Question

Line impedances for the power system shown in Figure 10.47 are${\mathit{Z}}_{\mathbf{12}}\mathbf{=}{\mathit{Z}}_{\mathbf{23}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{+}\mathit{j}\mathbf{40}\mathbf{.}\mathbf{0}$, and $\mathit{Z}\mathbf{24}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{+}\mathit{j}\mathbf{80}\mathbf{.}\mathbf{0}$. Reach for the zone 3B12 impedance a relay is set for 100% of line 1-2 plus 120% of line 2-4. (a) For a bolted three-phase fault at bus 4, show that the apparent primary impedance “seen” by the B12 relays is

${\mathit{Z}}_{\mathbf{apparent}}\mathbf{=}{\mathit{Z}}_{\mathbf{12}}\mathbf{+}{\mathit{Z}}_{\mathbf{24}}\mathbf{+}\mathbf{(}{\mathbf{I}}_{\mathbf{32}}\mathbf{/}{\mathbf{I}}_{\mathbf{12}}\mathbf{)}{\mathit{Z}}_{\mathbf{24}}$

Where $\mathbf{(}{\mathbf{I}}_{\mathbf{32}}\mathbf{/}{\mathbf{I}}_{\mathbf{12}}\mathbf{)}$is the line 2-3 to line 1-2 fault current ratio. (b) If $\mathbf{\left|}{\mathbf{I}}_{\mathbf{32}}\mathbf{\right|}\mathbf{\left|}{\mathbf{I}}_{\mathbf{12}}\mathbf{\right|}$, does the B12 relay see the fault at bus 4? Note: This problem illustrates the “infeed effect.” Fault currents from line 2-3 can cause the zone 3 B12relay to under reach. As such, remote backup of line 2-4 at B12is ineffective.

Figure 10.47

Short answer

Answer

(a) The value of apparent primary impedance seen by the B12 relay is calculated$Z_{apparent}=Z_{12}+Z_{24}+\left(\frac{I_{32}}{I_{12}}\right)\times Z_{24}$.

(b)

The value of secondary apparent impedance seen by B12 relay is$Z_{apparenr}=\frac{\left[9+6\times \left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\right]+j\left[120+80\times \left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\right]}{\left({\displaystyle \frac{n_v}{n_I}}\right)}$.

The value of secondary impedance $Z_{t3}$for B12 set relay is$Z_{t3}=\frac{10.2+j136}{\left({\displaystyle \frac{n_v}{n_I}}\right)}$.

Step-by-step solution

Write the given data from the question.

The value of line impedance, $Z_{12}=3+j40\Omega$.

The value of line impedance, $Z_{23}=3+j40\Omega$.

The value of line impedance, $Z_{24}=6+j80\Omega$.

Reach for zone 3 B12 impedance relay is 100% for line 1-2.

Reach for zone 3 B12 impedance relay is 120% for line 2-4.

Determine the formula of apparent primary impedance seen by the  relay is calculated.

Write the formula of apparent primary impedance seen by the B12 relay is calculated B12.

${\mathit{Z}}_{\mathbf{a}\mathbf{p}\mathbf{p}\mathbf{r}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{I}}_{\mathbf{12}}}$ …… (1)

Here,${\mathit{V}}_{\mathbf{1}}$is voltage and${\mathit{I}}_{\mathbf{12}}$is current.

Write the formula of secondary apparent impedance seen by B12relay.

${\mathit{Z}}_{\mathbf{apparent}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{apparent}}}{\left({\displaystyle \frac{n_v}{n_I}}\right)}$ …… (2)

Here, ${\mathit{Z}}_{\mathbf{apparent}}$is apparent primary impedance ${\mathit{n}}_{\mathbf{v}}$is VT ratio and ${\mathit{n}}_{\mathbf{I}}$is CT ratio.

(a) Determine the value of apparent primary impedance seen by the  relay is calculated.

Draw a circuit diagram of breakers and impedance relays for the system below.

Figure 1

Draw a circuit diagram for a three phase bolted three phase fault at bus 4:

Figure 2

Determine the apparent primary impedance seen by the B12 relay is calculated B12.

Substitute $V_1-V_2+V_2$for $V_1$and $I_{12}$for $I_{12}$into equation (1).

$$\begin{gathered} Z_{apparent}=\frac{V_1-V_2+V_2}{I_{12}} \\ =\frac{\left(V_1-V_2\right)}{I_{12}}+\frac{V_2}{I_{12}} \\ =Z_{12}+\frac{V_2}{I_{12}} \end{gathered}$$

From figure 2,

Determine the voltage$V_2$.

$V_2=Z_{12}\times I_{24}$

Determine the current$I_{24}$.

$I_{24}=I_{24}+I_{32}$

So, solve further as apparent impedance is determine below.

$$\begin{gathered} Z_{apparent}=Z_{12}+\frac{Z_{24}\times \left(I_{12}+I_{32}\right)}{I_{12}} \\ =Z_{12}+Z_{24}+\left(\frac{I_{32}}{I_{12}}\right)\times Z_{24} \end{gathered}$$

As a result, this is the same as the phrase that must be proven, hence it is proved.

(b) Determine the value of secondary apparent impedance seen by  relay.

Determine the secondary apparent impedance.

$$\begin{gathered} Z_{apparcent}=\frac{Z_{12}+Z_{24}+\left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\times Z_{24}}{\left({\displaystyle \frac{n_v}{n_I}}\right)} \\ =\frac{\left(3+j40\right)+\left(6+j80\right)+\left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\times Z_{24}}{\left({\displaystyle \frac{n_v}{n_I}}\right)}\Omega \\ =\frac{\left[9+6\times \left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\right]+j\left[120+80\times \left({\displaystyle \frac{I_{32}}{I_{12}}}\right)\right]}{\left({\displaystyle \frac{n_v}{n_I}}\right)}\Omega \end{gathered}$$ …… (3)

The relay is set B12 with secondary impedance .

$$\begin{gathered} Z_{t3}=\frac{\left(3+j40\right)+1.2\times \left(6+j80\right)}{\left({\displaystyle \frac{n_v}{n_I}}\right)} \\ =\frac{10.2+j136}{\left({\displaystyle \frac{n_v}{n_I}}\right)} \end{gathered}$$ …… (4)

When$\left(\frac{I_{32}}{I_{12}}\right)>0.20$, the relay impedance setting is larger than the apparent impedance, as shown in equations (3) and (4). When$\left(\frac{I_{32}}{I_{12}}\right)>0.20$, the apparent impedance is outside the trip zone.