Question

Question:Consider a protected bus that terminates four lines, as shown in Figure 10.51. Assume that the linear couplers have the standard and a three-phase fault externally located on line causes the fault currents shown in Figure 10.51. Note that the infeed current on line to the fault is . (a) Determine . (b) Let the fault be moved to an internal location on the protected bus between lines and . Find and discuss what happens. (c) By moving the external fault from line to a corresponding point on (i) line and (ii) line , determine in each case.

Short answer

Answer

(a) The value of voltage V₀ is 0 V .

(b) The value of voltage V₀ is 230 V .

(c)

(i) The value of voltage V₀ is 0 V .

(ii) The value of voltage V₀ is 0 V.

Step-by-step solution

Write the given data from the question.

Consider the impedance of linear couplers, $X_m=5m\mathit{\mho }$.

Consider the fault current on line 1 is -j16 k .

Consider the fault current on line 2 is -ji7 k .

Consider the fault current on line 3 is j36 KA .

Consider the fault current on line 3 is -J13 KA .

.Step 2: Determine the formula of voltage V0  .

Write the formula ofvoltage .

${\mathbf{V}}_{\mathbf{0}}={\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{V}}_{\mathbf{2}}\mathbf{+}{\mathbf{V}}_{\mathbf{3}}\mathbf{+}{\mathbf{V}}_{\mathbf{4}}$ …… (1)

Here, V₁ , V₂ , V₃ and V₄ are the output voltages of the linear couplers.

(a) Determine the value of voltage  V0

Draw a circuit diagram of the protected bus and the lines during a fault.

Figure 1

Determine the output voltages of the linear couplers are:

$V_1=j{X}_{\text{m}}I{\text{'}}_1$

Here, X_m is linear couplers, $I{\text{'}}_1$ is the fault current on line .

Similarly, determine the voltages due to other three lines.

Consider the voltage v₁ is as follows:

$$\begin{gathered} V_1=\left(j5\times {10}^{-3}\right)\times \left(-j16\times {10}^3\right) \\ =80\text{V} \end{gathered}$$

Consider the voltage v₂ is as follows:

$$\begin{gathered} V_2=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =35\text{V} \end{gathered}$$

Consider the voltage v₃ is as follows:

$$\begin{gathered} V_3=\left(j5\times {10}^{-3}\right)\times \left[j36\times {10}^3\right] \\ =-180 \text{V} \end{gathered}$$

Consider the voltage v₄ is as follows:

$$\begin{gathered} V_4=\left(j5\times {10}^{-3}\right)\times \left(-j13\times {10}^3\right) \\ =65\text{V} \end{gathered}$$

Determine the voltage as:

Substitute 80 for V₁ , 35 for V₂ , 180 for V₃ and 65 for V₄ into equation (1).

$$\begin{gathered} V_0=80+35-180+65 \\ =0\text{V} \end{gathered}$$

As a result, under the situation, the relay VR cannot be operated by voltage. As a result, the VR relay won't work.

(b) Determine the value of voltage V0 .

Draw the circuit diagram for fault location between line 3 and line 4

Figure 3

From Figure 3, determine the fault currents are:

$$\begin{gathered} {I\text{'}}_1=-j16\text{kA} \\ {I\text{'}}_2=-j7\text{kA} \\ {I\text{'}}_3=-j10\text{kA} \\ {I\text{'}}_4=-j13\text{kA} \end{gathered}$$

Determine the output voltages of the linear couplers are:

$V_1=j{X}_{\text{m}}I{\text{'}}_1$

Here, is linear couplers, is the fault current on line .

Similarly, determine the voltages due to other three lines.

$$\begin{gathered} V_1=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =80 \text{V} \end{gathered}$$

Consider the voltage V₂ is as follows:

$$\begin{gathered} V_2=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =35\text{V} \end{gathered}$$

Consider the voltage V₃ is as follows:

localid="1656156617650" $$\begin{gathered} V_3=\left(j5\times {10}^{-3}\right)\times \left(-j10\times {10}^3\right) \\ =50\text{V} \end{gathered}$$

Consider the voltage V₄ is as follows:

$$\begin{gathered} V_4=\left(j5\times {10}^{-7}\right)\times \left(-j13\times {10}^3\right) \\ =65\text{V} \end{gathered}$$

Determine the output voltage as:

Substitute 80 for V₁ , 35 for V₂, 50 for V₃ and 65 for V₄ into equation (1).

$$\begin{gathered} V_0=80+35+50+65 \\ =230\text{V} \end{gathered}$$

As a result, the relay VR will operate.

Draw a circuit diagram of voltage and current for external fault between line and line .

(c) Determine the value of voltage V0  .

(i) Draw the circuit diagram for the fault location on the line2 .

Figure 5

From Figure 4, determine the fault currents are:

$$\begin{gathered} {I\text{'}}_1=-j16\text{kA} \\ {I\text{'}}_2=-j39\text{kA} \\ {I\text{'}}_3=-j10\text{kA} \\ {I\text{'}}_4=-j13\text{kA} \end{gathered}$$

Determine the output voltages of the linear couplers are:

$V_1=j{X}_{\text{m}}I{\text{'}}_1$

Here, is linear couplers, is the fault current on line .

Similarly, determine the voltages due to other three lines.

$$\begin{gathered} V_1=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =80 \text{V} \end{gathered}$$

Consider the voltage V₂ is as follows:

$$\begin{gathered} V_2=\left(j5\times {10}^{-3}\right)\times \left(-j39\times {10}^3\right) \\ =-195\text{V} \end{gathered}$$

Consider the voltage V₃ is as follows:

$$\begin{gathered} V_3=\left(j5\times {10}^{-3}\right)\times \left(-j10\times {10}^3\right) \\ =50\text{V} \end{gathered}$$

Consider the voltage V₄ is as follows:

$$\begin{gathered} V_4=\left(j5\times {10}^{-3}\right)\times \left(-j13\times {10}^3\right) \\ =65\text{V} \end{gathered}$$

Determine the output voltage as:

Substitute 80 for V₁, -195for V₂,50 for V₃ and 65 for V₄ into equation (1).

$$\begin{gathered} V_0=80-195+50+65 \\ =0\text{V} \end{gathered}$$

As a result, the relay VR would not operate.

Draw a circuit diagram of voltage and current for external fault on bus

.

Figure 6

(ii) Draw the circuit diagram for the fault location on the line 4.

Figure 7

From Figure 7, determine the fault currents are:

From Figure 3, determine the fault currents are:

$$\begin{gathered} {I\text{'}}_1=-j16\text{kA} \\ {I\text{'}}_2=-j7\text{kA} \\ {I\text{'}}_3=-j10\text{kA} \\ {I\text{'}}_4=-j33\text{kA} \end{gathered}$$

Determine the output voltages of the linear couplers are:

$V_1=j{X}_{\text{m}}I{\text{'}}_1$

Here, is linear couplers, is the fault current on line .

Similarly, determine the voltages due to other three lines.

$$\begin{gathered} V_1=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =80 \text{V} \\ {V}_2=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =35\text{V} \\ {V}_3=\left(j5\times {10}^{-3}\right)\times \left(-j10\times {10}^3\right) \\ =50\text{V} \\ {V}_4=\left(j5\times {10}^{-7}\right)\times \left(-j33\times {10}^3\right) \\ =-165\text{V} \end{gathered}$$

Determine the output voltage as:

Substitute 80 for V₁ , 35 for V₂, 50 for V₃ and 65 for V₄ into equation (1).

$$\begin{gathered} V_1=\left(j5\times {10}^{-3}\right)\times \left(-j7\times {10}^3\right) \\ =80 \text{V} \end{gathered}$$

As a result, the relay VR would not operate.