Question

The CT of Problem 10.5 is utilized in conjunction with a current sensitive device that will operate at current levels of 8 A or above. Check whether the device will detect the 1300 A fault current for cases (b) and (d) in Problem 10.5.

Short answer

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

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

Step-by-step solution

Step 1; Write the given data from the question:

The CT current ratio, n = 500 : 5

The secondary leakage impedance,$Z_{2I}=4.9+j0.5\Omega$

The device operates equal or above the 8 A.

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, and are the constant.

Here E' is the secondary excitation voltage and I_e is the secondary excitation current.

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

Z_T = Z + Z_2I …… (2)

Here 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, is the primary current of the CT.

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

${\mathit{E}}_{\mathbf{T}}\mathbf{=}\left|I{\text{'}}_2\right|\left|Z_T\right|$ …… (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 current I₂ is 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 4.9+j0.5Ω 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 = 1200 A

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

Substitute 63 V for E' into equation (1).

$63=\frac{A}{B+1}$ …… (7)

Substitute 100 V for E' into equation (1).

$100=\frac{10A}{B+10}$ …… (8)

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

A = 107

B = 0.698

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_2I into equation (2).

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

Calculate the secondary input current.

Substitute 1200 A 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 12 A for I'₂ and$5.099\Omega$ for Z_T into equation (4).

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

Calculate the excitation current.

Substitute$\frac{107I_e}{0.698+I_e}$ for E₂ and 61.2 V 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 0.894. A

Calculate the excitation voltage.

Substitute 0.894 A 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 60.16 V for E' 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 8 A. Therefore, device will conduct.

Calculate the secondary output current of CT and CT error for the impedance of the terminating device is 14.9+j1.5Ω 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 = 1200 A

Calculate the total termination impedance.

Substitute$14.9+j1.5\Omega$for Z and$0.1+j0.5\Omega$ for Z_2I into equation (2).

$$\begin{gathered} Z_T=14.9+j1.5+0.1+j0.5 \\ {Z}_T=15+j2\Omega \\ {Z}_T=15.13\angle 7.{59}^{\circ }\Omega \end{gathered}$$

Calculate the secondary input current.

Substitute 1200 A 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 12 A I'₂ for and $15.13\Omega$ for Z_T into equation (4).

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

Calculate the excitation current.

Substitute$\frac{107I_e}{0.698+I_e}$ for E₂ and 181.56 V 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 9.21 A.

Calculate the excitation voltage.

Substitute 9.21 A 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 99.46 V for E' 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 8 A. Therefore, device will not conduct.