Question

The three-phase impedance load shown in Figure 8.7 has the following sequence impedance matrix:

${\mathbf{Z}}_{\mathbf{P}}=\left[\begin{array}{ccc}\left(\mathbf{6}\mathbf{+}\mathbf{j}\mathbf{10}\right)& 0& 0\\ 0& \mathbf{5}& 0\\ 0& 0& \mathbf{5}\end{array}\right]\mathbf{\Omega }$

Determine the phase impedance matrix ${\mathbf{Z}}_{\mathbf{P}}$for this load. Is the load symmetrical?

Short answer

The phase impedance matrix is$Z_P=\left[\begin{array}{ccc}\left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)\end{array}\right]$

Step-by-step solution

Write the given data from the question.

The three-phase impedance load has been shown in following diagram.

The sequence impedance matrix is given as:

.${\mathbf{Z}}_{\mathbf{P}}=\left[\begin{array}{ccc}\left(\mathbf{6}\mathbf{+}\mathbf{j}\mathbf{10}\right)& 0& 0\\ 0& \mathbf{5}& 0\\ 0& 0& \mathbf{5}\end{array}\right]\mathbf{\Omega }$

Determine the formula of phase impedance matrix.

Write the formula of phase impedance matrix.

${\mathbf{Z}}_{\mathbf{s}}={\mathbf{AZ}}_{\mathbf{P}}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}$…… (1)

Here, $\mathbf{A}$ is adjoin matrix, ${\mathbf{Z}}_{\mathbf{s}}$is sequence impedance matrix and ${\mathbf{A}}^{\mathbf{-}\mathbf{1}}$ is inverse of matrix.

Determine the phase impedance matrix.

The sequence impedance matrix in terms of phase impedance matrix is

$Z_s=A^{-1}{Z}_{P}A$…… (2)

Here, $A^{-1}$is inverse of matrix, $Z_P$is phase impedance matrix and A is adjoin matrix.

Multiply at both sides of equation (2).

$A{Z}_s=A{A}^{-1}{Z}_{P}A$…… (3)

Therefore, a multiplication of matrix and its inverse gives unit matrix of equation (3).

Now, multiply above equation by $A^{-1}$of equation (3).

$$\begin{gathered} A{Z}_s{A}^{-1}=I{Z}_{P}A{A}^{-1} \\ A{Z}_s{A}^{-1}=I{Z}_{P}I \\ {Z}_P=A{Z}_s{A}^{-1} \end{gathered}$$

Consider few identities are true for $a=1\angle 120°$.

$$\begin{gathered} a^4=a \\ {a}^3=1 \\ 1+a+a^2=0 \end{gathered}$$

Substitute localid="1656767520106" $\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]$for$A^{-1}$,$\left[\begin{array}{ccc}\left(\mathbf{6}\mathbf{+}\mathbf{j}\mathbf{10}\right)& 0& 0\\ 0& \mathbf{5}& 0\\ 0& 0& \mathbf{5}\end{array}\right]$ for Z and localid="1656767478246" $\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]$for into equation (1).

$$\begin{gathered} Z_P=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{ccc}\left(6+j10\right)& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{ccc}\left(6+j10\right)& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{ccc}\left(6+j10\right)+5+5& \left(1+j10\right)+5\left(1+a+a^2\right)& \left(1+j10\right)+5\left(1+a^2+a\right)\\ \left(1+j10\right)+5\left(1+a^2+a\right)& \left(6+j10\right)+5a^3+5a^3& \left(1+j10\right)+5\left(1+a^4+a^2\right)\\ \left(1+j10\right)+5\left(1+a+a^2\right)& \left(1+j10\right)+5\left(1+a^2+a^4\right)& \left(6+j10\right)+5a^3+5a^3\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{ccc}\left(16+j10\right)& \left(1+j10\right)& \left(1+j10\right)\\ \left(1+j10\right)& \left(16+j10\right)& \left(1+j10\right)\\ \left(1+j10\right)& \left(1+j10\right)& \left(16+j10\right)\end{array}\right] \end{gathered}$$

Solve further as,

$Z_P=\left[\begin{array}{ccc}\left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)\end{array}\right]$

Thus, the load is symmetrical and the phase impedance matrix is

$Z_P=\left[\begin{array}{ccc}\left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)\\ \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{1}{3}+j\frac{10}{3}\right)& \left(\frac{16}{3}+j\frac{10}{3}\right)\end{array}\right]$