Question

For the three-phase power system with single-line diagram shown in Figure 9.22, equipment ratings and per-unit reactances are given as follows:

Machine and: $\begin{array}{l}\mathbf{100}\mathbf{MVA}\\ {\mathbf{X}}_0=\mathbf{0}\mathbf{.}\mathbf{04}\end{array}$ $\begin{array}{l}\mathbf{20}\mathbf{kV}\\ {\mathbf{X}}_n=\mathbf{0}\mathbf{.}\mathbf{04}\end{array}$ ${\mathbf{X}}_1={\mathbf{X}}_2=\mathbf{0}\mathbf{.}\mathbf{2}$

Transformers and: $\begin{array}{l}\mathbf{100}\mathbf{MVA}\\ {\mathbf{X}}_1={\mathbf{X}}_2={\mathbf{X}}_0\end{array}$ $\mathbf{20}\mathbf{\Delta }/\mathbf{345}\mathbf{kV}$

Select a base of 100 MVA, 345 kV for the transmission line.On that base, the series reactances of the line are${\mathbf{X}}_1={\mathbf{X}}_2=\mathbf{0}\mathbf{.}\mathbf{15}$ and${\mathbf{X}}_0=\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{per}\mathbf{unit}$. With a nominal system voltage of 345 kV at bus 3, machine 2 is operating as a motor drawing 50 MVA at 0.8 power factor lagging. Compute the change in voltage at bus 3 when the transmission line undergoes(a) a one-conductor-open fault, (b) a two-conductor-open fault along its span between buses 2 and 3.

Short answer

(a) The value of change in voltage at bus 3,$\Delta V_3$when transmissions line under goes a one-conductor-open fault is $1.43\angle 53.14°\text{\hspace{0.33em}V}$.

(b) The change in voltage at bus 3,$\Delta V_3$ when transmission line under goes a two-conductor-open fault along its span between buses 2 and 3 is $0.259\angle 53.13°\text{\hspace{0.33em}V}$.

Step-by-step solution

Write the given data from the question.

Thevalue of base MVA is 100 MVA.

The value of series reactance of the line$X_1=X_2=0.15$and $X_0=0.5\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

The value of transmission line voltage is 345 kV.

The value of nominal system voltage at bus 3 is 345 kV.

The machine 2 is operating as a motor drawing 50 MVA.

The value of power factor lagging is 0.8

Determine the formula of transmissions line under goes a one-conductor-open fault and transmission line under goes a two-conductor-open fault along its span between buses 2 and 3.

Write the formula of transmissions line under goes a one-conductor-open fault.

${\mathbf{\Delta V}}_3={\mathbf{\Delta V}}_3^{\left(1\right)}\mathbf{+}{\mathbf{\Delta V}}_3^{\left(2\right)}\mathbf{+}{\mathbf{\Delta V}}_3^{\left(3\right)}$ …… (1)

Here, ${\mathbf{\Delta V}}_3^{\left(1\right)}\mathbf{,}\mathbf{}{\mathbf{\Delta V}}_3^{\left(2\right)}\mathbf{}\mathbf{\&}\mathbf{}{\mathbf{\Delta V}}_3^{\left(3\right)}$arechange in phase voltages of sequence components.

(a) Determine the transmissions line under goes a one-conductor-open fault.

Refer to the Figure 9.22 from the textbook.

Draw the circuit diagram of per unit positive sequence network.

Determine the Bus admittance matrix for positive-sequence network.

$Y_{\text{bus}}=j\left[\begin{array}{cccc}0.32& -0.08& 0& 0\\ -0.08& 0.23& -0.15& 0\\ 0& -0.15& 0.23& -0.08\\ 0& 0& -0.08& 0.32\end{array}\right]$

Determine the bus impedance matrix for the positive sequence network.

$\begin{array}{c}{Z}_{\text{bus}}=Y_{\text{bus}}^{-1}\\ =-j\left[\begin{array}{cccc}0.32& -0.08& 0& 0\\ -0.08& 0.23& -0.15& 0\\ 0& -0.15& 0.23& -0.08\\ 0& 0& -0.08& 0.32\end{array}\right]\\ =-j\left[\begin{array}{cccc}3.73& 2.43& 1.73& 0.43\\ 2.43& 9.72& 6.94& 1.73\\ 1.87& 7.51& 7.23& 3.66\\ 0.43& 1.73& 2.43& 3.73\end{array}\right]\end{array}$

Draw a circuit diagram of per unit negative sequence network.

Determine the Bus admittance matrix for negative-sequence network.

$Y_{\text{bus}}=j\left[\begin{array}{cccc}0.32& -0.08& 0& 0\\ -0.08& 0.23& -0.15& 0\\ 0& -0.15& 0.23& -0.08\\ 0& 0& -0.08& 0.32\end{array}\right]$

Determine the bus impedance matrix for the negative sequence network.

$\begin{array}{c}{Z}_{\text{bus}}=Y_{\text{bus}}^{-1}\\ =-j{\left[\begin{array}{cccc}0.32& -0.08& 0& 0\\ -0.08& 0.23& -0.15& 0\\ 0& -0.15& 0.23& -0.08\\ 0& 0& -0.08& 0.32\end{array}\right]}^{-1}\\ =-j\left[\begin{array}{cccc}3.73& 2.43& 1.73& 0.43\\ 2.43& 9.72& 6.94& 1.73\\ 1.87& 7.51& 7.23& 3.66\\ 0.43& 1.73& 2.43& 3.73\end{array}\right]\end{array}$

Draw the circuit diagram of per unit zero sequence network.

Determine the Bus admittance matrix for zero-sequence network.

$Y_{\text{bus}}=j\left[\begin{array}{cccc}0.04& 0& 0& 0\\ 0& 0.58& -0.5& 0\\ 0& -0.5& 0.58& 0\\ 0& 0& 0& 0.04\end{array}\right]$

Determine the bus impedance matrix for the zero sequence network.

$\begin{array}{c}{Z}_{\text{bus}}=Y_{\text{bus}}^{-1}\\ =-j{\left[\begin{array}{cccc}0.04& 0& 0& 0\\ 0& 0.58& -0.5& 0\\ 0& -0.5& 0.58& 0\\ 0& 0& 0& 0.04\end{array}\right]}^{-1}\\ =-j\left[\begin{array}{cccc}25& 0& 0& 0\\ 0& 6.713& 5.787& 0\\ 0& 5.787& 6.713& 0\\ 0& 0& 0& 25\end{array}\right]\end{array}$

Determine the line current in the system.

$\begin{array}{c}{I}_{23}=\frac{S}{V}\angle -\theta \\ =\frac{50}{100}\angle -{\cos }^{-1}\left(0.8\right)\\ =0.5\angle -36.86°\text{\hspace{0.33em}A}\end{array}$

Determine the positive-sequence impedance.

$\begin{array}{c}{Z}_{\text{PP}\left(\text{1}\right)}=Z_{\text{PP}\left(\text{2}\right)}\\ =\frac{-Z_1^2}{Z_{22}^{\left(1\right)}+Z_{33}^{\left(1\right)}-2Z_{23}^{\left(1\right)}-Z_1}\\ =\frac{-{\left(j1.903\right)}^2}{j\left(9.72+9.72-2\left(6.94\right)-1.903\right)}\\ =\frac{3.62}{j3.657}\end{array}$

Solve further as

$Z_{\text{PP}\left(\text{1}\right)}=-j1\text{\hspace{0.33em}pu}$

Determine the zero-sequence impedance.

$\begin{array}{c}{Z}_{\text{PP}\left(\text{0}\right)}=\frac{-Z_0^2}{Z_{22}^{\left(0\right)}+Z_{33}^{\left(0\right)}-2Z_{23}^{\left(0\right)}-Z_0}\\ =\frac{-{\left(j5.8\right)}^2}{j\left(6.713+6.713-2\left(5.78\right)-5.8\right)}\\ =\frac{33.64}{-j3.934}\\ =j8.55\text{\hspace{0.33em}pu}\end{array}$

For one conductor open fault occurs in a system:

Determine the Phase voltages of zero-sequence component.

$\begin{array}{c}{V}_{\text{a}}^{\left(0\right)}=V_{\text{b}}^{\left(\text{0}\right)}=V_c^{\left(0\right)}\\ =\frac{I_{23}}{\frac{1}{Z_{11}^{\left(1\right)}}+\frac{1}{Z_{11}^{\left(2\right)}}+\frac{1}{Z_{11}^{\left(0\right)}}}\\ =\frac{0.5\angle -36.86°}{1j+1j-0.117j}\\ =0.26\angle -126.87°\text{\hspace{0.33em}V}\end{array}$

For faulted at bus-3:

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_{3\left(0.26\right)}^{\left(1\right)}=\Delta V_3^{\left(2\right)}\\ =\frac{\left(Z_{32}^{\left(1\right)}-Z_{33}^{\left(1\right)}-Z_{34}^{\left(1\right)}\right)}{Z_1}{V}_{\text{a}}^{\left(0\right)}\\ =\left(\frac{j6.94-j9.72-j2.43}{j1.903}\right)\left(0.26\angle -126.87°\right)\\ =0.712\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

For during faulted at bus-3:

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_3^{\left(1\right)}=\Delta V_3^{\left(2\right)}\\ =\frac{\left(Z_{32}^{\left(1\right)}-Z_{23}^{\left(1\right)}-Z_{34}^{\left(1\right)}\right)}{Z_1}{V}_{\text{a}}^{\left(0\right)}\\ =\left(\frac{j6.94-j9.72-j2.43}{j1.903}\right)\left(0.26\angle -126.87°\right)\\ =0.712\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_3^{\left(0\right)}=\frac{\left(Z_{32}^{\left(0\right)}-Z_{33}^{\left(0\right)}-Z_{34}^{\left(0\right)}\right)}{Z_0}{V}_{\text{a}}^{\left(0\right)}\\ =\frac{\left(j5.787-j6.713-0\right)}{j5.8}\left(0.02\angle -126.87°\right)\\ =0.00319\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Now, determine the change in a phase voltage at bus-3.

Substitute$0.712\angle 53.13°$for $\Delta V_3^{\left(1\right)}$,$0.712\angle 53.13°$for$\Delta V_3^{\left(2\right)}$and$0.712\angle 53.13°$for$\Delta V_3^{\left(0\right)}$into equation (1).

$\begin{array}{c}\Delta V_3=0.712\angle 53.13°+0.712\angle 53.13°+0.00319\angle 53.13°\\ =1.43\angle 53.14°\text{\hspace{0.33em}V}\end{array}$

Therefore, the value of change in voltage at bus 3,$\Delta V_3$when transmissions line under goes a one-conductor-open fault is $1.43\angle 53.14°\text{\hspace{0.33em}V}$.

(b) Determine the transmission line under goes a two-conductor-open fault along its span between buses 2 and 3.

Determine the Positive-sequence phase voltage component.

$\begin{array}{c}{V}_{\text{a}}^{\left(1\right)}=I_{23}\left[\frac{Z_{\text{PP}}^{\left(1\right)}\left(Z_{\text{PP}}^{\left(2\right)}+Z_{\text{PP}}^{\left(0\right)}\right)}{Z_{\text{PP}}^{\left(1\right)}+\left(Z_{\text{PP}}^{\left(2\right)}+Z_{\text{PP}}^{\left(0\right)}\right)}\right]\\ =\left(0.5\angle -36.87°\right)\left[\frac{\left(-j1\right)\left(-j1+j8.55\right)}{-j1+\left(-j1+j8.55\right)}\right]\\ =\left(0.5\angle -36.87°\right)\left(-j1.161\right)\\ =0.058\angle -126.87°\text{\hspace{0.33em}V}\end{array}$

Determine the negative-sequence phase voltage component.

$\begin{array}{c}{V}_{\text{a}}^{\left(2\right)}=I_{23}\left[\frac{-Z_{\text{PP}}^{\left(1\right)}\left(Z_{\text{PP}}^{\left(2\right)}\right)}{Z_{\text{PP}}^{\left(1\right)}+\left(Z_{\text{PP}}^{\left(2\right)}+Z_{\text{PP}}^{\left(0\right)}\right)}\right]\\ =\left(0.5\angle -36.87°\right)\left[\frac{-\left(-j1\right)\left(-j1\right)}{-j1+\left(-j1+j8.55\right)}\right]\\ =\left(0.5\angle 36.87°\right)\left(-j0.153\right)\\ =0.07\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Determine the zero-sequence phase voltage component.

$\begin{array}{c}{V}_a^{\left(0\right)}=I_{23}\left[\frac{-Z_{\text{PP}}^{\left(1\right)}\left(Z_{\text{PP}}^{\left(0\right)}\right)}{Z_{\text{PP}}^{\left(1\right)}+\left(Z_{\text{PP}}^{\left(2\right)}+Z_{\text{PP}}^{\left(0\right)}\right)}\right]\\ =\left(0.5\angle -36.87°\right)\left[\frac{-\left(-j1\right)\left(j8.55\right)}{-j1+\left(-j1+j8.55\right)}\right]\\ =\left(0.5\angle -36.87°\right)\left(j1.3053\right)\\ =0.65\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_3^{\left(1\right)}=\frac{\left(Z_{32}^{\left(1\right)}-Z_{33}^{\left(1\right)}-Z_{34}^{\left(1\right)}\right)}{Z_1}{V}_{\text{a}}^{\left(1\right)}\\ =\left(\frac{-j7.51+j7.23+j3.66}{j1.903}\right)\left(0.058\angle 53.13°\right)\\ =\left(1.776\right)\left(0.058\angle 53.13°\right)\\ =0.103\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_3^{\left(2\right)}=\frac{\left(Z_{32}^{\left(2\right)}-Z_{33}^{\left(2\right)}-Z_{34}^{\left(2\right)}\right)}{Z_1}\\ =\left(\frac{-j7.51+j7.23+j3.66}{j1.903}\right)\left(0.07\angle 53.13°\right)\\ =\left(1.77\right)\left(0.07\angle 53.13°\right)\\ =0.124\angle 53.14°\text{\hspace{0.33em}V}\end{array}$

Determine the change in phase voltages of sequence components.

$\begin{array}{c}\Delta V_3^{\left(0\right)}=\frac{\left(Z_{32}^{\left(0\right)}-Z_{33}^{\left(0\right)}\right)}{Z_0}{V}_{\text{a}}^{\left(0\right)}\\ =\left(\frac{-j5.787+j6.713}{j1.903}\right)\left(0.65\angle 53.13°\right)\\ =\left(0.486\right)\left(0.65\angle 53.13°\right)\\ =0.032\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Now, determine the change in a phase voltage at bus-3.

Substitute $0.103\angle 53.13°$for $\Delta V_3^{\left(1\right)}$, $0.124\angle 53.13°$for $\Delta V_3^{\left(2\right)}$and$0.032\angle 53.13°$ for$\Delta V_3^{\left(0\right)}$into equation (1).

$\begin{array}{c}\Delta V_3=0.103\angle 53.13°+0.124\angle 53.13°+0.032\angle 53.13°\\ =0.259\angle 53.13°\text{\hspace{0.33em}V}\end{array}$

Therefore, the change in voltage at bus 3,$\Delta V_3$ when transmission line under goes a two-conductor-open fault along its span between buses 2 and 3 is $0.259\angle 53.13°\text{\hspace{0.33em}V}$.