Question

Reconsider case (b) of Problem 10.5. Let the load impedance be the input impedance to a CO-7 induction disc time-delay overcurrent relay. The CO-7 relay characteristic is shown in Figure 10.41. For a tap setting of and a time dial setting of , determine the relay operating time.

Short answer

The operating time of the relay is 0.9 sec.

Step-by-step solution

Write the given data from the question.

The CT current ratio n = 500:5.

The load impedance $Z_L=4.9+j0.5\Omega$.

Time dial setting TDS = 2.

Tap setting is 4 A.

The secondary leakage impedance $Z_{2l}=0.1+j0.5\Omega$.

Determine the equations to calculate the relay operating time.

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, A and B 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,

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

Here 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, I is the primary current of the CT.

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

${\mathit{E}}_{\mathbf{T}}\mathbf{=}\left|I{\text{'}}_2\right|\left|Z_T\right|$ …… (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 relay operating time.

Take two points (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 (8) and (9).

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 impedace.

Substituterole="math" localid="1655882078918" $4.9+j0.5\Omega$ 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 ninto 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 61.2V 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 0.894 A into equation (10).

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

Calculate the secondary output current.

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

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

The tap setting is equal to the pick-up current.

Therefore, Pick up current, I_p= 4 A

Calculate the relay input current multiple of tap setting.

$$\begin{gathered} \frac{I_{relay}}{I_p}=\frac{11.78}{4} \\ \frac{I_{relay}}{I_p}=2.94 \end{gathered}$$

The operating time of the relay corresponding to TDS = 2 and relay current input multiple of current tap setting 2.94 is 0.9 sec.

Hence, the operating time of the relay is 0.9 sec.