Question

Compute the 3 x 3 per-unit zero-, positive-, and negative-sequence bus impedance matrices for the power system given in Problem 4(a). Use a base of 1000 MVA and 500 kV in the zone of line 1-2.

Short answer

The value of per-unit positive-sequence bus impedance matrix,${\text{Z}}_{\text{bus}}$ is $-j\left[\begin{array}{ccc}1.87& 0.66& 0.406\\ 0.66& 1.81& 0.39\\ 0.406& 0.39& 1.12\end{array}\right]$.

The value of per-unit negative-sequence bus impedance matrix,${\text{Z}}_{\text{bus}}$ is $-j\left[\begin{array}{ccc}1.85& 0.65& 0.36\\ 0.65& 1.79& 0.35\\ 0.36& 0.35& 1.01\end{array}\right]$.

The value of per-unit zero-sequence impedance matrix,${\text{Z}}_{\text{bus}}$ is $-j\left[\begin{array}{ccc}2.93& 2.34& 2.05\\ 2.34& 2.93& 2.05\\ 2.05& 2.05& 2.56\end{array}\right]$.

Step-by-step solution

Write the given data from the question.

As reference of problem 4(a).

Thevalue of base MVA is 1000 MVA.

The value of system voltage in the zone of line 1-2 is 500 kV.

Determine the formula of per-unit positive-sequence bus impedance matrix, per-unit negative sequence bus impedance matrix and per-unit zero sequence bus impedance matrix.

Write the formula of per-unit positive-sequence, negative sequence and zero sequence bus impedance matrixes.

${\mathbf{Z}}_{\mathrm{bus}}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}}}$ …… (1)

Here,${\mathbf{Y}}_{\mathrm{bus}}$ is per-unit positive-sequence bus impedance matrix.

(a) Determine the per-unit positive-sequence bus impedance matrix.

Refer to the Figure 9.17 from the textbook.

Determine the base impedance value.

$\begin{array}{c}{\mathrm{Z}}_{\text{base}}=\frac{{\left(500\times {10}^3\right)}^2}{1000\times {10}^6}\\ =250\mathrm{\Omega }\end{array}$

Determine the per unit positive and negative quantities for line 1-2.

$\begin{array}{c}{X}_1=X_2\\ =\left(\frac{50}{250}\right)\\ =0.2\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit zero sequence reactance for lines 1-2.

$\begin{array}{c}{X}_0=\left(\frac{150}{250}\right)\\ =0.6\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit positive and negative quantities for lines 1-3 and 2-3.

$\begin{array}{c}{X}_1=X_2\\ =\left(\frac{40}{250}\right)\\ =0.16\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit Zero sequence reactance for line 1-3 and 2-3.

$\begin{array}{c}{X}_0=\left(\frac{100}{250}\right)\\ =0.4\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit positive-sequence bus admittance matrix.

$Y_{\text{bus}}=j\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1\end{array}\right]$

Determine the per unit positive-sequence bus impedance matrix.

Substitute$j\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1\end{array}\right]$for$Y_{\text{bus}}$into equation (1).

$\begin{array}{c}{Z}_{\text{bus}}=\frac{1}{j\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1\end{array}\right]}\\ =-j\left[\begin{array}{ccc}1.87& 0.66& 0.406\\ 0.66& 1.81& 0.39\\ 0.406& 0.39& 1.12\end{array}\right]\end{array}$

Therefore, the value of per-unit positive-sequence bus impedance matrix, ${\text{Z}}_{\text{bus}}$is $-j\left[\begin{array}{ccc}1.87& 0.66& 0.406\\ 0.66& 1.81& 0.39\\ 0.406& 0.39& 1.12\end{array}\right]$.

Draw the circuit diagram of per unit negative sequence single line diagram.

Determine the per unit negative-sequence bus admittance matrix.

$Y_{\text{bus}}=\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1.1\end{array}\right]$

Determine the per unit negative-sequence bus impedance matrix.

Substitute $\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1.1\end{array}\right]$for $Y_{\text{bus}}$into equation (2).

$\begin{array}{c}{Z}_{\text{bus}}=\frac{1}{\left[\begin{array}{ccc}0.64& -0.2& -0.16\\ -0.2& 0.66& -0.16\\ -0.16& -0.16& 1.1\end{array}\right]}\\ =-j\left[\begin{array}{ccc}1.85& 0.65& 0.36\\ 0.65& 1.79& 0.35\\ 0.36& 0.35& 1.01\end{array}\right]\end{array}$

Therefore, the value of per-unit negative-sequence bus impedance matrix, ${\text{Z}}_{\text{bus}}$is $-j\left[\begin{array}{ccc}1.85& 0.65& 0.36\\ 0.65& 1.79& 0.35\\ 0.36& 0.35& 1.01\end{array}\right]$.

Draw the circuit diagram of per unit zero sequence single line diagram.

Determine the per unit zero-sequence bus admittance matrix.

$Y_{\text{bus}}=\left[\begin{array}{ccc}1.1& -0.6& -0.4\\ -0.6& 1.1& -0.4\\ -0.4& -0.4& 1.03\end{array}\right]$

Determine the per unit zero-sequence bus impedance matrix.

Substitutelocalid="1656422391591" $\left[\begin{array}{ccc}1.1& -0.6& -0.4\\ -0.6& 1.1& -0.4\\ -0.4& -0.4& 1.03\end{array}\right]$for$Y_{\text{bus}}$into equation (3).

$\begin{array}{c}{Z}_{\text{bus}}=\frac{1}{\left[\begin{array}{ccc}1.1& -0.6& -0.4\\ -0.6& 1.1& -0.4\\ -0.4& -0.4& 1.03\end{array}\right]}\\ =-j\left[\begin{array}{ccc}2.93& 2.34& 2.05\\ 2.34& 2.93& 2.05\\ 2.05& 2.05& 2.56\end{array}\right]\end{array}$

Therefore, the value of per-unit zero-sequence impedance matrix${\text{Z}}_{\text{bus}}$is $-j\left[\begin{array}{ccc}2.93& 2.34& 2.05\\ 2.34& 2.93& 2.05\\ 2.05& 2.05& 2.56\end{array}\right]$.