Question

The bus impedance matrix of a 3-bus system (in pu) is: \[ Z_{{bus}} = \begin{bmatrix} j0.059 & j0.061 & j0.038
j0.061 & j0.093 & j0.066
j0.038 & j0.066 & j0.110 \end{bmatrix} \] A symmetrical fault (through a fault impedance of \( j0.007 \) pu) occurs at bus 2. Neglecting pre-fault loading conditions, the voltage at bus 1 during the fault is: \[ {(round off to three decimal places).} \]

Hint:

For a fault at a bus with fault impedance, compute the Thevenin impedance at that bus using the diagonal of the \( Z_{{bus}} \) matrix. Fault current is \( I_f = V / Z_{{th}} \), and voltages at other buses are adjusted using the mutual impedances.

Answer 1

Assume a prefault voltage of 1 pu at all buses. For a fault at bus 2 with fault impedance \( Z_f = j0.007 \), the Thevenin impedance at the fault point is: \[ Z_{{th}} = Z_{22} + Z_f = j0.093 + j0.007 = j0.100 \] Fault current: \[ I_f = \frac{V_{{prefault}}}{Z_{{th}}} = \frac{1}{j0.100} = -j10 \] Voltage at bus 1 during fault is: \[ V_1 = V_{{prefault}} - Z_{12} \cdot I_f = 1 - (j0.061)(-j10) = 1 - (-0.61) = 1 + 0.61 = 0.39 { pu} \]