|\begin{array}{c}I{\text{'}}_2\end{array}\right|\left|\begin{array}{c}{Z}_T\end{array}\right|$ …… (4)

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

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

The equation to calculate the percentage CT error is given as follows.

$\mathit{C}\mathit{T}\mathbf{}\mathit{e}\mathit{r}\mathit{r}\mathit{o}\mathit{r}\mathbf{=}\frac{\left|\begin{array}{c}I{\text{'}}_2\end{array}\right|\mathbf{-}\left|\begin{array}{c}{I}_2\end{array}\right|}{\left|\begin{array}{c}I{\text{'}}_2\end{array}\right|}\mathbf{\times }\mathbf{100}$ …… (7)

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 400 A.

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

Primary CT load current,$I=400A$

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

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

$63=\frac{A}{B+1}$ …… (8)

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

localid="1655726790842" $100-\frac{10A}{B+10}$ …… (9)

By solving the equation (8) and (9).

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

Calculate the excitation voltage.

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

$E\text{'}=\frac{107I_e}{0.698+I_e}$ …… (10)

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 $\mathbf{400}\mathbf{}\mathit{A}$for $I$and $500:5$for $n$into equation (3).

localid="1655727404542" $$\begin{gathered} I{\text{'}}_2=\frac{400}{500} \\ I{\text{'}}_2=\frac{2000}{500} \\ I{\text{'}}_2=4A \end{gathered}$$

Calculate the voltage across the termination impedance.

Substitute$4A$for $I{\text{'}}_2$and localid="1655727474180" $5.099\Omega$for $Z_T$into equation (4).

$$\begin{gathered} E_T=\left|\begin{array}{c}4\end{array}\right|\times \left|\begin{array}{c}5.099\end{array}\right| \\ {E}_T=20.4V \end{gathered}$$

Calculate the excitation current.

Substitutefor andforinto equation(5).