Question

A single-line diagram of a four-bus system is shown in Figure 7.20, for which$Z_{\mathrm{bus}}$is given below:

${\mathbf{Z}}_{\mathbf{bus}}\mathbf{=}\mathbf{j}\mathbf{\left[}\begin{array}{cccc}\mathbf{0}\mathbf{.}\mathbf{25}& \mathbf{0}\mathbf{.}\mathbf{2}& \mathbf{0}\mathbf{.}\mathbf{16}& \mathbf{0}\mathbf{.}\mathbf{14}\\ \mathbf{0}\mathbf{.}\mathbf{2}& \mathbf{0}\mathbf{.}\mathbf{23}& \mathbf{0}\mathbf{.}\mathbf{15}& \mathbf{0}\mathbf{.}\mathbf{151}\\ \mathbf{0}\mathbf{.}\mathbf{16}& \mathbf{0}\mathbf{.}\mathbf{15}& \mathbf{0}\mathbf{.}\mathbf{196}& \mathbf{0}\mathbf{.}\mathbf{1}\\ \mathbf{0}\mathbf{.}\mathbf{14}& \mathbf{0}\mathbf{.}\mathbf{151}& \mathbf{0}\mathbf{.}\mathbf{1}& \mathbf{0}\mathbf{.}\mathbf{195}\end{array}\mathbf{\right]}\mathbf{per}\mathbf{}\mathbf{unit}$

Let a three-phase fault occur at bus 2 of the network.

(a) Calculate the initial symmetrical rms current in the fault.

(b) Determine the voltages during the fault at buses 1, 3, and 4.

(c) Compute the fault currents contributed to bus 2 by the adjacent unfaulted buses 1, 3, and 4.

(d) Find the current flow in the line from bus 3 to bus 1. Assume the prefault voltage ${\mathbf{V}}_{\mathbf{f}}$at bus 2 to belocalid="1655403112082" $\mathbf{1}\mathbf{\angle }{\mathbf{0}}^{\mathbf{\circ }}$and neglect all prefault currents.

Short answer

(a)The value of initial symmetrical rms fault current at bus2 is$I_{f_2}^{"}=-j4.3478$$\mathrm{per}\mathrm{unit}$

(b)The voltage during fault at bus 1 is role="math" localid="1655402085619" $E,=0.1304\angle 0^{\circ }\mathrm{per}\mathrm{unit}$, at bus 2 is role="math" localid="1655402092099" $E_2=0\mathrm{per}\mathrm{unit}$, at bus 3 is role="math" localid="1655402101264" $E_3=0.3478\angle 0^{\circ }\mathrm{per}\mathrm{unit}$and at bus 4 is role="math" localid="1655402106250" $E_4=0.3435\angle 0^{\circ }\mathrm{per}\mathrm{unit}$.

(c)The fault current contributed to bus 2 by the adjacent unfaulted bus 1 isrole="math" localid="1655402112435" $I_{2-1}=-j1.0432\mathrm{per}\mathrm{unit}$, by bus 2 isrole="math" localid="1655402119806" $I_{3-2}=-j1.3912\mathrm{per}\mathrm{unit}$and by bus 3 is$I_{4-2}=-j1.7175\mathrm{per}\mathrm{unit}$

(d)The current flow in the line bus 3 to bus 1 is${/}_{3-1}=-j0.8696\mathrm{per}\mathrm{unit}$.

Step-by-step solution

Given data

The pre-fault voltage at bus 2 is $V_f=1\angle 0^{\circ }$.

The values of impedances are$$\begin{gathered} Z_{22}=j0.23\mathrm{per}\mathrm{unit}\mathrm{and}{\mathrm{Z}}_{12}=\mathrm{j}0.2\mathrm{per}\mathrm{unit}, \\ {\mathrm{Z}}_{32}=\mathrm{j}0.15\mathrm{per}\mathrm{unit}\mathrm{and}{\mathrm{Z}}_{42}=\mathrm{j}0.151\mathrm{per}\mathrm{unit} \end{gathered}$$.

The value of $Z_{2-1}isj0.125\mathrm{per}\mathrm{unit}and{Z}_{3-2}isj0.25\mathrm{per}\mathrm{unit}$

Calculate the value of initial symmetrical rms current in the fault

(a)

As given in the data the three phase fault has occurred in the bus 2,

The formula to calculate the rms value of the initial symmetrical fault current$I_{f_2}^{"}$is,

$I_{f_2}^{"}=\frac{V_f}{Z_{22}}$

Substitute$1\angle 0^{\circ }\mathrm{for}{\mathrm{V}}_{\mathrm{f}}\mathrm{and}j0.23perunit\mathrm{for}{\mathrm{Z}}_{22}$in the above equation.

$$\begin{gathered} I_{f_2}^{"}=\frac{1\angle 0}{j0.23} \\ =-j4.3478\mathrm{per}\mathrm{unit} \end{gathered}$$

The value of initial symmetrical rms fault current at bus 2 is$I_{f_2}^{"}=-j4.3478\mathrm{per}\mathrm{unit}$.

Calculate the value of fault voltages at buses 1, 3 and 4

(b)

The formula to calculate the voltage$E_1$for bus 1 is,

$E_1=\left(1-\frac{Z_{12}}{Z_{22}}\right)V_f$

Now, substitute the value$1\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{V}}_{\mathrm{f}},\mathrm{j}0.2\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{12}\mathrm{and}\mathrm{j}0.23\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{22}$in the above equation.

$$\begin{gathered} E_1=\left(1-\frac{j0.2}{j0.23}\right)\left(1\angle 0^{\circ }\right) \\ =\left(1-0.8696\right)\left(1\angle 0^{\circ }\right) \\ =\left(0.1304\right)\left(1\angle 0^{\circ }\right) \end{gathered}$$

During the fault, value of voltage$E_1$at bus 1 will be,

$E_1=0.1304\angle 0^{\circ }\mathrm{per}\mathrm{unit}$

The formula to calculate the voltage$E_2$for bus 3 during the fault is,

$E_2=\left(1-\frac{Z_{22}}{Z_{22}}\right)V_f$

Now, substitute the value $1\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{V}}_{\mathrm{f}},j0.23\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{22}$in the above equation.

$$\begin{gathered} E_2=\left(1-\frac{j0.23}{j0.23}\right)\left(1\angle 0^{\circ }\right) \\ =\left(1-1\right)\left(1\angle 0^{\circ }\right) \\ =0\mathrm{per}\mathrm{unit} \end{gathered}$$

The formula to calculate the voltage$E_3$, during the fault for bus 3 is,

$E_3=\left(1-\frac{Z_{32}}{Z_{22}}\right)V_f$

Now, substitute the value$1\angle 0^{\circ }\mathrm{per}\mathrm{unit}for{V}_f,j0.15\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{32}\mathrm{and}j0.23\mathrm{per}\mathrm{unit}fo{Z}_{22}$in the above equation.

$$\begin{gathered} E_3=\left(1-\frac{j0.15}{j0.23}\right)\left(1\angle 0^{\circ }\right) \\ =\left(1-0.6522\right)\left(1\angle 0^{\circ }\right) \\ =\left(0.3478\right)\left(1\angle 0^{\circ }\right) \end{gathered}$$

During the fault, value of voltage$E_3$at bus 3 will be,

$E_3=0.3478\angle 0^{\circ }\mathrm{per}\mathrm{unit}$

The formula to calculate the voltage$E_4$during the fault for bus 4 is,

$E_4=\left(1-\frac{Z_{42}}{Z_{22}}\right)V_f$

Substitute the value$1\angle 0^{\circ }\mathrm{for}{\mathrm{V}}_{\mathrm{f}},j0.151\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{42}\mathrm{and}j0.23\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{22}$in the above equation.

$$\begin{gathered} E_4=\left(1-\frac{j0.151}{j0.23}\right)\left(1\angle 0^{\circ }\right) \\ =\left(1-0.6565\right)\left(1\angle 0^{\circ }\right) \\ =\left(0.3435\right)\left(1\angle 0^{\circ }\right) \end{gathered}$$

During the fault, value of voltage$E_4$at bus 4 will be,

$E_4=0.3435\angle 0^{\circ }\mathrm{per}\mathrm{unt}$

Determine the values of fault rms currents contributed by buses 1, 3 and 4

(c)

The formula to calculate the value of the rms fault current between the bus 2 and bus 1 is,

$I_{2-1}=\frac{E_1-E_2}{Z_{2-1}}$

Substitute$0.1304\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{E}}_1,0\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{E}}_2\mathrm{an}\mathrm{j}0.125\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{2-1}$in the above equation.

$$\begin{gathered} I_{2-1}=\frac{0.1304-0}{j0.125} \\ =j1.0432\mathrm{per}\mathrm{unit} \end{gathered}$$

The formula to calculate the value of the rms fault current between the bus 3 and bus 2 is,

$I_{3-2}=\frac{E_3-E_2}{Z_{3-2}}$

Substitute$0.3478\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{E}_{3,}0\mathrm{per}\mathrm{unit}\mathrm{for}{E}_2\mathrm{and}j0.25\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{3-2}$in the above equation.

$$\begin{gathered} I_{3-2}=\frac{0.3478-0}{j0.25} \\ =j1.3912\mathrm{per}\mathrm{unit} \end{gathered}$$

The formula to calculate the value of the rms fault current between the bus 4 and bus 2 is

$I_{4-2}=\frac{E_4-E_2}{Z_{4-2}}$

Substitute$0.3435\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{E}_4,0\mathrm{per}\mathrm{unit}\mathrm{for}{E}_2\mathrm{and}j0.2\mathrm{per}\mathrm{unit}\mathrm{for}{\mathrm{Z}}_{4-2}$in the above equation.

$$\begin{gathered} I_{4-2}=\frac{0.3435-0}{j0.2} \\ =j1.7175\mathrm{per}\mathrm{unit} \end{gathered}$$

The fault current contributed to bus 2 by the adjacent unfaulted bus 1 is , by bus 2 is and by bus 3 is .

$$\begin{gathered} I_{2-1}=-j1.0432\mathrm{per}\mathrm{unit},\mathrm{by}\mathrm{bus}2\mathrm{is}{I}_{3-2}=-j1.3912\mathrm{per}\mathrm{unit}\mathrm{and}\mathrm{by}\mathrm{bus}3\mathrm{is} \\ {I}_{4-2}=-j1.7175\mathrm{per}\mathrm{unit} \end{gathered}$$

Determine the value of rms current flowing in the line from bus 3 to 1

The formula to calculate the value of the rms current flowing between the bus 3 and bus 1 is,

$I_{3-1}=\frac{E_3-E_1}{Z_{3-1}}$

Substitute$$\begin{gathered} 0.3478\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{E}_3,0.1304\angle 0^{\circ }\mathrm{per}\mathrm{unit}\mathrm{for}{E}_1\mathrm{and}j0.25\mathrm{per}\mathrm{unit} \\ \mathrm{for}{\mathrm{Z}}_{3-1} \end{gathered}$$in the above equation.

$$\begin{gathered} I_{3-1}=\frac{0.3478\angle 0^{\circ }-0.1304\angle 0^{\circ }}{j0.25} \\ =\frac{0.2174\angle 0^{\circ }}{j0.25} \\ =j0.8696\mathrm{per}\mathrm{unit} \end{gathered}$$

Thus, the current flow in the line bus 3 to bus 1 is $I_{3-1}=-j0.8696\mathrm{per}\mathrm{unit}$.